Soluções resistência dos materiais hibbeler_ 7ª edição

1. ISM for 2. Problem 1-1Determine the resultant internal normal force acting on the cross section through point A in eachcolumn. In (a), segment BC weighs 300 kg/m and segment CD weighs 400kg/m. In (b), the columnhas a mass of 200 kg/m.(a) Given: g 9.81ms2:= wBC 300kgm:=LBC := 3m wCA 400kgm:=FB := 5kN LCA := 1.2mFC := 3kNSolution:+↑Σ Fy = 0; FA wBC g − ( ⋅ )⋅LBC − (wCA⋅g)⋅LCA − FB − 2FC = 0FA := (wBC⋅g)⋅LBC + (wCA⋅g)⋅LCA + FB + 2FCFA = 24.5 kN Ans(b) Given: g 9.81ms2:= w 200kgm:=L := 3m F1 := 6kNFB := 8kN F2 := 4.5kNSolution:+↑Σ Fy = 0; FA w L − ( ⋅ )⋅g − FB − 2F1 − 2F2 = 0FA := (w⋅L)⋅g + FB + 2F1 + 2F2FA = 34.89 kN Ans 3. Problem 1-2Determine the resultant internal torque acting on the cross sections through points C and D of theshaft. The shaft is fixed at B.Given: TA := 250N⋅mTCD := 400N⋅mTDB := 300N⋅mSolution:Equations of equilibrium:+ TA − TC = 0TC := TATC = 250N⋅m Ans+ TA − TCD + TD = 0TD := TCD − TATD = 150N⋅m Ans 4. Problem 1-3Determine the resultant internal torque acting on the cross sections through points B and C.Given: TD := 500N⋅mTBC := 350N⋅mTAB := 600N⋅mSolution:Equations of equilibrium:Σ Mx = 0; TB + TBC − TD = 0TB := −TBC + TDTB = 150N⋅m AnsΣ Mx = 0; TC − TD = 0TC := TDTC = 500N⋅m Ans 5. Problem 1-4A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings actingon the section through point A.Given: P := 80Nθ := 30deg φ := 45dega := 0.3m b := 0.1mSolution:Equations of equilibrium:+ ΣFx'=0; NA − P⋅cos(φ − θ) = 0NA := P⋅cos(φ − θ)NA = 77.27N Ans+ ΣFy'=0; VA − P⋅ sin(φ − θ) = 0VA := P⋅ sin(φ − θ)VA = 20.71N Ans+ ΣΜA=0; MA + P⋅cos(φ)⋅a cos(θ) − P⋅ sin(φ)⋅ (b + a sin(θ)) = 0MA := −P⋅cos(φ)⋅a cos(θ) + P⋅ sin(φ)⋅ (b + a sin(θ))MA = −0.555N⋅m AnsNote: Negative sign indicates that MA acts in the opposite direction to that shown on FBD. 6. Problem 1-5Determine the resultant internal loadings acting on the cross section through point D of member AB.Given: ME := 70N⋅ma := 0.05m b := 0.3mSolution:Segment AB: Support Reactions+ ΣΜA=0; −ME − By⋅ (2⋅a + b) = 0By−ME2a + b:= By = −175N150200⎛⎜⎝⎞⎠At B: Bx := By ⋅ Bx = −131.25 N+Segment DB: NB := −Bx VB := −By+ ΣFx=0; ND + NB = 0ND := −NB ND = −131.25 N Ans+ ΣFy=0; VD + VB = 0VD := −VB VD = −175N Ans+ ΣΜD=0; −MD − ME − By⋅ (a + b) = 0MD := −ME − By⋅ (a + b)MD = −8.75 N⋅m Ans 7. Problem 1-6The beam AB is pin supported at A and supported by a cable BC. Determine the resultant internalloadings acting on the cross section at point D.Given: P := 5000Na := 0.8m b := 1.2m c := 0.6m d := 1.6me := 0.6mSolution:θ atanbd⎛⎜⎝⎞⎠:= θ = 36.87 degφ atana + bd⎛⎜⎝⎞⎠:= − θ φ = 14.47 degMember AB:+ ΣΜA=0; FBC⋅ sin(φ)⋅ (a + b) − P⋅ (b) = 0FBCP⋅ (b)sin(φ)⋅ (a + b):=FBC = 12.01 kNSegment BD:+ ΣFx=0; −ND − FBC⋅cos(φ) − P⋅cos(θ) = 0ND := −FBC⋅cos(φ) − P⋅cos(θ)ND = −15.63 kN Ans+ ΣFy=0; VD + FBC⋅ sin(φ) − P⋅ sin(θ) = 0VD := −FBC⋅ sin(φ) + P⋅ sin(θ)VD = 0 kN Ans+ ΣΜD=0; (FBC⋅ sin(φ) − P⋅ sin(θ)) d − csin(θ) ⋅ − MD = 0MD (FBC⋅ sin(φ) − P⋅ sin(θ)) d − csin(θ) := ⋅MD = 0 kN⋅m AnsNote: Member AB is the two-force member. Therefore the shear force and moment are zero. 8. Problem 1-7Solve Prob. 1-6 for the resultant internal loadings acting at point E.Given: P := 5000Na := 0.8m b := 1.2m c := 0.6m d := 1.6me := 0.6mSolution:θ atanbd⎛⎜⎝⎞⎠:= θ = 36.87 degφ atana + bd⎛⎜⎝⎞⎠:= − θ φ = 14.47 degMember AB:+ ΣΜA=0; FBC⋅ sin(φ)⋅ (a + b) − P⋅ (b) = 0FBCP⋅ (b)sin(φ)⋅ (a + b):=FBC = 12.01 kNSegment BE:+ ΣFx=0; −NE − FBC⋅cos(φ) − P⋅cos(θ) = 0NE := −FBC⋅cos(φ) − P⋅cos(θ)NE = −15.63 kN Ans+ ΣFy=0; VE + FBC⋅ sin(φ) − P⋅ sin(θ) = 0VE := −FBC⋅ sin(φ) + P⋅ sin(θ)VE = 0 kN Ans+ ΣΜE=0; (FBC⋅ sin(φ) − P⋅ sin(θ))⋅e − ME = 0ME := (FBC⋅ sin(φ) − P⋅ sin(θ))⋅eME = 0 kN⋅m AnsNote: Member AB is the two-force member. Therefore the shear force and moment are zero. 9. Problem 1-8The boom DF of the jib crane and the column DE have a uniform weight of 750 N/m. If the hoist andload weigh 1500 N, determine the resultant internal loadings in the crane on cross sections throughpoints A, B, and C.Given: P := 1500N w 750Nm:=a := 2.1m b := 1.5mc := 0.6m d := 2.4m e := 0.9mSolution:Equations of Equilibrium: For point A+ ΣFx=0; NA := 0 Ans+ΣF y=0;VA − w⋅e − P = 0 VA := w⋅e + P VA = 2.17 kN Ans+ ΣΜA=0; −MA − (w⋅e)⋅ (0.5⋅e) − P⋅ (e) = 0MA := −(w⋅e)⋅ (0.5⋅e) − P⋅ (e) MA = −1.654 kN⋅m AnsNote: Negative sign indicates that MA acts in the opposite direction to that shown on FBD.Equations of Equilibrium: For point B+ ΣFx=0; NB := 0 Ans+ΣF y=0;VB − w⋅ (d + e) − P = 0 VB := w⋅ (d + e) + P VB = 3.98 kN Ans+ ΣΜB=0; −MB − [w⋅ (d + e)]⋅ [0.5⋅ (d + e)] − P⋅ (d + e) = 0MB := −[w⋅ (d + e)]⋅ [0.5⋅ (d + e)] − P⋅ (d + e)MB = −9.034 kN⋅m AnsNote: Negative sign indicates that MB acts in the opposite direction to that shown on FBD.Equations of Equilibrium: For point C+ ΣFx=0; VC := 0 Ans+ΣF y=0;−NC − w⋅ (b + c + d + e) − P = 0 NC := −w⋅ (b + c + d + e) − PNC = −5.55 kN Ans+ ΣΜB=0; −MC − [w⋅ (c + d + e)]⋅ [0.5⋅ (c + d + e)] − P⋅ (c + d + e) = 0MC := −[w⋅ (c + d + e)]⋅ [0.5⋅ (c + d + e)] − P⋅ (c + d + e)MC = −11.554 kN⋅m AnsNote: Negative sign indicates that NC and MC acts in the opposite direction to that shownon FBD. 10. Problem 1-9The force F = 400 N acts on the gear tooth. Determine the resultant internal loadings on the root of thetooth, i.e., at the centroid point A of section a-a.Given: P := 400Nθ := 30deg φ := 45dega := 4mm b := 5.75mmSolution: α := φ − θEquations of equilibrium: For section a -a+ ΣFx'=0; VA − P⋅cos(α) = 0VA := P⋅cos(α)VA = 386.37 N Ans+ ΣFy'=0; NA − sin(α) = 0NA := P⋅ sin(α)NA = 103.53 N Ans+ ΣΜA=0; −MA − P⋅ sin(α)⋅a + P⋅cos(α)⋅b = 0MA := −P⋅ sin(α)⋅a + P⋅cos(α)⋅bMA = 1.808N⋅m Ans 11. Problem 1-10The beam supports the distributed load shown. Determine the resultant internal loadings on the crosssection through point C. Assume the reactions at the supports A and B are vertical.Given: w1 4.5kNm:= w2 6.0kNm:=a := 1.8m b := 1.8m c := 2.4md := 1.35m e := 1.35mSolution: L1 := a + b + cL2 := d + eSupport Reactions:+ ΣΜA=0; By⋅L1 − (w1⋅L1)(0.5⋅L1) (0.5w2⋅L2) L1L23+⎛⎜⎝⎞⎠− ⋅ = 0By (w1⋅L1)⋅ (0.5) (0.5w2⋅L2) 1L23⋅L1+⎛⎜⎝⎞⎠:= + ⋅ By = 22.82 kN+ ΣFy=0; Ay + By − w1⋅L1 − 0.5w2⋅L2 = 0Ay := −By + w1⋅L1 + 0.5w2⋅L2 Ay = 12.29 kNEquations of Equilibrium: For point C+ ΣFx=0; NC := 0 Ans+ ΣFy=0; Ay w1 a b + ( ) ⋅ − ⎡⎣⎤⎦− VC = 0:= − VC = 3.92 kN AnsVC Ay w1 a b + ( ) ⋅ − ⎡⎣⎤⎦+ ΣΜC=0; MC w1 a b + ( ) ⋅ ⎡⎣+ ⋅0.5⋅ (a + b) − Ay⋅ (a + b) = 0⎤⎦MC w1 a b + ( ) ⋅ ⎡⎣⎤⎦:= − ⋅0.5⋅ (a + b) + Ay⋅ (a + b)MC = 15.07 kN⋅m AnsNote: Negative sign indicates that VC acts in the opposite direction to that shown on FBD. 12. Problem 1-11The beam supports the distributed load shown. Determine the resultant internal loadings on the crosssections through points D and E. Assume the reactions at the supports A and B are vertical.Given: w1 4.5kNm:= w2 6.0kNm:=a := 1.8m b := 1.8m c := 2.4md := 1.35m e := 1.35mSolution: L1 := a + b + cL2 := d + eSupport Reactions:+ ΣΜA=0; By⋅L1 − (w1⋅L1)(0.5⋅L1) (0.5w2⋅L2) L1L23+⎛⎜⎝⎞⎠− ⋅ = 0By (w1⋅L1)⋅ (0.5) (0.5w2⋅L2) 1L23⋅L1+⎛⎜⎝⎞⎠:= + ⋅ By = 22.82 kN+ ΣFy=0; Ay + By − w1⋅L1 − 0.5w2⋅L2 = 0Ay := −By + w1⋅L1 + 0.5w2⋅L2 Ay = 12.29 kNEquations of Equilibrium: For point D+ ΣFx=0; ND := 0 Ans+ ΣFy=0; Ay w1 a ( ) ⋅ − ⎡⎣⎤⎦− VD = 0VD := Ay − w1⋅ (a) VD = 4.18 kN Ans+ ΣΜD=0; MD w1 a ( ) ⋅ ⎡⎣+ ⋅0.5⋅ (a) − Ay⋅ (a) = 0⎤⎦:= − ⋅0.5⋅ (a) + Ay⋅ (a) MD = 14.823 kN⋅m AnsMD w1 a ( ) ⋅ ⎡⎣⎤⎦Equations of Equilibrium: For point E+ ΣFx=0; NE := 0 Ans+ ΣFy=0; VE − 0.5w2⋅ (0.5⋅e) = 0VE := 0.5w2⋅ (0.5⋅e) VE = 2.03 kN Ans+ ΣΜD=0; ME − 0.5w2 0.5 e ⋅ ( ) ⋅ ⎡⎣⎤⎦e3⎛⎜⎝⎞⎠− ⋅ = 0ME 0.5 − w2 0.5 e ⋅ ( ) ⋅ ⎡⎣⎤⎦e3⎛⎜⎝⎞⎠:= ⋅ ME = −0.911 kN⋅m AnsNote: Negative sign indicates that ME acts in the opposite direction to that shown on FBD. 13. Problem 1-12Determine the resultant internal loadings acting on (a) section a-a and (b) section b-b. Each section islocated through the centroid, point C.Given: w 9kNm:= θ := 45dega := 1.2m b := 2.4mSolution: L := a + bSupport Reactions:+ ΣΜA=0; −Bx⋅L⋅ sin(θ) + (w⋅L)(0.5⋅L) = 0Bx(w⋅L)⋅ (0.5⋅L)L⋅ sin(θ):= Bx = 22.91 kN+ ΣFy=0; Ay − w⋅L⋅ sin(θ) = 0Ay := w⋅L⋅ sin(θ) Ay = 22.91 kN+ ΣFx=0; Bx − w⋅L⋅cos(θ) + Ax = 0Ax := w⋅L⋅cos(θ) − Bx Ax = −0 kN(a) Equations of equilibrium: For Section a - a :+ ΣFx=0; NC + Ay⋅ sin(θ) = 0NC := −Ay⋅ sin(θ)+ ΣFy=0; VC + Ay⋅cos(θ) − w⋅a = 0NC = −16.2 kN AnsVC := −Ay⋅cos(θ) + w⋅a VC = −5.4 kN Ans+ ΣΜA=0; −MC − (w⋅a)⋅ (0.5⋅a) + Ay cos(θ)⋅a = 0MC := (w⋅a)⋅ (0.5⋅a) − Ay cos(θ)⋅a MC = −12.96 kN⋅m Ans(b) Equations of equilibrium: For Section b - b :+ ΣFx=0; NC + w⋅a⋅cos(θ) = 0NC := −w⋅a⋅cos(θ) NC = −7.64 kN Ans+ ΣFy=0; VC − w⋅a⋅ sin(θ) + Ay = 0VC := w⋅a⋅ sin(θ) − Ay VC = −15.27 kN Ans+ ΣΜA=0; −MC − (w⋅a)⋅ (0.5⋅a) + Ay cos(θ)⋅a = 0MC := (w⋅a)⋅ (0.5⋅a) − Ay cos(θ)⋅a MC = −12.96 kN⋅m Ans 14. Problem 1-13Determine the resultant internal normal and shear forces in the member at (a) section a-a and (b)section b-b, each of which passes through point A. Take θ = 60 degree. The 650-N load is appliedalong the centroidal axis of the member.Given: P := 650N θ := 60deg(a) Equations of equilibrium: For Section a - a :+ ΣFy=0; P − Na_a = 0Na_a := PNa_a = 650N Ans+ ΣFx=0; Va_a := 0 Ans(b) Equations of equilibrium: For Section b - b :+ ΣFy=0; −Vb_b + P⋅cos(90deg − θ) = 0Vb_b := P⋅cos(90deg − θ)Vb_b = 562.92N Ans+ ΣFx=0; Nb_b − P⋅ sin(90deg − θ) = 0Nb_b := P⋅ sin(90deg − θ)Nb_b = 325N Ans 15. Problem 1-14Determine the resultant internal normal and shear forces in the member at section b-b, each as afunction of θ. Plot these results for 0o ≤ θ ≤ 90o. The 650-N load is applied along the centroidal axisof the member.Given: P := 650N θ := 0Equations of equilibrium: For Section b - b :+ ΣFx0; Nb_b − P⋅cos(θ) = 0Nb_b := P⋅cos(θ) Ans+ ΣFy=0; −Vb_b + P⋅cos(θ) = 0Vb_b := −P⋅cos(θ) Ans 16. Problem 1-15The 4000-N load is being hoisted at a constant speed using the motor M, which has a weight of 450 N.Determine the resultant internal loadings acting on the cross section through point B in the beam. Thebeam has a weight of 600 N/m and is fixed to the wall at A.Given:W1 := 4000N w 600Nm:=W2 := 450Na := 1.2m b := 1.2m c := 0.9md := 0.9m e := 1.2mf := 0.45m r := 0.075mSolution:Tension in rope: TW12:=T = 2.00 kNEquations of Equilibrium: For point B+ ΣFx=0; (−NB − T) = 0NB := −T NB = −2 kN Ans+ ΣFy=0; VB − w⋅ (e) − W1 = 0VB := w⋅ (e) + W1 VB = 4.72 kN Ans+ ΣΜB=0; −MB − [w⋅ (e)]⋅0.5⋅ (e) − W1⋅ (e + r) + T⋅ (f) = 0MB := −[w⋅ (e)]⋅0.5⋅ (e) − W1⋅ (e + r) + T⋅ (f)MB = −4.632 kN⋅m Ans 17. Problem 1-16Determine the resultant internal loadings acting on the cross section through points C and D of thebeam in Prob. 1-15.Given: W1 := 4000Nw 600Nm:=W2 := 450Na := 1.2m b := 1.2m c := 0.9md := 0.9m e := 1.2mf := 0.45m r := 0.075mSolution:Tension in rope: TW12:= T = 2.00 kNEquations of Equilibrium: For point C LC := d + e+ ΣFx=0; (−NC − T) = 0NC := −T NC = −2 kN Ans+ ΣFy=0; VC − w⋅ (LC) − W1 = 0VC := w⋅ (LC) + W1 VC = 5.26 kN Ans+ ΣΜC=0; MC − w LC ( ) ⋅ ⎡⎣− ⋅0.5⋅ (LC) − W1⋅ (LC + r) + T⋅ (f) = 0⎤⎦MC w LC ( ) ⋅ ⎡⎣:= − ⋅0.5⋅ (LC) − W1⋅ (LC + r) + T⋅ (f)⎤⎦MC = −9.123 kN⋅m AnsEquations of Equilibrium: For point D LD := b + c + d + e+ ΣFx=0; ND := 0 ND = 0 kN Ans+ ΣFy=0; VD − w⋅ (LD) − W1 − W2 = 0VD := w⋅ (LD) + W1 + W2 VD = 6.97 kN Ans+ ΣΜC=0; MD − w LD ( ) ⋅ ⎡⎣− ⋅0.5⋅ (LD) − W1⋅ (LD + r) − W2⋅ (b) = 0⎤⎦MD w LD ( ) ⋅ ⎡⎣:= − ⋅0.5⋅ (LD) − W1⋅ (LD + r) − W2⋅ (b)⎤⎦MD = −22.932 kN⋅m Ans 18. Problem 1-17Determine the resultant internal loadings acting on the cross section at point B.Given: w 900kNm:=a := 1m b := 4mSolution: L := a + bEquations of Equilibrium: For point B+ ΣFx=0; NB := 0NB = 0 kN Ans+ ΣFy=0; VB 0.5 wbL⎛⎜⎝⎞⎠⋅ ⎡⎢⎣⎤⎥⎦− ⋅ ⋅ (b) = 0VB 0.5 wbL⎛⎜⎝⎞⎠⋅ ⎡⎢⎣⎤⎥⎦:= ⋅ ⋅ (b)VB = 1440 kN Ans+ ΣΜB=0; −MB 0.5 wbL⎛⎜⎝⎞⎠⋅ ⎡⎢⎣⎤⎥⎦⋅ ⋅ (b)b3⎛⎜⎝⎞⎠− ⋅ = 0MB −0.5 wbL⎛⎜⎝⎞⎠⋅ ⎡⎢⎣⎤⎥⎦⋅ ⋅ (b)b3:= ⋅MB = −1920 kN⋅m Ans 19. Problem 1-18The beam supports the distributed load shown. Determine the resultant internal loadings acting on thecross section through point C. Assume the reactions at the supports A and B are vertical.Given: w1 0.5kNm:= a := 3mw2 1.5kNm:=Solution: L := 3⋅a w := w2 − w1Support Reactions:+ ΣΜA=0; By⋅L − (w1⋅L)(0.5⋅L) [0.5(w)⋅L]2L3⎛⎜⎝⎞⎠− ⋅ = 0By (w1⋅L)⋅ (0.5) [0.5(w)⋅L]23⎛⎜⎝⎞⎠:= + ⋅By = 5.25 kN+ ΣFy=0; Ay + By − w1⋅L − 0.5(w)⋅L = 0Ay := −By + w1⋅L + 0.5(w)⋅LAy = 3.75 kNEquations of Equilibrium: For point C+ ΣFx=0; NC := 0 NC = 0 kN Ans+ ΣFy=0; VC + w1⋅a 0.5 w⎛⎜⎝⎞⎠aL⋅ ⎡⎢⎣⎤⎥⎦+ ⋅ ⋅ (a) − Ay = 0VC −w1⋅a 0.5 waL⎛⎜⎝⎞⎠⋅ ⎡⎢⎣⎤⎥⎦:= − ⋅ ⋅ (a) + AyVC = 1.75 kN Ans+ ΣΜC=0; MC + (w1⋅a)(0.5⋅a) 0.5 waL⎛⎜⎝⎞⎠⋅ ⎡⎢⎣⎤⎥⎦⋅ ⋅ (a)a3⎛⎜⎝⎞⎠+ ⋅ − Ay⋅a = 0MC (−w1⋅a)(0.5⋅a) 0.5 waL⎛⎜⎝⎞⎠⋅ ⎡⎢⎣⎤⎥⎦⋅ ⋅ (a)a3⎛⎜⎝⎞⎠:= − ⋅ + Ay⋅aMC = 8.5 kN⋅m Ans 20. Problem 1-19Determine the resultant internal loadings acting on the cross section through point D in Prob. 1-18.Given: w1 0.5kNm:= a := 3mw2 1.5kNm:=Solution: L := 3⋅a w := w2 − w1Support Reactions:+ ΣΜA=0; By⋅L − (w1⋅L)(0.5⋅L) [0.5(w)⋅L]2L3⎛⎜⎝⎞⎠− ⋅ = 0By (w1⋅L)⋅ (0.5) [0.5(w)⋅L]23⎛⎜⎝⎞⎠:= + ⋅By = 5.25 kN+ ΣFy=0; Ay + By − w1⋅L − 0.5(w)⋅L = 0Ay := −By + w1⋅L + 0.5(w)⋅LAy = 3.75 kNEquations of Equilibrium: For point D+ ΣFx=0; ND := 0 ND = 0 kN Ans+ ΣFy=0; VD + w1⋅ (2a) 0.5 w2aL⎛⎜⎝⎞⎠⋅ ⎡⎢⎣⎤⎥⎦+ ⋅ ⋅ (2a) − Ay = 0⎞⎠⋅ ⎡⎢⎣VD −w1⋅ (2a) 0.5 w2⋅aL⎛⎜⎝⎤⎥⎦:= − ⋅ ⋅ (2a) + AyVD = −1.25 kN Ans+ ΣΜD=0; MD w1 2a ( ) ⋅ ⎡⎣⎤⎦a ( ) + 0.5 w2⋅aL⎛⎜⎝⎞⎠⋅ ⎡⎢⎣⎤⎥⎦⋅ ⋅ (2a)2a3⎛⎜⎝⎞⎠+ ⋅ − Ay⋅ (2a) = 0MD w1 − 2a ( ) ⋅ ⎡⎣⎤⎦(a) 0.5 w2⋅aL⎛⎜⎝⎞⎠⋅ ⎡⎢⎣⎤⎥⎦⋅ ⋅ (2a)2a3⎛⎜⎝⎞⎠:= − ⋅ + Ay⋅ (2a)MD = 9.5 kN⋅m Ans 21. Problem 1-20The wishbone construction of the power pole supports the three lines, each exerting a force of 4 kNon the bracing struts. If the struts are pin connected at A, B, and C, determine the resultant internalloadings at cross sections through points D, E, and F.Given: P := 4kN a := 1.2m b := 1.8mSolution:Support Reactions: FBD (a) and (b).Given+ ΣΜA=0; By⋅ (a) + Bx⋅ (0.5⋅b) − P⋅ (a) = 0 [1]+ ΣΜC=0; Bx⋅ (0.5⋅b) + P⋅ (a) − By⋅ (a) − P⋅ (a) = 0 [2]Solving [1] and [2]: Initial guess: Bx := 1kN By := 2kNBxBy⎛⎜⎜⎝⎞⎠:= Find(Bx , By)BxBy⎛⎜⎜⎝⎞⎠2.672⎛⎜⎝⎞⎠= kNFrom FBD (a):+ ΣFx=0; Bx − Ax = 0Ax := Bx Ax = 2.67 kN+ ΣFy=0; Ay − P − By = 0Ay := P + By Ay = 6 kNFrom FBD (b):+ ΣFx=0; Cx − Bx = 0Cx := Bx Cx = 2.67 kN+ ΣFy=0; Cy + By − P − P = 0Cy := 2P − By Cy = 6 kNEquations of Equilibrium: For point D [FBD (c)].+ ΣFx=0; VD := 0 VD = 0 kN Ans+ ΣFy=0; ND := 0 ND = 0 kN Ans+ ΣΜD=0; MD := 0 MD = 0 kN⋅m AnsFor point E [FBD (d)].+ ΣFx=0; Ax − VE = 0 VE := AxVE = 2.67 kN Ans+ ΣFy=0; NE − Ay = 0 NE := AyNE = 6 kN Ans+ ΣΜE=0; ME − Ax⋅ (0.5⋅b) = 0 ME := Ax⋅ (0.5⋅b) ME = 2.4 kN⋅m AnsFor point F [FBD (e)].+ ΣFx=0; VF + Ax − Cx = 0 VF := −Ax + Cx VF = 0 kN Ans+ ΣFy=0; NF − Ay ⋅ −Cy = 0 NF := Ay + Cy NF = 12 kN Ans+ ΣΜF=0; MF − (Ax + Cx)⋅ (0.5⋅b) = 0 MF := (Ax + Cx)⋅ (0.5⋅b) MF = 4.8 kN⋅m Ans 22. Problem 1-21The drum lifter suspends the 2.5-kN drum. The linkage is pin connected to the plate at A and B. Thegripping action on the drum chime is such that only horizontal and vertical forces are exerted on thedrum at G and H. Determine the resultant internal loadings on the cross section through point I.Given: P := 2.5 kN θ := 60dega := 200mm b := 125mm c := 75mmd := 125mm e := 125mm f := 50mmSolution:Equations of Equilibrium: Memeber Ac and BD aretwo-force members.ΣFy=0; P − 2⋅F sin(θ) = 0 [1]FP:= [2]2⋅ sin(θ)F = 1.443 kNEquations of Equilibrium: For point I.+ ΣFx=0; VI − F⋅cos(θ) = 0 VI := F⋅cos(θ) VI = 0.722 kN Ans+ ΣFy=0; −NI + F⋅ sin(θ) = 0 NI := F⋅ sin(θ) NI = 1.25 kN Ans=0; −MI + F⋅cos(θ)⋅ (a) = 0+ Σ ΜIMI := F⋅cos(θ)⋅ (a) MI = 0.144 kN⋅m Ans 23. Problem 1-22Determine the resultant internal loadings on the cross sections through points K and J on the drum lifterin Prob. 1-21.Given: P := 2.5 kN θ := 60dega := 200mm b := 125mm c := 75mmd := 125mm e := 125mm f := 50mmSolution:Equations of Equilibrium: Memeber Ac and BD aretwo-force members.ΣFy=0; P − 2⋅F sin(θ) = 0 [1]FP:= [2]2⋅ sin(θ)F = 1.443 kNEquations of Equilibrium: For point J.+ ΣFy'=0; VI := 0 VI = 0 kN Ans+ ΣFx'=0; NI + F = 0 NI := −F NI = −1.443 kN Ans+ Σ ΜJ=0; MJ := 0 MJ = 0 kN⋅m AnsNote: Negative sign indicates that NJ acts in the opposite direction to that shown on FBD.Support Reactions: For Member DFH :+ ΣΜH=0; FEF⋅ (c) − F cos(θ)⋅ (a + b + c) + F⋅ sin(θ)⋅ (f) = 0FEF F⋅cos(θ)a + b + cc⎛⎜⎝⎞⎠⋅ F⋅ sin(θ)fc⎛⎜⎝⎞⎠:= − ⋅FEF = 3.016 kNEquations of Equilibrium: For point K.+ ΣFx=0; NK + FEF = 0 NK := FEF NK = 3.016 kN Ans+ ΣFy=0; VK := 0 VK = 0 kN Ans+ ΣΜK=0; MK := 0 MK = 0 kN⋅m Ans 24. Problem 1-23The pipe has a mass of 12 kg/m. If it is fixed to the wall at A, determine the resultant internal loadingsacting on the cross section at B. Neglect the weight of the wrench CD.Given: g 9.81ms2:= ρ 12kgm:= P := 60Na := 0.150m b := 0.400mc := 0.200m d := 0.300mSolution: w := ρ⋅gΣFx=0; NBx := 0N AnsΣFy=0; VBy := 0N AnsΣFz=0; VBz − P + P − w⋅ (b + c) = 0VBz := P − P + w⋅ (b + c)VBz = 70.6 N AnsΣ Μx=0; TBx + P⋅ (b) − P⋅ (b) − (w⋅b)⋅ (0.5⋅b) = 0TBx := −P⋅ (b) + P⋅ (b) + (w⋅b)⋅ (0.5⋅b) TBx = 9.42N⋅m AnsΣ Μy=0; MBy − P⋅ (2a) + (w⋅b)⋅ (c) + (w⋅c)⋅ (0.5⋅c) = 0MBy := P⋅ (2⋅a) − (w⋅b)⋅ (c) − (w⋅c)⋅ (0.5⋅c) MBy = 6.23 N⋅m Ans=0; MBz := 0N⋅m AnsΣ Μz 25. Problem 1-24The main beam AB supports the load on the wing of the airplane. The loads consist of the wheelreaction of 175 kN at C, the 6-kN weight of fuel in the tank of the wing, having a center of gravity atD, and the 2-kN weight of the wing, having a center of gravity at E. If it is fixed to the fuselage at A,determine the resultant internal loadings on the beam at this point. Assume that the wing does nottransfer any of the loads to the fuselage, except through the beam.Given: PC := 175kN PE := 2kN PD := 6kNa := 1.8m b := 1.2m e := 0.3mc := 0.6m d := 0.45mSolution:ΣFx=0; VAx := 0kN AnsΣFy=0; NAy := 0kN AnsΣFz=0; VAz − PD − PE + PC = 0VAz := PD + PE − PCVAz = −167 kN AnsΣ Μx=0;MAx := PD⋅ (a) + PE⋅ (a + b + c) − PC⋅ (a + b) MAx = −507 kN⋅m AnsΣ Μy=0; TAy + PD⋅ (d) − PE⋅ (e) = 0TAy := −PD⋅ (d) + PE⋅ (e) TAy = −2.1 kN⋅m Ans=0; MAz := 0kN⋅m AnsΣ ΜzMAx − PD⋅ (a) − PE⋅ (a + b + c) + PC⋅ (a + b) = 0 26. Problem 1-25Determine the resultant internal loadings acting on the cross section through point B of the signpost.The post is fixed to the ground and a uniform pressure of 50 N/m2 acts perpendicular to the face of thesign.Given: a := 4m d := 2m p 50 Nm2:=b := 6m e := 3mc := 3mSolution: P := p⋅ (c)⋅ (d + e)ΣFx=0; VBx − P = 0VBx := PVBx = 750N AnsΣFy=0; VBy := 0N AnsΣFz=0; NBz := 0N AnsΣ Μx=0; MBx := 0N⋅m AnsΣ Μy=0; MBy − P⋅ (b + 0.5⋅c) = 0MBy := P⋅ (b + 0.5⋅c)MBy = 5625N⋅m AnsΣ Μz=0; TBz − P⋅ [e − 0.5⋅ (d + e)] = 0TBz := P⋅ [e − 0.5⋅ (d + e)]TBz = 375N⋅m Ans 27. Problem 1-26The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to thepulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section throughpoint D. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +ydirection. The journal bearings at A and B exert only y and z components of force on the shaft.Given: P1z := 400N P2y := 200N P3y := 80Na := 0.3m b := 0.4m c := 0.3m d := 0.4mSolution: L := a + b + c + dSupport Reactions:Σ Μz=0; 2P3y⋅ (d) + 2P2y⋅ (c + d) − Ay⋅ (L) = 0Ay 2P3ydL⋅ 2P2yc + dL:= + ⋅Ay = 245.71NΣFy=0; −Ay − By + 2⋅P2y + 2⋅P3y = 0By := −Ay + 2⋅P2y + 2⋅P3y By = 314.29 NΣ Μy=0; 2P1z⋅ (b + c + d) − Az⋅ (L) = 0Az 2P1zb + c + d:= ⋅ Az = 628.57 NLΣFz=0; Bz + Az − 2⋅P1z = 0Bz := −Az + 2⋅P1z Bz = 171.43NEquations of Equilibrium: For point D.ΣFx=0; NDx := 0N AnsΣFy=0; VDy − By + 2⋅P3y = 0VDy := By − 2⋅P3yVDy = 154.3N AnsΣFz=0; VDz + Bz = 0VDz := −BzVDz = −171.4N AnsΣ Μx=0; TDx := 0N⋅m AnsΣ Μy=0; MDy + Bz⋅ (d + 0.5⋅c) = 0:= − MDy = −94.29N⋅m AnsMDy Bz d 0.5 c ⋅ + ( ) ⋅ ⎡⎣⎤⎦Σ Μz=0; MDz + By⋅ (d + 0.5⋅c) − 2⋅P3y⋅ (0.5⋅c) = 0MDz := −By⋅ (d + 0.5⋅c) + 2⋅P3y⋅ (0.5⋅c) MDz = −148.86 N⋅m Ans 28. Problem 1-27The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to thepulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section throughpoint C. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +ydirection. The journal bearings at A and B exert only y and z components of force on the shaft.Given: P1z := 400N P2y := 200N P3y := 80Na := 0.3m b := 0.4m c := 0.3m d := 0.4mSolution: L := a + b + c + dSupport Reactions:Σ Μz=0; 2P3y⋅ (d) + 2P2y⋅ (c + d) − Ay⋅ (L) = 0Ay 2P3ydL⋅ 2P2yc + dL:= + ⋅ Ay = 245.71NΣFy=0; −Ay − By + 2⋅P2y + 2⋅P3y = 0By := −Ay + 2⋅P2y + 2⋅P3y By = 314.29 NΣ Μy=0; 2P1z⋅ (b + c + d) − Az⋅ (L) = 0Az 2P1zb + c + d:= ⋅ Az = 628.57 NLΣFz=0; Bz + Az − 2⋅P1z = 0Bz := −Az + 2⋅P1z Bz = 171.43NEquations of Equilibrium: For point C.ΣFx=0; NCx := 0N AnsΣFy=0; VCy − Ay = 0VCy := AyVCy = 245.7N AnsΣFz=0; VCz + Az − 2P1z = 0VCz := −Az + 2P1zVCz = 171.4N Ans=0; TCx := 0N⋅m AnsΣ ΜxΣ Μy=0; MCy − Az⋅ (a + 0.5⋅b) + 2⋅P1z⋅ (0.5⋅b) = 0MCy := Az⋅ (a + 0.5⋅b) − 2⋅P1z⋅ (0.5⋅b) MCy = 154.29 N⋅m AnsΣ Μz=0; MCz + Ay⋅ (a + 0.5⋅b) = 0MCz := −Ay⋅ (a + 0.5⋅b) MCz = −122.86 N⋅m Ans 29. Problem 1-28Determine the resultant internal loadings acting on the cross section of the frame at points F and G.The contact at E is smooth.Given: a := 1.2m b := 1.5m c := 0.9m P := 400Nd := 0.9m e := 1.2m θ := 30degSolution: L := d2 + e2Member DEF :+ ΣΜD=0; NE⋅ (b) − P⋅ (a + b) = 0NE Pa + bb:= ⋅ NE = 720NMember BCE :+ ΣΜB=0; FACeL⎛⎜⎝⎞⎠⋅ ⋅ (d) − NE⋅ sin(θ)⋅ (c + d) = 0FACLe⋅d⎛⎜⎝⎞⎠NE sin θ ( ) ⋅ c d + ( ) ⋅ ⎡⎣⎤⎦:= ⋅FAC = 900N+ ΣFx=0; Bx FACdL⎛⎜⎝⎞⎠+ ⋅ − NE⋅cos(θ) = 0Bx −FACdL⎛⎜⎝⎞⎠:= ⋅ + NE⋅cos(θ)Bx = 83.54N+ ΣFy=0; −By FACeL⎛⎜⎝⎞⎠+ ⋅ − NE⋅ sin(θ) = 0By FACeL⎛⎜⎝⎞⎠:= ⋅ − NE⋅ sin(θ)By = 360NEquations of Equilibrium: For point F.+ ΣFy'=0; NF := 0 NF = 0N Ans+ ΣFx'=0; VF − P = 0 VF := P VF = 400N Ans+ ΣΜF=0; MF − P⋅ (0.5⋅a) = 0 MF := P⋅ (0.5⋅a) MF = 240N⋅m AnsEquations of Equilibrium: For point G.+ ΣFx=0; Bx − NG = 0 NG := Bx NG = 83.54N Ans+ ΣFy=0; VG − By = 0 VG := By VG = 360N Ans+ ΣΜG=0; −MG + By⋅ (0.5⋅d) = 0 MG := By⋅ (0.5⋅d) MG = 162N⋅m Ans 30. Problem 1-29The bolt shank is subjected to a tension of 400 N. Determine the resultant internal loadings acting onthe cross section at point C.Given:P := 400Nr := 150mmθ := 90degSolution:Equations of Equilibrium: For segment AC.+ ΣFx=0; NC + P = 0 NC := P NC = 400N Ans+ ΣFy=0; VC := 0 VC = 0N Ans+ ΣΜG=0; MC + P⋅ (r) = 0 MC := −P⋅ (r) MC = −60N⋅m Ans 31. Problem 1-30The pipe has a mass of 12 kg/m. If it is fixed to the wall at A, determine the resultant internal loadingsacting on the cross section through B.Given: P := 750N MC := 800N⋅mρ 12kgm:= g 9.81ms2:=a := 1m b := 2m c := 2mSolution:Py45⎛⎜⎝⎞⎠:= ⋅P Pz35:= − ⋅PEquations of Equilibrium: For point B.ΣFx=0; VBx := 0kip AnsΣFy=0; NBy + Py = 0NBy := −Py AnsNBy = −600 N AnsΣFz=0; VBz + Pz − ρ⋅g⋅c − ρ⋅g⋅b = 0VBz := −Pz + ρ⋅g⋅c + ρ⋅g⋅bVBz = 920.9N AnsΣ Μx=0; MBx + Pz⋅ (b) − ρ⋅g⋅c⋅ (b) − ρ⋅g⋅b⋅ (0.5⋅b) = 0MBx := −Pz⋅ (b) + ρ⋅g⋅c⋅ (b) + ρ⋅g⋅b⋅ (0.5⋅b)MBx = 1606.3 N⋅m Ans=0; TBy := 0N⋅m AnsΣ ΜyΣ Μz=0; MBy + Mc = 0MBy := −MCMBy = −800N⋅m Ans 32. Problem 1-31The curved rod has a radius r and is fixed to the wall at B. Determine the resultant internal loadingsacting on the cross section through A which is located at an angle θ from the horizontal.Solution: P := kN θ := degEquations of Equilibrium: For point A.+ ΣFx=0; −NA + P⋅cos(θ) = 0NA := P⋅cos(θ) Ans+ ΣFy=0; VA − P⋅ sin(θ) = 0VA := P⋅ sin(θ) Ans+ ΣΜA=0; MA − P⋅ r⋅ (1 − cos(θ)) = 0MA := P⋅ r⋅ (1 − cos(θ)) Ans 33. Problem 1-32The curved rod AD of radius r has a weight per length of w. If it lies in the horizontal plane, determinethe resultant internal loadings acting on the cross section through point B. Hint: The distance from thecentroid C of segment AB to point O is CO = 0.9745r.Given: θ := 22.5deg r := m a := 0.9745r wkNm2:=Solution:Equations of Equilibrium: For point B.ΣFz=0; VBπ4− ⋅ r⋅w = 0 VB := 0.785⋅w⋅ r AnsΣFx=0; NB := 0 AnsΣ Μx=0; TBπ4− ⋅ r⋅w⋅ (0.09968r) = 0 TB := 0.0783w⋅ r2 AnsΣ Μy=0; MBπ4+ ⋅ r⋅w⋅ (0.37293r) = 0 MB := −0.293w⋅ r2 Ans 34. Problem 1-33A differential element taken from a curved bar is shown in the figure. Show that dN/dθ = V,dV/dθ = -N, dM/dθ = -T, and dT/dθ = M,Solution: 35. Problem 1-34The column is subjected to an axial force of 8 kN, which is applied through the centroid of thecross-sectional area. Determine the average normal stress acting at section a–a. Show this distributionof stress acting over the area’s cross section.Given: P := 8kNb := 150mm d := 140mm t := 10mmSolution:A := 2(b⋅ t) + d⋅ t A = 4400.00mm2σPA:= σ = 1.82MPa Ans 36. Problem 1-35The anchor shackle supports a cable force of 3.0 kN. If the pin has a diameter of 6 mm, determine theaverage shear stress in the pin.Given: P := 3.0kN d := 6mmSolution:+ ΣFy=0; 2⋅V − P = 0V := 0.5PV = 1500NAπ⋅d24:= A = 28.2743mm2τavgVA:= τavg = 53.05MPa Ans 37. Problem 1-36While running the foot of a 75-kg man is momentarily subjected to a force which is 5 times hisweight. Determine the average normal stress developed in the tibia T of his leg at the mid section a-a.The cross section can be assumed circular, having an outer diameter of 45 mm and an inner diameterof 25 mm. Assume the fibula F does not support a load.Given: g 9.81ms2=M := 75kgdo := 45mm di := 25mmSolution:Aπ4:= ⋅ A = 1099.5574mm2do2 − di2 ⎛⎝⎞⎠σ5M⋅gA:= σ = 3.345MPa Ans 38. Problem 1-37The thrust bearing is subjected to the loads shown. Determine the average normal stress developed oncross sections through points B, C, and D. Sketch the results on a differential volume element locatedat each section.Units Used: kPa := 103PaGiven: P := 500N Q := 200NdB := 65mm dC := 140mm dD := 100mmSolution:AB⋅ 24π dB:= AB = 3318.3mm2σBPAB:= σB = 150.7 kPa AnsAC⋅ 24π dC:= AC = 15393.8mm2σCPAC:= σC = 32.5 kPa AnsAD⋅ 24π dD:= AD = 7854.0mm2σDQAD:= σD = 25.5 kPa Ans 39. Problem 1-38The small block has a thickness of 5 mm. If the stress distribution at the support developed by the loadvaries as shown, determine the force F applied to the block, and the distance d to where it is applied.Given: a := 60mm b := 120mm t := 5mmσ1 := 0MPa σ2 := 40MPa σ3 := 60MPaSolution:⌠⎮⌡= dF σ AF := 0.5⋅σ2⋅ (a⋅ t) + σ2⋅ (b⋅ t) + 0.5⋅ (σ3 − σ2)⋅ (b⋅ t)F = 36.00 kN AnsRequire:⌠⎮⌡= dF⋅d x⋅σ Ad0.5 σ2 ⋅ a t ⋅ ( ) ⋅ ⎡⎣⎤⎦2a3a 0.5 b ⋅ + ( ) ⋅ + 0.5 σ3 σ2 − ( ) ⋅ b t ⋅ ( ) ⋅ ⎡⎣⋅ σ2 b t ⋅ ( ) ⋅ ⎡⎣⎤⎦⎤⎦a2⋅b3+ ⎛⎜⎝⎞⎠+ ⋅F:=d = 110mm Ans 40. Problem 1-39The lever is held to the fixed shaft using a tapered pin AB, which has a mean diameter of 6 mm. If acouple is applied to the lever, determine the average shear stress in the pin between the pin and lever.Given: a := 250mm b := 12mmd := 6mm P := 20NSolution:+ ΣΜO=0; V⋅b − P⋅ (2a) = 0V P2ab⎛⎜⎝⎞⎠:= ⋅V = 833.33 NAπ⋅d24:= A = 28.2743mm2τavgVA:= τavg = 29.47MPa Ans 41. Problem 1-40The cinder block has the dimensions shown. If the material fails when the average normal stressreaches 0.840 MPa, determine the largest centrally applied vertical load P it can support.Given: σallow := 0.840MPaao := 150mm ai := 100mmbo := [2⋅ (1 + 2 + 3) + 2]⋅mmbi := [2⋅ (1 + 3)]⋅mmSolution:A := ao⋅bo − ai⋅bi A = 1300mm2Pallow := σallow⋅ (A)Pallow = 1.092 kN Ans 42. Problem 1-41The cinder block has the dimensions shown. If it is subjected to a centrally applied force of P = 4 kN,determine the average normal stress in the material. Show the result acting on a differential volumeelement of the material.Given: P := 4kNao := 150mm ai := 100mmbo := [2⋅ (1 + 2 + 3) + 2]⋅mmbi := [2⋅ (1 + 3)]⋅mmSolution:A := ao⋅bo − ai⋅bi A = 1300mm2σPA:=σ = 3.08MPa Ans 43. Problem 1-42The 250-N lamp is supported by three steel rods connected by a ring at A. Determine which rod issubjected to the greater average normal stress and compute its value. Take θ = 30°. The diameter ofeach rod is given in the figure.Given: W := 250N θ := 30deg φ := 45degdB := 9mm dC := 6mm dD := 7.5mmSolution: Initial guess: FAC := 1N FAD := 1NGiven+ ΣFx=0; FAC⋅cos(θ) − FAD⋅cos(φ) = 0 [1]+ ΣFy=0; FAC⋅ sin(θ) + FAD⋅ sin(φ) − W = 0 [2]Solving [1] and [2]:FACFAD⎛⎜⎜⎝⎞⎠:= Find(FAC, FAD)FACFAD⎛⎜⎜⎝⎞⎠183.01224.14⎛⎜⎝⎞⎠= NRod AB:AAB⋅ 24π dB:= AAB = 63.61725mm2σABWAAB:= σAB = 3.93MPaRod AD :AAD⋅ 24π dD:= AAD = 44.17865mm2σADFADAAD:= σAD = 5.074MPaRod AC:AAC⋅ 24π dC:= AAC = 28.27433mm2σACFACAAC:= σAC = 6.473MPa Ans 44. Problem 1-43Solve Prob. 1-42 for θ = 45°.Given: W := 250N θ := 45deg φ := 45degdB := 9mm dC := 6mm dD := 7.5mmSolution: Initial guess: FAC := 1N FAD := 1NGiven+ ΣFx=0; FAC⋅cos(θ) − FAD⋅cos(φ) = 0 [1]+ ΣFy=0; FAC⋅ sin(θ) + FAD⋅ sin(φ) − W = 0 [2]Solving [1] and [2]:FACFAD⎛⎜⎜⎝⎞⎠:= Find(FAC, FAD)FACFAD⎛⎜⎜⎝⎞⎠176.78176.78⎛⎜⎝⎞⎠= NRod AB:AAB⋅ 24π dB:= AAB = 63.61725mm2σABWAAB:= σAB = 3.93MPaRod AD :AAD⋅ 24π dD:= AAD = 44.17865mm2σADFADAAD:= σAD = 4.001MPaRod AC:AAC⋅ 24π dC:= AAC = 28.27433mm2σACFACAAC:= σAC = 6.252MPa Ans 45. Problem 1-44The 250-N lamp is supported by three steel rods connected by a ring at A. Determine the angle oforientation θ of AC such that the average normal stress in rod AC is twice the average normal stress inrod AD. What is the magnitude of stress in each rod? The diameter of each rod is given in the figure.Given: W := 250N φ := 45degdB := 9mm dC := 6mm dD := 7.5mmSolution:Rod AB: AAB⋅ 24π dB:= AAB = 63.61725mm2Rod AD : AAD⋅ 24π dD:= AAD = 44.17865mm2Rod AC: AAC⋅ 24π dC:= AAC = 28.27433mm2Since σAC = 2σAD ThereforeFACAAC2FADAAD=Initial guess: FAC := 1N FAD := 2N θ := 30degGiven FACAAC2FADAAD= [1]+ ΣFx=0; FAC⋅cos(θ) − FAD⋅cos(φ) = 0 [2]+ ΣFy=0; FAC⋅ sin(θ) + FAD⋅ sin(φ) − W = 0 [3]Solving [1], [2] and [3]:FACFADθ⎛⎜⎜⎜⎝⎞⎟⎠:= Find(FAC, FAD, θ)FACFAD⎛⎜⎜⎝⎞⎠180.38140.92⎛⎜⎝⎞⎠= Nθ = 56.47 degσABWAAB:= σAB = 3.93MPa AnsσADFADAAD:= σAD = 3.19MPa AnsσACFACAAC:= σAC = 6.38MPa Ans 46. Problem 1-45The shaft is subjected to the axial force of 30 kN. If the shaft passes through the 53-mm diameter holein the fixed support A, determine the bearing stress acting on the collar C. Also, what is the averageshear stress acting along the inside surface of the collar where it is fixed connected tothe 52-mm diameter shaft?Given: P := 30kNdhole := 53mm dshaft := 52mmdcollar := 60mm hcollar := 10mmSolution:Bearing Stress:Abπ4dcollar2 − dhole2 ⎛⎝⎞⎠:= ⋅σbPAb:= σb = 48.3MPa AnsAverage Shear Stress:As := π⋅ (dshaft)⋅ (hcollar)τavgPAs:= τavg = 18.4MPa Ans 47. Problem 1-46The two steel members are joined together using a 60° scarf weld. Determine the average normal andaverage shear stress resisted in the plane of the weld.Given:P := 8kN θ := 60degb := 25mm h := 30mmSolution:Equations of Equilibrium:+ ΣFx=0; N − P⋅ sin(θ) = 0N := P⋅ sin(θ) N = 6.928 kN+ ΣFy=0; V − P⋅cos(θ) = 0V := P⋅cos(θ) V = 4 kNAh⋅bsin(θ) :=σNA:= σ = 8MPa AnsτavgVA:= τavg = 4.62MPa Ans 48. Problem 1-47The J hanger is used to support the pipe such that the force on the vertical bolt is 775 N. Determine theaverage normal stress developed in the bolt BC if the bolt has a diameter of 8 mm. Assume A is a pin.Given: P := 775Na := 40mm b := 30mmd := 8mm θ := 20degSolution:Support Reaction:ΣFA=0; P⋅ (a) − FBC⋅cos(θ)⋅ (a + b) = 0FBCP⋅a(a + b)⋅cos(θ):=FBC = 471.28NAverage Normal Stress:ABCπ⋅d24:=σFBCABC:=σ = 9.38MPa Ans 49. Problem 1-48The board is subjected to a tensile force of 425 N. Determine the average normal and average shearstress developed in the wood fibers that are oriented along section a-a at 15° with the axis of the board.Given: P := 425N θ := 15degb := 25mm h := 75mmSolution:Equations of Equilibrium:+ ΣFx=0; V − P⋅cos(θ) = 0V := P⋅cos(θ) V = 410.518N+ ΣFy=0; N − P⋅ sin(θ) = 0N := P⋅ sin(θ) N = 1NAverage Normal Stress:Ah⋅bsin(θ) :=σNA:= σ = 0.0152MPa AnsτavgVA:= τavg = 0.0567MPa Ans 50. Problem 1-49The open square butt joint is used to transmit a force of 250 kN from one plate to the other. Determinethe average normal and average shear stress components that this loading creates on the face of theweld, section AB.Given: P := 250kN θ := 30degb := 150mm h := 50mmSolution:Equations of Equilibrium:+ ΣFx=0; −V + P⋅ sin(θ) = 0V := P⋅ sin(θ) V = 125 kN+ ΣFy=0; N − P⋅cos(θ) = 0N := P⋅cos(θ) N = 216.506 kNAverage Normal and Shear Stress:Ah⋅bsin(2θ) :=σNA:= σ = 25MPa AnsτavgVA:= τavg = 14.434MPa Ans 51. Problem 1-50The specimen failed in a tension test at an angle of 52° when the axial load was 100 kN. If the diameterof the specimen is 12 mm, determine the average normal and average shear stress acting on the area ofthe inclined failure plane. Also, what is the average normal stress acting on the cross section whenfailure occurs?Given: P := 100kNd := 12mm θ := 52degSolution:Equations of Equilibrium:+ ΣFx=0; V − P⋅cos(θ) = 0V := P⋅cos(θ) V = 61.566 kN+ ΣFy=0; N − P⋅ sin(θ) = 0N := P⋅ sin(θ) N = 78.801 kNInclined plane:Aπ4d2sin(θ)⎛⎜⎝⎞⎠:=σNA:= σ = 549.05MPa AnsτavgVA:= τavg = 428.96MPa AnsCross section:Aπd24:=σPA:= σ = 884.19MPa Ansτavg := 0 τavg = 0MPa Ans 52. Problem 1-51A tension specimen having a cross-sectional area A is subjected to an axial force P. Determine themaximum average shear stress in the specimen and indicate the orientation θ of a section on which itoccurs.Solution:Equations of Equilibrium:+ ΣFy=0; V − P⋅cos(θ) = 0V = P⋅cos(θ)Inclined plane:AinclAsin(θ) =τVAincl= τP⋅cos(θ)⋅ sin(θ)= τAP⋅ sin(2θ)2A=dτdθP⋅cos(2θ)A=dτdθ= 0cos(2θ) = 02θ = 90degθ := 45deg AnsτmaxP⋅ sin(90°)= τmax2AP2A= Ans 53. Problem 1-52The joint is subjected to the axial member force of 5 kN. Determine the average normal stress actingon sections AB and BC. Assume the member is smooth and is 50-mm thick.Given: P := 5kN θ := 45deg φ := 60degdAB := 40mm dBC := 50mm t := 50mmSolution: α := 90deg − φ α = 30.00 degAAB := t⋅dABABC := t⋅dBC+ ΣFx=0; NAB⋅cos(α) − P⋅cos(θ) = 0NABP⋅cos(θ)cos(α) :=NAB = 4.082 kN+ ΣFy=0; −NAB⋅ sin(α) + P⋅ sin(θ) − NBC = 0NBC := −NAB⋅ sin(α) + P⋅ sin(θ)NBC = 1.494 kNσABNABAAB:= σAB = 2.041MPa AnsσBCNBCABC:= σBC = 0.598MPa Ans 54. Problem 1-53The yoke is subjected to the force and couple moment. Determine the average shear stress in the boltacting on the cross sections through A and B. The bolt has a diameter of 6 mm. Hint: The couplemoment is resisted by a set of couple forces developed in the shank of the bolt.Given: P := 2.5kN M := 120N⋅mho := 62mm hi := 50mmd := 6mm θ := 60degSolution:As a force on bolt shank is zero, thenτA := 0 AnsEquations od Equilibrium:ΣFz=0; P 2Fz− = 0Fz := 0.5P Fz = 1.25 kNΣMz=0; M − Fx⋅ (hi) = 0FxMhi:= Fx = 2.4 kNAverage Shear Stress:Aπd24:=:= + 22 FzThe bolt shank subjected to a shear force of VB FxτBVBA:= τB = 95.71MPa Ans 55. Problem 1-54The two members used in the construction of an aircraft fuselage are joined together using a 30°fish-mouth weld. Determine the average normal and average shear stress on the plane of each weld.Assume each inclined plane supports a horizontal force of 2 kN.Given:P := 4.0kNb := 37.5mm hhalf := 25mθ := 30degSolution:Equations of Equilibrium:+ ΣFx=0; −V + 0.5P⋅cos(θ) = 0V := 0.5P⋅cos(θ) V = 1.732 kN+ ΣFy=0; N − 0.5P⋅ sin(θ) = 0N := 0.5P⋅ sin(θ) N = 1 kNAverage Normal and Shear Stress:A(hhalf)⋅bsin(θ) :=σNA:= σ = 533.33 Pa AnsτavgVA:= τavg = 923.76 Pa Ans 56. Problem 1-55The row of staples AB contained in the stapler is glued together so that the maximum shear stress theglue can withstand is τ max = 84 kPa. Determine the minimum force F that must be placed on theplunger in order to shear off a staple from its row and allow it to exit undeformed through the grooveat C. The outer dimensions of the staple are shown in the figure. It has a thickness of 1.25 mmAssume all the other parts are rigid and neglect friction.Given: τmax := 0.084MPaa := 12.5mm b := 7.5mmt := 1.25mmSolution:Average Shear Stress:A := a⋅b − [(a − 2t)⋅ (b − t)]τmaxVA= V := (τmax)⋅AV = 2.63 NFmin := VFmin = 2.63 N Ans 57. Problem 1-56Rods AB and BC have diameters of 4mm and 6 mm, respectively. If the load of 8 kN is applied to thering at B, determine the average normal stress in each rod if θ = 60°.Given: W := 8kN θ := 60degdA := 4mm dC := 6mmSolution:Rod AB: AAB⋅ 24π dA:=Rod BC : ABC⋅ 24π dC:=+ ΣFy=0; FBC⋅ sin(θ) − W = 0FBCWsin(θ) :=FBC = 9.238 kN+ ΣFx=0; FBC⋅cos(θ) − FAB = 0FAB := FBC⋅cos(θ)FAB = 4.619 kNσABFABAAB:= σAB = 367.6MPa AnsσBCFBCABC:= σBC = 326.7MPa Ans 58. Problem 1-57Rods AB and BC have diameters of 4 mm and 6 mm, respectively. If the vertical load of 8 kN isapplied to the ring at B, determine the angle θ of rod BC so that the average normal stress in each rodis equivalent. What is this stress?Given: W := 8kNdA := 4mm dC := 6mmSolution:Rod AB: AAB⋅ 24π dA:=Rod BC : ABC⋅ 24π dC:=+ ΣFy=0; FBC⋅ sin(θ) − W = 0+ ΣFx=0; FBC⋅cos(θ) − FAB = 0Since FAB = σ⋅AABFBC = σ⋅ABCInitial guess: σ := 100MPa θ := 50degGivenσ⋅ABC⋅ sin(θ) − W = 0 [1]σ⋅ABC⋅cos(θ) − σ⋅AAB = 0 [2]Solving [1] and [2]:σθ⎛⎜⎝⎞⎠:= Find(σ , θ)θ = 63.61 deg Ansσ = 315.85MPa Ans 59. Problem 1-58The bars of the truss each have a cross-sectional area of 780 mm2. Determine the average normalstress in each member due to the loading P = 40 kN. State whether the stress is tensile or compressive.Given: P := 40kNa := 0.9m b := 1.2m A := 780mm2Solution: c := a2 + b2 c = 1.5mhbc:= vac:=Joint A:+ ΣFy=0; (v)⋅FAB − P = 0 FABPv:= FAB = 66.667 kN+ ΣFx=0; (h)⋅FAB − FAE = 0 FAE := (h)⋅FAB FAE = 53.333 kNσABFABA:= σAB = 85.47MPa (T) AnsσAEFAEA:= σAE = 68.376MPa (C) AnsJoint E:+ ΣFy=0; FEB − 0.75P = 0 FEB := 0.75P FEB = 30 kN+ ΣFx=0; FED − FAE = 0 FED := FAE FED = 53.333 kNσEBFEBA:= σEB = 38.462MPa (T) AnsσEDFEDA:= σED = 68.376MPa (C) AnsJoint B:+ ΣFy=0; (v)⋅FBD − (v)⋅FAB − FEB = 0 FBD FABFEBv⎛⎜⎝⎞⎠:= + FBD = 116.667 kN+ ΣFx=0; FBC − (h)FAB − (h)FBD = 0 FBC := (h)FAB + (h)FBD FBC = 146.667 kNσBCFBCA:= σBC = 188.034MPa (T) AnsσBDFBDA:= σBD = 149.573MPa (C) Ans 60. Problem 1-59The bars of the truss each have a cross-sectional area of 780 mm2. If the maximum average normalstress in any bar is not to exceed 140 MPa, determine the maximum magnitude P of the loads that canbe applied to the truss.Given: σallow := 140MPaa := 0.9m b := 1.2m A := 780mm2Solution: c := a2 + b2 c = 1.5mhbc:= vac:=For comparison purpose, set P := 1kNJoint A:+ ΣFy=0; (v)⋅FAB − P = 0 FABPv:= FAB = 1.667 kN+ ΣFx=0; (h)⋅FAB − FAE = 0 FAE := (h)⋅FAB FAE = 1.333 kNσABFABA:= σAB = 2.137MPa (T)σAEFAEA:= σAE = 1.709MPa (C)Joint E:+ ΣFy=0; FEB − 0.75P = 0 FEB := 0.75P FEB = 0.75 kN+ ΣFx=0; FED − FAE = 0 FED := FAE FED = 1.333 kNσEBFEBA:= σEB = 0.962MPa (T)σEDFEDA:= σED = 1.709MPa (C)Joint B:+ ΣFy=0; (v)⋅FBD − (v)⋅FAB − FEB = 0 FBD FABFEBv⎛⎜⎝⎞⎠:= + FBD = 2.917 kN+ ΣFx=0; FBC − (h)FAB − (h)FBD = 0 FBC := (h)FAB + (h)FBD FBC = 3.667 kNσBCFBCA:= σBC = 4.701MPa (T)σBDFBDA:= σBD = 3.739MPa (C)Since the cross-sectional areas are the same, the highest stress occurs in the member BC,which has the greatest forceFmax := max(FAB, FAE , FEB, FED, FBD, FBC) Fmax = 3.667 kNPallowPFmax⎛⎜⎝⎞⎠:= ⋅ (σallow⋅A) Pallow = 29.78 kN Ans 61. Problem 1-60The plug is used to close the end of the cylindrical tube that is subjected to an internal pressure of p =650 Pa. Determine the average shear stress which the glue exerts on the sides of the tube needed tohold the cap in place.Given: p := 650Pa a := 25mmdi := 35mm do := 40mmSolution:Ap⋅ 24π di:= As := (π⋅do)(a)P⋅ (a) − FBC⋅cos(θ)⋅ (a + b) = 0P p Ap:= ⋅ ( ) P = 0.625 NAverage Shear Stress:τavgPAs:=τavg = 199.1 Pa Ans 62. Problem 1-61The crimping tool is used to crimp the end of the wire E. If a force of 100 N is applied to the handles,determine the average shear stress in the pin at A. The pin is subjected to double shear and has adiameter of 5 mm. Only a vertical force is exerted on the wire.Given: P := 100Na := 37.5mm b := 50mm c := 25mmd := 125mm dpin := 5mmSolution:From FBD (a):+ ΣFx=0; Bx := 0Bx = 0N+ ΣΜD=0; P⋅ (d) − By⋅ (c) = 0By Pdc:= ⋅By = 500NFrom FBD (b):+ ΣFx=0; Ax := 0Ax = 0N+ ΣΜE=0; Ay⋅ (a) − By⋅ (a + b) = 0Ay Bya + ba:= ⋅Ay = 1166.67NAverage Shear Stress:Apin⋅ 24π dpin:=VA := 0.5⋅ (Ay) VA = 583.333NτavgVAApin:= τavg = 29.709MPa Ans 63. Problem 1-62Solve Prob. 1-61 for pin B. The pin is subjected to double shear and has a diameter of 5 mm.Given: a := 37.5mm b := 50mm c := 25mmd := 125mm dpin := 5mm P := 100NSolution:From FBD (a):+ ΣFx=0; Bx := 0Bx = 0N+ ΣΜD=0; P⋅ (d) − By⋅ (c) = 0By Pdc:= ⋅By = 500NAverage Shear Stress: Pin B is subjected to doule shearApin⋅ 24π dpin:=VB := 0.5⋅ (By) VB = 250NτavgVBApin:= τavg = 12.732MPa Ans 64. Problem 1-63The railcar docklight is supported by the 3-mm-diameter pin at A. If the lamp weighs 20 N, and theextension arm AB has a weight of 8 N/m, determine the average shear stress in the pin needed tosupport the lamp. Hint: The shear force in the pin is caused by the couple moment required forequilibrium at A.Given: w 8Nm:= P := 20Na := 900mm h := 32mmdpin := 3mmSolution:From FBD (a):+ ΣFx=0; Bx := 0+ ΣΜA=0; V⋅ (h) − (w⋅a)⋅ (0.5a) − P⋅ (a) = 0V (w⋅a) 0.5ah⎛⎜⎝⎞⎠⋅ Pah:= + ⋅ V = 663.75 NAverage Shear Stress:Apin⋅ 24π dpin:=τavgVApin:= τavg = 93.901MPa Ans 65. Problem 1-64The two-member frame is subjected to the distributed loading shown. Determine the average normalstress and average shear stress acting at sections a-a and b-b. Member CB has a square cross sectionof 35 mm on each side. Take w = 8 kN/m.Given: w 8kNm:=a := 3m b := 4m A := (0.0352)m2Solution: c := a2 + b2 c = 5mhac:= vbc:=Member AB:ΣMA=0; By⋅ (a) − (w⋅a)⋅ (0.5a) = 0By := 0.5w⋅a By = 12 kN+ ΣFy=0; (v)⋅FAB − By = 0FABByv:= FAB = 15 kNSection a-a:σa_aFABA:= σa_a = 12.24MPa Ansτa_a := 0 τa_a = 0MPa AnsSection b-b:+ ΣFx=0; N − FAB⋅ (h) = 0 N := FAB⋅ (h) N = 9 kN+ ΣFy=0; V − FAB⋅ (v) = 0 V := FAB⋅ (v) V = 12 kNAb_bAh:=σb_bNAb_b:= σb_b = 4.41MPa Ansτb_bVAb_b:= τb_b = 5.88MPa Ans 66. Problem 1-65Member A of the timber step joint for a truss is subjected to a compressive force of 5 kN. Determinethe average normal stress acting in the hanger rod C which has a diameter of 10 mm and in member Bwhich has a thickness of 30 mm.Given: P := 5kN θ := 60deg φ := 30degdrod := 10mm h := 40mm t := 30mmSolution:AB := t⋅hArodπ4:= ⋅ 2drod+ ΣFx=0; P⋅cos(θ) − FB = 0FB := P⋅cos(θ) FB = 2.5 kN+ ΣFy=0; Fc − P⋅ sin(θ) = 0FC := P⋅ sin(θ) FC = 4.33 kNAverage Normal Stress:σBFBAB:= σB = 2.083MPa AnsσCFCArod:= σC = 55.133MPa Ans 67. Problem 1-66Consider the general problem of a bar made from m segments, each having a constant cross-sectionalarea Am and length Lm. If there are n loads on the bar as shown, write a computer program that can beused to determine the average normal stress at any specified location x. Show an application of theprogram using the values L1 = 1.2 m, d1 = 0.6 m, P1 = 2 kN, A1 = 1875 mm2, L2 = 0.6 m, d2 = 1.8 m,P2 = -1.5 kN, A2 = 625 mm2. 68. Problem 1-67The beam is supported by a pin at A and a short link BC. If P = 15 kN, determine the average shearstress developed in the pins at A, B, and C. All pins are in double shear as shown, and each has adiameter of 18 mm.Given: P := 15kNa := 0.5m b := 1m c := 1.5md := 1.5m e := 0.5mθ := 30deg dpin := 18mmSolution: L := a + b + c + d + eSupport Reactions:ΣΜA=0; −By⋅ (L) + P⋅ (L − a) + 4P⋅ (c + d + e) + 4P⋅ (d + e) + 2P⋅ (e) = 0By PL − aL⋅ 4⋅Pc + d + e+ ⋅ 4⋅PLd + eL+ ⋅ 2⋅PeL:= + ⋅By = 82.5 kN+ ΣFy=0; −By + P + 4⋅P + 4⋅P + 2⋅P − Ay = 0Ay := −By + P + 4⋅P + 4P + 2⋅PAy = 82.5 kNFBCBysin(θ) := FBC = 165 kNAx := FBC⋅cos(θ) Ax = 142.89 kNAverage Shear Stress:Apin⋅ 24π dpin:=For Pins B and C:τB_and_C0.5FBCApin:= τB_and_C = 324.2MPa AnsFor Pin A::= + 2 FA = 165 kNFA Ax2 AyτA0.5FAApin:= τA = 324.2MPa Ans 69. Problem 1-68The beam is supported by a pin at A and a short link BC. Determine the maximum magnitude P of theloads the beam will support if the average shear stress in each pin is not to exceed 80 MPa. All pins arein double shear as shown, and each has a diameter of 18 mm.Given: τallow := 80MPaa := 0.5m b := 1m c := 1.5md := 1.5m e := 0.5mθ := 30deg dpin := 18mmSolution: L := a + b + c + d + eFor comparison purpose, set P := 1kNSupport Reactions:ΣΜA=0; −By⋅ (L) + P⋅ (L − a) + 4P⋅ (c + d + e) + 4P⋅ (d + e) + 2P⋅ (e) = 0By PL − aL⋅ 4⋅Pc + d + e+ ⋅ 4⋅PLd + eL+ ⋅ 2⋅PeL:= + ⋅ By = 5.5 kN+ ΣFy=0; −By + P + 4⋅P + 4⋅P + 2⋅P − Ay = 0Ay := −By + P + 4⋅P + 4P + 2⋅P Ay = 5.5 kNFBCBysin(θ) := FBC = 11 kNAx := FBC⋅cos(θ) Ax = 9.53 kNFA := Ax2 + Ay2 FA = 11 kNRequire:Fmax := max(FBC, FA) Fmax = 11 kNApin⋅ 24π dpin:=PallowPFmax⎛⎜⎝⎞⎠τallow 2Apin ( ) ⋅ ⎡⎣⎤⎦:= ⋅ Pallow = 3.70 kN Ans 70. Problem 1-69The frame is subjected to the load of 1 kN. Determine the average shear stress in the bolt at A as afunction of the bar angle θ. Plot this function, 0 ≤ θ ≤ 90o, and indicate the values of θ for which thisstress is a minimum. The bolt has a diameter of 6 mm and is subjected to single shear.Given: P := 1kN dbolt := 6mma := 0.6m b := 0.45m c := 0.15mSolution:Support Reactions:ΣΜC=0; FAB⋅cos(θ)⋅ (c) + FAB⋅ sin(θ)⋅ (a) − P⋅ (a + b) = 0FABP⋅ (a + b)cos(θ)⋅ (c) + sin(θ)⋅ (a)=Average Shear Stress: Pin B is subjected to doule shearτFABAbolt= Abolt⋅ 24π dbolt:=τ4P⋅ (a + b)2 ⋅ ⎛⎝⎡⎣cos(θ)⋅ (c) + sin(θ)⋅ (a)⎤⎦ π dbolt⎞⎠⋅=dτdθ4P⋅ (a + b)π ⋅ dbolt2sin(θ)⋅ (c) − cos(θ)⋅ (a)⎡⎣cos(θ)⋅ (c) + sin(θ)⋅ (a)⎤⎦2= ⋅dτdθ= 0 sin(θ)⋅ (c) − cos(θ)⋅ (a) = 0 tan(θ) ac=θ atanac⎛⎜⎝⎞⎠:=θ = 75.96 deg Ans 71. Problem 1-70The jib crane is pinned at A and supports a chain hoist that can travel along the bottom flange of thebeam, 1ft ≤ x ≤ 12ft. If the hoist is rated to support a maximum of 7.5 kN, determine the maximumaverage normal stress in the 18-mm-diameter tie rod BC and the maximum average shear stress in the16-mm-diameter pin at B.Given: P := 7.5kN xmax := 3.6ma := 3m θ := 30degdrod := 18mm dpin := 16mmSolution:Support Reactions:ΣΜC=0; FBC⋅ sin(θ)⋅ (a) − P⋅ (x) = 0FBCP⋅ (x)sin(θ)⋅ (a)=Maximum FBC occurs when x= xmax . Therefore,FBCP⋅ (xmax)sin(θ)⋅ (a):= FBC = 18.00 kNArod⋅ 24π drod:= Apin⋅ 24π dpin:=τpin0.5⋅FBCApin:= τpin = 44.762MPa AnsσrodFBCArod:= σrod = 70.736MPa Ans 72. Problem 1-71The bar has a cross-sectional area A and is subjected to the axial load P. Determine the average normaland average shear stresses acting over the shaded section, which is oriented at θ from the horizontal.Plot the variation of these stresses as a function of θ (0o ≤ θ ≤ 90o).Solution:Equations of Equilibrium:+ ΣFx=0; V − P⋅cos(θ) = 0V = P⋅cos(θ)+ ΣFy=0; N − P⋅ sin(θ) = 0N = P⋅ sin(θ)Inclined plane:AθAsin(θ) =σNA= σPA= ⋅ sin(θ)2 AnsτavgVA= τavgP2A= ⋅ sin(2θ) Ans 73. Problem 1-72The boom has a uniform weight of 3 kN and is hoisted into position using the cable BC. If the cable hasa diameter of 15 mm, plot the average normal stress in the cable as a function of the boom position θ for0o ≤ θ ≤ 90o.Given: W := 3kNa := 1mdo := 15mmSolution: Angle B: φB = 0.5(90deg + θ)φB = 45deg + 0.5θSupport Reactions:ΣΜA=0; FBC⋅ sin(φB)⋅ (a) − W⋅ (0.5a) cos(θ) = 0FBC0.5W⋅cos(θ)sin(45deg + 0.5θ) =Average Normal Stress:σBCFABABC= ABC⋅ 24π do:=σBC2Wπ do⋅ 2⎛⎜⎝⎞⎠cos(θ)sin(45deg + 0.5θ) = ⋅ Ans 74. Problem 1-73The bar has a cross-sectional area of 400 (10-6) m2. If it is subjected to a uniform axial distributedloading along its length and to two concentrated loads as shown, determine the average normal stress inthe bar as a function of for 0 < x ≤ 0.5m.Given: P1 := 3kN P2 := 6kN:= A := 400⋅ (10− 6)m2a := 0.5m b := 0.75mw 8kNmSolution: L := a + b+ ΣFx=0; −N + P1 + P2 + w⋅ (L − x) = 0N = P1 + P2 + w⋅ (L − x)Average Normal Stress:σNA=σP1 + P2 + w⋅ (L − x)= AnsA 75. Problem 1-74The bar has a cross-sectional area of 400 (10-6) m2. If it is subjected to a uniform axial distributedloading along its length and to two concentrated loads as shown, determine the average normal stress inthe bar as a function of for 0.5m < x ≤ 1.25m.Given: P1 := 3kN P2 := 6kN:= A := 400⋅ (10− 6)m2a := 0.5m b := 0.75mw 8kNmSolution: L := a + b+ ΣFx=0; −N + P1 + w⋅ (L − x) = 0N = P1 + w⋅ (L − x)Average Normal Stress:σNA=σP1 + w⋅ (L − x)= AnsA 76. Problem 1-75The column is made of concrete having a density of 2.30 Mg/m3. At its top B it is subjected to an axialcompressive force of 15 kN. Determine the average normal stress in the column as a function of thedistance z measured from its base. Note: The result will be useful only for finding the average normalstress at a section removed from the ends of the column, because of localized deformation at the ends.Given: P := 3kN ρ 2.3(103) kg:= g 9.81m3ms2:=r := 180mm h := 0.75mSolution:A := π⋅ r2 w := ρ⋅g⋅A+ ΣFz=0; N − P − w⋅ (h − z) = 0N = P + w⋅ (h − z)Average Normal Stress:σNA=σP + w⋅ (h − z)= AnsA 77. Problem 1-76The two-member frame is subjected to the distributed loading shown. Determine the largestintensity of the uniform loading that can be applied to the frame without causing either the averagenormal stress or the average shear stress at section b-b to exceed σ = 15 MPa, and τ = 16 MParespectively. Member CB has a square cross-section of 30 mm on each side.Given: σallow := 15MPa τallow := 16MPaa := 4m b := 3m A := (0.0302)m2Solution: c := a2 + b2 c = 5mvbcach:=:= Set wo 1kNm:=Member AB:ΣMA=0; −By⋅ (b) − (wo⋅b)⋅ (0.5b) = 0By := 0.5wo⋅bBy = 1.5 kNSection b-b:By = FBC⋅ (h) FBCByh:=FBC = 1.88 kN+ ΣFx=0; FBC⋅ (h) − Vb_b = 0 Vb_b := FBC⋅ (h)Vb_b = 1.5 kN+ ΣFy=0; −Nb_b + FBC⋅ (v) = 0 Nb_b := FBC⋅ (v)Nb_b = 1.125 kNAb_bAv:=σb_bNb_bAb_b:= σb_b = 0.75MPaτb_bVb_bAb_b:= τb_b = 1MPaAssume failure due to normal stress: wallow woσallowσb_b⎛⎜⎝⎞⎠:= ⋅ wallow 20.00kNm=Assume failure due to shear stress: wallow woτallowτb_b⎛⎜⎝⎞⎠:= ⋅ wallow 16.00kNm= AnsControls ! 78. Problem 1-77The pedestal supports a load P at its center. If the material has a mass density ρ, determine the radialdimension r as a function of z so that the average normal stress in the pedestal remains constant. Thecross section is circular.Solution:Require: σP + w1A= σP + w1 + dwA + dA=P⋅dA + w1⋅dA = A⋅dwdwdAP + w1A=dwdA= σ [1]dA = π (r + dr)2 − πr2dA = 2πr⋅drdw = πr2⋅ (ρ⋅g)⋅dzFrom Eq.[1],πr2⋅ (ρ⋅g)⋅dz2πr⋅dr= σr⋅ (ρ⋅g)⋅dz2dr= σz(1) z⌠⎮⌡ρ⋅g2σ 0drr1r1r⌠⎮⎮⌡= dρ⋅g⋅z2σlnrr1⎛⎜⎝⎞⎠= r r1 eρ⋅g2σ⎛⎜⎝⎞⎠⋅z= ⋅However,σPπ r1⋅ 2=Ansr r1 e⋅ 2⋅ρ⋅g2Pπ r1⎛⎜⎜⎝⎞⎠⋅z= ⋅ 79. Problem 1-78The radius of the pedestal is defined by r = (0.5e-0.08y2) m, where y is given in meters. If the materialhas a density of 2.5 Mg/m3, determine the average normal stress at the support.Given:ro := 0.5m h := 3m g 9.81ms2== ⋅ m ρ 2.5⋅ (103)r ro e− 0.08y2kgm3:=yunit := 1mSolution:dr π e− 0.08y2 ⎛⎝⎞⎠= ⋅dy:= 2 Ao = 0.7854 m2Ao πrodV = π(r2)⋅dy2 e 0.08 − y2 ⎛⎝dV πro⎞⎠2= ⋅dyV3πro y02 e 0.08 − y2 ⎛⎝⎞⎠2⋅ (yunit)⎡⎢⎣⎤⎥⎦⌠⎮⎮⌡:= dV = 1.584m3W := ρ⋅g⋅VW = 38.835 kNσWAo:=σ = 0.04945MPa Ans 80. Problem 1-79The uniform bar, having a cross-sectional area of A and mass per unit length of m, is pinned at itscenter. If it is rotating in the horizontal plane at a constant angular rate of ω, determine the averagenormal stress in the bar as a function of x.Solution:Equation of Motion :+ ΣFx=MaN=Mω r2 ;N mL− x 2⎛⎜⎝⎞⎠⋅ ω x12L− x 2⎛⎜⎝⎞⎠+ ⎡⎢⎣⎤⎥⎦= ⋅N= ⋅ (L2 − 4⋅x2)m⋅ω8Average Normal Stress:σNA=σ= ⋅ (L2 − 4⋅x2) Ansm⋅ω8A 81. Problem 1-80Member B is subjected to a compressive force of 4 kN. If A and B are both made of wood and are10mm. thick, determine to the nearest multiples of 5mm the smallest dimension h of the support so thatthe average shear stress does not exceed τallow = 2.1 MPa.Given: P := 4kNt := 10mm τallow := 2.1MPaa := 300mm b := 125mmSolution: c := a2 + b2 c = 325mmhac:= vbc:=V := P⋅ (v) V = 1.54 kNτallowVt⋅h=hVt⋅ τallow:= h = 73.26mmUse h := 75mm h = 75mm Ans 82. Problem 1-81The joint is fastened together using two bolts. Determine the required diameter of the bolts if the failureshear stress for the bolts is τfail = 350 MPa. Use a factor of safety for shear of F.S. = 2.5.Given: P := 80kN τfail := 350MPa γ := 2.5Solution:τallowτfailγ:= τallow = 140MPaVbolt 0.5P2⎛⎜⎝⎞⎠:= ⋅ Vbolt = 20 kNAboltVboltτallow=π4⎛⎜⎝⎞⎠⋅d2Vboltτallow=d4πVboltτallow⎛⎜⎝⎞⎠:= ⋅d = 13.49mm Ans 83. Problem 1-82The rods AB and CD are made of steel having a failure tensile stress of σfail = 510 MPa. Using a factorof safety of F.S. = 1.75 for tension, determine their smallest diameter so that they can support the loadshown. The beam is assumed to be pin connected at A and C.Given: P1 := 4kN P2 := 6kN P3 := 5kNa := 2m b := 2m c := 3m d := 3mγ := 1.75 σfail := 510MPaSolution: L := a + b + c + dSupport Reactions:ΣΜA=0; FCD⋅ (L) − P1⋅ (a) − P2⋅ (a + b) − P3⋅ (a + b + c) = 0FCD P1aL⎛⎜⎝⎞⎠:= ⋅ P2+ ⋅ FCD = 6.70 kNa + bL+ ⋅ P3a + b + cLΣΜC=0; −FAB⋅ (L) + P1⋅ (b + c + d) + P2⋅ (c + d) + P3⋅ (d) = 0FAB P1b + c + d⋅ P2Lc + dL+ ⋅ P3dL⎛⎜⎝⎞⎠:= + ⋅ FAB = 8.30 kNAverage Normal Stress: Design of rod sizesσallowσfailγ:= σallow = 291.43MPaFor Rod ABAboltFABσallow=π4⎛⎜⎝⎞⎠⋅ 2dABFABσallow=dAB4πFABσallow⎛⎜⎝⎞⎠:= ⋅ dAB = 6.02mm AnsFor Rod CDAboltFCDσallow=π4⎛⎜⎝⎞⎠⋅ 2dCDFCDσallow=dCD4πFCDσallow⎛⎜⎝⎞⎠:= ⋅ dCD = 5.41mm Ans 84. Problem 1-83The lever is attached to the shaft A using a key that has a width d and length of 25 mm. If the shaft isfixed and a vertical force of 200 N is applied perpendicular to the handle, determine the dimension d ifthe allowable shear stress for the key is τallow = 35 MPa.Given: P := 200N τallow := 35MPaL := 500mm a := 20mmb := 25mmSolution:ΣΜA=0; Fa_a⋅ (a) − P⋅ (L) = 0Fa_a PLa:= ⋅Fa_a = 5000NFor the keyAa_aFa_aτallow= b⋅dFa_aτallow=d1bFa_aτallow⎛⎜⎝⎞⎠:= ⋅d = 5.71mm Ans 85. Problem 1-84The fillet weld size a is determined by computing the average shear stress along the shaded plane,which has the smallest cross section. Determine the smallest size a of the two welds if the forceapplied to the plate is P = 100 kN. The allowable shear stress for the weld material is τallow = 100 MPa.Given: P := 100kN τallow := 100MPaL := 100mm θ := 45degSolution:Shear Plane in the Weld: Aweld = L⋅a⋅ sin(θ)Aweld0.5Pτallow=L⋅a⋅ sin(θ)0.5Pτallow=a1L⋅ sin(θ)0.5Pτallow⎛⎜⎝⎞⎠:= ⋅a = 7.071mm Ans 86. Problem 1-85The fillet weld size a = 8 mm. If the joint is assumed to fail by shear on both sides of the block alongthe shaded plane, which is the smallest cross section, determine the largest force P that can be appliedto the plate. The allowable shear stress for the weld material is τallow = 100 MPa.Given: a := 8mm L := 100mmθ := 45deg τallow := 100MPaSolution:Shear Plane in the Weld: Aweld = L⋅a⋅ sin(θ)P = τallow⋅ (2Aweld)P := τallow⋅⎡⎣2(L⋅a⋅ sin(θ))⎤⎦P = 113.14 kN Ans 87. Problem 1-86The eye bolt is used to support the load of 25 kN. Determine its diameter d to the nearest multiples of5mm and the required thickness h to the nearest multiples of 5mm of the support so that the washerwill not penetrate or shear through it. The allowable normal stress for the bolt is σallow = 150 MPa andthe allowable shear stress for the supporting material is τallow = 35 MPa.Given: P := 25kN dwasher := 25mmσallow := 150MPa τallow := 35MPaSolution:Allowable Normal Stress: Design of bolt sizeAboltPσallow=π4⎛⎜⎝⎞⎠⋅d2Pσallow=d4πPσallow⎛⎜⎝⎞⎠:= ⋅d = 14.567mmUse d := 15mm d = 15mm AnsAllowable Shear Stress: Design of support thicknessAsupportPτallow= π⋅ (dwasher)⋅hPτallow=h1π⋅ (dwasher)Pτallow⎛⎜⎝⎞⎠:= ⋅h = 9.095mmUse d := 10mm d = 10mm Ans 88. Problem 1-87The frame is subjected to the load of 8 kN. Determine the required diameter of the pins at A and B ifthe allowable shear stress for the material is τallow = 42 MPa. Pin A is subjected to double shear,whereas pin B is subjected to single shear.Given: P := 8kN τallow := 42MPaa := 1.5m b := 1.5m c := 1.5m d := 0.6mSolution: θBC := 45degSupport Reactions: From FBD (a),ΣΜD=0; FBC⋅ sin(θBC)⋅ (c) − P⋅ (c + d) = 0FBC Pc + dsin(θBC)⋅ (c):= ⋅FBC = 15.839 kNFrom FBD (b),ΣΜA=0; Dy⋅ (a + b) − P⋅ (c + d) = 0Dy Pc + da + b:= ⋅ Dy = 5.6 kN+ ΣFx=0; Ax − P = 0 Ax := P Ax = 8 kN+ ΣFy=0; Dy − Ay = 0 Ay := Dy:= + 2 FA = 9.77 kNFA Ax2 AyApin0.5FAτallow=π4⎛⎜⎝⎞⎠⋅d20.5FAτallow=d4π0.5FAτallow⎛⎜⎝⎞⎠:= ⋅ d = 12.166mm AnsFor pin B: Pin A is subjected to single shear, and FB := FBCApinFBτallow=π4⎛⎜⎝⎞⎠⋅d2FBτallow=d4πFBτallow⎛⎜⎝⎞⎠:= ⋅ d = 21.913mm Ans 89. Problem 1-88The two steel wires AB and AC are used to support the load. If both wires have an allowable tensilestress of σallow = 200 MPa, determine the required diameter of each wire if the applied load is P = 5kN.Given: P := 5kN σallow := 200MPaa := 4m b := 3m θ := 60degSolution:c := a2 + b2 hac:= vbc:=At joint A:Initial guess: FAB := 1kN FAC := 2kNGiven+ ΣFx=0; FAC⋅ (h) − FAB⋅ sin(θ) = 0 [1]+ ΣFy=0; FAC⋅ (v) + FAB⋅cos(θ) − P = 0 [2]Solving [1] and [2]:FABFAC⎛⎜⎜⎝⎞⎠:= Find(FAB, FAC)FABFAC⎛⎜⎜⎝⎞⎠4.34964.7086⎛⎜⎝⎞⎠= kNFor wire ABAABFABσallow=π4⎛⎜⎝⎞⎠⋅ 2dABFABσallow=dAB4πFABσallow⎛⎜⎝⎞⎠:= ⋅ dAB = 5.26mm AnsFor wire ACAACFACσallow=π4⎛⎜⎝⎞⎠⋅ 2dACFACσallow=dAC4πFACσallow⎛⎜⎝⎞⎠:= ⋅ dAC = 5.48mm Ans 90. Problem 1-89The two steel wires AB and AC are used to support the load. If both wires have an allowable tensilestress of σallow = 180 MPa, and wire AB has a diameter of 6 mm and AC has a diameter of 4 mm,determine the greatest force P that can be applied to the chain before one of the wires fails.Given: σallow := 180MPaa := 4m b := 3m θ := 60degdAB := 6mm dAC := 4mmSolution:c := a2 + b2 hac:= vbc:=Assume failure of AB: FAB = (AAB)⋅σallowFABπ4⎛⎜⎝⎞⎠σ:= ⋅ dAB2 ⋅ allow FAB = 5.09 kNAt joint A:Initial guess: P1 := 1kN FAC := 2kNGiven+ ΣFx=0; FAC⋅ (h) − FAB⋅ sin(θ) = 0 [1]+ ΣFy=0; FAC⋅ (v) + FAB⋅cos(θ) − P1 = 0 [2]Solving [1] and [2]:P1FAC⎛⎜⎜⎝⎞⎠:= Find(P1 , FAC)P1FAC⎛⎜⎜⎝⎞⎠⎛⎜⎝⎞5.85035.5094= kN⎠Assume failure of AC: FAC = (AAC)⋅σallowFACπ4⎛⎜⎝⎞⎠σ:= ⋅ dAC2 ⋅ allow FAC = 2.26 kNAt joint A:Initial guess: P2 := 1kN FAB := 2kNGiven+ ΣFx=0; FAC⋅ (h) − FAB⋅ sin(θ) = 0 [1]+ ΣFy=0; FAC⋅ (v) + FAB⋅cos(θ) − P2 = 0 [2]Solving [1] and [2]:FABP2⎛⎜⎜⎝⎞⎠:= Find(FAB, P2)FABP2⎛⎜⎜⎝⎞⎠2.08952.4019⎛⎜⎝⎞⎠= kNChosoe the smallest value: P := min(P1 , P2) P = 2.40 kN Ans 91. Problem 1-90The boom is supported by the winch cable that has a diameter of 6 mm and an allowable normal stressof σallow = 168 MPa. Determine the greatest load that can be supported without causing the cable to failwhen θ = 30° and φ = 45°. Neglect the size of the winch.Given: σallow := 168MPa do := 6mmθ := 30deg φ := 45degSolution:For the cable:Tcable = (Acable)⋅σallowTcableπ4⎛⎜⎝⎞⎠σ:= ⋅ do2 ⋅ allowTcable = 4.7501 kNAt joint B:Initial guess: FAB := 1 kN W := 2 kNGiven+ ΣFx=0; −Tcable cos(θ) + FAB⋅cos(φ) = 0 [1]+ ΣFy=0; −W + FAB⋅ sin(φ) − Tcable⋅ sin(θ) = 0 [2]Solving [1] and [2]:FABW⎛⎜⎝⎞⎠:= Find(FAB,W)FABW⎛⎜⎝⎞⎠⎛⎜⎝⎞5.8181.739= kN Ans⎠ 92. Problem 1-91The boom is supported by the winch cable that has an allowable normal stress of σallow = 168 MPa. Ifit is required that it be able to slowly lift 25 kN, from θ = 20° to θ = 50°, determine the smallestdiameter of the cable to the nearest multiples of 5mm. The boom AB has a length of 6 m. Neglect thesize of the winch. Set d = 3.6 m.Given: σallow := 168MPa W := 25 kNd := 3.6m a := 6mSolution:Maximum tension in canle occurs when θ := 20degsin(θ)asin(ψ)d= ψ asinda⎛⎜⎝⎞⎠⎤⎥⎦sin θ ( ) ⋅ ⎡⎢⎣:= ψ = 11.842 degAt joint B:Initial guess: FAB := 1 kN Tcable := 2 kNGiven φ := θ + ψ+ ΣFx=0; −Tcable cos(θ) + FAB⋅cos(φ) = 0 [1]+ ΣFy=0; −W + FAB⋅ sin(φ) − Tcable⋅ sin(θ) = 0 [2]Solving [1] and [2]:FABTcable⎛⎜⎜⎝⎞⎠:= Find(FAB, Tcable)FABTcable⎛⎜⎜⎝⎞⎠114.478103.491⎛⎜⎝⎞⎠= kNFor the cable:AcablePσallow=π4⎛⎜⎝⎞⎠⋅ 2doTcableσallow=do4πTcableσallow⎛⎜⎝⎞⎠:= ⋅do = 28.006mmUse do := 30mm do = 30mm Ans 93. Problem 1-92The frame is subjected to the distributed loading of 2 kN/m. Determine the required diameter of thepins at A and B if the allowable shear stress for the material is τallow = 100 MPa. Both pins aresubjected to double shear.Given: w 2kNm:= τallow := 100MPar := 3mSolution: Member AB is atwo-force memberθ := 45degSupport Reactions:ΣΜA=0; FBC⋅ sin(θ)⋅ (r) − (w⋅ r)⋅ (0.5r) = 0FBC0.5w⋅ rsin(θ) := FBC = 4.243 kN+ ΣFy=0; Ay + FBC⋅ sin(θ) − w⋅ r = 0Ay := −FBC⋅ sin(θ) + w⋅ r Ay = 3 kN+ ΣFx=0; Ax − FBC⋅cos(θ) = 0Ax := FBC⋅cos(θ) Ax = 3 kNAverage Shear Stress: Pin A and pin B are subjected to double shearFA := Ax2 + Ay2 FA = 4.243 kNFB := FBC FB = 4.243 kNSince both subjected to the same shear force V = 0.5 FA and V := 0.5FBApinVτallow=π4⎛⎜⎝⎞⎠⋅ 2dpinVτallow=dpin4πVτallow⎛⎜⎝⎞⎠:= ⋅dpin = 5.20mm Ans 94. Problem 1-93Determine the smallest dimensions of the circular shaft and circular end cap if the load it is required tosupport is P = 150 kN. The allowable tensile stress, bearing stress, and shear stress is (σt)allow = 175MPa, (σb)allow = 275 MPa, and τallow = 115 MPa.Given: P := 150kN σt_allow := 175MPaτallow := 115MPa σb_allow := 275MPad2 := 30mmSolution:Allowable Normal Stress: Design of end cap outer diameterAPσt_allow=π4⎛⎜⎝⎞⎠d12 − d22 ⎛⎝⎞⎠⋅Pσt_allow=d14πPσt_allow⎛⎜⎝⎞⎠:= ⋅ + d22 d1 = 44.62mm AnsAllowable Bearing Stress: Design of circular shaft diameterAPσb_allow=π4⎛⎜⎝⎞⎠⎛⎝d32 ⎞⎠⋅Pσb_allow=d34πPσb_allow⎛⎜⎝⎞⎠:= ⋅ d3 = 26.35mm AnsAllowable Shear Stress: Design of end cap thicknessAPτallow= (π⋅d3)⋅ tPτallow=t1π⋅d3Pτallow⎛⎜⎝⎞⎠:= ⋅ t = 15.75mm Ans 95. Problem 1-94If the allowable bearing stress for the material under the supports at A and B is (σb)allow = 2.8 MPa,determine the size of square bearing plates A' and B' required to support the loading. Take P = 7.5 kN.Dimension the plates to the nearest multiples of 10mm. The reactions at the supports are vertical.Given: σb_allow := 2.8MPa P := 7.5 kNP1 := 10 kN P2 := 10 kNP3 := 15 kN P4 := 10 kNa := 1.5m b := 2.5mSolution: L := 3⋅a + bSupport Reactions:ΣΜA=0; By 3a ( ) P2− ⋅ (a) − P3⋅ (2a) − P4⋅ (3a) − P⋅ (L) = 0By P2a3a⎛⎜⎝⎞⎠⋅ P32a3a⎛⎜⎝⎞⎠+ ⋅ P43a3a⎛⎜⎝⎞⎠+ ⋅ PL3a⎛⎜⎝⎞⎠:= + ⋅ By = 35 kNΣΜB=0; −Ay⋅ (3⋅a) + P1⋅ (3⋅a) + P2⋅ (2⋅a) + P3⋅ (a) − P⋅ (b) = 0Ay P13a3a⎛⎜⎝⎞⎠⋅ P22a3a⎛⎜⎝⎞⎠+ ⋅ P3a3a⎛⎜⎝⎞⎠+ ⋅ Pb3a⎛⎜⎝⎞⎠:= − ⋅ Ay = 17.5 kNFor Plate A:Aplate_AAyσb_allow2 Ay= aAσb_allow=aAAyσb_allow:=aA = 79.057mmUse aA x aA plate: aA = 80mm AnsFor Plate BAplate_BByσb_allow2 By= aBσb_allow=aBByσb_allow:=aB = 111.803mmUse aB x aB plate: aB = 120mm AnsRb = 35kN(b is in subscript) 96. Problem 1-95If the allowable bearing stress for the material under the supports at A and B is (σb)allow = 2.8 MPa,determine the maximum load P that can be applied to the beam. The bearing plates A' and B' havesquare cross sections of 50mm x 50mm and 100mm x 100mm, respectively.Given: σb_allow := 2.8MPaP1 := 10 kN P2 := 10 kNP3 := 15 kN P4 := 10 kNa := 1.5m b := 2.5maA := 50mm aB := 100mmSolution: L := 3⋅a + bSupport Reactions:ΣΜA=0; By 3a ( ) P2− ⋅ (a) − P3⋅ (2a) − P4⋅ (3⋅a) − P⋅ (L) = 0By P2a3a⎛⎜⎝⎞⎠⋅ P3 2a3a⋅ ⎛⎜⎝⎞⎠+ ⋅ P43⋅a3a⎛⎜⎝⎞⎠+ ⋅ PL3a⎛⎜⎝⎞⎠= + ⋅ΣΜB=0; −Ay⋅ (3⋅a) + P1⋅ (3⋅a) + P2⋅ (2⋅a) + P3⋅ (a) − P⋅ (b) = 0Ay P13a3a⎛⎜⎝⎞⎠⋅ P22a3a⎛⎜⎝⎞⎠+ ⋅ P3a3a⎛⎜⎝⎞⎠+ ⋅ Pb3a⎛⎜⎝⎞⎠= − ⋅2 ⎛⎝For Plate A: Ay aA⎞⎠:= ⋅σb_allow⎛⎝aA2 ⎞⎠⋅σb_allow P13a3a⎛⎜⎝⎞⎠⋅ P22a3a⎛⎜⎝⎞⎠+ ⋅ P3a3a⎛⎜⎝⎞⎠+ ⋅ Pb3a⎛⎜⎝⎞⎠= − ⋅P P13ab⎛⎜⎝⎞⎠⋅ P2⎛⎜⎝⎞⎠2ab+ ⋅ P3ab⎛⎜⎝⎞⎠2 ⎛⎝+ ⋅ aA⎞⎠⋅σb_allow3ab⎛⎜⎝⎞⎠:= − ⋅ P = 26.400 kNPcase_1 := P2 ⎛⎝ ⎞⎠For Plate B: By aB:= ⋅σb_allow⎛⎝aB2 ⎞⎠⋅σb_allow P2a3a⎛⎜⎝⎞⎠⋅ P3 2a3a⋅ ⎛⎜⎝⎞⎠+ ⋅ P43⋅a3a⎛⎜⎝⎞⎠+ ⋅ PL3a⎛⎜⎝⎞⎠= + ⋅P −P2aL⎛⎜⎝⎞⎠⋅ P3 2aL⋅ ⎛⎜⎝⎞⎠− ⋅ P43⋅aL⎛⎜⎝⎞⎠2 ⎛⎝− ⋅ aB⎞⎠⋅σb_allow3aL:= + ⋅ P = 3.000 kNPcase_2 := PPallow := min(Pcase_1 , Pcase_2) Pallow = 3 kN Ans 97. Problem 1-96Determine the required cross-sectional area of member BC and the diameter of the pins at A and B ifthe allowable normal stress is σallow = 21 MPa and the allowable shear stress is τallow = 28 MPa.Given: σallow := 21MPa τallow := 28MPaP := 7.5kip θ := 60dega := 0.6m b := 1.2m c := 0.6mSolution: L := a + b + cSupport Reactions:ΣΜA=0; By⋅ (L) − P⋅ (a) − P⋅ (a + b) = 0By PaL⋅ Pa + bL:= + ⋅FBCBysin(θ) := FBC = 38.523 kNBy = 33.362 kNBx := FBC⋅cos(θ) Bx = 19.261 kN+ ΣFy=0; −By + P + P − Ay = 0 Ay := −By + P + P Ay = 33.362 kN+ ΣFx=0; Bx − Ax = 0 Ax := Bx Ax = 19.261 kN:= + 2 FA = 38.523 kNFA Ax2 AyMember BC:ABCFBCσallow:= ABC = 1834.416mm2 AnsPin A:AAFAτallow=π4⎛⎜⎝⎞⎠⋅ 2dAFAτallow=dA4πFAτallow⎛⎜⎝⎞⎠:= ⋅ dA = 41.854mm AnsPin B:AB0.5FBCτallow=π4⎛⎜⎝⎞⎠⋅ 2dB0.5FBCτallow=dB4π0.5FBCτallow⎛⎜⎝⎞⎠:= ⋅ dB = 29.595mm Ans 98. Problem 1-97The assembly consists of three disks A, B, and C that are used to support the load of 140 kN.Determine the smallest diameter d1 of the top disk, the diameter d2 within the support space, and thediameter d3 of the hole in the bottom disk. The allowable bearing stress for the material is (τallow)b =350 MPa and allowable shear stress is τallow = 125 MPa.Given: P := 140kNτallow := 125MPa σb_allow := 350MPahB := 20mm hC := 10mmSolution:Allowable Shear Stress: Assume shear failure dor disk CAPτallow= (π⋅d2)⋅hCPτallow=d21π⋅hCPτallow⎛⎜⎝⎞⎠:= ⋅ d2 = 35.65mm AnsAllowable Bearing Stress: Assume bearing failure dor disk CAPσb_allow=π4⎛⎜⎝⎞⎠d22 − d32 ⎛⎝⎞⎠⋅Pσb_allow=2 4d3 d2πPσb_allow⎛⎜⎝⎞⎠:= − ⋅ d3 = 27.60mm AnsAllowable Bearing Stress: Assume bearing failure dor disk BAPσb_allow=π4⎛⎜⎝⎞⎠⎛⎝d12 ⎞⎠⋅Pσb_allow=d14πPσb_allow⎛⎜⎝⎞⎠:= ⋅ d1 = 22.57mmSince d3 > d1, disk B might fail due to shear.τPA= τP:= τ = 98.73MPa < τallow (O.K.!)π⋅d1⋅hBTherefore d1 = 22.57mm Ans 99. Problem 1-98Strips A and B are to be glued together using the two strips C and D. Determine the required thicknesst of C and D so that all strips will fail simultaneously. The width of strips A and B is 1.5 times that ofstrips C and D.Given: P := 40N t := 30mmbA := 1.5m bB := 1.5mbC := 1m bD := 1mSolution:Average Normal Stress: Requires,σA = σB σB = σC σC = σDN(bA)⋅ t0.5N(bC)⋅ (tC)=tC0.5(bA)⋅ tbC:= tC = 22.5mm Ans 100. Problem 1-99If the allowable bearing stress for the material under the supports at A and B is (σb)allow = 2.8 MPa,determine the size of square bearing plates A' and B' required to support the loading. Dimension theplates to the nearest multiples of 10mm. The reactions at the supports are vertical. Take P = 7.5 kN.Given: σb_allow := 2.8MPa P := 7.5kNw 10kNm:= a := 4.5m b := 2.25mSolution: L := a + bSupport Reactions:ΣΜA=0; By(a) − w⋅ (a)⋅ (0.5⋅a) − P⋅ (L) = 0By w⋅ (0.5⋅a) PLa⎛⎜⎝⎞⎠:= + ⋅ By = 33.75 kNΣΜB=0; −Ay⋅ (a) + w⋅ (a)⋅ (0.5⋅a) − P⋅ (L) = 0Ay w⋅ (0.5⋅a) Pba⎛⎜⎝⎞⎠:= − ⋅ Ay = 18.75 kNAllowable Bearing Stress: Design of bearing platesFor Plate A:AreaAyσb_allow2 Ay= aAσb_allow=aAAyσb_allow:=aA = 81.832mmUse aA x aA plate: aA := 90mm aA = 90mm AnsFor Plate BAreaByσb_allow2 By= aBσb_allow=aBByσb_allow:=aB = 109.789mmUse aB x aB plate: aB := 110mm aB = 110.00mm Ans 101. Problem 1-100If the allowable bearing stress for the material under the supports at A and B is (σb)allow = 2.8 MPa,determine the maximum load P that can be applied to the beam. The bearing plates A' and B' havesquare cross sections of 50mm x 50mm and 100mm x 100mm, respectively.Given: σb_allow := 2.8MPaw 10kNm:= a := 4.5m b := 2.25maA := 50mm aB := 100mmSolution: L := a + bSupport Reactions:ΣΜA=0;By⋅ (a) − w⋅ (a)⋅ (0.5⋅a) − P⋅ (L) = 0P ByaL⎛⎜⎝⎞⎠⋅ waL⎛⎜⎝⎞⎠= − ⋅ ⋅ (0.5⋅a)ΣΜB=0; −Ay⋅ (a) + w⋅ (a)⋅ (0.5⋅a) − P⋅ (b) = 0P −Ayab⎛⎜⎝⎞⎠⋅ wab⎛⎜⎝⎞⎠= + ⋅ ⋅ (0.5⋅a)Allowable Bearing Stress:Assume failure of material occurs under plate A. Ay aA2 ⎛⎝⎞⎠:= ⋅σb_allow2 ⎛⎝σb_allow ⋅ ⎡⎣P aA⎞⎠⎤⎦−ab⎛⎜⎝⎞⎠⋅ (w⋅a)0.5ab:= + ⋅ P = 31 kNPcase_1 := P2 ⎛⎝Assume failure of material occurs under plate B. By aB⎞⎠:= ⋅σb_allowP ByaL⎛⎜⎝⎞⎠⋅ waL⎛⎜⎝⎞⎠:= − ⋅ ⋅ (0.5⋅a) P = 3.67 kNPcase_2 := PPallow := min(Pcase_1 , Pcase_2) Pallow = 3.67 kN Ans 102. Problem 1-101The hanger assembly is used to support a distributed loading of w = 12 kN/m. Determine the averageshear stress in the 10-mm-diameter bolt at A and the average tensile stress in rod AB, which has adiameter of 12 mm. If the yield shear stress for the bolt is τy = 175 MPa, and the yield tensile stress forthe rod is σy = 266 MPa, determine the factor of safety with respect to yielding in each case.Given: τy := 175MPa w 12kNm:=σy := 266MPaa := 1.2m b := 0.6m e := 0.9mdo := 10mm drod := 12mmSolution:c := a2 + e2 hac:= vec:=Support Reactions: L := a + bΣΜC=0; FAB v ( ) ⋅ ⎡⎣⎤⎦⋅ (a) − w⋅ (L)⋅ (0.5⋅L) = 0FAB wLa⋅v⎛⎜⎝⎞⎠:= ⋅ ⋅ (0.5⋅L)FAB = 27 kNFor bolt A: Bolt A is subjected to double shear, and V := 0.5FAB V = 13.5 kNAπ4⎛⎜⎝⎞⎠:= ⋅ 2 τdoVA:= τ = 171.89MPa AnsFSτyτ:= FS = 1.02 AnsFor rod AB: N := FAB N = 27 kNAπ4⎛⎜⎝⎞⎠:= ⋅ 2 σdrodNA:= σ = 238.73MPa AnsFSσyσ:= FS = 1.11 Ans 103. Problem 1-102Determine the intensity w of the maximum distributed load that can be supported by the hangerassembly so that an allowable shear stress of τallow = 95 MPa is not exceeded in the 10-mm-diameterbolts at A and B, and an allowable tensile stress of σallow = 155 MPa is not exceeded in the12-mm-diameter rod AB.Given: τallow := 95MPa σallow := 155MPaa := 1.2m b := 0.6m e := 0.9mdo := 10mm drod := 12mmSolution: c := a2 + e2 hac:= vec:=Support Reactions: L := a + bΣΜC=0; FAB v ( ) ⋅ ⎡⎣⎤⎦⋅ (a) − w⋅ (L)⋅ (0.5⋅L) = 0FAB wLa⋅v⎛⎜⎝⎞⎠= ⋅ ⋅ (0.5⋅L)Assume failure of pin A or B:V = 0.5FAB V = τallow⋅A Aπ4⎛⎜⎝⎞⎠:= ⋅ 2do0.5⋅wLa⋅v⎛⎜⎝⎞⎠⋅ ⋅ (0.5⋅L) τallowπ⋅ 2 4⎡⎢⎣⎛⎜⎝⎞⎠do⎤⎥⎦= ⋅wa⋅v(0.5L)2⋅τallowπ⋅ 2 4⎡⎢⎣⎛⎜⎝⎞⎠do⎤⎥⎦:= ⋅w 6.632kNm= (controls!) AnsAssuming failure of rod AB:N = FAB N = σallow⋅A Aπ4⎛⎜⎝⎞⎠:= ⋅ 2drodwLa⋅v⎛⎜⎝⎞⎠⋅ ⋅ (0.5⋅L) σallowπ⋅ 2 4⎡⎢⎣⎛⎜⎝⎞⎠drod⎤⎥⎦= ⋅wa⋅v0.5L2⋅σallowπ⋅ 2 4⎡⎢⎣⎛⎜⎝⎞⎠drod⎤⎥⎦:= ⋅w 7.791kNm= 104. Problem 1-103The bar is supported by the pin. If the allowable tensile stress for the bar is (σt)allow = 150 MPa, andthe allowable shear stress for the pin is τallow = 85 MPa, determine the diameter of the pin for whichthe load P will be a maximum. What is this maximum load? Assume the hole in the bar has the samediameter d as the pin. Take t =6 mm and w = 50 mm.Given: τallow := 85MPaσt_allow := 150MPat := 6mm w := 50mmSolution: GivenAllowable Normal Stress: The effective cross-sectionalarea Ae for the bar must be considered here by takinginto account the reduction in cross-sectional areaintroduced by the hole. Here, effective area Ae is equalto (w - d) t, and σallow equals to P /Ae .σt_allowP= [1](w − d)⋅ tAllowable Shear Stress: The pin is subjected to doubleshear and therefore the allowable τ equals to 0.5P /Apin,and the area Apin is equal to ( π/4) d2.τallow2π⎛⎜⎝⎞⎠Pd2⎛⎜⎝⎞⎠= ⋅ [2]Solving [1] and [2]: Initial guess: d := 20mm P := 10kNPd⎛⎜⎝⎞⎠:= Find(P , d) P = 31.23 kN Ansd = 15.29mm Ans 105. Problem 1-104The bar is connected to the support using a pin having a diameter of d = 25 mm. If the allowabletensile stress for the bar is (σt)allow = 140 MPa, and the allowable bearing stress between the pin andthe bar is (σb)allow =210 MPA, determine the dimensions w and t such that the gross area of the crosssection is wt = 1250 mm2 and the load P is a maximum. What is this maximum load? Assume the holein the bar has the same diameter as the pin.Given: σt_allow := 140MPa σb_allow := 210MPaA := 1250mm2 d := 25mmSolution: A = w⋅ t GivenAllowable Normal Stress: The effective cross-sectionalarea Ae for the bar must be considered here by takinginto account the reduction in cross-sectional areaintroduced by the hole. Here, effective area Ae is equalto (w - d) t, that is (A- d t) and σallow equals to P /Ae .σt_allowPA − d⋅ t= [1]Allowable Bearing Stress: The projected area Ab is equalto (d t), and σallow equals to P /Ab .σb_allowPd⋅ t⎛⎜⎝⎞⎠= [2]Solving [1] and [2]: Initial guess: t := 0.5in P := 1kipPt⎛⎜⎝⎞⎠:= Find(P , t) P = 105.00 kN Anst = 20.00mm AnsAnd : wAt:= w = 62.50mm Ans 106. Problem 1-105The compound wooden beam is connected together by a bolt at B. Assuming that the connections atA, B, C, and D exert only vertical forces on the beam, determine the required diameter of the bolt at Band the required outer diameter of its washers if the allowable tensile stress for the bolt is (σt)allow =150 MPa. and the allowable bearing stress for the wood is (σb)allow = 28 MPa. Assume that the hole inthe washers has the same diameter as the bolt.Given: P1 := 3kN P2 := 1.5kN P3 := 2kNσt_allow := 150MPa a := 2mσb_allow := 28MPa b := 1.5mSolution:From FBD (a): GivenΣΜD=0; −Cy⋅ (4⋅b) + By⋅ (3b) + P2⋅ (2⋅b) + P3⋅ (b) = 0 [1]From FBD (b):ΣΜA=0; By 2 a ⋅ b + ( ) ⋅ Cy 2 a ⋅ ( ) ⋅ − ⎡⎣⎤⎦− P1⋅ (a) = 0 [2]Solving [1] and [2]: Initial guess: By := 1kN Cy := 2kNByCy⎛⎜⎜⎝⎞⎠:= Find(By , Cy)ByCy⎛⎜⎜⎝⎞⎠4.44.55⎛⎜⎝⎞⎠= kNFor bolt:AraeByσt_allow=π4⎛⎜⎝⎞⎠⎛⎝dB2 ⎞⎠⋅Byσt_allow=dB4πByσt_allow⎛⎜⎝⎞⎠:= ⋅dB = 6.11mm AnsFor washer:AreaByσb_allow=π4⎛⎜⎝⎞⎠dw2 − dB2 ⎞⎠⎛⎝⋅Byσb_allow=2 4dw dBπByσb_allow⎛⎜⎝⎞⎠:= + ⋅dw = 15.41mm Ans 107. Problem 1-106The bar is held in equilibrium by the pin supports at A and B. Note that the support at A has a singleleaf and therefore it involves single shear in the pin, and the support at B has a double leaf and thereforeit involves double shear. The allowable shear stress for both pins is τallow = 150 MPa. If a uniformdistributed load of w = 8 kN/m is placed on the bar, determine its minimum allowable position x fromB. Pins A and B each have a diameter of 8 mm. Neglect any axial force in the bar.Given: τallow := 150MPa do := 8mma := 2mw 8kNm:= b := 2mSolution:ΣΜA=0; By⋅ (a) − w⋅ (b − x)⋅ [a + x + 0.5⋅ (b − x)] = 0 [1]ΣΜB=0; Ay⋅ (a) − w⋅ (b − x)⋅ [x + 0.5⋅ (b − x)] = 0 [2]Assume failure of pin A:AraeAyτallow=π4⎛⎜⎝⎞⎠⎛⎝do2 ⎞⎠⋅Ayτallow=Ayπ4:= ⋅ ⋅ (τallow) Ay = 7.5398 kN⎛⎝do2 ⎞⎠Substitute value of force A into Eq [2],− ( ) ⋅ x1 0.5 b x1 − ( ) ⋅ + ⎡⎣− ⋅ = 0 [2]Given Ay a ( ) ⋅ w b x1⎤⎦Initial guess: x1 := 0.3m x1 := Find(x1) x1 = 0.480 mxcase_1 := x1Assume failure of pin B:Arae0.5Byτallow=π4⎛⎜⎝⎞⎠⎛⎝do2 ⎞⎠⋅0.5Byτallow=:= ⋅ ⋅ (τallow) By = 15.0796 kNBy 2π4⎛⎜⎝⎞⎠2 ⎛⎝⋅ do⎞⎠Substitute value of force A into Eq [1],− ( ) ⋅ a x2 + 0.5 b x2 − ( ) ⋅ + ⎡⎣− ⋅ = 0 [1]Given By a ( ) ⋅ w b x2⎤⎦Initial guess: x2 := 0.3m x2 := Find(x2) x2 = 0.909 mxcase_2 := x2Choose the larger x value: x := max(xcase_1 , xcase_2)x = 0.909 m Ans 108. Problem 1-107The bar is held in equilibrium by the pin supports at A and B. Note that the support at A has a singleleaf and therefore it involves single shear in the pin, and the support at B has a double leaf and thereforeit involves double shear. The allowable shear stress for both pins is τallow = 125 MPa. If x = 1 m,determine the maximum distributed load w the bar will support. Pins A and B each have a diameter of 8mm. Neglect any axial force in the bar.Given: τallow := 125MPa x := 1m do := 8mma := 2m b := 2mSolution:wokNm:=GivenΣΜA=0; Bw⋅ (a) − wo⋅ (b − x)⋅ [a + x + 0.5⋅ (b − x)] = 0 [1]ΣΜB=0; Aw⋅ (a) − wo⋅ (b − x)⋅ [x + 0.5⋅ (b − x)] = 0 [2]Initial guess: Bw := 1kN Aw := 1kNSolving [1] and [2]:BwAw⎛⎜⎜⎝⎞⎠:= Find(Bw,Aw)BwAw⎛⎜⎜⎝⎞⎠1.750.75⎛⎜⎝⎞⎠= kNFor pin A: Ay w1Awwo⎛⎜⎝⎞⎠= ⋅AraeAyτallow=π4⎛⎜⎝⎞⎠⎛⎝do2 ⎞⎠⋅w1τallow⎛⎜⎝⎞⎠Awwo⎛⎜⎝⎞⎠= ⋅w1π4⋅ ⋅ (τallow)⎛⎝do2 ⎞⎠woAw⎛⎜⎝⎞⎠:= ⋅ w1 8.378kNm=For pin B By wBwwo⎛⎜⎝⎞⎠= ⋅Arae0.5Byτallow=⎛⎜⎝⎞π2⎠⎛⎝do2 ⎞⎠⋅w2τallow⎛⎜⎝⎞⎠Bwwo⎛⎜⎝⎞⎠= ⋅w2π2⋅ ⋅ (τallow)⎛⎝do2 ⎞⎠woBw⎛⎜⎝⎞⎠:= ⋅ w2 7.181kNm=The smalleer w controls ! w := min(w1 ,w2)w 7.181kNm= Ans 109. Problem 1-108The bar is held in equilibrium by the pin supports at A and B. Note that the support at A has a singleleaf and therefore it involves single shear in the pin, and the support at B has a double leaf and thereforeit involves double shear. The allowable shear stress for both pins is τallow = 125 MPa. If x = 1 m and w= 12 kN/m, determine the smallest required diameter of pins A and B. Neglect any axial force in thebar.Given: τallow := 125MPa x := 1m do := 8mma := 2m b := 2mSolution:w 12kNm:=GivenΣΜA=0; By⋅ (a) − w⋅ (b − x)⋅ [a + x + 0.5⋅ (b − x)] = 0 [1]ΣΜB=0; Ay⋅ (a) − w⋅ (b − x)⋅ [x + 0.5⋅ (b − x)] = 0 [2]Initial guess: By := 1kN Ay := 1kNSolving [1] and [2]:ByAy⎛⎜⎜⎝⎞⎠:= Find(By ,Ay)ByAy⎛⎜⎜⎝⎞⎠219⎛⎜⎝⎞⎠= kNFor pin A:AraeAyτallow=π4⎛⎜⎝⎞⎠⎛⎝dA2 ⎞⎠⋅Ayτallow⎛⎜⎝⎞⎠=dA4πAyτallow⎛⎜⎝⎞⎠:= ⋅dA = 9.57mm AnsFor pin BArae0.5Byτallow=π2⎛⎜⎝⎞⎠dB2 ⎞⎠⎛⎝⋅Byτallow⎛⎜⎝⎞⎠=dB2πByτallow⎛⎜⎝⎞⎠:= ⋅dB = 10.34mm Ans 110. Problem 1-109The pin is subjected to double shear since it is used to connect the three links together. Due to wear,the load is distributed over the top and bottom of the pin as shown on the free-body diagram.Determine the diameter d of the pin if the allowable shear stress is τallow = 70 MPa and the load P = 40kN. Also, determine the load intensities w1 and w2 .Given: τallow := 70MPa P := 40kNa := 37.5mm b := 25mmSolution:Pin: + ΣFy=0; P − w1(a) = 0w1Pa:=w1 1066.67kNm= AnsLink: + ΣFy=0; P − 2(0.5w2)⋅ (b) = 0w2Pb:=w2 1600.00kNm= AnsShear StressArea0.5Pτallow=π2⎛⎜⎝⎞⎠⋅d2Pτallow=d2πPτallow⎛⎜⎝⎞⎠:= ⋅d = 19.073mm Ans 111. Problem 1-110The pin is subjected to double shear since it is used to connect the three links together. Due to wear,the load is distributed over the top and bottom of the pin as shown on the free-body diagram.Determine the maximum load P the connection can support if the allowable shear stress for thematerial is τallow = 56 MPa and the diameter of the pin is 12.5 mm. Also, determine the load intensitiesw1 and w2 .Given: τallow := 56MPa d := 12.5mma := 37.5mm b := 25mmSolution:Shear StressArea0.5Pτallow=π2⎛⎜⎝⎞⎠⋅d2Pτallow=Pπ2⎛⎜⎝⎞⎠:= ⋅d2⋅ (τallow)P = 13.7445 kN AnsPin:+ ΣFy=0; P − w1(a) = 0w1Pa:=w1 366.52kNm= AnsLink:+ ΣFy=0; P − 2(0.5w2)⋅ (b) = 0w2Pb:=w2 549.78kNm= Ans 112. Problem 1-111The cotter is used to hold the two rods together. Determine the smallest thickness t of the cotter andthe smallest diameter d of the rods. All parts are made of steel for which the failure tensile stress is σfail= 500 MPa and the failure shear stress is τfail = 375 MPa. Use a factor of safety of (F.S.)t = 2.50 intension and (F.S.)s = 1.75 in shear.Given: σfail := 500MPa τfail := 375MPa P := 30kNd2 := 40mm h := 10mmFSt := 2.50 FSs := 1.75Solution:Allowable Normal Stress : Design of rod sizeσallowσfailFSt:= σallow = 200MPaAreaPσallow=π4⎛⎜⎝⎞⎠⋅d2Pσallow⎛⎜⎝⎞⎠=d4πPσallow:= ⋅d = 13.82mm AnsAllowable Shear Stress : Design of cotter sizeτallowτfailFSs:= τallow = 214.29MPaArea0.5Pτallow= h⋅ t0.5Pτallow⎛⎜⎝⎞⎠=t1h0.5Pτallow:= ⋅t = 7mm Ans 113. Problem 1-112The long bolt passes through the 30-mm-thick plate. If the force in the bolt shank is 8 kN, determinethe average normal stress in the shank, the average shear stress along the cylindrical area of the platedefined by the section lines a-a, and the average shear stress in the bolt head along the cylindrical areadefined by the section lines b-b.Given: P := 8kN dshank := 7mmda_a := 18mm db_b := 7mmhhead := 8mm hplate := 30mmSolution:Average Normal Stress:Ashankπ4⎛⎜⎝⎞⎠⎛⎝dshank2 ⎞⎠:=σtPAshank:= σt = 207.9MPa AnsAverage Shear Stresses:Aa_a := π (da_a)hplateτa_avgPAa_a:= τa_avg = 4.72MPa AnsAb_b := π (db_b)hheadτb_avgPAb_b:= τb_avg = 45.47MPa Ans 114. Problem 1-113The bearing pad consists of a 150 mm by 150 mm block of aluminum that supports a compressiveload of 6 kN. Determine the average normal and shear stress acting on the plane through section aa.Show the results on a ifferential volume element located on the plane.Given: P := 6kNd := 150mm θ := 30degSolution: φ := 90deg − θEquations of Equilibrium:+ ΣFx=0; Va_a − P⋅cos(φ) = 0Va_a := P⋅cos(φ) Va_a = 3 kN+ ΣFy=0; Na_a − P⋅ sin(φ) = 0Na_a := P⋅ sin(φ) Na_a = 5.196 kNAt inclined plane:Ad2sin(φ) :=σa_aNa_aA:= σa_a = 0.2MPa Ansτa_aVa_aA:= τa_a = 0.115MPa Ans 115. Problem 1-114Determine the resultant internal loadings acting on the cross sections located through points D and E ofthe frame.Given: w 2.5kNm:= a := 1.2m b := 0.9mc := 1.5m e := 0.45mSolution: d := a2 + b2 d = 1.5mvad:= hbd:=Support Reactions: L := b + c+ ΣΜA=0; By⋅ (b) − (w⋅L)(0.5⋅L) = 0By (w⋅L) 0.5Lb⎛⎜⎝⎞⎠:= ⋅By = 8 kN+ ΣFy=0; Ay − By + w⋅L = 0Ay := By − w⋅LAy = 2 kNBy = FBC⋅ (v) FBCByv:=FBC = 10 kN+ ΣFx=0; FBC⋅ (h) − Ax = 0Ax := FBC⋅ (h)Ax = 6 kNSegment AD:+ ΣFx=0; ND − Ax = 0 ND := Ax ND = 6 kN Ans+ ΣFy=0; Ay + w⋅ (e) + VD = 0 VD := −Ay − w⋅ (e) VD = −3.13 kN Ans+ ΣΜD=0; MD + [w⋅ (e)]⋅ (0.5⋅e) + Ay⋅ (e) = 0MD := −[w⋅ (e)]⋅ (0.5⋅e) − Ay⋅ (e) MD = −1.153 kN⋅m AnsSegment CE:+ ΣFx=0; NE + FBC = 0 ND := −FBC ND = −10 kN Ans+ ΣFy=0; VD := 0 VD = 0 kN Ans+ ΣΜE=0; MD := 0 MD = 0 kN⋅m Ans 116. Problem 1-115The circular punch B exerts a force of 2 kN on the top of the plate A. Determine the average shearstress in the plate due to this loading.Given: P := 2kNdpunch := 4mm hplate := 2mmSolution:Average Shear Stresses:Aa_a := π (dpunch)hplateτa_avgPAa_a:=τa_avg = 79.58MPa Ans 117. Problem 1-116The cable has a specific weight γ (weight/volume) and cross-sectional area A. If the sag s is small, sothat its length is approximately L and its weight can be distributed uniformly along the horizontal axis,determine the average normal stress in the cable at its lowest point C.Solution:Equations of Equilibrium:ΣMA=0; T⋅ sγAL2⎛⎜⎝⎞⎠L4− ⋅ = 0γγ⋅A⋅L8⋅ s=Average Normal Stresses:σTA=σγ⋅L8⋅ s= Ans 118. Problem 1-117The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to holdup the frame. If AB weighs 2.0 kN/m and the column FC has a weight of 3.0 kN/m, determine theresultant internal loadings acting on cross sections located at points D and E. Neglect the thickness ofboth the beam and column in the calculation.Given: wb 2.0kNm:= L := 3.6m d := 1.8mwc 3.0kNm:= H := 4.8m e := 1.2ma := 3.6m b := 3.6m c := 1.2mSolution:Beam AB: Lc := L2 + c2 vbcLc:= hbLLc:=+ ΣΜA=0; By⋅ (L) − wAB⋅ (L)(0.5⋅L) = 0By wb⋅ (L) 0.5LL⎛⎜⎝⎞⎠:= ⋅By 7936.64ms2= lb+ ΣF Ay := −By + wb⋅ (L) Ay = 3.6 kN y=0; −Ay − By + wb⋅ (L) = 0By = FBC⋅ (vb) FBCByvb:= FBC = 11.38 kN+ ΣFx=0; −FBC⋅ (h) + Ax = 0 Ax := FBC⋅ (hb) Ax = 10.8 kNSegment AD:+ ΣFx=0; ND + Ax = 0 ND := −Ax ND = −10.8 kN Ans+ ΣFy=0; −Ay + wb⋅ (d) + VD = 0 VD := Ay − wb⋅ (d) VD = 0 kN Ans+ ΣΜD=0; MD wb d ( ) ⋅ ⎡⎣+ ⋅ (0.5⋅d) − Ay⋅ (d) = 0⎤⎦:= − ⋅ (0.5⋅d) + Ay⋅ (d) MD = 3.24 kN⋅m AnsMD wb d ( ) ⋅ ⎡⎣⎤⎦Member CG: Hb := H2 + b2 vcHHb:= hcbHb:=Column FC:+ ΣΜC=0; Fx⋅ (H) − Ax⋅c = 0 Fx AxcH⎛⎜⎝⎞⎠:= ⋅Fx = 2.7 kN+ ΣFx=0; FBC⋅ (hb) − Ax + Fx − FCG⋅ (hc) = 0+ ΣFy=0; FCGFBC⋅ (hb) − Ax + Fx:= FCG = 4.5 kNhc−Fy + By + wc⋅ (H) + FBC⋅ (vb) + FCG⋅ (vc) = 0Fy := By + wc⋅ (H) + FBC⋅ (vb) + FCG⋅ (vc)Fy = 25.2 kNSegment FE:+ ΣFx=0; VE − Fx = 0 VE := Fx VE = 2.7 kN Ans+ ΣFy=0; NE + wc⋅ (e) − Fy = 0 NE := −wc⋅ (e) + Fy NE = 21.6 kN Ans+ ΣΜE=0; −ME + Fy⋅ (e) = 0 ME := Fy⋅ (e) ME = 30.24 kN⋅m Ans 119. Problem 1-118The 3-Mg concrete pipe is suspended by the three wires. If BD and CD have a diameter of 10 mm andAD has a diameter of 7 mm, determine the average normal stress in each wire.Given: M := 3000kg g 9.81ms2:=h := 2m r := 1m α := 120degdBD := 10mm dCD := 10mm dAD := 7mmSolution: W := M⋅g W = 29.43 kNθ := 0.5α θ = 60 degL := r2 + h2 vhL:=Equations of Equilibrium:Σ Μx=0; 2⋅F⋅ (r⋅cos(θ)) − FAD⋅ (r) = 0FBD = FCD = F=0; FBD⋅ (r⋅ sin(θ)) − FCD⋅ (r⋅ sin(θ)) = 0Σ ΜyFAD = FΣFz=0; 3⋅ [F⋅ (v)] − W = 0FW3v:=F = 10.97 kNAllowable Normal Stress :σBD = σCD = σ1 σ1FArea= σ1F:= σ1 = 139.65MPa Ansπ4⎛⎝dBD2 ⎞⎠⋅σAD = σ2 σ2FArea= σ2F:= σ2 = 285MPa Ansπ4⎛⎝dAD2 ⎞⎠⋅ 120. Problem 1-119The yoke-and-rod connection is subjected to a tensile force of 5 kN. Determine the average normalstress in each rod and the average shear stress in the pin A between the members.Given: P := 5kN dpin := 25mmd30 := 30mm d40 := 40mmSolution:Average Normal Stress:A40π4⎛⎜⎝⎞⎠⎛⎝d402 ⎞⎠:=σ40PA40:= σ40 = 3.979MPa AnsA30π4⎛⎜⎝⎞⎠⎛⎝d302 ⎞⎠:=σ30PA30:= σ30 = 7.074MPa AnsAverage Shear Stresses:Apinπ4⎛⎜⎝⎞⎠⎛⎝dpin2 ⎞⎠:=τavg0.5PApin:= τavg = 5.093MPa Ans 121. Problem 2-1An air-filled rubber ball has a diameter of 150 mm. If the air pressure within it is increased until theball's diameter becomes 175 mm, determine the average normal strain in the rubber.Given: d0 := 150mm d := 175mmSolution:επd − πd0πd0:=ε 0.1667mmmm= Ans 122. Problem 2-2A thin strip of rubber has an unstretched length of 375 mm. If it is stretched around a pipe having anouter diameter of 125 mm, determine the average normal strain in the strip.Given: L0 := 375mmSolution:L := π⋅ (125) mmεπL − πL0πL0:=ε 0.0472mmmm= Ans 123. Problem 2-3The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes theend C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD.Given: a := 3m LCE := 4mb := 4m LBD := 4mΔLCE := 10mmSolution:ΔLBDaa + b⎛⎜⎝⎞⎠:= ⋅ΔLCE ΔLBD = 4.2857mmεCEΔLCELCE:= εCE 0.00250mmmm= AnsεBDΔLBDLBD:= εBD 0.00107mmmm= Ans 124. Problem 2-4The center portion of the rubber balloon has a diameter of d = 100 mm. If the air pressure within itcauses the balloon's diameter to become d = 125 mm, determine the average normal strain in therubber.Given: d0 := 100mm d := 125mmSolution:επd − πd0πd0:=ε 0.2500mmmm= Ans 125. Problem 2-5The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam is displace10 mm downward, determine the normal strain developed in wires CE and BD.Given: a := 3m b := 2m c := 2mLCE := 4m LBD := 3mΔtip := 10mmSolution:ΔLBDaΔLCEa + b=ΔLCEa + bΔ tipa + b + c=ΔLCEa + ba + b + c⎛⎜⎝⎞⎠:= ⋅Δtip ΔLCE = 7.1429mmΔLBDaa + b + c⎛⎜⎝⎞⎠:= ⋅Δtip ΔLBD = 4.2857mmAverage Normal Strain:εCEΔLCELCE:= εCE 0.00179mmmm= AnsεBDΔLBDLBD:= εBD 0.00143mmmm= Ans 126. Problem 2-6The rigid beam is supported by a pin at A and wires BD and CE. If the maximum allowable normalstrain in each wire is εmax = 0.002 mm/mm, determine the maximum vertical displacement of the loadP.Given: a := 3m b := 2m c := 2mLCE := 4m LBD := 3mεallow 0.002mmmm:=Solution:ΔLBDaΔLCEa + b=ΔLCEa + bΔ tipa + b + c=Average Elongation/Vertical Displacement:ΔLBD := LBD⋅ εallow ΔLBD = 6.00mmΔ tipa + b + ca⎛⎜⎝⎞⎠:= ⋅ΔLBDΔtip = 14.00mmΔLCE := LCE⋅ εallow ΔLCE = 8.00mmΔ tipa + b + ca + b⎛⎜⎝⎞⎠:= ⋅ΔLCEΔtip = 11.20mm (Controls !) Ans 127. Problem 2-7The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2mm, determine the normal strain developed in each wire.Given:a := 300mm θ := 30degΔA := 2mmSolution:Consider the triangle CAA':φA := 180deg − θ φA = 150 deg+ 2 2⋅a Δ:= − ⋅ ( A)⋅cos(φA)LCA' a2 ΔALCA' = 301.734mmΔCALCA' − aa:=ΔCA 0.00578mmmm= Ans 128. Problem 2-8Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If aforce is applied to the end D of the member and causes it to rotate by θ = 0.3°, determine the normalstrain in the cable. Originally the cable is unstretched.Given:a := 400mm b := 300mm c := 300mmθ := 0.3degSolution:LAB := a2 + b2 LAB = 500mmConsider the triangle ACB':φC := 90deg + θ φC = 90.3 degLAB' := a2 + b2 − 2⋅a⋅b⋅cos(φC)LAB' = 501.255mmεABLAB' − LABLAB:=εAB 0.00251mmmm= Ans 129. Problem 2-9Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If aforce is applied to the end D of the member and causes a normal strain in the cable of 0.0035 mm/mmdetermine the displacement of point D. Originally the cable is unstretched.Given: a := 400mm b := 300mm c := 300mmεAB 0.0035mmmm:=Solution:LAB := a2 + b2 LAB = 500mmLAB' := LAB⋅ (1 + εAB) LAB' = 501.750mmConsider the triangle ACB':φC = 90deg + θLAB' = a2 + b2 − 2⋅a⋅b⋅cos(φC)φC acos(a2 + b2) LAB'− 22⋅a⋅b⎡⎢⎣⎤⎥⎦:=φC = 90.419 degθ := φC − 90deg θ = 0.41852 degθ = 0.00730 radΔD := (b + c)⋅θ ΔD = 4.383mm Ans 130. Problem 2-10The wire AB is unstretched when θ = 45°. If a vertical load is applied to bar AC, which causes θ =47°, determine the normal strain in the wire.Given:θ := 45deg Δθ := 2degSolution:LAB = L2 + L2 LAB = 2LLCB = (2L)2 + L2 LCB = 5LFrom the triangle CAB:φA := 180deg − θ φA = 135.00 degsin(φB)Lsin(φA)LCB=φB asinL⋅ sin(φA)5L⎛⎜⎝⎞⎠:= φB = 18.435 degFrom the triangle CA'B:φ'B := φB + Δθ φ'B = 20.435 degsin(φ'B)Lsin(180deg − φA')LCB=φA' 180deg asin5⋅L⋅ sin(φ'B)L⎛⎜⎝⎞⎠:= −φA' = 128.674 degφ'C := 180deg − φ'B − φA'φ'C = 30.891 degsin(φ'B)Lsin(φ'C)LA'B= LA'Bsin(φ'C)sin(φ'B) := ⋅L LAB := 2LεABLA'B − LABLAB:= εAB = 0.03977 Ans 131. Problem 2-11If a load applied to bar AC causes point A to be displaced to the left by an amount ΔL, determine thenormal strain in wire AB. Originally, θ = 45°.Given: θ := 45degSolution: εACΔLL=LAB = L2 + L2 LAB = 2LLCB = (2L)2 + L2 LCB = 5LFrom the triangle A'AB:φA := 180deg − θ φA = 135.00 degLA'B ΔL2 LAB+ 2 2⋅ (ΔL) L= − ⋅ ( AB)⋅cos(φA)LA'B ΔL2 = + 2⋅L2 + 2⋅ (ΔL)⋅LεABLA'B − LABLAB=εABΔL2 + 2⋅L2 + 2⋅ (ΔL)⋅L − 2L2L=εAB12ΔLL⎛⎜⎝⎞⎠⋅ ⎡⎢⎣⎤⎥⎦2+ 1ΔLL⎛⎜⎝⎞⎠= + − 1Neglecting the higher-order terms,εAB 1ΔLL+ ⎛⎜⎝⎞⎠0.5= − 1+ ⋅ + ..... ⎡⎢⎣εAB 112ΔLL⎛⎜⎝⎞⎠⎤⎥⎦= − 1 (Binomial expansion)εAB12ΔLL⎛⎜⎝⎞⎠= ⋅ AnsAlternatively,εABLA'B − LABLAB= εABΔL⋅ sin(θ)2L=εAB12ΔLL⎛⎜⎝⎞⎠= ⋅ Ans 132. Problem 2-12The piece of plastic is originally rectangular. Determine the shear strain γxy at corners A and B if theplastic distorts as shown by the dashed lines.Given:a := 400mm b := 300mmΔAx := 3mm ΔAy := 2mmΔCx := 2mm ΔCy := 2mmΔBx := 5mm ΔBy := 4mmSolution:Geometry : For small angles,αΔCxb + ΔCy:= α = 0.00662252 radβΔBy − ΔCya + (ΔBx − ΔCx):= β = 0.00496278 radψΔBx − ΔAxb + (ΔBy − ΔAy):= ψ = 0.00662252 radθΔAya + ΔAx:= θ = 0.00496278 radShear Strain :γxy_B := β + ψ γxy_B = 11.585 × 10− 3 rad Ansγxy_A := −(θ + ψ) γxy_A = −11.585 × 10− 3 rad Ans 133. Problem 2-13The piece of plastic is originally rectangular. Determine the shear strain γxy at corners D and C if theplastic distorts as shown by the dashed lines.Given:a := 400mm b := 300mmΔAx := 3mm ΔAy := 2mmΔCx := 2mm ΔCy := 2mmΔBx := 5mm ΔBy := 4mmSolution:Geometry : For small angles,αΔCxb + ΔCy:= α = 0.00662252 radβΔBy − ΔCya + (ΔBx − ΔCx):= β = 0.00496278 radψΔBx − ΔAxb + (ΔBy − ΔAy):= ψ = 0.00662252 radθΔAya + ΔAx:= θ = 0.00496278 radShear Strain :γxy_D := α + θ γxy_D = 11.585 × 10− 3 rad Ansγxy_C := −(α + β) γxy_C = −11.585 × 10− 3 rad Ans 134. Problem 2-14The piece of plastic is originally rectangular. Determine the average normal strain that occurs along thdiagonals AC and DB.Given:a := 400mm b := 300mmΔAx := 3mm ΔAy := 2mmΔCx := 2mm ΔCy := 2mmΔBx := 5mm ΔBy := 4mmSolution:Geometry :LAC := a2 + b2 LAC = 500mmLDB := a2 + b2 LDB = 500mmLA'C' (a + ΔAx − ΔCx)2 := + (b + ΔCy − ΔAy)2LA'C' = 500.8mmLDB' (a + ΔBx)2 := + (b + ΔBy)2LDB' = 506.4mmAverage Normal Strain :εACLA'C' − LAC:= εAC 1.601 × 10− 3LACmmmm= AnsεBDLDB' − LDBLDB:= εBD 12.800 × 10− 3mmmm= Ans 135. Problem 2-15The guy wire AB of a building frame is originally unstretched. Due to an earthquake, the two columnsof the frame tilt θ = 2°. Determine the approximate normal strain in the wire when the frame is in thiposition. Assume the columns are rigid and rotate about their lower supports.Given:a := 4m b := 3m c := 1m θ := 2degSolution:θθ180⎛⎜⎝⎞⎠= ⋅π θ = 0.03490659 radGeometry : The vertical dosplacement is negligible.ΔAx := c⋅θ ΔAx = 34.907mmΔBx := (b + c)⋅θ ΔBx = 139.626mmLAB := a2 + b2 LAB = 5000mmLA'B' := (a + ΔBx − ΔAx)2 + b2LA'B' = 5084.16mmAverage Normal Strain :LA'B' − LABεABLAB:=εAB 16.833 × 10− 3mmmm= Ans 136. Problem 2-16The corners of the square plate are given the displacements indicated. Determine the shear strain alongthe edges of the plate at A and B.Given: ax := 250mm ay := 250mmΔv := 5mm Δh := 7.5mmSolution:At A :tanθ'A2⎛⎜⎝⎞⎠ax − Δhay + Δv=θ'A 2 atanax − Δhay + Δv⎛⎜⎝⎞⎠:= ⋅ θ'A = 1.52056 radγnt_Aπ2⎛⎜⎝⎞⎠:= − θ'A γnt_A = 0.05024 rad AnsAt B :tanφ'B2⎛⎜⎝⎞⎠ay + Δvax − Δh=φ'B 2 atanay + Δvax − Δh⎛⎜⎝⎞⎠:= ⋅ φ'B = 1.62104 radγnt_B φ'Bπ2:= − γnt_B = 0.05024 rad Ans 137. Problem 2-17The corners of the square plate are given the displacements indicated. Determine the average normalstrains along side AB and diagonals AC and DB.Given: ax := 250mm ay := 250mmΔv := 5mm Δh := 7.5mmSolution:For AB :LAB ax:= + 22 ayLA'B' (ax − Δh)2 := + (ay + Δv)2εABLA'B' − LABLAB:=LAB = 353.55339mmLA'B' = 351.89665mmεAB −4.686 × 10− 3mmmm= AnsFor AC :LAC := 2(ay) LAC = 500mmLA'C' := 2⋅ (ay + Δv) LA'C' = 510mmεACLA'C' − LAC:= εAC 20.000 × 10− 3LACmmmm= AnsFor DB :LDB := 2(ax) LDB = 500mmLD'B' := 2⋅ (ax − Δh) LD'B' = 485mmεDBLD'B' − LDB:= εDB −30.000 × 10− 3LDBmmmm= Ans 138. Problem 2-18The square deforms into the position shown by the dashed lines. Determine the average normal strainalong each diagonal, AB and CD. Side D'B' remains horizontal.Given:a := 50mm b := 50mmΔBx := −3mmΔCx := 8mm ΔCy := 0mmθ'A := 91.5deg LAD' := 53mmSolution:For AB :ΔBy := LAD'⋅cos(θ'A − 90deg) − b ΔBy = 2.9818mmLAB := a2 + b2 LAB = 70.7107mmLAB' (a + ΔBx)2 := + (b + ΔBy)2 LAB' = 70.8243mmεABLAB' − LABLAB:=εAB 1.606 × 10− 3mmmm= AnsFor CD :ΔDy := ΔBy ΔDy = 2.9818mmLCD := a2 + b2 LCD = 70.7107mmLC'D' (a + ΔCx)2 := + (b + ΔDy)2 − 2⋅ (a + ΔCx)⋅ (b + ΔDy)⋅cos(θ'A)LC'D' = 79.5736mmεCDLC'D' − LCDLCD:=εCD 125.340 × 10− 3mmmm= Ans 139. Problem 2-19The square deforms into the position shown by the dashed lines. Determine the shear strain at each ofits corners, A, B, C, and D. Side D'B' remains horizontal.Given: a := 50mm b := 50mmΔBx := −3mmΔCx := 8mm ΔCy := 0mmθ'A := 91.5deg LAD' := 53mmSolution: θ'A = 1.597 radGeometry :ΔBy := LAD'⋅cos(θ'A − 90deg) − b ΔBy = 2.9818mmΔDy := ΔBy ΔDy = 2.9818mmΔDx := −LAD'⋅ sin(θ'A − 90deg) ΔDx = −1.3874mmIn triangle C'B'D' :LC'B' := (ΔCx − ΔBx)2 + (b + ΔBy)2LC'B' = 54.1117mmLD'B' := a + ΔBx − ΔDx LD'B' = 48.3874mmLC'D' := (a + ΔCx − ΔDx)2 + (b + ΔDy)2LC'D' = 79.586mm2 cos(θB') LC'B'+ LD'B'2 − LC'D'22⋅ (LC'B')⋅ (LD'B')=θB' acosLC'B'2 LD'B'+ 2 LC'D'− 22⋅ (LC'B')⋅ (LD'B')⎡⎢⎢⎣⎤⎥⎥⎦:= θB' = 101.729 deg θB' = 1.7755 radθD' := 180deg − θ'A θD' = 88.500 deg θD' = 1.5446 radθC' := 180deg − θB' θC' = 78.271 deg θC' = 1.3661 radShear Strain :γxy_A := 0.5π − (θ'A) γxy_A = −26.180 × 10− 3 rad Ansγxy_B := 0.5π − (θB') γxy_B = −204.710 × 10− 3 rad Ansγxy_C := 0.5π − (θC') γxy_C = 204.710 × 10− 3 rad Ansγxy_D := 0.5π − (θD') γxy_D = 26.180 × 10− 3 rad Ans 140. Problem 2-20The block is deformed into the position shown by the dashed lines. Determine the average normalstrain along line AB.Given: ΔxBA := (70 − 30)mm ΔyBA := 100mmΔxB'A := (55 − 30)mmΔyB'A := ( 1102 − 152)mmSolution:For AB :LAB ΔxBA:= 2 + ΔyBA2 LAB = 107.7033mm:= + 2 LAB' = 111.8034mmLAB' ΔxB'A2 ΔyB'AεABLAB' − LABLAB:=εAB 38.068 × 10− 3mmmm= Ans 141. Problem 2-21A thin wire, lying along the x axis, is strained such that each point on the wire is displaced Δx = k x2along the x axis. If k is constant, what is the normal strain at any point P along the wire?Given: Δx kx 2= ⋅Solution:εd Δxdx=ε = 2⋅k⋅x Ans 142. Problem 2-22The rectangular plate is subjected to the deformation shown by the dashed line. Determine the averagshear strain γxy of the plate.Given: a := 150mm b := 200mmΔa := 0mm Δb := −3mmSolution:Δθ atanΔba⎛⎜⎝⎞⎠:=Δθ = −1.146 degΔθ = −19.9973 × 10− 3 radShear Strain :γxy := Δθγxy = −19.997 × 10− 3 rad Ans 143. Problem 2-23The rectangular plate is subjected to the deformation shown by the dashed lines. Determine the averagshear strain γxy of the plate.Given: a := 200mm Δa := 3mmb := 150mm Δb := 0mmSolution:Δθ atanΔab⎛⎜⎝⎞⎠:=Δθ = 1.146 degΔθ = 19.9973 × 10− 3 radShear Strain :γxy := Δθγxy = 19.997 × 10− 3 rad Ans 144. Problem 2-24The rectangular plate is subjected to the deformation shown by the dashed lines. Determine theaverage normal strains along the diagonal AC and side AB.Given: a := 200mm b := 150mmΔAx := −3mm ΔAy := 0mmΔBx := 0mm ΔBy := 0mmΔCx := 0mm ΔCy := 0mmΔDx := −3mm ΔDy := 0mmSolution:Geometry :LAC := a2 + b2 LAC = 250mmLAB := b LAB = 150mmLA'C := (a + ΔCx − ΔAx)2 + (b + ΔCy − ΔAy)2LA'C = 252.41mmLA'B := ΔAx2 + b2LA'B = 150.03mmAverage Normal Strain :εACLA'C − LACLAC:= εAC 9.626 × 10− 3mmmm= AnsεABLA'B − LABLAB:= εAB 199.980 × 10− 6mmmm= Ans 145. Problem 2-25The piece of rubber is originally rectangular. Determine the average shear strain γxy if the corners Band D are subjected to the displacements that cause the rubber to distort as shown by the dashed linesGiven: a := 300mm b := 400mmΔAx := 0mm ΔAy := 0mmΔBx := 0mm ΔBy := 2mmΔDx := 3mm ΔDy := 0mmSolution:ΔθAB atanΔBya⎛⎜⎝⎞⎠:=ΔθAB = 0.38197 degΔθAB = 6.6666 × 10− 3 radΔθAD atanΔDxb⎛⎜⎝⎞⎠:=ΔθAD = 0.42971 degΔθAD = 7.4999 × 10− 3 radShear Strain :γxy_A := ΔθAB + ΔθADγxy_A = 14.166 × 10− 3 rad Ans 146. Problem 2-26The piece of rubber is originally rectangular and subjected to the deformation shown by the dashedlines. Determine the average normal strain along the diagonal DB and side AD.Given: a := 300mm b := 400mmΔAx := 0mm ΔAy := 0mmΔBx := 0mm ΔBy := 2mmΔDx := 3mm ΔDy := 0mmSolution:Geometry :LDB := a2 + b2 LDB = 500mmLAD := b LAD = 400mmLD'B' := (a + ΔBx − ΔDx)2 + (b + ΔDy − ΔBy)2 LD'B' = 496.6mmLAD' := ΔDx2 + (b + ΔDy)2 LAD' = 400.01mmAverage Normal Strain :LD'B' − LDBεBD:= εBD −6.797 × 10− 3LDBmmmm= AnsεADLAD' − LAD:= εAD 28.125 × 10− 6LADmmmm= Ans 147. Problem 2-27The material distorts into the dashed position shown. Determine (a) the average normal strains εxand εy , the shear strain γxy at A, and (b) the average normal strain along line BE.Given: a := 80mm Bx := 0mm Ex := 80mmb := 125mm By := 100mm Ey := 50mmΔAx := 0mm ΔCx := 10mm ΔDx := 15mmΔAy := 0mm ΔCy := 0mm ΔDy := 0mmSolution:ΔxAC := ΔCx − ΔAx ΔxAC = 10.00mmΔyAC := ΔCy − ΔAy ΔyAC = 0.00mmΔθAC atanΔCxb⎛⎜⎝⎞⎠:= ΔθAC = 79.8300 × 10− 3 radSince there is no deformation occuring along the y- and x-axis,εx_A := ΔyAC εx_A = 0 Ansεy_A2 + b2 − bbΔxAC:=εy_A = 0.00319 Ansγxy_A := ΔθACγxy_A = 79.830 × 10− 3 rad AnsGeometry :ΔBxByBy= ΔBxΔCxbb⎛⎜⎝⎞⎠:= ⋅ΔCx ΔBx = 8mmΔBy := 0mmΔExΔDxEyb= ΔExEyb⎛⎜⎝⎞⎠:= ⋅ΔDx ΔEx = 6mmΔEy := 0mmLBE := (Ex − Bx)2 + (Ey − By)2 LBE = 94.34mmLB'E' (a + ΔEx − ΔBx)2 := + (Ey + ΔEy − ΔBy)2 LB'E' = 92.65mmεBELB'E' − LBE:= εBE −17.913 × 10− 3LBEmmmm= AnsNote: Negative sign indicates shortening of BE. 148. Problem 2-28The material distorts into the dashed position shown. Determine the average normal strain that occursalong the diagonals AD and CF.Given: a := 80mm Bx := 0mm Ex := 80mmb := 125mm By := 100mm Ey := 50mmΔAx := 0mm ΔCx := 10mm ΔDx := 15mmΔAy := 0mm ΔCy := 0mm ΔDy := 0mmSolution:ΔxAC := ΔCx − ΔAx ΔxAC = 10.00mmΔyAC := ΔCy − ΔAy ΔyAC = 0.00mmΔθAC atanΔCxb⎛⎜⎝⎞⎠:= ΔθAC = 79.8300 × 10− 3 radGeometry :LAD := a2 + b2 LAD = 148.41mmLCF := a2 + b2 LCF = 148.41mmLA'D' := (a + ΔDx − ΔAx)2 + (b + ΔDy − ΔAy)2LA'D' = 157.00mmLC'F := (a − ΔCx)2 + (b − ΔCy)2LC'F = 143.27mmAverage Normal Strain :LA'D' − LADεAD:= εAD 57.914 × 10− 3LADmmmm= AnsεCFLC'F − LCFLCF:= εCF −34.653 × 10− 3mmmm= Ans 149. Problem 2-29The block is deformed into the position shown by the dashed lines. Determine the shear strain atcorners C and D.Given: a := 100mm ΔAx := −15mmb := 100mm ΔBx := −15mmLCA' := 110mmSolution:Geometry :ΔθC asinΔAxLCA'⎛⎜⎝⎞⎠:= ΔθC = −7.84 degΔθC = −0.1368 radΔθD asinΔBxLCA'⎛⎜⎝⎞⎠:= ΔθD = −7.84 degΔθD = −0.1368 radShear Strain :γxy_C := ΔθCγxy_C = −136.790 × 10− 3 rad Ansγxy_D := −ΔθDγxy_D = 136.790 × 10− 3 rad Ans 150. Problem 2-30The bar is originally 30 mm long when it is flat. If it is subjected to a shear strain defined by γxy = 0.0x, where x is in millimeters, determine the displacement Δy at the end of its bottom edge. It is distorteinto the shape shown, where no elongation of the bar occurs in the x direction.Given: L := 300mm γxy = 0.02⋅xunit := 1mmSolution:dydx= tan(γxy)dydx= tan(0.02⋅x)Δy01 y⌠⎮⌡dLtan(0.02⋅x)⋅ (unit) x⌠⎮⌡= d0Δy30tan(0.02x)⋅ (unit) x⌠⎮⌡:= d0Δy = 9.60mm Ans 151. Problem 2-31The curved pipe has an original radius of 0.6 m. If it is heated nonuniformly, so that the normal strainalong its length is ε = 0.05 cos θ, determine the increase in length of the pipe.Given: r := 0.6m ε = 0.05⋅cos(θ)Solution:⌠⎮⌡= dΔL ε LΔL90deg00.05⋅cos(θ) (rθ)⌠⎮⌡= dΔL90deg00.05⋅ r⋅cos(θ) θ⌠⎮⌡:= dΔL = 30.00mm Ans 152. Problem 2-32Solve Prob. 2-31 if ε = 0.08 sin θ .Given: r := 0.6m ε = 0.08⋅ sin(θ)Solution:⌠⎮⌡= dΔL ε LΔL90deg00.08⋅ sin(θ) (rθ)⌠⎮⌡= dΔL90deg00.08⋅ r⋅ sin(θ) θ⌠⎮⌡:= dΔL = 0.0480m Ans 153. Problem 2-33A thin wire is wrapped along a surface having the form y = 0.02 x2, where x and y are in mm. Originathe end B is at x = 250 mm. If the wire undergoes a normal strain along its length of ε = 0.0002x,determine the change in length of the wire. Hint: For the curve, y = f (x), ds = 1dydx⎛⎜⎝⎞⎠2+ ⋅dx .Given: Bx := 250mm ε = 0.0002⋅x unit := 1mmSolution:y = 0.02x2dydx= 0.04xds 1dydx⎛⎜⎝⎞⎠2= + ⋅dxds = 1 + (0.04x)2⋅dx⌠⎮⌡= d ΔLΔL ε sBx0(0.0002x)⋅ 1 + (0.04x)2 x⌠⎮⌡= dΔL (unit)2500(0.0002x)⋅ 1 + (0.04x)2 x⌠⎮⌡:= ⋅ dΔL = 42.252mm Ans 154. Problem 2-34The fiber AB has a length L and orientation If its ends A and B undergo very small displacements uAand vB, respectively, determine the normal strain in the fiber when it is in position A'B'.Solution:Geometry:LA'B' = (L⋅cos(θ) − uA)2 + (L⋅ sin(θ) − vB)2LA'B' v= L2 + uA2 + vB2 + 2L ⋅ ( B⋅ sin(θ) − uA⋅cos(θ))Average Normal Strain:LA'B' ⋅ −LεABL== + − 1εAB 12 vBuA+ 2L2+2⋅ (vB⋅ sin(θ) − uA⋅cos(θ))L2 and vBNeglecting higher-order terms uA2 ,= + − 1εAB 12⋅ (vB⋅ sin(θ) − uA⋅cos(θ))LUsing the binomial theorem:εAB 1122⋅ (vB⋅ sin(θ) − uA⋅cos(θ))L⎡⎢⎣⎤⎥⎦= + + .. − 1εABvB⋅ sin(θ)= − AnsLuA⋅cos(θ)L 155. Problem 2-35If the normal strain is defined in reference to the final length, that is,ε 'n =p p'Δs' − ΔsΔ s'⎛⎜⎝⎞⎠lim→instead of in reference to the original length, Eq.2-2, show that the difference in these strains isrepresented as a second-order term, namely, εn - ε'n = εn ε'n.Solution:ε nΔS' − ΔSΔS=ε n − ε'nΔS' − ΔSΔS⎛⎜⎝⎞⎠ΔS' − ΔSΔ S'⎛⎜⎝⎞⎠= −ε n − ε'nΔ S'2 − 2⋅ (ΔS)⋅ (ΔS') ΔS+ 2(ΔS)⋅ (ΔS')=ε n − ε'n(ΔS' − ΔS)2(ΔS)⋅ (ΔS')=ε n − ε'nΔS' − ΔSΔS⎛⎜⎝⎞⎠ΔS' − ΔSΔ S'⎛⎜⎝⎞⎠= ⋅ε n − ε'n = (ε n)⋅ (ε'n) (Q.E.D.) 156. Problem 3-1A concrete cylinder having a diameter of 150 mm. and gauge length of 300 mm is tested incompression. The results of the test are reported in the table as load versus contraction. Draw thestress-strain diagram using scales of 10 mm = 2 MPa and 10 mm = 0.1(10-3) mm/mm. From thediagram, determine approximately the modulus of elasticity.0 1.10 4 2 .10 4 3 .10 4 4 .10 4 5 .10 4 6 .10 4 7 .10 4161412108642σY(x)ε , xδ0.00000.01500.03000.05000.06500.08500.10000.11250.12500.15500.17500.1875:=Given: d := 150 L := 300Solution: Aπ4⎛⎜⎝⎞⎠:= ⋅d2 * σ103PA:= εδL:=σ=P/A(MPa)ε=δ/L(mm/mm)Regression x := min(ε) , 0.00005 .. max(ε)0.001.412.694.675.807.228.499.7610.8913.1614.1515.00⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟σ curve: Coeff := loess(ε , σ , 1.5)= ε⎠0.0000000.0000500.0001000.0001670.0002170.0002830.0003330.0003750.0004170.0005170.0005830.000625⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠=Y(x) := interp(Coeff , ε , σ , x)Modulus of Elasticity:From the stress-strain diagram,Δε (0.0003 − 0)mmmm:= *Δσ := (8.0 − 0)MPa *EapproxΔσΔε:=Eapprox = 26.67GPa AnsContractionP (mm)0.025.047.582.5102.5127.5150.0172.5192.5232.5250.0265.0:=Load(kN) 157. Problem 3-2Data taken from a stress-strain test for a ceramic are given in the table. The curve is linear between theorigin and the first point. Plot the diagram, and determine the modulus of elasticity and the modulus ofresilience.Unit used: MJ := (106)Jσ:= ε0.0232.4318.5345.8360.5373.80.00000.00060.00100.00140.00180.0022Solution: := Regression curve:x := min(ε) , 0.00005 .. max(ε)Coeff := loess(ε , σ , 0.9)Y(x) := interp(Coeff , ε , σ , x)0 5.10 4 0.001 0.0015 0.002 0.00254003603202802402001601208040σY(x)ε, xModulus of Elasticity:From the stress-strain diagram,Δε (0.0006 − 0)mmmm:=Δσ := (232.4 − 0)MPaEapproxΔσΔε:=Eapprox = 387.3GPa AnsModulus of Resilience:The modulus of resilience is equal to the areaunder the initial linear portion of the curve.Δε (0.0006 − 0)mmmm:=Δσ (232.4 − 0)103 kNm2:=ur12:= ⋅Δε⋅Δσur 0.0697MJm3= Ansσ=P/A(MPa)ε=δ/L(mm/mm) 158. rProblem 3-σ3Data taken from a stress-strain test for a ceramic are given in the table. The curve is linear between theorigin and the first point. Plot the diagram, and determine approximately the modulus of toughness. Therupture stress is = 373.8 MPa.Unit used: MJ := (106)Jσ:= ε0.0232.4318.5345.8360.5373.80.00000.00060.00100.00140.00180.0022:=Solution: Regression curve:x := min(ε) , 0.00005 .. max(ε)Coeff := loess(ε , σ , 0.9)Y(x) := interp(Coeff , ε , σ , x)0 5.10 4 0.001 0.0015 0.002 0.00254003603202802402001601208040σY(x)ε , xModulus of Resilience:The modulus of resilience is equalto the area under the curve.A112:= ⋅ (232.4)⋅ (0.0004 + 0.0010)A2 := 318.5⋅ (0.0022 − 0.0010)A312:= ⋅ (373.8 − 318.5)⋅ (0.0022 − 0.0010)A412:= ⋅ (318.5 − 232.4)⋅ (0.0010 − 0.0006)Atotal := A1 + A2 + A3 + A4ut Atotal⋅106Jm3:= ⋅ *ut 0.595MJm3= * Ansσ=P/A(MPA)ε=δ/L(mm/mm) 159. Problem 3-4A tension test was performed on a steel specimen having an original diameter of 13 mm and gaugelength of 50 mm. The data is listed in the table. Plot the stress-strain diagram and determineapproximately the modulus of elasticity, the yield stress, the ultimate stress, and the rupture stress. Usea scale of 10 mm = 209 MPa and 10 mm = 0.05 mm/mm. Redraw the elastic region, using the samestress scale but a strain scale of 10 mm = 0.001 mm/mm.LoadGiven: d := 12.5* L := 50* P:= (kN)δ0.07.523.040.055.059.059.060.083.0100.0107.597.592.5Elongation0.00000.01250.03750.06250.08750.12500.20000.50001.00002.50007.000010.000011.5000:=Solution: Aπ4⎛⎜⎝⎞⎠:= ⋅d2 * σ103⋅PA:= *εδL:= *σ=P/A(MPa)ε=δ/L(mm/mm.)σ0.0061.12187.42325.95448.18480.78480.78488.92676.34814.87875.99794.50753.76⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟= * ε⎠0.000000.000250.000750.001250.001750.002500.004000.010000.020000.050000.140000.200000.23000⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟= * Curve Fit:⎠Use: F(x)xx0.3x0.6⎛⎜⎜⎜⎝⎞⎟⎠:= Fit := linfit(ε , σ , F)Fit−4189.6962285.571596.538⎛⎜⎜⎝⎞⎠=x := min(ε) , 0.00005 .. max(ε)Y(x) := F(x)⋅Fit0 0.001 0.002 0.003 0.004 0.0051000800600400200σε0 0.05 0.1 0.15 0.2 0.251000800600400200σY(x)ε , xModulus of Elasticity: From the stress-strain diagram,(mm) 160. From the stress-strain diagram, σY = 448MPa AnsΔε (0.00125 − 0)mmmm:= * σult 890MPa = AnsΔσ := (326 − 0)MPa * σR = 753.8MPa AnsEapproxΔσΔε:= *Eapprox = 260.8GPa* Ans 161. Problem 3-5The stress-strain diagram for a steel alloy having an original diameter of 12mm and a gauge length of50 mm is given in the figure. Determine approximately the modulus of elasticity for the material, theload on the specimen that causes yielding, and the ultimate load the specimen will support.Given: d := 12mm L := 50mmSolution:Aπ4⎛⎜⎝⎞⎠:= ⋅d2Modulus of Elasticity:From the stress-strain diagram,Δε (0.001 − 0)mmmm:=Δσ := (290 − 0)MPaEΔσΔε:= *E = 290GPa* AnsFrom the stress-strain diagram, σY := 290MPaYield Load: PY := (σY)⋅APY = 32.80 kN AnsFrom th stress-strain diagram, σu := 550MPaUltimate Load: Pu := (σu)⋅APu = 62.20 kN Ans 162. Problem 3-6The stress-strain diagram for a steel alloy having an original diameter of 12 mm and a gauge length of50 mm is given in the figure. If the specimen is loaded until it is stressed to 500MPa, determine theapproximate amount of elastic recovery and the increase in the gauge length after it is unloaded.Given: d := 12mm L := 50mmσmax := 500MPaSolution:Aπ4⎛⎜⎝⎞⎠:= ⋅d2Modulus of Elasticity:From the stress-strain diagram,Δε (0.001 − 0)mmmm:=Δσ := (290 − 0)MPaEΔσΔε:= *E = 290GPa* AnsElastic Recovery: rReσmaxE:= rRe 0.00172mmmm=AmountRe := rRe⋅ (L) AmountRe = 0.08621mm AnsFrom the stress-strain diagram, εmax 0.08mmmm:=Permanent set: (Permanent elongation)rps := εmax − rRe rps 0.07828mmmm=Amountps := rps⋅ (L) Amountps = 3.91379mm Ans 163. Problem 3-7The stress-strain diagram for a steel alloy having an original diameter of 12 mm and a gauge length of50 mm is given in the figure. Determine approximately the modulus of resilience and the modulus oftoughness for the material.Unit used: MJ := (106)JGiven: d := 12mm L := 50mmSolution:Modulus of Resilience:The modulus of resilience is equal to the areaunder the initial linear portion of the curve.Δε (0.001 − 0)mmmm:=Δσ := (290 − 0)MPaur12:= ⋅Δε⋅Δσur = 0.145MPa AnsModulus of Toughness:The modulus of toughness is equal to the area under the curve, and could be approximated bycounting the number of sqaures. the total number of squares is:n := 33Δεsq (0.04)mmmm:=Δσsq := 100MPaut := n⋅ (Δεsq)⋅ (Δσsq)ut = 132MPa Ans 164. Problem 3-8The stress-strain diagram for a steel bar is shown in the figure. Determine approximately the modulusof elasticity, the proportional limit, the ultimate stress, and the modulus of resilience. If the bar isloaded until it is stressed to 450 MPa, determine the amount of elastic strain recovery and thepermanent set or strain in the bar when it is unloaded.Unit used: kJ := 103JGiven: σmax := 450MPaSolution:Modulus of Elasticity:From th stress-strain diagram,Δε (0.0015 − 0)mmmm:=Δσ := (325 − 0)⋅MPaEΔσΔε:=E = 216.67GPa AnsModulus of Resilience:The modulus of resilience is equal to the area under theinitial linear portion of the curve.Δε (0.0015 − 0)mmmm:=Δσ := (325 − 0)⋅MPaur12:= ⋅Δε⋅Δσur 243.75kJm3= AnsElastic Recovery: rReσmaxE:= rRe 0.00208mmmm= AnsFrom th stress-strain diagram, εmax 0.0750mmmm:=Permanent set: (Permanent elongation)rps := εmax − rRe rps 0.07292mmmm= Ans 165. Problem 3-9The σ-ε diagram for elastic fibers that make up human skin and muscle is shown. Determine themodulus of elasticity of the fibers and estimate their modulus of toughness and modulus of resilience.Modulus of Elasticity:From the stress-strain diagram,Δε (2.00 − 0)mmmm:=Δσ := (77 − 0)MPaEΔσΔε:= E = 38.5MPa AnsModulus of Resilience:The modulus of resilience is equal to the area under the initial linear portion of the curve.Δε (2.00 − 0)mmmm:=Δσ := (77 − 0)MPaur12:= ⋅Δε⋅Δσur = 77.00MPa AnsModulus of Toughness:The modulus of toughness is equal to the area under the curve. ε0 := 0 ε1 := 2.00 ε2 := 2.25σ0 := 0 σ1 := 77MPa σ2 := 385MPaA1 := urA2 := 0.5(σ1 + σ2)⋅ (ε2 − ε1)ut := A1 + A2ut = 134.75MPa Ans 166. Problem 3-10An A-36 steel bar has a length of 1250 mm and cross-sectional area of 430 mm2. Determine the lengthof the bar if it is subjected to an axial tension of 25 kN. The material has linear-elastic behavior.Given: A := 430mm2 L0 := 1250mm P := 25kNσY := 250MPa Est := 200GPaSolution:Normal Stress:σPA:= σ = 58.140MPa [less than yield stress σy]Hence Hook's law is still valid.Normal Strain:εσEst:= ε 290.6977 × 10− 6mmmm=Thus,δL := ε⋅L0 δL = 0.36337mmL := L0 + δL L = 1250.363mm Ans 167. Problem 3-11The stress-strain diagram for polyethylene, which is used to sheath coaxial cables, is determined fromtesting a specimen that has a gauge length of 250 mm. If a load P on the specimen develops a strain ofε = 0.024 mm/mm, determine the approximate length of the specimen, measured between the gaugepoints, when the load is removed. Assume the specimen recovers elastically.Given: L0 := 250mmSolution:Modulus of Elasticity:From th stress-strain diagram,Δε (0.004 − 0)mmmm:=Δσ := (14.0 − 0)⋅MPaEΔσΔε:=E = 3500.00MPaElastic Recovery: From the stress-strain diagram, εmax 0.024mmmm:= σmax := 26MParReσmaxE:= rRe 0.00743mmmm=Permanent set:rps := εmax − rRe rps 0.01657mmmm=Permanent elongation:ΔL := rps⋅L0 ΔL = 4.14286mmL := L0 + ΔL L = 254.143mm Ans 168. Problem 3-12Fiberglass has a stress-strain diagram as shown. If a 50-mm-diameter bar of length 2 m made fromthis material is subjected to an axial tensile load of 60 kN, determine its elongation.Given: L0 := 2m d := 50mmP := 60kN unit := 1PaSolution:Aπ4⎛⎜⎝⎞⎠:= ⋅d2σPA:= σ = 30.558 × 106 Paσ'σunitGiven: σ' = 300(106)⋅ε0.5 :=εσ'29(1016)⎡⎢⎢⎣⎤⎥⎥⎦:=ε 0.010375mmmm=ΔL := ε⋅ (L0) ΔL = 20.75mm Ans 169. Problem 3-13The change in weight of an airplane is determined from reading the strain gauge A mounted in theplane's aluminum wheel strut. Before the plane is loaded, the strain gauge reading in a strut is ε1 =0.00100 mm/mm, whereas after loading ε2 = 0.00243 mm/mm. Determine the change in the force onthe strut if the cross-sectional area of the strut is 2200 mm2. Eal = 70 GPa.Given: A := 2200mm2 Eal := 70⋅GPaε1 0.00100mmmm:= ε2 0.00243mmmm:=Solution:Stress-strain Relationship: Applying Hook's law σ = Eε .σ1 := Eal⋅ ε1 σ1 = 70.00MPaσ2 := Eal⋅ ε2 σ2 = 170.10MPaNormal Force: Applying equation σ = P / A .P1 := A⋅σ1 P1 = 154.00 kNP2 := A⋅σ2 P2 = 374.22 kNThus,ΔP := P2 − P1 ΔP = 220.22 kN Ans 170. Problem 3-14A specimen is originally 300 mm long, has a diameter of 12 mm, and is subjected to a force of 2.5 kN.When the force is increased to 9 kN, the specimen elongates 22.5 mm. Determine the modulus ofelasticity for the material if it remains elastic.Given: d := 12mm L0 := 300mmP1 := 2.5kN P2 := 9kN ΔL := 22.5mmSolution:Aπ4⎛⎜⎝⎞⎠:= ⋅d2Normal Force: Applying equation σ = P / A .σ1P1A:= σ1 = 22.105MPaσ2P2A:= σ2 = 79.577MPaThus,Δσ := σ2 − σ1 Δσ = 57.473MPaΔεΔLL0:=EΔσΔε:= E = 766.3MPa Ans 171. Problem 3-15A structural member in a nuclear reactor is made from a zirconium alloy. If an axial load of 20 kN is tobe supported by the member, determine its required cross-sectional area. Use a factor of safety of 3with respect to yielding. What is the load on the member if it is 1-m long and its elongation is 0.5 mm?Ezr = 100 GPa, σY = 400 MPa. The material has elastic behavior.Given: P := 20kN L0 := 1m ΔL := 0.5mmEzr := 100⋅GPa σY := 400⋅MPa FoS := 3Solution:σallowσYFoS:= σallow = 133.33MPaAreqPσallow:= Areq = 150mm2 AnsA := AreqεΔLL0:= ε 0.0005mmmm=σ := Ezr⋅ ε σ = 50MPaP := σ⋅A P = 7.5 kN Ans 172. Problem 3-16The pole is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 5 mm,determine how much it stretches when a horizontal force of 15 kN acts on the pole.Given: P := 15kN Est := 200⋅GPaa := 1.2m b := 1md := 5mm θ := 30degSolution:Aπ4⎛⎜⎝⎞⎠:= ⋅d2 L := a + bSupport Reactions:ΣΜB=0; Cx⋅ (L) + P⋅ (b) = 0Cx −PbL:= ⋅Cx = −6.82 kN+ ΣFx=0; Cx + P + Bx = 0Bx := −Cx − P−BxBx = −8.18 kNFAB:= sin(θ) FAB = 16.36 kN LABLcos(θ) := LAB = 2.54mσABFABA:= σAB = 833.393MPaεABσABEst:= εAB 0.0041670mmmm=δAB := εAB⋅LAB δAB = 10.586mm Ans 173. Problem 3-17By adding plasticizers to polyvinyl chloride, it is possible to reduce its stiffness. The stress-straindiagrams for three types of this material showing this effect are given below. Specify the type thatshould be used in the manufacture of a rod having a length of 125 mm and a diameter of 50 mm, thatis required to support at least an axial load of 100 kN and also be able to stretch at most 6 mm.Given: d := 50mm L0 := 125mmP := 100kN δL := 6mmSolution:Aπ4⎛⎜⎝⎞⎠:= ⋅d2Normal Stress:σPA:= σ = 50.930MPaNormal Strain:εδLL0:= ε 0.048000mmmm=From the stress-strain diagram,the copolymer will satisfy bothstress and strain requirements. Ans 174. Problem 3-18The steel wires AB and AC support the 200-kg mass. If the allowable axial stress for the wires is σallow= 130 MPa, determine the required diameter of each wire. Also, what is the new length of wire ABafter the load is applied? Take the unstretched length of AB to be 750 mm. Est = 200 GPa.Given: g 9.81ms2:= m := 200kgθ := 60deg v45:= h35:=L0 := 750mmEst := 200GPa σallow := 130MPaSolution: W := m⋅gAxial force in steel wires AB and AC:Initial guess: FAC := 1N FAB := 2NGiven+ ΣFx=0; −FAB⋅cos(θ) + FAC⋅ (h) = 0 [1]+ ΣFy=0; FAB⋅ sin(θ) + FAC⋅ (v) − W = 0 [2]Solving [1] and [2]:FACFAB⎛⎜⎜⎝⎞⎠:= Find(FAC, FAB) FACFAB⎛⎜⎜⎝⎞⎠1066.751280.10⎛⎜⎝⎞⎠= NWire AB := ⋅ 2FAB σallowπ4⎛⎜⎝⎞⎠⋅ dABdAB4πFABσallow⎛⎜⎝⎞⎠:= ⋅ dAB = 3.54mm AnsWire AC:= ⋅ 2FAC σallowπ4⎛⎜⎝⎞⎠⋅ dACdAC4πFACσallow⎛⎜⎝⎞⎠:= ⋅ dAC = 3.23mm AnsStress-strain Relationship: Applying Hook's law σ = Eε .εABσallowEst:= εAB 0.000650mmmm=Thus, LAB := L0⋅ (1 + εAB) LAB = 750.487mm Ans 175. Problem 3-19The two bars are made of polystyrene, which has the stress-strain diagram shown. If thecross-sectional area of bar AB is 950 mm2 and BC is 2500 mm2, determine the largest force P that canbe supported before any member ruptures. Assume that buckling does not occur.AnsGiven: a := 1.2m b := 0.9mAAB := 950mm2 ABC := 2500mm2Solution:c := a2 + b2 hac:= vbc:=+ ΣFy=0; FAB⋅ (v) − P = 0 [1]+ ΣFx=0; FAB⋅ (h) − FBC = 0 [2]Solving Eqs.[1] and [2]:FAB53= P [1a]FBC43= P [2a]Assume tension failure of BC:From the stress-strain diagram, σR_t := 35MPaFBC := (ABC)⋅ (σR_t) FBC = 87.50 kNFrom Eq.[2a], P := 0.75⋅FBC P = 65.63 kNPcase_1 := PAssume compression failure of AB:From the stress-strain diagram, σR_c := 175MPaFAB := (AAB)⋅ (σR_c) FAB = 166.25 kNFrom Eq.[1a], P := 0.60⋅FAB P = 99.75 kNPcase_2 := PChosoe the smallest value: P := min(Pcase_1 , Pcase_2) P = 65.63 kN 176. Problem 3-20The two bars are made of polystyrene, which has the stress-strain diagram shown. Determine thecross-sectional area of each bar so that the bars rupture simultaneously when the load P = 15 kN.Assume that buckling does not occur.AnsGiven: a := 1.2m b := 0.9m P := 15kNSolution:c := a2 + b2 hac:= vbc:=+ ΣFy=0; FAB⋅ (v) − P = 0 [1]+ ΣFx=0; FAB⋅ (h) − FBC = 0 [2]Solving Eqs.[1] and [2]:FAB53:= P FAB = 25.00 kNFBC43:= P FBC = 20.00 kNFor member BC:From the stress-strain diagram, σR_t := 35MPaABCFBCσR_t⎛⎜⎝⎞⎠:= ABC = 571.43mm2 AnsFor member AB:From the stress-strain diagram, σR_c := 175MPaAABFABσR_c⎛⎜⎝⎞⎠:= AAB = 142.86mm2 177. Problem 3-21The stress-strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by astrut AB and post CD, both made from this material, and subjected to a load of P = 80 kN, determinethe angle of tilt of the beam when the load is applied. The diameter of the strut is 40 mm and thediameter of the post is 80 mm.Given: P := 80kNLAB := 2m LCD := 0.5m LAC := 1.5mdAB := 40mm dCD := 80mmSolution:AyP2:= CyP2Support Reactions: :=FAB := Ay FCD := CyAreaABπ4⎛⎜⎝⎞⎠:= ⋅ dAB2 AreaCDπ4⎛⎜⎝⎞⎠:= ⋅ 2dCDFrom the stress-strain diagram, E32.2MPa0.01:=E = 3220.00MPaσABFABAreaAB:= σAB = 31.831MPaεABσABE:= εAB 0.0098854mmmm=δAB := εAB⋅LAB δAB = 19.771mmσCDFCDAreaCD:= σCD = 7.958MPaεCDσCDE:= εCD 0.0024714mmmm=δCD := εCD⋅LCD δCD = 1.236mmAngle of tilt α: tan(α)δAB − δCDLAC:=α atanδAB − δCDLAC⎛⎜⎝⎞⎠:= α = 0.708 deg Ans 178. Problem 3-22The stress-strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by astrut AB and post CD made from this material, determine the largest load P that can be applied to thebeam before it ruptures. The diameter of the strut is 12 mm and the diameter of the post is 40 mm.Given:LAB := 2m LCD := 0.5m LAC := 1.5mdAB := 12mm dCD := 40mmSolution:Support Reactions: AyP2= CyP2=FAB = Ay FCD = CyAreaABπ4⎛⎜⎝⎞⎠:= ⋅ dAB2 AreaCDπ4⎛⎜⎝⎞⎠:= ⋅ 2dCDFor rupture of strut AB:From the stress-strain diagram, σR_t := 50.0MPaFAB = AreaAB⋅ (σR_t)P := 2AreaAB⋅ (σR_t)P = 11.31 kN Ans(Controls!)For rupture of post CD:From the stress-strain diagram, σR_c := 95.0MPaFCD = AreaCD⋅ (σR_c)P := 2AreaCD⋅ (σR_c)P = 238.76 kN 179. Problem 3-23The beam is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 5mm, determine how much it stretches when a distributed load of w = 1.5 kN/m acts on the pipe. Thematerial remains elastic.Given: L := 3m w 1.5kNm:=dAB := 5mm θ := 30degEst := 200⋅GPaSolution:Support Reactions:ΣΜC=0; FAB⋅ sin(θ)⋅ (L) − w⋅ (L)⋅ (0.5⋅L) = 0FABw⋅L2 sin(θ) :=AreaABπ4⎛⎜⎝⎞⎠:= ⋅ dAB2 LABLcos(θ) :=σABFABAreaAB:= σAB = 229.183MPaεABσABEst:= εAB 0.0011459mmmm=δAB := εAB⋅LAB δAB = 3.970mm Ans 180. Problem 3-24The beam is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 5mm, determine the distributed load w if the end B is displaced 18 mm downward.Given: L := 3m δBy := 18mmdAB := 5mm θB := 30degEst := 200⋅GPaSolution:Consider triangle BB'C: L⋅ sin(θC) δB= yθC asinδByL⎛⎜⎝⎞⎠:=Consider triangle AB'C: θ'C := 90deg + θC LAC := L⋅ tan(θB)2 + L2 2 L:= − ⋅ ( AC)⋅L⋅cos(θ'C)L'AB LACL'AB = 3.4731mLAreaABLAB:= cos(θB) π4⎛⎜⎝⎞⎠:= ⋅ 2dABδAB := L'AB − LAB δAB = 8.9883mmεABδABLAB:= εAB 0.0025947mmmm=σAB := εAB⋅Est σAB = 518.942MPaFAB := σAB⋅ (AreaAB) FAB = 10.19 kNSupport Reactions:ΣΜC=0; FAB⋅ sin(θB)⋅ (L) − w⋅ (L)⋅ (0.5⋅L) = 0w2 sin(θB)L⎛⎜⎝⎞⎠:= ⋅FAB w 3.40kNm= Ans 181. Problem 3-25Direct tension indicators are sometimes used instead of torque wrenches to insure that a bolt has aprescribed tension when used for connections. If a nut on the bolt is tightened so that the six heads ofthe indicator that were originally 3 mm high are crushed 0.3 mm, leaving a contact area on each headof 1.5 mm2, determine the tension in the bolt shank. The material has the stressstrain diagram shown.Given: h := 3mm δh := 0.3mmArea := 1.5mm2number := 6 unit := 1MPaSolution:Stress-strain Relationship:εδhh:= ε 0.1000mmmm=From the stress-strain diagram, σ − 450600 − 450ε − 0.00150.3 − 0.0015=σ 450 (600 − 450)ε − 0.00150.3 − 0.0015⎛⎜⎝⎞⎠+ ⎡⎢⎣⎤⎥⎦:= ⋅unitσ = 499.50MPaAxial Force: For each head P := σ⋅ (Area)P = 0.749 kNThus, the tension in the bolt is T := (number)⋅PT = 4.50 kN Ans 182. pProblem 3-ν26The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it,determine the change in its length and the change in its diameter. Ep = 2.70 GPa, = 0.4.Given: P := 300NL := 200mm d := 15mmEP := 2.70⋅GPa ν := 0.4Solution:Aπ4⎛⎜⎝⎞⎠:= ⋅d2σPA:= σ = 1.698MPaεlongσEP:= εlong 0.0006288mmmm=δ := εlong⋅L δ = 0.126mm Ansεlat := −ν⋅εlong εlat −0.0002515mmmm=Δd := εlat⋅ (d) Δd = −0.003773mm Ans 183. xyεεProblem 3-27The block is made of titanium Ti-6A1-4V and is subjected to a compression of 1.5 mm along the yaxis, and its shape is given a tilt of θ = 89.7°.Determine , , and γxy.Given: Lx := 125mm Ly := 100mmδy := −1.5mm θ := 89.7degν := 0.36Solution:Normal Strain:εyδyLy:= εy −0.01500mmmm= AnsPoisson's Ratio: The lateral and longitudinal strain can be related using Poisson's ratio.εx := −ν⋅ (εy) εx 0.00540mmmm= AnsShear Strain:β := 180deg − θ β = 90.30 degβ = 1.58 radThus, γxyπ2:= − β γxy = −0.00524 rad Ans 184. Problem 3-28A short cylindrical block of bronze C86100, having an original diameter of 38 mm and a length of 75mm, is placed in a compression machine and squeezed until its length becomes 74.5 mm. Deterinmethe new diameter of the block.Given: L := 75mm L' := 74.5mmd := 38mm ν := 0.34Solution:Aπ4⎛⎜⎝⎞⎠:= ⋅d2εlongL' − LL:= εlong −0.0066667mmmm=εlat := −ν⋅εlong εlat 0.0022667mmmm=Δd := εlat⋅ (d) Δd = 0.086133mmd' := d + Δd d' = 38.0861mm Ans 185. Problem 3-29The elastic portion of the stress-strain diagram for a steel alloy is shown in the figure. The specimenfrom which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. When theapplied load on the specimen is 50 kN, the diameter is 12.99265 mm. Determine Poisson's ratio for thematerial.Given: d := 13mm d' := 12.99265mmL := 50mm P := 50kNSolution:Normal Stress:Aπ4⎛⎜⎝⎞⎠:= ⋅d2σPA:= σ = 376.698MPaNormal Strain: From the stress-strain diagram, the modulus of elasticity isE400MPa0.002:= E = 200GPaApplying Hook's law σ = Eε .εlongσE:= εlong 0.0018835mmmm=εlatd' − dd:= εlat −0.0005654mmmm=Poisson's Ratio: The lateral and longitudinal strain can be related using Poisson's ratio.ν−εlatεlong:= ν = 0.30018 Ans 186. Problem 3-30The elastic portion of the stress-strain diagram for a steel alloy is shown in the figure. The specimenfrom which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. If a loadof P = 20 kN is applied to the specimen, determine its diameter and gauge length. Take ν = 0.4.Given:L := 50mm d := 13mmP := 20kN ν := 0.4Solution:Normal Stress:Aπ4⎛⎜⎝⎞⎠:= ⋅d2σPA:= σ = 150.679MPaNormal Strain: From the stress-strain diagram, the modulus of elasticity isE400MPa0.002:= E = 200GPaApplying Hook's law σ = Eε .εlongσE:= εlong 0.0007534mmmm=Thus,δL := εlong⋅ (L) δL = 0.037670mmL' := L + δL L' = 50.0377mm AnsPoisson's Ratio: The lateral and longitudinal strain can be related using Poisson's ratio.εlat := −ν⋅ (εlong) εlat −0.00030136mmmm=Δd := εlat⋅ (d) Δd = −0.003918mmd' := d + Δd d' = 12.99608mm Ans 187. Problem 3-31The shear stress-strain diagram for a steel alloy is shown in the figure. If a bolt having a diameter of 6mm is made of this material and used in the lap joint, determine the modulus of elasticity E and theforce P required to cause the material to yield. Take ν = 0.3.Given:d := 6mm ν := 0.3Solution:Modulus of Rigidity:From the stress-strain diagram,G350MPa0.004:= G = 87500MPaModulus of Elasticity: GE2⋅ (1 + ν)=E := 2G⋅ (1 + ν)E = 227500MPa AnsYielding Stress:From the stress-strain diagram, τY := 350MPaAπ4⎛⎜⎝⎞⎠:= ⋅d2 τYPA=P := (τY)⋅A P = 9.896 kN Ans 188. Problem 3-32The brake pads for a bicycle tire are made of rubber. If a frictional force of 50 N is applied to eachside of the tires, determine the average shear strain in the rubber. Each pad has cross-sectionaldimensions of 20 mm and 50 mm. Gr = 0.20 MPa.Given: a := 20mm b := 50mmV := 50N G := 0.20MPaSolution:A := a⋅b A = 1000.00mm2Average Shear Stress:The shear force is V.τVA:= τ = 0.050MPaShear Stress-strain Relationship:Applying Hooke's law for shear: τ = G⋅ γγτG:= γ = 0.250 rad Ans 189. Problem 3-33The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm.Both the plug and the sleeve are 50 mm long. Determine the axial pressure p that must be applied to thetop of the plug to cause it to contact the sides of the sleeve. Also, how far must the plug becompressed downward in order to do this? The plug is made from a material for which E = 5 MPa, ν= 0.45.Given:d := 30mm d' := 32mm L := 50mmE := 5MPa ν := 0.45Solution:εlatd' − dd:= εlat 0.0666667mmmm=ν−εlatεlong= εlong−εlatν:= εlong −0.14815mmmm=Applying Hook's law σ = Eε .p := E⋅ (εlong) p = −0.741MPa AnsδL := (εlong)⋅L δL = −7.41mm Ans 190. rProblem 3-34The rubber block yxνεεis subjected to an elongation of 0.75 mm. along the x axis, and its vertical faces aregiven a tilt so that θ = 89.3°. Deterinme the strains , and γxy. Take = 0.5.Given: Lx := 100mm Ly := 75mmδx := 0.75mm θ := 89.3degν := 0.5Solution:Normal Strain:εxδxLx:= εx 0.00750mmmm= AnsPoisson's Ratio: The lateral and longitudinal strain can be related using Poisson's ratio.εy := −ν⋅εx εy −0.00375mmmm= AnsShear Strain:θ = 89.30 degθ = 1.56 radThus, γxyπ2:= − θ γxy = 0.01222 rad Ans 191. Problem 3-35The elastic portion of the tension stress-strain diagram for an aluinmum alloy is shown in the figure.The specimen used for the test has a gauge length of 50 mm. and a diameter of 12.5 mm. When theapplied load is 45 kN, the new diameter of the specimen is 12.48375 mm. Compute shear modulus Galfor the aluinmum.Given: d := 12.5mm d' := 12.48375mmL := 50mm P := 45kNSolution:Normal Stress:Aπ4⎛⎜⎝⎞⎠:= ⋅d2σPA:= σ = 366.693MPaNormal Strain: From the stress-strain diagram, the modulus of elasticity isE500MPa0.00614:= E = 81433.22MPaApplying Hook's law σ = Eε .εlongσE:= εlong 0.0045030mmmm=εlatd' − dd:= εlat −0.0013000mmmm=Poisson's Ratio: The lateral and longitudinal strain can be related using Poisson's ratio.ν−εlatεlong:= ν = 0.28870GE:= G = 31.60GPa Ans2⋅ (1 + ν) 192. Problem 3-36The elastic portion of the tension stress-strain diagram for an aluinmum alloy is shown in the figure. Thespecimen used for the test has a gauge length of 50 mm and a diameter of 12.5 mm. If the applied load is50 kN deterinme the new diameter of the specimen. The shear modulus is Gal = 28 GPa.Given: d := 12.5mm L := 50mmP := 50kN G := 28⋅GPaSolution:Normal Stress:Aπ4⎛⎜⎝⎞⎠:= ⋅d2σPA:= σ = 407.44MPaNormal Strain: From the stress-strain diagram, the modulus of elasticity isE500MPa0.00614:= E = 81433.22MPaApplying Hook's law σ = Eε .εlongσE:= εlong 0.0050033mmmm=Poisson's Ratio: GE2⋅ (1 + ν)=Thus, νE2⋅G⎛⎜⎝⎞⎠:= − 1 ν = 0.454The lateral and longitudinal strain can be related using Poisson's ratio.εlat := −ν⋅ (εlong) εlat −0.00227233mmmm=Δd := εlat⋅ (d) Δd = −0.028404mmd' := d + Δd d' = 12.4716mm Ans 193. Problem 3-37The head H is connected to the cylinder of a compressor using six steel bolts. If the clamping force ineach bolt is 4 kN, deterinme the normal strain in the bolts. Each bolt has a diameter of 5 mm. If σY =280 MPa and Est = 200 GPa, what is the strain in each bolt when the nut is unscrewed so that theclamping force is released?Given: d := 5mm σY := 28MPaP := 4kN E := 200⋅GPaSolution:Normal Stress:Aπ4⎛⎜⎝⎞⎠:= ⋅d2σPA:= σ = 203.7183MPa ( < σY = 280 MPa )Normal Strain: Since σ < σY, Hook's law is still valid.εσE:= ε 0.0010186mmmm= AnsIf the nut is unscrewed, the load is zero. Therefore, the strain ε = 0. Ans 194. Problem 3-38The rigid pipe is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 5mm., determine how much it stretches when a load of P = 1.5 kN acts on the pipe. The materialremains elastic.Given: P := 1.5kN Est := 200⋅GPaLBC := 2.4m d := 5mmθ := 60degSolution:Support Reactions:ΣΜC=0; −FAB⋅cos(θ)⋅ (LAB) + P⋅ (LAB) = 0FABPcos(θ) := FAB = 3 kNNormal Stress:Areaπ4⎛⎜⎝⎞⎠:= ⋅d2σABFABArea:= σAB = 152.789MPaNormal Strain: LABLBCsin(θ) := LAB = 2.771mApplying Hook's law σ = Eε .εABσABEst:= εAB 0.0007639mmmm=Thus, δLAB := εAB⋅ (LAB)δLAB = 2.1171mm Ans 195. Problem 3-39The rigid pipe is supported by a pin at C and an A-36 guy wire AB. If the wire has a diameter of 5mm., determine the load P if the end B is displaced 2.5 mm. to the right. Est = 200 GPa.Given: L := 2.4m d := 5mmθ := 60deg δBx := 2.5mmEst := 200⋅GPaSolution:Consider triangle BB'C: L⋅ sin(θC) δB= xθC asinδBxL⎛⎜⎝⎞⎠:=Consider triangle AB'C: θ'C := 90deg + θC LAC := L⋅cot(θ)2 + L2 2 L:= − ⋅ ( AC)⋅L⋅cos(θ'C)L'AB LACL'AB = 2.7725mNormal Strain:LABLsin(θ) :=εABL'AB − LABLAB:= εAB 0.0004510mmmm=Normal Stress:Areaπ4⎛⎜⎝⎞⎠:= ⋅d2Applying Hook's law σ = Eε .σAB := εAB⋅Est σAB = 90.191MPaThus, FAB := σAB⋅ (Area) FAB = 1.771 kNSupport Reactions:ΣΜC=0; FAB⋅cos(θ)⋅ (L) − P⋅L = 0P := FAB⋅cos(θ) P = 0.885 kN Ans 196. Problem 3-40While undergoing a tension test, a copper-alloy specimen having a gauge length of 50 mm. is subjectedto a strain of 0.40 mm./mm. when the stress is 490 MPa. If σY = 315 MPa when εY = 0.0025mm./mm., determine the distance between the gauge points when the load is released.Given: L0 := 50mm ε1 0.40mmmm:= σ1 := 490⋅MPaεY 0.0025mmmm:= σY := 315MPaSolution:Modulus of Elasticity:EσYεY:=E = 126.00GPaElastic Recovery:rReσ1E:= rRe 0.0038889mmmm=Permanent set:rps := ε1 − rRe rps 0.39611mmmm=Permanent elongation:ΔL := rps⋅L0 ΔL = 19.806mmL := L0 + ΔL L = 69.806mm Ans 197. Problem 3-41The 8-mm-diameter bolt is made of an aluminum alloy. It fits through a magnesium sleeve that has aninner diameter of 12 mm and an outer diameter of 20 mm. If the original lengths of the bolt and sleeveare 80 mm and 50 mm, respectively, determine the strains in the sleeve and the bolt if the nut on thebolt is tightened so that the tension in the bolt is 8 kN. Assume the material at A is rigid. Eal = 70 GPa,Emg = 45 GPa.Given: Lb := 80mm db := 8mmLs := 30mm ds_o := 20mm ds_i := 12mmP := 8kN Eal := 70GPa Emg := 45GPaSolution:Abπ4⎛⎜⎝⎞⎠:= ⋅ db2 Asπ4⎛⎜⎝⎞⎠ds_2 2 o− ds_i⎛⎝⎞⎠:= ⋅Normal Stress:σbPAb:= σb = 159.15MPaσsPAs:= σs = 39.79MPaNormal Strain:εbσbEal:= εb 0.002274mmmm= AnsεsσsEmg:= εs 0.000884mmmm= Ans 198. Problem 3-42A tension test was performed on a steel specimen having an original diameter of 12.5mm and a gaugelength of 50 mm. The data is listed in the table. Plot the stress-strain diagram and determineapproximately the modulus of elasticity, the ultimate stress, and the rupture stress. Use a scale of 20mm = 50 MPa and 20 mm = 0.05 mm/mm. Redraw the linear-elastic region, using the same stressscale but a strain scale of 20 mm = 0.001 mm/mm.Given: d := 12.5* L := 50* P:= δ0.011.131.937.840.943.653.462.364.562.358.80.00000.01750.06000.10200.16500.24901.01603.04806.35008.890011.9380:=Solution: Aπ4⎛⎜⎝⎞⎠:= ⋅d2 * σP⋅103A:= *εδL:= *σ=P/A(MPa)ε=δ/L(mm/mm)UseF(x)xx0.3x0.6⎛⎜⎜⎜⎝⎞⎟⎠0.000090.4509259.9446308.0221333.2832355.2848435.1423507.6661525.5933507.6661479.1455⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟σ :== * ε⎠0.0000000.0003500.0012000.0020400.0033000.0049800.0203200.0609600.1270000.1778000.238760⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟= *⎠Fit107.3252123.113−2190.474⎛⎜⎜⎝⎞⎠=x := min(ε) , 0.005 .. max(ε)Y(x) := F(x)⋅Fit0 0.001 0.002 0.003 0.004 0.00555050045040035030025020015010050σε0 0.05 0.1 0.15 0.2 0.2555050045040035030025020015010050σY(x)ε , xLoad(kN)Elongation(mm)Fit := linfit(ε , σ , F) 199. Modulus of Elasticity: From th stress-strain diagram,From th stress-strain diagram, σult = 530MPa AnsΔε (0.0005 − 0)mmmm:= * σR = 479MPa AnsΔσ := (125 − 0)MPa *EapproxΔσΔε:= *Eapprox = 250GPa* Ans 200. Problem 3-43A tension test was performed on a steel specimen having an original diameter of 12.5 mm and a gaugelength of 50 mm. Using the data listed in the table, plot the stress-strain diagram and determineapproximately the modulus of toughness. Use a scale of 20 mm = 50 MPa and 20 mm = 0.05 mm/mm.Given: d := 12.5* L := 50* P:= δ0.011.131.937.840.943.653.462.364.562.358.80.00000.01750.06000.10200.16500.24901.01603.04806.35008.890011.9380:=Solution: Aπ4⎛⎜⎝⎞⎠:= ⋅d2 * σP⋅103A:= *εδL:= *σ=P/A(MPa)ε=δ/L(mm/mm)UseF(x)xx0.3x0.6⎛⎜⎜⎜⎝⎞⎟⎠0.000090.4509259.9446308.0221333.2832355.2848435.1423507.6661525.5933507.6661479.1455⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟σ :== * ε⎠0.0000000.0003500.0012000.0020400.0033000.0049800.0203200.0609600.1270000.1778000.238760⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟= * Fit := linfit(ε , σ , F)⎠Fit107.3252123.113−2190.474⎛⎜⎜⎝⎞⎠=x := min(ε) , 0.005 .. max(ε)Y(x) := F(x)⋅Fit0 0.05 0.1 0.15 0.2 0.2555050045040035030025020015010050σY(x)ε , xModulus of Toughness:The modulus of toughness is equal to the area under thecurve, and could be approximated by counting thenumber of sqaures. the total number of squares is:n := 188.5Δεsq (0.025)mmmm:=Δσsq := 25MPaut := n⋅ (Δεsq)⋅ (Δσsq)ut = 117.81MPa AnsElongation(mm)Load(kN) 201. Problem 3-44An 8-mm-diameter brass rod has a modulus of elasticity of Ebr = 100 GPa. If it is 3 m long andsubjected to an axial load of 2 kN, determine its elongation. What is its elongation under the same loadif its diameter is 6 mm?Given: P := 2kNL := 3m d1 := 8mm d2 := 6mmEbr := 100⋅GPaSolution:Case 1: A1π4⎛⎜⎝⎞⎠:= ⋅ 2d1σ1PA1:= σ1 = 39.789MPaε1σ1Ebr:= ε1 0.0003979mmmm=δ1 := ε1⋅L δ1 = 1.194mm AnsCase 2: A2π4⎛⎜⎝⎞⎠:= ⋅ 2d2σ2PA2:= σ2 = 70.736MPaε2σ2Ebr:= ε2 0.0007074mmmm=δ2 := ε2⋅L δ2 = 2.122mm Ans 202. Problem 4-1The ship is pushed through the water using an A-36 steel propeller shaft that is 8 m long, measuredfrom the propeller to the thrust bearing D at the engine. If it has an outer diameter of 400 mm and awall thickness of 50 mm, determine the amount of axial contraction of the shaft when the propellerexerts a force on the shaft of 5 kN. The bearings at B and C are journal bearings.Given: F := 5kNL := 8m do := 400mm t := 50mmE := 200⋅GPaSolution:di := do − 2tAπ4⎛⎜⎝⎞⎠do2 − di2 ⎛⎝⎞⎠:= ⋅Internal Force: As shown on FBD.P := −FDisplacement:δAP⋅LA⋅E:= δA = −0.00364mm AnsNote: Negative sign indicates that end A moves towards end D. 203. Problem 4-2The A-36 steel column is used to support the symmetric loads from the two floors of a building.Determine the vertical displacement of its top, A, if P1 = 200 kN, P2 = 310 kN, and the column has across-sectional area of 14625 mm2.Given: L := 3.6m A := 14625mm2P1 := 200kN P2 := 310kNE := 200⋅GPaSolution:Internal Force: As shown on FBD.Displacement:δAB−2P1⋅LA⋅E:=δBC−2(P1 + P2)⋅LA⋅E:=δA := δAB + δBC δA = −1.74769mm AnsNote: Negative sign indicates that end A moves towards end C. 204. Problem 4-3The A-36 steel column is used to support the symmetric loads from the two floors of a building.Determine the loads P1 and P2 if A moves downward 3 mm and B moves downward 2.25 mm whenthe loads are applied. The column has a cross-sectional area of 14625 mm2.Given: L := 3.6m A := 14625mm2δA := −3mm δB := −2.25mmE := 200⋅GPaSolution:Internal Force: As shown on FBD.Displacement:Initial guess: P1 := 1kN P2 := 2kNGivenFor AB: δA − δB−2P1⋅LA⋅E= [1]For BC: δB−2(P1 + P2)⋅L= [2]A⋅ESolving [1] and [2]:P1P2⎛⎜⎜⎝⎞⎠:= Find(P1 , P2)P1P2⎛⎜⎜⎝⎞⎠304.69609.38⎛⎜⎝⎞⎠= kN Ans 205. Problem 4-4The copper shaft is subjected to the axial loads shown. Determine the displacement of end A withrespect to end D if the diameters of each segment are dAB = 20 mm, dBC = 25 mm, and dCD = 12 mm.Take Ecu = 126 GPa.Given:LAB := 2m dAB := 20mmLBC := 3.75m dBC := 25mmLCD := 2.5m dCD := 12mm E := 126⋅GPaPA := −40kN PB := 25kN PC := 10kN PD := −30kNSolution:Internal Force: As shown on FBD.Displacement:AABπ4⎛⎜⎝⎞⎠:= ⋅ dAB2 δABPA⋅ (LAB)E⋅ (AAB):=ABCπ4⎛⎜⎝⎞⎠:= ⋅ dBC2 δBC(PA + 2PB)⋅ (LBC)E⋅ (ABC):=ACDπ4⎛⎜⎝⎞⎠:= ⋅ dCD2 δCD(PA + 2PB + 2PC)⋅ (LCD)E⋅ (ACD):=δA_D := δAB + δBC + δCDδA_D = 3.8483mm AnsNote: The positive sign indicates that end A moves away from end D. 206. Problem 4-5The A-36 steel rod is subjected to the loading shown. If the cross-sectional area of the rod is 60 mm2,Determine the displacement of B and A. Neglect the size of the couplings at B, C, and D.Given:LAB := 0.50m LBC := 1.50m LCD := 0.75mPA := 8kN PB := 2kN PC := 3.3kN45hB:= vBθC := 60deg 35:=A := 60mm2 E := 200⋅GPaSolution:Internal Force: As shown on FBD.Displacement:δABPA⋅ (LAB)E⋅A:=δBCPA 2PB vB ( ) ⋅ + ⎡⎣⎤⎦⋅ (LBC)E⋅A:=δCDPA 2PB vB ( ) ⋅ + 2PC sin θC ( ) ⋅ + ⎡⎣⎤⎦⋅ (LCD)E⋅A:=δB := δBC + δCD δB = 2.307mm AnsδA := δAB + δBC + δCD δA = 2.641mm Ans 207. Problem 4-6The assembly consists of an A-36 steel rod CB and a 6061-T6 aluminum rod BA, each having adiameter of 25 mm. Determine the applied loads P1 and P2 if A is displaced 2 mm to the right and B isdisplaced 0.5 mm to the left when the loads are applied. The unstretched length of each segment isshown in the figure. Neglect the size of the connections at B and C, and assume that they are rigid.Given: LBA := 1.2m LCB := 0.6mδA := 2mm δB := −0.5mmd := 25mm ECB := 200⋅GPaEBA := 68.9⋅GPaSolution:Internal Force: As shown on FBD.Displacement:Aπ4⎛⎜⎝⎞⎠:= ⋅d2Initial guess: P1 := 1kN P2 := 2kNGivenFor BA: δA − δBP1⋅LBAA⋅EBA= [1]For CB: δB(P1 − P2)⋅LCB= [2]A⋅ECBSolving [1] and [2]:P1P2⎛⎜⎜⎝⎞⎠:= Find(P1 , P2)P1P2⎛⎜⎜⎝⎞⎠70.46152.27⎛⎜⎝⎞⎠= kN Ans 208. Problem 4-7The 15-mm-diameter A-36 steel shaft AC is supported by a rigid collar, which is fixed to the shaft atB. If it is subjected to an axial load of 80 kN at its end, determine the uniform pressure distribution pon the collar required for equilibrium.Also, what is the elongation on segment BC and segment BA?Given:LAB := 200mm LBC := 500mm d := 15mmPC := 80kN E := 200⋅GPa rc := 35mmSolution:Acollarπ4⎛⎜⎝⎞⎠2rc ( )2 d2 − ⎡⎣⎤⎦:= ⋅Equations of equilibrium:+ ΣFy=0; p⋅ (Acollar) − PC = 0pPCAcollar:= p = 21.79MPa AnsInternal Force: As shown on FBD.Displacement:π4⎛⎜⎝⎞⎠A:= ⋅d2PBC := PC PBA := 0 δBCPBC⋅ (LBC):= δBC = 1.132mm AnsE⋅AδBAPBA⋅ (LAB):= δBA = 0mm AnsE⋅A 209. Problem 4-8The load is supported by the four 304 stainless steel wires that are connected to the rigid members ABand DC. Determine the vertical displacement of the 2.5-kN load if the members were horizontal whenthe load was originally applied. Each wire has a cross-sectional area of 16 mm2.Given: LDE := 0.9m LCF := 0.9m P := 2.5kNLDH := 0.3m LHC := 0.6m A := 16mm2LAH := 0.54m LBG := 1.5mE := 193⋅GPa LAI := 0.9m LIB := 0.3mSolution: LDC := LDH + LHC LAB := LAI + LIBInternal Forces in the wires :From FBD (b):ΣΜA=0; FBG⋅ (LAB) − P⋅ (LAI) = 0+ ΣFy=0; FAH + FBG − P = 0From FBD (a):FBG PLAILAB⎛⎜⎝⎞⎠:= ⋅ FBG = 1.875 kN ΣΜD=0; FCF⋅ (LDC) − FAH⋅ (LDH) = 0FAH := P − FBG FAH = 0.625 kN +ΣFy=0; FCF + FDE − FAH = 0FCF FAHLDHLDC⎛⎜⎝ ⎞:= ⋅ FCF = 0.2083 kN⎠FDE := FAH − FCF FDE = 0.4167 kNDisplacement:δDFDE⋅ (LDE):= δD = 0.12143782mmE⋅AδCFCF⋅ (LCF):= δC = 0.06071891mmE⋅AδH δCLHCLDC⎛⎜⎝⎞⎠:= + ⋅ (δD − δC) δH = 0.10119819mmδA δHFAH⋅ (LAH)E⋅A⎡⎢⎣⎤⎥⎦:= +δBFBG⋅ (LBG)E⋅A:=δI δALAILAB⎛⎜⎝⎞⎠:= + ⋅ (δB − δA) δI = 0.736mm Ans 210. Problem 4-9The load is supported by the four 304 stainless steel wires that are connected to the rigid members ABand DC. Determine the angle of tilt of each member after the 2.5-kN load is applied. The memberswere originally horizontal, and each wire has a cross-sectional area of 16 mm2.Given: P := 2.5kN LDE := 0.9m LCF := 0.9mLHC := 0.6mLDH := 0.3mSolution:A := 16mm2LAH := 0.54m LBG := 1.5mE := 193⋅GPa LAI := 0.9m LIB := 0.3mFAH := P − FBG FAH = 0.625 kN ΣFy=0; FCF + FDE − FAH = 0AnsLDC := LDH + LHC LAB := LAI + LIBInternal Forces in the wires :From FBD (b):ΣΜA=0; FBG⋅ (LAB) − P⋅ (LAI) = 0+ ΣFy=0; FAH + FBG − P = 0From FBD (a):FBG PLAILAB⎛⎜⎝⎞⎠:= ⋅ FBG = 1.875 kN ΣΜD=0; FCF⋅ (LDC) − FAH⋅ (LDH) = 0+FCF FAHLDHLDC⎛⎜⎝⎞⎠:= ⋅ FCF = 0.2083 kNFDE := FAH − FCF FDE = 0.4167 kNDisplacement:δDFDE⋅ (LDE):= E⋅AδD = 0.12143782mmFCF⋅ (LCF)δC:= δC = 0.06071891mmE⋅AδH δCLHCLDC⎛⎜⎝⎞⎠:= + ⋅ (δD − δC) δH = 0.10119819mmtan(αDC) δD − δC:= αDC atanLDCδD − δCLDC⎛⎜⎝⎞⎠:= αDC = 0.0039 deg AnsδA δHFAH⋅ (LAH)E⋅A⎡⎢⎣⎤⎥⎦:= + δA = 0.21049223mmδBFBG⋅ (LBG):= = 0.91078368mmE⋅AδB tan(βAB) δB − δA:= βAB atanLABδB − δALAB⎛⎜⎝⎞⎠:= βAB = 0.0334 deg 211. Problem 4-10The bar has a cross-sectional area of 1800 mm2, and E = 250 GPa. Determine the displacement of itsend A when it is subjected to the distributed loading.Given: L := 1.5m E := 250⋅GPaA := 1800mm21⋅ 3w 500 xNm=Solution:Internal Force: As shown on FBD.Pxxw⋅x x⌠⎮⌡= d Px0x013⎛⎜⎝⎞⎤⎥⎦⋅ ⎠⋅x⎡⎢⎣500 x x⌠⎮⎮⌡= dPx15004x43⎛⎜⎝⎞= ⋅ ⎠Displacement: unit := 1N⋅mδAL0xPxE⋅A⌠⎮⎮⌡= d L = 1.50mδAunitA⋅E48⌠⎮⎮⎮⌡d0x⎛⎜⎜⎝ ⎞150044⋅ 3x⎠⎡⎢⎢⎢⎣⎤⎥⎥⎥⎦:=δA = 2.990mm Ans 212. Problem 4-11The assembly consists of three titanium (Ti-6A1-4V) rods and a rigid bar AC. The cross-sectional areaof each rod is given in the figure. If a force of 30 kN is applied to the ring F, determine the horizontaldisplacement of point F.Given: LCD := 1.2m LEC := 0.6m LEF := 0.3mLAB := 1.8m LAE := 0.3mP := 30kN E := 120⋅GPaACD := 600mm2 AAB := 900mm2AEF := 1200mm2Solution: LAC := LAE + LECInternal Forces in the rods :From FBD :FCD PLAELAC⎛⎜⎝⎞⎠ΣΜ := ⋅ A=0; FCD⋅ (LAC) − P⋅ (LAE) = 0 FCD = 10.00 kN+ ΣFy=0; FCD + FAB − P = 0 FAB := P − FCD FAB = 20.00 kNDisplacement: δCFCD⋅ (LCD)E⋅ACD:= δC = 0.1667mmδAFAB⋅ (LAB)E⋅AAB:= δA = 0.3333mmδE δCLECLAC⎛⎜⎝⎞⎠:= + ⋅ (δA − δC) δE = 0.2778mmδF_EP⋅ (LEF)E⋅AEF:= δF_E = 0.0625mmδF := δE + δF_E δF = 0.340278mm Ans 213. Problem 4-12The assembly consists of three titanium (Ti-6A1-4V) rods and a rigid bar AC.The cross-sectional areaof each rod is given in the figure. If a force of 30 kN is applied to the ring F, determine the angle of tiltof bar AC.Given: LCD := 1.2m LEC := 0.6m LEF := 0.3mLAB := 1.8m LAE := 0.3mP := 30kN E := 120⋅GPaACD := 600mm2 AAB := 900mm2AEF := 1200mm2Solution: LAC := LAE + LECInternal Forces in the rods :From FBD :FCD PLAELAC⎛⎜⎝⎞⎠ΣΜ := ⋅ A=0; FCD⋅ (LAC) − P⋅ (LAE) = 0 FCD = 10.00 kN+ ΣFy=0; FCD + FAB − P = 0 FAB := P − FCD FAB = 20.00 kNDisplacement: δCFCD⋅ (LCD)E⋅ACD:= δC = 0.1667mmδAFAB⋅ (LAB)E⋅AAB:= δA = 0.3333mmtan(αAC) δA − δC:= αAC atanLACδA − δCLAC⎛⎜⎝⎞⎠:= αAC = 0.01061 deg Ans 214. Problem 4-13A spring-supported pipe hanger consists of two springs which are originally unstretched and have astiffness of k = 60 kN/m, three 304 stainless steel rods, AB and CD, which have a diameter of 5 mm,and EF, which has a diameter of 12 mm, and a rigid beam GH. If the pipe and the fluid it carries havea total weight of 4 kN, determine the displacement of the pipe when it is attached to the support.Given: LAB := 0.75m LCD := 0.75m LEF := 0.75mdAB := 5mm dCD := 5mm dEF := 12mmLGE := 0.25m LEH := 0.25m k 60kNm:=P := 4kN E := 193⋅GPaSolution: LGH := LGE + LEHInternal Forces in the rods :From FBD (a) :ΣΜA=0; FCD⋅ (LGH) − P⋅ (LGE) = 0+ ΣFy=0; FCD + FAB − P = 0FCD PLGELGH⎛⎜⎝⎞⎠:= ⋅ FCD = 2.00 kNFAB := P − FCD FAB = 2.00 kNFrom FBD (b) :+ ΣFy=0; FEF ⋅ −(FCD + FAB) = 0FEF := FCD + FAB FEF = 4.00 kNDisplacement:AABπ4⎛⎜⎝⎞⎠:= ⋅ dAB2 ACDπ4⎛⎜⎝⎞⎠:= ⋅ dCD2 AEFπ4⎛⎜⎝⎞⎠:= ⋅ 2dEFδEFEF⋅ (LEF)E⋅ (AEF):= δE = 0.13744mmδD δEFCDk⎛⎜⎝⎞⎠:= + δD = 33.47077mmDue to symmetry, δB := δDδC_DFCD⋅ (LCD)E⋅ (ACD):= δC_D = 0.39583mmDue to symmetry, δB_A := δC_Dδtotal := δD + δC_D δtotal = 33.8666mm Ans 215. Problem 4-14A spring-supported pipe hanger consists of two springs, which are originally unstretched and have astiffness of k = 60 kN/m, three 304 stainless steel rods, AB and CD, which have a diameter of 5 mm,and EF, which has a diameter of 12 mm, and a rigid beam GH. If the pipe is displaced 82 mm when itis filled with fluid, determine the weight of the fluid.Given: LAB := 0.75m LCD := 0.75m LEF := 0.75mdAB := 5mm dCD := 5mm dEF := 12mmLGE := 0.25m LEH := 0.25m k 60kNm:=δtotal := 82mm E := 193⋅GPaSolution: LGH := LGE + LEHInternal Forces in the rods : Iniatlly set: P := 1kNFrom FBD (a) :ΣΜA=0; FCD⋅ (LGH) − P⋅ (LGE) = 0+ ΣFy=0; FCD + FAB − P = 0FCD PLGELGH⎛⎜⎝⎞⎠:= ⋅ FCD = 0.50 kNFAB := P − FCD FAB = 0.50 kNFrom FBD (b) :+ ΣFy=0; FEF ⋅ −(FCD + FAB) = 0FEF := FCD + FAB FEF = 1.00 kNDisplacement:AABπ4⎛⎜⎝⎞⎠:= ⋅ dAB2 ACDπ4⎛⎜⎝⎞⎠:= ⋅ dCD2 AEFπ4⎛⎜⎝⎞⎠:= ⋅ 2dEFδEFEF⋅ (LEF)E⋅ (AEF):= δE = 0.03436mm:= + δD = 8.36769mmδD δEFCDk⎛⎜⎝⎞⎠Due to symmetry, δB := δDδC_DFCD⋅ (LCD)E⋅ (ACD):= δC_D = 0.09896mmDue to symmetry, δB_A := δC_Dδ'total := δD + δC_D δ'total = 8.4666mmWPδtotalδ'total= W Pδtotalδ'total⎛⎜⎝⎞⎠:= ⋅ W = 9.69 kN Ans 216. Problem 4-15The assembly consists of three titanium rods and a rigid bar AC.The cross-sectional area of each rod isgiven in the figure. If a vertical force P = 20 kN is applied to the ring F, determine the verticaldisplacement of point F. Eti = 350 GPa.Given: LBA := 2m LDC := 2m LEF := 1.5mLAE := 0.5m LEC := 0.75mABA := 60mm2 ADC := 45mm2 AEF := 75mm2P := 20kN E := 350⋅GPaSolution: LAC := LAE + LECInternal Forces in the rods :From FBD :FDC PLAELAC⎛⎜⎝⎞⎠ΣΜ := ⋅ A=0; FDC⋅ (LAC) − P⋅ (LAE) = 0 FDC = 8.00 kN+ ΣFy=0; FDC + FBA − P = 0 FBA := P − FDC FBA = 12.00 kNDisplacement: δCFDC⋅ (LDC)E⋅ADC:= δC = 1.0159mmδAFBA⋅ (LBA)E⋅ABA:= δA = 1.1429mmδE δCLECLAC⎛⎜⎝⎞⎠:= + ⋅ (δA − δC) δE = 1.0921mmδF_EP⋅ (LEF)E⋅AEF:= δF_E = 1.1429mmδF := δE + δF_E δF = 2.2349mm Ans 217. Problem 4-16The linkage is made of three pin-connected A-36 steel members, each having a cross-sectional area of500 mm2. If a vertical force of P = 250 kN is applied to the end B of member AB, determine thevertical displacement of point B.Given: a := 1.5m b := 2mLAD := a2 + b2 LAD = 2.5m LAB := 3mLAC := a2 + b2 LAC = 2.5mE := 200⋅GPa A := 500mm2 P := 250kNSolution: c := a2 + b2 hac:= vbc:=θ atanab⎛⎜⎝⎞⎠:= θ = 36.869898 degEquation of equilibrium :For AB :+ ΣFy=0; (FAD + FAC)⋅v − P = 0Since FAD = FAC FACP2v:= FAC = 156.25 kNDisplacement:δA_CFAC⋅ (LAC):= * δA_C = 3.9063mmE⋅AδB_AP⋅ (LAB)E⋅A:= δB_A = 7.5mmConsider triangle AA'C: LA'C := LAC + δA_CLAA' = δAθ'A := 180deg − θsin(φA')LACsin(θ'A)LA'C= φA' asin LACsin(θ'A)LA'C⋅⎛⎜⎝⎞⎠:= φA' = 36.80289 degθ'C := 180deg − θ'A − φA' θ'C = 0.0670094 degsin(θ'C)δAsin(θ'A)LA'Csin(θ'C)sin(θ'A) := ⋅ δA = 4.8731mm= δA LACδB := δA + δB_A δB = 12.37mm Ans 218. Problem 4-17The linkage is made of three pin-connected A-36 steel members, each having a cross-sectional area of500 mm2. Determine the magnitude of the force P needed to displace point B 2.5 mm downward.Given: a := 1.5m b := 2mLAD := a2 + b2 LAD = 2.5m LAB := 3mLAC := a2 + b2 LAC = 2.5mE := 200⋅GPa A := 500mm2 δB := 2.5mmSolution: c := a2 + b2 hac:= vbc:=θ atanab⎛⎜⎝⎞⎠:= θ = 36.869898 degEquation of equilibrium :For AB :+ ΣFy=0; (FAD + FAC)⋅v − P = 0Since FAD = FAC FACP2v= FADP2v=Displacement:δA_CFAC⋅ (LAC)= δA_CE⋅AP⋅ (LAC)2v(E⋅A)=δB_AP⋅ (LAB)E⋅A=δB = δA + δB_A δA δBP⋅ (LAB)E⋅A= −Consider triangle AA'C: LA'C = LAC + δA_CLAA' = δAθ'A := 180deg − θ θ'A = 143.130102 degLA'Cδ= 2 + LAC2 − 2 ( A)⋅ (LAC)⋅cos(θ'A)2 δAInitial guess: P := 1kNGiven⎡⎢⎣⎤LACP⋅ (LAC)2v(E⋅A)+⎥⎦2= δB− ⋅ (LAC)⋅cos(θ'A)P⋅ (LAB)E⋅A−⎡⎢⎣⎤⎥⎦2LAC+ 2 2 δBP⋅ (LAB)E⋅A−⎡⎢⎣⎤⎥⎦Solving : P := Find(P) P = 50.47 kN Ans 219. Problem 4-18Consider the general problem of a bar made from m segments, each having a constant cross-sectionalarea Am and length Lm. If there are n loads on the bar as shown, write a computer program that can beused to determine the displacement of the bar at any specified location x. Show an application of theprogram using the values L1 = 1.2 m, d1 = 0.6 m, P1 = 2 kN, A1 = 1875 mm2, L2 = 0.6 m, d2 = 1.8 m,P2 = -1.5 kN, A2 = 625 mm2. 220. Problem 4-19The rigid bar is supported by the pin-connected rod CB that has a cross-sectional area of 14 mm2 andis made from 6061-T6 aluminum. Determine the vertical deflection of the bar at D when the distributedload is applied.Given: L := 4m w 300Nm:=a := 1.5m b := 2mA := 14mm2 E := 68.9⋅GPaSolution:LBC := a2 + b2 vaLBC:= hbLBC:=Support Reactions:ΣΜA=0; FBC⋅ (v)⋅ (b) − w⋅ (L)⋅ (0.5⋅L) = 0 FBCw⋅L22⋅v⋅b:=FBC = 2 kNDisplacement:δB_CFBC⋅ (LBC):= δB_C = 5.183mmE⋅AConsider triangle AB'C: LB'C := LBC + δB_C LB'C = 2.505183m2 a2 + b2 2(a)⋅ (b) cos θ'= − ⋅ ( A)LB'Cθ'A acosa2 + b2 LB'C− 22⋅a⋅b⎛⎜⎝⎞⎠:= θ'A = 90.24775 degφ := θ'A − 90deg φ = 0.24775 degφ = 0.0043241 radδD := φ⋅L δD = 17.30mm Ans 221. Problem 4-20The rigid beam is supported at its ends by two A-36 steel tie rods. If the allowable stress for the steel isσallow = 115 MPa, the load w = 50 kN/m, and x = 1.2 m, determine the diameter of each rod so that thebeam remains in the horizontal position when it is loaded.Given: LCD := 1.8m LAC := 2.4mLAB := 1.8m x := 1.2mw 50kNm:= σallow := 115⋅MPaSolution:Internal Forces in the rods :From FBD :FCDw⋅x22⋅LACΣΜ := A=0; FCD⋅ (LAC) − (w⋅x)⋅ (0.5x) = 0 FCD = 15.00 kN+ ΣFy=0; FCD + FAB − w⋅x = 0 FAB := w⋅x − FCD FAB = 45.00 kNDisplacement: To maintain the rigid beam in the horizontal position, the elongation of bothrods AB and CD must be the same. δA = δCAABπ4⎛⎜⎝⎞⎠= ⋅ dAB2 ACDπ4⎛⎜⎝⎞⎠= ⋅ 2dCDδAFAB⋅ (LAB)E⋅AAB= δCFCD⋅ (LCD)E⋅ACD=Thus,FAB⋅ (LAB)Eπ⋅ 2 4⎡⎢⎣⎛⎜⎝⎞⎠dAB⎤⎥⎦⋅FCD⋅ (LCD)E= dCDπ2 ⎥⎦⎤⋅⋅ 4⎡⎢⎣⎛⎜⎝⎞⎠dCDdABFCD⋅ (LCD)FAB⋅ (LAB)=dAB = 3⋅dCDAllowable Normal Stress:Assume failure of rods AB.σallowFABπ4= dAB⎛⎜⎝⎞⎠⋅ 2dAB4⋅FAB(π)⋅σallow:= dAB = 22.321mmdCDdAB3:= dCD = 12.887mmAssume failure of rods CD.σallowFCDπ4= dCD⎛⎜⎝⎞⎠⋅ 2dCD4⋅FCD(π)⋅σallow:= dCD = 12.887mm AnsdAB := 3⋅dCD dAB = 22.321mm Ans 222. Problem 4-21The rigid beam is supported at its ends by two A-36 steel tie rods. The rods have diameters dAB = 12mm and dCD = 7.5 mm. If the allowable stress for the steel is σallow = 115 MPa, determine the intensityof the distributed load w and its length x on the beam so that the beam remains in the horizontal positionwhen it is loaded.Given: LCD := 1.8m LAC := 2.4m LAB := 1.8mdAB := 12mm dCD := 7.5mmσallow := 115⋅MPaSolution:AABπ4⎛⎜⎝⎞⎠= ⋅ dAB2 ACDπ4⎛⎜⎝⎞⎠= ⋅ 2dCDAllowable Normal Stress:Assume failure of rods AB.σallowFABAAB⋅ 2 ⎡⎢⎣= FAB (σallow) π4⎛⎜⎝⎞⎠dAB⎤⎥⎦:= ⋅ FAB = 13.006 kNDisplacement: To maintain the rigid beam in the horizontal position, the elongation of bothrods AB and CD must be the same. δA = δCδAFAB⋅ (LAB)E⋅AAB= δCFCD⋅ (LCD)E⋅ACD=Thus,FAB⋅ (LAB)Eπ⋅ 2 4⎡⎢⎣⎛⎜⎝⎞⎠dAB⎤⎥⎦⋅FCD⋅ (LCD)E⋅ 2 ⎡⎢⎣π⎤⎥⎦⋅4⎛⎜⎝⎞⎠dCD=FCDFAB2dCD= FCD2dAB2dCD:= ⋅FAB FCD = 5.081 kN2dAB⎛⎜⎜⎝⎞⎠Internal Forces in the rods :GivenFrom FBD : ΣΜA=0; FCD⋅ (LAC) − (w⋅x)⋅ (0.5x) = 0+ ΣFy=0; FCD + FAB − w⋅x = 0Initial guess: w 1kNm:= x := 1mSolving :wx⎛⎜⎝⎞⎠:= Find(w, x) w 13.41kNm=Ansx = 1.35m 223. Allowable Normal Stress:Assume failure of rods CD.σallowFCDACD⋅ 2 ⎡⎢⎣= FCD (σallow) π4⎛⎜⎝⎞⎠dCD⎤⎥⎦:= ⋅ FCD = 5.081 kNDisplacement: To maintain the rigid beam in the horizontal position, the elongation of bothrods AB and CD must be the same. δA = δCδAFAB⋅ (LAB)E⋅AAB= δCFCD⋅ (LCD)E⋅ACD=Thus,FAB⋅ (LAB)Eπ⋅ 2 4⎡⎢⎣⎛⎜⎝⎞⎠dAB⎤⎥⎦⋅FCD⋅ (LCD)Eπ⋅ 2 4⎡⎢⎣⎛⎜⎝⎞⎠dCD⎤⎥⎦⋅=FCDFAB2dCD= FAB2dAB2dAB:= ⋅FCD FAB = 13.006 kN2dCD⎛⎜⎜⎝⎞⎠The results are the same as assuming failure of rod AB.Therefore, rods AB and CD fail simultaneously. Ans 224. Problem 4-22The post is made of Douglas fir and has a diameter of 60 mm. If it is subjected to the load of 20 kNand the soil provides a frictional resistance that is uniformly distributed along its sides of w = 4 kN/m,determine the force F at its bottom needed for equilibrium. Also, what is the displacement of the top ofthe post A with respect to its bottom B? Neglect the weight of the post.Given: L := 2m d := 60mmP := 20kN w 4kNm:= E := 13.1⋅GPaSolution:Equation od Equilibrium : For entire post [FBD (a)]+ ΣFy=0; −P + w⋅L + F = 0F := P − w⋅L F = 12 kNInternal Force: FBD (b).+ ΣFy=0; −P + w⋅y − Fy = 0Fy = −P + w⋅yDisplacement: Aπ4⎛⎜⎝⎞⎠:= ⋅d2unit := 1kNδA_BL0yFyA⋅E⌠⎮⎮⌡= dδA_BunitA⋅EL(−P + w⋅y) y01kN⋅⌠⎮⎮⌡d⎡⎢⎢⎣⎤⎥⎥⎦:=δA_B = −0.864mm AnsNote: Negative sign indicates that ens A moves towards end B. 225. Problem 4-23The post is made of Douglas fir and has a diameter of 60 mm. If it is subjected to the load of 20 kNand the soil provides a frictional resistance that is distributed along its length and varies linearly from w= 0 at y = 0 to w = 3 kN/m at y = 2 m, determine the force F at its bottom needed for equilibrium.Also, what is the displacement of the top of the post A with respect to its bottom B? Neglect theweight of the post.Given: L := 2m d := 60mmP := 20kN wo 3kNm:= E := 13.1⋅GPaSolution:Equation od Equilibrium : For entire post [FBD (a)]+ ΣFy=0; −P + 0.5wo⋅L + F = 0F := P − 0.5wo⋅L F = 17 kNInternal Force: FBD (b).+ ΣFy=0; −P 0.5 woyL⋅ ⎛⎜⎝⎞⎠+ ⋅ ⋅y − Fy = 0Fy −Pwo2⋅L⎛⎜⎝⎞⎠= + ⋅y2Displacement: Aπ4⎛⎜⎝⎞⎠:= ⋅d2unit := 1kNδA_BL0yFyA⋅E⌠⎮⎮⌡= dδA_BunitA⋅EL0wo2⋅L⎛⎜⎝⎞⎠⎤+ ⋅y2⎡⎢⎣⎥⎦1kN−P ⋅y⌠⎮⎮⌡d⎡⎢⎢⎢⎣⎤⎥⎥⎥⎦:=δA_B = −1.026mm AnsNote: Negative sign indicates that ens A moves towards end B. 226. Problem 4-24The rod has a slight taper and length L. It is suspended from the ceiling and supports a load P at itsend. Show that the displacement of its end due to this load is δ = PL/(πEr2r1). Neglect the weight ofthe material. The modulus of elasticity is E.Solution:Geometry :L + xor2xor1= xoL⋅ r1r2 − r1=Thus, rx r1r2 − r1L= + ⋅x rxr1⋅L + (r2 − r1)⋅xL=2 ⎛⎝Ax π rx⎞⎠= ⋅ AxπL2r1 L ⋅ r2 r1 − ( ) x ⋅ + ⎡⎣= ⋅ 2⎤⎦Displacement:δL0xPAx⋅E⌠⎮⎮⌡= dδP⋅L2π⋅EL0x1r1 L ⋅ r2 r1 − ( ) x ⋅ + ⎡⎣⎤⎦2⌠⎮⎮⎮⌡d⎡⎢⎢⎢⎣⎤⎥⎥⎥⎦= ⋅L0δP⋅L2π⋅E1r2 r1 − ( ) r1 L ⋅ r2 r1 − ( ) x ⋅ + ⎡⎣⎤⎦⋅= ⋅δ−P⋅L2π⋅E⋅ (r2 − r1)1r1⋅L + (r2 − r1)⋅L1r1⋅L− ⎡⎢⎣⎤⎥⎦= ⋅δ−P⋅L2π⋅E⋅ (r2 − r1)1r2⋅L1r1⋅L− ⎛⎜⎝⎞⎠= ⋅δ−P⋅L2π⋅E⋅ (r2 − r1)r2 − r1r1⋅ r2⋅L⎛⎜⎝⎞⎠= ⋅δ−P⋅Lπ⋅E⋅ r1⋅ r2= Ans 227. Problem 4-25Solve Prob. 4-24 by including both P and the weight of the material, considering its specific weight tobe γ (weight per volume).Solution:Internal Force: FBD (b).+ ΣFy=0; Px − wx − P = 0 Px = Wx + PGeometry :L + xor2xor1= xoL⋅ r1r2 − r1=Thus, rx r1r2 − r1L= + ⋅x rxr1⋅L + (r2 − r1)⋅xL=2 ⎛⎝Ax π rx⎞⎠= ⋅ AxπL2r1 L ⋅ r2 r1 − ( ) x ⋅ + ⎡⎣= ⋅ 2⎤⎦Wxγ3⎛⎜⎝⎞⎠= ⋅Ax⋅ (x0 + x)− ⋅ ⋅ (x0)γ⋅π3⎛⎜⎝⎞⎠⎛⎝r12 ⎞⎠Wxγ⋅π3⋅L2⎛⎜⎝⎞⎠r1 L ⋅ r2 r1 − ( ) x ⋅ + ⎡⎣⋅ 2⎤⎦L⋅ r1r2 − r1+ x⎛⎜⎝⎞⎠⋅γ⋅π3⎛⎜⎝⎞⎠⎛⎝r12 ⎞⎠⋅L⋅ r1r2 − r1⎛⎜⎝⎞⎠= − ⋅Wxγ⋅π3⋅L2⋅ (r2 − r1)r1 L ⋅ r2 r1 − ( ) x ⋅ + ⎡⎣⎤⎦3 r13 L3 ⋅ ⎛⎝⎞⎠− ⎡⎣⎤⎦= ⋅Displacement:δ1γ3⋅E⋅ (r2 − r1)L⌠⎮⎮⎮⌡ d0xr1 L ⋅ r2 r1 − ( ) x ⋅ + ⎡⎣⎤⎦3 r13 L3 ⋅ ⎛⎝⎞⎠−r1 L ⋅ r2 r1 − ( ) x ⋅ + ⎡⎣⎤⎦2⎡⎢⎢⎢⎢⎣⎤⎥⎥⎥⎥⎦Lδ = ⋅ 10xWxAx⋅E⌠⎮⎮⎮⌡= dδ1γ3⋅E⋅ (r2 − r1)L0x r1 L ⋅ r2 r1 − ( ) x ⋅ + ⎡⎣⎤⎦⌠⎮⌡3 L3 ⋅ ⎛⎝d r1⎞⎠L0x1r1 L ⋅ r2 r1 − ( ) x ⋅ + ⎡⎣⎤⎦2⌠⎮⎮⎮⌡− ⋅ d⎡⎢⎢⎢⎣⎤⎥⎥⎥⎦= ⋅Using the result of Prob.(4-24) for the 2nd Integral, we haveδ1γ3⋅E⋅ (r2 − r1)r1⋅L2 (r2 − r1) L22+ ⋅⎡⎢⎣⎤⎥⎦r13 ⋅ L3 ⎛⎝⎞⎠1L⋅ r1⋅ r2⎛⎜⎝⎞⎠− ⋅⎡⎢⎣⎤⎥⎦= ⋅δ1γ⋅L2⋅ (r2 + r1)6⋅E⋅ (r2 − r1)γ⋅L2 r1⋅ 23⋅E⋅ (r2 − r1)⋅ r2= ⋅−Using the result of Prob.(4-24) for the 2nd Integral, we haveδ δ1L0xPAx⋅E⌠⎮⎮⌡= + d δγ⋅L2⋅ (r2 + r1)6⋅E⋅ (r2 − r1)γ⋅L2 r1⋅ 23⋅E⋅ (r2 − r1)⋅ r2⋅−P⋅Lπ⋅E⋅ r1⋅ r2= + Ans 228. Problem 4-26The support is made by cutting off the two opposite sides of a sphere that has a radius r0. If theoriginal height of the support is r0 /2, determine how far it shortens when it supports a load P.Themodulus of elasticity is E.Solution:Geometry : r = ro⋅cos(θ) y = ro⋅ sin(θ)Ay = π⋅ r2 dy = ro⋅cos(θ)⋅dθ= ⋅ 2⋅cos(θ)2 h 0.25 r= ⋅ oAy π roDisplacement: h = ro⋅ sin(θo)0.25⋅ ro = ro⋅ sin(θo)δ 2h0yPAy⋅E⌠⎮⎮⌡d⎛⎜⎜⎜⎝⎞⎟⎠= ⋅ δθo0θ2P⋅ ro⋅cos(θ)π ro⋅ 2⋅cos(θ)2⋅E⌠⎮⎮⎮⌡= d θo := asin(0.25)θo = 14.48 degδ2Pπ⋅ ro⋅E⎛⎜⎝14.48o⎞⎠ 0θ1cos(θ)⌠⎮⎮⌡d⎛⎜⎜⎜⎝⎞⎟⎠= ⋅14.48o2⋅Pπ⋅ ro⋅E⎛⎜⎝⎞⎠= ⋅ (ln(sec(θ) + tan(θ)))δ 0δ0.511⋅Pπ⋅ ro⋅E= AnsAlternatively,Geometry : Ay = π⋅x2 Ay π ro2 y2 − ⎛⎝⎞⎠= ⋅Displacement:δ0.25⋅ro0θ2⋅P2 y2 − ⎛⎝π ro⎞⎠⋅ ⋅E⌠⎮⎮⎮⌡hδ 2 = d0yPAy⋅E⌠⎮⎮⌡d⎛⎜⎜⎜⎝⎞⎟⎠= ⋅0.25ro0 δ2⋅Pπ⋅E⎛⎜⎝⎞⎠12⋅ ro⎛⎜⎝⎞⎠⋅ lnro + yro − y⎛⎜⎝⎞⎠= ⋅δ0.511⋅Pπ⋅ ro⋅E= Ans 229. Problem 4-27The ball is truncated at its ends and is used to support the bearing load P. If the modulus of elasticityfor the material is E, determine the decrease in its height when the load is applied.Solution:Geometry : x2 = r2 − y2Ay = π⋅x2 Ay = π⋅ (r2 − y2)Displacement:δ 2h0yPAy⋅E⌠⎮⎮⌡d⎛⎜⎜⎜⎝⎞⎟⎠= ⋅r2⎛⎜⎝⎞⎠2= r2 − h2δ0.866⋅r0θ2⋅Pπ⋅ (r2 − y2)⋅E⌠⎮⎮⌡= d h32= ⋅ r0.866 r0h = 0.866⋅ rδ2⋅Pπ⋅E⎛⎜⎝⎞⎠12⋅ r⎛⎜⎝⎞⎠⋅ lnr + yr − y⎛⎜⎝⎞⎠= ⋅δ2.63⋅Pπ⋅ ro⋅E= Ans 230. Problem 4-28Determine the elongation of the aluminum strap when it is subjected to an axial force of 30 kN. Eal =70 GPa.Given: L1 := 250mmL2 := 800mmd1 := 15mmd2 := 50mm t := 6mmP := 30kN E := 70⋅GPaSolution:Displacement:δ 2P⋅L1⋅ lnE⋅ t⋅ (d2 − d1)d2d1⎛⎜⎝⎞⎠⋅P⋅L2E⋅ t⋅ (d2):= +δ = 2.371mm Ans 231. Problem 4-29The casting is made of a material that has a specific weight γ and modulus of elasticity E. If it isformed into a pyramid having the dimensions shown, determine how far its end is displaced due togravity when it is suspended in the vertical position.Solution:Internal Force: FBD (b).+ ΣFy=0; Pz13− γ⋅A⋅z = 0Pz13= γ⋅A⋅zDisplacement:δL0yPzA⋅E⌠⎮⎮⌡d⎛⎜⎜⎜⎝⎞⎟⎠=δL0yγ⋅A⋅z3A⋅E⌠⎮⎮⌡= dδLz y⌠⎮⌡γ3⋅E 0d⎛⎜⎜⎝⎞⎠=δγ⋅L26⋅E= Ans 232. Problem 4-30The pedestal is made in a shape that has a radius defined by the function r = 2/(2 + y1/2) m, where y isin meter. If the modulus of elasticity for the material is E = 100 MPa, determine the displacement of itstop when it supports the 5-kN load.Given: H := 4m do := 1m d1 := 0.5mP := 5kN E := 100⋅ (106)Par22 + y0.5=Solution:Ay4⋅π(2 + y0.5)2Displacement: A = y = π⋅ r2unit := 1m H := 4δH0yPAy⋅E⌠⎮⎮⌡= dδPE⎛⎜⎝⎞⎠1unit⎛⎜⎝⎞⎠⋅H0y(2 + y0.5)24⋅π⌠⎮⎮⎮⌡d⎡⎢⎢⎢⎣⎤⎥⎥⎥⎦ :=δ = 0.1804mm Ans 233. Problem 4-31The column is constructed from high-strength concrete and six A-36 steel reinforcing rods. If it issubjected to an axial force of 150 kN, determine the average normal stress in the concrete and in eachrod. Each rod has a diameter of 20 mm.Given: L := 1.2m rconc := 100mm dst := 20mmP := 150kN Est := 200⋅GPa Econc := 29⋅GPaSolution:Ast 6π4⎛⎜⎝⎞⎠⎛⎝dst2 ⎞⎠2 ⎛⎝:= ⋅ Aconc π rconc⎞⎠:= ⋅ − AstCompatibility: δst = δconcGiven (Pst)⋅LAst⋅Est(Pconc)⋅LAconc⋅Econc= [1]Equations of equilibrium:Σ+Fy=0; Pst + Pconc − P = 0 [2]Initial guess: Pconc := 1kN Pst := 2kNSolving [1] and [2]:PconcPst⎛⎜⎜⎝⎞⎠:= Find(Pconc , Pst)PconcPst⎛⎜⎜⎝⎞⎠104.15245.848⎛⎜⎝⎞⎠= kNAverage Normal Stress:σstPstAst:= σst = 24.323MPa AnsσconcPconcAconc:= σconc = 3.527MPa Ans 234. Problem 4-32The column is constructed from high-strength concrete and six A-36 steel reinforcing rods. If it issubjected to an axial force of 150 kN, determine the required diameter of each rod so that one-fourthof the load is carried by the concrete and three-fourths by the steel.Given: L := 1.2m rconc := 100mmP := 150kN Pst := 0.75P Pconc := 0.25PEst := 200⋅GPa Econc := 29⋅GPaSolution:Pst = 112.50 kN Pconc = 37.50 kNCompatibility: δst = δconc(Pst)⋅LAst⋅Est(Pconc)⋅LAconc⋅Econc=Ast 6π4⎛⎜⎝⎞⎠⎛⎝dst2 ⎞⎠2 ⎛⎝= ⋅ Aconc π rconc⎞⎠= ⋅ − AstThus,2 ⎛⎝π rconc⎞⎠⋅ − AstAstEstPstPconcEconc= ⋅2 ⎛⎝π rconc⎞⎠⋅AstEstPstPconcEconc⋅ + 1⎛⎜⎝⎞⎠=Ast2 ⎛⎝π rconc⎞⎠⋅EstPst:= Ast = 9523.2948mm2PconcEconc⋅ + 12 2dst= Ast3⋅πdst23⋅π:= Astdst = 44.95mm Ans 235. Problem 4-33The A-36 steel pipe has a 6061-T6 aluminum core. It is subjected to a tensile force of 200 kN.Determine the average normal stress in the aluminum and the steel due to this loading. The pipe has anouter diameter of 80 mm and an inner diameter of 70 mm.Given: L := 400mm do := 80mmP := 200kN di := 70mmEst := 200GPa Eal := 68.9GPaSolution:Aalπ4⎛⎝di2 ⎞⎠:= ⋅ Astπ4do2 − di2 ⎛⎝⎞⎠:= ⋅Compatibility: δst = δalGiven (Pst)⋅LAst⋅Est(Pal)⋅LAal⋅Eal= [1]Equations of equilibrium:+ ΣFx=0; Pst + Pal − P = 0 [2]Initial guess: Pal := 1kN Pst := 2kNSolving [1] and [2]:PalPst⎛⎜⎜⎝⎞⎠:= Find(Pal , Pst)PalPst⎛⎜⎜⎝⎞⎠105.89994.101⎛⎜⎝⎞⎠= kNAverage Normal Stress:σstPstAst:= σst = 79.88MPa AnsσalPalAal:= σal = 27.52MPa Ans 236. Problem 4-34The concrete column is reinforced using four steel reinforcing rods,each having a diameter of 18 mm.Determine the stress in the concrete and the steel if the column is subjected to an axial load of 800 kN.Est = 200 GPa, Ec = 25 GPa.Given: P := 800kN bc := 300mm dst := 18mmEst := 200⋅GPa Ec := 25⋅GPaSolution:Ast 4π4⎛⎜⎝⎞⎠⎛⎝dst2 ⎞⎠2 ⎛⎝:= ⋅ Ac bc⎞⎠:= − AstSet: L := 1mCompatibility: δst = δconcGiven (Pst)⋅LAst⋅Est(Pc)⋅LAc⋅Ec= [1]Equations of equilibrium:+ ΣFy=0; Pst + Pc − P = 0 [2]Initial guess: Pc := 1kN Pst := 2kNSolving [1] and [2]:PcPst⎛⎜⎜⎝⎞⎠:= Find(Pc , Pst)PcPst⎛⎜⎜⎝⎞⎠732.92867.072⎛⎜⎝⎞⎠= kNAverage Normal Stress:σstPstAst:= σst = 65.89MPa AnsσcPcAc:= σc = 8.24MPa Ans 237. Problem 4-35The column is constructed from high-strength concrete and four A-36 steel reinforcing rods. If it issubjected to an axial force of 800 kN, determine the required diameter of each rod so that one-fourthof the load is carried by the steel and three-fourths by the concrete. Est = 200 GPa, Ec = 25 GPa.Given: P := 800kN Ec := 25⋅GPa bc := 300mmPc := 0.75P Est := 200⋅GPaPst := 0.25PSolution:Pst = 200 kN Pc = 600 kNCompatibility: δst = δc(Pst)⋅LAst⋅Est(Pc)⋅LAc⋅Ec== ⋅ 2 Ac bcAst 4π4⎛⎜⎝⎞⎠dst2 ⎛⎝⎞⎠= − AstThus,2 − AstAstbcEstPstPcEc= ⋅bc2AstEstPstPcEc⋅ + 1⎛⎜⎝⎞⎠=Ast2bc:= Ast = 3600mm2EstPstPcEc⋅ + 12 1dst= Astπdst1π:= Astdst = 33.85mm Ans 238. Problem 4-36The A-36 steel pipe has an outer radius of 20 mm and an inner radius of 15 mm. If it fits snuglybetween the fixed walls before it is loaded, determine the reaction at the walls when it is subjected tothe load shown.Given: LAB := 300mm ro := 20mmLBC := 700mm ri := 15mmP := 8kN Est := 200GPa⎛⎝2 − ri2 Solution: A π ro⎞⎠:= ⋅ L := LAB + LBCBy superposition :+ −ΔC + δC = 0(−2P)⋅LABA⋅E(FC)⋅LA⋅E+ = 0FC 2⋅PLABL⎛⎜⎝⎞⎠:= ⋅ FC = 4.80 kN AnsEquations of equilibrium:+ ΣFx=0; FA + FC − 2P = 0FA := 2P − FC FA = 11.20 kN Ans 239. Problem 4-37The 304 stainless steel post A has a diameter of d = 50 mm and is surrounded by a red brass C83400tube B. Both rest on the rigid surface. If a force of 25 kN is applied to the rigid cap, determine theaverage normal stress developed in the post and the tube.Given: L := 200mm dst := 50mm ro := 75mm t := 12mmP := 25kN Est := 193⋅GPa Ebr := 101⋅GPaSolution: ri := ro − tAstπ4⎛⎝dst2 ⎞⎠⎛⎝2 − ri2 := ⋅ Abr π ro⎞⎠:= ⋅Compatibility: δst = δbrGiven (Pst)⋅LAst⋅Est(Pbr)⋅LAbr⋅Ebr= [1]Equations of equilibrium:+ ΣFy=0; Pst + Pbr − P = 0 [2]Initial guess: Pbr := 1kN Pst := 2kNSolving [1] and [2]:PbrPst⎛⎜⎜⎝⎞⎠:= Find(Pbr , Pst)PbrPst⎛⎜⎜⎝⎞⎠14.52510.475⎛⎜⎝⎞⎠= kNAverage Normal Stress:σstPstAst:= σst = 5.335MPa AnsσbrPbrAbr:= σbr = 2.792MPa Ans 240. Problem 4-38The 304 stainless steel post A is surrounded by a red brass C83400 tube B. Both rest on the rigidsurface. If a force of 25 kN is applied to the rigid cap, determine the required diameter d of the steelpost so that the load is shared equally between the post and tube.Given:L := 200mm ro := 75mm t := 12mmP := 25kN Est := 193⋅GPa Ebr := 101⋅GPaPst := 0.5P Pbr := 0.5PSolution: ri := ro − tPst = 12.5 kN Pbr = 12.5 kNCompatibility: δst = δbr(Pst)⋅LAst⋅Est(Pbr)⋅LAbr⋅Ebr=Astπ4⎛⎜⎝⎞⎠= ⋅ dst2 Abr π ro⎛⎝2 − ri2 ⎞⎠:= ⋅Thus,2 − ⎛⎝π ro2 ri⎞⎠⋅AstEstPstPbrEbr= ⋅Ast2 − ⎛⎝π ro2 ri⎞⎠⋅EstPst:= Ast = 2722.54mm2PbrEbr⋅2 4dst= Astπdst4π:= Astdst = 58.88mm Ans 241. Problem 4-39The load of 7.5 kN is to be supported by the two vertical steel wires for which σY = 500 MPa. If,originally, wire AB is 1250 mm long and wire AC is 1252.5 mm long, determine the force developed ineach wire after the load is suspended. Each wire has a cross-sectional area of 12.5 mm2.Given: LAB := 1250mm LAC := 1252.5mm A := 12.5mm2W := 7.5kN E := 200⋅GPa EY := 70500⋅MPaSolution: ΔL := LAC − LAB ΔL = 2.50mmCompatibility:+ δAC + ΔL = δABGiven(TAC)⋅LACA⋅E+ ΔL(TAB)⋅LAB= [1]A⋅EEquations of equilibrium:+ ΣFy=0; TAC + TAB − W = 0 [2]Initial guess: TAC := 1kN TAB := 2kNSolving [1] and [2]:TACTAB⎛⎜⎜⎝⎞⎠:= Find(TAC, TAB)TACTAB⎛⎜⎜⎝⎞⎠1.2496.251⎛⎜⎝⎞⎠= kN AnsCheck Average Normal Stress:σACTACA:= σAC = 99.9MPa [ < σY = 500 MPa]σABTABA:= σAB = 500.1MPa [ < σY = 500 MPa] 242. Problem 4-40The load of 4 kN is to be supported by the two vertical steel wires for which σY = 560 MPa. If,originally, wire AB is 1250 mm long and wire AC is 1252.5 mm long, determine the cross-sectionalarea of AB if the load is to be shared equally between both wires. Wire AC has a cross-sectional areaof 13 mm2.Given: W := 4kN LAB := 1250mm AAC := 13mm2TAC := 0.5W LAC := 1252.5mm EY := 560⋅MPaTAB := 0.5W E := 200⋅GPaSolution: ΔL := LAC − LAB ΔL = 2.50mmTAC = 2 kN TAB = 2 kNCompatibility:+ δAC + ΔL = δAB(TAC)⋅LAC(AAC)⋅E+ ΔL(TAB)⋅LAB(AAB)⋅E=Thus,AAB(TAB)⋅LAB(TAC)⋅LACAAC+ (ΔL)⋅E:=AAB = 3.60911mm2 AnsCheck Average Normal Stress:σACTACAAC:= σAC = 153.846MPa [ < σY = 560 MPa]σABTABAAB:= σAB = 554.15MPa [ < σY = 560 MPa] 243. Problem 4-41The support consists of a solid red brass C83400 post surrounded by a 304 stainless steel tube. Beforethe load is applied, the gap between these two parts is 1 mm. Given the dimensions shown, determinethe greatest axial load that can be applied to the rigid cap A without causing yielding of any one of thematerials.Given: Lbr := 0.251m Ebr := 101⋅GPa σY_br := 70⋅MPaLst := 0.250m Est := 193⋅GPadbr := 60mm di := 80mm t := 10mmSolution: ΔL := Lbr − Lst ΔL = 1mm do := di + 2tAbrπ4⎛⎝dbr2 ⎞⎠:= ⋅ Astπ4do2 − di2 ⎛⎝⎞⎠:= ⋅Compatibility: + δbr + ΔL = δst(Fbr)⋅LbrAbr⋅Ebr+ ΔL(Fst)⋅LstAst⋅Est= [1]Equations of equilibrium:+ ΣFy=0; Fst + Fbr − P = 0 [2]Assume brass yields, thenFbr := (σY_br)Abr Fbr = 197.92 kNεbrσY_brEbr:= εbr 0.00069307mmmm=δbr := (εbr)Lbr δbr = 0.17mm [ < ΔL = 1 mm ]Thus, only the brass is loaded. P := FbrP = 197.92 kN Ans 244. Problem 4-42Two A-36 steel wires are used to support the 3.25-kN (~325-kg) engine. Originally, AB is 800 mmlong and A'B' is 800.2 mm long. Determine the force supported by each wire when the engine issuspended from them. Each wire has a cross-sectional area of 6.25 mm2.Given: LAB := 800mm A := 6.25mm2LA'B' := 800.2mm W := 3.25kN:= ⋅ ( )⋅ksiE 29 10 3Solution: ΔL := LA'B' − LAB ΔL = 0.200mmCompatibility:+ δAB + ΔL = δA'B'Given(TAB)⋅LAB [1]A⋅E+ ΔL(TA'B')⋅LA'B'A⋅E=Equations of equilibrium:+ ΣFy=0; TAB + TA'B' − W = 0 [2]Initial guess: TAB := 1kN TA'B' := 2kNSolving [1] and [2]:TABTA'B'⎛⎜⎜⎝⎞⎠:= Find(TAB, TA'B')TABTA'B'⎛⎜⎜⎝⎞⎠1.4691.781⎛⎜⎝⎞⎠= kN Ans 245. Problem 4-43The bolt AB has a diameter of 20 mm and passes through a sleeve that has an inner diameter of 40 mmandan outer diameter of 50 mm. The bolt and sleeve are made of A-36 steel and are secured to the rigidbrackets as shown. If the bolt length is 220 mm and the sleeve length is 200 mm, determine the tensionin the bolt when a force of 50 kN is applied to the brackets.Given: Lb := 220mm db := 20mm do := 50mmLs := 200mm P := 25kN di := 40mmE := 200GPaSolution:Abπ4⎛⎝db2 ⎞⎠:= ⋅ Asπ4do2 − di2 ⎛⎝⎞⎠:= ⋅Compatibility: δb = δsGiven (Pb)⋅Lb [1]Ab⋅E(Ps)⋅LsAs⋅E=Equations of equilibrium:+ ΣFx=0; Pb + Ps − 2P = 0 [2]Initial guess: Pb := 1kN Ps := 2kNSolving [1] and [2]:PbPs⎛⎜⎜⎝⎞⎠:= Find(Pb , Ps)PbPs⎛⎜⎜⎝⎞⎠14.38835.612⎛⎜⎝⎞⎠= kN 246. Problem 4-44The specimen represents a filament-reinforced matrix system made from plastic (matrix) and glass(fiber). If there are n fibers, each having a cross-sectional area of Af and modulus of Ef , embedded ina matrix having a cross-sectional area of Am and modulus of Em, determine the stress in the matrix andeach fiber when the force P is imposed on the specimen.Solution:Compatibility: δm = δf(Pm)⋅LAm⋅Em(Pf)⋅Ln⋅Af⋅Ef=PmAm⋅Emn⋅Af⋅Ef⎛⎜⎝⎞⎠= ⋅Pf [1]Equations of equilibrium:+ ΣFy=0; P − Pm − Pf = 0 [2]Solving [1] and [2]:PmAm⋅Emn⋅Af⋅Ef + Am⋅Em⎛⎜⎝⎞⎠= ⋅PPfn⋅Af⋅Efn⋅Af⋅Ef + Am⋅Em⎛⎜⎝⎞⎠= ⋅PAverage Normal Stress:σmPmAm= σmEmn⋅Af⋅Ef + Am⋅Em⎛⎜⎝⎞⎠= ⋅P AnsσfPfn⋅Af= σfEf⎛⎜⎝ ⎞= ⋅P Ansn⋅Af⋅Ef + Am⋅Em⎠ 247. Problem 4-45The distributed loading is supported by the three suspender bars. AB and EF are made from aluminumandCD is made from steel. If each bar has a cross-sectional area of 450 mm2, determine the maximumintensity w of the distributed loading so that an allowable stress of (σallow)st = 180 MPa in the steel and(σallow)al = 94 MPa in the aluminum is not exceeded. Est = 200 GPa, Eal = 70 GPa.Given: L := 2m b := 1.5m A := 450mm2Est := 200⋅GPa Eal := 70⋅GPaσal_allow := 94MPaσst_allow := 180MPaSolution:Compatibility:+ δA = δC(FAB)⋅L(Eal)⋅A(FCD)⋅L(Est)⋅A= FABEalEst⎛⎜⎝⎞⎠= ⋅FCD [1]Equations of equilibrium:ΣΜC=0; FEF⋅ (b) − FAB⋅ (b) = 0 FAB = FEF+ ΣFy=0; FAB + FCD + FEF − w⋅ (2b) = 02⋅FAB = w⋅ (2b) − FCD [2]Assume failure of AB and EF:FAB := (σal_allow)⋅A FAB = 42.30 kNFrom [1]: FCDEstEal⎛⎜⎝⎞⎠:= ⋅FAB FCD = 120.86 kNFrom [2]: wFABbFCD2⋅b:= + w 68.49kNm=Assume failure of CD:FCD := (σst_allow)⋅A FCD = 81.00 kNFrom [1]: FABEalEst⎛⎜⎝ ⎞:= ⋅FCD FAB = 28.35 kN⎠From [2]: wFABbFCD2⋅b:= + w 45.90kNm= [Controls !] Ans 248. Problem 4-46The rigid link is supported by a pin at A, a steel wire BC having an unstretched length of 200 mm andcross-sectional area of 22.5 mm2, and a short aluminum block having an unloaded length of 50 mmand cross-sectional area of 40 mm2. If the link is subjected to the vertical load shown, determine theaverage normal stress in the wire and the block. Est = 200 GPa, Eal = 70 GPa.Given: Lk := 200mm Ak := 22.5mm2eD := 50mm AD := 40mm2 P := 450Na := 100mm b := 150mm c := 150mmEst := 200GPa Eal := 70GPaSolution:Compatibility:θBC = θADδBCbδDc=Given(FBC)⋅Lk (FD)⋅eD[1]=Ak⋅Est⋅bAD⋅Eal⋅cEquations of equilibrium:ΣΜA=0; −FBC⋅ (b) − FD⋅ (c) + P⋅ (a + b) = 0 [2]Initial guess: FBC := 1kN FD := 2kNSolving [1] and [2]:FBCFD⎛⎜⎜⎝⎞⎠:= Find(FBC, FD)FBCFD⎛⎜⎜⎝⎞⎠214.968535.032⎛⎜⎝⎞⎠= NAverage Normal Stress:σBCFBCAk:= σBC = 9.554MPa AnsσDFDAD:= σD = 13.376MPa Ans 249. Problem 4-47The rigid link is supported by a pin at A, a steel wire BC having an unstretched length of 200 mm andcross-sectional area of 22.5 mm2, and a short aluminum block having an unloaded length of 50 mmand cross-sectional area of 40 mm2. If the link is subjected to the vertical load shown, determine therotation of the link about the pin A. Report the answer in radians. Est = 200 GPa, Eal = 70 GPa.Given: Lk := 200mm Ak := 22.5mm2eD := 50mm AD := 40mm2 P := 450Na := 100mm b := 150mm c := 150mmEst := 200GPa Eal := 70GPaSolution:Compatibility:θBC = θAD δBCbδDc=Given(FBC)⋅Lk (FD)⋅eD[1]=Ak⋅Est⋅bAD⋅Eal⋅cEquations of equilibrium:ΣΜA=0; −FBC⋅ (b) − FD⋅ (c) + P⋅ (a + b) = 0 [2]Initial guess: FBC := 1kN FD := 2kNSolving [1] and [2]:FBCFD⎛⎜⎜⎝⎞⎠:= Find(FBC, FD)FBCFD⎛⎜⎜⎝⎞⎠214.968535.032⎛⎜⎝⎞⎠= NDisplacement:δD(FD)⋅eDAD⋅Eal:= δD = 0.009554mmθADδDc:= θAD = 63.69 × 10− 6 rad Ans 250. Problem 4-48The three A-36 steel wires each have a diameter of 2 mm and unloaded lengths of LAC = 1.60 m andLAB = LAD = 2.00 m. Determine the force in each wire after the 150-kg mass is suspended from thering at A.Given: LAC := 1.60m a := 3 d := 2mmLAB := 2.00m b := 4 M := 150kgLAD := 2.00m c := 5 g 9.81ms2=Solution: hac:= vbc:= Aπ4⎛⎜⎝⎞⎠:= d2W := M⋅gCompatibility: In triangle AA'B, since the displacement δA is very small, cos(θA') = vδAC⋅cos(θA') = δADδAC⋅ (v) = δAD(FAC)⋅LACA⋅E⋅ (v)(FAD)⋅LAD= [1]A⋅EEquations of equilibrium:+ ΣFx=0; FAB⋅ (h) − FAD⋅ (h) = 0 FAB = FAD+ ΣFy=0; FAB⋅ (v) + FAC + FAD⋅ (v) − W = 0Given 2⋅FAD⋅ (v) W F= − AC [2]From [1]: FAC⋅LAC⋅ (v) = FAD⋅LAD [3]Initial guess: FAD := 1N FAC := 2NSolving [2] and [3]:FADFAC⎛⎜⎜⎝ ⎞⎠:= Find(FAD, FAC)FADFAC⎛⎜⎜⎝⎞⎠465.1726.8⎛⎜⎝⎞⎠= N AnsFAB = FAD 251. Problem 4-49The A-36 steel wires AB and AD each have a diameter of 2 mm and the unloaded lengths of each wireare LAC = 1.60 m and LAB = LAD = 2.00 m. Determine the required diameter of wire AC so that eachwire is subjected to the same force caused by the 150-kg mass suspended from the ring at A.Given: LAC := 1.60m a := 3 dAB := 2mmLAB := 2.00m b := 4 dAD := 2mmLAD := 2.00m c := 5 g 9.81ms2=M := 150kgSolution: hac:= vbc:= W := M⋅gAABπ4⎛⎜⎝⎞⎠:= dAB2 AADπ4⎛⎜⎝⎞⎠:= 2dADEquations of equilibrium:Since each wire is required to carry the same amount of load.Hence,FAB = F FAC = F FAD = FCompatibility: In triangle AA'B, since the displacement δA is very small, cos(θA') = vδAC⋅cos(θA') = δADδAC⋅ (v) = δAD(FAC)⋅LAC⋅ (v)AAC⋅E(FAD)⋅LADAAD⋅E= AACLACLAD:= ⋅ (v)⋅AADAAC = 2.010619mm2AACπ4⎛⎜⎝⎞⎠= dAC2 dAC4π:= (AAC) dAC = 1.60mm Ans 252. Problem 4-50The three suspender bars are made of the same material and have equal cross-sectional areas A.Determinethe average normal stress in each bar if the rigid beam ACE is subjected to the force P.Solution:Compatibility: δC − δEdδA − δE2d=2δC = δA + δE2(FCD)⋅LA⋅E(FAB)⋅LA⋅E(FEF)⋅LA⋅E= +2FCD = FAB + FEF [1]Equations of equilibrium:ΣΜA=0; FCD⋅ (d) + FEF⋅ (2⋅d) Pd2⎛⎜⎝⎞⎠− ⋅ = 0FCD + 2FEF = 0.5P [2]+ ΣFy=0; FAB + FCD + FEF − P = 0 [3]Solving [1], [2] and [3]:FAB712= P FCD13= P FEF112= PσAB7P12A= σABP3A= σABP12A= Ans 253. Problem 4-51The assembly consists of an A-36 steel bolt and a C83400 red brass tube. If the nut is drawn up snugagainst the tube so that L = 75 mm, then turned an additional amount so that it advances 0.02 mm onthe bolt, determine the force in the bolt and the tube. The bolt has a diameter of 7 mm and the tube hasa cross-sectional area of 100 mm2.Given: L := 75mm Ebr := 101⋅GPaΔL := 0.02mm Est := 200⋅GPadst := 7mm Abr := 100mm2Solution: Astπ4⎛⎜⎝⎞⎠:= ⋅ dst2 Kst := (Est)⋅Ast Kbr := (Ebr)⋅AbrEquations of equilibrium: Since no external load is applie, the force acting on thetube and the bolt is the same.Compatibility: ΔL = δst + δbrΔLP⋅LKstP⋅LKbr= +ΔL P⋅LKst + KbrKst⋅Kbr⎛⎜⎝⎞⎠= ⋅PΔLLKst⋅KbrKst + Kbr⎛⎜⎝⎞⎠:=P = 1.165 kN Ans 254. Problem 4-52The assembly consists of an A-36 steel bolt and a C83400 red brass tube. The nut is drawn up snugagainst the tube so that L = 75 mm. Determine the maximum additional amount of advance of the nuton the bolt so that none of the material will yield. The bolt has a diameter of 7 mm and the tube has across-sectional area of 100 mm2.Given: L := 75mm Ebr := 101⋅GPadst := 7mm Est := 200⋅GPaAbr := 100mm2 σY_st := 250MPaσY_br := 70MPaSolution: Astπ4⎛⎜⎝⎞⎠:= ⋅ dst2 Kst := (Est)⋅Ast Kbr := (Ebr)⋅AbrAllowable Normal Stress:Pst := (σY_st)⋅Ast Pst = 9.621 kNPbr := (σY_br)⋅Abr Pbr = 7.000 kNSince Pst > Pbr , by comparison, the brass will yield first.P := min(Pst , Pbr)P = 7.00 kNCompatibility: ΔL = δst + δbrΔLP⋅LKstP⋅LKbr= +ΔLP⋅LKstP⋅LKbr:= +ΔL = 0.120mm Ans 255. Problem 4-53The 10-mm-diameter steel bolt is surrounded by a bronze sleeve. The outer diameter of this sleeve is20 mm, and its inner diameter is 10 mm. If the bolt is subjected to a compressive force of P = 20 kN,determine the average normal stress in the steel and the bronze. Est = 200 GPa, Ebr = 100 GPa.Given: ds := 10mm do := 20mm Est := 200GPaP := 20kN di := 10mm Ebr := 100GPaSolution:Astπ4⎛⎝ds2 ⎞⎠:= ⋅ Abrπ4do2 − di2 ⎛⎝⎞⎠:= ⋅Compatibility:δb = δs(Pbr)⋅LAbr⋅Ebr(Pst)⋅LAst⋅Est=Given PbrAbr⋅EbrPstAst⋅Est= [1]Equations of equilibrium:+ ΣFy=0; Pbr + Pst − P = 0 [2]Initial guess: Pbr := 1kN Pst := 2kNSolving [1] and [2]:PbrPst⎛⎜⎜⎝⎞⎠:= Find(Pbr , Pst)PbrPst⎛⎜⎜⎝⎞⎠12.0008.000⎛⎜⎝⎞⎠= kNσstPstAst:= σst = 101.86MPa AnsσbrPbrAbr:= σbr = 50.93MPa Ans 256. Problem 4-54The 10-mm-diameter steel bolt is surrounded by a bronze sleeve.The outer diameter of this sleeve is 20mm,and its inner diameter is 10 mm. If the yield stress for the steel is (σY)st = 640 MPa, and for thebronze (σY)br = 520 MPa, determine the magnitude of the largest elastic load P that can be applied tothe assembly. Est = 200 GPa, Ebr = 100 GPa.Given: do := 20mm Est := 200GPa σY_st := 640MPadi := 10mm Ebr := 100GPa σY_br := 520MPads := 10mmSolution:Astπ4⎛⎝ds2 ⎞⎠:= ⋅ Abrπ4do2 − di2 ⎛⎝⎞⎠:= ⋅Equations of equilibrium:+ ΣFy=0; Pbr + Pst − P = 0 [1]Compatibility:δb = δs(Pbr)⋅LAbr⋅Ebr(Pst)⋅LAst⋅Est=PbrAbr⋅EbrPstAst⋅Est= [2]Assume failure of bolt, thenPst := (σY_st)Ast Pst = 50.265 kNFrom [2]: PbrAbr⋅EbrAst⋅Est⎛⎜⎝⎞⎠:= ⋅Pst Pbr = 75.398 kNFrom [1]: P := Pbr + Pst P = 125.66 kN (Controls!): AnsAssume failure of sleeve, thenPbr := σY_brAbr Pbr = 122.522 kNFrom [2]: PstAst⋅EstAbr⋅Ebr⎛⎜⎝ ⎞:= ⋅Pbr Pst = 81.681 kN⎠From [1]: P := Pbr + Pst P = 204.20 kN 257. Problem 4-55The rigid member is held in the position shown by three A-36 steel tie rods. Each rod has anunstretched length of 0.75 m and a cross-sectional area of 125 mm2. Determine the forces in the rodsif a turnbuckle on rod EF undergoes one full turn. The lead of the screw is 1.5 mm. Neglect the size ofthe turnbuckle and assume that it is rigid. Note: The lead would cause the rod, when unloaded, toshorten 1.5 mm when the turnbuckle is rotated one revolution.Given: L := 0.75m b := 0.5m A := 125mm2ΔL := 1.5mm E := 200⋅GPaSolution:Equations of equilibrium:ΣΜE=0; TCD⋅ (b) + TAB⋅ (b) = 0TCD = TAB [1]+ ΣFy=0; TAB + TCD − TEF = 0 [1]TEF = 2TAB [2]Compatibility:Rod EF shortens 1.5mm causing AB (and CD) to elongate. Thus,:+ ΔL = δA + δCΔL(TAB)⋅LE⋅A(TEF)⋅LE⋅A= +ΔL(TAB)⋅LE⋅A(2TAB)⋅LE⋅A= +TABE⋅A3L:= ⋅ΔL TAB = 16.667 kN AnsFrom [1]: TCD := TAB TCD = 16.667 kN AnsFrom [2]: TEF := 2TAB TEF = 33.333 kN Ans 258. Problem 4-56The bar is pinned at A and supported by two aluminum rods, each having a diameter of 25 mm and amodulus of elasticity Eal = 70 GPa. If the bar is assumed to be rigid and initially vertical, determine thedisplacement of the end B when the force of 10 kN is applied.Given: LCD := 0.6m LEF := 0.3m d := 25mma := 0.3m P := 10kN Eal := 70GPaSolution: Aπ4:= ⋅d2Compatibility: δE3⋅aδCa=Given(FEF)⋅LEF3A⋅Eal(FCD)⋅LCDA⋅Eal=FEF⋅LEF = 3FCD⋅LCD [1]Equations of equilibrium:ΣΜA=0; −FEF⋅ (3⋅a) − FCD⋅ (a) + P⋅ (2a) = 0−3FEF − FCD + 2P = 0 [2]Initial guess: FEF := 1kN FCD := 2kNSolving [1] and [2]:FEFFCD⎛⎜⎜⎝⎞⎠:= Find(FEF, FCD)FEFFCD⎛⎜⎜⎝⎞⎠6.315791.05263⎛⎜⎝⎞⎠= kNDisplacement:δE(FEF)⋅LEFA⋅Eal:= δE = 0.055142mmδB4⋅aδE3a= δB43:= δE δB = 0.073522mm Ans 259. Problem 4-57The bar is pinned at A and supported by two aluminum rods, each having a diameter of 25 mm and amodulus of elasticity Eal = 70 GPa. If the bar is assumed to be rigid and initially vertical, determine theforce in each rod when the 10-kN load is applied.Given: LCD := 0.6m LEF := 0.3m d := 25mma := 0.3m P := 10kN Eal := 70GPaSolution: Aπ4:= ⋅d2Compatibility: δE3⋅aδCa=Given(FEF)⋅LEF3A⋅Eal(FCD)⋅LCDA⋅Eal=FEF⋅LEF = 3FCD⋅LCD [1]Equations of equilibrium:ΣΜA=0; −FEF⋅ (3⋅a) − FCD⋅ (a) + P⋅ (2a) = 0−3FEF − FCD + 2P = 0 [2]Initial guess: FEF := 1kN FCD := 2kNSolving [1] and [2]:FEFFCD⎛⎜⎜⎝⎞⎠:= Find(FEF, FCD)FEFFCD⎛⎜⎜⎝⎞⎠6.3161.053⎛⎜⎝⎞⎠= kN 260. Problem 4-58The assembly consists of two posts made from material 1 having a modulus of elasticity of E1 andeach a cross-sectional area A1 , and a material 2 having a modulus of elasticity E2 and cross-sectionalarea A2. If a central load P is applied to the rigid cap, determine the force in each material.Solution:FAB = F1 FEF = F2Equations of equilibrium:ΣΜE=0; FCD⋅ (d) − FAB⋅ (d) = 0FCD = FABFCD = F1+ ΣFy=0; FAB + FCD + FEF − P = 02F1 + F2 = P [1]Compatibility: δA = δB δB = δC(F1)⋅LA1⋅E1(F2)⋅LA2⋅E2=F1A1⋅E1A2⋅E2⎛⎜⎝⎞⎠= ⋅F2 [2]Solving [1] and [2]:F1A1⋅E12A1⋅E1 + A2⋅E2⎛⎜⎝⎞⎠= P AnsF2A2⋅E22A1⋅E1 + A2⋅E2⎛⎜⎝⎞⎠= P Ans 261. Problem 4-59The assembly consists of two posts AB and CD made from material 1 having a modulus of elasticityE1 ofand each a cross-sectional area A1, and a central post EF made from material 2 having a modulusof elasticity E2 and a cross-sectional area A2. If posts AB and CD are to be replaced by those having amaterial 2, determine the required cross-sectional area of these new posts so that both assembliesdeform the same amount when loaded.Solution:FAB = F1 FEF = F2Equations of equilibrium:ΣΜE=0; FCD⋅ (d) − FAB⋅ (d) = 0FCD = FABFCD = F1+ ΣFy=0; FAB + FCD + FEF − P = 02F1 + F2 = P [1]Compatibility:δA = δB δB = δC δA = δ(F1)⋅LA1⋅E1(F2)⋅LA2⋅E2= F1A1⋅E1A2⋅E2⎛⎜⎝⎞⎠= ⋅F2 [2]Solving [1] and [2]:F1A1⋅E12A1⋅E1 + A2⋅E2⎛⎜⎝⎞⎠= P F2A2⋅E2⎛⎜⎝ ⎞= P2A1⋅E1 + A2⋅E2⎠When material 1 has been replaced by material 2 for two side posts, thenEquations of equilibrium: [1] becomes 2(F'1) + F'2 = P [1']Compatibility: [2] becomes F'1A'1A2⎛⎜⎝⎞⎠= ⋅F'2 [2']Solving [1'] and [2']:F'1A'12A'1 + A2⎛⎜⎝⎞⎠= P F'2A22A'1 + A2⎛⎜⎝⎞⎠= PRequires, δB = δ'B(F2)⋅L(F'2)⋅L=A2⋅E2A2⋅E2A2⋅E22A1⋅E1 + A2⋅E2⎛⎜⎝⎞⎠PA22A'1 + A2⎛⎜⎝⎞⎠= P A'1E1E2⎛⎜⎝⎞⎠= ⋅A1 Ans 262. Problem 4-60The assembly consists of two posts AB and CD made from material 1 having a modulus of elasticityE1 ofand each a cross-sectional area A1, and a central post EF made from material 2 having a modulusof elasticity E2 and a cross-sectional area A2. If post EF is to be replaced by one having a material 1,determine the required cross-sectional area of this new post so that both assemblies deform the sameamount when loaded.Solution:FAB = F1 FEF = F2Equations of equilibrium:ΣΜE=0; FCD⋅ (d) − FAB⋅ (d) = 0FCD = FABFCD = F1+ ΣFy=0; FAB + FCD + FEF − P = 02F1 + F2 = P [1]Compatibility:δA = δB δB = δC δA = δ(F1)⋅L(F2)⋅L= A1⋅E1A2⋅E2F2A2⋅E2A1⋅E1⎛⎜⎝⎞⎠= ⋅F1 [2]Solving [1] and [2]:F1A1⋅E12A1⋅E1 + A2⋅E2⎛⎜⎝⎞⎠= P F2A2⋅E22A1⋅E1 + A2⋅E2⎛⎜⎝⎞⎠= PWhen material 2 has been replaced by material 1 for central post, thenEquations of equilibrium: [1] becomes 2(F'1) + F'2 = P [1']Compatibility: [2] becomes F'2A'2A1⎛⎜⎝⎞⎠= ⋅F'1 [2']Solving [1'] and [2']:F'1A12A1 + A'2⎛⎜⎝⎞⎠= P F'2A'22A1 + A'2⎛⎜⎝⎞⎠= PRequires, δA = δ'A(F1)⋅L(F'1)⋅L=A1⋅E1A1⋅E1A1⋅E12A1⋅E1 + A2⋅E2⎛⎜⎝⎞⎠PA12A1 + A'2⎛⎜⎝⎞⎠= P A'2E2E1⎛⎜⎝⎞⎠= ⋅A2 Ans 263. Problem 4-61The bracket is held to the wall using three A-36 steel bolts at B, C, and D. Each bolt has a diameter of12.5 mm and an unstretched length of 50 mm. If a force of 4 kN is placed on the bracket as shown,determine the force developed in each bolt. For the calculation, assume that the bolts carry no shear;rather, the vertical force of 4 kN is supported by the toe at A. Also, assume that the wall and bracketare rigid. A greatly exaggerated deformation of the bolts is shown.Given: a := 12.5mm b := 25mm c := 50mmd := 12.5mm e := 50mmL := 50mm P := 4kN E := 200GPaSolution: LAD := a + b + cLAC := a + b A π4d2:= ⋅LAB := aCompatibility:δDLADδCLAC=(FD)⋅L(LAD)⋅A⋅E(FC)⋅L(LAC)⋅A⋅E=δBLABδCLAC=(FB)⋅L(LAB)⋅A⋅E(FC)⋅L(LAC)⋅A⋅E=Given FDLADLAC⎛⎜⎝⎞⎠= ⋅FC [1]FBLABLAC⎛⎜⎝⎞⎠= ⋅FC [2]Equations of equilibrium:ΣΜA=0; FD⋅ (LAD) + FC⋅ (LAC) + FB⋅ (LAB) − P⋅ (e) = 0 [3]Initial guess: FB := 10kN FC := 20kN FD := 30kNSolving [1], [2] and [3]:FBFCFD⎛⎜⎜⎜⎝⎞⎟⎠:= Find(FB, FC, FD)FBFCFD⎛⎜⎜⎜⎝⎞⎟⎠0.27120.81361.8983⎛⎜⎜⎝⎞= kN Ans⎠ 264. Problem 4-62The bracket is held to the wall using three A-36 steel bolts at B, C, and D. Each bolt has a diameter of12.5 mm and an unstretched length of 50 mm. If a force of 4 kN is placed on the bracket as shown,determine how far, s, the top bracket at bolt D moves away from the wall. For the calculation, assumethat the bolts carry no shear; rather, the vertical force of 4 kN is supported by the toe at A. Also,assume that the wall and bracket are rigid. A greatly exaggerated deformation of the bolts is shown.Given: a := 12.5mm b := 25mm c := 50mmd := 12.5mm e := 50mmL := 50mm P := 4kN E := 200GPaSolution: LAD := a + b + cLAC := a + bLAB := a Aπ4:= ⋅d2Compatibility:δDLADδCLAC=(FD)⋅L(LAD)⋅A⋅E(FC)⋅L(LAC)⋅A⋅E=δBLABδCLAC=(FB)⋅L(LAB)⋅A⋅E(FC)⋅L(LAC)⋅A⋅E=Given FDLADLAC⎛⎜⎝⎞⎠= ⋅FC [1]FBLABLAC⎛⎜⎝⎞⎠= ⋅FC [2]Equations of equilibrium:ΣΜA=0; FD⋅ (LAD) + FC⋅ (LAC) + FB⋅ (LAB) − P⋅ (e) = 0 [3]Initial guess: FB := 10kN FC := 20kN FD := 30kNSolving [1], [2] and [3]:FBFCFD⎛⎜⎜⎜⎝⎞⎟⎠:= Find(FB, FC, FD)FBFCFD⎛⎜⎜⎜⎝⎞⎟⎠0.27120.81361.8983⎛⎜⎜⎝⎞= kN⎠Displacement: δD(FD)⋅LA⋅E:= δD = 0.003867mm Ans 265. Problem 4-63The rigid bar is supported by the two short white spruce wooden posts and a spring. If each of theposts has an unloaded length of 1 m and a cross-sectional area of 600 mm2, and the spring has astiffness of k = 2 MN/m and an unstretched length of 1.02 m, determine the force in each post afterthe load is applied to the bar.Given: L := 1m Lk := 1.02m b := 2mw 50kNm⋅ ( )kNm:= k := 2 1⋅30 A := 600mm2 E := 9.65GPaSolution: ΔLk := Lk − L ΔLk = 0.02mCompatibility:+ δA + ΔLk = δk [1]Equations of equilibrium:ΣΜC=0; FB⋅ (b) − FA⋅ (b) = 0FA = FB+ ΣFy=0; FA + FB + Fsp − w⋅ (b) = 0Given 2FA + Fsp − w⋅ (b) = 0 [2](FA)⋅LE⋅AFrom [1]: [1']+ ΔLkFspk=Initial guess: FA := 1kN Fsp := 2kNSolving [2] and [1']:FAFsp⎛⎜⎜⎝⎞⎠:= Find(FA, Fsp)FAFsp⎛⎜⎜⎝⎞⎠25.5848.84⎛⎜⎝⎞⎠= kN AnsFB := FA FB = 25.58 kN Ans 266. Problem 4-64The rigid bar is supported by the two short white spruce wooden posts and a spring. If each of theposts has an unloaded length of 1 m and a cross-sectional area of 600 mm2, and the spring has astiffness of k = 2 MN/m and an unstretched length of 1.02 m, determine the vertical displacement of Aand B after the load is applied to the bar.Given: L := 1m Lk := 1.02m b := 2mw 50kNm⋅ ( )kNm:= k := 2 1⋅30 A := 600mm2 E := 9.65GPaSolution: ΔLk := Lk − L ΔLk = 0.02mCompatibility:+ δA + ΔLk = δk [1]Equations of equilibrium:ΣΜC=0; FB⋅ (b) − FA⋅ (b) = 0FA = FB+ ΣFy=0; FA + FB + Fsp − w⋅ (b) = 0Given 2FA + Fsp − w⋅ (b) = 0 [2](FA)⋅LE⋅AFrom [1]: [1']+ ΔLkFspk=Initial guess: FA := 1kN Fsp := 2kNSolving [2] and [1']:FAFsp⎛⎜⎜⎝⎞⎠:= Find(FA, Fsp)FAFsp⎛⎜⎜⎝⎞⎠25.5848.84⎛⎜⎝⎞⎠= kNFB := FA FB = 25.58 kNDisplacement:δA(FA)⋅LA⋅E:= δA = 4.418mm AnsδB(FB)⋅LA⋅E:= δB = 4.418mm Ans 267. Problem 4-65The wheel is subjected to a force of 18 kN from the axle. Determine the force in each of the threespokes. Assume the rim is rigid and the spokes are made of the same material, and each has the samecross-sectional area.Given: L := 0.4m θ := 120deg P := 18kNSolution: φ := 180deg − θCompatibility:δAB⋅cos(φ) = δAC(FAB)⋅LE⋅A⋅cos(φ)(FAC)⋅LE⋅A= [1]Equations of equilibrium: At A :+ ΣFx=0; FAC⋅ sin(φ) − FAD⋅ sin(b) = 0FAC = FAD+ ΣFy=0; FAB + FAC⋅cos(φ) + FAD⋅cos(φ) − P = 0Given FAB + 2(FAC⋅cos(φ)) − P = 0 [2]From [1]: FAB⋅cos(φ) = FAC [1']Initial guess: FAB := 1kN FAC := 2kNSolving [2] and [1']:FABFAC⎛⎜⎜⎝⎞⎠:= Find(FAB, FAC)FABFAC⎛⎜⎜⎝⎞⎠126⎛⎜⎝⎞⎠= kN AnsFAD := FAC FAD = 6 kN Ans 268. Problem 4-66The post is made from 6061-T6 aluminum and has a diameter of 50 mm. It is fixed supported at A andB, and at its center C there is a coiled spring attached to the rigid collar. If the spring is originallyuncompressed, determine the reactions at A and B when the force P = 40 kN is applied to the collar.Given: LCA := 0.25m P := 40kN d := 50mmLBC := 0.25m E := 68.9GPa k 200(103) kNm:=Solution:Aπ4:= ⋅d2 kCAE⋅ALCA:= kBCE⋅ALBC:=Equations of equilibrium:+ ΣFy=0; FA − P + Fk + FB = 0 [1]Compatibility: Considera a combined force F (=FB+Fk) acted at free end B .ΔB = δP − δF ΔB = 0Given0PkBCFB + FkkCAFB + FkkBC + k+⎛⎜⎝⎞⎠= − [2]Also, Δsp=ΔBCFkkFB + FkkBC + k= [3]Initial guess: FB := 1kN Fk := 2kNSolving [1] and [2]:FBFk⎛⎜⎜⎝⎞⎠:= Find(FB, Fk)FBFk⎛⎜⎜⎝ ⎞⎠16.886.24⎛⎜⎝⎞⎠= kN AnsFrom [1]: FA := P − Fk − FB FA = 16.88 kN Ans 269. Problem 4-67The post is made from 6061-T6 aluminum and has a diameter of 50 mm. It is fixed supported at A andB, and at its center C there is a coiled spring attached to the rigid collar. If the spring is originallyuncompressed, determine the compression in the spring when the load of P = 50 kN is applied to thecollar.Given: LCA := 0.25m P := 50kN d := 50mmLBC := 0.25m E := 68.9GPa k 200(103) kNm:=Solution:Aπ4:= ⋅d2 kCAE⋅ALCA:= kBCE⋅ALBC:=Equations of equilibrium:+ ΣFy=0; FA − P + Fk + FB = 0 [1]Compatibility: Considera a combined force F (=FB+Fk) acted at free end B .ΔB = δP − δF ΔB = 0Given0PkBCFB + FkkCAFB + FkkBC + k+⎛⎜⎝⎞⎠= − [2]Also, Δsp=ΔBCFkkFB + FkkBC + k= [3]Initial guess: FB := 1kN Fk := 2kNSolving [1] and [2]:FBFk⎛⎜⎜⎝⎞⎠:= Find(FB, Fk)FBFk⎛⎜⎜⎝⎞⎠21.1017.799⎛⎜⎝⎞⎠= kNThus, ΔspFkk:=Δsp = 0.03899mm Ans 270. Problem 4-68The rigid bar supports the uniform distributed load of 90 kN/m. Determine the force in each cable ifeach cable has a cross-sectional area of 36 mm2, and E = 200 GPa.Given: a := 1m b := 2mA := 36mm2 E := 200GPaw 90kNm:=Solution: c := a2 + b2 vbc:= hac:=LAC := 4a2 + b2 L := 3aLBC := c LDC := cEquations of equilibrium:ΣΜA=0; TBC⋅ (v)⋅a + TDC⋅ (v)⋅L − w⋅L⋅ (0.5L) = 0 [1]Compatibility:2 a2 LACIn triangle AB'C: LB'C+ 2 2⋅a L= − ⋅ ( AC)⋅cos(θ'A)In triangle AD'C: LD'C+ 2 2⋅L L= − ⋅ ( AC)⋅cos(θ'A)2 L2 LACThus, eliminating cos(θ'A):2LLD'C2aLB'C− ⎛⎜⎝ ⎞⎠2 1− L − a LACL1a= + ⋅LD'C2 3LB'C− 2 6a2 2LAC= − 22 LD'C3LB'C− 2 = 2a2 + 2b22 LD'C3LB'C− 2 = 2c2But,LB'C = LBC + ΔBC LD'C = LDC + ΔDCLB'C = c + ΔBC LD'C = c + ΔDCThus, 3 (c + ΔBC)2 − (c + ΔDC)2 = 2c2Neglect squares of Δ's since small strain occurs:3 c2 2cΔBC + ⎛⎝⎞⎠− = 2c2c2 2cΔDC + ⎛⎝⎞⎠3 c2 2cΔBC + ⎛⎝⎞⎠− = 2c2c2 2cΔDC + ⎛⎝⎞⎠3ΔBC − ΔDC = 0 271. 3TBC⋅ (c)E⋅ATDC⋅ (c)E⋅A− = 0TDC = 3TBC [2]Substituting [2] into [1]:TBC⋅ (v)⋅a + 3TBC⋅ (v)⋅L − w⋅L⋅ (0.5L) = 0TBC9a⋅c20⋅b⎛⎜⎝⎞⎠:= ⋅w TBC = 45.2804 kN AnsFrom [2] : TDC := 3TBC TDC = 135.8411 kN Ans 272. Problem 4-69The rigid bar is originally horizontal and is supported by two cables each having a cross-sectional areaof 36 mm2, and E = 200 GPa. Determine the slight rotation of the bar when the uniform load is applied.Given: a := 1m b := 2mA := 36mm2 E := 200GPaw 90kNm:=Solution: c := a2 + b2 vbc:= hac:=LAC := 4a2 + b2 L := 3aLBC := c LDC := cEquations of equilibrium:ΣΜA=0; TBC⋅ (v)⋅a + TDC⋅ (v)⋅L − w⋅L⋅ (0.5L) = 0 [1]Compatibility: See solution of Prob. 4-68.TDC = 3TBC [2]Solving [1] and [2]: TDC := 27.1682 kNΔDCTDC⋅ (c)E⋅A:= ΔDC = 8.4374919mmGeometry: tan(θA) b= θA atan2ab2a⎛⎜⎝⎞⎠:= θA = 45.000 degIn triangle AD'C: LD'C+ 2 2⋅L L= − ⋅ ( AC)⋅cos(θ'A)2 L2 LAC(LDC + ΔDC)2 L2 LAC+ 2 2⋅L L= − ⋅ ( AC)⋅cos(θ'A)θ'A acos+ 2 L( DC + ΔDC)− 2L2 LAC2⋅L⋅ (LAC)⎡⎢⎢⎣⎤⎥⎥⎦:= θ'A = 45.180 degΔθ := θ'A − θA Δθ = 0.180 deg Ans 273. Problem 4-70The electrical switch closes when the linkage rods CD and AB heat up, causing the rigid arm BDEboth to translate and rotate until contact is made at F. Originally, BDE is vertical, and the temperatureis 20°C. If AB is made of bronze C86100 and CD is made of aluminum 6061-T6, determine the gap srequired so that the switch will close when the temperature becomes 110°C.Unit used: °C := degGiven: LBD := 400mm LDE := 200mm L := 300mmT1 := 20°C T2 := 110°Cαcu 17.0⋅ (10− 6):= αst 24.0⋅ (10− 6)1°C1°C:=Solution:ΔT := T2 − T1 ΔT = 90 °CThermal Expansion:δAB := αcu⋅ (ΔT)⋅L δAB = 0.4590mmδCD := αst⋅ (ΔT)⋅L δCD = 0.6480mmGeomotry:θδCD − δABLBD:=s := δAB + θ⋅ (LBD + LDE)s = 0.7425mm Ans 274. Problem 4-71A steel surveyor's tape is to be used to measure the length of a line. The tape has a rectangular crosssection of 1.25 mm by 5 mm and a length of 30 m when T1 = 20°C and the tension or pull on the tapeis 100 N. Determine the true length of the line if the tape shows the reading to be 139 m when usedwith a pull of 175 N at T2 = 40°C. The ground on which it is placed is flat. αst = 17(10-6)/°C, Est =200 GPa.Unit used: °C := degGiven: a := 5mm b := 1.25mmT1 := 20°C P1 := 100N L1 := 30mT2 := 40°C P2 := 175N L2 := 139mαst 17⋅ (10− 6)1°C:= Est := 200GPaSolution: A := a⋅bΔT := T2 − T1 ΔT = 20 °CThermal Expansion:δT := αst⋅ (ΔT)⋅L2 δT = 47.2600mm δT = 0.0473mδP(P2 − P1)⋅L2:= δP = 8.3400mm δP = 0.0083mA⋅EstL' := L2 + δT + δPL' = 139.056m Ans 275. Problem 4-72The assembly has the diameters and material make-up indicated. If it fits securely between its fixedsupports when the temperature is T1 = 20°C, determine the average normal stress in each materialwhen the temperature reaches T2 = 40°C.Unit used: °C := degGiven: T1 := 20°C T2 := 40°CL1 := 1.2m d1 := 300mmL2 := 1.8m d2 := 200mmL3 := 0.9m d3 := 100mmα1 23⋅ (10− 6):= α2 17⋅ (10− 6)1°C:= α3 17⋅ (10− 6)1°C1°C:=E1 := 73.1GPa E2 := 103GPa E3 := 193GPaSolution: ΔT := T2 − T1 ΔT = 20 °CA1π4⎛⎜⎝⎞⎠:= ⋅ d12 A2π4⎛⎜⎝⎞⎠:= ⋅ d22 A3π4⎛⎜⎝⎞⎠:= ⋅ 2d3Equations of equilibrium:+ ΣFx=0; FA − FD = 0 FA = FDLet FA=F. Then, FA = FAB = FBC = FCD = FD = FThermal Expansion:δT := α1⋅ (ΔT)⋅L1 + α2⋅ (ΔT)⋅L2 + α3⋅ (ΔT)⋅L3 δT = 1.470000mmElongation: negative sign indicates shorteningδFF⋅L1A1⋅E1−F⋅L2A2⋅E2−F⋅L3A3⋅E3= −Compatibility:0 = δT + δF Given 0 δTF⋅L1A1⋅E1−F⋅L2A2⋅E2−F⋅L3A3⋅E3= − [1]Initial guess: F := 1kN Solving [1] : F := Find(F) F = 1063.49 kNAverage Normal Stress: σalFA1:= σal = 15.05MPa AnsσbrFA2:= σbr = 33.85MPa AnsσstFA3:= σst = 135.41MPa Ans 276. Problem 4-73A high-strength concrete driveway slab has a length of 6 m when its temperature is 10°C. If there is agap of 3 mm on one side before it touches its fixed abutment, determine the temperature required toclose the gap. What is the compressive stress in the concrete if the temperature becomes 60°C?Unit used: °C := degGiven: α 11⋅ (10− 6)1°C:= L := 6m T1 := 10°CΔgap := 3mm T2 := 60°C E := 29GPaSolution:Require, Δgap = α⋅ (T' − T1)⋅LT'1Δgapα⋅L⎛⎜⎝⎞⎠:= + T1 T'1 = 55.45 °C AnsFor ΔT := T2 − T1 ΔT = 50 °CΔgap = δT − δFΔgap α⋅ (ΔT)⋅LF⋅LA⋅E= −FAα⋅ (ΔT)⋅E ΔgapEL⎛⎜⎝⎞⎠= − ⋅σFA⎞⎠⋅ − := σ 1.45MPa = Ans= σ α⋅ (ΔT)⋅E ΔgapEL⎛⎜⎝ 277. Problem 4-74A 1.8-m-long steam pipe is made of steel with σY = 280 MPa. It is connected directly to two turbinesA and B as shown.The pipe has an outer diameter of 100 mm and a wall thickness of 6 mm.Theconnection was made at T1 = 20°C. If the turbines' points of attachment are assumed rigid, determinethe force the pipe exerts on the turbines when the steam and thus the pipe reach a temperature of T2 =135°C.Unit used: °C := degGiven:L := 1.8m T1 := 20°C T2 := 135°CσY := 280MPa do := 100mm t := 6mmE := 200GPa α 12⋅ (10− 6)1°C:=Solution:di := do − 2t Aπ4do2 − di2 ⎛⎝⎞⎠:= ⋅ΔT := T2 − T1 ΔT = 115 °CCompatibility:0 = δT + δF0 α⋅ (ΔT)⋅LF⋅LE⋅A= −F := α⋅ (ΔT)⋅E⋅AF = 489.03 kN AnsCheck stress:σFA:= σ = 276.00MPa (< σY = 280 MPa) ok. 278. Problem 4-75A 1.8-m-long steam pipe is made of steel with σY = 280 MPa. It is connected directly to two turbinesA and B as shown.The pipe has an outer diameter of 100 mm and a wall thickness of 6 mm. Theconnection was made at T1 = 20°C. If the turbines' points of attachment are assumed to have astiffness of k = 16 MN/m., determine the force the pipe exerts on the turbines when the steam andthus the pipe reach a temperature of T2 = 135°C.Unit used: °C := degGiven: L := 1.8m T1 := 20°C T2 := 135°CσY := 280MPa do := 100mm t := 6mmE := 200GPa α 12⋅ (10− 6)1°C⋅ ( )kNmm30 := k :=16 1Solution:di := do − 2t Aπ4do2 − di2 ⎛⎝⎞⎠:= ⋅ΔT := T2 − T1 ΔT = 115 °CCompatibility:2x = δT + δF2x α⋅ (ΔT)⋅L(k⋅x)⋅LE⋅A= −xα⋅ (ΔT)⋅E⋅A⋅Lk⋅L + 2E⋅A:=x = 0.029830mmF := k⋅x F = 477.29 kN AnsCheck stress:σFA:= σ = 269.37MPa (< σY = 280 MPa) ok. 279. Problem 4-76The 12-m-long A-36 steel rails on a train track are laid with a small gap δ between them to allowforthermal expansion. Determine the required gap so that the rails just touch one another when thetemperature is increased from T1 = -30°C to T2 = 30°C. Using this gap, what would be the axial forcein the rails if the temperature were to rise to T3 = 40°C? The cross-sectional area of each rail is 3200mm2.Unit used: °C := degGiven: L := 12m A := 3200mm2T1 := −30°C T2 := 30°C T3 := 40°CE := 200GPa α 12⋅ (10− 6)1°C:=Solution:Require, Δgap := α⋅ (T2 − T1)⋅LΔgap = 8.640mm AnsFor ΔT := T3 − T1 ΔT = 70 °CCompatibility:Δgap = δT + δFΔgap α⋅ (ΔT)⋅LF⋅LE⋅A= −F α ΔT ( ) ⋅ L ⋅ Δgap − ⎡⎣⎤⎦E⋅AL:= ⋅F = 76.80 kN Ans 280. Problem 4-77The two circular rod segments, one of aluminum and the other of copper, are fixed to the rigid wallssuch that there is a gap of 0.2 mm between them when T1 = 15°C. What larger temperature T2 isrequired in order to just close the gap? Each rod has a diameter of 30 mm, αal = 24(10-6)/°C, Eal = 70GPa, αcu = 17(10-6)/°C, Ecu = 126 GPa. Determine the average normal stress in each rod if T2 = 95°C.Unit used: °C := degGiven: Lcu := 100mm Lal := 200mm d := 30mmT1 := 15°C T2 := 95°C Δgap := 0.2mmEcu := 126GPa αcu 17⋅ (10− 6)1°C:=Eal := 70GPa αal 24⋅ (10− 6)1°C:=Solution:Aπ4⎛⎜⎝⎞⎠:= ⋅d2Require, Δgap = αcu⋅ (T' − T1)⋅Lcu + αal⋅ (T' − T1)⋅LalT' T1Δgapαcu⋅Lcu + αal⋅Lal:= +T' = 45.77 °C AnsFor ΔT := T2 − T1 ΔT = 80 °CCompatibility: δcu = δ1T + δ1F δal = δ2T + δ2FΔgap = δcu + δalΔgap αcu⋅ (ΔT)⋅LcuF⋅LcuEcu⋅A− + αal⋅ (ΔT)⋅LalF⋅LalEal⋅A= −Fαcu⋅ (ΔT)⋅Lcu + αal⋅ (ΔT)⋅Lal − ΔgapLcuEcu⋅ALalEal⋅A+:=F = 61.958 kN AnsAverage Normal Stress:σFA:= σ = 87.65MPa Ans 281. Problem 4-78The two circular rod segments, one of aluminum and the other of copper, are fixed to the rigid wallssuch that there is a gap of 0.2 mm between them when T1 = 15°C. Each rod has a diameter of 30 mm,αal = 24(10-6)/°C, Eal = 70 GPa, αcu = 17(10-6)/°C, Ecu = 126 GPa. Determine the average normalstress in each rod if T2 = 150°C, and also calculate the new length of the aluminum segment.Unit used: °C := degGiven: Lcu := 100mm Lal := 200mm d := 30mmT1 := 15°C T2 := 150°C Δgap := 0.2mmEcu := 126GPa αcu 17⋅ (10− 6)1°C:=Eal := 70GPa αal 24⋅ (10− 6)1°C:=Solution: Aπ4⎛⎜⎝⎞⎠:= ⋅d2For ΔT := T2 − T1 ΔT = 135 °CCompatibility:δcu = δ1T + δ1F δal = δ2T + δ2FΔgap = δcu + δalΔgap αcu⋅ (ΔT)⋅LcuF⋅LcuEcu⋅A− + αal⋅ (ΔT)⋅LalF⋅LalEal⋅A= −Fαcu⋅ (ΔT)⋅Lcu + αal⋅ (ΔT)⋅Lal − Δgap:= F = 131.176 kNLcuEcu⋅ALalEal⋅A+Average Normal Stress:σFA:= σ = 185.58MPa AnsDisplacement:δal αal⋅ (ΔT)⋅LalF⋅LalEal⋅A:= − δal = 0.117783mmL'al := Lal + δal L'al = 200.117783mm Ans 282. Problem 4-79Two bars, each made of a different material, are connected and placed between two walls when thetemperature is T1 = 10°C. Determine the force exerted on the (rigid) supports when the temperaturebecomes T2 = 20°C. The material properties and cross-sectional area of each bar are given in thefigure.Unit used: °C := degGiven: T1 := 10°C T2 := 20°C L := 300mmAst := 200mm2 Abr := 450mm2αst 12⋅ (10− 6):= αbr 21⋅ (10− 6)1°C1°C:=Est := 200GPa Ebr := 100GPaSolution: ΔT := T2 − T1 ΔT = 10 °CEquations of equilibrium:+ ΣFx=0; FA − FC = 0 FA = FCLet FA=F. Then, FA = FAB = FBC = FC = FCompatibility: 0 = δT + δF0 αst⋅ (ΔT)⋅L + αbr⋅ (ΔT)⋅LF⋅LAst⋅Est−F⋅LAbr⋅Ebr= −F(αst + αbr)⋅ΔT1Ast⋅Est:= F = 6.988 kN Ans1Abr⋅Ebr+ 283. Problem 4-80The center rod CD of the assembly is heated from T1 = 30°C to T2 = 180°C using electrical resistanceheating. At the lower temperature T1 the gap between C and the rigid bar is 0.7 mm. Determine theforce in rods AB and EF caused by the increase in temperature. Rods AB and EF are made of steel,and each has a cross-sectional area of 125 mm2. CD is made of aluminum and has a cross-sectionalarea of and 375 mm2. Est = 200 GPa, Eal = 70 GPa, αst = 12(10-6)/°C, αal = 23(10-6)/°C.Unit used: °C := degGiven: T1 := 30°C T2 := 180°C Δgap := 0.7mmAst := 125mm2 Aal := 375mm2αst 12⋅ (10− 6):= αal 23⋅ (10− 6)1°C1°C:=Est := 200GPa Eal := 70GPaLst := 300mm Lal := 240mmSolution: ΔT := T2 − T1 ΔT = 150 °CEquations of equilibrium:ΣΜC=0; FAB⋅ (b) − FEF⋅ (b) = 0FAB = FEFLet FAB= Fst. Then, FAB = FEF = Fst+ ΣFy=0; FAB + FEF − Fal = 0Fal = 2Fst [1]Compatibility:δst = δal − ΔgapFst⋅LstEst⋅Astαal⋅ (ΔT)⋅LalFal⋅LalEal⋅Aal= −− Δgap [2]⎡⎢⎣⎤⎥⎦Substituting [1] into [2]:Fstαal⋅ (ΔT) Lal − ΔgapLstAst⋅Est:= Fst = 4.226 kN Ans2LalAal⋅Eal+From [1]: Fal := 2Fst Fal = 8.453 kN Ans 284. Problem 4-81The center rod CD of the assembly is heated from T1 = 30°C to T2 = 180°C using electrical resistanceheating. Also, the two end rods AB and EF are heated from T1 = 30°C to T2 = 50°C. At the lowertemperature T1 the gap between C and the rigid bar is 0.7 mm. Determine the force in rods AB and EFcaused by the increase in temperature. Rods AB and EF are made of steel, and each has across-sectional area of 125 mm2. CD is made of aluminum and has a cross-sectional area of and 375mm2. Est = 200 GPa, Eal = 70 GPa, αst = 12(10-6)/°C, αal = 23(10-6)/°C.Unit used: °C := degGiven: T1 := 30°C T2 := 180°C T'2 := 50°CAst := 125mm2 Aal := 375mm2αst 12⋅ (10− 6):= αal 23⋅ (10− 6)1°C1°C:=Est := 200GPa Eal := 70GPaΔgap := 0.7mmLst := 300mm Lal := 240mmSolution: ΔT := T2 − T1 ΔT = 150 °CΔT' := T'2 − T1 ΔT' = 20 °CEquations of equilibrium:ΣΜC=0; FAB⋅ (b) − FEF⋅ (b) = 0FAB = FEFLet FAB= Fst. Then, FAB = FEF = Fst+ ΣFy=0; FAB + FEF − Fal = 0Fal = 2Fst [1]Compatibility:δst = δal − Δgapαst⋅ (ΔT')⋅LstFst⋅LstEst⋅Ast⎛⎜⎝⎞⎠+ αal⋅ (ΔT)⋅LalFal⋅LalEal⋅Aal= −− Δgap [2]⎡⎢⎣⎤⎥⎦Substituting [1] into [2]:Fstαal⋅ (ΔT) Lal − Δgap − αst⋅ (ΔT')⋅Lst:= Fst = 1.849 kN AnsLstAst⋅Est2LalAal⋅Eal+From [1]: Fal := 2Fst Fal = 3.698 kN Ans 285. Problem 4-82The pipe is made of A-36 steel and is connected to the collars at A and B.When the temperature is 15°C, there is no axial load in the pipe. If hot gas traveling through the pipe causes its temperature to riseby ΔT = (20 + 30x)°C, where x is in meter, determine the average normal stress in the pipe. The innerdiameter is 50 mm, the wall thickness is 4 mm.Unit used: °C := degGiven: L := 2.4m di := 50mT1 := 15°C t := 4mmΔT = (20 + 30x)°CE := 200GPa α 12⋅ (10− 6)1°C:=Solution: do := di + 2t Aπ4do2 − di2 ⎛⎝⎞⎠:= ⋅Compatibility:0 = δT + δF0Lα⋅ (ΔT) x⌠⎮⌡0dF⋅LE⋅A= −unit := 1m⋅ °C LxLm:=Fα⋅E⋅AL⎛⎜⎝⎞⎠⋅ (unit)Lx0(20 + 30x) x⌠⎮⌡:= ⋅ dF = 84452.766 kNAverage Normal Stress:σFA:= σ = 134.40MPa Ans 286. Problem 4-83The bronze 86100 pipe has an inner radius of 12.5 mm and a wall thickness of 5 mm. If the gasflowing through it changes the temperature of the pipe uniformly from TA = 60°C at A to TB = 15°C atB, determine the axial force it exerts on the walls. The pipe was fitted between the walls when T =15°C.Unit used: °C := degGiven: L := 2.4m ri := 12.5mmTA := 60°C t := 5mmTB := 15°C To := 15°CE := 103GPa α 17⋅ (10− 6)1°C:=⎛⎝2 − ri2 Solution: ro := ri + t A π ro⎞⎠:= ⋅Tx 60TB − TA+ ⋅xL⎛⎜⎝⎞⎠= °C Tx = (60 − 18.75x)°CΔT = (Tx − To)ΔT = (45 − 18.75x)°CCompatibility:0 = δT + δF0Lα⋅ (ΔT) x⌠⎮⌡0dF⋅LE⋅A= −unit := 1m⋅ °C LxLm:=Fα⋅E⋅AL⎛⎜⎝⎞⎠⋅ (unit)Lx0(45 − 18.75x) x⌠⎮⌡:= ⋅ dF = 18.566 kN Ans 287. Problem 4-84The rigid block has a weight of 400 kN and is to be supported by posts A and B, which are made ofA-36 steel, and the post C, which is made of C83400 red brass. If all the posts have the same originallength before they are loaded, determine the average normal stress developed in each post when post C isheated so that its temperature is increased by 10°C. Each post has a cross-sectional area of 5000 mm2.Unit used: °C := degGiven: ΔT := 10°C A := 5000mm2 b := 1mEst := 200GPa Ebr := 101GPaW := 400kN αbr 18⋅ (10− 6)1°C:=Solution: Set: L := 1mEquations of equilibrium:ΣΜC=0; FA⋅ (b) − FB⋅ (b) = 0FA = FBLet FA= Fst. Then, FA = FB = Fst+ ΣFy=0; FA + FB + Fbr − W = 0Given Fbr = W − 2Fst [1]Compatibility: δst = δbrFst⋅LEst⋅AFbr⋅LEbr⋅A= − αbr⋅ (ΔT)⋅L [2]Initial guess: Fst := 1kN Fbr := 2kNSolving [1] and [2]:FstFbr⎛⎜⎜⎝⎞⎠:= Find(Fst , Fbr)FstFbr⎛⎜⎜⎝⎞⎠123.393153.214⎛⎜⎝⎞⎠= kN AnsAverage Normal Stress: σstFstA:= σst = 24.68MPa AnsσbrFbrA:= σbr = 30.64MPa Ans 288. Problem 4-85The bar has a cross-sectional area A, length L, modulus of elasticity E, and coefficient of thermalexpansion α. The temperature of the bar changes uniformly along its length from TA at A to TB at B sothat at any point x along the bar T = TA + x (TB - TA)/L. Determine the force the bar exerts on the rigidwalls. Initially no axial force is in the bar.Solution:Compatibility:0 = δT + δF [1]= + ⋅xTx TATB − TALΔT = (Tx − TA)ΔT (TB − TA) x= ⋅ d(ΔT) TB − TAL= ⋅xLHowever,d(δT) = α⋅ (ΔT) dxd(δT) α (TB − TA) xL⋅ ⎡⎢⎣⎤⎥⎦= ⋅ dxδTLα (TB − TA) x x0L⋅ ⎡⎢⎣⎤⎥⎦⋅⌠⎮⎮⌡= dδTα⋅L2= ⋅ (TB − TA)From [1]: 0α⋅L2⋅ (TB − TA)F⋅LE⋅A= −Fα⋅E⋅A2= ⋅ (TB − TA) Ans 289. Problem 4-86The rod is made of A-36 steel and has a diameter of 6 mm. If the springs are compressed 12 mmwhen the temperature of the rod is T = 10°C, determine the force in the rod when its temperature is T= 75°C.Unit used: °C := degGiven: L := 1.2m d := 6mm k 200Nmm:=T1 := 10°C T2 := 75°C xo := 12mmE := 200GPa α 12⋅ (10− 6)1°C:=Solution: Aπ4⎛⎜⎝⎞⎠:= ⋅d2For ΔT := T2 − T1 ΔT = 65 °C= ⋅ ( + )F k x xoCompatibility:2x = δT + δF2x α⋅ (ΔT)⋅LF⋅LE⋅A= −x α⋅ (ΔT)L2⋅⋅ ( + )⋅L2E⋅Ak x xo= −xα⋅ (ΔT)⋅E⋅A⋅L − k⋅xo⋅L:= x = 0.20892mm2E⋅A + k⋅L:= ⋅ ( + ) F = 2.442 kN AnsF k x xo 290. Problem 4-87Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8kN.Given: P := 8kN d := 20mmw := 40mm h := 20mmt := 5mm r := 10mmSolution:For the fillet: A := h⋅ twh= 2rh= 0.5From Fig. 4-24, K := 1.4σmax = K⋅σavgσmax KPA:= ⋅ σmax = 112MPa AnsFor the hole: Ao := (w − d)⋅ t rod2:=row= 0.25From Fig. 4-25, Ko := 2.375σmax = K⋅σavgσmax KoPAo:= ⋅ σmax = 190MPa Ans 291. Problem 4-88If the allowable normal stress for the bar is σallow = 120 MPa, determine the maximum axial force Pthat can be applied to the bar.Given: d := 20mm t := 5mm r := 10mmw := 40mm h := 20mmσallow := 120MPaSolution:Assume failure of the fillet: A := h⋅ twh= 2rh= 0.5From Fig. 4-24, K := 1.4σallow = σmax σmax = K⋅σavg σallow KPA= ⋅P(σallow)⋅A:= P = 8.57 kNKAssume failure of the hole: Ao := (w − d)⋅ t rod2:=row= 0.25From Fig. 4-25, Ko := 2.375σallow = σmax σmax = K⋅σavg σmax KoPAo:= ⋅P(σallow)⋅Ao:= P = 5.05 kN AnsKo(Controls!) 292. Problem 4-89The steel bar has the dimensions shown. Determine the maximum axial force P that can be applied soas not to exceed an allowable tensile stress of σallow =150 MPa.Given: d := 24mm t := 20mm r := 15mmw := 60mm h := 30mmσallow := 150MPaSolution:Assume failure occurs at the fillet: A := h⋅ twh= 2rh= 0.5From Fig. 4-24, K := 1.4σallow = σmax σmax = K⋅σavg σallow KPA= ⋅P(σallow)⋅A:= P = 64.29 kNKAssume failure occrs at the hole: Ao := (w − d)⋅ t rod2:=row= 0.2From Fig. 4-25, Ko := 2.45σallow = σmax σmax = K⋅σavg σmax KoPAo:= ⋅P(σallow)⋅Ao:= P = 44.08 kN AnsKo(Controls!) 293. Problem 4-90Determine the maximum axial force P that can be applied to the bar. The bar is made from steel andhas an allowable stress of σallow = 147 MPa.Given: d := 15mm t := 4mm r := 5mmw := 37.5mm h := 25mmσallow := 147MPaSolution:Assume failure of the fillet: A := h⋅ twh= 1.5rh= 0.2From Fig. 4-24, K := 1.73σallow = σmax σmax = K⋅σavg σallow KPA= ⋅P(σallow)⋅A:= P = 8.497 kNKAssume failure of the hole: Ao := (w − d)⋅ t rod2:=row= 0.2From Fig. 4-25, Ko := 2.45σallow = σmax σmax = K⋅σavg σmax KoPAo:= ⋅P(σallow)⋅Ao:= P = 5.4 kN AnsKo(Controls!) 294. Problem 4-91Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8kN.Given: d := 15mm t := 4mm r := 5mmw := 37.5mm h := 25mmP := 8kNSolution:For the fillet: A := h⋅ twh= 1.5rh= 0.2From Fig. 4-24, K := 1.73σmax = K⋅σavgσmax KPA:= ⋅ σmax = 138.4MPaFor the hole: Ao := (w − d)⋅ t rod2:=row= 0.20From Fig. 4-25, Ko := 2.45σmax = K⋅σavgσmax KoPAo:= ⋅ σmax = 217.78MPa Ans(Controls!) 295. Problem 4-92Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8kN.Given: P := 8kN d := 12mmw := 60mm h := 30mmt := 5mm r := 15mmSolution:At the fillet: A := h⋅ twh= 2rh= 0.5From Fig. 4-24, K := 1.4σmax = K⋅σavgσmax KPA:= ⋅ σmax = 74.67MPaAt the hole: Ao := (w − d)⋅ t rod2:=row= 0.1From Fig. 4-25, Ko := 2.65σmax = K⋅σavgσmax KoPAo:= ⋅ σmax = 88.33MPa Ans(Controls!) 296. Problem 4-93The resulting stress distribution along section AB for the bar is shown. From this distribution,determine the approximate resultant axial force P applied to the bar. Also, what is thestress-concentration factor for this geometry?Given: h := 80mm t := 10mmu := 5MPa v := 20mmSolution:⌠⎮⌡= dP σ AP = Volume under curveNumber of sqaures, n := 19P := n⋅ (u⋅v)⋅ t P = 19.00 kN AnsσavgPh⋅ t:= σavg = 23.75MPa AnsFrom the Figure, σmax := 30MPaKσmaxσavg:= K = 1.26 Ans 297. Problem 4-94The resulting stress distribution along section AB for the bar is shown. From this distribution,determine the approximate resultant axial force P applied to the bar.Also, what is thestress-concentration factor for this geometry?Given: h := 80mm t := 20mmu := 18MPa v := 20mm⌠⎮⌡= dP σ AP = Volume under curveNumber of sqaures, n := 10P := n⋅ (u⋅v)⋅ t P = 72.00 kN AnsσavgPh⋅ t:= σavg = 45.00MPa AnsFrom the Figure, σmax := 72MPaKσmaxσavg:= K = 1.60 AnsSolution: 298. Problem 4-95The A-36 steel plate has a thickness of 12 mm. If there are shoulder fillets at B and C, and σallow = 150MPa, determine the maximum axial load P that it can support. Compute its elongation neglecting theeffect of the fillets.Given: t := 12mm r := 30mmw := 120mm h := 60mmL1 := 200mm L2 := 800mmσallow := 150MPaE := 200GPaSolution: A1 := h⋅ t A2 := w⋅ tMaximum Normal Stress at fillet:wh= 2rh= 0.5From Fig. 4-23, K := 1.4σallow = σmax σmax = K⋅σavg σallow KPA1= ⋅P(σallow)⋅A1:= P = 77.14 kN AnsKDisplacement:δ = δ1 + δ2 + δ1δ 2P⋅L1E⋅A1P⋅L2E⋅A2:= + δ = 0.429mm Ans 299. Problem 4-96The 1500-kN weight is slowly set on the top of a post made of 2014-T6 aluminum with an A-36 steelcore. If both materials can be considered elastic perfectly plastic, determine the stress in each material.Given: ro := 50mm Est := 200⋅GPari := 25mm Eal := 73.1⋅GPars := 25mm σY_st := 250MPaP := 1500kN σY_al := 414MPaSolution:2 ⎛⎝Ast π rs⎞⎠⎛⎝2 − ri2 := ⋅ Aal π ro⎞⎠:= ⋅Equations of equilibrium:Given + ΣFy=0; Pal + Pst − P = 0 [1]Compatibility: δst = δal (Pal)⋅LAal⋅Eal(Pst)⋅LAst⋅Est=PalAal⋅EalPstAst⋅Est= [2]Initial guess: Pal := 1kN Pst := 2kNSolving [1] and [2]:PalPst⎛⎜⎜⎝⎞⎠:= Find(Pal , Pst)PalPst⎛⎜⎜⎝⎞⎠784.5218715.4782⎛⎜⎝⎞⎠= kNAverage Normal stress:σalPalAal:= σal = 133.18MPa ( < σY_al = 414MPa ) o.k.!σstPstAst:= σst = 364.39MPa ( > σY_st = 250MPa )Thererfore, the steel core yields and so the elastic analysis is invalid.Plastic Analysis: The stress in the steel is σst := σY_stσst = 250MPa AnsPst := (σY_st)Ast Pst = 490.87 kNFrom [1]: Pal := P − Pst Pal = 1009.13 kNσalPalAal:= σal = 171.31MPa ( < σY_al = 414MPa ) o.k.! Ans 300. Problem 4-97The 10-mm-diameter shank of the steel bolt has a bronze sleeve bonded to it. The outer diameter ofthis sleeve is 20 mm. If the yield stress for the steel is (σY)st = 640 MPa, and for the bronze (σY)br =520 MPa, determine the magnitude of the largest elastic load P that can be applied to the assembly. Est= 200 GPa, Ebr = 100 GPa.Given: do := 20mm Est := 200GPa σY_st := 640MPadi := 10mm Ebr := 100GPa σY_br := 520MPads := 10mmSolution:Astπ4⎛⎝ds2 ⎞⎠:= ⋅ Abrπ4do2 − di2 ⎛⎝⎞⎠:= ⋅Equations of equilibrium:+ ΣFy=0; Pbr + Pst − P = 0 [1]Compatibility:δb = δs(Pbr)⋅LAbr⋅Ebr(Pst)⋅LAst⋅Est=PbrAbr⋅EbrPstAst⋅Est= [2]Assume yielding of bolt, thenPst := (σY_st)Ast Pst = 50.265 kNFrom [2]: PbrAbr⋅EbrAst⋅Est⎛⎜⎝⎞⎠:= ⋅Pst Pbr = 75.398 kNFrom [1]: P := Pbr + Pst P = 125.66 kN (Controls!): AnsAssume yielding of sleeve, thenPbr := σY_brAbr Pbr = 122.522 kNFrom [2]: PstAst⋅EstAbr⋅Ebr⎛⎜⎝⎞⎠:= ⋅Pbr Pst = 81.681 kNFrom [1]: P := Pbr + Pst P = 204.20 kN 301. Problem 4-98The weight is suspended from steel and aluminum wires, each having the same initial length of 3 m andcross-sectional area of 4 mm2. If the materials can be assumed to be elastic perfectly plastic, with(σY)st = 120 MPa and (σY)al = 70 MPa, determine the force in each wire if the weight is (a) 600 N and(b) 720 N. Eal = 70 GPa, Est = 200 GPa.Given: L := 3m Est := 200GPa σY_st := 120MPaA 4mm 2:= Eal := 70GPa σY_al := 70MPaSolution: (a) W := 600NEquations of equilibrium:Given + ΣFy=0; Pal + Pst − W = 0 [1]Compatibility: δst = δal (Pal)⋅LA⋅Eal(Pst)⋅LA⋅Est=PalA⋅EalPstA⋅Est= [2]Initial guess: Pal := 1N Pst := 2NSolving [1] and [2]:PalPst⎛⎜⎜⎝⎞⎠:= Find(Pal , Pst)PalPst⎛⎜⎜⎝⎞⎠155.56444.44⎛⎜⎝⎞⎠= N AnsAverage Normal stress:σalPalA:= σal = 38.89MPa ( < σY_al = 70MPa ) o.k.!σstPstA:= σst = 111.11MPa ( < σY_st = 120MPa ) o.k.!The average normal stress for both wires do not exceed their respective yieldstress. Thererfore, the elastic analysis is valid for both wires. 302. Solution: (b) W := 720NEquations of equilibrium:Given + ΣFy=0; Pal + Pst − W = 0 [1]Compatibility: δst = δal (Pal)⋅LA⋅Eal(Pst)⋅LA⋅Est=PalA⋅EalPstA⋅Est= [2]Initial guess: Pal := 1N Pst := 2NSolving [1] and [2]:PalPst⎛⎜⎜⎝⎞⎠:= Find(Pal , Pst)PalPst⎛⎜⎜⎝⎞⎠186.67533.33⎛⎜⎝⎞⎠= NAverage Normal stress:σalPalA:= σal = 46.67MPa ( < σY_al = 70MPa ) o.k.!σstPstA:= σst = 133.33MPa ( > σY_st = 120MPa )Thererfore, the steel wire yields and so the elastic analysis is invalid.Plastic Analysis: The stress in the steel is σst := σY_stσst = 120MPaPst := (σY_st)A Pst = 480.00N AnsFrom [1]: Pal := W − Pst Pal = 240.00 N AnsσalPalA:= σal = 60MPa ( < σY_al = 70MPa ) o.k.! 303. Problem 4-99The bar has a cross-sectional area of 625 mm2. If a force of P = 225 kN is applied at B and thenremoved, determine the residual stress in sections AB and BC. σY = 210 MPa.Given: L := 1m LAB := 0.75L LBC := 0.25LA := 625mm2 σY := 210MPa P := 225kNSolution:By superposition :+ΔC − δC = 0(P)⋅LABA⋅E(FC)⋅LA⋅E− = 0 FC PLABL⎛⎜⎝⎞⎠:= ⋅ FC = 168.75 kNEquations of equilibrium:+ ΣFy=0; FA + FC − P = 0 FA := P − FC FA = 56.25 kN [1]Average Normal stress:σABFAA:= σAB = 90MPa ( < σY = 210MPa ) o.k.!σBCFCA:= σBC = 270MPa ( > σY = 210MPa )Thererfore, the segment BC yields and so the elastic analysis is invalid.Plastic Analysis: The stress in the BC is σBC := σYσBC = 210MPaFC := (σY)A FC = 131.25 kNFrom [1]: FA := P − FC FA = 93.75 kNσABFAA:= σAB = 150MPa ( < σY = 210MPa ) o.k.!A reversal of force of 45kip applied results in a reversed FC=270 MPa and FA=90 MPa,which produceσ'AB := 90 MPa (T)σ'BC := 270 MPa (C)Hence,ΔσAB := σAB − σ'AB ΔσAB = 60MPa (T) AnsΔσBC := −(σBC − σ'BC) ΔσBC = 60MPa (T) Ans 304. Problem 4-100The bar has a cross-sectional area of 300 mm2 and is made of a material that has a stressstrain diagramthat can be approximated by the two line segments shown. Determine the elongation of the bar due tothe applied loading.Given: LAB := 1.5m LBC := 0.6m A := 300mm2PB := 40kN PC := 25kN σ1 := 140MPaσ2 := 280MPa ε2 0.021mmmm:= ε1 0.001mmmm:=Solution: PBC := PC PAB := PC + PBAverage Normal stress and Strain: For segment BCσBCPBCA:= σBC = 83.33MPa (< 140MPa )εBCσBCσ1⎛⎜⎝⎞⎠:= ⋅ε1 εBC 0.00060mmmm=Average Normal stress and Strain: For segment ABσABPABA:= σAB = 216.67MPa (> 140MPa )εAB ε1σAB − σ1σ2 − σ1⎛⎜⎝⎞⎠:= + ⋅ (ε2 − ε1) εAB 0.01195mmmm=Elongation:δAB := εAB⋅LAB δAB = 17.92857mmδBC := εBC⋅LBC δBC = 0.35714mmδTotal := δAB + δBCδTotal = 18.286mm Ans 305. Problem 4-101The rigid bar is supported by a pin at A and two steel wires, each having a diameter of 4 mm. If theyield stress for the wires is σY = 530 MPa, and Est = 200 GPa, determine the intensity of thedistributed load w that can be placed on the beam and will just cause wire EB to yield. What is thedisplacement of point G for this case? For the calculation, assume that the steel is elastic perfectlyplastic.Given: a := 400mm b := 250mm c := 150mmd := 4mm Lwire := 800mmEst := 200GPa σY := 530MPaSolution: L := a + b + c Aπ4:= ⋅d2Plastic Analysis:Wire CD will yield first followed by wire BE. Whenboth wores yield,FBE := (σY)A FBE = 6.660 kNFCD := (σY)A FCD = 6.660 kNEquations of equilibrium:ΣΜA=0; FBE⋅ (a) + FCD⋅ (a + b) − w⋅L⋅ (0.5L) = 0w FBE2aL2⎛⎜⎝⎞⎠⋅ FCD2(a + b)L2:= + ⋅w 21.85kNm= AnsDisplacement:When wire BE achieves yield stress, the correspondingyield strain isεYσYEst:= εY 0.002650mmmm=δBE := εY⋅ (Lwire) δBE = 2.120mmGeometry:δBEaδGL=δGLa:= ⋅ (δBE) δG = 4.24mm Ans 306. Problem 4-102The rigid bar is supported by a pin at A and two steel wires, each having a diameter of 4 mm. If theyield stress for the wires is σY = 530 MPa, and Est = 200 GPa, determine (a) the intensity of thedistributed load w that can be placed on the beam that will cause only one of the wires to start to yieldand (b) the smallest intensity of the distributed load that will cause both wires to yield. For thecalculation, assume that the steel is elastic perfectly plastic.Given: a := 400mm b := 250mm c := 150mmd := 4mm Lwire := 800mmEst := 200GPa σY := 530MPaSolution: L := a + b + c Aπ4:= ⋅d2Equations of equilibrium:ΣΜA=0; FBE⋅ (a) + FCD⋅ (a + b) − w⋅L⋅ (0.5L) = 0 [1]a) By observation, wire CD will yield first. Then,FCD := (σY)A FCD = 6.660 kNGeometry:δBEaδCDa + b= δBEaa + b⎛⎜⎝⎞⎠= ⋅δCDFBE⋅LwireEst⋅Aaa + bFCD⋅LwireEst⋅A⎛⎜⎝⎞⎠= ⋅FBEaa + b⎛⎜⎝⎞⎠:= ⋅FCD FBE = 4.0986 kNFrom [1]: w FBE2aL2⎛⎜⎝⎞⎠⋅ FCD2(a + b)L2:= + ⋅w 18.65kNm= Ansb) When both wires yield,FBE := (σY)A FBE = 6.660 kNFCD := (σY)A FCD = 6.660 kNFrom [1]: w FBE2aL2⎛⎜⎝⎞⎠⋅ FCD2(a + b)L2:= + ⋅w 21.85kNm= Ans 307. Problem 4-103The rigid beam is supported by the three posts A,B, and C of equal length. Posts A and C have adiameter of 75 mm and are made of aluminum, for which Eal = 70 GPa and (σY)al = 20 MPa. Post Bhas a diameter of 20 mm and is made of brass, for which Ebr = 100 GPa and (σY)br = 590 MPa.Determine the smallest magnitude of P so that (a) only rods A and C yield and (b) all the posts yield.Given: dA := 75mm dB := 20mm dC := 75mmEal := 70GPa σY_al := 20MPa L := 4mEbr := 100GPa σY_br := 590MPaSolution: AAπ4:= ⋅ dA2 ABπ4:= ⋅ dB2 ACπ4:= ⋅ 2dCEquations of equilibrium:ΣΜB=0; FC⋅ (L) − P⋅ (0.5L) − FA⋅ (L) + P⋅ (0.5L) = 0FA = FCLet FA= Fal. Then, FA = FC = Fal+ ΣFy=0; FA + FC + Fbr − 2P = 0 2Fal + Fbr − 2P = 0 [1]a) Post A and C will yield,Fal := (σY_al)AA Fal = 88.357 kNεalσY_alEal:= εal 0.0002857mmmm=Compatibility condition: δal = δbrFal⋅LpostEal⋅AAFbr⋅LpostEbr⋅AB= FbrEbr⋅ABEal⋅AA⎛⎜⎝⎞⎠:= ⋅Fal Fbr = 8.976 kNσbrFbrAB:=σbr = 28.57MPa (< σY ) o.k.!From [1]: P := Fal + 0.5Fbr P = 92.85 kN Ansb) All the posts yield. Then,Fal := (σY_al)AA Fal = 88.357 kNFbr := (σY_br)AB Fbr = 185.354 kNFrom [1]: P := Fal + 0.5Fbr P = 181.0 kN Ans 308. Problem 4-104The rigid beam is supported by the three posts A,B, and C. Posts A and C have a diameter of 60 mmand are made of aluminum, for which Eal = 70 GPa and (σY)al = 20 MPa. Post B is made of brass, forwhich Ebr = 100 GPa and (σY)br = 590 MPa. If P = 130 kN, determine the largest diameter of post Bso that all the posts yield at the same time.Given: dA := 60mm dC := 60mm P := 130kNEal := 70GPa σY_al := 20MPa L := 4mEbr := 100GPa σY_br := 590MPaSolution: AAπ4:= ⋅ dA2 ACπ4:= ⋅ 2dCEquations of equilibrium:ΣΜB=0; FC⋅ (L) − P⋅ (0.5L) − FA⋅ (L) + P⋅ (0.5L) = 0FA = FCLet FA= Fal. Then, FA = FC = Fal+ ΣFy=0; FA + FC + Fbr − 2P = 0 2Fal + Fbr − 2P = 0 [1]When all the posts yield,Fal := (σY_al)AA Fal = 56.549 kNFrom [1]: Fbr := 2P − 2Fal Fbr = 146.90 kNAlso, Fbr = (σY_br)AB ABπ4= ⋅ 2dB⋅ 2 ⎛⎜⎝Fbr (σY_br) π4dB⎞⎠=dB4Fbrπ⋅σY_br:=dB = 17.805mm Ans 309. Problem 4-105The rigid beam is supported by three A-36 steel wires, each having a length of 1.2 m. The cross-sectionalarea of AB and EF is 10 mm2, and the cross-sectional area of CD is 4 mm2. Determine thelargest distributed load w that can be supported by the beam before any of the wires begin to yield. Ifthe steel is assumed to be elastic perfectly plastic, determine how far the beam is displaced downwardjust before all the wires begin to yield.Given: L := 1.2m b := 1.5m AAB := 10mm2AEF := 10mm2 ACD := 4mm2E := 200⋅GPa σY := 250MPaSolution:Compatibility:Beam remains horizontal after the displacementsince the loading and the system are symmetrical.δAB = δCD(FAB)⋅LE⋅AAB(FCD)⋅LE⋅ACD= FCDACDAAB⎛⎜⎝⎞⎠= ⋅FAB [1]Equations of equilibrium:ΣΜC=0; FEF⋅ (b) − FAB⋅ (b) = 0 FAB = FEF+ ΣFy=0; FAB + FCD + FEF − w⋅ (2b) = 02⋅FAB = w⋅ (2b) − FCD [2]Plastic Analysis: Assume wires AB and EF yield.FAB := (σY)⋅AAB FAB = 2.500 kNFrom [1]: FCDACDAAB⎛⎜⎝⎞⎠:= ⋅FAB FCD = 1.000 kNFrom [2]: wFABbFCD2⋅b:= + w 2.0000kNm=Plastic Analysis: Assume failure of CD.FCD := (σY)⋅ACD FCD = 1.000 kNFrom [1]: FABAABACD⎛⎜⎝⎞⎠:= ⋅FCD FAB = 2.500 kNFrom [2]: wFABbFCD2⋅b:= + w 2.0000kNm=The three wires AB, CD and EF yield simultaneously. Hence, w 2.00kNm= AnsDisplacement:δFCD⋅LE⋅ACD:= δ = 1.500mm Ans 310. Problem 4-106A material has a stressstrain diagram that can be described by the curve σ = cε 1/2. Determine thedeflection δ of the end of a rod made from this material if it has a length L, cross-sectional area A, anda specific weight γ.Solution:Stress-strain relationship :σ = c⋅ εσ2 = c2⋅εHowever, σPA= and εdδdx=Thus,P2A2c2 dδdx= ⋅dδdxP2c2 A2=Since P = γ⋅A⋅x dδdxγ2c2⎛⎜⎜⎝⎞⎠= ⋅x2Displacement:δL0xγ2c2⎛⎜⎜⎝⎞⎠⋅x2⌠⎮⎮⎮⌡= d δLx2 x ⌠⎮⌡γ2c2 0d⎛⎜⎜⎝⎞⎠= ⋅δγ2 L3 ⋅3⋅c2= Ans 311. Problem 4-107Solve Prob. 4-106 if the stressstrain diagram is defined by σ = cε 3/2.Solution:Stress-strain relationship :σ = c⋅ ε3σ2 = c2⋅ε3However, σPA= and εdδdx=Thus,P2A2c2 dδdx⎛⎜⎝⎞⎠3= ⋅dδdxPc⋅A⎛⎜⎝⎞⎠23=Since P = γ⋅A⋅x dδdxγc⎛⎜⎝⎞⎠23x2= ⋅ 3Displacement:δL0xγc⎛⎜⎝⎞⎠23x2⋅ 3⌠⎮⎮⎮⎮⌡= d δγc⎛⎜⎝⎞⎠23L023x x⌠⎮⎮⌡d⎛⎜⎜⎜⎝⎞⎟⎠= ⋅δ35γc⎛⎜⎝⎞⎠23⋅ L5= ⋅ 3 Ans 312. Problem 4-108The bar having a diameter of 50 mm is fixed connected at its ends and supports the axial load P. Ifthe material is elastic perfectly plastic as shown by the stressstrain diagram, determine the smallest loadP needed to cause segment AC to yield. If this load is released, determine the permanent displacementof point C.Given: LAC := 0.6m LCB := 0.9m d := 50mmσY := 140MPa εY 0.001mmmm:=Solution: Aπ⋅d2 4⎛⎜⎝⎞⎠:=When P is increased, the segment AC will becomeplastic first, then CB will become plastic. Thus,FA := (σY)A FA = 274.889 kNFB := (σY)A FB = 274.889 kNEquations of equilibrium:+ ΣFx=0; FA + FB − P = 0 [1]From the figure, EσYεY:= E = 140 × 103MPa P := FA + FBP = 549.78 kN AnsThe deflection of point C is,δC := (εY)⋅LCBδC = 0.900mmConsider the reverse of P on the bar.(F'A)⋅LAC F'A = 1.5F'B [2]A⋅E(F'B)⋅LCBA⋅E=Equations of equilibrium:+ ΣFx=0; F'A + F'B − P = 0 [1']Substituting [2] into [1']: F'A := 0.6P F'A = 329.87 kNF'B := 0.4P F'B = 219.91 kNThe deflection of point C is,δ'C(F'B)⋅LCB:= δ'C = 0.72000mmA⋅EHence, Δδ := δC − δ'C Δδ = 0.180mm Ans 313. Problem 4-109Determine the elongation of the bar in Prob. 4-108 when both the load P and the supports are removed.Given: LAC := 0.6m LCB := 0.9m d := 50mmσY := 140MPa εY 0.001mmmm:=Solution: Aπ⋅d2 4⎛⎜⎝⎞⎠:= L := LAC + LCBWhen P is increased, the segment AC will becomeplastic first, then CB will become plastic. Thus,FA := (σY)A FA = 274.889 kNFB := (σY)A FB = 274.889 kNEquations of equilibrium:+ ΣFx=0; FA + FB − P = 0 [1]From the figure, EσYεY:= E = 140 × 103MPa P := FA + FBP = 549.78 kN AnsThe deflection of point C is,δC := (εY)⋅LCBδC = 0.900mmConsider the reverse of P on the bar.(F'A)⋅LAC F'A = 1.5F'B [2]A⋅E(F'B)⋅LCBA⋅E=Equations of equilibrium:+ ΣFx=0; F'A + F'B − P = 0 [1']Substituting [2] into [1']: F'A := 0.6P F'A = 329.87 kNF'B := 0.4P F'B = 219.91 kNThe resultant reactions are: F''A := F'A − FA F''A = 54.978 kNF''B := −(F'B − FB) F''B = 54.978 kNWhen the supports are removed, the elongation will beδ''CF''A⋅LA⋅E:= δ''C = 0.300mm Ans 314. Problem 4-110A 6-mm-diameter steel rivet having a temperature of 800°C is secured between two plates such that atthis temperature it is 50 mm long and exerts a clamping force of 1.25 kN between the plates.Determine the approximate clamping force between the plates when the rivet cools to 5°C. For thecalculation, assume that the heads of the rivet and the plates are rigid. Take αst = 14(10-6)/°C, Est =200 GPa. Is the result a conservative estimate of the actual answer? Why or why not?Unit used: °C := degGiven: T1 := 800°C T2 := 5°C P := 1.250kNL := 50mm d := 6mmαst 14⋅ (10− 6)1°C:= Est := 200GPaSolution: ΔT := T1 − T2 ΔT = 795 °CAπ4⎛⎜⎝⎞⎠:= ⋅d2By superposition:+ 0 = δT + δF0 αst⋅ (ΔT)⋅L(FT)⋅LA⋅Est= −FT := (αst)⋅ΔT⋅ (A⋅Est) FT = 62.939 kN:= + F = 64.189 kN AnsF P FTYes. Because as the rivet cools, the plates and the rivet head will also deform.Consequently, the force FT on the rivets will not be as great. 315. Problem 4-111Determine the maximum axial force P that can be applied to the steel plate. The allowable stress isσallow = 150 MPa.Given: d := 24mm t := 6mm r := 6mmw := 120mm h := 60mmσallow := 150MPaSolution:Assume failure of the fillet: A := h⋅ twh= 2rh= 0.1From Fig. 4-24, K := 2.4σallow = σmax σmax = K⋅σavg σallow KPA= ⋅P(σallow)⋅A:= P = 22.5 kN (Controls!) AnsKAssume failure of the hole: Ao := (w − d)⋅ t rod2:=row= 0.1From Fig. 4-25, Ko := 2.65σallow = σmax σmax = K⋅σavg σmax KoPAo:= ⋅P(σallow)⋅Ao:= P = 32.604 kNKo 316. Problem 4-112The rigid link is supported by a pin at A and two A-36 steel wires, each having an unstretched length of300 mm and cross-sectional area of 7.8 mm2. Determine the force developed in the wires when thelink supports the vertical load of 1.75 kN.Given: L := 300mm A := 7.8mm2 P := 1.75kNa := 150mm b := 100mm c := 125mmSolution:Compatibility: δBbδCb + c=(FB)⋅Lb⋅ (A⋅E)(FC)⋅L(b + c)⋅ (A⋅E)=Given FBbFCb + c= [1]Equations of equilibrium:ΣΜA=0; −FC⋅ (b + c) − FB⋅ (b) + P⋅ (a) = 0 [2]Initial guess: FC := 1kN FB := 2kNSolving [1] and [2]:FCFB⎛⎜⎜⎝⎞⎠:= Find(FC, FB)FCFB⎛⎜⎜⎝⎞⎠0.9740.433⎛⎜⎝⎞⎠= kN Ans 317. Problem 4-113The force P is applied to the bar, which is composed of an elastic perfectly plastic material. Constructa graph to show how the force in each section AB and BC (ordinate) varies as P (abscissa) isincreased. The bar has cross-sectional areas of 625 mm2 in region AB and 2500 mm2 in region BC,and σY = 210 MPa.Given: LAB := 150mm AAB := 625mm2LBC := 50mm ABC := 2500mm2σY := 210MPaSolution:Equations of equilibrium:+ ΣFx=0; P − FA − FC = 0 [1]Elastic behavior:+ 0 = ΔC − δC0(P)⋅LABE⋅AAB(FC)⋅LBCE⋅ABC(FC)⋅LABE⋅AAB+⎡⎢⎣⎤⎥⎦= −= − ( )(0.5 + 6) FC0 6P FC1213= P [2]Substituting [2] into [1]: FA113= P [3]By comparison, segment BC will yield first. Hence,FC := (σY)⋅ABC FC = 525 kNFrom [2]: P1312:= ⋅FC P = 568.75 kNFrom [3]: FA113:= ⋅P FA = 43.75 kNWhen segment AB yields,FA := (σY)⋅AAB FA = 131.25 kNFC := (σY)⋅ABC FC = 525 kNFrom [1]: P := FA + FC P = 656.25 kN 318. Problem 4-114The 2014-T6 aluminum rod has a diameter of 12 mm and is lightly attached to the rigid supports at Aand B when T1 = 25°C. If the temperature becomes T2 = -20°C, and an axial force of P = 80 kN isapplied to the rigid collar as shown, determine the reactions at A and B.Unit used: °C := degGiven: LAC := 125mm T1 := 25°C d := 12mmLCB := 200mm T2 := −20°C P := 80NE := 73.1⋅GPa α 23⋅ (10− 6)1°C:=Solution: ΔT := T2 − T1 ΔT = −45 °CL := LAC + LCB A:= ⋅ (d2)π4By superposition :+ 0 = ΔB + ΔT + δB0(P)⋅LACA⋅E+ α⋅ (ΔT)⋅L(FB)⋅LA⋅E= +FB −PLACL⎛⎜⎝⎞⎠:= ⋅ − α⋅ (ΔT)⋅ (A⋅E) FB = 8.526 kN AnsEquations of equilibrium:+ ΣFx=0; −FA + P + FB = 0FA := P + FB FA = 8.606 kN Ans 319. Problem 4-115The 2014-T6 aluminum rod has a diameter of 12 mm and is lightly attached to the rigid supports at Aand B when T1 = 40°C. Determine the force P that must be applied to the collar so that, when T = 0°C,the reaction at B is zero.Unit used: °C := degGiven: LAC := 125mm T1 := 40°C d := 12mmLCB := 200mm T2 := 0°C FB := 0kNE := 73.1⋅GPa α 23⋅ (10− 6)1°C:=Solution: ΔT := T2 − T1 ΔT = −40 °CL := LAC + LCB A:= ⋅ (d2)π4By superposition :+ 0 = ΔB + ΔT + δB0(P)⋅LACA⋅E+ α⋅ (ΔT)⋅L(FB)⋅LA⋅E= +PLLACα − ΔT ( ) ⋅ A E ⋅ ( ) ⋅ FB − ⎡⎣:= ⋅ P = 19.776 kN Ans⎤⎦ 320. Problem 4-116The A-36 steel column, having a cross-sectional area of 11250 mm2, is encased in high-strengthconcrete as shown. If an axial force of 300 kN is applied to the column, determine the averagecompressive stress in the concrete and in the steel. How far does the column shorten? It has an originallength of 2.4 m.Given: ac := 225mm bc := 400mm Ast := 11250mm2 L := 2.4mP := 300kN Est := 200⋅GPa Ec := 29⋅GPaSolution: Ac := (ac⋅bc) − AstCompatibility: δst = δconcGiven (Pst)⋅LAst⋅Est(Pc)⋅LAc⋅Ec= [1]Equations of equilibrium:+ ΣFy=0; Pst + Pc − P = 0 [2]Initial guess: Pc := 1kN Pst := 2kNSolving [1] and [2]:PcPst⎛⎜⎜⎝⎞⎠:= Find(Pc , Pst)PcPst⎛⎜⎜⎝⎞⎠151.117148.883⎛⎜⎝⎞⎠= kNAverage Normal Stress:σstPstAst:= σst = 13.23MPa AnsσcPcAc:= σc = 1.92MPa AnsDisplacement: Either the concrete or steel can be used for the calculation.δ(Pst)⋅LAst⋅Est:= δ = 0.15881mm Ans 321. Problem 4-117The A-36 steel column is encased in high-strength concrete as shown. If an axial force of 300 kN isapplied to the column, determine the required area of the steel so that the force is shared equallybetween the steel and concrete. How far does the column shorten? It has an original length of 2.4 m.Given: ac := 225mm bc := 400mm L := 2.4mP := 300kN Est := 200⋅GPa Ec := 29⋅GPaPc := 0.5P Pst := 0.5PSolution: Ac = (ac⋅bc) − AstCompatibility: δst = δconc(Pst)⋅LAst⋅Est(Pc)⋅LAc⋅Ec=AstEcEc + Est:= ⋅ (ac⋅bc)Ast = 11397.38mm2 AnsDisplacement: Either the concrete or steel can be used for the calculation.δ(Pst)⋅LAst⋅Est:= δ = 0.15793mm Ans 322. Problem 4-118The assembly consists of a 30-mm-diameter aluminum bar ABC with fixed collar at B and a10-mm-diameter steel rod CD. Determine the displacement of point D when the assembly is loaded asshown. Neglect the size of the collar at B and the connection at C. Est = 200 GPa, Eal = 70 GPa.Given:LAB := 300mm dAB := 30mmLBC := 500mm dBC := 30mm Eal := 70GPaLCD := 700mm dCD := 10mm Est := 200GPaPB := −8kN PD := 20kNSolution:Internal Force: As shown on FBD.Displacement:AABπ4⎛⎜⎝⎞⎠:= ⋅ dAB2 δAB(PD + PB)⋅ (LAB)Eal⋅ (AAB):=ABCπ4⎛⎜⎝⎞⎠:= ⋅ dBC2 δBC(PD)⋅ (LBC)Eal⋅ (ABC):=ACDπ4⎛⎜⎝⎞⎠:= ⋅ dCD2 δCD(PD)⋅ (LCD)Est⋅ (ACD):=δD := δAB + δBC + δCDδD = 1.166mm Ans 323. Problem 4-119The joint is made from three A-36 steel plates that are bonded together at their seams. Determine thedisplacement of end A with respect to end B when the joint is subjected to the axial loads shown. Eachplate has a thickness of 5 mm.Given: b := 100mm t := 5mmLAB := 600mm AAB := t⋅bLBC := 200mm ABC := 3t⋅bLCD := 800mm ACD := 2t⋅bP := 46kN E := 200GPaSolution:Internal Force: As shown on FBD.δA_D = δAB + δBC + δCDδA_DP⋅ (LAB)E⋅ (AAB)P⋅ (LBC)E⋅ (ABC)+P⋅ (LCD)E⋅ (ACD):= +δA_D = 0.491mm Ans 324. Problem 5-1A shaft is made of a steel alloy having an allowable shear stress of τallow = 84 MPa. If the diameter ofthe shaft is 37.5 mm, determine the maximum torque T that can be transmitted. What would be themaximum torque T' if a 25-mm-diameter hole is bored through the shaft? Sketch the shear-stressdistribution along a radial line in each case.Given: do := 37.5mm di := 25mmτallow := 84MPaSolution: cdo2:=a) Allowable shear srtess : Aplying the torsion formulaτallowT⋅cJπ2⎛⎜⎝⎞⎠do2⎛⎜⎝⎞⎠4J := ⋅=T(τallow)⋅ Jc:=T = 0.87 kN⋅m Ansb) Allowable shear srtess : Aplying the torsion formulaτallowT'⋅cJ'π2⎛⎜⎝⎞⎠do2⎛⎜⎝⎞⎠4 diJ' =2⎛⎜⎝⎞⎠4−⎡⎢⎣⎤⎥⎦:= ⋅T'(τallow)⋅ Jc:=T' = 0.87 kN⋅m AnsShera stress at ρ := 0.5⋅diτρT'⋅ρJ':= τρ = 69.78MPa 325. Problem 5-2The solid shaft of radius r is subjected to a torque T. Determine the radius r' of the inner core of theshaft that resists one-half of the applied torque (T/2). Solve the problem two ways: (a) by using thetorsion formula, (b) by finding the resultant of the shear-stress distribution. 326. Problem 5-3The solid shaft of radius r is subjected to a torque T. Determine the radius r' of the inner core of theshaft that resists one-quarter of the applied torque (T/4). Solve the problem two ways: (a) by using thetorsion formula, (b) by finding the resultant of the shear-stress distribution. 327. Problem 5-4The tube is subjected to a torque of 750 N·m. Determine the amount of this torque that is resisted bythe gray shaded section. Solve the problem two ways: (a) by using the torsion formula, (b) by findingthe resultant of the shear-stress distribution.Given: ro := 100mm ri := 25mm T := 750N⋅mrc := 75mmSolution: c := roa) Aplying the torsion formula:Jπ2⎛⎜⎝⎞⎠ro4 − ri4 ⎛⎝⎞⎠:= ⋅ J'π2⎛⎜⎝⎞⎠ro4 − rc4 ⎛⎝⎞⎠:= ⋅τmaxT⋅cJ:= τmaxT'⋅cJ'=τmax = 0.479MPaT'τmax⋅ J'c:=T' = 0.515 kN⋅m Ansb) Integartion Method :dT' = ρ⋅τ⋅dA dA = 2π ρ⋅dρ τρc⎛⎜⎝⎞⎠= ⋅ τmaxdT' ρρc⎛⎜⎝⎞⎠= ⋅ ⋅τmax⋅ (2π ρ⋅dρ)dT'2πc⋅ τmaxρ3 = ⋅dρT'2πc⋅ τmaxroρ3 ρ ⌠⎮⌡rcd⎛⎜⎜⎝⎞⎠:=T' = 0.515 kN⋅m Ans 328. Problem 5-5The solid 30-mm-diameter shaft is used to transmit the torques applied to the gears. Determine theabsolute maximum shear stress on the shaft.Given: a := 300mm TA := −300N⋅mb := 400mm TC := 500N⋅mc := 500mm TD := 200N⋅mdo := 30mm TB := −400N⋅mSolution: c'do2:=Internal Torque : As shown in the torque diagram.Allowable shear srtess : Aplying the torsion formulaFrom the torque diagram, Tmax := TBJπ2⎛⎜⎝⎞⎠do2⎛⎜⎝⎞⎠4:= ⋅ τmaxTmax⋅c'J:=τmax = 75.45MPa Ansx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cT1(x1) TA1N⋅m:= ⋅ T2(x2) (TA + TC) 1:= ⋅ T3(x3) (TA + TC + TD) 1N⋅mN⋅m:= ⋅0 0.2 0.4 0.6 0.8 15000500Distance (m)Torque (Nm)T1(x1)T2(x2)T3(x3)x1, x2, x3 329. Problem 5-6The solid 32-mm-diameter shaft is used to transmit the torques applied to the gears. If it is supportedby smooth bearings at A and B, which do not resist torque, determine the shear stress developed in theshaft at points C and D. Indicate the shear stress on volume elements located at these points.Given: do := 32mm T1 := 185N⋅mT2 := −260N⋅m T3 := 75N⋅mSolution: cdo2:= Jπ2⎛⎜⎝⎞⎠do2⎛⎜⎝⎞⎠4:= ⋅TC := T1TD := T1 + T2τC(TC)⋅cJ:= τC = 28.75MPa AnsτD(TD)⋅cJ:= τD = −11.66MPa Ans 330. Problem 5-7The shaft has an outer diameter of 32 mm and an inner diameter of 25 mm. If it is subjected to theapplied torques as shown, determine the absolute maximum shear stress developed in the shaft. Thesmooth bearings at A and B do not resist torque.Given: do := 32mm di := 25mmT1 := 185N⋅mT2 := −260N⋅m T3 := 75N⋅mSolution: cdo2:=Jπ2⎛⎜⎝⎞⎠do2⎛⎜⎝⎞⎠4 di2⎛⎜⎝⎞⎠4−⎡⎢⎣⎤⎥⎦:= ⋅Tmax := T1τmax(Tmax)⋅c:= τmax = 45.82MPa AnsJ 331. Problem 5-8The shaft has an outer diameter of 32 mm and an inner diameter of 25 mm. If it is subjected to theapplied torques as shown, plot the shear-stress distribution acting along a radial line lying within regionEA of the shaft. The smooth bearings at A and B do not resist torque.Given: do := 32mm di := 25mmT1 := 185N⋅mT2 := −260N⋅m T3 := 75N⋅mSolution: cdo2:=Jπ2⎛⎜⎝⎞⎠do2⎛⎜⎝⎞⎠4 di2⎛⎜⎝⎞⎠4−⎡⎢⎣⎤⎥⎦:= ⋅TEA := T1τmax(TEA)⋅c:= τmax = 45.82MPa AnsJShera stress at ρ := 0.5⋅diτρTEA⋅ρJ:= τρ = 35.80MPa Ans 332. Problem 5-9The assembly consists of two sections of galvanized steel pipe connected together using a reducingcoupling at B. The smaller pipe has an outer diameter of 18.75 mm and an inner diameter of 17 mm,whereas the larger pipe has an outer diameter of 25 mm and an inner diameter of 21.5 mm. If the pipeis tightly secured to the wall at C, determine the maximum shear stress developed in each section of thepipe when the couple shown is applied to the handles of the wrench.Given: d1o := 18.75mm d1i := 17mm F := 75Nd2o := 25mm d2i := 21.5mmaL := 150mm aR := 200mmSolution: T FaL:= ⋅ + F⋅aRT = 26.25N⋅mSegment AB :c1d1o2:=J1π2⎛⎜⎝⎞⎠d1o2⎛⎜⎝⎞⎠4 d1i2⎛⎜⎝⎞⎠4−⎡⎢⎣⎤⎥⎦:= ⋅τABT⋅c1J1:= τAB = 62.55MPa AnsSegment BC :c2d2o2:=J2π2⎛⎜⎝⎞⎠d2o2⎛⎜⎝⎞⎠4 d2i2⎛⎜⎝⎞⎠4−⎡⎢⎣⎤⎥⎦:= ⋅τBCT⋅c2J2:= τBC = 18.89MPa Ans 333. Problem 5-10The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has aninner diameter of 25 mm and a wall thickness of 5 mm, determine the maximum shear stress in thetube when the cable force of 600 N is applied to the cables. Also, sketch the shear-stress distributionover the cross section.Given: di := 25mm t := 5mm P := 600Na := 75mmSolution: do := di + 2t cdo2:=Jπ2⎛⎜⎝⎞⎠do2⎛⎜⎝⎞⎠4 di2⎛⎜⎝⎞⎠4−⎡⎢⎣⎤⎥⎦:= ⋅T := P⋅ (2a) T = 90.00N⋅mτmaxT⋅cJ:= τmax = 14.45MPa AnsShera stress at ρ := 0.5⋅diτρT⋅ρJ:= τρ = 10.32MPa Ans 334. Problem 5-11The shaft consists of three concentric tubes, each made from the same material and having the innerand outer radii shown. If a torque of T = 800 N·m is applied to the rigid disk fixed to its end, determinthe maximum shear stress in the shaft.Given: ri1 := 20mm ro1 := 25mmri2 := 26mm ro2 := 30mmri3 := 32mm ro3 := 38mmT := 800N⋅m L := 2mSolution: cmax := ro3J1π2⎛⎜⎝⎞⎠ro14 − ri14 ⎛⎝⎞⎠:= ⋅J2π2⎛⎜⎝⎞⎠ro24 − ri24 ⎛⎝⎞⎠:= ⋅J3π2⎛⎜⎝⎞⎠ro34 − ri34 ⎛⎝⎞⎠:= ⋅J := J1 + J2 + J3τmaxT⋅cmax:= τmax = 11.94MPa AnsJ 335. Problem 5-12The solid shaft is fixed to the support at C and subjected to the torsional loadings shown. Determinethe shear stress at points A and B and sketch the shear stress on volume elements located at thesepoints.Given: ro := 35mmρA := 35mm ρB := 20mmTD := 800N⋅m TB := −300N⋅m:= ⋅ 4TBA := TD + TB TAB := TDSolution: Jπ2roτATBA⋅ρA:= τBJTAB⋅ρBJ:=τA = 7.42MPa τB = 6.79MPa Ans 336. Problem 5-13A steel tube having an outer diameter of 62.5 mm is used to transmit 3 kW when turning at 27 rev/min.Determine the inner diameter d of the tube to the nearest multiples of 5mm if the allowable shear stressis τallow = 70 MPa.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: do := 62.5mm ω := 27⋅ rpmP := 3kW τallow := 70MPaSolution: ω 2.83rads=TPω:= T = 1061.03N⋅mMax. stress : cdo2:=τallowT⋅cJ= Jπ2⎛⎜⎝⎞⎠do2⎛⎜⎝⎞⎠4 di2⎛⎜⎝⎞⎠4−⎡⎢⎣⎤⎥⎦= ⋅di 2do2⎛⎜⎝⎞⎠42π⎛⎜⎝⎞⎠T⋅cτallow⎛⎜⎝⎞⎠− ⋅⎡⎢⎢⎣⎤⎥⎥⎦0.25:=di = 56.83mmUse di = 60mm Ans 337. Problem 5-14The solid aluminum shaft has a diameter of 50 mm and an allowable shear stress of τallow = 6 MPa.Determine the largest torque T1 that can be applied to the shaft if it is also subjected to the othertorsional loadings. It is required that T1 act in the direction shown. Also, determine the maximum shestress within regions CD and DE.Given: do := 50mm τallow := 6MPaTA := 68N⋅m TC := 49N⋅m TD := 35N⋅mSolution: Jπ32:= ⋅ 4 cdodo2:=Assume failure at region BC. TBC = −(T1 + TA)Aplying the torsion formula(TBC)⋅cτallow= τallowJ−(T1 + TA)⋅cJ=T1τallow⋅ J:= − − TA T1 = −215.26 N⋅m AnscInternal Torque : Maximum torque occurs ithin region BC as indicated on the torque diagram.Maximum shear srtesses at Other Regions :TCD := TA + T1 + TC TCD = −98.26N⋅mτmax.CD(TCD)⋅c:= τmax.CD = −4.00MPa AnsJTDE := TCD + TD TDE = −63.26N⋅mτmax.DE(TDE)⋅c:= τmax.DE = −2.58MPa AnsJLet a := 1mx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 2ax3 := 2a , 1.01⋅ (2a) .. 3a x4 := 3a , 1.01⋅ (3a) .. 4aT'1(x1) TA1N⋅m:= ⋅ T'2(x2) (TA + T1) 1N⋅m:= ⋅T'3(x3) (TA + T1 + TC) 1N⋅m:= ⋅T'4(x3) (TA + T1 + TC + TD) 1N⋅m:= ⋅ 338. 0 1 2 3 41000100200Distance (m)Torque (Nm)T'1(x1)T'2(x2)T'3(x3)T'4(x4)x1, x2, x3, x4 339. Problem 5-15The solid aluminum shaft has a diameter of 50 mm. Determine the absolute maximum shear stress inthe shaft and sketch the shear-stress distribution along a radial line of the shaft where the shear stressis maximum. Set T1 = 20 N·m.Given: do := 50mm T1 := −20N⋅mTA := 68N⋅m TC := 49N⋅m TD := 35N⋅mSolution: Jπ32:= ⋅ 4 cdodo2:=Maximum Torque :Maximum torque occurs ithin region DEas indicated on the torque diagram.TDE := TA + T1 + TC + TD TDE = 132.00 N⋅mTmax := TDEAplying the torsion formula τmax(Tmax)⋅c:= τmax = 5.38MPa AnsJLet a := 1mx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 2ax3 := 2a , 1.01⋅ (2a) .. 3a x4 := 3a , 1.01⋅ (3a) .. 4aT'1(x1) TA1N⋅m:= ⋅ T'2(x2) (TA + T1) 1N⋅m:= ⋅T'3(x3) (TA + T1 + TC) 1N⋅m:= ⋅T'4(x3) (TA + T1 + TC + TD) 1N⋅m:= ⋅ 340. 0 1 2 3 4150100500Distance (m)Torque (Nm)T'1(x1)T'2(x2)T'3(x3)T'4(x4)x1, x2, x3, x4 341. Problem 5-16The motor delivers a torque of 50 N·m to the shaft AB. This torque is transmitted to shaft CD usingthe gears at E and F. Determine the equilibrium torque T' on shaft CD and the maximum shear stresseach shaft. The bearings B, C, and D allow free rotation of the shafts.Given: dAB := 30mm RE := 50mm TAB := 50N⋅mdCD := 35mm RF := 125mmSolution:Equilibrium :ΣΜE=0; TAB − F⋅ (RE) = 0 FTABRE:=ΣΜF=0; T' − F⋅RF = 0T' := F⋅RFT' = 125N⋅m AnsInternal Torque : As shown in FBD.JABπ32:= ⋅ dAB4 cAB := 0.5dABJCDπ32:= ⋅ dCD4 cCD := 0.5dCDMaximum shear srtesses:τmax.AB(TAB)⋅cABJAB:= τmax.AB = 9.43MPa Ansτmax.CDT'⋅cCDJCD:= τmax.CD = 14.85MPa Ans 342. Problem 5-17If the applied torque on shaft CD is T' = 75 N·m, determine the absolute maximum shear stress in eacshaft. The bearings B, C, and D allow free rotation of the shafts, and the motor holds the shafts fixedfrom rotating.Given: dEA := 30mm RE := 50mm TAB := 50N⋅mdCD := 35mm RF := 125mm T' := 75N⋅mSolution:Equilibrium :ΣΜF=0; T' − F⋅RF = 0 FT'RF:=ΣΜE=0; TAB − F⋅ (RE) − TA = 0TA := TAB − F⋅RETA = 20N⋅m AnsInternal Torque : As shown in FBD.TEA := TAB − TA TEA = 30.00N⋅mJEAπ32:= ⋅ dEA4 cEA := 0.5dEAJCDπ32:= ⋅ dCD4 cCD := 0.5dCDMaximum shear srtesses:τmax.EA(TEA)⋅cEAJEA:= τmax.EA = 5.66MPa Ansτmax.CDT'⋅cCDJCD:= τmax.CD = 8.91MPa Ans 343. Problem 5-18The copper pipe has an outer diameter of 62.5 mm and an inner diameter of 57.5 mm. If it is tightlysecured to the wall at C and a uniformly distributed torque is applied to it as shown, determine theshear stress developed at points A and B. These points lie on the pipe's outer surface. Sketch the shearstress on volume elements located at A and B.Given: do := 62.5mm di := 57.5mm q 625N⋅mm:=LOA := 300mm LAB := 225mm LBC := 100mmSolution:Internal Torque : As shown on FBDTA := q⋅(LOA) TA = 187.50N⋅mTB := q⋅(LOA + LAB) TB = 328.13N⋅mMax. shear stress : τ T⋅cJ=cdo2:= Jπ2⎛⎜⎝⎞⎠do2⎛⎜⎝⎞⎠4 di2⎛⎜⎝⎞⎠4−⎡⎢⎣⎤⎥⎦:= ⋅τATA⋅cJ:= τA = 13.79MPa AnsτBTB⋅cJ:= τB = 24.14MPa Ans 344. Problem 5-19The copper pipe has an outer diameter of 62.5 mm and an inner diameter of 57.5 mm. If it is tightlysecured to the wall at C and it is subjected to the uniformly distributed torque along its entire length,determine the absolute maximum shear stress in the pipe. Discuss the validity of this result.Given: do := 62.5mm di := 57.5mm q 625N⋅mm:=LOA := 300mm LAB := 225mm LBC := 100mmSolution:Internal Torque : The maximum torque occurs at thesupport CTC := q⋅ (LOA + LAB + LBC) TC = 390.63 N⋅mMax. shear stress : τT⋅cJ=cdo2:= Jπ2⎛⎜⎝⎞⎠do2⎛⎜⎝⎞⎠4 di2⎛⎜⎝⎞⎠4−⎡⎢⎣⎤⎥⎦:= ⋅τCTC⋅cJ:= τC = 28.73MPa AnsAccording to Saint-Venant's principle, application of the torsion formula should be at pointssufficiently removed from the supports or points of concentrated loading. 345. Problem 5-20The 60-mm-diameter solid shaft is subjected to the distributed and concentrated torsional loadingsshown. Determine the shear stress at points A and B, and sketch the shear stress on volume elementslocated at these points.Given: L1 := 0.3m L2 := 0.4mT1 := 400N⋅m T2 := −600N⋅mdo := 60mm q 2kN⋅mm:=Solution:Internal Torque : As shown on FBD.TA := T1 TA = 400.00N⋅mTB := T1 + T2 + q⋅L2 TB = 600.00 N⋅mMaximum shear stress : τT⋅cJ=cdo2:= Jπ2do2⎛⎜⎝⎞⎠4:= ⋅τATA⋅cJ:= τA = 9.43MPa AnsτBTB⋅cJ:= τB = 14.15MPa Ans 346. Problem 5-21The 60-mm diameter solid shaft is subjected to the distributed and concentrated torsional loadingsshown. Determine the absolute maximum and minimum shear stresses in the shaft and specify theirlocations, measured from the fixed end.Given: L1 := 0.3m L2 := 0.4mT1 := 400N⋅m T2 := −600N⋅mdo := 60mm q 2kN⋅mm:=Solution:Internal Torque :TC T1 T2 + q 2L2:= + ⋅ ( ) TC = 1400.00N⋅mThe maximum torque occurs at the fixed support C.Tmax := TC Tmax = 1400.00N⋅mThe minimum torque occurs in segment loaded with q. Tmin := 0xo2L2TCq 2L2= xo⋅ ( )TCq:= xo = 0.700mShear stress : τT⋅cJ= cdo2:= Jπ32⎛⎜⎝⎞⎠:= ⋅ 4doτabs.minTmin⋅cJ:= τabs.min = 0.00MPa Ansτabs.maxTmax⋅c:= τabs.max = 33.01MPa AnsJAccording to Saint-Venant's principle, application of the torsion formula should be at pointssufficiently removed from the supports or points of concentrated loading. Therefore, theabsolute τmax is not valid.x1 := 0 , 0.01⋅2L2 .. (2L2) x2 := 2L2 , 1.01⋅ (2L2) .. (2L2 + 2L1)Ta(x1) (−TC + q⋅x1) 1:= ⋅ Tb(x2) (−TC + 2q⋅L2 + T2) 1N⋅mN⋅m:= ⋅ 347. 0 0.2 0.4 0.6 0.8 1 1.2050010001500Distance (m)Torque (Nm)Ta(x1)Tb(x2)x1, x2 348. Problem 5-22The solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine therequired diameter d of the shaft if the allowable shear stress for the material is τallow = 175 MPa.Given: L1 := 0.3m L2 := 0.4mT1 := 400N⋅m T2 := −600N⋅mτallow := 175MPa q 2kN⋅mm:=Solution:Internal Torque :TC T1 T2 + q 2L2:= + ⋅ ( ) TC = 1400.00N⋅mThe maximum torque occurs at the fixed support C.Tmax := TC Tmax = 1400.00N⋅mAllowable Shear stress : τT⋅cJ= cdo2= Jπ32⎛⎜⎝⎞⎠= ⋅ 4doτallow16Tmaxπ ⋅ do3=do3 16Tmaxπ⋅τallow:= do = 34.41mm AnsAccording to Saint-Venant's principle, application of the torsion formula should be at pointssufficiently removed from the supports or points of concentrated loading. Therefore, theabove analysis is not valid.x1 := 0 , 0.01⋅2L2 .. (2L2) x2 := 2L2 , 1.01⋅ (2L2) .. (2L2 + 2L1)Ta(x1) (−TC + q⋅x1) 1:= ⋅ Tb(x2) (−TC + 2q⋅L2 + T2) 1N⋅mN⋅m:= ⋅0 0.2 0.4 0.6 0.8 1 1.2050010001500Distance (m)Torque (Nm)Ta(x1)Tb(x2)x1, x2 349. Problem 5-23The steel shafts are connected together using a fillet weld as shown. Determine the average shearstress in the weld along section aa if the torque applied to the shafts is T = 60 N·m. Note: The criticalsection where the weld fails is along section aa.Given: do := 50mm a := 12mmT := 60N⋅m θa := 45degSolution:The maen radius of weld is:rwelddo + a2:=rweld = 31.00mmShear stress :VTrweld:= Aweld := 2π⋅ rweld⋅ (a⋅ sin(θa))τavgVAweld:=τavg = 1.17MPa Ans 350. Problem 5-24The rod has a diameter of 12 mm and a weight of 80 N/m. Determine the maximum torsional stress inthe rod at a section located at A due to the rod's weight.Given: d 12mm := w 80 N:= LA := 0.3mmLx := 0.9m Ly := 0.9m Lz := 0.3mSolution:Equilibrium :Σ Mx = 0; TA − w⋅Ly⋅ (0.5Ly) − w⋅Lz⋅ (Ly) = 0TA := w⋅Ly⋅ (0.5⋅Ly) + w⋅Lz⋅ (Ly)TA = 54.00N⋅mMax. shear stress : τT⋅cJ=cd2:= Jπ2d2⎛⎜⎝⎞⎠4:= ⋅τATA⋅cJ:= τA = 159.15MPa Ans 351. Problem 5-25Solve Prob. 5-24 for the maximum torsional stress at B.Given: d 12mm := w 80 N:= LA := 0.3mmLx := 0.9m Ly := 0.9m Lz := 0.3mSolution:Equilibrium :Σ Mx = 0; TB − w⋅Ly⋅ (0.5Ly) − w⋅Lz⋅ (Ly) = 0TB := w⋅Ly⋅ (0.5⋅Ly) + w⋅Lz⋅ (Ly)TB = 54.00N⋅mMax. shear stress : τT⋅cJ=cd2:= Jπ2d2⎛⎜⎝⎞⎠4:= ⋅τBTB⋅cJ:= τB = 159.15MPa Ans 352. Problem 5-26Consider the general problem of a circular shaft made from m segments each having a radius of cm. Ifthere are n torques on the shaft as shown, write a computer program that can be used to determine themaximum shearing stress at any specified location x along the shaft. Show an application of theprogram using the values L1 = 0.6 m, c1 = 50 mm, L2 = 1.2 m, c2 = 25 mm, T1 = 1200 N·m, d1 = 0, T2= -900 N·m, d2 = 1.5 m. 353. Problem 5-27The wooden post, which is half buried in the ground, is subjected to a torsional moment of 50 N·mthat causes the post to rotate at constant angular velocity.This moment is resisted by a lineardistribution of torque developed by soil friction, which varies from zero at the ground to t0 N·m/m atits base. Determine the equilibrium value for t0 , and then calculate the shear stress at points A and B,which lie on the outer surface of the post.Given: do := 100mm L := 0.75m LAB := 0.5⋅mT := 50N⋅mSolution:Equilibrium :Σ Mz = 0; (0.5to)⋅L − T = 0to 2TL:= to 133.33N⋅mm=Internal Torque : As shown on FBD.TA := T TA = 50.00N⋅mTB T 0.5toLABL⎛⎜⎝⎞⎠:= − ⋅ ⋅LAB TB = 27.78N⋅mMaximum Shear Stress : τT⋅cJ=cdo2:= Jπ32:= ⋅ 4doτATA⋅cJ:= τA = 0.255MPa AnsτBTB⋅cJ:= τB = 0.141MPa Ans 354. Problem 5-28A cylindrical spring consists of a rubber annulus bonded to a rigid ring and shaft. If the ring is heldfixed and a torque T is applied to the shaft, determine the maximum shear stress in the rubber.Solution:Shear stress :τFA= FTr= A = 2π⋅ r⋅hτT2π⋅ r2⋅h=Shear stress is maximm when r is the smallest, i.e. r = ri.Hence,Tτmax= Ans⋅ 2⋅h2π ri 355. Problem 5-29The shaft has a diameter of 80 mm and due to friction at its surface within the hole, it is subjected to avariable torque described by the function t = (25x ex2) N·m/m, where x is in meters. Determine theminimum torque T0 needed to overcome friction and cause it to twist. Also, determine the absolutemaximum stress in the shaft.Given: do := 80mm L := 2m t 25x ⋅ex2N⋅mm=Solution: unit := N⋅mEquilibrium :Σ Mz = 0; To unitLm25x ex x 20⋅⌠⎮⎮⌡:= ⋅ dTo = 669.98 N⋅m AnsMaximum Shear Stress : τT⋅cJ=cdo2:= Jπ32:= ⋅ 4doτabs.maxTo⋅cJ:=τabs.max = 6.664MPa Ans 356. Problem 5-30The solid shaft has a linear taper from rA at one end to rB at the other. Derive an equation that gives thmaximum shear stress in the shaft at a location x along the shaft's axis. 357. Problem 5-31When drilling a well at constant angular velocity, the bottom end of the drill pipe encounters a torsionresistance TA. Also, soil along the sides of the pipe creates a distributed frictional torque along itslength, varying uniformly from zero at the surface B to tA at A. Determine the minimum torque TB thamust be supplied by the drive unit to overcome the resisting torques, and compute the maximum sheastress in the pipe. The pipe has an outer radius ro and an inner radius ri . 358. Problem 5-32The drive shaft AB of an automobile is made of a steel having an allowable shear stress of τallow= 56MPa. If the outer diameter of the shaft is 62.5 mm and the engine delivers 165 kW to the shaft whenis turning at 1140 rev/min, determine the minimum required thickness of the shaft's wall.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: do := 62.5mm ω := 1140⋅ rpmP := 165kW τallow := 56MPaSolution: ω 119.38rads=TPω:= T = 1382.14 JMax. shear stress : cdo2:=τallowT⋅cJ= Jπ2⎛⎜⎝⎞⎠do2⎛⎜⎝⎞⎠4 di2⎛⎜⎝⎞⎠4−⎡⎢⎣⎤⎥⎦= ⋅di 2do2⎛⎜⎝⎞⎠42π⎛⎜⎝⎞⎠T⋅cτallow⎛⎜⎝⎞⎠− ⋅⎡⎢⎢⎣⎤⎥⎥⎦0.25:=di = 52.16mmtdo − di2:=t = 5.17mm Ans 359. Problem 5-33The drive shaft AB of an automobile is to be designed as a thin-walled tube. The engine delivers 125kW when the shaft is turning at 1500 rev/min. Determine the minimum thickness of the shaft's wall ifthe shaft's outer diameter is 62.5 mm.The material has an allowable shear stress of τallow = 50 MPa.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: do := 62.5mm ω := 1500⋅ rpmP := 125kW τallow := 50MPaSolution: ω 157.08rads=TPω:= T = 795.77 JMax. shear stress : cdo2:=τallowT⋅cJ= Jπ2⎛⎜⎝⎞⎠do2⎛⎜⎝⎞⎠4 di2⎛⎜⎝⎞⎠4−⎡⎢⎣⎤⎥⎦= ⋅di 2do2⎛⎜⎝⎞⎠42π⎛⎜⎝⎞⎠T⋅cτallow⎛⎜⎝⎞⎠− ⋅⎡⎢⎢⎣⎤⎥⎥⎦0.25:=di = 56.50mmtdo − di2:=t = 2.998mm Ans 360. Problem 5-34The gear motor can develop 100 W when it turns at 300 rev/min. If the shaft has a diameter of 12 m,determine the maximum shear stress that will be developed in the shaft.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: d := 12mm ω := 300⋅ rpmP := 100WSolution: ω 31.42rads=TPω:= T = 3.183N⋅mMax. shear stress :cd2:= Jπ2d2⎛⎜⎝⎞⎠4:= ⋅τmaxT⋅cJ:=τmax = 9.382MPa Ans 361. Problem 5-35The gear motor can develop 100 W when it turns at 80 rev/min. If the allowable shear stress for theshaft is τallow = 28 MPa, determine the smallest diameter of the shaft to the nearest multiples of 5mmthat can be used.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: P := 100W ω := 80⋅ rpmτallow := 28MPaSolution: ω 8.38rads=TPω:= T = 11.937N⋅mMax. shear stress : cd2= Jπ2d2⎛⎜⎝⎞⎠4= ⋅τallowT⋅cJ=π2d2⎛⎜⎝⎞⎠4⋅T⋅d2⋅τallow=d16Tπ⋅τallow⎛⎜⎝⎞⎠13:=d = 12.95mmUse d = 15mm Ans 362. Problem 5-36The drive shaft of a tractor is made of a steel tube having an allowable shear stress of τallow = 42 MPaIf the outer diameter is 75 mm and the engine delivers 145 kW to the shaft when it is turning at 1250rev/min., determine the minimum required thickness of the shaft's wall.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: do := 75mm ω := 1250⋅ rpmP := 145kW τallow := 42MPaSolution: ω 130.90rads=TPω:= T = 1107.718N⋅mMax. shear stress : cdo2:=τallowT⋅cJ= Jπ2⎛⎜⎝⎞⎠do2⎛⎜⎝⎞⎠4 di2⎛⎜⎝⎞⎠4−⎡⎢⎣⎤⎥⎦= ⋅di 2do2⎛⎜⎝⎞⎠42π⎛⎜⎝⎞⎠T⋅cτallow⎛⎜⎝⎞⎠− ⋅⎡⎢⎢⎣⎤⎥⎥⎦0.25:=di = 68.15mmtdo − di2:=t = 3.427mm Ans 363. Problem 5-37The 2.5-kW reducer motor can turn at 330 rev/min. If the shaft has a diameter of 20 mm, determinethe maximum shear stress that will be developed in the shaft.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: d := 20mm ω := 330⋅ rpmP := 2.5kWSolution: ω 34.56rads=TPω:= T = 72.343N⋅mMax. shear stress :cd2:= Jπ2d2⎛⎜⎝⎞⎠4:= ⋅τmaxT⋅cJ:=τmax = 46.055MPa Ans 364. Problem 5-38The 2.5-kW reducer motor can turn at 330 rev/min. If the allowable shear stress for the shaft is τallow= 56 MPa, determine the smallest diameter of the shaft to the nearest multiples of 5mm that can beused.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: d := 20mm ω := 330⋅ rpmP := 2.5kW τallow := 56MPaSolution: ω 34.56rads=TPω:= T = 72.343N⋅mMax. shear stress : cd2= Jπ2d2⎛⎜⎝⎞⎠4= ⋅τallowT⋅cJ=π2d2⎛⎜⎝⎞⎠4⋅T⋅d2⋅τallow=d16Tπ⋅τallow⎛⎜⎝⎞⎠13:=d = 18.74mmUse d = 20mm Ans 365. Problem 5-39The solid steel shaft AC has a diameter of 25 mm and is supported by smooth bearings at D and E. Itis coupled to a motor at C, which delivers 3 kW of power to the shaft while it is turning at 50 rev/s. Ifgears A and B remove 1 kW and 2 kW, respectively, determine the maximum shear stress developed ithe shaft within regions AB and BC. The shaft is free to turn in its support bearings D and E.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: do := 25mm ω := 50⋅ (60)⋅ rpmPA := −1kW Po := 3kWPB := −2kWSolution: ω 314.16rads=TCPoω:= TC = 9.549N⋅mTAPAPo:= ⋅TC TA = 3.183N⋅mMaximum Shear Stress :cdo2:= Jπ2do2⎛⎜⎝⎞⎠4:= ⋅τAB.maxTA⋅cJ:= τAB.max = 1.038MPa AnsτBC.maxTC⋅cJ:= τBC.max = 3.113MPa Ans 366. Problem 5-40A ship has a propeller drive shaft that is turning at 1500 rev/min. while developing 1500 kW. If it is 2m long and has a diameter of 100 mm, determine the maximum shear stress in the shaft caused bytorsion.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: d := 100mm ω := 1500⋅ rpmP := 1500kW L := 2.4mSolution: ω 157.08rads=TPω:= T = 9549.297N⋅mMax. shear stress :cd2:= Jπ2d2⎛⎜⎝⎞⎠4:= ⋅τmaxT⋅cJ:=τmax = 48.634MPa Ans 367. Problem 5-41The motor A develops a power of 300 W and turns its connected pulley at 90 rev/min. Determine therequired diameters of the steel shafts on the pulleys at A and B if the allowable shear stress is τallow =85 MPa.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: rA := 60mm ωA := 90⋅ rpmrB := 150mm P := 300Wτallow := 85MPaSolution:ωB ωArArB⎛⎜⎝⎞⎠:= ⋅ ωB 3.77rads=TAPωA:= TA = 31.831N⋅mTBPωB:= TB = 79.577 N⋅mAllowable Shear Stress :For shaft A:τallowTA⋅cJ= cdA2= Jπ32= ⋅ 4dAτallow16TAπ dA⋅ 3=dA3 16TAπ⋅τallow:= dA = 12.40mm AnsFor shaft B:τallowTB⋅cJ= cdB2= Jπ32= ⋅ 4dBτallow16TBπ dB⋅ 3=dB3 16TBπ⋅τallow:= dB = 16.83mm Ans 368. Problem 5-42The motor delivers 400 kW to the steel shaft AB, which is tubular and has an outer diameter of 50 mmand an inner diameter of 46 mm. Determine the smallest angular velocity at which it can rotate if theallowable shear stress for material is τallow = 175 MPa.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: do := 50mm di := 46mmP := 400kW τallow := 175MPaSolution:cdo2:=Jπ2⎛⎜⎝⎞⎠do2⎛⎜⎝⎞⎠4 di2⎛⎜⎝⎞⎠4−⎡⎢⎣⎤⎥⎦:= ⋅τallowT⋅cJ=TJ⋅ (τallow):= T = 1218.13N⋅mcωPT:=ω 328.37rads=ω = 3135.714 rpm Ans 369. Problem 5-43The motor delivers 40 kW while turning at a constant rate of 1350 rpm at A. Using the belt and pulleysystem this loading is delivered to the steel blower shaft BC. Determine to the nearest multiples of 5mmthe smallest diameter of this shaft if the allowable shear stress for steel if τallow = 84 MPa.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: rA := 100mm rB := 200mmP := 40kW ω := 1350⋅ rpmτallow := 84MPaSolution: ω 141.37rads=TAPω:= TA = 282.942N⋅mTA = 2rA⋅ (F' − F)TB = 2rB⋅ (F − F')TBrBrA⎛⎜⎝⎞⎠:= ⋅TA TB = 565.88 N⋅mMax. shear stress : cd2= Jπ2d2⎛⎜⎝⎞⎠4= ⋅τallowTB⋅cJ=π2d2⎛⎜⎝⎞⎠4⋅TB⋅d2⋅τallow=d16TBπ⋅τallow⎛⎜⎝ ⎞⎠13:=d = 32.49mmUse d = 35mm Ans 370. Problem 5-44The propellers of a ship are connected to a solid A-36 steel shaft that is 60 m long and has an outerdiameter of 340 mm and inner diameter of 260 mm. If the power output is 4.5 MW when the shaftrotates at 20 rad/s, determine the maximum torsional stress in the shaft and its angle of twist.Given: do := 340mm di := 260mm L := 60mP := 4500kW G := 75GPa ω 20rads:=Solution:TPω:= T = 225 kN⋅mMaximum Shear Stress :cdo2:= Jπ32do4 − di4 ⎛⎝⎞⎠:= ⋅τmaxT⋅cJ:= τmax = 44.31MPa AnsAngle of Twist :φT⋅LG⋅ J:=φ = 0.2085 radφ = 11.946 deg Ans 371. Problem 5-45A shaft is subjected to a torque T. Compare the effectiveness of using the tube shown in the figurewith that of a solid section of radius c. To do this, compute the percent increase in torsional stress andangle of twist per unit length for the tube versus the solid section.Given: ro = c ri = 0.5cSolution:Maximum Shear Stress :For solid shaft::τs.maxT⋅cJ= c = ro Jsπ2= ⋅ 4roτs.max2Tπ ro⋅ 3=For the tube::τt.maxT⋅cJ= c = ro Jtπ2ro4 − ri4 ⎛⎝⎞⎠= ⋅ Jt15π32= ⋅ 4roτt.max32T15π ro⋅ 3=% increase in shear stress: τ%τt.max − τs.max= ⋅100τs.maxτ%3215− 22:= ⋅100 τ% = 6.67 AnsAngle of Twist :For solid shaft:: φsT⋅LG⋅ Js=For the tube:: φtT⋅LG⋅ Jt=% increase in angle of twist: φ φ%= ⋅100 %1Js1Jt−1Jsφt − τs.maxτs.max= ⋅100φ%π215π32−15π32:= ⋅100 φ% = 6.67 Ans 372. Problem 5-46The tubular drive shaft for the propeller of a hover-craft is 6 m long. If the motor delivers 4 MW ofpower to the shaft when the propellers rotate at 25 rad/s, determine the required inner diameter of theshaft if the outer diameter is 250 mm.What is the angle of twist of the shaft when it is operating? Takτallow = 90 MPa and G = 75 GPa.Given: do := 250mm L := 6mP := 4000kW τallow := 90MPaG := 75GPa ω 25rads:=Solution:TPω:= T = 160 kN⋅mMaximum Shear Stress :τallowT⋅cJ= cdo2:= Jπ32do4 − di4 ⎛⎝⎞⎠= ⋅τallow16T⋅do4 − ⎛⎝π do4 di⎞⎠⋅=di4do4 16T⋅do:= − di = 201.3mm Ansπ⋅τallowAngle of Twist : Jπ32do4 − di4 ⎛⎝⎞⎠:= ⋅φT⋅LG⋅ J:=φ = 0.05760 radφ = 3.30 deg Ans 373. Problem 5-47The A-36 steel axle is made from tubes AB and CD and a solid section BC. It is supported on smoothbearings that allow it to rotate freely. If the gears, fixed to its ends, are subjected to 85-N·m torques,determine the angle of twist of gear A relative to gear D. The tubes have an outer diameter of 30 mmand an inner diameter of 20 mm. The solid section has a diameter of 40 mm.Given: do := 30mm di := 20mm ds := 40mmLAB := 0.4m LBC := 0.25m LCD := 0.4mT := 85N⋅m G := 75GPaSolution:TAB := TTBC := TTCD := TAngle of Twist :JABπ32do4 − di4 ⎛⎝⎞⎠:= ⋅ JBCπ32:= ⋅ ds4 JCD := JABφ 2TAB⋅LABG⋅ JABTBC⋅LBCG⋅ JBC:= +φ = 0.01534 radφ = 0.879 deg Ans 374. Problem 5-48The A-36 steel axle is made from tubes AB and CD and a solid section BC. It is supported on smoothbearings that allow it to rotate freely. If the gears, fixed to its ends, are subjected to 85-N·m torques,determine the angle of twist of the end B of the solid section relative to end C. The tubes have an outediameter of 30 mm and an inner diameter of 20 mm.The solid section has a diameter of 40 mm.Given: do := 30mm di := 20mm ds := 40mmLAB := 0.4m LBC := 0.25m LCD := 0.4mT := 85N⋅m G := 75GPaSolution:TAB := TTBC := TTCD := TAngle of Twist : JBCπ32:= ⋅ 4dsφTBC⋅LBCG⋅ JBC:=φ = 0.001127 radφ = 0.0646 deg Ans 375. Problem 5-49The hydrofoil boat has an A-36 steel propeller shaft that is 30 m long. It is connected to an in-linediesel engine that delivers a maximum power of 2000 kW and causes the shaft to rotate at 1700 rpm. Ifthe outer diameter of the shaft is 200 mm and the wall thickness is 10 mm, determine the maximumshear stress developed in the shaft. Also, what is the “wind up,” or angle of twist in the shaft at fullpower?Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: do := 200mm t := 10mmP := 2000kW ω := 1700rpmG := 75GPa L := 30mSolution: ω 178.02rads=TPω:= T = 11234.467N⋅mMax. shear stress :di := do − 2t cdo2:=Jπ2⎛⎜⎝⎞⎠do2⎛⎜⎝⎞⎠4 di2⎛⎜⎝⎞⎠4−⎡⎢⎣⎤⎥⎦:= ⋅τmaxT⋅cJ:=τmax = 20.797MPa AnsAngle of Twist :φT⋅LG⋅ J:=φ = 0.0832 radφ = 4.766 deg Ans 376. Problem 5-50The splined ends and gears attached to the A-36 steel shaft are subjected to the torques shown.Determine the angle of twist of gear C with respect to gear D. The shaft has a diameter of 40 mm.Given: LAC := 0.3m LCD := 0.4m LDE := 0.5mdo := 40mm G := 75GPaTA := −300N⋅m TC := 500N⋅mTD := 200N⋅m TE := −400N⋅mSolution:TAC := TATCD := TA + TCTDE := TA + TC + TDAngle of Twist : Jπ32:= ⋅ 4doφC_DTCD⋅LCDG⋅ J:=φC_D = 0.004244 radφC_D = 0.243 deg Ans 377. Problem 5-51The 20-mm-diameter A-36 steel shaft is subjected to the torques shown. Determine the angle of twistof the end B.Given: LAD := 0.2m LDC := 0.6m LCB := 0.8mTB := 80N⋅m TC := −20N⋅m TD := 30N⋅mdo := 20mm G := 75GPaSolution:TCB := TBTDC := TB + TCTAD := TB + TC + TDAngle of Twist : Jπ32:= ⋅ 4doφTCB⋅LCBG⋅ JTDC⋅LDCG⋅ J+TAD⋅LADG⋅ J:= +φ = 0.100162 radφ = 5.739 deg Ans 378. Problem 5-52The 8-mm-diameter A-36 bolt is screwed tightly into a block at A. Determine the couple forces F thatshould be applied to the wrench so that the maximum shear stress in the bolt becomes 18 MPa. Also,compute the corresponding displacement of each force F needed to cause this stress. Assume that thewrench is rigid.Given: do := 8mm L := 80mm a := 150mmG := 75GPa τallow := 18MPaSolution:Allowable Shear Stress :cdo2:= Jπ32⎛⎜⎝⎞⎠:= ⋅ 4doτallowT⋅cJ= Tτallow⋅ J:= T = 1.8096N⋅mcEquilibrium :T − F(2a) = 0 FT2a:= F = 6.03 N AnsAngle of Twist :φT⋅LG⋅ J:=φ = 0.00480 radDisplacement :s' := a⋅φs' = 0.720mm Ans 379. Problem 5-53The turbine develops 150 kW of power, which is transmitted to the gears such that C receives 70%and D receives 30%. If the rotation of the 100-mm-diameter A-36 steel shaft is ω = 800 rev/min.,determine the absolute maximum shear stress in the shaft and the angle of twist of end E of the shaftrelative to B. The journal bearing at E allows the shaft to turn freely about its axis.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: do := 100mm ω := 800⋅ rpmLBC := 3m LCD := 4mLDE := 2m P := 150kWG := 75GPaTC = 0.7T TD = 0.3TSolution: ω 83.78rads=TPω:= T = 1.790 kN⋅mTC := 0.7T TC = 1.253 kN⋅mTD := 0.3T TD = 0.537 kN⋅mTBC := TTCD := 0.3TTDE := 0Maximum Shear Stress :Maximum torque occurs in region BC.cdo2:= Jπ32:= ⋅ 4doτmaxT⋅cJ:= τmax = 9.119MPa AnsAngle of Twist :φE_BTBC⋅LBCG⋅ JTCD⋅LCDG⋅ J+TDE⋅LDEG⋅ J:= +φE_B = 0.010213 radφE_B = 0.5852 deg Ans 380. Problem 5-54The turbine develops 150 kW of power, which is transmitted to the gears such that both C and Dreceive an equal amount. If the rotation of the 100-mm-diameter A-36 steel shaft is ω = 500 rev/min.determine the absolute maximum shear stress in the shaft and the rotation of end B of the shaft relativto E. The journal bearing at C allows the shaft to turn freely about its axis.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: do := 100mm ω := 500⋅ rpmLBC := 3m LCD := 4mLDE := 2m P := 150kWG := 75GPaTC = 0.5T TD = 0.5TSolution: ω 52.36rads=TPω:= T = 2.865 kN⋅mTC := 0.5T TC = 1.432 kN⋅mTD := 0.5T TD = 1.432 kN⋅mTBC := TTCD := 0.5TTDE := 0Maximum Shear Stress :Maximum torque occurs in region BC.cdo2:= Jπ32:= ⋅ 4doτmaxT⋅cJ:= τmax = 14.59MPa AnsAngle of Twist :φB_ETBC⋅LBCG⋅ JTCD⋅LCDG⋅ J+TDE⋅LDEG⋅ J:= +φB_E = 0.019454 radφB_E = 1.1146 deg Ans 381. Problem 5-55The motor delivers 33 kW to the 304 stainless steel shaft while it rotates at 20 Hz. The shaft issupported on smooth bearings at A and B, which allow free rotation of the shaft. The gears C and Dfixed to the shaft remove 20 kW and 12 kW, respectively. Determine the diameter of the shaft to thenearest mm if the allowable shear stress is τallow = 56 MPa and the allowable angle of twist of C withrespect to D is 0.20°.Given: LAC := 250mm LCD := 200mm LDB := 150mmP := 32kW PC := −20kW PD := −12kWf := 20Hz G := 75GPa φallow := 0.20degτallow := 56MPaSolution: ω := 2⋅π⋅ f ω 125.6637rads=TACPω:= TAC = 254.648N⋅mTCDP + PCω:= TCD = 95.493N⋅mTmax := max(TAC, TCD) Tmax = 254.6479 N⋅mMax. shear stress : Assume failure due to shear stresscd2= Jπ2d2⎛⎜⎝⎞⎠4= ⋅ τallowTmax⋅cJ=π2d2⎛⎜⎝⎞⎠4⋅Tmax⋅d2⋅ τallow= d16Tmaxπ⋅τallow⎛⎜⎝⎞⎠13:= d = 28.5041mmAngle of Twist : Assume failure due to angle of twist limitation (occured between C and D)Jπ2d2⎛⎜⎝⎞⎠4= ⋅ φallowTCD⋅LCD= φallow = 0.003491 radG⋅ Jπ2d2⎛⎜⎝⎞⎠4⋅TCD⋅LCDG⋅φallow= d 22TCD⋅LCDπ⋅G⋅φallow⎛⎜⎝⎞⎠0.25:= d = 29.3601mmUse d = 30mm Ans 382. Problem 5-56The motor delivers 33 kW to the 304 stainless steel solid shaft while it rotates at 20 Hz. The shaft has adiameter of 37.5 mm and is supported on smooth bearings at A and B, which allow free rotation of theshaft. The gears C and D fixed to the shaft remove 20 kW and 12 kW, respectively. Determine theabsolute maximum stress in the shaft and the angle of twist of gear C with respect to gear D.Given: LAC := 250mm LCD := 200mm LDB := 150mmP := 32kW PC := −20kW PD := −12kWf := 20Hz G := 75GPa d := 37.5mmSolution: ω := 2⋅π⋅ f ω 125.66rads=TACPω:= TAC = 254.648N⋅mTCDP + PCω:= TCD = 95.493 N⋅mTmax := max(TAC, TCD) Tmax = 254.65N⋅mMax. shear stress : Maximum shear stress occured between A and Ccd2:= Jπ2d2⎛⎜⎝⎞⎠4:= ⋅ τmaxTmax⋅cJ:=τmax = 24.59MPa AnsAngle of Twist : φCDTCD⋅LCDG⋅ J:=φCD = 0.001312 radφCD = 0.075152 deg Ans 383. Problem 5-57The motor produces a torque of T = 20 N·m on gear A. If gear C is suddenly locked so it does notturn, yet B can freely turn, determine the angle of twist of F with respect to E and F with respect to Dof the L2-steel shaft, which has an inner diameter of 30 mm and an outer diameter of 50 mm. Also,calculate the absolute maximum shear stress in the shaft. The shaft is supported on journal bearings atG and H.Given: do := 50mm di := 30mmrA := 30mm rF := 100mmG := 75GPa T := 20N⋅mLEF := 0.6mSolution:Equilibrium :− ⋅ = 0 FT FrATrA:= F = 666.67 NT' − F⋅ rF = 0 T' := F⋅ rF T' = 66.67N⋅mAngle of Twist :Jπ32do4 − di4 ⎛⎝⎞⎠:= ⋅Since shaft is held fixed at C, the torque is only in region EF of the shaft.φF_ET'⋅LEFG⋅ J:= φF_E = 9.986 × 10− 4 rad AnsSince the torque in region ED is zero,φF_D := φF_E φF_D = 9.986 × 10− 4 rad AnsMaximum Shear Stress :cdo2:=τmaxT'⋅cJ:= τmax = 3.121MPa Ans 384. Problem 5-58The two shafts are made of A-36 steel. Each has a diameter of 25 mm, and they are supported bybearings at A, B, and C, which allow free rotation. If the support at D is fixed, determine the angle oftwist of end B when the torques are applied to the assembly as shown.Given: LDH := 250mm LHE := 750mmLAG := 200mm LGF := 250mmLFB := 300mm d := 25mmrE := 150mm rF := 100mmTH := 120N⋅m TG := 60N⋅mG := 75GPaSolution:Internal Torque : AS shown on FBDAt F : TF=TGPTGrF:= P = 600 NAt E : TE −P r:= ⋅ ( E) TE = −90N⋅mAngle of Twist :Jπ2d2⎛⎜⎝⎞⎠4:= ⋅φETH⋅LDHG⋅ JTE⋅ (LDH + LHE)G⋅ J:= +φE = −0.020861 radφE⋅ (rE) = −φF⋅ (rF) φFrErF:= − ⋅φE φF = 0.031291 radSince there is no torque applied between F and B,φB := φF φB = 0.031291 radAnsφB = 1.793 deg 385. Problem 5-59The two shafts are made of A-36 steel. Each has a diameter of 25 mm, and they are supported bybearings at A, B, and C, which allow free rotation. If the support at D is fixed, determine the angle oftwist of end A when the torques are applied to the assembly as shown.Given: LDH := 250mm LHE := 750mmLAG := 200mm LGF := 250mmLFB := 300mm d := 25mmrE := 150mm rF := 100mmTH := 120N⋅m TG := 60N⋅mG := 75GPaSolution:Internal Torque : AS shown on FBDAt F : TF=TGPTGrF:= P = 600 NAt E : TE −P r:= ⋅ ( E) TE = −90N⋅mAngle of Twist :Jπ2d2⎛⎜⎝⎞⎠4:= ⋅φETH⋅LDHG⋅ JTE⋅ (LDH + LHE)G⋅ J:= +φE = −0.020861 radφE⋅ (rE) = −φF⋅ (rF) φFrErF:= − ⋅φEφF = 0.031291 radSince there is no torque applied between A and G,φA_FTG⋅LGFG⋅ J:= φA_F = 0.005215 radφA := φA_F + φF φA = 0.036506 radφA = 2.092 deg Ans 386. Problem 5-60Consider the general problem of a circular shaft made from m segments, each having a radius of cmand shearing modulus Gm. If there are n torques on the shaft as shown, write a computer program thatcan be used to determine the angle of twist of its end A. Show an application of the program using thevalues L1 = 0.5 m, c1 = 0.02 m, G1 = 30 GPa, L2 = 1.5 m, c2 = 0.05 m, G2 = 15 GPa, T1 = -450 N·m,d1 = 0.25 m, T2 = 600 N·m , d2 = 0.8 m. 387. Problem 5-61The 30-mm-diameter shafts are made of L2 tool steel and are supported on journal bearings that allowthe shaft to rotate freely. If the motor at A develops a torque of T = 45 N·m on the shaft AB, while theturbine at E is fixed from turning, determine the amount of rotation of gears B and C.Given: LAB := 1.5m LDC := 0.5m LCE := 0.75mTA := 45N⋅m rB := 50mm rC := 75mmdo := 30mm G := 75GPaSolution:TAB := TAEquilibrium :TAB − F⋅ rB = 0 FTABrB:= F = 900.00 NT' − F⋅ rC = 0 T' := F⋅ rC T' = 67.50N⋅mTCE := T'Angle of Twist : Jπ32:= ⋅ 4doφBTAB⋅LAB:= φCG⋅ JTCE⋅LCEG⋅ J:=φB = 0.011318 rad φC = 0.008488 radφB = 0.648 deg Ans φC = 0.486 deg Ans 388. Problem 5-62The 60-mm-diameter solid shaft is made of A-36 steel and is subjected to the distributed andconcentrated torsional loadings shown. Determine the angle of twist at the free end A of the shaft dueto these loadings.Given: LAC := 0.6m LCB := 0.8m do := 60mmTA := 400N⋅m TC := −600N⋅mG := 75GPa q 2000N⋅mm:=Solution:Internal Torque : As shown in the torque diagram.TAC := TATCB = TA + TC + q⋅xAngle of Twist : Jπ32:= ⋅ 4doφATAC⋅LACG⋅ J1G⋅ JLCB0TCB x⌠⎮⌡= + ⋅ dφATAC⋅LACG⋅ J1G⋅ JLCB0TA + TC + q⋅x x⌠⎮⌡:= + ⋅ dφA = 0.007545 radφA = 0.432 deg Ans 389. Problem 5-63When drilling a well, the deep end of the drill pipe is assumed to encounter a torsional resistance TA.Furthermore, soil friction along the sides of the pipe creates a linear distribution of torque per unitlength, varying from zero at the surface B to t0 at A. Determine the necessary torque TB that must besupplied by the drive unit to turn the pipe. Also, what is the relative angle of twist of one end of thepipe with respect to the other end at the instant the pipe is about to turn? The pipe has an outer radiusro and an inner radius ri . The shear modulus is G. 390. Problem 5-64The assembly is made of A-36 steel and consists of a solid rod 15 mm in diameter connected to theinside of a tube using a rigid disk at B. Determine the angle of twist at A. The tube has an outerdiameter of 30 mm and wall thickness of 3 mm.Given: do := 30mm t := 3mm dr := 15mmLAB := 0.3m LBC := 0.3m G := 75GPaTA := 50N⋅m TB := 30N⋅mSolution: di := do − 2tTAB := TATBC := TA + TBAngle of Twist :JABπ32:= ⋅ dr4 JBCπ32do4 − di4 ⎛⎝⎞⎠:= ⋅φATAB⋅LABG⋅ JABTBC⋅LBCG⋅ JBC:= +φA = 0.04706 radφA = 2.696 deg Ans 391. Problem 5-65The device serves as a compact torsional spring. It is made of A-36 steel and consists of a solid innershaft CB which is surrounded by and attached to a tube AB using a rigid ring at B. The ring at A canalso be assumed rigid and is fixed from rotating. If a torque of T = 0.25 N·m. is applied to the shaft,determine the angle of twist at the end C and the maximum shear stress in the tube and shaft.Given: LBC := 600mm LBA := 300mmrti := 18.75mm rto := 25mmr := 12.5mm TC := 0.25kN⋅mG := 75GPaSolution:Internal Torque : AS shown on FBDInner shaft : TBC := TCOuter tube : TBA := TCMax. Shear Stress :Inner shaft : cBC := r JBCπ2:= ⋅ r4τBCTBC⋅ (cBC)JBC:= τBC = 81.49MPa AnscBA := rto JBAπ2rto4 − rti4 ⎛⎝⎞⎠Outer tube : := ⋅τBATBC⋅ (cBA)JBA:= τBA = 14.9MPa AnsAngle of Twist : φBTBA⋅LBAG⋅ JBA:= φB = 0.002384 radφC_BTBC⋅LBCG⋅ JBC:= φC_B = 0.052152 radφC := φC_B + φB φC = 0.054536 radφC = 3.125 deg Ans 392. Problem 5-66The device serves as a compact torsion spring. It is made of A-36 steel and consists of a solid innershaft CB which is surrounded by and attached to a tube AB using a rigid ring at B. The ring at A canalso be assumed rigid and is fixed from rotating. If the allowable shear stress for the material is τallow =84 MPa and the angle of twist at C is limited to φallow = 3°, determine the maximum torque T that canbe applied at the end C.Given: LBC := 600mm LBA := 300mmrti := 18.75mm rto := 25mmτallow := 84MPa r := 12.5mmφallow := 3deg G := 75GPaSolution:Internal Torque : AS shown on FBDInner shaft : TBC=TCOuter tube : TBA=TCAllowable Shear Stress : Assume failure due to shear stress.Inner shaft : cBC := r JBCπ2:= ⋅ r4TBCτallow⋅ (JBC):= TBC = 257.71 N⋅mcBCcBA := rto JBAπ2rto4 − rti4 ⎛⎝⎞⎠Outer tube : := ⋅TBAτallow⋅ (JBA):= TBA = 1409.34N⋅mcBAAngle of Twist : Assume failure due to angle of twist limitation (maximum occurred at C).φC = φC_B + φB φC_BTC⋅LBCG⋅ JBC= φBTC⋅LBAG⋅ JBA=Thus, φallowTC⋅LBCG⋅ JBCTC⋅LBAG⋅ JBA= +TCG⋅φallowLBCJBC:= TC = 240.02 N⋅m AnsLBAJBA+(controls !) 393. Problem 5-67The shaft has a radius c and is subjected to a torque per unit length of t0 , which is distributeduniformly over the shaft's entire length L. If it is fixed at its far end A, determine the angle of twist φ oend B. The shear modulus is G.Solution:Internal Torque : As shown in the torque diagram.= − ⋅xT x ( ) toAngle of Twist : Jπ2:= ⋅c4φBLT(x) x⌠⎮⌡1G⋅ J 0= ⋅ dφBLx x−toG⋅ J 0⌠⎮⌡= ⋅ dφB−to⋅L22G⋅ J=φB−to⋅L2πG⋅c4=φBto⋅L2πG⋅c4= Ans 394. Problem 5-68The A-36 bolt is tightened within a hole so that the reactive torque on the shank AB can be expressedby the equation t = (k x2) N·m /m, where x is in meters. If a torque of T = 50 N·m is applied to the bolhead, determine the constant k and the amount of twist in the 50-mm length of the shank. Assume theshank has a constant radius of 4 mm.Given: ro := 4mm L := 50mm2TA := x 50N⋅m G := 75GPat k⋅N⋅mm=Solution: c := roInternal Torque :As shown in the torque diagram.t x ( ) k x 2= ⋅Equilibrium :TALt(x) x⌠⎮⌡− d = 0 TA0Lk⋅x2 x⌠⎮⌡− d = 00TAk⋅L33− = 0 k3TAL3:= k = 1.200MPa Ans⌠⎮⎮⌡Hence, T(x) = TA − k⋅x2 dxT(x) TAk⋅x33= −Angle of Twist : Jπ2:= ⋅c4φLT(x) x⌠⎮⌡1G⋅ J 0= ⋅ d φ1G⋅ JLTA x0k⋅x33−⎛⎜⎝⎞⎠⌠⎮⎮⌡:= ⋅ dφ = 0.06217 radφ = 3.562 deg Ans 395. Problem 5-69Solve Prob. 5-68 if the distributed torque is t = (k x2/3) N·m /m.Given: ro := 4mm L := 50mmTA := 50N⋅m G := 75GPa2⋅ 3t kxN⋅mm=Solution: c := roInternal Torque :As shown in the torque diagram.Equilibrium :LTAt(x) x⌠⎮⌡− d = 0 TA0L02⋅ 3k x x⌠⎮⎮⌡− d = 0Let unit m− 4:= 3 TA3k L53⎛⎜⎝⎞⋅ ⎠5− = 0k5TA3⎛⎜⎝⎞⎠L− 5:= ⋅ 3 ⋅unitk = 0.01228MPa Ans2⋅ 3⌠⎮⎮⎮⌡Hence, T(x) = TA − k x dxT(x) TA3k x53⎛⎜⎝⎞⋅ ⎠5= −Angle of Twist : Jπ2:= ⋅c4φLT(x) x⌠⎮⌡1G⋅ J 0= ⋅ d φ⌠⎮⎮⎮⎮⌡d ⋅ :=1G⋅ JL0⎛⎜⎝⋅ ⎠5⋅unit3k x53⎞TA −xφ = 0.05181 radφ = 2.968 deg Ans 396. Problem 5-70The contour of the surface of the shaft is defined by the equation y = e ax, where a is a constant. If theshaft is subjected to a torque T at its ends, determine the angle of twist of end A with respect to end B.The shear modulus is G. 397. Problem 5-71The A-36 steel shaft has a diameter of 50 mm and is subjected to the distributed and concentratedloadings shown. Determine the absolute maximum shear stress in the shaft and plot a graph of theangle of twist of the shaft in radians versus x.Given: do := 50mm L := 0.5mTC := −250N⋅m G := 75GPaq 200N⋅mm:=Solution:Support Reaction:RA := −TC − q⋅L RA = 150.00N⋅mInternal Torque : As shown in the torque diagram.T(x) = RA + q⋅xThe maximum torque occurs at C.Tmax := RA + q⋅L Tmax = 250.00N⋅mMaximum Shear Stress : c := 0.5⋅do Jπ2:= ⋅c4τmaxTmax⋅c:= τmax = 10.19MPa AnsJAngle of Twist :φxxT(x) x⌠⎮⌡1G⋅ J 0= ⋅ d φx1G⋅ Jx(RA + q⋅x) x ⌠⎮⌡= ⋅ d0φx1G⋅ JRA⋅xq⋅x22+⎛⎜⎝⎞⎠= ⋅At C, x = L φC1G⋅ JRA⋅Lq⋅L22+⎛⎜⎝⎞⎠:= ⋅ φC = 0.00217 radφC = 0.125 deg AnsFor L < x < 2L Since T(x) = 0, then, φx := φCx1 := 0 , 0.01⋅L .. L x2 := L , 1.01⋅L .. 2Lφ1(x1) 1G⋅ JRA⋅x1⋅ 22q x1+⎛⎜⎝⎞⎠⋅1N⋅m:= ⋅ φ2(x2) (φC) 1N⋅m:= ⋅ 398. 0 0.2 0.4 0.6 0.80.0020.0010Distance (m)Twist (rad)φ1(x1)φ2(x2)x1, x2 399. Problem 5-72A cylindrical spring consists of a rubber annulus bonded to a rigid ring and shaft. If the ring is heldfixed and a torque T is applied to the rigid shaft, determine the angle of twist of the shaft. The shearmodulus of the rubber is G. Hint: As shown in the figure, the deformation of the element at radius rcan be determined from rdθ = drγ. Use this expression along with τ = T/(2π r2h) from Prob. 5-28, toobtain the result. 400. Problem 5-73The A-36 steel shaft has a diameter of 50 mm and is fixed at its ends A and B. If it is subjected to thecouple, determine the maximum shear stress in regions AC and CB of the shaft.Given: do := 50mm LAC := 0.4m LBC := 0.8mG := 75GPa TC := 300N⋅mSolution:c := 0.5⋅do Jπ2:= ⋅c4Equilibrium : GivenTA + TB − TC = 0 (1)Compatibility : φC/A = φC/BTA⋅LACG⋅ JTB⋅LBCG⋅ J= (2)Solving Eqs. (1) and (2): Guess TA := 1N⋅m TB := 1N⋅mTATB⎛⎜⎜⎝⎞⎠:= Find(TA, TB)TATB⎛⎜⎜⎝⎞⎠200100⎛⎜⎝⎞⎠= N⋅mMaximum Shear Stress :τAC.maxTA⋅cJ:= τAC.max = 8.15MPa AnsτBC.maxTB⋅cJ:= τBC.max = 4.07MPa Ans 401. Problem 5-74The bronze C86100 pipe has an outer diameter of 37.5 mm and a thickness of 3 mm. The coupling onit at C is being tightened using a wrench. If the torque developed at A is 16 N·m, determine themagnitude F of the couple forces. The pipe is fixed supported at end B.Given: LCB := 200mm LCA := 250mmLw := 300mm do := 37.5mm t := 3mmTA := 16N⋅m G := 38GPaSolution: ro := 0.5do ri := ro − tCompatibility : φC_B = φC_ATB⋅LCBG⋅ JTB⋅LCAG⋅ J=TB⋅LCB = TA⋅LCATBLCALCB⎛⎜⎝⎞⎠:= ⋅TA TB = 20.00N⋅mEquilibrium: F⋅ (Lw) − TB − TA = 0FTB + TALw:=F = 120.00 N Ans 402. Problem 5-75The bronze C86100 pipe has an outer diameter of 37.5 mm and a thickness of 3 mm. The coupling onit at C is being tightened using a wrench. If the applied force is F = 100 N, determine the maximumshear stress in the pipe.Given: LCB := 200mm LCA := 250mmLw := 300mm do := 37.5mm t := 3mmF := 100N G := 38GPaSolution: ro := 0.5do ri := ro − tCompatibility : φC_B = φC_ATB⋅LCBG⋅ JTB⋅LCAG⋅ J=Given TB⋅LCB = TA⋅LCA [1]Equilibrium: F⋅ (Lw) − TB − TA = 0 [2]Initial guess: TA := 1N⋅m TB := 2N⋅mSolving [1] and [2]:TATB⎛⎜⎜⎝⎞⎠:= Find(TA, TB)TATB⎛⎜⎜⎝⎞⎠⎛⎜⎝⎞13.3316.67= N⋅m⎠Max. Shear Stress :c := ro Jπ2ro4 − ri4 ⎛⎝⎞⎠:= ⋅τmaxTB⋅ (c)J:=τmax = 3.21MPa Ans 403. Problem 5-76The steel shaft is made from two segments: AC has a diameter of 12 mm, and CB has a diameter of 25mm. If it is fixed at its ends A and B and subjected to a torque of 750 N·m, determine the maximumshear stress in the shaft. Gst = 75 GPa.Given: LAC := 125mm LCD := 200mmLDB := 300mm dAC := 12mmdCD := 25mm dDB := 25mmTD := 750N⋅m G := 75GPaSolution:JACπ2dAC2⎛⎜⎝⎞⎠4:= ⋅JCDπ2dCD2⎛⎜⎝⎞⎠4:= ⋅ JDBπ2dDB2⎛⎜⎝⎞⎠4:= ⋅Compatibility : φD_A = φD_BGivenTA⋅LACG⋅ JACTA⋅LCDG⋅ JCD+TB⋅LDBG⋅ JDB= [1]Equilibrium: TD − TB − TA = 0 [2]Initial guess: TA := 1N⋅m TB := 2N⋅mSolving [1] and [2]:TATB⎛⎜⎜⎝⎞⎠:= Find(TA, TB)TATB⎛⎜⎜⎝ ⎞⎠78.82671.18⎛⎜⎝⎞⎠= N⋅mMax. Shear Stress :cACdAC2:= τACTA⋅ (cAC)JAC:= τAC = 232.3MPacDBdDB2:= τDBTB⋅ (cDB)JDB:= τDB = 218.77MPaτmax := max(τAC, τDB) τmax = 232.30MPa Ans 404. Problem 5-77The shaft is made of L2 tool steel, has a diameter of 40 mm, and is fixed at its ends A and B. If it issubjected to the couple, determine the maximum shear stress in regions AC and CB.Given: do := 40mm LAC := 0.4m LBC := 0.6mG := 75GPa PC := 2kN rC := 50mmSolution:c:= 0.5⋅do Jπ2:= ⋅c4Equilibrium : GivenTA + TB − PC⋅ (2rC) = 0 (1)Compatibility : φC/A = φC/BTA⋅LACG⋅ JTB⋅LBCG⋅ J= (2)Solving Eqs. (1) and (2): Guess TA := 1N⋅m TB := 1N⋅mTATB⎛⎜⎜⎝⎞⎠:= Find(TA, TB)TATB⎛⎜⎜⎝⎞⎠12080⎛⎜⎝⎞⎠= N⋅mMaximum Shear Stress :τAC.maxTA⋅cJ:= τAC.max = 9.55MPa AnsτBC.maxTB⋅cJ:= τBC.max = 6.37MPa Ans 405. Problem 5-78The composite shaft consists of a mid-section that includes the 20-mm-diameter solid shaft and a tubethat is welded to the rigid flanges at A and B. Neglect the thickness of the flanges and determine theangle of twist of end C of the shaft relative to end D. The shaft is subjected to a torque of 800 N·m.The material is A-36 steel.Given: LCA := 100mm LAB := 150mmLBD := 100mm ds := 20mmdto := 60mm t := 5mmT := 800N⋅m G := 75GPaSolution: dti := dto − 2tJsπ2ds2⎛⎜⎝⎞⎠4:= ⋅ Jtπ2dto2⎛⎜⎝⎞⎠4 dti2⎛⎜⎝⎞⎠4−⎡⎢⎣⎤⎥⎦:= ⋅Compatibility : φs = φtTs⋅LABG⋅ JsTt⋅LABG⋅ Jt=GivenTsJsTtJt= [1]Equilibrium: T − Ts − Tt = 0 [2]Initial guess: Ts := 1N⋅m Tt := 2N⋅mSolving [1] and [2]:TsTt⎛⎜⎜⎝⎞⎠:= Find(Ts, Tt)TsTt⎛⎜⎜⎝⎞⎠18.63781.37⎛⎜⎝⎞⎠= N⋅mAngle of Twist:φC_DTs⋅LCAG⋅ JsTt⋅LABG⋅ Jt+Ts⋅LBDG⋅ Js:= +φC_D = 0.005535 radφC_D = 0.317 deg Ans 406. Problem 5-79The shaft is made from a solid steel section AB and a tubular portion made of steel and having a brasscore. If it is fixed to a rigid support at A, and a torque of T = 50 N·m is applied to it at C, determine theangle of twist that occurs at C and compute the maximum shear stress and maximum shear strain inthe brass and steel. Take Gst = 80 GPa, Gbr = 40 GPa.Given: LAB := 1.5m LBC := 1mrs := 20mm rto := 20mm rti := 10mmT := 50N⋅m Gst := 80GPa Gbr := 40GPaSolution: Jsπ2:= ⋅ rs4 Jtπ2rto4 − rti4 ⎛⎝⎞⎠:= ⋅Jbrπ2:= ⋅ 4rtiCompatibility : φst = φbrTst⋅LBCGst⋅ JtTbr⋅LBCGbr⋅ Jbr=GivenTstGst⋅ JtTbrGbr⋅ Jbr= [1]Equilibrium: T − Tst − Tbr = 0 [2]Initial guess: Tst := 1N⋅m Tbr := 2N⋅mSolving [1] and [2]:TstTbr⎛⎜⎜⎝⎞⎠:= Find(Tst , Tbr)TstTbr⎛⎜⎜⎝⎞⎠48.391.61⎛⎜⎝⎞⎠= N⋅mAngle of Twist:φCT⋅LABGst⋅ JsTbr⋅LBCGbr⋅ Jbr:= + φC = 0.006297 rad φC = 0.361 deg AnsMax. Shear Stress :cAB := rs τABT⋅ (cAB)Js:= τAB = 3.98MPacBC := rto τBCTst⋅ (cBC):= τBC = 4.11MPaJtτst := max(τAB, τBC) τst = 4.11MPa AnsγstτstGst:= γst = 51.34 × 10− 6 rad AnsτbrTbr⋅ (rti)Jbr:= τbr = 1.03MPa AnsγbrτbrGbr:= γbr = 25.67 × 10− 6 rad Ans 407. Problem 5-80The two 1-m-long shafts are made of 2014-T6 aluminum. Each has a diameter of 30 mm and they areconnected using the gears fixed to their ends. Their other ends are attached to fixed supports at A andB. They are also supported by bearings at C and D, which allow free rotation of the shafts along theiraxes. If a torque of 900 N·m is applied to the top gear as shown, determine the maximum shear stressin each shaft.Given: L := 1000mm d := 30mmrA := 80mm rB := 40mmT := 900N⋅m G := 27GPaSolution: Jπ2d2⎛⎜⎝⎞⎠4:= ⋅Compatibility : φE⋅ (rA) = φF⋅ (rB)TA⋅LG⋅ J⋅ (rA)TB⋅LG⋅ J= ⋅ (rB)Given TA⋅ (rA) = TB⋅ (rB) [1]Equilibrium: T − TA − F⋅ (rA) = 0 [2]TB − F⋅ (rB) = 0 [3]Initial guess: TA := 1N⋅m TB := 2N⋅m F := 1NSolving [1], [2] and [3]:TATBF⎛⎜⎜⎜⎝⎞⎟⎠:= Find(TA, TB, F) TATB⎛⎜⎜⎝⎞⎠180360⎛⎜⎝⎞⎠= N⋅mF = 9000.00 NMax. Shear Stress :cd2:= τACTA⋅ (c)J:= τAC = 33.95MPa AnsτBDTB⋅ (c)J:= τBD = 67.91MPa Ans 408. Problem 5-81The two shafts are made of A-36 steel. Each has a diameter of 25 mm and they are connected usingthe gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are alsosupported by journal bearings at C and D, which allow free rotation of the shafts along their axes. If atorque of 500 N·m is applied to the gear at E as shown, determine the reactions at A and B.Given: LAE := 1.5m LBF := 0.75m do := 25mmrE := 100mm rF := 50mmTE := 500N⋅m G := 75GPaSolution: Jπ32:= ⋅ 4doCompatibility : φE⋅ (rE) = φF⋅ (rF)TA⋅LAEG⋅ J⋅ (rE)TB⋅LBFG⋅ J= ⋅ (rF)Given TA⋅LAE⋅ (rE) = TB⋅LBF⋅ (rF) [1]Equilibrium: TA + F⋅ (rE) − TE = 0 [2]TB − F⋅ (rF) = 0 [3]Initial guess: TA := 1N⋅m TB := 2N⋅m F := 1NSolving [1], [2] and [3]:TATBF⎛⎜⎜⎜⎝⎞⎟⎠:= Find(TA, TB, F) TA⎛⎜⎜⎝ ⎞TB⎠55.56222.22⎛⎜⎝⎞⎠= N⋅m AF = 4444.44 N 409. Problem 5-82Determine the rotation of the gear at E in Prob. 5-81.Given: LAE := 1.5m LBF := 0.75m do := 25mmrE := 100mm rF := 50mmTE := 500N⋅m G := 75GPaSolution: Jπ32:= ⋅ 4doCompatibility : φE⋅ (rE) = φF⋅ (rF)TA⋅LAEG⋅ J⋅ (rE)TB⋅LBFG⋅ J= ⋅ (rF)Given TA⋅LAE⋅ (rE) = TB⋅LBF⋅ (rF) [1]Equilibrium: TA + F⋅ (rE) − TE = 0 [2]TB − F⋅ (rF) = 0 [3]Initial guess: TA := 1N⋅m TB := 2N⋅m F := 1NSolving [1], [2] and [3]:TATBF⎛⎜⎜⎜⎝⎞⎟⎠:= Find(TA, TB, F) TATB⎛⎜⎜⎝⎞⎠55.56222.22⎛⎜⎝⎞⎠= N⋅mF = 4444.44 NAngle of Twist :φETA⋅LAEG⋅ J:=φE = 0.02897 radφE = 1.660 deg Ans 410. Problem 5-83The A-36 steel shaft is made from two segments: AC has a diameter of 10 mm and CB has a diameterof 20 mm. If the shaft is fixed at its ends A and B and subjected to a uniform distributed torque of 300N·m/m along segment CB, determine the absolute maximum shear stress in the shaft.Given: LAC := 0.1m LCB := 0.4mdAC := 10mm dCB := 20mmG := 75GPa q 300N⋅mm:=Solution: JACπ2dAC2⎛⎜⎝⎞⎠4:= ⋅ JCBπ2dCB2⎛⎜⎝⎞⎠4:= ⋅Compatibility : φC_A = φC_BGivenTA⋅LACG⋅ JAC− ⋅ 2G⋅ JCBTB⋅LCB 0.5q LCB= [1]Equilibrium: q⋅LCB − TA − TB = 0 [2]Initial guess: TA := 1N⋅m TB := 2N⋅mSolving [1] and [2]: TATB⎛⎜⎜⎝⎞⎠:= Find(TA, TB)TATB⎛⎜⎜⎝⎞⎠12108⎛⎜⎝⎞⎠= N⋅mMax. Shear Stress :cACdAC2:= τACTA⋅ (cAC)JAC:= τAC = 61.12MPa AnscCBdCB2:= τCBTB⋅ (cCB)JCB:= τCB = 68.75MPa Ansτmax := max(τAC, τCB) τmax = 68.75MPa Ans 411. Problem 5-84The tapered shaft is confined by the fixed supports at A and B. If a torque T is applied at its mid-poindetermine the reactions at the supports. 412. Problem 5-85A portion of the A-36 steel shaft is subjected to a linearly distributed torsional loading. If the shaft hasthe dimensions shown, determine the reactions at the fixed supports A and C. Segment AB has adiameter of 30 mm and segment BC has a diameter of 15 mm.Given: LAB := 1.2m LBC := 0.96mdAB := 30mm dBC := 15mmG := 75GPa q 1.5kN⋅mm:=Solution: JABπ2dAB2⎛⎜⎝⎞⎠4:= ⋅ JBCπ2dBC2⎛⎜⎝⎞⎠4:= ⋅qx 1xLab− ⎛⎜⎝⎞⎠= ⋅q TR = (qx)⋅x + 0.5(q − qx)⋅xTR 1xLab− ⎛⎜⎝⎞⎠⋅q⋅x 0.5xLab⎛⎜⎝⎞⎠= + ⋅q⋅xTR 10.5xLab− ⎛⎜⎝⎞⎠= ⋅q⋅xβLAB00.5xLAB− ⎛⎜⎝⎞⎠1 ⋅xx⌠⎮⎮⌡:= d β = 0.48 m2Compatibility : φB_A = φB_CGivenTA⋅LAB − q⋅βG⋅ JABTC⋅LBCG⋅ JBC= [1]Equilibrium: 0.5q⋅LAB − TA − TC = 0 [2]Initial guess: TA := 1kN⋅m TC := 2kN⋅mSolving [1] and [2]: TATC⎛⎜⎜⎝⎞⎠:= Find(TA, TC)TATC⎛⎜⎜⎝⎞⎠878.2621.74⎛⎜⎝⎞⎠= N⋅m Ans 413. Problem 5-86Determine the rotation of joint B and the absolute maximum shear stress in the shaft in Prob. 5-85.Given: LAB := 1.2m LBC := 0.96mdAB := 30mm dBC := 15mmG := 75GPa q 1.5kN⋅mm:=Solution: JABπ2dAB2⎛⎜⎝⎞⎠4:= ⋅ JBCπ2dBC2⎛⎜⎝⎞⎠4:= ⋅qx 1xLab− ⎛⎜⎝⎞⎠= ⋅q TR = (qx)⋅x + 0.5(q − qx)⋅xTR 1xLab− ⎛⎜⎝⎞⎠⋅q⋅x 0.5xLab⎛⎜⎝⎞⎠= + ⋅q⋅x TR 10.5xLab− ⎛⎜⎝⎞⎠= ⋅q⋅xβLAB00.5xLAB− ⎛⎜⎝⎞⎠1 ⋅xx⌠⎮⎮⌡:= d β = 0.48 m2Compatibility : φB_A = φB_CGivenTA⋅LAB − q⋅βG⋅ JABTC⋅LBCG⋅ JBC= [1]Equilibrium: 0.5q⋅LAB − TA − TC = 0 [2]Initial guess: TA := 1kN⋅m TC := 2kN⋅mSolving [1] and [2]:TATC⎛⎜⎜⎝⎞⎠:= Find(TA, TC)TATC⎛⎜⎜⎝⎞⎠878.2621.74⎛⎜⎝⎞⎠= N⋅mAngle of Twist:φBTC⋅LBCG⋅ JBC:= φB = 0.055987 radφB = 3.208 deg AnsMax. Shear Stress :cABdAB2:= τABTA⋅ (cAB)JAB:= τAB = 165.66MPacBCdBC2:= τBCTC⋅ (cBC)JBC:= τBC = 32.8MPaτmax := max(τAB, τBC) τmax = 165.66MPa Ans 414. Problem 5-87The shaft of radius c is subjected to a distributed torque t, measured as torque/length of shaft.Determine the reactions at the fixed supports A and B. 415. Problem 5-88Compare the values of the maximum elastic shear stress and the angle of twist developed in 304stainless steel shafts having circular and square cross sections. Each shaft has the same cross-sectionaarea of 5600 mm2, length of 900 mm, and is subjected to a torque of 500 N·m.Given: L := 900mm A := 5600mm2G := 75GPa T := 500N⋅mSolution:Max. Shear Stress :For circular shaft: A = π⋅ r2 rAπ:=c := r Jπ2:= ⋅ r4 τC_maxT⋅ (c)J:=τC_max = 4.23MPa AnsFor squarr shaft: A = a2 a := AτS_maxT⋅ (4.81)a3:=τS_max = 5.74MPa AnsAngle of Twist:For circular shaft:φCT⋅LG⋅ J:= φC = 0.001202 rad φC = 0.0689 deg AnsFor sqaure shaft:φS(7.10)T⋅LG⋅a4:= φS = 0.001358 rad φS = 0.0778 deg AnsNote: The sqaure shaft has a greater maximum shear stress and angle of twist. 416. Problem 5-89The shaft is made of red brass C83400 and has an elliptical cross section. If it is subjected to thetorsional loading shown, determine the maximum shear stress within regions AC and BC, and the angof twist φ of end B relative to end A.Given: LAC := 2m LCB := 1.5ma := 50mm b := 20mmTA := 50N⋅m TB := 30N⋅mG := 37GPa TC := 20N⋅mSolution:Max. Shear Stress :τBC2TBπ⋅a⋅b2:= τBC = 0.955MPa AnsτAC2TAπ⋅a⋅b2:= τAC = 1.592MPa AnsAngle of Twist: φB_A(a2 + b2)Tn⋅Lnπ⋅a3⋅b3⋅G =ΣnφB_A(a2 + b2)(−TB⋅LCB − TA⋅LAC)π⋅a3⋅b3⋅G:=φB_A = −0.003618 radφB_A = 0.2073 deg Ans 417. Problem 5-90Solve Prob. 5-89 for the maximum shear stress within regions AC and BC, and the angle of twist φ ofend B relative to C.Given: LAC := 2m LCB := 1.5ma := 50mm b := 20mmTA := 50N⋅m TB := 30N⋅mG := 37GPa TC := 20N⋅mSolution:Max. Shear Stress :τBC2TBπ⋅a⋅b2:= τBC = 0.955MPa AnsτAC2TAπ⋅a⋅b2:= τAC = 1.592MPa AnsAngle of Twist: φB_C(a2 + b2)Tn⋅Lnπ⋅a3⋅b3⋅G =ΣnφB_C(a2 + b2)(−TB⋅LCB)π⋅a3⋅b3⋅G:=φB_C = −0.001123 radφB_C = 0.0643 deg Ans 418. Problem 5-91The steel shaft is 300 mm long and is screwed into the wall using a wrench. Determine the largestcouple forces F that can be applied to the shaft without causing the steel to yield. τY = 56 MPa.Given: L := 300mm Lw := 400mma := 25mm τY := 56MPaSolution:Max. Shear Stress :τmax4.81Ta3= Ta3τY4.81:=T = 181.91N⋅mEquilibrium:F⋅ (Lw) − T = 0 FTLw:=F = 454.78 N Ans 419. Problem 5-92The steel shaft is 300 mm long and is screwed into the wall using a wrench. Determine the maximumshear stress in the shaft and the amount of displacement that each couple force undergoes if the coupleforces have a magnitude of F = 150 N. Gst = 75 GPa.Given: L := 300mm Lw := 400mma := 25mm F := 150NGst := 75GPaSolution:Equilibrium:F Lw ( ) ⋅ T − 0 = T F Lw:= ⋅ ( )T = 60N⋅mMax. Shear Stress :τmax4.81Ta3:= τmax = 18.47MPa AnsAngle of Twist:φ(7.10)T⋅LGst⋅a4:= φ = 0.004362 radδF φLw2:= ⋅ δF = 0.872mm Ans 420. Problem 5-93The shaft is made of plastic and has an elliptical cross-section. If it is subjected to the torsional loadinshown, determine the shear stress at point A and show the shear stress on a volume element located atthis point. Also, determine the angle of twist φ at the end B. Gp = 15 GPa.Given: LOC := 2m LCB := 1.5ma := 50mm b := 20mmTB := 50N⋅m TC := 40N⋅mG := 15GPaSolution: TOC := TB + TCShear Stress :τA2TOCπ⋅a⋅b2:= τA = 2.865MPa AnsAngle of Twist: φB(a2 + b2)Tn⋅Lnπ⋅a3⋅b3⋅G =ΣnφB(a2 + b2)(TB⋅LCB + TOC⋅LOC)π⋅a3⋅b3⋅G:=φB = 0.015693 radφB = 0.8991 deg Ans 421. Problem 5-94The square shaft is used at the end of a drive cable in order to register the rotation of the cable on agauge. If it has the dimensions shown and is subjected to a torque of 8 N·m, determine the shear stresin the shaft at point A. Sketch the shear stress on a volume element located at this point.Given: a := 5mm T := 8N⋅mSolution:Maximum Shear Stress :τA.max4.81Ta3:=τA.max = 307.8MPa Ans 422. Problem 5-95The brass wire has a triangular cross section, 2 mm on a side. If the yield stress for brass is τY = 205MPa, determine the maximum torque T to which it can be subjected so that the wire will not yield. Ifthis torque is applied to a segment 4 m long, determine the greatest angle of twist of one end of thewire relative to the other end that will not cause permanent damage to the wire. Gbr = 37 GPa.Given: a := 2mm L := 4mG := 37GPa τY := 205MPaSolution:Allowable Shear Stress :τallow20Ta3= τallow := τYTτY⋅a320:=T = 0.0820N⋅m AnsAngle of Twist :φ46T⋅LG⋅a4:=φ = 25.49 rad Ans 423. Problem 5-96It is intended to manufacture a circular bar to resist torque; however, the bar is made elliptical in theprocess of manufacturing, with one dimension smaller than the other by a factor k as shown.Determine the factor by which the maximum shear stress is increased.Given: a = 0.5⋅do b = 0.5⋅k⋅doSolution:Maximum Shear Stress :For the circular shaft::τc.maxT⋅cJ= c = 0.5⋅do Jπ32= ⋅ 4doτc.max16Tπ do⋅ 3=For the elliptical shaft:τe.max2Tπ⋅a⋅b2= τe.max2Tπ⋅ (0.5⋅do)⋅ (0.5⋅k⋅do)2=τt.max16Tπ⋅k2 do⋅ 3=Factor of increase in shear stress:Fττe.maxτc.max= Fτ16Tπ⋅k2 do⋅ 316Tπ do= Fτ⋅ 31k2= Ans 424. Problem 5-97The 2014-T6 aluminum strut is fixed between the two walls at A and B. If it has a 50 mm by 50 mmsquare cross section, and it is subjected to the torsional loading shown, determine the reactions at thefixed supports. Also, what is the angle of twist at C?Given: LAC := 600mm LCD := 600mm a := 50mmLDB := 600mm TC := 60N⋅mTD := 30N⋅m G := 27GPaSolution: T := TC + TD T = 90N⋅mL := LAC + LCD + LDBCompatibility : φT − φB = 07.10(T⋅LAC)G⋅a47.10(TD⋅LCD)G⋅a4+7.10(TB⋅L)− = 0G⋅a4TBT⋅LAC + TD⋅LCDL:=TB = 40N⋅m AnsEquilibrium:TA + TB − T = 0TA := T − TB TA = 50N⋅m AnsAngle of Twist:φC7.10 ( )TA⋅LAC:= φC = 0.001262 radG⋅a4φC = 0.0723 deg Ans 425. Problem 5-98The 304 stainless steel tube has a thickness of 10 mm. If the allowable shear stress is τallow = 80 MPa,determine the maximum torque T that it can transmit. Also, what is the angle of twist of one end of thtube with respect to the other if the tube is 4 m long? Neglect the stress concentrations at the corners.The mean dimensions are shown.Given: a := 70mm b := 30mmL := 4m τallow := 80⋅MPat := 10mm G := 75GPaSolution:Section properties : S = ΣdsAm := a⋅b Am = 2100mm2S := 2a + 2⋅b S = 200mmAverage shear stress:τavgT2t⋅Am= τavg := τallowT := 2⋅ t⋅Am⋅ τallowT = 3.36 kN⋅m AnsAngle of Twist :φT⋅L4Am2⋅Gs1t⌠⎮⎮⌡= ⋅ dφT⋅L4Am2⋅GSt⎛⎜⎝⎞⎠:= ⋅φ = 0.203175 radφ = 11.641 deg Ans 426. Problem 5-99The 304 stainless steel tube has a thickness of 10 mm. If the applied torque is T = 50 N·m, determinethe average shear stress in the tube. Neglect the stress concentrations at the corners. The meandimensions are shown.Given: a := 70mm b := 30mmt := 10mm T := 50N⋅mG := 75GPaSolution:Section properties :Am := a⋅b Am = 2100mm2Average shear stress:τavgT2t⋅Am:=τavg = 1.19MPa Ans 427. Problem 5-100Determine the constant thickness of the rectangular tube if the average shear stress is not to exceed84 MPa when a torque of T = 2.5 kN·m is applied to the tube. Neglect stress concentrations at thecorners. The mean dimensions of the tube are shown.Given: a := 100mm b := 50mmT := 2.5kN⋅m τallow := 84MPaSolution:Section properties :Am := a⋅b Am = 5000mm2Average shear stress:τavgT2t⋅Am=tT:= t = 2.98mm Ans2⋅Am⋅ τallow 428. Problem 5-101Determine the torque T that can be applied to the rectangular tube if the average shear stress is not toexceed 84 MPa. Neglect stress concentrations at the corners. The mean dimensions of the tube areshown and the tube has a thickness of 3 mm.Given: a := 100mm b := 50mmt := 3mm τallow := 84MPaSolution:Section properties :Am := a⋅b Am = 5000mm2Average shear stress:τavgT2t⋅Am=T := 2⋅ t⋅Am⋅ τallowT = 2.52 kN⋅m Ans 429. Problem 5-102A tube having the dimensions shown is subjected to a torque of T = 50 N·m. Neglecting the stressconcentrations at its corners, determine the average shear stress in the tube at points A and B. Showthe shear stress on volume elements located at these points.Given: ao := 50mm bo := 50mmta := 3mm tb := 5mmT := 50N⋅mSolution:Section properties :a := ao − ta b := bo − tbAm := a⋅b Am = 2115mm2Average shear stress:τA.avgT2ta⋅Am:= τA.avg = 3.94MPa AnsτB.avgT2tb⋅Am:= τB.avg = 2.36MPa Ans 430. Problem 5-103The tube is made of plastic, is 5 mm thick, and has the mean dimensions shown. Determine theaverage shear stress at points A and B if it is subjected to the torque of T = 5 N·m. Show the shearstress on volume elements located at these points.Given: a := 80mm b := 110mm t := 5mmc1 := 40mm c2 := 30mmT := 5N⋅mSolution:Section properties :c := c12 + c22 c = 50mmAm a⋅b12:= + ⋅a⋅c2 Am = 10000mm2Average shear stress:τA.avgT2t⋅Am:= τA.avg = 0.05MPa AnsτB.avgT2t⋅Am:= τB.avg = 0.05MPa Ans 431. Problem 5-104The steel tube has an elliptical cross section of mean dimensions shown and a constant thickness oft = 5 mm. If the allowable shear stress is τallow = 56 MPa, and the tube is to resist a torque of T = 375N·m, determine the necessary dimension b. The mean area for the ellipse is Am = π b (0.5b).Given: T := 375N⋅m t := 5mmτallow := 56MPaSolution:Section properties :Am = π⋅b⋅ (0.5⋅b)Average shear stress:τavgT2t⋅Am=τavgT2t⋅⎡⎣π⋅b⋅ (0.5⋅b)⎤⎦=bTt⋅π⋅τallow:=b = 20.65mm Ans 432. Problem 5-105The tube is made of plastic, is 5 mm thick, and has the mean dimensions shown. Determine theaverage shear stress at points A and B if the tube is subjected to the torque of T = 500 N·m. Show theshear stress on volume elements located at these points. Neglect stress concentrations at the corners.Given: a := 40mm b := 100mm t := 5mmc1 := 20mm c2 := 30mmT := 500N⋅mSolution:Section properties :c := c12 + c22 c = 36.06mmAm a⋅b 21⋅ a ⋅ 2c2 ⎛⎜⎝⎞⎠:= + Am = 5200mm2Average shear stress:τA.avgT2t⋅Am:= τA.avg = 9.62MPa AnsτB.avgT2t⋅Am:= τB.avg = 9.62MPa Ans 433. Problem 5-106The steel tube has an elliptical cross section of mean dimensions shown and a constant thickness oft = 5 mm. If the allowable shear stress is τallow = 56 MPa, determine the necessary dimension b neededto resist the torque shown.. The mean area for the ellipse is Am = π b (0.5b).Given: T1 := 75N⋅m τallow := 56MPaT2 := −120N⋅m t := 5mmT3 := 450N⋅mSolution:Section properties :Am = π⋅b⋅ (0.5⋅b)Internal torque :Tmax := T1 + T2 + T3 Tmax = 405N⋅mAverage shear stress:τavgT2t⋅Am=τavgT2t⋅⎡⎣π⋅b⋅ (0.5⋅b)⎤⎦=bTmaxt⋅π⋅ τallow:=b = 21.46mm Ans 434. Problem 5-107The symmetric tube is made from a high-strength steel, having the mean dimensions shown and athickness of 5 mm. If it is subjected to a torque of T = 40 N·m, determine the average shear stressdeveloped at points A and B. Indicate the shear stress on volume elements located at these points.Given: a := 40mm b := 60mm t := 5mmT := 40N⋅mSolution:Section properties :Am := a⋅a + 4(a⋅b) Am = 11200mm2Average shear stress:τA.avgT2t⋅Am:= τA.avg = 0.357MPa AnsτB.avgT2t⋅Am:= τB.avg = 0.357MPa Ans 435. Problem 5-108Due to a fabrication error the inner circle of the tube is eccentric with respect to the outer circle. Bywhat percentage is the torsional strength reduced when the eccentricity e is one-fourth of thedifference in the radii?Given: ea − b4=Solution:For the aligned tube::Section properties :t = a − b Am πa + b2⎛⎜⎝⎞⎠2= ⋅Average shear stress:τA.avgT2t⋅Am= T = τA.avg⋅ (2t⋅Am)T τA.avg⋅ (2)(a − b)⋅πa + b2⎛⎜⎝⎞⎠2= ⋅For the eccentric tube:Section properties : A'm πa + b2⎛⎜⎝⎞⎠2= ⋅t' ae2−e+ b 2⎛⎜⎝⎞⎠= − t' = a − e − b t' aa − b4= − − b t'34= (a − b)Average shear stress:τA.avgT'2t'⋅A'm= T' = τA.avg⋅ (2t'⋅A'm)T' τA.avg⋅ (2)34(a − b)⋅πa + b2⎛⎜⎝⎞⎠2= ⋅Factor of increase in shear stress:FTT'T= FττA.avg⋅ (2)34(a − b)⋅πa + b2⎛⎜⎝ ⎞⎠2⋅= FTτA.avg⋅ (2)(a − b)⋅πa + b2⎛⎜⎝⎞⎠2⋅34:=% reduction in strength : T% := (1 − FT)⋅100T% = 25.00 Ans 436. Problem 5-109For a given average shear stress, determine the factor by which the torque-carrying capacity isincreased if the half-circular sections are reversed from the dashed-line positions to the section shown.The tube is 2.5 mm thick.Given: a := 30mm b := 45mmr := 15mm t := 2.5mmSolution:Section properties :am := a − t rm := r − 0.5tA'm (am)⋅b 2π⋅ 2rm⎛⎜⎝2 ⎞⎠:= − ⋅ A'm = 643.54mm2Am (am)⋅b 2π⋅ 2rm⎛⎜⎝2 ⎞⎠:= + ⋅ Am = 1831.46mm2Average shear stress:τavgT2t⋅Am= T = τavg⋅ (2t⋅Am)T' = τavg⋅ (2t⋅A'm)Hence, the factor of increase is:αTT'= αAmA'm:= α = 2.85 Ans 437. Problem 5-110For a given maximum shear stress, determine the factor by which the torque carrying capacity isincreased if the half-circular section is reversed from the dashed-line position to the section shown.The tube is 2.5 mm thick.Given: a := 30mm b := 45mmr := 15mm t := 2.5mmSolution:Section properties :am := a − t bm := b − 0.5t rm := r − 0.5tA'm (am)⋅bmπ⋅ 2rm⎛⎜⎝2 ⎞⎠:= − A'm = 906.15mm2Am (am)⋅bmπ⋅ 2rm⎛⎜⎝2 ⎞⎠:= + Am = 1500.10mm2Average shear stress:τavgT2t⋅Am= T = τavg⋅ (2t⋅Am)T' = τavg⋅ (2t⋅A'm)Hence, the factor of increase is:αTT'= αAmA'm:= α = 1.66 Ans 438. Problem 5-111The steel used for the shaft has an allowable shear stress of τallow = 8 MPa. If the members areconnected with a fillet weld of radius r = 4 mm, determine the maximum torque T that can be appliedGiven: D := 50mm d := 20mmr := 4mm τallow := 8MPaSolution:Stress Concentration Factor :Dd= 2.50rd= 0.20From Fig. 5-36, K := 1.25Allowable Shear Stress: cd2:= Jπ32:= ⋅d4τallow K0.5T⋅cJ⎛⎜⎝⎞⎠= ⋅Tτallow⋅ J0.5K⋅c:=T = 20.11N⋅m Ans 439. Problem 5-112The shaft is used to transmit 660 W while turning at 450 rpm. Determine the maximum shear stress inthe shaft. The segments are connected together using a fillet weld having a radius of 1.875 mm.Unit used: rpm2π60rads:=Given: D := 25mm d := 12.5mmr := 1.875mm ω := 450rpm P := 660WSolution:TPω:= T = 14.01N⋅mStress Concentration Factor :Dd= 2.00rd= 0.15From Fig. 5-36, K := 1.30Max. shear stress: cd2:= Jπ2d2⎛⎜⎝⎞⎠4:= ⋅τmax KT⋅cJ⎛⎜⎝⎞⎠:= ⋅ τmax = 47.48MPa Ans 440. Problem 5-113The shaft is fixed to the wall at A and is subjected to the torques shown. Determine the maximumshear stress in the shaft. A fillet weld having a radius of 4.5 mm is used to connect the shafts at B.Given: D := 60mm d := 30mm r := 4.5mmTC := 250N⋅m TD := −300N⋅mTE := 800N⋅mSolution:Internal Torque : As shown in the torque diagram.TCD := TCTDB := TC + TDTBE := TC + TDTEA := TC + TD + TEMaximum Shear Stress :For segment CD: cd2:= Jπ32:= ⋅d4τCDTCD⋅ (c)J:=τCD = 47.16MPa Ans(Max.)For segment EA: c'D2:= J'π32:= ⋅D4τEATEA⋅ (c')J':=τEA = 17.68MPa AnsFor the fillet: Stress Concentration Factor :Dd= 2.00rd= 0.15From Fig. 5-36, K := 1.30Max. shear stress:τmax KTDB⋅cJ⎛⎜⎝⎞⎠:= ⋅ τmax = 12.26MPa Ans 441. Problem 5-114The built-up shaft is to be designed to rotate at 720 rpm while transmitting 30 kW of power. Is thispossible? The allowable shear stress is τallow = 12 MPa.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: d := 60mm D := 75mm ω := 720rpmP := 30kW τallow := 12MPaSolution:ω 75.40rads= TPω:=T = 397.89N⋅mMaximum Shear Stress : cd2:= Jπ32:= ⋅d4τallow KT⋅cJ= ⋅ Kτallow⋅ JT⋅c:= K = 1.28Stress Concentration Factor :Dd= 1.25 K = 1.28From Fig. 5-36,rd= 0.133 r := 0.133d r = 7.98mmCheck:D − d2= 7.50mm < r = 7.98mmNo. It si impossible. 442. Problem 5-115The built-up shaft is designed to rotate at 540 rpm. If the radius of the fillet weld connecting the shaftis r = 7.20 mm, and the allowable shear stress for the material is τallow = 55 MPa, determine themaximum power the shaft can transmit.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: d := 60mm D := 75mm ω := 540rpmr := 7.20mm τallow := 55MPaSolution:Stress Concentration Factor :Dd= 1.25rd= 0.12From Fig. 5-36, K := 1.30Maximum Shear Stress : cd2:= Jπ32:= ⋅d4τallow KT⋅cJ= ⋅ Tτallow⋅ JK⋅c:= T = 1.7943 kN⋅mMaximum Power :ω 56.55rads=P := T⋅ωP = 101.47 kW Ans 443. Problem 5-116The steel used for the shaft has an allowable shear stress of τallow = 8 MPa. If the members areconnected together with a fillet weld of radius r = 2.25 mm, determine the maximum torque T that canbe applied.Given: d := 15mm D := 30mmr := 2.25mm τallow := 8MPaSolution:Stress Concentration Factor :Dd= 2.00rd= 0.15From Fig. 5-36, K := 1.30Maximum Shear Stress : cd2:= Jπ32:= ⋅d4τallow K0.5T⋅cJ= ⋅ Tτallow⋅ J0.5K⋅c:=T = 8.156N⋅m Ans 444. Problem 5-117A solid shaft is subjected to the torque T, which causes the material to yield. If the material iselastic-plastic, show that the torque can be expressed in terms of the angle of twist φ of the shaft asT = 4/3 TY (1 - φY3/ 4φ3), where TY and φY are the torque and angle of twist when the material beginsto yield. 445. Problem 5-118A solid shaft having a diameter of 50 mm is made of elastic-plastic material having a yield stress of τY= 112 MPa and shear modulus of G = 84 GPa. Determine the torque required to develop an elastic corein the shaft having a diameter of 25 mm. Also, what is the plastic torque?Given: d := 50mm G := 84GPadY := 25mm τY := 112MPaSolution: cd2:= ρYdY2:=Elastic-plastic torque:Use Eq. 5-26 from the text :Tπ⋅ (τY)6⎛⎝4c3 − ρY3 ⎞⎠:= ⋅ T = 3.551 kN⋅m AnsPlastic torque:Use Eq. 5-27 from the text :2π⋅ (τY)TP:= ⋅ (c3) TP = 3.665 kN⋅m Ans3 446. Problem 5-119Determine the torque needed to twist a short 3-mm-diameter steel wire through several revolutions if itis made from steel assumed to be elastic-plastic and having a yield stress of τY = 80 MPa. Assume thatthe material becomes fully plastic.Given: d := 3mmτY := 80MPaSolution: cd2:=Plastic torque:Use Eq. 5-27 from the text :2π⋅ (τY)TP:= ⋅ (c3)3TP = 0.565N⋅m Ans 447. Problem 5-120A solid shaft has a diameter of 40 mm and length of 1 m. It is made from an elastic-plastic materialhaving a yield stress of τY = 100 MPa. Determine the maximum elastic torque TY and thecorresponding angle of twist. What is the angle of twist if the torque is increased to T = 1.2TY?G = 80 GPa.Given: d := 40mm G := 80GPaL := 1m τY := 100MPaSolution: cd2:= Jπ32:= ⋅d4Maximum Elastic Torque:τYTY⋅cJ= TY(τY)⋅ Jc:=TY = 1.257 kN⋅m AnsAngle of Twist:γYτYG:= γY = 0.00125 radφ(γY)⋅Lc:= φ = 0.0625 radφ = 3.581 deg AnsElastic-plastic torque: T = 1.2TYUse Eq. 5-26 from the text :Tπ⋅ (τY)6⎛⎝4c3 − ρY3 ⎞⎠= ⋅1.2TYπ⋅ (τY)6⎛⎝4c3 − ρY3 ⎞⎠= ⋅ρY34c3 7.2 TY ⋅π⋅ (τY):= −ρY = 14.74mmAngle of Twist:φ'(γY)⋅LρY:= φ' = 0.084826 radφ' = 4.86 deg Ans 448. Problem 5-121The stepped shaft is subjected to a torque T that produces yielding on the surface of the larger diametsegment. Determine the radius of the elastic core produced in the smaller diameter segment. Neglectthe stress concentration at the fillet.Given: do := 60mm ds := 55mmSolution: codo2:= Joπ32:= ⋅ 4docsds2:= Jsπ32:= ⋅ 4dsSet τY := 1MPaMaximum Elastic Torque: For the larger diameter segment.τYTY⋅cJ= TY(τY)⋅ Joco:= TY = 42.41N⋅mElastic-plastic torque: For the smaller diameter segment.2π⋅ (τY)TP3⎛⎝cs3 ⎞⎠:= ⋅ TP = 43.56N⋅m > TYApplying Eq. 5-26 from the text :Tπ⋅ (τY)64cs3 − ρY3 ⎛⎝⎞⎠= ⋅TYπ⋅ (τY)64cs3 − ρY3 ⎛⎝⎞⎠= ⋅ρY34cs3 6⋅TYπ⋅ (τY):= −ρY = 12.98mm Ans 449. Problem 5-122A bar having a circular cross section of 75 mm diameter is subjected to a torque of 12 kN·m. If thematerial is elastic-plastic, with τY = 120 MPa, determine the radius of the elastic core.Given: d := 75mm T := 12kN⋅mτY := 120MPaSolution: cd2:=Elastic-plastic torque:Use Eq. 5-26 from the text :Tπ⋅ (τY)6⎛⎝4c3 − ρY3 ⎞⎠= ⋅ρY34c3 6Tπ⋅ (τY):= −ρY = 27.12mm Ans 450. Problem 5-123A tubular shaft has an inner diameter of 20 mm, an outer diameter of 40 mm, and a length of 1 m. It imade of an elastic perfectly plastic material having a yield stress of τY = 100 MPa. Determine themaximum torque it can transmit.What is the angle of twist of one end with respect to the other end ifthe shear strain on the inner surface of the tube is about to yield? G = 80 GPa.Given: do := 40mm G := 80GPa L := 1mdi := 20mm τY := 100MPaSolution: ρYdi2:=Plastic Torque: co := 0.5do ci := 0.5diTP 2πco(τY)⋅ρ2 ρ⌠⎮⎮⌡:= ⋅ dciTP = 1.466 kN⋅m AnsAngle of Twist:γYτYG:= γY = 0.00125 radφ(γY)⋅LρY:= φ = 0.125 radφ = 7.162 deg Ans 451. Problem 5-124The 2-m-long tube is made from an elastic-plastic material as shown. Determine the applied torque T,which subjects the material of the tube's outer edge to a shearing strain of γmax = 0.008 rad. Whatwould be the permanent angle of twist of the tube when the torque is removed? Sketch the residualstress distribution of the tube.Given: ro := 45mm ri := 40mm L := 2mγY := 0.003 τY := 240MPa γmax := 0.008Solution:Determine if it is fully plastic :φmaxγmax⋅Lro:= φmax = 0.35556 radHowever,φmaxγY⋅LρY= ρYγY⋅Lφmax:=ρY = 16.88mm < ri = 40mmTherefore, the tube is filly plastic.Also, at r = ri :γrriroγmax γr=:= ⋅ γmax γr = 0.00711 > γY = 0.003roriAgain, the tube is filly plastic.Plastic Torque :Tp 2πro(τY)⋅ρ2 ρ⌠⎮⎮⌡:= ⋅ d Tp = 13.63 kN⋅m AnsriAngle of Twist:When the torque is removed: The equal but opposite torque TP is applied.GτYγY:= Jπ2ro4 − ri4 ⎛⎝⎞⎠:= ⋅φ'Tp⋅LG⋅ J:= φ' = 0.14085 radPermanent angle of twist: φr := φmax − φ'φr = 0.21470 rad φr = 12.30 deg AnsShear Stresses :At r = ro : τ'poTp⋅ roJ:= τ'po = 253.5MPaAt r = ri : τ'piTp⋅ riJ:= τ'pi = 225.4MPa 452. Problem 5-125The tube has a length of 2 m and is made of an elastic-plastic material as shown. Determine the torquneeded to just cause the material to become fully plastic. What is the permanent angle of twist of thetube when this torque is removed?Given: ro := 100mm ri := 60mm L := 2mγY := 0.007 τY := 350MPaSolution:At Just Fully Plastic : ρY := riφpγY⋅LρY:= φp = 0.23333 radPlastic Torque :Tp 2πro(τY)⋅ρ2 ρ⌠⎮⎮⌡:= ⋅ d Tp = 574.70 kN⋅m AnsriAngle of Twist:When the torque is removed: The equal but opposite torque TP is applied.GτYγY:= Jπ2ro4 − ri4 ⎛⎝⎞⎠:= ⋅φ'Tp⋅LG⋅ J:= φ' = 0.16814 radPermanent angle of twist: φr := φp − φ'φr = 0.06520 radφr = 3.74 deg Ans 453. cProblem 5-ρ126The shaft is made from a strain-hardening material having a τ -γ diagram as shown. Determine thetorque T that must be applied to the shaft in order to create an elastic core in the shaft having a radiusof = 12.5 mm.Given: r := 15mm ρc := 12.5mmγa := 0.005 τa := 70MPaγb := 0.010 τb := 105MPaSolution: c := r ρa := ρcFor linear strain variations against ρ :γργaρa= γγaρa⎛⎜⎝⎞⎠= ⋅ρPath 1: τ1τaγa⎛⎜⎝⎞⎠= ⋅ γ τ1τaρa⎛⎜⎝⎞⎠= ⋅ρPath 2:τ2 − τaγ − γaτb − τaγb − γa= τ2τb − τaγb − γa⎛⎜⎝⎞⎠= ⋅ (γ ⋅ −γa) + τaτ2τb − τaγb − γa⎛⎜⎝⎞⎠ρ− 1 ρa⎛⎜⎝⎞⎠= ⋅ ⋅ γa + τa= ⋅ dT 2πcτ⋅ρ2 ρ⌠⎮⌡0= + dT 2πρaτ1⋅ρ2 ρ⌠⎮⌡⋅ d 2π0cτ2⋅ρ2 ρ⌠⎮⎮⌡ρa:= + ⋅ dT 2πτaρa⎛⎜⎝⎞⎠⋅ρaρ3 ρ ⌠⎮⌡⋅ d 2π⋅ γa0τb − τaγb − γa⎛⎜⎝⎞⎠⋅cρaρρ− 1 ρa⎛⎜⎝⎞⎠⋅ρ2⌠⎮⎮⎮⌡+ d 2π⋅τacρ2 ρ ⌠⎮⌡ρaT = 434.27N⋅m Ans 454. Problem 5-127The 2-m-long tube is made of an elastic perfectly plastic material as shown. Determine the appliedtorque T that subjects the material at the tube's outer edge to a shear strain of γmax = 0.006 rad. Whatwould be the permanent angle of twist of the tube when this torque is removed? Sketch the residualstress distribution in the tube.Given: ro := 35mm ri := 30mm L := 2mγY := 0.003 τY := 210MPa γmax := 0.006Solution:At Fully Plastic : ρY := riφpγmax⋅Lro:= φp = 0.34286 radAlso, at r = ri :γrriroγmax γr=:= ⋅ γmax γr = 0.00514rori> γY = 0.003 Confirm that tube is filly plastic.Plastic Torque :Tp 2πro(τY)⋅ρ2 ρ⌠⎮⎮⌡:= ⋅ d Tp = 6.98 kN⋅m AnsriAngle of Twist:When the torque is removed: The equal but opposite torque TP is applied.GτYγY:= Jπ2ro4 − ri4 ⎛⎝⎞⎠:= ⋅φ'Tp⋅LG⋅ J:= φ' = 0.18389 radPermanent angle of twist: φr := φp − φ'φr = 0.15897 radφr = 9.11 deg AnsResidual Shear Stresses :At r = ro : τ'poTp⋅ roJ:= τ'po = 225.3MPaτro := −τY + τ'po τro = 15.3MPaAt r = ri : τ'piTp⋅ riJ:= τ'pi = 193.1MPaτri := −τY + τ'pi τri = −16.9MPa 455. Problem 5-128The shear stressstrain diagram for a solid 50-mm diameter shaft can be approximated as shown in thefigure. Determine the torque required to cause a maximum shear stress in the shaft of 125 MPa. If theshaft is 3 m long, what is the corresponding angle of twist?Given: do := 50mm L := 3mγ1 := 0.0025 τ1 := 50MPaγmax := 0.010 τmax := 125MPaSolution: ro := 0.5⋅doτ-ρ Function :At r = ρ1 :γmaxroγ1ρ1= ρ1γ1γmax:= ⋅ ro ρ1 = 6.25mmFor the region 0 < r < ρ1:k1τ1 − 0ρ1 − 0:= k1 8.00MPamm=k1τ − 0ρ − 0= τ = k1⋅ρFor the region ρ1 < r < r0:k2τmax − τ1ro − ρ1:= k2 4.00MPamm=k2τ' − τ1ρ' − ρ1= τ' = τ1 + k2⋅ (ρ' − ρ1)Plastic Torque :Tp 2πρ1(τ') ρ' ρ' ⋅ 2⌠⎮⌡= ⋅ d + 2π⋅ d0ro(τ)⋅ρ2 ρ⌠⎮⌡ρ1Tp 2πρ1(k1⋅ρ)⋅ρ2 ρ⌠⎮⌡⋅ d 2π0roρ1ρ' τ1 k2 ρ' ρ1 − ( ) ⋅ + ⎡⎣⎤⎦ρ'⋅ 2⌠⎮⎮⌡:= + ⋅ dTp = 3.27 kN⋅m AnsAngle of Twist:φmaxγmax⋅Lro:= φmax = 1.20000 radφmax = 68.75 deg Ans 456. Problem 5-129The shaft consists of two sections that are rigidly connected. If the material is elastic perfectly plasticas shown, determine the largest torque T that can be applied to the shaft. Also, draw the shear-stressdistribution over a radial line for each section. Neglect the effect of stress concentration.Given: r1 := 20mm r2 := 25mmγY := 0.002 τY := 70MPaSolution:Plastic Torque : For the smaller-diameter segmentc := r1Tp 2πc(τY)⋅ρ2 ρ⌠⎮⌡:= ⋅ d0Tp = 1.173 kN⋅m AnsMax. Shear Stress : For the bigger-diameter segmentc := r2 Jπ2:= ⋅ 4r2τmaxTp⋅ (c)J:=τmax = 47.79MPa Ans 457. Problem 5-130The shaft is made of an elastic-perfectly plastic material as shown. Plot the shear-stress distributionacting along a radial line if it is subjected to a torque of T = 2 kN·m. What is the residual stressdistribution in the shaft when the torque is removed?Given: ro := 20mm T := 2kN⋅mγY := 0.001875 τY := 150MPaSolution: c := ro Jπ2:= ⋅ 4roMaximum Elastic Torque:τYTY⋅cJ= TY(τY)⋅ Jc:=TY = 1.885 kN⋅m < T = 2 kNmPlastic Torque:TP2π⋅τY3:= ⋅ (c3) TP = 2.513 kN⋅m > T = 2 kNmTherefore, it is elastic-plastic.Elastic-plastic Torque:Applying Eq. 5-26 from the text :Tπ⋅ (τY)6⎛⎝4c3 − ρY3 ⎞⎠= ⋅ρY34c3 6 T ⋅π⋅ (τY):= −ρY = 18.70mmResidual Shear Stresses :When the torque is removed:The equal but opposite torque T is applied.At r = ro : τ'poT⋅ roJ:= τ'po = 159.15MPaτro := −τY + τ'po τro = 9.15MPaAt r = ρY : τ'piT⋅ρYJ:= τ'pi = 148.78MPaτri := −τY + τ'pi τri = −1.22MPa 458. Problem 5-131A 40-mm-diameter shaft is made from an elasticplastic material as shown. Determine the radius of itselastic core if it is subjected to a torque of T = 300 N·m. If the shaft is 250 mm long, determine theangle of twist.Given: d := 40mm L := 250mm T := 300N⋅mγY := 0.006 τY := 21MPaSolution: cd2:=Elastic-plastic torque:Use Eq. 5-26 from the text :Tπ⋅ (τY)6⎛⎝4c3 − ρY3 ⎞⎠= ⋅ρY34c3 6T:= − ρY = 16.77mm Ansπ⋅ (τY)Angle of Twist:φ(γY)⋅LρY:= φ = 0.089445 radφ = 5.1248 deg Ans 459. Problem 5-132A torque is applied to the shaft having a radius of 100 mm. If the material obeys a shear stress-strainrelation of τ = 20γ 1/3 MPa, determine the torque that must be applied to the shaft so that the maximumshear strain becomes 0.005 rad.Given: ro := 100mmγmax := 0.005 τ 20 = 3 γ⋅unitSolution: c := roτ-ρ Function : unit := MPaγρc= ⋅ γmax τ 20 = 3 γ⋅unitτ 203 ρc= ⋅ γmax⋅unitUltimate Torque := ⋅ dT 2πcτ⋅ρ2 ρ⌠⎮⌡0T 2π⋅unitc20 ρ03 ρc⋅ γmax⋅ (ρ2)⌠⎮⎮⌡:= ⋅ dT = 6.446 kN⋅m Ans 460. Problem 5-133The shaft is made of an elastic-perfectly plastic material as shown. Determine the torque that the shafcan transmit if the allowable angle of twist is 0.375 rad. Also, determine the permanent angle of twistonce the torque is removed. The shaft is 2-m-long.Given: ro := 20mm φallow := 0.375radγY := 0.001875 τY := 150MPa L := 2mSolution: c := roAngle of Twist:γmaxφallow⋅c:= γmax = 0.00375 radLγmaxcγYρY= ρYγYγmax:= ⋅c ρY = 10.00mmElastic-plastic Torque:Applying Eq. 5-26 from the text :Tπ⋅ (τY)6⎛⎝4c3 − ρY3 ⎞⎠:= ⋅T = 2.435 kN⋅m AnsPermanent Angle of Twist:When the torque is removed: The equal but opposite torque T is applied.GτYγY:= Jπ2:= ⋅ 4roφ'T⋅LG⋅ J:= φ' = 0.24219 radPermanent angle of twist: φr := φallow − φ'φr = 0.13281 radφr = 7.61 deg Ans 461. Problem 5-134Consider a thin-walled tube of mean radius r and thickness t. Show that the maximum shear stress inthe tube due to an applied torque T approaches the average shear stress computed from Eq. 5-18 as r/tapproaches infinity. 462. Problem 5-135The 304 stainless steel shaft is 3 m long and has an outer diameter of 60 mm. When it is rotating at 60rad/s, it transmits 30 kW of power from the engine E to the generator G. Determine the smallestthickness of the shaft if the allowable shear stress is τallow = 150 MPa and the shaft is restricted not totwist more than 0.08 rad.Given: do := 60mm L := 3m ω 60rads:= ⋅P := 30kW G := 75GPaτallow := 150MPa φallow := 0.08radSolution:TPω:= T = 500N⋅mAllowable Shear Stress : Assume failure due to shear strss.rodo2:= c := ro Jπ2ro4 − ri4 ⎛⎝⎞⎠= ⋅τallowT⋅cJ= JT⋅cτallow=π2ro4 − ri4 ⎛⎝⎞⎠⋅T⋅ roτallow= ri4ro4 2T⋅ ro:= − ri = 29.392mmπ⋅τallowAngle of Twist : Assume failure due to angle of twist limitation.φT⋅LG⋅ J= J'T⋅LG⋅φallow:=π2ro4 − r'i4 ⎛⎝⎞⎠⋅T⋅LG⋅φallow= r'i4ro4 2T⋅L:= − r'i = 28.403mmπ⋅G⋅φallowChoose the smallest value of ri : ri := min(ri , r'i) ri = 28.403mmt := ro − rit = 1.597mm Ans 463. Problem 5-136The 304 stainless solid steel shaft is 3 m long and has a diameter of 50 mm. It is required to transmit40 kW of power from the engine E to the generator G. Determine the smallest angular velocity theshaft can have if it is restricted not to twist more than 1.5°.Given: do := 50mm L := 3mP := 40kW G := 75GPaφallow := 1.5degSolution:Angle of Twist :rodo2:= c := ro Jπ2:= ⋅ 4roφT⋅LG⋅ J= TG⋅ JL:= ⋅φallowT = 401.60N⋅mAngular Velocity:TPω= ωPT:=ω 99.6rads= Ans 464. Problem 5-137The drilling pipe on an oil rig is made from steel pipe having an outside diameter of 112 mm and athickness of 6 mm. If the pipe is turning at 650 rev/min while being powered by a 12-kW motor,determine the maximum shear stress in the pipe.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: do := 112mm t := 6mm ω := 650⋅ rpmP := 12kW τallow := 70MPaSolution: ω 68.07rads=TPω:= T = 176.29N⋅mMax. stress :cdo2:= di := do − 2t Jπ2⎛⎜⎝⎞⎠do2⎛⎜⎝⎞⎠4 di2⎛⎜⎝⎞⎠4−⎡⎢⎣⎤⎥⎦:= ⋅τmaxT⋅cJ:=τmax = 1.75MPa Ans 465. Problem 5-138The tapered shaft is made from 2014-T6 aluminum alloy, and has a radius which can be described bythe function r = 0.02 (1 + x3/2) m, where x is in meters. Determine the angle of twist of its end A if it isubjected to a torque of 450 N·m.Given: T := 450N⋅m L := 4mG := 27GPa r = 0.02(1 + x3)⋅unitSolution: unit := mAngle of Twist :J= ⋅ r4 r = 0.02(1 + x3)⋅unitπ2φL0xTG⋅ J⌠⎮⎮⌡= dφ unitLm0x2TG π ⋅ 0.02 1 x3 + ( ) unit ⋅ ⎡⎣⎤⎦4⋅⌠⎮⎮⎮⌡:= ⋅ dφ = 0.02771 radφ = 1.588 deg Ans 466. Problem 5-139The engine of the helicopter is delivering 660 kW to the rotor shaft AB when the blade is rotating at1500 rev/min. Determine to the nearest multiples of 5mm the diameter of the shaft AB if the allowablshear stress is τallow = 56 MPa and the vibrations limit the angle of twist of the shaft to 0.05 rad. Theshaft is 0.6 m long and made of L2 tool steel.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: L := 600mm ω := 1500⋅ rpmP := 660kW τallow := 56MPaG := 75GPa φallow := 0.05radSolution: ω 157.08rads=TPω:= T = 4.202 kN⋅mAllowable shear stress : Assume failure due to shear stresscd2= Jπ2d2⎛⎜⎝⎞⎠4= ⋅ τallowT⋅cJ=Thus, d 22π⎛⎜⎝⎞⎠Tτallow⎛⎜⎝⎞⎠⋅ ⎡⎢⎣⎤⎥⎦13:=d = 72.57mmAngle of Twist : Assume failure due to angle of twist limitationJπ2d2⎛⎜⎝⎞⎠4= ⋅ φT⋅LG⋅ J=Thus, d 2⎛⎜⎝⎞2π⎠T⋅LG⋅φallow⎛⎜⎝⎞⎠⋅ ⎡⎢⎣⎤⎥⎦0.25:=d = 51.15mmShear stress failure controls the design. Hence,Use d = 75mm Ans 467. Problem 5-140The engine of the helicopter is delivering 660 kW to the rotor shaft AB when the blade is rotating at1500 rev/min. Determine to the nearest multiples of 5mm the diameter of the shaft AB if the allowablshear stress is τallow = 75 MPa and the vibrations limit the angle of twist of the shaft to 0.03 rad. Theshaft is 0.6 m long and made of L2 tool steel.Unit used: rpm2π60⎛⎜⎝⎞⎠rads:=Given: L := 600mm ω := 1500⋅ rpmP := 660kW τallow := 75MPaG := 75GPa φallow := 0.03radSolution: ω 157.08rads=TPω:= T = 4.202 kN⋅mAllowable shear stress : Assume failure due to shear stresscd2= Jπ2d2⎛⎜⎝⎞⎠4= ⋅ τallowT⋅cJ=Thus, d 22π⎛⎜⎝⎞⎠Tτallow⎛⎜⎝⎞⎠⋅ ⎡⎢⎣⎤⎥⎦13:=d = 65.83mmAngle of Twist : Assume failure due to angle of twist limitationJπ2d2⎛⎜⎝⎞⎠4= ⋅ φT⋅LG⋅ J=Thus, d 2⋅ ⎡⎢⎣⎤⎛⎜⎝⎞2π⎠T⋅LG⋅φallow⎛⎜⎝⎞⎠⎥⎦0.25:=d = 58.12mmShear stress failure controls the design. Hence,Use d = 70mm Ans 468. Problem 5-141The material of which each of three shafts is made has a yield stress of τY and a shear modulus of G.Determine which shaft geometry will resist the largest torque without yielding. What percentage of thtorque can be carried by the other two shafts? Assume that each shaft is made of the same amount ofmaterial and that it has the same cross-sectional area A.Given: AΟ = A Asq = A AΔ = ASolution:For circular shaft::AΟ = πc2 cAπ=Jπ2= ⋅c4 JA22π=τmaxT⋅cJ= τmax2⋅ π TA A= TA A2 π= ⋅ τYαΟ12 π:= αΟ = 0.2821For square shaft:Asq = a2 a = Aτmax4.81Ta3= τmax4.81TA⋅ A= TA A4.81= ⋅ τYαsq14.81:= αsq = 0.2079For triangular shaft: θ := 60degAΔa2= ⋅a⋅ sin(θ) a2 A4 3=τmax20Ta3= τmax5⋅4 27T2A⋅ A= T2A A5⋅4 27= ⋅τYαΔ25⋅4 27:= αΔ = 0.1755The circular shaft will carry the largest torque. AnsFor square shaft: %sqαsqαΟ:= ⋅100 %sq = 73.7 AnsFor triangular shaft: %ΔαΔαΟ:= ⋅100 %Δ = 62.2 Ans 469. Problem 5-142The A-36 steel circular tube is subjected to a torque of 10 kN·m. Determine the shear stress at themean radius ρ = 60 mm and compute the angle of twist of the tube if it is 4 m long and fixed at its farend. Solve the problem using Eqs. 5-7 and 5-15 and by using Eqs. 5-18 and 5-20.Given: ρ := 60mm t := 5mm L := 4mT := 10kN⋅m G := 75GPaSolution:Section properties :ro := ρ + 0.5t ri := ρ − 0.5tΣds := 2π ρ Σds = 376.99mmAm := π ρ2 Am = 11309.73mm2Jπ2:= ⋅ J = 6797621.10mm4ro4 − ri4 ⎛⎝⎞⎠Shear Stress:Applying Eq. 5-7,c := ρ τT⋅cJ:= τ = 88.27MPa AnsApplying Eq. 5-18,τavgT2⋅ tAm:= τavg = 88.42MPa AnsAngle of Twist :Applying Eq. 5-20,φ'T⋅LJ⋅G:= φ' = 0.07846 rad φ' = 4.495 deg AnsApplying Eq. 5-20,φT⋅L4Am2⋅Gs1t⌠⎮⎮⌡= ⋅ dφT⋅L4Am2⋅GΣdst⎛⎜⎝⎞⎠:= ⋅ φ = 0.07860 rad φ = 4.503 deg Ans 470. Problem 5-143The aluminum tube has a thickness of 5 mm and the outer cross-sectional dimensions shown.Determine the maximum average shear stress in the tube. If the tube has a length of 5 m, determine thangle of twist. Gal = 28 GPa.Given: ao := 150mm bo := 100mmL := 4m t := 5mmTA := 280N⋅m TB := 135N⋅mLAB := 2m LBC := 3mG := 28GPaSolution:Section properties : S = Σdsa := ao − t b := bo − tAm := a⋅b Am = 13775mm2S := 2a + 2⋅b S = 480mmMaximum Average shear stress:TAB := TATBC := TA − TBTmax := max(TAB, TBC) Tmax = 280N⋅mτavg_maxTmax2t⋅Am:=τavg_max = 2.03MPa AnsAngle of Twist :φT⋅L4Am2⋅Gs1t⌠⎮⎮⌡= ⋅ dφTAB⋅LAB + TBC⋅LBC4Am2⋅GSt⎛⎜⎝⎞⎠:= ⋅φ = 0.004495 radφ = 0.258 deg Ans 471. Problem 6-1Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only verticalreactions on the shaft.Given: a := 250mm b := 800mmF := 24kNSolution:GivenEquilibrium :+ ΣFy=0; A + B − F = 0ΣΜA=0; −F⋅a − B⋅b = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠31.50−7.50⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + bV1(x1) −F1kN:= ⋅ V2(x2) (−F + A)1kN:= ⋅M1(x1) −F⋅x1:= M2 x2 ( ) F − x2 ( ) ⋅ A x2 a − ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅0 0.5 12002040Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 0.5 105Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 472. Problem 6-2The load binder is used to support a load. If the force applied to the handle is 250 N, determine thetensions T1 and T2 in each end of the chain and then draw the shear and moment diagrams for the armABC.Given: a := 300mm b := 75mm F := 250NSolution: GivenEquilibrium :+ ΣFy=0; T1 − T2 − F = 0ΣΜB=0; −F⋅a + T2⋅b = 0Guess T1 := 1N T2 := 1NT1T2⎛⎜⎜⎝⎞⎠:= Find(T1 , T2)T1T2⎛⎜⎜⎝⎞⎠1.251.00⎛⎜⎝⎞⎠= kN Ansx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + bV1(x1) −F1kN:= ⋅ V2(x2) (−F + T1) 1kN:= ⋅M1(x1) −F⋅x1:= M2 x2 ( ) F − x2 ( ) ⋅ T1 x2 a − ( ) ⋅ + ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅0 0.2101Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 0.2050100Distane (m)Moment (N-m)M1(x1)M2(x2)x1, x2 473. Problem 6-3Draw the shear and moment diagrams for the shaft. The bearings at A and D exert only verticalreactions on the shaft. The loading is applied to the pulleys at B and C and E.Given: a := 350mm b := 500mmc := 375mm d := 300mmB := 400N C := 550NSolution: E := 175NEquilibrium : Given+ ΣFy=0; A + D − B − C − E = 0ΣΜD=0; A⋅ (a + b + c) − B⋅ (b + c) − C⋅c + E⋅d = 0Guess A := 1N D := 1NAD⎛⎜⎝⎞⎠:= Find(A,D)AD⎛⎜⎝⎞⎠411.22713.78⎛⎜⎝⎞⎠= Nx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cx4 := a + b + c , 1.01⋅ (a + b + c) .. a + b + c + dV1(x1) A1N:= ⋅ V2(x2) (A − B)1N:= ⋅ V3(x3) (A − B − C)1N:= ⋅V4(x4) (A − B − C + D)1N:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ B x2 a − ( ) ⋅ − ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ B x3 a − ( ) ⋅ − C x3 a − b − ( ) ⋅ − ⎡⎣⎤⎦1N⋅m:= ⋅ 474. M4 x4 ( ) A x4 ( ) ⋅ B x4 a − ( ) ⋅ − C x4 a − b − ( ) ⋅ − D x4 a − b − c − ( ) ⋅ + ⎡⎣⎤⎦1N⋅m:= ⋅0 0.2 0.4 0.6 0.8 1 1.2 1.450005001000Distance (m)Shear (N)V1(x1)V2(x2)V3(x3)V4(x4)x1, x2, x3, x40 0.2 0.4 0.6 0.8 1 1.2 1.42001000100Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)M4(x4)x1, x2, x3, x4 475. Problem 6-4Draw the shear and moment diagrams for the beam.Given: a := 1m F := 10kNSolution:Equilibrium : Given+ ΣFy=0; A − 4F + B = 0ΣΜB=0; A⋅ (5a) − F⋅ (4a + 3a + 2a + a) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠2020⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 2a x3 := 2a , 1.01⋅ (2a) .. 3ax4 := 3a , 1.01⋅ (3a) .. 4a x5 := 4a , 1.01⋅ (4a) .. 5aV1(x1) A1kN:= ⋅ V2(x2) (A − F)1kN:= ⋅ V3(x3) (A − 2F)1kN:= ⋅V4(x4) (A − 3F)1kN:= ⋅ V5(x5) (A − 4F)1kN:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ F x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ F x3 a − ( ) ⋅ − F x3 2a − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅M4 x4 ( ) A x4 ( ) ⋅ F x4 a − ( ) ⋅ − F x4 2a − ( ) ⋅ − F x4 3a − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅ 476. M5 x5 ( ) A x5 ( ) ⋅ F x5 a − ( ) ⋅ − F x5 2a − ( ) ⋅ − F x5 3a − ( ) ⋅ − F x5 4a − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 3 4 520020Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)V4(x4)V5(x5)x1, x2, x3, x4, x50 1 2 3 4 53020100Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)M4(x4)M5(x5)x1, x2, x3, x4, x5 477. Problem 6-5A reinforced concrete pier is used to support the stringers for a bridge deck. Draw the shear andmoment diagrams for the pier when it is subjected to the stringer loads shown. Assume the columns aA and B exert only vertical reactions on the pier.Given: a := 1m b := 1.5m F1 := 60kN F2 := 35kNSolution: L := 4a + 2⋅bEquilibrium : Given+ ΣFy=0; A + B − 2F1 − 3F2 = 0ΣΜB=0; A⋅ (2a + 2⋅b) − F1⋅ (L − 2a) − F2⋅ (3⋅b + 3a) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠112.5112.5⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 2a x3 := 2a , 1.01⋅ (2a) .. (2a + b)x4 := (2a + b) , 1.01⋅ (2a + b) .. (2a + 2⋅b)x5 := (2a + 2⋅b) , 1.01⋅ (2a + 2⋅b) .. (3a + 2⋅b)x6 := (3a + 2⋅b) , 1.01⋅ (3a + 2⋅b) .. (4a + 2⋅b)V1(x1) −F11kN:= ⋅ V2(x2) (A − F1) 1:= ⋅ V3(x3) (A − F1 − F2) 1kNkN:= ⋅V4(x4) (A − F1 − 2F2) 1:= ⋅ V5(x5) (A − F1 − 3F2) 1kNkN:= ⋅V6(x6) (A − F1 − 3F2 + B) 1kN:= ⋅M1(x1) −F1⋅x1:= M2 x2 ( ) A x2 a − ( ) ⋅ F1 x2 ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) A x3 a − ( ) ⋅ F1 x3 ( ) ⋅ − F2 x3 2a − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅M4 x4 ( ) A x4 a − ( ) ⋅ F1 x4 ( ) ⋅ − F2 2 x4 2 a ⋅ − ( ) b − ⎡⎣⎤⎦⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅M5 x5 ( ) A x5 a − ( ) ⋅ F1 x5 ( ) ⋅ − F2 3 x5 2 a ⋅ − ( ) 3 b ⋅ − ⎡⎣⎤⎦⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅M6 x6 ( ) A x6 a − ( ) ⋅ F1 x6 ( ) ⋅ − F2 3 x6 2 a ⋅ − ( ) ⋅ 3 b ⋅ − ⎡⎣− ⋅ + B ⋅ ⎤⎦⎡⎣x6 − ( L − a ⎡⎣) ⎤⎦⎤⎦1kN⋅:= ⋅ 478. 0 1 2 3 4 5 6 750050Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)V4(x4)V5(x5)V6(x6)x1, x2, x3, x4, x5, x60 1 2 3 4 5 6 7200204060Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)M4(x4)M5(x5)M6(x6)x1, x2, x3, x4, x5, x6 479. Problem 6-6Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only verticalreactions on the shaft. Also, express the shear and moment in the shaft as a function of x within theregion 125 mm < x < 725 mm.Given: a := 125mm b := 600mm c := 75mmF1 := 0.8kN F2 := 1.5kNSolution: L := a + b + cEquilibrium : Given+ ΣFy=0; A − F1 − F2 + B = 0ΣΜB=0; A⋅ (L) − F1⋅ (b + c) − F2⋅ (c) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠0.81561.4844⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cV1(x1) A1kN:= ⋅ V2(x2) (A − F1) 1:= ⋅ V3(x3) (A − F1 − F2) 1kNkN:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ F1 x2 a − ( ) ⋅ − ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ F1 x3 a − ( ) ⋅ − F2 x3 a − b − ( ) ⋅ − ⎡⎣⎤⎦1N⋅m:= ⋅0 0.1 0.2 0.3 0.4 0.5 0.6 0.71012Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3 480. 0 0.2 0.4 0.6150100500Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 481. Problem 6-7Draw the shear and moment diagrams for the shaft and determine the shear and moment throughoutthe shaft as a function of x. The bearings at A and B exert only vertical reactions on the shaft.Given: a := 0.9m b := 0.6mc := 0.3m d := 0.15mF1 := 4kN F2 := 2.5kNSolution:Equilibrium : Given+ ΣFy=0; A − F1 + B − F2 = 0ΣΜB=0 ;A⋅ (a + b) − F1⋅ (b) + F2⋅ (c + d) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠0.855.65⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cV1(x1) := A⋅ V2(x2) (F1) 1A − 1kN:= ⋅ V3(x3) (A − F1 + B) 1kNkN:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ F1 x2 a − ( ) ⋅ − ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ F1 x3 a − ( ) ⋅ − B x3 a − b − ( ) ⋅ + ⎡⎣⎤⎦1N⋅m:= ⋅0 0.5 1 1.542024Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3 482. 0 0.5 1 1.5100050005001000Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 483. Problem 6-8Draw the shear and moment diagrams for the pipe. The end screw is subjected to a horizontal force of5 kN. Hint: The reactions at the pin C must be replaced by equivalent loadings at point B on the axisthe pipe.Given: a := 400mm h := 80mm F := 5kNSolution: GivenEquilibrium :+ ΣFy=0; A + C = 0ΣΜC=0; A⋅a + F⋅h = 0Guess A := 1N C := 1NAC⎛⎜⎝⎞⎠:= Find(A, C)AC⎛⎜⎝⎞⎠−1.001.00⎛⎜⎝⎞⎠= kN Ansx1 := 0 , 0.01⋅a .. aV1(x1) A1kN:= ⋅M1(x1) A⋅x1N⋅m:=0 0.2 0.41012Distance (m)Shear (kN)V1(x1)x10 0.2 00200400600Distane (m)Moment (N-m)M1(x1)x1 484. Problem 6-9Draw the shear and moment diagrams for the beam. Hint: The 100-kN load must be replaced byequivalent loadings at point C on the axis of the beam.Given: a := 1m b := 1mc := 1m d := 0.25mF1 := 75kN F2 := 100kNSolution:Equilibrium : Given+ ΣFy=0; A − F1 + B = 0ΣΜC=0; A⋅ (a + b + c) − F1⋅ (b + c) − F2⋅ (d) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠58.3316.67⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cV1(x1) := A⋅ V2(x2) (F1) 1A − 1kN:= ⋅ V3(x3) (A − F1) 1kNkN:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ F1 x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ F1 x3 a − ( ) ⋅ − F2 d ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅Distance (m) Shear (kN)0 0.5 1 1.5 2 2.5 3500V1(x1)V2(x2)V3(x3)x1, x2, x3 485. 0 0.5 1 1.5 2 2.5 36040200Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 486. Problem 6-10The engine crane is used to support the engine, which has a weight of 6 kN. Draw the shear andmoment diagrams of the boom ABC when it is in the horizontal position shown.Given: a := 0.9m b := 1.5mc := 1.2m W := 6kNSolution: d := a2 + c2vcd:= had:=Equilibrium : Given+ΣFy=0; −Ay + B⋅v − W = 0ΣΜA=0; (−B⋅v)⋅a + W⋅ (a + b) = 0+ ΣFx=0; Ax − B⋅h = 0Guess Ax := 1kN Ay := 1kN B := 1kNAxAyB⎛⎜⎜⎜⎝⎞⎟⎠:= Find(Ax ,Ay , B)AxAyB⎛⎜⎜⎜⎝⎞⎟⎠121020⎛⎜⎜⎝⎞= kN⎠x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + bV1(x1) −Ay1kN:= ⋅ V2(x2) (−Ay + B⋅v) 1kN:= ⋅M1(x1) −Ay⋅x1:= M2 x2 ( ) Ay − x2 ( ) ⋅ B v ⋅ ( ) x2a − ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅0 1 210010Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 1 20510Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 487. Problem 6-11Draw the shear and moment diagrams for the compound beam. It is supported by a smooth plate at Awhich slides within the groove and so it cannot support a vertical force, although it can support amoment and axial load.Set: a := 1m P := 1kNSolution:Equilibrium : Given+ ΣFy=0; A − P + C − P = 0ΣΜB=0; P⋅ (2a) − C⋅a = 0Guess A := 1kN C := 1kNAC⎛⎜⎝⎞⎠:= Find(A, C)AC⎛⎜⎝⎞⎠02⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 2a x3 := 2a , 1.01⋅ (2a) .. 3ax4 := 3a , 1.01⋅ (3a) .. 4aV1(x1) A1kN:= ⋅ V2(x2) (A − P)1kN:= ⋅ V3(x3) (A − P)1kN:= ⋅V4(x4) (A − P + C)1kN:= ⋅0 0.5 1 1.5 2 2.5 3 3.5 421012Distance (m)Shear (P kN)V1(x1)V2(x2)V3(x3)V4(x4)x1, x2, x3, x4 488. MA := C⋅ (3a) − P⋅ (4a) − P⋅ (a) MA = 1.00 kN⋅mM1(x1) MA + A⋅x1:= M2 x2 ( ) MA A x2 ( ) ⋅ + P x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) A P − ( ) x32a − ( ) ⋅ ⎡⎣⎤⎦1kN⋅m:= ⋅M4 x4 ( ) A P − ( ) x42a − ( ) ⋅ C x4 3a − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 3 4101Distance (m)Moment (P kN-m)M1(x1)M2(x2)M3(x3)M4(x4)x1, x2, x3, x4 489. Problem 6-12Draw the shear and moment diagrams for the compound beam which is pin connected at B.Given: a := 1m b := 1.5mc := 1m d := 1mF1 := 30kN F2 := 40kNSolution:Equilibrium : Given+ ΣFy=0; −F1 + A − F2 + C = 0ΣΜB=0; −F1⋅ (a + b) + A⋅b = 0Guess A := 1kN C := 1kNAC⎛⎜⎝⎞⎠:= Find(A, C)AC⎛⎜⎝⎞⎠5020⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cx4 := a + b + c , 1.01⋅ (a + b + c) .. a + b + c + d:= ⋅ V2(x2) (−F1 + A) 1V1(x1) −F11kN:= ⋅ V3(x3) (−F1 + A) 1kNkN:= ⋅V4(x4) (−F1 + A − F2) 1kN:= ⋅0 1 2 3 42002040Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)V4(x4)x1, x2, x3, x4 490. M1(x1) −F1⋅x1:= M2 x2 ( ) F1 − x2 ( ) ⋅ A x2 a − ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) F1 − A + ( ) x3 a − b − ( ) ⋅ ⎡⎣⎤⎦1kN⋅m:= ⋅M4 x4 ( ) F1 − A + ( ) x4 a − b − ( ) ⋅ F2 x4 a − b − c − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 3 420100102030Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)M4(x4)x1, x2, x3, x4 491. Problem 6-13Draw the shear and moment diagrams for the beam.Set: a := 1m Mo := 1kN⋅mSolution:Equilibrium : Given+ ΣFy=0; A + B = 0ΣΜB=0; A⋅ (3a) + 2Mo − Mo = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠−0.330.33⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 2a x3 := 2a , 1.01⋅ (2a) .. 3aV1(x1) A1kN:= ⋅ V2(x2) (A)1kN:= ⋅ V3(x3) (A)1kN:= ⋅M1(x1) Mo + A⋅x1:= M2 x2 ( ) 2 Mo ⋅ A x2 ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) 2 Mo ⋅ A x3 ( ) ⋅ + Mo − ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 30.500.5Distance (m)Shear (Mo/a kN)V1(x1)V2(x2)V3(x3)210x1, x2, x3 0 1 2Distance (m)Moment (Mo kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 492. Problem 6-14Consider the general problem of a simply supported beam subjected to n concentrated loads. Write acomputer program that can be used to determine the internal shear and moment at any specifiedlocation x along the beam, and plot the shear and moment diagrams for the beam. Show an applicatioof the program using the values P1 = 2.5 kN, d1 = 1.5 m, P2 = 4 kN, d2 = 4.5 m, L1 = 3 m, L = 4.5 m. 493. Problem 6-15The beam is subjected to the uniformly distributed moment m (Moment/length). Draw the shear andmoment diagrams for the beam.Set: L := 1m mo 1kN⋅mm:=Solution: GivenEquilibrium :+ ΣFy=0; A := 0ΣΜA=0; MA := mo⋅Lx1 := 0 , 0.01⋅L .. LV1(x1) A1kN:= ⋅M1(x1) (MA + A⋅x1 − mo⋅x1) 1kN⋅m:= ⋅0 0.5 11.510.50Distance (m)Shear (kN)V1(x1)x1Distane (m) Moment m*L (kN-m)0 0.5 110.50M1(x1)x1 494. Problem 6-16Draw the shear and moment diagrams for the beam.Given: a := 2.5m w 10kNm:=b := 2.5mSolution:Equilibrium :+ΣF A := w⋅a − w⋅a y=0;A = 0 kNΣΜA=0; MA := (w⋅a)⋅ (0.5a) − (w⋅b)⋅ (a + 0.5b)MA = −62.50 kN⋅mx1x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b( ) ( ) 1V1 x1 A − w ⋅ ⋅ := V2 x2 ( ) A w a ⋅ − w x2 a − ( ) ⋅ + ⎡⎣kN⎤⎦1kN:= ⋅M1(x1) −MA + A⋅x1 0.5w x12 ⋅ − ⎛⎝⎞⎠1kN⋅m:= ⋅M2 x2 ( ) MA − A x2 ⋅ + w a ⋅ ( ) x20.5a − ( ) ⋅ − 0.5w x2 a − ( )2 ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 2 4020Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 2 4500Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 495. Problem 6-17The 75-kg man sits in the center of the boat, which has a uniform width and a weight per linear foot of50 N/m. Determine the maximum bending moment exerted on the boat. Assume that the water exerts auniform distributed load upward on the bottom of the boat.Given: a := 2.5m Mw := 75kgb := 2.5m w 50 Nm:=Solution: W := Mw⋅gEquilibrium :+ ΣFy=0; qW + w⋅ (2a)2a:=q 197.1Nm=x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + bV1 x1 ( ) q w − ( ) x1⋅ ⎡⎣⎤⎦1N:= ⋅V2 x2 ( ) q w − ( ) x2⋅ W − ⎡⎣⎤⎦1N:= ⋅M1 x1 ( ) 0.5 q w − ( ) x12 ⋅ ⎡⎣⎤⎦1N⋅m:= ⋅M2 x2 ( ) 0.5 q w − ( ) x2⋅ 2 W x− ⋅ ( 2 − a) ⎡⎣⎤⎦1N⋅m:= ⋅0 2 45000500Distance (m)Shear (N)V1(x1)V2(x2)x1, x20 2 44002000Distane (m)Moment (N-m)M1(x1)M2(x2)x1, x2 496. Problem 6-18Draw the shear and moment diagrams for the beam. It is supported by a smooth plate at A whichslides within the groove and so it cannot support a vertical force, although it can support a moment anaxial load.Set: L := 1m w 1kNm:= A := 0Solution: GivenEquilibrium :+ ΣFy=0; A + B − w⋅L = 0 B := w⋅LΣΜA=0; MA + A⋅L (w⋅L)L2− ⋅ = 0 MAw⋅L22:=x1 := 0 , 0.01⋅L .. LV1 x1 ( ) A w x1( − ⋅ ) 1kN:= ⋅2 ⋅ − ⎛⎜⎝M1(x1) MA + A⋅x1w2x1⎞⎠1kN⋅m:= ⋅0 0.5 101Distance (m)Shear w*L (kN)V1(x1)x10 0.5 10.60.40.20Distane (m)Moment w*L*L (kN-m)M1(x1)x1 497. Problem 6-19Draw the shear and moment diagrams for the beam.Given: a := 1.5m w 30kNm:=Mo := 45kN⋅mSolution:Equilibrium : Given+ ΣFy=0; −w⋅a + A + B = 0ΣΜA=0; −(w⋅a)⋅ (0.5a) + Mo − B⋅ (2a) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠41.253.75⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 2a x3 := 2a , 1.01⋅ (2a) .. 3aV1(x1) −w⋅x1:= V2(x2) (−w⋅a + A)kN1kN:= ⋅ V3(x3) (−w⋅a + A)1kN:= ⋅⋅ 2M1(x1) −0.5w x1:= M2 x2 ( ) w a ⋅ ( ) − x2 0.5a − ( ) ⋅ A x2 a − ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) w a ⋅ ( ) − x3 0.5a − ( ) ⋅ A x3 a − ( ) ⋅ + Mo + ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 3 402040Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3 498. 0 1 2 3 402040Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 499. Problem 6-20Draw the shear and moment diagrams for the beam, and determine the shear and moment throughoutthe beam as functions of x.Given: a := 2.4m b := 1.2mP1 := 50kNw 30kNm:= P2 := 40kNM2 := 60kN⋅mSolution:Equilibrium :+ΣF A := w⋅a + P1 + P2 A = 162 kN y=0;ΣΜ MA := (w⋅a)⋅ (0.5a) + P1⋅a + P2⋅ (a + b) + M2 A=0;MA = 410.40 kN⋅mx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + bV1 x1 ( ) A w x111:= ( − ⋅ ) ⋅ V2(x2) (A − w⋅a − P1) kNkN:= ⋅M1(x1) −MA + A⋅x1 0.5w x12 ⋅ − ⎛⎝⎞⎠1kN⋅m:= ⋅M2 x2 ( ) MA − A x2 ⋅ + w a ⋅ ( ) x20.5 a ⋅ − ( ) ⋅ − P1 x2 a − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 3200150100500Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 1 2 30200400Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 500. Problem 6-21Draw the shear and moment diagrams for the beam and determine the shear and moment in the beamas functions of x, where 1.2 m < x < 3 m.Given: a := 1.2m Mo := 300N⋅mb := 1.8mw 2.5kNm:= c := 1.2mSolution:Equilibrium : Given+ ΣFy=0; −w⋅b + A + B = 0ΣΜB=0; −Mo + A⋅b − (w⋅b)⋅ (0.5b) + Mo = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠2.252.25⎛⎜⎝⎞⎠= kNx2x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (a + b) x3 := (a + b) , 1.01⋅ (a + b) .. (a + b + c)V1(x1) 0:= V2 ( x2 ) A w kNa − ( ) ⋅ − ⎡⎣⎤⎦1kN:= ⋅ V3(x3) (A − w⋅b + B)1kN:= ⋅M1(x1) −Mo:= M2 x2 ( ) Mo − A x2 a − ( ) ⋅ + 0.5w x2 a − ( )2 ⋅ − ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅M3 x3 ( ) Mo − A x3 a − ( ) ⋅ + w b ⋅ ( ) x3a − 0.5 b ⋅ − ( ) ⋅ − B x3 a − b − ( ) ⋅ + ⎡⎣⎤⎦1N⋅m:= ⋅0 1 2 3 4202Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3 501. 0 1 2 3 48006004002000200Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 502. Problem 6-22Draw the shear and moment diagrams for the compound beam.The three segments are connected bypins at B and E.Given: a := 2m b := 1m F := 3kNLBE := a + 2⋅bw 0.8kNm:= LAB := a + bSolution: L := 3a + 4⋅bEquilibrium :Consider segment AB:ΣΜB=0; A⋅ (a + b) − F⋅ (b) = 0 Aba + b:= ⋅F A = 1.00 kN+ ΣFy=0; A + B − F = 0 B := F − A B = 2.00 kNConsider segment BE:By symmetry, E = B D = C+ ΣFy=0; 2C − 2B − w⋅ (a + 2⋅b) = 0 C Bw2:= + ⋅ (a + 2⋅b) C = 3.60 kND := C D = 3.60 kNE := B E = 2.00 kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (a + b) x3 := (a + b) , 1.01⋅ (a + b) .. (a + 2⋅x4 := (a + 2⋅b) , 1.01⋅ (a + 2⋅b) .. (2a + 2⋅b)x5 := (2a + 2⋅b) , 1.01⋅ (2a + 2⋅b) .. (2a + 3⋅b)x6 := (2a + 3⋅b) , 1.01⋅ (2a + 3⋅b) .. (2a + 4⋅b)x7 := (2a + 4⋅b) , 1.01⋅ (2a + 4⋅b) .. (3a + 4⋅b)V1(x1) A1kN:= ⋅ V2(x2) (A − F)1kN⋅ := V3 x3 ( ) A F − w x3 LAB − ( ) ⋅ − ⎡⎣⎤⎦1kN:= ⋅V4 x4 ( ) A F − C + w x4 LAB − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅ := V5 x5 ( ) A F − C + D + w x5 LAB − ( ) ⋅ − ⎡⎣⎤⎦:= ⋅V6 x6 ( ) A F − C + D + w LBE ( ) ⋅ − ⎡⎣⎤⎦1kN⋅ := V7 x7 ( ) A 2F − C + D + w LBE ( ) ⋅ − ⎡⎣⎤⎦1kN:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ F x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) B − x3 LAB − ( ) ⋅ 0.5w x3 LAB − ( )2 ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅M4 x4 ( ) B − x4 LAB − ( ) ⋅ 0.5w x4 LAB − ( )2 ⋅ − C x4 LAB − b − ( ) ⋅ + ⎡⎣ ⎤⎦1kN⋅m:= ⋅M5 x5 ( ) B − x5 LAB − ( ) ⋅ 0.5w x5 LAB − ( )2 ⋅ − C x5 LAB − b − ( ) ⋅ + D x5 LAB − b − a − ( ) ⋅ + ⎡⎣⎤⎦:= ⋅M6 x6 ( ) E x6 LAB − LBE − ( ) ⋅ ⎡⎣⎤⎦1kN⋅m:= ⋅ 503. M7 x7 ( ) E x7 LAB − LBE − ( ) ⋅ F x7 LAB − LBE − b − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 2 4 6 8 1042024Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)V4(x4)V5(x5)V6(x6)V7(x7)x1, x2, x3, x4, x5, x6, x70 2 4 6 8 10210123Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)M4(x4)M5(x5)M6(x6)M7(x7)x1, x2, x3, x4, x5, x6, x7 504. Problem 6-23Draw the shear and moment diagrams for the beam.Given: a := 1.5m Mo := 30kN⋅mw 30kNm:=Solution:Equilibrium : Given+ ΣFy=0; −w⋅a + A − w⋅a + B = 0ΣΜB=0; Mo − (w⋅a)⋅ (2.5a) + A⋅ (2a) − (w⋅a)⋅ (0.5a) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠57.5032.50⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 2a x3 := 2a , 1.01⋅2a .. 3aV1(x1) −w⋅x1:= V2(x2) [−(w⋅a) + A]kN1kN:= ⋅V3 x3 ( ) w a ⋅ ( ) − A + w x3 2a − ( ) ⋅ − ⎡⎣⎤⎦1kN:= ⋅− ⋅ 2kN⋅mM1(x1) Mo 0.5w x1:= M2 x2 ( ) Mo w a ⋅ ( ) x20.5a − ( ) ⋅ − A x2 a − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅M3 x3 ( ) Mo w a ⋅ ( ) x30.5a − ( ) ⋅ − A x3 a − ( ) ⋅ + 0.5w x3 2a − ( )2 ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅ 505. 0 1 2 3 42002040Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x30 1 2 3 4302010010Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 506. Problem 6-24The beam is bolted or pinned at A and rests on a bearing pad at B that exerts a uniform distributedloading on the beam over its 0.6-m length. Draw the shear and moment diagrams for the beam if itsupports a uniform loading of 30 kN/m.Given: a := 0.3m c := 0.6mb := 2.4m w 30kNm:=Solution:Equilibrium : Given+ ΣFy=0; A − w⋅b + (qB)⋅c = 0ΣΜA=0; (w⋅b)⋅ (a + 0.5b) − (qB⋅c)⋅ (a + b + 0.5c) = 0Guess A := 1kN qB 1kNm:=AqB⎛⎜⎝⎞⎠:= Find(A, qB) A = 36.00 kN qB 60.00kNm=x2x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (a + b) x3 := (a + b) , 1.01⋅ (a + b) .. (a + b + c)V1(x1) A:= V2 ( x2 ) A w kNa − ( ) ⋅ − ⎡⎣⎤⎦1kN:= ⋅V3 x3 ( ) A w b ⋅ − qB x3 a − b − ( ) ⋅ + ⎡⎣⎤⎦1kN:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ 0.5w x2 a − ( )2 ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ w b ⋅ ( ) x3a − 0.5 b ⋅ − ( ) ⋅ − 0.5qB x3 a − b − ( )2 ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.5 1 1.5 2 2.5 3402002040Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3 507. 0 0.5 1 1.5 2 2.5 3403020100Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 508. Problem 6-25Draw the shear and moment diagrams for the beam. The two segments are joined together at B.Given: a := 0.9m P := 40kNb := 1.5mw 50kNm:= c := 2.4mSolution:Equilibrium : Given+ ΣFy=0; A − P − w⋅c + C = 0ΣΜB=0; (w⋅c)⋅ (0.5c) − C⋅ (c) = 0Guess A := 1kN C := 1kNAC⎛⎜⎝⎞⎠:= Find(A, C)AC⎛⎜⎝⎞⎠10060⎛⎜⎝⎞⎠= kNMA := P⋅a − (C − w⋅c)⋅ (a + b) MA = 180 kN⋅mx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (a + b) x3 := (a + b) , 1.01⋅ (a + b) .. (a + b + c)V1(x1) A:= V2(x2) (A − P)kN1kN⋅ := V3 x3 ( ) A P − w x3 a − b − ( ) ⋅ − ⎡⎣⎤⎦1kN:= ⋅M1(x1) −MA + A⋅x1:= M2 x2 ( ) MA − A x2 ⋅ + P x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) MA − A x3 ⋅ + P x3 a − ( ) ⋅ − 0.5w x3 a − b − ( )2 ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 3 410050050Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3 509. 0 1 2 3 41000100200Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 510. Problem 6-26Consider the general problem of a cantilevered beam subjected to n concentrated loads and a constantdistributed loading w. Write a computer program that can be used to determine the internal shear andmoment at any specified location x along the beam, and plot the shear and moment diagrams for thebeam. Show an application of the program using the values P1 = 4 kN, d1 = 2 m, w = 800 N/m, a1 = 2m, a2 = 4 m, L = 4 m. 511. Problem 6-27Determine the placement distance a of the roller support so that the largest absolute value of themoment is a minimum. Draw the shear and moment diagrams for this condition. 512. Problem 6-28Draw the shear and moment diagrams for the rod. Only vertical reactions occur at its ends A and B.Given: a := 900mm A := 360NB := 720N w' 2.4kNm:=Solution:+ ΣFy=0; A 0.5w'xoa⎛⎜⎝⎞⎠− ⋅ ⋅xo = 0xoA⋅a0.5w':=xo = 519.62mmΣΜ Mmax A⋅xo 0.5w'xoa⎛⎜⎝⎞⎠⋅ ⋅xoxo3⎛⎜⎝⎞⎠:= − ⋅Mmax = 124.71 N⋅mx := 0 , 0.01⋅a .. a V(x) A 0.5w'xa⎛⎜⎝⎞⎠⎤⎥⎦⋅ x ⋅ − ⎡⎢⎣1N:= ⋅ M(x) A⋅x 0.5w'xa⎛⎜⎝⎞⎠⋅ ⋅xx3⎛⎜⎝⎞⎠⋅ − ⎡⎢⎣⎤⎥⎦1N⋅m:=0 0.2 0.4 0.6 0.80500Distance (m)Shear (N)V(x)x0 0.2 0.4 0.6 0.8100500Distance m)Moment (N-m)M(x)x 513. Problem 6-29Draw the shear and moment diagrams for the beam.Given: Set L := 1m wo 1kNm:=aL3:=Solution:Equilibrium : Given+ ΣFy=0; A − 2(0.5wo)⋅a − wo⋅a + B = 0ΣΜB=0; A⋅ (3⋅a) (0.5⋅wo⋅a) 2aa3+ ⎛⎜⎝⎞⎠− ⋅ − (wo⋅a)⋅ (1.5a) (0.5⋅wo⋅a) 2a3⎛⎜⎝⎞⎠− ⋅ = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠0.330.33⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (2a) x3 := (2a) , 1.01⋅ (2a) .. (3aV1(x1) Awo2x1a⎛⎜⎝⎞⎠− ⋅ ⋅x1⎡⎢⎣⎤⎥⎦1kN⋅ := V2 x2 ( ) A 0.5 wo ⋅ a ⋅ − wo x2 a − ( ) ⋅ − ⎡⎣⎤⎦1kN:= ⋅V3(x3) A − 0.5⋅wo⋅a − wo⋅a wo⋅ (x3 − 2a) 1 0.5x3 − 2aa− ⋅⎛⎜⎝⎞⎠− ⋅⎡⎢⎣⎤⎥⎦1kN:= ⋅Distance (m) Shear (kN)0 0.2 0.4 0.6 0.80.40.200.20.4V1(x1)V2(x2)V3(x3)x1, x2, x3 514. M1(x1) A⋅x1wo2x1a⎛⎜⎝⎞⎠⋅ ⋅x1x13− ⋅⎡⎢⎣⎤⎥⎦1N⋅m:= ⋅M2(x2) A⋅x2wo⋅a2x22a3− ⎛⎜⎝⎞⎠− ⋅ − 0.5wo⋅ (x2 − a)2⎡⎢⎣⎤⎥⎦1N⋅m:= ⋅M'3(x3) wo2⋅ (x3 − 2⋅a)2 1x3 − 2⋅aa⎛⎜⎝⎞⎠13− ⋅⎡⎢⎣⎤⎥⎦:= ⋅M3(x3) A⋅x3wo⋅a2x32⋅a3− ⎛⎜⎝⎞⎠− ⋅ − (wo⋅a)⋅ (x3 − 1.5⋅a) − M'3(x3)⎡⎢⎣⎤⎥⎦1N⋅m:= ⋅0 0.2 0.4 0.6 0.8120100806040200Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 515. Problem 6-30Draw the shear and moment diagrams for the beam.Set: L := 1m wo 1kNm:=Solution:Equilibrium : Given+ ΣFy=0; A + B − 0.5⋅wo⋅L = 0ΣΜB=0; A2⋅L3⋅wo2⋅L⎛⎜⎝⎞⎠L3⎛⎜⎝⎞⎠− ⋅ = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠0.250.25⎛⎜⎝⎞⎠= kNLet aL3:=x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 3aV1(x1) wo2−x1L⋅ ⋅ (x1)⎡⎢⎣⎤⎥⎦1kN:= ⋅ V2(x2) Awo2x2L− ⋅ ⋅ (x2)⎡⎢⎣⎤⎥⎦1kN:= ⋅M1(x1) wo2−x1L⋅ ⋅ (x1)x13⎛⎜⎝⎞⎠⋅⎡⎢⎣⎤⎥⎦1kN⋅m:= ⋅M2(x2) A⋅ (x2 − a)wo2x2L⋅ ⋅ (x2)x23⎛⎜⎝⎞⎠− ⋅⎡⎢⎣⎤⎥⎦1kN⋅m:= ⋅0 0.2 0.4 0.6 0.80.200.2Distance (m)Shear Wo (kN)V1(x1)V2(x2)x1, x2 516. 0 0.2 0.4 0.6 0.80.040.020Distance (m)Moment Wo*L*L (kN-m)M1(x1)M2(x2)x1, x2 517. Problem 6-31The T-beam is subjected to the loading shown. Draw the shear and moment diagrams.Given: a := 2m P := 10kNb := 3mw 3kNm:= c := 3mSolution:Equilibrium : Given+ ΣFy=0; −P + A − w⋅c + B = 0ΣΜB=0; −P⋅ (a + b + c) + A⋅ (b + c) − (w⋅c)⋅ (0.5c) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠15.583.42⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (a + b) x3 := (a + b) , 1.01⋅ (a + b) .. (a + b + c)V1(x1) −P:= V2(x2) (−P + A)kN1kN⋅ := V3 x3 ( ) P − A + w x3 a − b − ( ) ⋅ − ⎡⎣⎤⎦1kN:= ⋅M1(x1) −P⋅x1:= M2 x2 ( ) P − x2 ⋅ A x2 a − ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) P − x3 ⋅ A x3 a − ( ) ⋅ + 0.5 w ⋅ x3 a − b − ( )2 ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 3 4 5 6 7 850510Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3 518. 0 2 4 6 850510152025Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 519. Problem 6-32The ski supports the 900-N (~90-kg) weight of the man. If the snow loading on its bottom surface istrapezoidal as shown, determine the intensity w, and then draw the shear and moment diagrams for theski.Given: a := 0.5m P := 900NSolution:Equilibrium :+ ΣFy=0; −Pw2+ ⋅ (2a + 4a) = 0wP3a:= w 600Nm=x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 2ax3 := 2a , 1.01⋅ (2a) .. 3a x4 := 3a , 1.01⋅ (3a) .. 4aV1(x1) w2x1a⎛⎜⎝⎞⎠⋅ ⋅x1⎡⎢⎣⎤⎥⎦1N:= ⋅V2 x2 ( ) 0.5 w ⋅ a ⋅ w x2 a − ( ) ⋅ + ⎡⎣⎤⎦1N:= ⋅V3 x3 ( ) 0.5 w ⋅ a ⋅ w a ⋅ + P − w x3 2a − ( ) ⋅ + ⎡⎣⎤⎦1N:= ⋅V4(x4) 0.5⋅w⋅a + w⋅a − P + wa w⋅ (x4 − 3a) 1 0.5x4 − 3aa− ⋅⎛⎜⎝⎞⎠+ ⋅⎡⎢⎣⎤⎥⎦1N:= ⋅0 0.5 1 1.5 24002000200400Distance (m)Shear (N)V1(x1)V2(x2)V3(x3)V4(x4)x1, x2, x3, x4 520. M1(x1) w2x1a⎛⎜⎝⎞⎠⋅ ⋅x1x13⋅⎡⎢⎣⎤⎥⎦1N⋅m:= ⋅x2 a − ( )2 ⋅ + ⎡⎢⎣M2(x2) w⋅a2x22a3− ⎛⎜⎝⎞⎠⋅w2⎤⎥⎦1N⋅m:= ⋅x3 a − ( )2 ⋅ + P x3 2a − ( ) ⋅ − ⎡⎢⎣M3(x3) w⋅a2x32a3− ⎛⎜⎝⎞⎠⋅w2⎤⎥⎦1N⋅m:= ⋅M'4(x4) w2⋅ (x4 − 3⋅a)2 1x4 − 3⋅aa⎛⎜⎝⎞⎠13− ⋅⎡⎢⎣⎤⎥⎦:= ⋅2 a ⋅ − ( ) ⋅ + P x4 2a − ( ) ⋅ − M'4 x4 ( ) + ⎡⎢⎣M4(x4) w⋅a2x42⋅a3− ⎛⎜⎝⎞⎠⋅ 2w a ⋅ ( ) x4⎤⎥⎦1N⋅m:= ⋅0 0.5 1 1.5 2150100500Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)M4(x4)x1, x2, x3, x4 521. Problem 6-33Draw the shear and moment diagrams for the beam.Given: L := 9mwo 50kNm:= a := 0.5LSolution:Equilibrium : Given+ ΣFy=0; A − 2(0.5wo)⋅a + B = 0ΣΜB=0; A⋅L (0.5⋅wo⋅a) aa3+ ⎛⎜⎝⎞⎠− ⋅ (0.5⋅wo⋅a) 2a3⎛⎜⎝⎞⎠− ⋅ = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠112.50112.50⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (2a)V1(x1) A wo⋅x1 1 0.5x1a− ⋅⎛⎜⎝⎞⎠− ⋅⎡⎢⎣⎤⎥⎦1kN:= ⋅V2(x2) A − 0.5⋅wo⋅awo2x2 − aa⎛⎜⎝⎞⎠− ⋅ ⋅ (x2 − a)⎡⎢⎣⎤⎥⎦1kN:= ⋅M1(x1) A⋅x1wo2⋅ 2 1x1x1a⎛⎜⎝⎞⎠13− ⋅⎡⎢⎣⎤⎥⎦− ⋅⎡⎢⎣⎤⎥⎦1kN⋅m:= ⋅M'2(x2) wo2x2 − aa⎛⎜⎝⎞⎠⋅ ⋅ (x2 − a)x2 − a3⎛⎜⎝⎞⎠:= ⋅M2(x2) A⋅x2wo⋅a2x2a3− ⎛⎜⎝⎞⎠− ⋅ − M'2(x2)⎡⎢⎣⎤⎥⎦1kN⋅m:= ⋅0 51000100Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 52001000Distance (m)Moment (N-m)M1(x1)M2(x2)x1, x2 522. Problem 6-34Draw the shear and moment diagrams for the wood beam, and determine the shear and momentthroughout the beam as functions of x.Given: a := 1m P := 1kNb := 1.5mw 2kNm:= c := 1mSolution:Equilibrium : Given+ ΣFy=0; A − w⋅b + B − 2P = 0ΣΜB=0; −P⋅ (a + b) + A⋅b − (w⋅b)⋅ (0.5b) + P⋅c = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠2.502.50⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (a + b) x3 := (a + b) , 1.01⋅ (a + b) .. (a + b + c)V1(x1) −P:= V2 x2 ( ) P − A + w x2 a − ( ) ⋅ − ⎡⎣kN⎤⎦1kN:= ⋅ V3(x3) (−P + A − w⋅b + B)1kN:= ⋅M1(x1) −P⋅x1:= M2 x2 ( ) P − x2 ⋅ A x2 a − ( ) ⋅ + 0.5w x2 a − ( )2 ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) P − x3 ⋅ A x3 a − ( ) ⋅ + w b ⋅ ( ) x3a − 0.5 b ⋅ − ( ) ⋅ − B x3 a − b − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.5 1 1.5 2 2.5 3 3.521012Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3 523. 0 0.5 1 1.5 2 2.5 3 3.500.20.40.60.81Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 524. Problem 6-35The smooth pin is supported by two leaves A and B and subjected to a compressive load of 0.4 kN/mcaused by bar C. Determine the intensity of the distributed load w0 of the leaves on the pin and drawthe shear and moment diagrams for the pin.Given: L := 100mm w 0.4kNm:=a := 0.2LSolution:Equilibrium :+ ΣFy=0; 2(0.5wo)⋅a − w⋅ (3a) = 0wo := 3w wo 1.20kNm=x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (4a) x3 := (4a) , 1.01⋅ (4a) .. LV1(x1) wo:= ⋅ V2(x2) wo2x1a⎛⎜⎝⎞⎠⋅ ⋅x1⎡⎢⎣⎤⎥⎦1N2⋅a − w⋅ (x2 − a)⎡⎢⎣⎤⎥⎦1N:= ⋅V3(x3) wo2⋅a − w⋅ (3a) wo⋅ (x3 − 4a) 1 0.5x3 − 4aa− ⋅⎛⎜⎝⎞⎠+ ⋅⎡⎢⎣⎤⎥⎦1N:= ⋅0 0.02 0.04 0.06 0.0810010Distance (m)Shear (N)V1(x1)V2(x2)V3(x3)x1, x2, x3 525. M1(x1) wo2x1a⎛⎜⎝⎞⎠⋅ ⋅x1x13⋅⎡⎢⎣⎤⎥⎦1N⋅m:= ⋅M2(x2) wo⋅a2x22a3− ⎛⎜⎝⎞⎠⋅ − 0.5w⋅ (x2 − a)2⎡⎢⎣⎤⎥⎦1N⋅m:= ⋅M'3(x3) wo2⋅ (x3 − 4⋅a)2 1x3 − 4⋅aa⎛⎜⎝⎞⎠13− ⋅⎡⎢⎣⎤⎥⎦:= ⋅M3(x3) wo⋅a2x32⋅a3− ⎛⎜⎝⎞⎠⋅ − w⋅ (3a)⋅ (x3 − 2.5⋅a)⎡⎢⎣⎤⎥⎦+ M'3(x3)⎡⎢⎣⎤⎥⎦1N⋅m:= ⋅0 0.02 0.04 0.06 0.080.20.10Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 526. Problem 6-36Draw the shear and moment diagrams for the beam.Given: a := 3.6m b := 1.8mMA := 2.25kN⋅m w 45kNm:=Solution:Equilibrium : Given+ ΣFy=0; A + B − 0.5w⋅b = 0ΣΜB=0; MA + A⋅a (0.5w⋅b)b3⎛⎜⎝⎞⎠+ ⋅ = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠−7.3847.88⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + bV1(x1) A:= kNV2(x2) A + B w⋅ (x2 − a) 1 0.5x2 − ab− ⋅⎛⎜⎝⎞⎠− ⋅⎡⎢⎣⎤⎥⎦1kN:= ⋅M1(x1) (MA + A⋅x1) 1kN⋅m:= ⋅M2(x2) MA + A⋅x2 + B⋅ (x2 − a)w2⋅ (x2 − a)2 1x2 − ab⎛⎜⎝⎞⎠13− ⋅⎡⎢⎣⎤⎥⎦− ⋅⎡⎢⎣⎤⎥⎦1kN⋅m:= ⋅0 1 2 3 4 54020020Distance (m)Shear (kN)V1(x1)V2(x2)x1, x2 527. 0 1 2 3 4 50102030Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 528. Problem 6-37The compound beam consists of two segments that are pinned together at B. Draw the shear andmoment diagrams if it supports the distributed loading shown.Set: L := 1m aL3:= w 1kNm:=Solution:Consider segment AB.ΣMB=0; A⋅ (2a)w22aL⎛⎜⎝⎞⎠⋅ ⋅ (2⋅a)2a3⎛⎜⎝⎞⎠− ⋅ = 0A2w⋅L27:= A = 0.0741 kNConsider whole beam ABC.MC A⋅Lw⋅L2L3⎛⎜⎝⎞⎠= − ⋅MC−5w⋅L254:=MC = −0.09259 kN⋅mx1 := 0 , 0.01⋅ (2a) .. (2a)V(x) Aw2xL⎛⎜⎝⎞⎠⎤⎥⎦⋅ x ⋅ − ⎡⎢⎣1kN⎞⎠⋅ − ⎡⎢⎣:= ⋅ M(x) A⋅xw2xL⎛⎜⎝⎞⎠⋅ ⋅xx3⎛⎜⎝⎤⎥⎦1kN⋅m:=0 0.5 10.200.20.4Distance (m)Shear W*L (kN)V(x)x0 0.5 100.050.1Distance m)Moment w*L*L (kN-m)M(x)x 529. Problem 6-38Draw the shear and moment diagrams for the beam.Given: L := 3m wo 12kNm:= w1 18kNm:=Solution:Equilibrium :+ ΣFy=0; B (wo + w1) L2:= ⋅ΣΜB=0; MB (wo⋅L) L⋅ (w1 − wo) L22⋅L3⎛⎜⎝⎞⎠:= + ⋅B = 45.00 kN MB = 63.00 kN⋅moww' := w1 − wox1 := 0 , 0.01⋅L .. LV ( x ) − ⋅xw'2xL⎛⎜⎝⎞⎠⎤⎥⎦⋅ x ⋅ − ⎡⎢⎣1kN⋅ := M x ( ) wo− ⋅xx2⎛⎜⎝⎞⎠⋅w'2xL⎛⎜⎝⎞⎠⋅ ⋅xx3⎛⎜⎝⎞⎠⋅ − ⎡⎢⎣⎤⎥⎦1kN⋅m:= ⋅0 1 2 302040Distance (m)Shear (kN)V(x)x0 1 2 3050Distance (m)Moment (kN-m)M(x)x 530. Problem 6-39Draw the shear and moment diagrams for the beam and determine the shear and moment as functionsof x.Given: a := 3m wo 200Nm:= w1 400Nm:=Solution: L := 2a w' := w1 − woEquilibrium : Given+ ΣFy=0; A (wo + w1) a− ⋅ + B = 02+ ΣΜB=0; A⋅ (2⋅a) (wo⋅a) a− ⋅ w'2a2⋅a3⎛⎜⎝⎞⎠− ⋅ = 0Guess A := 1N B := 1NAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠200700⎛⎜⎝⎞⎠= NShear and Moment Functions :For 0 < x < 3m,V := A V = 200 N AnsM = A⋅x M = (200x)⋅N⋅m AnsFor 3m < x < 6m,V' A − wo⋅ (x − a)w'2x − aa⎛⎜⎝⎞⎠= − ⋅ ⋅ (x − a)x2 ⋅ − ⎛⎜⎝V' 5001003⎞⎠= ⋅N AnsM' A⋅x wo⋅ (x − a)x − a2− ⋅w'2x − aa⎛⎜⎝⎞⎠⋅ ⋅ (x − a)⎛⎜⎝⎞⎠x − a3= − ⋅x3 ⋅ + ⎛⎜⎝M' 600 − 500x1009⎞⎠= ⋅N⋅m Ansx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. LV1(x1) A:= V2(x2) A − wo⋅ (x2 − a)Nw'2x2 − aa⎛⎜⎝⎞⎠− ⋅ ⋅ (x2 − a)⎡⎢⎣⎤⎥⎦1N:= ⋅M1(x1) (A⋅x1) 1N⋅m:= ⋅M2(x2) A⋅x2 wo⋅ (x2 − a)x2 − a2− ⋅w'2x2 − aL⎛⎜⎝⎞⎠⋅ ⋅ (x2 − a)x2 − a3⎛⎜⎝⎞⎠− ⋅⎡⎢⎣⎤⎥⎦1N⋅m:= ⋅ 531. 0 2 4 65000500Distance (m)Shear (N)V1(x1)V2(x2)x1, x20 2 4 66004002000Distance (m)Moment (N-m)M1(x1)M2(x2)x1, x2 532. Problem 6-40Determine the placement distance a of the roller support so that the largest absolute value of themoment is a minimum. Draw the shear and moment diagrams for this condition.Solution:Equilibrium :ΣFy=0; A − 2P + B = 0+ΣΜA=0; PL2⋅ − B⋅a + P⋅L = 0+ B3⋅L2a= ⋅P A4a − 3L2a= ⋅PInternal Moment :For positive moment, Mmax AL2= ⋅For negative moment, Mmin = −P⋅ (L − a)When Mmax = MminAL2⋅ = P⋅ (L − a)4a − 3L⋅ P 2a⎛⎜⎝⎞⎠L2⋅ = P⋅ (L − a)(4a − 3L)⋅L = 4a(L − a)a32= L AnsSet: L := 1m P := 1kNa32:= L A4a − 3L2a:= ⋅P B3⋅L2a:= ⋅Pa' := 0.5L b' := a − a' c' := L − ax1 := 0 , 0.01⋅a' .. a' x2 := a' , 1.01⋅a' .. (a' + b') x3 := (a' + b') , 1.01⋅ (a' + b') .. LV1(x1) A1kN:= ⋅ V2(x2) (A − P)1kN:= ⋅ V3(x3) (A − P + B)1kN:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ P x2 a' − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ P x3 a' − ( ) ⋅ − B x3 a' − b' − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅ 533. 0 0.2 0.4 0.6 0.810.500.51Distance (m)Shear P (kN)V1(x1)V2(x2)V3(x3)x1, x2, x30 0.2 0.4 0.6 0.80.100.1Distance (m)Moment P*L (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 534. Problem 6-41Draw the shear and moment diagrams for the beam.Given: a := 2m w0 8kNm:=Solution: unitkNm3:= w = (unit)⋅2x2Wxxw x⌠⎮⌡= d0Wx (unit)x2x2 x ⌠⎮⌡0d⎛⎜⎜⎝⎞⎠= ⋅ Wx (unit)23= ⋅ ⋅x3Wa (unit)23= ⋅ ⋅a3+ ΣFy=0; Aaw x⌠⎮⌡− d = 0 A (unit)0a2x2 x ⌠⎮⌡:= ⋅ d A = 5.33 kN0xcaw⋅x x⌠⎮⌡0d= xcAa2⋅x3 x⌠⎮⌡unitA 0:= d xc = 1.500mMA := A⋅ (xc) MA = 8.00 kN⋅m⋅ x3 − ⎡⎢⎣x := 0 , 0.01⋅a .. a V(x) A (unit)23⎤⎥⎦1kN:= ⋅M(x) −MA + A⋅x (unit)23⋅ ⋅x3⋅x 1xca−⎛⎜⎝⎞⎠− ⋅⎡⎢⎣⎤⎥⎦1kN⋅m:= ⋅0 0.5 1 1.5 26420Distance (m)Shear (kN)V(x)x 535. 0 0.5 1 1.5 202468Distance m)Moment (kN-m)M(x)x 536. Problem 6-42The truck is to be used to transport the concrete column. If the column has a uniform weight of w(force/length), determine the equal placement a of the supports from the ends so that the absolutemaximum bending moment in the column is as small as possible. Also, draw the shear and momentdiagrams for the column.Solution:Support Reactions: By symmetry, A=B=R+ ΣFy=0; 2R − wL = 0R = 0.5w⋅LInternal Moment :For negative moment, Mmin = −0.5w⋅a2For positive moment, Mmax wL2⋅ ⎛⎜⎝⎞⎠L4⋅ RL− a 2⎛⎜⎝⎞⎠= −(at mid-span)Mmaxw⋅L8= ⋅ (4a − L)For optimal minimum: Mmax = Mminw⋅L8⋅ (4a − L)12= − ⋅w⋅a2 (4a − L)⋅L = −4a2Let αaL= α2 + α − 0.25 = 0α121 − 12 4 0.25 − ( ) ⋅ − + ⎡⎣⎤⎦:= ⋅α = 0.2071a = 0.2071L AnsSet: L := 1m w 1kNm:=a := αL R := 0.5w⋅L b := L − 2ax1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (a + b) x3 := (a + b) , 1.01⋅ (a + b) .. LV1(x1) −w⋅x1:= ⋅ V3(x3) (2R − w⋅x3) 1:= V2 x2 ( ) R w x2kN( − ⋅ ) 1kNkN:= ⋅2 ⋅ ⎛⎜⎝M1(x1) −w2 ⋅ − ⎡⎢⎣:= ⋅ M2(x2) R⋅ (x2 − a)2x1⎞⎠1kN⋅mw2x2⎤⎥⎦1kN⋅m:= ⋅2 ⋅ − ⎡⎢⎣M3(x3) R⋅ (x3 − a) + R⋅ (x3 − a − b)w2x3⎤⎥⎦1kN⋅m:= ⋅ 537. 0 0.2 0.4 0.6 0.80.40.200.20.4Distance (m)Shear w*L (kN)V1(x1)V2(x2)V3(x3)x1, x2, x30 0.2 0.4 0.6 0.80.040.0200.020.04Distance (m)Moment w*L*L (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 538. Problem 6-43A member having the dimensions shown is to be used to resist an internal bending moment of M = 2kN·m. Determine the maximum stress in the member if the moment is applied (a) about the z axis, (b)about the y axis. Sketch the stress distribution for each case.Given: d := 120mm b := 60mmMz := 2kN⋅m My := 2kN⋅mSolution:Iz112:= ⋅b⋅d3 Iy112:= ⋅d⋅b3Maximum Stress: σ McI= ⋅(a) About the z axisymax:= σmax (Mz) ymaxd2Iz:= ⋅σmax = 13.89MPa Ans(b) About the y axiszmax:= σmax (My) zmaxb2Iy:= ⋅σmax = 27.78MPa Ans 539. Problem 6-44The steel rod having a diameter of 20 mm is subjected to an internal moment of M = 300 N·m.Determine the stress created at points A and B. Also, sketch a threedimensional view of the stressdistribution acting over the cross section.Given: d := 20mm M := 300N⋅mθ := 45degSolution: Iπ4d2⎛⎜⎝⎞⎠4:= ⋅ σ MyI= ⋅yAd2:= σA MyAI:= ⋅σA = 381.97MPa AnsyBd2⎛⎜⎝⎞⎠:= ⋅ sin(θ) σB MyBI:= ⋅σB = 270.09MPa Ans 540. Problem 6-45The beam is subjected to a moment M. Determine the percentage of this moment that is resisted by thstresses acting on both the top and bottom boards,A and B, of the beam.Given: bf := 200mm tf := 25mmtw := 25mm dw := 150mmSolution: D := dw + 2tfI112bf⋅D3 (bf − 2tw) dw3 ⋅ − ⎡⎣⎤⎦:= ⋅Bending Stress: σ McI= ⋅Set M := 1kN⋅mco := 0.5D σo McoI:= ⋅ σo = 1.097143MPaci := 0.5dw σi MciI:= ⋅ σi = 0.822857MPaResultant Force and Moment: For board A or B.F12:= ⋅ (σo + σi)⋅bf⋅ tf F = 4.800 kNCentroid of force: σi⋅ (bf⋅ tf)tf2⋅12⋅ (σo − σi)⋅ (bf⋅ tf)tf3+ ⋅ = F⋅ycyc1Fσi⋅ (bf⋅ tf)tf2⋅12⋅ (σo − σi)⋅ (bf⋅ tf)tf3+ ⋅⎡⎢⎣⎤⎥⎦:=yc = 11.905mmM' := F⋅ (D − 2yc)M' = 0.8457 kN⋅mHence, %MM'M:= ⋅100%M = 84.57 Ans 541. Problem 6-46Determine the moment M that should be applied to the beam in order to create a compressive stress atpoint D of σD = 30 MPa. Also sketch the stress distribution acting over the cross section and computethe maximum stress developed in the beam.Given: bf := 200mm tf := 25mm dw := 150mmtw := 25mm σD := 30MPaSolution: D := dw + 2tfI112bf⋅D3 (bf − 2tw) dw3 ⋅ − ⎡⎣⎤⎦:= ⋅I = 91145833.33mm4Bending Stress: σ McI= ⋅cD := 0.5dw σD McDI= ⋅ MσD⋅ IcD:=M = 36.46 kN⋅m Anscmax := 0.5D σmax McmaxI:= ⋅σmax = 40.00MPa Ans 542. Problem 6-47The slab of marble, which can be assumed a linear elastic brittle material, has a specific weight of 24kN/m3 and a thickness of 20 mm. Calculate the maximum bending stress in the slab if it is supported(a) on its side and (b) on its edges. If the fracture stress is σ f = 1.5 MPa, explain the consequences ofsupporting the slab in each position.Given: t := 20mm L := 1.5m d := 0.5mγ 24kNm3:= σf := 1.5MPaSolution: w := γ⋅d⋅ t w 0.24kNm=Mmax18:= ⋅w⋅L2Is112:= ⋅ t⋅d3 Ie112:= ⋅d⋅ t3Maximum Stress: σ McI= ⋅(a) Supported on its sidec1:= σmax (Mmax) c1d2Is:= ⋅σmax = 0.081MPa Ans(b) Supported on its edgesc2:= σmax (Mmax) c2t2Ie:= ⋅σmax = 2.025MPa Ans> σf = 1.5 MPaThe marble slab will break if it is supported as in case (b). 543. Problem 6-48The slab of marble, which can be assumed a linear elastic brittle material, has a specific weight of 24kN/m3. If it is supported on its edges as shown in (b), determine the minimum thickness it should havewithout causing it to break.The fracture stress is σ f = 1.5 MPa.Given: L := 1.5m d := 0.5mγ 24kNm3:= σf := 1.5MPaSolution: w = γ⋅d⋅ tMmax18= ⋅w⋅L2 Mmax18= ⋅ (γ⋅d⋅ t)⋅L2Maximum Stress: ct2= Ie112= ⋅d⋅ t3σmax MmaxcIe= ⋅ σmax Mmax6d⋅ t2= ⋅Thus, σmax18⋅ (γ⋅d⋅ t)⋅L26d⋅ t2= ⋅t3⋅ γ⋅L24⋅σf:=t = 27mm Ans 544. Problem 6-49A beam has the cross section shown. If it is made of steel that has an allowable stress of σallow = 170MPa, determine the largest internal moment the beam can resist if the moment is applied (a) about the zaxis, (b) about the y axis.Given: bf := 120mm tf := 5mm d := 120mmtw := 5mm σallow := 170MPafSolution: D := d +2tIz112bf D3 ⋅ bf tw − ( ) d3 ⋅ − ⎡⎣⎤⎦:= ⋅3 ⋅ ⎛⎜⎝:= + ⋅ 3Iy 2112⋅ tf bf⎞⎠112⋅d twBending Stress: σallow McI= ⋅(a) About the z axiscz:= Mz (σallow) IzD2:= ⋅czMz = 14.15 kN⋅m Ans(b) About the y axiscybf2:= My (σallow) Iy:= ⋅cyMy = 4.08 kN⋅m Ans 545. Problem 6-50Two considerations have been proposed for the design of a beam. Determine which one will support amoment of with the least amount of M = 150 kN·m bending stress. What is that stress? By whatpercentage is it more effective?Given: bf := 200mm dw := 300mmtf.a := 15mm tw.a := 30mmtf.b := 30mm tw.b := 15mmM := 150kN⋅mSolution:Section Property:For section (a):Da := dw + 2tf.a Ia112⋅ 3 b( f − tw.a) dw⎡⎣bf Da− ⋅ 3 ⎤⎦:= ⋅For section (b):Db := dw + 2tf.b Ib112⋅ 3 b( f − tw.b) dw⎡⎣bf Db− ⋅ 3 ⎤⎦:= ⋅Maximum Bending Stress: σ McI= ⋅For section (a):cmax := 0.5Da σmax McmaxIa:= ⋅σmax = 114.35MPaFor section (b):c'max := 0.5Db σ'max Mc'maxIb:= ⋅σ'max = 74.72MPa AnsBy comparison, section (b) will have the least amount of bending stress.%effσmax − σ'max:= ⋅100σ'max%eff = 53.03 Ans 546. Problem 6-51The aluminum machine part is subjected to a moment of Determine the bending stress M = 75 N·m.created at points B and C on the cross section. Sketch the results on a volume element located at eachof these points.Given: bf := 80mm tf := 10mmtw := 10mm dw := 40mmM := 75N⋅mSolution: D := dw + tf⎯ Σ yi ⎯y⋅ ( ⋅Ai)Σ⋅ (Ai)=yc(bf⋅ tf)⋅0.5tf + 2(dw⋅ tw)⋅ (0.5dw + tf)bf⋅ tf + 2dw⋅ tw:=yc = 17.50mmIf112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (yc − 0.5tf)2Iw112⋅ ⋅ 3 dw tw ⋅ ( ) yc 0.5dw tf + ( ) − ⎡⎣:= + ⋅ 2tw dw⎤⎦I := If + 2IwBending Stress: σ McI= ⋅At B: cB := yc σB McBI:= ⋅ σB = 3.612MPa AnsAt C: cC := yc − tf σC McCI:= ⋅ σC = 1.548MPa Ans 547. Problem 6-52The aluminum machine part is subjected to a moment of M = 75 N·m. Determine the maximum tensiland compressive bending stresses in the part.Given: bf := 80mm tf := 10mmtw := 10mm dw := 40mmM := 75N⋅mSolution: D := dw + tf⎯ Σ yi ⎯y⋅ ( ⋅Ai)Σ⋅ (Ai)=yc(bf⋅ tf)⋅0.5tf + 2(dw⋅ tw)⋅ (0.5dw + tf)bf⋅ tf + 2dw⋅ tw:=yc = 17.50mmIf112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (yc − 0.5tf)2Iw112⋅ ⋅ 3 dw tw ⋅ ( ) yc 0.5dw tf + ( ) − ⎡⎣:= + ⋅ 2tw dw⎤⎦I := If + 2IwBending Stress: σ McI= ⋅For compression:cc := yc σc_max MccI:= ⋅ σc_max = 3.612MPa AnsFor tension:ct := D − yc σt_max MctI:= ⋅ σt_max = 6.709MPa Ans 548. Problem 6-53A beam is constructed from four pieces of wood, glued together as shown. If the moment acting onthe cross section is M = 450 N·m, determine the resultant force the bending stress produces on the topboard A and on the side board B.Given: bf := 240mm tf := 15mmtw := 20mm dw := 200mmM := 450N⋅mSolution: D := dw + 2tfIy112D ⋅ bf3 − dw ⋅ ⎡⎣(bf − 2tw)3 ⎤⎦:= ⋅Bending Stress: σ McI= ⋅co := 0.5bf σo McoIy:= ⋅ σo = 0.410251MPaci := 0.5bf − tw σi MciIy:= ⋅ σi = 0.341876MPaResultant Force : For board A or B.FAσo2bf2⋅ tf⎛⎜⎝⎞⎠⋅σo2bf2⋅ tf⎛⎜⎝⎞⎠:= − ⋅ FA = 0 kN AnsFB12:= ⋅ (σo + σi)⋅dw⋅ tw FB = 1.504 kN Ans 549. Problem 6-54The aluminum strut has a cross-sectional area in the form of a cross. If it is subjected to the momentM = 8 kN·m, determine the bending stress acting at points A and B, and show the results acting onvolume elements located at these points.Given: b'f := 50mm tf := 20mmtw := 20mm dw := 220mmM := 8kN⋅mSolution:I112⋅ ⋅ 3 2tw dw1⋅⋅ 3 12b'f tf⎛⎜⎝⎞⎠:= +I = 17813333.33mm4Bending Stress: σ McI= ⋅At A: cA := 0.5dw σA McAI:= ⋅ σA = 49.401MPa AnsAt B: cB := 0.5tf σB McBI:= ⋅ σB = 4.491MPa Ans 550. Problem 6-55The aluminum strut has a cross-sectional area in the form of a cross. If it is subjected to the momentM = 8 kN·m, determine the maximum bending stress in the beam, and sketch a three-dimensional viewof the stress distribution acting over the entire cross-sectional area.Given: b'f := 50mm tf := 20mmtw := 20mm dw := 220mmM := 8kN⋅mSolution:I112⋅ ⋅ 3 2tw dw1⋅⋅ 3 12b'f tf⎛⎜⎝⎞⎠:= +I = 17813333.33mm4Bending Stress: σ McI= ⋅cmax := 0.5dw σmax McmaxI:= ⋅ σmax = 49.401MPa AnsAt B:cB := 0.5tf σB McBI:= ⋅ σB = 4.491MPa 551. Problem 6-56The beam is made from three boards nailed together as shown. If the moment acting on the crosssection is M = 1.5 kN·m, determine the maximum bending stress in the beam. Sketch athree-dimensional view of the stress distribution acting over the cross section.Given: bf := 250mm b'f := 150mm tf := 38mmtw := 25mm d := 300mmM := 1.5kN⋅mfSolution: D := d +2t⎯ Σ yi ⎯y⋅ ( ⋅Ai)Σ⋅ (Ai)=yc(bf⋅ tf)⋅0.5tf + (d⋅ tw)⋅ (0.5d + tf) + (b'f⋅ tf)⋅ (D − 0.5tf)bf⋅ tf + d⋅ tw + b'f⋅ tf:=yc = 159.71mmIf112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (yc − 0.5tf)2Iw112tw ⋅ d3 ⋅ d tw ⋅ ( ) yc 0.5d tf + ( ) − ⎡⎣:= + ⋅ 2⎤⎦I'f112⋅ 3 b'f tf ⋅ ( ) yc D 0.5tf − ( ) − ⎡⎣:= + ⋅ 2⋅b'f tf⎤⎦I := If + Iw + I'fBending Stress: σ McI= ⋅At B: cmax := D − yc σmax McmaxI:= ⋅ σmax = 0.684MPa AnsAt A: cA := cmax − tf σA McAI:= ⋅ σA = 0.564MPaAt C: cC := yc − tf σC McCI:= ⋅ σC = 0.385MPaAt D: cD := yc σD McDI:= ⋅ σD = 0.505MPa 552. Problem 6-57Determine the resultant force the bending stresses produce on the top board A of the beam if M = 1.5kN·m.Given: bf := 250mm b'f := 150mm tf := 38mmtw := 25mm d := 300mmM := 1.5kN⋅mfSolution: D := d +2t⎯ Σ yi ⎯y⋅ ( ⋅Ai)Σ⋅ (Ai)=yc(bf⋅ tf)⋅0.5tf + (d⋅ tw)⋅ (0.5d + tf) + (b'f⋅ tf)⋅ (D − 0.5tf)bf⋅ tf + d⋅ tw + b'f⋅ tf:=yc = 159.71mmIf112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (yc − 0.5tf)2Iw112tw ⋅ d3 ⋅ d tw ⋅ ( ) yc 0.5d tf + ( ) − ⎡⎣:= + ⋅ 2⎤⎦I'f112⋅ 3 b'f tf ⋅ ( ) yc D 0.5tf − ( ) − ⎡⎣:= + ⋅ 2⋅b'f tf⎤⎦I := If + Iw + I'fBending Stress: σ McI= ⋅At C: cC := yc − tf σC McCI:= ⋅ σC = 0.385MPaAt D: cD := yc σD McDI:= ⋅ σD = 0.505MPaThe resultant Force: For top board AF := 0.5(σC + σD)⋅ (bf⋅ tf) F = 4.23 kN Ans 553. Problem 6-58The control level is used on a riding lawn mower. Determine the maximum bending stress in the leverat section a-a if a force of 100 N is applied to the handle. The lever is supported by a pin at A and awire at B. Section a-a is square, 6 mm by 6 mm.Given: L := 50mm b := 6mmd := 6mm F := 100NSolution: I:= ⋅ (b⋅d3)112M := F⋅LM = 5.00 N⋅mBending Stress: σ McI= ⋅cd2:= σmax McI:= ⋅σmax = 138.89MPa Ans 554. Problem 6-59Determine the largest bending stress developed in the member if it is subjected to an internal bendingmoment of M = 40 kN·m.Given: bf := 100mm tf := 10mm rf := 30mmtw := 10mm dw := 180mmM := 40kN⋅mSolution: D := dw + tf + 2rf⎯ Σ yi ⎯y⋅ ( ⋅Ai)Σ⋅ (Ai)=yc2 ⋅ ⎛⎝(bf⋅ tf)⋅0.5tf + (dw⋅ tw)⋅ (0.5dw + tf) π rf⎞⎠+ ⋅ (rf + dw + tf)+ ⋅ 2bf⋅ tf + dw⋅ tw π rf:=yc = 143.41mmIf112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (yc − 0.5tf)2Iw112⋅ 3 dw tw ⋅ ( ) yc 0.5dw tf + ( ) − ⎡⎣:= + ⋅ 2⋅ tw dw⎤⎦I'fπ4:= + ⋅ 2⋅ rf4 π rf2 ⋅ ⎛⎝⎞⎠yc rf dw + tf + ( ) − ⎡⎣⎤⎦I := If + Iw + I'fBending Stress: σ McI= ⋅Maximum stress occurs at the bottom fibre.cmax := yc σmax McmaxI:= ⋅ σmax = 128.51MPa Ans 555. Problem 6-60The tapered casting supports the loading shown. Determine the bending stress at points A and B. Thecross section at section a-a is given in the figure.Given: La := 250mm P := 750NLb := 375mm Lc := 125mmb := 100mm t := 25mmd := 75mmSolution:Equilibrium :ΣΜC=0; F1⋅ (2Lb + Lc) − P⋅ (Lc + Lb) − P⋅ (Lb) = 0F1P⋅ (Lc + Lb) + P⋅ (Lb):=F1 = 750.00 N2⋅Lb + LcSection a-a : D := d + 2tM := F1⋅La M = 187.50 N⋅mI:= ⋅ (b⋅D3 − b⋅d3)112Bending Stress: σ McI= ⋅cAD2:= σA McAI:= ⋅ σA = 0.918MPa AnscBd2:= σB McBI:= ⋅ σB = 0.551MPa Ans 556. Problem 6-61If the shaft in Prob. 6-1 has a diameter of 100 mm, determine the absolute maximum bending stress inthe shaft.Given: a := 250mm b := 800mmF := 24kN do := 100mmSolution:Equilibrium : Given+ ΣFy=0; A + B − F = 0ΣΜA=0; −F⋅a − B⋅b = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠31.50−7.50⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + bM1(x1) −F⋅x1:= M2 x2 ( ) F − x2 ( ) ⋅ A x2 a − ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅0 0.5 105Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2Max. Moment : unit := kN⋅mM1(a) = −6.00Bending Stress: M := M1(a)⋅unit I⋅ 464π do:= c'do2:=σ Mc'I= ⋅ σmax Mc'I:= ⋅ σmax = 61.12MPa Ans 557. Problem 6-62If the shaft in Prob. 6-3 has a diameter of 40 mm, determine the absolute maximum bending stress inthe shaft.Given: a := 350mm b := 500mm c := 375mmd := 300mm do := 40mmB := 400N C := 550N E := 175NSolution:Equilibrium : Given+ ΣFy=0; A + D − B − C − E = 0ΣΜD=0; A⋅ (a + b + c) − B⋅ (b + c) − C⋅c + E⋅d = 0Guess A := 1N D := 1NAD⎛⎜⎝⎞⎠:= Find(A,D)AD⎛⎜⎝⎞⎠411.22713.78⎛⎜⎝⎞⎠= Nx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cx4 := a + b + c , 1.01⋅ (a + b + c) .. a + b + c + dM1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ B x2 a − ( ) ⋅ − ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ B x3 a − ( ) ⋅ − C x3 a − b − ( ) ⋅ − ⎡⎣⎤⎦1N⋅m:= ⋅M4 x4 ( ) A x4 ( ) ⋅ B x4 a − ( ) ⋅ − C x4 a − b − ( ) ⋅ − D x4 a − b − c − ( ) ⋅ + ⎡⎣⎤⎦1N⋅m:= ⋅0 0.2 0.4 0.6 0.8 1 1.2 1.42001000100Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)M4(x4)x1, x2, x3, x4Max. Moment : unit := N⋅mM3(a + b) = 149.54Bending Stress: M := M3(a + b)⋅unit I⋅ 464π do:= c'do2:=σ Mc'I= ⋅ σmax Mc'I:= ⋅ σmax = 23.8MPa Ans 558. Problem 6-63If the shaft in Prob. 6-6 has a diameter of 50 mm, determine the absolute maximum bending stress inthe shaft.Given: a := 125mm b := 600mm c := 75mmF1 := 0.8kN F2 := 1.5kN do := 50mmSolution: L := a + b + cEquilibrium : Given+ ΣFy=0; A − F1 − F2 + B = 0ΣΜB=0; A⋅ (L) − F1⋅ (b + c) − F2⋅ (c) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠0.81561.4844⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cM1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ F1 x2 a − ( ) ⋅ − ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ F1 x3 a − ( ) ⋅ − F2 x3 a − b − ( ) ⋅ − ⎡⎣⎤⎦1N⋅m:= ⋅0 0.2 0.4 0.6150100500Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3Max. Moment : unit := N⋅mM2(a + b) = 111.33Bending Stress: M := M2(a + b)⋅unit I⋅ 464π do:= c'do2:=σ Mc'I= ⋅ σmax Mc'I:= ⋅ σmax = 9.072MPa Ans 559. Problem 6-64If the shaft in Prob. 6-8 has a diameter of 30 mm and thickness of 10 mm, determine the absolutemaximum bending stress in the shaft.Given: a := 400mm h := 80mm F := 5kNdo := 30mm t := 10mmSolution: Given di := do − 2tEquilibrium :+ ΣFy=0; A + C = 0ΣΜC=0; A⋅a + F⋅h = 0Guess A := 1N C := 1NAC⎛⎜⎝⎞⎠:= Find(A, C)AC⎛⎜⎝⎞⎠−1.001.00⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a M1(x1) A⋅x1N⋅m:=0 0.2 00200400600Distane (m)Moment (N-m)M1(x1)x1Max. Moment : unit := N⋅mM1(a) = −400.00Bending Stress: M := M1(a)⋅unit Iπ64do4 − di4 ⎛⎝⎞⎠:= ⋅ c'do2:=σ Mc'I= ⋅ σmax Mc'I:= ⋅ σmax = 152.8MPa Ans 560. Problem 6-65If the beam ACB in Prob. 6-9 has a square cross section, 150 mm by 150 mm, determine the absolutemaximum bending stress in the beam.Given: a := 1m b := 1mc := 1m d := 0.25mao := 150mmF1 := 75kN F2 := 100kNSolution:Equilibrium : Given+ ΣFy=0; A − F1 + B = 0ΣΜC=0; A⋅ (a + b + c) − F1⋅ (b + c) − F2⋅ (d) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠58.3316.67⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cM1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ F1 x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ F1 x3 a − ( ) ⋅ − F2 d ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.5 1 1.5 2 2.5 3100500Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3Max. Moment : unit := kN⋅mM1(a) = 58.33Bending Stress: M := M1(a)⋅unit I412ao:= c'ao2:=σ Mc'I= ⋅ σmax Mc'I:= ⋅ σmax = 103.7MPa Ans 561. Problem 6-66If the crane boom ABC in Prob. 6-10 has a rectangular cross section with a base of 60 mm, determineits required height h to the nearest multiples of 5 mm if the allowable bending stress is σallow = 170 MPa.Given: a := 0.9m b := 1.5m bo := 60mmc := 1.2m W := 6kN σallow := 170MPaSolution: d := a2 + c2vcd:= had:=Equilibrium : Given+ΣFy=0; −Ay + B⋅v − W = 0ΣΜA=0; (−B⋅v)⋅a + W⋅ (a + b) = 0+ ΣFx=0; Ax − B⋅h = 0Guess Ax := 1kN Ay := 1kN B := 1kNAxAyB⎛⎜⎜⎜⎝⎞⎟⎠:= Find(Ax ,Ay , B)AxAyB⎛⎜⎜⎜⎝⎞⎟⎠121020⎛⎜⎜⎝⎞= kN⎠x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + bM1(x1) −Ay⋅x1:= M2 x2 ( ) Ay − x2 ( ) ⋅ B v ⋅ ( ) x2kN⋅ma − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 20510Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2Max. Moment : unit := kN⋅mM1(a) = −9.00Bending Stress: M := M1(a)⋅unitI⋅ 312bo ho= c'ho2=σ Mc'I= ⋅ho6 Mbo⋅ (σallow):= ho = 72.76mmUse ho = 75mm Ans 562. Problem 6-67If the crane boom ABC in Prob. 6-10 has a rectangular cross section with a base of 50 mm and aheight of 75 mm, determine the absolute maximum bending stress in the boom.Given: a := 0.9m b := 1.5m bo := 50mmc := 1.2m W := 6kN ho := 75mmSolution: d := a2 + c2vcd:= had:=Equilibrium : Given+ΣFy=0; −Ay + B⋅v − W = 0ΣΜA=0; (−B⋅v)⋅a + W⋅ (a + b) = 0+ ΣFx=0; Ax − B⋅h = 0Guess Ax := 1kN Ay := 1kN B := 1kNAxAyB⎛⎜⎜⎜⎝⎞⎟⎠:= Find(Ax ,Ay , B)AxAyB⎛⎜⎜⎜⎝⎞⎟⎠121020⎛⎜⎜⎝⎞= kN⎠x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + bM1(x1) −Ay⋅x1:= M2 x2 ( ) Ay − x2 ( ) ⋅ B v ⋅ ( ) x2kN⋅ma − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 20510Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2Max. Moment : unit := kN⋅mM1(a) = −9.00Bending Stress: M := M1(a)⋅unitI⋅ 312bo ho:= c'ho2:=σ Mc'I= ⋅ σmax Mc'I:= ⋅σmax = 192MPa Ans 563. Problem 6-68Determine the absolute maximum bending stress in the beam in Prob. 6-24. The cross section isrectangular with a base of 75 mm and height of 100 mm.Given: a := 0.3m b := 2.4m c := 0.6mbo := 75mm ho := 100mm w 30kNm:=Solution:Equilibrium : Given+ ΣFy=0; A − w⋅b + (qB)⋅c = 0ΣΜA=0; (w⋅b)⋅ (a + 0.5b) − (qB⋅c)⋅ (a + b + 0.5c) = 0Guess A := 1kN qB 1kNm:=AqB⎛⎜⎝⎞⎠:= Find(A, qB) A = 36.00 kN qB 60.00kNm=x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (a + b) x3 := (a + b) , 1.01⋅ (a + b) .. (a + b + c)M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ 0.5w x2 a − ( )2 ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ w b ⋅ ( ) x3⎤⎦1a − 0.5 b ⋅ − ( ) ⋅ − 0.5qB x3 a − b − ( )2 ⋅ + ⎡⎣kN⋅m:= ⋅0 1 2 340200Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3Max. Moment : unit := kN⋅mb' := 0.5⋅bM2(a + b') = 32.40Bending Stress:M := M2(a + b')⋅unitI⋅ 312bo ho:= c'ho2:=σ Mc'I= ⋅ σmax Mc'I:= ⋅σmax = 259.2MPa Ans 564. Problem 6-69Determine the absolute maximum bending stress in the beam in Prob. 6-25. Each segment has arectangular cross section with a base of 100 mm and height of 200 mm.Given: a := 0.9m b := 1.5m c := 2.4mbo := 100mm ho := 200mmP := 40kN w 50kNm:=Solution:Equilibrium : Given+ ΣFy=0; A − P − w⋅c + C = 0ΣΜB=0; (w⋅c)⋅ (0.5c) − C⋅ (c) = 0Guess A := 1kN C := 1kNAC⎛⎜⎝⎞⎠:= Find(A, C)AC⎛⎜⎝⎞⎠10060⎛⎜⎝⎞⎠= kNMA := P⋅a − (C − w⋅c)⋅ (a + b) MA = 180 kN⋅mx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (a + b) x3 := (a + b) , 1.01⋅ (a + b) .. (a + b + c)M1(x1) −MA + A⋅x1:= M2 x2 ( ) MA − A x2 ⋅ + P x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) MA − A x3 ⋅ + P x3 a − ( ) ⋅ − 0.5w x3 a − b − ( )2 ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 2 40200Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3Max. Moment : unit := kN⋅mc' := 0.5⋅cM3(a + b + c') = 36.00Bending Stress:M := M2(a + b + c')⋅unitI⋅ 312bo ho:= coho2:=σ Mc'I= ⋅ σmax McoI:= ⋅σmax = 108MPa Ans 565. Problem 6-70Determine the absolute maximum bending stress in the 20-mm-diameter pin in Prob. 6-35.Given: L := 100mm w 0.4kNm:=a := 0.2L do := 20mmSolution:Equilibrium :+ ΣFy=0; 2(0.5wo)⋅a − w⋅ (3a) = 0wo := 3w wo 1.20kNm=x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (4a) x3 := (4a) , 1.01⋅ (4a) .. LM1(x1) wo:= ⋅ M2(x2) wo⋅a2x1a⎛⎜⎝⎞⎠⋅ ⋅x1x13⋅⎡⎢⎣⎤⎥⎦1N⋅m2x22a3− ⎛⎜⎝⎞⎠⋅ − 0.5w⋅ (x2 − a)2⎡⎢⎣⎤⎥⎦1N⋅m:= ⋅M'3(x3) wo2⋅ (x3 − 4⋅a)2 1x3 − 4⋅aa⎛⎜⎝⎞⎠13− ⋅⎡⎢⎣⎤⎥⎦:= ⋅M3(x3) wo⋅a2x32⋅a3− ⎛⎜⎝⎞⎠⋅ − w⋅ (3a)⋅ (x3 − 2.5⋅a)⎡⎢⎣⎤⎥⎦+ M'3(x3)⎡⎢⎣⎤⎥⎦1N⋅m:= ⋅0 0.02 0.04 0.06 0.080.20.10Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3Max. Moment : unit := N⋅mM2(.5L) = 0.260Bending Stress: M := M2(.5L)⋅unit Iπ64:= ⋅ 4 c'dodo2:=σ Mc'I= ⋅ σmax Mc'I:= ⋅ σmax = 0.331MPa Ans 566. Problem 6-71The member has a cross section with the dimensions shown. Determine the largest internal moment Mthat can be applied without exceeding allowable tensile and compressive stresses of (σ t )allow = 150MPa and (σ c )allow = 100 MPa, respectively.Given: bf := 100mm tf := 10mm rf := 30mmtw := 10mm dw := 180mmσt.allow := 150MPa σc.allow := 100MPaSolution: D := dw + tf + 2rf⎯ Σ yi ⎯y⋅ ( ⋅Ai)Σ⋅ (Ai)=yc2 ⋅ ⎛⎝(bf⋅ tf)⋅0.5tf + (dw⋅ tw)⋅ (0.5dw + tf) π rf⎞⎠+ ⋅ (rf + dw + tf)+ ⋅ 2bf⋅ tf + dw⋅ tw π rf:=yc = 143.41mmIf112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (yc − 0.5tf)2Iw112⋅ 3 dw tw ⋅ ( ) yc 0.5dw tf + ( ) − ⎡⎣:= + ⋅ 2⋅ tw dw⎤⎦I'fπ4:= + ⋅ 2⋅ rf4 π rf2 ⋅ ⎛⎝⎞⎠yc rf dw + tf + ( ) − ⎡⎣⎤⎦I := If + Iw + I'f I = 44639608.23mm4Maximum Bending Stress: σ McI= ⋅Assume failure due to tensile stress.ct.max := yc σt.max Mct.maxI= ⋅ Mtσt.allow⋅ Ict.max:= Mt = 46.69 kN⋅mAssume failure due to compressive stress.cc.max := D − yc σc.max Mcc.maxI= ⋅ Mcσc.allow⋅ Icc.max:= Mc = 41.88 kN⋅mMallow := min(Mt ,Mc)Mallow = 41.88 kN⋅m Ans 567. Problem 6-72Determine the absolute maximum bending stress in the 30-mm-diameter shaft which is subjected to thconcentrated forces. The sleeve bearings at A and B support only vertical forces.Given: a := 0.8m b := 1.2m c := 0.6mF1 := 0.6kN F2 := 0.4kNdo := 30mmSolution: L := a + b + cEquilibrium : Given+ΣFy=0; A − F1 − F2 + B = 0ΣΜB=0; A⋅ (b) − F1⋅ (a + b) + F2⋅ (c) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠0.80.2⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cM1(x1) −F1⋅x1:= M2 x2 ( ) F1 − x2 ( ) ⋅ A x2 a − ( ) ⋅ + ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅M3 x3 ( ) F1 − x3 ( ) ⋅ A x3 a − ( ) ⋅ + B x3 a − b − ( ) ⋅ + ⎡⎣⎤⎦1N⋅m:= ⋅0 0.5 1 1.5 2 2.50200400600Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3Max. Moment : unit := N⋅mM1(a) = −480.000Bending Stress: M := M1(a)⋅unit Iπ64:= ⋅ 4 c'dodo2:=σ Mc'I= ⋅ σmax Mc'I:= ⋅ σmax = 181.1MPa Ans 568. Problem 6-73Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces.The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is σallow160 MPa.Given: a := 0.8m b := 1.2m c := 0.6mF1 := 0.6kN F2 := 0.4kNσallow := 160MPaSolution: L := a + b + cEquilibrium : Given+ΣFy=0; A − F1 − F2 + B = 0ΣΜB=0; A⋅ (b) − F1⋅ (a + b) + F2⋅ (c) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠0.80.2⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cM1(x1) −F1⋅x1:= M2 x2 ( ) F1 − x2 ( ) ⋅ A x2 a − ( ) ⋅ + ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅M3 x3 ( ) F1 − x3 ( ) ⋅ A x3 a − ( ) ⋅ + B x3 a − b − ( ) ⋅ + ⎡⎣⎤⎦1N⋅m:= ⋅0 0.5 1 1.5 2 2.50200400600Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3Max. Moment : unit := N⋅mM1(a) = −480.000Bending Stress: M := M1(a)⋅unit Iπ64= ⋅ 4 c'dodo2=σ Mc'I= ⋅ σallow M32π do= ⋅ do⋅ 33 32 Mπ σallow:= do = 31.26mm Ans 569. Problem 6-74Determine the absolute maximum bending stress in the 40-mm-diameter shaft which is subjected to theconcentrated forces. The sleeve bearings at A and B support only vertical forces.Given: a := 300mm b := 450mmc := 375mm do := 40mmF1 := 2kN F2 := 1.5kNSolution:Equilibrium : Given+ ΣFy=0; A − F1 + B − F2 = 0ΣΜB=0; A⋅ (a + b) − F1⋅ (b) + F2⋅ (c) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠0.453.05⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cV1(x1) A1kN:= ⋅ V2(x2) (A − F1) 1:= ⋅ V3(x3) (A − F1 + B) 1kNkN:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ F1 x2 a − ( ) ⋅ − ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ F1 x3 a − ( ) ⋅ − B x3 a − b − ( ) ⋅ + ⎡⎣⎤⎦1N⋅m:= ⋅Distance (m) Shear (kN)0 0.2 0.4 0.6 0.8 121012V1(x1)V2(x2)V3(x3)x1, x2, x3 570. 0 0.2 0.4 0.6 0.8 150005001000Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3Max. Moment : unit := N⋅mM3(a + b) = −562.50Bending Stress:M := M2(a + b)⋅unit I⋅ 464π do:= codo2:=σ Mc'I= ⋅ σmax McoI:= ⋅σmax = 89.52MPa Ans 571. Problem 6-75Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces.The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is σallow =150 MPa.Given: a := 300mm b := 450mmc := 375mm σallow := 150MPaF1 := 2kN F2 := 1.5kNSolution:Equilibrium : Given+ ΣFy=0; A − F1 + B − F2 = 0ΣΜB=0; A⋅ (a + b) − F1⋅ (b) + F2⋅ (c) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠0.453.05⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cM1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ F1 x2 a − ( ) ⋅ − ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ F1 x3 a − ( ) ⋅ − B x3 a − b − ( ) ⋅ + ⎡⎣⎤⎦1N⋅m:= ⋅0 0.5 150005001000Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3Max. Moment : unit := N⋅mM3(a + b) = −562.50Bending Stress:M := −M3(a + b)⋅unitI⋅ 464π do= codo2=σ McoI= ⋅do3 32Mπ⋅σallow:= do = 33.68mm Ans 572. Problem 6-76The bolster or main supporting girder of a truck body is subjected to the uniform distributed load.Determine the bending stress at points A and B.Given: L1 := 2.4m L2 := 3.6mb := 150mm tf := 20mmd := 300mm tw := 12mmw 25kNm:=Solution:By symmetry : F1 = R F2 = REquilibrium :+ ΣFy=0; −w⋅ (L1 + L2) + 2R = 0R := 0.5⋅w⋅ (L1 + L2):= − ⋅ 2MAB R⋅L1 0.5w L1ftSection properties : D := d + 2⋅I112b D3 ⋅ b tw − ( ) d3 ⋅ − ⎡⎣⎤⎦:= ⋅Bending Stress: σ McI= ⋅cBd2:= σB MABcBI:= ⋅σB = 89.6MPa AnscAd2:= − tf σA MABcAI:= ⋅σA = 77.65MPa Ans 573. Problem 6-77A portion of the femur can be modeled as a tube having an inner diameter of 9.5 mm and an outerdiameter of 32 mm. Determine the maximum elastic static force P that can be applied to its centerwithout causing failure. Assume the bone to be roller supported at its ends. The σ-ε diagram for thebone mass is shown and is the same in tension as in compression.Given: L1 := 100mm L2 := 100mmdi := 9.5mm do := 32mmεe 0.02mmmm:= σe := 8.75MPaεr 0.06mmmm:= σr := 16.1MPaSolution:By symmetry : R = 0.5PMmax = R⋅L1 Mmax = 0.5P⋅L1Section properties :Iπ64do4 − di4 ⎛⎝⎞⎠:= ⋅Bending Stress: σ McI= ⋅ cdo2=M2σ⋅ Ido=Requires: σmax = σeP2σe⋅ I(0.5⋅L1)⋅do:=P = 558.6N Ans 574. Problem 6-78If the beam in Prob. 6-20 has a rectangular cross section with a width of 200 mm and a height of 400mm, determine the absolute maximum bending stress in the beam.Given: a := 2.4m b := 1.2mP1 := 50kNw 30kNm:= M2 := 60kN⋅m P2 := 40kNbo := 200mm do := 400mmSolution:Equilibrium :+ΣF A := w⋅a + P1 + P2 A = 162 kN y=0;ΣΜ MA := (w⋅a)⋅ (0.5a) + P1⋅a + P2⋅ (a + b) + M2 A=0;MA = 410.40 kN⋅mAs indicated in the moment diagram, the maximum moment is MA.Section properties :I112:= ⋅ 3⋅bo doBending Stress: σ McI= ⋅codo2:= σmax MAcoI:= ⋅σmax = 76.95MPa Ans 575. Problem 6-79If the shaft has a diameter of 37.5 mm, determine the absolute maximum bending stress in the shaft.Given: a := 450mm b := 600mm c := 300mmdo := 37.5mm F1 := 1000N F2 := 750NSolution:Equilibrium : Given+ ΣFy=0; A 2F1− + B − 2F2 = 0ΣΜB=0; −2F1⋅ (a + b) + A⋅ (b) + 2F2⋅ (c) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠2.750.75⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cM1(x1) −2F1⋅x1:= M2 x2 ( ) 2 − F1 x2 ( ) ⋅ A x2 a − ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) 2 − F1 x3 ( ) ⋅ A x3 a − ( ) ⋅ + B x3 a − b − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.2 0.4 0.6 0.8 1 1.200.51Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3Max. Moment : unit := kN⋅mM1(a) = −0.900Bending Stress:Mmax := M1(a)⋅unit I⋅ 464π do:= codo2:=σ Mc'I= ⋅ σmax MmaxcoI:= ⋅ σmax = 173.84MPa Ans 576. Problem 6-80If the beam has a square cross section of 225 mm on each side, determine the absolute maximumbending stress in the beam.Given: a := 2.5m b := 2.5mP := 6kN w 15kNm:=bo := 225mm do := 225mmSolution:Equilibrium :+ΣF A := w⋅a + P A = 43.5 kN y=0;ΣΜ MA := (w⋅a)⋅ (0.5a) + P⋅ (a + b) A=0;MA = 76.88 kN⋅mx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + bM1(x1) −MA + A⋅x1 0.5w x12 ⋅ − ⎛⎝⎞⎠1kN⋅m:= ⋅M2 x2 ( ) MA − A x2 ⋅ + w a ⋅ ( ) x20.5 a ⋅ − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 2 4050Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2As indicated in the moment diagram, themaximum moment is MA.Section properties :I112:= ⋅ 3⋅bo doBending Stress: σ McI= ⋅codo2:= σmax MAcoI:= ⋅σmax = 40.49MPa Ans 577. Problem 6-81The beam is subjected to the load P at its center. Determine the placement a of the supports so that thabsolute maximum bending stress in the beam is as large as possible. What is this stress?Solution:Equilibrium : By symmetry, A=B=R+ ΣFy=0; 2R − P = 0R = 0.5PMax. Moment :Mmax = R⋅ (0.5L − a)Mmax = 0.5P⋅ (0.5L − a)For the largest Mmax require, a := 0 AnsMmaxP⋅L4=Bending Stress: σ Mc'I= ⋅Ib⋅d312= c'd2= σmax Mmaxc'I= ⋅σmaxP⋅L23b⋅d2= ⋅ Ans 578. Problem 6-82If the beam in Prob. 6-23 has a cross section as shown, determine the absolute maximum bendingstress in the beam.Given: Mo := 30kN⋅m w 30kNm:=a := 1.5m b := 100mm tf := 12mmd := 168mm tw := 6mmSolution:Equilibrium : Given+ ΣFy=0; −w⋅a + A − w⋅a + B = 0ΣΜB=0; Mo − (w⋅a)⋅ (2.5a) + A⋅ (2a) − (w⋅a)⋅ (0.5a) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠57.5032.50⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 2a x3 := 2a , 1.01⋅2a .. 3aM1(x1) Mo 0.5w x1− ⋅ 2kN⋅m:= M2 x2 ( ) Mo w a ⋅ ( ) x20.5a − ( ) ⋅ − A x2 a − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅M3 x3 ( ) Mo w a ⋅ ( ) x30.5a − ( ) ⋅ − A x3 a − ( ) ⋅ + 0.5w x3 2a − ( )2 ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅Distance (m) Moment (kN-m)0 1 2 3 4200M1(x1)M2(x2)M3(x3)x1, x2, x3As indicated in the momentdiagram, the maximummoment is Mo.Section properties : D d 2tf:= + ⋅ I112b D3 ⋅ b tw − ( ) d3 ⋅ − ⎡⎣⎤⎦:= ⋅Bending Stress: σ McI= ⋅coD2:= σmax MocoI:= ⋅ σmax = 131.87MPa Ans 579. Problem 6-83The pin is used to connect the three links together. Due to wear, the load is distributed over the top andbottom of the pin as shown on the free-body diagram. If the diameter of the pin is 10 mm, determinethe maximum bending stress on the cross-sectional area at the center section a-a. For the solution it isfirst necessary to determine the load intensities w1 and w2 .Given: P := 2kNa := 25mm b := 37.5mmdo := 10mmSolution:Maa −Pa3b2+ ⎛⎜⎝⎞⎠⋅ Pb2⎛⎜⎝⎞⎠:= + ⋅Maa = −16.6667N⋅mBending Stress:I⋅ 464π do:= codo2:=σ Mc'I= ⋅ σmax MaacoI:= ⋅σmax = 169.77MPa Ans 580. Problem 6-84A shaft is made of a polymer having an elliptical cross-section. If it resists an internal moment of M =50 N·m determine the maximum bending stress developed in the material (a) using the flexure formulwhere Iz = 1/4 π (0.08 m)(0.04 m)3, (b) using integration. Sketch a three-dimensional view of thestress distribution acting over the cross-sectional area.Given: a := 80mm b := 40mmMz := 50N⋅mSolution:a) Using the flexure formula, cmax := bIzπ⋅a⋅b34:= Iz = 4021238.60mm4σmax MzcmaxIz:= ⋅σmax = 0.497MPa Ansb) Using integration,y2b2z2a2+ = 1 zab= ⋅ b2 − y2Izoy2 A ⌠⎮⌡Ad⎛⎜⎜⎝⎞⎠= Izoby2⋅ (2z) y⌠⎮⌡= d− bIzob2y2 y a− b⋅ b2 − y2 b⎛⎜⎝⎞⎠⋅⌠⎮⎮⌡:= dIzo = 4021631.98mm4Bending Stress: σ'max MzcmaxIzo:= ⋅σ'max = 0.497MPa Ans 581. Problem 6-85Solve Prob. 6-84 if the moment is M = 50 N·m, applied about the y axis instead of the z axis. HereIy = 1/4 π (0.04 m)(0.08 m)3.Given: a := 80mm b := 40mmMy := 50N⋅mSolution:a) Using the flexure formula, cmax := aIyπ⋅b⋅a34:= Iy = 16084954.39mm4σmax MycmaxIy:= ⋅σmax = 0.249MPa Ansb) Using integration,y2b2z2a2+ = 1 yba= ⋅ a2 − z2Iyoz2 A ⌠⎮⌡Ad⎛⎜⎜⎝⎞⎠= Iyobz2⋅ (2y) z⌠⎮⌡= d− bIyoa2z2 z b− a⋅ a2 − z2 a⎛⎜⎝⎞⎠⋅⌠⎮⎮⌡:= dIyo = 16086527.94mm4Bending Stress: σ'max MycmaxIyo:= ⋅σ'max = 0.249MPa Ans 582. Problem 6-86The simply supported beam is made from four 16-mm-diameter rods, which are bundled as shown.Determine the maximum bending stress in the beam due to the loading shown.Given: L1 := 0.5m L2 := 1.5mdo := 16mm P := 400NSolution:By symmetry : F1 = R F2 = REquilibrium :+ ΣFy=0; −2P + 2⋅R = 0 R := PMmax := R⋅ (L1 + 0.5L2) − P⋅ (0.5L2)Section properties : Aπ4:= ⋅ 2doI 4π64⋅ 4 Adodo2⎛⎜⎝⎞⎠2+ ⋅⎡⎢⎣⎤⎥⎦:= ⋅Bending Stress: σ McI= ⋅cmax := do σmax MmaxcmaxI:= ⋅σmax = 49.74MPa Ans 583. Problem 6-87Solve Prob. 6-86 if the bundle is rotated 45° and set on the supports.Given: L1 := 0.5m L2 := 1.5mdo := 16mm P := 400NSolution:By symmetry : F1 = R F2 = REquilibrium :+ ΣFy=0; −2P + 2⋅R = 0 R := PMmax := R⋅ (L1 + 0.5L2) − P⋅ (0.5L2)Section properties : Aπ4:= ⋅ do2 d' 0.5 do:= ⋅ + 22 doI 2π⋅ 64do⎛⎜⎝4 ⎞⎠⋅ 2π⋅ 64do4 + A⋅d'2 ⎛⎜⎝⎞⎠:= + ⋅Bending Stress: σ McI= ⋅cmax d'do2:= + σmax MmaxcmaxI:= ⋅σmax = 60.04MPa Ans 584. Problem 6-88The steel beam has the cross-sectional area shown. Determine the largest intensity of distributed loadw0 that it can support so that the maximum bending stress in the beam does not exceed σ max = 150MPa.Given: L1 := 4m L2 := 4mb := 200mm t := 8mmd := 250mm σmax := 150MPaSolution:By symmetry : F1 = R F2 = REquilibrium :+ ΣFy=0; −0.5wo⋅ (L1 + L2) + 2R = 0R = 0.25⋅wo⋅ (L1 + L2)Mmax R⋅L1 (0.5wo⋅L1) L13= − ⋅Mmax 0.25⋅wo⋅ (L1 + L2)⋅L1 (0.5wo⋅L1) L13= − ⋅Section properties : D := d + 2⋅ tI112b D3 ⋅ b t − ( ) d 3⋅ − ⎡⎣⎤⎦:= ⋅Bending Stress: σ McI= ⋅ cmaxD2:=σmax 0.25⋅wo⋅ (L1 + L2)⋅L1 (0.5wo⋅L1) L13− ⋅⎡⎢⎣⎤⎥⎦D2I= ⋅woσmax⋅ (2I)D10.25⋅ (L1 + L2)⋅L1 (0.5L1) L13− ⋅:= ⋅wo 13.47kNm= Ans 585. Problem 6-89The steel beam has the cross-sectional area shown. If w0 = 10 kN/m, determine the maximum bendingstress in the beam.Given: L1 := 4m L2 := 4mb := 200mm t := 8mmd := 250mm wo 10kNm:=Solution:By symmetry : F1 = R F2 = REquilibrium :+ ΣFy=0; −0.5wo⋅ (L1 + L2) + 2R = 0R := 0.25⋅wo⋅ (L1 + L2)Mmax R⋅L1 (0.5wo⋅L1) L13:= − ⋅Section properties : D := d + 2⋅ tI112b D3 ⋅ b t − ( ) d 3⋅ − ⎡⎣⎤⎦:= ⋅Bending Stress: σ McI= ⋅cmaxD2:= σmax MmaxcmaxI:= ⋅σmax = 111.38MPa Ans 586. Problem 6-90The beam has a rectangular cross section as shown. Determine the largest load P that can besupported on its overhanging ends so that the bending stress in the beam does not exceed σ max = 10MPa.Given: a := 0.5m bo := 50mmσallow := 10MPa do := 100mmSolution:By symmetry : A = R B = REquilibrium :+ ΣFy=0; −2P + 2⋅R = 0 R = PMmax = −P⋅ (1.5a) + R⋅ (0.5a) Mmax = −P⋅aSection properties :I112⎛⎝bo ⋅ do3 ⎞⎠:= ⋅Bending Stress: σ McI= ⋅ cmaxdo2:=σ (P⋅a)do2I= ⋅ Pσallow⋅ (2I):= P = 1.67 kN Ansa⋅doR := Px1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 2a x3 := 2a , 1.01⋅ (2a) .. 3aM1(x1) −P⋅x1:= M2 x2 ( ) P − x2 ( ) ⋅ R x2 a − ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) P − x3 ( ) ⋅ R x3 a − ( ) ⋅ + R x3 2a − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.5 1 1.500.51Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 587. Problem 6-91The beam has the rectangular cross section shown. If P = 1.5 kN, determine the maximum bendingstress in the beam. Sketch the stress distribution acting over the cross section.Given: a := 0.5m bo := 50mmP := 1.5kN do := 100mmSolution:By symmetry : A = R B = REquilibrium :+ ΣFy=0; −2P + 2⋅R = 0 R := PMmax = −P⋅ (1.5a) + R⋅ (0.5a) Mmax := −P⋅a Mmax = −0.75 kN⋅mSection properties :I112⎛⎝bo ⋅ do3 ⎞⎠:= ⋅Bending Stress:cmaxdo2:= σmax MmaxcmaxI:= ⋅ σmax = 9.00MPa Ansx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 2a x3 := 2a , 1.01⋅ (2a) .. 3aM1(x1) −P⋅x1:= M2 x2 ( ) P − x2 ( ) ⋅ R x2 a − ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) P − x3 ( ) ⋅ R x3 a − ( ) ⋅ + R x3 2a − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.5 1 1.500.51Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 588. Problem 6-92The beam is subjected to the loading shown. If its cross-sectional dimension a = 180 mm, determinethe absolute maximum bending stress in the beam.Given: L1 := 2m L2 := 1mP := 60kN a := 180mmw 40kNm:=Solution:Equilibrium : Given+ ΣFy=0; A + B − P − w⋅L1 = 0ΣΜA=0; P⋅ (L1 + L2) − B⋅L1 0.5w L1+ ⋅ 2 = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠10.00130.00⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. L1 + L2M1(x1) A⋅x1 0.5w x12 ⋅ − ⎛⎝⎞⎠1kN⋅m:= ⋅M2 x2 ( ) A x2 ⋅ w L1 ⋅ ( ) x2 0.5 L1 ⋅ − ( ) ⋅ − B x2 L1 − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 30204060Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2Max. Moment : unit := kN⋅mM1(L1) = −60.000Mmax := −M1(L1)⋅unitMmax = 60.00 kN⋅mSection properties :ycΣ yi ⎯⋅ ( ⋅Ai)Σ⋅ (Ai)=ycaa3⋅ ⎛⎜⎝⎞⎠a6⋅a22a3⋅ ⎛⎜⎝⎞⎠2a3⎛⎜⎝⎞⎠+ ⋅:= yc = 75.00mmaa3⋅ ⎛⎜⎝⎞⎠a22a3⋅ ⎛⎜⎝⎞⎠+Iw112a2⎛⎜⎝⎞⎠⋅2a3⎛⎜⎝⎞⎠3⋅a22a3⋅ ⎛⎜⎝⎞⎠yc2a3⎛⎜⎝⎞⎠− ⎡⎢⎣⎤⎥⎦2:= + ⋅If112⋅ (a)a3⎛⎜⎝⎞⎠3⋅ aa3⋅ ⎛⎜⎝⎞⎠yca6⎛⎜⎝⎞⎠− ⎡⎢⎣⎤⎥⎦2:= + ⋅ 589. I := If + IwBending Stress: σ McI= ⋅ cmax := a − ycσmax (Mmax) cmaxI:= ⋅σmax = 105.11MPa Ans 590. Problem 6-93The beam is subjected to the loading shown. Determine its required cross-sectional dimension a, if theallowable bending stress for the material is σallo w = 150 MPa.Given: L1 := 2m L2 := 1mP := 60kN σallow := 150MPaw 40kNm:=Solution:Equilibrium : Given+ ΣFy=0; A + B − P − w⋅L1 = 0ΣΜA=0; P⋅ (L1 + L2) − B⋅L1 0.5w L1+ ⋅ 2 = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠10.00130.00⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. L1 + L2M1(x1) A⋅x1 0.5w x12 ⋅ − ⎛⎝⎞⎠1kN⋅m:= ⋅M2 x2 ( ) A x2 ⋅ w L1 ⋅ ( ) x2 0.5 L1 ⋅ − ( ) ⋅ − B x2 L1 − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 30204060Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2Max. Moment : unit := kN⋅mM1(L1) = −60.000Mmax := −M1(L1)⋅unitMmax = 60.00 kN⋅mSection properties :ycΣ yi ⎯⋅ ( ⋅Ai)Σ⋅ (Ai)=ycaa3⋅ ⎛⎜⎝⎞⎠a6⋅a22a3⋅ ⎛⎜⎝⎞⎠2a3⎛⎜⎝⎞⎠+ ⋅= ycaa3⋅ ⎛⎜⎝⎞⎠a22a3⋅ ⎛⎜⎝⎞⎠+5a12=Iw112⎞⎠⋅a2⎛⎜⎝2a3⎛⎜⎝⎞⎠3⋅a22a3⋅ ⎛⎜⎝⎞⎠yc2a3⎛⎜⎝⎞⎠− ⎡⎢⎣⎤⎥⎦2= + ⋅ Iw43a41296=If112⋅ (a)a3⎛⎜⎝⎞⎠3⋅ aa3⋅ ⎛⎜⎝⎞⎠yca6⎛⎜⎝⎞⎠− ⎡⎢⎣⎤⎥⎦2= + ⋅ If31a41296= 591. I = If + Iw I37a4648=Bending Stress: σ McI= ⋅ cmax = a − yc cmax7a12=I (Mmax) cmaxσallow= ⋅a3648Mmax712⎛⎜⎝⎞⎠⋅37σallow:=a = 159.88mm Ans 592. Problem 6-94The wing spar ABD of a light plane is made from 2014T6 aluminum and has a cross-sectional area of1000 mm2, a depth of 80 mm, and a moment of inertia about its neutral axis of 1.662 (106) mm4.Determine the absolute maximum bending stress in the spar if the anticipated loading is to be as shown.Assume A, B, and C are pins. Connection is made along the central longitudinal axis of the spar.Given: a := 1m b := 2mA := 1000m2 wo 15kNm:=I := 1.662⋅ (106)mm4do := 80mmSolution: L := a + bEquilibrium : Given+ ΣFy=0; A − B + 0.5wo⋅ (L) = 0ΣΜA=0; B⋅a (0.5wo⋅L) L3⎛⎜⎝⎞⎠− ⋅ = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠−0.0022.50⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + bM1(x1) A⋅x1wo2⋅ 2 1x1x1L⎛⎜⎝⎞⎠13− ⋅⎡⎢⎣⎤⎥⎦+ ⋅⎡⎢⎣⎤⎥⎦1kN⋅m:= ⋅M2 x2 ( ) A x2 ⋅ B x2 a − ( ) ⋅ − ⎡⎣⎤⎦wo2⎡⎢⎣ ⎤⎥⎦⋅ 2 1x2x2L⎛⎜⎝⎞⎠13− ⋅+ ⋅⎡⎢⎣⎤⎥⎦1kN⋅m:= ⋅0 1 2 350Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2Max. Moment : unit := kN⋅mM1(a) = 6.667Mmax := M1(a)⋅unitBending Stress:σ McI= ⋅ cmaxdo2:=σmax MmaxcmaxI:= ⋅σmax = 160.45MPa Ans( < σY = 414 MPa) 593. Problem 6-95The boat has a weight of 11.5 kN and a center of gravity at G. If it rests on the trailer at the smoothcontact A and can be considered pinned at B, determine the absolute maximum bending stressdeveloped in the main strut of the trailer. Consider the strut to be a box-beam having the dimensionsshown and pinned at C.Given: a := 0.9m b := 1.8mc := 1.2m d := 0.3mbo := 45mm do := 75mmbi := 38mm di := 45mmSolution: W := 11.5kNEquilibrium (for boat) : Given+ ΣFy=0; A − W + B = 0ΣΜB=0; A⋅ (a + b) − W⋅ (b − d) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠6.3895.111⎛⎜⎝⎞⎠= kNEquilibrium (for assembly) : Given+ ΣFy=0; D − A − B + C = 0ΣΜC=0; −A⋅ (a + b + c) + D⋅ (b + c) − B⋅c = 0Guess C := 1kN D := 1kNCD⎛⎜⎝⎞⎠:= Find(C,D)CD⎛⎜⎝⎞⎠1.1510.35⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cM1(x1) −A⋅x1:= M2 x2 ( ) A − x2 ( ) ⋅ D x2 a − ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅ 594. M3 x3 ( ) A − x3 ( ) ⋅ D x3 a − ( ) ⋅ + B x3 a − b − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 305Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3Max. Moment : unit := kN⋅mM1(a) = −5.750Mmax := −M1(a)⋅unitSection properties :I112bo ⋅ do3 − bi ⋅ di3 ⎛⎝⎞⎠:= ⋅Bending Stress:σ McI= ⋅ cmaxdo2:=σmax MmaxcmaxI:= ⋅σmax = 166.7MPa Ans 595. Problem 6-96The beam supports the load of 25 kN. Determine the absolute maximum bending stress in the beam ifthe sides of its triangular cross section are a = 150 mm.Given: a := 150mm L := 0.6m P := 25kNSolution:Mmax := P⋅LSection Property :I136:= ⋅a⋅ (a⋅ sin(60deg))3Maximum Bending Stress: σ McI= ⋅cmax23:= ⋅a⋅ sin(60deg)σmax (Mmax) cmaxI:= ⋅σmax = 142.2MPa Ans 596. Problem 6-97The beam supports the load of 25 kN. Determine the required size a of the sides of its triangular crosssection if the allowable bending stress is σ allow = 126 MPa.Given: L := 0.6m P := 25kN σallow := 126MPaSolution:Mmax := P⋅LSection Property :I136= ⋅a⋅ (a⋅ sin(60deg))3Maximum Bending Stress: σ McI= ⋅cmax23= ⋅a⋅ sin(60deg)σ (Mmax) cmaxI= ⋅I (Mmax) cmaxσallow= ⋅a24sin(60deg)2Mmaxσallow⋅⎛⎜⎝⎞⎠13:=a = 156.2mm Ans 597. Problem 6-98The wood beam is subjected to the uniform load of w = 3 kN/m. If the allowable bending stress for thematerial is σ allow = 10 MPa, determine the required dimension b of its cross section. Assume thesupport at A is a pin and B is a roller.Given: L1 := 2m σallow := 10MPaL2 := 1m w 3kNm:=Solution:Equilibrium : Given+ ΣFy=0; A + B − w⋅L1 = 0ΣΜA=0; −B⋅ (L1 + L2) 0.5w L1+ ⋅ 2 = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠4.002.00⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. L1 + L2M1(x1) A⋅x1 0.5w x12 ⋅ − ⎛⎝⎞⎠1kN⋅m:= ⋅M2 x2 ( ) A x2 ⋅ w L1 ⋅ ( ) x2 0.5 L1 ⋅ − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 320Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2As indicated in the moment diagram, themaximum moment occurs in L1 such thatV1=0:V1 A w xc= − ⋅ ( )xcAw:= xc = 1.333 mMax. Moment : unit := kN⋅mM1(xc) = 2.667Mmax := M1(xc)⋅unitMmax = 2.667 kN⋅mSection properties : I112= ⋅b⋅ (1.5⋅b)3Bending Stress: σ McoI= ⋅ co1.5⋅b2=I Mmaxcoσallow= ⋅ b6⋅Mmax2.25⋅σallow⎛⎜⎝⎞⎠13:= b = 89.3mm Ans 598. Problem 6-99The wood beam has a rectangular cross section in the proportion shown. Determine its requireddimension b if the allowable bending stress is σ allow = 10 MPa.Given: L1 := 2m σallow := 10MPaL2 := 2m w 0.5kNm:=Solution:Equilibrium : Given+ ΣFy=0; A + B − w⋅L1 = 0ΣΜA=0; −B⋅ (L1 + L2) 0.5w L1+ ⋅ 2 = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠0.750.25⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. L1 + L2M1(x1) A⋅x1 0.5w x12 ⋅ − ⎛⎝⎞⎠1kN⋅m:= ⋅M2 x2 ( ) A x2 ⋅ w L1 ⋅ ( ) x2 0.5 L1 ⋅ − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 210.50Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2As indicated in the moment diagram, themaximum moment occurs in L1 such thatV1=0:V1 A w xc= − ⋅ ( )xcAw:= xc = 1.500 mMax. Moment : unit := kN⋅mM1(xc) = 0.5625Mmax := M1(xc)⋅unitMmax = 0.5625 kN⋅mSection properties : I112= ⋅b⋅ (1.5⋅b)3Bending Stress: σ McoI= ⋅ co1.5⋅b2=I Mmaxcoσallow= ⋅ b3 6⋅Mmax2.25⋅σallow:= b = 53.1mm Ans 599. Problem 6-100A beam is made of a material that has a modulus of elasticity in compression different from that givenfor tension. Determine the location c of the neutral axis, and derive an expression for the maximumtensile stress in the beam having the dimensions shown if it is subjected to the bending moment M. 600. Problem 6-101The beam has a rectangular cross section and is subjected to a bending moment M. If the materialfrom which it is made has a different modulus of elasticity for tension and compression as shown,determine the location c of the neutral axis and the maximum compressive stress in the beam. 601. Problem 6-102The box beam is subjected to a bending moment of M = 25 kN·m directed as shown. Determine themaximum bending stress in the beam and the orientation of the neutral axis.Given: ay := −3 az := 4 ar := 5 M := 25kN⋅mbo := 150mm do := 150mmbi := 100mm di := 100mmSolution:Internal Moment Components :Myayar:= ⋅M Mzazar:= ⋅MSection Property :Iy112do ⋅ bo3 − di ⋅ bi3 ⎛⎝⎞⎠:= ⋅ Iz112bo ⋅ do3 − bi ⋅ di3 ⎛⎝⎞⎠:= ⋅Maximum Bending Stress: By inspection, maximum bending stress occurs at B and D.σMz⋅yIz−My⋅zIy= +At B : yB := 0.5do zB := 0.5boσBMz⋅yBIz−My⋅zBIy:= +σB = −77.5MPa (C) AnsAt D : yD := −0.5do zD := −0.5boσDMz⋅yDIz−My⋅zDIy:= +σD = 77.5MPa (T) AnsOrientation of Neutral Axis : tan(α) Iz= ⋅ tan(θ)Iyθ atanayaz⎛⎜⎝⎞⎠:= α atanIzIy⋅ tan(θ)⎛⎜⎝⎞⎠:= α = −36.87 deg Ansy' := bo⋅ tan(α) y' = −112.50mm 602. Problem 6-103Determine the maximum magnitude of the bending moment M so that the bending stress in the memberdoes not exceed 100 MPa.Given: ay := −3 az := 4 ar := 5 σallow := 100MPabo := 150mm do := 150mmbi := 100mm di := 100mmSolution:Internal Moment Components :Myayar= ⋅M Mzazar= ⋅MSection Property :Iy112do ⋅ bo3 − di ⋅ bi3 ⎛⎝⎞⎠:= ⋅ Iz112bo ⋅ do3 − bi ⋅ di3 ⎛⎝⎞⎠:= ⋅Maximum Bending Stress: By inspection, maximum bending stress occurs at B and D.Apply the flexure formula for biaxial bending at either point B or D. σMz⋅yIz−My⋅zIy= +At B : yB := 0.5do zB := 0.5boσBMz⋅yBIz−My⋅zBIy= +σallowazar⋅M⎛⎜⎝⎞⎠−yBIz⋅ayar⋅M⎛⎜⎝⎞⎠zBIy= + ⋅Mσallowazar−yBIz⋅ayar⎛⎜⎝ ⎞⎠zBIy+ ⋅:=M = 32.24 kN⋅m Ans 603. Problem 6-104The beam has a rectangular cross section. If it is subjected to a bending moment of M = 3500 N·mdirected as shown, determine the maximum bending stress in the beam and the orientation of theneutral axis.Given: M := 3.5kN⋅m θ' := 30degb := 150mm d := 300mmSolution: θ := (180deg − θ') θ = 150 degInternal Moment Components :My := M⋅ sin(θ) Mz := M⋅cos(θ)Section Property :Iy112:= ⋅d⋅b3 Iz112:= ⋅b⋅d3Maximum Bending Stress: σMz⋅yIz−My⋅zIy= +At A : yA := 0.5d zA := 0.5bσAMz⋅yAIz−My⋅zAIy:= + σA = 2.903MPa (T) AnsAt B : yB := −0.5d zB := −0.5bσBMz⋅yBIz−My⋅zBIy:= + σB = −2.903MPa (C) AnsAt C : yC := 0.5d zC := −0.5bσCMz⋅yCIz−My⋅zCIy:= + σC = −0.208MPa (C)At D : yD := −0.5d zD := 0.5bσDMz⋅yDIz−My⋅zDIy:= + σD = 0.208MPa (T)Orientation of Neutral Axis : tan(α) Iz= ⋅ tan(θ)Iyα atanIzIy⋅ tan(θ)⎛⎜⎝⎞⎠:= α = −66.59 deg Ans0.5dtan(α) := − z' = 10.05mmz' 0.5⋅b 604. Problem 6-105The T-beam is subjected to a bending moment of M = 15 kN·m. directed as shown. Determine themaximum bending stress in the beam and the orientation of the neutral axis. The location of thecentroid, C, must be determined.Given: M := 15kN⋅m θ' := 60degbf := 300mm tf := 50mmtw := 50mm dw := 200mmSolution: θ := (180deg − θ') θ = 120 degInternal Moment Components :My := M⋅ sin(θ) Mz := M⋅cos(θ)Section Property :Iy112:= ⋅ Iy = 110416666.67mm4tf ⋅ bf3 − dw ⋅ tw3 ⎛⎝⎞⎠ycΣ yi ⎯⋅ ( ⋅Ai)Σ⋅ (Ai)= yc(bf⋅ tf)⋅ (0.5tf) + (tw⋅dw)⋅ (0.5dw + tf):= yc = 75.00mm(bf⋅ tf) + (tw⋅dw)Iz112⋅ 3 b( f⋅ tf) (0.5tf − yc)+ ⋅ 2⋅bf tf1⋅ ⋅ 12tw dw3 + ⎡⎢⎣t( w⋅dw) ⋅ (0.5dw + tf − yc)2 ⎤⎥⎦:= +Iz = 130208333.33mm4Maximum Bending Stress: σMz⋅yIz−My⋅zIy= +At A : yA := yc zA := 0.5bfσAMz⋅yAIz−My⋅zAIy:= + σA = 21.97MPa (T) AnsAt B : yB := yc zB := −0.5bf 605. σBMz⋅yBIz−My⋅zBIy:= + σB = −13.33MPa (C)At D : yD := −(tf + dw − yc) zD := −0.5twσDMz⋅yDIz−My⋅zDIy:= + σD = −13.02MPa (C)Orientation of Neutral Axis : tan(α) Iz= ⋅ tan(θ)Iyα atanIzIy⋅ tan(θ)⎛⎜⎝⎞⎠:= α = −63.91 deg Ansyctan(α) := − z' = 186.72mmz' 0.5bf 606. Problem 6-106If the resultant internal moment acting on the cross section of the aluminum strut has a magnitude ofM = 520 N·m and is directed as shown, determine the bending stress at points A and B. The location yof the centroid C of the strut's cross-sectional area must be determined.Also, specify the orientation othe neutral axis.θ' atan512⎛⎜⎝⎞⎠Given: M := 520N⋅m :=bf := 400mm tf := 20mmtw := 20mm dw := 180mmSolution: D := dw + tf θ := (180deg − θ')Internal Moment Components :My := M⋅ sin(θ) Mz := M⋅cos(θ)Section Property :Iy112⋅ 3 dw (bf − 2tw)3 ⋅ − ⎡⎣:= ⋅ D bfIy = 366826666.67mm4⎤⎦ycΣ yi ⎯⋅ ( ⋅Ai)Σ⋅ (Ai)= yc(bf⋅ tf)⋅ (0.5tf) + 2(tw⋅dw)⋅ (0.5dw + tf):= yc = 57.37mm(bf⋅ tf) + 2(tw⋅dw)Iz112⋅ 3 b( f⋅ tf) (0.5tf − yc)+ ⋅ 2 2⋅bf tf1⋅ ⋅ 12tw dw3 + ⎡⎢⎣t( w⋅dw) ⋅ (0.5dw + tf − yc)2 ⎤⎥⎦:= + ⋅Iz = 57601403.51mm4Maximum Bending Stress: σMz⋅yIz−My⋅zIy= +At A : yA := yc − D zA := −0.5bfσAMz⋅yAIz−My⋅zAIy:= + σA = −1.298MPa (C) AnsAt B : yB := yc zB := 0.5bfσBMz⋅yBIz−My⋅zBIy:= + σB = 0.587MPa (T)Orientation of Neutral Axis : tan(α) Iz= ⋅ tan(θ)Iyα atanIzIy⋅ tan(θ)⎛⎜⎝⎞⎠:= α = −3.74 deg Ans 607. Problem 6-107The resultant internal moment acting on the cross section of the aluminum strut has a magnitude ofM = 520 N·m and is directed as shown. Determine the maximum bending stress in the strut. Thelocation y of the centroid C of the strut's cross-sectional area must be determined.Also, specify theorientation of the neutral axis.θ' atan512⎛⎜⎝⎞⎠Given: M := 520N⋅m :=bf := 400mm tf := 20mmtw := 20mm dw := 180mmSolution: D := dw + tf θ := (180deg − θ')Internal Moment Components :My := M⋅ sin(θ) Mz := M⋅cos(θ)Section Property :Iy112⋅ 3 dw (bf − 2tw)3 ⋅ − ⎡⎣:= ⋅ D bfIy = 366826666.67mm4⎤⎦ycΣ yi ⎯⋅ ( ⋅Ai)Σ⋅ (Ai)= yc(bf⋅ tf)⋅ (0.5tf) + 2(tw⋅dw)⋅ (0.5dw + tf):= yc = 57.37mm(bf⋅ tf) + 2(tw⋅dw)Iz112⋅ 3 b( f⋅ tf) (0.5tf − yc)+ ⋅ 2 2⋅bf tf1⋅ ⋅ 12tw dw3 + ⎡⎢⎣t( w⋅dw) ⋅ (0.5dw + tf − yc)2 ⎤⎥⎦:= + ⋅Iz = 57601403.51mm4Maximum Bending Stress: σMz⋅yIz−My⋅zIy= +By inspection, the maximum bending stress can occur at either point A or B.At A : yA := yc − D zA := −0.5bfσAMz⋅yAIz−My⋅zAIy:= + σA = −1.298MPa (C) AnsAt B : yB := yc zB := 0.5bfσBMz⋅yBIz−My⋅zBIy:= + σB = 0.587MPa (T)Orientation of Neutral Axis : tan(α) Iz= ⋅ tan(θ)Iyα atanIzIy⋅ tan(θ)⎛⎜⎝⎞⎠:= α = −3.74 deg Ans 608. Problem 6-108The 30-mm-diameter shaft is subjected to the vertical and horizontal loadings of two pulleys as shownIt is supported on two journal bearings at A and B which offer no resistance to axial loading.Furthermore, the coupling to the motor at C can be assumed not to offer any support to the shaft.Determine the maximum bending stress developed in the shaft.Given: a := 1m Fy := 150N Fz := 400N do := 30mmSolution:Equilibrium : In x-y plane. Given+ ΣFy=0; Ay + By − 2Fy = 0ΣΜA=0; −2Fy⋅a − By⋅ (2a) = 0Guess Ay := 1N By := 1NAyBy⎛⎜⎜⎝⎞⎠:= Find(Ay , By)AyBy⎛⎜⎜⎝⎞⎠0.45−0.15⎛⎜⎝⎞⎠= kNEquilibrium : In x-z plane, by symmetry: Az = Bz = Rz.+ ΣFz=0; 2Rz − 2Fz = 0Rz := Fz Rz = 400 NInternal Moment Components:The shaft is subjected to two bending moment components My and Mz. The moment disgramfor each component is drawn.x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (2a) x3 := (2a) , 1.01⋅ (2a) .. 3aMz1(x1) −2Fy⋅x1:= Mz2 x2 ( ) 2 − Fy x2 ( ) ⋅ Ay x2 a − ( ) ⋅ + ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅Mz3 x3 ( ) 2 − Fy x3 ( ) ⋅ Ay x3 a − ( ) ⋅ + ⎡⎣⎤⎦1N⋅m:= ⋅My1(x1) := 0 My2(x2) Rz⋅ (x2 − a):= My3 x3 ( ) Rz x3 a − ( ) ⋅ 2Fz x3 2a − ( ) ⋅ − ⎡⎣N⋅m⎤⎦1N⋅:= ⋅0 1 2 30200400Distane (m)Moment (N-m)Mz1(x1)Mz2(x2)Mz3(x3)x1, x2, x30 1 25000Distane (m)Moment (N-m)My2(x2)My3(x3)x2, x3 609. Maximum Bending Stress: unit := N⋅mSince all the axes through the circle's center for circular shaft are principal axes, then theresultant moment M = (My2 + Mz2)0.5 can be used to determine the maximum bending stress.The maximum bending stress moment occurs at E (x=2a).Mmax := Mz2(2a)2 + My2(2a)2⋅unit Mmax = 427.20 N⋅mHence,cmaxdo2:= Iπ64:= ⋅ do4 σmax MmaxcmaxI:= ⋅σmax = 161.2MPa AnsM1(x1) := Mz1(x1)2 + My1(x1)2M2(x2) := Mz2(x2)2 + My2(x2)2M3(x3) := Mz3(x3)2 + My3(x3)20 0.5 1 1.5 2 2.5 34002000Distane (m)Moment (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 610. Problem 6-109The shaft is subjected to the vertical and horizontal loadings of two pulleys D and E as shown. It issupported on two journal bearings at A and B which offer no resistance to axial loading.Furthermore,the coupling to the motor at C can be assumed not to offer any support to the shaft.Determine the required diameter d of the shaft if the allowable bending stress for the material is σ allow= 180 MPa.Given: a := 1m Fy := 150N Fz := 400N σallow := 180MPaSolution:Equilibrium : In x-y plane. Given+ ΣFy=0; Ay + By − 2Fy = 0ΣΜA=0; −2Fy⋅a − By⋅ (2a) = 0Guess Ay := 1N By := 1NAyBy⎛⎜⎜⎝⎞⎠:= Find(Ay , By)AyBy⎛⎜⎜⎝⎞⎠0.45−0.15⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 3aMz1(x1) −2Fy⋅x1:= Mz2 x2 ( ) 2 − Fy x2 ( ) ⋅ Ay x2 a − ( ) ⋅ + ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅Equilibrium : In x-z plane, by symmetry: Az = Bz = Rz.+ ΣFz=0; 2Rz − 2Fz = 0Rz := Fz Rz = 400 Nx'1 := a , 1.01⋅a .. (2a) x'2 := (2a) , 1.01⋅ (2a) .. 3aMy1(x'1) Rz⋅ (x'1 − a):= My2 x'2 ( ) Rz x'2 a − ( ) ⋅ 2Fz x'2 2a − ( ) ⋅ − ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅Internal Moment Components:The shaft is subjected to two bending moment components My and Mz. The moment disgramfor each component is drawn.0 1 2 30200400Distane (m)Moment (N-m)Mz1(x1)Mz2(x2)x1, x20 1 25000Distane (m)Moment (N-m)My1(x'1)My2(x'2)x'1, x'2 611. Maximum Bending Stress: unit := N⋅mSince all the axes through the circle's center for circular shaft are principal axes, then theresultant moment M = (My2 + Mz2)0.5 can be used to determine the maximum bending stress.The maximum bending stress moment occurs at E (x=2a).Mmax := Mz2(2a)2 + My1(2a)2⋅unit Mmax = 427.20 N⋅mHence,cmaxdo2= Iπ64= ⋅ do4 σallow MmaxcmaxI= ⋅π64⋅ 4doMmaxσallow⎛⎜⎝⎞⎠do2= ⋅do3 32Mmaxπ⋅σallow:=do = 28.91mm Ans 612. Problem 6-110The board is used as a simply supported floor joist. If a bending moment of M = 1.2 kN·m is applied3° from the z axis, determine the stress developed in the board at the corner A. Compare this stresswith that developed by the same moment applied along the z axis (θ = 0°). What is the angle a for theneutral axis when θ = 3°? Comment: Normally, floor boards would be nailed to the top of the beam sothat θ = 0° (nearly) and the high stress due to misalignment would not occur.Given: M := 1.2kN⋅m θ := 3degb := 50mm d := 150mmSolution:Internal Moment Components :My := −M⋅ sin(θ) Mz := M⋅cos(θ)Section Property : yc := 0.5dIy112:= ⋅d⋅b3 Iz112:= ⋅b⋅d3Maximum Bending Stress: σMz⋅yIz−My⋅zIy= +At A : yA := −yc zA := −0.5bσAMz⋅yAIz−My⋅zAIy:= + σA = 7.40MPa (T) AnsOrientation of Neutral Axis : tan(α) Iz= ⋅ tan(θ)Iyα atanIzIy⋅ tan(θ)⎛⎜⎝⎞⎠:= α = 25.25 deg AnsWhen θ = 0 : M'y := 0 M'z := Mσ'AM'z⋅yAIz−M'y⋅zAIy:= + σ'A = 6.40MPa (T) Ans 613. Problem 6-111Consider the general case of a prismatic beam subjected to bending-moment components My and Mz,as shown, when the x, y, z axes pass through the centroid of the cross section. If the material islinear-elastic, the normal stress in the beam is a linear function of position such that σ = a + by + cz.Using the equilibrium conditions 0 σ dA , M zσ dA , M yσ dA, determine theconstants a, b, and c, and show that the normal stress can be determined from the equationσ = [- (Mz Iy + My Iyz) y + (My Iz + Mz Iyz) z] / (Iy Iz - Iyz2), where the moments and products ofinertia are defined in Appendix A.= ∫ = ∫ = ∫ −A y A z AGiven: Linear function: σx = a + by + zSolution:Equilibrium Conditios :∫ ∫0 = σ dA , 0 = (a + by +cz)dAA x A∫ ∫ ∫0 = a dA + b y dA +c z dAA A AM = zσ dA , M = z (a + by +cz)dAy A x y A∫ ∫ ∫∫ ∫M = a z dA + b yz dA +c z dAA2y A AM = - yσ dA , M = - y (a + by +cz)dA∫ ∫ ∫∫ ∫2z A x z AM = − a y dA − b y dA −c yz dAA Az A∫ = ∫ =A AThe integrals are defined in Appendix A. Note that ydA z dA 0Thus, = ∫ + ⋅ + ⋅0 a dA b 0 c 0Ay yz y M = a ⋅ 0 + b ⋅ I + c ⋅ Iz y yz M = −a ⋅ 0 − b ⋅ I − c ⋅ ISolving for a, b and c : a = 0 (since A ≠ 0) Ansb = - 1 ⋅ + ⋅ = ⋅ 2Ansy yz z y y z yz (M I M I ) where D I I - IDc 1 y z z yz = ⋅ + ⋅ Ans(M I M I )DBending Stress:σ 1 x z y y yz z yz y z = ⋅ + ⋅ ⋅ − + ⋅ + ⋅ ⋅ QED(M I M I ) ( y) 1D(M I M I ) zDIn matrix form,σ1D(Mz My )IyIyzIyzIz⎛⎜⎜⎝⎞⎠⋅−yz⎛⎜⎝⎞⎠= ⋅ 614. Problem 6-112The 65-mm-diameter steel shaft is subjected to the two loads that act in the directions shown. If thejournal bearings at A and B do not exert an axial force on the shaft, determine the absolute maximumbending stress developed in the shaft.Given: a := 1.25m b := 1m F := 4kNL := 2a + b do := 65mm θ := 30degSolution:Equilibrium : In x-y plane, by symmetry: Ay = By = Ry.+ ΣFy=0; 2Ry − 2F⋅ cos(θ) = 0Ry := F⋅cos(θ) Ry = 3.464 kNEquilibrium :In x-z plane, by anti-symmetry: Az = -Bz = Rz.ΣΜB=0; Az⋅L − F⋅ sin(θ)⋅ (b) = 0Rz F⋅ sin(θ)bL⎛⎜⎝⎞⎠:= ⋅ Rz = 0.571 kNInternal Moment Components:The shaft is subjected to two bending moment components My and Mz. The moment disgramfor each component is drawn.x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (a + b) x3 := (a + b) , 1.01⋅ (a + b) .. LMz1(x1) Ry⋅x1:= Mz2 x2 ( ) Ry x2 ( ) ⋅ F cos θ ( ) ⋅ x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅Mz3 x3 ( ) Ry x3 ( ) ⋅ F cos θ ( ) ⋅ x3 a − ( ) ⋅ − F cos θ ( ) ⋅ x3 a − b − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅My1(x1) Rz⋅x1:= My2 x2 ( ) Rz x2 ( ) ⋅ F sin θ ( ) ⋅ x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅My3 x3 ( ) Rz x3 ( ) ⋅ F sin θ ( ) ⋅ x3 a − ( ) ⋅ − F sin θ ( ) ⋅ x3 a − b − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 250Distane (m)Moment (kN-m)Mz1(x1)Mz2(x2)Mz3(x3)x1, x2, x30 2101Distane (m)Moment (kN-m)My1(x1)My2(x2)My3(x3)x1, x2, x3 615. Maximum Bending Stress: unit := kN⋅mSince all the axes through the circle's center for circular shaft are principal axes, then theresultant moment M = (My2 + Mz2)0.5 can be used to determine the maximum bending stress.The maximum bending stress moment occurs at x=a.Mmax := Mz1(a)2 + My1(a)2⋅unit Mmax = 4.389 kN⋅mHence,cmaxdo2:= Iπ64:= ⋅ do4 σmax MmaxcmaxI:= ⋅σmax = 162.8MPa AnsM1(x1) := Mz1(x1)2 + My1(x1)2M2(x2) := Mz2(x2)2 + My2(x2)2M3(x3) := Mz3(x3)2 + My3(x3)20 1 2 350Distane (m)Moment (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 616. Problem 6-113The steel shaft is subjected to the two loads that act in the directions shown. If the journal bearings atA and B do not exert an axial force on the shaft, determine the required diameter of the shaft if theallowable bending stress is σ allow = 180 MPa.Given: a := 1.25m b := 1m F := 4kNL := 2a + b θ := 30deg σallow := 180MPaSolution:Equilibrium : In x-y plane, by symmetry: Ay = By = Ry.+ ΣFy=0; 2Ry − 2F⋅ cos(θ) = 0Ry := F⋅cos(θ) Ry = 3.464 kNEquilibrium :In x-z plane, by anti-symmetry: Az = -Bz = Rz.ΣΜB=0; Az⋅L − F⋅ sin(θ)⋅ (b) = 0Rz F⋅ sin(θ)bL⎛⎜⎝⎞⎠:= ⋅ Rz = 0.571 kNInternal Moment Components:The shaft is subjected to two bending moment components My and Mz. The moment disgramfor each component is drawn.x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (a + b) x3 := (a + b) , 1.01⋅ (a + b) .. LMz1(x1) Ry⋅x1:= Mz2 x2 ( ) Ry x2 ( ) ⋅ F cos θ ( ) ⋅ x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅Mz3 x3 ( ) Ry x3 ( ) ⋅ F cos θ ( ) ⋅ x3 a − ( ) ⋅ − F cos θ ( ) ⋅ x3 a − b − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅My1(x1) Rz⋅x1:= My2 x2 ( ) Rz x2 ( ) ⋅ F sin θ ( ) ⋅ x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅My3 x3 ( ) Rz x3 ( ) ⋅ F sin θ ( ) ⋅ x3 a − ( ) ⋅ − F sin θ ( ) ⋅ x3 a − b − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 250Distane (m)Moment (kN-m)Mz1(x1)Mz2(x2)Mz3(x3)x1, x2, x30 2101Distane (m)Moment (kN-m)My1(x1)My2(x2)My3(x3)x1, x2, x3 617. Maximum Bending Stress: unit := kN⋅mSince all the axes through the circle's center for circular shaft are principal axes, then theresultant moment M = (My2 + Mz2)0.5 can be used to determine the maximum bending stress.The maximum bending stress moment occurs at x=a.Mmax := Mz1(a)2 + My1(a)2⋅unit Mmax = 4.389 kN⋅mHence,cmaxdo2= Iπ64= ⋅ do4 σallow MmaxcmaxI= ⋅π64⋅ 4doMmaxσallow⎛⎜⎝⎞⎠do2= ⋅do3 32Mmaxπ⋅σallow:=do = 62.86mm Ans 618. Problem 6-114Using the techniques outlined in Appendix A, Example A.4 or A.5, the Z section has principal momeof inertia of Iy = 0.060(10-3) m4 and Iz = 0.471(10-3) m4, computed about the principal axes of inertia yand z, respectively. If the section is subjected to an internal moment of M = 250 N·m directedhorizontally as shown, determine the stress produced at point A. Solve the problem using Eq. 6-17.Given: M := 250N⋅m θ := 32.9degbf := 300mm tf := 50mmtw := 50mm dw := 150mmIy := 0.060(10− 3)m4 Iz := 0.471(10− 3)m4Solution:Internal Moment Components :My := M⋅cos(θ) Mz := M⋅ sin(θ)Coordinates of Point A :y'A := 0.5bf z'A := −(dw + 0.5tf)yAzA⎛⎜⎜⎝⎞⎠cos(θ)sin(θ)−sin(θ)cos(θ)⎛⎜⎝⎞⎠y'Az'A⎛⎜⎜⎝⎞⎠:= ⋅yAzA⎛⎜⎜⎝⎞⎠221.00−65.46⎛⎜⎝⎞⎠= mmBending Stress:σAMz⋅yAIz−My⋅zAIy+⎛⎜⎝⎞⎠:=σA = −0.293MPa (C) Ans 619. Problem 6-115Solve Prob. 6-114 using the equation developed in Prob. 6-111.Given: M := 250N⋅m θ := 32.9degbf := 300mm tf := 50mmtw := 50mm dw := 150mmIy := 0.060(10− 3)m4 Iz := 0.471(10− 3)m4Solution:Internal Moment Components :My' := M Mz' := 0Section Property : cz.w := 0.5dw + 0.5tf cy.w := 0.5bf − 0.5twIy'1122 ⋅ + ⎛⎜⎝⋅ 3 2⋅bf tf112⋅ 3 tw⋅dw cz.w⋅ tw dw⎞⎠:= + Iy' = 181.25 × 10− 6 m4Iz'1122 ⋅ + ⎛⎜⎝⋅ 3 2⋅ tf bf112⋅ 3 dw⋅ tw cy.w⋅dw tw⎞⎠:= + Iz' = 350.00 × 10− 6 m4Iy'z' := tw⋅dw⋅ (cz.w)⋅ (−cy.w) + tw⋅dw⋅ (−cz.w)⋅ (cy.w) Iy'z' = −187.50 × 10− 6 m4Coordinates of Point A :y'A := 0.5bf z'A := −(dw + 0.5tf)y'A = 150mm z'A = −175mmBending Stress: Using formula developed in Prob. 6-111.:= − 2D Iy'⋅ Iz' Iy'z'σA1D(Mz' My' )Iy'Iy'z'Iy'z'Iz'⎛⎜⎜⎝⎞⎠⋅−y'Az'A⎛⎜⎜⎝ ⎞⎠:= ⋅σA = −0.293MPa (C) Ans 620. Problem 6-116Using the techniques outlined in Appendix A, Example A.4 or A.5, the Z section has principal momeof inertia of Iy = 0.060(10-3) m4 and Iz = 0.471(10-3) m4, computed about the principal axes of inertia yand z, respectively. If the section is subjected to an internal moment of M = 250 N·m directedhorizontally as shown, determine the stress produced at point B. Solve the problem using Eq. 6-17.Given: M := 250N⋅m θ := 32.9degbf := 300mm tf := 50mmtw := 50mm dw := 150mmIy := 0.060(10− 3)m4 Iz := 0.471(10− 3)m4Solution:Internal Moment Components :My := M⋅cos(θ) Mz := M⋅ sin(θ)Coordinates of Point B :y'B := −0.5bf z'B := dw + 0.5tfyBzB⎛⎜⎜⎝⎞⎠cos(θ)sin(θ)−sin(θ)cos(θ)⎛⎜⎝⎞⎠y'Bz'B⎛⎜⎜⎝⎞⎠:= ⋅yBzB⎛⎜⎜⎝⎞⎠−221.0065.46⎛⎜⎝⎞⎠= mmBending Stress:σBMz⋅yBIz−My⋅zBIy+⎛⎜⎝⎞⎠:=σB = 0.293MPa (T) Ans 621. Problem 6-117For the section, Iy' = 31.7110-62 m4, Iz' = 114(10-6) m4, Iy'z' = 15.1(10-6) m4. Using the techniquesoutlined in Appendix A, the member's cross-sectional area has principal moments of inertia of Iy =29.0(10-6) m4 and Iz = 117(10-6) m4, computed about the principal axes of inertia y and z, respectivelyIf the section is subjected to a moment of M = 2500 N·m directed as shown, determinethe stress produced at point A, using Eq. 6-17.Given: M := 2500N⋅m θ' := 10.10degh1 := 80mm h2 := 140mmb1 := 60mm b2 := 60mmIy := 29.0(10− 6)m4 Iz := 117(10− 6)m4Solution: θ := −θ'Internal Moment Components :My := M⋅ sin(θ) Mz := M⋅ cos(θ)Coordinates of Point A :y'A := −h2 z'A := −b2yAzA⎛⎜⎜⎝⎞⎠cos(θ)sin(θ)−sin(θ)cos(θ)⎛⎜⎝⎞⎠y'Az'A⎛⎜⎜⎝⎞⎠:= ⋅yAzA⎛⎜⎜⎝⎞⎠−148.35−34.52⎛⎜⎝⎞⎠= mmBending Stress:σAMz⋅yAIz−My⋅zAIy+⎛⎜⎝⎞⎠:=σA = 2.599MPa (T) Ans 622. Problem 6-118Solve Prob. 6-117 using the equation developed in Prob. 6-111.Given: M := 2500N⋅m θ' := 10.10degh1 := 80mm h2 := 140mmb1 := 60mm b2 := 60mmIy' := 31.7(10− 6)m4 Iz' := 114(10− 6)m4Iy'z' := 15.1(10− 6)m4Solution: θ := −θ'Internal Moment Components :My' := 0 Mz' := MCoordinates of Point A :y'A := −h2 z'A := −b2yAzA⎛⎜⎜⎝⎞⎠cos(θ)sin(θ)−sin(θ)cos(θ)⎛⎜⎝⎞⎠y'Az'A⎛⎜⎜⎝⎞⎠:= ⋅yAzA⎛⎜⎜⎝⎞⎠−148.35−34.52⎛⎜⎝⎞⎠= mmBending Stress: Using formula developed in Prob. 6-111.:= − 2D Iy'⋅ Iz' Iy'z'σA1D(Mz' My' )Iy'Iy'z'Iy'z'Iz'⎛⎜⎜⎝⎞⎠⋅−y'Az'A⎛⎜⎜⎝⎞⎠:= ⋅σA = 2.608MPa (T) Ans 623. Problem 6-119The composite beam is made of 6061-T6 aluminum (A) and C83400 red brass (B). Determine thedimension h of the brass strip so that the neutral axis of the beam is located at the seam of the twometals. What maximum moment will this beam support if the allowable bending stress for thealuminum is (σ allow) al = 128 MPa and for the brass (σ allow) br = 35 MPa?Given: b := 150mm hal := 50mmEal := 68.9GPa Ebr := 101GPaσal_allow := 128MPa σbr_allow := 35MPaSolution:Section Property : nEalEbr:= n = 0.682178Abr = b⋅h A'al := (n⋅b)⋅halycΣyi⋅AiΣA= ycA'al⋅ (0.5⋅hal) + Abr⋅ (hal + 0.5⋅h)A'al + Abr=Given yc := halhalA'al⋅ (0.5⋅hal) + b⋅h⋅ (hal + 0.5⋅h)A'al + b⋅h=Guess h := 10mmh := Find(h) h = 41.30mm AnsAbr := b⋅hIbr112:= ⋅b⋅h3 + Abr⋅ (0.5h)2I'al112:= ⋅ (n⋅b) ⋅ hal3 + A'al ⋅ (yc − 0.5hal)2I := Ibr + I'al I = 7785108.17mm4Allowable Bending Stress: σM⋅yI=Assume failure of red brass:ybr := h Mbr(σbr_allow)⋅ I:= Mbr = 6.60 kN⋅mybrAssume failure of aluminum:yal := hal Mal(σal_allow)⋅ I:= Mal = 29.22 kN⋅mn⋅yalMallow := min(Mbr ,Mal)Mallow = 6.60 kN⋅m Ans 624. Problem 6-120The composite beam is made of 6061-T6 aluminum (A) and C83400 red brass (B). If the height h = 4mm, determine the maximum moment that can be applied to the beam if the allowable bending stressfor the aluminum is (σ allow) al = 128 MPa and for the brass (σ allow) br = 35 MPa?Given: b := 150mm hal := 50mm hbr := 40mmEal := 68.9GPa Ebr := 101GPaσal_allow := 128MPa σbr_allow := 35MPaSolution:Section Property : nEalEbr:= n = 0.682178Abr := b⋅hbr A'al := (n⋅b)⋅halycΣyi⋅AiΣA= ycA'al⋅ (0.5⋅hal) + Abr⋅ (hal + 0.5⋅hbr):= yc = 49.289mmA'al + AbrIbr:= ⋅ 3 + Abr ⋅ (hal + 0.5hbr − yc)2I'al112⋅b hbr112:= ⋅ (n⋅b) ⋅ hal3 + A'al ⋅ (yc − 0.5hal)2I := Ibr + I'al I = 7457987.63mm4Allowable Bending Stress: σM⋅yI=Assume failure of red brass:ybr := hal + hbr − yc Mbr(σbr_allow)⋅ I:= Mbr = 6.41 kN⋅mybrAssume failure of aluminum:yal := yc Mal(σal_allow)⋅ I:= Mal = 28.39 kN⋅mn⋅yalMallow := min(Mbr ,Mal)Mallow = 6.41 kN⋅m Ans 625. Problem 6-121A wood beam is reinforced with steel straps at its top and bottom as shown. Determine the maximumbending stress developed in the wood and steel if the beam is subjected to a bending moment of M =kN·m. Sketch the stress distribution acting over the cross section. Take Ew = 11 GPa, Est = 200 GPa.Given: b := 200mm hw := 300mm tst := 20mmEw := 11GPa Est := 200GPaM := 5kN⋅mSolution:Section Property :nEstEw:= n = 18.181818Aw := b⋅hw A'st := (n⋅b)⋅ tstI'st112:= ⋅ (n⋅b) ⋅ tst3 + A'st ⋅ (0.5hw + 0.5tst)2I1122I':= ⋅b ⋅ hw3 + 4st I 4178484848.48mm= Maximum Bending Stress: σM⋅yI=For wood beam,yw := 0.5hw σw.maxM⋅ywI:= σw.max = 0.179MPa AnsFor steel straps,yst := 0.5hw + tst σst.maxn⋅M⋅yst:= σst.max = 3.699MPa AnsIAt y'st := 0.5hw σ'stn⋅M⋅y'st:= σ'st = 3.263MPaI 626. Problem 6-122The Douglas Fir beam is reinforced with A-36 steel straps at its center and sides. Determine themaximum stress developed in the wood and steel if the beam is subjected to a bending moment of Mz =10 kN·m. Sketch the stress distribution acting over the cross section.Given: bst := 12mm d := 150mmbw := 50mm Mz := 10kN⋅mEw := 13.1GPa Est := 200GPaSolution:Section Property : nEwEst:= n = 0.0655Iz1123bst ( ) d3 ⋅ n 2bw ( ) ⋅ d3 ⋅ + ⎡⎣⎤⎦:= ⋅Maximum Bending Stress: σMz⋅yIz=ymax := 0.5dσstMz⋅ymaxIz:= σst = 62.7MPa Ansσw := n⋅ (σst) σw = 4.10MPa Ans 627. Problem 6-123The steel channel is used to reinforce the wood beam. Determine the maximum stress in the steel andin the wood if the beam is subjected to a moment of M = 1.2 kN·m. Est = 200 GPa, Ew = 12GPa.Given: bst := 399mm dst := 100mm tst := 12mmbw := 375mm dw := 88mm M := 1.2kN⋅mEw := 12GPa Est := 200GPaSolution: df := dwnEwEstSection Property : := n = 0.06As1 := 2tst⋅df As2 := bst⋅ tst A'w := n⋅bw⋅dwyc(A'w + As1)⋅ (0.5dw + tst) + As2⋅ (0.5tst):= yc = 29.04mmA'w + As1 + As2Iw:= ⋅ + A'w⋅ (0.5dw + tst − yc)2Is1112n⋅bw ⋅ dw3 ⎛⎝⎞⎠112:= ⋅ 2tst ⋅ df3 + As1⋅ (0.5dw + tst − yc)2Is2⎛⎝⎞⎠112:= ⋅ + As2⋅ (0.5tst − yc)2⎛⎝bst ⋅ tst3 ⎞⎠I := Iw + Is1 + Is2Maximum Bending Stress: σM⋅cI=cmax := dst − ycσstM⋅cmax:= σst = 10.37MPa AnsIσw := n⋅ (σst) σw = 0.62MPa Ans 628. Problem 6-124The Douglas Fir beam is reinforced with A-36 steel straps at its sides. Determine the maximum stressdeveloped in the wood and steel if the beam is subjected to a bending moment of Mz = 4 kN·m. Sketcthe stress distribution acting over the cross section.Given: bst := 15mm d := 350mmbw := 200mm Mz := 4kN⋅mEw := 13.1GPa Est := 200GPaSolution:Section Property : nEwEst:= n = 0.0655Iz1122bst ( ) d3 ⋅ n bw ⋅ d3 ⋅ + ⎡⎣⎤⎦:= ⋅Maximum Bending Stress: σMz⋅yIz=ymax := 0.5dσstMz⋅ymaxIz:= σst = 4.546MPa Ansσw := n⋅ (σst) σw = 0.298MPa Ans 629. Problem 6-125The composite beam is made of A-36 steel (A) bonded to C83400 red brass (B) and has the crosssection shown. If it is subjected to a moment of M = 6.5 kN·m, determine the maximum stress in thebrass and steel. Also, what is the stress in each material at the seam where they are bonded together?Given: b := 125mm hst := 100mm hbr := 100mmEst := 200GPa Ebr := 101GPaM := 6.5kN⋅mSolution:Section Property : nEbrEst:= n = 0.505Ast := b⋅hst A'br := (n⋅b)⋅hbrycΣyi⋅AiΣA= ycA'br⋅ (0.5⋅hbr) + Ast⋅ (hbr + 0.5⋅hst):= yc = 116.445mmA'br + AstIst:= ⋅ 3 + Ast ⋅ (hbr + 0.5hst − yc)2I'br112⋅b hst112:= ⋅ (n⋅b) ⋅ hbr3 + A'br ⋅ (yc − 0.5hbr)2I := Ist + I'br I = 57620604.93mm4Maximum Bending Stress: σM⋅yI=For steel,yst := hbr + hst − yc σst.maxM⋅ystI:= σst.max = 9.426MPa AnsFor red brass,ybr := yc σbr.maxn⋅M⋅ybr:= σbr.max = 6.634MPa AnsIBending Stress at the Seam:yseam := yc − hbr σ'stM⋅yseam:= σ'st = 1.855MPa AnsIσ'br := n⋅σ'st σ'br = 0.937MPa Ans 630. Problem 6-126The composite beam is made of A-36 steel (A) bonded to C83400 red brass (B) and has the crosssection shown. If the allowable bending stress for the steel is (σ allow)st = 180 MPa and for the brass(σ allow)br = 60 MPa, determine the maximum moment M that can be applied to the beam.Given: b := 125mm hst := 100mm hbr := 100mmEst := 200GPa Ebr := 101GPaσst_allow := 180MPa σbr_allow := 60MPaSolution:Section Property : nEbrEst:= n = 0.505Ast := b⋅hst A'br := (n⋅b)⋅hbrycΣyi⋅AiΣA= ycA'br⋅ (0.5⋅hbr) + Ast⋅ (hbr + 0.5⋅hst):= yc = 116.445mmA'br + AstIst:= ⋅ 3 + Ast ⋅ (hbr + 0.5hst − yc)2I'br112⋅b hst112:= ⋅ (n⋅b) ⋅ hbr3 + A'br ⋅ (yc − 0.5hbr)2I := Ist + I'br I = 57620604.93mm4Allowable Bending Stress: σM⋅yI=Assume failure of red brass:ybr := yc Mbr(σbr_allow)⋅ I:= Mbr = 58.79 kN⋅mn⋅ybrAssume failure of steel:yst := hbr + hst − yc Mst(σst_allow)⋅ I:= Mst = 124.13 kN⋅mystMallow := min(Mbr ,Mst)Mallow = 58.79 kN⋅m Ans 631. Problem 6-127The reinforced concrete beam is made using two steel reinforcing rods. If the allowable tensile stressfor the steel is (σ st) allow = 280 MPa and the allowable compressive stress for the concrete is (σ conc)allow = 21 MPa, determine the maximum moment M that can be applied to the section. Assume theconcrete cannot support a tensile stress. Est = 200 GPa, Econc = 26.5 GPa.Given: bf := 550mm df := 100mmbw := 150mm dw := 450mmdr := 25mm hr := 50mmEconc := 26.5GPa Est := 200GPaσc_allow := 21MPa σst_allow := 280MPaSolution:nEstEconcSection Property : := n = 7.54717A'st := n⋅2π⋅ (0.5dr)2Given A'st⋅ (dw − h'w − hr) − bf⋅df⋅ (0.5df + h'w) − bw⋅h'w⋅ (0.5h'w) = 0Guess h'w := 10mmh'w := Find(h'w) h'w = 3.41mmIst := A'st⋅ (dw − h'w − hr)2If112:= ⋅ bf ⋅ df3 + (bf⋅df)⋅ (0.5df + h'w)2I'w⎛⎝⎞⎠112:= ⋅ + (bw⋅h'w)⋅ (0.5h'w)2⎛⎝bw ⋅ h'w3 ⎞⎠I := Ist + If + I'wσM⋅yIAssume concrete fails: =ymax := df + h'w Mconc(σc_allow)⋅ Iymax:= Mconc = 277.83 kN⋅mAssume steel fails:yst := dw − h'w − hr Mst(σst_allow)⋅ In⋅yst:=Mst = 127.98 kN⋅mMallow := min(Mconc ,Mst)Mallow = 127.98 kN⋅m Ans 632. Problem 6-128Determine the maximum uniform distributed load w0 that can be supported by the reinforced concretebeam if the allowable tensile stress for the steel is (σ st) allow = 200 MPa and the allowable compressivestress for the concrete is (σ conc) allow = 20 MPa. Assume the concrete cannot support a tensile stress.Take Est = 200 GPa, Econc = 25 GPa.Given: b := 250mm d := 500mmdr := 16mm hr := 50mmEconc := 25GPa Est := 200GPaσc_allow := 20MPa L := 2.5mσst_allow := 200MPaSolution:nEstEconcSection Property : := n = 8A'st := n⋅2π⋅ (0.5dr)2Given A'st⋅ (d − h' − hr) − b⋅h'⋅ (0.5h') = 0Guess h' := 10mmh' := Find(h') h' = 95.51mmI A'st⋅ (d − h' − hr)21⋅ ( b ⋅ h'3 ) + ( b ⋅ h' ) ⋅ ( 0.5h' )2 12⎡⎢⎣⎤⎥⎦:= +σM⋅yIAssume concrete fails: =ymax := h' Mconc(σc_allow)⋅ Iymax:= Mconc = 99.85 kN⋅mAssume steel fails:yst := d − h' − hr Mst(σst_allow)⋅ I:= Mst = 33.63 kN⋅mn⋅ystThus, steel fails first.Maximum moment ocurs over the middle support: Mmaxwo⋅L22=wo2⋅MstL2:=wo 10.76kNm= Ans 633. Problem 6-129A bimetallic strip is made from pieces of 2014-T6 aluminum and C83400 red brass, having the crosssection shown. A temperature increase causes its neutral surface to be bent into a circular arc having aradius of 400 mm. Determine the moment that must be acting on its cross section due to the thermalstress. Take Eal = 74 GPa, Ebr = 102 GPa.Given: bbr := 6mm dbr := 2mmbal := 6mm dal := 2mmEal := 74GPa ρ := 400mmEbr := 102GPaSolution: Transform the section to brassSection Property : nEalEbr:= n = 0.72549A'al := n⋅bal⋅dal Abr := bbr⋅dbrycA'al⋅ (0.5dal) + Abr⋅ (0.5dbr + dal):= yc = 2.16mmA'al + AbrIal:= ⋅ + A'al⋅ (0.5dal − yc)2Ibr112n⋅bal ⋅ dal3 ⎛⎝⎞⎠112:= ⋅ + Abr⋅ (0.5dbr + dal − yc)2⎛⎝bbr ⋅ dbr3 ⎞⎠I := Ial + IbrMaximum Bending Stress: 1ρME⋅ I=M(Ebr)⋅ Iρ:=M = 6.91 N⋅m Ans 634. Problem 6-130The fork is used as part of a nosewheel assembly for an airplane. If the maximum wheel reaction at theend of the fork is 4.5 kN, determine the maximum bending stress in the curved portion of the fork atsection a-a. There the cross-sectional area is circular, having a diameter of 50 mm.Given: rc := 250mm a := 150mm F := 4.5kNθ := 30deg do := 50mmSolution:Internal Moment :− ⋅ ( − ⋅ sin(θ)) = 0M F a rcΣΜC=0; M F a rc:= ⋅ ( − ⋅ sin(θ))M = 112.5N⋅mSection Property : ro := 0.5do:= ⋅ 2A π roI Σ dA IA_r 2⋅π rc rc= ∫A_r A r⎛⎝− 2 − ro2 ⎞⎠= ⋅RAIA_r= R⋅ 2π ro⎛⎝− 2 − ro2 2⋅π rc rc⎞⎠⋅:=R = 249.373mmBending Stress: σM⋅ (R − r)A⋅ r⋅ (rc − R)=rA := rc − ro σAM⋅ (R − rA)A⋅ rA⋅ (rc − R):= σA = 9.91MParB := rc + ro σBM⋅ (R − rB)A⋅ rB⋅ (rc − R):= σB = −8.52MPaσmax := max( σA , σB )σmax = 9.91MPa (T) Ans 635. Problem 6-131Determine the greatest magnitude of the applied forces P if the allowable bending stress is (σ allow)c =50 MPa in compression and (σ allow)t = 120 MPa in tension.Given: bf := 75mm b'f := 150mm tf := 10mmtw := 10mm dw := 150mm ri := 250mmσc.allow := −50MPa σt.allow := 120MPaSolution:Internal Moment :M = P(dw+tf ) kN-m is positive since it tendsto increase the beam's radius of curvature.Section Property : re := ri + dw + 2tfA := bf⋅ tf + dw⋅ tw + b'f⋅ tf A = 3750mm2⎯ Σ ri ⎯r⋅ ( ⋅Ai)Σ⋅ (Ai)= rc(b'f⋅ tf)⋅ (ri + 0.5tf) + (dw⋅ tw)⋅ (ri + 0.5dw + tf) + (bf⋅ tf)⋅ (re − 0.5tf)A:=rc = 319.000mmI Σ dA IA_r b'f ln= ∫A_r A rri + tfri⎛⎜⎝⎞⎠⋅ tw lnre − tfri + tf⎛⎜⎝⎞⎠+ ⋅ bf lnrere − tf⎛⎜⎝⎞⎠:= + ⋅IA_r = 12.245mmRAIA_r:= R = 306.243mmNormal Stress: σM⋅ (R − r)A⋅ r⋅ (rc − R)= Mσ⋅A⋅ r⋅ (rc − R)R − r=Assume tension failure.r := ri P⋅ (dw + tf)σt.allow⋅A⋅ r⋅ (rc − R)= PR − rσt.allow⋅A⋅ r⋅ (rc − R)(dw + tf)⋅ (R − r):=P = 159.48 kNAssume compression failure.r' := re P'⋅ (dw + tf)σc.allow⋅A⋅ r'⋅ (rc − R)= P'R − r'σc.allow⋅A⋅ r'⋅ (rc − R)(dw + tf)⋅ (R − r'):=P' = 55.2 kNPallow := min(P , P') Pallow = 55.20 kN Ans 636. Problem 6-132If P = 6 kN, determine the maximum tensile and compressive bending stresses in the beam.Given: bf := 75mm b'f := 150mm tf := 10mmtw := 10mm dw := 150mm ri := 250mmP := 6kNSolution:Internal Moment :M P dw:= ⋅ ( + tf) M = 0.960 kN⋅mM is positive since it tends to increase thebeam's radius of curvature.Section Property : re := ri + dw + 2tfA := bf⋅ tf + dw⋅ tw + b'f⋅ tf A = 3750mm2⎯ Σ ri ⎯r⋅ ( ⋅Ai)Σ⋅ (Ai)= rc(b'f⋅ tf)⋅ (ri + 0.5tf) + (dw⋅ tw)⋅ (ri + 0.5dw + tf) + (bf⋅ tf)⋅ (re − 0.5tf)A:=rc = 319.000mmI Σ dA IA_r b'f ln= ∫A_r A rri + tfri⎛⎜⎝⎞⎠⋅ tw lnre − tfri + tf⎛⎜⎝⎞⎠+ ⋅ bf lnrere − tf⎛⎜⎝⎞⎠:= + ⋅IA_r = 12.245mmRAIA_r:= R = 306.243mmNormal Stress: σM⋅ (R − r)A⋅ r⋅ (rc − R)=Maximum tensile stress: r := riσt_maxM⋅ (R − r)A⋅ r⋅ (rc − R):= σt_max = 4.51MPa (T) AnsMaximum compressive stress: r := reσc_maxM⋅ (R − r)A⋅ r⋅ (rc − R):= σc_max = −5.44MPa (C) Ans 637. Problem 6-133The curved beam is subjected to a bending moment of M = 900 N·m as shown. Determine the stressat points A and B, and show the stress on a volume element located at each of these points.Given: bf := 100mm tf := 20mm ri := 400mmtw := 15mm dw := 150mmM := −900N⋅m θ := 30degSolution:Internal Moment :M is negative since it tends to decrease thebeam's radius of curvature.Section Property : re := ri + dw + tfA := bf⋅ tf + dw⋅ tw A = 4250mm2⎯ Σ ri ⎯r⋅ ( ⋅Ai)Σ⋅ (Ai)= rc(dw⋅ tw)⋅ (ri + 0.5dw) + (bf⋅ tf)⋅ (re − 0.5tf)A:=rc = 515.000mmI Σ dA IA_r tw ln= ∫A_r A rre − tfri⎛⎜⎝⎞⎠⋅ bf lnrere − tf⎛⎜⎝⎞⎠:= + ⋅IA_r = 8.349mmRAIA_r:= R = 509.067mmNormal Stress: σM⋅ (R − r)A⋅ r⋅ (rc − R)=At A: rA := re σAM⋅ (R − rA)A⋅ rA⋅ (rc − R):= σA = 3.82MPa (T) AnsAt B: rB := ri σBM⋅ (R − rB)A⋅ rB⋅ (rc − R):= σB = −9.73MPa (C) Ans 638. Problem 6-134The curved beam is subjected to a bending moment of M = 900 N·m. Determine the stress at point C.Given: bf := 100mm tf := 20mm ri := 400mmtw := 15mm dw := 150mmM := −900N⋅m θ := 30degSolution:Internal Moment :M is negative since it tends to decrease thebeam's radius of curvature.Section Property : re := ri + dw + tfA := bf⋅ tf + dw⋅ tw A = 4250mm2⎯ Σ ri ⎯r⋅ ( ⋅Ai)Σ⋅ (Ai)= rc(dw⋅ tw)⋅ (ri + 0.5dw) + (bf⋅ tf)⋅ (re − 0.5tf)A:=rc = 515.000mmI Σ dA IA_r tw ln= ∫A_r A rre − tfri⎛⎜⎝⎞⎠⋅ bf lnrere − tf⎛⎜⎝⎞⎠:= + ⋅IA_r = 8.349mmRAIA_r:= R = 509.067mmNormal Stress: σM⋅ (R − r)A⋅ r⋅ (rc − R)=At C: rC := re − tf σCM⋅ (R − rC)A⋅ rC⋅ (rc − R):= σC = 2.66MPa (T) Ans 639. Problem 6-135The curved bar used on a machine has a rectangular cross section. If the bar is subjected to a coupleas shown, determine the maximum tensile and compressive stress acting at section a-a. Sketch thestress distribution on the section in three dimensions.Given: b := 50mm h := 75mm ri := 162.5mmP := 250N θ := 60deg a := 150mmSolution:Internal Moment :M := P⋅ (a⋅ sin(θ) + h⋅cos(θ))M = 41.851 N⋅mM is positive since it tends to increase thebeam's radius of curvature.Section Property : re := ri + hA := b⋅h A = 3750mm2rc := ri + 0.5⋅h rc = 200mmI Σ dA IA_r b ln= ∫A_r A rreri⎛⎜⎝⎞⎠:= ⋅ IA_r = 18.974mmRAIA_r:= R = 197.634mmNormal Stress: σM⋅ (R − r)A⋅ r⋅ (rc − R)=At A: rA := re σAM⋅ (R − rA)A⋅ rA⋅ (rc − R):= σA = −0.792MPa (C) AnsAt B: rB := ri σBM⋅ (R − rB)A⋅ rB⋅ (rc − R):= σB = 1.020MPa (T) Ans 640. Problem 6-136The circular spring clamp produces a compressive force of 3 N on the plates. Determine the maximumbending stress produced in the spring at A. The spring has a rectangular cross section as shown.Given: b := 20mm h := 10mm ri := 200mmP := 3N a := 220mmSolution:Internal Moment :M := P⋅aM = 0.660N⋅mM is positive since it tends to increase thebeam's radius of curvature.Section Property : re := ri + hA := b⋅h A = 200mm2rc := ri + 0.5⋅h rc = 205mmI Σ dA IA_r b ln= ∫A_r A rreri⎛⎜⎝⎞⎠:= ⋅ IA_r = 0.976mmRAIA_r:= R = 204.959mmNormal Stress: σM⋅ (R − r)A⋅ r⋅ (rc − R)=Maximum tensile stress: rA := riσt_maxM⋅ (R − rA)A⋅ rA⋅ (rc − R):= σt_max = 2.01MPa (T) AnsMaximum compressive stress: r'A := reσc_maxM⋅ (R − r'A)A⋅ r'A⋅ (rc − R):= σc_max = −1.95MPa (C) Ans 641. Problem 6-137Determine the maximum compressive force the spring clamp can exert on the plates if the allowablebending stress for the clamp is σ allow = 4 MPa.Given: b := 20mm h := 10mm ri := 200mm a := 220mmσt.allow := 4MPa σc.allow := −4MPaSolution:Section Property : re := ri + hA := b⋅h A = 200mm2rc := ri + 0.5⋅h rc = 205mmI Σ dA IA_r b ln= ∫A_r A rreri⎛⎜⎝⎞⎠:= ⋅ IA_r = 0.976mmRAIA_r:= R = 204.959mmInternal Moment :Mmax = P⋅ (a + R)M is positive since it tends to increase thebeam's radius of curvature.Normal Stress: σM⋅ (R − r)A⋅ r⋅ (rc − R)= Mσ⋅A⋅ r⋅ (rc − R)R − r=Assume tension failure.r := ri P⋅ (a + R)σt.allow⋅A⋅ r⋅ (rc − R)= PR − rσt.allow⋅A⋅ r⋅ (rc − R)(a + R)⋅ (R − r):=P = 3.087NAssume compression failure.r' := re P'⋅ (a + R)σc.allow⋅A⋅ r'⋅ (rc − R)= P'R − r'σc.allow⋅A⋅ r'⋅ (rc − R)(a + R)⋅ (R − r'):=P' = 3.189NPallow := min(P , P') Pallow = 3.087N Ans 642. Problem 6-138While in flight, the curved rib on the jet plane is subjected to an anticipated moment of M = 16 N·m athe section. Determine the maximum bending stress in the rib at this section, and sketch a two-dimensionalview of the stress distribution.Given: bf := 30mm tf := 5mmtw := 5mm dw := 20mmM := 16N⋅m ri := 600mmSolution:Internal Moment :M is positive since it tends to increase thebeam's radius of curvature.Section Property : re := ri + dw + 2tf:= ⋅ tf + dw⋅ tw A = 400mm2A 2bf⎯ Σ ri ⎯r⋅ ( ⋅Ai)Σ⋅ (Ai)= rc(bf⋅ tf)⋅ (ri + 0.5tf) + (dw⋅ tw)⋅ (ri + 0.5dw + tf) + (bf⋅ tf)⋅ (re − 0.5tf)A:=rc = 615.000mmI Σ dA IA_r bf ln= ∫A_r A rri + tfri⎛⎜⎝⎞⎠⋅ tw lnre − tfri + tf⎛⎜⎝⎞⎠+ ⋅ bf lnrere − tf⎛⎜⎝⎞⎠:= + ⋅IA_r = 0.651mmRAIA_r:= R = 614.793mmNormal Stress: σM⋅ (R − r)A⋅ r⋅ (rc − R)=Maximum tensile stress: r := riσt_maxM⋅ (R − r)A⋅ r⋅ (rc − R):= σt_max = 4.77MPa (T) AnsMaximum compressive stress: r' := reσc_maxM⋅ (R − r')A⋅ r'⋅ (rc − R):= σc_max = −4.67MPa (C) Ans 643. Problem 6-139The steel rod has a circular cross section. If it is gripped at its ends and a couple moment of M = 1.5N·m is developed at each grip, determine the stress acting at points A and B and at the centroid C.Given: rci := 50mm rce := 75mmro := 12mm M := 1.5N⋅mSolution:Internal Moment :M = 1.5Nm is positive since it tends toincrease the beam's radius of curvature.Section Property :A π ro:= ⋅ 2 rc := 0.5(rci + rce)= ∫I Σ dA IA_r 2⋅π rc rcA_r A r⎛⎝− 2 − ro2 ⎞⎠:= ⋅RAIA_r:= R = 61.919mmNormal Stress: σM⋅ (R − r)A⋅ r⋅ (rc − R)=rA := rci σAM⋅ (R − rA)A⋅ rA⋅ (rc − R):= σA = 1.3594MPa (T) AnsrB := rce σBM⋅ (R − rB)A⋅ rB⋅ (rc − R):= σB = −0.9947MPa (C) AnsrC := rc σCM⋅ (R − rC)A⋅ rC⋅ (rc − R):= σC = −0.0531MPa (C) Ans 644. Problem 6-140The curved bar used on a machine has a rectangular cross section. If the bar is subjected to a coupleas shown, determine the maximum tensile and compressive stresses acting at section a-a. Sketch thestress distribution on the section in three dimensions.Given: b := 50mm h := 75mmri := 100mm P := 250Nd' := 50mm d := 150mmSolution:Internal Moment :M := P⋅dM = 37.5 N⋅mM is positive since it tends to increase thebeam's radius of curvature.Section Property : re := ri + hA := b⋅h A = 3750mm2rc := ri + 0.5⋅h rc = 137.5mmI Σ dA IA_r b ln= ∫A_r A rreri⎛⎜⎝⎞⎠:= ⋅ IA_r = 27.981mmRAIA_r:= R = 134.021mmNormal Stress: σM⋅ (R − r)A⋅ r⋅ (rc − R)=Maximum tensile stress: r := riσt_maxM⋅ (R − r)A⋅ r⋅ (rc − R):= σt_max = 0.978MPa (T) AnsMaximum compressive stress: r' := reσc_maxM⋅ (R − r')A⋅ r'⋅ (rc − R):= σc_max = −0.673MPa (C) Ans 645. Problem 6-141The member has an elliptical cross section. If it is subjected to a moment of M = 50 N·m, determinethe stress at points A and B. Is the stress at point A', which is located on the member near the wall, thesame as that at A? Explain.Given: ri := 100mm La := 150mm Lb := 75mmM := 50N⋅mSolution:Internal Moment :M is positive since it tends to increase themember's radius of curvature.Section Property : ao := 0.5La bo := 0.5Lbre := ri + LaA := π⋅ao⋅bo A = 8835.73mm2rc := ri + ao rc = 175mmI Σ dA IA_r= ∫A_r A r2π⋅boaorc − rc2 ⎛⎝− ao2 ⎞⎠= ⋅RAIA_r= Rπ⋅ao⋅bo⋅ (ao)⎛⎝− 2 − ao2 2⋅π⋅bo rc rc⎞⎠⋅:=R = 166.557mmBending Stress: σM⋅ (R − r)A⋅ r⋅ (rc − R)=rA := ri σAM⋅ (R − rA)A⋅ rA⋅ (rc − R):= σA = 0.446MPa (T) AnsrB := re σBM⋅ (R − rB)A⋅ rB⋅ (rc − R):= σB = −0.224MPa (C) AnsNo. The stress at point A' is not the same as that at A,because of localized stress concentration. Ans 646. Problem 6-142The member has an elliptical cross section. If the allowable bending stress is σ allow = 125 MPa,determine the maximum moment M that can be applied to the member.Given: ri := 100mm La := 150mm Lb := 75mmσt.allow := 125MPa σc.allow := −125MPaSolution:Internal Moment :M is positive since it tends to increase themember's radius of curvature.Section Property : ao := 0.5La bo := 0.5Lbre := ri + LaA := π⋅ao⋅bo A = 8835.73mm2rc := ri + ao rc = 175mmI Σ dA IA_r= ∫A_r A r2π⋅boaorc − rc2 ⎛⎝− ao2 ⎞⎠= ⋅RAIA_r= Rπ⋅ao⋅bo⋅ (ao)⎛⎝− 2 − ao2 2⋅π⋅bo rc rc⎞⎠⋅:=R = 166.557mmNormal Stress: σM⋅ (R − r)A⋅ r⋅ (rc − R)= Mσ⋅A⋅ r⋅ (rc − R)R − r=Assume tension failure.r := ri Mσt.allow⋅A⋅ r⋅ (rc − R)= MR − rσt.allow⋅A⋅ r⋅ (rc − R)R − r:=M = 14.01 kN⋅mAssume compression failure.r' := re M'σc.allow⋅A⋅ r'⋅ (rc − R)= M'R − r'σc.allow⋅A⋅ r'⋅ (rc − R)R − r':=M' = 27.94 kN⋅mMallow := min(M,M') Mallow = 14.01 kN⋅m Ans 647. Problem 6-143The bar has a thickness of 6.25 mm and is made of a material having an allowable bending stress ofσ allow = 126 MPa. Determine the maximum moment M that can be applied.Given: t := 6.25mm r := 6.25mmw := 100mm h := 25mmσallow := 126MPaSolution:Section Property:I112:= ⋅ t⋅h3 I = 8138.02mm4Stress Concentration Factor :wr= 4= 0.25hhFrom Fig. 6-48, K := 1.45Maximum Moment : σ KM⋅cI= ⋅c := 0.5⋅h Mmax(σallow)⋅ IK⋅c:=Mmax = 56.57N⋅m Ans 648. Problem 6-144The bar has a thickness of 12.5 mm and is subjected to a moment of 90 N·m. Determine the maximumbending stress in the bar.Given: t := 12.5mm r := 6.25mmw := 100mm h := 25mmM := 90N⋅mSolution:Section Property:I112:= ⋅ t⋅h3 I = 16276.04mm4Stress Concentration Factor :wr= 4= 0.25hhFrom Fig. 6-48, K := 1.45Maximum Bending Stress :c := 0.5⋅h σmax KM⋅cI:= ⋅σmax = 100.2MPa Ans 649. Problem 6-145The bar is subjected to a moment of M = 40 N·m. Determine the smallest radius r of the fillets so thatan allowable bending stress of σ allow = 124 MPa is not exceeded.Given: w := 80mm h := 20mm t := 7mmM := 40N⋅m σallow := 124MPaSolution:Section Property:I112:= ⋅ t⋅h3 I = 4666.67mm4Allowable Bending Stress : σ KM⋅cI= ⋅c := 0.5⋅h Kσallow⋅ IM⋅c:=K = 1.45Stress Concentration Factor :From Fig. 6-48, with K = 1.45wh= 4then,rh= 0.25r := 0.25⋅hr = 5.00mm Ans 650. Problem 6-146The bar is subjected to a moment of M = 17.5 N · m. If r = 5 mm, determine the maximum bendingstress in the material.Given: w := 80mm h := 20mmt := 7mm r := 5mmM := 17.5N⋅mSolution:Section Property:I112:= ⋅ t⋅h3 I = 4666.67mm4Stress Concentration Factor :wr= 4= 0.25hhFrom Fig. 6-48, K := 1.45Maximum Bending Stress :c := 0.5⋅h σmax KM⋅cI:= ⋅σmax = 54.4MPa Ans 651. Problem 6-147The bar is subjected to a moment of M = 20 N·m. Determine the maximum bending stress in the barand sketch, approximately, how the stress varies over the critical section.Given: w := 30mm h := 10mmt := 5mm r := 1.5mmM := 20N⋅mSolution:Section Property:I112:= ⋅ t⋅h3 I = 416.67mm4Stress Concentration Factor :wr= 3= 0.15hhFrom Fig. 6-48, K := 1.6Maximum Bending Stress :c := 0.5⋅h σmax KM⋅cI:= ⋅σmax = 384MPa Ans 652. Problem 6-148The allowable bending stress for the bar is σ allow = 175 MPa. Determine the maximum moment Mthat can be applied to the bar.Given: w := 30mm h := 10mmt := 5mm r := 1.5mmσallow := 175MPaSolution:Section Property:I112:= ⋅ t⋅h3 I = 416.67mm4Stress Concentration Factor :wr= 3= 0.15hhFrom Fig. 6-48, K := 1.6Maximum Moment : σ KM⋅cI= ⋅c := 0.5⋅h M(σallow)⋅ IK⋅c:=M = 9.11 N⋅m Ans 653. Problem 6-149Determine the maximum bending stress developed in the bar if it is subjected to the couples shown.The bar has a thickness of 6 mm.Given: t := 6mm w := 108mmh1 := 72mm h2 := 36mmr1 := 7.2mm r2 := 27mmM1 := 20N⋅m M2 := 7.5N⋅mMw := 12.5N⋅mSolution:Section Property:For the larger section 1: For the smaller section 2:I1112:= ⋅ t ⋅ h13 I1 = 186624mm4 I2112:= ⋅ t ⋅ h23 I2 = 23328mm4Stress Concentration Factor :For the larger section 1: For the smaller section 2:wh1= 1.5r1h1= 0.1wh2= 3r2h2= 0.75From Fig. 6-48, K1 := 1.755 From Fig. 6-48, K2 := 1.15Maximum Moment : σ KM⋅cI= ⋅For the larger section 1: For the smaller section 2:c1 := 0.5⋅h1 σ1 K1M1⋅c1I1:= ⋅ c2 := 0.5⋅h2 σ2 K2M2⋅c2I2:= ⋅σ1 = 6.77MPa σ2 = 6.66MPaσmax := max(σ1 , σ2)σmax = 6.77MPa Ans 654. Problem 6-150Determine the length L of the center portion of the bar so that the maximum bending stress at A, B,and C is the same.The bar has a thickness of 10 mm.Given: w := 60mm t := 10mmh := 40mm r := 7mma := 200mm P := 350NSolution:Section Property:I112:= ⋅ t⋅h3 I = 53333.33mm4 Support Reaction : By symmetry, A =B = R+ ΣFy=0; 2R − P = 0Stress Concentration Factor :R := 0.5P R = 175Nwh= 1.5rh= 0.175 Internal Moment :From Fig. 6-48, K := 1.5 MA := R⋅a MA = 35.00N⋅mMB := MAMaximum Bending Stresses at A and B :MC = R⋅ (a + 0.5L)c := 0.5⋅h σA.max KMA⋅cI:= ⋅σA.max = 19.688MPaσB.max := σA.maxAt Section C-C:Require, σC.max := σA.maxI'112:= ⋅ t⋅w3 I' = 180000mm4Maximum Bending Stress :c' := 0.5⋅w σC.maxMC⋅c'I'=σA.maxR⋅ (a + 0.5L)⋅c'I'=L⎛⎜⎝⎞2σA.max⋅ I'R⋅c'− 2a⎠:=L = 950mm Ans 655. Problem 6-151If the radius of each notch on the plate is r = 10 mm, determine the largest moment M that can beapplied. The allowable bending stress for the material is σ allow = 180 MPa.Given: w := 165mm h := 125mmt := 20mm r := 10mmσallow := 180MPaSolution:Section Property:I112:= ⋅ t⋅h3 I = 3255208.33mm4Stress Concentration Factor :b := 0.5⋅ (w − h)br= 2= 0.08rhFrom Fig. 6-50, K := 2.1Maximum Moment : σ KM⋅cI= ⋅c := 0.5⋅h M(σallow)⋅ IK⋅c:=M = 4.464 kN⋅m Ans 656. Problem 6-152The stepped bar has a thickness of 15 mm. Determine the maximum moment that can be applied to itsends if it is made of a material having an allowable bending stress of σ allow = 200 MPa.Given: w := 45mm t := 15mmh1 := 30mm h2 := 10mmr1 := 3mm r2 := 6mmσallow := 200MPaSolution:Section Property:For the larger section 1: For the smaller section 2:I1112:= ⋅ t ⋅ h13 I1 = 33750mm4 I2112:= ⋅ t ⋅ h23 I2 = 1250mm4Stress Concentration Factor :For the larger section 1: For the smaller section 2:wh1= 1.5r1h1= 0.1h1h2= 3r2h2= 0.6From Fig. 6-48, K1 := 1.75 From Fig. 6-48, K2 := 1.2Maximum Moment : σ KM⋅cI= ⋅For the larger section 1: For the smaller section 2:c1 := 0.5⋅h1 M1(σallow)⋅ I1K1⋅c1:= c2 := 0.5⋅h2 M2(σallow)⋅ I2K2⋅c2:=M1 = 257.14N⋅m M2 = 41.67N⋅mMallow := min(M1 ,M2)Mallow = 41.67N⋅m Ans 657. Problem 6-153The bar has a thickness of 12.5 mm and is made of a material having an allowable bending stress ofσ allow = 140 MPa. Determine the maximum moment M that can be applied.Given: t := 12.5mm r := 7.5mmw := 150mm h := 50mmσallow := 140MPaSolution:Section Property:I112:= ⋅ t⋅h3 I = 130208.33mm4Stress Concentration Factor :wr= 3= 0.15hhFrom Fig. 6-48, K := 1.6Maximum Moment : σ KM⋅cI= ⋅c := 0.5⋅h M(σallow)⋅ IK⋅c:=M = 455.73 N⋅m Ans 658. Problem 6-154The bar has a thickness of 12.5 mm and is subjected to a moment of 900 N·m. Determine themaximum bending stress in the bar.Given: t := 12.5mm r := 7.5mmw := 150mm h := 50mmM := 900N⋅mSolution:Section Property:I112:= ⋅ t⋅h3 I = 130208.33mm4Stress Concentration Factor :wr= 3= 0.15hhFrom Fig. 6-48, K := 1.6Maximum Bending Stress :c := 0.5⋅h σmax KM⋅cI:= ⋅σmax = 276.5MPa Ans 659. Problem 6-155The simply supported notched bar is subjected to two forces P. Determine the largest magnitude of Pthat can be applied without causing the material to yield. The material is A-36 steel. Each notch has aradius of r = 3 mm.Given: t := 12mm r := 3mmw := 42mm h := 30mma := 500mm σY := 250MPaSolution:Section Property:I112:= ⋅ t⋅h3 I = 27000.00mm4Support Reaction : By symmstry, R1=R2=R+Stress Concentration Factor : ΣFy=0; 2R − 2P = 0 R = Pb := 0.5⋅ (w − h)b rInternal Moment : At mid-span,= 2rh= 0.1 MC = R⋅aFrom Fig. 6-50, K := 1.92 MC = P⋅aMaximum Moment : σY KMC⋅cI= ⋅c := 0.5⋅h MC(σY)⋅ IK⋅c=P⋅a(σY)⋅ IK⋅c=P(σY)⋅ Ia⋅K⋅c:=P = 468.75 N Ans 660. Problem 6-156The simply supported notched bar is subjected to the two loads, each having a magnitude of P = 500N. Determine the maximum bending stress developed in the bar, and sketch the bending-stressdistribution acting over the cross section at the center of the bar. Each notch has a radius of r = 3 mm.Given: t := 12mm r := 3mmw := 42mm h := 30mma := 500mm P := 500NSolution:Section Property:I112:= ⋅ t⋅h3 I = 27000.00mm4Support Reaction : By symmstry, R1=R2=R+Stress Concentration Factor : ΣFy=0; 2R − 2P = 0 R := Pb := 0.5⋅ (w − h)b rInternal Moment : At mid-span,= 2rh= 0.1 MC := R⋅aFrom Fig. 6-50, K := 1.92 MC = 250N⋅mMaximum Bending Stress :c := 0.5⋅h σmax KMC⋅cI:= ⋅σmax = 266.7MPa Ans 661. Problem 6-157A rectangular A-36 steel bar has a width of 25 mm and height of 75 mm. Determine the momentapplied about the horizontal axis that will cause half the bar to yield.Given: b := 25mm σY := 250MPad := 75mmSolution: de := 0.5d dp := 0.5dElastic-plastic Moment:M σY bde2⋅⎛⎜⎝⎞⎠⋅2de3⋅ σY bdp2⋅⎛⎜⎝⎞⎠⋅ dedp2+⎛⎜⎝⎞⎠:= + ⋅M = 9.52 kN⋅m Ans 662. Problem 6-158The box beam is made of an elastic perfectly plastic material for which σY = 250 MPa. Determine theresidual stress in the top and bottom of the beam after the plastic moment Mp is applied and thenreleased.Given: bo := 200mm do := 200mmbi := 150mm di := 150mmσY := 250MPaSolution:Section Property:tb := 0.5(do − di) td := 0.5(bo − bi)I112:= ⋅ I = 91145833.33mm4bo ⋅ do3 − bi ⋅ di3 ⎛⎝⎞⎠Plastic Moment:Mp σY⋅ (bo⋅ tb)⋅ (do − tb) σY 2tddi2⋅⎛⎜⎝⎞⎠⋅di2⎛⎜⎝⎞⎠:= + ⋅Mp = 289062.50N⋅mModulus of Rupture:The modulus of rupture σr can be determined using the flexure formulawith the application of reverse plastic moment Mp.c := 0.5do σrMp⋅cI:= σr = 317.14MPaResidul Bending Stress:σ't := σr − σY σ't = 67.14MPa Ansσ'b := σr − σY σ'b = 67.14MPa Ans 663. Problem 6-159The box beam is made of an elastic plastic material for which σY = 250 MPa. Determine the residualstress in the top and bottom of the beam after the plastic moment Mp is applied and then released.Given: bf := 200mm dw := 200mmtf := 15mm tw := 20mmσY := 250MPaSolution:Section Property:D := dw + 2tf I112bf⋅D3 (bf − tw) dw3 ⋅ − ⎡⎣⎤⎦:= ⋅I = 82783333.33mm4Plastic Moment:Mp σY⋅ (bf⋅ tf)⋅ (D − tf) σY twdw2⋅⎛⎜⎝⎞⎠⋅dw2⎛⎜⎝⎞⎠:= + ⋅Mp = 211.25 kN⋅mModulus of Rupture:The modulus of rupture σr can be determined using the flexure formulawith the application of reverse plastic moment Mp.c := 0.5D σrMp⋅cI:= σr = 293.46MPaResidul Bending Stress:σ't := σr − σY σ't = 43.46MPa Ansσ'b := σr − σY σ'b = 43.46MPa Ans 664. Problem 6-160Determine the plastic section modulus and the shape factor of the beam's cross section.Set a := mmGiven: bf := 2a tf := adw := 2a tw := aSolution:Section Property :A := bf⋅ tf + dw⋅ twyc⋅ ( ⋅Ai)Σ⋅ (Ai)Σ yi ⎯= yc(bf⋅ tf)⋅ (0.5tf) + (dw⋅ tw)⋅ (0.5dw + tf)A:=yc = 1.25 aI112⋅ 3 b( f⋅ tf) (0.5tf − yc)+ ⋅ 2⋅bf tf1⋅ ⋅ 12tw dw3 + ⎡⎢⎣t( w⋅dw) ⋅ (0.5dw + tf − yc)2 ⎤⎥⎦:= +I = 3.08 a4Maximum Elastic Moment :c := dw + tf − yc c = 1.75 aσYMY⋅cI=MYσYIc=Ic⎛⎜⎝⎞⎠= 1.7619 a3MY = (1.7619 a3)⋅σYPlastic Moment :∫ = σY⋅ tw⋅ (d) − σY⋅ tw⋅ (dw − d) − σY⋅ (bf⋅ tf) = 0σdA 0Adtw⋅dw + bf⋅ tf:= d = 2 a2twdarm := 0.5tf + dw − 0.5d darm = 1.50 aMp = σY⋅ tw⋅ (d)⋅darmMpσY= tw⋅ (d)⋅darm tw⋅ (d)⋅darm = 3.00 a3Mp = (3.00 a3)⋅σYShape Factor : kMpMY= k3.00 a31.7619 a3:= k = 1.70 AnsPlastic Section Modulus : zMpσY= z = 3.00 a3 Ans 665. Problem 6-161The beam is made of elastic perfectly plastic material. Determine the maximum elastic moment and thplastic moment that can be applied to the cross section. Take a = 50 mmand σY = 230 MPa.Given: a := 50mm bf := 2a tf := aσY := 230MPa dw := 2a tw := aSolution:Section Property :A := bf⋅ tf + dw⋅ twyc⋅ ( ⋅Ai)Σ⋅ (Ai)Σ yi ⎯= yc(bf⋅ tf)⋅ (0.5tf) + (dw⋅ tw)⋅ (0.5dw + tf)A:=yc = 62.5mmI112⋅ 3 b( f⋅ tf) (0.5tf − yc)+ ⋅ 2⋅bf tf1⋅ ⋅ 12tw dw3 + ⎡⎢⎣t( w⋅dw) ⋅ (0.5dw + tf − yc)2 ⎤⎥⎦:= +I = 19270833.33mm4Maximum Elastic Moment :c := dw + tf − yc c = 87.50mmσYMY⋅cI= MYσY⋅ Ic:= MY = 50.65 kN⋅m AnsPlastic Moment :∫ = σY⋅ tw⋅ (d) − σY⋅ tw⋅ (dw − d) − σY⋅ (bf⋅ tf) = 0σdA 0Adtw⋅dw + bf⋅ tf:= d = 100mm2twdarm := 0.5tf + dw − 0.5d darm = 75.00mmMp := σY⋅ tw⋅ (d)⋅darmMp = 86.25 kN⋅m Ans 666. Problem 6-162The rod has a circular cross section. If it is made of an elastic plastic material, determine the shapefactor and the plastic section modulus Z.Set r := mmSolution:Section Property :A := πr2 I:= (r4)π4Maximum Elastic Moment :c := r σYMY⋅cI=MYσYIc=Ic⎛⎜⎝⎞⎠= 0.7854 r3MY = (0.7854 r3)⋅σYPlastic Moment :darm 24r3π⎛⎜⎝⎞⎠:=Mp σYA2⎛⎜⎝⎞⎠= ⋅ ⋅darmMpσYA2⎛⎜⎝⎞⎠= ⋅darmA2⎛⎜⎝⎞⎠⋅darm = 1.3333 r3Mp = (1.3333 r3)⋅σYShape Factor : kMpMY= k1.3333 r30.7854 r3:= k = 1.70 AnsPlastic Section Modulus : zMpσY= z = 1.333 r3 Ans 667. Problem 6-163The rod has a circular cross section. If it is made of an elastic plastic material, determine the maximumelastic moment and plastic moment that can be applied to the cross section.Take r = 75 mm, σY = 250MPa.Given: r := 75mm σY := 250MPaSolution:Section Property :A := πr2 I:= (r4)π4Maximum Elastic Moment :c := r σYMY⋅cI= MYσY⋅ Ic:=MY = 82.83 kN⋅m AnsPlastic Moment :darm 24r3π⎛⎜⎝⎞⎠:=Mp σYA2⎛⎜⎝⎞⎠:= ⋅ ⋅darm Mp = 140.63 kN⋅m Ans 668. Problem 6-164Determine the plastic section modulus and the shape factor of the cross section.Set a := mmGiven: bf := 3a tf := adw := 3a tw := a2Solution:Section a Property :A := bf ⋅ tf + ( dw − tf ) ⋅ tw A =5I112:= ⋅ 3+ ⋅ I = 2.41667 a4⋅bf tf112⎛⎝3 − tf3 ⋅ tw dw⎞⎠Maximum Elastic Moment :c := 0.5dw c = 1.5 aσYMY⋅cI=MYσYIc=Ic⎛⎜⎝⎞⎠= 1.61111 a3MY = (1.61111 a3)⋅σYPlastic Moment :dw.arm := tf + 0.5(dw − tf) dw.arm = 2.00 adf.arm := 0.5tf df.arm = 0.50 aMp σY bftf2⋅⎛⎜⎝⎞⎠⋅ ⋅df.arm σY⋅ twdw − tf2⎛⎜⎝⎞⎠= + ⋅ ⋅dw.armMptfbf⋅σY2⎛⎜⎝⎞⎠⋅df.arm twdw − tf2⎛⎜⎝⎞⎠= + ⋅ ⋅dw.armbftf2⋅⎛⎜⎝⎞⎠⋅df.arm twdw − tf2⎛⎜⎝⎞⎠+ ⋅ ⋅dw.arm = 2.75 a3Mp = (2.75 a3)⋅σYShape Factor : kMpMY= k2.75 a31.61111 a3:= k = 1.71 AnsPlastic Section Modulus : zMpσY= z = 2.75 a3 Ans 669. Problem 6-165The beam is made of elastic perfectly plastic material. Determine the maximum elastic moment and thplastic moment that can be applied to the cross section. Take a = 50 mm and σY = 250 MPa.Given: a := 50mm σY := 250MPabf := 3a tf := adw := 3a tw := a2Solution:Section a Property :A := bf ⋅ tf + ( dw − tf ) ⋅ tw A =5I112:= ⋅ 3+ ⋅ I = 2.41667 a4⋅bf tf112⎛⎝3 − tf3 ⋅ tw dw⎞⎠Maximum Elastic Moment :c := 0.5dw c = 1.5 aσYMY⋅cI= MYσY⋅ Ic:= MY = 50.35 kN⋅m AnsPlastic Moment :dw.arm := tf + 0.5(dw − tf) dw.arm = 2.00 adf.arm := 0.5tf df.arm = 0.50 aMp σY bftf2⋅⎛⎜⎝⎞⎠⋅ ⋅df.arm σY⋅ twdw − tf2⎛⎜⎝⎞⎠:= + ⋅ ⋅dw.armMp = 85.94 kN⋅m Ans 670. Problem 6-166The beam is made of an elastic perfectly plastic material. Determine the plastic moment Mp that can besupported by a beam having the cross section shown. σY = 210 MPa.Given: ro := 50mm tw := 25mm σY := 210MPari := 25mm dw := 250mmSolution:Plastic Moment :A1 π ro⎛⎝2 − ri2 ⎞⎠:= ⋅ darm1 := dw + 2roA2 := tw(0.5dw) darm2 := 0.5dwMp := (σY⋅A1)⋅darm1 + (σY⋅A2)⋅darm2Mp = 515 kN⋅m Ans 671. Problem 6-167Determine the plastic moment Mp that can be supported by a beam having the cross section shown.σY = 210 MPa.Given: ro := 50mm tw := 25mm σY := 210MPari := 25mm dw := 250mmSolution:⎛⎝2 − ri2 A1 π ro⎞⎠= ⋅A2 = tw(dw − d')A3 = twd'σdA 0A∫ = σY⋅A1 + σY⋅A2 − σY⋅A3 = 0A1 + A2 − A3 = 0⋅ + tw(dw − d') − twd' = 02 − ⎛⎝π ro2 ri⎞⎠d'2 − ⎛⎝π ro2 ri⎞⎠⋅ + tw(dw)2tw:=d' = 242.81mmPlastic Moment :A1 π ro:= ⋅ 2 − ri2 darm1 := ro + (dw − d')A2 := tw(dw − d') darm2 := 0.5(dw − d')A3 := twd' darm3 := 0.5d'Mp := (σY⋅A1)⋅darm1 + (σY⋅A2)⋅darm2 + (σY⋅A3)⋅darm3Mp ⎛⎝= 225.6 kN⋅m Ans⎞⎠ 672. Problem 6-168Determine the plastic section modulus and the shape factor for the member having the tubular crosssection.Set d := mmSolution:Section Property :Aπ4:= A = 2.35619 d22d ( )2 d2 − ⎡⎣⎤⎦Iπ64:= I = 0.73631 d42d ( )4 d4 − ⎡⎣⎤⎦Maximum Elastic Moment :c := d σYMY⋅cI=MYσYIc=Ic⎛⎜⎝⎞⎠= 0.73631 d3MY = (0.73631 d3)⋅σYPlastic Moment :yc⋅ ( ⋅Ai)Σ⋅ (Ai)Σ yi ⎯= yc0.5π4⎛⎜⎝⎞⎠(2d)2 4d3π⎛⎜⎝⎞⎠⋅ 0.5π4⎛⎜⎝⎞⎠d2 12⋅4d3π⎛⎜⎝⎞⎠−0.5A:=yc = 0.49515 ddarm := 2ycMp σYA2⎛⎜⎝⎞⎠= ⋅ ⋅darmMpσYA2⎛⎜⎝⎞⎠= ⋅darmA2⎛⎜⎝⎞⎠⋅darm = 1.16667 d3Mp = (1.16667 d3)⋅σYShape Factor : kMpMY= k1.16667 d30.73631 d3:= k = 1.58 AnsPlastic Section Modulus : zMpσY= z = 1.16667 d3 Ans 673. Problem 6-169Determine the plastic section modulus and the shape factor for the member.Solution: Set b := mm h := mmSection Property :A12= (b⋅h) I= (b⋅h3)136Maximum Elastic Moment :c23:= ⋅h σYMY⋅cI=MYσYIc=Ic⎛⎜⎝⎞⎠124= ⋅b⋅h2MY1⋅ b ⋅ h2 24⎛⎜⎝⎞⎠Plastic Moment : = ⋅σYFrom the geometry, b'dh= ⋅bAΔ12= ⋅b'⋅d Atrp12= ⋅ (b + b')⋅ (h − d)AΔ12d⋅ b h⎛⎜⎝⎞⎠= ⋅ ⋅d Atrp12db+ ⋅ b h⎛⎜⎝⎞⎠= ⋅ ⋅ (h − d)∫ σdA = 0σY⋅ (AΔ) − σY⋅ (Atrp) = 0AAΔ = Atrp12d⋅ b h⎛⎜⎝⎞⎠⋅ ⋅d12db+ ⋅ b h⎛⎜⎝⎞⎠= ⋅ ⋅ (h − d) dh2= b'b2=Note: The centroid of a trapezoidal area was usedin the calculation.hch − d32⋅b' + bb' + b= ⋅darm13− ⎡⎣⎤⎦= +d ⋅ h d − ( ) hcdarm13⋅d (h − d)b' + 2⋅b3(b' + b)= +darm13h2⋅ hh2− ⎛⎜⎝⎞⎠b + 2 2⋅b3(b + 2⋅b) = + ⋅ darm4(2 − 2)⋅h6=Mp σYA2⎛⎜⎝⎞⎠= ⋅ ⋅darmMpσYA2⎛⎜⎝⎞⎠= ⋅darmA2⋅darm2 − 26= ⋅b⋅h2Mp2 − 2⋅b⋅h2 6⎛⎜⎝⎞⎠= ⋅σYShape Factor : kMpMY= k2 − 26⋅b⋅h2:= k = 2.34 Ans124⋅b⋅h2Plastic Section Modulus : zMpσY= z2 − 26= ⋅b⋅h2 Ans 674. Problem 6-170The member is made of elastic perfectly plastic material for which σY = 230 MPa. Determine themaximum elastic moment and the plastic moment that can be applied to the cross section.Take b = 50mm and h = 80 mm.Given: b := 50mm h := 80mm σY := 230MPaSolution:Section Property :A12:= (b⋅h) A = 2000mm2I:= (b⋅h3) I = 711111.11mm4136Maximum Elastic Moment :c23:= ⋅h σYMY⋅cI= MYσY⋅ Ic:= MY = 3.07 kN⋅m AnsPlastic Moment :From the geometry, b'dh= ⋅b AΔ12= ⋅b'⋅d Atrap12= ⋅ (b + b')⋅ (h − d)AΔ12d⋅ b h⎛⎜⎝⎞⎠= ⋅ ⋅d Atrap12db+ ⋅ b h⎛⎜⎝⎞⎠= ⋅ ⋅ (h − d)∫ = σY⋅ (AΔ) − σY⋅ (Atrap) = 0 AΔ = AtrapσdA 0A12d⋅ b h⎛⎜⎝⎞⎠⋅ ⋅d12db+ ⋅ b h⎛⎜⎝⎞⎠= ⋅ ⋅ (h − d) dh2= b'b2=Note: The centroid of a trapezoidal area was usedin the calculation.hch − d32⋅b' + bb' + b= ⋅darm13− ⎡⎣⎤⎦= +d ⋅ h d − ( ) hcdarm13⋅d (h − d)b' + 2⋅b3(b' + b)= +darm13h2⋅ hh2− ⎛⎜⎝⎞⎠b + 2 2⋅b3(b + 2⋅b) = + ⋅darm4(2 − 2)⋅h:= darm = 31.24mm6Mp σY⎛⎜⎝⎞⎠A2:= ⋅ ⋅darmMp = 7.19 kN⋅m Ans 675. Problem 6-171The wide-flange member is made from an elasticplastic material. Determine the shape factor and theplastic section modulus Z.Set b := mm h := mm t := mmGiven: bf := b tf := tD := h tw := tSolution: dw D 2tf= − ⋅ dw := h − 2tSection Property :A = bf⋅D − (bf − tw)⋅dw A := b⋅h − (b − t)⋅ (h − 2t)I112⋅bf⋅D3112⋅ (bf − tw) dw= − ⋅ 3I112⋅b⋅h3112Maximum Elastic Moment : := − ⋅ (b − t)⋅ (h − 2t)3c := 0.5D c := 0.5hσYMY⋅cI=MYσYIc=Ic16⋅b⋅h216⋅h= − ⋅ (b − t)⋅ (h − 2t)3MY16⋅hb h3 ⋅ b t − ( ) h 2 t ⋅ − ( )3 ⋅ − ⎡⎣⎤⎦= ⋅ ⋅σYPlastic Moment :dw.arm = 0.5dw dw.armh − 2t2=df.arm = D − tf df.arm = h − tMp σY⋅ (bf⋅ tf)⋅df.arm σY⋅ twdw2⎛⎜⎝⎞⎠= + ⋅ ⋅dw.armMph − 2t(b⋅ t)⋅ (h − t) tσY2⎛⎜⎝⎞⎠⋅h − 2t2= + ⋅h 2t − ( )2 ⋅ + ⎡⎢⎣Mp (b⋅ t)⋅ (h − t)t4⎤⎥⎦= ⋅σYShape Factor : kMpMY= k(b⋅ t)⋅ (h − t)t4+ ⋅ (h − 2t)216⋅hb h3 ⋅ b t − ( ) h 2 t ⋅ − ( )3 ⋅ − ⎡⎣⎤⎦⋅=k3⋅h24b⋅ t⋅ (h − t) + t⋅ (h − 2t)2b⋅h3 − (b − t)⋅ (h − 2⋅ t)3⎡⎢⎢⎣⎤⎥⎥⎦= AnsPlastic Section Modulus : zMpσY= z (b⋅ t)⋅ (h − t)t4= + ⋅ (h − 2t)2 Ans 676. Problem 6-172The beam is made of an elastic-plastic material for which σY = 200 MPa. If the largest moment in thebeam occurs within the center section a-a, determine the magnitude of each force P that causes thismoment to be (a) the largest elastic moment and (b) the largest plastic moment.Given: a := 2m b := 100mm h := 200mmσY := 200MPaSolution:Section Property :A := b⋅h A = 20000mm2I112:= ⋅b⋅h3 I = 66666666.67mm4a) Maximum Elastic Moment :c := 0.5⋅h MY = P⋅a σYMY⋅cI=σYP⋅a⋅cI= PσY⋅ Ia⋅c:=P = 66.67 kN Ansb) Plastic Moment :darmh2:= Mp σYA2⎛⎜⎝⎞⎠:= ⋅ ⋅darm Mp = 200.00 kN⋅mP'Mpa:= P' = 100.00 kN Ans 677. Problem 6-173The beam is made of phenolic, a structural plastic, that has the stress-strain curve shown. If a portionof the curve can be represented by the equation σ = (5(106)ε )1/2 MPa, determine the magnitude of wthe distributed load that can be applied to the beam without causing the maximum strain in its fibers athe critical section to exceed εmax = 0.005 mm/mm.Given: b := 150mm h := 150mmσ2 = 5(106)εL := 2m εmax 0.005mmmm:=Solution:Stress-strain Relationship : unit := MPaWhen εmax = 0.005,σmax := unit⋅ 5(106)εmaxσmax = 158.11MPaResultant Internal Forces :The resultant internal forces T and C can be evaluated from the volume ofstress block which is a paraboloid, T = C.T23⋅σmaxb⋅h2⎛⎜⎝⎞⎠:= ⋅ T = 1185.85 kNdarm 235h2⎛⎜⎝⎞⎠⋅ ⎡⎢⎣⎤⎥⎦:= ⋅ darm = 90mmMaximum Internal Moment :Mmax := T⋅darm Mmax = 106.73 kN⋅mBy observation, the maximum moment occurs over the middle support.Mmax = w⋅L wMmaxL:=w 53.36 mkNm= Ans 678. Problem 6-174The box beam is made from an elastic plastic material for which σY = 175 MPa. Determine theintensity of the distributed load w0 that will cause the moment to be (a) the largest elastic moment and(b) the largest plastic moment.Given: bo := 200mm do := 400mmbi := 150mm di := 300mmL := 3m σY := 175MPaSolution:Support Reaction : By symmstry, R1=R2=R+ ΣFy=0; 2R − 2(0.5⋅wo⋅L) = 0R = 0.5⋅wo⋅LMaximum Moment :M R⋅LL3= − (0.5⋅wo⋅L) Mwo⋅L23=a) Elastic Analysis : σM⋅cI=c := 0.5do I112bo ⋅ do3 − bi ⋅ di3 ⎛⎝⎞⎠:= ⋅MYσY⋅ Ic:= MY = 638.02 kN⋅mwo3⋅MYL2:= wo 212.67kNm= Ansb) Plastic Analysis : tw := 0.5(bo − bi) tf := 0.5(do − di)Af := bo⋅ tf darm1 := do − tfAw := di⋅ tw darm2 := 0.5diMp := (σY⋅Af)⋅darm1 + (σY⋅Aw)⋅darm2Mp = 809.4 kN⋅mw'o3⋅MpL2:= w'o 269.79kNm= Ans 679. Problem 6-175The beam is made of a polyester that has the stress-strain curve shown. If the curve can berepresented by the equation σ = [140 tan-1(15ε )] MPa, where tan-1(15ε ) is in radians, determine themagnitude of the force P that can be applied to the beam without causing the maximum strain in itsfibers at the critical section to exceed εmax = 0.003 mm/mm.Given: b := 50mm h := 100mmσ = 140 atan(15ε)L := 2.4m εmax 0.003mmmm:=Solution:Support Reaction : By symmstry, R1=R2=R+ ΣFy=0; 2R − 2P = 0R = PMaximum Moment :M = R⋅L M = P⋅LStress-strain Relationship : unit := MPaThe bending stress can bs expressedin terms of y usingεεmax0.5h= ⋅yσ 140 atan 15εmax0.5h⋅y⎛⎜⎝⎞⎠= ⋅unitWhen εmax = 0.003, ymax := 0.5hσmax 140 atan 15εmax0.5h⋅ymax⎛⎜⎝⎞⎠:= ⋅ ⋅unitσmax = 6.30MPa= ∫Resultant Internal Moment : M yσdAAM 20.5h0εmax0.5hy 140 atan 15 ⋅yy⎛⎜⎝⎞⎠⋅unit⎛⎜⎝⎞⎠⋅ ⋅b⌠⎮⎮⌡:= dM = 524.79 N⋅mPML:= P = 218.66 N Ans 680. Problem 6-176The stress-strain diagram for a titanium alloy can be approximated by the two straight lines. If a strutmade of this material is subjected to bending, determine the moment resisted by the strut if themaximum stress reaches a value of (a) σA and (b) σB.Given: b := 50mm d := 75mmσA := 980MPa σB := 1260MPaεA 0.01mmmm:= εB 0.04mmmm:=Solution:Maximum Elastic Moment :Since the stress is linearly related to strain upto point A, the flexure formula can be applied.σM⋅cI=3d c := 0.5d I := b ⋅MYσA⋅ Ic:= MY = 551.25 kN⋅m AnsUItimate Moment :yAεAεB:= ⋅ (0.5d) h := 0.5d − yAC1 = T1 T1σA + σB:= ⋅ (b⋅h)2C2 = T2 T2σA2:= ⋅ (b⋅yA)Note: The centroid of a trapezoidal area was usedin the calculation of moment.hch32σB + σAσB + σA:= ⋅darm1 := 2(yA + hc)darm223:= (2yA)M := (T1)⋅darm1 + (T2)⋅darm2M = 78.54 kN⋅m Ansσnot zero. 681. Problem 6-177A beam is made from polypropylene plastic and has a stress-strain diagram that can be approximatedby the curve shown. If the beam is subjected to a maximum tensile and compressive strain of ε = 0.02mm/mm, determine the maximum moment M.Given: b := 30mm h := 100mmσ 10 = 4 εMPaεmax 0.02mmmm:=Solution:Stress-strain Relationship : unit := MPaThe bending stress can bs expressedin terms of y usingεεmax0.5h= ⋅yσ 104 εmax0.5h= ⋅y⋅unit= ∫Resultant Internal Moment : M yσdAAM 20.5h04 εmax0.5hy 10 ⋅y⋅unity⎛⎜⎜⎝⎞⎠⋅ ⋅b⌠⎮⎮⎮⌡:= dM = 0.251 kN⋅m Ans 682. Problem 6-178The bar is made of an aluminum alloy having a stress-strain diagram that can be approximated by thestraight line segments shown. Assuming that this diagram is the same for both tension andcompression, determine the moment the bar will support if the maximum strain at the top and bottomfibers of the beam is εmax = 0.03.Given: b := 75mm d := 100mmσA := 420MPa σB := 560MPaεA 0.006mmmm:= εB 0.025mmmm:=σC := 630MPa εC 0.05mmmm:=εmax 0.03mmmm:=Solution:Maximum Stress :σmax − σBεmax − εBσC − σBεC − εB= σmaxσC − σBεC − εB:= ⋅ (εmax − εB) + σB σmax = 574MPaMaximum Moment :yAεAεmax:= ⋅ (0.5d) yBεBεmax:= ⋅ (0.5d) h1 := 0.5d − yB h2 := yB − yAC1 = T1 T1σB + σmax:= ⋅ (b⋅h1)2C2 = T2 T2σA + σB:= ⋅ (b⋅h2)2C3 = T3 T3σA2:= ⋅ (b⋅yA)Note: The centroid of a trapezoidal area was usedin the calculation of moment.hc1h132σmax + σBσmax + σB:= ⋅ hc2h232σB + σAσB + σA:= ⋅darm1 := 2(yB + hc1)darm2 := 2(yA + hc2)darm323:= (2yA)M := (T1)⋅darm1 + (T2)⋅darm2 + (T3)⋅darm3M = 96.48 kN⋅m Ans 683. Problem 6-179The bar is made of an aluminum alloy having a stress-strain diagram that can be approximated by thestraight line segments shown. Assuming that this diagram is the same for both tension andcompression, determine the moment the bar will support if the maximum strain at the top and bottomfibers of the beam is εmax = 0.05.Given: b := 75mm d := 100mmσA := 420MPa σB := 560MPaεA 0.006mmmm:= εB 0.025mmmm:=σC := 630MPa εC 0.05mmmm:=Solution: εmax := εCStress-strain Relationship :σ1εσAεA= σ1σAεA= ⋅ εσ2 − σAε − εAσB − σAεB − εA= σ2σB − σAεB − εA= ⋅ (ε − εA) + σAσ3 − σBε − εBσC − σBεC − εB= σ3σC − σBεC − εB= ⋅ (ε − εB) + σByAεAεmax:= ⋅ (0.5d) yBεBεmaxεmax0.5dStrain : ε := ⋅ (0.5d)= ⋅yσ1σAεAεmax0.5d⋅y⎛⎜⎝⎞⎠= ⋅ for 0 < y < yAσ2σB − σAεB − εAεmax0.5d⋅y − εA⎛⎜⎝⎞⎠= ⋅ + σA for yA < y < yBσ3σC − σBεC − εBεmax0.5d⋅y − εB⎛⎜⎝⎞⎠= ⋅ + σB for yB < y < 0.5d= ∫Resultant Moment : M yσdAAM1 2yA0σAεA⎛⎜⎝⎞εmax0.5dy ⋅yy⎠⋅⎡⎢⎣⎤⎥⎦⋅ ⋅b⌠⎮⎮⌡:= d M1 = 0.76 kN⋅mM2 2yByAσB − σAεB − εAεmax0.5dy y⋅y − εA⎛⎜⎝⎞⎠⋅ + σA⎡⎢⎣⎤⎥⎦⋅ ⋅b⌠⎮⎮⎮⌡:= d M2 = 22.28 kN⋅mM3 2yByAσC − σBεC − εBεmax0.5dy y⋅y − εB⎛⎜⎝⎞⎠⋅ + σB⎡⎢⎣⎤⎥⎦⋅ ⋅b⌠⎮⎮⎮⌡:= d M3 = 23.8 kN⋅m 684. M := M1 + M2 + M3M = 46.84 kN⋅m AnsNote: The solution can also be obtained from stress blocks as in Prob, 6-178 685. Problem 6-180The beam is made of a material that can be assumed perfectly plastic in tension and elastic perfectlyplastic in compression. Determine the maximum bending moment M that can be supported by the beaso that the compressive material at the outer edge starts to yield.Solution:A1 = a⋅d C12= σY⋅A1A2 = a(h − d) T = σY⋅A2σdA 0A∫ = C − T = 012σY⋅A1 − σY⋅A2 = 012A1 − A2 = 012a⋅d − a(h − d) = 0d2⋅h3=Plastic Moment :darm23d12= + (h − d) darm232⋅h312h2⋅h3− ⎛⎜⎝⎞⎠= + darm11⋅h18=Mp = (σY⋅A2)⋅darmMp σY⋅a⋅ (h − d)11⋅h18= ⋅Mp11⋅a⋅h254= ⋅σY Ans 686. Problem 6-181The plexiglass bar has a stress-strain curve that can be approximated by the straight-line segmentsshown. Determine the largest moment M that can be applied to the bar before it fails.Given: b := 20mm h := 20mmσt1 := 40MPa σt2 := 60MPaεt1 0.02mmmm:= εt2 0.04mmmm:=σc1 := −80MPa σc2 := −100MPaεc1 −0.04mmmm:= εc2 −0.06mmmm:=Solution:Assume failure due to tension and εc < εc1A1 = b(h − d) C12= σc⋅A1A212= ⋅b⋅d T112= σt1⋅A2A312= ⋅b⋅d T212= (σt1 + σt2)⋅A3∫ = C − T1 − T2 = 0σdA 0A12σc⋅ [b(h − d)]12σt11⋅ b ⋅ d 2⎛⎜⎝⎞⎠− ⋅12b ⋅ d ⋅ ⎛⎜⎝(σt1 + σt2) 12⎞⎠− ⋅ = 0σch− 1 d⎛⎜⎝⎞⎠⋅ = σt.1 + 0.5σt.2Try σc := 74.833MPa then dh⋅σc:= d = 10.334mmσt1 + 0.5σt2 + σcσc < σc1Check :From the strain diagram, εch − dd:= ⋅ εt2 εc 0.037417mmmm= O.K! εc < εc1From the σ-ε diagram, σεcεc1:= ⋅σc1 σ = 74.833MPa O.K! Close to assumed value.Hence,C12:= σc⋅ [b(h − d)] C = 7.2336 kNT1⎞⎠⋅ := T1 2.0667 kN =12σt11⋅ b ⋅ d 2⎛⎜⎝T212b ⋅ d ⋅ ⎛⎜⎝(σt1 + σt2) 12⎞⎠:= ⋅ T2 = 5.1668 kN 687. Ultimate Moment :darm123:= (h − d) darm1 = 6.4442mmdarm223d2⎛⎜⎝⎞⎠:= darm2 = 3.4446mmNote: The centroid of a trapezoidal area was usedin the calculation.hc0.5d32σt1 + σt2σt1 + σt2:= ⋅ hc = 2.4112mmdarm3 := d − hc darm3 = 7.9225mmMult := C⋅darm1 + T1⋅darm2 + T2⋅darm3Mult = 94.67N⋅m Ans 688. Problem 6-182The beam is made from three boards nailed together as shown. If the moment acting on the crosssection is M = 650 N·m, determine the resultant force the bending stress produces on the top board.Given: bf := 290mm tf := 15mmtw := 20mm dw := 125mmM := 650N⋅mSolution: D := dw + tf⎯ Σ yi ⎯y⋅ ( ⋅Ai)Σ⋅ (Ai)=yc(bf⋅ tf)⋅0.5tf + 2(dw⋅ tw)⋅ (0.5dw + tf)bf⋅ tf + 2dw⋅ tw:=yc = 44.933mmIf112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (yc − 0.5tf)2Iw112⋅ (2tw) dw⋅ 3 2tw dw ⋅ ( ) yc 0.5dw tf + ( ) − ⎡⎣:= + ⋅ 2⎤⎦I := If + Iw I = 17990374.89mm4Bending Stress: σ McI= ⋅At B: cB := yc − tf σB McBI:= ⋅ σB = 1.0815MPaAt A: cA := yc σA McAI:= ⋅ σA = 1.6235MPaResultant Force: For the top board.F := 0.5(σA + σB)⋅ (bf⋅ tf) F = 5.883 kN Ans 689. Problem 6-183The beam is made from three boards nailed together as shown. Determine the maximum tensile andcompressive stresses in the beam.Given: bf := 290mm tf := 15mmtw := 20mm dw := 125mmM := 650N⋅mSolution: D := dw + tf⎯ Σ yi ⎯y⋅ ( ⋅Ai)Σ⋅ (Ai)=yc(bf⋅ tf)⋅0.5tf + 2(dw⋅ tw)⋅ (0.5dw + tf)bf⋅ tf + 2dw⋅ tw:=yc = 44.933mmIf112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (yc − 0.5tf)2Iw112⋅ (2tw) dw⋅ 3 2tw dw ⋅ ( ) yc 0.5dw tf + ( ) − ⎡⎣:= + ⋅ 2⎤⎦I := If + Iw I = 17990374.89mm4Maximum Bending Stress: σ McI= ⋅For compression:cc := yc σc_max MccI:= ⋅ σc_max = 1.623MPa AnsFor tension:ct := D − yc σt_max MctI:= ⋅ σt_max = 3.435MPa Ans 690. Problem 6-184Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam asfunctions of x, where 0 ≤ x < 1.8 m.Given: a := 2.4m b := 1.2mP := 40kN w 30kNm:=M := 75kN⋅mSolution:Equilibrium :+A := w⋅a + P A = 112 kN 305.4kN.m112kN30xΣFy=0;ΣΜ MA := (w⋅a)⋅ (0.5a) + P⋅ (a + b) + M A=0;MA = 305.40 kN⋅mx1x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + bV1 ( 1x1 ) := ( A − w ⋅ ) ⋅ V2(x2) (A − w⋅a)kN1kN:= ⋅ AnsM1(x1) −MA + A⋅x1 0.5w x12 ⋅ − ⎛⎝⎞⎠1kN⋅m:= ⋅ AnsM2 x2 ( ) MA − A x2 ⋅ + w a ⋅ ( ) x20.5 a ⋅ − ( ) ⋅ − M + ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 3150100500Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 1 2 30200400Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 691. Problem 6-185Draw the shear and moment diagrams for the beam. Hint: The 100-kN load must be replaced byequivalent loadings at point C on the axis of the beam.Given: a := 1.2m b := 1.2mc := 1.2m d := 0.3mF1 := 75kN F2 := 100kNSolution:Equilibrium : Given+ ΣFy=0; A − F1 + B = 0ΣΜC=0; A⋅ (a + b + c) − F1⋅ (b + c) − F2⋅ (d) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠58.3316.67⎛⎜⎝⎞⎠= kNx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cV1(x1) := A⋅ V2(x2) (F1) 1A − 1kN:= ⋅ V3(x3) (A − F1) 1kNkN:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ F1 x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ F1 x3 a − ( ) ⋅ − F2 d ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 3500Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x30 1 2 3100500Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 692. Problem 6-186Determine the plastic section modulus and the shape factor for the wide-flange beam.Given: bf := 180mm tf := 20mmdw := 180mm tw := 30mmSolution: D := dw + 2⋅ tfSection Property :A := bf⋅D − (bf − tw)⋅dw A = 12600mm2I1⋅D31− ⋅ (12bf⋅12bf − tw) ⋅ dw3 ⎡⎢⎣⎤⎥⎦:= I = 86820000mm4Set σY := MPaMaximum Elastic Moment :c := 0.5D σYMY⋅cI= MYIc⎛⎜⎝⎞⎠:= ⋅σYMYσY= 789272.73mm3Plastic Moment :dw.arm := 0.5dw dw.arm = 90mmdf.arm := D − tf df.arm = 200mmMp (bf⋅ tf)⋅df.arm twdw2⎛⎜⎝⎞⎠+ ⋅ ⋅dw.arm⎡⎢⎣⎤⎥⎦:= ⋅σYMp= 963000.00mm3σYPlastic Section Modulus : zMpσY:= z = 963 × 10− 6m3 AnsShape Factor : kMpMY:= k = 1.22 Ans 693. Problem 6-187Draw the shear and moment diagrams for the shaft if it is subjected to the vertical loadings of the belt,gear, and flywheel. The bearings at A and B exert only vertical reactions on the shaft.Given: a := 200mm b := 400mmc := 300mm d := 200mmC := 450N D := −300NSolution: E := 150NEquilibrium : Given+ ΣFy=0; A + B − C − D − E = 0ΣΜB=0; A⋅ (a + b + c) − C⋅ (b + c) − D⋅c + E⋅d = 0Guess A := 1N B := 1NAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠216.6783.33⎛⎜⎝⎞⎠= Nx1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cx4 := a + b + c , 1.01⋅ (a + b + c) .. a + b + c + dV1(x1) A1N:= ⋅ V2(x2) (A − C)1N:= ⋅ V3(x3) (A − C − D)1N:= ⋅V4(x4) (A − C − D + B)1N:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ C x2 a − ( ) ⋅ − ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ C x3 a − ( ) ⋅ − D x3 a − b − ( ) ⋅ − ⎡⎣⎤⎦1N⋅m:= ⋅M4 x4 ( ) A x4 ( ) ⋅ C x4 a − ( ) ⋅ − D x4 a − b − ( ) ⋅ − B x4 a − b − c − ( ) ⋅ + ⎡⎣⎤⎦1N⋅m:= ⋅0 0.2 0.4 0.6 0.8 14002000200400Distance (m)Shear (N)V1(x1)V2(x2)V3(x3)V4(x4)x1, x2, x3, x4 694. 0 0.2 0.4 0.6 0.8 150050Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)M4(x4)x1, x2, x3, x4 695. Problem 6-188The beam is constructed from four pieces of wood, glued together as shown. If the internal bendingmoment is M = 120 kN·m, determine the maximum bending stress in the beam. Sketch athree-dimensional view of the stress distribution acting over the cross section.Given: bo := 300mm do := 300mmbi := 250mm di := 250mmM := 120kN⋅mSolution:Section Property :I112bo ⋅ do3 − bi ⋅ di3 ⎛⎝⎞⎠:= ⋅Maximum Bending Stress: σM⋅cI=cmax := 0.5doσmaxM⋅cmaxI:=σmax = 51.51MPa Ans 696. Problem 6-189The beam is constructed from four pieces of wood, glued together as shown. If the internal bendingmoment is M = 120 kN·m, determine the resultant force the bending moment exerts on the top andbottom boards of the beam.Given: bo := 300mm do := 300mmbi := 250mm di := 250mmM := 120kN⋅mSolution:Section Property : t := 0.5(do − di)I112bo ⋅ do3 − bi ⋅ di3 ⎛⎝⎞⎠:= ⋅Maximum Bending Stress: σM⋅cI=co := 0.5do σoM⋅coI:= σo = 51.505MPaci := 0.5di σiM⋅ciI:= σi = 42.921MPaResultant Force :F12:= ⋅ (σo + σi)⋅ (t⋅bo)F = 354.10 kN Ans 697. Problem 6-190For the section, Iy =31.7(10-6) m4, Iz = 114(10-6) m4, Iyz = 15.1(10-6) m4. Using the techniques outlinein Appendix A, the member's cross-sectional area has principal moments of inertia of Iy' = 29(10-6) mand Iz' = 117(10-6) m4, computed about the principal axes of inertia y' and z', respectively. If the sectiois subjected to a moment of M = 2 kN·m directed as shown, determine the stress produced at point A,(a) using Eq. 6-11 and (b) using the equation developed in Prob. 6-111.Given: M := 2000N⋅m θ := 10.10degb1 := 80mm b2 := 140mmh1 := 60mm h2 := 60mmIy := 31.7(10− 6)m4 Iz := 114(10− 6)m4Iyz := 15.1(10− 6)m4Iy' := 29.0(10− 6)m4 Iz' := 117(10− 6)m4Solution: θ' := −θCoordinates of Point A : yA := b2 zA := h2y'Az'A⎛⎜⎜⎝⎞⎠cos(θ')sin(θ')−sin(θ')cos(θ')⎛⎜⎝⎞⎠yAzA⎛⎜⎜⎝⎞⎠:= ⋅y'Az'A⎛⎜⎜⎝⎞⎠148.3534.52⎛⎜⎝⎞⎠= mma) Using Eq. 6-11Internal Moment Components :My' := M⋅ sin(θ) Mz' := M⋅ cos(θ)Bending Stress:σAMz'⋅y'AIz'−My'⋅z'AIy'+⎛⎜⎝⎞⎠:= σA = −2.079MPa (C) Ansb) Using the equation developed in Prob. 6-111Internal Moment Components :My := 0 Mz := MBending Stress: Using formula developed in Prob. 6-111.:= − 2D Iy⋅ Iz IyzσA1D(Mz My )IyIyzIyzIz⎛⎜⎜⎝⎞⎠⋅−yAzA⎛⎜⎜⎝⎞⎠:= ⋅σA = −2.086MPa (C) Ans 698. Problem 6-191The strut has a square cross section a by a and is subjected to the bending moment M applied at anangle θ as shown. Determine the maximum bending stress in terms of a, M, and θ. What angle θ wilgive the largest bending stress in the strut? Specify the orientation of the neutral axis for this case.Solution:Internal Moment Components :My = −M⋅ sin(θ) Mz = −M⋅cos(θ)Section Property :Iya412= Iza412=Maximum Bending Stress:By inspection, maximum bending stress occurs at A (and B).At A : yA = 0.5a zA = −0.5aσMz⋅yIz−My⋅zIy= +σ−12M⋅cos(θ)⋅ (0.5a)a4−−12M⋅ sin(θ)⋅ (−0.5a)a4= +σ6⋅Ma3= ⋅ (cos(θ) + sin(θ)) Ansd σdθ6⋅Ma3= ⋅ (−sin(θ) + cos(θ))d σdθ= 0 −sin(θ) + cos(θ) = 0tan(θ) = 1 θ := 45deg AnsOrientation of Neutral Axis : θ' := −θtan(α) Iz= ⋅ tan(θ')IyIzIy= 1α := atan(1⋅ tan(θ'))α = −45.00 deg Ans 699. Problem 7-1If the beam is subjected to a shear of V = 15 kN, determine the web's shear stress at A and B. Indicatethe shear-stress components on a volume element located at these points. Set w = 125 mm. Show thatthe neutral axis is located at y = 0.1747 m from the bottom and INA = 0.2182(10-3) m4.Given: bf := 200mm b'f := 125mm tf := 30mmtw := 25mm dw := 250mmV := 15kNSolution:Section Property : D := dw + 2tfA := bf⋅ tf + dw⋅ tw + b'f⋅ tf A = 16000mm2⎯ Σ yi ⎯y⋅ ( ⋅Ai)Σ⋅ (Ai)= yc(b'f⋅ tf)⋅ (0.5tf) + (dw⋅ tw)⋅ (0.5dw + tf) + (bf⋅ tf)⋅ (D − 0.5tf)A:=yc = 174.69mmI'f:= ⋅ 3 + b'( f⋅ tf) ⋅ (0.5tf − yc)2Iw112⋅b'f tf112:= ⋅ tw ⋅ dw3 + t( w⋅dw) ⋅ (0.5dw + tf − yc)2If112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (D − 0.5tf − yc)2I := If + Iw + I'f I = 218.18 × 10− 6m4Q Σ yi ⎯= ⋅ ⋅A1 QA := (D − 0.5tf − yc)⋅ (bf⋅ tf) QA = 721875.00mm3QB := (yc − 0.5tf)⋅ (b'f⋅ tf) QB = 598828.12mm3Shear Stress: τV⋅QI⋅ t=τAV⋅QAI⋅ tw:= τA = 1.99MPa AnsτBV⋅QBI⋅ tw:= τB = 1.65MPa Ans 700. Problem 7-2If the wide-flange beam is subjected to a shear of V = 30 kN, determine the maximum shear stress inthe beam. Set w = 200 mm.Given: bf := 200mm tf := 30mmtw := 25mm dw := 250mmV := 30kNSolution:Section Property : D := dw + 2tfI112bf⋅D3 (bf − tw) dw3 ⋅ − ⎡⎣⎤⎦:= ⋅I = 268.65 × 10− 6m4Q Σ yi ⎯= ⋅ ⋅A1 Qmax (bf⋅ tf) D2tf2−⎛⎜⎝⎞⎠⋅dw2⋅ tw⎛⎜⎝⎞⎠dw4⎛⎜⎝⎞⎠:= + ⋅Qmax = 1035312.50mm3Shear Stress: τV⋅QI⋅ t=Maximum shear stress occurs at the point where the neutral axis passes through the section.τmaxV⋅QmaxI⋅ tw:= τmax = 4.62MPa Ans 701. Problem 7-3If the wide-flange beam is subjected to a shear of V = 30 kN, determine the shear force resisted by theweb of the beam. Set w = 200 mm.Given: bf := 200mm dw := 250mmtf := 30mm tw := 25mmV := 30kNSolution:Section Property : D := dw + 2tfI112bf⋅D3 (bf − tw) dw3 ⋅ − ⎡⎣⎤⎦:= ⋅I = 268.65 × 10− 6m4Af 0.5D y − ( ) bf= ⋅ yf = 0.5(0.5D − y) + yyf = 0.5(0.5D + y)Qf = Af⋅yfQf 0.5 0.5D y − ( ) bf= ⋅ ⋅ (0.5D + y)Qf = 0.5bf(0.25D2 − y2)Shear Stress: τV⋅QI⋅ t=τfVI⋅bf⎛⎜⎝⎞⎠= ⋅QfτfVI⋅bf⎛⎜⎝⎞⎠0.5bf 0.25D2 y2 − ( ) ⎡⎣⎤⎦= ⋅= ∫ =Resultant Shear Force: V τ dA 0AFor the flange,VfVI⋅bf0.5D0.5dwy 0.5 bf ⋅ 0.25 D2 ⋅ y2 − ( ) ⋅ ⎡⎣⎤⎦⋅bf⌠⎮⌡d⎡⎢⎢⎣⎤⎥⎥⎦:= ⋅Vf = 1.46 kNVw V 2Vf:= −Vw = 27.09 kN Ans 702. Problem 7-4If the wide-flange beam is subjected to a shear of V = 125 kN, determine the maximum shear stress inthe beam.Given: bf := 200mm dw := 250mmtf := 25mm tw := 25mmV := 125kNSolution:Section Property : D := dw + 2tfI112bf⋅D3 (bf − tw) dw3 ⋅ − ⎡⎣⎤⎦:= ⋅Qmax (bf⋅ tf) D2tf2−⎛⎜⎝⎞⎠⋅dw2⋅ tw⎛⎜⎝⎞⎠dw4⎛⎜⎝⎞⎠:= + ⋅Maximum Shear Stress: τV⋅QI⋅ t=Maximum shear stress occurs at the point where the neutral axis passes through the section.τmaxV⋅QmaxI⋅ tw:=τmax = 19.87MPa Ans 703. Problem 7-5If the wide-flange beam is subjected to a shear of V = 125 kN, determine the shear force resisted bythe web of the beam.Given: bf := 200mm dw := 250mmtf := 25mm tw := 25mmV := 125kNSolution:Section Property : D := dw + 2tfI112bf⋅D3 (bf − tw) dw3 ⋅ − ⎡⎣⎤⎦:= ⋅A1 = bf⋅ tf y1c = 0.5(D − tf)A2 = (0.5dw − y)⋅ tw y2c = 0.5(0.5dw − y) + yy2c = 0.5(0.5dw + y)Qw = A1⋅ (y1c) + A2⋅ (y2c)Qw = 0.5bf⋅ tf⋅ (D − tf) + 0.5(0.5dw − y)⋅ tw⋅ (0.5dw. + y)Qw 0.5bf⋅ tf⋅ (D − tf) 0.5tw 0.25dw2 y2 − ⎛⎝⎞⎠= +Shear Stress: τV⋅QI⋅ t=τwVI⋅ tw⎛⎜⎝⎞⎠= ⋅QwτwVI⋅ tw⎛⎜⎝⎞⎠0.5⋅bf⋅ tf⋅ (D − tf) 0.5⋅ tw 0.25 dw2 ⋅ y2 − ⎛⎝⎞⎠⋅ + ⎡⎣⎤⎦= ⋅= ∫ =Resultant Shear Force: For the web. V τ dA 0AVw0.5dw− 0.5dwyVI⋅ tw⎛⎜⎝⎞⎠0.5⋅bf⋅ tf⋅ (D − tf) 0.5⋅ tw 0.25 dw2 ⋅ y2 − ⎛⎝⎞⎠⋅ + ⎡⎣⎤⎦⋅ ⋅ tw⌠⎮⎮⌡:= dVw = 115.04 kN Ans 704. Problem 7-6The beam has a rectangular cross section and is made of wood having an allowable shear stress ofτallow = 11.2 MPa. If it is subjected to a shear of V = 20 kN, determine the smallest dimension a of itsbottom and 1.5a of its sides.Given: V := 20kN τallow := 11.2MPaSolution:Section Property :I112= ⋅a⋅ (1.5a)3 t = aQmax1.5a⋅ a 2⎛⎜⎝⎞⎠1.5a4⎛⎜⎝⎞⎠= ⋅Allowablwe Shear Stress: τV⋅QI⋅ t=I⋅ tV⋅Qmaxτallow=112⋅a⋅ (1.5a)3⋅aVτallow⎛⎜⎝⎞⎠1.5a⋅ a 2⎛⎜⎝⎞⎠⋅1.5a4⎛⎜⎝⎞⎠= ⋅aVτallow:=a = 42.26mm Ans 705. Problem 7-7The beam has a rectangular cross section and is made of wood. If it is subjected to a shear of V = 20kN, and a = 250 mm, determine the maximum shear stress and plot the shearstress variation over thecross section. Sketch the result in three dimensions.Given: a := 250mm V := 20kNSolution:Section Property : b := a d := 1.5aI112:= ⋅b⋅d3Qmaxd⋅ b 2⎛⎜⎝⎞⎠d4⎛⎜⎝⎞⎠:= ⋅Maximum Shear Stress: τV⋅QI⋅b=Maximum shear stress occurs at the point wherethe neutral axis passes through the section.τmaxV⋅QmaxI⋅b:=τmax = 0.320MPa Ans 706. Problem 7-8Determine the maximum shear stress in the strut if it is subjected to a shear force of V = 20 kN.Given: bf := 120mm tf := 12mmtw := 80mm dw := 60mmV := 20kNSolution:Section Property : D := dw + 2tfI112bf⋅D3 (bf − tw) dw3 ⋅ − ⎡⎣⎤⎦:= ⋅I = 5.21 × 10− 6m4Q Σ yi ⎯= ⋅ ⋅A1 Qmax (bf⋅ tf) D2tf2−⎛⎜⎝⎞⎠⋅dw2⋅ tw⎛⎜⎝⎞⎠dw4⎛⎜⎝⎞⎠:= + ⋅Qmax = 87840.00mm3Shear Stress: τV⋅QI⋅ t=Maximum shear stress occurs at the point where the neutral axis passes through the section.τmaxV⋅QmaxI⋅ tw:=τmax = 4.22MPa Ans 707. Problem 7-9Determine the maximum shear force V that the strut can support if the allowable shear stress for thematerial is τallow = 40 MPa.Given: bf := 120mm tf := 12mmtw := 80mm dw := 60mmτallow := 40MPaSolution:Section Property : D := dw + 2tfI112bf⋅D3 (bf − tw) dw3 ⋅ − ⎡⎣⎤⎦:= ⋅I = 5.21 × 10− 6m4Q Σ yi ⎯= ⋅ ⋅A1 Qmax (bf⋅ tf) D2tf2−⎛⎜⎝⎞⎠⋅dw2⋅ tw⎛⎜⎝⎞⎠dw4⎛⎜⎝⎞⎠:= + ⋅Qmax = 87840.00mm3Shear Stress: τV⋅QI⋅ t=Maximum shear stress occurs at the point where the neutral axis passes through the section.τallowV⋅QmaxI⋅ tw= VI⋅ tw⋅ τallowQmax:=V = 189.69 kN Ans 708. Problem 7-10Plot the intensity of the shear stress distributed over the cross section of the strut if it is subjected to ashear force of V = 15 kN.Given: bf := 120mm tf := 12mmtw := 80mm dw := 60mmV := 15kNSolution:Section Property : D := dw + 2tfI112bf⋅D3 (bf − tw) dw3 ⋅ − ⎡⎣⎤⎦:= ⋅I = 5.21 × 10− 6m4Q Σ yi ⎯= ⋅ ⋅A1QA (bf⋅ tf) D2tf2−⎛⎜⎝⎞⎠:= ⋅ QA = 51840.00mm3Qmax (bf⋅ tf) D2tf2−⎛⎜⎝⎞⎠⋅dw2⋅ tw⎛⎜⎝⎞⎠dw4⎛⎜⎝⎞⎠:= + ⋅Qmax = 87840.00mm3Shear Stress: τV⋅QI⋅ t=Maximum shear stress occurs at the point where the neutral axis passes through the section.τmaxV⋅QmaxI⋅ tw:= τmax = 3.16MPa Ansτw_AV⋅QAI⋅ tw:= τw_A = 1.87MPa Ansτf_AV⋅QAI⋅bf:= τf_A = 1.24MPa Ans 709. Problem 7-11If the pipe is subjected to a shear of V = 75 kN, determine the maximum shear stress in the pipe.Given: ro := 60mm ri := 50mm V := 75kNSolution:Section Property : t := ro − riIπ4ro4 − ri4 ⎛⎝⎞⎠:= ⋅Qmax4ro3π⋅ 22π ro⎛⎜⎝⎞⎠⋅4ri3π⋅ 22π ri⎛⎜⎝⎞⎠:= − ⋅Maximum Shear Stress: τV⋅QI⋅b=Maximum shear stress occurs at the point where the neutral axis passes through the section.τmaxV⋅QmaxI⋅ (2t):=τmax = 43.17MPa Ans 710. Problem 7-12The strut is subjected to a vertical shear of V = 130 kN. Plot the intensity of the shear-stressdistribution acting over the cross-sectional area, and compute the resultant shear force developed in thevertical segment AB.Given: bf := 350mm tf := 50mmtw := 50mm dw := 350mmV := 130kNSolution:Section Property : a := 0.5(bf − tw)I112bf ⋅ tf3 + tw ⋅ dw3 − tw ⋅ tf3 ⎛⎝⎞⎠:= ⋅I = 181.77 × 10− 6m4Q Σ yi ⎯= ⋅ ⋅A1QC (a⋅ tw) a2tf2+⎛⎜⎝⎞⎠:= ⋅ QC = 750000mm3QD (a⋅ tw) a2tf2+⎛⎜⎝⎞⎠⋅tf2⋅bf⎛⎜⎝⎞⎠tf4⎛⎜⎝⎞⎠:= + ⋅ QD = 859375mm3τV⋅QI⋅ tShear Stress: =τw_CV⋅QCI⋅ tw:= τf_CV⋅QCI⋅bf:= τDV⋅QDI⋅bf:=τw_C = 10.73MPa τf_C = 1.53MPa τD = 1.76MPa= ∫ =Resultant Shear Force: V τ dA 0AAw = (0.5dw − y)⋅ tw yw = 0.5(0.5dw − y) + yyw = 0.5(0.5dw + y)Qw = Aw⋅ywQw = 0.5(0.5dw − y)⋅ tw⋅ (0.5dw + y)Q 0.5tw 0.25dw2 y2 − ⎛⎝⎞⎠=τwVI⋅ tw⎛⎜⎝⎞⎠= ⋅Qw τwVI⋅ tw⎛⎜⎝⎞⎠2 y2 − ⎛⎝0.5tw 0.25dw⎞⎠⎡⎣⎤⎦= ⋅VABVI⋅ tw0.5dw0.5tf2 ⋅ y2 − ⎛⎝⎞⎠⋅ ⎡⎣⎤⎦0.5⋅ tw 0.25 dw ⋅ twy⌠⎮⎮⌡d⎡⎢⎢⎣⎤⎥⎥⎦:= ⋅ VAB = 50.29 kN Ans 711. Problem 7-13The steel rod has a radius of 30 mm. If it is subjected to a shear of V = 25 kN, determine themaximum shear stress.Given: r := 30mm V := 25kNSolution:Section Property :Iπ4:= ⋅ r4Qmax4r3ππ⋅ r22⎛⎜⎝⎞⎠:= ⋅Maximum Shear Stress: τV⋅QI⋅b=Maximum shear stress occurs at the point where the neutral axis passes through the section.τmaxV⋅QmaxI⋅ (2r):=τmax = 11.79MPa Ans 712. Problem 7-14If the T-beam is subjected to a vertical shear of V = 60 kN, determine the maximum shear stress in thebeam. Also, compute the shear-stress jump at the flange-web junction AB. Sketch the variation of theshear-stress intensity over the entire cross section.Given: bf := 300mm dw := 150mmtf := 75mm tw := 100mmV := 60kNSolution:Section Property : D := dw + tfyc0.5tf(bf⋅ tf) + (0.5dw + tf)(dw⋅ tw):= yc = 82.50mmbf⋅ tf + dw⋅ twI1:= ⋅ 3 + b( f⋅ tf) ⋅ (0.5tf − yc)2I2112⋅bf tf112:= ⋅ tw ⋅ dw3 + d( w⋅ tw) ⋅ (0.5dw + tf − yc)2I := I1 + I2Qmax tw⋅ (D − yc)D − yc2:= ⋅QAB := (bf⋅ tf)⋅ (yc − 0.5tf)Shear Stress: τV⋅QI⋅b=τmaxV⋅QmaxI⋅ tw:= τmax = 3.993MPa Ansτf_ABV⋅QABI⋅bf:= τf_AB = 1.327MPa Ansτw_ABV⋅QABI⋅ tw:= τw_AB = 3.982MPa Ans 713. Problem 7-15If the T-beam is subjected to a vertical shear of V = 60 kN, determine the vertical shear force resistedby the flange.Given: bf := 300mm dw := 150mmtf := 75mm tw := 100mmV := 60kNSolution:Section Property : D := dw + tfyc0.5tf(bf⋅ tf) + (0.5dw + tf)(dw⋅ tw):= yc = 82.50mmbf⋅ tf + dw⋅ twI1:= ⋅ 3 + b( f⋅ tf) ⋅ (0.5tf − yc)2I2112⋅bf tf112:= ⋅ tw ⋅ dw3 + d( w⋅ tw) ⋅ (0.5dw + tf − yc)2I := I1 + I2A'f = (yc − y)⋅bf yfc = 0.5(yc − y) + yyfc = 0.5(yc + y)Q = Af⋅yfcQ = 0.5(yc − y)⋅bf(yc + y)Q 0.5bf yc'2 y2 − ⎛⎝⎞⎠=Shear Stress: τV⋅QI⋅b=τfVI⋅bf⎛⎜⎝⎞⎠= ⋅QτfVI⋅bf⎛⎜⎝⎞⎠2 y2 − ⎛⎝⎞⎠⎡⎣0.5bf yc⎤⎦= ⋅= ∫ =Resultant Shear Force: For the flange. yo := yc − tf V τ dA 0AVfycyoyVI⋅bf⎛⎜⎝⎞⎠2 y2 − ⎛⎝0.5bf yc⎞⎠⎡⎣⎤⎦⋅ ⋅bf⌠⎮⎮⌡:= dVf = 19.08 kN Ans 714. Problem 7-16The T-beam is subjected to the loading shown. Determine the maximum transverse shear stress in thebeam at the critical section.Given: L1 := 2m L2 := 2m L3 := 3mbf := 100mm dw := 100mmtf := 20mm tw := 20mmP := 20kN w 8kNm:=Solution: L := L1 + L2 + L3Support Reaction :Equilibrium : Given+ ΣFy=0; A − P − w⋅L3 + B = 0ΣΜB=0; AL − P⋅ (L2 + L3) − (w⋅L3)⋅ (0.5L3) = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠19.4324.57⎛⎜⎝⎞⎠= kNSection Property : D := dw + tfyc0.5tf(bf⋅ tf) + (0.5dw + tf)(dw⋅ tw):= yc = 40.00mmbf⋅ tf + dw⋅ twI1:= ⋅ 3 + b( f⋅ tf) ⋅ (0.5tf − yc)2I2112⋅bf tf112:= ⋅ tw ⋅ dw3 + d( w⋅ tw) ⋅ (0.5dw + tf − yc)2I := I1 + I2 I = 5333333.33mm4Qmax tw⋅ (D − yc)D − yc2:= ⋅ Qmax = 64000mm3Maximum Shear Stress: τV⋅QI⋅b= Vmax := BτmaxVmax⋅Qmax(I)⋅ tw:=τmax = 14.74MPa Ans 715. Shear Force Diagram:x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2) x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. (L)V1(x1) A:= V2(x2) (A − P)kN1kN⋅ := V3 x3 ( ) A P − w x3 L1 − L2 − ( ) ⋅ − ⎡⎣⎤⎦1kN:= ⋅0 2 4 620020Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3 716. Problem 7-17Determine the largest end forces P that the member can support if the allowable shear stress is τallow =70 MPa. The supports at A and B only exert vertical reactions on the beam.Given: L1 := 1m τallow := 70MPaL2 := 2mw 3kNm:= L3 := 1mdo := 100mm bo := 160mmdi := 60mm bi := 80mmSolution:Section Property :yc0.5do(bo⋅do) − (0.5di)(bi⋅di):= yc = 58.57mmbo⋅do − bi⋅diI1:= ⋅ 3 + b( o⋅do) ⋅ (0.5do − yc)2I2112⋅bo do112:= ⋅bi ⋅ di3 + b( i⋅di) ⋅ (0.5di − yc)2 I I:= 1 − I2Qmax = A'f⋅yfc A'f = yc⋅ (bo − bi) y'c = 0.5yc Qmax 0.5yc= 2 × (bo − bi)Maximum Shear Stress: τV⋅QI⋅b= Vmax = PτallowPI⋅ (bo − bi)⎡⎢⎣⎤⎥⎦0.5yc2 ( bo − bi ) ⎡⎣⎤⎦= ⋅τallowPI⎛⎜⎝⎞⎠⎛⎝0.5yc2 ⎞⎠= ⋅PI0.5yc2⎛⎜⎝⎞⎠:= ⋅ (τallow)P = 373.42 kN AnsShear Force Diagram: L := L1 + L2 + L3Equilibrium : Given+ ΣFy=0; A wL2− ⋅ + B − 2P = 0ΣΜB=0; −P⋅ (L1 + L2) + A⋅L2 − (w⋅L2)⋅ (0.5L2) + P⋅L3 = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠376.42376.42⎛⎜⎝⎞⎠= kN 717. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2) x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. (L)V1(x1) −P:= ⋅ V3(x3) (−P + A − w⋅L2 + B) 1:= V2 x2 ( ) P − A + w x2 L1 − ( ) ⋅ − ⎡⎣kN⎤⎦1kNkN:= ⋅0 1 2 3 40Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3 718. Problem 7-18If the force P = 4 kN, determine the maximum shear stress in the beam at the critical section. Thesupports at A and B only exert vertical reactions on the beam.Given: L1 := 1m P := 4kNL2 := 2mw 3kNm:= L3 := 1mdo := 100mm bo := 160mmdi := 60mm bi := 80mmSolution:Section Property :yc0.5do(bo⋅do) − (0.5di)(bi⋅di):= yc = 58.57mmbo⋅do − bi⋅diI1:= ⋅ 3 + b( o⋅do) ⋅ (0.5do − yc)2I2112⋅bo do112:= ⋅bi ⋅ di3 + b( i⋅di) ⋅ (0.5di − yc)2 I I:= 1 − I2Qmax = A'f⋅yfc A'f = yc⋅ (bo − bi) y'c = 0.5yc Qmax 0.5yc= 2 × (bo − bi)Maximum Shear Stress: τV⋅QI⋅b= Vmax = PτmaxPI⋅ (bo − bi)⎡⎢⎣⎤⎥⎦0.5yc2 ( bo − bi ) ⎡⎣⎤⎦= ⋅τmaxPI⎛⎜⎝⎞⎠⎛⎝0.5yc2 ⎞⎠:= ⋅τmax = 0.750MPa AnsShear Force Diagram: L := L1 + L2 + L3Equilibrium : Given+ ΣFy=0; A wL2− ⋅ + B − 2P = 0ΣΜB=0; −P⋅ (L1 + L2) + A⋅L2 − (w⋅L2)⋅ (0.5L2) + P⋅L3 = 0Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠7.007.00⎛⎜⎝⎞⎠= kN 719. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2) x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. (L)V1(x1) −P:= ⋅ V3(x3) (−P + A − w⋅L2 + B) 1:= V2 x2 ( ) P − A + w x2 L1 − ( ) ⋅ − ⎡⎣kN⎤⎦1kNkN:= ⋅0 1 2 3 410010Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3 720. Problem 7-19Plot the shear-stress distribution over the cross section of a rod that has a radius c. By what factor isthe maximum shear stress greater than the average shear stress acting over the cross section? 721. Problem 7-20Develop an expression for the average vertical component of shear stress acting on the horizontal planethrough the shaft, located a distance y from the neutral axis. 722. Problem 7-21Railroad ties must be designed to resist large shear loadings. If the tie is subjected to the 150-kN railloadings and the gravel bed exerts a distributed reaction as shown, determine the intensity w forequilibrium, and find the maximum shear stress in the tie.Given: L1 := 0.45m P := 150kNL2 := 0.90mw 3kNm:= L3 := 0.45md := 150mm b := 200mmSolution:Equilibrium :+ ΣFy=0; 0.5w⋅L1 + w⋅L2 + 0.5w⋅L3 − 2P = 0w2P0.5L1 + L2 + 0.5L3:=w 222.22kNm=Section Property : I112:= ⋅b⋅d3 Qmax := (0.5⋅b⋅d)⋅0.25dShear Force Diagram: L := L1 + L2 + L3x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2) x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. (L)V1(x1) 0.5⋅x1w⋅x1L1⎛⎜⎝⎞⎠1kN⋅ := V2 x2 ( ) 0.5w L1 ⋅ P − w x2 L1 − ( ) ⋅ + ⎡⎣⎤⎦1kN:= ⋅V3(x3) 0.5⋅w⋅L1 − 2⋅P + w⋅L2 w⋅ (x3 − L1 − L2) 1x3 − L1 − L22L3−⎛⎜⎝⎞⎠+ ⋅⎡⎢⎣⎤⎥⎦1kN:= ⋅0 0.5 1 1.51000100Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3Maximum Shear Stress:τV⋅QI⋅b=Vmax := V2(L1) ⋅kNτmaxVmaxI⋅b⎛⎜⎝⎞⎠:= ⋅Qmaxτmax = 5MPa Ans 723. Problem 7-22The beam is subjected to a uniform load w. Determine the placement a of the supports so that the shearstress in the beam is as small as possible. What is this stress?Set wkNm:= a := m L := 5mGiven: L1 := a L3 := aL2 := L − 2aSolution:Equilibrium : By equilibrium, A = B = RΣFy=0; 2R − w⋅L = 0+R := 0.5w⋅LShear Force Diagram:x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2) x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. (L)V1(x1) (w⋅x1) 1:= ⋅ V2(x2) (w⋅x2 − R) 1kN:= ⋅ V3(x3) (w⋅x3 − 2⋅R) 1kNkN:= ⋅Distance (m) Shear (kN)0 1 2 3 4 50V1(x1)V2(x2)V3(x3)x1, x2, x3Require,V1 a ( ) V2= − (a)w⋅a = −(w⋅a − R)2w⋅a = 0.5w⋅La = 0.25LVmax = w⋅a Vmax := 0.25w⋅LSection Property : I112= ⋅b⋅d3 Qmax bd2⋅ ⎛⎜⎝⎞⎠d4= ⋅Maximum Shear Stress: τV⋅QI⋅b=τmaxVmaxI⋅b⎛⎜⎝⎞⎠= ⋅Qmax τmax0.25w⋅L112⋅b⋅d3⋅b⎛⎜⎜⎝⎞⎠bd2⋅ ⎛⎜⎝⎞⎠d4⋅ ⎡⎢⎣⎤⎥⎦= ⋅τmax3w⋅L8⋅b⋅d= Ans 724. Problem 7-23The timber beam is to be notched at its ends as shown. If it is to support the loading shown, determinethe smallest depth d of the beam at the notch if the allowable shear stress is τallow = 3 MPa. The beamhas a width of 200 mm.Given: L1 := 1.2m P1 := 12.5kNL2 := 1.8m P2 := 25.0kNL3 := 1.8m P3 := 12.5kNL4 := 1.2m b := 200mmτallow := 3MPaSolution:Equilibrium : By symmetry, R1=R , R2=R+ ΣFy=0; P1 + P2 + P3 − 2R = 0R := 0.5(P1 + P2 + P3) R = 25.00 kNSection Property : I112= ⋅b⋅d3Qmax = (0.5⋅b⋅d)⋅0.25d Qmax = 0.125⋅b⋅d2Maximum Shear Stress:τV⋅QI⋅b= Vmax := RτallowR⋅ (0.125b⋅d2)112⎞⎠b ⋅ d3 ⋅ ⎛⎜⎝⋅b=d12⋅ (0.125)⋅R(τallow)⋅b:=d = 62.5mm Ans 725. Problem 7-24The beam is made from three boards glued together at the seams A and B. If it is subjected to theloading shown, determine the shear stress developed in the glued joints at section a-a. The supports atC and D exert only vertical reactions on the beam.Given: bf := 150mm dw := 200mmtf := 40mm tw := 50mmP := 25kNSolution:Equilibrium : By symmetry, RC=R , RD=R+ ΣFy=0; 3P − 2R = 0R := 1.5PR = 37.50 kNSection Property : D := dw + 2tfI112:= ⋅bf⋅D3− ⋅ 3QA := (bf⋅ tf)⋅ (0.5D − 0.5tf)QB := QA112⋅ tw dwShear Stress: τV⋅QI⋅b= Vaa := R − PτAVaa⋅QAI⋅ tw:= τA = 0.747MPa AnsτBVaa⋅QBI⋅ tw:= τB = 0.747MPa Ans 726. Problem 7-25The beam is made from three boards glued together at the seams A and B. If it is subjected to theloading shown, determine the maximum shear stress developed in the glued joints. The supports at Cand D exert only vertical reactions on the beam.Given: bf := 150mm dw := 200mmtf := 40mm tw := 50mmP := 25kNSolution:Equilibrium : By symmetry, RC=R , RD=R+ ΣFy=0; 3P − 2R = 0R := 1.5PR = 37.50 kNSection Property : D := dw + 2tfI112:= ⋅bf⋅D3− ⋅ 3QA := (bf⋅ tf)⋅ (0.5D − 0.5tf)QB := QA112⋅ tw dwShear Stress: τV⋅QI⋅b= Vmax := RτAVmax⋅QAI⋅ tw:= τA = 2.24MPa AnsτBVmax⋅QBI⋅ tw:= τB = 2.24MPa Ans 727. Problem 7-26The beam is made from three boards glued together at the seams A and B. If it is subjected to theloading shown, determine the maximum vertical shear force resisted by the top flange of the beam. Thesupports at C and D exert only vertical reactions on the beam.Given: bf := 150mm dw := 200mmtf := 40mm tw := 50mm P := 25kNSolution:Equilibrium : By symmetry, RC=R , RD=R+ ΣFy=0; 3P − 2R = 0R := 1.5PSection Property : D := dw + 2tfI112:= ⋅bf⋅D3− ⋅ 3yc := 0.5DA'f = (yc − y)⋅bf yfc = 0.5(yc − y) + y112⋅ tw dwyfc = 0.5(yc + y)Q = Af⋅yfcQ = 0.5(yc − y)⋅bf(yc + y)Q 0.5bf yc'2 y2 − ⎛⎝⎞⎠=Shear Stress in flange: τV⋅QI⋅b= V = RτRI⋅bf⎛⎜⎝⎞⎠= ⋅QτRI⎛⎜⎝⎞⎠2 y2 − ⎛⎝0.5 yc⎞⎠⎡⎣⎤⎦= ⋅Resultant Shear Force: For the flange. yo := yc − tf VfAτ A⌠⎮⌡= dVfycyoy⎛⎜⎝⎞⎠RI2 y2 − ⎛⎝0.5 yc⎞⎠⎡⎣⎤⎦⋅ ⋅bf⌠⎮⎮⌡:= dVf = 2.36 kN Ans 728. Problem 7-27Determine the shear stress at points B and C located on the web of the fiberglass beam.Given: bf := 100mm dw := 150mmtf := 18mm tw := 12mmL1 := 2m L2 := 0.6mL3 := 2mwo 2.5kNm:= w1 3kNm:=Solution: L := L1 + L2 + L3Equilibrium : Given+ ΣFy=0; A − wo⋅L1 − 0.5⋅w1⋅L3 + D = 0ΣΜD=0; A⋅L − wo⋅L1⋅ (L − 0.5L1) 0.5⋅w1⋅L32L33⎛⎜⎝⎞⎠− ⋅ = 0Guess A := 1kN D := 1kNAD⎛⎜⎝⎞⎠:= Find(A,D)AD⎛⎜⎝⎞⎠4.7833.217⎛⎜⎝⎞⎠= kNSection Property : D := dw + 2tfI112:= ⋅bf⋅D3− ⋅ 3QB := (bf⋅ tf)⋅ (0.5D − 0.5tf)QC := QB112⋅ tw dwShear Stress: τV⋅QI⋅b100mm84mm= VBC := A − wo⋅ (0.5L1)τBVBC⋅QBI⋅ tw:= τB = 0.572MPa AnsτCVBC⋅QCI⋅ tw:= τC = 0.572MPa Ans18mm18mm12mm75mm75mm 729. Problem 7-28Determine the maximum shear stress acting in the fiberglass beam at the critical section.Given: bf := 100mm dw := 150mmtf := 18mm tw := 12mmL1 := 2m L2 := 0.6mL3 := 2mwo 2.5kNm:= w1 3kNm:=Solution: L := L1 + L2 + L3Equilibrium : Given+ ΣFy=0; A − wo⋅L1 − 0.5⋅w1⋅L3 + D = 0ΣΜD=0; A⋅L − wo⋅L1⋅ (L − 0.5L1) 0.5⋅w1⋅L32L33⎛⎜⎝⎞⎠− ⋅ = 0Guess A := 1kN D := 1kNAD⎛⎜⎝⎞⎠:= Find(A,D)AD⎛⎜⎝⎞⎠4.7833.217⎛⎜⎝⎞⎠= kNSection Property : D := dw + 2tfI112:= ⋅bf⋅D3− ⋅ 3Qmax := (bf⋅ tf)⋅ (0.5D − 0.5tf) + (0.5dw⋅ tw)⋅ (0.25dw)112⋅ tw dwShear Stress: τV⋅QI⋅b= Vmax := AτmaxVmax⋅Qmax:= τmax = 1.467MPa AnsI⋅ twShear Force Diagram: L := L1 + L2 + L3x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2) x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. (L)V1(x1) (A − wo⋅x1) 1:= ⋅ V2(x2) (A − wo⋅L1) 1kNkN:= ⋅V3(x3) A − wo⋅L1 w1⋅ (x3 − L1 − L2) 1x3 − L1 − L22L3−⎛⎜⎝⎞⎠− ⋅⎡⎢⎣⎤⎥⎦1kN:= ⋅0 1 2 3 4505Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3 730. Problem 7-29The beam is made from three plastic pieces glued together at the seams A and B. If it is subjected tothe loading shown, determine the shear stress developed in the glued joints at the critical section. Thesupports at C and D exert only vertical reactions on the beam.Given: bf := 200mm dw := 200mmtf := 50mm tw := 50mmwo 3kNm:= L := 2.5mSolution:Equilibrium : By symmetry, RC=R , RD=R+ ΣFy=0; wo⋅L − 2R = 0R := 0.5(wo⋅L)R = 3.75 kNSection Property : D := dw + 2tfI112:= ⋅bf⋅D3− ⋅ 3QA := (bf⋅ tf)⋅ (0.5D − 0.5tf)QB := QA112⋅ tw dwShear Stress: τV⋅QI⋅b= Vmax := RτAVmax⋅QAI⋅ tw:= τA = 0.225MPa AnsτBVmax⋅QBI⋅ tw:= τB = 0.225MPa Ans 731. Problem 7-30The beam is made from three plastic pieces glued together at the seams A and B. If it is subjected tothe loading shown, determine the vertical shear force resisted by the top flange of the beam at thecritical section. The supports at C and D exert only vertical reactions on the beam.Given: bf := 200mm dw := 200mmtf := 50mm tw := 50mmL := 2.5m wo 3kNm:=Solution:Equilibrium : By symmetry, RC=R , RD=R+ ΣFy=0; wo⋅L − 2R = 0R := 0.5(wo⋅L)R = 3.75 kNSection Property : D := dw + 2tfI112:= ⋅bf⋅D3− ⋅ 3yc := 0.5DA'f = (yc − y)⋅bf yfc = 0.5(yc − y) + y112⋅ tw dwyfc = 0.5(yc + y)Q = Af⋅yfcQ = 0.5(yc − y)⋅bf(yc + y)Q 0.5bf yc'2 y2 − ⎛⎝⎞⎠=Shear Stress in flange: τV⋅QI⋅b= Vmax = RτRI⋅bf⎛⎜⎝⎞⎠= ⋅QτRI⎛⎜⎝⎞⎠2 y2 − ⎛⎝⎤⎦⋅ =0.5 yc⎞⎠⎡⎣Resultant Shear Force: For the flange. yo := yc − tf VfAτ A⌠⎮⌡= dVfycyoyRI⎛⎜⎝⎞⎠2 y2 − ⎛⎝0.5 yc⎞⎠⎡⎣⎤⎦⋅ ⋅bf⌠⎮⎮⌡:= dVf = 0.30 kN Ans 732. Problem 7-31Determine the variation of the shear stress over the cross section of a hollow rivet. What is themaximum shear stress in the rivet? Also, show that if then 1 0 τ max = 2( V/A ). r → r 733. Problem 7-32The beam has a square cross section and is subjected to the shear force V. Sketch the shear-stressdistribution over the cross section and specify the maximum shear stress. Also, from the neutral axis,locate where a crack along the member will first start to appear due to shear. 734. Problem 7-33Write a computer program that can be used to determine the maximum shear stress in the beam thathas the cross section shown, and is subjected to a specified constant distributed load w andconcentrated force P. Show an application of the program using the values L = 4 m, a = 2 m, P = 1.5kN, d1 = 0, d2 = 2 m, w = 400 N/m, t1 = 15 mm, t2 = 20 mm, b = 50 mm, and h = 150 mm. 735. Problem 7-34The beam has a rectangular cross section and is subjected to a load P that is just large enough todevelop a fully plastic moment Mp = PL at the fixed support. If the material is elastic-plastic, then at adistance x < L the moment M = P x creates a region of plastic yielding with an associated elastic corehaving a height 2y'. This situation has been described by Eq. 6-30 and the moment M is distributedover the cross section as shown in Fig. 6-54e. Prove that the maximum shear stress developed in thebeam is given by τ max = 3/2 (P/A'),where A' = 2y'b, the cross-sectional area of the elastic core. 736. Problem 7-35The beam in Fig. 6-54f is subjected to a fully plastic moment Mp. Prove that the longitudinal andtransverse shear stresses in the beam are zero. Hint: Consider an element of the beam as shown in Fig.7-4d. 737. Problem 7-36The beam is constructed from two boards fastened together at the top and bottom with two rows ofnails spaced every 150 mm. If each nail can support a 2.5-kN shear force, determine the maximumshear force V that can be applied to the beam.Given: b := 150mm d1 := 50mmd2 := 50mm sn := 150mmFallow := 2.5kNSolution:Section Property : D := d1 + d2I112:= ⋅b⋅D3:= ( ⋅ )⋅ (0.5d1)Q bd1Shear Flow : qV⋅QI=There are two rows of nails. Hence, the allowable shear flow is qallow2Fallowsn:=2FallowsnVmax⋅QI=Vmax2⋅ I FallowQ⋅ sn:=Vmax = 2.222 kN Ans 738. Problem 7-37The beam is constructed from two boards fastened together at the top and bottom with two rows ofnails spaced every 150 mm. If an internal shear force of V = 3 kN is applied to the boards, determinethe shear force resisted by each nail.Given: b := 150mm d1 := 50mmd2 := 50mm sn := 150mmV := 3kNSolution:Section Property : D := d1 + d2I112:= ⋅b⋅D3:= ( ⋅ )⋅ (0.5d1)Q bd1Shear Flow : qV⋅QI:= q 45.00kNm=There are two rows of nails. Hence, the shear force resisted by each nail is Fq2:= ⋅ snF = 3.37 kN Ans 739. Problem 7-38A beam is constructed from five boards bolted together as shown. Determine the maximum shearforce developed in each bolt if the bolts are spaced s = 250 mm apart and the applied shear is V = 35kN.Given: d1 := 250mm d2 := 350mmt := 25mm a := 100mms := 250mm V := 35kNSolution: a' := d2 − (d1 − a) a' = 200mmh := a' + 0.5d1 h = 325mmSection Property :yc2(d1⋅ t)⋅ (0.5d1 + a') + 3(d2⋅ t)⋅ (0.5d2)2(d1⋅ t) + 3(d2⋅ t):=yc = 223.39mmI1112:= ⋅ (2t) ⋅ d13 + d( 2t ⋅ 1) ⋅ (0.5d1 + a' − yc)2I2112:= ⋅ (3⋅ t) ⋅ d23 + d( 3⋅ t ⋅ 2) ⋅ (0.5d2 − yc)2I := I1 + I2 I = 523597110.22mm4Q := d1⋅ (2t)⋅ (h − yc) Q = 1270161.29mm3Shear Flow :qV⋅QI:= q 84.90kNm=There are four planes on the bolt. Hencs, the shearforce resisted by each shear plane of the bolt isFq⋅ s 4⎛⎜⎝⎞⎠:= F = 5.31 kN Ans 740. Problem 7-39A beam is constructed from five boards bolted together as shown. Determine the maximum spacing sof the bolts if they can each resist a shear of 20 kN and the applied shear is V = 45 kN.Given: d1 := 250mm d2 := 350mmt := 25mm a := 100mmV := 45kN Fallow := 20kNSolution: a' := d2 − (d1 − a) a' = 200mmh := a' + 0.5d1 h = 325mmSection Property :yc2(d1⋅ t)⋅ (0.5d1 + a') + 3(d2⋅ t)⋅ (0.5d2)2(d1⋅ t) + 3(d2⋅ t):=yc = 223.39mmI1112:= ⋅ (2t) ⋅ d13 + d( 2t ⋅ 1) ⋅ (0.5d1 + a' − yc)2I2112:= ⋅ (3⋅ t) ⋅ d23 + d( 3⋅ t ⋅ 2) ⋅ (0.5d2 − yc)2I := I1 + I2 I = 523597110.22mm4Q := d1⋅ (2t)⋅ (h − yc) Q = 1270161.29mm3Shear Flow : qV⋅QI:= q 109.16kNm=Since there are four planes on the bolt, the allowable shear flow is qallow4Fallows=s4Fallowq:=s = 732.9mm Ans 741. Problem 7-40The beam is subjected to a shear of V = 800 N. Determine the average shear stress developed in thenails along the sides A and B if the nails are spaced s = 100 mm apart. Each nail has a diameter of 2mm.Given: bf := 250mm dw := 150mmt := 30mm a := 100mms := 100mm V := 800Ndo := 2mmSolution: h' := 0.5t h' = 15mmSection Property :ycbf⋅ t⋅ (0.5t) + dw⋅ (2t)⋅ (0.5dw):=yc = 47.73mmI1bf⋅ t + 2(dw⋅ t)112:= ⋅bf⋅ t3 + (bf⋅ t)⋅ (0.5t − yc)2I2112:= ⋅ (2⋅ t) ⋅ dw3 + d( 2⋅ t ⋅ w) ⋅ (0.5dw − yc)2I := I1 + I2 I = 32164772.73mm4Q := bf⋅ t⋅ (yc − h') Q = 245454.55mm3Shear Flow :qV⋅QI:= q 6.105kNm=F := q⋅ s F = 0.6105 kNSince each side of the beam resists this shear force, thenAoπ4:= ⋅ do2 τavgF2Ao:=τavg = 97.16MPa Ans 742. Problem 7-41The beam is fabricated from two equivalent structural tees and two plates. Each plate has a height of150 mm and a thickness of 12 mm. If a shear of V = 250 kN is applied to the cross section, determinethe maximum spacing of the bolts. Each bolt can resist a shear force of 75 kN.Given: bf := 75mm dw := 75mm t := 12mmdp := 150mm dg := 50mmV := 250kN Fallow := 75kNSolution:Section Property : D 2dw:= + 2t + dgIT112⋅bf⋅D3112− ⋅ 3⋅ t dg112:= − ⋅ (bf − t)⋅ (D − 2t)3IP112:= ⋅ 3⋅ (2t) dpI := IT + IPQ := (bf⋅ t)⋅ (0.5D − 0.5t) + (t⋅dw)⋅ (0.5dw + 0.5dg)Shear Flow : qV⋅QI=Since there are two shear planes on the bolt, the allowable shear flow is q2Fsn=2FallowsnV⋅QI=sn2⋅ I FallowV⋅Q:=sn = 138.0mm Ans 743. Problem 7-42The beam is fabricated from two equivalent structural tees and two plates. Each plate has a height of150 mm and a thickness of 12 mm. If the bolts are spaced at s = 200 mm, determine the maximumshear force V that can be applied to the cross section. Each bolt can resist a shear force of 75 kN.Given: bf := 75mm dw := 75mm t := 12mmdp := 150mm dg := 50mmsn := 200mm Fallow := 75kNSolution:Section Property : D 2dw:= + 2t + dgIT112⋅bf⋅D3112− ⋅ 3⋅ t dg112:= − ⋅ (bf − t)⋅ (D − 2t)3IP112:= ⋅ 3⋅ (2t) dpI := IT + IPQ := (bf⋅ t)⋅ (0.5D − 0.5t) + (t⋅dw)⋅ (0.5dw + 0.5dg)Shear Flow : V⋅Qq=ISince there are two shear planes on the bolt, the allowable shear flow is q2Fsn=2FallowsnV⋅QI=V2⋅ I Fallowsn⋅Q:=V = 172.5 kN Ans 744. Problem 7-43The double-web girder is constructed from two plywood sheets that are secured to wood members atits top and bottom. If each fastener can support 3 kN in single shear, determine the required spacing sof the fasteners needed to support the loading P = 15 kN. Assume A is pinned and B is a roller.Given: bi := 150mm di := 250mmt := 12mm do := 350mmP := 15kN Fallow := 3kNSolution:Equilibrium : By symmetry, A=R , B=R+ ΣFy=0; 2R − P = 0R := 0.5PSection Property : bo := bi + 2t d' := 0.5(do − di)I112:= ⋅ 3− ⋅ 3⋅bo do112⋅bi diQ := (bi⋅d')⋅ (0.5do − 0.5d')Shear Flow : qV⋅QI= Vmax := RSince there are two shear planes on the bolt, the allowable shear flow is q2Fsn=2FallowsnVmax⋅QI=sn2⋅ I FallowVmax⋅Q:=sn = 303.2mm Ans 745. Problem 7-44The double-web girder is constructed from two plywood sheets that are secured to wood members atits top and bottom. The allowable bending stress for the wood is σallow = 56 MPa and the allowableshear stress is τallow = 21 MPa. If the fasteners are spaced s = 150 mm and each fastener can support3 kN in single shear, determine the maximum load P that can be applied to the beam.Given: bi := 150mm di := 250mmt := 12mm do := 350mmsn := 150mm σallow := 56MPaFallow := 3kN τallow := 21MPaSolution:Equilibrium : By symmetry, A=R , B=R+ ΣFy=0; 2R − P = 0R = 0.5PSection Property : bo := bi + 2td' := 0.5(do − di)I112:= ⋅ 3− ⋅ 3⋅bo do112⋅bi diQ := (bi⋅d')⋅ (0.5do − 0.5d')Shear Flow : qV⋅QI= Vmax = R q2Fsn=Since there are two shear planes on the bolt, the allowable shear flow is2FallowsnVmax⋅QI=2Fallowsn0.5P⋅QI=P4⋅ I Fallowsn⋅Q:=P = 30.32 kN Ans 746. Problem 7-45The beam is made from three polystyrene strips that are glued together as shown. If the glue has ashear strength of 80 kPa, determine the maximum load P that can be applied without causing the glueto lose its bond.Given: bf := 30mm tf := 40mmtw := 20mm dw := 60mmτallow := 0.080MPaSolution:Equilibrium : By equilibrium, A = B = R+P ⋅ ⎛⎜⎝ΣFy=0; 2R − P 214⎞⎠− = 0R = 0.75PMaximum Shear : Vmax = R Vmax34= PSection Property : D := dw + 2tfI112bf⋅D3 (bf − tw) dw3 ⋅ − ⎡⎣⎤⎦:= ⋅I = 6.68 × 10− 6m4Q Σ yi ⎯= ⋅ ⋅A1 Q (bf⋅ tf) D2tf2−⎛⎜⎝⎞⎠:= ⋅ Q = 60000mm3Shear Stress: τV⋅QI⋅ t=τallowVmax⋅QI⋅ tw= Vmaxτallow⋅ I⋅ twQ=34Pτallow⋅ I⋅ twQ=P4τallow⋅ I⋅ tw3Q:=P = 0.238 kN Ans 747. Problem 7-46The beam is made from four boards nailed together as shown. If the nails can each support a shearforce of 500 N., determine their required spacings s' and s if the beam is subjected to a shear of V =3.5 kN.Given: bf := 250mm dw := 250mmtf := 25mm tw := 40mmtb := 25mm db := 75mmV := 3.5kN Fallow := 0.5kNSolution:Section Property : D := dw + tfyc0.5tf(bf⋅ tf) + (0.5dw + tf)(dw⋅ tw) + 0.5db(2tb⋅db)bf⋅ tf + dw⋅ tw + 2(tb⋅db):=yc = 85.94mmI1112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (0.5tf − yc)2I2112:= ⋅ tw ⋅ dw3 + d( w⋅ tw) ⋅ (0.5dw + tf − yc)2I3112:= ⋅ (2tb) ⋅ db3 + 2t( b⋅db) ⋅ (0.5db − yc)2I := I1 + I2 + I3QC := (tb⋅db)⋅ (yc − 0.5db)QD := (dw⋅ tw)⋅ (D − yc − 0.5dw)Shear Flow : qV⋅QI=The allowable shear flow at points C and D are : qCFsn= qDFs'n=FallowsnV⋅QCI=Fallows'nV⋅QDI=snI FallowV⋅QC:= s'nI FallowV⋅QD:=sn = 216.6mm s'n = 30.7mm Ans 748. Problem 7-47The beam is fabricated from two equivalent channels and two plates. Each plate has a height of 150mm and a thickness of 12 mm. If a shear of V = 250 kN is applied to the cross section, determine themaximum spacing of the bolts. Each bolt can resist a shear force of 75 kN.Given: bf := 300mm dw := 88mm t := 12mmdp := 150mm dg := 50mmV := 250kN Fallow := 75kNSolution:Section Property : D 2dw:= + 2t + dgIU112⋅bf⋅D3112− ⋅ 3⋅ (2t) dg112:= − ⋅ (bf − 2t)⋅ (D − 2t)3IP112:= ⋅ 3⋅ (2t) dpI := IU + IPQ := (bf⋅ t)⋅ (0.5D − 0.5t) + (2t⋅dw)⋅ (0.5dw + 0.5dg)Shear Flow : qV⋅QI=Since there are two rows of bolts, the allowable shear flow is q2Fsn=2FallowsnV⋅QI=sn2⋅ I FallowV⋅Q:=sn = 137.6mm Ans 749. Problem 7-48A built-up timber beam is made from n boards, each having a rectangular cross section. Write acomputer program that can be used to determine the maximum shear stress in the beam when it issubjected to any shear V. Show an application of the program using a cross section that is inthe form of a “T” and a box. 750. Problem 7-49The timber T-beam is subjected to a load consisting of n concentrated forces Pn ., If the allowableshear Vnail for each of the nails is known, write a computer program that will specify the nail spacingbetween each load. Show an application of the program using the values L = 4.5 m, a1 = 1.2 m, P1 = 3kN, a2 = 2.4 m, P2 = 7.5 kN, b1 = 37.5 mm, h1 = 250 mm, b2 = 200 mm, h2 = 25 mm, and Vnail = 1kN. 751. Problem 7-50The strut is constructed from three pieces of plastic that are glued together as shown. If the allowableshear stress for the plastic is τallow = 5.6 MPa and each glue joint can withstand 50 kN/m, determinethe largest allowable distributed loading w that can be applied to the strut.Given: L1 := 1m τallow := 5.6MPaL2 := 2m qallow 50kNm:=L3 := 1mbf := 74mm tf := 25mmdw := 75mm tw := 12mmSolution:Section Property : D := dw + tfyc0.5tf(bf⋅ tf) + (0.5dw + tf)(2tw⋅dw)bf⋅ tf + 2tw⋅dw:=yc = 37.16mmI1112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (0.5tf − yc)2I2112:= ⋅ (2tw) ⋅ dw3 + 2t( w⋅dw) ⋅ (0.5dw + tf − yc)2I := I1 + I2Qmax (2tw)⋅ (D − yc)D − yc2:= ⋅QA := (bf⋅ tf)⋅ (yc − 0.5tf)Allowable Shear Stress: τV⋅QI⋅ t= Vmax = w⋅L1τalloww⋅L1I 2tw⋅ ( )⎡⎢⎣⎤⎥⎦(2tw)⋅ (D − yc)D − yc2⋅⎡⎢⎣⎤⎥⎦= ⋅τalloww⋅L12I= ⋅ (D − yc)2 w2I:= ⋅ (τallow) w 9.13L1⋅ (D − yc)2kNm=Shear Flow : Assume the beam fails at the glue joint and the allowable shear flow is 2⋅qallow2qallowV⋅QI= 2qallow(w⋅L1)⋅QAI=w2⋅ I qallowL1⋅QA:= w 7.06kNm= (Controls !) Ans 752. Shear Force Diagram: L := L1 + L2 + L3Equilibrium : By symmetry, R1 = R R2 = R+ ΣFy=0; 2R − w⋅L = 0R := 0.5w⋅Lx1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2) x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. (L)V1(x1) −w⋅x1:= V2(x2) (−w⋅x2 + R) 1kN:= ⋅ V3(x3) (−w⋅x3 + 2R) 1kNkN:= ⋅0 1 2 3 410010Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3 753. Problem 7-51The strut is constructed from three pieces of plastic that are glued together as shown. If the distributedload w = 3 kN/m, determine the shear flow that must be resisted by each glue joint.Given: L1 := 1mL2 := 2m w 3kNm:=L3 := 1mbf := 74mm tf := 25mmdw := 75mm tw := 12mmSolution:Section Property : D := dw + tfyc0.5tf(bf⋅ tf) + (0.5dw + tf)(2tw⋅dw)bf⋅ tf + 2tw⋅dw:=yc = 37.16mmI1112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (0.5tf − yc)2I2112:= ⋅ (2tw) ⋅ dw3 + 2t( w⋅dw) ⋅ (0.5dw + tf − yc)2I := I1 + I2QA := (bf⋅ tf)⋅ (yc − 0.5tf)Shear Flow : Since there are two glue joints, hence 2qV⋅QI=Vmax := w⋅L1qVmax⋅QA2I:=q 21.24kNm= Ans 754. Shear Force Diagram: L := L1 + L2 + L3Equilibrium : By symmetry, R1 = R R2 = R+ ΣFy=0; 2R − w⋅L = 0R := 0.5w⋅Lx1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2) x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. (L)V1(x1) −w⋅x1:= V2(x2) (−w⋅x2 + R) 1kN:= ⋅ V3(x3) (−w⋅x3 + 2R) 1kNkN:= ⋅0 1 2 3 410010Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3 755. Problem 7-52The beam is subjected to the loading shown, where P = 7 kN. Determine the average shear stressdeveloped in the nails within region AB of the beam. The nails are located on each side of the beam andare spaced 100 mm apart. Each nail has a diameter of 5 mm.Given: bf := 250mm dw := 150mmt := 30mm a := 2ms := 100mm do := 5mmP' := 3kN P := 7kNSolution:Section Property :I112(bf + 2⋅ t) ⋅ dw3 − bf ⋅ (dw − 2⋅ t)3 ⎡⎣⎤⎦:= ⋅I = 72000000mm4Q := bf⋅ t⋅ (0.5dw − 0.5t)Q = 450000mm3Maximum Shear : Vmax := P' + P Vmax = 10 kNShear Flow :Vmax⋅Qq:= q 62.500IkNm=There are two rows of nails. Hence, the sher forceresisted by each nail isFq2:= ⋅ s F = 3.125 kNAoπ4:= ⋅ do2 τavgFAo:=τavg = 159.2MPa Ans 756. Problem 7-53The beam is constructed from four boards which are nailed together. If the nails are on both sides ofthe beam and each can resist a shear of 3 kN, determine the maximum load P that can be applied to theend of the beam.Given: bf := 250mm dw := 150mmt := 30mm a := 2ms := 100mm P' := 3kNFallow := 3kNSolution:Section Property :I112(bf + 2⋅ t) ⋅ dw3 − bf ⋅ (dw − 2⋅ t)3 ⎡⎣⎤⎦:= ⋅I = 72000000mm4Q := bf⋅ t⋅ (0.5dw − 0.5t)Q = 450000mm3Maximum Shear : Vmax = P' + PThere are two rows of nails. Hence, the allowable sher flow is2Fallowqallows:=qallow 60.00kNm=Shear Flow :qallowVmax⋅Q= qallowI(P' + P)⋅QI=Pqallow⋅ I⎛⎜⎝⎞:= − P'Q⎠P = 6.60 kN Ans 757. Problem 7-54The member consists of two plastic channel strips 12 mm thick, bonded together at A and B. If theglue can support an allowable shear stress of τallow = 4.2 MPa, determine the maximum intensity w0 ofthe triangular distributed loading that can be applied to the member based on the strength of the glue.Given: bo := 150mm t := 12mmdo := 150mm L := 4mτallow := 4.2MPaSolution:Equilibrium : By symmetry, A=R , B=R+ ΣFy=0; 2R − 0.5wo⋅L = 0R = 0.25wo⋅LSection Property : bi := bo − 2t di := do − 2tI112:= ⋅ 3− ⋅ 3⋅bo do112⋅bi diQ bo t ⋅ ( ) 0.5do 0.5t − ( ) ⋅ 2 t ⋅ 0.5 di ⋅ ( ) ⋅ ⎡⎣:= + ⋅ (0.5di)⎤⎦Shear Flow : qV⋅QI= Vmax = RSince there are two planes of glue, the allowable shear flow is 2t⋅τallow(2t)⋅τallowVmax⋅QI=(2⋅ t)τallow(0.25wo⋅L)⋅QI=wo8t⋅ IτallowL⋅Q:=wo 9.73kNm= Ans 758. Problem 7-55The member consists of two plastic channel strips 12 mm thick, glued together at A and B. If thedistributed load has a maximum intensity of w0 = 50 kN/m, determine the maximum shear stressresisted by the glue.Given: bo := 150mm do := 150mm L := 4mt := 12mm wo 50kNm:=Solution:Equilibrium : By symmetry, A=R , B=R+ ΣFy=0; 2R − 0.5wo⋅L = 0R := 0.25wo⋅LSection Property : bi := bo − 2t di := do − 2tI112:= ⋅ 3− ⋅ 3⋅bo do112⋅bi diQ bo t ⋅ ( ) 0.5do 0.5t − ( ) ⋅ 2 t ⋅ 0.5 di ⋅ ( ) ⋅ ⎡⎣:= + ⋅ (0.5di)⎤⎦Allowable Shear Stress: τV⋅QI⋅b= Vmax := RτmaxVmax⋅QI⋅ (2t):=τmax = 21.58MPa Ans 759. Problem 7-56A shear force of V = 18 kN is applied to the symmetric box girder. Determine the shear flow at A andB.Given: bf := 125mm dw := 300mm t := 10mmdm := 200mm dg := 30mm V := 18kNSolution:Section Property :I1112:= ⋅bf⋅ t3 + (bf⋅ t)⋅ (0.5dw − 0.5t)2I2112:= ⋅bf⋅ t3 + (bf⋅ t)⋅ (0.5t + 0.5dm)2I3112:= ⋅ 3⋅ t dwI := 21+ 2I2 + 2I3 I = 125166666.67mm4QA := (bf⋅ t)⋅ (0.5dw − 0.5t) QA = 181250mm3QB := (bf⋅ t)⋅ (0.5t + 0.5dm) QB = 131250mm3IShear Flow :qA12V⋅QAI:= qA 13.03kNm= AnsqB12V⋅QBI:= qB 9.44kNm= Ans 760. Problem 7-57A shear force of V = 18 kN is applied to the box girder. Determine the shear flow at C.Given: bf := 125mm dw := 300mm t := 10mmdm := 200mm dg := 30mm V := 18kNSolution:Section Property :I1112:= ⋅bf⋅ t3 + (bf⋅ t)⋅ (0.5dw − 0.5t)2I2112:= ⋅bf⋅ t3 + (bf⋅ t)⋅ (0.5t + 0.5dm)2I3112:= ⋅ 3⋅ t dwI := 21+ 2I2 + 2I3I = 125166666.67mm4QC := (bf⋅ t)⋅ (0.5dw − 0.5t) + (bf⋅ t)⋅ (0.5t + 0.5dm) + 2(0.5dw⋅ t)⋅ (0.25dw)QC = 537500mm3Shear Flow :qCI12V⋅QCI:= qC 38.65kNm= Ans 761. Problem 7-58The channel is subjected to a shear of V = 75 kN. Determine the shear flow developed at point A.Given: bf := 400mm dw := 200mmtf := 30mm tw := 30mmV := 75kNSolution:Section Property : D := dw + tfyc0.5tf(bf⋅ tf) + (0.5dw + tf)(2tw⋅dw)bf⋅ tf + 2tw⋅dw:=yc = 72.50mmI1112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (0.5tf − yc)2I2112:= ⋅ (2tw) ⋅ dw3 + 2t( w⋅dw) ⋅ (0.5dw + tf − yc)2I := I1 + I2I = 120250000mm4QA := (bf⋅ tf)⋅ (yc − 0.5tf)QA = 690000mm3Shear Flow :qA12V⋅QAI:= qA 215.2kNm= Ans 762. Problem 7-59The channel is subjected to a shear of V = 75 kN. Determine the maximum shear flow in the channel.Given: bf := 400mm dw := 200mmtf := 30mm tw := 30mmV := 75kNSolution:Section Property : D := dw + tfyc0.5tf(bf⋅ tf) + (0.5dw + tf)(2tw⋅dw)bf⋅ tf + 2tw⋅dw:=yc = 72.50mmI1112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (0.5tf − yc)2I2112:= ⋅ (2tw) ⋅ dw3 + 2t( w⋅dw) ⋅ (0.5dw + tf − yc)2I := I1 + I2I = 120250000mm4Qmax tw⋅ (D − yc)12:= ⋅ ⋅ (D − yc)Qmax = 372093.75mm3Shear Flow :V⋅Qmaxqmax:= qmax 232.1IkNm= Ans 763. Problem 7-60The beam supports a vertical shear of V = 35 kN. Determine the resultant force developed in segmentAB of the beam.Given: bf := 125mm dw := 250mmtf := 12mm tw := 12mmV := 35kNSolution:Section Property : D := dw + 2tff'I:= + ⋅ 3Q A112⋅ (2tf) bf⋅ 3112⋅dw tw= ⋅yfcQ tf 0.5bf y − ( ) ⋅ 0.5 0.5bf y − ( ) y + ⎡⎣⎤⎦= ⋅Q 0.5tf 0.25bf2 y2 − ⎛⎝⎞⎠=Shear Flow: qV⋅QI=qV⋅ tf2I0.25bf2 − y2 ⎛⎝⎞⎠= ⋅Resultant Shear Force: For AB: yo := 0.5tw VABAq y⌠⎮⌡= dVAB0.5bfyoyV⋅ tf2I0.25bf2 − y2 ⎛⎝⎞⎠⋅⌠⎮⎮⌡:= dVAB = 7.43 kN Ans 764. Problem 7-61The aluminum strut is 10 mm thick and has the cross section shown. If it is subjected to a shear of V= 150 N, determine the shear flow at points A and B.Given: bf := 60mm b'f := 80mmtf := 10mm dw := 40mmtw := 10mm V := 150NSolution:Section Property : D := dw + 2tfA := bf⋅ tf + 2dw⋅ tw + b'f⋅ tfA = 2200mm2⎯ Σ yi ⎯y⋅ ( ⋅Ai)Σ⋅ (Ai)= yc(b'f⋅ tf)⋅ (0.5tf) + 2(dw⋅ tw)⋅ (0.5dw + tf) + (bf⋅ tf)⋅ (D − 0.5tf)A:=yc = 27.73mmI'f:= ⋅ 3 + b'( f⋅ tf) ⋅ (0.5tf − yc)2Iw112⋅b'f tf112:= ⋅ tw ⋅ dw3 + t( w⋅dw) ⋅ (0.5dw + tf − yc)2If112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (D − 0.5tf − yc)2I := If + 2Iw + I'fI = 981.9697 × 10− 9m4QA := (0.5b'f⋅ tf)⋅ (yc − 0.5tf) QA = 9090.91mm3QB := (bf⋅ tf)⋅ (D − 0.5tf − yc) QB = 16363.64mm3Shear Flow :qAV⋅QAI:= qA 1.39kNm= AnsqB12V⋅QBI:= qB 1.25kNm= Ans 765. Problem 7-62The aluminum strut is 10 mm thick and has the cross section shown. If it is subjected to a shear of V= 150 N, determine the maximum shear flow in the strut.Given: bf := 60mm b'f := 80mmtf := 10mm dw := 40mmtw := 10mm V := 150NSolution:Section Property : D := dw + 2tfA := bf⋅ tf + 2dw⋅ tw + b'f⋅ tfA = 2200mm2⎯ Σ yi ⎯y⋅ ( ⋅Ai)Σ⋅ (Ai)= yc(b'f⋅ tf)⋅ (0.5tf) + 2(dw⋅ tw)⋅ (0.5dw + tf) + (bf⋅ tf)⋅ (D − 0.5tf)A:=yc = 27.73mmI'f:= ⋅ 3 + b'( f⋅ tf) ⋅ (0.5tf − yc)2Iw112⋅b'f tf112:= ⋅ tw ⋅ dw3 + t( w⋅dw) ⋅ (0.5dw + tf − yc)2If112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (D − 0.5tf − yc)2I := If + 2Iw + I'fI = 981.9697 × 10− 9m4Qmax (bf⋅ tf)⋅ (D − 0.5tf − yc) 2⋅ tw⋅ (D − yc − tf)12:= + ⋅ ⋅ (D − yc − tf)Qmax = 21324.38mm3Shear Flow :qmax12V⋅Qmax:= qmax 1.63IkNm= Ans 766. Problem 7-63The angle is subjected to a shear of V = 10 kN. Sketch the distribution of shear flow along the leg AB.Indicate numerical values at all peaks.Given: L := 125mm t := 6mmθ := 45deg V := 10kNSolution:Section Property :h := L⋅cos(θ) btsin(θ) :=I112:= ⋅ (2b)⋅h3= ⋅Q A'y'cQ t0.5h − ysin(θ)⎛⎜⎝⎞⎠= ⋅ ⋅ [0.5(0.5h − y) + y]Qt= (0.25h2 y2)2 sin(θ) − Shear Flow: qV⋅QI=q= ⋅ (0.25h2 − y2)V⋅ t2I⋅ sin(θ)At y = 0, q = qmaxqmax:= ⋅ (0.25h2)V⋅ t2I⋅ sin(θ)qmax 84.85kNm= Ans 767. Problem 7-64The beam is subjected to a shear force of V = 25 kN. Determine the shear flow at points A and B.Given: bf := 274mm tf := 12mmb'f := 250mm t'f := 12mmdw := 200mm tw := 12mmd'w := 50mm V := 25kNSolution:Section Property : D := dw + tfD' := D − d'wyc0.5tf(bf⋅ tf) + (0.5dw + tf)(2tw⋅dw) + (D' − 0.5t'f)(b'f⋅ t'f)bf⋅ tf + 2tw⋅dw + b'f⋅ t'f:=yc = 92.47mmI1112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (0.5tf − yc)2I2112:= ⋅ (2tw) ⋅ dw3 + 2t( w⋅dw) ⋅ (0.5dw + tf − yc)2I3112:= ⋅b'f ⋅ t'f3 + b'( f⋅ t'f) ⋅ (D' − 0.5t'f − yc)2I := I1 + I2 + I3QA := (bf⋅ tf)⋅ (yc − 0.5tf)QB := (b'f⋅ t'f)⋅ (D' − 0.5t'f − yc)Shear Flow :qAV⋅QA2I:= qA 65.09kNm= AnsqBV⋅QB2I:= qB 43.63kNm= Ans 768. Problem 7-65The beam is constructed from four plates and is subjected to a shear force of V = 25 kN. Determinethe maximum shear flow in the cross section.Given: bf := 274mm tf := 12mmb'f := 250mm t'f := 12mmdw := 200mm tw := 12mmd'w := 50mm V := 25kNSolution:Section Property : D := dw + tfD' := D − d'wyc0.5tf(bf⋅ tf) + (0.5dw + tf)(2tw⋅dw) + (D' − 0.5t'f)(b'f⋅ t'f)bf⋅ tf + 2tw⋅dw + b'f⋅ t'f:=yc = 92.47mmI1112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (0.5tf − yc)2I2112:= ⋅ (2tw) ⋅ dw3 + 2t( w⋅dw) ⋅ (0.5dw + tf − yc)2I3112:= ⋅b'f ⋅ t'f3 + b'( f⋅ t'f) ⋅ (D' − 0.5t'f − yc)2I := I1 + I2 + I3Qmax (bf⋅ tf)⋅ (yc − 0.5tf) 2twyc − tf2⎛⎜⎝⎞⎠:= + ⋅ ⋅ (yc − tf)Maximum Shear Flow :qmaxV⋅Qmax2I:= qmax 82.88kNm= Ans 769. Problem 7-66A shear force of V = 18 kN is applied to the box girder. Determine the position d of the stiffener platesBE and FG so that the shear flow at A is twice as great as the shear flow at B. Use the centerlinedimensions for the calculation. All plates are 10 mm thick.Given: t := 10mm bf := 135mm − tV := 18kN dw := t + 290mmSolution:Section Property :QA := (bf⋅ t)⋅ (0.5dw − 0.5t) QA = 181250mm3QB = (bf⋅ t)⋅ (d)Shear Flow :qA12V⋅QAI= qB12V⋅QBI=Require, qA = 2qB12V⋅QAI212V⋅QBI⎛⎜⎝⎞⎠= ⋅QA = 2QBQA = 2(bf⋅ t)⋅ (d)dQA2bf⋅ t:=d = 72.50mm Ans 770. Problem 7-67The pipe is subjected to a shear force of V = 40 kN. Determine the shear flow in the pipe at points Aand B.Given: ri := 150mm t := 5mmV := 40kNSolution:Section Property : ro := ri + tIπ4ro4 − ri4 ⎛⎝⎞⎠:= ⋅Since a' -> 0, then QA := 0QB4ro3π⋅ 22π ro⎛⎜⎝⎞⎠4ri3π⋅ 22π ri⎛⎜⎝⎞⎠:= −Shear Flow :qAV⋅QAI:= qA 0.00kNm= AnsqBV⋅QB2I:= qB 83.48kNm= Ans 771. Problem 7-68Determine the location e of the shear center, point O, for the thin-walled member having the crosssection shown where b2 > b1. The member segments have the same thickness t. 772. Problem 7-69Determine the location e of the shear center, point O, for the thin-walled member having the crosssection shown.The member segments have the same thickness t. 773. Problem 7-70Determine the location e of the shear center, point O, for the thin-walled member having the crosssection shown. The member segments have the same thickness t. 774. Problem 7-71Determine the location e of the shear center, point O, for the thin-walled member having the crosssection shown. The member segments have the same thickness t. 775. Problem 7-72Determine the location e of the shear center, point O, for the thin-walled member having the crosssection shown. The member segments have the same thickness t. 776. Problem 7-73Determine the location e of the shear center, point O, for the thin-walled member having the crosssection shown. The member segments have the same thickness t. 777. Problem 7-74Determine the location e of the shear center, point O, for the thin-walled member having the crosssection shown. The member segments have the same thickness t.Given: dθ := 150mm dv := 150mmθ := 30degSolution: Set t := 1mmSection Property :hθ := dθ⋅ sin(θ) bθtsin(θ) :=Iθ112⋅ 3 d( θ⋅ t) dv + hθ⋅bθ hθ2⎛⎜⎝⎞⎠2:= + ⋅Iv112:= ⋅ 3⋅ t dvI := 2c'+ Ivy'c = 0.5dv + hθ − 0.5x⋅ sin(θ)θIQ A'y= ⋅ Q = (x⋅ t)⋅ (0.5dv + hθ − 0.5x⋅ sin(θ))Shear Flow Resultant:qV⋅QI= V = PqPx⋅ tI= ⋅ (0.5dv + hθ − 0.5x⋅ sin(θ))F1dθP0xx⋅ tI⋅ (0.5dv + hθ − 0.5x⋅ sin(θ))⌠⎮⎮⌡= dShear Center: Summing moment about point AP⋅e = F1⋅dv⋅cos(θ)eF1P= ⋅dv⋅cos(θ)edθdv⋅cos(θ) x⌠⎮⎮⌡d :=0x⋅ tI⎛⎜⎝⎞⎠⋅ ⋅ (0.5dv + hθ − 0.5x⋅ sin(θ))e = 43.30mm Ans 778. Problem 7-75Determine the location e of the shear center, point O, for the thin-walled member having a slit along itsside.Given: a := 100mm b := 100mmSolution:Set t := mm P := kNSection Property :h := 2aI112:= ⋅ 2a ( 2t ) ⋅ h3 + 2 ( b ⋅ t ) ⋅ I = 3.3333 a3tQ1 (y⋅ t)12= ⋅ (y) Q1t2= y2Q2 (a⋅ t)12= ⋅ (a) + (x⋅ t)⋅ (a) Q1a⋅ t2= (a + 2x)Shear Flow Resultant:q1V⋅Q1I= q1P⋅ t⋅y22I=q2V⋅Q2I= q2P⋅ (a⋅ t)⋅ (a + 2x)2I=Fwaq1 y⌠⎮⌡= d Fw0a0yP⋅ t⋅y22I⌠⎮⎮⌡:= d Fw = 0.05 PFfbq2 x⌠⎮⌡= d Ff0b0xP⋅ (a⋅ t)⋅ (a + 2x)2I⌠⎮⎮⌡:= d Ff = 0.3 PShear Center: Summing moment about point AP⋅e = 2Fw⋅b + Ff⋅he1P:= ⋅ (2Fw⋅b + Ff⋅h)e = 70mm Ans 779. Problem 7-76Determine the location e of the shear center, point O, for the thin-walled member having a slit along itsside. Each element has aconstant thickness t.Given: a := mm b := aSolution:Set t := mm P := kNSection Property :h := 2aI112:= ⋅ 2a ( 2t ) ⋅ h3 + 2 ( b ⋅ t ) ⋅ I = 3.3333 a3tQ1 (y⋅ t)12= ⋅ (y) Q1t2= y2Q2 (a⋅ t)12= ⋅ (a) + (x⋅ t)⋅ (a) Q1a⋅ t2= (a + 2x)Shear Flow Resultant:q1V⋅Q1I= q1P⋅ t⋅y22I=q2V⋅Q2I= q2P⋅ (a⋅ t)⋅ (a + 2x)2I=Fwaq1 y⌠⎮⌡= d Fw0a0yP⋅ t⋅y22I⌠⎮⎮⌡:= dFw = 0.05 PFfbq2 x⌠⎮⌡= d Ff0b0xP⋅ (a⋅ t)⋅ (a + 2x)2I⌠⎮⎮⌡:= dFf = 0.3 PShear Center: Summing moment about point AP⋅e = 2Fw⋅b + Ff⋅he1P:= ⋅ (2Fw⋅b + Ff⋅h)e = 0.7 a Ans 780. Problem 7-77Determine the location e of the shear center, point O, for the thin-walled member having the crosssection shown.Given: a := mm θ := 60degSolution:Set t := mm P := kNSection Property :b := a⋅ sin(θ) b = 0.86603 at'tcos(θ) := t' = 2 t ha2:=I:= + ⋅ I = 0.25 a3tQ1 (y'⋅ t')112⋅ (t)⋅a3112t' ( ) a 312= ⋅ (y') Q1t'2= y'2Q2 (h⋅ t')12h y − ( ) + ⎡⎢⎣⋅ (h) (h − y)⋅ t y12⎤⎥⎦= +Q212t'h2 t h2 y2 − ( ) + ⎡⎣⎤⎦=Shear Flow Resultant:q1V⋅Q1I= q1P⋅ t'⋅y'22I=q2V⋅Q2I= q2t h2 y2 − ( ) + ⎡⎣P t'h 2⎤⎦⋅2I=F'hq1 y⌠⎮⌡= d F'0h0y'P⋅ t'⋅y'22I⌠⎮⎮⌡:= d F' = 0.1667 PFhq2 y⌠⎮⌡= d F0h0yt h2 y2 − ( ) + ⎡⎣P t'h 2⎤⎦⋅2I⌠⎮⎮⌡:= d F = 0.6667 PShear Center: Summing moment about point AP⋅e = 2(F⋅b + F'⋅0)e1P:= ⋅ (2F⋅b)e = 1.1547 a Ans 781. Problem 7-78If the angle has a thickness of 3 mm, a height h = 100 mm, and it is subjected to a shear of V = 50 N,determine the shear flow at point A and the maximum shear flow in the angle.Given: h := 100mm t := 3mmθ := 45deg V := 50NSolution:Section Property :tt':= sin(θ) t' = 4.2426mmI112:= ⋅ (2t')⋅h3= ⋅Q A'y'cQ t'⋅ (0.5h − y)1( 0.5h − y ) + y 2⎡⎢⎣⎤⎥⎦= ⋅Q= (0.25h2 − y2)t'2Shear Flow: qV⋅QI=q= ⋅ (0.25h2 − y2)V⋅ t'2IAt A, yA := 0.5⋅h qA := 0 AnsAt y = 0, q = qmaxqmax:= ⋅ (0.25h2)V⋅ t'2Iqmax 375Nm= Ans 782. Problem 7-79The angle is subjected to a shear of V = 10 kN. Sketch the distribution of shear flow along the leg AB.Indicate numerical values at all peaks. The thickness is 6 mm and the legs (AB) are 125 mm.Given: L := 125mm t := 6mmθ := 45deg V := 10kNSolution:Section Property :h := L⋅cos(θ) t'tsin(θ) := t' = 8.4853mmI112:= ⋅ (2t')⋅h3= ⋅Q A'y'cQ t'⋅ (0.5h − y)1( 0.5h − y ) + y 2⎡⎢⎣⎤⎥⎦= ⋅Q= (0.25h2 − y2)t'2Shear Flow: qV⋅QI=q= ⋅ (0.25h2 − y2) AnsV⋅ t'2IAt y = 0, q = qmaxqmax:= ⋅ (0.25h2)V⋅ t'2Iqmax 84.85kNm= Ans 783. Problem 7-80Determine the placement e for the force P so that the beam bends downward without twisting. Takeh = 200 mm.Given: h1 := 100mm bf := 300mmh2 := 200mmSolution:Set t := mm P := kNSection Property :I112bf⋅ t3 + t ⋅ h13 + t ⋅ h23 ⎛⎝⎞⎠:= ⋅0.5h2 y − ( ) + ⎡⎢⎣Qw2 t⋅ (0.5h2 − y) y12⎤⎥⎦= ⋅Qw2t20.25h22 − y2 ⎛⎝⎞⎠= ⋅Shear Flow Resultant:qw2V⋅Qw2I= qw22 y2 − ⎛⎝P⋅ t 0.25h2⎞⎠⋅2I=Fw20.5h2− 0.5h2qw2 x⌠⎮⎮⌡= dFw20.5h2− 0.5h2y2 y2 − ⎛⎝P⋅ t 0.25h2⎞⎠⋅2I⌠⎮⎮⎮⌡:= dFw2 = 0.8889 PShear Center: Summing moment about point AP⋅e = Fw2⋅ (bf + t)e1PFw2 bf t + ( ) ⋅ ⎡⎣⎤⎦:= ⋅e = 267.5mm Ans 784. Problem 7-81A force P is applied to the web of the beam as shown. If e = 250 mm, determine the height h of theright flange so that the beam will deflect downward without twisting. The member segments have thesame thickness t.Given: h1 := 100mm bf := 300mme := 250mmSolution:Set t := mm P := kNShear Center: Summing moment about point AP⋅e = Fw2⋅ (bf + t) Fw2ebf + t:= ⋅PAssume bf+t equal to bf : Fw2ebf:= ⋅PSection Property : Assume bf t3 negligible.I112t ⋅ h13 + t ⋅ h23 ⎛⎝⎞⎠= ⋅0.5h2 y − ( ) + ⎡⎢⎣Qw2 t⋅ (0.5h2 − y) y12⎤⎥⎦= ⋅Qw2t20.25h22 − y2 ⎛⎝⎞⎠= ⋅Shear Flow Resultant:qw2V⋅Qw2I= qw22 y2 − ⎛⎝⎞⎠⋅P⋅ t 0.25h22I=Fw20.5h2− 0.5h2qw2 x⌠⎮⎮⌡= debf⋅P0.5h2− 0.5h2y2 y2 − ⎛⎝6P⋅ t 0.25h2⎞⎠⋅t h1⋅ 3 t h2+ ⋅ 3⌠⎮⎮⎮⌡= dGiven⎛⎝3 + h23 e h1⎞⎠⋅0.5h26bf − 0.5h22 y2 − ⎛⎝⎞⎠0.25h2 y⌠⎮⎮⌡= dGuess h2 := 10mm h2 := Find(h2)h2 = 171.0mm Ans 785. Problem 7-82Determine the location e of the shear center, point O, for the thin-walled member having the crosssection shown. 786. Problem 7-83Determine the location e of the shear center, point O, for the tube having a slit along its length. 787. Problem 7-84The beam is fabricated from four boards nailed together as shown. Determine the shear force each nailalong the sides C and the top D must resist if the nails are uniformly spaced s = 75 mm. The beam issubjected to a shear of V = 22.5 kN.Given: bf := 250mm dw := 300mmtf := 25mm tw := 25mmtb := 25mm db := 100mmV := 22.5kN sn := 75mmSolution:Section Property : D := dw + tfyc0.5tf(bf⋅ tf) + (0.5dw + tf)(dw⋅ tw) + 0.5db(2tb⋅db)bf⋅ tf + dw⋅ tw + 2(tb⋅db):=yc = 87.50mmI1112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (0.5tf − yc)2I2112:= ⋅ tw ⋅ dw3 + d( w⋅ tw) ⋅ (0.5dw + tf − yc)2I3112:= ⋅ (2tb) ⋅ db3 + 2t( b⋅db) ⋅ (0.5db − yc)2I := I1 + I2 + I3QC := (tb⋅db)⋅ (yc − 0.5db)QD := (dw⋅ tw)⋅ (D − yc − 0.5dw)Shear Flow : qV⋅QI=The allowable shear flow at points C and D are : qCFCsn= qDFDsn=FCsnV⋅QCI=FDsnV⋅QDI=FCV⋅QC⋅ sn:= FDIV⋅QD⋅ snI:=FC = 0.987 kN FD = 6.906 kN Ans 788. Problem 7-85The beam is constructed from four boards glued together at their seams. If the glue can withstand 15kN/m, what is the maximum vertical shear V that the beam can support?Given: bf := 100mm dw := 249mmtf := 12mm tw := 12mm di := 75mmqallow 15kNm:=Solution:Section Property :I212⋅ 3⋅ tw dw212+ ⋅bf ⋅ tf3 2bf⋅ tfdi + tf2⎛⎜⎝⎞⎠2:= + ⋅Q (bf⋅ tf) di + tf2:= ⋅Shear Flow : Since there are two glue joints, hence 2qV⋅QI=Vmax2I⋅qallowQ:=Vmax = 20.37 kN Ans 789. Problem 7-86Solve Prob. 7-85 if the beam is rotated 90° from the position shown.Given: bf := 100mm dw := 249mmtf := 12mm tw := 12mm di := 75mmqallow 15kNm:=Solution:Section Property :I212⋅ 3⋅ tf bf212+ ⋅dw ⋅ tw3 2dw⋅ twbf + tw2⎛⎜⎝⎞⎠2:= + ⋅Q (dw⋅ tw) bf + tw2:= ⋅Shear Flow : Since there are two glue joints, hence 2qV⋅QI=Vmax2I⋅qallowQ:=Vmax = 3.731 kN Ans 790. Problem 7-87The member is subjected to a shear force of V = 2 kN. Determine the shear flow at points A, B, and C.The thickness of each thin-walled segment is 15 mm.Given: bf := 200mm dw := 300mmtf := 15mm tw := 15mmtb := 15mm db := 115mmV := 2kNSolution:Section Property : D := dw + tfyc0.5tf(bf⋅ tf) + (0.5dw + tf)(dw⋅ tw) + 0.5db(2tb⋅db)bf⋅ tf + dw⋅ tw + 2(tb⋅db):=yc = 87.98mmI1112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (0.5tf − yc)2I2112:= ⋅ tw ⋅ dw3 + d( w⋅ tw) ⋅ (0.5dw + tf − yc)2I3112:= ⋅ (2tb) ⋅ db3 + 2t( b⋅db) ⋅ (0.5db − yc)2I := I1 + I2 + I3I = 86939045.38mm4QA := 0QB := (tb⋅db)⋅ (yc − 0.5db) QB = 52577.05mm3QC := QB + tf⋅ (0.5bf − 0.5tw)⋅ (yc − 0.5tf) QC = 164242.29mm3Shear Flow : qV⋅QI=qA := 0 AnsqBV⋅QBI:= qB 1.210kNm= AnsqCV⋅QCI:= qC 3.778kNm= Ans 791. Problem 7-88The member is subjected to a shear force of V = 2 kN. Determine the maximum shear flow in themember. All segments of the cross section are 15 mm thick.Given: bf := 200mm dw := 300mmtf := 15mm tw := 15mmtb := 15mm db := 115mmV := 2kNSolution:Section Property : D := dw + tfyc0.5tf(bf⋅ tf) + (0.5dw + tf)(dw⋅ tw) + 0.5db(2tb⋅db)bf⋅ tf + dw⋅ tw + 2(tb⋅db):=yc = 87.98mmI1112:= ⋅bf ⋅ tf3 + b( f⋅ tf) ⋅ (0.5tf − yc)2I2112:= ⋅ tw ⋅ dw3 + d( w⋅ tw) ⋅ (0.5dw + tf − yc)2I3112:= ⋅ (2tb) ⋅ db3 + 2t( b⋅db) ⋅ (0.5db − yc)2I := I1 + I2 + I3I = 86939045.38mm4Qmax tw⋅ (D − yc)12:= ⋅ ⋅ (D − yc)Qmax = 386537.47mm3Shear Flow : qV⋅QI=Maximum shear flow occurs at the point where the neutral axis passes through the section.V⋅Qmaxqmax:= qmax 8.892IkNm= Ans 792. Problem 7-89The beam is made from three thin plates welded together as shown. If it is subjected to a shear ofV = 48 kN, determine the shear flow at points A and B. Also, calculate the maximum shear stress inthe beam.Given: bf := 215mm tf := 15mmtw := 15mm dw := 315mmV := 48kN h := 200mmSolution:Section Property : a := 0.5(bf − tw)ycbf tw − ( ) tf ⋅ ⎡⎣(h + 0.5tf) + (dw⋅ tw)(0.5dw)(bf − tw)⋅ tf + dw⋅ tw⎤⎦:=yc = 176.92mmIf112:= ⋅ (bf − tw) ⋅ tf3 + b( f − tw)⋅ tf ⋅ (h + 0.5⋅ tf − yc)2Iw112:= ⋅ tw ⋅ dw3 + t( w⋅dw) ⋅ (yc − 0.5dw)2I := If + Iw I = 43.71347 × 10− 6m4QA := (a⋅ tw)⋅ (dw − yc − 0.5a) QA = 132123.79mm3QB := (a⋅ tf)⋅ (h + 0.5⋅ tf − yc) QB = 45873.79mm3Qmax tw⋅ (yc)12:= ⋅ ⋅ (yc) Qmax = 234748.45mm3Shear Flow : qV⋅QI=qAV⋅QAI:= qA 145.1kNm= AnsqBV⋅QBI:= qB 50.37kNm= AnsMaximum Shear Stress: τV⋅QI⋅b=Maximum shear stress occurs at the point where the neutral axis passes through the section.τmaxV⋅QmaxI⋅ (tw):= τmax = 17.18MPa Ans 793. Problem 7-90A steel plate having a thickness of 6 mm is formed into the thin-walled section shown. If it is subjectedto a shear force of V = 1.25 kN, determine the shear stress at points A and C. Indicate the results onvolume elements located at these points.Given: bf := 100mm dw := 50mm b'f := 25mmt := 6mm V := 1.25kNSolution:Section Property : D := dw + 2tmmmmyc0.5t fb( 2b'f ⋅ t ) + ( 0.5dw + t ) ( 2dw ⋅ t ) + ( D − 0.5t ) ( ⋅ t)2b'f⋅ t + 2dw⋅ t + bf⋅ t:=yc = 36.60mmI1112:= ⋅b'f⋅ t3 + (b'f⋅ t)⋅ (0.5t − yc)2I2112:= ⋅ t ⋅ dw3 + d( w⋅ t) ⋅ (0.5dw + t − yc)2I3112:= ⋅bf⋅ t3 + (bf⋅ t)⋅ (D − 0.5t − yc)2I 2I1:= + 2I2 + I3QA := (b'f⋅ t)⋅ (yc − 0.5t)QC := (b'f⋅ t)⋅ (yc − 0.5t) + (dw⋅ t)⋅ 0.5dw + t − yc − (0.5bf⋅ t)⋅ (D − 0.5t − yc)QC = 0.00mm3 (since A' = 0)Shear Stress : τV⋅QI⋅ t=τAV⋅QAI⋅ t:= τA = 1.335MPa AnsτCV⋅QCI⋅ t:= τC = 0MPa Ans 794. Problem 7-91A steel plate having a thickness of 6 mm is formed into the thin-walled section shown. If it is subjectedto a shear force of V = 1.25 kN, determine the shear stress at point B.Given: bf := 100mm dw := 50mm b'f := 25mmt := 6mm V := 1.25kNSolution:Section Property : D := dw + 2tyc0.5t fb( 2b'f ⋅ t ) + ( 0.5dw + t ) ( 2dw ⋅ t ) + ( D − 0.5t ) ( ⋅ t)2b'f⋅ t + 2dw⋅ t + bf⋅ t:=yc = 36.60mmI1112:= ⋅b'f⋅ t3 + (b'f⋅ t)⋅ (0.5t − yc)2I2112:= ⋅ t ⋅ dw3 + d( w⋅ t) ⋅ (0.5dw + t − yc)2I3112:= ⋅bf⋅ t3 + (bf⋅ t)⋅ (D − 0.5t − yc)2I 2I1:= + 2I2 + I3QB := (bf⋅ t)⋅ (D − 0.5t − yc)Shear Stress : τV⋅QI⋅b=τBV⋅QBI⋅ (2t):= τB = 1.781MPa Ans 795. Problem 7-92Determine the location e of the shear center, point O, for the thin-walled member having the crosssection shown. 796. Problem 7-93Sketch the intensity of the shear-stress distribution acting over the beam's cross-sectional area, anddetermine the resultant shear force acting on the segment AB. The shear acting at the section is V =175 kN. Show that INA = 340.82(106) mm4.Given: b1 := 200mm d1 := 200mm V := 175kNb2 := 50mm d2 := 150mmSolution:Section Property : D := d1 + d2yc0.5d1(b1⋅d1) + (0.5d2 + d1)(b2⋅d2):= yc = 127.63mmb1⋅d1 + b2⋅d2y'c := D − ycI1:= ⋅ 3 + b( 1⋅d1) ⋅ (0.5d1 − yc)2I2112⋅b1 d1112:= ⋅b2 ⋅ d23 + b( 2⋅d2) ⋅ (D − 0.5d2 − yc)2I := I1 + I2 I = 340.82 × 106mm4 (Q.E.D)A'1 = (yc − y1)⋅b1 y1c = 0.5(yc − y1) + y1 y1c = 0.5(yc + y1)Q1 = A'1⋅y'1c Q1 = 0.5(yc − y1)⋅b1(yc + y1) Q1 0.5b1 yc⎛⎝2 − y12 ⎞⎠=A'2 = (y'c − y2)⋅b2 y2c = 0.5(y'c − y2) + y2 y2c = 0.5(y'c + y2)Q2 = A'2⋅y'2c Q2 = 0.5(y'c − y2)⋅b2(y'c + y2) Q2 0.5b2 y'c⎛⎝2 − y22 ⎞⎠=Shear Stress : τV⋅QI⋅b= τCBVI⎛⎜⎝⎞⎠⎛⎝2 − y12 0.5 yc⎞⎠⎡⎣⎤⎦= ⋅τ1BVI⎛⎜⎝⎞⎠⎛⎝0.5yc2 ⎞⎠At B: y := ⋅ 1 = 0τ1B = 4.18MPaAt C: y1 := yc − d1 τ1CVI⎛⎜⎝⎞⎠⎛⎝2 − y12 0.5 yc⎞⎠⎡⎣⎤⎦:= ⋅τ1C = 2.84MPaτABVINA⎛⎜⎝⎞⎠⎛⎝2 − y22 0.5 y'c⎞⎠⎡⎣⎤⎦= ⋅At C: y2 := y'c − d2 τ2CVI⎛⎜⎝⎞⎠⎛⎝2 − y22 0.5 y'c⎞⎠⎡⎣⎤⎦:= ⋅τ2C = 11.35MPaResultant Shear Force: For segment AB. yo := y'c − d2 VABAτAB A⌠⎮⌡= dVABy'cyoyVI⎛⎜⎝⎞⎠2 y2 − ⎛⎝0.5 y'c⎞⎠⎡⎣⎤⎦⋅ ⋅b2⌠⎮⎮⌡:= dVAB = 49.78 kN Ans 797. Problem 8-1A spherical gas tank has an inner radius of r = 1.5 m. If it is subjected to an internal pressure of p =300 kPa, determine its required thickness if the maximum normal stress is not to exceed 12 MPa.Given: r := 1.5m p := 0.3MPa σallow := 12MPaSolution:Normal Stress : σallowp⋅ r2⋅ t=tp⋅ r2⋅σallow:=t = 18.75mm Ans 798. Problem 8-2A pressurized spherical tank is to be made of 125mm-thick steel. If it is subjected to an internalpressure of p = 1.4 MPa, determine its outer radius if the maximum normal stress is not to exceed 105MPa.Given: t := 125mm p := 1.4MPa σallow := 105MPaSolution:Normal Stress : σp⋅ r2⋅ t=ri2⋅ tσallow:= = 18.750mpri ro := ri + t ro = 18.875 m Ans 799. Problem 8-3The thin-walled cylinder can be supported in one of two ways as shown. Determine the state of stressin the wall of the cylinder for both cases if the piston P causes the internal pressure to be 0.5 MPa.The wall has a thickness of 6 mm and the inner diameter of the cylinder is 200 mm.Given: t := 6mm p := 0.5MPari := 200mmSolution:Case (a) :Hoop Stress : σ1p⋅ rit:= σ1 = 16.67MPa AnsNormal Stress : σ2 := 0 AnsCase (b) :Hoop Stress : σ1p⋅ rit:= σ1 = 16.67MPa AnsNormal Stress : σ2p⋅ ri2⋅ t:= σ2 = 8.33MPa Ans 800. Problem 8-4The tank of the air compressor is subjected to an internal pressure of 0.63 MPa. If the internal diameterof the tank is 550 mm, and the wall thickness is 6 mm, determine the stress components acting at pointA. Draw a volume element of the material at this point, and show the results on the element.Given: t := 6mm p := 0.63MPadi := 550mmSolution: ri := 0.5diHoop Stress : αrit:= α = 45.83Since α > 10. then thin-wall analysis can be used.σ1p⋅ rit:= σ1 = 28.88MPa AnsLongitudinal Stress :σ2p⋅ ri2⋅ t:= σ2 = 14.44MPa Ans 801. Problem 8-5The open-ended pipe has a wall thickness of 2 mm and an internal diameter of 40 mm. Calculate thepressure that ice exerted on the interior wall of the pipe to cause it to burst in the manner shown. Themaximum stress that the material can support at freezing temperatures is σmax = 360 MPa. Show thestress acting on a small element of material just before the pipe fails.Given: t := 2mm σallow := 360MPadi := 40mmSolution: ri := 0.5diHoop Stress : αrit:= α = 10.00Since α > 10. then thin-wall analysis can be used.σ1 := σallowσ1p⋅ rit= pσallow⋅ tri:=p = 36.0MPa AnsLongitudinal Stress :Since the pipe is open at both neds, thenσ2 := 0 Ans 802. Problem 8-6The open-ended polyvinyl chloride pipe has an inner diameter of 100 mm and thickness of 5 mm. If itcarries flowing water at 0.42 MPa pressure, determine the state of stress in the walls of the pipe.Given: t := 5mm p := 0.42MPadi := 100mmSolution: ri := 0.5diHoop Stress : σ1p⋅ rit:= σ1 = 4.2MPa AnsNormal Stress : σ2 := 0 AnsThere is no stress componenet in the longitudinal direction since pipe has open ends. 803. Problem 8-7If the flow of water within the pipe in Prob. 8-6 is stopped due to the closing of a valve, determine thestate of stress in the walls of the pipe. Neglect the weight of the water. Assume the supports only exertvertical forces on the pipe.Given: t := 5mm p := 0.42MPadi := 100mmSolution: ri := 0.5diHoop Stress : σ1p⋅ rit:= σ1 = 4.2MPa AnsNormal Stress : σ2p⋅ ri2⋅ t:= σ2 = 2.1MPa Ans 804. Problem 8-8The A-36-steel band is 50 mm wide and is secured around the smooth rigid cylinder. If the bolts aretightened so that the tension in them is 2 kN, determine the normal stress in the band, the pressureexerted on the cylinder, and the distance half the band stretches.Given: t := 3mm b := 50mmr := 200mm F := 2kNE := 200GPaSolution: rb := r + 0.5t Lb := π⋅ rbTensile Stress in the Band :σ1Fb⋅ t:= σ1 = 13.33MPa AnsHoop Stress : σp⋅ rt=ptσ1rb:= p = 0.199MPa AnsStectch :δ = ε1⋅Lb ε1σ1E=δσ1⋅LbE:= δ = 0.0422mm Ans 805. Problem 8-9The 304 stainless steel band initially fits snugly around the smooth rigid cylinder. If the band is thensubjected to a nonlinear temperature drop of ΔT = 12 sin2θ °C, where θ is in radians, determine thecircumferential stress in the band.Unit used: °C := degGiven: t := 0.4mm b := 25mmr' := 250mm E := 193GPaΔT = 12⋅ sin(θ)2 α 17⋅ (10− 6)1°C:=Solution:Compatibility: Since the band is fitted to a rigid cylinder(which does not deform under load), thenδF − δT = 0P⋅ (2π⋅ r)A⋅E2π0α⋅ (ΔT)⋅ r θ⌠⎮⌡− d = 02π⋅ rEPA⎛⎜⎝⎞⎠⋅ 122π0α⋅ sin(θ)2⋅ r θ⌠⎮⌡− d = 0However,PA= σc2π⋅ rE= d⋅σc 12⋅α⋅ r2πsin(θ)2 θ ⌠⎮⌡0σc2πsin(θ)2 θ ⌠⎮⌡6⋅αEπ 0d⎛⎜⎜⎝⎞⎠:= ⋅ ⋅ °Cσc = 19.69MPa Ans 806. Problem 8-10The barrel is filled to the top with water. Determine the distance s that the top hoop should be placedfrom the bottom hoop so that the tensile force in each hoop is the same. Also, what is the force in eachhoop? The barrel has an inner diameter of 1.2 m. Neglect its wall thickness. Assume that only thehoops resist the water pressure. Note: Water develops pressure in the barrel according to Pascal's law,p = (0.01z) MPa, where z is the depth from the surface of the water in meter.Given: d := 1.2m p = 0.01z⋅MPah := 2.4m h' := 0.6mSolution: r := 0.5dPhp⋅ (2r) z⌠⎮⌡= d0P 2rh0.01z z⌠⎮⌡0d⎛⎜⎜⎝⎞⎠⋅MPam:= ⋅P = 34.56 kNEquilibrium for the Steel Hoop :ΣFy=0; P − 4F = 0 F := 0.25P F = 8.64 kNΣΜBase=0; Ph3⋅ − 2F⋅h' − 2F⋅ (h' + s) = 0sP⋅h6F:= − h' − h's = 400mm Ans 807. 30 Problem 8-11A wood pipe having an inner diameter of 0.9 m is bound together using steel hoops having across-sectional area of 125 mm2. If the allowable stress for the hoops is σallow = 84 MPa, determinetheir maximum spacing s along the section of pipe so that the pipe can resist an internal gauge pressureof 28 kPa. Assume each hoop supports the pressure loading acting along the length s of the pipe.Given: d := 0.9m p := 28 ⋅ ( 1− )⋅MPaAs := 125mm2 σallow := 84MPaSolution: r := 0.5dP = p⋅ (2r⋅ s) F = σallow⋅ (As)Equilibrium for the Steel Hoop :From the FBD,+ ΣFy=0; P − 2F = 0p⋅ (2r⋅ s) − 2σallow⋅ (As) = 0σallow⋅ (As)sp⋅ r:=s = 833.33mm Ans 808. Problem 8-12A boiler is constructed of 8-mm thick steel plates that are fastened together at their ends using a buttjoint consisting of two 8-mm cover plates and rivets having a diameter of 10 mm and spaced 50 mmapart as shown. If the steam pressure in the boiler is 1.35 MPa, determine (a) the circumferential stressin the boiler's plate apart from the seam, (b) the circumferential stress in the outer cover plate along therivet line a-a, and (c) the shear stress in the rivets.Given: to := 8mm ri := 750mm p := 1.35MPatc := 8mm db := 10mm s := 50mmSolution:a) Hoop Stress :σ1p⋅ rito:= σ1 = 126.6MPa Ansb) Hoop Stress in cover plate along line a-a :Consider a width of s (mm), Fo (in boiler plate) = F'o (in cover plates)σ1⋅ (s⋅ to) = σ'1⋅ (s − db)⋅ (2⋅ tc)σ'1σ1⋅ (s⋅ to)(s − db)⋅ (2⋅ tc):=σ'1 = 79.1MPa Ansc) Shear Stress in Rivet : rb := 0.5dbFrom the FBD,+ ΣFy=0; Fb − σ1⋅ (s⋅ to) = 0Fb := σ1⋅ (s⋅ to)τavg12Fbπ rb⋅ 2⎛⎜⎜⎝⎞⎠:= ⋅τavg = 322.3MPa Ans 809. Problem 8-13The ring, having the dimensions shown, is placed over a flexible membrane which is pumped up with apressure p. Determine the change in the internal radius of the ring after this pressure is applied. Themodulus of elasticity for the ring is E.Solution:Equilibrium for the Ring :+ ΣFy=0; 2P − 2p⋅ ri⋅w = 0= ⋅ ⋅wP priHoop Stress and Strain for the Ring :σ1PA= σ1p⋅ ri⋅w(ro − ri)⋅w= σ1p⋅ riro − ri=Using Hooke's Law,ε1σ1E= ε1p⋅ ri(ro − ri)⋅E= (1)However,ε1⋅ − 2⋅π⋅ ri2⋅π⋅ r2⋅π ri1= ε1ri1− rir= ε1δrir=Then, from Eq.(1)δrirp⋅ ri(ro − ri)⋅E=δrip ⋅ ri2(ro − ri)⋅E= Ans 810. 0σProblem 8-14A closed-ended pressure vessel is fabricated by cross-winding glass filaments over a mandrel, so thatthe wall thickness t of the vessel is composed entirely of filament and an epoxy binder as shown in thefigure. Consider a segment of the vessel of width w and wrapped at an angle θ. If the vessel is subjectedto an internal pressure p, show that the force in the segment is Fθ = σ0 wt , where is the stress in thefilaments. Also, show that the stresses in the hoop and longitudinal directions are σh= σ0sin2 θ and σl= σ0cos2 θ , respectively. At what angle θ (optimum winding angle) would the filaments have to bewound so that the hoop and longitudinal stresses are equivalent? 811. Problem 8-15The steel bracket is used to connect the ends of two cables. If the allowable normal stress for the steelis σallow = 168 MPa, determine the largest tensile force P that can be applied to the cables. The brackethas a thickness of 12 mm and a width of 18 mm.Given: b := 18mm a := 50mmt := 12mm σallow := 168MPaSolution:Internal Force and Moment :ao := a + 0.5⋅bN = PM Pao= ⋅Section Property :A := t⋅b I112:= ⋅ t⋅b3Alowable Normal Stress: σNAM⋅cI= +The maximum normal stress occurs at the bottom of the steel bracket.cmax := 0.5⋅b σallowPAP⋅ao⋅cmaxI= +Pσallow1Aao⋅cmaxI+:=P = 1.756 kN Ans 812. Problem 8-16The steel bracket is used to connect the ends of two cables. If the applied force P = 2.5 kN, determinethe maximum normal stress in the bracket. The bracket has a thickness of 12 mm and a width of 18mm.Given: b := 18mm a := 50mmt := 12mm P := 2.5kNSolution:Internal Force and Moment :oao := aa + 0.5⋅bN := PM := P⋅Section Property :A := t⋅b I112:= ⋅ t⋅b3Alowable Normal Stress: σNAM⋅cI= +The maximum normal stress occurs at the bottom of the steel bracket.cmax := 0.5⋅b σmaxNAM⋅cmaxI:= +σmax = 239.2MPa Ans 813. Problem 8-17The joint is subjected to a force of 1.25 kN as shown. Sketch the normal-stress distribution acting oversection a-a if the member has a rectangular cross section of width 12 mm and thickness 18 mm.Given: b := 18mm ah := 50mm av := 32mmt := 12mm P := 1.25kNv := 3 h := 4 r := 5Solution:Internal Force and Moment :+ ΣFx=0; Phr⎛⎜⎝⎞⎠⋅ − N = 0 N Phr⎛⎜⎝⎞⎠:= ⋅+ ΣFy=0; V Pvr⎛⎜⎝⎞⎠− ⋅ = 0 V Pvr⎛⎜⎝⎞⎠:= ⋅+ ΣΜA=0; M Phr⎛⎜⎝⎞⎠+ ⋅ ⋅av Pvr⎛⎜⎝⎞⎠− ⋅ ⋅ah = 0M Pvr⎛⎜⎝⎞⎠⋅ ⋅ah Phr⎛⎜⎝⎞⎠:= − ⋅ ⋅avSection Property :A := b⋅ t I112:= ⋅b⋅ t3Normal Stress: σNAM⋅cI= +ctop := 0.5⋅ t σtopNAM⋅ctopI:= + σtop = 17.36MPa (T) Anscbot := −0.5⋅ t σbotNAM⋅cbotI:= + σbot = −8.10MPa (C) AnsLocation of zero stress:σtopσbotyot − yo=yot⋅ σtopσbot + σtop:=yo = 8.18mm 814. Problem 8-18The joint is subjected to a force of 1.25 kN as shown. Determine the state of stress at points A and B,and sketch the results on differential elements located at these points. The member has a rectangularcross-sectional area of width 12 mm and thickness 18 mm.Given: b := 18mm ah := 50mm av := 32mmt := 12mm P := 1.25kNv := 3 h := 4 r := 5Solution:Internal Force and Moment :+ ΣFx=0; Phr⎛⎜⎝⎞⎠⋅ − N = 0 N Phr⎛⎜⎝⎞⎠:= ⋅+ ΣFy=0; V Pvr⎛⎜⎝⎞⎠− ⋅ = 0 V Pvr⎛⎜⎝⎞⎠:= ⋅+ ΣΜA=0; M Phr⎛⎜⎝⎞⎠+ ⋅ ⋅av Pvr⎛⎜⎝⎞⎠− ⋅ ⋅ah = 0M Pvr⎛⎜⎝⎞⎠⋅ ⋅ah Phr⎛⎜⎝⎞⎠:= − ⋅ ⋅avSection Property :A := b⋅ t I112:= ⋅b⋅ t3QA := (0.5⋅ t⋅b)⋅ (0.25t)QB := 0 (since A' = 0)Normal Stress: σNAM⋅cI= +cA := 0 σANAM⋅cAI:= + σA = 4.63MPa (T) AnscB := −0.5⋅ t σBNAM⋅cBI:= + σB = −8.10MPa (C) AnsShear Stress : τV⋅QI⋅b=τAV⋅QAI⋅b:= τA = 5.21MPa AnsτBV⋅QBI⋅b:= τB = 0MPa Ans 815. Problem 8-19The coping saw has an adjustable blade that is tightened with a tension of 40 N. Determine the state ofstress in the frame at points A and B.Given: t := 3mm a := 100mmh := 8mm P := 40NSolution:Internal Force and Moment :At A: NA := −P MA := P⋅aAt B: NB := 0 MB := P⋅ (0.5a)Section Property :A := t⋅h I112:= ⋅ t⋅h3State of Stress: σNAM⋅cI= +At A:cA := 0.5⋅h σANAAMA⋅cA:= +IσA = 123.3MPa (T) AnsAt B:cB := 0.5⋅h σBNBAMB⋅cBI:= +σB = 62.5MPa (T) Ans 816. Problem 8-20Determine the maximum and minimum normal stress in the bracket at section a when the load isapplied at x = 0.Given: a := 30mm b := 20mmxe := 0.5a P := 4kNSolution:Internal Force and Moment :N := −PM Pxe:= ⋅ M = 0.060 kN⋅mSection Property :A := a⋅b A = 600mm2I112:= ⋅b⋅a3 I = 45000mm4Normal Stress: σNAM⋅cI= +cmax := 0.5a σtNAM⋅cmax:= + σt = 13.33MPa (T)I:= + σc = −26.67MPa (C)σmax := max( σt , σc ) σmax = 26.67MPa Ansσmin := min( σt , σc ) σmin = 13.33MPa Anscmin := −0.5⋅a σcNAM⋅cminI 817. Problem 8-21Determine the maximum and minimum normal stress in the bracket at section a when the load isapplied at x = 50 mm.Given: a := 30mm b := 20mmP := −4kN xe := 0.5a − 50mmSolution:Internal Force and Moment :N := PM Pxe:= ⋅ M = 0.140 kN⋅mSection Property :A := a⋅b A = 600mm2I112:= ⋅b⋅a3 I = 45000mm4Normal Stress: σNAM⋅cI= +cmax := 0.5a σtNAM⋅cmax:= + σt = 40.00MPa (T)I:= + σc = −53.33MPa (C)σmax := max( σt , σc ) σmax = 53.33MPa Ansσmin := min( σt , σc ) σmin = 40.00MPa Anscmin := −0.5⋅a σcNAM⋅cminI 818. Problem 8-22The vertical force P acts on the bottom of the plate having a negligible weight. Determine the maximumdistance d to the edge of the plate at which it can be applied so that it produces no compressivestresses on the plate at section a-a. The plate has a thickness of 10 mm and P acts along the centerlineof this thickness.Given: t := 10mm b := 150mmSolution:Internal Force and Moment :N = Pxexe = d − 0.5b M = P ⋅Section Property :A := t⋅b A = 1500mm2I112:= ⋅ t⋅b3 I = 2812500mm4Normal Stress: Require σmin := 0σ = N ±McAIcmin := −0.5⋅b 0PA(P⋅xe)⋅cminI= +0PAP⋅ (d − 0.5b)⋅ (−0.5⋅b)I= +d2IA⋅bb2:= +d = 100mm Ans 819. Problem 8-23The vertical force P = 600 N acts on the bottom of the plate having a negligible weight. The plate has athickness of 10 mm and P acts along the centerline of this thickness such that d = 100 mm. Plot thedistribution of normal stress acting along section a-a.Given: t := 10mm b := 150mmd := 100mm P := 0.6kNSolution:Internal Force and Moment :N := Pxe d 0.5b − := M P xe:= ⋅ M = 0.015 kN⋅mSection Property :A := t⋅b A = 1500mm2I112:= ⋅ t⋅b3 I = 2812500mm4Normal Stress: σNAM⋅cI= +cA := −0.5b σANAM⋅cAI:= + σA = 0MPa (C) AnscB := 0.5⋅b σBNAM⋅cBI:= + σB = 0.800MPa (T) Ans 820. Problem 8-24The gondola and passengers have a weight of 7.5 kN and center of gravity at G. The suspender armAE has a square cross-sectional area of 38 mm by 38 mm, and is pin connected at its ends A and E.Determine the largest tensile stress developed in regions AB and DC of the arm.Given: b := 38mm d := 38mm ah := 375mmL1 := 1.2m L2 := 1.65m W := 7.5kNSolution:Section Property :A := b⋅d I112:= ⋅b⋅d3Segment AB :NAB := W MAB := 0Maximum Normal Stress: σNAM⋅cI= +σABNABA:= σAB = 5.19MPa (T) AnsSegment DC :NDC := W MDC := W⋅ahMaximum Normal Stress: σNAM⋅cI= +cmax := 0.5d σDCNDCAMDC⋅cmaxI:= +σDC = 312.73MPa (T) Ans 821. Problem 8-25The stepped support is subjected to the bearing load of 50 kN. Determine the maximum and minimumcompressive stress in the material.Given: a := 100mm b := 100mmP := 50kN xe := 0.5a − 30mmSolution:Internal Force and Moment :N := −PM Pxe:= ⋅ M = 1.000 kN⋅mSection Property :For the bottom portion of the stepped support.A := a⋅b A = 10000mm2I112:= ⋅b⋅a3 I = 8333333.33mm4Normal Stress: σNAM⋅cI= +cmax := 0.5a σtNAM⋅cmax:= + σt = 1MPa (T)I:= + σc = −11.00MPa (C)σc_max := max(0 , σc ) σc_max = 11MPa Ansσc_min := min(0 , σc ) σc_min = 0MPa Anscmin := −0.5⋅a σcNAM⋅cminI 822. Problem 8-26The bar has a diameter of 40 mm. If it is subjected to a force of 800 N as shown, determine the stresscomponents that act at point A and show the results on a volume element located at this point.Given: do := 40mm a := 200mmP := 0.8kN θ := 30degSolution:Internal Force and Moment : At section AB.N := P⋅ sin(θ) N = 0.400 kNV := P⋅cos(θ) V = 0.693 kNM := V⋅a M = 0.1386 kN⋅mSection Property : ro := 0.5doA π ro:= ⋅ 2 A = 1256.64mm2Iπ4:= ⋅ ro4 I = 125663.71mm4QA4ro3πA2⎛⎜⎝⎞⎠:= ⋅ QA = 5333.33mm3Normal Stress: σNAM⋅cI= +cA := 0 σANAM⋅cAI:= +σA = 0.318MPa (T) AnsShear Stress : τV⋅QI⋅b=τAV⋅QAI⋅do:= τA = 0.735MPa Ans 823. Problem 8-27Solve Prob. 8-26 for point B.Given: do := 40mm a := 200mmP := 0.8kN θ := 30degSolution:Internal Force and Moment : At section AB.N := P⋅ sin(θ) N = 0.400 kNV := P⋅cos(θ) V = 0.693 kNM := V⋅a M = 0.1386 kN⋅mSection Property : ro := 0.5doA π ro:= ⋅ 2 A = 1256.64mm2Iπ4:= ⋅ ro4 I = 125663.71mm4QB := 0 (since A'=0 )Normal Stress: σNAM⋅cI= +cB := −ro σBNAM⋅cBI:= +σB = −21.73MPa (C) AnsShear Stress : τV⋅QI⋅b=τAV⋅QBI⋅do:= τA = 0MPa Ans 824. Problem 8-28Since concrete can support little or no tension, this problem can be avoided by using wires or rods toprestress the concrete once it is formed. Consider the simply supported beam shown, which has arectangular cross section of 450 mm by 300 mm. If concrete has a specific weight of 24 kN/m3,determine the required tension in rod AB, which runs through the beam so that no tensile stress isdeveloped in the concrete at its center section a-a. Neglect the size of the rod and any deflection of thebeam.Given: b := 300mm d := 450mmd' := 400mm γ 24kNm3:=L := 2.4mSolution: a := d − d'w := γ⋅b⋅dSupport Reactions : By symmetry, RA= R ; RA= R+ 2R − w⋅L = 0 R := 0.5w⋅LInternal Force and Moment :+ ΣFx=0; T − N = 0 N = T+ ΣΜO=0; M + T⋅ (0.5d − a) − R⋅ (0.5L) + (0.5w⋅L)⋅ (0.25L) = 0M = R⋅ (0.25⋅L) − T⋅ (0.5⋅d − a)Section Property :A := b⋅d I112:= ⋅b⋅d3Normal Stress: σaNAM⋅cI= +Requires σa= 0 0−TAM⋅caI= +ca := 0.5d 0−TA[R⋅ (0.25⋅L) − T⋅ (0.5⋅d − a)]⋅caI= +TR⋅ (0.25⋅L)(0.5⋅d − a)IA⋅ca+:=T = 9.331 kN Ans 825. Problem 8-29Solve Prob. 8-28 if the rod has a diameter of 12 mm. Use the transformed area method discussed inSec. 6.6. Est = 200 GPa, Ec = 25 GPa.Given: b := 300mm d := 450mmd' := 400mm do := 12mmEst := 200GPa Ec := 25GPaL := 2.4m γ 24kNm3:=Solution: a := d − d'w := γ⋅b⋅dSupport Reactions : By symmetry, RA= R ; RA= R+ 2R − w⋅L = 0 R := 0.5w⋅LInternal Force and Moment :+ ΣFx=0; T − N = 0 N = T+ ΣΜO=0; M T d ycyc+ ⋅ ( − − a) − R⋅ (0.5L) + (0.5w⋅L)⋅ (0.25L) = 0M = R ⋅ ( 0.25 ⋅ L ) − T ⋅ ( d − − a)Section Property :nEstEc:= A'conc (n − 1)π⋅ 4do⎛⎜⎝2 ⎞⎠:= ⋅A := b⋅d + A'concycb⋅d⋅ (0.5d) + A'conc⋅d'A:=I112:= ⋅b⋅d3 + b⋅d⋅ (0.5d − yc)2 + A'conc⋅ (d' − yc)2Normal Stress: σaNAM⋅cI= +Requires σa= 0 0−TAM⋅caI= +ca := d − yc 0−TAR yc⋅ ( 0.25 ⋅ L ) − T ⋅ ( d − − a ⎡⎣) ⎤⎦⋅caI= +TR⋅ (0.25⋅L)(d − yc − a) IA⋅ca+:=T = 9.343 kN Ans 826. Problem 8-30The block is subjected to the two axial loads shown. Determine the normal stress developed at points Aand B. Neglect the weight of the block.Given: b := 50mm d := 75mmP1 := 250N P2 := 500NSolution:Internal Force and Moment :+ ΣFx=0; −N − P1 − P2 = 0 N := −P1 − P2+ Σ Μz=0; Mz + P1⋅ (0.5d) − P2⋅ (0.5d) = 0Mz := 0.5⋅d⋅ (P2 − P1)+ Σ Μy=0; My + P1⋅ (0.5b) − P2⋅ (0.5b) = 0My := 0.5⋅b⋅ (P2 − P1)Section Property :A := b⋅d Iz112:= ⋅b⋅d3 Iy112:= ⋅d⋅b3Normal Stress: σNAMz⋅yIz−My⋅zIy= +At A: yA := 0.5d zA := 0.5bσANAMz⋅yAIz−My⋅zAIy:= + σA = −0.200MPa (C) AnsAt B: yB := 0.5d zB := −0.5bσBNAMz⋅yBIz−My⋅zBIy:= + σB = −0.600MPa (C) Ans 827. Problem 8-31The block is subjected to the two axial loads shown. Sketch the normal stress distribution acting overthe cross section at section a-a. Neglect the weight of the block.Solution:(T)Given: b := 50mm d := 75mmP1 := 250N P2 := 500NInternal Force and Moment :+ ΣFx=0; −N − P1 − P2 = 0 N := −P1 − P2+Σ Μz=0;AnsMz + P1⋅ (0.5d) − P2⋅ (0.5d) = 0Mz := 0.5⋅d⋅ (P2 − P1)+ Σ Μy=0; My + P1⋅ (0.5b) − P2⋅ (0.5b) = 0My := 0.5⋅b⋅ (P2 − P1)Section Property :A := b⋅d Iz112:= ⋅b⋅d3 Iy112:= ⋅d⋅b3Normal Stress: σNAMz⋅yIz−My⋅zIy= +At A: yA := 0.5d zA := 0.5bσANAMz⋅yAIz−My⋅zAIy:= + σA = −0.200MPa (C) AnsAt B: yB := 0.5d zB := −0.5bσBNAMz⋅yBIz−My⋅zBIy:= + σB = −0.600MPa (C) AnsAt C: yC := −0.5d zC := −0.5bσCNAMz⋅yCIz−My⋅zCIy:= + σC = −0.200MPa (C) AnsAt D: yD := −0.5d zD := 0.5bσDNAMz⋅yDIz−My⋅zDIy:= + σD = 0.200MPa 828. Problem 8-32A bar having a square cross section of 30 mm by 30 mm is 2 m long and is held upward. If it has amass of 5 kg/m, determine the largest angle θ, measured from the vertical, at which it can besupported before it is subjected to a tensile stress near the grip.Given: b := 30mm L := 2mt := 30mm mo 5kgm:=Solution: W := mo⋅g⋅L W = 0.0981 kNInternal Force and Moment :+ ΣFy=0; −N − W⋅cos(θ) = 0 N = −W⋅cos(θ)+ ΣΜO=0; M − W⋅ sin(θ)⋅ (0.5⋅L) = 0 M = 0.5W⋅L⋅ sin(θ)Section Property :A := b⋅ t I112:= ⋅b⋅ t3Normal Stress: Require σmax := 0σ = N ±McAIcmax := 0.5⋅b 0NAM⋅cmaxI= +0−W⋅cos(θ)b⋅ t0.5W⋅L⋅ sin(θ)⋅ (0.5⋅b)112⋅b⋅ t3= +tan(θ) t23⋅L⋅b=θ atant23⋅L⋅b⎛⎜⎝⎞⎠:=θ = 0.00500 rad θ = 0.286 deg Ans 829. Problem 8-33Solve Prob. 8-32 if the bar has a circular cross section of 30-mm diameter.Given: do := 30mm L := 2mmo 5kgm:=Solution: W := mo⋅g⋅L W = 0.0981 kNInternal Force and Moment :+ ΣFy=0; −N − W⋅cos(θ) = 0 N = −W⋅cos(θ)+ ΣΜO=0; M − W⋅ sin(θ)⋅ (0.5⋅L) = 0 M = 0.5W⋅L⋅ sin(θ)Section Property : ro := 0.5doA := π ⋅ ro2 A = 706.86mm2Iπ4:= ⋅ 4 I = 39760.78mm4roNormal Stress: Require σmax := 0σ = N ±McAIcmax := ro 0NAM⋅cmaxI= +0−W⋅cos(θ)A0.5W⋅L⋅ sin(θ)⋅ (ro)I= +tan(θ) 2IA⋅L⋅ ro=θ atan2IA⋅L⋅ ro⎛⎜⎝⎞⎠:=θ = 0.00375 rad θ = 0.215 deg Ans 830. Problem 8-34The wide-flange beam is subjected to the loading shown. Determine the stress components at points Aand B and show the results on a volume element at each of these points. Use the shear formula tocompute the shear stress.Given: b := 100mm d := 150mmt := 12mm d'B := 50mmP1 := 2.5kN P2 := 12.5kNP3 := 15kN L1 := 0.5mL2 := 1m L3 := 1.5mSolution: L 3L1:= ⋅ + L2 + L3Support Reactions : Given+ ΣFy=0; R1 + R2 − P1 − P2 − P3 = 0+ ΣΜR2=0; −R1⋅L + P1⋅ (L − L1) + P2⋅ (L2 + L3) + P3⋅L3 = 0Guess R1 := 1N R2 := 1NR1R2⎛⎜⎜⎝⎞⎠:= Find(R1 , R2)R1R2⎛⎜⎜⎝⎞⎠15.6314.38⎛⎜⎝⎞⎠= kNAt Section A-B:M := R1⋅ (2L1) V := R1 − P1Section Property : D := d + 2tA := 2⋅b⋅ t + d⋅ t I112:= ⋅b⋅D3− ⋅ (b − t)⋅d3112QB := b⋅ t⋅ (0.5D − 0.5t) + d'B⋅ t⋅ (0.5D − t − 0.5d'B)QA := 0 (since A' = 0)Normal Stress: σM⋅yI=yA := 0.5D σA−M⋅yAI:= σA = −70.98MPa (C) AnsyB := 0.5d − d'B σBM⋅yBI:= σB = 20.4MPa (T) AnsShear Stress : τV⋅QI⋅ t=τAV⋅QAI⋅b:= τA = 0.00MPa AnsτBV⋅QBI⋅ t:= τB = 7.265MPa Ans 831. Problem 8-35The cantilevered beam is used to support the load of 8 kN. Determine the state of stress at points Aand B, and sketch the results on differential elements located at each of these points.Given: bo := 100mm d := 100mmt := 10mm L := 3mdA := 25mm dB := 45mmP := 8kNSolution:Internal Force and Moment : At Section A-B:V := P V = 8 kNM := P⋅L M = 24 kN⋅mSection Property : bi := bo − 2tA := 2⋅d⋅ t + bi⋅ t A = 2800mm2I112⋅bi⋅ t3112:= + ⋅ (2t)⋅d3 I = 1673333.33mm4QA := (dA⋅ t)⋅ (0.5d − 0.5dA) QA = 9375mm3QB := (dB⋅ t)⋅ (0.5d − 0.5dB) QB = 12375mm3Normal Stress: σM⋅yI=yA := 0.5d − dA σAM⋅yAI:= σA = 358.6MPa (T) AnsyB := 0.5d − dB σBM⋅yBI:= σB = 71.7MPa (T) AnsShear Stress : τV⋅QI⋅ t=τAV⋅QAI⋅ t:= τA = 4.48MPa AnsτBV⋅QBI⋅ t:= τB = 5.92MPa Ans 832. Problem 8-36The cylinder of negligible weight rests on a smooth floor. Determine the eccentric distance ey at whichthe load can be placed so that the normal stress at point A is zero.Solution:Internal Force and Moment :V = −PM P ey= ⋅ ( )Section Property :A = π⋅ r2Iπ4= ⋅ r4Normal Stress: Require σmax := 0McIσ = N ±AcA = r 0NAM⋅crI= +0−Pπ⋅ r24P⋅ (ey)⋅ rπ⋅ r4= +eyr4= Ans 833. Problem 8-37The beam supports the loading shown. Determine the state of stress at points E and F at section a-a,and represent the results on a differential volume element located at each of these points.Given: bf := 150mm dw := 200mmtf := 10mm tw := 15mma := 1m b := 2myB := 3.3mw 10kNm:= yD := 0.3mSolution: L := 2⋅a + bSupport Reactions : GivenDxLyB − yD⎛⎜⎝⎞⎠= ⋅Dy (1)+ ΣFy=0; Cy + Dy − w⋅ (2a) = 0 (2)+ ΣΜC=0; −w⋅ (2a)⋅a + Dy⋅ (L) + Dx⋅ (yD) = 0 (3)Solving Eqs.(1), (2) and (3). Guess Cy := 1N Dx := 1N Dy := 1NCyDxDy⎛⎜⎜⎜⎝⎞⎟⎠:= Find(Cy ,Dx ,Dy)CyDxDy⎛⎜⎜⎜⎝⎞⎟⎠15.45456.06064.5455⎛⎜⎜⎝⎞= kN⎠Internal Force and Moment : At Section a-a:N := −Dx N = −6.0606 kNV := w⋅a − Dy V = 5.4545 kNM := −Dy⋅ (a + b) − Dx⋅yD + w⋅a⋅ (0.5a) M = −10.4545 kN⋅mSection Property : do := dw + 2tfA 2bf:= ⋅ ⋅ tf + dw⋅ tw A = 6000mm2I112⋅ 3⋅bf do112⋅ (bf − tw) dw:= − ⋅ 3 I = 43100000mm4QE := bf⋅ tf⋅ (0.5do − 0.5tf) + 0.5dw⋅ tw⋅ (0.25dw) QE = 232500mm3QF := 0 (since A' = 0)McINormal Stress:σ = N ± AcE := 0 σENAM⋅cEI:= + σE = −1.01MPa (C) Ans 834. cF := 0.5⋅do σFNAM⋅cFI:= + σF = −27.69MPa (C) AnsShear Stress : τV⋅QI⋅ t=τEV⋅QEI⋅ tw:= τE = 1.96MPa AnsτFV⋅QFI⋅bf:= τF = 0MPa Ans 835. Problem 8-38The metal link is subjected to the axial force of P = 7 kN. Its original cross section is to be altered bycutting a circular groove into one side. Determine the distance a the groove can penetrate into the crosssection so that the tensile stress does not exceed σallow = 175 MPa. Offer a better way to remove thisdepth of material from the cross section and calculate the tensile stress for this case. Neglect theeffects of stress concentration.Given: h := 80mm t := 25mmP := 7kN σallow := 175MPaSolution:Internal Force and Moment : At narrow section.+ ΣFx=0; P − N = 0 N := P+ ΣΜbase=0; M Nh − a2+ ⋅ Ph2− ⋅ = 0 M = 0.5P⋅aSection Property :A = (h − a)⋅ t It12= ⋅ (h − a)3Normal Stress: Require σmax := σallowcmax = 0.5(d − a)σmaxNAM⋅cmaxI= +σallowP(h − a)⋅ t0.5P⋅a⋅ [0.5(h − a)]t12⋅ (h − a)3= +Given (σallow⋅ t)⋅ (h − a)2 − P⋅ (h − a) − 3P⋅a = 0 (1)Solving Eq.(1),. Guess a := 1mma := Find(a)a = 61.94mm AnsBetter way: To remove material equally from both sides such that M=0.A := (h − a)⋅ tσ'maxNA:= + 0σ'max = 15.50MPa Ans 836. Problem 8-39Determine the state of stress at point A when the beam is subjected to the cable force of 4 kN. Indicatethe result as a differential volume element.Given: bf := 150mm dw := 200mmtf := 20mm tw := 15mma := 2m b := 0.75m c := 1mhG := 375mm rG := 250mm P := 4kNSolution: L := a + b + cSupport Reactions : Given+ ΣFx=0; Cx − P = 0 Cx := P+ ΣΜD=0; P⋅ (hg + rG)⋅a + Cy⋅ (L) = 0 CyhG + rG:= PLInternal Force and Moment : At Section A-B:N := Cx N = 4 kNV := Cy V = 0.6667 kNM := Cy⋅c M = 0.6667 kN⋅mSection Property : do := dw + 2tfA 2bf:= ⋅ ⋅ tf + dw⋅ tw A = 9000mm2I112⋅ 3⋅bf do112⋅ (bf − tw) dw:= − ⋅ 3 I = 82800000mm4QA := bf⋅ tf⋅ (0.5do − 0.5tf) + 0.5dw⋅ tw⋅ (0.25dw) QA = 405000mm3McINormal Stress:σ = N ± AcA := 0 σANAM⋅cAI:= +σA = 0.444MPa (T) AnsShear Stress : τV⋅QI⋅ t=τAV⋅QAI⋅ tw:= τA = 0.217MPa Ans 837. Problem 8-40Determine the state of stress at point B when the beam is subjected to the cable force of 4 kN. Indicatethe result as a differential volume element.Given: bf := 150mm dw := 200mmtf := 20mm tw := 15mma := 2m b := 0.75m c := 1mhG := 375mm rG := 250mm P := 4kNSolution: L := a + b + cSupport Reactions : Given+ ΣFx=0; Cx − P = 0 Cx := P+ ΣΜD=0; P⋅ (hg + rG)⋅a + Cy⋅ (L) = 0 CyhG + rG:= PLInternal Force and Moment : At Section A-B:N := Cx N = 4 kNV := Cy V = 0.6667 kNM := Cy⋅c M = 0.6667 kN⋅mSection Property : do := dw + 2tfA 2bf:= ⋅ ⋅ tf + dw⋅ tw A = 9000mm2I112⋅ 3⋅bf do112⋅ (bf − tw) dw:= − ⋅ 3 I = 82800000mm4QB := 0 (since A' = 0)McINormal Stress:σ = N ± AcB := −0.5do σBNAM⋅cBI:= +σB = −0.522MPa (C) AnsShear Stress : τV⋅QI⋅ t=τBV⋅QBI⋅ tw:= τB = 0MPa Ans 838. Problem 8-41The bearing pin supports the load of 3.5 kN. Determine the stress components in the support memberat point A. The support is 12 mm thick.Given: b := 12mm d := 18mm W := 3.5kNL1 := 50mm L2 := 75mm a := 32mmθ := 30degSolution:Internal Force and Moment :ΣFx=0; N − W⋅cos(θ) = 0 N := W⋅cos(θ)ΣFy=0; V − W⋅ sin(θ) = 0 V := W⋅ sin(θ)+ ΣΜ=0; M W a L1− ⋅ ( − ⋅ sin(θ)) = 0:= ⋅ ( − ⋅ sin(θ))M W a L1Section Property :A := b⋅d I112:= ⋅b⋅d3QA := 0 (since A' = 0)Normal Stress: σNAM⋅yI= +At A: yA := −0.5dσANAM⋅yAI:= +σA = −23.78MPa (C) AnsShear Stress : τV⋅QI⋅ t=τAV⋅QAI⋅b:=τA = 0MPa Ans 839. Problem 8-42The bearing pin supports the load of 3.5 kN. Determine the stress components in the support memberat point B. The support is 12 mm thick.Given: b := 12mm d := 18mm W := 3.5kNL1 := 50mm L2 := 75mm a := 32mmθ := 30degSolution:Internal Force and Moment :ΣFx=0; N − W⋅cos(θ) = 0 N := W⋅cos(θ)ΣFy=0; V − W⋅ sin(θ) = 0 V := W⋅ sin(θ)+ ΣΜ=0; M W a L1− ⋅ ( − ⋅ sin(θ)) = 0:= ⋅ ( − ⋅ sin(θ))M W a L1Section Property :A := b⋅d I112:= ⋅b⋅d3QB := 0 (since A' = 0)Normal Stress: σNAM⋅yI= +At B: yB := 0.5dσBNAM⋅yBI:= +σB = 51.84MPa (T) AnsShear Stress : τV⋅QI⋅ t=τBV⋅QBI⋅b:=τB = 0MPa Ans 840. Problem 8-43The uniform sign has a weight of 7.5 kN and is supported by the pipe AB, which has an inner radius of68 mm and an outer radius of 75 mm. If the face of the sign is subjected to a uniform wind pressureof p = 8 kN/m2, determine the state of stress at points C and D. Show the results on a differentialvolume element located at each of these points. Neglect the thickness of the sign, and assume that it issupported along the outside edge of the pipe.p 8kNm2:=Given: b := 3.6m h := 1.8mro := 75mm ri := 68mmW := 7.5kN d := 0.9mSolution: P := p⋅b⋅hInternal Force and Moment :W + Nx = 0ΣFx=0;AnsNx := −WΣFy=0; Vy − P = 0 Vy := PΣFz=0; Vz := 0Σ Μx=0; Mx − P⋅ (0.5b) = 0 Mx := 0.5⋅P⋅bΣ Μy=0; My − W⋅ (0.5b) = 0 My := 0.5⋅W⋅bΣ Μz=0; Mz + P⋅ (0.5h + d) = 0 Mz := −P⋅ (0.5h + d)Nx = −7.5 kN Vy = 51.84 kN Vz = 0 kNMx = 93.312 kN⋅m My = 13.5 kN⋅m Mz = −93.312 kN⋅m2 − ⎛⎝Section Property : A π ro2 ri⎞⎠:= ⋅Iyπ4ro4 − ri4 ⎛⎝⎞⎠:= ⋅ Iz := IyQC_y4ro3ππ⋅ 2ro⎛⎜⎝2 ⎞⎠⋅4ri3ππ⋅ 2ri⎛⎜⎝2 ⎞⎠:= − ⋅QD_z := QC_y QD_y := 0π2QC_z := 0 Jro4 − ri4 ⎛⎝⎞⎠:= ⋅Normal Stress: σNAMz⋅yIz−My⋅zIy= +At C: yC := 0 zC := riσCNAMz⋅yCIz−My⋅zCIy:= + σC = 113.9MPa (T) AnsAt D: yD := ro zD := 0σDNAMz⋅yDIz−My⋅zDIy:= + σD = 868.5MPa (T) 841. Shear Stress :The transverse shear stress in the z and y directions and the torsional shear stress can beobtained using the shear formula for τv and the torsional formula for τt respectively.τvV⋅QI⋅ t= τtT⋅ρJ=:= ⋅ ( − ri) ρ := riAt C: t 2 roτv_yVy⋅QC_yIz⋅ t:= τtMx⋅ρJ:=τC_xy := τv_y − τt τC_xy = −360.8MPa AnsτC_xz := 0 AnsτC_yz := 0 Ans:= ⋅ ( − ri) ρ' := roAt D: t 2 roτv_zVz⋅QD_zIy⋅ t:= τ'tMx⋅ρ'J:=τD_xz := τv_z + τ't τD_xz = 434.3MPa AnsτD_xy := 0 AnsτD_yz := 0 Ans 842. Problem 8-44Solve Prob. 8-43 for points E and F.Given: b := 3.6m h := 1.8mro := 75mm ri := 68mmp 8kNm2:= W := 7.5kN d := 0.9mSolution: P := p⋅b⋅hInternal Force and Moment :ΣFx=0; W + Nx = 0 Nx := −WΣFy=0; Vy − P = 0 Vy := PΣFz=0; Vz := 0Σ Μx=0; Mx − P⋅ (0.5b) = 0 Mx := 0.5⋅P⋅bΣ Μy=0; My − W⋅ (0.5b) = 0 My := 0.5⋅W⋅bΣ Μz=0; Mz + P⋅ (0.5h + d) = 0 Mz := −P⋅ (0.5h + d)Nx = −7.5 kN Vy = 51.84 kN Vz = 0 kNMx = 93.312 kN⋅m My = 13.5 kN⋅m Mz = −93.312 kN⋅m⎛⎝2 − ri2 Section Property : A π ro⎞⎠:= ⋅Iyπ4ro4 − ri4 ⎛⎝⎞⎠:= ⋅ Iz := IyQF_y4ro3ππ⋅ 2ro⎛⎜⎝2 ⎞⎠⋅4ri3ππ⋅ 2ri⎛⎜⎝2 ⎞⎠:= − ⋅QE_z := QF_y QE_y := 0π2QF_z := 0 Jro4 − ri4 ⎛⎝⎞⎠:= ⋅Normal Stress: σNAMz⋅yIz−My⋅zIy= +At F: yF := 0 zF := −roσFNAMz⋅yFIz−My⋅zFIy:= + σF = −125.7MPa (C) AnsAt E: yE := −ro zE := 0σENAMz⋅yEIz−My⋅zEIy:= + σE = −868.5MPa (C) Ans 843. Shear Stress :The transverse shear stress in the z and y directions and the torsional shear stress can beobtained using the shear formula for τv and the torsional formula for τt respectively.τvV⋅QI⋅ t= τtT⋅ρJ=:= ⋅ ( − ri) ρ := roAt F: t 2 roτv_yVy⋅QF_yIz⋅ t:= τtMx⋅ρJ:=τF_xy := τv_y + τt τF_xy = 467.2MPa AnsτF_xz := 0 AnsτF_yz := 0 Ans:= ⋅ ( − ri) ρ' := roAt E: t 2 roτv_zVz⋅QE_zIy⋅ t:= τ'tMx⋅ρ'J:=τE_xz := τv_z − τ't τE_xz = −434.3MPa AnsτE_xy := 0 AnsτE_yz := 0 Ans 844. Problem 8-45The bar has a diameter of 40 mm. If it is subjected to the two force components at its end as shown,determine the state of stress at point A and show the results on a differential volume element located atthis point.Given: a := 100mm b := 150mmdo := 40mm Py := 0.3kNPz := −0.5kNSolution:Internal Force and Moment :ΣFx=0; Nx := 0ΣFy=0; Vy + Py = 0 Vy := −PyΣFz=0; Vz + Pz = 0 Vz := −PzΣ Μx=0; Mx := 0Σ Μy=0; My − Pz⋅ (−b) = 0 My := −Pz⋅bΣ Μz=0; Mz + Py⋅ (−b) = 0 Mz := Py⋅bNx = 0 kN Vy = −0.3 kN Vz = 0.5 kNMx = 0 kN⋅m My = 0.075 kN⋅m Mz = 0.045 kN⋅mSection Property : ro := 0.5doA := π ⋅ ro2 Jπ2:= ⋅ 4roIy:= ⋅ 4 Iz := IyQA_z := 0 QA_yπ64do4ro3π⎞⎠:= ⋅A2⎛⎜⎝Normal Stress: σNxAMz⋅yIz−My⋅zIy= +At A: yA := 0 zA := roσANxAMz⋅yAIz−My⋅zAIy:= + σA = 11.9MPa (T) AnsShear Stress :The transverse shear stress in the z and y directions and the torsional shear stress can beobtained using the shear formula for τv and the torsional formula for τt respectively.τvV⋅QI⋅ t= τtT⋅ρJ=At A: ty := do ρ := ro 845. τv_yVy⋅QA_yIz⋅ ty:= τtMx⋅ρJ:=τv_y = −0.318MPa τt = 0MPaτA_xy := τv_y − τt τA_xy = −0.318MPa AnsτA_xz := 0 AnsτA_yz := 0 Ans 846. Problem 8-46Solve Prob. 8-45 for point B.Given: a := 100mm b := 150mmdo := 40mm Py := 0.3kNPz := −0.5kNSolution:Internal Force and Moment :ΣFx=0; Nx := 0ΣFy=0; Vy + Py = 0 Vy := −PyΣFz=0; Vz + Pz = 0 Vz := −PzΣ Μx=0; Mx := 0Σ Μy=0; My − Pz⋅ (−b) = 0 My := −Pz⋅bΣ Μz=0; Mz + Py⋅ (−b) = 0 Mz := Py⋅bNx = 0 kN Vy = −0.3 kN Vz = 0.5 kNMx = 0 kN⋅m My = 0.075 kN⋅m Mz = 0.045 kN⋅mSection Property : ro := 0.5doA := π ⋅ ro2 Jπ2:= ⋅ 4roIy:= ⋅ 4 Iz := IyQB_y := 0 QB_zπ64do4ro3πA2⎛⎜⎝⎞⎠:= ⋅Normal Stress: σNxAMz⋅yIz−My⋅zIy= +At B: yB := ro zB := 0σBNxAMz⋅yBIz−My⋅zBIy:= + σB = −7.16MPa (C) AnsShear Stress :The transverse shear stress in the z and y directions and the torsional shear stress can beobtained using the shear formula for τv and the torsional formula for τt respectively.τvV⋅QI⋅ t= τtT⋅ρJ=At B: tz := do ρ := roτv_zVz⋅QB_zIy⋅ tz:= τtMx⋅ρJ:= 847. τv_z = 0.531MPa τt = 0MPaτB_xz := τv_z + τt τB_xz = 0.531MPa AnsτB_xy := 0 AnsτB_yz := 0 Ans 848. Problem 8-47The strongback AB consists of a pipe that is used to lift the bundle of rods having a total mass of 3Mgand center of mass at G. If the pipe has an outer diameter of 70 mm and a wall thickness of 10mm,determine the state of stress acting at point C. Show the results on a differential volume elementlocated at this point. Neglect the weight of the pipe.Given: do := 70mm t := 10mmhA := 75mm L := 3mMo := 3000kg θ := 45degSolution: W := Mo⋅g W = 29.42 kNEquilibrium :+ ΣFy=0; P − W = 0 P := WSupport Reactions : By symmetry, Ay = By = RP + 2R = 0R := −0.5P R = −14.71 kNAlso, Ax = -Bx and tan(θ) AyAx=AxRtan(θ) := Ax = −14.71 kNInternal Force and Moment : At Section C-D:N := Ax N = −14.71 kNV := R + 0.5W V = 0 kNM := V⋅ (0.5⋅L) + Ax⋅ (hA) M = −1.103 kN⋅mSection Property : di := do − 2tAπ4:= ⋅ A = 1884.96mm2do2 − di2 ⎛⎝⎞⎠Iπ64:= ⋅ I = 871791.96mm4do4 − di4 ⎛⎝⎞⎠McINormal Stress:σ = N ± AcC := 0.5do σCNAM⋅cCI:= + σC = −52.10MPa (C) AnsShear Stress : τV⋅QI⋅ t=τC := 0 (since V = 0) Ans 849. Problem 8-48The strongback AB consists of a pipe that is used to lift the bundle of rods having a total mass of 3Mgand center of mass at G. If the pipe has an outer diameter of 70 mm and a wall thickness of 10mm,determine the state of stress acting at point D. Show the results on a differential volume elementlocated at this point. Neglect the weight of the pipe.Given: do := 70mm t := 10mmhA := 75mm L := 3mMo := 3000kg θ := 45degSolution: W := Mo⋅g W = 29.42 kNEquilibrium :+ ΣFy=0; P − W = 0 P := WSupport Reactions : By symmetry, Ay = By = RP + 2R = 0R := −0.5P R = −14.71 kNAlso, Ax = -Bx and tan(θ) AyAx=AxRtan(θ) := Ax = −14.71 kNInternal Force and Moment : At Section C-D:N := Ax N = −14.71 kNV := R + 0.5W V = 0 kNM := V⋅ (0.5⋅L) + Ax⋅ (hA) M = −1.103 kN⋅mSection Property : di := do − 2tAπ4:= ⋅ A = 1884.96mm2do2 − di2 ⎛⎝⎞⎠Iπ64:= ⋅ I = 871791.96mm4do4 − di4 ⎛⎝⎞⎠McINormal Stress:σ = N ± AcD := 0 σDNAM⋅cDI:= + σD = −7.80MPa (C) AnsShear Stress : τV⋅QI⋅ t=τD := 0 (since V = 0) Ans 850. Problem 8-49The sign is subjected to the uniform wind loading. Determine the stress components at points A and Bon the 100-mm-diameter supporting post. Show the results on a volume element located at each ofthese points.Given: do := 100mm a := 3m po := 1500Pabo := 2m ho := 1mSolution:Px := −po⋅bo⋅ho Px = −3.00 kNInternal Force and Moment :ΣFx=0; Vx + Px = 0 Vx := −PxΣFy=0; Vy := 0ΣFz=0; Nz := 0Σ Μx=0; Mx := 0=0; My + Px⋅ (a + 0.5ho) = 0 My := −Px⋅ (a + 0.5ho)Σ Μy=0; Mz − Px⋅ (0.5bo) Mz := Px⋅ (0.5bo)Vx = 3 kN Vy = 0 kN Nz = 0 kNMx = 0 kN⋅m My = 10.5 kN⋅m Mz = −3 kN⋅mΣ ΜzSection Property : ro := 0.5⋅doA π ro:= ⋅ ⎛⎝2 A = 7853.98mm2I⎞⎠π4:= ⋅ ro4 I ⎛⎝= 4908738.52mm4J⎞⎠π2:= ⋅ ro4 J = 9817477.04mm4QA := 0 (since A' = 0)QB⎛⎝⎞⎠4ro3πA2⎛⎜⎝⎞⎠:= ⋅At A: xA := ro At B: xB := 0ρA := ro ρB := roNormal Stress:σ = N ± McσAIANzAMy⋅xAI:= + σBNzAMy⋅xBI:= +σA = 107.0MPa (T) Ans σB = 0.0MPa AnsShear Stress :τT⋅ρJV⋅QI⋅ t= + τAMz⋅ρAJ:= (since QA = 0) τBMz⋅ρBJVx⋅QBI⋅do:= +τA = 15.28MPa Ans τB = 14.77MPa Ans 851. Problem 8-50The sign is subjected to the uniform wind loading. Determine the stress components at points C and Don the 100-mm-diameter supporting post. Show the results on a volume element located at each ofthese points.Given: do := 100mm a := 3m po := 1500Pabo := 2m ho := 1mSolution:Px := −po⋅bo⋅ho Px = −3.00 kNInternal Force and Moment :ΣFx=0; Vx + Px = 0 Vx := −PxΣFy=0; Vy := 0ΣFz=0; Nz := 0Σ Μx=0; Mx := 0=0; My + Px⋅ (a + 0.5ho) = 0 My := −Px⋅ (a + 0.5ho)Σ Μy=0; Mz − Px⋅ (0.5bo) Mz := Px⋅ (0.5bo)Vx = 3 kN Vy = 0 kN Nz = 0 kNMx = 0 kN⋅m My = 10.5 kN⋅m Mz = −3 kN⋅mΣ ΜzSection Property : ro := 0.5⋅doA π ro:= ⋅ ⎛⎝2 A = 7853.98mm2I⎞⎠π4:= ⋅ I = 4908738.52mm4⎛⎝ro4 ⎞⎠J:= ⋅ J = 9817477.04mm4QC := 0 (since A' = 0)QDπ2⎛⎝ro4 ⎞⎠4ro3πA2⎛⎜⎝⎞⎠:= ⋅At C: xC := −ro At D: xD := 0ρC := ro ρD := roNormal Stress:σ = N ± McσCIANzAMy⋅xCI:= + σDNzAMy⋅xDI:= +σC = −107.0MPa (C) Ans σD = 0MPa AnsShear Stress :τT⋅ρJV⋅QI⋅ t= + τCMz⋅ρCJ:= (since QC = 0) τD−Mz⋅ρDJVx⋅QDI⋅do:= +τC = 15.28MPa Ans τD = 15.79MPa Ans 852. Problem 8-51The 18-mm-diameter shaft is subjected to the loading shown. Determine the stress components atpoint A. Sketch the results on a volume element located at this point. The journal bearing at C can exertonly force components Cy and Cz on the shaft, and the thrust bearing at D can exert force componentsDx, Dy, and Dz on the shaft.Given: do := 18mm L := 500mmax := 50mm ay := 200mmP := 600N a := 250mmSolution: ro := 0.5doSupport Reactions at C :Tx := 0 My := 0 Mz := 0Cx := 0 Cy := 0 Cz := PInternal Force and Moment at A :N := 0 Vy := 0 Vz := CzTx := 0 My := −Cz⋅a Mz := 0Section Property :A π roIy:= ⋅ 4 Iz := Iy π4ro:= ⋅ 2QA := 0 Jπ2:= ⋅ 4roNormal Stress: σNAMz⋅yIz−My⋅zIy= +At A: yA := 0 zA := roσANAMz⋅yAIz−My⋅zAIy:= + σA = −262.0MPa (C) AnsShear Stress : τV⋅QI⋅b=At A: b := 0 τAVz⋅QAIy⋅b:= τA = 0.00 Ans 853. Problem 8-52Solve Prob. 8-51 for the stress components at point B.Given: do := 18mm L := 500mmax := 50mm ay := 200mmP := 600N a := 250mmSolution: ro := 0.5doSupport Reactions at C :Tx := 0 My := 0 Mz := 0Cx := 0 Cy := 0 Cz := PInternal Force and Moment at B :N := 0 Vy := 0 Vz := CzTx := 0 My := −Cz⋅a Mz := 0Section Property :A π roIy:= ⋅ 4 Iz := Iy π4ro:= ⋅ 2:= ⋅ 4 QBJπ2ro4ro3ππ⋅ 2ro⎛⎜⎝2 ⎞⎠:= ⋅Normal Stress: σNAMz⋅yIz−My⋅zIy= +At B: yB := ro zB := 0σANAMz⋅yBIz−My⋅zBIy:= + σA = 0.0MPa AnsShear Stress : τV⋅QI⋅b=At B: b 2ro:= ⋅ τBVz⋅QBIy⋅b:= τB = 3.14MPa Ans 854. Problem 8-53The solid rod is subjected to the loading shown. Determine the state of stress developed in the materialat point A, and show the results on a differential volume element at this point.Given: ro := 30mm a := 150mmPx := −10kN Py := 1kNPz := 15kN Tx := 0.2kN⋅mSolution:Internal Force and Moment : At Section AΣFx=0; Nx + Px = 0 Nx := −PxΣFy=0; Vy := 0ΣF Nx = 10 kN z=0; Vz := 0Vy 0 kN = Σ Μx=0; Mx + Tx = 0 Mx := −TxVz 0 kN = Σ Μy=0; My := 0Mx 0.2 − kN m ⋅ = Σ Μz=0; Mz − Px⋅ (−ro) = 0 Mz := −Px⋅ roMy = 0 kN⋅mSection Property : ρ := ro Mz = 0.3 kN⋅mA π ro:= ⋅ ⎛⎝2 A = 2827.43mm2I⎞⎠π4:= ⋅ I = 636172.51mm4⎛⎝ro4 ⎞⎠J:= ⋅ J = 1272345.02mm4QA_zπ2⎛⎝ro4 ⎞⎠4ro3πA2⎛⎜⎝⎞⎠:= ⋅ QA_y := 0 (since A' = 0)M zyyM yNormal Stress: σ = x − z+zIINAyA := −ro zA := 0 σANxAMz⋅yAI−My⋅zAI:= + σA = 17.7MPa (T) AnsShear Stress :The transverse shear stress in the z and y directions and the torsional shear stress can beobtained using the shear formula for τv and the torsional formula for τt respectively.τvV⋅QI⋅ t= τv_z := 0 (since Vz= 0)τtT⋅ρJ= τtMx⋅ρJ:= τt = −4.72MPaτxz := τv_z − τt τxz = 4.716MPa Ansτxy := 0 Ansτyz := 0 Ans 855. Problem 8-54The solid rod is subjected to the loading shown. Determine the state of stress at point B, and show theresults on a differential volume element at this point.Given: ro := 30mm a := 150mmPx := −10kN Py := 10kNPz := 15kN Tx := 0.2kN⋅mSolution:Internal Force and Moment : At Section BΣFx=0; Nx + Px = 0 Nx := −PxΣFy=0; Vy + Py = 0 Vy := −PyΣFz=0; Vz := 0 Nx = 10 kNΣ Μx=0; Mx − Py⋅ ro + Tx = 0 Mx := Py⋅ ro − Tx Vy = −10 kNΣ Μy=0; My := 0 Vz = 0 kN=0; Mz − Px⋅ (−ro) + Py⋅ (−a) = 0 Mz := −Px⋅ ro + Py⋅a Mx = 0.1 kN⋅mΣ ΜzSection Property : ρ r My = 0 kN⋅m := oMz = 1.8 kN⋅mA := π ⋅ ro2 A = 2827.43mm2⎛⎝I⎞⎠π4:= ⋅ I = 636172.51mm4⎛⎝ro4 ⎞⎠J:= ⋅ J = 1272345.02mm4QB_zπ2⎛⎝ro4 ⎞⎠4ro3πA2⎛⎜⎝⎞⎠:= ⋅ QB_y := 0 (since A' = 0)M zyyM yNormal Stress: σ = x − z+zIINAyB := ro zB := 0 σBNxAMz⋅yBI−My⋅zBI:= + σB = −81.3MPa (C) AnsShear Stress :The transverse shear stress in the z and y directions and the torsional shear stress can beobtained using the shear formula for τv and the torsional formula for τt respectively.τvV⋅QI⋅ t= τv_z := 0 (since Vz= 0)τtT⋅ρJ= τtMx⋅ρJ:= τt = 2.36MPaτxz := τv_z + τt τxz = 2.358MPa Ansτxy := 0 Ansτyz := 0 Ans 856. Problem 8-55The solid rod is subjected to the loading shown. Determine the state of stress at point C, and show theresults on a differential volume element at this point.Given: ro := 30mm a := 150mmPx := −10kN Py := 10kNPz := 15kN Tx := 0.2kN⋅mSolution:Internal Force and Moment : At Section AΣFx=0; Nx + Px = 0 Nx := −PxΣFy=0; Vy + Py = 0 Vy := −PyΣFz=0; Vz + Pz = 0 Vz := −Pz Nx = 10 kNΣ Μx=0; Mx − Py⋅ ro + Pz⋅ ro + Tx = 0 Mx := Py⋅ ro − Pz⋅ ro − Tx Vy = −10 kNΣ Μy=0; My − Pz⋅ (−a) = 0 My := −Pz⋅a Vz = −15 kN=0; Mz − Px⋅ (−ro) + Py⋅ (−3a) = 0 Mz := −Px⋅ ro + 3Py⋅a Mx = −0.35 kN⋅mΣ ΜzSection Property : ρ := ro My = −2.25 kN⋅mMz = 4.8 kN⋅m A π ro:= ⋅ ⎛⎝2 A = 2827.43mm2I⎞⎠π4:= ⋅ I = 636172.51mm4⎛⎝ro4 ⎞⎠J:= ⋅ J = 1272345.02mm4QC_yπ2⎛⎝ro4 ⎞⎠4ro3πA2⎛⎜⎝⎞⎠:= ⋅ QC_z := 0 (since A' = 0)M zyyM yNormal Stress: σ = x − z+zIINAyC := 0 zC := ro σCNxAMz⋅yCI−My⋅zCI:= + σC = −102.6MPa (C) AnsShear Stress :The transverse shear stress in the z and y directions and the torsional shear stress can beobtained using the shear formula for τv and the torsional formula for τt respectively.τvV⋅QI⋅ t= τv_yVy⋅QC_yI 2ro⋅ ( ):= τv_y = −4.72MPaτtT⋅ρJ= τtMx⋅ρJ:= τt = −8.25MPaτxy := τv_y − τt τxy = 3.54MPa Ansτxz := 0 Ansτyz := 0 Ans 857. Problem 8-56The 25-mm-diameter rod is subjected to the loads shown. Determine the state of stress at point A, andshow the results on a differential element located at this point.Given: ax := −200mm az := 75mm do := 25mmPx := −375N Py := −400N Pz := 500NSolution: ro := 0.5doInternal Force and Moment :ΣFx=0; Nx + Px = 0 Nx := −PxΣFy=0; Vy + Py = 0 Vy := −PyΣFz=0; Vz + Pz = 0 Vz := −PzΣ Μx=0; Tx − Py⋅ (az) = 0 Tx := Py⋅ (az)=0; My − Pz⋅ (ax) + Px⋅ (az) = 0 My := Pz⋅ (ax) − Px⋅ (az)Σ Μy=0; Mz + Py⋅ (ax) = 0 Mz := −Py⋅ (ax)Σ Μz:= ⋅ 2Section Property : A π roIyπ4⎛⎝ro4 ⎞⎠:= ⋅ Iz := IyQA_y := 0 QA_z4ro3ππ⋅ 2ro⎛⎜⎝2 ⎞⎠:= ⋅Jπ2⎛⎝ro4 ⎞⎠:= ⋅Normal Stress: σNAMz⋅yIz−My⋅zIy= +At A: yA := ro zA := 0σANxAMz⋅yAIz−My⋅zAIy:= + σA = 52.9MPa (T) AnsShear Stress : τvV⋅QI⋅b= τtT⋅ρJ=At A: bz := 2⋅ ro ρ := roτv_zVz⋅QA_zIy⋅bz:= τtTx⋅ρJ:=τA_xz := τv_z + τt τA_xz = 11.14MPa AnsτA_xy := 0 Ans 858. Problem 8-57The 25-mm-diameter rod is subjected to the loads shown. Determine the state of stress at point B, andshow the results on a differential element located at this point.Given: ax := −200mm az := 75mm do := 25mmPx := −375N Py := −400N Pz := 500NSolution: ro := 0.5doInternal Force and Moment :ΣFx=0; Nx + Px = 0 Nx := −PxΣFy=0; Vy + Py = 0 Vy := −PyΣFz=0; Vz + Pz = 0 Vz := −PzΣ Μx=0; Tx − Py⋅ (az) = 0 Tx := Py⋅ (az)=0; My − Pz⋅ (ax) + Px⋅ (az) = 0 My := Pz⋅ (ax) − Px⋅ (az)Σ Μy=0; Mz + Py⋅ (ax) = 0 Mz := −Py⋅ (ax)Σ Μz:= ⋅ 2Section Property : A π roIyπ4⎛⎝ro4 ⎞⎠:= ⋅ Iz := IyQB_z := 0 QB_y4ro3ππ⋅ 2ro⎛⎜⎝2 ⎞⎠:= ⋅Jπ2⎛⎝ro4 ⎞⎠:= ⋅Normal Stress: σNAMz⋅yIz−My⋅zIy= +At B: yB := 0 zB := roσBNxAMz⋅yBIz−My⋅zBIy:= + σB = −46.1MPa (C) AnsShear Stress : τvV⋅QI⋅b= τtT⋅ρJ=At B: by := 2⋅ ro ρ := roτv_yVy⋅QB_yIz⋅by:= τtTx⋅ρJ:=τB_xy := τv_y − τt τB_xy = 10.86MPa AnsτB_xz := 0 Ans 859. Problem 8-58The crane boom is subjected to the load of 2.5 kN. Determine the state of stress at points A and B.Show the results on a differential volume element located at each of these points.Given: b := 75mm d := 76mm t := 12mmah := 1.5m av := 2.4m P := 2.5kNv := 4 h := 3 r := 5Solution:Internal Force and Moment :+ ΣFx=0; −V Phr⎛⎜⎝⎞⎠+ ⋅ = 0 V Phr⎛⎜⎝⎞⎠:= ⋅+ ΣFy=0; −N Pvr⎛⎜⎝⎞⎠− ⋅ = 0 N −Pvr⎛⎜⎝⎞⎠:= ⋅+ ΣΜO=0; M Phr⎛⎜⎝⎞⎠− ⋅ ⋅av Pvr⎛⎜⎝⎞⎠− ⋅ ⋅ah = 0M Pvr⎛⎜⎝⎞⎠⋅ ⋅ah Phr⎛⎜⎝⎞⎠:= + ⋅ ⋅avSection Property : D := d + 2tA := b⋅D − t⋅d I112⋅b⋅D3112:= − ⋅ (b − t)⋅d3QA := 0 QB := 0 (since A' = 0)Normal Stress: σNAM⋅cI= +cA := 0.5D σANAM⋅cAI:= + σA = 83.34MPa (T) AnscB := −0.5D σBNAM⋅cBI:= + σB = −83.95MPa (C) AnsShear Stress : τV⋅QI⋅b=τAV⋅QAI⋅b:= τA = 0MPa AnsτBV⋅QBI⋅b:= τB = 0MPa Ans 860. Problem 8-59The masonry pier is subjected to the 800-kN load. Determine the equation of the line y = f (x) alongwhich the load can be placed without causing a tensile stress in the pier. Neglect the weight of the pier.Given: a := 1.5m b := 2.25m Pz := −800kNxA := −a yA := −bSolution:Section Property :A := (2a)⋅ (2⋅b) A = 13.5m2Ix112:= ⋅ (2a) (2⋅b)3 Ix = 22.78125m4Iy112:= ⋅ (2b) (2⋅a)3 Iy = 10.125m4Force and Moment :Mx = Pz⋅yMy = −Pz⋅xNormal Stress: Require σA := 0σAPzAMx⋅yAIx+My⋅xAIy= −0PzA(Pz⋅y)⋅yAIx+(Pz⋅x)⋅xAIy= +01AyAIx⎛⎜⎝⎞⎠+ ⋅yxAIy⎛⎜⎝⎞⎠= + ⋅x014a⋅b34a⋅b2⎛⎜⎝⎞⎠− ⋅y34b⋅a2⎛⎜⎝⎞⎠= − ⋅xyb3ba= − ⋅xy = 0.75 − 1.5⋅x Ans 861. Problem 8-60The masonry pier is subjected to the 800-kN load. If x = 0.25 m and y = 0.5 m, determine the normalstress at each corner A, B, C, D (not shown) and plot the stress distribution over the cross section.Neglect the weight of the pier.Unit Used: kPa := 103PaGiven: a := 1.5m b := 2.25m Pz := −800kNxA := −a xB := a xC := a xD := −ayA := −b yB := −b yC := b yD := bx := 0.25m y := 0.5mSolution:Section Property :A := (2a)⋅ (2⋅b) A = 13.5m2Ix112:= ⋅ (2a) (2⋅b)3 Ix = 22.78125m4Iy112:= ⋅ (2b) (2⋅a)3 Iy = 10.125m4Force and Moment :Mx := Pz⋅y Mx = −400 kN⋅mMy := −Pz⋅x My = 200 kN⋅mNormal Stress: σPzAMx⋅yIx+My⋅xIy= −σAPzAMx⋅yAIx+My⋅xAIy:= − σA = 9.877 kPa (T) AnsσBPzAMx⋅yBIx+My⋅xBIy:= − σB = −49.38 kPa (C) AnsσCPzAMx⋅yCIx+My⋅xCIy:= − σC = −128.4 kPa (C) AnsσDPzAMx⋅yDIx+My⋅xDIy:= − σD = −69.1 kPa (C) Ans 862. Problem 8-61The symmetrically loaded spreader bar is used to lift the 10-kN (~1-tonne) tank. Determine the state ofstress at points A and B, and indicate the results on a differential volume elements.Given: b := 25mm d := 50mm a := 0.45mL := 1.2m W := 10kN θ := 30degSolution:Support Reactions :+ ΣFy=0; −W + 2F⋅cos(θ) = 0 FW2⋅cos(θ):=Internal Force and Moment :+ ΣFx=0; F⋅ sin(θ) − N = 0 N := F⋅ sin(θ)+ ΣFy=0; V − F⋅cos(θ) = 0 V := F⋅cos(θ)+ ΣΜB=0; M − F⋅cos(θ)⋅a = 0 M := F⋅cos(θ)⋅aSection Property :A := b⋅d I112:= ⋅b⋅d3QB := (0.5⋅d⋅b)⋅ (0.25d)QA := 0 (since A' = 0)Normal Stress: σNAM⋅cI= +cA := 0.5d σANAM⋅cAI:= + σA = 218.31MPa (T) AnscB := 0 σBNAM⋅cBI:= + σB = 2.31MPa (T) AnsShear Stress : τV⋅QI⋅b=τAV⋅QAI⋅b:= τA = 0MPa AnsτBV⋅QBI⋅b:= τB = 6.00MPa Ans 863. Problem 8-62A post having the dimensions shown is subjected to the bearing load P. Specify the region to whichthis load can be applied without causing tensile stress to be developed at points A, B, C, and D. 864. Problem 8-63The man has a mass of 100 kg and center of mass at G. If he holds himself in the position shown,determine the maximum tensile and compressive stress developed in the curved bar at section a-a. Heis supported uniformly by two bars, each having a diameter of 25 mm. Assume the floor is smooth.Given: Ri := 150mm do := 25mm mo := 100kge := 0.3m a := 0.35m b := 1mSolution:Equilibrium: For the man.+ ΣΜtoe=0; (mo⋅g)⋅b − 2P⋅ (a + b) = 0P (mo⋅g) b:= ⋅P = 0.3632 kN2(a + b)Section Property : ro := 0.5do rc := Ri + ro:= ⋅ 2 A = 490.87mm2= ∫A π roI Σ dA IA_r 2π rc rcA_r A r⎛⎝− 2 − ro2 ⎞⎠:= ⋅IA_r = 3.0252mmRAIA_r:= R = 162.259mmInternal Force and Moment :As shown on BFBD. The internal moment must becomputed about the neutral axis. M is negative since ittends to decrease the bar's radius of curvature.N := −P N = −0.3632 kNM := −P⋅ (R + e) M = −0.16790 kN⋅mMaximum Normal Stress:For tensile stress, σtNA⋅ ( − )M R r2A⋅ r1⋅ (rc − R)= +r2 := rc + ro σtNA⋅ ( − )M R r2:= + σt = 102.7MPa (T) AnsA⋅ r2⋅ (rc − R)For compressive stress, σcNA⋅ ( − )M R r1A⋅ r1⋅ (rc − R)= +r1 := rc − ro σcNA⋅ ( − )M R r1:= + σc = −116.9MPa (C) AnsA⋅ r1⋅ (rc − R) 865. Problem 8-64The block is subjected to the three axial loads shown. Determine the normal stress developed at pointsA and B. Neglect the weight of the block.Given: b := 100mm b' := 50mmd := 75mm d' := 125mmP1 := 500N P2 := 1250N P3 := 250NSolution: B := b + 2b' D := d + 2d'Internal Force and Moment :+ ΣFz=0; −N − P1 − P2 − P3 = 0N := −(P1 + P2 + P3)+ Σ Μx=0; Mx − P1⋅ (0.5d) − P2⋅ (0.5d) + P3⋅ (0.5D) = 0Mx P1 0.5 d ⋅ ( ) ⋅ P2 0.5 d ⋅ ( ) ⋅ + ⎡⎣⎤⎦:= − P3⋅ (0.5⋅D)+ Σ Μy=0; My + P1⋅ (0.5B) − P2⋅ (0.5B) − P3⋅ (0.5⋅b) = 0My := −P1⋅ (0.5⋅B) + P2⋅ (0.5⋅B) + P3⋅ (0.5⋅b)Section Property :A := B⋅D − 4b'⋅d' Ix112⋅b⋅D3212:= + ⋅b'⋅d3Iy112⋅d⋅B3212:= + ⋅d'⋅b3Normal Stress: σNAMx⋅yIx−My⋅xIy= −At A: xA := 0.5B yA := −0.5dσANAMx⋅yAIx−My⋅xAIy:= − σA = −0.1703MPa (C) AnsAt B: xB := 0.5b yB := −0.5DσBNAMx⋅yBIx−My⋅xBIy:= − σB = −0.0977MPa (C) Ans 866. Problem 8-65If P = 15 kN, plot the distribution of stress acting over the cross section a-a of the offset link.Given: ho := 50mm P := 15kNto := 10mm a := 30mmSolution:Section Property :A := ho⋅ to A = 500mm2I112:= 3 I = 104166.67mm4⋅ to hoMoment :M := P⋅ (a + 0.5ho) M = 0.825 kN⋅mNormal Stress:σ = N ± McσAIAPAM⋅ (0.5ho):= + σA = 228MPa (T) AnsIσBPAM⋅ (0.5ho):= − σB = −168MPa (C) AnsIyσAho − yσB= y hoσAσA + σB:= ⋅ y = 28.79mm 867. Problem 8-66Determine the magnitude of the load P that will cause a maximum normal stress σmax = 200 MPa of inthe link at section a-a.Given: ho := 50mm a := 30mmto := 10mm σallow := 200MPaSolution:Section Property :A := ho⋅ to A = 500mm2I112:= 3 I = 104166.67mm4⋅ to hoMoment :M = P⋅ (a + 0.5ho)Normal Stress: The maximum normal stress occurs at A.σ = N ± McσAIAPAM⋅ (0.5ho)I= +σallowPAP⋅ (a + 0.5ho)⋅ (0.5ho)I= +P (σallow⋅A) II + A⋅ (a + 0.5ho)⋅ (0.5ho):= ⋅P = 13.16 kN Ans 868. Problem 8-67Air pressure in the cylinder is increased by exerting forces P = 2 kN on the two pistons, each having aradius of 45 mm. If the cylinder has a wall thickness of 2 mm, determine the state of stress in the wallof the cylinder.Given: t := 2mm P := 2kNri := 45mmSolution::= 2 pA πriPA:=p = 0.3144MPaHoop Stress : αrit:= α = 22.50Since α > 10. then thin-wall analysis can be used.σ1p⋅ rit:= σ1 = 7.07MPa AnsLongitudinal Stress :σ2 := 0 AnsThe pressure p is supported by the surface of thepistons in the longitudinal direction. 869. Problem 8-68Determine the maximum force P that can be exerted on each of the two pistons so that thecircumferential stress component in the cylinder does not exceed 3 MPa. Each piston has a radius of45 mm and the cylinder has a wall thickness of 2 mm.Given: t := 2mm σallow := 3MPari := 45mmSolution::= 2 pA πriPA=Hoop Stress : αrit:= α = 22.50Since α > 10. then thin-wall analysis can be used.σ1p⋅ rit= σallowP⋅ riA⋅ t=Pσallowri:= ⋅A⋅ tP = 0.848 kN Ans 870. Problem 8-69The screw of the clamp exerts a compressive force of 2.5 kN on the wood blocks. Determine themaximum normal stress developed along section a-a. The cross section there is rectangular, 18 mm by12 mm.Given: b := 12mm d := 18mma := 100mm P := 2.5kNSolution:Internal Force and Moment :N := PM := P⋅aSection Property :A := b⋅d I112:= ⋅b⋅d3Normal Stress: σNAM⋅cI= +cmax := 0.5d σmaxNAM⋅cmaxI:= +σmax = 397.4MPa (T) Ans 871. Problem 8-70The wall hanger has a thickness of 6 mm and is used to support the vertical reactions of the beam thatis loaded as shown. If the load is transferred uniformly to each strap of the hanger, determine the stateof stress at points C and D of the strap at B. Assume the vertical reaction F at this end acts in thecenter and on the edge of the bracket as shown.Given: t := 6mm d := 50mm P := 50kNL1 := 1.2m L2 := 1.8m w 30kNm:=Solution: L := L1 + L2Support Reactions : Given+ ΣFy=0; FA − P − w⋅L2 + FB = 0ΣΜB=0; FA⋅L − P⋅ (L − 0.5L1) − w⋅L2⋅ (0.5L2) = 0Guess FA := 1kN FB := 1kNFAFB⎛⎜⎜⎝⎞⎠:= Find(FA, FB)FAFB⎛⎜⎜⎝⎞⎠56.2047.80⎛⎜⎝⎞⎠= kNSection Property :A := 2t⋅d I212:= ⋅ t⋅d3At Section CD: P := FB M := P⋅ (0.5d) V := 0Stresses: σPAM⋅yI= + τV⋅QI⋅b=At C: yC := 0 σCPAM⋅yCI:= +σC = 79.67MPa (T) AnsτC := 0 AnsAt D: yD := −0.5dσDPAM⋅yDI:= +σD = −159.33MPa (C) AnsτD := 0 Ans 872. Problem 8-71The support is subjected to the compressive load P. Determine the absolute maximum and minimumnormal stress acting in the material. 873. Problem 8-72The support has a circular cross section with a radius that increases linearly with depth. If it issubjected to the compressive load P, determine the maximum and minimum normal stress acting in thematerial. 874. Problem 8-73The cap on the cylindrical tank is bolted to the tank along the flanges. The tank has an inner diameterof 1.5 m and a wall thickness of 18 mm. If the largest normal stress is not to exceed 150 MPa,determine the maximum pressure the tank can sustain. Also, compute the number of bolts required toattach the cap to the tank if each bolt has a diameter of 20 mm. The allowable stress for the bolts is(σallow)b = 180 MPa.Given: t := 18mm di := 1.5m σallow := 150MPadb := 20mm σb.allow := 180MPaSolution: ri := 0.5diHoop Stress : αrit:= α = 41.67Since α > 10. then thin-wall analysis can be used.σ1p⋅ rit= σallowp⋅ rit=pσallow⋅ tri:=p = 3.6MPa Ans:= 2Force Equilibrium for the Cap : A πri+ ΣFy=0; p⋅A − Fb = 0Fb := p⋅AFb = 6361.73 kNAllowable Normal Stress for Bolts : Abπ4:= 2dbσb.allowFbn⋅Ab=nFb(σb.allow)⋅Ab:=n = 112.5Use n := 113 Ans 875. Problem 8-74The cap on the cylindrical tank is bolted to the tank along the flanges. The tank has an inner diameterof 1.5 m and a wall thickness of 18 mm. If the pressure in the tank is p = 1.20 MPa, determine theforce in the 16 bolts that are used to attach the cap to the tank. Also, specify the state of stress in thewall of the tank.Given: t := 18mm di := 1.5m p := 1.20MPan := 16Solution: ri := 0.5diHoop Stress : αrit:= α = 41.67Since α > 10. then thin-wall analysis can be used.σ1p⋅ rit:=σ1 = 50MPa AnsLongitudinal Stress :σ2p⋅ ri2⋅ t:=σ2 = 25MPa Ans:= 2Force Equilibrium for the Cap : A πri+ ΣFy=0; p⋅A − 16⋅Fb = 0Fbp⋅A16:=Fb = 132.5 kN Ans 876. Problem 8-75The crowbar is used to pull out the nail at A. If a force of 40 N is required, determine the stresscomponents in the bar at points D and E. Show the results on a differential volume element located ateach of these points. The bar has a circular cross section with a diameter of 12 mm. No slippingoccurs at B.Given: do := 12mm dA := 60mm hA := 75mma := 125mm dP := 300mmF := 40N hP := 300mm:= + 2Solution: rp dP2 hPSupport Reactions :+ ΣΜB=0; F⋅hA − P⋅ (rp) = 0PF⋅hArp:=Internal Force and Moment :+ ΣFx=0; N := 0+ ΣFy=0; V − P = 0 V := P+ ΣΜO=0; M − P⋅a = 0 M := P⋅aSection Property : ro := 0.5doA := π ⋅ ro2 Iπ4:= ⋅ 4roQE4ro3π⎛⎝0.5π ⋅ ro2 ⎞⎠:= ⋅QD := 0 (since A' = 0)Normal Stress: σNAM⋅cI= +cD := ro σDNAM⋅cDI:= + σD = 5.21MPa (T) AnscE := 0 σENAM⋅cEI:= + σE = 0.00MPa AnsShear Stress : τV⋅QI⋅b=bE := 2⋅ ro τEV⋅QEI⋅bE:= τE = 0.0834MPa AnsτD := 0 Ans 877. Problem 8-76The screw of the clamp exerts a compressive force of 2.5 kN on the wood blocks. Sketch the stressdistribution along section a-a of the clamp. The cross section there is rectangular, 18 mm by 12 mm.Given: b := 12mm d := 18mma := 100mm P := 2.5kNSolution:Internal Force and Moment :N := PM := P⋅aSection Property :A := b⋅d I112:= ⋅b⋅d3Normal Stress: σNAM⋅cI= +cmax := 0.5d σmaxNAM⋅cmaxI:= +σmax = 397.4MPa (T) Anscmin := 0.5d σminNAM⋅cminI:= −σmin = −374.2MPa (C) Ansyd − yσminσmax=yd σminσmin + σmax:=y = 8.73mm 878. Problem 8-77The clamp is made from members AB and AC, which are pin connected at A. If the compressive forceat C and B is 180 N, determine the state of stress at point F, and indicate the results on a differentialvolume element. The screw DE is subjected only to a tensile force along its axis.Given: h := 15mm t := 15mm P := 180Na := 30mm b := 40mmSolution:Support Reactions :+ ΣΜO=0; P⋅ (b + a) − FDE⋅ (a) = 0FDEb + aa:= ⋅PFDE = 0.420 kNInternal Force and Moment :+ ΣFy=0; N' := 0+ ΣFx=0; V + FDE − P = 0 V := P − FDE+ ΣΜO=0; M + P⋅ (b + 0.5a) − FDE⋅ (0.5a) = 0 M := FDE⋅ (0.5a) − P⋅ (b + 0.5a)M = −3.60 N⋅mSection Property:A := h⋅ t A = 225mm2I112:= ⋅ t⋅h3 I = 4218.75mm4QF := 0 (since A' = 0)Normal Stress: σN'AM⋅cI= +cF := 0.5⋅h σFN'AM⋅cFI:= +σF = −6.40MPa (C) AnsShear Stress : τV⋅QI⋅b=τFV⋅QFI⋅ t:= τF = 0MPa Ans 879. Problem 8-78The eye is subjected to the force of 250 N. Determine the maximum tensile and compressive stressesat section a-a. The cross section is circular and has a diameter of 6 mm. Use the curved-beam formulato compute the bending stress.Given: Ri := 30mm do := 6mm P := 0.250kNSolution: ro := 0.5do rc := Ri + roSection Property ::= ⋅ 2 A = 28.27mm2A π roI Σ dA IA_r 2π rc rc= ∫A_r A r⎛⎝− 2 − ro2 ⎞⎠:= ⋅IA_r = 0.8586mmRAIA_r:= R = 32.932mmInternal Force and Moment :As shown on BFBD. The internal moment must becomputed about the neutral axis. M is positive since ittends to increase the beam's radius of curvature.N := PM := P⋅RMaximum Normal Stress:For tensile stress, σtNA⋅ ( − )M R r1A⋅ r1⋅ (rc − R)= +r1 := rc − ro σtNA⋅ ( − )M R r1A⋅ r1⋅ (rc − R):= +σt = 425.3MPa (T) AnsFor compressive stress, σcNA⋅ ( − )M R r2A⋅ r1⋅ (rc − R)= +r2 := rc + ro σcNA⋅ ( − )M R r2A⋅ r2⋅ (rc − R):= +σc = −354.4MPa (C) Ans 880. Problem 8-79Solve Prob. 8-78 if the cross section is square, having dimensions of 6 mm by 6 mm.Given: Ri := 30mm do := 6mm P := 0.250kNSolution: ro := 0.5do rc := Ri + roSection Property ::= 2 A = 36.00mm2A doI Σ dA IA_r (do) ln= ∫A_r A rrc + rorc − ro⎛⎜⎝⎞⎠:= ⋅IA_r = 1.0939mmRAIA_r:= R = 32.91mmInternal Force and Moment :As shown on BFBD. The internal moment must becomputed about the neutral axis. M is positive since ittends to increase the beam's radius of curvature.N := PM := P⋅RMaximum Normal Stress:For tensile stress, σtNA⋅ ( − )M R r1A⋅ r1⋅ (rc − R)= +r1 := rc − ro σtNA⋅ ( − )M R r1A⋅ r1⋅ (rc − R):= +σt = 250.2MPa (T) AnsFor compressive stress, σcNA⋅ ( − )M R r2A⋅ r1⋅ (rc − R)= +r2 := rc + ro σcNA⋅ ( − )M R r2A⋅ r2⋅ (rc − R):= +σc = −208.4MPa (C) Ans 881. Problem 9-1Prove that the sum of the normal stresses σx+ σy= σx' + σy' is constant. See Figs. 9-2a and 9-2b.Solution:Stress Transformation Equations: Applying Eqs. 9-1 and 9-3 of the text.σx'σx + σy= + ⋅cos(2θ) + τxy⋅ sin(2θ) (1)2σx − σy2σy'σx + σy= − ⋅cos(2θ) − τxy⋅ sin(2θ) (2)2σx − σy2(1) + (2) :LHS = σx' + σy'RHSσx + σy= + RHS = σx + σy2σx + σy2Hence,σx' + σy' = σx + σy (Q.E.D.) 882. Problem 9-2The state of stress at a point in a member is shown on the element. Determine the stress componentsacting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.9.1.Given: σx := 3MPa σy := 5MPa τxy := −8MPa φ := 40degSolution: Set ΔA := m2 θ := 180deg + φForce Equilibrium: For the sectioned element,ΔAx := ΔA⋅cos(φ) ΔAy := ΔA⋅ sin(φ)Fxx := σx⋅ΔAx Fxy := τxy⋅ΔAxFyy := σy⋅ΔAy Fyx := τxy⋅ΔAyGiven+ ΣFx'=0; ΔFx' + Fxy⋅ sin(θ) + Fxx⋅cos(θ) + Fyx⋅cos(θ) + Fyy⋅ sin(θ) = 0+ ΣFy'=0; ΔFy' + Fxy⋅cos(θ) − Fxx⋅ sin(θ) − Fyx⋅ sin(θ) + Fyy⋅cos(θ) = 0Guess ΔFx' := 1kN ΔFy' := 1kNΔ Fx'Δ Fy'⎛⎜⎜⎝⎞⎠:= Find(ΔFx' , ΔFy')Δ Fx'Δ Fy'⎛⎜⎜⎝⎞⎠⎛⎜⎝⎞−4052.11−404.38= kN⎠Normal and Shear Stress: σA 0FA⎛⎜⎝⎞⎠lim→=σx'Δ Fx'ΔA:= σx' = −4.052MPa Ansτx'y'Δ Fy'ΔA:= τx'y' = −0.404MPa AnsThe negative signs indicate that the sense of σx' and τx'y' are opposite to that shown in FBD. 883. Problem 9-3The state of stress at a point in a member is shown on the element. Determine the stress componentsacting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.9.1.Given: σx := −0.200MPa σy := 0.350MPaφ := 50deg τxy := 0MPaSolution: Set ΔA := m2 θ := 180deg + φForce Equilibrium: For the sectioned element,ΔAx := ΔA⋅cos(φ) ΔAy := ΔA⋅ sin(φ)Fxx := σx⋅ΔAx Fxy := τxy⋅ΔAx Fxy = 0.00Fyy := σy⋅ΔAy Fyx := τxy⋅ΔAy Fyx = 0.00Given+ ΣFx'=0; ΔFx' + Fxy⋅ sin(θ) + Fxx⋅cos(θ) + Fyx⋅cos(θ) + Fyy⋅ sin(θ) = 0+ ΣFy'=0; ΔFy' + Fxy⋅cos(θ) − Fxx⋅ sin(θ) − Fyx⋅ sin(φ) + Fyy⋅cos(θ) = 0Guess ΔFx' := 1kN ΔFy' := 1kNΔ Fx'Δ Fy'⎛⎜⎜⎝⎞⎠:= Find(ΔFx' , ΔFy')Δ Fx'Δ Fy'⎛⎜⎜⎝⎞⎠122.75270.82⎛⎜⎝⎞⎠= kNNormal and Shear Stress: σA 0FA⎛⎜⎝⎞⎠lim→=σx'Δ Fx'ΔA:= σx' = 0.123MPa Ansτx'y'Δ Fy'ΔA:= τx'y' = 0.271MPa Ans 884. Problem 9-4The state of stress at a point in a member is shown on the element. Determine the stress componentsacting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.9.1.Given: σx := −0.650MPa σy := 0.400MPaφ := 60deg τxy := 0MPaSolution: Set ΔA := m2 θ := −(90deg + φ)Force Equilibrium: For the sectioned element,ΔAx := ΔA⋅ sin(φ) ΔAy := ΔA⋅cos(φ)Fxx := σx⋅ΔAx Fxy := τxy⋅ΔAx Fxy = 0.00Fyy := σy⋅ΔAy Fyx := τxy⋅ΔAy Fyx = 0.00Given+ ΣFx'=0; ΔFx' + Fxy⋅ sin(θ) + Fxx⋅cos(θ) + Fyx⋅cos(θ) + Fyy⋅ sin(θ) = 0+ ΣFy'=0; ΔFy' + Fxy⋅cos(θ) − Fxx⋅ sin(θ) − Fyx⋅ sin(θ) + Fyy⋅cos(θ) = 0Guess ΔFx' := 1kN ΔFy' := 1kNΔ Fx'Δ Fy'⎛⎜⎜⎝⎞⎠:= Find(ΔFx' , ΔFy')Δ Fx'Δ Fy'⎛⎜⎜⎝⎞⎠−387.50454.66⎛⎜⎝⎞⎠= kNNormal and Shear Stress: σA 0FA⎛⎜⎝⎞⎠lim→=σx'Δ Fx'ΔA:= σx' = −0.387MPa Ansτx'y'Δ Fy'ΔA:= τx'y' = 0.455MPa AnsThe negative signs indicate that the sense of σx' is opposite to that shown in FBD. 885. Problem 9-5The state of stress at a point in a member is shown on the element. Determine the stress componentsacting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.9.1.Given: σx := −60MPa σy := −50MPaφ := 30deg τxy := 28MPaSolution: Set ΔA := m2 θ := 180deg + φForce Equilibrium: For the sectioned element,ΔAx := ΔA⋅cos(φ) ΔAy := ΔA⋅ sin(φ)Fxx := σx⋅ΔAx Fxy := τxy⋅ΔAx Fxy = 24248.71 kNFyy := σy⋅ΔAy Fyx := τxy⋅ΔAy Fyx = 14000.00 kNGiven+ ΣFx'=0; ΔFx' + Fxy⋅ sin(θ) + Fxx⋅cos(θ) + Fyx⋅cos(θ) + Fyy⋅ sin(θ) = 0+ ΣFy'=0; ΔFy' + Fxy⋅cos(θ) − Fxx⋅ sin(θ) − Fyx⋅ sin(θ) + Fyy⋅cos(θ) = 0Guess ΔFx' := 1kN ΔFy' := 1kNΔ Fx'Δ Fy'⎛⎜⎜⎝⎞⎠:= Find(ΔFx' , ΔFy')Δ Fx'Δ Fy'⎛⎜⎜⎝⎞⎠−33251.2918330.13⎛⎜⎝⎞⎠= kNNormal and Shear Stress: σA 0FA⎛⎜⎝⎞⎠lim→=σx'Δ Fx'ΔA:= σx' = −33.251MPa Ansτx'y'Δ Fy'ΔA:= τx'y' = 18.330MPa AnsThe negative signs indicate that the sense of σx' is opposite to that shown in FBD. 886. Problem 9-6The state of stress at a point in a member is shown on the element. Determine the stress componentsacting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.9.1.Given: σx := 90MPa σy := 50MPaτxy := −35MPa φ := 60degSolution: Set ΔA := m2 θ := −(90deg + φ)Force Equilibrium: For the sectioned element,ΔAx := ΔA⋅ sin(φ) ΔAy := ΔA⋅cos(φ)Fxx := σx⋅ΔAx Fxy := τxy⋅ΔAxFyy := σy⋅ΔAy Fyx := τxy⋅ΔAyGiven+ ΣFx'=0; ΔFx' + Fxy⋅ sin(θ) + Fxx⋅cos(θ) + Fyx⋅cos(θ) + Fyy⋅ sin(θ) = 0+ ΣFy'=0; ΔFy' + Fxy⋅cos(θ) − Fxx⋅ sin(θ) − Fyx⋅ sin(θ) + Fyy⋅cos(θ) = 0Guess ΔFx' := 1kN ΔFy' := 1kNΔ Fx'Δ Fy'⎛⎜⎜⎝⎞⎠:= Find(ΔFx' , ΔFy')Δ Fx'Δ Fy'⎛⎜⎜⎝⎞⎠⎛⎜⎝⎞49689.11−34820.51= kN⎠Normal and Shear Stress: σA 0FA⎛⎜⎝⎞⎠lim→=σx'Δ Fx'ΔA:= σx' = 49.69MPa Ansτx'y'Δ Fy'ΔA:= τx'y' = −34.82MPa AnsThe negative signs indicate that the sense of τx'y' isopposite to that shown in FBD. 887. Problem 9-7Solve Prob. 9-2 using the stress-transformation equations developed in Sec. 9.2.Given: σx := 5MPa σy := 3MPa τxy := 8MPa φ := 40degSolution: θ := 90deg + φNormal Stress:σx':= + ⋅cos(2θ) + τxy⋅ sin(2θ)σx' = −4.05MPa AnsThe negative signs indicate that the sense of σx' is a compressive stress.Shear Stress:σx + σy2σx − σy2τx'y'σx − σy:= − ⋅ sin(2θ) + τxy⋅cos(2θ)2τx'y' = −0.404MPa AnsThe negative signs indicate that the sense of τx'y' is in the -y' direction. 888. Problem 9-8Solve Prob. 9-4 using the stress-transformation equations developed in Sec. 9.2.Given: σx := −0.650MPa σy := 0.400MPaφ := 60deg τxy := 0MPaSolution: θ := 90deg − φNormal Stress:σx'σx + σy:= + ⋅cos(2θ) + sin(2θ)2τxy⋅ σx' = −0.387MPa AnsThe negative signs indicate that the sense of σx' is a compressive stress.Shear Stress:σx − σy2τx'y'σx − σy:= − ⋅ sin(2θ) + τxy⋅cos(2θ)2τx'y' = 0.455MPa Ans 889. Problem 9-9Solve Prob. 9-6 using the stress-transformation equations developed in Sec. 9.2. Show the result on asketch.Given: σx := 90MPa σy := 50MPaτxy := −35MPa φ := 60degSolution: θ := −(90deg + φ)Normal Stress:σx':= + ⋅cos(2θ) + τxy⋅ sin(2θ)σx' = 49.69MPa Ansσx + σy2σx − σy2Shear Stress:τx'y'σx − σy:= − ⋅ sin(2θ) + τxy⋅cos(2θ)2τx'y' = −34.82MPa AnsThe negative signs indicate that the sense of τx'y' is in the -y' direction. 890. Problem 9-10Determine the equivalent state of stress on an element if the element is oriented 30° counterclockwisefrom the element shown. Use the stress-transformation equations.Unit Used: kPa := 1000PaGiven: σx := 0kPa σy := −300kPaθ := 30deg τxy := 950kPaSolution:Normal Stress:σx'σx + σy:= + ⋅cos(2θ) + τxy⋅ sin(2θ)2σx' = 747.7 kPa Ansσx − σy2σy':= − ⋅cos(2θ) − τxy⋅ sin(2θ)σy' = −1047.7 kPa Ansσx + σy2σx − σy2Shear Stress:τx'y'σx − σy:= − ⋅ sin(2θ) + τxy⋅cos(2θ)2τx'y' = 345.1 kPa Ans 891. Problem 9-11Determine the equivalent state of stress on an element if the element is oriented 60° clockwise from theelement shown.Given: σx := 0.300MPa σy := 0MPaθ := −60deg τxy := 0.120MPaSolution:Normal Stress:σx'σx + σy:= + ⋅cos(2θ) + τxy⋅ sin(2θ)2σx' = −0.0289MPa Ansσx − σy2σy':= − ⋅cos(2θ) − τxy⋅ sin(2θ)σy' = 0.329MPa Ansσx + σy2σx − σy2Shear Stress:τx'y'σx − σy:= − ⋅ sin(2θ) + τxy⋅cos(2θ)2τx'y' = 0.0699MPa Ans 892. Problem 9-12Solve Prob. 9-6 using the stress-transformation equations.Given: σx := 90MPa σy := 50MPaφ := 60deg τxy := −35MPaSolution: θ := −(90deg + φ)Normal Stress:σx':= + ⋅cos(2θ) + τxy⋅ sin(2θ)σx' = 49.69MPa Ansσx + σy2σx − σy2Shear Stress:τx'y'σx − σy:= − ⋅ sin(2θ) + τxy⋅cos(2θ)2τx'y' = −34.82MPa Ans 893. Problem 9-13The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) themaximum in-plane shear stress and average normal stress at the point. Specify the orientation of theelement in each case.Given: σx := 45MPa σy := −60MPa τxy := 30MPaSolution:(a) Principal Stress:σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2 σ1 = 52.97MPaAnsσ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2 σ2 = −67.97MPaAnsOrientation of Principal Stress:tan(2θp) 2τxy= θpσx − σy12atan2τxyσx − σy⎛⎜⎝⎞⎠:= θ'p := θp − 90degθp = 14.87 deg θ'p = −75.13 degUse Eq. 9-1 to determine the principal plane of σ1 and σ2.σx + σyσx':= + ⋅cos(2θp) + sin(2θp)2τxy⋅ σx' = 52.97MPaTherefore, θp1 := θp θp1 = 14.87 deg Ansσx − σy2θp2 := θ'p θp2 = −75.13 deg Ans(b)τmaxσx − σy2⎛⎜⎝⎞⎠2τxy:= + 2 τmax = 60.47MPa Ansσavgσx + σy:= σavg = −7.50MPa Ans2Orientation of Maximum In-plane Shear Stress:tan(2θs) σx − σy= − θs2τxy12atanσx − σy2τxy−⎛⎜⎝⎞⎠:= θ's := θs + 90degθs = −30.13 deg θ's = 59.87 degBy observation, in order to preserve equilibrium along AB, τmax has to act in thedirection shown in the figure. 894. Problem 9-14The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) themaximum in-plane shear stress and average normal stress at the point. Specify the orientation of theelement in each case.Given: σx := 180MPa σy := 0MPa τxy := −150MPaSolution:(a) Principal Stress:σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2 σ1 = 264.93MPa Ansσ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2 σ2 = −84.93MPa AnsOrientation of Principal Stress:tan(2θp) 2τxy= θpσx − σy12atan2τxyσx − σy⎛⎜⎝⎞⎠:= θ'p := θp + 90degθp = −29.52 deg θ'p = 60.48 degUse Eq. 9-1 to determine the principal plane of σ1 and σ2.σx + σyσx':= + ⋅cos(2θp) + sin(2θp)2τxy⋅ σx' = 264.93MPaTherefore, θp1 := θp θp1 = −29.52 deg Ansσx − σy2θp2 := θ'p θp2 = 60.48 deg Ans(b)τmaxσx − σy2⎛⎜⎝⎞⎠2τxy:= + 2 τmax = 174.93MPa Ansσavgσx + σy:= σavg = 90.00MPa Ans2Orientation of Maximum In-plane Shear Stress:tan(2θs) σx − σy= − θs2τxy12atanσx − σy2τxy−⎛⎜⎝⎞⎠:= θ's := θs − 90degθs = 15.48 deg θ's = −74.52 degBy observation, in order to preserve equilibrium along AB, τmax has to act in thedirection shown in the figure. 895. Problem 9-15The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) themaximum in-plane shear stress and average normal stress at the point. Specify the orientation of theelement in each case.Given: σx := −30MPa σy := 0MPa τxy := −12MPaSolution:(a) Principal Stress:σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2 σ1 = 4.21MPa Ansσ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2 σ2 = −34.21MPa AnsOrientation of Principal Stress:tan(2θp) 2τxy= θpσx − σy12atan2τxyσx − σy⎛⎜⎝⎞⎠:= θ'p := θp − 90degθp = 19.33 deg θ'p = −70.67 degUse Eq. 9-1 to determine the principal plane of σ1 and σ2.σx + σyσx':= + ⋅cos(2θp) + sin(2θp)2τxy⋅ σx' = −34.21MPaTherefore, θp1 := θ'p θp1 = −70.67 deg Ansσx − σy2θp2 := θp θp2 = 19.33 deg Ans(b)τmaxσx − σy2⎛⎜⎝⎞⎠2τxy:= + 2 τmax = 19.21MPa Ansσavgσx + σy:= σavg = −15.00MPa Ans2Orientation of Maximum In-plane Shear Stress:tan(2θs) σx − σy= − θs2τxy12atanσx − σy2τxy−⎛⎜⎝⎞⎠:= θ's := θs + 90degθs = −25.67 deg θ's = 64.33 degBy observation, in order to preserve equilibrium along AB, τmax has to act in thedirection shown in the figure. 896. Problem 9-16The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) themaximum in-plane shear stress and average normal stress at the point. Specify the orientation of theelement in each case.Given: σx := −200MPa σy := 250MPa τxy := 175MPaSolution:(a) Principal Stress:σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2 σ1 = 310.04MPaAnsσ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2 σ2 = −260.04MPaAnsOrientation of Principal Stress:tan(2θp) 2τxy= θpσx − σy12atan2τxyσx − σy⎛⎜⎝⎞⎠:= θ'p := θp + 90degθp = −18.94 deg θ'p = 71.06 degUse Eq. 9-1 to determine the principal plane of σ1 and σ2.σx + σyσx':= + ⋅cos(2θp) + sin(2θp)2τxy⋅ σx' = −260.04MPaTherefore, θp1 := θ'p θp1 = 71.06 deg Ansσx − σy2θp2 := θp θp2 = −18.94 deg Ans(b)τmaxσx − σy2⎛⎜⎝⎞⎠2τxy:= + 2 τmax = 285.04MPa Ansσavgσx + σy:= σavg = 25.00MPa Ans2Orientation of Maximum In-plane Shear Stress:tan(2θs) σx − σy= − θs2τxy12atanσx − σy2τxy−⎛⎜⎝⎞⎠:= θ's := θs − 90degθs = 26.06 deg θ's = −63.94 degBy observation, in order to preserve equilibrium along AB, τmax has to act in thedirection shown in the figure. 897. Problem 9-17A point on a thin plate is subjected to the two successive states of stress shown. Determine theresultant state of stress represented on the element oriented as shown on the right.Given:(a) σx'_a := −200MPa τx'y'_a := 0MPaσy'_a := −350MPa θa := −30deg(a) σx'_b := 0 τx'y'_b := 58MPaσy'_b := 0 θb := 25degSolution:Stress Transformation Equations: Applying Eqs. 9-1, 9-2, and 9-3 of the text.For element (a):σx'_a + σy'_aσx_a2σx'_a − σy'_a+ ⋅cos(2θa) + τx'y'_a⋅ sin(2θa)2⎛⎜⎝⎞⎠:= σx_a = −237.50MPaσy_aσx'_a + σy'_a2σx'_a − σy'_a− ⋅cos(2θa) − τx'y'_a⋅ sin(2θa)2⎛⎜⎝⎞⎠:= σy_a = −312.50MPaτxy_aσx'_a − σy'_a− ⋅ sin(2θa) + τx'y'_a⋅cos(2θa)2⎛⎜⎝⎞⎠:= τxy_a = 64.95MPaFor element (b):σx_bσx'_b + σy'_b2σx'_b − σy'_b+ ⋅cos(2θb) + τx'y'_b⋅ sin(2θb)2⎛⎜⎝⎞⎠:= σx_b = 44.43MPaσy_bσx'_b + σy'_b2σx'_b − σy'_b− ⋅cos(2θb) − τx'y'_b⋅ sin(2θb)2⎛⎜⎝⎞⎠:= σy_b = −44.43MPa 898. τxy_bσx'_b − σy'_b− ⋅ sin(2θb) + τx'y'_b⋅cos(2θb)2⎛⎜⎝⎞⎠:= τxy_b = 37.28MPaCombining the stress componenets of two elements yieldsσx := σx_a + σx_b σx = −193.1MPa Ansσy := σy_a + σy_b σy = −356.9MPa Ansτxy := τxy_a + τxy_b τxy = 102.2MPa Ans 899. Problem 9-18The steel bar has a thickness of 12 mm and is subjected to the edge loading shown. Determine theprincipal stresses developed in the bar.Given: d := 50mm t := 12mmq 4kNm:= L := 500mmSolution:Normal and Shear Stress:In accordance with the established sign convention.σx := 0MPa σy := 0MPaτxyqt:= τxy = 0.333MPaIn-plane Principal Stress: Apply Eq. 9-5.σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2 σ1 = 0.333MPa Ansσ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2 σ2 = −0.333MPa Ans 900. Problem 9-19The steel plate has a thickness of 10 mm and is subjected to the edge loading shown. Determine themaximum in-plane shear stress and the average normal stress developed in the steel.Given: ax := 300mm ay := 100mm t := 10mmqx 30kNm:= qy 40kNm:=Solution:Normal and Shear Stress:In accordance with the established sign convention.σxqxt:= σx = 3.00MPaσyqyt:= σy = 4.00MPaτxy := 0Maximum In-plane Shear Stress: Apply Eq. 9-7.τmaxσx − σy2⎛⎜⎝⎞⎠2τxy:= + 2 τmax = 0.500MPa AnsAverage Normal Stress: Apply Eq. 9-8.σavgσx + σy:= σavg = 3.50MPa Ans2 901. Problem 9-20The stress acting on two planes at a point is indicated. Determine the shear stress on plane a-a and theprincipal stresses at the point.Given: σa := 80MPa σb := 60MPaθ := 45deg β := 60degSolution:σx := σb⋅ sin(β)τxy := σb⋅cos(β)Givenσx + σyσa= + ⋅cos(2θ) + τxy⋅ sin(2θ)2σx − σy2τa= − ⋅ sin(2θ) + τxy⋅cos(2θ)Guess σy := 1MPa τa := 1MPaσyτaσx − σy2⎛⎜⎜⎝⎞⎠:= Find(σy , τa)σyτa⎛⎜⎜⎝⎞⎠48.04−1.96⎛⎜⎝⎞⎠= MPa AnsPrincipal Stress:σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2 σ1 = 80.06MPa Ansσ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2 σ2 = 19.94MPa Ans 902. Problem 9-21The stress acting on two planes at a point is indicated. Determine the normal stress σband the principalstresses at the point.Given: σa := 4MPa τx'y' := −2MPaφbb := 45deg β := 60degSolution:Stress Transformation Equations :Applying Eqs. 9-3 and 9-1 with θ := −φbb − 90degσy := σa⋅ sin(β) τxy := σa⋅cos(β) σx' = σbGivenσx + σyσx'= + ⋅cos(2θ) + τxy⋅ sin(2θ)2σx − σy2τx'y'σx − σy= − ⋅ sin(2θ) + τxy⋅cos(2θ)2Guess σx := 1MPa σx' := 1MPaσxσx'⎛⎜⎜⎝⎞⎠:= Find(σx , σx')σxσx'⎛⎜⎜⎝⎞⎠7.4647.464⎛⎜⎝⎞⎠= MPaσb := σx'σb = 7.464MPa AnsIn-plane Principal Stress: Applying Eq. 9-5,σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2 σ1 = 8.29MPa Ansσ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2 σ2 = 2.64MPa Ans 903. Problem 9-22The clamp bears down on the smooth surface at E by tightening the bolt. If the tensile force in the boltis 40 kN, determine the principal stresses at points A and B and show the results on elements located ateach of these points. The cross-sectional area at A and B is shown in the adjacent figure.Given: h := 50mm t := 30mm P := 40kNa := 100mm b := 200mmSolution: L := 3a + bSupport Reactions :+ ΣΜO=0; E'⋅L − P⋅ (a + b) = 0E'P⋅ (a + b):= E' = 24 kNLInternal Force and Moment : At Section A-B:+ ΣFx=0; E + V = 0 V := −E'+ ΣΜO=0; M − E⋅ (a) = 0 M := E'⋅aSection Property :A := h⋅ t A = 1500mm2I112:= ⋅ t⋅h3 I = 312500mm4QB := (0.5⋅h⋅ t)(0.25⋅h) QB = 9375mm3QA := 0 (since A' = 0)Normal Stress: σM⋅cI=cA := −0.5h σAM⋅cAI:= σA = −192.00MPacB := 0 σBM⋅cBI:= σB = 0MPaShear Stress : τV⋅QI⋅ t=τAV⋅QAI⋅ t:= τA = 0.00MPaτBV⋅QBI⋅ t:= τB = −24.00MPaIn-plane Principal Stress:At A: σxA := σA σyA := 0 τxy := τASince no shear stress acts upon the element,σA1 := σyA σA1 = 0MPa AnsσA2 := σxA σA2 = −192MPa Ans 904. At B: σxB := σB σyB := 0 τxy := τBσ σ σ σ⎞⎛ −+x y x y τ22σ + ⎟ ⎟⎠⎜ ⎜⎝±=1.2 2 2 xyσB1 := −τxy σB1 = 24MPa AnsσB2 := τxy σB2 = −24MPa AnsOrientation of Principal Plane: Applying Eq. 9-4 for point B,tan(2θp) 2τB= − θpσxB − σyB12:= (90deg) θ'p := θp − 90degθp = 45 deg θ'p = −45 degUse Eq. 9-1 to determine the principal plane of σ1 and σ2.σxB + σyBσx'_B:= + ⋅cos(2θp) + τB⋅ sin(2θp)2σxB − σyB2σx'_B = −24.00MPaTherefore, θp1 := θ'p θp1 = −45.00 deg Ansθp2 := θp θp2 = 45.00 deg Ans 905. Problem 9-23Solve Prob. 9-22 for points C and D.Given: h := 50mm t := 30mm P := 40kNa := 100mm b := 200mm hD := 10mmSolution: L := 3a + bSupport Reactions :+ ΣΜE=0; P⋅ (2a) − R⋅L = 0R2P⋅aL:= R = 16 kNInternal Force and Moment : At Section C-D:+ ΣFx=0; R − V = 0 V := R+ ΣΜO=0; M − R⋅ (b) = 0 M := R⋅ (b)Section Property :A := h⋅ t A = 1500mm2I112:= ⋅ t⋅h3 I = 312500mm4QD := (hD⋅ t)(0.5⋅h − 0.5⋅hD) QD = 6000mm3QC := 0 (since A' = 0)Normal Stress: σM⋅cI=cC := 0.5h σCM⋅cCI:= σC = 256.00MPacD := −0.5⋅h + hD σDM⋅cDI:= σD = −153.6MPaShear Stress : τV⋅QI⋅ t=τCV⋅QCI⋅ t:= τC = 0.00MPaτDV⋅QDI⋅ t:= τD = 10.24MPaIn-plane Principal Stress:At C: σxC := 0 σyC := σC τxy := τCSince no shear stress acts upon the element,σC1 := σyC σC1 = 256MPa AnsσC2 := σxC σC2 = 0MPa AnsAt D: σxD := 0 σyD := σD τxy := τD 906. σD1σxD + σyD2σxD − σyD2⎛⎜⎝⎞⎠2τD:= + + 2 σD1 = 0.680MPa AnsσD2σxD + σyD2σxD − σyD2⎛⎜⎝⎞⎠2τD:= − + 2 σD2 = −154.28MPa AnsOrientation of Principal Plane: Applying Eq. 9-4 for point D,tan(2θp) 2τD= θpσxD − σyD12atan2τDσxD − σyD⎛⎜⎝⎞⎠:= θ'p := θp − 90degθp = 3.797 deg θ'p = −86.203 degUse Eq. 9-1 to determine the principal plane of σ1 and σ2.σxD + σyDσx'_D:= + ⋅cos(2θp) + τD⋅ sin(2θp)2σxD − σyD2σx'_D = 0.680MPaTherefore, θp1 := θp θp1 = 3.80 deg Ansθp2 := θ'p θp2 = −86.20 deg Ans 907. Problem 9-24The grains of wood in the board make an angle of 20° with the horizontal as shown. Determine thenormal and shear stress that act perpendicular to the grains if the board is subjected to an axial load of250 N.Unit Used: kPa := 1000PaGiven: P := 250N φ := 20degh := 60mm t := 25mmSolution:σxPh⋅ t:= σx = 0.1667MPaσy := 0τxy := 0θ := 90deg + φσx'σx + σy:= + ⋅cos(2θ) + τxy⋅ sin(2θ) σx' = 19.50 kPa Ans2σx − σy2τx'y'σx − σy:= − ⋅ sin(2θ) + τxy⋅cos(2θ) τx'y' = 53.57 kPa Ans2 908. Problem 9-25The wooden block will fail if the shear stress acting along the grain is 3.85 MPa. If the normal stressσx= 2.8 MPa, determine the necessary compressive stress σythat will cause failure.Given: σx := 2.8MPa τxy := 0MPaθgrain := 58deg τx'y' := 3.85MPaSolution: θ := θgrain + 90degShear Stress:τx'y'σx − σy= − ⋅ sin(2θ) + τxy⋅cos(2θ)2τx'y'σx − σy= − ⋅ sin(2θ)22τx'y'σy:= sin(2θ) + σxσy = −5.767MPa Ans 909. Problem 9-26The T-beam is subjected to the distributed loading that is applied along its centerline. Determine theprincipal stresses at points A and B and show the results on elements located at each of these points.Given: bf := 150mm tf := 20mmdw := 150mm tw := 20mma := 2m b := 1mhB := 50mm w 12kNm:=Solution:Internal Force and Moment : At Section A-B:+ ΣFy=0; V − w⋅a = 0 V := w⋅a+ ΣΜA=0; M − w⋅a⋅ (0.5a + b) = 0 M := w⋅a⋅ (0.5a + b)Section Property : D := dw + tfyc0.5tf(bf⋅ tf) + (0.5dw + tf)(dw⋅ tw):= yc = 52.50mmbf⋅ tf + dw⋅ twI1:= ⋅ 3 + b( f⋅ tf) ⋅ (0.5tf − yc)2I2112⋅bf tf112:= ⋅ tw ⋅ dw3 + d( w⋅ tw) ⋅ (0.5dw + tf − yc)2I := I1 + I2 I = 16562500.00mm4QA := 0 (since A' = 0)QB := (hB⋅ tw)(D − yc − 0.5⋅hB) QB = 92500mm3Normal Stress: σM⋅cI=cA := yc σAM⋅cAI:= σA = 152.15MPacB := −(D − yc − hB) σBM⋅cBI:= σB = −195.62MPaShear Stress : τV⋅QI⋅ t= τAV⋅QAI⋅bf:= τA = 0.00MPaτBV⋅QBI⋅ tw:= τB = 6.702MPaIn-plane Principal Stress:At A: σxA := σA σyA := 0 τxy := τASince no shear stress acts upon the element,σA1 := σxA σA1 = 152.15MPa Ans 910. σA2 := σyA σA2 = 0MPa AnsAt B: σxB := σB σyB := 0 τxy := τBσB1σxB + σyB2σxB − σyB2⎛⎜⎝⎞⎠2τB:= + + 2 σB1 = 0.229MPa AnsσB2σxB + σyB2σxB − σyB2⎛⎜⎝⎞⎠2τB:= − + 2 σB2 = −195.852MPa AnsOrientation of Principal Plane: Applying Eq. 9-4 for point B,tan(2θp) 2τB= − θpσxB − σyB12atan−2τBσxB − σyB⎛⎜⎝⎞⎠:= θ'p := θp − 90degθp = 1.96 deg θ'p = −88.04 degUse Eq. 9-1 to determine the principal plane of σ1 and σ2.σxB + σyBσx'_B:= + ⋅cos(2θp) + τB⋅ sin(2θp)2σxB − σyB2σx'_B = −194.94MPaTherefore, θp1 := θ'p θp1 = −88.04 deg Ansθp2 := θp θp2 = 1.96 deg Ans 911. Problem 9-27The bent rod has a diameter of 15 mm and is subjected to the force of 600 N. Determine the principalstresses and the maximum in-plane shear stress that are developed at point A and point B. Show theresults on properly oriented elements located at these points.Given: do := 15mm a := 50mm P := 0.6kNSolution:Internal Force and Moment : At Section A-B:+ ΣFx=0; N − P = 0 N := P+ ΣΜO=0; M − P⋅ (a) = 0 M := P⋅aSection Property :A⋅ 24π do:= A = 176.71mm2I⋅ 464π do:= I = 2485.05mm4McI= N ± 1.2 Normal Stress: σAcA := 0.5do σANAM⋅cAI:= − σA = −87.15MPacB := 0.5do σBNAM⋅cBI:= + σB = 93.94MPaIn-plane Principal Stress:At A: σxA := σA σyA := 0 τxy := 0Since no shear stress acts upon the element,σA1 := σyA σA1 = 0MPa AnsσA2 := σxA σA2 = −87.15MPa AnsAt B: σxB := σB σyB := 0 τxy := 0Since no shear stress acts upon the element,σB1 := σxB σB1 = 93.94MPa AnsσB2 := σyB σB2 = 0MPa AnsMaximum In-plane Shear Stress: Applying Eq. 9-7τA.maxσxA − σyA2⎛⎜⎝⎞⎠2τxy:= + 2 τA.max = 43.6MPa AnsτB.maxσxB − σyB2⎛⎜⎝⎞⎠2τxy:= + 2 τB.max = 47.0MPa Ans 912. Orientation of Plane for Maximum In-plane Shear Stress: Applying Eq. 9-6tan(2θ ) = ∞ S_Aθs_A := 45deg Anstan(2θs_A) σxA − σyA2τxy= −Ans θ's_A := θs_A − 90deg θ's_A = −45 deg Anstan(2θ ) = ∞ tan(2θs_B) S_B σxB − σyB= − θs_B := −45deg Ans2τxyAns θ's_B := θs_B + 90deg θ's_B = 45 deg AnsBy observation, in order to preserve equilibrium along AB, τmax has to act in thedirection shown in the figure.Average Normal Stress: Applying Eq. 9-8σxA + σyAσavg_A:= σavg_A = −43.57MPa2σavg_BσxB + σyB:= σavg_B = 46.97MPa2 913. Problem 9-28The simply supported beam is subjected to the traction stress τ0 on its top surface. Determine theprincipal stresses at points A and B. 914. Problem 9-29The beam has a rectangular cross section and is subjected to the loadings shown. Determine theprincipal stresses and the maximum in-plane shear stress that are developed at point A and point B.These points are just to the left of the 10-kN load. Show the results on properly oriented elementslocated at these points.Given: b := 150mm d := 375mmF := 10kN P := 5kN L := 1.2mSolution:Support Reactions : By symmetry, R1=R ; R2= R+ ΣFy=0; 2R − F = 0 R := 0.5F+ ΣFx=0; H1 − P = 0 H1 := PInternal Force and Moment : At Section A-B:+ ΣFx=0; H1 + N = 0 N := −H1+ ΣFy=0; R + V = 0 V := −R+ ΣΜO=0; M − R⋅ (0.5L) = 0 M := 0.5R⋅LSection Property :A := b⋅d I112:= ⋅b⋅d3QA := 0 QB := 0 (since A' = 0)Normal Stress: σNAM⋅cI= +cA := −0.5d σANAM⋅cAI:= + σA = −0.942MPacB := 0.5d σBNAM⋅cBI:= + σB = 0.764MPaShear Stress : Since QA = QB = 0, τA := 0 τB := 0In-plane Principal Stress:At A: σxA := σA σyA := 0 τxy := 0Since no shear stress acts upon the element,σA1 := σyA σA1 = 0MPa AnsσA2 := σxA σA2 = −0.942MPa AnsAt B: σxB := σB σyB := 0 τxy := 0Since no shear stress acts upon the element,σB1 := σxB σB1 = 0.764MPa AnsσB2 := σyB σB2 = 0MPa Ans 915. Maximum In-plane Shear Stress: Applying Eq. 9-7,τmax_AσxA − σyA2⎛⎜⎝⎞⎠2τxy:= + 2 τmax_A = 0.471MPa Ansτmax_BσxB − σyB2⎛⎜⎝⎞⎠2τxy:= + 2 τmax_B = 0.382MPa AnsOrientation of Plane for Maximum In-plane Shear Stress: Applying Eq. 9-6tan(2θ ) = ∞ S_Aθs_A := 45deg Anstan(2θs_A) σxA − σyA2τxy= −Ans θ's_A := θs_A − 90deg θ's_A = −45 deg Anstan(2θ ) = ∞ tan(2θs_B) S_B σxB − σyB= − θs_B := −45deg Ans2τxyAns θ's_B := θs_B + 90deg θ's_B = 45 deg AnsBy observation, in order to preserve equilibrium along AB, τmax has to act in thedirection shown in the figure.Average Normal Stress: Applying Eq. 9-8,σxA + σyAσavg_A:= σavg_A = −0.471MPa Ans2σavg_BσxB + σyB:= σavg_B = 0.382MPa Ans2 916. Problem 9-30The wide-flange beam is subjected to the loading shown. Determine the principal stress in the beam atpoint A and at point B. These points are located at the top and bottom of the web, respectively.Although it is not very accurate, use the shear formula to compute the shear stress.Given: bf := 200mm tf := 10mmtw := 10mm dw := 200mmP := 25kN θ := 30dega := 3m w 8kNm:=Solution:Internal Force and Moment : At Section A-B:+ ΣFx=0; P⋅cos(θ) − N = 0 N := P⋅cos(θ)+ ΣFy=0; V − P⋅ sin(θ) − w⋅a = 0V := P⋅ sin(θ) + w⋅a+ ΣΜO=0; M − P⋅ sin(θ)⋅ (a) − (w⋅a)⋅ (0.5a) = 0M := P⋅a⋅ sin(θ) + 0.5w⋅a2Section Property : D := dw + 2tfA := bf⋅D − (bf − tw)⋅dw A = 6000mm2I112bf⋅D3 (bf − tw) dw:= ⋅ ⎡⎣− ⋅ 3 I = 50.80 × 10− 6m4(tf) DQA bf⋅ ⎤⎦2tf2−⎛⎜⎝⎞⎠:= ⋅ QA = 210000mm3 QB := QANormal Stress: σNAM⋅cI= +cA := 0.5dw σANAM⋅cAI:= + σA = 148.293MPacB := −0.5dw σBNAM⋅cBI:= + σB = −141.077MPaShear Stress : τV⋅QI⋅ t=τAV⋅QAI⋅ tw:= τA = 15.09MPaτBV⋅QBI⋅ tw:= τB = 15.09MPa 917. In-plane Principal Stress:At A: σxA := σA σyA := 0 τxy := τAσA1σxA + σyA2σxA − σyA2⎛⎜⎝⎞⎠2τA:= + + 2 σA1 = 149.8MPa AnsσA2σxA + σyA2σxA − σyA2⎛⎜⎝⎞⎠2τA:= − + 2 σA2 = −1.52MPa AnsAt B: σxB := σB σyB := 0 τxy := τBσB1σxB + σyB2σxB − σyB2⎛⎜⎝⎞⎠2τB:= + + 2 σB1 = 1.60MPa AnsσB2σxB + σyB2σxB − σyB2⎛⎜⎝⎞⎠2τB:= − + 2 σB2 = −142.7MPa Ans 918. Problem 9-31The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stresses andthe maximum in-plane shear stress that is developed anywhere on the surface of the shaft. 919. Problem 9-32A paper tube is formed by rolling a paper strip in a spiral and then gluing the edges together as shown.Determine the shear stress acting along the seam, which is at 30° from the vertical, when the tube issubjected to an axial force of 10 N. The paper is 1 mm thick and the tube has an outer diameter of 30mm.Unit used: kPa := 1000PaGiven: do := 30mm t := 1mmθ := 30deg P := 10NSolution:Section Property : di := do − 2tAπ4:= ⋅ A = 91.11mm2do2 − di2 ⎛⎝⎞⎠Normal Stress:σxPA:= σx = 109.76 kPaσy := 0τxy := 0Shear stress along the seam:τx'y'σx − σy:= − ⋅ sin(2θ) + τxy⋅cos(2θ)2τx'y' = −47.53 kPa Ans 920. Problem 9-33Solve Prob. 9-32 for the normal stress acting perpendicular to the seam.Unit used: kPa := 1000PaGiven: do := 30mm t := 1mmθ := 30deg P := 10NSolution:Section Property : di := do − 2tAπ4:= ⋅ A = 91.11mm2do2 − di2 ⎛⎝⎞⎠Normal Stress:σxPA:= σx = 109.76 kPaσy := 0τxy := 0Normal stress perpendicular to the seam:σx'σx + σy:= + ⋅cos(2θ) + τxy⋅ sin(2θ)2σx − σy2σx' = 82.32 kPa Ans 921. Problem 9-34The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stresses andthe maximum in-plane shear stress that is developed at point A. The bearings only support verticalreactions. 922. Problem 9-35The drill pipe has an outer diameter of 75 mm, a wall thickness of 6 mm, and a weight of 0.8 kN/m. Ifit is subjected to a torque and axial load as shown, determine (a) the principal stresses and (b) themaximum in-plane shear stress at a point on its surface at section a.Given: do := 75mm t := 6mm L := 6mP := 7.5kN Mx := 1.2kN⋅m w 0.8kNm:=Solution:Internal Force and Moment : At section a:ΣFx=0; N + P + w⋅L = 0 N := −P − w⋅LΣΜx=0; T − Mx = 0 T := MxSection Property : di := do − 2tAπ4do2 − di2 ⎛⎝⎞⎠:= ⋅ Jπ32do4 − di4 ⎛⎝⎞⎠:= ⋅Normal Stress: σNA:= σ = −9.457MPaShear Stress :c := 0.5do τT⋅cJ:= τ = 28.850MPaa) In-plane Principal Stresses:σx := 0 σy := σ τxy := τfor any point on the shaft's surface. Applying Eq. 9-5,σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2 σ1 = 24.51MPa Ansσ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2 σ2 = −33.96MPa Ansb) Maximum In-plane Shear Stress: Applying Eq. 9-7,τmaxσx − σy2⎛⎜⎝⎞⎠2τxy:= + 2 τmax = 29.24MPa Ans 923. Problem 9-36The internal loadings at a section of the beam are shown. Determine the principal stresses at point A.Also compute the maximum in-plane shear stress at this point.Given: bf := 200mm tf := 50mmtw := 50mm dw := 200mmPx := −500kN My := −30kN⋅mPy := −800kN Mz := 40kN⋅mSolution:Section Property : D := dw + 2tfA := bf⋅D − (bf − tw)⋅dw A = 30000mm2Iz112bf⋅D3 (bf − tw) dw:= ⋅ Iz = 350.00 × 10− 6m43 ⋅ − ⎡⎣⎤⎦Iy:= ⋅ Iy = 68.75 × 10− 6m4QA := 0 (since A' = 0)Normal Stress: yA := 0.5D zA := 0.5bf1122tf ⋅ bf3 + dw ⋅ tw3 ⎛⎝⎞⎠σAPxAMz⋅yAIz−My⋅zAIy:= + σA = −77.446MPaShear Stress : Since QA = 0, τA := 0In-plane Principal Stress:σx := σA σy := 0 τxy := 0Since no shear stress acts upon the element,σ1 := σy σ1 = 0MPa Ansσ2 := σx σ2 = −77.45MPa AnsMaximum In-plane Shear Stress: Applying Eq. 9-7,τmaxσx − σy2⎛⎜⎝⎞⎠2τxy:= + 2 τmax = 38.72MPa Ans 924. Problem 9-37Solve Prob. 9-36 for point B.Given: bf := 200mm tf := 50mmtw := 50mm dw := 200mmPx := −500kN My := −30kN⋅mPy := −800kN Mz := 40kN⋅mSolution:Section Property : D := dw + 2tfA := bf⋅D − (bf − tw)⋅dw A = 30000mm2Iz112bf⋅D3 (bf − tw) dw:= ⋅ Iz = 350.00 × 10− 6m43 ⋅ − ⎡⎣⎤⎦Iy:= ⋅ Iy = 68.75 × 10− 6m4QB := 0 (since A' = 0)Normal Stress: yB := −0.5D zB := −0.5bf1122tf ⋅ bf3 + dw ⋅ tw3 ⎛⎝⎞⎠σBPxAMz⋅yBIz−My⋅zBIy:= + σB = 44.113MPaShear Stress : Since QB = 0, τB := 0In-plane Principal Stress:σx := σB σy := 0 τxy := 0Since no shear stress acts upon the element,σ1 := σx σ1 = 44.113MPa Ansσ2 := σy σ2 = 0.00MPa AnsMaximum In-plane Shear Stress: Applying Eq. 9-7,τmaxσx − σy2⎛⎜⎝⎞⎠2τxy:= + 2 τmax = 22.06MPa Ans 925. Problem 9-38Solve Prob. 9-36 for point C, located in the center on the bottom of the web.Given: bf := 200mm tf := 50mmtw := 50mm dw := 200mmPx := −500kN My := −30kN⋅mPy := −800kN Mz := 40kN⋅mSolution:Section Property : D := dw + 2tfA := bf⋅D − (bf − tw)⋅dw A = 30000mm2Iz112bf⋅D3 (bf − tw) dw:= ⋅ ⎡⎣− ⋅ 3 Iz = 350.00 × 10− 6m4Iy⎤⎦112:= ⋅ Iy = 68.75 × 10− 6m42tf ⋅ bf3 + dw ⋅ tw3 ⎛⎝⎞⎠QC (bf⋅ tf) D2tf2−⎛⎜⎝⎞⎠:= ⋅ QC = 1250000mm3Normal Stress: yC := −0.5dw zC := 0σCPxAMz⋅yCIz−My⋅zCIy:= + σC = −5.238MPaShear Stress : τV⋅QI⋅ t=τCPy⋅QCIz⋅ tw:= τC = −57.14MPaIn-plane Principal Stress:σx := σC σy := 0 τxy := τCσ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2 σ1 = 54.58MPa Ansσ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2 σ2 = −59.82MPa AnsMaximum In-plane Shear Stress: Applying Eq. 9-7,τmaxσx − σy2⎛⎜⎝⎞⎠2τxy:= + 2 τmax = 57.20MPa Ans 926. Problem 9-39The wide-flange beam is subjected to the 50-kN force. Determine the principal stresses in the beam atpoint A located on the web at the bottom of the upper flange. Although it is not very accurate, use theshear formula to calculate the shear stress.Given: bf := 200mm tf := 12mmtw := 10mm dw := 250mmP := 50kN a := 3mSolution:Internal Force and Moment : At Section A-B:+ ΣFy=0; V − P = 0 V := P+ ΣΜO=0; M − P⋅ (a) = 0 M := P⋅aSection Property : D := dw + 2tfA := bf⋅D − (bf − tw)⋅dw A = 7300mm2I112bf⋅D3 (bf − tw) dw:= ⋅ I = 95.45 × 10− 6m43 ⋅ − ⎡⎣⎤⎦QA (bf⋅ tf) D2tf2−⎛⎜⎝⎞⎠:= ⋅ QA = 314400mm3Normal Stress: σM⋅cI=cA := 0.5dw σAM⋅cAI:= σA = 196.435MPaShear Stress : τV⋅QI⋅ t=τAV⋅QAI⋅ tw:= τA = 16.47MPaIn-plane Principal Stress:σx := σA σy := 0 τxy := τAσ1σx + σy2⎛⎜⎝⎞σx − σy2⎠2τxy:= + + 2 σ1 = 197.81MPa Ansσ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2 σ2 = −1.37MPa Ans 927. Problem 9-40Solve Prob. 9-39 for point B located on the web at the top of the bottom flange.Given: bf := 200mm tf := 12mmtw := 10mm dw := 250mmP := 50kN a := 3mSolution:Internal Force and Moment : At Section A-B:+ ΣFy=0; V − P = 0 V := P+ ΣΜO=0; M − P⋅ (a) = 0 M := P⋅aSection Property : D := dw + 2tfA := bf⋅D − (bf − tw)⋅dw A = 7300mm2I112bf⋅D3 (bf − tw) dw:= ⋅ I = 95.45 × 10− 6m43 ⋅ − ⎡⎣⎤⎦QB (bf⋅ tf) D2tf2−⎛⎜⎝⎞⎠:= ⋅ QB = 314400mm3Normal Stress: σM⋅cI=cB := −0.5dw σBM⋅cBI:= σB = −196.435MPaShear Stress : τV⋅QI⋅ t=τBV⋅QBI⋅ tw:= τB = 16.47MPaIn-plane Principal Stress:σx := σB σy := 0 τxy := τBσ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2 σ1 = 1.37MPa Ansσ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2 σ2 = −197.81MPa Ans 928. Problem 9-41The bolt is fixed to its support at C. If a force of 90 N is applied to the wrench to tighten it, determinethe principal stresses developed in the bolt shank at point A. Represent the results on an element locatedat this point. The shank has a diameter of 6 mm.Given: do := 6mm a := 150mm L := 50mmP := 90NSolution:Internal Force and Moment : At section AB:Mx := P⋅LTy := P⋅aSection Property :Iπ64:= ⋅ 4 Jdoπ32:= ⋅ do4 cA := 0.5doNormal Stress: σAMx⋅cAI:= σA = 212.21MPaShear Stress : τATy⋅cAJ:= τA = 318.31MPaIn-plane Principal Stresses: Applying Eq. 9-5,σx := σA σy := 0 τxy := τAσ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2 σ1 = 441.63MPa Ansσ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2 σ2 = −229.42MPa AnsOrientation of Principal Stress:tan(2θp) 2τxy= θpσx − σy12atan2τxyσx − σy⎛⎜⎝⎞⎠:= θ'p := θp − 90degθp = 35.783 deg θ'p = −54.217 degUse Eq. 9-1 to determine the principal plane of σ1 and σ2.σx + σyσx':= + ⋅cos(2θp) + sin(2θp)2τxy⋅ σx' = 441.63MPaTherefore, θp1 := θp θp1 = 35.78 deg Ansσx − σy2θp2 := θ'p θp2 = −54.22 deg Ans 929. Problem 9-42Solve Prob. 9-41 for point B.Given: do := 6mm a := 150mm L := 50mmP := 90NSolution:Internal Force and Moment : At section AB:Mx := P⋅LTy := P⋅aVz := PSection Property : ro := 0.5doIπ64:= ⋅ 4 Jdoπ32:= ⋅ 4doQB4ro3π⋅ 22π ro⎛⎜⎝⎞⎠:= ⋅Normal Stress: cBσ := 0 σBMx⋅cBσ:= σB = 0MPaIShear Stress : bB := do cBτ := roτBVz⋅QBI⋅bBTy⋅cBτJ:= − τB = −314.07MPaIn-plane Principal Stresses: Applying Eq. 9-5,σx := σB σy := 0 τxy := τBσ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2 σ1 = 314.07MPa Ansσ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2 σ2 = −314.07MPa AnsOrientation of Principal Stress:tan(2θ ) = ∞ tan(2θp) P 2τxy= θp1 := 45deg Ansσx − σyθp2 := θp1 − 90deg θp2 = −45 deg Ans 930. Problem 9-43The beam has a rectangular cross section and is subjected to the loadings shown. Determine theprincipal stresses that are developed at point A and point B, which are located just to the left of the20-kN load. Show the results on elements located at these points.Given: b := 100mm d := 200mmF := 20kN P := 10kNL := 4mSolution:Support Reactions : By symmetry, R1=R ; R2= R+ ΣFy=0; 2R − F = 0 R := 0.5F+ ΣFx=0; H1 − P = 0 H1 := PInternal Force and Moment : At Section A-B:+ ΣFx=0; H1 + N = 0 N := −H1+ ΣFy=0; R + V = 0 V := −R+ ΣΜO=0; M − R⋅ (0.5L) = 0 M := 0.5R⋅LSection Property :A := b⋅d I112:= ⋅b⋅d3QA := 0 (since A' = 0) QB := b⋅ (0.5d)⋅ (0.25d)Normal Stress: σNAM⋅cI= +cA := −0.5d σANAM⋅cAI:= + σA = −30.5MPacB := 0 σBNAM⋅cBI:= + σB = −0.5MPaShear Stress :Since QA = 0, τA := 0τBV⋅QBI⋅b:= τB = −0.75MPaIn-plane Principal Stress: Applying Eq. 9-5At A: σxA := σA σyA := 0 τxy := 0Since no shear stress acts upon the element,σA1 := σyA σA1 = 0MPa AnsσA2 := σxA σA2 = −30.50MPa AnsAt B: σxB := σB σyB := 0 τxy := τB 931. σB1σxB + σyB2σxB − σyB2⎛⎜⎝⎞⎠2τB:= + + 2 σB1 = 0.541MPa AnsσB2σxB + σyB2σxB − σyB2⎛⎜⎝⎞⎠2τB:= − + 2 σB2 = −1.041MPa AnsOrientation of Principal Plane: Applying Eq. 9-4 for point B,tan(2θp) 2τB= θpσxB − σyB12atan2τBσxB − σyB⎛⎜⎝⎞⎠:= θp = 35.783 degθ'p := θp − 90deg θ'p = −54.217 degUse Eq. 9-1 to determine the principal plane of σ1 and σ2.σxB + σyBσx'_B:= + ⋅cos(2θp) + τB⋅ sin(2θp)2σxB − σyB2σx'_B = −1.04MPaTherefore, θp1 := θ'p θp1 = −54.22 deg Ansθp2 := θp θp2 = 35.78 deg Ans 932. Problem 9-44The solid propeller shaft on a ship extends outward from the hull. During operation it turns at ω = 15rad/s when the engine develops 900 kW of power. This causes a thrust of F = 1.23 MN on the shaft.If the shaft has an outer diameter of 250 mm, determine the principal stresses at any point located onthe surface of the shaft.Given: do := 250mm L := 0.75mF := 1230kN P := 900kW ω 15rads:=Solution:Internal Force and Moment : As shown on FBDToPω:= To = 60.00 kN⋅mN := −FSection Property :A⋅ 24π do:= A = 49087.39mm2J⋅ 432π do:= J = 383495196.97mm4Normal Stress:σaNA:= σa = −25.06MPaShear Stress : cmax := 0.5⋅doTo⋅cmaxτo:= τo = 19.56MPaJIn-plane Principal Stresses: Applying Eq. 9-5,σx := σa σy := 0 τxy := τoσ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2 σ1 = 10.70MPa Ansσ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2 σ2 = −35.75MPa Ans 933. Problem 9-45The solid propeller shaft on a ship extends outward from the hull. During operation it turns at ω = 15rad/s when the engine develops 900 kW of power. This causes a thrust of F = 1.23 MN on the shaft.If the shaft has a diameter of 250 mm, determine the maximum in-plane shear stress at any pointlocated on the surface of the shaft.Given: do := 250mm L := 0.75mF := 1230kN P := 900kW ω 15rads:=Solution:Internal Force and Moment : As shown on FBDToPω:= To = 60.00 kN⋅mN := −FSection Property :A⋅ 24π do:= A = 49087.39mm2J⋅ 432π do:= J = 383495196.97mm4Normal Stress:σaNA:= σa = −25.06MPaShear Stress : cmax := 0.5⋅doTo⋅cmaxτo:= τo = 19.56MPaJMaximum In-plane Shear Stress: Applying Eq. 9-7σx := σa σy := 0 τxy := τoτmaxσx − σy2⎛⎜⎝⎞⎠2τxy:= + 2 τmax = 23.2MPa Ans 934. Problem 9-46The steel pipe has an inner diameter of 68mm and an outer diameter of 75 mm. If it is fixed at C andsubjected to the horizontal 100-N force acting on the handle of the pipe wrench at its end, determinethe principal stresses in the pipe at point A which is located on the surface of the pipe.Given: do := 75mm di := 68mm L := 250mmP := 100N a := 300mmAt section AB:Solution:Internal Force and Moment :AnsTx := P⋅aMy := P⋅LVz := PSection Property : ro := 0.5do ri := 0.5diIπ64do4 − di4 ⎛⎝⎞⎠:= ⋅ Jπ32do4 − di4 ⎛⎝⎞⎠:= ⋅QAz4ro3π⋅ 22π ro⎛⎜⎝⎞⎠⋅4ri3π⋅ 22π ri⎛⎜⎝⎞⎠:= − ⋅Normal Stress: cAσ := 0 σAMy⋅cAσ:= σA = 0MPaIShear Stress : bA := do − di cAτ := roτAVz⋅QAzI⋅bATx⋅cAτ:= − JτA = −0.863MPaIn-plane Principal Stresses: Applying Eq. 9-5,σx := σA σz := 0 τxz := τAσ1σx + σz2σx − σz2⎛⎜⎝⎞⎠2τxz:= + + 2 σ1 = 0.863MPa Ansσ2σx + σz2⎛⎜⎝⎞σx − σz2⎠2τxz:= − + 2 σ2 = −0.863MPa 935. Problem 9-47Solve Prob. 9-46 for point B, which is located on the surface of the pipe.Given: do := 75mm di := 68mm L := 250mmP := 100N a := 300mmAt section AB:Solution:Internal Force and Moment :AnsTx := P⋅aMy := P⋅LVz := PSection Property : ro := 0.5do ri := 0.5diIπ64do4 − di4 ⎛⎝⎞⎠:= ⋅ Jπ32do4 − di4 ⎛⎝⎞⎠:= ⋅QBz := 0 (Since A'=0)Normal Stress: cBσ := ro σBMy⋅cBσ:=IσB = 1.862MPaShear Stress : cBτ := ro τBTx⋅cBτJ:= −τB = −1.117MPaIn-plane Principal Stresses: Applying Eq. 9-5,σx := σB σz := 0 τxz := τBσ1σx + σz2σx − σz2⎛⎜⎝⎞⎠2τxz:= + + 2 σ1 = 2.385MPa Ansσ2σx + σz2σx − σz2⎛⎜⎝⎞⎠2τxz:= − + 2 σ2 = −0.523MPa 936. Problem 9-48The cantilevered beam is subjected to the load at its end. Determine the principal stresses in the beam atpoints A and B.Given: b := 120mm h := 150mm P := 15kN L := 1.2myA := 45mm zA := 60mm yθ−45:= zθ35:=yB := 75mm zB := −20mmSolution:Internal Force and Moment : At Section A-B:+ ΣFz0; Vz + P⋅zθ = 0 Vz := −P⋅zθ+ ΣFy=0; Vy + P⋅yθ = 0 Vy := −P⋅yθMz := P⋅yθ⋅LMy := −P⋅zθ⋅LSection Property :Iz112:= ⋅b⋅h3 Iz = 33.75 × 10− 6m4Iy112:= ⋅h⋅b3 Iy = 21.60 × 10− 6m4QA.y b 0.5h yA − ( ) ⋅ yA 0.5 0.5h yA − ( ) + ⎡⎣:= ⋅ QA.y = 216000mm3⎤⎦QB.z h 0.5b zB − ( ) ⋅ zB 0.5 0.5b zB − ( ) + ⎡⎣:= ⋅ QB.z = 240000mm3⎤⎦QA.z := 0 (since A' = 0)QB.y := 0 (since A' = 0)Normal Stress:σAMz⋅yAIz−My⋅zAIy:= + σA = −10.8MPaσBMz⋅yBIz−My⋅zBIy:= + σB = 42.0MPaShear Stress : τV⋅QI⋅ t=τAVy⋅QA.yIz⋅b:= τA = 0.640MPaτBVz⋅QB.zIy⋅h:= τB = −0.667MPaIn-plane Principal Stress: Applying Eq. 9-5At A: σxA := σA σyA := 0 τxy := 0 937. σA1σxA + σyA2σxA − σyA2⎛⎜⎝⎞⎠2τA:= + + 2 σA1 = 0.0378MPa AnsσA2σxA + σyA2σxA − σyA2⎛⎜⎝⎞⎠2τA:= − + 2 σA2 = −10.84MPa AnsAt B: σxB := σB σzB := 0 τxz := τBσB1σxB + σzB2σxB − σzB2⎛⎜⎝⎞⎠2τB:= + + 2 σB1 = 42.01MPa AnsσB2σxB + σzB2σxB − σzB2⎛⎜⎝⎞⎠2τB:= − + 2 σB2 = −0.0106MPa Ans 938. Problem 9-49The box beam is subjected to the loading shown. Determine the principal stresses in the beam at pointsA and B.Given: bo := 200mm bi := 150mmdo := 200mm di := 150mmL1 := 0.9m L2 := 1.5mP1 := 4kN P2 := 6kNSolution: L := L1 + 2L2Support Reactions : Given+ ΣFy=0; R1 + R2 − P1 − P2 = 0+ ΣΜR2=0;P1⋅L − R1⋅ (L − L1) + P2⋅L2 = 0Guess R1 := 1kN R2 := 1kNR1R2Ans⎛⎜⎜⎝⎞⎠:= Find(R1 , R2)R1R2⎛⎜⎜⎝⎞⎠8.21.8⎛⎜⎝⎞⎠= kNInternal Force and Moment : At section A-B:N := 0V := P1 − R1M := P1⋅ (L1 + 0.5L2) − R1⋅ (0.5L2)Section Property :I112bo ⋅ do3 − bi ⋅ di3 ⎛⎝⎞⎠:= ⋅QA := 0 QB := 0 (Since A'=0)For point A: τA := 0cA := 0.5⋅do σAM⋅cAI:= σA = 0.494MPaσ1 := σA σ1 = 0.494MPa Ansσ2 := 0 σ2 = 0.000MPa AnsFor point B: τB := 0cB := −0.5⋅di σBM⋅cBI:= σB = −0.37MPaσ1 := 0 σ1 = 0.000MPa Ansσ2 := σB σ2 = −0.370MPa 939. Problem 9-50A bar has a circular cross section with a diameter of 25 mm. It is subjected to a torque and a bendingmoment. At the point of maximum bending stress the principal stresses are 140 MPa and -70 MPa.Determine the torque and the bending moment.Given: do := 6mmσ1 := 140MPa σ2 := −70MPaSolution:In-plane Principal Stresses: Applying Eq. 9-5,Given σy := 0σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy= + + 2 (1)σ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy= − + 2 (2)Solving Eqs. (1) and (2):Guess σx := 2MPa τxy := 1MPaσxτxy⎛⎜⎜⎝⎞⎠:= Find(σx , τxy)σxτxy⎛⎜⎜⎝⎞⎠70.0098.99⎛⎜⎝⎞⎠= MPaSection Property : ro := 0.5doIπ64:= ⋅ 4 Jdoπ32:= ⋅ 4doNormal Stress: σM⋅cI=c := ro Mσx⋅ Ic:= M = 1.484N⋅m AnsShear Stress: τT⋅cJ=c := ro Tτxy⋅ Jc:= T = 4.199N⋅m Ans 940. Problem 9-51The internal loadings at a section of the beam consist of an axial force of 500 N, a shear force of 800N, and two moment components of 30 N·m and 40 N·m. Determine the principal stresses at point A.Also calculate the maximum in-plane shear stress at this point.Unit Used: kPa := 1000PaGiven: b := 100mm h := 200mm Px := 0.5kNyA := 100mm zA := 0mm Py := 0.8kNMy := −0.04kN⋅m Mz := 0.03kN⋅mSolution:Section Property :A := b⋅h A = 20000mm2Iz112:= ⋅b⋅h3 Iz = 66.67 × 10− 6m4Iy112:= ⋅h⋅b3 Iy = 16.67 × 10− 6m4QA.y := 0 (since A' = 0)Normal Stress:σAPxAMz⋅yAIz−My⋅zAIy:= + σA = −20.0 kPaShear Stress : Since QA = 0, τA := 0In-plane Principal Stress:σx := σA σy := 0 τxy := 0Since no shear stress acts upon the element,σ1 := σy σ1 = 0 kPa Ansσ2 := σx σ2 = −20 kPa AnsMaximum In-plane Shear Stress: Applying Eq. 9-7,τmaxσx − σy2⎛⎜⎝⎞⎠2τxy:= + 2 τmax = 10 kPa Ans 941. Problem 9-52The internal loadings at a section of the beam consist of an axial force of 500 N, a shear force of 800N, and two moment components of 30 N·m and 40 N·m. Determine the principal stresses at point B.Also calculate the maximum in-plane shear stress at this point.Unit Used: kPa := 1000PaGiven: b := 100mm h := 200mm Px := 0.5kNyB := 0mm zB := −50mm Py := 0.8kNMy := −0.04kN⋅m Mz := 0.03kN⋅mSolution:Section Property :A := b⋅h A = 20000mm2Iz112:= ⋅b⋅h3 Iz = 66.67 × 10− 6m4Iy112:= ⋅h⋅b3 Iy = 16.67 × 10− 6m4QB.y := b⋅ (0.5h)⋅ (0.25h)Normal Stress:σBPxAMz⋅yBIz−My⋅zBIy:= + σB = 145.0 kPaShear Stress :τBPy⋅QB.yIz⋅b:= τB = 60.00 kPaIn-plane Principal Stress:σxB := σB σyB := 0 τxy := τBσB1σxB + σyB2σxB − σyB2⎛⎜⎝⎞⎠2τB:= + + 2 σB1 = 166.6 kPa AnsσB2σxB + σyB2⎛⎜⎝⎞σxB − σyB2⎠2τB:= − + 2 σB2 = −21.61 kPa AnsMaximum In-plane Shear Stress: Applying Eq. 9-7,τmax⎛⎜⎝⎞σxB − σyB2⎠2τxy:= + 2 τmax = 94.11 kPa Ans 942. Problem 9-53The internal loadings at a section of the beam consist of an axial force of 500 N, a shear force of 800N, and two moment components of 30 N·m and 40 N·m. Determine the principal stresses at point C.Also calculate the maximum in-plane shear stress at this point.Unit Used: kPa := 1000PaGiven: b := 100mm h := 200mm Px := 0.5kNyC := −50mm zC := 0mm Py := 0.8kNMy := −0.04kN⋅m Mz := 0.03kN⋅mSolution:Section Property :A := b⋅h A = 20000mm2Iz112:= ⋅b⋅h3 Iz = 66.67 × 10− 6m4Iy112:= ⋅h⋅b3 Iy = 16.67 × 10− 6m4QC.y b 0.5h yC − ( ) ⋅ yC 0.5 0.5h yC − ( ) + ⎡⎣:= ⋅ QC.y = 375000mm3⎤⎦Normal Stress:σCPxAMz⋅yCIz−My⋅zCIy:= + σC = 47.5 kPaShear Stress :τCPy⋅QC.yIz⋅b:= τC = 45.00 kPaIn-plane Principal Stress:σxC := σC σyC := 0 τxy := τCσC1σxC + σyC2σxC − σyC2⎛⎜⎝⎞⎠2τC:= + + 2 σC1 = 74.63 kPa AnsσC2σxC + σyC2⎛⎜⎝⎞σxC − σyC2⎠2τC:= − + 2 σC2 = −27.13 kPa AnsMaximum In-plane Shear Stress: Applying Eq. 9-7,τmax⎛⎜⎝⎞σxC − σyC2⎠2τxy:= + 2 τmax = 50.88 kPa Ans 943. Problem 9-54The beam has a rectangular cross section and is subjected to the loads shown. Write a computerprogram that can be used to determine the principal stresses at points A, B, C, and D. Show anapplication of the program using the values h = 300 mm, b = 200 mm, Nx = 2 kN, Vy = 1.5 kN, Vz = 0,My = 0, and Mz = -225 kN·m. 944. Problem 9-55The member has a rectangular cross section and is subjected to the loading shown. Write a computerprogram that can be used to determine the principal stresses at points A, B, and C. Show an applicationof the program using the values b = 150 mm, h = 200 mm, P = 1.5 kN, x = 75 mm, z = -50 mm, Vx =300 N, and Vz = 600 N. 945. Problem 9-56Solve Prob. 9-4 using Mohr's circle.Given: σx := −0.650MPa σy := 0.400MPaφ' := 60deg τxy := 0MPaSolution: θ := 90deg − φ' θ = 30.00 degCenter :σcσx + σy:= σc = −0.125MPa2Radius :R := σx − σc R = 0.525MPaCoordinates:A(σx , 0) B(σy , 0) C(σc , 0)Stresses:σx' := σc − R⋅cos(2θ) σx' = −0.387MPa Ansτx'y' := R⋅ sin(2θ) τx'y' = 0.455MPa Ans 946. Problem 9-57Solve Prob. 9-2 using Mohr's circle.Given: σx := 5MPa σy := 3MPa τxy := 8MPa φ' := 40degSolution:Center :σcσx + σy:= σc = 4MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 8.062MPaAngles: θ := 90deg + φ' θ = 130 degφ atanτxyσx − σc⎛⎜⎝⎞⎠:= φ = 82.875 degα := 180deg − (2θ − φ) α = 2.875 degStresses:σx' := σc − R⋅cos(α) σx' = −4.052MPa Ansτx'y' := −R⋅ sin(α) τx'y' = −0.404MPa Ans 947. Problem 9-58Solve Prob. 9-3 using Mohr's circle.Given: σx := 0.350MPa σy := −0.200MPaφ' := 50deg τxy := 0MPaSolution: θ := 90deg + φ' θ = 140 degCenter :σcσx + σy:= σc = 0.075MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 0.275MPaCoordinates:A(σx , 0) C(σc , 0)Angles:α := 360deg − 2θ α = 80 degStresses: (represented by coordinates of point P)σx' := σc + R⋅cos(α) σx' = 0.123MPa Ansτx'y' := R⋅ sin(α) τx'y' = 0.271MPa Ans 948. Problem 9-59Solve Prob. 9-10 using Mohr's circle.Given: σx := 0MPa σy := −0.300MPaθ := 30deg τxy := 0.950MPaSolution:Center :σcσx + σy:= σc = −0.15MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 0.962MPaAngles:φ atanτxyσx − σc⎛⎜⎝⎞⎠:= φ = 81.027 degα := 2θ − φ α = −21.027 degStresses:σx' := σc + R⋅cos(α) σx' = 0.748MPa Ansσy' := σc − R⋅cos(α) σy' = −1.048MPa Ansτx'y' := −R⋅ sin(α) τx'y' = 0.345MPa Ans 949. Problem 9-60Solve Prob. 9-6 using Mohr's circle.Given: σx := 90MPa σy := 50MPa τxy := −35MPaφ' := 60degSolution:Center :σcσx + σy:= σc = 70MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 40.311MPaAngles: θ := 90deg − φ'φ atanτxyσx − σc⎛⎜⎝⎞⎠:= φ = −60.255 degα := 180deg − (2θ − φ) α = 59.745 degStresses:σx' := σc − R⋅cos(α) σx' = 49.689MPa Ansτx'y' := −R⋅ sin(α) τx'y' = −34.821MPa Ans 950. Problem 9-61Solve Prob. 9-11 using Mohr's circle.Given: σx := 0.300MPa σy := 0MPaθ := −60deg τxy := 0.120MPaSolution:Center :σcσx + σy:= σc = 0.15MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 0.1921MPaCoordinates:A(σx , τxy) C(σc , 0)Angles:φ atanτxyσx − σc⎛⎜⎝⎞⎠:= φ = 38.660 degα := 180deg + 2θ − φ α = 21.340 degStresses: (represented by coordinates of points P and Q)σx' := σc − R⋅cos(α) σx' = −0.0289MPa Ansσy' := σc + R⋅cos(α) σy' = 0.329MPa Ansτx'y' := R⋅ sin(α) τx'y' = 0.0699MPa Ans 951. Problem 9-62Solve Prob. 9-13 using Mohr's circle.Given: σx := 45MPa σy := −60MPa τxy := 30MPaSolution:Center :σcσx + σy:= σc = −7.5MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 60.467MPaCoordinates:A(σx , τxy) C(σc , 0)Stresses:σ1 := σc + R σ1 = 52.97MPa Ansσ2 := σc − R σ2 = −67.97MPa Ansτmax := R τmax = 60.47MPa Ansσavg := σc σavg = −7.5MPa AnsAngles:θp112atanτxyσx − σc⎛⎜⎝⎞⎠:=θp1 = 14.872 deg (Counter-clockwise) Ans2θs1 = 90deg − 2θp1θs1 := 45deg − θp1θs1 = 30.128 deg (Clockwise) Ans 952. Problem 9-63Solve Prob. 9-14 using Mohr's circle.Given: σx := 180MPa σy := 0MPa τxy := −150MPaSolution:Center :σcσx + σy:= σc = 90MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 174.929MPaCoordinates:A(σx , τxy) B(σy , −τxy) C(σc , 0)Stresses:σ1 := σc + R σ1 = 264.93MPa Ansσ2 := σc − R σ2 = −84.93MPa Ansτmax := R τmax = 174.93MPa Ansσavg := σc σavg = 90MPa AnsAngles:θp12atanτxyσx − σc⎛⎜⎝⎞⎠:=θp = 29.518 deg (Clockwise) Ans2θs = 90deg − 2θpθs := 45deg − θpθs = 15.482 deg (Counter-clockwise) Ans 953. Problem 9-64Solve Prob. 9-16 using Mohr's circle.Given: σx := −200MPa σy := 250MPa τxy := 175MPaSolution:Center :σcσx + σy:= σc = 25MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 285.044MPaCoordinates:A(σx , τxy) B(σy , −τxy) C(σc , 0)Stresses:σ1 := σc + R σ1 = 310.04MPa Ansσ2 := σc − R σ2 = −260.04MPa Ansτmax := R τmax = 285.04MPa Ansσavg := σc σavg = 25MPa AnsAngles:θp12atanτxyσx − σc⎛⎜⎝⎞⎠:=θp = 18.937 deg (Clockwise) Ans2θs = 90deg − 2θpθs := 45deg − θpθs = 26.063 deg (Counter-clockwise) Ans 954. Problem 9-65Solve Prob. 9-15 using Mohr's circle.Given: σx := −30MPa σy := 0MPa τxy := −12MPaSolution:Center :σcσx + σy:= σc = −15MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 19.209MPaCoordinates:A(σx , τxy) C(σc , 0)Stresses:σ1 := σc + R σ1 = 4.21MPa Ansσ2 := σc − R σ2 = −34.21MPa Ansτmax := R τmax = 19.21MPa Ansσavg := σc σavg = −15MPa AnsAngles:θp212atanτxyσx − σc⎛⎜⎝⎞⎠:=θp2 = 19.330 deg (Countr-clockwise) Ans2θs2 = 2θp2 + 90degθs2 := θp2 + 45degθs2 = 64.330 deg (Countr-clockwise) Ans 955. Problem 9-66Determine the equivalent state of stress if an element is oriented 20° clockwise from the elementshown. Show the result on the element.Given: σx := 3MPa σy := −2MPaθ := −20deg τxy := −4MPaSolution:Center :σcσx + σy:= σc = 0.5MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 4.717MPaCoordinates:A(σx , τxy) C(σc , 0)Angles:φ atanτxyσx − σc⎛⎜⎝⎞⎠:= φ = 57.995 degα := φ + 2θ α = 17.995 degStresses: (represented by coordinates of points P and Q)σx' := σc + R⋅cos(α) σx' = 4.986MPa Ansσy' := σc − R⋅cos(α) σy' = −3.986MPa Ansτx'y' := −R⋅ sin(α) τx'y' = −1.457MPa Ans 956. Problem 9-67Determine the equivalent state of stress if an element is oriented 60° counterclockwise from theelement shown.Given: σx := 0.750MPa σy := −0.800MPaθ := 60deg τxy := 0.450MPaSolution:Center :σcσx + σy:= σc = −0.025MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 0.8962MPaCoordinates:A(σx , τxy) B(σy , −τxy) C(σc , 0)Angles:φ atanτxyσx − σc⎛⎜⎝⎞⎠:= φ = 30.141 degα := 2θ − φ α = 89.859 degStresses:σx' := σc + R⋅cos(α) σx' = −0.0228MPa Ansσy' := σc − R⋅cos(α) σy' = −0.0272MPa Ansτx'y' := −R⋅ sin(α) τx'y' = −0.896MPa Ans 957. Problem 9-68Determine the equivalent state of stress if an element is oriented 30° clockwise from the elementshown.Given: σx := 350MPa σy := 230MPaθ := 30deg τxy := −480MPaSolution:Center :σcσx + σy:= σc = 290MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 483.7355MPaCoordinates:A(σx , τxy) B(σy , −τxy) C(σc , 0)Angles:φ atanτxyσx − σc⎛⎜⎝⎞⎠:= φ = −82.875 degα := −2θ − φ α = 22.875 degStresses:σx' := σc + R⋅cos(α) σx' = 735.6922MPa Ansσy' := σc − R⋅cos(α) σy' = −155.6922MPa Ansτx'y' := −R⋅ sin(α) τx'y' = −188.038MPa Ans 958. Problem 9-69Determine the equivalent state of stress if an element is oriented 30° clockwise from the elementshown. Show the result on the element.Given: σx := 7MPa σy := 8MPaθ := −30deg τxy := 15MPaSolution:Center :σcσx + σy:= σc = 7.5MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 15.0083MPaCoordinates:A(σx , τxy) C(σc , 0)Angles:φ atanτxyσx − σc⎛⎜⎝⎞⎠:= φ = 88.091 degα := 180deg − (2 θ + φ) α = 31.909 degStresses: (represented by coordinates of points P and Q)σx' := σc − R⋅cos(α) σx' = −5.240MPa Ansσy' := σc + R⋅cos(α) σy' = 20.240MPa Ansτx'y' := R⋅ sin(α) τx'y' = 7.933MPa Ans 959. Problem 9-70Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normalstress. Specify the orientation of the element in each case.Given: σx := 350MPa σy := −200MPa τxy := 500MPaSolution:Center :σcσx + σy:= σc = 75MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 570.636MPaCoordinates:A(σx , τxy) C(σc , 0)a) Principal Stresses:σ1 := σc + R σ1 = 645.64MPa Ansσ2 := σc − R σ2 = −495.64MPa AnsAngles:θp112atanτxyσx − σc⎛⎜⎝⎞⎠:=θp1 = 30.595 deg (Counter-clockwise) Ansb) Maximum In-plane Shear Stress: (represented by coordinates of point E)τmax := R τmax = 570.64MPa Ansσavg := σc σavg = 75MPa Ans2θs = 90deg − 2⋅θp1θs := 45deg − θp1θs = 14.405 deg (Clockwise) Ans 960. Problem 9-71Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normalstress. Specify the orientation of the element in each case.Given: σx := 10MPa σy := 80MPa τxy := −60MPaSolution:Center :σcσx + σy:= σc = 45MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 69.462MPaCoordinates:A(σx , τxy) C(σc , 0)a) Principal Stresses:σ1 := σc + R σ1 = 114.46MPa Ansσ2 := σc − R σ2 = −24.46MPa AnsAngles:θp212atanτxyσx − σc⎛⎜⎝⎞⎠:=θp1 := 90deg − θp2θp1 = 60.128 deg (Clockwise) Ansb) Maximum In-plane Shear Stress: (represented by coordinates of point E)τmax := −R τmax = −69.46MPa Ansσavg := σc σavg = 45MPa Ans2θs2 = 90deg − 2⋅θp2θs2 := 45deg − θp2θs2 = 15.128 deg (Clockwise) Ans 961. Problem 9-72Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normalstress. Specify the orientation of the element in each case.Given: σx := 0MPa σy := 50MPa τxy := −30MPaSolution:Center :σcσx + σy:= σc = 25MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 39.051MPaCoordinates:A(σx , τxy) B(σy , −τxy) C(σc , 0)a) Principal Stresses:σ1 := σc + R σ1 = 64.05MPa Ansσ2 := σc − R σ2 = −14.05MPa AnsAngles:θp12atanτxyσx − σc⎛⎜⎝⎞⎠:=θp = 25.097 deg (Clockwise)b) Maximum In-plane Shear Stress: (represented by coordinates of point E)τmax := R τmax = 39.05MPa Ansσavg := σc σavg = 25MPa Ans2θs2 = 90deg − 2⋅θpθs2 := 45deg − θpθs2 = 19.903 deg (Counter-clockwise) 962. Problem 9-73Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normalstress. Specify the orientation of the element in each case.Given: σx := −12MPa σy := −8MPa τxy := 4MPaSolution:Center :σcσx + σy:= σc = −10MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 4.472MPaCoordinates:A(σx , τxy) B(σy , −τxy) C(σc , 0)a) Principal Stresses:σ1 := σc + R σ1 = −5.53MPa Ansσ2 := σc − R σ2 = −14.47MPa AnsAngles:θp12atanτxyσx − σc⎛⎜⎝⎞⎠:=θp = 31.717 deg (Clockwise)b) Maximum In-plane Shear Stress: (represented by coordinates of point E)τmax := R τmax = 4.47MPa Ansσavg := σc σavg = −10MPa Ans2θs2 = 90deg − 2⋅θpθs2 := 45deg − θpθs2 = 13.283 deg (Counter-clockwise) 963. Problem 9-74Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normalstress. Specify the orientation of the element in each case.Given: σx := 45MPa σy := 30MPa τxy := −50MPaSolution:Center :σcσx + σy:= σc = 37.5MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 50.559MPaCoordinates:A(σx , τxy) B(σy , −τxy) C(σc , 0)a) Principal Stresses:σ1 := σc + R σ1 = 88.06MPa Ansσ2 := σc − R σ2 = −13.06MPa AnsAngles:θp12atanτxyσx − σc⎛⎜⎝⎞⎠:=θp = 40.735 deg (Clockwise)b) Maximum In-plane Shear Stress: (represented by coordinates of point E)τmax := R τmax = 50.56MPa Ansσavg := σc σavg = 37.5MPa Ans2θs2 = 90deg − 2⋅θpθs2 := 45deg − θpθs2 = 4.265 deg (Counter-clockwise) 964. Problem 9-75The square steel plate has a thickness of 12 mm and is subjected to the edge loading shown. Determinethe principal stresses developed in the steel.Given: σx := 0MPa σy := 0MPa qxy 3.2kNm:=L := 100mm t := 12mmSolution:τxyqxyt:= τxy = 0.267MPaCenter :σcσx + σy:= σc = 0MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 0.267MPaCoordinates:A(σx , τxy) C(σc , 0)In-plane Principal Stresses:The coordinates of points B and D represent σ1 and σ2, respectively.σ1 := σc + R σ1 = 0.267MPa Ansσ2 := σc − R σ2 = −0.267MPa Ans 965. Problem 9-76Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normalstress. Specify the orientation of the element in each case.Given: σx := 105MPa σy := 0MPa τxy := −35MPaSolution:Center ::= σc = 52.5MPaσ1 = 115.60MPaσcσx + σy2Radius :R (σx − σc)2 τxy:= + 2 R = 63.097MPaCoordinates:A(σx , τxy ) C(σc , 0)a) In-planePrincipal Stresses:The coordinates of points B and D represent σ1 and σ2, respectively.σ1 := σc + RAnsσ2 := σc − R σ2 = −10.60MPa AnsAngles:θpAns12atanτxyσx − σc⎛⎜⎝⎞⎠:=θp = 16.845 deg (Clockwise)b) Maximum In-plane Shear Stresses:Represented by the coordinates of point E on the circle.τmax := −R τmax = −63.10MPa Ans2θs = 90deg − 2⋅θpθs := 45deg − θpθs = 28.155 deg (Counter-clockwise) 966. Problem 9-77Draw Mohr's circle that describes each of the following states of stress.Given: σx := −30MPa σy := 30MPaτxy := 0MPaSolution:Cases (a) and (b) are the same.Center :σcσx + σy:= σc = 0MPa2Radius :R (σx − σc)2 τxy:= + 2 R = 30MPaCoordinates:A(σx , τxy) C(σc , 0) 967. Problem 9-78Draw Mohr's circle that describes each of the following states of stress.a) Given: σx := 0.8MPa τxy := 0MPaσy := −0.6MPaSolution:Center : σcσx + σy:=2σc = 0.1MPaRadius : R (σx − σc)2 τxy:= + 2R = 0.7MPaCoordinates: A(σx , τxy) C(σc , 0)b) Given: σx := 0MPa τxy := 0MPaσy := −2MPaSolution:Center : σcσx + σy:=2σc = −1MPaRadius : R (σx − σc)2 τxy:= + 2R = 1MPaCoordinates: A(σx , τxy) C(σc , 0)c) Given: σx := 0MPa τxy := 20MPaσy := 0MPaSolution:Center : σcσx + σy:=2σc = 0MPaRadius : R (σx − σc)2 τxy:= + 2R = 20MPaCoordinates: A(σx , τxy) C(σc , 0) 968. Problem 9-79A point on a thin plate is subjected to two successive states of stress as shown. Determine the resultingstate of stress with reference to an element oriented as shown on the bottom.Unit used: kPa := 1000PaGiven: σx_a := 50kPa σy_a := 0 τxy_a := 0σy_b := −18kPa σx_b := 0 τxy_b := −45kPaθa := 30deg βb := 50degSolution:For element a:Center : σc_aσx_a + σy_a:= σc_a = 25 kPa2Radius : Ra (σx_a − σc_a)2 τxy_a:= + 2 Ra = 25 kPaCoordinates: A(σx_a , τxy_a) B(σx_a , −τxy_a) C(σc_a , 0)σ'x_a := σc_a + Ra⋅cos(2θa)σ'y_a := σc_a − Ra⋅cos(2θa)τ'xy_a := Ra⋅ sin(2θa)For element b: θb := −(90deg − βb)σx_b + σy_Center : bσc_b:= σc_b = −9 kPa2Radius : Rb (σx_b − σc_b)2 τxy_b:= + 2 Rb = 45.891 kPaCoordinates: A(σx_b , τxy_b) B(σx_a , −τxy_b) C(σc_b , 0)Angles :φ atanτxy_bσx_b − σc_b⎛⎜⎝⎞⎠:= φ = 78.69 degαb := 2θb + φαb = −1.310 degσ'x_b := σc_b + Rb⋅cos(αb)σ'y_b := σc_b − Rb⋅cos(αb)τ'xy_b := Rb⋅ sin( αb )Resultants:σ'x := σ'x_a + σ'x_b σ'x = 74.38 kPa Ansσ'y := σ'y_a + σ'y_b σ'y = −42.38 kPa Ansτ'xy := τ'xy_a + τ'xy_b τ'xy = 22.70 kPa Ans 969. Problem 9-80Mohr's circle for the state of stress in Fig. 9-15a is shown in Fig. 9-15b. Show that finding thecoordinates of point P(σx' , τx'y' ) on the circle gives the same value as the stress-transformation Eqs.9-1 and 9-2. 970. Problem 9-81The cantilevered rectangular bar is subjected to the force of 25 kN. Determine the principal stresses atpoint A.Given: b := 80mm d := 160mm cA := 40mmP := 25kN L := 0.4m bA := −40mmrv35:= rh45:=Solution:Internal Force and Moment : At Section A-B:hr+ ΣFx=0; − P ⋅ rh + N = 0 N := P ⋅vr+ ΣFy=0; − P ⋅ rv + V = 0 V := P ⋅⋅ ( ) L ⋅ − 0 = M P rv+ ΣΜO=0; M Prv:= ( ⋅ )⋅LSection Property : dA := 0.5d − cAA := b⋅d I112:= ⋅b⋅d3QA := b⋅dA⋅ (0.5dA + cA)Normal Stress:σANAM⋅cAI:= + σA = 10.352MPaShear Stress :τAV⋅QAI⋅b:= τA = 1.318MPaConstruction of Mohr's Circle :σx := σA σy := 0MPa τxy := −τACenter : σcσx + σy:= σc = 5.176MPa2Radius : R (σx − σc)2 τxy:= + 2R = 5.341MPaCoordinates: A(σx , τxy) C(σc , 0)In-plane Principal Stresses:The coordinates of points B and D represent σ1 and σ2, respectively.σ1 := σc + R σ1 = 10.52MPa Ansσ2 := σc − R σ2 = −0.165MPa Ans 971. Problem 9-82Solve Prob. 9-81 for the principal stresses at point B.Given: b := 80mm d := 160mm cB := −25mmP := 25kN L := 0.4m bB := 25mmrv35:= rh45:=Solution:Internal Force and Moment : At Section A-B:hr+ ΣFx=0; − P ⋅ rh + N = 0 N := P ⋅vr+ ΣFy=0; − P ⋅ rv + V = 0 V := P ⋅⋅ ( ) L ⋅ − 0 = M P rv+ ΣΜO=0; M Prv:= ( ⋅ )⋅LSection Property : dB := 0.5d − cBA := b⋅d I112:= ⋅b⋅d3QB := b⋅dB⋅ (0.5dB + cB )Normal Stress:σBNAM⋅cBI:= + σB = −3.931MPaShear Stress :τBV⋅QBI⋅b:= τB = 1.586MPaConstruction of Mohr's Circle :σx := σB σy := 0MPa τxy := −τBCenter : σcσx + σy:= σc = −1.965MPa2Radius : R (σx − σc)2 τxy:= + 2R = 2.526MPaCoordinates: A(σx , τxy) C(σc , 0)In-plane Principal Stresses:The coordinates of points B and D represent σ1 and σ2, respectively.σ1 := σc + R σ1 = 0.560MPa Ansσ2 := σc − R σ2 = −4.491MPa Ans 972. Problem 9-83The stair tread of the escalator is supported on two of its sides by the moving pin at A and the roller atB. If a man having a weight of 1500 N (~150 kg) stands in the center of the tread, determine theprincipal stresses developed in the supporting truck on the cross section at point C. The stairs move atconstant velocity.Given: b := 12mm d := 50mm cC := 0mmW := 1.5kN hC := 150mm θ := 30degvA := 450mm hA := 375mm hB := 150mmSolution:Support Reactions :+ ΣΜA=0; W⋅hA − By⋅hB = 0 ByW⋅hAhB:=Bx := −By⋅ tan(θ)Internal Force and Moment : At Section C:+ ΣFx=0; Bx + V = 0 V := −Bx+ ΣFy=0; By + N = 0 N := −By+ ΣΜO=0; Bx⋅hC − M = 0 M := Bx⋅hCSection Property : dC := 0.5dA := b⋅d I112:= ⋅b⋅d3 QC := b⋅dC⋅ (0.5dC)Normal Stress:σCNAM⋅cCI:= + σC = −6.250MPaShear Stress :τCV⋅QCI⋅b:= τC = 5.413MPaConstruction of Mohr's Circle :σx := 0MPa σy := σC τxy := τCσx + σyCenter : σc:= σc = −3.125MPa2Radius : R (σx − σc)2 τxy:= + 2 R = 6.25MPaCoordinates: A(σx , τxy) C(σc , 0)In-plane Principal Stresses:The coordinates of points B and D represent σ1 and σ2, respectively.σ1 := σc + R σ1 = 3.125MPa Ansσ2 := σc − R σ2 = −9.375MPa Ans 973. Problem 9-84The pedal crank for a bicycle has the cross section shown. If it is fixed to the gear at B and does notrotate while subjected to a force of 400 N, determine the principal stresses in the material on the crosssection at point C.Given: b := 7.5mm d := 20mm cC := 5mmP := 0.4kN L := 100mmSolution:Internal Force and Moment : At Section C:+ ΣFy=0; V − P = 0 V := P+ ΣΜO=0; M + P⋅L = 0 M := −P⋅LSection Property : dC := 0.5d − cCA := b⋅d I112:= ⋅b⋅d3QC := b⋅dC⋅ (cC + 0.5dC)Normal Stress:σCM⋅cCI:= − σC = 40.000MPaShear Stress :τCV⋅QCI⋅b:= τC = 3.000MPaConstruction of Mohr's Circle :σx := σC σy := 0MPa τxy := τCσx + σyCenter : σc:= σc = 20MPa2Radius : R (σx − σc)2 τxy:= + 2 R = 20.224MPaCoordinates: A(σx , τxy) C(σc , 0)In-plane Principal Stresses:The coordinates of points B and D represent σ1 and σ2, respectively.σ1 := σc + R σ1 = 40.224MPa Ansσ2 := σc − R σ2 = −0.224MPa Ans 974. Problem 9-85The frame supports the distributed loading of 200 N/m. Determine the normal and shear stresses atpoint D that act perpendicular and parallel, respectively, to the grains. The grains at this point make anangle of 30° with the horizontal as shown.Unit Used: kPa := 1000PaGiven: b := 100mm a := 1m w 0.2kNm:=h := 200mm L := 2.5mhD := 75mm θ := 60degSolution:Support Reactions: Due to symmetry, B=C=R+ ΣFy=0; 2R − w⋅L = 0 R := 0.5w⋅LInternal Force and Moment : At Section D:+ ΣFy=0; −V + R − w⋅a = 0V := R − w⋅a+ ΣΜO=0; M − R⋅a + wa⋅ (0.5a) = 0M := R⋅a − w⋅a⋅ (0.5a)Section Property : cD := 0.5h − hDA := b⋅h I112:= ⋅b⋅h3QD := b⋅hD⋅ (cD + 0.5hD)Normal Stress: σDM⋅ (−cD):= − σD = 56.25 kPaIShear Stress : τDV⋅QDI⋅b:= τD = 3.516 kPaConstruction of Mohr's Circle :σx := σD σy := 0MPa τxy := τDσx + σyCenter : σc:= σc = 28.125 kPa2Radius : R (σx − σc)2 τxy:= + 2 R = 28.344 kPaCoordinates: A(σx , −τxy) C(σc , 0)Angles:φ atanτxyσx − σc⎛⎜⎝⎞⎠:= φ = 7.125 degα := 180deg − (2 θ + φ) α = 52.875 degStresses on the Rotated Element: (represented by coordinates of point P)σx' := σc − R⋅cos(α) σx' = 11.02 kPa Ansτx'y' := −R⋅ sin(α) τx'y' = −22.60 kPa Ans 975. Problem 9-86The frame supports the distributed loading of 200 N/m. Determine the normal and shear stresses atpoint E that act perpendicular and parallel, respectively, to the grains. The grains at this point make anangle of 60° with the horizontal as shown.Unit Used: kPa := 1000PaGiven: b := 50mm w 0.2kNm:=h := 100mmL := 2.5m θ := 60degSolution:Support Reactions: Due to symmetry, B=C=R+ ΣFy=0; 2R − w⋅L = 0 R := 0.5w⋅LInternal Force: At Section E:+ ΣFy=0; −N − R = 0 N := −RSection Property :A := b⋅h A = 5000mm2Normal Stress: σENA:= σE = −50 kPaShear Stress : τD := 0Construction of Mohr's Circle :σx := σE σy := 0MPa τxy := τDσx + σyCenter : σc:= σc = −25 kPa2Radius : R (σx − σc)2 τxy:= + 2 R = 25 kPaCoordinates: A(σx , τxy) C(σc , 0)Angles:φ atanτxyσx − σc⎛⎜⎝⎞⎠:= φ = 0.000 degα := 180deg − (2 θ + φ) α = 60.000 degStresses on the Rotated Element: (represented by coordinates of point P)σx' := σc + R⋅cos(α) σx' = −12.50 kPa Ansτx'y' := R⋅ sin(α) τx'y' = 21.65 kPa Ans 976. Problem 9-87The bent rod has a diameter of 15 mm and is subjected to the force of 600 N. Determine the principalstresses and the maximum in-plane shear stress that are developed at point A and point B. Show theresults on elements located at these points.Given: do := 15mm P := 0.6kNa := 50mm L := 0.1mSolution:Internal Force: At Section AB:N := P N = 0.600 kNM := P⋅a M = 0.030 kN⋅mSection Property :Aπ4:= ⋅ 2 A = 176.71mm2doIπ644 ⋅ := I 0 m 2do= mm2QA := 0 (since A' = 0)QB := 0 (since A' = 0)Normal Stress:cA := 0.5do σANAM⋅cAI:= − σA = −87.15MPacB := 0.5do σBNAM⋅cBI:= + σB = 93.94MPaShear Stress :τA := 0 (since QA = 0)τB := 0 (since QB = 0)For A:Construction of Mohr's Circle :σx := σA σy := 0MPa τxy := τAσx + σyCenter : σc:= σc = −43.573MPa2Radius : R (σx − σc)2 τxy:= + 2 R = 43.573MPaCoordinates: A(σx , τxy) C(σc , 0)In-plane Principal Stresses:The coordinates of points B and A represent σ1 and σ2, respectively.σ1 := σc + R σ1 = 0MPa Ansσ2 := σc − R σ2 = −87.15MPa Ans 977. Maximum In-plane Shear Stresses:Represented by the coordinates of point E on the circle.τmax := R τmax = 43.57MPa Ans2θs = 90deg θs := 45deg (Counter-clockwise) AnsFor B:Construction of Mohr's Circle :σx := σB σy := 0MPa τxy := τBσx + σyCenter : σc:= σc = 46.968MPa2Radius : R (σx − σc)2 τxy:= + 2 R = 46.968MPaCoordinates: A(σx , τxy) C(σc , 0)In-plane Principal Stresses:The coordinates of points A and B represent σ1 and σ2, respectively.σ1 := σc + R σ1 = 93.94MPa Ansσ2 := σc − R σ2 = 0MPa AnsMaximum In-plane Shear Stresses:Represented by the coordinates of point E on the circle.τmax := R τmax = 46.97MPa Ans2θs = 90deg θs := 45deg (Clockwise) Ans 978. Problem 9-88Draw the three Mohr's circles that describe each of the following states of stress.a) σmax := 6MPa σint := 0σmin := 0b) σmax := 50MPa σint := 0σmin := −40MPac) Unit used: kPa := 1000Paσmax := 600kPa σint := 200kPaσmin := 100kPad) σmax := 0 σint := −7MPaσmin := −9MPae) σmax := −30MPa σint := −30MPaσmin := −30MPa 979. Problem 9-89Draw the three Mohr's circles that describe each of the following states of stress.a) σ1 := 15MPa σ2 := 0σ3 := −15MPaτmax := 15MPab) σ1 := 65MPa σ2 := −65MPaσ3 := −65MPaτmax := 65MPa 980. Problem 9-90The stress at a point is shown on the element. Determine the principal stresses and the absolutemaximum shear stress.Given: σx := −100MPa σy := 90MPa σz := −80MPaτxy := 0MPa τyz := 40MPa τxz := 0MPaSolution:Construction of Mohr's Circle in y-z Plane :Center : σcσy + σz:= σc = 5MPa2Radius : R (σy − σc)2 τyz:= + 2 R = 93.941MPaCoordinates: A(σy , τyz) C(σc , 0)In-plane Principal Stresses:The coordinates of points A and B represent σ1 and σ2, respectively.σ1 := σc + R σ1 = 98.941MPaσ2 := σc − R σ2 = −88.941MPaConstruction of Three Circles :From the results obtained above,σmax := σ1 σint := σ2 σmin := σxσmax = 98.94MPa Ansσint = −88.94MPa Ansσmin = −100.00MPa AnsAbsolute Maximum Shear Stress :From the three Mohr's circles,σmax − σminτabs.max2:=τabs.max = 99.47MPa Ans 981. Problem 9-91The stress at a point is shown on the element. Determine the principal stresses and the absolutemaximum shear stress.Given: σx := 0MPa σy := 7MPa σz := 0MPaτxy := 0MPa τyz := 50MPa τxz := 5MPaSolution:Construction of Mohr's Circle in x-z Plane :Center : σcσx + σz:= σc = 0MPa2Radius : R (σx − σc)2 τxz:= + 2 R = 5MPaCoordinates: A(σy , τxz) C(σc , 0)In-plane Principal Stresses:The coordinates of points A and B represent σ1 and σ2, respectively.σ1 := σc + R σ1 = 5MPaσ2 := σc − R σ2 = −5MPaConstruction of Three Circles :From the results obtained above,σmax := σy σint := σ1 σmin := σ2σmax = 7MPa Ansσint = 5MPa Ansσmin = −5MPa AnsAbsolute Maximum Shear Stress :From the three Mohr's circles,σmax − σminτmax2:=τmax = 6MPa Ans 982. Problem 9-92The stress at a point is shown on the element. Determine the principal stresses and the absolutemaximum shear stress.Given: σx := 150MPa σy := 0MPa σz := 90MPaτxy := 0MPa τyz := −80MPa τxz := 0MPaSolution:Construction of Mohr's Circle in y-z Plane :Center : σcσy + σz:= σc = 45MPa2Radius : R (σz − σc)2 τyz:= + 2 R = 91.788MPaCoordinates: A(σy , τyz) B(σz , −τyz) C(σc , 0)In-plane Principal Stresses:The coordinates of points A and B represent σ1 and σ2, respectively.σ1 := σc + R σ1 = 136.788MPaσ2 := σc − R σ2 = −46.788MPaConstruction of Three Circles :From the results obtained above,σmax := σx σint := σ1 σmin := σ2σmax = 150.00MPa Ansσint = 136.79MPa Ansσmin = −46.79MPa AnsAbsolute Maximum Shear Stress :From the three Mohr's circles,σmax − σminτmax2:=τmax = 98.39MPa Ans 983. Problem 9-93The principal stresses acting at a point in a body are shown. Draw the three Mohr's circles thatdescribe this state of stress, and find the maximum in-plane shear stresses and associated averagenormal stresses for the x-y, y-z, and x-z planes. For each case, show the results on the element orientedin the appropriate direction.Given: σx := 40MPa σy := −40MPa σz := −40MPaτxy := 0MPa τyz := 0MPa τxz := 0MPaSolution:Construction of Three Circles :σmax := σx σint := σy σmin := σzσmax = 40.00MPaσint = −40.00MPaσmin = −40.00MPaFor x-y Plane :σavgσx + σy:= σavg = 0MPa Ans2τmaxσx − σy:= τmax = 40MPa Ans2For y-z Plane :σ'avgσy + σz:= σ'avg = −40MPa Ans2τ'maxσy − σz:= τ'max = 0MPa Ans2For x-y Plane :σ''avgσx + σz:= σ''avg = 0MPa Ans2τ''maxσx − σz:= τ''max = 40MPa Ans2 984. Problem 9-94Consider the general case of plane stress as shown. Write a computer program that will show a plot ofthe three Mohr's circles for the element, and will also calculate the maximum in-plane shear stress andthe absolute maximum shear stress. 985. Problem 9-95The solid shaft is subjected to a torque, bending moment, and shear force as shown. Determine theprincipal stresses acting at points A and B and the absolute maximum shear stress.Given: ro := 25mm L := 450mmP := 0.8kN To := 0.045kN⋅m Mo := 0.3kN⋅mSolution: ρ := roInternal Force and Moment : At Section AB:Vy := P Tx := To Mz := Mo − P⋅LSection Property ::= ⋅ 4 A π roJπ2ro:= ⋅ 2 Izπ4:= ⋅ 4roQA := 0 (Since A'=0) QB4⋅ ro3⋅π:= ⋅ (0.5A)Normal Stress: cA := ro cB := 0σAMz⋅cAIz:= − σA = 4.889MPaσBMz⋅cBIz:= − σB = 0MPaShear Stress : bB := 2⋅ roτATx⋅ρJ:= τA = 1.833MPaτBVy⋅QBIz⋅bBTx⋅ρJ:= − τB = −1.290MPaFor Point A:Construction of Mohr's Circle:σx := σA σz := 0MPa τxz := −τAσx + σzCenter : σc:= σc = 2.445MPa2Radius : R (σz − σc)2 τxz:= + 2 R = 3.056MPaCoordinates: A(σz , τxz) C(σc , 0)In-plane Principal Stresses:The coordinates of points B and D represent σ1 and σ2, respectively.σ1 := σc + R σ1 = 5.500MPaσ2 := σc − R σ2 = −0.611MPa 986. Three Mohr's Circlesσ: y := 0From the results obtained above,σmax := σ1 σint := σy σmin := σ2σmax = 5.50MPa Ansσint = 0.00MPa Ansσmin = −0.611MPa AnsAbsolute Maximum Shear Stress :From the three Mohr's circles,σmax − σminτmax:= τmax = 3.06MPa Ans2For Point B:Construction of Mohr's Circle:σx := σB σz := 0MPa τxz := τBσx + σzCenter : σc:= σc = 0MPa2Radius : R (σz − σc)2 τxz:= + 2 R = 1.290MPaCoordinates: A(σz , τxz) C(σc , 0)In-plane Principal Stresses:The coordinates of points B and D represent σ1 and σ2, respectively.σ1 := σc + R σ1 = 1.290MPaσ2 := σc − R σ2 = −1.290MPaThree Mohr's Circlesσ: y := 0From the results obtained above,σmax := σ1 σint := σy σmin := σ2σmax = 1.29MPa Ansσint = 0.00MPa Ansσmin = −1.29MPa AnsAbsolute Maximum Shear Stress :From the three Mohr's circles,σmax − σminτmax:= τmax = 1.29MPa Ans2 987. Problem 9-96The bolt is fixed to its support at C. If a force of 90 kN is applied to the wrench to tighten it, determinethe principal stresses and the absolute maximum shear stress developed in the bolt shank at point A.Represent the results on an element located at this point. The shank has a diameter of 6 mm.Given: do := 6mm a := 50mmP := 90N L := 150mmSolution: ρ := 0.5doInternal Force and Moment : At Section AB:Vy := P Mz := −P⋅a Tx := P⋅LSection Property :Aπ4:= ⋅ do2 Izπ64:= ⋅ 4 Jdoπ32:= ⋅ 4doQA := 0 (Since A'=0)Normal Stress: cA := 0.5doσAMz⋅cAIz:= − σA = 212.21MPaShear Stress :τATx⋅ρJ:= τA = 318.31MPaConstruction of Mohr's Circle in x-z Plane :σx := σA σz := 0MPa τxz := −τAσx + σzCenter : σc:= σc = 106.1MPa2Radius : R (σz − σc)2 τxz:= + 2 R = 335.53MPaCoordinates: A(σz , τxz) C(σc , 0)In-plane Principal Stresses:The coordinates of points B and D represent σ1 and σ2, respectively.σ1 := σc + R σ1 = 441.631MPaσ2 := σc − R σ2 = −229.425MPaAngles:θp212atanτxzσz − σc⎛⎜⎝⎞⎠:= θp2 = 35.78 deg2θp1 = 180deg − 2⋅θp2 θp1 := 90deg − θp2 θp1 = 54.22 deg (Clockwise)Construction of Three Circles : σy := 0From the results obtained above, 988. σmax := σ1 σint := σy σmin := σ2σmax = 441.63MPa Ansσint = 0.00MPa Ansσmin = −229.42MPa AnsAbsolute Maximum Shear Stress :From the three Mohr's circles,σmax − σminτmax2:=τmax = 335.53MPa AnsAnd the orientation is,θs12atanσz − σcτxz⎛⎜⎝⎞⎠:= θs = 9.22 deg 989. Problem 9-97Solve Prob. 9-96 for point B.Given: do := 6mm a := 50mmP := 90N L := 150mmSolution: ρ := 0.5doInternal Force and Moment : At Section AB:Vy := P Mz := −P⋅a Tx := P⋅LSection Property :Aπ4:= ⋅ do2 Izπ64:= ⋅ 4 Jdoπ32:= ⋅ 4doQB4ρ3πA2⎛⎜⎝⎞⎠:= ⋅Normal Stress: cB := 0σBMz⋅cBIz:= − σB = 0MPaShear Stress : bB := doτBVy⋅QBIz⋅bBTx⋅ρJ:= − τB = −314.07MPaConstruction of Mohr's Circle in x-z Plane :σx := σB σz := 0MPa τxz := τBσx + σzCenter : σc:= σc = 0MPa2Radius : R (σz − σc)2 τxz:= + 2 R = 314.07MPaCoordinates: A(σz , τxz) C(σc , 0)In-plane Principal Stresses:The coordinates of points B and D represent σ1 and σ2, respectively.σ1 := σc + R σ1 = 314.07MPaσ2 := σc − R σ2 = −314.07MPaAngles:θp112:= (90deg) θp1 = 45.00 deg (Clockwise) 990. Construction of Three Circles : σy := 0From the results obtained above,σmax := σ1 σint := σy σmin := σ2σmax = 314.07MPa Ansσint = 0.00MPa Ansσmin = −314.07MPa AnsAbsolute Maximum Shear Stress :From the three Mohr's circles,σmax − σminτmax2:=τmax = 314.07MPa Ans 991. Problem 9-98The stress at a point is shown on the element. Determine the principal stresses and the absolutemaximum shear stress.Given: σx := −90MPa σy := 70MPa σz := −120MPaτxy := 30MPa τyz := 0MPa τxz := 0MPaSolution:Construction of Mohr's Circle in x-y Plane :Center : σcσx + σy:= σc = −10MPa2Radius : R (σx − σc)2 τxy:= + 2 R = 85.44MPaCoordinates: A(σy , −τxy) B(σx , τxy) C(σc , 0)In-plane Principal Stresses:The coordinates of points representing σ1 and σ2, respectively, areσ1 := σc + R σ1 = 75.440MPaσ2 := σc − R σ2 = −95.440MPaConstruction of Three Circles :From the results obtained above,σmax := σ1 σint := σ2 σmin := σzσmax = 75.44MPa Ansσint = −95.44MPa Ansσmin = −120.00MPa AnsAbsolute Maximum Shear Stress :From the three Mohr's circles,σmax − σminτmax2:=τmax = 97.72MPa Ans 992. Problem 9-99The cylindrical pressure vessel has an inner radius of 1.25 m and a wall thickness of 15 mm. It is madefrom steel plates that are welded along a 45° seam with the horizontal. Determine the normal and shearstress components along this seam if the vessel is subjected to an internal pressure of 3 MPa.Given: t := 15mm ri := 1250mm p := 3MPaθ := 45degSolution:Hoop Stress : αrit:= α = 83.33Since α > 10. then thin-wall analysis can be used.σ1p⋅ rit:= σ1 = 250MPaLongitudinal Stress :σ2p⋅ ri2⋅ t:= σ2 = 125MPaConstruction of Mohr's Circle:σx := σ2 σy := σ1 τxy := 0σx + σyCenter : σc:= σc = 187.5MPa2Radius : R (σy − σc)2 τxy:= + 2 R = 62.5MPaCoordinates: A(σx , τxz) C(σc , 0)Stresses: (represented by coordinates of point P on the circle)σx' := σc σx' = 187.5MPa Ansτx'y' := R τx'y' = 62.5MPa Ans 993. Problem 9-100Determine the equivalent state of stress if an element is oriented 40° clockwise from the elementshown. Use Mohr's circle.Given: σx := 6MPa σy := −10MPaθ := 40deg τxy := 0MPaSolution:Center :σcσx + σy:= σc = −2MPa2Radius :R := σx − σc R = 8MPaCoordinates:A(σx , 0) B(σy , 0) C(σc , 0)Stresses:σx' := σc + R⋅cos(2θ) σx' = −0.611MPa Ansτx'y' := R⋅ sin(2θ) τx'y' = 7.878MPa Ansσy' := σc − R⋅cos(2θ) σy' = −3.389MPa Ans 994. Problem 9-101The internal loadings at a cross section through the 150-mm-diameter drive shaft of a turbine consistof an axial force of 12.5 kN, a bending moment of 1.2 kN·m, and a torsional moment of 2.25 kN·m.Determine the principal stresses at point A. Also compute the maximum in-plane shear stress at thispoint.Given: do := 150mm N := −12.5kNMz := 1.2kN⋅m Tx := 2.25kN⋅mSolution: ρ := 0.5doSection Property :Aπ4:= ⋅ do2 Izπ64:= ⋅ 4 Jdoπ32:= ⋅ 4doNormal Stress: cA := ρσANAMz⋅cAIz:= − σA = −4.33MPaShear Stress :τATx⋅ρJ:= τA = 3.4MPaConstruction of Mohr's Circle in x-y Plane :σx := σA σy := 0MPa τxy := τAσx + σyCenter : σc:= σc = −2.16MPa2Radius : R (σx − σc)2 τxy:= + 2 R = 4.03MPaCoordinates: A(σx , τxy) C(σc , 0)In-plane Principal Stresses:The coordinates of points B and D represent σ1 and σ2, respectively.σ1 := σc + R σ1 = 1.862MPaσ2 := σc − R σ2 = −6.191MPaMaximum In-plane Shear Stress :Represented by the coordinates of point E on the circle.τmax := R τmax = 4.03MPa Ans 995. Problem 9-102The internal loadings at a cross section through the 150-mm-diameter drive shaft of a turbine consistof an axial force of 12.5 kN, a bending moment of 1.2 kN·m, and a torsional moment of 2.25 kN·m.Determine the principal stresses at point B. Also compute the maximum in-plane shear stress at thispoint.Given: do := 150mm N := −12.5kNMz := 1.2kN⋅m Tx := 2.25kN⋅mSolution: ρ := 0.5doSection Property :Aπ4:= ⋅ do2 Izπ64:= ⋅ 4 Jdoπ32:= ⋅ 4doNormal Stress: cB := 0σBNAMz⋅cBIz:= + σB = −0.707MPaShear Stress :τBTx⋅ρJ:= τB = 3.395MPaConstruction of Mohr's Circle in x-y Plane :σx := σB σy := 0MPa τxy := τBσx + σyCenter : σc:= σc = −0.354MPa2Radius : R (σx − σc )2 τxy:= + 2 R = 3.414MPaCoordinates: A(σx , τxy) C(σc , 0)In-plane Principal Stresses:The coordinates of points B and D represent σ1 and σ2, respectively.σ1 := σc + R σ1 = 3.060MPaσ2 := σc − R σ2 = −3.767MPaMaximum In-plane Shear Stress :Represented by the coordinates of point E on the circle.τmax := R τmax = 3.41MPa Ans 996. Problem 9-103Determine the equivalent state of stress on an element if it is oriented 30° clockwise from the elementshown. Use the stress-transformation equations.Given: σx := 0MPa σy := −0.300MPaθ := −30deg τxy := 0.950MPaSolution:Normal Stress:σx'σx + σy:= + ⋅cos(2θ) + τxy⋅ sin(2θ)2σx' = −0.898MPa Ansσx − σy2σy':= − ⋅cos(2θ) − τxy⋅ sin(2θ)σy' = 0.598MPa Ansσx + σy2σx − σy2Shear Stress:τx'y'σx − σy:= − ⋅ sin(2θ) + τxy⋅cos(2θ)2τx'y' = 0.605MPa Ans 997. Problem 9-104The state of stress at a point in a member is shown on the element. Determine the stress componentsacting on the inclined plane AB.AnsGiven: σx := −50MPa σy := −100MPaφ' := 30deg τxy := 28MPaSolution:Construction of Mohr's Circle: θ := 90deg + φ'Center : σcσx + σy:= σc = −75MPa2Radius : R (σy − σc)2 τxy:= + 2 R = 37.54MPaCoordinates: A(σx , τxz) C(σc , 0)Angles:φ atanτxyσx − σc⎛⎜⎝⎞⎠:= φ = 48.240 degα := 360deg − 2θ − φ α = 71.760 degStresses: (represented by coordinates of point P on the circle)σx' := σc + R⋅cos(α) σx' = −63.25MPa Ansτx'y' := R⋅ sin(α) τx'y' = 35.65MPa 998. Problem 10-1Prove that the sum of the normal strains in perpendicular directions is constant.Solution:Stress Transformation Equations: Applying Eqs. 10-5 and 10-7 of the text.εx'εx + εy2εx − εy2+ ⋅cos(2θ)γxy2= + ⋅ sin(2θ) (1)εy'εx + εy2εx − εy2− ⋅cos(2θ)γxy2= − ⋅ sin(2θ) (2)(1) + (2) :LHS = εx' + εy'RHSεx + εy2εx + εy2= + RHS = εx + εyHence,εx' + εy' = εx + εy (Q.E.D.) 999. Problem 10-2The state of strain at the point on the leaf of the caster assembly has components of εx= -400(10-6),= 860(10-6), and γxy = 375(10-6). Use the strain-transformation equations to determine the equivalentin-plane strains on an element oriented at an angle of θ = 30° counterclockwise from the originalposition. Sketch the deformed element due to these strains within the x-y plane.Given: εx := −400⋅ (10− 6) γxy := 375⋅ (10− 6)εyεy := 860⋅ (10− 6) θ := 30degSolution:Stress Transformation Equations:εx'εx + εy2εx − εy2+ ⋅cos(2θ)γxy2:= + ⋅ sin(2θ)εx' = 77.38 × 10− 6 Ansεy'εx + εy2εx − εy2− ⋅cos(2θ)γxy2:= − ⋅ sin(2θ)εy' = 382.62 × 10− 6 Ansγx'y' 2εx − εy2− ⋅ sin(2θ)γxy2+ ⋅cos (2θ)⎛⎜⎝⎞⎠:=γx'y' = 1.279 × 10− 3 Ans 1000. Problem 10-3The state of strain at the point on the pin leaf has components of εx= 200(10-6), εy= 180(10-6), and γxy= -300(10-6). Use the strain-transformation equations to determine the equivalent in-plane strains on anelement oriented at an angle of θ = 60° counterclockwise from the original position. Sketch thedeformed element due to these strains within the x-y plane.Given: εx := 200⋅ (10− 6) γxy := −300⋅ (10− 6)εy := 180⋅ (10− 6) θ := 60degSolution:Stress Transformation Equations:εx'εx + εy2εx − εy2+ ⋅cos(2θ)γxy2:= + ⋅ sin(2θ)εx' = 55.10 × 10− 6 Ansεy'εx + εy2εx − εy2− ⋅cos(2θ)γxy2:= − ⋅ sin(2θ)εy' = 324.90 × 10− 6 Ansγx'y' 2εx − εy2− ⋅ sin(2θ)γxy2+ ⋅cos(2θ)⎛⎜⎝⎞⎠:=γx'y' = 132.679 × 10− 6 Ans 1001. Problem 10-4Solve Prob. 10-3 for an element oriented θ = 30° clockwise.Given: εx := 200⋅ (10− 6) γxy := −300⋅ (10− 6)εy := 180⋅ (10− 6) θ := −30degSolution:Stress Transformation Equations:εx'εx + εy2εx − εy2+ ⋅cos(2θ)γxy2:= + ⋅ sin(2θ)εx' = 324.90 × 10− 6 Ansεy'εx + εy2εx − εy2− ⋅cos(2θ)γxy2:= − ⋅ sin(2θ)εy' = 55.10 × 10− 6 Ansγx'y' 2εx − εy2− ⋅ sin(2θ)γxy2+ ⋅cos(2θ)⎛⎜⎝⎞⎠:=γx'y' = −132.679 × 10− 6 Ans 1002. Problem 10-5Due to the load P, the state of strain at the point on the bracket has components of εx= 500(10-6),= 350(10-6), and γxy = -430(10-6). Use the strain-transformation equations to determine the equivalentin-plane strains on an element oriented at an angle of θ = 30° clockwise from the original position.Sketch the deformed element due to these strains within the x-y plane.Given: εx := 500⋅ (10− 6) γxy := −430⋅ (10− 6)εyεy := 350⋅ (10− 6) θ := −30degSolution:Stress Transformation Equations:εx'εx + εy2εx − εy2+ ⋅cos(2θ)γxy2:= + ⋅ sin(2θ)εx' = 648.70 × 10− 6 Ansεy'εx + εy2εx − εy2− ⋅cos(2θ)γxy2:= − ⋅ sin(2θ)εy' = 201.30 × 10− 6 Ansγx'y' 2εx − εy2− ⋅ sin(2θ)γxy2+ ⋅cos(2θ)⎛⎜⎝⎞⎠:=γx'y' = −85.096 × 10− 6 Ans 1003. Problem 10-6The state of strain at the point on a wrench has components εx= 120(10-6), εy= -180(10-6), γxy =150(10-6). Use the strain-transformation equations to determine (a) the in-plane principal strains and (b)the maximum in-plane shear strain and average normal strain. In each case specify the orientation ofthe element and show how the strains deform the element within the x-y plane.Given: εx := 120⋅ (10− 6) εy := −180⋅ (10− 6) γxy := 150⋅ (10− 6)Solution:a) In-plane Principal Strains: Applying Eq. 10-9,ε1εx + εy2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= + + ε1 = 137.71 × 10− 6 Ansε2εx + εy2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= − + ε2 = −197.71 × 10− 6 AnsOrientation of Principal Strain:tan(2θp) γxy= θpεx − εy12atanγxyεx − εy⎛⎜⎝⎞⎠:= θ'p := θp − 90degθp = 13.283 deg θ'p = −76.717 degUse Eq. 10-5 to determine the direction of ε1 and ε2.εx'εx + εy2εx − εy2+ ⋅cos(2θp)γxy2:= + ⋅ sin(2θp)εx' = 137.71 × 10− 6Therefore, θp1 := θp θp1 = 13.28 deg Ansθp2 := θ'p θp2 = −76.72 deg Ansb) Maximum In-plane Shear Strain: Applying Eq. 10-11,γmax 2εx − εy2⎛⎜⎝⎞⎠2 γxy⎛⎜⎝⎞2⎠2:= +γmax = 335.41 × 10− 6 Ansεavgεx + εy2:= εavg = −30 × 10− 6 AnsOrientation of Principal Strain:tan(2θs) εx − εy= − θsγxy12atanεx − εyγxy−⎛⎜⎝⎞⎠:= θ's := θs + 90deg 1004. θs = −31.717 deg θ'.s = 58.283 degUse Eq. 10-6 to determine the sign of γmax.γx'y' 2εx − εy2− ⋅ sin(2θs)γxy2+ ⋅cos(2θs)⎛⎜⎝⎞⎠:=γx'y' = 335.41 × 10− 6Therefore, θs1 := θs θs1 = −31.72 deg Ansθs2 := θ's θs2 = 58.28 deg Ans 1005. Problem 10-7The state of strain at the point on the gear tooth has components εx= 850(10-6), εy= 480(10-6), γxy =650(10-6). Use the strain-transformation equations to determine (a) the in-plane principal strains and (b)the maximum in-plane shear strain and average normal strain. In each case specify the orientation ofthe element and show how the strains deform the element within the x-y plane.Given: εx := 850⋅ (10− 6) εy := 480⋅ (10− 6) γxy := 650⋅ (10− 6)Solution:a) In-plane Principal Strains: Applying Eq. 10-9,ε1εx + εy2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= + + ε1 = 1.039 × 10− 3Ansε2εx + εy2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= − + ε2 = 291.03 × 10− 6AnsOrientation of Principal Strain:tan(2θp) γxy= θpεx − εy12atanγxyεx − εy⎛⎜⎝⎞⎠:= θ'p := θp + 90degθp = 30.175 deg θ'p = 120.175 degUse Eq. 10-5 to determine the direction of ε1 and ε2.εx'εx + εy2εx − εy2+ ⋅cos(2θp)γxy2:= + ⋅ sin(2θp)εx' = 1.039 × 10− 3Therefore, θp1 := θp θp1 = 30.18 deg Ansθp2 := θ'p θp2 = 120.18 deg Ansb) Maximum In-plane Shear Strain: Applying Eq. 10-11,γmax 2εx − εy2⎛⎜⎝⎞⎠2 γxy⎛⎜⎝⎞2⎠2:= +γmax = 747.93 × 10− 6 Ansεavgεx + εy2:= εavg = 665 × 10− 6 Ans 1006. Orientation of Principal Strain:tan(2θs) εx − εy= − θsγxy12atanεx − εyγxy−⎛⎜⎝⎞⎠:= θ's := θs + 90degθs = −14.825 deg θ's = 75.175 degUse Eq. 10-6 to determine the sign of γmax.γx'y' 2εx − εy2− ⋅ sin(2θs)γxy2+ ⋅cos(2θs)⎛⎜⎝⎞⎠:=γx'y' = 747.93 × 10− 6Therefore, θs1 := θs θs1 = −14.82 deg Ansθs2 := θ's θs2 = 75.18 deg Ans 1007. Problem 10-8The state of strain at the point on the gear tooth has the components εx= 520(10-6), εy= -760(10-6),γxy = -750(10-6). Use the strain-transformation equations to determine (a) the in-plane principal strainsand (b) the maximum in-plane shear strain and average normal strain. In each case specify theorientation of the element and show how the strains deform the element within the x-y plane.Given: εx := 520⋅ (10− 6) εy := −760⋅ (10− 6) γxy := −750⋅ (10− 6)Solution:a) In-plane Principal Strains: Applying Eq. 10-9,ε1εx + εy2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= + + ε1 = 621.772 × 10− 6Ansε2εx + εy2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= − + ε2 = −861.77 × 10− 6AnsOrientation of Principal Strain:tan(2θp) γxy= θpεx − εy12atanγxyεx − εy⎛⎜⎝⎞⎠:= θ'p := θp + 90degθp = −15.184 deg θ'p = 74.816 degUse Eq. 10-5 to determine the direction of ε1 and ε2.εx'εx + εy2εx − εy2+ ⋅cos(2θp)γxy2:= + ⋅ sin(2θp)εx' = 621.772 × 10− 6Therefore, θp1 := θp θp1 = −15.18 deg Ansθp2 := θ'p θp2 = 74.82 deg Ansb) Maximum In-plane Shear Strain: Applying Eq. 10-11,γmax 2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= +γmax = 1.48 × 10− 3 Ansεavgεx + εy2:= εavg = −120 × 10− 6 AnsOrientation of Principal Strain:tan(2θs) εx − εy= − θsγxy12atanεx − εyγxy−⎛⎜⎝⎞⎠:= θ's := θs − 90degθs = 29.816 deg θ's = −60.184 degUse Eq. 10-6 to determine the sign of γmax. 1008. γx'y' 2εx − εy2− ⋅ sin(2θs)γxy2+ ⋅cos(2θs)⎛⎜⎝⎞⎠:=γx'y' = −1.48 × 10− 3Therefore, θs1 := θs θs1 = 29.82 deg Ansθs2 := θ's θs2 = −60.18 deg Ans 1009. Problem 10-9The state of strain at the point on the spanner wrench has components εx= 260(10-6), εy= 320(10-6),γxy = 180(10-6). Use the strain-transformation equations to determine (a) the in-plane principal strainsand (b) the maximum in-plane shear strain and average normal strain. In each case specify theorientation of the element and show how the strains deform the element within the x-y plane.Given: εx := 260⋅ (10− 6) εy := 320⋅ (10− 6)γxy := 180⋅ (10− 6)Solution:a) In-plane Principal Strains: Applying Eq. 10-9,ε1εx + εy2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= + + ε1 = 384.87 × 10− 6 Ansε2εx + εy2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= − + ε2 = 195.13 × 10− 6 AnsOrientation of Principal Strain:tan(2θp) γxy= θpεx − εy12atanγxyεx − εy⎛⎜⎝⎞⎠:= θ'p := θp + 90degθp = −35.783 deg θ'p = 54.217 degUse Eq. 10-5 to determine the direction of ε1 and ε2.εx'εx + εy2εx − εy2+ ⋅cos(2θp)γxy2:= + ⋅ sin(2θp)εx' = 195.132 × 10− 6Therefore, θp1 := θ'p θp1 = 54.22 deg Ansθp2 := θp θp2 = −35.78 deg Ansb) Maximum In-plane Shear Strain: Applying Eq. 10-11,γmax 2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= + γmax = 189.74 × 10− 6 Ansεavgεx + εy2:= εavg = 290 × 10− 6 AnsOrientation of Principal Strain:tan(2θs) εx − εy= − θsγxy12atanεx − εyγxy−⎛⎜⎝⎞⎠:= θ's := θs − 90degθs = 9.217 deg θ's = −80.783 deg 1010. Use Eq. 10-6 to determine the sign of γmax.γx'y' 2εx − εy2− ⋅ sin(2θs)γxy2+ ⋅cos(2θs)⎛⎜⎝⎞⎠:=γx'y' = 189.74 × 10− 6Therefore, θs1 := θs θs1 = 9.22 deg Ansθs2 := θ's θs2 = −80.78 deg Ans 1011. Problem 10-10The state of strain at the point on the arm has components εx= 250(10-6), εy= -450(10-6), γxy= -825(10-6). Use the strain-transformation equations to determine (a) the in-plane principal strains and(b) the maximum in-plane shear strain and average normal strain. In each case specify the orientationof the element and show how the strains deform the element within the x-y plane.Given: εx := 250⋅ (10− 6) εy := −450⋅ (10− 6)γxy := −825⋅ (10− 6)Solution:a) In-plane Principal Strains: Applying Eq. 10-9,ε1εx + εy2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= + + ε1 = 440.98 × 10− 6Ansε2εx + εy2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= − + ε2 = −640.98 × 10− 6AnsOrientation of Principal Strain:tan(2θp) γxy= θpεx − εy12atanγxyεx − εy⎛⎜⎝⎞⎠:= θ'p := θp + 90degθp = −24.843 deg θ'p = 65.157 degUse Eq. 10-5 to determine the direction of ε1 and ε2.εx'εx + εy2εx − εy2+ ⋅cos(2θp)γxy2:= + ⋅ sin(2θp)εx' = 440.977 × 10− 6Therefore, θp1 := θp θp1 = −24.84 deg Ansθp2 := θ'p θp2 = 65.16 deg Ansb) Maximum In-plane Shear Strain: Applying Eq. 10-11,γmax 2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= +γmax = 1.082 × 10− 3 Ansεavgεx + εy2:= εavg = −100 × 10− 6 Ans 1012. Orientation of Principal Strain:tan(2θs) εx − εy= − θsγxy12atanεx − εyγxy−⎛⎜⎝⎞⎠:= θ's := θs − 90degθs = 20.157 deg θ's = −69.843 degUse Eq. 10-6 to determine the sign of γmax.γx'y' 2εx − εy2− ⋅ sin(2θs)γxy2+ ⋅cos(2θs)⎛⎜⎝⎞⎠:=γx'y' = −1.082 × 10− 3Therefore, θs1 := θs θs1 = 20.16 deg Ansθs2 := θ's θs2 = −69.84 deg Ans 1013. Problem 10-11The state of strain at the point on the fan blade has components εx= 250(10-6), εy= -450(10-6), γxy= -825(10-6). Use the strain-transformation equations to determine (a) the in-plane principal strains and(b) the maximum in-plane shear strain and average normal strain. In each case specify the orientationof the element and show how the strains deform the element within the x-y plane.Given: εx := 250⋅ (10− 6) εy := −450⋅ (10− 6)γxy := −825⋅ (10− 6)Solution:a) In-plane Principal Strains: Applying Eq. 10-9,ε1εx + εy2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= + + ε1 = 440.98 × 10− 6Ansε2εx + εy2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= − + ε2 = −640.98 × 10− 6AnsOrientation of Principal Strain:tan(2θp) γxy= θpεx − εy12atanγxyεx − εy⎛⎜⎝⎞⎠:= θ'p := θp + 90degθp = −24.843 deg θ'p = 65.157 degUse Eq. 10-5 to determine the direction of ε1 and ε2.εx'εx + εy2εx − εy2+ ⋅cos(2θp)γxy2:= + ⋅ sin(2θp)εx' = 440.977 × 10− 6Therefore, θp1 := θp θp1 = −24.84 deg Ansθp2 := θ'p θp2 = 65.16 deg Ansb) Maximum In-plane Shear Strain: Applying Eq. 10-11,γmax 2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= + γmax = 1.082 × 10− 3 Ansεavgεx + εy2:= εavg = −100 × 10− 6 AnsOrientation of Principal Strain:tan(2θs) εx − εy= − θsγxy12atanεx − εyγxy−⎛⎜⎝⎞⎠:= θ's := θs − 90degθs = 20.157 deg θ's = −69.843 deg 1014. Use Eq. 10-6 to determine the sign of γmax.γx'y' 2εx − εy2− ⋅ sin(2θs)γxy2+ ⋅cos(2θs)⎛⎜⎝⎞⎠:=γx'y' = −1.082 × 10− 3Therefore, θs1 := θs θs1 = 20.16 deg Ansθs2 := θ's θs2 = −69.84 deg Ans 1015. Problem 10-12A strain gauge is mounted on the 25-mm-diameter A-36 steel shaft in the manner shown. When theshaft isrotating with an angular velocity of ω = 1760 rev/min, using a slip ring the reading on the straingauge is ε = 800(10-6). Determine the power output of the motor. Assume the shaft is only subjected toa torque.rpm(2π)rad60secUnit used: :=Given: εx' := 800⋅ (10− 6) ω := 1760rpmεy := 0 εx := 0 θ := 60degG := 75GPa do := 25mmSolution:Section Property :ρ := 0.5do Jπ32:= ⋅ 4doStress Transformation Equations:εx'εx + εy2εx − εy2+ ⋅cos(2θ)γxy2= + ⋅ sin(2θ)γxy2sin(2θ) εx'εx + εy2−εx − εy2− ⋅cos(2θ)⎛⎜⎝⎞⎠:= γxy = 1.848 × 10− 3Shear Stress :τT⋅ρJ= τ := G⋅ γxyTG⋅ J⋅ γxy:= T = 0.4251 kN⋅mρP := T⋅ω P = 78.35 kW Ans 1016. Problem 10-13The state of strain at the point on the support has components εx= 350(10-6), εy= 400(10-6), γxy= -675(10-6). Use the strain-transformation equations to determine (a) the in-plane principal strains and(b) the maximum in-plane shear strain and average normal strain. In each case specify the orientationof the element and show how the strains deform the element within the x-y plane.Given: εx := 350⋅ (10− 6) εy := 400⋅ (10− 6)γxy := −675⋅ (10− 6)Solution:a) In-plane Principal Strains: Applying Eq. 10-9,ε1εx + εy2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= + + ε1 = 713.42 × 10− 6Ansε2εx + εy2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= − + ε2 = 36.58 × 10− 6AnsOrientation of Principal Strain:tan(2θp) γxy= θpεx − εy12atanγxyεx − εy⎛⎜⎝⎞⎠:= θ'p := θp − 90degθp = 42.882 deg θ'p = −47.118 degUse Eq. 10-5 to determine the direction of ε1 and ε2.εx'εx + εy2εx − εy2+ ⋅cos(2θp)γxy2:= + ⋅ sin(2θp)εx' = 36.575 × 10− 6Therefore, θp1 := θ'p θp1 = −47.12 deg Ansθp2 := θp θp2 = 42.88 deg Ansb) Maximum In-plane Shear Strain: Applying Eq. 10-11,γmax 2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= + γmax = 676.849 × 10− 6 Ansεavgεx + εy2:= εavg = 375 × 10− 6 AnsOrientation of Principal Strain:tan(2θs) εx − εy= − θsγxy12atanεx − εyγxy−⎛⎜⎝⎞⎠:= θ's := θs + 90degθs = −2.118 deg θ's = 87.882 deg 1017. Use Eq. 10-6 to determine the sign of γmax.γx'y' 2εx − εy2− ⋅ sin(2θs)γxy2+ ⋅cos(2θs)⎛⎜⎝⎞⎠:=γx'y' = −676.849 × 10− 6Therefore, θs1 := θs θs1 = −2.12 deg Ansθs2 := θ's θs2 = 87.88 deg Ans 1018. Problem 10-14Consider the general case of plane strain where εx, εy, γxy and are known. Write a computer programthat can be used to determine the normal and shear strain, εx' and γx'y' , on the plane of an elementoriented θ from the horizontal. Also, compute the principal strains and the element's orientation, and themaximum in-plane shear strain, the average normal strain, and the element's orientation. 1019. Problem 10-15Solve Prob. 10-2 using Mohr's circle.Given: εx := −400⋅ (10− 6) γxy := 375⋅ (10− 6)εy := 860⋅ (10− 6) θ := 30degSolution:Construction of Mohr's Circle :Center :εcεx + εy2:= εc = 230 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 657.31 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)Angles:φ atan−0.5γxyεy − εc⎛⎜⎝⎞⎠:= φ = −16.574 degα := 2θ − φ α = 76.574 degStrain on the inclined Element: (represented by coordinates of points P and Q)εx' := εc − R⋅cos(α) εx' = 77.38 × 10− 6 Ansεy' := εc + R⋅cos(α) εy' = 382.62 × 10− 6 Ansγx'y' := 2R⋅ sin(α) γx'y' = 1.279 × 10− 3 Ans 1020. Problem 10-16Solve Prob. 10-4 using Mohr's circle.Given: εx := 200⋅ (10− 6) γxy := −300⋅ (10− 6)εy := 180⋅ (10− 6) θ := −30degSolution:Construction of Mohr's Circle :Center :εcεx + εy2:= εc = 190 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 150.33 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)Angles:φ atan−0.5γxyεy − εc⎛⎜⎝⎞⎠:= φ = −86.186 degα := 2θ − φ α = 26.186 degStrain on the inclined Element: (represented by coordinates of points P and Q)εx' := εc + R⋅cos(α) εx' = 324.9 × 10− 6 Ansεy' := εc − R⋅cos(α) εy' = 55.1 × 10− 6 Ansγx'y' := −2R⋅ sin(α) γx'y' = −132.7 × 10− 6 Ans 1021. Problem 10-17Solve Prob. 10-3 using Mohr's circle.Given: εx := 200⋅ (10− 6) γxy := −300⋅ (10− 6)εy := 180⋅ (10− 6) θ := 60degSolution:Construction of Mohr's Circle :Center :εcεx + εy2:= εc = 190 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 150.33 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)Angles:φ atan−0.5γxyεy − εc⎛⎜⎝⎞⎠:= φ = −86.186 degα := 2θ − φ α = 206.186 degStrain on the inclined Element: (represented by coordinates of points P and Q)εx' := εc + R⋅cos(α) εx' = 55.1 × 10− 6 Ansεy' := εc − R⋅cos(α) εy' = 324.9 × 10− 6 Ansγx'y' := −2R⋅ sin(α) γx'y' = 132.7 × 10− 6 Ans 1022. Problem 10-18Solve Prob. 10-5 using Mohr's circle.Given: εx := 500⋅ (10− 6) γxy := −430⋅ (10− 6)εy := 350⋅ (10− 6) θ := −30degSolution:Construction of Mohr's Circle :Center :εcεx + εy2:= εc = 425 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 227.71 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)Angles:φ atan−0.5γxyεy − εc⎛⎜⎝⎞⎠:= φ = −70.769 degα := 2θ − φ α = 10.769 degStrain on the inclined Element: (represented by coordinates of points P and Q)εx' := εc + R⋅cos(α) εx' = 648.7 × 10− 6 Ansεy' := εc − R⋅cos(α) εy' = 201.3 × 10− 6 Ansγx'y' := −2R⋅ sin(α) γx'y' = −85.1 × 10− 6 Ans 1023. Problem 10-19Solve Prob. 10-6 using Mohr's circle.Given: εx := 120⋅ (10− 6) εy := −180⋅ (10− 6) γxy := 150⋅ (10− 6)Solution:Construction of Mohr's Circle :Center :εcεx + εy2:= εc = −30 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 167.71 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)In-plane Principal Strains: (represented by coordinates of points B and D)ε1 := εc + R ε1 = 137.71 × 10− 6 Ansε2 := εc − R ε2 = −197.71 × 10− 6 AnsOrientation of Principal Strain:tan(2θp1) 0.5γxy= θp1εx − εc12atan0.5γxyεx − εc⎛⎜⎝⎞⎠:= θp1 = 13.28 deg Ansθp2 := θp1 − 90deg θp2 = −76.72 deg AnsMaximum In-plane Shear Strain: (represented by coordinates of point E)γmax := −2R γmax = −335.41 × 10− 6 AnsOrientation of Maximum In-plane Shear Strain:tan(2θs) εx − εc= − θs0.5γxy12atanεx − εc0.5γxy−⎛⎜⎝⎞⎠:=θs = −31.717 deg (Clockwise) Ans 1024. Problem 10-20Solve Prob. 10-8 using Mohr's circle.Given: εx := 520⋅ (10− 6) εy := −760⋅ (10− 6) γxy := −750⋅ (10− 6)Solution:Construction of Mohr's Circle :Center :εcεx + εy2:= εc = −120 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 741.77 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)a) In-plane Principal Strains:(represented by coordinates of points B and D)ε1 := εc + R ε1 = 621.77 × 10− 6 Ansε2 := εc − R ε2 = −861.77 × 10− 6 AnsOrientation of Principal Strain:tan(2θp1) 0.5γxy= θp1εx − εc12atan0.5γxyεx − εc⎛⎜⎝⎞⎠:= θp1 = −15.18 deg Ansθp2 := θp1 − 90deg θp2 = −105.18 deg Ansb) Maximum In-plane Shear Strain: (represented by coordinates of point E)γmax := −2R γmax = −1.484 × 10− 3 AnsOrientation of Maximum In-plane Shear Strain:tan(2θs) εx − εc= − θs0.5γxy12atanεx − εc0.5γxy−⎛⎜⎝⎞⎠:=θs = 29.816 deg (Clockwise) Ans 1025. Problem 10-21Solve Prob. 10-7 using Mohr's circle.Given: εx := 850⋅ (10− 6) εy := 480⋅ (10− 6) γxy := 650⋅ (10− 6)Solution:Construction of Mohr's Circle :Center :εcεx + εy2:= εc = 665 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 373.97 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)a) In-plane Principal Strains:(represented by coordinates of points B and D)ε1 := εc + R ε1 = 1.039 × 10− 3 Ansε2 := εc − R ε2 = 291.03 × 10− 6 AnsOrientation of Principal Strain:tan(2θp1) 0.5γxy= θp1εx − εc12atan0.5γxyεx − εc⎛⎜⎝⎞⎠:= θp1 = 30.18 deg Ansθp2 := θp1 − 90deg θp2 = −59.82 deg Ansb) Maximum In-plane Shear Strain: (represented by coordinates of point E)γmax := −2R γmax = −747.93 × 10− 6 AnsOrientation of Maximum In-plane Shear Strain:tan(2θs) εx − εc= − θs0.5γxy12atanεx − εc0.5γxy−⎛⎜⎝⎞⎠:=θs = −14.825 deg (Clockwise) Ans 1026. Problem 10-22Solve Prob. 10-9 using Mohr's circle.Given: εx := 260⋅ (10− 6) εy := 320⋅ (10− 6)γxy := 180⋅ (10− 6)Solution:Construction of Mohr's Circle :Center :εcεx + εy2:= εc = 290 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 94.87 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)a) In-plane Principal Strains:(represented by coordinates of points B and D)ε1 := εc + R ε1 = 384.87 × 10− 6 Ansε2 := εc − R ε2 = 195.13 × 10− 6 AnsOrientation of Principal Strain:tan(2θp2) −0.5γxy= θp2εx − εc12:= atanθp2 = 35.78 deg Ans⎛⎜⎝ ⎞−0.5γxyεx − εc⎠θp1 := 90deg − θp2 θp1 = 54.22 deg Ansb) Maximum In-plane Shear Strain: (represented by coordinates of point E)γmax := −2R γmax = −189.74 × 10− 6 AnsOrientation of Maximum In-plane Shear Strain:tan(2θs) εx − εc= − θs0.5γxy12atanεx − εc0.5γxy−⎛⎜⎝⎞⎠:=θs = 9.217 deg (Counter-clockwise) Ans 1027. Problem 10-23The strain at point A on the bracket has components εx= 300(10-6), εy= 550(10-6), γxy = -650(10-6), εz= 0. Determine (a) the principal strains at A, (b) the maximum shear strain in the x-y plane, and (c) theabsolute maximum shear strain.Given: εx := 300⋅ (10− 6) εy := 550⋅ (10− 6) γxy := −650⋅ (10− 6)εz := 0Solution:Construction of Mohr's Circle for x-y plane :Center :εcεx + εy2:= εc = 425 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 348.21 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)a) In-plane Principal Strains:ε1 := εc + R ε1 = 773.21 × 10− 6 Ansε2 := εc − R ε2 = 76.79 × 10− 6 Ansb) Maximum In-plane Shear Strain:γmax := 2R γmax = 696.42 × 10− 6 Ansc) Absolute Maximum Shear Strain:Construction of Three Circles :From the results obtained above,εmax := ε1 εint := ε2 εmin := 0Absolute Maximum Shear Strain :From the three Mohr's circles,γabs.max := εmax − εminγabs.max = 773.21 × 10− 6 Ans 1028. Problem 10-24The strain at a point has components of εx= -480(10-6), εy= 650(10-6), γxy = 780(10-6), εz= 0.Determine (a) the principal strains, (b) the maximum shear strain in the x-y plane, and (c) the absolutemaximum shear strain.Given: εx := −480⋅ (10− 6) εy := 650⋅ (10− 6) γxy := 780⋅ (10− 6) εz := 0Solution:Construction of Mohr's Circle for x-y plane :Center :εcεx + εy2:= εc = 85 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 686.53 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)a) In-plane Principal Strains: (represented by coordinates of points B and D)ε1 := εc + R ε1 = 771.53 × 10− 6 Ansε2 := εc − R ε2 = −601.53 × 10− 6 Ansb) Maximum In-plane Shear Strain: (represented by coordinates of point E)γmax := 2R γmax = 1.373 × 10− 3 Ansc) Absolute Maximum Shear Strain:Construction of Three Circles :From the results obtained above,εmax := ε1 εint := 0 εmin := ε2Absolute Maximum Shear Strain :From the three Mohr's circles,γabs.max := εmax − εminγabs.max = 1.373 × 10− 3 Ans 1029. Problem 10-25The strain at a point on a pressure-vessel wall has components of εx= 350(10-6), εy= -460(10-6), γxy= -560(10-6), εz= 0. Determine (a) the principal strains at the point, (b) the maximum shear strain inthe x-y plane, and (c) the absolute maximum shear strain.Given: εx := 350⋅ (10− 6) εy := −460⋅ (10− 6) γxy := −560⋅ (10− 6) εz := 0Solution:Construction of Mohr's Circle for x-y plane :Center :εcεx + εy2:= εc = −55 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 492.37 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)a) In-plane Principal Strains: (represented by coordinates of points B and D)ε1 := εc + R ε1 = 437.37 × 10− 6 Ansε2 := εc − R ε2 = −547.37 × 10− 6 Ansb) Maximum In-plane Shear Strain: (represented by coordinates of point E)γmax := −2R γmax = −984.7 × 10− 6 Ansc) Absolute Maximum Shear Strain:Construction of Three Circles :From the results obtained above,εmax := ε1 εint := 0 εmin := ε2Absolute Maximum Shear Strain :From the three Mohr's circles,γabs.max := εmax − εminγabs.max = 984.7 × 10− 6 Ans 1030. Problem 10-26The strain at point A on the leg of the angle has components of εx= -140(10-6), εy= 180(10-6), γxy= -125(10-6), εz= 0. Determine (a) the principal strains at A, (b) the maximum shear strain in the x-yplane, and (c) the absolute maximum shear strain.Given: εx := −140⋅ (10− 6) εy := 180⋅ (10− 6)γxy := −125⋅ (10− 6) εz := 0Solution:Construction of Mohr's Circle for x-y plane :Center :εcεx + εy2:= εc = 20 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 171.77 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)a) In-plane Principal Strains:ε1 := εc + R ε1 = 191.77 × 10− 6 Ansε2 := εc − R ε2 = −151.77 × 10− 6 Ansb) Maximum In-plane Shear Strain:γmax := −2R γmax = −343.5 × 10− 6 Ansc) Absolute Maximum Shear Strain:Construction of Three Circles :From the results obtained above,εmax := ε1 εint := 0 εmin := ε2Absolute Maximum Shear Strain :From the three Mohr's circles,γabs.max := εmax − εminγabs.max = 343.5 × 10− 6 Ans 1031. Problem 10-27The steel bar is subjected to the tensile load of 2.5 kN. If it is 12 mm thick determine the absolutemaximum shear strain. E = 200 GPa, ν = 0.3.Given: d := 50mm t := 12mmL := 375mm N := 2.5kNE := 200GPa ν := 0.3Solution:Stress: σxNd⋅ t:= σy := 0 σz := 0Strain:εxσxE:= εx = 2.0833 × 10− 5εy := −ν⋅ (εx) εy = −6.25 × 10− 6εz := −ν⋅ (εx) εz = −6.25 × 10− 6γxy := 0Construction of Mohr's Circle in x-y Plane :Center : εcεx + εy2:= εc = 7.2917 × 10− 6Radius : R (εx − εc)2 := + (0.5γxy)2 R = 1.3542 × 10− 5Coordinates: A(εx , 0) C(εc , 0)In-plane Principal Strains:ε1 := εc + R ε1 = 2.0833 × 10− 5ε2 := εc − R ε2 = −6.25 × 10− 6Similarly, from Mohr's Circle in x-z Plane :ε3 := εc − R ε3 = −6.25 × 10− 6Absolute Maximum In-plane Shear Strain :γabs.max := ε1 − ε2γabs.max = 2.7083 × 10− 5 Ans 1032. Problem 10-28The 45° strain rosette is mounted on the surface of an aluminum plate. The following readings areobtained for each gauge: εa = 475(10-6), εb = 250(10-6), and εc= -360(10-6). Determine the in-planeprincipal strains.Given: εa := 475⋅ (10− 6) εb := 250⋅ (10− 6) εc := −360⋅ (10− 6)θa := 0deg θb := −45deg θc := −90degSolution:Strain Rosettes (450): Applying Eq. 10-16,Givenεa = εx⋅cos(θa)2 + εy⋅ sin(θa)2 + γxy⋅ sin(θa)⋅cos(θa) (1)εb = εx⋅cos(θb)2 + εy⋅ sin(θb)2 + γxy⋅ sin(θb)⋅cos(θb) (2)εc = εx⋅cos(θc)2 + εy⋅ sin(θc)2 + γxy⋅ sin(θc)⋅cos(θc) (3)Solving Eqs.(1), (2) and (3), Guess εx := 10− 6 εy := 10− 6 γxy := 10− 6εxεyγxy⎛⎜⎜⎜⎝⎞⎟⎠:= Find(εx , εy , γxy)εxεyγxy⎛⎜⎜⎜⎝⎞⎟⎠475 × 10− 6−360 × 10− 6−385 × 10− 6⎛⎜⎜⎜⎜⎝⎞⎟⎟⎠=Construction of Mohr's Circle :Center :εcεx + εy2:= εc = 57.5 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 459.74 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)In-plane Principal Strains: (represented by coordinates of points B and D)ε1 := εc + R ε1 = 517.24 × 10− 6 Ansε2 := εc − R ε2 = −402.24 × 10− 6 Ans 1033. Problem 10-29The 60° strain rosette is mounted on the surface of the bracket. The following readings are obtainedfor each gauge: εa = -780(10-6), εb = 400(10-6), and εc= 500(10-6).Determine (a) the principal strainsand (b) the maximum in-plane shear strain and associated average normal strain. In each case show thedeformed element due to these strains.Given: εa := −780⋅ (10− 6) θa := 0degεb := 400⋅ (10− 6) θb := 60degεc := 500⋅ (10− 6) θc := 120degSolution:Strain Rosettes (600): Applying Eq. 10-16,Givenεa = εx⋅cos(θa)2 + εy⋅ sin(θa)2 + γxy⋅ sin(θa)⋅cos(θa) (1)εb = εx⋅cos(θb)2 + εy⋅ sin(θb)2 + γxy⋅ sin(θb)⋅cos(θb) (2)εc = εx⋅cos(θc)2 + εy⋅ sin(θc)2 + γxy⋅ sin(θc)⋅cos(θc) (3)Solving Eqs.(1), (2) and (3),Guess εx := 10− 6 εy := 10− 6 γxy := 10− 6εxεyγxy⎛⎜⎜⎜⎝⎞⎟⎠:= Find(εx , εy , γxy)εxεyγxy⎛⎜⎜⎜⎝⎞⎟⎠−780 × 10− 6860 × 10− 6−115.47 × 10− 6⎛⎜⎜⎜⎜⎝⎞⎟⎟⎠=Construction of Mohr's Circle :Center :εcεx + εy2:= εc = 40 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 822.03 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)a) In-plane Principal Strains:(represented by coordinates of points B and D)ε1 := εc + R ε1 = 862.03 × 10− 6 Ansε2 := εc − R ε2 = −782.03 × 10− 6 Ans 1034. Orientation of Principal Strain:tan(2θp2) 0.5γxy= θp2εx − εc12atan0.5γxyεx − εc⎛⎜⎝⎞⎠:= θp2 = 2.01 deg Ansθp1 := 90deg − θp2 θp1 = 87.99 deg Ansb) Maximum In-plane Shear Strain: (represented by coordinates of point E)γmax := −2R γmax = −1.644 × 10− 3 AnsOrientation of Maximum In-plane Shear Strain:tan(2θs) εx − εc= θs0.5γxy12atanεx − εc0.5γxy⎛⎜⎝⎞⎠:=θs = 42.986 deg (Clockwise) Ans 1035. cProblem 10-ε30The 45° strain rosette is mounted near the tooth of the wrench. The following readings are obtained foreach gauge: εa = 800(10-6), εb = 520(10-6), and = -450(10-6). Determine (a) the in-plane principalstrains and (b) the maximum in-plane shear strain and associated average normal strain. In each caseshow the deformed element due to these strains.Given: εa := 800⋅ (10− 6) θa := −135degεb := 520⋅ (10− 6) θb := −90degεc := −450⋅ (10− 6) θc := −45degSolution:Strain Rosettes (450): Applying Eq. 10-16,Givenεa = εx⋅cos(θa)2 + εy⋅ sin(θa)2 + γxy⋅ sin(θa)⋅cos(θa) (1)εb = εx⋅cos(θb)2 + εy⋅ sin(θb)2 + γxy⋅ sin(θb)⋅cos(θb) (2)εc = εx⋅cos(θc)2 + εy⋅ sin(θc)2 + γxy⋅ sin(θc)⋅cos(θc) (3)Solving Eqs.(1), (2) and (3),Guess εx := 10− 6 εy := 10− 6 γxy := 10− 6εxεyγxy⎛⎜⎜⎜⎝⎞⎟⎠:= Find(εx , εy , γxy)εxεyγxy⎛⎜⎜⎜⎝⎞⎟⎠−170 × 10− 6520 × 10− 61.25 × 10− 3⎛⎜⎜⎜⎜⎝⎞⎟⎟⎠=Construction of Mohr's Circle :Center :εcεx + εy2:= εc = 175 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 713.9 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)a) In-plane Principal Strains:(represented by coordinates of points B and D)ε1 := εc + R ε1 = 888.90 × 10− 6 Ansε2 := εc − R ε2 = −538.90 × 10− 6 Ans 1036. Orientation of Principal Strain:tan(2θp2) −0.5γxy= θp2εx − εc12atan−0.5γxyεx − εc⎛⎜⎝⎞⎠:= θp2 = 30.55 deg Ansθp1 := 90deg − θp2 θp1 = 59.45 deg Ansb) Maximum In-plane Shear Strain: (represented by coordinates of point E)γmax := 2R γmax = 1.428 × 10− 3 AnsOrientation of Maximum In-plane Shear Strain:tan(2θs) εx − εc= − θs0.5γxy12atanεx − εc0.5γxy−⎛⎜⎝⎞⎠:=θs = 14.449 deg (Counter-clockwise) Ans 1037. Problem 10-31The 60° strain rosette is mounted on a beam. The following readings are obtained from each gauge: εa= 150(10-6), εb = -330(10-6), and εc= 400(10-6). Determine (a) the in-plane principal strains and (b) themaximum in-plane shear strain and average normal strain. In each case show the deformed element dueto these strains.Given: εa := 150⋅ (10− 6) θa := −30degεb := −330⋅ (10− 6) θb := 30degεc := 400⋅ (10− 6) θc := 90degSolution:Strain Rosettes (600): Applying Eq. 10-16,Givenεa = εx⋅cos(θa)2 + εy⋅ sin(θa)2 + γxy⋅ sin(θa)⋅cos(θa) (1)εb = εx⋅cos(θb)2 + εy⋅ sin(θb)2 + γxy⋅ sin(θb)⋅cos(θb) (2)εc = εx⋅cos(θc)2 + εy⋅ sin(θc)2 + γxy⋅ sin(θc)⋅cos(θc) (3)Solving Eqs.(1), (2) and (3),Guess εx := 10− 6 εy := 10− 6 γxy := 10− 6εxεyγxy⎛⎜⎜⎜⎝⎞⎟⎠:= Find(εx , εy , γxy)εxεyγxy⎛⎜⎜⎜⎝⎞⎟⎠−253.33 × 10− 6400 × 10− 6−554.26 × 10− 6⎛⎜⎜⎜⎜⎝⎞⎟⎟⎠=Construction of Mohr's Circle :Center :εcεx + εy2:= εc = 73.33 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 428.38 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)a) In-plane Principal Strains:(represented by coordinates of points B and D)ε1 := εc + R ε1 = 501.72 × 10− 6 Ansε2 := εc − R ε2 = −355.05 × 10− 6 Ans 1038. Orientation of Principal Strain:tan(2θp2) 0.5γxy= θp2εx − εc12atan0.5γxyεx − εc⎛⎜⎝⎞⎠:= θp2 = 20.15 deg Ansθp1 := 90deg − θp2 θp1 = 69.85 deg Ansb) Maximum In-plane Shear Strain: (represented by coordinates of point E)γmax := −2R γmax = −856.764 × 10− 6 AnsOrientation of Maximum In-plane Shear Strain:tan(2θs) εx − εc= θs0.5γxy12atanεx − εc0.5γxy⎛⎜⎝⎞⎠:=θs = 24.845 deg (Clockwise) Ans 1039. Problem 10-32The 45° strain rosette is mounted on a steel shaft. The following readings are obtained from eachgauge: εa = 800(10-6), εb = 520(10-6), and εc= -450(10-6). Determine the in-plane principal strains andtheir orientation.Given: εa := 800⋅ (10− 6) εb := 520⋅ (10− 6) εc := −450⋅ (10− 6)θa := −45deg θb := 0deg θc := 45degSolution:Strain Rosettes (450): Applying Eq. 10-16,Givenεa = εx⋅cos(θa)2 + εy⋅ sin(θa)2 + γxy⋅ sin(θa)⋅cos(θa) (1)εb = εx⋅cos(θb)2 + εy⋅ sin(θb)2 + γxy⋅ sin(θb)⋅cos(θb) (2)εc = εx⋅cos(θc)2 + εy⋅ sin(θc)2 + γxy⋅ sin(θc)⋅cos(θc) (3)Solving Eqs.(1), (2) and (3),Guess εx := 10− 6 εy := 10− 6 γxy := 10− 6εxεyγxy⎛⎜⎜⎜⎝⎞⎟⎠:= Find(εx , εy , γxy)εxεyγxy⎛⎜⎜⎜⎝⎞⎟⎠520 × 10− 6−170 × 10− 6−1.25 × 10− 3⎛⎜⎜⎜⎜⎝⎞⎟⎟ ⎠=Construction of Mohr's Circle :Center :εcεx + εy2:= εc = 175 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 713.9 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)a) In-plane Principal Strains:ε1 := εc + R ε1 = 888.90 × 10− 6 Ansε2 := εc − R ε2 = −538.90 × 10− 6 AnsOrientation of Principal Strain:tan(2θp) 0.5γxy= θpεx − εc12atan0.5γxyεx − εc⎛⎜⎝⎞⎠:= θp = −30.55 deg Ans 1040. baProblem 10-33Consider θθthe general orientation of three strain gauges at a point as shown. Write a computer programthat can be used to determine the principal in-plane strains and the maximum in-plane shear strain at thepoint. Show an application of the program using the values = 40°, εεa = 160(10-6), = 125°, b =100(10-6), θc= 220°, εc= 80(10-6). 1041. Problem 10-34For the case of plane stress, show that Hooke's law can be written asE ε νενE ε νεν= ( )( )σ +x (1 −2 ) x yσ +y (1 −2 ) y x= 1042. Problem 10-35Use Hooke's law, Eq. 10-18, to develop the strain-transformation equations, Eqs. 105 and 106, fromthe stress-transformation equations, Eqs. 9-1 and 9-2. 1043. Problem 10-36Abar of copper alloy is loaded in a tension machine and it is determined that εx= 940(10-6) and σx=100 MPa, σy= 0, σz= 0. Determine the modulus of elasticity, Ecu , and the dilatation, ecu , of thecopper. νcu = 0.35.Given: σx := 100MPa σy := 0 σz := 0εx := 940⋅ (10− 6) νcu := 0.35Solution:εx1Eσx ν σy σz + ( ) ⋅ − ⎡⎣⎤⎦=Ecuσx − νcu⋅ (σy + σz):= Ecu = 106.38 GPa Ansεxεcu1 − 2νcuEcu:= ⋅ (σx + σy + σz) εcu = 2.820 × 10− 4 Ans 1044. Problem 10-37The principal plane stresses and associated strains in a plane at a point are σ1= 250 MPa, σ2= 112MPa, ε1 = 1.02(10-3), ε2 = 0.180(10-3). Determine the modulus of elasticity and Poisson's ratio.Given: σ1 := 250MPa σ2 := 112MPaε1 := 1.02⋅10− 3 ε2 := 0.180⋅10− 3Solution: σ3 := 0MPaGivenε11Eσ1 ν σ2 σ3 + ( ) ⋅ − ⎡⎣= (1)ε2⎤⎦1Eσ2 ν σ1 σ3 + ( ) ⋅ − ⎡⎣= (2)⎤⎦Solving (1) and (2): Guess E := 1GPa ν := 10− 6Eν⎛⎜⎝⎞⎠:= Find(E , ν)E = 212.77GPa Ansν = 0.295 Ans 1045. Problem 10-38Determine the bulk modulus for hard rubber if Er = 5 GPa, νr= 0.43.Given: Er := 5GPa νr := 0.43Solution:Bulk Modulus: Applying Eq. 10-25κEr3⋅ (1 − 2νr):=κ = 11.90GPa Ans 1046. Problem 10-39The principal strains at a point on the aluminum fuselage of a jet aircraft are ε1 = 780(10-6) andε2 = 400(10-6). Determine the associated principal stresses at the point in the same plane. Eal = 70GPa, νal= 0.33. Hint: See Prob. 10-34.Given: ε1 := 780⋅ (10− 6) E := 70GPaε2 := 400⋅ (10− 6) ν := 0.33Solution:Plane stress: σ3 := 0MPaUse the formula developed in Prob. 10-34,σ1E1 − ν2:= (ε1 + ν⋅ε2) σ1 = 71.64MPa Ansσ2E1 − ν2:= (ε2 + ν⋅ε1) σ2 = 51.64MPa Ans 1047. Problem 10-40The rod is made of aluminum 2014-T6. If it is subjected to the tensile load of 700 N and has a diameterof 20 mm, determine the absolute maximum shear strain in the rod at a point on its surface.Given: Px := 700N do := 20mmσy := 0 σz := 0E := 73.1GPa ν := 0.35Solution:σx4Pxπ do:= σx = 2.228MPa⋅ 2Normal Strains: Apply the general Hooke's Law,εx1Eσx ν σy σz + ( ) ⋅ − ⎡⎣:= εx = 30.481 × 10− 6⎤⎦εy1Eσy ν σx σz + ( ) ⋅ − ⎡⎣:= εy = −10.668 × 10− 6⎤⎦εz1Eσz ν σx σy + ( ) ⋅ − ⎡⎣:= εz = −10.668 × 10− 6⎤⎦Principal Strains: From the results obtained above,εmax := εx εmax = 30.481 × 10− 6εmin := εz εmin = −10.668 × 10− 6Absolute Maximum Shear Strain :From the three Mohr's circles,γabs.max := εmax − εminγabs.max = 41.149 × 10− 6 Ans 1048. Problem 10-41The rod is made of aluminum 2014-T6. If it is subjected to the tensile load of 700 N and has a diameterof 20 mm, determine the principal strains at a point on the surface of the rod.Given: Px := 700N do := 20mmσy := 0 σz := 0E := 73.1GPa ν := 0.35Solution:σx4Pxπ do:= σx = 2.228MPa⋅ 2Normal Strains: Apply the general Hooke's Law,εx1Eσx ν σy σz + ( ) ⋅ − ⎡⎣:= εx = 30.481 × 10− 6⎤⎦εy1Eσy ν σx σz + ( ) ⋅ − ⎡⎣:= εy = −10.668 × 10− 6⎤⎦εz1Eσz ν σx σy + ( ) ⋅ − ⎡⎣:= εz = −10.668 × 10− 6⎤⎦Principal Strains: From the results obtained above,εmax := εx εmax = 30.481 × 10− 6 Ansεint := εy εint = −10.668 × 10− 6 Ansεmin := εz εmin = −10.668 × 10− 6 Ans 1049. xProblem 10-ε42A rod has a radius of 10 mm. If it is subjected to an axial load of 15 N such that the axial strain in therod is = 2.75(10-6), determine the modulus of elasticity E and the change in its diameter. ν = 0.23.Given: Px := 15N σy := 0ro := 10mm σz := 0ν := 0.23 εx := 2.75⋅ (10− 6)Solution:Normal Stresses :σxPxπ ro:= σx = 0.0477MPa⋅ 2Normal Strains: Appling the generalized Hooke's Law,εx1Eσx ν σy σz + ( ) ⋅ − ⎡⎣⎤⎦=E:= E = 17.36GPa Ansεy := −ν⋅εx εy = −632.50 × 10− 9εz := −ν⋅εx εz = −632.50 × 10− 9Thus,Δd := εy⋅ (2⋅ ro) Δd = −12.65 × 10− 6mm Ans1εxσx ν σy σz + ( ) ⋅ − ⎡⎣⎤⎦ 1050. Problem 10-43The principal strains at a point on the aluminum surface of a tank are ε1 = 630(10-6) and ε2 = 350(10-6).If this is a case of plane stress, determine the associated principal stresses at the point in the sameplane. Eal = 70 GPa, νal= 0.33. Hint: See Prob. 10-34.Given: ε1 := 630⋅ (10− 6) Eal := 70GPaε2 := 350⋅ (10− 6) νal := 0.33Solution:Plane stress: σ3 := 0MPaUse the formula developed in Prob. 10-34,σ1Eal1 νal:= (ε1 + νal⋅ ε2) σ1 = 58.56MPa Ans− 2σ2Eal1 νal:= (ε2 + νal⋅ ε1) σ2 = 43.83MPa Ans− 2 1051. pνProblem 10-44A uniform edge load of 100 kN/m and 70 kN/m is applied to the polystyrene specimen. If the specimenis originally square and has dimensions of a = 50 mm, b = 50 mm, and a thickness of t = 6 mm,determine its new dimensions a', b', and t' after the load is applied. Ep = 4 GPa, = 0.25.Given: a := 50mm b := 50mm t := 6mmqa 100kNm:= qb 70kNm:=E := 4GPa ν := 0.25Solution:Plane stress: σz := 0MPaσxqat:= σx = 16.67MPaσyqbt:= σy = 11.67MPaNormal Strains: Apply the general Hooke's Law,εx1Eσx ν σy σz + ( ) ⋅ − ⎡⎣:= εx = 3.438 × 10− 3⎤⎦εy1Eσy ν σx σz + ( ) ⋅ − ⎡⎣:= εy = 1.875 × 10− 3⎤⎦εz1Eσz ν σx σy + ( ) ⋅ − ⎡⎣:= εz = −1.771 × 10− 3⎤⎦The new dimensions for the new specimen are,a' := a⋅ (1 + εy) a' = 50.09mm Ansb' := b⋅ (1 + εx) b' = 50.17mm Anst' := t⋅ (1 + εz) t' = 5.99mm Ans 1052. Problem 10-45The principal stresses at a point are shown. If the material is graphite for which Eg = 5.6 GPa, νg=0.23, determine the principal strains.Given: σx := 70MPaσy := −105MPaσz := −182MPaE := 5.6GPa ν := 0.23Solution:Normal Strains: Apply the general Hooke's Law,εx1Eσx ν σy σz + ( ) ⋅ − ⎡⎣⎤⎦:=εy1Eσy ν σx σz + ( ) ⋅ − ⎡⎣εx = 24.287 × 10− 3:= εy = −14.150 × 10− 3⎤⎦εz1Eσz ν σx σy + ( ) ⋅ − ⎡⎣:= εz = −31.063 × 10− 3⎤⎦Principal Strains: From the results obtained above,εmax := εx εmax = 0.0243 Ansεint := εy εint = −0.0142 Ansεmin := εz εmin = −0.0311 Ans 1053. Problem 10-46The shaft has a radius of 15 mm and is made of L2 tool steel. Determine the strains in the x' and y'directions if a torque T = 2 kN·m is applied to the shaft.Given: ro := 15mm T := 2kN⋅m G := 75GPaθ := 45degSolution:Section Property : J⋅ 42π ro:=Shear Stress: τT⋅ roJ:=Shear Stress-strain Relationship:Applying Hooke's Law, γxyτG:=γxy = 5.03 × 10− 3Strain Rosettes: For pure shear, εx := 0 εy := 0Applying Eq. 10-15, θx' := θ θy' := θ + 90degεx' := εx⋅cos(θx')2 + εy⋅ sin(θx')2 + γxy⋅ sin(θx')⋅cos(θx') εx' = 2.52 × 10− 3 Ansεy' := εx⋅cos(θy')2 + εy⋅ sin(θy')2 + γxy⋅ sin(θy')⋅cos(θy') εy' = −2.52 × 10− 3 Ans 1054. Problem 10-47The cross section of the rectangular beam is subjected to the bending moment M. Determine anexpression for the increase in length of lines AB and CD. The material has a modulus of elasticity Eand Poisson's ratio is ν. 1055. Problem 10-48The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gaugehaving a length of 20mm is attached to it, and it is observed to increase in length by 0.012 mm whenthe vessel is pressurized. Determine the pressure causing this deformation, and find the maximumin-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of thevessel. The material is steel, for which Est = 200 GPa and νst = 0.3.Given: t := 10mm di := 2m L := 20mmE := 200GPa ν := 0.3 ΔL := 0.012mmSolution:Normal Strains: Appling the generalized Hooke's Law withεmaxΔLL:= εmax = 600 × 10− 6εint := εmax εint = 600 × 10− 6εmax1Eσmax ν σint σmin + ( ) ⋅ − ⎡⎣⎤⎦=σmax = E⋅ εmax + ν⋅ (σint + σmin)50 p = E⋅εmax + ν⋅ (50 p + 0)pE⋅ εmax50(1 − ν) := p = 3.43MPa AnsNormal Stresses :ri := 0.5di αrit:= α = 100Since α > 10. then thin-wall analysis can be used.This is a plane stress problem where σmin := 0since there is no load acting on the outer surface of the wall.σmaxp⋅ ri2⋅ t:= σmax = 171.43MPaσint := σmax σint = 171.43MPaMaximum In-plane Shear Stress (Sperical Surface):Mohr's circle is simply a dot. As the result, the state of stress is the same consisting of twonormal stresses with zero shear stress regardless of the orientation of the element.τmax := 0 Ansσmax − σminAbsolute Maximum Shear Stress : τabs.max2:=τabs.max = 85.71MPa Ans 1056. xProblem 10-ε49A rod has a radius of 10 mm. If it is subjected to an axial load of 15 N such that the axial strain in therod is = 2.75(10-6), determine the modulus of elasticity E and the change in its diameter. ν = 0.23.Given: Px := 15N σy := 0ro := 10mm σz := 0ν := 0.23 εx := 2.75⋅ (10− 6)Solution:Normal Stresses :σxPxπ ro:= σx = 0.0477MPa⋅ 2Normal Strains: Appling the generalized Hooke's Law,εx1Eσx ν σy σz + ( ) ⋅ − ⎡⎣⎤⎦=E:= E = 17.36GPa Ansεy := −ν⋅εx εy = −632.50 × 10− 9εz := −ν⋅εx εz = −632.50 × 10− 9Thus,Δd := εy⋅ (2⋅ ro) Δd = −12.65 × 10− 6mm Ans1εxσx ν σy σz + ( ) ⋅ − ⎡⎣⎤⎦ 1057. Problem 10-50A single strain gauge, placed in the vertical plane on the outer surface and at an angle of 60° to the axisof the pipe, gives a reading at point A of εA = -250(10-6). Determine the vertical force P if the pipe hasan outer diameter of 25 mm and an inner diameter of 15 mm.The pipe is made of C86100 bronze.Given: do := 25mm di := 15mm G := 38GPaa := 200mm L := 150mmθ := 60deg εA := −250⋅ (10− 6)Solution: ρo := 0.5do ρi := 0.5diStrain Rosettes: For pure shear, εx := 0 εy := 0Applying Eq. 10-15,εA = εx + εy + γxy⋅ sin(θ)⋅cos(θ)γxyεA − εx − εysin(θ)⋅cos(θ):=Shear Stress-strain Relationship:Applying Hooke's Law, τA := G⋅ γxyInternal Force and Moment : At Section A:Vy = P Mz = −P⋅a Tx = P⋅LSection Property : Aπ4do2 − di2 ⎛⎝⎞⎠:= ⋅Izπ64do4 − di4 ⎛⎝⎞⎠:= ⋅ Jπ32do4 − di4 ⎛⎝⎞⎠:= ⋅QA4ρo3π⋅ 22π ρo⎛⎜⎜⎝⎞⎠⋅4ρi3π⋅ 22π ρi⎛⎜⎜⎝⎞⎠:= − ⋅Normal Stress: cA := 0σAMz⋅cAIz= − σA := 0MPaShear Stress in x-y plane : bA := do − diτAVy⋅QAIz⋅bATx⋅ρoJ= −G⋅ γxyP⋅QAIz⋅bA(P⋅L)⋅ρoJ= −PG⋅ γxyQAIz⋅bA:= P = 0.438 kN AnsL⋅ρoJ− 1058. Problem 10-51A single strain gauge, placed in the vertical plane on the outer surface and at an angle of 60° to the axisof the pipe, gives a reading at point A of εA = -250(10-6). Determine the principal strains in the pipe atpoint A. The pipe has an outer diameter of 25 mm and an inner diameter of 15 mm and is made ofC86100 bronze.Given: do := 25mm di := 15mm G := 38GPaa := 200mmL := 150mmθ := 60deg εA := −250⋅ (10− 6)ρi := 0.5diSolution: ρo := 0.5doStrain Rosettes: For pure shear, εx := 0 εy := 0Applying Eq. 10-15,AnsεA = εx + εy + γxy⋅ sin(θ)⋅cos(θ)γxyεA − εx − εysin(θ)⋅cos(θ):= γxy = −577.35 × 10− 6Construction of Mohr's Circle in x-y Plane :Center : εcεx + εy2:= εc = 0Radius : R (εx − εc)2 := + (0.5γxy)2 R = 288.675 × 10− 6Coordinates: A(εx , 0.5γxy) C(εc , 0)In-plane Principal Strains:The coordinates of points B and D represent ε1and ε2, respectively.ε1 := εc + R ε1 = 288.675 × 10− 6ε2 := εc − R ε2 = −288.675 × 10− 6Principal Stresses:Since σx=σy=σz=0, then from the generalized Hooke's Law,εz := 0From the results obtained above, we haveεmax := ε1 εmax = 288.7 × 10− 6 Ansεint := εz εint = 0 Ansεmin := ε2 εmin = −288.7 × 10− 6 1059. Problem 10-52A material is subjected to principal stresses σx and σy. Determine the orientation θ of a strain gaugeplaced at the point so that its reading of normal strain responds only to σyand not σx.The materialconstants are E and ν. 1060. Problem 10-53The principal stresses at a point are shown in the figure. If the material is aluminum for which Eal =70 GPa and νal = 0.33, determine the principal strains.Given: σx := 70MPaσy := −105MPaσz := −182MPaE := 70GPa ν := 0.33Solution:Apply the general Hooke's Law,εx1Eσx ν σy σz + ( ) ⋅ − ⎡⎣:= εx = 2.353 × 10− 3 Ans⎤⎦εy1Eσy ν σx σz + ( ) ⋅ − ⎡⎣:= εy = −9.720 × 10− 4 Ans⎤⎦εz1Eσz ν σx σy + ( ) ⋅ − ⎡⎣:= εz = −2.435 × 10− 3 Ans⎤⎦Ans 1061. Problem 10-54A thin-walled cylindrical pressure vessel has an inner radius r, thickness t, and length L. If it issubjected to an internal pressure p, show that the increase in its inner radius is dr = rε1= pr2(1 - ½ ν) /Et and the increase in its length is ΔL = pLr ( ½ - ν) /Et. Using these results, show thatthe change in internal volume becomes dV = π r2(1 + ε1)2 (1 + ε2)L - π r2L. Since ε1and ε2 are smallquantities, show further that the change in volume per unit volume, called volumetric strain, can bewritten as dV /V = pr ( 2.5 - 2ν) /Et. 1062. Problem 10-55The cylindrical pressure vessel is fabricated using hemispherical end caps in order to reduce thebending stress that would occur if flat ends were used. The bending stresses at the seam where thecaps are attached can be eliminated by proper choice of the thickness th and tc of the caps and cylinder,respectively. This requires the radial expansion to be the same for both the hemispheres and cylinder.Show that this ratio is tc / th = (2 - ν) / (1 - ν). Assume that the vessel is made of the same material andboth the cylinder and hemispheres have the same inner radius. If the cylinder is to have a thickness of12 mm, what is the required thickness of the hemispheres? Take ν = 0.3.Given: tc := 12mm ν := 0.3Solution:For cylindrical vessel:ο1p⋅ rtc= ο2p⋅ r2tc= ο3 = 0ε11Eσ1 ν σ2 σ3 + ( ) ⋅ − ⎡⎣⎤⎦=ε11Ep⋅ rtc= νε1p⋅ r2tc⎛⎜⎝⎞⎠⋅ − ⎡⎢⎣⎤⎥⎦p⋅ rE⋅ tc1ν2− ⎛⎜⎝⎞⎠=dr = ε1⋅ r drp⋅ r2E⋅ tc1ν2− ⎛⎜⎝⎞⎠= (1)For hemispherical end cap:ο1p⋅ r2th= ο2p⋅ r2th= ο3 = 0ε11Eσ1 ν σ2 σ3 + ( ) ⋅ − ⎡⎣⎤⎦=ε11Ep⋅ r2th= νε1p⋅ r2th⎛⎜⎝⎞⎠⋅ − ⎡⎢⎣⎤⎥⎦p⋅ r2E⋅ th= (1 − ν)dr = ε1⋅ r drp⋅ r22E⋅ th= (1 − ν) (2)Equate Eqs. (1) and (2):p⋅ r2ν⎞p⋅ r21− E⋅ tc2⎠2E⋅ th⎛⎜⎝= (1 − ν)1tc1ν2− ⎛⎜⎝⎞⎠12th= (1 − ν)tcth2 − ν1 − ν= QEDHence, th1 − ν2 − ν⎛⎜⎝⎞⎠:= ⋅ tc th = 4.94mm Ans 1063. Problem 10-56TheA-36 steel pipe is subjected to the axial loading of 60 kN. Determine the change in volume of thematerial after the load is applied.Given: do := 40mm di := 30mmL := 0.5m P := 60kNE := 200GPa ν := 0.32Solution:Section Property : Aπ4do2 − di2 ⎛⎝⎞⎠:= ⋅Normal Stress: The pipe is subjected to uniaxial load. Therefore,σxPA:= σx = 109.13MPaσy := 0 σz := 0Dilation: Apply Eq. 10-23,δV1 − 2ν= (σx + σy + σz)VEδV1 − 2νE:= (σx + σy + σz)⋅ (A⋅L)δV = 54.00mm3 Ans 1064. Problem 10-57The smooth rigid-body cavity is filled with liquid 6061-T6 aluminum. When cooled it is 0.3 mm fromthe top of the cavity. If the top of the cavity is covered and the temperature is increased by 110°C,determine the stress components σxUnit used: °C := degGiven: H := 150mm ΔH := 0.3mmE := 68.9GPa ν := 0.35ΔT := 110°C α 24 × 10− 6Given1°C:=Solution:Normal Strains: Since the aluminum is confined at thesides by a rigid container and allowed to expand in thez-direction,εx := 0 εy := 0 εzΔHH:=Applying the generalized Hooke's Law with the additional thernal strain, εT := α⋅ΔTE' := E⋅ (10− 3) (Scale down to avoid floating-point error during calculation)= + εT (3)Ansεx1E'σx ν σy σz + ( ) ⋅ − ⎡⎣⎤⎦= + εT (1)εy1E'σy ν σx σz + ( ) ⋅ − ⎡⎣⎤⎦= + εT (2)εz1E'σz ν σy σx + ( ) ⋅ − ⎡⎣⎤⎦Solving (1), (2) and (3):Guess σx := 1MPa σy := 2MPa σz := 3MPaσxσyσz⎛⎜⎜⎜⎝⎞⎟⎠:= Find(σx , σy , σz)σxσyσz⎛⎜⎜⎜⎝⎞⎟⎠103σxσyσz⎛⎜⎜⎜⎝⎞⎟⎠:= ⋅ (Scale back up)σxσyσz⎛⎜⎜⎜⎝⎞⎟⎠−487.2−487.2−385.2⎛⎜⎜⎝⎞= MPa⎠, σy, and σzin the aluminum. Hint: Use Eqs. 10-18 with anadditional strain term of αΔT (Eq. 4-4). 1065. zyxεεεProblem 10-58The smooth rigid-body cavity is filled with liquid 6061-T6 aluminum. When cooled it is 0.3 mm fromthe top of the cavity. If the top of the cavity is not covered and the temperature is increased by 110°C,determine the strain components , , and in the aluminum. Hint: Use Eqs. 10-18 with anadditional strain term of αΔT (Eq. 4-4).Unit used: °C := degGiven: H := 150mm ΔH := 0.3mmE := 68.9GPa ν := 0.35ΔT := 110°C α 24 × 10− 61°C:=Solution:Normal Strains: Since the aluminum is confined at thesides by a rigid container, thenεx := 0 εy := 0 Ansand since it is not restrained in z-direction, σz := 0Applying the generalized Hooke's Law with the additional thernal strain, εT := α⋅ΔTGivenεx1Eσx ν σy σz + ( ) ⋅ − ⎡⎣= + εT (1)⎤⎦εy1Eσy ν σx σz + ( ) ⋅ − ⎡⎣= + εT (2)⎤⎦Solving (1) and (2):Guess σx := 1MPa σy := 2MPaσxσy⎛⎜⎜⎝⎞⎠:= Find(σx , σy)σxσy⎛⎜⎜⎝⎞⎠−279.8−279.8⎛⎜⎝⎞⎠= MPaεz1σz − ν ⋅ ( Eσy + σx ⎡⎣⎡⎢⎣) + εT ⎤⎦⎤⎥⎦:=εz = 5.48 × 10− 3 Ans 1066. Problem 10-59The thin-walled cylindrical pressure vessel of inner radius r and thickness t is subjected to an internalpressure p. If the material constants are E and ν, determine the strains in the circumferential andlongitudinal directions. Using these results, compute the increase in both the diameter and the length ofa steel pressure vessel filled with air and having an internal gauge pressure of 15 MPa. The vessel is3 m long, and has an inner radius of 0.5 m and a wall thick of 10 mm. Est = 200 GPa, νst = 0.3.Given: t := 10mm r := 0.5m L := 3mE := 200GPa ν := 0.3 p := 15MPaSolution:Normal Stresses : αrt:= α = 50Since α > 10. then thin-wall analysis can be used.σ1p⋅ rt:= σ2p⋅ r2⋅ t:= σ3 := 0Normal Strains: Appling the generalized Hooke's Law,εcir1Eσ1 ν σ2 σ3 + ( ) ⋅ − ⎡⎣⎤⎦=εcir1Ep⋅ rtνp⋅ r+ 0 2⋅ t⎛⎜⎝⎞⎠⋅ − ⎡⎢⎣⎤⎥⎦=εcirp⋅ r2E⋅ t:= (2 − ν) εcir = 3.1875 × 10− 3 Ansεlong1Eσ2 ν σ1 σ3 + ( ) ⋅ − ⎡⎣⎤⎦=εlong1Ep⋅ r2tνp⋅ r+ 0 t⎛⎜⎝⎞⎠⋅ − ⎡⎢⎣⎤⎥⎦=εlongp⋅ r2E⋅ t:= (1 − 2ν) εlong 750 10 6 = × − AnsDeformations :Δd := εcir⋅ (2r) Δd = 3.19mm AnsΔL := εlong⋅L ΔL = 2.25mm Ans 1067. Problem 10-60Estimate the increase in volume of the tank in Prob. 10-59. Suggestion: Use the results of Prob. 10-54as a check.Given: t := 10mm r := 0.5m L := 3mE := 200GPa ν := 0.3 p := 15MPaSolution:Section Property : V := π⋅ r2⋅LNormal Stresses : αrt:= α = 50Since α > 10. then thin-wall analysis can be used.σ1p⋅ rt:= σ2p⋅ r2⋅ t:= σ3 := 0Normal Strains: Appling the generalized Hooke's Law,εcir1Eσ1 ν σ2 σ3 + ( ) ⋅ − ⎡⎣:= εcir = 3.1875 × 10− 3⎤⎦εlong1Eσ2 ν σ1 σ3 + ( ) ⋅ − ⎡⎣:= εlong = 750 × 10− 6⎤⎦Deformations :Δr := εcir⋅ (r) Δr = 1.59mmΔL := εlong⋅L ΔL = 2.25mmΔV π (r + Δr)2 := ⋅ (L + ΔL) − πr2⋅LΔV = 0.0168m3 AnsOr, appling the result of Prob. 10-54,ΔVp⋅ r= (2.5 − 2ν)VE⋅ tΔV:= (2.5 − 2ν)⋅ (π⋅ r2⋅L)p⋅ rE⋅ tΔV = 0.0168m3 Ans 1068. yProblem 10-x61A εεsoft material is placed within the confines of a rigid cylinder which rests on a rigid support.Assuming that = 0 and = 0, determine the factor by which the modulus of elasticity will beincreased when a load is applied if ν = 0.3 for the material.Given: ν := 0.3Solution:Normal Strains: Since the material is confined in a rigid cylinder,εx := 0 εy := 0Appling the generalized Hooke's Law,εx1E= ⎡⎣σx − ν ⋅ ( σy + σz ) σx = ν⋅ (σy + σz) (1)εy⎤⎦1E= σy = ν⋅ (σx + σz) (2)σy ν σx σz + ( ) ⋅ − ⎡⎣⎤⎦Solving Eqs. (1) and (2): σxν1 − ν⎛⎜⎝⎞⎠= ⋅σzσyν1 − ν⎛⎜⎝⎞⎠= ⋅σzThus,εz1Eσz ν σx σy + ( ) ⋅ − ⎡⎣⎤⎦=εz1Eνσz − ν⋅ ⋅ (σz + σz) 1 − ν⎡⎢⎣⎛⎜⎝⎞⎠⎤⎥⎦=εzσzE12ν21 − ν−⎛⎜⎝⎞⎠=εzσzE(1 + ν)⋅ (1 − 2ν)1 − ν⎡⎢⎣⎤⎥⎦=Hence, when the material is not being confined and undergoes the same normalstrain of εz, then the required modulus of elasticity isE'σzεz= E'1 − ν= E(1 + ν)⋅ (1 − 2ν)The increased factor is κE'E=κ1 − ν(1 + ν)⋅ (1 − 2ν):=κ = 1.35 Ans 1069. Problem 10-62A thin-walled spherical pressure vessel having an inner radius r and thickness t is subjected to aninternal pressure p. Show that the increase in the volume within the vessel is ΔV = (2pπ r4/Et)(1-ν).Use a small-strain analysis. 1070. Problem 10-63A material is subjected to plane stress. Express the distortion-energy theory of failure in terms of σx,σy, and τxy . 1071. Problem 10-64A material is subjected to plane stress. Express the maximum-shear-stress theory of failure in terms ofσx, σy, and τxy . Assume that the principal stresses are of different algebraic signs. 1072. Problem 10-65The components of plane stress at a critical point on an A-36 structural steel shell are shown.Determine if failure (yielding) has occurred on the basis of the maximum-shear-stress theory.Given: σx := −75MPa σy := 125MPa τxy := −80MPaSolution:In-plane Principal Stress: Applying Eq. 9-5,σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2σ1 = 153.06MPaσ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2σ2 = −103.06MPaMaximum Shear Stress Theory:σ1 and σ2 have opposite signs. Therefore,fσ := σ1 − σ2 fσ = 256.12MPa > σY (= 250 MPa) AnsBased on the result obtained above, the material yields accordingto the maximum shear stress theory. Ans 1073. Problem 10-66The components of plane stress at a critical point on an A-36 structural steel shell are shown.Determine if failure (yielding) has occurred on the basis of the maximum-distortion-energy theory.Given: σx := −75MPa σy := 125MPa τxy := −80MPaSolution:In-plane Principal Stress: Applying Eq. 9-5,σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2σ1 = 153.06MPaσ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2σ2 = −103.06MPaMaximum Distortion Energy Theory:2 − σ1⋅σ2 σ2σ1+ 2 σY2 =:= + 2 fσ = 49825MPa2 < σYfσ σ12 − σ1⋅σ2 σ22(= 62500 MPa2) AnsBased on the result obtained above, the material does not yield according tothe maximum distortion energy theory. Ans 1074. 21σσProblem 10-67The yield stress for a zirconium-magnesium alloy is σY = 107 MPa. If a machine part is made of thismaterial and a critical point in the material is subjected to in-plane principal stresses and = -0.5 σ1,determine the magnitude of σ1that will cause yielding according to the maximum-shear-stress theory.Given: σY := 107MPaσ2 = −0.5σ1Solution:σ1 − σ2 = σYσ1 − (−0.5σ1) = σY1.5σ1 = σYσ1σY1.5:=σ1 = 71.33MPa Ans 1075. Problem 10-68Solve Prob. 10-67 using the maximum-distortion-energy theory.Given: σY := 107MPaσ2 = −0.5σ1Solution:2 − σ1⋅σ2 σ2σ1+ 2 σY2 =2 σ1 −0.5σ− ⋅ ( 1) (−0.5σ1)+ 2 σYσ12 =1.75σ12 σY2 =σ12σY1.75:=σ1 = 80.88MPa Ans 1076. Problem 10-69If a shaft is made of a material of which σY = 350 MPa, determine the maximum torsional shear stressrequired to cause yielding using the maximum-distortion-energy theory.Given: σY := 350MPaSolution:σ1 = τ σ2 = −τσ12 − σ1⋅σ2 σ2+ 2 σY2 =τ2 − τ⋅ (−τ) + (−τ)2 σY2 =3τ2 σY2 =τ23σY:=τ = 202.1MPa Ans 1077. Problem 10-70Solve Prob. 10-69 using the maximum-shear-stress theory. Both principal stresses have opposite signs.Given: σY := 350MPaSolution:σ1 = τ σ2 = −τσ1 − σ2 = σYτ − (−τ) = σY2τ = σYτσY2:=τ = 175.0MPa Ans 1078. Problem 10-71The yield stress for a plastic material is σY =110 MPa. If this material is subjected to plane stress andelastic failure occurs when one principal stress is 120 MPa, what is the smallest magnitude of the otherprincipal stress? Use the maximum-distortion-energy theory.Given: σY := 110MPaσ1 := 120MPaSolution:Given2 − σ1⋅σ2 σ2σ1+ 2 σY2 = (1)Solving Eq. (1), Guess σ2 := 1MPaσ2 := Find(σ2)σ2 = 23.94MPa Ans 1079. Problem 10-72Solve Prob. 10-71 using the maximum-shear-stress theory. Both principal stresses have the same sign.Given: σY := 110MPaσ1 := 120MPaSolution:Since σ1 > σY (= 110 MPa) ,the material will fail for any σ2. Ans 1080. yxσσProblem 10-73The plate is made of Tobin bronze, which yields at σY = 175 MPa. Using the maximum-shear-stresstheory, determine the maximum tensile stress that can be applied to the plate if a tensile stress =0.75 σxis also applied.Given: σY := 175MPaσy = 0.75σxSolution: σ1 = σx σ2 = 0.75σxσ1 and σ1 have the same signs, soσ2 = σY0.75σx = σY0.75σx = σY σxσY0.75:= σx = 233.3MPaOr,σ1 = σYσx = σY σx := σY σx = 175.0MPa (Controls !) Ans 1081. Problem 10-74The plate is made of Tobin bronze, which yields at σY = 175 MPa. Using the maximum-distortion-energytheory, determine the maximum tensile stress σxthat can be applied to the plate if a tensilestress σy= 0.75 σxis also applied.Given: σY := 175MPaσy = 0.75σxSolution: σ1 = σx σ2 = 0.75σx2 − σ1⋅σ2 σ2σ1+ 2 σY2 =2 − σx⋅ (0.75σx) 0.75σ( x)+ 2 σYσx2 =0.8125σx2 σY2 =σ12σY0.8175:=σ1 = 193.55MPa Ans 1082. Problem 10-75An aluminum alloy 6061-T6 is to be used for a solid drive shaft such that it transmits 33 kW at 2400rev/min. Using a factor of safety of 2 with respect to yielding, determine the smallest-diameter shaftthat can be selected based on the maximum-shear-stress theory.Unit Used: rpm2π60rads:=Given: P := 33kW ω := 2400rpmσY := 255MPa Fsafety := 2Solution:Torsion :TPω:= T = 0.1313 kN⋅mSection Property : Jπ2= ⋅ρ4Shear Stress :τT⋅ρJ= τ2Tπ⋅ρ3=Principal Stresses :σ1 = τ σ2 = −τMaximum shear stress theory: Both principal stresses have opposite sign, henceσ1 − σ2σYFsafety=τ − (−τ)σYFsafety=4Tπ⋅ρ3σYFsafety=ρ34TπFsafetyσY⎛⎜⎝⎞⎠:= ⋅do := 2ρdo = 21.89mm Ans 1083. Problem 10-76Solve Prob. 10-75 using the maximum-distortion-energy theory.Unit Used: rpm2π60rads:=Given: P := 33kW ω := 2400rpmσY := 255MPa Fsafety := 2Solution:Torsion :TPω:= T = 0.1313 kN⋅mSection Property : Jπ2= ⋅ρ4Shear Stress :τT⋅ρJ= τ2Tπ⋅ρ3=Principal Stresses :σ1 = τ σ2 = −τMaximum distortion energy theory:2 − σ1⋅σ2 σ2σ1+ 2σYFsafety⎛⎜⎝⎞⎠2=τ2 − τ⋅ (−τ) + (−τ)2σYFsafety⎛⎜⎝⎞⎠2=3⋅ τσYFsafety=32Tπ⋅ρ3⎛⎜⎝⎞⎠⋅σYFsafety=ρ32⋅ 3TπFsafetyσY⎛⎜⎝⎞⎠:= ⋅do := 2ρdo = 20.87mm Ans 1084. Problem 10-77An aluminum alloy is to be used for a drive shaft such that it transmits 20 kW at 1500 rev/min. Using afactor of safety of 2.5 with respect to yielding, determine the smallest-diameter shaft that can beselected based on the maximum-distortion-energy theory. σY = 25 MPa.Unit Used: rpm2π60rads:=Given: P := 30kW ω := 1500rpmσY := 25MPa Fsafety := 2.5Solution:Torsion :TPω:= T = 0.1910 kN⋅mSection Property : Jπ2= ⋅ρ4Shear Stress :τT⋅ρJ= τ2Tπ⋅ρ3=Principal Stresses :σ1 = τ σ2 = −τMaximum distortion energy theory:2 − σ1⋅σ2 σ2σ1+ 2σYFsafety⎛⎜⎝⎞⎠2=τ2 − τ⋅ (−τ) + (−τ)2σYFsafety⎛⎜⎝⎞⎠2=3⋅ τσYFsafety=32Tπ⋅ρ3⎛⎜⎝⎞⎠⋅σYFsafety=ρ32⋅ 3TπFsafetyσY⎛⎜⎝⎞⎠:= ⋅do := 2ρdo = 55.23mm Ans 1085. Problem 10-78A bar with a square cross-sectional area is made of a material having a yield stress of σY = 840 MPa. Ifthe bar is subjected to a bending moment of 10 kN·m., determine the required size of the bar accordingto the maximum-distortion-energy theory. Use a factor of safety of 1.5 with respect to yielding.Given: M := 10kN⋅m σY := 840MPa Fsafety := 1.5Solution:Section Property : Ia412=Normal Stresses : σy := 0ca2= σxMcI= σx6Ma3=In-plane Principal Stresses :Since no shaer stress acts on the element,σ1 = σx σ2 = σyMaximum distortion energy theory:σ12 − σ1⋅σ2 + σ22σYFsafety⎛⎜⎝⎞⎠2=2 − σx⋅ (0) + 02σxσYFsafety⎛⎜⎝⎞⎠2=σxσYFsafety=6Ma3σYFsafety=a36MFsafetyσY⎛⎜⎝⎞⎠:= ⋅a = 47.50mm Ans 1086. Problem 10-79Solve Prob. 10-78 using the maximum-shear-stress theory.Given: M := 10kN⋅m σY := 840MPa Fsafety := 1.5Solution:Section Property : Ia412=Normal Stresses : σy := 0ca2= σxMcI= σx6Ma3=In-plane Principal Stresses :Since no shaer stress acts on the element,σ1 = σx σ2 = σyMaximum shear stress theory:σ2 = 0 σ2σYFsafety< (O.K.!)σ1σYFsafety=6Ma3σYFsafety= a36MFsafetyσY⎛⎜⎝⎞⎠:= ⋅a = 47.50mm Ans 1087. Problem 10-80The principal plane stresses acting on a differential element are shown. If the material is machine steelhaving a yield stress of σY = 700 MPa, determine the factor of safety with respect to yielding using themaximum-distortion-energy theory.Given: σx := −480MPa σy := −475MPa τxy := 0θ := 30deg σY := 700MPaSolution:Principal Stresses:σ1 := σy σ2 := σxMaximum Distortion Eenergy Theory:σ12 − σ1⋅σ2 + σ22 =σallow2 := + 22 − σ1⋅σ2 σ2σallow σ1σallow = 477.5MPaFsafetyσYσallow:= Fsafety = 1.47 Ans 1088. Problem 10-81The principal plane stresses acting on a differential element are shown. If the material is machine steelhaving a yield stress of σY = 700 MPa, determine the factor of safety with respect to yielding if themaximum-shear-stress theory is considered.Given: σx := 80MPa σy := −50MPa τxy := 0θ := 0deg σY := 700MPaSolution:Principal Stresses:σmax := σx σmin := σyMaximum Shear Stress Theory:σmax − σminτabs.max:= τabs.max = 65MPa2τmaxσY2:= τmax = 350MPaFsafetyτmaxτabs.max:= Fsafety = 5.38 Ans 1089. Problem 10-82The state of stress acting at a critical point on a machine element is shown in the figure. Determine thesmallest yield stress for a steel that might be selected for the part, based on the maximum-shear-stresstheory.Given: σx := 56MPa σy := −70MPaτxy := 28MPaSolution:Principal Stresses:σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2σ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2σ1 = 61.94MPa σ2 = −75.94MPaMaximum shear stress theory: Both principal stresses have opposite sign, henceσ1 − σ2 = σYσY := σ1 − σ2σY = 137.9MPa Ans 1090. 11Problem 10-832The yield stress for a σσσ1σuranium alloy is σY = 160 MPa. If a machine part is made of this materialand a critical point in the material is subjected to plane stress, such that the principal stresses are and = 0.25 , determine the magnitude of that will cause yielding according to the maximum-distortion-energy theory.Given: σ2 = 0.25σ1 σY := 160MPaSolution:Maximum Distortion Eenergy Theory:2 − σ1⋅σ2 σ2σ1+ 2 σY2 =2 − σ1⋅ (0.25σ1) 0.25σ( 1)+ 2 σYσ12 =⋅ 2 σY0.8125 σ12 =Principal Stress:σ1σY0.8125:=σ1 = 177.5MPa Ans 1091. Problem 10-84Solve Prob. 10-83 using the maximum-shear-stress theory.Given: σ2 = 0.25σ1 σY := 160MPaSolution:Principal Stresses: This is a plane stress case.σmax = σ1 σint = 0.25σ1 σmin = 0Maximum Shear Stress Theory:τallowσY2:= τallow = 80MPaτabs.maxσmax − σmin= τabs.max2σ12=τabs.max = τallowσ1= 80 MPa2σ1 := 160MPa Ans 1092. Problem 10-85An aluminum alloy is to be used for a solid drive shaft such that it transmits 25 kW at 1200 rev/min.Using a factor of safety of 2.5 with respect to yielding, determine the smallest-diameter shaft that canbe selected based on the maximum-shear-stress theory. σY = 70 MPa.Unit Used: rpm2π60rads:=Given: P := 25kW ω := 1200rpmσY := 70MPa Fsafety := 2.5Solution:Torsion :TPω:= T = 0.1989 kN⋅mSection Property : Jπ2= ⋅ρ4Shear Stress :τT⋅ρJ= τ2Tπ⋅ρ3=Principal Stresses :σ1 = τ σ2 = −τMaximum shear stress theory: Both principal stresses have opposite sign, henceσ1 − σ2σYFsafety=τ − (−τ)σYFsafety=4Tπ⋅ρ3σYFsafety=ρ34TπFsafetyσY⎛⎜⎝⎞⎠:= ⋅do := 2ρdo = 41.67mm Ans 1093. Problem 10-86The state of stress acting at a critical point on the seat frame of an automobile during a crash is shownin the figure. Determine the smallest yield stress for a steel that can be selected for the member, basedon the maximum-shear-stress theory.Given: σx := 560MPa σy := 0MPaτxy := 175MPaSolution:Principal Stresses:σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2σ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2σ1 = 610.19MPa σ2 = −50.19MPaMaximum shear stress theory: Both principal stresses have opposite sign, henceσ1 − σ2 = σYσY := σ1 − σ2σY = 660.4MPa Ans 1094. Problem 10-87Solve Prob. 10-86 using the maximum-distortion-energy theory.Given: σx := 560MPa σy := 0MPaτxy := 175MPaSolution:Principal Stresses:σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2σ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2σ1 = 610.19MPa σ2 = −50.19MPaMaximum distortion energy theory:σ12 − σ1⋅σ2 + σ22 =σY2 σY := σ12 − σ1⋅σ2 + σ22σY = 636.8MPa Ans 1095. 1Problem 10-8821If a machine σσσpart is made of titanium (Ti-6A1-4V) and a critical point in the material is subjectedto plane stress, such that the principal stresses are and = 0.5 , determine the magnitude of σ1in MPa that will cause yielding according to (a) the maximum-shear-stress theory, and (b) themaximum-distortion-energy theory.Given: σ2 = 0.5σ1 σY := 924MPaSolution:a) Maximum shear stress theory:Both principal stresses have the same signs, soσ1 = σY (Controls !)σ2 = σY 0.5σ1 = σY σ1 = 2σYHence σ1 := σYσ1 = 924.0MPa Ansb) Maximum Distortion Eenergy Theory:2 − σ1⋅σ2 σ2σ1+ 2 σY2 =2 − σ1⋅ (0.5σ1) 0.5σ( 1)+ 2 σYσ12 =⋅ 2 σY0.75 σ12 =Principal Stress:σ1σY0.75:=σ1 = 1066.9MPa Ans 1096. Problem 10-89Derive an expression for an equivalent torque Te that, if applied alone to a solid bar with a circularcross section, would cause the same energy of distortion as the combination of an applied bendingmoment M and torque T. 1097. Problem 10-90An aluminum alloy 6061-T6 is to be used for a drive shaft such that it transmits 40 kW at 1800rev/min. Using a factor of safety of F.S. = 2, with respect to yielding, determine the smallest-diametershaft that can be selected based on the maximum-distortion-energy theory.Unit Used: rpm2π60rads:=Given: P := 40kW ω := 1800rpmσY := 255MPa Fsafety := 2Solution:Torsion :TPω:= T = 0.2122 kN⋅mSection Property : Jπ2= ⋅ρ4Shear Stress :τT⋅ρJ= τ2Tπ⋅ρ3=Principal Stresses :σ1 = τ σ2 = −τMaximum distortion energy theory:2 − σ1⋅σ2 σ2σ1+ 2σYFsafety⎛⎜⎝⎞⎠2=τ2 − τ⋅ (−τ) + (−τ)2σYFsafety⎛⎜⎝⎞⎠2=3 τσYFsafety=2⋅ 3Tπ⋅ρ3σYFsafety=ρ32⋅ 3TπFsafetyσY⎛⎜⎝⎞⎠:= ⋅do := 2ρdo = 24.49mm Ans 1098. Problem 10-91Derive an expression for an equivalent bending moment Me that, if applied alone to a solid bar with acircular cross section, would cause the same energy of distortion as the combination of an appliedbending moment M and torque T. 1099. Problem 10-92The internal loadings at a critical section along the steel drive shaft of a ship are calculated to be atorque of 3.45 kN·m, a bending moment of 2.25 kN·m, and an axial thrust of 12.5 kN. If the yieldpoints for tension and shear are σY = 700 MPa and τY = 350 MPa, respectively, determine the requireddiameter of the shaft using the maximum-shear-stress theory.Given: Nx := −12.5kN M := −2.25kN⋅m T := 3.45kN⋅mσY := 700MPa τY := 350MPaSolution:Section Property : Aπ2= ⋅ρ2 Iπ4= ⋅ρ4 Jπ2= ⋅ρ4Normal Stress : σy := 0σxNxAM⋅ρI= + σx2Nxπ⋅ρ24Mπ⋅ρ3= +Shear Stress : τxyT⋅ρJ= τxy2Tπ⋅ρ3=Principal Stresses :σx + σyσ12σx − σy2⎛⎜⎝⎞⎠2τxy= + + 2 σ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy= − + 2Maximum shear stress theory: Assume σ1 and σ2 have opposite sign, hence σ1 − σ2 = σY2σx − σy2⎛⎜⎝⎞⎠2τxy+ 2 σ= Yσx − σy2⎛⎜⎝⎞⎠2τxy+ 224σY=142Nxπ⋅ρ24Mπ⋅ρ3+⎛⎜⎜⎝⎞⎠22Tπ⋅ρ3⎛⎜⎝⎞⎠2+24σY=(1)Given (2ρ⋅Nx + 4M)2 4T ( )2 + π ρ3 ⋅ σY ⋅ ⎛⎝⎞⎠2=Solving Eq. (1): Guess ρ := 10mmρ := Find(ρ) ρ = 19.68mmτxy2Tπ⋅ρ32Nxπ⋅ρ24Mπ⋅ρ3Check signs: σ := x+⎛⎜⎜⎝⎞⎠:=σ1σx2σx2⎛⎜⎝⎞⎠2τxy:= + + 2 σ2σx2σx2⎛⎜⎝⎞⎠2τxy:= − + 2σ1 = 151.65MPa σ2 = −548.35MPaσ1 and σ2 are of opposite sign. (O.K.!)Therefore, do := 2ρ do = 39.35mm Ans 1100. Problem 10-93The element is subjected to the stresses shown. If σY = 350 ksi, determine the factor of safety for thisloading based on (a) the maximum-shear-stress theory and (b) the maximum-distortion-energy theory.Given: σx := 84MPa σy := −56MPaτxy := 49MPa σY := 350MPaSolution:Principal Stresses:σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2σ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2σ1 = 99.45MPa σ2 = −71.45MPaa) Maximum shear stress theory: Both principal stresses have opposite sign, henceσallow := σ1 − σ2 σallow = 170.9MPaFactor of safety is, FsafetyσYσallow:= Fsafety = 2.05 Ansb) Maximum distortion energy theory:2 − σ1⋅σ2 σ2σ1+ 2 σ'allow2 =:= 2 − σ1⋅σ2 + σ22 σ'allow = 148.7MPaσ'allow σ1Factor of safety is, FsafetyσYσ'allow:= Fsafety = 2.35 Ans 1101. Problem 10-94The state of stress acting at a critical point on a wrench is shown in the figure. Determine the smallestyield stress for steel that might be selected for the part, based on the maximum-distortion-energytheory.Given: σx := 175MPa σy := 0MPaτxy := 70MPaSolution:Principal Stresses:σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2σ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2σ1 = 199.55MPa σ2 = −24.55MPaMaximum distortion energy theory:σ12 − σ1⋅σ2 + σ22 =σY2 σY := σ12 − σ1⋅σ2 + σ22σY = 212.9MPa Ans 1102. Problem 10-95The state of stress acting at a critical point on a wrench is shown in the figure. Determine the smallestyield stress for steel that might be selected for the part, based on the maximum-shear-stress theory.Given: σx := 175MPa σy := 0MPaτxy := 70MPaSolution:Principal Stresses:σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2σ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2σ1 = 199.55MPa σ2 = −24.55MPaMaximum shear stress theory: Both principal stresses have opposite sign, henceσY := σ1 − σ2σY = 224.1MPa Ans 1103. Problem 10-96The short concrete cylinder having a diameter of 50 mm is subjected to a torque of 500 N·m and anaxial compressive force of 2 kN. Determine if it fails according to the maximum-normal-stress theory.The ultimate stress of the concrete is σult = 28 MPa.Given: do := 50mm σult := 28MPaP := 2kN T := 0.5kN⋅mSolution:Section Property :Aπ4:= ⋅ 2 Jdoπ32:= ⋅ 4doNormal Stress: σPA:= σ = 1.019MPaShear Stress :ρ := 0.5do τT⋅ρJ:= τ = 20.37MPaIn-plane Principal Stresses:σx := 0 σy := −σ τxy := τfor any point on the shaft's surface. Applying Eq. 9-5,σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2 σ1 = 19.87MPa < σult (=28 MPa)σ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2 σ2 = −20.89MPa < σult (=28 MPa)Failure Criteria:σ1 < σult OKσ2 < σult OKBased on the result obtained above, the material does not fail according to themaximum normal-stress theory.Ans 1104. Problem 10-97If a solid shaft having a diameter d is subjected to a torque T and moment M, show that by themaximum-normal-stress theory the maximum allowable principal stress is(16 / d 3 ) (M M 2 T 2 ). allow σ = π + + 1105. Problem 10-98The principal stresses acting at a point on a thin-walled cylindrical pressure vessel are σ1= pr/t, σ2=pr/2t, and σ3= 0. If the yield stress is σy , determine the maximum value of p based on (a) themaximum-shear-stress theory and (b) the maximum-distortion-energy theory.Given: σ1p⋅ rt= σ2p⋅ r2t= σ3 := 0Solution:a) Maximum Shear Stress Theory:σ2 = 0.5σ1Both principal stresses have the same signs, soσ1 = σY (Controls !)σ2 = σY 0.5σ1 = σY σ1 = 2σYHencep⋅ rt= σYptr= ⋅σY Ansb) Maximum Distortion Eenergy Theory:2 − σ1⋅σ2 σ2σ1+ 2 σY2 =2 − σ1⋅ (0.5σ1) 0.5σ( 1)+ 2 σYσ12 =⋅ 2 σY0.75 σ12 =Principal Stress:σ1σY0.75=Hencep⋅ rtσY0.75=p2t3⋅ r= ⋅σY Ans 1106. Problem 10-99A thin-walled spherical pressure vessel has an inner radius r, thickness t, and is subjected to an internalpressure p. If the material constants are E and v, determine the strain in the circumferential direction interms of the stated parameters.Solution:Normal Stresses :This is a plane stress problem where σmin := 0since there is no load acting on the outer surface of the wall.σ1p⋅ r2⋅ t=σ2 = σ1Normal Strains: Appling the generalized Hooke's Law,ε11Eσ1 ν σ2 σ3 + ( ) ⋅ − ⎡⎣⎤⎦= ε11E= (σ1 − ν⋅σ2) ε1σ1E= (1 − ν)ε21Eσ2 ν σ1 σ3 + ( ) ⋅ − ⎡⎣⎤⎦= ε21E= (σ2 − ν⋅σ1) ε2σ1E= (1 − ν)Hence,εcirσ1E= (1 − ν)εcirp⋅ r2E⋅ t= (1 − ν) Ans 1107. Problem 10-100The strain at point A on the shell has components εx= 250(10-6), εy= 400(10-6), γxy = 275(10-6), εz=0. Determine (a) the principal strains at A, (b) the maximum shear strain in the x-y plane, and (c) theabsolute maximum shear strain.Given: εx := 250⋅ (10− 6) εy := 400⋅ (10− 6)γxy := 275⋅ (10− 6) εz := 0Solution:Construction of Mohr's Circle :Center :εcεx + εy2:= εc = 325 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 156.62 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)a) In-plane Principal Strains:ε1 := εc + R ε1 = 481.62 × 10− 6 Ansε2 := εc − R ε2 = 168.38 × 10− 6 Ansb) Maximum In-plane Shear Strain:γmax := 2R γmax = 313.25 × 10− 6 Ansc) Absolute Maximum Shear Strain :From the results obtained above,εmax := ε1 εmax = 481.625 × 10− 6εmin := εz εmin = 0γabs.max := εmax − εminγabs.max = 481.62 × 10− 6 Ans 1108. Problem 10-101A differential element is subjected to plane strain that has the following components: εx= 950(10-6), εy= 420(10-6), γxy = -325(10-6). Use the strain-transformation equations and determine (a) the principalstrains and (b) the maximum in-plane shear strain and the associated average strain. In each casespecify the orientation of the element and show how the strains deform the element.Given: εx := 950⋅ (10− 6) εy := 420⋅ (10− 6)γxy := −325⋅ (10− 6)Solution:a) In-plane Principal Strains: Applying Eq. 10-9,ε1εx + εy2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= + + ε1 = 995.86 × 10− 6 Ansε2εx + εy2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= − + ε2 = 374.14 × 10− 6 AnsOrientation of Principal Strain:tan(2θp) γxy= θpεx − εy12atanγxyεx − εy⎛⎜⎝⎞⎠:= θ'p := θp + 90degθp = −15.758 deg θ'p = 74.242 degUse Eq. 10-5 to determine the direction of ε1 and ε2.εx'εx + εy2εx − εy2+ ⋅cos(2θp)γxy2:= + ⋅ sin(2θp)εx' = 995.856 × 10− 6Therefore, θp1 := θp θp1 = −15.76 deg Ansθp2 := θ'p θp2 = 74.24 deg Ansb) Maximum In-plane Shear Strain: Applying Eq. 10-11,γmax 2εx − εy2⎛⎜⎝⎞⎠2 γxy2⎛⎜⎝⎞⎠2:= + γmax = 621.711 × 10− 6 Ansεavgεx + εy2:= εavg = 685 × 10− 6 AnsOrientation of Principal Strain:tan(2θs) εx − εy= − θsγxy12atanεx − εyγxy−⎛⎜⎝⎞⎠:= θ's := θs + 90degθs = 29.242 deg θ's = 119.242 deg 1109. Use Eq. 10-6 to determine the sign of γmax.γx'y' 2εx − εy2− ⋅ sin(2θs)γxy2+ ⋅cos(2θs)⎛⎜⎝⎞⎠:=γx'y' = −621.711 × 10− 6Therefore, θs1 := θs θs1 = 29.24 deg Ansθs2 := θ's θs2 = 119.24 deg Ans 1110. Problem 10-102The components of plane stress at a critical point on a thin steel shell are shown. Determine if failure(yielding) has occurred on the basis of the maximum-distortion-energy theory. The yield stress for thesteel is σY = 650 MPa.Given: σx := −55MPa σy := 340MPaτxy := 65MPa σY := 650MPaSolution:Principal Stresses:σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2σ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2σ1 = 350.42MPa σ2 = −65.42MPaMaximum Distortion Energy Theory:σ12 − σ1⋅σ2 + σ22 =σY2 := + 2 fσ = 150000MPa2 < σYfσ σ12 − σ1⋅σ2 σ22(= 422500 MPa2) AnsBased on the result obtained above, the material does not yield according tothe maximum distortion energy theory. Ans 1111. Problem 10-103Solve Prob. 10-102 using the maximum-shear-stress theory.Given: σx := −55MPa σy := 340MPaτxy := 65MPa σY := 650MPaSolution:Principal Stresses:σ1σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= + + 2σ2σx + σy2σx − σy2⎛⎜⎝⎞⎠2τxy:= − + 2σ1 = 350.42MPa σ2 = −65.42MPaMaximum Shear Stress Theory: Both principal stresses have opposite sign, henceσ1 − σ2 = σYfσ := σ1 − σ2 fσ = 415.84MPa < σY (= 650 MPa) AnsBased on the result obtained above, the material does not yield according tothe maximum shear stress theory. Ans 1112. cProblem 10-ε104The 60° strain rosette is mounted on a beam. The following readings are obtained for each gauge: εa =600(10-6), εb = -700(10-6), and = 350(10-6). Determine (a) the in-plane principal strains and (b) themaximum in-plane shear strain and average normal strain. In each case show the deformed element dueto these strains.Given: εa := 600⋅ (10− 6) θa := 150degεb := −700⋅ (10− 6) θb := −150degεc := 350⋅ (10− 6) θc := −90degSolution:Strain Rosettes (600): Applying Eq. 10-16,Givenεa = εx⋅cos(θa)2 + εy⋅ sin(θa)2 + γxy⋅ sin(θa)⋅cos(θa) (1)εb = εx⋅cos(θb)2 + εy⋅ sin(θb)2 + γxy⋅ sin(θb)⋅cos(θb) (2)εc = εx⋅cos(θc)2 + εy⋅ sin(θc)2 + γxy⋅ sin(θc)⋅cos(θc) (3)Solving Eqs.(1), (2) and (3),Guess εx := 10− 6 εy := 10− 6 γxy := 10− 6εxεyγxy⎛⎜⎜⎜⎝⎞⎟⎠:= Find(εx , εy , γxy)εxεyγxy⎛⎜⎜⎜⎝⎞⎟⎠−183.333 × 10− 6350 × 10− 6−1.501 × 10− 3⎛⎜⎜⎜⎜⎝⎞⎟⎟⎠=Construction of Mohr's Circle :Center :εcεx + εy2:= εc = 83.33 × 10− 6Radius :R (εy − εc)2 γxy2⎛⎜⎝⎞⎠2:= + R = 796.52 × 10− 6Coordinates:A(εx , 0.5⋅ γxy) C(εc , 0)a) In-plane Principal Strains:(represented by coordinates of points B and D)ε1 := εc + R ε1 = 879.85 × 10− 6 Ansε2 := εc − R ε2 = −713.19 × 10− 6 Ans 1113. Orientation of Principal Strain:tan(2θp2) 0.5γxy= θp2εx − εc12atan0.5γxyεx − εc⎛⎜⎝⎞⎠:= θp2 = 35.22 deg Ansθp1 := 90deg − θp2 θp1 = 54.78 deg Ansb) Maximum In-plane Shear Strain: (represented by coordinates of point E)γmax := −2R γmax = −1.593 × 10− 3 AnsOrientation of Maximum In-plane Shear Strain:tan(2θs) εx − εc= θs0.5γxy12atanεx − εc0.5γxy⎛⎜⎝⎞⎠:=θs = 9.78 deg (Clockwise) Ans 1114. Problem 10-105The aluminum beam has the rectangular cross section shown. If it is subjected to a bending moment ofM = 7.5 kN·m., determine the increase in the 50-mm dimension at the top of the beam and thedecrease in this dimension at the bottom. Eal = 70 GPa, νal= 0.3.Given: M := 7.5kN⋅m b := 50mm h := 75mmE := 70GPa ν := 0.3Solution:Section Property : Ib⋅h312:=Normal Stresses :ch2:= σzM⋅cI:=Lateral Strain and deformation :εxν⋅σzE:= εx = 685.71 × 10− 6At the top:Δb := εx⋅b Δb = 0.03429mmAt the bottom:Δb := −εx⋅b Δb = −0.03429mmThe negative sign indicates shortening. 1115. Problem 11-01The simply supported beam is made of timber that has an allowable bending stress of σallow = 6.5 MPaand an allowable shear stress of τallow = 500 kPa. Determine its dimensions if it is to be rectangularand have a height-to-width ratio of 1.25.Given: σallow := 6.5MPa La := 2m Lb := 4mτallow := 0.5MPa wo 8kNm:=h = 1.25⋅bSolution: L 2La:= + LbSupport Reactions : By symmetry, RL=RR=R+ ΣFy=0; 2R − wo⋅L = 0 R := 0.5wo⋅L R = 32 kNMaximum Moment and Shear:Vmax := wo⋅La Vmax = 16 kNMmax := wo⋅La⋅ (0.5⋅La) Mmax = 16 kN⋅mSection Property : Ib⋅h312=SxI0.5h= Sxb⋅h26= Sxb⋅ (1.25⋅b)2= Sx625⋅b396=Bending Stress:Sreq'dMmaxσallow=25⋅b396Mmaxσallow= b3 96Mmax25σallow:= b = 211.4mm Ansh := 1.25⋅b h = 264.3mm AnsShear Check : Ib⋅h312:= Qmax := (0.5b⋅h)⋅0.25hτmaxVmax⋅Qmax:= τmax = 0.429MPaI⋅b< τallow =0.5 MPa (O.K.!) 1116. x1 := 0 , 0.01⋅La .. La x2 := La , 1.01⋅La .. (La + Lb) x3 := (La + Lb) , 1.01⋅ (La + Lb) .. LV1(x1) −wo⋅x1:= V2(x2) (−wo⋅x2 + R) 1kN:= ⋅ V3(x3) (−wo⋅x3 + 2R) 1kNkN:= ⋅⋅ 2M1(x1) −0.5⋅wo x1:= M2(x2) −0.5⋅wo x2kN⋅m⋅ 2 R x+ ⋅ ( 2 − La) ⎡⎣⎤⎦1kN⋅m:= ⋅M3(x3) −0.5⋅wo x3⋅ 2 R x+ ⋅ ( 3 − La) + R⋅ (x3 − La − Lb) ⎡⎣⎤⎦1kN⋅m:= ⋅0 2 4 6 8201001020Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x30 2 4 6 801020Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 1117. Problem 11-02The joists of a floor in a warehouse are to be selected using square timber beams made of oak. If eachbeam is to be designed to carry 1.5 kN/m over a simply supported span of 7.5 m, determine thedimension a of its square cross section to the nearest multiples of 5mm. The allowable bending stressis σallow = 32 MPa and the allowable shear stress is τallow = 0.875 MPa.Given: σallow := 32MPa L := 7.5mτallow := 0.875MPa w 1.5kNm:=Solution:Support Reactions : By symmetry, RL=RR=R+ ΣFy=0; 2R − w⋅L = 0 R := 0.5w⋅LMaximum Moment and Shear:Vmax := R Vmax = 5.63 kNMmax := R⋅ (0.5L) − w⋅ (0.5L)⋅ (0.25L) Mmax = 10.55 kN⋅mSection Property : Ia412= Qmax = (0.5a⋅ a)⋅0.25aBending Stress: cmax = 0.5aa3 6MmaxσallowM⋅cmaxσ := max= σallowI12Mmax⋅ (0.5a)a4=a = 125.52mm (Use 130mm) AnsShear Stress : Ia412:= Qmax := (0.5a⋅a)⋅0.25aτmaxVmax⋅Qmax:= τmax = 0.536MPaI⋅a< τallow =0.875 MPa (O.K.!)x := 0 , 0.01⋅L .. L V(x) (R − w⋅x)1kN:= ⋅ M(x) [R⋅x − w⋅x⋅ (0.5x)]1kN⋅m:=0 2 4 6Distance (m)Shear (kN)V(x)x0 2 4 6105Distance m)Moment (kN-m)M(x)x 1118. Problem 11-03The timber beam is to be loaded as shown. If the ends support only vertical forces, determine thegreatest magnitude of P that can be applied. σallow = 25 MPa, τallow = 700 kPa.Given: σallow := 25MPa a := 4mτallow := 0.7MPabf := 150mm df := 30mmtw := 40mm dw := 120mmSolution: L := 2aSection Property : h := df + dwycΣ yi ⎯⋅ ( ⋅Ai)Σ⋅ (Ai)=yc(bf⋅df)⋅ (0.5df) + (tw⋅dw)⋅ (0.5dw + df):= yc = 53.71mm(bf⋅df) + (tw⋅dw)I112⋅ 3 b( f⋅df) (0.5df − yc)+ ⋅ 2⋅bf df1⋅ ⋅ 12tw dw3 + ⎡⎢⎣t( w⋅dw) ⋅ (0.5dw + df − yc)2 ⎤⎥⎦:= +I = 19162016.13mm4Qmax h yc − ( ) tw ⋅ 0.5 h yc − ( ) ⋅ ⎡⎣:= ⋅ Qmax = 185436.52mm3⎤⎦Support Reactions : By symmetry, A=B=R+ ΣFy=0; 2R − P = 0 R = 0.5PMaximum Load: Assume failure due to bending moment.Mmax = R⋅a Mmax = 0.5P⋅a cmax := h − ycσallowMmax⋅cmax= σallowI0.5P⋅a⋅ (h − yc)= PI2I(σallow)a h yc⋅ ( − ):=P = 2.49 kN AnsCheck Shear : R := 0.5P Vmax := R Mmax := R⋅aτmaxVmax⋅Qmax:= τmax = 0.301MPaI⋅ tw< τallow =0.7 MPa (O.K.!) 1119. Problem 11-04Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the machineloading shown. The allowable bending stress is σallow = 168 MPa and the allowable shear stress is τallow= 98 MPa.Given: σallow := 168MPa a := 0.6mτallow := 98MPa P := 25kNSolution: L := 5aSupport Reactions : By symmetry, RL=RR=R+ ΣFy=0; 2R − 4P = 0 R := 2PMaximum Moment and Shear:Vmax := R Vmax = 50 kNMmax := R⋅ (2a) − P⋅a Mmax = 45 kN⋅mBending Stress:Sreq'dMmaxσallow:= Sreq'd = 267.86 × 103mm3Select W 310x24 : Sx := 281⋅ (103)mm3 d := 305mm tw := 5.59mmShear Stress : Provide a shear stress check.τmaxVmaxtw⋅d:= τmax = 29.33MPa< τallow =98 MPa (O.K.!)Hence, Use W 310x24 Ans 1120. x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 2a x3 := 2a , 1.01⋅ (2a) .. 3ax4 := 3a , 1.01⋅ (3a) .. 4a x5 := 4a , 1.01⋅ (4a) .. 5aV1(x1) R1kN:= ⋅ V2(x2) (R − P)1kN:= ⋅ V3(x3) (R − 2P)1kN:= ⋅V4(x4) (R − 3P)1kN:= ⋅ V5(x5) (R − 4P)1kN:= ⋅M1(x1) R⋅x1:= M2 x2 ( ) R x2 ( ) ⋅ P x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) R x3 ( ) ⋅ P x3 a − ( ) ⋅ − P x3 2a − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅M4 x4 ( ) R x4 ( ) ⋅ P x4 a − ( ) ⋅ − P x4 2a − ( ) ⋅ − P x4 3a − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅M5 x5 ( ) R x5 ( ) ⋅ P x5 a − ( ) ⋅ − P x5 2a − ( ) ⋅ − P x5 3a − ( ) ⋅ − P x5 4a − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.5 1 1.5 2 2.5 35050Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)V4(x4)V5(x5)x1, x2, x3, x4, x50 0.5 1 1.5 2 2.5 34020Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)M4(x4)M5(x5)x1, x2, x3, x4, x5 1121. Problem 11-05The simply supported beam is made of timber that has an allowable bending stress of σallow = 7 MPaand an allowable shear stress of τallow = 0.5 MPa. Determine its dimensions if it is to be rectangular andhave a height-to-width ratio of 1.25.Given: σallow := 7MPa L := 2mτallow := 0.5MPa wo 75kNm:=h = 1.25⋅bSolution:Support Reactions : By symmetry, RL=RR=R+ ΣFy=0; 2R − 0.5wo⋅ (2L) = 0 R := 0.5(wo)⋅LMaximum Moment and Shear:Vmax := R Vmax = 75 kNMmax R⋅L (0.5wo⋅L) L:= − ⋅ Mmax = 100 kN⋅m3Section Property : Ib⋅h312=SxI0.5h= Sxb⋅h26= Sxb⋅ (1.25⋅b)2= Sx625⋅b396=Bending Stress:Sreq'dMmaxσallow=25⋅b396Mmaxσallow= b3 96Mmax25σallow:= b = 380.0mmShear Stress : Provide a shear stress check.h := 1.25⋅b τmax1.5Vmaxb⋅h:= τmax = 0.62MPa> τallow =0.5 MPa (NG.!)Shear controls :τallow1.5Vmaxb⋅h= τallow1.5Vmaxb⋅ (1.25b)=b1.5Vmax1.25τallow:=b = 424.3mm Ans 1122. x1 := 0 , 0.01⋅L .. L x2 := L , 1.01⋅ (L) .. (2L)V1(x1) Rwo2x1L⎛⎜⎝⎞⎠− ⋅ ⋅x1⎡⎢⎣⎤⎥⎦1kN:= ⋅V2(x2) R − 0.5⋅wo⋅L wo⋅ (x2 − L) 1 0.5x2 − LL− ⋅⎛⎜⎝⎞⎠− ⋅⎡⎢⎣⎤⎥⎦1kN:= ⋅M1(x1) R⋅x1wo2x1L⎛⎜⎝⎞⎠⋅ ⋅x1x13− ⋅⎡⎢⎣⎤⎥⎦1kN⋅m:= ⋅M2(x2) R⋅x2wo⋅L2x22⋅L3− ⎛⎜⎝⎞⎠− ⋅wo2⋅ (x2 − L)2 1x2 − LL⎛⎜⎝⎞⎠13− ⋅⎡⎢⎣⎤⎥⎦− ⋅⎡⎢⎣⎤⎥⎦1kN⋅m:= ⋅0 1 2 3 40Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 1 2 3 4100500Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 1123. Problem 11-06The wooden beam has a rectangular cross section and is used to support a load of 6 kN. If theallowable bending stress is σallow = 14 MPa and the allowable shear stress is τallow = 5 MPa, determinethe height h of the cross section to the nearest multiples of 5mm if it is to be rectangular and have awidth of b = 75 mm. Assume the supports at A and B only exert vertical reactions on the beam.Given: σallow := 14MPa L1 := 1.2mτallow := 5MPa L2 := 1.8mP := 6kN b := 75mmSolution: L := L1 + L2Support Reactions :+ ΣFy=0; A + B − P = 0 (1)ΣΜB=0; A⋅L − P⋅L2 = 0 (2)Solving Eqs. (1) and (2): AP⋅L2L:= BP⋅L1L:=Maximum Moment and Shear:Vmax := max(A, B) Vmax = 3.60 kNMmax := A⋅L1 Mmax = 4.32 kN⋅mSection Property : Ib⋅h312= Qmax = (0.5h⋅b)⋅0.25hBending Stress: cmax = 0.5hh6Mmaxb⋅σallowM⋅cmaxσ := max= σallowI12Mmax⋅ (0.5h)b⋅h3=h = 157.12mm (Use 160mm) AnsShear Stress : Ib⋅h312:= Qmax := (0.5h⋅b)⋅0.25hτmaxVmax⋅Qmax:= τmax = 0.458MPaI⋅b< τallow =5 MPa (O.K.!) 1124. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. LV1(x1) A1kN:= ⋅ V2(x2) (A − P)1kN:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ P x2 L1 − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅0 1 2 3505Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 1 2 350Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 1125. Problem 11-07Solve Prob. 11-6 if the cross section has an unknown width but is to be square, i.e., h = b.Given: σallow := 14MPa L1 := 1.2mτallow := 5MPa L2 := 1.8mP := 6kN b = hSolution: L := L1 + L2Support Reactions :+ ΣFy=0; A + B − P = 0 (1)ΣΜB=0; A⋅L − P⋅L2 = 0 (2)Solving Eqs. (1) and (2): AP⋅L2L:= BP⋅L1L:=Maximum Moment and Shear:Vmax := max(A, B) Vmax = 3.60 kNMmax := A⋅L1 Mmax = 4.32 kN⋅mSection Property : Ib⋅h312= Qmax = (0.5h⋅b)⋅0.25hBending Stress: cmax = 0.5h b = hh3 6MmaxσallowM⋅cmaxσ := max= σallowI12Mmax⋅ (0.5h)h⋅ (h3)=h = 122.79mm (Use 125mm) AnsShear Stress : b := h Ib⋅h312:= Qmax := (0.5h⋅b)⋅0.25hτmaxVmax⋅Qmax:= τmax = 0.358MPaI⋅b< τallow =5 MPa (O.K.!) 1126. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. LV1(x1) A1kN:= ⋅ V2(x2) (A − P)1kN:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ P x2 L1 − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅0 1 2 3505Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 1 2 350Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 1127. Problem 11-08The simply supported beam is composed of two W310 X 33 sections built up as shown. Determine themaximum uniform loading w the beam will support if the allowable bending stress is σallow = 160 MPaand the allowable shear stress is τallow = 100 MPa.Given: σallow := 160MPa L := 8mτallow := 100MPa Use W 310x33Solution:Support Reactions : By symmetry, RL=RR=R+ ΣFy=0; 2R − w⋅L = 0 R = 0.5w⋅LMaximum Moment and Shear:Vmax = RMmax = R⋅ (0.5L) − w⋅ (0.5L)⋅ (0.25L)Mmax = (0.5w⋅L)⋅ (0.5L) − w⋅ (0.5L)⋅ (0.25L) Mmax = 0.125⋅w⋅L2Section Property : One W 310x33tw := 6.60mm Ix := 65.0⋅ (106)mm4 A := 4180mm2 d := 313mmTwo W 310x33 : Ic 2 Ix A 0.5d ( )2 + ⎡⎣⎤⎦:= ScIcd:=Maximum Loading: Assume moment controls.Mmax⋅cmaxcmax := d σmax= σallowIc0.125⋅w⋅L2⋅ (d)2 Ix A 0.5d ( )2 + ⎡⎣⎤⎦=w16σallowL2⋅dIx A 0.5d ( )2 + ⎡⎣⎤⎦⋅ := w 21.39kNm= AnsCheck Shear : (Neglect area of flanges.) Aw := 2d⋅ tw R := 0.5w⋅L Vmax := RτmaxVmaxAw:= τmax = 20.71MPa < τallow =100 MPa (O.K.!)x := 0 , 0.01⋅L .. L V(x) (R − w⋅x)1kN:= ⋅ M(x) [R⋅x − w⋅x⋅ (0.5x)]1kN⋅m:=0 51000100Distance (m)Shear (kN)V(x)x0 52001000Distance m)Moment (kN-m)M(x)x 1128. Problem 11-09The simply supported beam is composed of two W310 X 33 sections built up as shown. Determine ifthe beam will safely support a loading of w = 30 kN/m. The allowable bending stress is σallow = 160MPa and the allowable shear stress is τallow = 100 MPa.Given: σallow := 160MPa L := 8m Use W 310x33τallow := 100MPa w 30kNm:=Solution:Support Reactions : By symmetry, RL=RR=R+ ΣFy=0; 2R − w⋅L = 0 R := 0.5w⋅LMaximum Moment and Shear:Vmax := RMmax = R⋅ (0.5L) − w⋅ (0.5L)⋅ (0.25L)Mmax = (0.5w⋅L)⋅ (0.5L) − w⋅ (0.5L)⋅ (0.25L) Mmax := 0.125⋅w⋅L2Section Property : One W 310x33tw := 6.60mm Ix := 65.0⋅ (106)mm4 A := 4180mm2 d := 313mmTwo W 310x33 : Ic 2 Ix A 0.5d ( )2 + ⎡⎣⎤⎦:= ScIcd:=Bending Stress:cmax := d σmaxMmax⋅cmaxIc:=σmax = 224.4MPa > σallow =160 MPa (Not O.K.!)The beam fails due to bending stress criteria.. AnsCheck Shear : (Neglect area of flanges.) Aw := 2d⋅ twτmaxVmaxAw:= τmax = 29.04MPa < τallow =100 MPa (O.K.!)x := 0 , 0.01⋅L .. L V(x) (R − w⋅x)1kN:= ⋅ M(x) [R⋅x − w⋅x⋅ (0.5x)]1kN⋅m:=0 50Distance (m)Shear (kN)V(x)x0 52000Distance m)Moment (kN-m)M(x)x 1129. Problem 11-10Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the loadingshown, where w = 100 kN/m and P = 25 kN. The allowable bending stress is σallow = 160 MPa, andthe allowable shear stress is τallow = 100 MPa.Given: σallow := 160MPa L1 := 2.4mτallow := 100MPa L2 := 1.8mP := 25kN w 100kNm:=Solution: L := L1 + L2Support Reactions : Given+ ΣFy=0; A + B − P − w⋅L1 = 0 (1)− ⋅ 2 = 0 (2)ΣΜB=0; A⋅L1 + P⋅L2 0.5w L1Solving Eqs. (1) and (2): Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠101.25163.75⎛⎜⎝⎞⎠= kNMaximum Moment and Shear:Vmax := P − B Vmax = −138.75 kNWhen V=0,xoL1AA + B= xoA⋅L1A + B:= xo = 0.917 m:= − ⋅ 2 Mmax = 50.8 kN⋅mMmax A⋅xo 0.5w xoBending Stress: Assume bending controls the design.Sreq'dMmaxσallow:= Sreq'd = 317510.1mm3Select W 310x33 : Sx := 415⋅ (103)mm3 d := 313mm tw := 6.60mmShear Stress : Provide a shear stress check.τmaxVmaxtw⋅d:= τmax = 67.17MPa< τallow =100 MPa (O.K.!)Hence, Use W 310x33 Ans 1130. x1x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. L( ) ( ) 1V1 x1 A − w ⋅ ⋅ := V2 x2 ( ) A w L1kN( − ⋅ + B) 1kN:= ⋅2 ⋅ − ⎛⎝M1(x1) A⋅x1 0.5w x1⎞⎠1kN⋅m:= ⋅M2 x2 ( ) A x2 ⋅ w L1 ⋅ ( ) x2 0.5 L1 ⋅ − ( ) ⋅ − B x2 L1 − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 2 41000100Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 2 450050Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 1131. Problem 11-11Select the lightest-weight steel wide-flange beam having the shortest height from Appendix B that willsafely support the loading shown, where w = 0 and P = 50 kN. The allowable bending stress is σallow =168 MPa, and the allowable shear stress is τallow = 100 MPa.Given: σallow := 168MPa L1 := 2.4mτallow := 100MPa L2 := 1.8mP := 50kN w 0kNm:=Solution: L := L1 + L2Support Reactions : Given+ ΣFy=0; A + B − P = 0 (1)ΣΜB=0; A⋅L1 + P⋅L2 = 0 (2)Solving Eqs. (1) and (2): Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠−37.5087.50⎛⎜⎝⎞⎠= kNMaximum Moment and Shear:Vmax := P Vmax = 50 kNMmax := P⋅L2 Mmax = 90 kN⋅mBending Stress: Assume bending controls the design.Sreq'dMmaxσallow:= Sreq'd = 535714.3mm3Three choices of wide flange section having the weight of 39 kg/m can bemade. W 310x39 , W 360x39 , and W 410x39. However, the shortest isthe W 310x39.Select W 310x39 : Sx := 547⋅ (103)mm3 d := 310mm tw := 5.84mmShear Stress : Provide a shear stress check.τmaxVmaxtw⋅d:= τmax = 27.62MPa< τallow =100 MPa (O.K.!)Hence, Use W 310x39 Ans 1132. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. LV1(x1) (A)1kN:= ⋅ V2(x2) (A + B)1kN:= ⋅M1(x1) (A⋅x1) 1kN⋅m:= ⋅M2 x2 ( ) A x2 ⋅ B x2 L1 − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 2 450050Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 2 4050100Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 1133. Problem 11-12Determine the minimum width of the beam to the nearest multiples of 5mm that will safely support theloading of P = 40 kN. The allowable bending stress is σallow = 168 MPa, and the allowable shear stressis τallow = 105 MPa.Given: σallow := 168MPa L1 := 2mτallow := 105MPa L2 := 2mP := 40kN h := 150mmSolution: L := L1 + L2Support Reactions : Given+ ΣFy=0; A + B − P = 0 (1)ΣΜB=0; A⋅L1 − P⋅L = 0 (2)Solving Eqs. (1) and (2): Guess A := 1kN B := 1kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠80−40⎛⎜⎝⎞⎠= kNMaximum Moment and Shear:Vmax := P Vmax = 40 kNMmax := P⋅L1 Mmax = 80 kN⋅mSection Property : Ib⋅h312=SxI0.5h= Sxb⋅h26=Qmax = (0.5h⋅b)⋅0.25hBending Stress: Assume bending controls the design.Sreq'dMmaxσallow=b⋅h26Mmaxσallow= b6Mmax(h2)σallow:=b = 126.98mm (Use 130mm) AnsCheck Shear : Ib⋅h312:= Qmax := (0.5h⋅b)⋅0.25hτmaxVmax⋅Qmax:= τmax = 3.150MPaI⋅b< τallow =105 MPa (O.K.!) 1134. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. LV1(x1) (−P)1kN:= ⋅ V2(x2) (−P + A)1kN:= ⋅M1(x1) (−P⋅x1) 1kN⋅m:= ⋅M2 x2 ( ) P − x2 ⋅ A x2 L1 − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 2 450050Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 2 4050100Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 1135. Problem 11-13Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the loadingshown. The allowable bending stress is σallow = 168 MPa, and the allowable shear stress is τallow = 100MPa.Given: σallow := 168MPa L1 := 2mτallow := 100MPa L2 := 3mP := 75kN wo 75kNm:=Solution: L := L1 + L2Support Reactions : Given+ ΣFy=0; A + B − P − 0.5wo⋅L2 = 0 (1)ΣΜA=0; −P⋅L1 0.5wo⋅L2L23⎛⎜⎝⎞⎠+ ⋅ − B⋅L2 = 0 (2)Solving Eqs. (1) and (2): Guess A := 1kN B := 2kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠200.00−12.50⎛⎜⎝⎞⎠= kNMaximum Moment and Shear:Vmax := −P + A Vmax = 125 kNMmax := −P⋅L1 Mmax = −150 kN⋅mBending Stress: Assume bending controls the design.Sreq'dMmaxσallow:= Sreq'd = 892857.1mm3Select W 410x53 : Sx := 923⋅ (103)mm3 d := 403mm tw := 7.49mmShear Stress : Provide a shear stress check.τmaxVmaxtw⋅d:= τmax = 41.41MPa< τallow =100 MPa (O.K.!)Hence, Use W 410x53 Ans 1136. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. LV1(x1) −P:= kNV2(x2) −P + A wo⋅ (x2 − L1) 1 0.5x2 − L1L2− ⋅⎛⎜⎝⎞⎠− ⋅⎡⎢⎣⎤⎥⎦1kN:= ⋅M1(x1) (−P⋅x1) 1kN⋅m:= ⋅M2(x2) −P⋅x2 + A⋅ (x2 − L1)wo2⋅ (x2 − L1)2 1x2 − L1L2⎛⎜⎝⎞⎠13− ⋅⎡⎢⎣⎤⎥⎦− ⋅⎡⎢⎣⎤⎥⎦1kN⋅m:= ⋅0 1 2 3 4 51000100Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 1 2 3 4 5050100150Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 1137. Problem 11-14Select the lightest-weight steel structural wide-flange beam with the shortest depth from Appendix Bthat will safely support the loading shown. The allowable bending stress is σallow = 168 MPa, and theallowable shear stress is τallow = 100 MPa.Given: σallow := 168MPa L := 2mτallow := 100MPa wo 120kNm:=Solution:Support Reactions : Given+ ΣFy=0; R − 0.5wo⋅L = 0 R := 0.5wo⋅LΣΜA=0; M 0.5wo⋅LL3⎛⎜⎝⎞⎠+ ⋅ = 0 M−wo⋅L26:=Maximum Moment and Shear:Vmax := R Vmax = 120 kNMmax := M Mmax = −80 kN⋅mBending Stress: Assume bending controls the design.Sreq'dMmaxσallow:= Sreq'd = 476190.5mm3Select W 310x39 : Sx := 547⋅ (103)mm3 d := 310mm tw := 5.84mmShear Stress : Provide a shear stress check.τmaxVmaxtw⋅d:= τmax = 66.28MPa< τallow =100 MPa (O.K.!)Hence, Use W 310x39 Ans 1138. Problem 11-15Select the shortest and lightest-weight steel wide-flange beam from Appendix B that will safely supportthe loading shown. The allowable bending stress is σallow = 160 MPa, and the allowable shear stress isτallow = 84 MPa.Given: σallow := 160MPa a := 1.2mτallow := 84MPa P1 := 20kNP2 := 50kN P3 := 30kNSolution: L := 4aSupport Reactions : Given+ ΣFy=0; A + B − P1 − P2 − P3 = 0 (1)ΣΜB=0; A⋅L − P1⋅ (3a) − P2⋅ (2⋅a) − P3⋅ (a) = 0 (2)Solving Eqs. (1) and (2): Guess A := 1kN B := 2kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠47.5052.50⎛⎜⎝⎞⎠= kNMaximum Moment and Shear:Vmax := max(A, B) Vmax = 52.5 kNMmax := A⋅ (2a) − P1⋅ (a) Mmax = 90 kN⋅mBending Stress: Assume bending controls the design.Sreq'dMmaxσallow:= Sreq'd = 562500mm3Select W 250x58 : Sx := 693⋅ (103)mm3 d := 252mm tw := 8.00mmShear Stress : Provide a shear stress check.τmaxVmaxtw⋅d:= τmax = 26.04MPa< τallow =84 MPa (O.K.!)Hence, Use W 250x58 Ans 1139. x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 2a x3 := 2a , 1.01⋅ (2a) .. 3ax4 := 3a , 1.01⋅ (3a) .. 4aV1(x1) A1kN:= ⋅ V2(x2) (A − P1) 1:= ⋅ V3(x3) (A − P1 − P2) 1kNkN:= ⋅V4 x4 ( ) A P1 − P2 − ( ) P3 − ⎡⎣⎤⎦1kN:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ P1 x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ P1 x3 a − ( ) ⋅ − P2 x3 2a − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅M4 x4 ( ) A x4 ( ) ⋅ P1 x4 a − ( ) ⋅ − P2 x4 2a − ( ) ⋅ − ⎡⎣P3 x4 3a − ( ) ⋅ − ⎡⎣⎤⎦⎤⎦1kN⋅m:= ⋅0 1 2 3 450050Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)V4(x4)x1, x2, x3, x40 1 2 3 4100500Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)M4(x4)x1, x2, x3, x4 1140. Problem 11-16The beam is made of a ceramic material having an allowable bending stress of σallow = 5 MPa and anallowable shear stress of τallow = 2.8 MPa. Determine the width b of the beam if the height h = 2b.Given: σallow := 5MPa L1 := 50mmτallow := 2.8MPa Lo := 150mmP1 := 75N w 1.2kNm:=P2 := 50N h = 2bSolution: L 2L1:= + LoSupport Reactions : Given+ ΣFy=0; A + B − P1 − P2 − w⋅Lo = 0 (1)ΣΜB=0; A⋅ (Lo) − P1⋅ (L1 + Lo) − w⋅Lo⋅ (0.5Lo) + P2⋅L1 = 0 (2)Solving Eqs. (1) and (2): Guess A := 1N B := 1NAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠173.33131.67⎛⎜⎝⎞⎠= NMaximum Moment and Shear:Vmax := −P1 + A Vmax = 98.33NMmax := P1⋅L1 Mmax = 3.75 N⋅mSection Property : Ib⋅h312= h = 2bSxI0.5h= Sxb⋅h26= Sx4b36=Bending Stress: Assume bending controls the design.Sreq'dMmaxσallow=4b36Mmaxσallow= b3 6Mmax4σallow:=b = 10.40mm AnsCheck Shear : h := 2⋅bτmax1.5Vmaxh⋅b:= τmax = 0.682MPa< τallow =2.8 MPa (O.K.!) 1141. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + Lo) x3 := (L1 + Lo) , 1.01⋅ (L1 + Lo) .. LV1(x1) −P1:= ⋅ V3(x3) (−P1 + A − w⋅Lo + B) 1:= V2 x2 ( ) P1 − A + w x2 L1 − ( ) ⋅ − ⎡⎣N⎤⎦1NN:= ⋅M1(x1) −P1⋅x1:= M2 x2 ( ) P1 − x2 ⋅ A x2 L1 − ( ) ⋅ + 0.5w x2 L1 − ( )2 ⋅ − ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅M3 x3 ( ) P1 − x3 ⋅ A x3 L1 − ( ) ⋅ + w Lo ⋅ ( ) x3 L1 − 0.5 Lo ⋅ − ( ) ⋅ − B x3 L1 − Lo − ( ) ⋅ + ⎡⎣⎤⎦1N⋅m:= ⋅0 0.05 0.1 0.15 0.21000100Distance (m)Shear (N)V1(x1)V2(x2)V3(x3)x1, x2, x30 0.05 0.1 0.15 0.22024Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 1142. Problem 11-17The steel cantilevered T-beam is made from two plates welded together as shown. Determine themaximum loads P that can be safely supported on the beam if the allowable bending stress is σallow =170 MPa and the allowable shear stress is τallow = 95 MPa.Given: σallow := 170MPa a := 2mτallow := 95MPabf := 150mm df := 15mmtw := 15mm dw := 150mmSolution: L := 2aSection Property : h := df + dwycΣ yi ⎯⋅ ( ⋅Ai)Σ⋅ (Ai)= yc(bf⋅df)⋅ (0.5df) + (tw⋅dw)⋅ (0.5dw + df):= yc = 48.75mm(bf⋅df) + (tw⋅dw)I112⋅ 3 b( f⋅df) (0.5df − yc)+ ⋅ 2⋅bf df1⋅ ⋅ 12tw dw3 + ⎡⎢⎣t( w⋅dw) ⋅ (0.5dw + df − yc)2 ⎤⎥⎦:= +I = 11917968.75mm4Qmax h yc − ( ) tw ⋅ 0.5 h yc − ( ) ⋅ ⎡⎣:= ⋅ Qmax = 101355.47mm3⎤⎦Support Reactions :+ ΣFy=0; R − P − P = 0 R = 2PΣΜA=0; MA + P⋅a + P⋅ (2a) = 0 MA = 3P⋅aMaximum Load: Assume failure due to bending moment.Mmax = MA cmax := h − ycσallowMmax⋅cmax= σallowI3P⋅a⋅ (h − yc)= PII(σallow)3a⋅ (h − yc):=P = 2.90 kN AnsCheck Shear : R := 2P Vmax := R MA := 3P⋅aτmaxVmax⋅Qmax:= τmax = 3.294MPaI⋅ tw< τallow =95 MPa (O.K.!) 1143. x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. LV1(x1) (R)1kN:= ⋅ V2(x2) (R − P)1kN:= ⋅M1(x1) (−MA + R⋅x1) 1kN⋅m:= ⋅M2 x2 ( ) MA − R x2 ⋅ + P x2 a − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 2 46420Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 2 401020Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 1144. Problem 11-18Draw the shear and moment diagrams for the W310 X 21 beam and check if the beam will safelysupport the loading. The allowable bending stress is σallow = 160 MPa and the allowable shear stress isτallow = 84 MPa.Given: σallow := 160MPa L1 := 1mτallow := 84MPa L2 := 4mMo := 75kN⋅m w 25kNm:=Solution: L := L1 + L2Support Reactions : Given+ ΣFy=0; A + B − w⋅L2 = 0 (1)+ ⋅ 2 B L− ⋅ 2 = 0 (2)ΣΜA=0; −Mo 0.5w L2Solving Eqs. (1) and (2): Guess A := 1kN B := 2kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠68.7531.25⎛⎜⎝⎞⎠= kNMaximum Moment and Shear:Vmax := max(A, B) Vmax = 68.75 kNMmax := Mo Mmax = 75 kN⋅mBending Stress: Assume bending controls the design.Use W 310x21 : Sx := 244⋅ (103)mm3 d := 303mm tw := 5.08mmSreq'dMmaxσallow:= Sreq'd = 468750mm3> Sx =244⋅ (103)mm3 (No Good.!)Shear Stress : Provide a shear stress check.τmaxVmaxtw⋅d:= τmax = 44.66MPa< τallow =84 MPa (O.K.!)Hence, the wide flange section W 310x21 fails due to the bending stressand will not safely support the loading. Ans 1145. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. LV1(x1) 0x2⎡⎣:= kNV2 ( x2 ) A − w ⋅ ( − L1 ) ⎤⎦1kN:= ⋅M1(x1) (−Mo) 1kN⋅m:= ⋅x2 L1 − ( )2 ⋅ − ⎡⎢⎣M2(x2) −Mo + A⋅ (x2 − L1)w2⎤⎥⎦1kN⋅m:= ⋅0 1 2 3 4 5500Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 1 2 3 4 5050100Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 1146. Problem 11-19Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the loadingshown. The allowable bending stress is σallow = 160 MPa and the allowable shear stress is τallow = 84MPa.Given: σallow := 160MPa L1 := 1mτallow := 84MPa L2 := 4mMo := 75kN⋅m w 25kNm:=Solution: L := L1 + L2Support Reactions : Given+ ΣFy=0; A + B − w⋅L2 = 0 (1)+ ⋅ 2 B L− ⋅ 2 = 0 (2)ΣΜA=0; −Mo 0.5w L2Solving Eqs. (1) and (2): Guess A := 1kN B := 2kNAB⎛⎜⎝⎞⎠:= Find(A, B)AB⎛⎜⎝⎞⎠68.7531.25⎛⎜⎝⎞⎠= kNMaximum Moment and Shear:Vmax := max(A, B) Vmax = 68.75 kNMmax := Mo Mmax = 75 kN⋅mBending Stress: Assume bending controls the design.Sreq'dMmaxσallow:= Sreq'd = 468750mm3Select W 360x33 : Sx := 475⋅ (103)mm3 d := 349mm tw := 5.84mmShear Stress : Provide a shear stress check.τmaxVmaxtw⋅d:= τmax = 33.73MPa< τallow =84 MPa (O.K.!)Hence, Use W 360x33 Ans 1147. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. LV1(x1) 0x2⎡⎣:= kNV2 ( x2 ) A − w ⋅ ( − L1 ) ⎤⎦1kN:= ⋅M1(x1) (−Mo) 1kN⋅m:= ⋅x2 L1 − ( )2 ⋅ − ⎡⎢⎣M2(x2) −Mo + A⋅ (x2 − L1)w2⎤⎥⎦1kN⋅m:= ⋅0 1 2 3 4 5500Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 1 2 3 4 5050100Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 1148. Problem 11-20The compound beam is made from two sections, which are pinned together at B. Use Appendix B andselect the light wide-flange beam that would be safe for each section if the allowable bending stress isσallow = 168 MPa and the allowable shear stress is τallow = 100 MPa. The beam supports a pipe loadingof 6 kN and 9 kN as shown.Given: σallow := 168MPa L1 := 3.6mτallow := 100MPa L2 := 2.4mP1 := 6kN L3 := 3mP2 := 9kNSolution: L := L1 + L2 + L3Support Reactions : GivenFor segment BC :+ ΣFy=0; B + C − P2 = 0 (1)ΣΜB=0; −C⋅ (L2 + L3) + P2⋅L2 = 0 (2)Solving Eqs. (1) and (2): Guess C := 1kN B := 2kNCB⎛⎜⎝⎞⎠:= Find(C, B)CB⎛⎜⎝⎞⎠45⎛⎜⎝⎞⎠= kNFor segment AB :+ A := P1 + B A = 11 kNMA := −P1⋅ (0.5L1) − B⋅L1 MA = −28.80 kN⋅mMaximum Moment and Shear:For segment AB : Vmax := max(A, B) Vmax = 11 kNMmax := MA Mmax = −28.8 kN⋅mFor segment BC : V'max := max(B, C) V'max = 5 kNM'max := C⋅L3 M'max = 12 kN⋅mBending Stress: Assume bending controls the design.For segment AB : Sreq'dMmaxσallow:= Sreq'd = 171428.6mm3Select W 250x18 :Sx := 179⋅ (103)mm3 d := 251mm tw := 4.83mmFor segment BC :S'req'dM'maxσallow:= S'req'd = 71428.6mm3Select W 150x14 :S'x := 91.2⋅ (103)mm3 d' := 150mm t'w := 4.32mm 1149. Shear Stress : Provide a shear stress check.For segment AB :τmaxVmaxtw⋅d:= τmax = 9.07MPa< τallow =100 MPa (O.K.!)Hence, Use W 250x18 AnsFor segment BC :τ'maxV'maxt'w⋅d':= τ'max = 7.72MPa< τallow =100 MPa (O.K.!)Hence, Use W 150x14 AnsSet a := 0.5L1 b := 0.5L1 c := L2 d := L3x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. a + b x3 := a + b , 1.01⋅ (a + b) .. a + b + cx4 := a + b + c , 1.01⋅ (a + b + c) .. a + b + c + dV1(x1) A1kN:= ⋅ V2(x2) (A − P1) 1:= ⋅ V3(x3) (A − P1) 1kNkN:= ⋅V4(x4) (A − P1 − P2) 1kN:= ⋅M1(x1) MA + A⋅x1:= M2 x2 ( ) MA A x2 ⋅ + P1 x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅⎤⎦1M3 x3 ( ) B x3 a − b − ( ) ⋅ ⎡⎣kN⋅m:= ⋅M4 x4 ( ) B x4 a − b − ( ) ⋅ P2 x4 a − b − c − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 2 4 6 810010Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)V4(x4)x1, x2, x3, x4 1150. 0 2 4 6 820020Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)M4(x4)x1, x2, x3, x4 1151. Problem 11-21The steel beam has an allowable bending stress σallow = 140 MPa and an allowable shear stress ofτallow = 90 MPa. Determine the maximum load that can safely be supported.Given: σallow := 140MPa a := 2mτallow := 90MPabf := 120mm df := 20mmtw := 20mm dw := 150mmSolution: L := 2aSection Property : h := df + dwycΣ yi ⎯⋅ ( ⋅Ai)Σ⋅ (Ai)=yc(bf⋅df)⋅ (0.5df) + (tw⋅dw)⋅ (0.5dw + df):= yc = 57.22mm(bf⋅df) + (tw⋅dw)I112⋅ 3 b( f⋅df) (0.5df − yc)+ ⋅ 2⋅bf df1⋅ ⋅ 12tw dw3 + ⎡⎢⎣t( w⋅dw) ⋅ (0.5dw + df − yc)2 ⎤⎥⎦:= +I = 15338333.33mm4Qmax h yc − ( ) tw ⋅ 0.5 h yc − ( ) ⋅ ⎡⎣:= ⋅ Qmax = 127188.27mm3⎤⎦Support Reactions : By symmetry, RR= - P+ ΣFy=0; Rc − P − P = 0 Rc = 2PMaximum Load: Assume failure due to bending moment.Mmax = P⋅a cmax := h − ycσallowMmax⋅cmax= σallowIP⋅a⋅ (h − yc)= PII(σallow)a h yc⋅ ( − ):=P = 9.52 kN AnsCheck Shear : Vmax := P Rc := 2PτmaxVmax⋅Qmax:= τmax = 3.947MPaI⋅ tw< τallow =90 MPa (O.K.!) 1152. x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. LV1(x1) (−P)1kN:= ⋅ V2(x2) (−P + Rc) 1kN:= ⋅M1(x1) (−P⋅x1) 1kN⋅m:= ⋅M2 x2 ( ) P − x2 ⋅ Rc x2 a − ( ) ⋅ + ⎡⎣⎤⎦1kN⋅m:= ⋅0 2 410010Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 2 401020Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 1153. Problem 11-22The timber beam has a rectangular cross section. If the width of the beam is 150 mm, determine itsheight h so that it simultaneously reaches its allowable bending stress of σallow = 10 MPa and anallowable shear stress of τallow = 0.35 MPa. Also, what is the maximum load P that the beam can thensupport?Given: σallow := 10MPa L := 1.5mτallow := 0.35MPa b := 150mmSolution:Support Reactions : By symmetry, A=B=R+ ΣFy=0; 2R − P = 0 R = 0.5PSection Property :Ib⋅h312= SxI0.5h= Sxb⋅h26=Qmax = (0.5h⋅b)⋅0.25h Qmax = 0.125b⋅h2Maximum Moment and Shear:Vmax = R Vmax = 0.5PMmax = R⋅L Mmax = 0.5P⋅LIf shear conrols : τallowVmax⋅QmaxI⋅b=I⋅bQmaxVmaxτallow=(b⋅h3)⋅b12⋅ (0.125b⋅h2)0.5Pτallow= b⋅h3⋅P4τallow= (1)If bending conrols : Sreq'dMmaxσallow=b⋅h260.5P⋅Lσallow= (2)Solving Eqs. (1) and (2): h4τallow⋅Lσallow:= h = 210mm AnsFrom Eq. (1): P:= ⋅ ( h) τallow P = 14.70 kN Ans43⎯bR := 0.5P R = 7.35 kN 1154. x1 := 0 , 0.01⋅L .. L x2 := L , 1.01⋅L .. 2LV1(x1) (R)1kN:= ⋅ V2(x2) (R − P)1kN:= ⋅M1(x1) (R⋅x1) 1kN⋅m:= ⋅M2 x2 ( ) R x2 ⋅ P x2 L − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 30Distance (m)Shear (kN)V1(x1)V2(x2)20100x1, x2 0 1 2 3Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 1155. Problem 11-23The beam is to be used to support the machine, which has a weight of 80 kN and a center of gravity atG. If the maximum bending stress is not to exceed σallow = 160 MPa, determine the required width b ofthe flanges. The supports at B and C are smooth.Given: W := 80kN σallow := 160MPaL1 := 1.8m L2 := 0.9mL3 := 1.5m L4 := 1.8mdw := 175mm df := 12mmtw := 12mmSolution: L := L1 + L2 + L3 + L4Support Reactions : Given+For machine BC :ΣFy=0; B + C − W = 0 (1)ΣΜC=0;B⋅ (L2 + L3) − W⋅L3 = 0 (2)Solving Eqs. (1) and (2): Guess B := 1kN C := 2kNBC⎛⎜⎝⎞⎠:= Find(B, C)BC⎛⎜⎝⎞⎠5030⎛⎜⎝⎞⎠= kN+For beam AD :ΣFy=0; A + D − W = 0 (3)ΣΜD=0;A⋅L − W⋅ (L3 + L4) = 0 (4)Solving Eq. (4) : AW⋅ (L3 + L4):= A = 44 kNLfdFrom Eq. (3) : D := W − A D = 36 kNSection Property : h := 2+ dwI112b⋅h3 (b − tw) dw3 ⋅ − ⎡⎣⎤⎦= ⋅ I112b h3 − dw3 + ⋅ ⎛⎝⎡⎣3 ⎞⎠⋅ tw dw⎤⎦= ⋅Bending Stress:Mmax = A⋅L1 cmax := 0.5h σallowMmax⋅cmax= IIMmax⋅cmaxσallow=112b ⋅ h3 − dw3 + tw ⋅ dw⎞⎠⎛⎝3 ⎡⎣⎤⎦⋅A⋅L1⋅ (0.5h)σallow= b12A⋅L1⋅ (0.5h)σallow− ⋅ 3tw dwh3 dw− 3:=b = 208.9mm Ans 1156. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2 + L3)x3 := (L1 + L2 + L3) , 1.01⋅ (L1 + L2 + L3) .. LV1(x1) A:= V2(x2) (A − B)kN1kN:= ⋅ V3(x3) (A − B − C)1kN:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ⋅ B x2 L1 − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) A x3 ⋅ B x3 L1 − ( ) ⋅ − C x3 L1 − L2 − L3 − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 3 4 5 650050Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x30 1 2 3 4 5 6100500Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 1157. Problem 11-24The beam has a flange width b = 200 mm. If the maximum bending stress is not to exceed σallow = 160MPa, determine the greatest weight of the machine that the beam can support. The center of gravityfor the machine is at G, and the supports at B and C are smooth.Given: L1 := 1.8m L2 := 0.9mL3 := 1.5m L4 := 1.8mdw := 175mm df := 12mmtw := 12mm b := 200mmσallow := 160MPaSolution: L := L1 + L2 + L3 + L4Support Reactions :+For beam AD :ΣFy=0; A + D − W = 0 (1)ΣΜD=0;A⋅L − W⋅ (L3 + L4) = 0 (2)Solving Eq. (2) : AW⋅ (L3 + L4)L=From Eq. (1) : D = W − ASection Property : h 2df:= + dwI112b⋅h3 (b − tw) dw:= ⋅ I = 47379775.00mm43 ⋅ − ⎡⎣⎤⎦Bending Stress:Mmax = A⋅L1 cmax := 0.5h σallowMmax⋅cmax= MmaxII⋅σallowcmax=WI⋅σallowcmax⎛⎜⎝⎞⎠LL1⋅ (L3 + L4)W⋅ (L3 + L4) := ⋅L⎡⎢⎣⎤⎥⎦⋅L1I⋅σallowcmax=W = 76.96 kN AnsEvaluate Support Reactions :For beam AD : For machine BC : GivenFrom Eqs. (1) and (2) : + ΣFy=0; B + C − W = 0 (3)AW⋅ (L3 + L4):= ΣΜC=0L;B⋅ (L2 + L3) − W⋅L3 = 0 (4)D := W − A Solving Eqs. (3) and (4): Guess B := 1kN C := 2kNA = 42.33 kN BC⎛⎜⎝⎞⎠:= Find(B, C)BC⎛⎜⎝⎞⎠48.128.86⎛⎜⎝⎞⎠= kND = 34.63 kN 1158. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2 + L3)x3 := (L1 + L2 + L3) , 1.01⋅ (L1 + L2 + L3) .. LV1(x1) A:= V2(x2) (A − B)kN1kN:= ⋅ V3(x3) (A − B − C)1kN:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ⋅ B x2 L1 − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) A x3 ⋅ B x3 L1 − ( ) ⋅ − C x3 L1 − L2 − L3 − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 1 2 3 4 5 650050Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x30 1 2 3 4 5 6100500Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 1159. Problem 11-25The box beam has an allowable bending stress of σallow = 10 MPa and an allowable shear stress ofτallow = 775 kPa. Determine the maximum intensity w of the distributed loading that it can safelysupport. Also, determine the maximum safe nail spacing for each third of the length of the beam. Eachnail can resist a shear force of 200 N.Given: Vallow := 0.2kN L := 6mbf := 210mm df := 30mmtw := 30mm dw := 190mmσallow := 10MPaτallow := 0.775MPaSolution:Section Property : h 2df:= + dwI112⋅bf⋅h3112⋅ (bf − 2tw) dw:= − ⋅ 3 I = 187700000mm4QA := (0.5⋅h − 0.5df)⋅ (bf − 2tw)⋅df QA = 495000mm3Qmax18⋅bf⋅h218⋅ (bf − 2tw) dw:= − ⋅ 2 Qmax = 963750mm3Support Reactions : By symmetry, RL=RR=R+ ΣFy=0; 2R − wo⋅L = 0 R = 0.5wo⋅LMaximum Moment and Shear:Vmax = R Vmax = 0.5⋅wo⋅LMmax = R⋅ (0.5L) − wo⋅ (0.5L)⋅ (0.25⋅L) Mmax = 0.125⋅wo⋅L2Maximum Load:Assume failure due to bending: cmax := 0.5⋅hMmax⋅cmaxσallowI=σallow0.125⋅wo⋅L2⋅ (0.5⋅h)= woI16(I)(σallow):= wo 3.337h⋅L2kNm=Assume failure due to shear:τallowVmax⋅QmaxI 2tw⋅ ( )=τallow0.5⋅wo⋅L⋅Qmax= w'o⋅ ( )I 2tw4(I)(τallow)⋅ tw:= w'o 3.019L⋅QmaxkNm=w := min(wo ,w'o) w 3.019kNm= AnsShear Flow: Since there are two rows of nails, the allowable shear flow is q = 2V/sRegion AB (0 < x < 2m) and CD (4m < x < 6m): 1160. Vmax := 0.5⋅w⋅L q1Vmax ⋅QA:= q1 23.88IkNm=s12Vallowq1:= s1 = 16.7mm AnsRegion BC (2m < x < 4m) :V'max Vmax wL3:= − ⋅ q2V'max ⋅QA:= q2 7.96IkNm=s22Vallowq2:= s2 = 50.2mm AnsR := 0.5w⋅Lx := 0 , 0.01⋅L .. L V(x) (R − w⋅x)1kN:= ⋅ M(x) [R⋅x − w⋅x⋅ (0.5x)]1kN⋅m:=0 2 4 610010Distance (m)Shear (kN)V(x)x0 2 4 620100Distance m)Moment (kN-m)M(x)x 1161. Problem 11-26The beam is constructed from three boards as shown. If each nail can support a shear force of 250 N,determine the maximum spacing of the nails, s, s', and s”, for regions AB, BC, and CD, respectively.Given: Vallow := 250N a := 1.5mbf := 200mm df := 25mmtw := 25mm dw := 150mmP1 := 4kN P2 := 6kNSolution: L := 3aSection Property : h := df + dwycΣ yi ⎯⋅ ( ⋅Ai)Σ⋅ (Ai)=yc(bf⋅df)⋅ (0.5df) + 2(tw⋅dw)⋅ (0.5dw + df):= yc = 65.00mm(bf⋅df) + 2(tw⋅dw)I112⋅ 3 t( w⋅dw) (0.5dw + df − yc)2 ⋅ + ⎡⎢⎣⋅ 3 b( f⋅df) (0.5df − yc)+ ⋅ 2 2⋅bf df112⋅ tw dw⎤⎥⎦:= +Q := (yc − 0.5df)⋅bf⋅df I = 37291666.67mm4Q = 262500.00mm3Support Reactions : Given+ ΣFy=0; B + D P− 1 − P2 = 0 (1)ΣΜD=0;−P1⋅ (3a) + B⋅ (2a) − P2⋅a = 0 (2)Solving Eqs. (1) and (2): Guess B := 1kN D := 2kNBD⎛⎜⎝⎞⎠:= Find(B,D)BD⎛⎜⎝⎞⎠91⎛⎜⎝⎞⎠= kNRegion AB : VAB := −P1VAB ⋅Qq:= q 28.16IkNm= sVallow0.5q:= s = 17.76mm AnsRegion BC : VBC := −P1 + BVBC ⋅Qq':= q' 35.20IkNm= s'Vallow0.5q':= s' = 14.21mm AnsRegion CD : VCD := −P1 + B − P2q''VCD ⋅Q:= q'' 7.04IkNm= s''Vallow0.5q'':= s'' = 71.03mm Ans 1162. x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (2a) x3 := (2a) , 1.01⋅ (2a) .. LV1(x1) −P1:= V2(x2) (−P1 + B) 1kN:= ⋅ V3(x3) (−P1 + B − P2) 1kNkN:= ⋅0 1 2 3 41050510Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3 1163. Problem 11-27The beam is constructed from two boards as shown. If each nail can support a shear force of 1 kN,determine the maximum spacing of the nails, s, s', and s”, to the nearest multiples of 5 mm for regionsAB,BC, and CD, respectively.Given: Vallow := 1kN a := 1.5mbf := 200mm df := 25mmtw := 25mm dw := 150mmP1 := 2.5kN P2 := 7.5kNSolution: L := 3aSection Property : h := df + dwycΣ yi ⎯⋅ ( ⋅Ai)Σ⋅ (Ai)=yc(bf⋅df)⋅ (0.5df) + (tw⋅dw)⋅ (0.5dw + df):= yc = 50.00mm(bf⋅df) + (tw⋅dw)I:= + ⋅ 3 + t( w⋅dw) ⋅ (0.5dw + df − yc)2Q := (yc − 0.5df)⋅bf⋅df I = 23697916.67mm4112⋅ 3 b( f⋅df) (0.5df − yc)+ ⋅ 2⋅bf df112⋅ tw dwQ = 187500.00mm3Support Reactions : Given+ΣFy=0; B + D − P1 − P2 = 0 (1)ΣΜD=0;−P1⋅ (3a) + B⋅ (2a) − P2⋅a = 0 (2)Solving Eqs. (1) and (2): Guess B := 1kN D := 2kNBD⎛⎜⎝⎞⎠:= Find(B,D)BD⎛⎜⎝⎞⎠7.52.5⎛⎜⎝⎞⎠= kNRegion AB : VAB := −P1VAB ⋅Qq:= q 19.78IkNm= sVallowq:= s = 50.56mmUse s = 55mm AnsRegion BC : VBC := −P1 + BVBC ⋅Qq':= q' 39.56IkNm= s'Vallowq':= s' = 25.28mmUse s' = 30mm AnsRegion CD : VCD := −P1 + B − P2q''VCD ⋅Q:= q'' 19.78IkNm= s''Vallowq'':= s'' = 50.56mmUse s'' = 55mm Ans 1164. x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. (2a) x3 := (2a) , 1.01⋅ (2a) .. LV1(x1) −P1:= V2(x2) (−P1 + B) 1kN:= ⋅ V3(x3) (−P1 + B − P2) 1kNkN:= ⋅0 1 2 3 41050510Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x3 1165. Problem 11-28Draw the shear and moment diagrams for the shaft, and determine its required diameter to the nearestmultiples of 5mm. if σallow = 50 kN and τallow = 20 kN. The bearings at A and D exert only verticalreactions on the shaft. The loading is applied to the pulleys at B, C, and E. Take P = 550 N.Given: σallow := 50MPa L1 := 350mmτallow := 20MPa L2 := 500mmPB := 400N L3 := 375mmPC := 550N L4 := 300mmPE := 175NSolution: Lo := L1 + L2 + L3 L := Lo + L4Support Reactions : Given+ ΣFy=0; A + D − PB − PC − PE = 0 (1)ΣΜD=0;A⋅Lo − PB⋅ (L2 + L3) − PC⋅L3 + PE⋅L4 = 0 (2)Solving Eqs. (1) and (2): Guess A := 1N D := 1NAD⎛⎜⎝⎞⎠:= Find(A,D)AD⎛⎜⎝⎞⎠411.22713.78⎛⎜⎝⎞⎠= NMaximum Moment and Shear:Vmax := A − PB − PC Vmax = −538.78 NMmax := A⋅ (L1 + L2) − PB⋅L2 Mmax = 149.54 N⋅mSection Property :I⋅ 464π do= SxI0.5⋅do= Sx⋅ 332π do=Qmax4⋅ (0.5⋅do)= ⋅ Qmax3π12⋅⋅ 24⎛⎜⎝ ⎞π do⎠312do=Bending Stress: Assume bending controls the design.Sreq'dMmaxσallow=⋅ 332π doMmaxσallow3 32Mmaxπ(σallow) :=do = 31.23mm (Use 35mm) Ans= doCheck Shear : I⋅ 464π do:= Qmax312do:=τmaxVmax ⋅Qmax:= τmax = 0.938MPaI⋅do< τallow =20 MPa (O.K.!) 1166. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2)x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. (L1 + L2 + L3)x4 := (L1 + L2 + L3) , 1.01⋅ (L1 + L2 + L3) .. LV1(x1) A1N:= ⋅ V2(x2) (A − PB) 1:= ⋅ V3(x3) (A − PB − PC) 1kNN:= ⋅V4(x4) (A − PB − PC + D) 1N:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ PB x2 L1 − ( ) ⋅ − ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ PB x3 L1 − ( ) ⋅ − PC x3 L1 − L2 − ( ) ⋅ − ⎡⎣⎤⎦1N⋅m:= ⋅M4 x4 ( ) A x4 ( ) ⋅ PB x4 L1 − ( ) ⋅ − PC x4 L1 − L2 − ( ) ⋅ − D L4 x4 L − − ( ) ⋅ + ⎡⎣⎤⎦1N⋅m:= ⋅0 0.2 0.4 0.6 0.8 1 1.2 1.45000500Distance (m)Shear (N)V1(x1)V2(x2)V3(x3)V4(x4)x1, x2, x3, x40 0.2 0.4 0.6 0.8 1 1.2 1.42001000100Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)M4(x4)x1, x2, x3, x4 1167. Problem 11-29Draw the shear and moment diagrams for the shaft, and determine its required diameter to the nearestmultiples of 5mm if σallow = 50 kN and τallow = 20 kN. The bearings at A and D exert only verticalreactions on the shaft. The loading is applied to the pulleys at B, C, and E. Take P = 400 N.Given: σallow := 50MPa L1 := 350mmτallow := 20MPa L2 := 500mmPB := 400N L3 := 375mmPC := 400N L4 := 300mmPE := 175NSolution: Lo := L1 + L2 + L3 L := Lo + L4Support Reactions : Given+ ΣFy=0; A + D − PB − PC − PE = 0 (1)ΣΜD=0;A⋅Lo − PB⋅ (L2 + L3) − PC⋅L3 + PE⋅L4 = 0 (2)Solving Eqs. (1) and (2): Guess A := 1N D := 1NAD⎛⎜⎝⎞⎠:= Find(A,D)AD⎛⎜⎝⎞⎠365.31609.69⎛⎜⎝⎞⎠= NMaximum Moment and Shear:Vmax := A − PB − PC Vmax = −434.69NMmax := A⋅L1 Mmax = 127.86N⋅mSection Property :I⋅ 464π do= SxI0.5⋅do= Sx⋅ 332π do=Qmax4⋅ (0.5⋅do)3π12⋅⋅ 24π do⎛⎜⎝⎞⎠= ⋅ Qmax312do=Bending Stress: Assume bending controls the design.Sreq'dMmaxσallow=⋅ 332π doMmaxσallow3 32Mmaxπ(σallow) :=do = 29.64mm (Use 30mm) Ans= doCheck Shear : I⋅ 464π do:= Qmax312do:=τmaxVmax ⋅Qmax:= τmax = 0.840MPaI⋅do< τallow =20 MPa (O.K.!)x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2)x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. (L1 + L2 + L3)x4 := (L1 + L2 + L3) , 1.01⋅ (L1 + L2 + L3) .. L 1168. V1(x1) A1N:= ⋅ V2(x2) (A − PB) 1:= ⋅ V3(x3) (A − PB − PC) 1kNN:= ⋅V4(x4) (A − PB − PC + D) 1N:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ( ) ⋅ PB x2 L1 − ( ) ⋅ − ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅M3 x3 ( ) A x3 ( ) ⋅ PB x3 L1 − ( ) ⋅ − PC x3 L1 − L2 − ( ) ⋅ − ⎡⎣⎤⎦1N⋅m:= ⋅M4 x4 ( ) A x4 ( ) ⋅ PB x4 L1 − ( ) ⋅ − PC x4 L1 − L2 − ( ) ⋅ − D L4 x4 L − − ( ) ⋅ + ⎡⎣⎤⎦1N⋅m:= ⋅0 0.2 0.4 0.6 0.8 1 1.2 1.45000500Distance (m)Shear (N)V1(x1)V2(x2)V3(x3)V4(x4)x1, x2, x3, x40 0.2 0.4 0.6 0.8 1 1.2 1.42001000100Distance (m)Moment (N-m)M1(x1)M2(x2)M3(x3)M4(x4)x1, x2, x3, x4 1169. Problem 11-30The overhang beam is constructed using two 50-mm by 100-mm pieces of wood braced as shown. Ifthe allowable bending stress is σallow = 4.2 MPa, determine the largest load P that can be applied. Also,determine the associated maximum spacing of nails, s, along the beam section AC if each nail can resista shear force of 4 kN. Assume the beam is pin-connected at A, B, and D. Neglect the axial forcedeveloped in the beam along DA.Given: Vallow := 4kN L := 0.9mb := 100mm ho := 50mmσallow := 4.2MPaoSolution: h :=2hMaximum Moment and Shear:Vmax = PMmax = P⋅LSection Property : Ib⋅h312:= SxI0.5h:=Qmax = (0.5h⋅b)⋅0.25h Qmax = 0.125b h2Maximum Load :Sreq'dMmaxσallow= Mmax = Sx⋅ (σallow) P⋅L = Sx⋅ (σallow)PSx⋅ (σallow):= P = 777.8N AnsLNail Spacing : Vmax := P Qmax := 0.125b h2qVmax ⋅Qmax:= q 11.7IkNm= smaxVallowq:=smax = 342.9mm Ans 1170. Problem 11-31The tapered beam supports a concentrated force P at its center. If it is made from a plate that has aconstant width b, determine the absolute maximum bending stress in the beam. 1171. Problem 11-32Determine the variation of the radius r of the cantilevered beam that supports the uniform distributedload so that it has a constant maximum bending stress σmax throughout its length. 1172. Problem 11-33Determine the variation in the depth d of a cantilevered beam that supports a concentrated force P atits end so that it has a constant maximum bending stress σallow throughout its length. The beam has aconstant width b0 . 1173. Problem 11-34The beam is made into the shape of a frustum and has a diameter of 12 mm at A and a diameter of 300mm at B. If it supports a force of 750 N at A, determine the absolute maximum bending stress in thebeam and specify its location x.Given: P := 750N L := 900mmdo := 150mm d1 := 300mmSolution:Section Property : δrd1 − do2:=r − δrxδrL= rδr⋅ (L + x)L=Iπ⋅ r44= SIr= Sπ⋅ r34= S⋅ 3⋅ (L + x)3π δr4⋅L3=Bendiug Stress : M = P⋅xσMS= ⋅ (1)= σ (P⋅x)4⋅L3⋅ 3⋅ (L + x)3π δrIn order to have the absolute maximum bending stress,d σdx= 0Differentiate Eq. (1): 4P⋅L3⎡⎢⎣⋅ 3 xπ δrx(L + x)3dd⎤⎥⎦⋅ = 0(L + x)3d xdx⎛⎜⎝⎞⎠⋅ xd (L + x)3dx⎡⎢⎣⎤⎥⎦⋅ − 0 =L x + ( )3 x 3 L x + ( )2 ⋅ ⎡⎣⎤⎦− ⋅ = 0(L + x)2⋅ (L − 2x) = 0xL2:= x = 450mm AnsSubstituting into Eq. (1):σmax (P⋅x)4⋅L3:= ⋅ σmax = 0.3018MPa Ans⋅ 3⋅ (L + x)3π δr 1174. Problem 11-35The beam has a width w and a depth that varies as shown. If it supports a concentrated force P at itsend, determine the absolute maximum bending stress in the beam and specify its location x. 1175. Problem 11-36The tapered beam supports a uniform distributed load w. If it is made from a plate and has a constantwidth b, determine the absolute maximum bending stress in the beam. 1176. Problem 11-37The tapered simply supported beam supports the concentrated force P at its center. Determine theabsolute maximum bending stress in the beam. 1177. Problem 11-38t0ta0t0nt0The bearings at A and D exert only y and z components of force on the shaft. If τallow = 60MPa, determine to the nearest millimeter the smallest-diameter shaft that will support the loading. Usethe maximum-shear-stress theory of failure.Given: a := 300mm r := 50mm τallow := 60MPaPB := 5kN PC := 5kNSolution: L := 3aSupport Reactions :In x-z plane : + ΣFz=0; Az + DZ − PB = 0 (1)ΣΜD=0; Az⋅ (3a) − PB⋅ (2a) = 0 (2)Solving Eqs. (1) and (2):Az23:= PB Az = 3.3333 kNDz := PB − Az Dz = 1.6667 kNIn x-y plane : ΣFy=0; (Ay + Dy) − PC = 0 (3)ΣΜD=0; −PC⋅a + Ay⋅ (3a) = 0 (4)Solving Eqs. (3) and (4):Ay13:= PC Ay = 1.6667 kNDy := PC − Ay Dy = 3.3333 kNTorsion occurs in segment BC : TBC := (PB)⋅ r TBC = 0.250 kN⋅mCritical Section : Located just to the left of gear C and just to the right of gear B, where:= 2 + Mz2 M = 1.118 kN⋅mT := TBC T = 0.250 kN⋅mMy := Dz⋅a Mz := Dy⋅a M MyMaximum Shear Stress Theory :c3 2π⋅τallow:= ⋅ M2 + T2 c = 22.99mmdo := 2c do = 45.99mm Use do = 46mm Ans 1178. x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 2a x3 := 2a , 1.01⋅ (2a) .. 3aMy1(x1) Az⋅ (x1):= My2 x2 ( ) Az x2 ( ) ⋅ PB x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅My3 x3 ( ) Az x3 ( ) ⋅ PB x3 a − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.2 0.4 0.6 0.810.50Distance (m)Moment (kN-m)My1(x1)My2(x2)My3(x3)x1, x2, x3Mz1(x1) −Ay⋅ (x1):= Mz2(x2) −Ay⋅ (x2)kN⋅m:= Mz3 x3 ( ) Ay − x3 ( ) ⋅ PC x3 2a − ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅0 0.2 0.4 0.6 0.800.51Distance (m)Moment (kN-m)Mz1(x1)Mz2(x2)Mz3(x3)x1, x2, x3Mx1(x1) := 0 Mx2(x2) −r⋅PB:= Mx3(x3) := 0kN⋅m0 0.2 0.4 0.6 0.800.5Distance (m)Moment (kN-m)Mx1(x1)Mx2(x2)Mx3(x3)x1, x2, x3 1179. Problem 11-39Solve Prob. 11-38 using the maximum-distortion-energy theory of failure with σallow = 180 MPa.Given: a := 300mm r := 50mm σallow := 180MPaPB := 5kN PC := 5kNSolution: L := 3aSupport Reactions :In x-z plane : + ΣFz=0; Az + DZ − PB = 0 (1)ΣΜD=0; Az⋅ (3a) − PB⋅ (2a) = 0 (2)Solving Eqs. (1) and (2):Az23:= PB Az = 3.3333 kNDz := PB − Az Dz = 1.6667 kNIn x-y plane : ΣFy=0; (Ay + Dy) − PC = 0 (3)ΣΜD=0; −PC⋅a + Ay⋅ (3a) = 0 (4)Solving Eqs. (3) and (4):Ay13:= PC Ay = 1.6667 kNDy := PC − Ay Dy = 3.3333 kNTorsion occurs in segment BC : TBC := (PB)⋅ r TBC = 0.250 kN⋅mCritical Section : Located just to the left of gear C and just to the right of gear B, where:= + 2 M = 1.118 kN⋅mT := TBC T = 0.250 kN⋅mMy := Dz⋅a Mz := Dy⋅a M My2 MzMaximum Distortion Energy Theory : Applying Eq. 9-5:σ1 0.5 σx' σy' + ( ) 0.5 σx' σy' + ( ) ⎡⎣⎤⎦⎡⎣= + 2 + τx'y'2σ2 0.5 ( σx' + σy' ) 0.5 ⋅ ( σx' + σy' ) = − + 2⎤⎦2 τx'y'where σy' := 0σx'M⋅cI=M⋅cπ4⋅c4=4M⋅cπ⋅c4=τx'y'T⋅cJ=T⋅cπ2⋅c4=2T⋅cπ⋅c4= 1180. Let a' = 0.5σx' and b' (0.5σx')2 τx'y'= + 22 = (a' + b')2 σ2Then σ12 = (a − 'b')2σ1⋅σ2 = (a' + b')⋅ (a' − b') = a'2 − b'2σ1+ 2 = (a' + b')2 − (a'2 − b'2) + (a' − b')2 = a'2 + 3b'22 − σ1⋅σ2 σ22 − σ1⋅σ2 σ2Hence σ1+ 2 σallow2 =2 + ⎡⎣(0.5σx')2 3 (0.5σx')2 τx'y'⎤⎦22 =+ σallow2 3τx'y'σx'+ 2 σallow2 =4M⋅cπ⋅c4⎛⎜⎝⎞⎠232T⋅cπ⋅c4⎛⎜⎝⎞⎠22 =+ σallowc616M2 + 12T2π2 σallow:= c = 20.05mm⋅ 2do := 2c do = 40.09mm Use do = 41mm Ans 1181. x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. 2a x3 := 2a , 1.01⋅ (2a) .. 3aMy1(x1) Az⋅ (x1):= My2 x2 ( ) Az x2 ( ) ⋅ PB x2 a − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅My3 x3 ( ) Az x3 ( ) ⋅ PB x3 a − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.2 0.4 0.6 0.810.50Distance (m)Moment (kN-m)My1(x1)My2(x2)My3(x3)x1, x2, x3Mz1(x1) −Ay⋅ (x1):= Mz2(x2) −Ay⋅ (x2)kN⋅m:= Mz3 x3 ( ) Ay − x3 ( ) ⋅ PC x3 2a − ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅0 0.2 0.4 0.6 0.800.51Distance (m)Moment (kN-m)Mz1(x1)Mz2(x2)Mz3(x3)x1, x2, x3Mx1(x1) := 0 Mx2(x2) −r⋅PB:= Mx3(x3) := 0kN⋅m0 0.2 0.4 0.6 0.800.5Distance (m)Moment (kN-m)Mx1(x1)Mx2(x2)Mx3(x3)x1, x2, x3 1182. Problem 11-40The bearings at A and D exert only y and z components of force on the shaft. If τallow = 60 MPa,determine to the nearest millimeter the smallest-diameter shaft that will support the loading. Use themaximum-shear-stress theory of failure.Given: L1 := 200mm L2 := 400mm L3 := 350mm τallow := 60MParB := 50mm rC := 75mm PB := 3kN PC := 2kNSolution: L := L1 + L2 + L3Support Reactions :In x-z plane : + ΣFz=0; Az + DZ − PC = 0 (1)ΣΜD=0; Az⋅ (L) − PC⋅ (L3) = 0 (2)Solving Eqs. (1) and (2):AzL3L:= PC Az = 0.7368 kNDz := PB − Az Dz = 2.2632 kNIn x-y plane : ΣFy=0; (Ay + Dy) − PB = 0 (3)ΣΜD=0; −PB⋅ (L2 + L3) + Ay⋅ (L) =(40)Solving Eqs. (3) and (4):L2 + L3Ay:= = 2.3684 kNLPB Ay Dy := PB − Ay Dy = 0.6316 kNTorsion occurs in segment BC : TBC := (PB)⋅ rB TBC = 0.150 kN⋅mCritical Section : Located just to right of gear B, whereMy := Az⋅L1 Mz := Ay⋅L1 M My:= 2 + Mz2 M = 0.496 kN⋅mT := TBC T = 0.150 kN⋅mMaximum Shear Stress Theory :c3 2π⋅τallow:= ⋅ M2 + T2 c = 17.65mmdo := 2c do = 35.30mm Use do = 36mm Ans 1183. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2) x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. LMy1(x1) Az⋅ (x1):= My2(x2) Az⋅ (x2)kN⋅m:= My3 x3 ( ) Az x3 ( ) ⋅ PC x3 L1 − L2 − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅0 0.2 0.4 0.6 0.80.40.20Distance (m)Moment (kN-m)My1(x1)My2(x2)My3(x3)x1, x2, x3Mz1(x1) Ay⋅ (x1):= Mz2 x2 ( ) Ay x2 ( ) ⋅ PB x2 L1 − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅Mz3 x3 ( ) Ay x3 ( ) ⋅ PB x3 L1 − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.2 0.4 0.6 0.80.40.20Distance (m)Moment (kN-m)Mz1(x1)Mz2(x2)Mz3(x3)x1, x2, x3Mx1(x1) := 0 Mx2(x2) rB⋅PB:= Mx3(x3) := 0kN⋅m0 0.2 0.4 0.6 0.80.10Distance (m)Moment (kN-m)Mx1(x1)Mx2(x2)Mx3(x3)x1, x2, x3 1184. Problem 11-41The bearings at A and D exert only y and z components of force on the shaft. If τallow = 60 MPa,determine to the nearest millimeter the smallest-diameter shaft that will support the loading. Use themaximum-distortion-energy theory of failure. σallow = 130 MPa.Given: L1 := 200mm L2 := 400mm L3 := 350mm σallow := 130MParB := 50mm rC := 75mm PB := 3kN PC := 2kNSolution: L := L1 + L2 + L3Support Reactions :In x-z plane : + ΣFz=0; Az + DZ − PC = 0 (1)ΣΜD=0; Az⋅ (L) − PC⋅ (L3) = 0 (2)Solving Eqs. (1) and (2):AzL3L:= PC Az = 0.7368 kNDz := PB − Az Dz = 2.2632 kNIn x-y plane : ΣFy=0; (Ay + Dy) − PB = 0 (3)ΣΜD=0; −PB⋅ (L2 + L3) + Ay⋅ (L) =(40)Solving Eqs. (3) and (4):L2 + L3Ay:= = 2.3684 kNLPB Ay Dy := PB − Ay Dy = 0.6316 kNTorsion occurs in segment BC : TBC := (PB)⋅ rB TBC = 0.150 kN⋅mCritical Section : Located just to right of gear B, whereMy := Az⋅L1 Mz := Ay⋅L1 M My:= 2 + Mz2 M = 0.496 kN⋅mT := TBC T = 0.150 kN⋅mMaximum Distortion Energy Theory : Applying Eq. 9-5:σ1 0.5 σx' σy' + ( ) 0.5 σx' σy' + ( ) ⎡⎣= + + 2σ2 0.5 σx' σy' + ( ) 0.5 σx' σy' + ( ) ⋅ ⎡⎣⎤⎦2 τx'y'= − + 2⎤⎦2 τx'y'where σy' := 0σx'M⋅cI=M⋅cπ4⋅c4=4M⋅cπ⋅c4=τx'y'T⋅cJ=T⋅cπ2⋅c4=2T⋅cπ⋅c4= 1185. Let a' = 0.5σx' and b' (0.5σx')2 τx'y'= + 2Then σ12 = (a' + b')2 σ22 = (a − 'b')2σ1⋅σ2 = (a' + b')⋅ (a' − b') = a'2 − b'2σ1+ 2 = (a' + b')2 − (a'2 − b'2) + (a' − b')2 = a'2 + 3b'22 − σ1⋅σ2 σ2Hence σ12 − σ1⋅σ2 σ2+ 2 σallow2 =2 + ⎡⎣(0.5σx')2 3 (0.5σx')2 τx'y'⎤⎦22 =+ σallow2 3τx'y'σx'+ 2 σallow2 =4M⋅cπ⋅c4⎛⎜⎝⎞⎠232T⋅cπ⋅c4⎛⎜⎝⎞⎠22 =+ σallowc616M2 + 12T2π2 σallow:= c = 17.13mm⋅ 2do := 2c do = 34.25mm Use do = 35mm Ans 1186. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2) x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. LMy1(x1) Az⋅ (x1):= My2(x2) Az⋅ (x2)kN⋅m:= My3 x3 ( ) Az x3 ( ) ⋅ PC x3 L1 − L2 − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅0 0.2 0.4 0.6 0.80.40.20Distance (m)Moment (kN-m)My1(x1)My2(x2)My3(x3)x1, x2, x3Mz1(x1) Ay⋅ (x1):= Mz2 x2 ( ) Ay x2 ( ) ⋅ PB x2 L1 − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅Mz3 x3 ( ) Ay x3 ( ) ⋅ PB x3 L1 − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.2 0.4 0.6 0.80.40.20Distance (m)Moment (kN-m)Mz1(x1)Mz2(x2)Mz3(x3)x1, x2, x3Mx1(x1) := 0 Mx2(x2) rB⋅PB:= Mx3(x3) := 0kN⋅mDistance (m) Moment (kN-m)0 0.2 0.4 0.6 0.80.10Mx1(x1)Mx2(x2)Mx3(x3)x1, x2, x3 1187. Problem 11-42The pulleys attached to the shaft are loaded as shown. If the bearings at A and B exert only horizontaland vertical forces on the shaft, determine the required diameter of the shaft to the nearest mm. usingthe maximum-shear-stress theory of failure. τallow = 84 MPa.Given: P1 := 1250N P2 := 750N r := 150mmL2 := 1.5mL1 := 0.3m+L3 := 0.6mτallow := 84MPaSolution: L := L1 + L2 + L3Support Reactions : Ps := P1 + P2AnsIn y-z plane : ΣFz=0; Az + BZ − P1 − P2 = 0 (1)ΣΜB=0; Az⋅L − (P1 + P2)⋅ (L2 + L3) = 0 (2)Solving Eqs. (1) and (2):Az1L:= (P1 + P2)⋅ (L2 + L3) Az = 1750 NBz := P1 + P2 − Az Bz = 250 NIn x-y plane : ΣFx=0; Ax + Bx − P1 − P2 = 0 (3)ΣΜB=0; Ax⋅L − (P1 + P2)⋅ (L3) = 0 (4)Solving Eqs. (3) and (4):Ax1L:= (P1 + P2)⋅ (L3) Ax = 500 NBx := P1 + P2 − Ax Bx = 1500 NTorsion occurs in segment DC : T := (P1 − P2)⋅ r T = 75N⋅mCritical Section : Located just to the left of point C.:= + 2 M = 912.41 N⋅mMx := Bx⋅L3 Mz := Bz⋅L3 M Mx2 MzMaximum Shear Stress Theory :c3 2π⋅τallow:= ⋅ M2 + T2 c = 19.07mmdo := 2c do = 38.15mm Use do = 39mm 1188. y1 := 0 , 0.01⋅L1 .. L1 y2 := L1 , 1.01⋅L1 .. L1 + L2 y3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. LMz1(y1) (Az⋅y1) 1⎡⎣:= ⋅ N⋅mMz2 ( y2 ) Az ⋅ y2 − Ps ⋅ ( y2 − L1 ) ⎤⎦1N⋅m:= ⋅Mz3 y3 ( ) Az y3 ⋅ Ps y3 L1 − ( ) ⋅ − ⎡⎣⎤⎦1N⋅m:= ⋅0 1 25000Distance (m)Mz (N-m)Mz1(y1)Mz2(y2)Mz3(y3)y1, y2, y3Mx1(y1) Ax⋅y1:= Mx2(y2) Ax⋅y2N⋅m:= Mx3 y3 ( ) Ax y3 ⋅ Ps y3 L1 − L2 − ( ) ⋅ − ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅0 1 210005000Distance (m)Mx (N-m)Mx1(y1)Mx2(y2)Mx3(y3)y1, y2, y3My1(y1) := 0 My2(y2) T:= My3(y3) := 0N⋅m0 1 2100500Distance (m)My (N-m)My1(y1)My2(y2)My3(y3)y1, y2, y3 1189. Problem 11-43The pulleys attached to the shaft are loaded as shown. If the bearings at A and B exert only horizontaland vertical forces on the shaft, determine the required diameter of the shaft to the nearest mm. Usethe maximum-ditortion-energy theory of failure. σallow = 140 MPaGiven: P1 := 1250N P2 := 750N r := 150mmL1 := 0.3m L2 := 1.5m L3 := 0.6mσallow := 140MPaSolution: L := L1 + L2 + L3Support Reactions : Ps := P1 + P2In y-z plane : +ΣFz=0; Az + BZ − P1 − P2 = 0 (1)ΣΜB=0; Az⋅L − (P1 + P2)⋅ (L2 + L3) = 0 (2)Solving Eqs. (1) and (2):Az1L:= (P1 + P2)⋅ (L2 + L3) Az = 1750 NBz := P1 + P2 − Az Bz = 250 NIn x-y plane : ΣFx=0; Ax + Bx − P1 − P2 = 0 (3)ΣΜB=0; Ax⋅L − (P1 + P2)⋅ (L3) = 0 (4)Solving Eqs. (3) and (4):Ax1L:= (P1 + P2)⋅ (L3) Ax = 500 NBx := P1 + P2 − Ax Bx = 1500 NTorsion occurs in segment DC : T := (P1 − P2)⋅ r T = 75N⋅mCritical Section : Located just to the left of point C.:= + 2 M = 912.41N⋅m2 MzMx := Bx⋅L3 Mz := Bz⋅L3 M MxMaximum Distortion Energy Theory :Both states of stress will yield the same result.σ1 = 0.5σ + (0.5σ)2 + τ2σ2 = 0.5σ − (0.5σ)2 + τ2Let a' = 0.5σ and b' = (0.5σ)2 + τ2Then σ12 = (a' + b')2 σ22 = (a − 'b')2σ1⋅σ2 = (a' + b')⋅ (a' − b') = a'2 − b'2σ1+ 2 = (a' + b')2 − (a'2 − b'2) + (a' − b')2 = a'2 + 3b'22 − σ1⋅σ2 σ2 1190. Hence σ12 − σ1⋅σ2 σ2+ 2 σallow2 =(0.5σ)2 3 0.5σ ( )2 τ2 + ⎡⎣⎤⎦22 =+ σallowσ2 + 3τ2 σallow2 = (5)σM⋅cI=M⋅cπ4⋅c4=4M⋅cπ⋅c4=τT⋅cJ=T⋅cπ2⋅c4=2T⋅cπ⋅c4=From Eq. (5),4M⋅cπ⋅c4⎛⎜⎝⎞⎠232T⋅cπ⋅c4⎛⎜⎝⎞⎠22 =+ σallowc616M2 + 12T2π2 σallow:= c = 20.26mm⋅ 2do := 2c do = 40.52mm Use do = 41mm Ans 1191. y1 := 0 , 0.01⋅L1 .. L1 y2 := L1 , 1.01⋅L1 .. L1 + L2 y3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. LMz1(y1) (Az⋅y1) 1⎡⎣:= ⋅ N⋅mMz2 ( y2 ) Az ⋅ y2 − Ps ⋅ ( y2 − L1 ) ⎤⎦1N⋅m:= ⋅Mz3 y3 ( ) Az y3 ⋅ Ps y3 L1 − ( ) ⋅ − ⎡⎣⎤⎦1N⋅m:= ⋅0 1 25000Distance (m)Mz (N-m)Mz1(y1)Mz2(y2)Mz3(y3)y1, y2, y3Mx1(y1) Ax⋅y1:= Mx2(y2) Ax⋅y2N⋅m:= Mx3 y3 ( ) Ax y3 ⋅ Ps y3 L1 − L2 − ( ) ⋅ − ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅0 1 210005000Distance (m)Mx (N-m)Mx1(y1)Mx2(y2)Mx3(y3)y1, y2, y3My1(y1) := 0 My2(y2) T:= My3(y3) := 0N⋅m0 1 2100500Distance (m)My (N-m)My1(y1)My2(y2)My3(y3)y1, y2, y3 1192. Problem 11-44The shaft is supported on journal bearings that do not offer resistance to axial load. If the allowablenormal stress for the shaft is σallow = 80 MPa, determine to the nearest millimeter the smallest diameterof the shaft that will support the loading. Use the maximum-distortion-energy theory of failure.Given: La := 250mm Lb := 500mm σallow := 80MParD := 150mm rC := 100mm θ := 30degPD := 0.20kN PC := 0.35kNΔPD := 0.10kN ΔPC := 0.15kNSolution:L := La + Lb + LaSupport Reactions :In y-z plane : + ΣFz=0; Az + BZ − PC⋅ sin(θ) − PD⋅ sin(θ) = 0 (1)ΣΜA=0; Bz⋅ (L) − PC⋅ sin(θ)⋅ (L − La) − PD⋅ sin(θ)⋅La = 0 (2)Solving Eqs. (1) and (2):Bz PCL − LaL⋅ PDLaL+ ⋅⎛⎜⎝⎞⎠:= sin(θ) Bz = 0.15625 kNAz := (PC + PD)⋅ sin(θ) − Bz Az = 0.11875 kNIn x-y plane : ΣFx=0; Ax + Bx − PC⋅cos(θ) + PD⋅cos(θ) = 0 (3)ΣΜA=0; PC⋅cos(θ)⋅ (L − La) − PD⋅cos(θ)⋅La − Bx⋅L = 0 (4)Solving Eqs. (3) and (4):Bx PCL − LaL⋅ PDLaL− ⋅⎛⎜⎝⎞⎠:= cos(θ) Bx = 0.18403 kNAx := (PC − PD)⋅cos(θ) − Bx Ax = −0.05413 kNTorsion occurs in segment CD : TCD := (ΔPC)⋅ rC TCD = 0.015 kN⋅mCritical Section : Located just to the left of gear C, where.:= 2 + Mz2 M = 0.060354 kN⋅mT := TCD T = 0.015 kN⋅mMx := Bz⋅La Mz := Bx⋅La M MxMaximum Distortion Energy Theory : Applying Eq. 9-5:σ1 0.5 σx' σy' + ( ) 0.5 σx' σy' + ( ) ⎡⎣= + + 2σ2 0.5 σx' σy' + ( ) 0.5 σx' σy' + ( ) ⋅ ⎡⎣⎤⎦2 τx'y'= − + 2⎤⎦2 τx'y' 1193. where σy' := 0σx'M⋅cI=M⋅cπ4⋅c4=4M⋅cπ⋅c4=τx'y'T⋅cJ=T⋅cπ2⋅c4=2T⋅cπ⋅c4=Let a' = 0.5σx' and b' (0.5σx')2 τx'y'= + 2Then σ12 = (a' + b')2 σ22 = (a − 'b')2σ1⋅σ2 = (a' + b')⋅ (a' − b') = a'2 − b'2σ1+ 2 = (a' + b')2 − (a'2 − b'2) + (a' − b')2 = a'2 + 3b'22 − σ1⋅σ2 σ2Hence σ12 − σ1⋅σ2 σ2+ 2 σallow2 =2 + ⎡⎣(0.5σx')2 3 (0.5σx')2 τx'y'⎤⎦22 =+ σallow2 3τx'y'σx'+ 2 σallow2 =4M⋅cπ⋅c4⎛⎜⎝⎞⎠232T⋅cπ⋅c4⎛⎜⎝⎞⎠22 =+ σallowc616M2 + 12T2π2 σallow:= c = 9.94mm⋅ 2do := 2c do = 19.88mm Use do = 20mm Ans 1194. y1 := 0 , 0.01⋅La .. La y2 := La , 1.01⋅La .. (La + Lb) y3 := (La + Lb) , 1.01⋅ (La + Lb) .. LMx1(y1) Az⋅ (y1):= Mx2 y2 ( ) Az y2 ⋅ PD sin θ ( ) ⋅ y2 La − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:=Mx3 y3 ( ) Az y3 ⋅ PD sin θ ( ) ⋅ y3 La − ( ) ⋅ − PC sin θ ( ) ⋅ y3 La − Lb − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.2 0.4 0.6 0.80.040.020Distance (m)Moment (kN-m)Mx1(y1)Mx2(y2)Mx3(y3)y1, y2, y3Mz1(y1) Ax⋅ (y1):= Mz2 y2 ( ) Ax y2 ⋅ PD cos θ ( ) ⋅ y2 La − ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:=⎤⎦1Mz3 y3 ( ) Ax y3 ⋅ PD cos θ ( ) ⋅ y3 La − ( ) ⋅ + PC cos θ ( ) ⋅ y3 La − Lb − ( ) ⋅ − ⎡⎣kN⋅m:= ⋅0 0.2 0.4 0.6 0.80.050Distance (m)Moment (kN-m)Mz1(y1)Mz2(y2)Mz3(y3)y1, y2, y3My1(y1) := 0 My2(y2) T:= My3(y3) := 0kN⋅m0 0.2 0.4 0.6 0.80.020Distance (m)Moment (kN-m)My1(y1)My2(y2)My3(y3)y1, y2, y3 1195. Problem 11-45The shaft is supported on journal bearings that do not offer resistance to axial load. If the allowableshear stress for the shaft is τallow = 35 MPa, determine to the nearest millimeter the smallest diameter ofthe shaft that will support the loading. Use the maximum-shear-stress theory of failure.Given: La := 250mm Lb := 500mm τallow := 35MParD := 150mm rC := 100mm θ := 30degPD := 0.20kN PC := 0.35kNΔPD := 0.10kN ΔPC := 0.15kNSolution: L := La + Lb + LaSupport Reactions :In y-z plane : + ΣFz=0; Az + BZ − PC⋅ sin(θ) − PD⋅ sin(θ) = 0 (1)ΣΜA=0; Bz⋅ (L) − PC⋅ sin(θ)⋅ (L − La) − PD⋅ sin(θ)⋅La = 0 (2)Solving Eqs. (1) and (2):Bz PCL − LaL⋅ PDLaL+ ⋅⎛⎜⎝⎞⎠:= sin(θ) Bz = 0.15625 kNAz := (PC + PD)⋅ sin(θ) − Bz Az = 0.11875 kNIn x-y plane : ΣFx=0; Ax + Bx − PC⋅cos(θ) + PD⋅cos(θ) = 0 (3)ΣΜA=0; PC⋅cos(θ)⋅ (L − La) − PD⋅cos(θ)⋅La − Bx⋅L = 0 (4)Solving Eqs. (3) and (4):Bx PCL − LaL⋅ PDLaL− ⋅⎛⎜⎝⎞⎠:= cos(θ) Bx = 0.18403 kNAx := (PC − PD)⋅cos(θ) − Bx Ax = −0.05413 kNTorsion occurs in segment CD : TCD := (ΔPC)⋅ rC TCD = 0.015 kN⋅mCritical Section : Located just to the left of gear C, where.:= 2 + Mz2 M = 0.060354 kN⋅mT := TCD T = 0.015 kN⋅mMx := Bz⋅La Mz := Bx⋅La M MxMaximum Shear Stress Theory :c3 2π⋅τallow:= ⋅ M2 + T2 c = 10.42mmdo := 2c do = 20.84mm Use do = 21mm Ans 1196. y1 := 0 , 0.01⋅La .. La y2 := La , 1.01⋅La .. (La + Lb) y3 := (La + Lb) , 1.01⋅ (La + Lb) .. LMx1(y1) Az⋅ (y1):= Mx2 y2 ( ) Az y2 ⋅ PD sin θ ( ) ⋅ y2 La − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:=Mx3 y3 ( ) Az y3 ⋅ PD sin θ ( ) ⋅ y3 La − ( ) ⋅ − PC sin θ ( ) ⋅ y3 La − Lb − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.2 0.4 0.6 0.80.040.020Distance (m)Moment (kN-m)Mx1(y1)Mx2(y2)Mx3(y3)y1, y2, y3Mz1(y1) Ax⋅ (y1):= Mz2 y2 ( ) Ax y2 ⋅ PD cos θ ( ) ⋅ y2 La − ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:=Mz3 y3 ( ) Ax y3 ⋅ PD cos θ ( ) ⋅ y3 La − ( ) ⋅ + PC cos θ ( ) ⋅ y3 La − Lb − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.2 0.4 0.6 0.80.050Distance (m)Moment (kN-m)Mz1(y1)Mz2(y2)Mz3(y3)y1, y2, y3My1(y1) := 0 My2(y2) T:= My3(y3) := 0kN⋅m0 0.2 0.4 0.6 0.80.020Distance (m)Moment (kN-m)My1(y1)My2(y2)My3(y3)y1, y2, y3 1197. Problem 11-46The shaft is supported by bearings at A and B that exert force components only in the x and zdirections on the shaft. If the allowable normal stress for the shaft is σallow = 105 MPa, determine tothe nearest mm the smallest diameter of the shaft that will support the gear loading. Use themaximum-distortion-energy theory of failure.Given: La := 200mm Lb := 100mmPC := 1kN r := 100mmPD := 0.25kN PE := 1.25kNσallow := 105MPaSolution:L 3La:= + LbSupport Reactions :In y-z plane : + ΣFz=0; Az + BZ − PD = 0 (1)ΣΜB=0; Az⋅ (2La + Lb) − PD⋅ (La + Lb) = 0 (2)Solving Eqs. (1) and (2):AzLa + Lb2La + Lb:= PD Az = 150 NBz := PD − Az Bz = 100 NIn x-y plane : ΣFx=0; (Ax + Bx) + PC − PE = 0 (3)ΣΜB=0; PC⋅L + Ax⋅ (2La + Lb) − PE⋅ (Lb) = 0 (4)Solving Eqs. (3) and (4):Ax12La + Lb:= (PE⋅Lb − PC⋅L) Ax = −1150 NBx := PE − PC − Ax Bx = 1400 NTorsion occurs in segment DC : TDC := (PC)⋅ r TDC = 100N⋅min segment DE : TDE := (PE)⋅ r TDE = 125N⋅mCritical Section : Located at support A (within segment DC).:= + 2 M = 200N⋅mT := TDC T = 100N⋅mMx := PC⋅La Mz := 0 M Mx2 Mz 1198. Maximum Distortion Energy Theory : Applying Eq. 9-5:σ1 0.5 σx' σy' + ( ) 0.5 σx' σy' + ( ) ⎡⎣= + + 2σ2 0.5 σx' σy' + ( ) 0.5 σx' σy' + ( ) ⋅ ⎡⎣⎤⎦2 τx'y'= − + 2⎤⎦2 τx'y'where σy' := 0σx'M⋅cI=M⋅cπ4⋅c4=4M⋅cπ⋅c4=τx'y'T⋅cJ=T⋅cπ2⋅c4=2T⋅cπ⋅c4=Let a' = 0.5σx' and b' (0.5σx')2 τx'y'= + 22 = (a' + b')2 σ2Then σ12 = (a − 'b')2σ1⋅σ2 = (a' + b')⋅ (a' − b') = a'2 − b'2σ1+ 2 = (a' + b')2 − (a'2 − b'2) + (a' − b')2 = a'2 + 3b'22 − σ1⋅σ2 σ22 − σ1⋅σ2 σ2Hence σ1+ 2 σallow2 =2 + ⎡⎣(0.5σx')2 3 (0.5σx')2 τx'y'⎤⎦22 =+ σallow2 3τx'y'σx'+ 2 σallow2 =4M⋅cπ⋅c4⎛⎜⎝⎞⎠232T⋅cπ⋅c4⎛⎜⎝⎞⎠22 =+ σallowc616M2 + 12T2π2 σallow:= c = 13.83mm⋅ 2do := 2c do = 27.65mm Use do = 28mm Ans 1199. y1 := 0 , 0.01⋅La .. La y2 := La , 1.01⋅La .. 2La y3 := 2La , 1.01⋅ (2La) .. 3Lay4 := 3La , 1.01⋅ (3La) .. LMz1(y1) 0⋅ (y1):= Mz2 y2 ( ) Az y2 La − ( ) ⋅ ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅Mz3 y3 ( ) Az y3 La − ( ) ⋅ PD y3 2La − ( ) ⋅ − ⎡⎣⎤⎦1N⋅m:= ⋅Mz4 y4 ( ) Az y4 La − ( ) ⋅ PD y4 2La − ( ) ⋅ − ⎡⎣⎤⎦1N⋅m:= ⋅0 0.2 0.4 0.6200Distance (m)Moment (N-m)Mz1(y1)Mz2(y2)Mz3(y3)Mz4(y4)y1, y2, y3, y4Mx1(y1) PC⋅ (y1):= Mx2 y2 ( ) PC y2 ( ) ⋅ Ax y2 La − ( ) ⋅ + ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅Mx3 y3 ( ) PC y3 ( ) ⋅ Ax y3 La − ( ) ⋅ + ⎡⎣⎤⎦1N⋅m:= ⋅Mx4 y4 ( ) PC y4 ( ) ⋅ Ax y4 La − ( ) ⋅ + PE y4 3La − ( ) ⋅ − ⎡⎣⎤⎦1N⋅m:= ⋅0 0.2 0.4 0.62001000Distance (m)Moment (N-m)Mx1(y1)Mx2(y2)Mx3(y3)Mx4(y4)y1, y2, y3, y4 1200. My1(y1) −r⋅PC:= My2(y2) −r⋅PCN⋅mN⋅m:=My3(y3) (−r⋅PC − r⋅PD) 1N⋅m:= ⋅My4(y4) (−r⋅PC − r⋅PD + r⋅PE) 1N⋅m:= ⋅0 0.2 0.4 0.6050100150Distance (m)Moment (N-m)My1(y1)My2(y2)My3(y3)My4(y4)y1, y2, y3, y4 1201. Problem 11-47The shaft is supported by bearings at A and B that exert force components only in the x and zdirections on the shaft. If the allowable normal stress for the shaft is σallow = 105 MPa, determine tothe nearest mm the smallest diameter of the shaft that will support the gear loading. Use themaximum-shear-stress theory of failure with τallow = 42 MPa.Given: La := 200mm Lb := 100mmPC := 1kN r := 100mmPD := 0.25kN PE := 1.25kNτallow := 42MPaSolution: L 3La:= + LbSupport Reactions :In y-z plane : +ΣFz=0; Az + BZ − PD = 0 (1)ΣΜB=0; Az⋅ (2La + Lb) − PD⋅ (La + Lb) = 0 (2)Solving Eqs. (1) and (2):AzLa + Lb2La + Lb:= PD Az = 150 NBz := PD − Az Bz = 100 NIn x-y plane : ΣFx=0; (Ax + Bx) + PC − PE = 0 (3)ΣΜB=0; PC⋅L + Ax⋅ (2La + Lb) − PE⋅ (Lb) = 0 (4)Solving Eqs. (3) and (4):Ax12La + Lb:= (PE⋅Lb − PC⋅L) Ax = −1150 NBx := PE − PC − Ax Bx = 1400 NTorsion occurs in segment DC : TDC := (PC)⋅ r TDC = 100N⋅min segment DE : TDE := (PE)⋅ r TDE = 125N⋅mCritical Section : Located at support A (within segment DC).:= + 2 M = 200.00 N⋅mT := TDC T = 100.00N⋅mMx := PC⋅La Mz := 0 M Mx2 MzMaximum Shear Stress Theory :c3 2π⋅τallow:= ⋅ M2 + T2 c = 15.0mmdo := 2c do = 30.0mm Use do = 31mm Ans 1202. y1 := 0 , 0.01⋅La .. La y2 := La , 1.01⋅La .. 2La y3 := 2La , 1.01⋅ (2La) .. 3Lay4 := 3La , 1.01⋅ (3La) .. LMz1(y1) 0⋅ (y1):= Mz2 y2 ( ) Az y2 La − ( ) ⋅ ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅Mz3 y3 ( ) Az y3 La − ( ) ⋅ PD y3 2La − ( ) ⋅ − ⎡⎣⎤⎦1N⋅m:= ⋅Mz4 y4 ( ) Az y4 La − ( ) ⋅ PD y4 2La − ( ) ⋅ − ⎡⎣⎤⎦1N⋅m:= ⋅0 0.2 0.4 0.6200Distance (m)Moment (N-m)Mz1(y1)Mz2(y2)Mz3(y3)Mz4(y4)y1, y2, y3, y4Mx1(y1) PC⋅ (y1):= Mx2 y2 ( ) PC y2 ( ) ⋅ Ax y2 La − ( ) ⋅ + ⎡⎣N⋅m⎤⎦1N⋅m:= ⋅Mx3 y3 ( ) PC y3 ( ) ⋅ Ax y3 La − ( ) ⋅ + ⎡⎣⎤⎦1N⋅m:= ⋅Mx4 y4 ( ) PC y4 ( ) ⋅ Ax y4 La − ( ) ⋅ + PE y4 3La − ( ) ⋅ − ⎡⎣⎤⎦1N⋅m:= ⋅0 0.2 0.4 0.62001000Distance (m)Moment (N-m)Mx1(y1)Mx2(y2)Mx3(y3)Mx4(y4)y1, y2, y3, y4 1203. My1(y1) −r⋅PC:= My2(y2) −r⋅PCN⋅mN⋅m:=My3(y3) (−r⋅PC − r⋅PD) 1N⋅m:= ⋅My4(y4) (−r⋅PC − r⋅PD + r⋅PE) 1N⋅m:= ⋅0 0.2 0.4 0.6050100150Distance (m)Moment (N-m)My1(y1)My2(y2)My3(y3)My4(y4)y1, y2, y3, y4 1204. Problem 11-48The end gear connected to the shaft is subjected to the loading shown. If the bearings at A and B exertonly y and z components of force on the shaft, determine the equilibrium torque T at gear C and thendetermine the smallest diameter of the shaft to the nearest millimeter that will support the loading. Usethe maximum-shear-stress theory of failure with τallow = 60 MPa.Given: L1 := 150mm L2 := 250mm L3 := 100mm τallow := 60MParD := 100mm rC := 75mm rC' := 50mm PD := 1.5kNSolutio.n: L L:= 1 + L2 + L3Equilibrium Torque at C :Σ Μx=0; −PD⋅ rD + FC⋅ rC = 0FCrDrC:= ⋅PD FC = 2.00 kNThus, TC := FC⋅ rC' TC = 0.100 kN⋅m AnsSupport Reactions :In x-z plane : + ΣFz=0; Az + Bz − PD = 0 (1)ΣΜB=0; Az⋅ (L − L1) − PD⋅L = 0 (2)Solving Eqs. (1) and (2):AzLL − L1:= PD Az = 2.1429 kNBz := PD − Az Bz = −0.6429 kNIn x-y plane : ΣFy=0; (Ay + By) − FC = 0 (3)ΣΜB=0; Ay⋅ (L − L1) − FC⋅L3 = 0 (4)Solving Eqs. (3) and (4):AyL3L − L1:= FC Ay = 0.5714 kNBy := FC − Ay By = 1.4286 kNTorsion occurs in segment D-C : TDC := PD⋅ rD TDC = 0.150 kN⋅mCritical Section : Located just to right of gear A, whereMy := PD⋅L1 Mz := 0 M My:= 2 + Mz2 M = 0.225 kN⋅mT := TDC T = 0.150 kN⋅mMaximum Shear Stress Theory :c3 2π⋅τallow:= ⋅ M2 + T2 c = 14.21mmdo := 2c do = 28.42mm Use do = 29mm Ans 1205. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2) x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. LMy1(x1) PD⋅ (x1):= My2 x2 ( ) PD x2 ( ) ⋅ Az x2 L1 − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅My3 x3 ( ) PD x3 ( ) ⋅ Az x3 L1 − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.1 0.2 0.3 0.40.20Distance (m)Moment (kN-m)My1(x1)My2(x2)My3(x3)x1, x2, x3Mz1 x1 ( ) 0 := Mz2 x2 ( ) Ay x2 L1 − ( ) ⋅ ⎡⎣⎤⎦1kN⋅m:= ⋅⎤⎦1Mz3 x3 ( ) Ay x3 L1 − ( ) ⋅ FC x3 L1 − L2 − ( ) ⋅ − ⎡⎣kN⋅m:= ⋅0 0.1 0.2 0.3 0.40.20.10Distance (m)Moment (kN-m)Mz1(x1)Mz2(x2)Mz3(x3)x1, x2, x3Mx1(x1) T:= Mx2(x2) TkN⋅m:= Mx3(x3) := 0kN⋅m0 0.1 0.2 0.3 0.40.20Distance (m)Moment (kN-m)Mx1(x1)Mx2(x2)Mx3(x3)x1, x2, x3 1206. Problem 11-49The end gear connected to the shaft is subjected to the loading shown. If the bearings at A and B exertonly y and z components of force on the shaft, determine the equilibrium torque T at gear C and thendetermine the smallest diameter of the shaft to the nearest millimeter that will support the loading. Usethe maximum-distortion-energy theory of failure with σallow = 80 MPa.Given: L1 := 150mm L2 := 250mm L3 := 100mm σallow := 80MParD := 100mm rC := 75mm rC' := 50mm PD := 1.5kNSolutio.n: L L:= 1 + L2 + L3Equilibrium Torque at C :Σ Μx=0; −PD⋅ rD + FC⋅ rC = 0FCrDrC:= ⋅PD FC = 2.00 kNThus, TC := FC⋅ rC' TC = 0.100 kN⋅m AnsSupport Reactions :In x-z plane : + ΣFz=0; Az + Bz − PD = 0 (1)ΣΜB=0; Az⋅ (L − L1) − PD⋅L = 0 (2)Solving Eqs. (1) and (2):AzLL − L1:= PD Az = 2.1429 kNBz := PD − Az Bz = −0.6429 kNIn x-y plane : ΣFy=0; (Ay + By) − FC = 0 (3)ΣΜB=0; Ay⋅ (L − L1) − FC⋅L3 = 0 (4)Solving Eqs. (3) and (4):AyL3L − L1:= FC Ay = 0.5714 kNBy := FC − Ay By = 1.4286 kNTorsion occurs in segment D-C : TDC := PD⋅ rD TDC = 0.150 kN⋅mCritical Section : Located just to right of gear A, whereMy := PD⋅L1 Mz := 0 M My:= 2 + Mz2 M = 0.225 kN⋅mT := TDC T = 0.150 kN⋅mMaximum Distortion Energy Theory : Applying Eq. 9-5:σ1 0.5 σx' σy' + ( ) 0.5 σx' σy' + ( ) ⎡⎣= + + 2σ2 0.5 σx' σy' + ( ) 0.5 σx' σy' + ( ) ⋅ ⎡⎣⎤⎦2 τx'y'= − + 2⎤⎦2 τx'y' 1207. where σy' := 0σx'M⋅cI=M⋅cπ4⋅c4=4M⋅cπ⋅c4=τx'y'T⋅cJ=T⋅cπ2⋅c4=2T⋅cπ⋅c4=Let a' = 0.5σx' and b' (0.5σx')2 τx'y'= + 22 = (a' + b')2 σ2Then σ12 = (a − 'b')2σ1⋅σ2 = (a' + b')⋅ (a' − b') = a'2 − b'2σ1+ 2 = (a' + b')2 − (a'2 − b'2) + (a' − b')2 = a'2 + 3b'22 − σ1⋅σ2 σ22 − σ1⋅σ2 σ2Hence σ1+ 2 σallow2 =2 + ⎡⎣(0.5σx')2 3 (0.5σx')2 τx'y'⎤⎦22 =+ σallow2 3τx'y'σx'+ 2 σallow2 =4M⋅cπ⋅c4⎛⎜⎝⎞⎠232T⋅cπ⋅c4⎛⎜⎝⎞⎠22 =+ σallowc616M2 + 12T2π2 σallow:= c = 16.05mm⋅ 2do := 2c do = 32.10mm Use do = 33mm Ansx1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2) x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. LMy1(x1) PD⋅ (x1):= My2 x2 ( ) PD x2 ( ) ⋅ Az x2 L1 − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅My3 x3 ( ) PD x3 ( ) ⋅ Az x3 L1 − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅ 1208. 0 0.1 0.2 0.3 0.40.20Distance (m)Moment (kN-m)My1(x1)My2(x2)My3(x3)x1, x2, x3Mz1 x1 ( ) 0 := Mz2 x2 ( ) Ay x2 L1 − ( ) ⋅ ⎡⎣⎤⎦1kN⋅m:= ⋅Mz3 x3 ( ) Ay x3 L1 − ( ) ⋅ FC x3 L1 − L2 − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.1 0.2 0.3 0.40.20.10Distance (m)Moment (kN-m)Mz1(x1)Mz2(x2)Mz3(x3)x1, x2, x3Mx1(x1) T:= Mx2(x2) TkN⋅m:= Mx3(x3) := 0kN⋅m0 0.1 0.2 0.3 0.40.20Distance (m)Moment (kN-m)Mx1(x1)Mx2(x2)Mx3(x3)x1, x2, x3 1209. Problem 11-50Draw the shear and moment diagrams for the shaft, and then determine its required diameter to thenearest millimeter if σallow = 140 MPa and τallow = 80 MPa. The bearings at A and B exert only verticalreactions on the shaft.Given: L1 := 125mm L2 := 600mm L3 := 75mmP1 := 0.8kN P2 := 1.5kNτallow := 80MPa σallow := 140MPaSolution: L := L1 + L2 + L3Support Reactions :+ΣFz=0; A + B − P1 − P2 = 0 (1)ΣΜB=0; A⋅L − P1⋅ (L − L1) − P2⋅ (L3) = 0 (2)Solving Eqs. (1) and (2):A1LP1 L L1 − ( ) ⋅ P2 L3 ( ) ⋅ + ⎡⎣⎤⎦:= ⋅ A = 0.815625 kNB := P1 + P2 − A B = 1.484375 kNMaximum Moment and Shear:Vmax := B Vmax = 1.484375 kNMmax := B⋅L3 Mmax = 0.111328 kN⋅mSection Property : I⋅ 464π do=SI0.5do= S⋅ 332π do=Bending Stress:Sreq'dMmaxσallow=⋅ 332π doMmaxσallow= do3 32Mmaxπ σallow:= do = 20.1mmUse do := 21mm AnsShear Check : Ao⋅ 24π do:= I⋅ 464π do:= Qmax2⋅do3π⎛⎜⎝⎞⎠Ao2:= ⋅τmaxVmax⋅Qmax:= τmax = 5.714MPaI⋅do< τallow =80 MPa (O.K.!) 1210. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2) x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. LV1(x1) A:= V2(x2) (A − P1) 1kN:= ⋅ V3(x3) (A − P1 − P2) 1kNkN:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ⋅ P1 x2 L1 − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) A x3 ⋅ P1 x3 L1 − ( ) ⋅ − P2 x3 L1 − L2 − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.2 0.4 0.61012Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x30 0.2 0.4 0.60.10.050Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 1211. Problem 11-51The cantilevered beam has a circular cross section. If it supports a force P at its end, determine itsradius y as a function of x so that it is subjected to a constant maximum bending stress σallowthroughout its length. 1212. Problem 11-52The simply supported beam is made of timber that has an allowable bending stress of σallow = 8 MPaand an allowable shear stress of τallow = 750 kPa. Determine its dimensions if it is to be rectangular andhave a height-to-width ratio of h/b = 1.25.Given: σallow := 8MPa a := 3mτallow := 0.75MPa wo 0.3kNm:=h = 1.25⋅bSolution: L := 2aSupport Reactions :+ ΣFy=0; A + B − 0.5⋅wo⋅a = 0 (1)ΣΜA=0; (0.5⋅wo⋅a) 2a3⎛⎜⎝⎞⎠⋅ − B⋅L = 0 (2)Solving Eqs. (1) and (2): Awo⋅a3:= Bwo⋅a6:=Maximum Moment and Shear:Mmax occurs at x', where V(x')=0.V A x' w⋅ − = 0 A x' wo2⋅x'ao2⋅x'a= − ⋅ x'2awo⎛⎜⎝⎞⎠wo⋅a3= ⋅ x'23:= ⋅aVmax := max(A, B) Vmax = 0.300 kNMmax A⋅x' 0.5⋅x'⋅wox'a⋅ ⎛⎜⎝⎞⎠x'3⎛⎜⎝⎞⎠:= − ⋅ Mmax = 0.4899 kN⋅mSection Property : Ib⋅h312=SxI0.5h= Sxb⋅h26= Sxb⋅ (1.25⋅b)2= Sx625⋅b396=Bending Stress:Sreq'dMmaxσallow=25⋅b396Mmaxσallow= b3 96Mmax25σallow:= b = 61.7mm Ansh := 1.25⋅b h = 77.2mm AnsShear Check : Ib⋅h312:= Qmax := (0.5b⋅h)⋅0.25hτmaxVmax⋅Qmax:= τmax = 0.094MPaI⋅b< τallow =0.75 MPa (O.K.!) 1213. x1 := 0 , 0.01⋅a .. a x2 := a , 1.01⋅a .. LV1(x1) Awo2x1a⎛⎜⎝⎞⎠− ⋅ ⋅x1⎡⎢⎣⎤⎥⎦1kN:= ⋅V2(x2) (A − 0.5⋅wo⋅a) 1kN:= ⋅M1(x1) A⋅x1wo2x1a⎛⎜⎝⎞⎠⋅ ⋅x1x13− ⋅⎡⎢⎣⎤⎥⎦1kN⋅m:= ⋅M2(x2) A⋅x2wo⋅a2x22⋅a3− ⎛⎜⎝⎞⎠− ⋅⎡⎢⎣⎤⎥⎦1kN⋅m:= ⋅0 1 2 3 4 5 60.500.5Distance (m)Shear (kN)V1(x1)V2(x2)x1, x20 1 2 3 4 5 60.50Distane (m)Moment (kN-m)M1(x1)M2(x2)x1, x2 1214. Problem 11-53The beam is made in the shape of a frustum that has a diameter of 0.3 m at A and a diameter of 0.6 mat B. If it supports a couple moment of 12 kN·m at its end, determine the absolute maximum bendingstress in the beam and specify its location x.Given: M := 12kN⋅m L := 1.8mdo := 0.3m d1 := 0.6mSolution:Section Property : δrd1 − do2:=r − δrxδrL= rδr⋅ (L + x)L=Iπ⋅ r44= SIr= Sπ⋅ r34= S⋅ 3⋅ (L + x)3π δr4⋅L3=Bendiug Stress :σMS= ⋅ (1)= σ (M)4⋅L3⋅ 3⋅ (L + x)3π δrSince σ is a decreasing function, the maximum bending stress occurs at x := 0 AnsSubstituting x=0 into Eq. (1)::= ⋅ σmax = 4.527MPa Ansσmax (M)4⋅L3⋅ 3⋅ (L + x)3π δr 1215. Problem 11-54Select the lightest-weight steel wide-flange overhanging beam from Appendix B that will safely supportthe loading. Assume the support at A is a pin and the support at B is a roller. The allowable bendingstress is σallow = 168 MPa and the allowable shear stress is τallow = 100 MPa.Given: σallow := 168MPa P := 10kNτallow := 100MPa L1 := 2.4mL2 := 0.6m L3 := 1.2mSolution: L := L1 + L2 + L3Support Reactions :+ ΣFy=0; A + B − 2P = 0 (1)ΣΜB=0; A⋅L1 + P⋅L2 + P⋅ (L2 + L3) = 0 (2)Solving Eqs. (1) and (2):A2L2 + L3L1:= − P A = −10 kNB := 2P − A B = 30 kNMaximum Moment and Shear:Vmax := A + B Vmax = 20 kNMmax := A⋅L1 Mmax = −24 kN⋅mBending Stress:Sreq'dMmaxσallow:= Sreq'd = 142857.14mm3Select W 250x18 : Sx := 179⋅ (103)mm3 d := 251mm tw := 4.83mmShear Stress : Provide a shear stress check.τmaxVmaxtw⋅d:= τmax = 16.5MPa< τallow =100 MPa (O.K.!)Hence, Use W 250x18 Ans 1216. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2)x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. LV1(x1) A:= V2(x2) (A + B)kN1kN:= ⋅ V3(x3) (A + B − P)1kN:= ⋅M1(x1) A⋅x1:= M2 x2 ( ) A x2 ⋅ B x2 L1 − ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) A x3 ⋅ B x3 L1 − ( ) ⋅ + ⎡⎣P x3 L1 − L2 − ( ) ⋅ − ⎡⎣⎤⎦⎤⎦1kN⋅m:= ⋅0 1 2 3 4200Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)x1, x2, x30 1 2 3 4020Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)x1, x2, x3 1217. Problem 11-55The bearings at A and B exert only x and z components of force on the steel shaft. Determine theshaft's diameter to the nearest millimeter so that it can resist the loadings of the gears withoutexceeding an allowable shear stress of τallow = 80 MPa. Use the maximum-shear-stress theory offailure.Given: PC := 7.5kN PD := 5kN rC := 50mmL1 := 150mm L2 := 350mm L3 := 250mmτallow := 80MPa rD := 75mmSolution: L := L1 + L2 + L3Support Reactions :In y-z plane : + ΣFz=0; Az + BZ − PC = 0 (1)ΣΜB=0; Az⋅L − PC⋅L3 = 0 (2)Solving Eqs. (1) and (2):AzL3L:= ⋅PC Az = 2.5 kNBz := PC − Az Bz = 5 kNIn x-y plane : ΣFx=0; Ax + Bx − PD = 0 (3)ΣΜB=0; Ax⋅L − PD⋅ (L − L1) = 0 (4)Solving Eqs. (3) and (4):AxL − L1L:= ⋅PD Ax = 4 kNBx := PD − Ax Bx = 1 kNTorsion occurs in segment DC : T := PC⋅ rC T = 0.375 kN⋅mCritical Section : Located just to the left of gear C.:= + 2 M = 1.27475 kN⋅mMx := Bx⋅L3 Mz := Bz⋅L3 M Mx2 MzMaximum Shear Stress Theory :c3 2π⋅τallow:= ⋅ M2 + T2 c = 21.95mmdo := 2c do = 43.90mm Use do = 44mm Ans 1218. y1 := 0 , 0.01⋅L1 .. L1 y2 := L1 , 1.01⋅L1 .. L1 + L2 y3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. LMz1(y1) Az⋅y1:= Mz2(y2) Az⋅y2kN⋅m:= Mz3 y3 ( ) Az y3 ⋅ PC y3 L1 − L2 − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅0 0.510Distance (m)Mz (kN-m)Mz1(y1)Mz2(y2)Mz3(y3)y1, y2, y3Mx1(y1) Ax⋅y1:= Mx2 y2 ( ) Ax y2 ⋅ PD y2 L1 − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅Mx3 y3 ( ) Ax y3 ⋅ PD y3 L1 − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.50.50Distance (m)Mx (kN-m)Mx1(y1)Mx2(y2)Mx3(y3)y1, y2, y3My1(y1) := 0 My2(y2) T:= My3(y3) := 0kN⋅m0 0.50.50Distance (m)My (kN-m)My1(y1)My2(y2)My3(y3)y1, y2, y3 1219. Problem 11-56The bearings at A and B exert only x and z components of force on the steel shaft. Determine theshaft's diameter to the nearest millimeter so that it can resist the loadings of the gears withoutexceeding an allowable shear stress of τallow = 80 MPa. Use the maximum-distortion-energy theory offailure with σallow = 200 MPa.Given: PC := 7.5kN PD := 5kN rC := 50mmL1 := 150mm L2 := 350mm L3 := 250mmσallow := 200MPa rD := 75mmSolution: L := L1 + L2 + L3Support Reactions :In y-z plane : + ΣFz=0; Az + BZ − PC = 0 (1)ΣΜB=0; Az⋅L − PC⋅L3 = 0 (2)Solving Eqs. (1) and (2):AzL3L:= ⋅PC Az = 2.5 kNBz := PC − Az Bz = 5 kNIn x-y plane : ΣFx=0; Ax + Bx − PD = 0 (3)ΣΜB=0; Ax⋅L − PD⋅ (L − L1) = 0 (4)Solving Eqs. (3) and (4):AxL − L1L:= ⋅PD Ax = 4 kNBx := PD − Ax Bx = 1 kNTorsion occurs in segment DC : T := PC⋅ rC T = 0.375 kN⋅mCritical Section : Located just to the left of gear C.:= + 2 M = 1.27475 kN⋅m2 MzMx := Bx⋅L3 Mz := Bz⋅L3 M MxMaximum Distortion Energy Theory : Applying Eq. 9-5:σ1 0.5 σx' σy' + ( ) 0.5 σx' σy' + ( ) ⎡⎣= + + 2σ2 0.5 σx' σy' + ( ) 0.5 σx' σy' + ( ) ⋅ ⎡⎣⎤⎦⎤⎦2 τx'y'= − + 22 τx'y'where σy' := 0σx'M⋅cI=M⋅cπ4⋅c4=4M⋅cπ⋅c4=τx'y'T⋅cJ=T⋅cπ2⋅c4=2T⋅cπ⋅c4= 1220. Let a' = 0.5σx' and b' (0.5σx')2 τx'y'= + 2Then σ12 = (a' + b')2 σ22 = (a − 'b')2σ1⋅σ2 = (a' + b')⋅ (a' − b') = a'2 − b'2σ1+ 2 = (a' + b')2 − (a'2 − b'2) + (a' − b')2 = a'2 + 3b'22 − σ1⋅σ2 σ2Hence σ12 − σ1⋅σ2 σ2+ 2 σallow2 =2 + ⎡⎣(0.5σx')2 3 (0.5σx')2 τx'y'⎤⎦22 =+ σallow2 3τx'y'σx'+ 2 σallow2 =4M⋅cπ⋅c4⎛⎜⎝⎞⎠232T⋅cπ⋅c4⎛⎜⎝⎞⎠22 =+ σallowc616M2 + 12T2π2 σallow:= c = 20.31mm⋅ 2do := 2c do = 40.61mm Use do = 41mm Ans 1221. y1 := 0 , 0.01⋅L1 .. L1 y2 := L1 , 1.01⋅L1 .. L1 + L2 y3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. LMz1(y1) Az⋅y1:= Mz2(y2) Az⋅y2kN⋅m:= Mz3 y3 ( ) Az y3 ⋅ PC y3 L1 − L2 − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅0 0.510Distance (m)Mz (kN-m)Mz1(y1)Mz2(y2)Mz3(y3)y1, y2, y3Mx1(y1) Ax⋅y1:= Mx2 y2 ( ) Ax y2 ⋅ PD y2 L1 − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅Mx3 y3 ( ) Ax y3 ⋅ PD y3 L1 − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅0 0.50.50Distance (m)Mx (kN-m)Mx1(y1)Mx2(y2)Mx3(y3)y1, y2, y3My1(y1) := 0 My2(y2) T:= My3(y3) := 0kN⋅m0 0.50.50Distance (m)My (kN-m)My1(y1)My2(y2)My3(y3)y1, y2, y3 1222. Problem 11-57Select the lightest-weight steel wide-flange beam from Appendix B that will safely support the loadingshown. The allowable bending stress is σallow = 160 MPa and the allowable shear stress is τallow = 84MPa.Given: σallow := 160MPa L1 := 3mτallow := 84MPa L2 := 1.5mP1 := 40kN P2 := 50kN:= ( + L2)Solution: L 2L1Support Reactions : By symmetry, RA=RB=R+ ΣFy=0; 2R − 2P1 − P2 = 0 R := P1 + 0.5P2Maximum Moment and Shear:Vmax := R Vmax = 65 kNMmax := R⋅ (L1 + L2) − P1⋅L2 Mmax = 232.5 kN⋅mBending Stress:Sreq'dMmaxσallow:= Sreq'd = 1453125.00mm3Select W 460x74 : Sx := 1460⋅ (103)mm3 d := 457mm tw := 9.02mmShear Stress : Provide a shear stress check.τmaxVmaxtw⋅d:= τmax = 15.77MPa< τallow =98 MPa (O.K.!)Hence, Use W 460x74 Ans 1223. x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. (L1 + L2)x3 := (L1 + L2) , 1.01⋅ (L1 + L2) .. (L1 + 2L2)x4 := (L1 + 2L2) , 1.01⋅ (L1 + 2L2) .. LV1(x1) R1kN:= ⋅ V2(x2) (R − P1) 1:= ⋅ V3(x3) (R − P1 − P2) 1kNkN:= ⋅V4(x4) (R − P1 − P2 − P1) 1kN:= ⋅M1(x1) R⋅x1:= M2 x2 ( ) R x2 ( ) ⋅ P1 x2 L1 − ( ) ⋅ − ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅M3 x3 ( ) R x3 ( ) ⋅ P1 x3 L1 − ( ) ⋅ − P2 x3 L1 − L2 − ( ) ⋅ − ⎡⎣⎤⎦1kN⋅m:= ⋅M4 x4 ( ) R x4 ( ) ⋅ P1 x4 L1 − ( ) ⋅ − P2 x4 L1 − L2 − ( ) ⋅ − ⎡⎣P1 L1 x4 L − − ( ) ⋅ − ⎡⎣⎤⎦⎤⎦1kN⋅m:= ⋅0 2 4 6 81000100Distance (m)Shear (kN)V1(x1)V2(x2)V3(x3)V4(x4)x1, x2, x3, x40 2 4 6 83002001000Distance (m)Moment (kN-m)M1(x1)M2(x2)M3(x3)M4(x4)x1, x2, x3, x4 1224. Problem 12-01An L2 steel strap having a thickness of 3 mm and a width of 50 mm is bent into a circular arc of radius15 m. Determine the maximum bending stress in the strap.Given: b := 50mm t := 3mm ρ := 15mE := 200GPaSolution:Section Property :ct2:= Ib⋅ t312:=Moment - Curvature Relationship :1ρME⋅ I= ME⋅ Iρ:=Bendiug Stress :σM⋅cI= σE⋅ IρcI⎛⎜⎝⎞⎠= ⋅ σE⋅cρ:=σ = 20MPa Ans 1225. Problem 12-02A picture is taken of a man performing a pole vault, and the minimum radius of curvature of the pole isestimated by measurement to be 4.5 m. If the pole is 40 mm in diameter and it is made of a glass-reinforcedplastic for which Eg = 131 GPa, determine the maximum bending stress in the pole.Given: do := 40mm ρ := 4.5mE := 131GPaSolution:Section Property :cdo2:= I⋅ 464π do:=Moment - Curvature Relationship :1ρME⋅ I= ME⋅ Iρ:=Bendiug Stress :σM⋅cI= σE⋅ IρcI⎛⎜⎝⎞⎠= ⋅ σE⋅cρ:=σ = 582.2MPa Ans 1226. Problem 12-03Determine the equation of the elastic curve for the beam using the x coordinate that is valid forSpecify the slope at A and the beam's maximum 0 ≤ x < L / 2 deflection. EI is constant. 1227. Problem 12-04Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates. Specify thebeam's maximum deflection. EI is constant. 1228. Problem 12-05Determine the equations of the elastic curve using the x1 and x2 coordinates. EI is constant. 1229. Problem 12-06Determine the equations of the elastic curve for the beam using the x1 and x3 coordinates. Specify thebeam's maximum deflection. EI is constant. 1230. Problem 12-07Determine the equations of the elastic curve for the shaft using the x1 and x2 coordinates. Specify theslope at A and the displacement at the center of the shaft. EI is constant. 1231. Problem 12-08Determine the equations of the elastic curve for the shaft using the x1 and x3 coordinates. Specify theslope at A and the deflection at the center of the shaft. EI is constant. 1232. Problem 12-09The beam is made of two rods and is subjected to the concentrated load P. Determine the maximumdeflection of the beam if the moments of inertia of the rods are IAB and IBC, and the modulus ofelasticity is E. 1233. Problem 12-10The beam is made of two rods and is subjected to the concentrated load P. Determine the slope at C.The moments of inertia of the rods are IAB and IBC, and the modulus of elasticity is E. 1234. Problem 12-11The bar is supported by a roller constraint at B, which allows vertical displacement but resists axialload and moment. If the bar is subjected to the loading shown, determine the slope at A and thedeflection at C. EI is constant. 1235. Problem 12-12Determine the deflection at B of the bar in Prob. 12-11. 1236. Problem 12-13The fence board weaves between the three smooth fixed posts. If the posts remain along the same line,determine the maximum bending stress in the board. The board has a width of 150 mm. and athickness of 12 mm. E = 12 GPa. Assume the displacement of each end of the board relative to itscenter is 75 mm.Given: b := 150mm t := 12mm L := 2.4mE := 12GPa δ := 75mmSolution:Support Reactions : By symmetry, RA=RC=R+ ΣFy=0; 2R − P = 0 R = 0.5PMoment Function :M(x) = R⋅x M(x) = 0.5⋅P⋅xMmax = R⋅ (0.5L) Mmax = 0.25P⋅LSection Property :ct2:= Ib⋅ t312:=Slope and Elastic Curve :E⋅ Id2⋅vdx2⋅ = M(x) E⋅ Id2vdx2⋅P⋅x2=E⋅ Idvdx⋅P⋅x24= + C1 (1)E⋅ I⋅vP⋅x312= + C1⋅x + C2 (2)Boundary Conditions : Due to symmetry, dv/dx=0 at x=L/2. Also, v=0 at x=0.From Eq. (1): 0P4L2⎛⎜⎝⎞⎠2= ⋅ + C1 C1P⋅L216= −From Eq. (2): 0 = 0 + 0 + C2 C2 := 0The Elastic Curve : Substitute the values of C1 and C2 into Eq. (2),E⋅ I⋅vP⋅x312P⋅L216= − ⋅x v= ⋅ (4x2 − 3L2) (3)P⋅x48E⋅ IRequire at x=L/2, v=-δ. From Eq. (3),−δP48E⋅ IL2⎛⎜⎝⎞⎠⋅ 4⎛⎜⎝⎞⎠ 2L2− 3L2⎡⎢⎣⎤⎥⎦= ⋅ −δ−P⋅L348⋅E⋅ I= P48⋅E⋅ IL3:= ⋅δBending Stress: Mmax := 0.25P⋅Lcmaxt2:= σmaxMmax⋅cmax:= σmax = 11.25MPa AnsI 1237. Problem 12-14Determine the equation of the elastic curve for the beam using the x coordinate. Specify the slope at Aand the maximum deflection. EI is constant. 1238. Problem 12-15Determine the deflection at the center of the beam and the slope at B. EI is constant. 1239. Problem 12-16A torque wrench is used to tighten the nut on a bolt. If the dial indicates that a torque of 90 N·m isapplied when the bolt is fully tightened, determine the force P acting at the handle and the distance s theneedle moves along the scale. Assume only the portion AB of the beam distorts. The cross section issquare having dimensions of 12 mm by 12 mm. E = 200 GPa.Given: b := 12mm h := 12mm L := 0.45mE := 200GPa δ := 75mm R := 0.3mTz := 90N⋅mSolution:Equations of Equilibrium :+ΣFy=0; Ay − P = 0 (1)ΣΜB=0; Tz − P⋅L = 0 (2)Solving Eqs. (1) and (2): PTzL:= Ay := P Ay = 200NP = 200 N AnsMoment Function : M x ( ) Ay= ⋅x − TzSection Property : Ib⋅h312:=Slope and Elastic Curve :E⋅ Id2⋅vdx2⋅ = M(x)E⋅ Id2vdx2⋅ = Ay⋅x − TzE⋅ Idvdx⋅Ay⋅x22= − Tz⋅x + C1 (1)E⋅ I⋅vAy⋅x36Tz⋅x22= − + C1⋅x + C2 (2)Boundary Conditions : Due to symmetry, dv/dx=0 at x=0, and v=0 at x=0.From Eq. (1): 0 = 0 − 0 + C1 C1 := 0From Eq. (2): 0 = 0 − 0 + 0 + C2 C2 := 0The Elastic Curve : Substitute the values of C1 and C2 into Eq. (2),E⋅ I⋅vAy⋅x36Tz⋅x22= − vx26E⋅ I= ⋅ ( − ) (3)Ayx 3TzAt x=R, v=-s. From Eq. (3),s−R26E⋅ I:= ⋅ ( AyR − 3Tz) s = 9.11mm Ans 1240. Problem 12-17The shaft is supported at A by a journal bearing that exerts only vertical reactions on the shaft and at Bby a thrust bearing that exerts horizontal and vertical reactions on the shaft. Draw the bending-momentdiagram for the shaft and then, from this diagram, sketch the deflection or elastic curve for the shaft'scenterline. Determine the equations of the elastic curve using the coordinates x1 and x2 . EI is constant.Given: h := 60mm L1 := 150mmP := 5kN L2 := 400mmSolution: L := L1 + L2Support Reactions:+ΣFy=0; A + B = 0 (1)ΣΜB=0; P⋅h + A⋅L2 = 0 (2)Solving Eqs. (1) and (2): A−P⋅hL2:= A = −0.75 kNB := −A B = 0.75 kNMoment Function : M1(x1) := P⋅hM2(x2) := B⋅x2Section Property : EI := kN⋅m2Slope and Elastic Curve :⋅ = M1(x1) EIEId2⋅v1dx12d2⋅v2dx2⋅ = M2(x2)2⋅ = P⋅h EIEId2v1dx12d2v2dx2⋅ = B⋅x22EIdv1dx1= ⋅ + C1 (1) EI⋅ P h ⋅ ( ) x1dv2dx2⋅B2C= ⋅ x22 + 3 (3)EI⋅v1⋅ 22P⋅h x1= + C1⋅x1 + C2 (2) EI⋅v2⋅ 36B x2= + C3⋅x2 + C4 (4)Boundary Conditions :v1=0 at x1=0.15m, From Eq. (2): 0P⋅h⋅ (0.15m)2= + C1⋅ (0.15m) + C2 (5)2v2=0 at x2=0, From Eq. (4): 0 = 0 + 0 + C4 C4 := 0 AnsB⋅ (0.4m)3v2=0 at x2=0.4m, From Eq. (4): 0= + C3⋅ (0.4m) + 6C4C3B6:= − ⋅ (0.4m)2 C3 = −0.02 kN⋅m2 Ans 1241. Continuity Condition:dv1/dx1= - dv2/dx2 at A (x1=0.15m and x2=0.4m)From Eqs. (1) and (3), P⋅h⋅ (0.15m) + C1B2= ⋅ (0.4m)2 + C3C1B2:= ⋅ (0.4m)2 + C3 − P⋅h⋅ (0.15m) C1 = −0.005 kN⋅m2 AnsFrom Eq. (5): C2P⋅h⋅ (0.15m)2:= − − C1⋅ (0.15m) C2 = −0.00263 kN⋅m3 Ans2The Elastic Curve : Substitute the values of C1 and C2 into Eq. (2), and C3 and C4 into Eq. (4),v11EIP⋅h⋅ 2x12 C+ 1⋅x1 + C2 ⎛⎜⎝⎞⎠= Ansv21EIB⋅ 6x2⎛⎜⎝3 C+ 3⋅x2 + C4 ⎞⎠= AnsBMD :x'1 := 0 , 0.01⋅L1 .. L1 x'2 := L1 , 1.01⋅L1 .. LM'1(x'1) P⋅h:= M'2 x'2 ( ) P h ⋅ A x'2L1 − ( ) ⋅ + ⎡⎣kN⋅m⎤⎦1kN⋅m:= ⋅0 0.2 0.40.40.20Distane (m)Moment (kN-m)M'1(x'1)M'2(x'2)x'1, x'2 1242. Problem 12-18Determine the equations of the elastic curve using the coordinates x1 and x2 , and specify the slope anddeflection at C. EI is constant. 1243. Problem 12-19Determine the equations of the elastic curve using the coordinates x1 and x2 , and specify the slope atA. EI is constant. 1244. Problem 12-20Determine the equations of the elastic curve using the coordinates x1 and x2 , and specify the slope anddeflection at B. EI is constant. 1245. Problem 12-21Determine the equations of the elastic curve using the coordinates x1 and x3 , and specify the slope anddeflection at B. EI is constant. 1246. Problem 12-22Determine the maximum slope and maximum deflection of the simply-supported beam which issubjected to the couple moment M0 . EI is constant. 1247. Problem 12-23The two wooden meter sticks are separated at their centers by a smooth rigid cylinder having adiameter of 50 mm. Determine the force F that must be applied at each end in order to just make theirends touch. Each stick has a width of 20 mm and a thickness of 5 mm. Ew = 11 GPa.Given: do := 50mm L := 0.5mb := 20mm t := 5mm E := 11GPaSolution:Section Property: Ib⋅ t312:=Moment Function : M(x) = −F⋅xSlope and Elastic Curve :E⋅ Id2⋅vdx2⋅ = M(x)E⋅ Id2vdx2⋅ = −F⋅xE⋅ Idvdx⋅F⋅x22= − + C1 (1)E⋅ I⋅vF⋅x36= − + C1⋅x + C2 (2)Boundary Conditions :dv/dx=0 at x=L. From Eq. (1): 0F⋅L22= − + C1 C1F⋅L22:=v=0 at x=L. From Eq. (2): 0F⋅L36= − + C1⋅L + C2C2F⋅L36= − C1⋅L C2F⋅L33:= −Require : v := −0.5do at x=0.From Eq. (2): −0.5do⋅E⋅ I −0 0F⋅L33= + −F1.5do⋅E⋅ IL3:=F = 1.375N Ans 1248. Problem 12-24The pipe can be assumed roller supported at its ends and by a rigid saddle C at its center. The saddlerests on a cable that is connected to the supports. Determine the force that should be developed in thecable if the saddle keeps the pipe from sagging or deflecting at its center. The pipe and fluid within ithave a combined weight of 2 kN/m. EI is constant.Given:e := 0.3m L := 3.75m w2kNm:=Solution:Moment Function : M(x) P⋅x12= − ⋅w⋅x2Slope and Elastic Curve :E⋅ Id2⋅vdx2⋅ = M(x)E⋅ Id2vdx2⋅ P⋅xw2= − ⋅x2E⋅ Idvdx⋅P⋅x22w6= − ⋅x3 + C1 (1)E⋅ I⋅vP⋅x36w24= − ⋅x4 + C1⋅x + C2 (2)Boundary Conditions : v=0 at x=0 and at x=L.From Eq. (2): 0 = 0 − 0 + 0 + C2 C2 := 00P⋅L36w24= − ⋅L4 + C1⋅L (3)Also, dv/dx=0 at x=L.From Eq. (1): 0P⋅L22w6= − ⋅L3 + C1 (4)Solving Eqs. (3) and (4) for P,P⋅L26w24− ⋅L3 + C1P⋅L22w6= − ⋅L3 + C1P⋅L23w8= ⋅L3 P3w⋅L8:= P = 2.813 kNEquations of Equilibrium :+ ΣFy=0; 2P + F − w⋅ (2L) = 0F := 2w⋅L − 2P F = 9375 NAt C : + ΣFy=0; 2Tcableee2 + L2⋅ − F = 0 Tcablee2 + L22e:= ⋅FTcable = 58.78 kN Ans 1249. Problem 12-25Determine the equations of the elastic curve using the coordinates x1 and x2 , and specify the slope at Cand displacement at B. EI is constant. 1250. Problem 12-26Determine the equations of the elastic curve using the coordinates x1 and x3 , and specify the slope at Band deflection at C. EI is constant. 1251. Problem 12-27Determine the elastic curve for the simply supported beam using the x coordinate 0 ≤ x ≤ L / . 2Also,determine the slope at A and the maximum deflection of the beam. EI is constant. 1252. Problem 12-28Determine the elastic curve for the cantilevered beam using the x coordinate. Also determine themaximum slope and maximum deflection. EI is constant. 1253. Problem 12-29The beam is made of a material having a specific weight γ. Determine the displacement and slope at itsend A due to its weight. The modulus of elasticity for the material is E. 1254. Problem 12-30The beam is made of a material having a specific weight γ. Determine the displacement and slope at itsend A due to its weight. The modulus of elasticity for the material is E. 1255. Problem 12-31The leaf spring assembly is designed so that it is subjected to the same maximum stress throughout itslength. If the plates of each leaf have a thickness t and can slide freely between each other, show thatthe spring must be in the form of a circular arc in order that the entire spring becomes flat when alarge enough load P is applied. What is the maximum normal stress in the spring? Consider the springto be made by cutting the n strips from the diamond-shaped plate of thickness t and width b. Themodulus of elasticity for the material is E. Hint: Show that the radius of curvature of the spring isconstant. 1256. Problem 12-32The beam has a constant width b and is tapered as shown. If it supports a load P at its end, determinethe deflection at B.The load P is applied a short distance s from the tapered end B, where s << L. EI isconstant. 1257. Problem 12-33A thin flexible 6-m-long rod having a weight of 10 N/m rests on the smooth surface. If a force of 15 Nis applied at its end to lift it, determine the suspended length x and the maximum moment developed inthe rod.Given: P := 15N w10Nm:=Lmax := 6mSolution:Since the horizontal section has no curvature,the moment in the rod is zero. Hence, R actsat the end of the suspended portion and thisportion acts like a simply supported beam.Thus,Equations of Equilibrium :+ ΣFy=0; P + R − w⋅x = 0 (1)Σ Μ0=0; P⋅xw2− ⋅x2 = 0 (2)From Eqs. (2) :x2Pw:= x = 3m AnsMaximum moment occurs at center.Mmax Px2⎛⎜⎝⎞⎠⋅w2x2⎛⎜⎝⎞⎠2:= − ⋅Mmax = 11.25N⋅m Ans 1258. Problem 12-34The shaft supports the two pulley loads shown. Determine the equation of the elastic curve. Thebearings at A and B exert only vertical reactions on the shaft. EI is constant.Use Macaulay Function: Ψ(z) := Φ(z)⋅ (z)Given: a := m P := kN EI := kN⋅m2 EIo := 1Solution:Support Reactions :+ΣFy=0; A + B − P − 2P = 0 (1)ΣΜB=0; A⋅ (2a) − P⋅a + 2P⋅a = 0 (2)Solving Eqs. (1) and (2): A−P2:= B7P2:=Moment Function :M(x) = A⋅Ψ(x − 0) − P⋅Ψ(x − a) + B⋅Ψ(x − 2a)Slope and Elastic Curve :EId2⋅vdx2⋅ = M(x)EId2vdx2⋅ = A⋅Ψ(x − 0) − P⋅Ψ(x − a) + B⋅Ψ(x − 2a)EIdvdx⋅A2Ψ(x − 0)⋅ 2P2Ψ(x − a)− ⋅ 2B2Ψ(x − 2a)+ ⋅ 2 C= + 1EI⋅vA6Ψ(x − 0)⋅ 3P6Ψ(x − a)− ⋅ 3B6Ψ(x − 2a)+ ⋅ 3 C= + 1⋅x + C2EI⋅vA6⋅x3P6Ψ(x − a)− ⋅ 3B6Ψ(x − 2a)+ ⋅ 3 C= + 1⋅x + C2 (1)Boundary Conditions : v=0 at x=0From Eq. (1): 0 = 0 − 0 + 0 + 0 + C2 C2 := 0Also v=0 at x=2a.From Eq. (1): 0A6⋅ (2a)3P6Ψ(2a − a)− ⋅ 3 + 0 C= + 1⋅ (2a)C1A12a− ⋅ (2a)3P12a:= + ⋅a3 C1512= P⋅a2Equation of Elastic Curve :v1EIAx3PB⋅− ⋅ Ψ(x − a)3+ ⋅ Ψ(x − 2a)3 C+ x 661⋅⎞⎠6⎛⎜⎝= mv(x)1EIoPx3P37P5− ⋅− ⋅ Ψ(x − a)+ ⋅ Ψ(x − 2a)3+ P ⋅ a2 ⋅ x 1261212⎛⎜⎝⎞⎠:= m Ans 1259. Problem 12-35Determine the equation of the elastic curve. Specify the slopes at A and B. EI is constant.Use Macaulay Function: Ψ(z) := Φ(z)⋅ (z)Given:a := m wkNm:=Solution: L := 2aSupport Reactions :+ΣFy=0; A + B − w⋅a = 0 (1)ΣΜA=0; (wa)⋅ (0.5a) − B⋅ (2⋅a) = 0 (2)Solving Eqs. (1) and (2): B14:= ⋅ (wa) A := w⋅a − BB = 0.25 w⋅a A = 0.75w⋅aMoment Function :M(x) −0.5w Ψ(x − 0)⋅ 2 (−0.5)w Ψ(x − a)= − ⋅ 2 + A⋅Ψ(x − 0)M(x) −0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 + A⋅xSlope and Elastic Curve : EI := kN⋅m2 EIo := 1EId2⋅vdx2⋅ = M(x)EId2vdx2⋅ −0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 + A⋅xEIdvdx⋅16− w⋅x316w Ψ(x − a)+ ⋅ 3A2= + ⋅x2 + C1 (3)EI⋅v124− w⋅x4124w Ψ(x − a)+ ⋅ 4A6= + ⋅x3 + C1⋅x + C2 (4)Boundary Conditions : v=0 at x=0 and x=2a. From Eq. (4):0 = −0 + 0 + 0 + 0 + C2 C2 := 00124− w⋅ (2a)4124⎛⎜⎝⎞⎠+ ⋅w⋅a4A6= + ⋅ (2a)3 + C1⋅ (2a)C113w⋅a3148− w⋅a32A3:= − ⋅a2 C1316= − w⋅a3Equations of Elastic Curve and Slope :v(x)1EIo124− w⋅x4124w Ψ(x − a)+ ⋅ 43w⋅a24+ ⋅x3316⎞⎠w a3 ⋅ x ⋅ − ⎛⎜⎝:= Ansθ(x)1EIo16− w⋅x316w Ψ(x − a)+ ⋅ 3A2⎞⎠x2 ⋅ + C1 + ⎛⎜⎝:=Slope at A: Substitute x=0 into Eq.(3). θ(0)316−w⋅a3EIo⎛⎜⎝⎞⎠= AnsSlope at B: Substitute x=L into Eq.(3). θ(L)748w⋅a3EIo⎛⎜⎝⎞⎠= Ans 1260. Problem 12-36The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant.Use Macaulay Function: Ψ(z) := Φ(z)⋅ (z)Given: a := 1.5m P := 20kN w 6kNm:=b := 3mSolution: L := 2a + bSupport Reactions :+ΣFy=0; A + B − P − w⋅a = 0 (1)ΣΜA=0; −(wa)⋅ (0.5a) + P⋅ (a + b) − B⋅b = 0 (2)Solving Eqs. (1) and (2): B1b:= ⋅ [−(wa)⋅ (0.5a) + P⋅ (a + b)] A := P + w⋅a − BB = 27.75 kN A = 1.25 kNMoment Function :M(x) −0.5w Ψ(x − 0)⋅ 2 (−0.5)w Ψ(x − a)= − ⋅ 2 + A⋅Ψ(x − a) + B⋅Ψ(x − a − b)M(x) −0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 + A⋅Ψ(x − a) + B⋅Ψ(x − a − b)Slope and Elastic Curve :E⋅ Id2⋅vdx2⋅ = M(x)E⋅ Id2vdx2⋅ −0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 + A⋅Ψ(x − a) + B⋅Ψ(x − a − b)E⋅ Idvdx⋅16− w⋅x316w Ψ(x − a)+ ⋅ 3A2Ψ(x − a)+ ⋅ 2B2Ψ(x − a − b)+ ⋅ 2 C= + 1 (3)E⋅ I⋅v124− w⋅x4124w Ψ(x − a)+ ⋅ 4A6Ψ(x − a)+ ⋅ 3B6Ψ(x − a − b)+ ⋅ 3 C= + 1⋅x + C2 (4)Boundary Conditions : v=0 at x=a and x=a+b. From Eq. (4):0124= − w⋅a4 + 0 + 0 + 0 + C1⋅a + C2 (5)0124− w⋅ (a + b)4124+ w⋅b4A6= + ⋅b3 + 0 + C1⋅ (a + b) + C2 (6)(6)-(5): C11b124w⋅ (a + b)4124− w⋅b4A6− ⋅b3124⎤⎥⎦w ⋅ a4 ⋅ − ⎡⎢⎣:= C1 = 25.125 kN⋅m2From Eq. (5): C2124:= ⋅w⋅a4 − C1⋅a C2 = −36.422 kN⋅m3Equation of Elastic Curve : aow24:= a1A6:= a2B6:= a3 := C1 a4 := C2ao 0.25kNm= a1 = 0.2083 kN a2 = 4.625 kN a3 = 25.13 kN⋅m2 a4 = −36.42 kN⋅m3 Ansv1E⋅ I−ao x4 ao Ψ(x − a)+ ⋅ 4 a1 Ψ(x − a)+ ⋅ 3 a2 Ψ(x − a − b)+ ⋅ 3 a+ 3⋅x + a4 ⎛⎝⎞⎠= Ans 1261. Problem 12-37The shaft supports the two pulley loads shown. Determine the equation of the elastic curve. Thebearings at A and B exert only vertical reactions on the shaft. EI is constant.Use Macaulay Function: Ψ(z) := Φ(z)⋅ (z)Given: a := 0.5m P1 := 200N P2 := 300NSolution:Support Reactions :+ΣFy=0; A + B − P1 − P2 = 0 (1)ΣΜB=0; A⋅ (2a) − P1⋅a + P2⋅a = 0 (2)P1 − P2Solving Eqs. (1) and (2): A:= B2P1 + 3P22:=A = −50 N B = 550 NMoment Function :M(x) = A⋅Ψ(x − 0) − P1⋅Ψ(x − a) + B⋅Ψ(x − 2a)Slope and Elastic Curve :E⋅ Id2⋅vdx2⋅ = M(x)E⋅ Id2vdx2⋅ = A⋅Ψ(x − 0) − P1⋅Ψ(x − a) + B⋅Ψ(x − 2a)E⋅ Idvdx⋅A2Ψ(x − 0)⋅ 2P12Ψ(x − a)− ⋅ 2B2Ψ(x − 2a)+ ⋅ 2 C= + 1E⋅ I⋅vA6Ψ(x − 0)⋅ 3P16Ψ(x − a)− ⋅ 3B6Ψ(x − 2a)+ ⋅ 3 C= + 1⋅x + C2 (1)Boundary Conditions : v=0 at x=0From Eq. (1): 0 = 0 − 0 + 0 + 0 + C2 C2 := 0Also v=0 at x=2a.From Eq. (1): 0A6Ψ(2a − 0)⋅ 3P16Ψ(2a − a)− ⋅ 3 + 0 C= + 1⋅ (2a)C1A12a− ⋅ (2a)3P112a:= + ⋅a3 C1 = 12.50N⋅m2Equation of Elastic Curve : aoA6:= a1P16:= − a2B6:= a3 := C1ao = −8.33 N a1 = −33.33 N a2 = 91.67 N a3 = 12.50N⋅m2 Ansv1E⋅ Iao Ψ(x − 0)⋅ 3 a1 Ψ(x − a)+ ⋅ 3 a2 Ψ(x − 2a)+ ⋅ 3 a+ 3⋅x ⎛⎝⎞⎠= Ans 1262. Problem 12-38The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant.Use Macaulay Function: Ψ(z) := Φ(z)⋅ (z)Given: a := 4m P := 50kN w 3kNm:=b := 3mSolution: L := a + 2⋅bSupport Reactions :+ΣFy=0; A + B − P − w⋅a = 0 (1)ΣΜA=0; (wa)⋅ (0.5a) + P⋅ (a + b) − B⋅L = 0 (2)Solving Eqs. (1) and (2): B1L:= ⋅ [(wa)⋅ (0.5a) + P⋅ (a + b)] A := P + w⋅a − BMoment Function : B = 37.4 kN A = 24.6 kNAnsM(x) A⋅Ψ(x − 0) 0.5w Ψ(x − 0)− ⋅ 2 (−0.5)w Ψ(x − a)= − ⋅ 2 − P⋅Ψ(x − a − b)M(x) A⋅x − 0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 − P⋅Ψ(x − a − b)Slope and Elastic Curve : EI := kN⋅m2 EIo := 1EId2⋅vdx2⋅ = M(x)EId2vdx2⋅ A⋅x − 0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 − P⋅Ψ(x − a − b)EIdvdx⋅A2⋅x2w6− ⋅x3w6Ψ(x − a)+ ⋅ 3P2Ψ(x − a − b)− ⋅ 2 C= + 1 (3)EI⋅vA6⋅x3w24− ⋅x4w24Ψ(x − a)+ ⋅ 4P6Ψ(x − a − b)− ⋅ 3 C= + 1⋅x + C2 (4)Boundary Conditions :v=0 at x=0. From Eq. (4): 0 = 0 − 0 + 0 + 0 + 0 + C2 C2 := 0v=0 at x=L. From Eq. (4): 0A6⋅L3w24− ⋅L4w24+ ⋅ (L − a)4P6= − ⋅ (L − a − b)3 + C1⋅LC1A⋅L26−w24+ ⋅L3w24L− ⋅ (L − a)4P6L:= + ⋅ (L − a − b)3C1 = −278.70 kN⋅m2Equation of Elastic Curve : aoA6:= a1w24:= a2P6:= a3 := C1 a3 := C2ao = 4.10 kN a1 0.125kNm= a2 = 8.33 kN a3 = 0.00 N⋅m2 Ansv1EIao⋅x3 − a1⋅x4 a1 Ψ(x − a)+ ⋅ 4 a2 Ψ(x − a − b)− ⋅ 3 a+ 3⋅x + a4 ⎛⎝⎞⎠= 1263. Problem 12-39The beam is subjected to the load shown. Determine the displacement at x = 7 m and the slope at A.EI is constant.Use Macaulay Function: Ψ(z) := Φ(z)⋅ (z)Given: a := 4m P := 50kN w 3kNm:=b := 3mSolution: L := a + 2⋅bSupport Reactions :+ΣFy=0; A + B − P − w⋅a = 0 (1)ΣΜA=0; (wa)⋅ (0.5a) + P⋅ (a + b) − B⋅L = 0 (2)Solving Eqs. (1) and (2): B1L:= ⋅ [(wa)⋅ (0.5a) + P⋅ (a + b)] A := P + w⋅a − BB = 37.4 kN A = 24.6 kNMoment Function :AnsM(x) A⋅Ψ(x − 0) 0.5w Ψ(x − 0)− ⋅ 2 (−0.5)w Ψ(x − a)= − ⋅ 2 − P⋅Ψ(x − a − b)M(x) A⋅x − 0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 − P⋅Ψ(x − a − b)Slope and Elastic Curve : EI := kN⋅m2 EIo := 1EId2⋅vdx2⋅ = M(x)EId2vdx2⋅ A⋅x − 0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 − P⋅Ψ(x − a − b)EIdvdx⋅A2⋅x2w6− ⋅x3w6Ψ(x − a)+ ⋅ 3P2Ψ(x − a − b)− ⋅ 2 C= + 1 (3)EI⋅vA6⋅x3w24− ⋅x4w24Ψ(x − a)+ ⋅ 4P6Ψ(x − a − b)− ⋅ 3 C= + 1⋅x + C2 (4)Boundary Conditions :v=0 at x=0. From Eq. (4): 0 = 0 − 0 + 0 + 0 + 0 + C2 C2 := 0v=0 at x=L. From Eq. (4): 0A6⋅L3w24− ⋅L4w24+ ⋅ (L − a)4P6= − ⋅ (L − a − b)3 + C1⋅LC1A⋅L26−w24+ ⋅L3w24L− ⋅ (L − a)4P6L:= + ⋅ (L − a − b)3C1 = −278.70 kN⋅m2Equations of Elastic Curve and Slope :v(x)1EIA⋅x3w− ⋅x4wP+ ⋅ Ψ(x − a)4− ⋅ Ψ(x − a − b)3 C+ x + 6⎛⎜⎝242461⋅C2 ⎞⎠:=Displacement at x=7m.: Substitute x=7m into Eq.(4). v(7m) −834.60mEIo= Ansθ(x)1EIA⎞⎠x2wwP⋅− ⋅x3+ ⋅ Ψ(x − a)3− ⋅ Ψ(x − a − b)2 C+ 26⎛⎜⎝621 :=Slope at A: Substitute x=0 into Eq.(3). θ(0) −278.701EIo= 1264. Problem 12-40The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant.Use Macaulay Function: Ψ(z) := Φ(z)⋅ (z)Given: a := 2.4m P1 := 10kNMB := 6kN⋅m P2 := 20kNSolution:Support Reactions :+ΣFy=0; A + B − P1 − P2 = 0 (1)ΣΜB=0; A⋅ (3a) − P1⋅ (2a) − P2⋅a + MB = 0 (2)2P1 + P2Solving Eqs. (1) and (2): A3MB3a:= − B := P1 + P2 − AA = 12.5 kN B = 17.5 kNMoment Function :M(x) = A⋅Ψ(x − 0) − P1⋅Ψ(x − a) − P2⋅Ψ(x − 2a)Slope and Elastic Curve :E⋅ Id2⋅vdx2⋅ = M(x)E⋅ Id2vdx2⋅ = A⋅Ψ(x − 0) − P1⋅Ψ(x − a) − P2⋅Ψ(x − 2a)E⋅ Idvdx⋅A2Ψ(x − 0)⋅ 2P12Ψ(x − a)− ⋅ 2P22Ψ(x − 2a)− ⋅ 2 C= + 1E⋅ I⋅vA6Ψ(x − 0)⋅ 3P16Ψ(x − a)− ⋅ 3P26Ψ(x − 2a)− ⋅ 3 C= + 1⋅x + C2 (1)Boundary Conditions : v=0 at x=0From Eq. (1): 0 = 0 − 0 + 0 + C2 C2 := 0Also v=0 at x=3a.From Eq. (1): 0A6Ψ(3a − 0)⋅ 3P16Ψ(3a − a)− ⋅ 3P26Ψ(3a − 2a)− ⋅ 3 C= + 1⋅ (3a)C1A18a− ⋅ (3a)3P118a+ ⋅ (2a)3P218a:= + ⋅a3 C1 = −76.00 kN⋅m2Equation of Elastic Curve : aoA6:= a1P16:= − a2P26:= − a3 := C1ao = 2.08 kN a1 = −1.67 kN a2 = −3.33 kN a3 = −76.00 kN⋅m2 Ansv1E⋅ Iao Ψ(x − 0)⋅ 3 a1 Ψ(x − a)+ ⋅ 3 a2 Ψ(x − 2a)+ ⋅ 3 a+ 3⋅x ⎛⎝⎞⎠= Ans 1265. Problem 12-41The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant.Use Macaulay Function: Ψ(z) := Φ(z)⋅ (z)Given: a := 1.5m b := 3mw 6kNm:=P := 20kNSolution: L := 2a + bSupport Reactions :+ΣFy=0; A + B − P − w⋅a = 0 (1)ΣΜA=0; −(wa)⋅ (0.5a) + P⋅ (a + b) − B⋅b = 0 (2)Solving Eqs. (1) and (2): B1b:= ⋅ [−(wa)⋅ (0.5a) + P⋅ (a + b)] A := P + w⋅a − BB = 27.75 kN A = 1.25 kNMoment Function :M(x) −0.5w Ψ(x − 0)⋅ 2 (−0.5)w Ψ(x − a)= − ⋅ 2 + A⋅Ψ(x − a) + B⋅Ψ(x − a − b)M(x) −0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 + A⋅Ψ(x − a) + B⋅Ψ(x − a − b)Slope and Elastic Curve :E⋅ Id2⋅vdx2⋅ = M(x)E⋅ Id2vdx2⋅ −0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 + A⋅Ψ(x − a) + B⋅Ψ(x − a − b)E⋅ Idvdx⋅16− w⋅x316w Ψ(x − a)+ ⋅ 3A2Ψ(x − a)+ ⋅ 2B2Ψ(x − a − b)+ ⋅ 2 C= + 1 (3)E⋅ I⋅v124− w⋅x4124w Ψ(x − a)+ ⋅ 4A6Ψ(x − a)+ ⋅ 3B6Ψ(x − a − b)+ ⋅ 3 C= + 1⋅x + C2 (4)Boundary Conditions : v=0 at x=a and x=a+b. From Eq. (4):0124= − w⋅a4 + 0 + 0 + 0 + C1⋅a + C2 (5)0124− w⋅ (a + b)4124+ w⋅b4A6= + ⋅b3 + 0 + C1⋅ (a + b) + C2 (6)(6)-(5): C11b11A1w⋅ (a + b)4− w⋅b4− ⋅b3− ⋅ w ⋅ a4 2424624⎡⎢⎣⎤⎥⎦:= C1 = 25.125 kN⋅m2From Eq. (5): C2124:= ⋅w⋅a4 − C1⋅a C2 = −36.422 kN⋅m3Equation of Elastic Curve : aow24:= a1A6:= a2B6:= a3 := C1 a4 := C2ao 0.25kNm= a1 = 0.2083 kN a2 = 4.625 kN a3 = 25.13 kN⋅m2 a4 = −36.42 kN⋅m3 Ansv1E⋅ I−ao x4 ao Ψ(x − a)+ ⋅ 4 a1 Ψ(x − a)+ ⋅ 3 a2 Ψ(x − a − b)+ ⋅ 3 a+ 3⋅x + a4 ⎛⎝⎞⎠= Ans 1266. Problem 12-42The beam is subjected to the load shown. Determine the equation of the slope and elastic curve. EI isconstant.Use Macaulay Function: Ψ(z) := Φ(z)⋅ (z)Given: a := 5m b := 3mw 3kNm:=Mo := 15kN⋅mSolution: L := a + bSupport Reactions :+ΣFy=0; A + B − w⋅a = 0 (1)ΣΜA=0; (wa)⋅ (0.5a) − B⋅a + Mo = 0 (2)Solving Eqs. (1) and (2): B1awa ( ) 0.5a ( ) ⋅ Mo + ⎡⎣⎤⎦:= ⋅ A := w⋅a − BB = 10.5 kN A = 4.5 kNMoment Function :M(x) −0.5w Ψ(x − 0)⋅ 2 (−0.5)w Ψ(x − a)= − ⋅ 2 + A⋅Ψ(x − 0) + B⋅Ψ(x − a)M(x) −0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 + A⋅x + B⋅Ψ(x − a)Slope and Elastic Curve :E⋅ Id2⋅vdx2⋅ = M(x)E⋅ Id2vdx2⋅ −0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 + A⋅x + B⋅Ψ(x − a)E⋅ Idvdx⋅16− w⋅x316w Ψ(x − a)+ ⋅ 3A2+ ⋅x2B2Ψ(x − a)+ ⋅ 2 C= + 1 (3)E⋅ I⋅v124− w⋅x4124w Ψ(x − a)+ ⋅ 4A6+ ⋅x3B6Ψ(x − a)+ ⋅ 3 C= + 1⋅x + C2 (4)Boundary Conditions : v=0 at x=0 and x=a. From Eq. (4):0 = −0 + 0 + 0 + 0 + 0 + C2 C2 := 00124− w⋅a4 + 0A6= + ⋅a3 + 0 + C1⋅ (a) C1124w⋅a3A6:= − ⋅a2 C1 = −3.125 kN⋅m2Equation of Elastic Curve and slope : aow24:= a1A6:= a2B6:= a3 := C1ao 0.125kNm= a1 = 0.75 kN a2 = 1.75 kN a3 = −3.125 kN⋅m2 Ansv1E⋅ I−ao x4 ao Ψ(x − a)+ ⋅ 4 a+ 1⋅x3 a2 Ψ(x − a)+ ⋅ 3 a+ 3⋅x ⎛⎝⎞⎠= Ansθ1E⋅ I−4⋅ao x3 4⋅ao Ψ(x − a)+ ⋅ 3 3a+ 1⋅x2 3a2 Ψ(x − a)+ ⋅ 2 a+ 3 ⎛⎝⎞⎠= Ans 1267. Problem 12-43Determine the equation of the elastic curve. Specify the slope at A and the displacement at C. EI isconstant.Use Macaulay Function: Ψ(z) := Φ(z)⋅ (z)Given: a := m wkNm:=Solution: L := 2aSupport Reactions :+ΣFy=0; A + B − w⋅a = 0 (1)ΣΜA=0; (wa)⋅ (0.5a) − B⋅ (2⋅a) = 0 (2)Solving Eqs. (1) and (2): B14:= ⋅ (wa) A := w⋅a − BB = 0.25 w⋅a A = 0.75w⋅aMoment Function :M(x) −0.5w Ψ(x − 0)⋅ 2 (−0.5)w Ψ(x − a)= − ⋅ 2 + A⋅Ψ(x − 0)M(x) −0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 + A⋅xSlope and Elastic Curve : EI := kN⋅m2 EIo := 1EId2⋅vdx2⋅ = M(x)EId2vdx2⋅ −0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 + A⋅xEIdvdx⋅16− w⋅x316w Ψ(x − a)+ ⋅ 3A2= + ⋅x2 + C1 (3)EI⋅v124− w⋅x4124w Ψ(x − a)+ ⋅ 4A6= + ⋅x3 + C1⋅x + C2 (4)Boundary Conditions : v=0 at x=0 and x=2a. From Eq. (4):0 = −0 + 0 + 0 + 0 + C2 C2 := 00124− w⋅ (2a)4124⎛⎜⎝⎞⎠+ ⋅w⋅a4A6= + ⋅ (2a)3 + C1⋅ (2a)C113w⋅a3148− w⋅a32A3:= − ⋅a2 C1316= − w⋅a3Equations of Elastic Curve and Slope :v(x)1EIo124− w⋅x4124w Ψ(x − a)+ ⋅ 43w⋅a24+ ⋅x3316⎞⎠w a3 ⋅ x ⋅ − ⎛⎜⎝:=Displacement at C: Substitute x=a into Eq.(4). v(a)548−w⋅a4EIo⎛⎜⎝⎞⎠= Ansθ(x)1EIo16− w⋅x316w Ψ(x − a)+ ⋅ 3A2⎞⎠x2 ⋅ + C1 + ⎛⎜⎝:= mSlope at A: Substitute x=0 into Eq.(3). θ(0)316− mw⋅a3EIo⎛⎜⎝⎞⎠= Ans 1268. Problem 12-44Determine the equation of the elastic curve. Specify the slope at A and B. EI is constant.Use Macaulay Function: Ψ(z) := Φ(z)⋅ (z)Given:a := m wkNm:=Solution: L := 2aSupport Reactions :+ΣFy=0; A + B − w⋅a = 0 (1)ΣΜA=0; (wa)⋅ (0.5a) − B⋅ (2⋅a) = 0 (2)Solving Eqs. (1) and (2): B14:= ⋅ (wa) A := w⋅a − BB = 0.25 w⋅a A = 0.75w⋅aMoment Function :M(x) −0.5w Ψ(x − 0)⋅ 2 (−0.5)w Ψ(x − a)= − ⋅ 2 + A⋅Ψ(x − 0)M(x) −0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 + A⋅xSlope and Elastic Curve : EI := kN⋅m2 EIo := 1EId2⋅vdx2⋅ = M(x)EId2vdx2⋅ −0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 + A⋅xEIdvdx⋅16− w⋅x316w Ψ(x − a)+ ⋅ 3A2= + ⋅x2 + C1 (3)EI⋅v124− w⋅x4124w Ψ(x − a)+ ⋅ 4A6= + ⋅x3 + C1⋅x + C2 (4)Boundary Conditions : v=0 at x=0 and x=2a. From Eq. (4):0 = −0 + 0 + 0 + 0 + C2 C2 := 00124− w⋅ (2a)4124⎛⎜⎝⎞⎠+ ⋅w⋅a4A6= + ⋅ (2a)3 + C1⋅ (2a)C113w⋅a3148− w⋅a32A3:= − ⋅a2 C1316= − w⋅a3Equations of Elastic Curve and Slope :v(x)1EIo1x413w⋅a3− w⋅+ w ⋅ Ψ(x − a)4+ ⋅x3− w ⋅ a3 ⋅ x 24242416⎛⎜⎝⎞⎠:= Ansθ(x)1EIo1− w⋅x31A+ w ⋅ Ψ(x − a)3+ ⋅ x2 + 662C1 ⎛⎜⎝⎞⎠:=Slope at A: Substitute x=0 into Eq.(3). θ(0)316−w⋅a3EIo⎛⎜⎝⎞⎠= AnsSlope at B: Substitute x=L into Eq.(3). θ(L)748w⋅a3EIo⎛⎜⎝⎞⎠= Ans 1269. Problem 12-45The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant.Use Macaulay Function: Ψ(z) := Φ(z)⋅ (z)Given: a := 1.5m P := 20kNSolution: L := 4aSupport Reactions :+ΣFy=0; A + B − 2P = 0 (1)ΣΜB=0; A⋅ (2a) − P⋅ (3a) + P⋅a = 0 (2)Solving Eqs. (1) and (2): A := P B := PMoment Function :M(x) = −P⋅Ψ(x − 0) + A⋅Ψ(x − a) + B⋅Ψ(x − 3a)Slope and Elastic Curve : EI := kN⋅m2EId2⋅vdx2⋅ = M(x)EId2vdx2⋅ = −P⋅Ψ(x − 0) + A⋅Ψ(x − a) + B⋅Ψ(x − 3a)EIdvdx⋅P2− Ψ(x − 0)⋅ 2A2Ψ(x − a)+ ⋅ 2B2Ψ(x − 3a)+ ⋅ 2 C= + 1 (1)EI⋅vP6− Ψ(x − 0)⋅ 3A6Ψ(x − a)+ ⋅ 3B6Ψ(x − 3a)+ ⋅ 3 C= + 1⋅x + C2EI⋅vP6− ⋅x3A6Ψ(x − a)+ ⋅ 3B6Ψ(x − 3a)+ ⋅ 3 C= + 1⋅x + C2 (2)Boundary Conditions : Due to symmetry, dv/dx=0 at x=2aFrom Eq. (1): 0P2− ⋅ (2a)2A2Ψ(2a − a)+ ⋅ 2 + 0 C= + 1 C1 2⋅PA2− ⎛⎜⎝⎞⎠:= ⋅a2Also v=0 at x=a. C1 = 67.50 kN⋅m2From Eq. (2): 0P6= − ⋅a3 + 0 + 0 + C1⋅ (a) + C2 C2P6:= ⋅a3 − C1⋅aC2 = −90 kN⋅m3Equation of Elastic Curve : aoP6:= a1A6:= a2 := C1 a3 := C2ao = 3333.33 N a1 = 3.3333 kN a2 = 67.5 kN⋅m2 a3 = −90 kN⋅m3 Ansv1E⋅ I−ao⋅x3 a1 Ψ(x − a)+ ⋅ 3 a1 Ψ(x − 3a)+ ⋅ 3 a+ 2⋅x + a3 ⎛⎝⎞⎠= Ans 1270. Problem 12-46The beam is subjected to the load shown. Determine the equation of the slope and elastic curve. EI isconstant.Use Macaulay Function: Ψ(z) := Φ(z)⋅ (z)Given: a := 5m b := 3mw 2kNm:=P := 8kNSolution: L := a + bSupport Reactions :+ΣFy=0; A + B − w⋅a − P = 0 (1)ΣΜA=0; (wa)⋅ (0.5a) − B⋅a + P⋅L = 0 (2)Solving Eqs. (1) and (2): B1a:= ⋅ [(wa)⋅ (0.5a) + P⋅L]B = 17.8 kNA := w⋅a + P − B A = 0.2 kNMoment Function :M(x) −0.5w Ψ(x − 0)⋅ 2 (−0.5)w Ψ(x − a)= − ⋅ 2 + A⋅Ψ(x − 0) + B⋅Ψ(x − a)M(x) −0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 + A⋅x + B⋅Ψ(x − a)Slope and Elastic Curve :EId2⋅vdx2⋅ = M(x)EId2vdx2⋅ −0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 + A⋅x + B⋅Ψ(x − a)EIdvdx⋅16− w⋅x316w Ψ(x − a)+ ⋅ 3A2+ ⋅x2B2Ψ(x − a)+ ⋅ 2 C= + 1 (3)EI⋅v124− w⋅x4124w Ψ(x − a)+ ⋅ 4A6+ ⋅x3B6Ψ(x − a)+ ⋅ 3 C= + 1⋅x + C2 (4)Boundary Conditions : v=0 at x=0 and x=a. From Eq. (4):0 = −0 + 0 + 0 + 0 + 0 + C2 C2 := 00124− w⋅a4 + 0A6= + ⋅a3 + 0 + C1⋅ (a) C1124w⋅a3A6:= − ⋅a2 C1 = 9.583 kN⋅m2Equation of Elastic Curve and slope : aow24:= a1A6:= a2B6:= a3 := C1ao 0.0833kNm= a1 = 0.0333 kN a2 = 2.9667 kN a3 = 9.583 kN⋅m2 Ansv1EI−ao x4 ao Ψ(x − a)+ ⋅ 4 a+ 1⋅x3 a2 Ψ(x − a)+ ⋅ 3 a+ 3⋅x ⎛⎝⎞⎠= Ansθ1EI−4⋅ao x3 4⋅ao Ψ(x − a)+ ⋅ 3 3a+ 1⋅x2 3a2 Ψ(x − a)+ ⋅ 2 a+ 3 ⎛⎝⎞⎠= Ans 1271. Problem 12-47The beam is subjected to the load shown. Determine the slope at A and the displacement at C. EI isconstant.Use Macaulay Function: Ψ(z) := Φ(z)⋅ (z)Given: a := 5m b := 3mSolution: L := a + bSupport Reactions :+ΣFy=0; A + B − w⋅a − P = 0 (1)ΣΜA=0; (wa)⋅ (0.5a) − B⋅a + P⋅L = 0 (2)Solving Eqs. (1) and (2): B:= ⋅ [(wa)⋅ (0.5a) + P⋅L] A := w⋅a + P − BB = 17.8 kN Moment Function : A = 0.2 kNM(x) −0.5w Ψ(x − 0)⋅ 2 (−0.5)w Ψ(x − a)= − ⋅ 2 + A⋅Ψ(x − 0) + B⋅Ψ(x − a)M(x) −0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 + A⋅x + B⋅Ψ(x − a)Slope and Elastic Curve : EI := kN⋅m2d2⋅vdx2EIo := 1 EI⋅ = M(x)EIP := 8kNd2vdx2w 2kNm:=1a⋅ −0.5w⋅x2 0.5w Ψ(x − a)= + ⋅ 2 + A⋅x + B⋅Ψ(x − a)EIdvdx⋅16− w⋅x316w Ψ(x − a)+ ⋅ 3A2+ ⋅x2B2Ψ(x − a)+ ⋅ 2 C= + 1 (3)EI⋅v124− w⋅x4124w Ψ(x − a)+ ⋅ 4A6+ ⋅x3B6Ψ(x − a)+ ⋅ 3 C= + 1⋅x + C2 (4)Boundary Conditions : v=0 at x=0 and x=a. From Eq. (4):0 = −0 + 0 + 0 + 0 + 0 + C2 C2 := 00124− w⋅a4 + 0A6= + ⋅a3 + 0 + C1⋅ (a) C1124w⋅a3A6:= − ⋅a2 C1 = 9.583 kN⋅m2Equation of Elastic Curve and slope :v(x)1EI1x41AB− w⋅+ w ⋅ Ψ(x − a)4+ ⋅x3+ ⋅ Ψ(x − a)3 C+ x 24246⎛⎜⎝61⋅⎞⎠:=Displacement at C: Substitute x=L into Eq.(4). v(L) −160.75mEIo= Ansθ(x)1EI1x31AB− w⋅+ w ⋅ Ψ(x − a)3+ ⋅x2+ ⋅ Ψ(x − a)2 C+ 6622⎛⎜⎝1 ⎞⎠:=Slope at A: Substitute x=0 into Eq.(3). θ(0) 9.581EIo= Ans 1272. Problem 12-48The beam is subjected to the load shown. Determine the equation of the elastic curve.Use Macaulay Function: Ψ(x) := Φ(x)⋅xGiven: L1 := 2m L2 := 3m wo 150kNm:=Solution: L := L1 + L2Support Reactions :+ ΣFy=0; A + B − wo⋅L1wo2− ⋅L2 = 0 (1)ΣΜA=0;wo2⋅ 2− L1wo2⋅L2L23⎛⎜⎝⎞⎠+ ⋅ − B⋅L2 = 0 (2)Solving Eqs. (1) and (2): B+ 26L2−3L12 L2:= ⋅wo A := wo⋅ (L1 + 0.5L2) − BB = −25 kN A = 550 kNMoment Function :M(x)wo2− Ψ(x − 0)⋅ 2wo6L2= + ⋅Ψ(x − L1)3 + A⋅Ψ(x − L1)Slope and Elastic Curve : EId2⋅vdx2⋅ = M(x) EI := kN⋅m2 EIo := 1EId2vdx2⋅wo2− Ψ(x − 0)⋅ 2wo6L2= + ⋅Ψ(x − L1)3 + A⋅Ψ(x − L1)EIdvdx⋅wo6− Ψ(x − 0)⋅ 3wo24⋅L2+ ⋅Ψ(x − L1)4A2= + ⋅Ψ(x − L1)2 + C1 (3)EI⋅vwo24− Ψ(x − 0)⋅ 4wo120⋅L2+ ⋅Ψ(x − L1)5A6= + ⋅Ψ(x − L1)3 + C1⋅x + C2 (4)Boundary Conditions : v=0 at x=L1 GivenFrom Eq. (4): (5) 0wo24C= − ⋅ L14 + 0 + 0 + 1⋅L1 + C2Also v=0 at x=L.From Eq. (4): 0wo24− ⋅L4wo120⋅L2+ ⋅ (L − L1)5A6= + ⋅ (L − L1)3 + C1⋅L + C2 (6)Solving Eqs. (5) and (6): Guess C1 := 1kN⋅m2 C2 := 1kN⋅m3C1C2⎛⎜⎜⎝⎞⎠:= Find(C1 , C2) C1 = 410 kN⋅m2 C2 = −720 kN⋅m3 1273. Equation of Elastic Curve : aowo24:= − a1wo120⋅L2:= a2A6:= a3 := C1 a4 := C2ao −6.251m= kN a1 0.421m2= kN a2 = 91.67 kN Ansa3 = 410.00 kN⋅m2 a4 = −720.00 kN⋅m3 Ansv1E⋅ Iao Ψ(x − 0)⋅ 4 a1 Ψ(x − L1)+ ⋅ 5 a2 Ψ(x − L1)+ ⋅ 3 a+ 3⋅x + a4 ⎛⎝⎞⎠= Ans 1274. Problem 12-49Determine the displacement at C and the slope at A of the beam.Use Macaulay Function: Ψ(x) := Φ(x)⋅xGiven: kNL1 := 2m L2 := 3mwo :=150mSolution: L := L1 + L2Support Reactions :+ ΣFy=0; (1)A + B − wo⋅L1wo2− ⋅L2 = 0ΣΜA=0;wo2⋅ 2− L1wo2⋅L2L23⎛⎜⎝⎞⎠+ ⋅ − B⋅L2 = 0 (2)Solving Eqs. (1) and (2): B+ 26L2−3L12 L2:= ⋅wo A := wo⋅ (L1 + 0.5L2) − BB = −25 kN A = 550 kNMoment Function :M(x)wo2− Ψ(x − 0)⋅ 2wo6L2= + ⋅Ψ(x − L1)3 + A⋅Ψ(x − L1)Slope and Elastic Curve : EId2⋅vdx2⋅ = M(x) EI := kN⋅m2 EIo := 1EId2vdx2⋅wo2− Ψ(x − 0)⋅ 2wo6L2= + ⋅Ψ(x − L1)3 + A⋅Ψ(x − L1)EIdvdx⋅wo6− Ψ(x − 0)⋅ 3wo24⋅L2+ ⋅Ψ(x − L1)4A2= + ⋅Ψ(x − L1)2 + C1 (3)EI⋅vwo24− Ψ(x − 0)⋅ 4wo120⋅L2+ ⋅Ψ(x − L1)5A6= + ⋅Ψ(x − L1)3 + C1⋅x + C2 (4)Boundary Conditions : v=0 at x=L1 GivenFrom Eq. (4): (5) 0wo24C= − ⋅ L14 + 0 + 0 + 1⋅L1 + C2Also v=0 at x=L.From Eq. (4): 0wo24− ⋅L4wo120⋅L2+ ⋅ (L − L1)5A6= + ⋅ (L − L1)3 + C1⋅L + C2 (6)Solving Eqs. (5) and (6): Guess C1 := 1kN⋅m2 C2 := 1kN⋅m3C1C2⎛⎜⎜⎝⎞⎠:= Find(C1 , C2) C1 = 410 kN⋅m2 C2 = −720 kN⋅m3 1275. Equation of Elastic Curve and Slope :v(x)1EIwo24− Ψ(x − 0)⋅ 4wo120⋅L2+ ⋅Ψ(x − L1)5A6+ ⋅Ψ(x − L1)3 + C1⋅x + C2⎛⎜⎝⎞⎠:= (7)Displacement at C: Substitute x=0 into Eq.(7). v(0m) −720mEIo= Ansθ(x)1EIwo6− Ψ(x − 0)⋅ 3wo24⋅L2+ ⋅Ψ(x − L1)4A2+ ⋅Ψ(x − L1)2 + C1⎛⎜⎝⎞⎠:=Slope at A: Substitute x=L1 into Eq.(3). θ(L1) 2101EIo= Ans 1276. Problem 12-50Determine the equation of the elastic curve. Specify the slope at A. EI is constant.Use Macaulay Function: Ψ(x) := Φ(x)⋅xGiven: L := m EIo := 1A + B − w⋅L = 0wkNm:=Solution:Support Reactions :+ΣFy=0; (1)ΣΜA=0; −(0.5L)⋅wL − B⋅L = 0 (2)Solving Eqs. (1) and (2): B := −0.5wLA := w⋅L − B A = 1.5w⋅LMoment Function :M(x) −(0.5w) Ψ(x − 0)⋅ 2 (−0.5w) Ψ(x − L)= − ⋅ 2 + A⋅Ψ(x − L)Slope and Elastic Curve :w L4 ⋅ − ⎛⎜⎝Ψ(x − L)+ ⋅ 2 C+ 1 ⎛⎜⎝AnsEI := kN⋅m2EId2vdx2⋅ −(0.5w) Ψ(x − 0)⋅ 2 (−0.5w) Ψ(x − L)= − ⋅ 2 + A⋅Ψ(x − L)EIdvdx⋅w6− Ψ(x − 0)⋅ 3w6Ψ(x − L)+ ⋅ 3A2Ψ(x − L)+ ⋅ 2 C= + 1 (3)EI⋅vw24− Ψ(x − 0)⋅ 4w24Ψ(x − L)+ ⋅ 4A6Ψ(x − L)+ ⋅ 3 C= + 1⋅x + C2 (4)Boundary Conditions : Givenv=0 at x=L From Eq. (4): 0w24= − ⋅L4 + 0 + 0 + C1⋅L + C2 (5)v=0 at x=2L From Eq. (4): 0w24− ⋅ (2L)4w24+ ⋅L4A6= + ⋅L3 + C1⋅ (2L) + C2 (6)Solving Eqs. (5) and (6): Guess C1 := 1kN⋅m2 C2 := 1kN⋅m3C1C2⎛⎜⎜⎝⎞⎠:= Find(C1 , C2) C113= w⋅L3 C2724= − w⋅L4Equation of Elastic Curve and Slope :v(x)1EIw24− ⋅x4w24Ψ(x − L)+ ⋅ 43w⋅L12Ψ(x − L)+ ⋅ 313+ w⋅L3⋅x724⎞⎠:= Ansθ(x)1EIw6− Ψ(x − 0)⋅ 3w6Ψ(x − L)+ ⋅ 39w⋅L12⎞⎠:=Slope at A: Substitute x=L into Eq.(3). θ(L)16w⋅L3EI=EId2⋅vdx2⋅ = M(x) 1277. Problem 12-51Determine the equation of the elastic curve. Specify the deflection at C. EI is constant.Use Macaulay Function: Ψ(x) := Φ(x)⋅xGiven: L := mwkNm:=Solution:Support Reactions :+ΣFy=0; A + B − w⋅L = 0 (1)ΣΜA=0; −(0.5L)⋅wL − B⋅L = 0 (2)Solving Eqs. (1) and (2): B := −0.5wLA := w⋅L − B A = 1.5w⋅LMoment Function :M(x) −(0.5w) Ψ(x − 0)⋅ 2 (−0.5w) Ψ(x − L)= − ⋅ 2 + A⋅Ψ(x − L)Slope and Elastic Curve :w L4 ⋅ − ⎛⎜⎝AnsEI := kN⋅m2 EIo := 1EId2vdx2⋅ −(0.5w) Ψ(x − 0)⋅ 2 (−0.5w) Ψ(x − L)= − ⋅ 2 + A⋅Ψ(x − L)EIdvdx⋅w6− Ψ(x − 0)⋅ 3w6Ψ(x − L)+ ⋅ 3A2Ψ(x − L)+ ⋅ 2 C= + 1 (3)EI⋅vw24− Ψ(x − 0)⋅ 4w24Ψ(x − L)+ ⋅ 4A6Ψ(x − L)+ ⋅ 3 C= + 1⋅x + C2 (4)Boundary Conditions : Givenv=0 at x=L From Eq. (4): 0w24= − ⋅L4 + 0 + 0 + C1⋅L + C2 (5)v=0 at x=2L From Eq. (4): 0w24− ⋅ (2L)4w24+ ⋅L4A6= + ⋅L3 + C1⋅ (2L) + C2 (6)Solving Eqs. (5) and (6): Guess C1 := 1kN⋅m2 C2 := 1kN⋅m3C1C2⎛⎜⎜⎝⎞⎠:= Find(C1 , C2) C113= w⋅L3 C2724= − w⋅L4Equation of Elastic Curve :v(x)1EIw24− ⋅x4w24Ψ(x − L)+ ⋅ 43w⋅L12Ψ(x − L)+ ⋅ 313+ w⋅L3⋅x724⎞⎠:= Ansv(0)724−w⋅L4EI=EId2⋅vdx2⋅ = M(x) 1278. Problem 12-52Determine the equation of the elastic curve. Specify the slope at B. EI is constant.Use Macaulay Function: Ψ(x) := Φ(x)⋅xGiven: kNL := mw:=mSolution:Support Reactions :+ΣFy=0; A + B − w⋅L = 0 (1)ΣΜA=0; −(0.5L)⋅wL − B⋅L = 0 (2)Solving Eqs. (1) and (2): B := −0.5wLA := w⋅L − B A = 1.5w⋅LMoment Function :M(x) −(0.5w) Ψ(x − 0)⋅ 2 (−0.5w) Ψ(x − L)= − ⋅ 2 + A⋅Ψ(x − L)Slope and Elastic Curve :EI := kN⋅m2 EIo := 1w L4 ⋅ − ⎛⎜⎝Ψ(x − L)+ ⋅ 2 C+ 1 ⎛⎜⎝AnsEId2vdx2⋅ −(0.5w) Ψ(x − 0)⋅ 2 (−0.5w) Ψ(x − L)= − ⋅ 2 + A⋅Ψ(x − L)EIdvdx⋅w6− Ψ(x − 0)⋅ 3w6Ψ(x − L)+ ⋅ 3A2Ψ(x − L)+ ⋅ 2 C= + 1 (3)EI⋅vw24− Ψ(x − 0)⋅ 4w24Ψ(x − L)+ ⋅ 4A6Ψ(x − L)+ ⋅ 3 C= + 1⋅x + C2 (4)Boundary Conditions : Givenv=0 at x=L From Eq. (4): 0w24= − ⋅L4 + 0 + 0 + C1⋅L + C2 (5)v=0 at x=2L From Eq. (4): 0w24− ⋅ (2L)4w24+ ⋅L4A6= + ⋅L3 + C1⋅ (2L) + C2 (6)Solving Eqs. (5) and (6): Guess C1 := 1kN⋅m2 C2 := 1kN⋅m3C1C2⎛⎜⎜⎝⎞⎠:= Find(C1 , C2) C113= w⋅L3 C2724= − w⋅L4Equation of Elastic Curve and Slope :v(x)1EIw24− ⋅x4w24Ψ(x − L)+ ⋅ 43w⋅L12Ψ(x − L)+ ⋅ 313+ w⋅L3⋅x724⎞⎠:= Ansθ(x)1EIw6− Ψ(x − 0)⋅ 3w6Ψ(x − L)+ ⋅ 39w⋅L12⎞⎠:=Slope at A: Substitute x=2L into Eq.(3). θ(2L)112−w⋅L3EI=EId2⋅vdx2⋅ = M(x) 1279. Problem 12-53The shaft is made of steel and has a diameter of 15 mm. Determine its maximum deflection. Thebearings at A and B exert only vertical reactions on the shaft. Est = 200 GPa.Use Macaulay Function: Ψ(x) := Φ(x)⋅xGiven: a := 200mm P1 := 250Nb := 300mm P2 := 80Ndo := 15mm E := 200GPaSolution: L := 2a + bSection Property: I464πdo:=Support Reactions :+ΣFy=0; A + B − P1 − P2 = 0 (1)ΣΜB=0; A⋅L − P1⋅ (a + b) − P2⋅b = 0 (2)P1⋅ (a + b) + P2⋅aSolving Eqs. (1) and (2): A:= A = 201.429NLB := P1 + P2 − A B = 128.571NMoment Function :M(x) = A⋅Ψ(x − 0) − P1⋅Ψ(x − a) − P2⋅Ψ(x − a − b)Slope and Elastic Curve :E⋅ Id2vdx2E⋅ Id2⋅vdx2⋅ = M(x)⋅ = A⋅Ψ(x − 0) − P1⋅Ψ(x − a) − P2⋅Ψ(x − a − b)E⋅ Idvdx⋅A2Ψ(x − 0)⋅ 2P12Ψ(x − a)− ⋅ 2P22Ψ(x − a − b)− ⋅ 2 C= + 1 (1)E⋅ I⋅vA6Ψ(x − 0)⋅ 3P16Ψ(x − a)− ⋅ 3P26Ψ(x − a − b)− ⋅ 3 C= + 1⋅x + C2 (2)Boundary Conditions :v=0 at x=0 From Eq. (2): 0 = 0 − 0 + 0 + C2 C2 := 0v=0 at x=L From Eq. (2): 0A6⋅L3P16− ⋅ (L − a)3P26= − ⋅a3 + C1⋅ (L)C1A6− ⋅L2P16L+ ⋅ (L − a)3P26L:= + ⋅a3 C1 = −8.857N⋅m2Equation of Elastic Curve :v(x)1E⋅ IA6⋅x3P16Ψ(x − a)− ⋅ 3P26Ψ(x − a − b)− ⋅ 3 C+ 1⋅x⎛⎜⎝⎞⎠:= (3)Maximum Deflection: Assume vmax occurs at a < x < a+b. Given 1280. dv/dx=0 at x', From Eq. (1): 0A2⋅x'2P12= − ⋅ (x' − a)2 − 0 + C1 (4)Solving Eq.(3): Guess x' := 300⋅mm x' := Find(x')x' = 330.05mmFor vmax, substitute x' into Eq.(3). v(x') = −3.64mm AnsNote: The negative sign indicates downward displacement. 1281. Problem 12-54Determine the slope and deflection at C. EI is constant.Given: L1 := 8m L2 := 4m P := 75kNSolution:Support Reactions : L := L1 + L2+ ΣFy=0; A + B − P = 0 (1)ΣΜB=0; A⋅L1 + P⋅L2 = 0 (2)Solving Eqs. (1) and (2): A −PL2L1:= ⋅ B PL1 + L2L1:= ⋅A = −37.5 kN B = 112.5 kNM/EI Diagram:Set EI := kN⋅m2 EIo := 1x1 := 0 , 0.01⋅L1 .. L1 x2 := L1 , 1.01⋅L1 .. LM'1(x1) (A⋅x1) 1⋅ := M'2 x2 ( ) A x2 ⋅ B x2 L1 − ( ) ⋅ + ⎡⎣EI⎤⎦1EI:= ⋅0 5 100200Distance (m)M/EI (1/m)M'1(x1)M'2(x2)x1, x2Moment-Area Theorems :Slopes :tB.A12M'1(L1)