ENGC6353 Dr.Mohammed Arafa Page 1 Design Example 3 Rectangular Silo Design a single rectangular concrete silo for storing peas. The bottom is a symmetrical pyramidal Hopper. The silo walls rest on the Hopper base which is supported by four columns. The Roof load ( DL = 150 kg/m 2 and LL= 100 kg/m 2 Use ). ' 2 2 350 / , 4200 / c y f kg cm f kg cm = = Solution For Peas 3 ' 800 / 25 0.296 o kg m γ φ µ = = = b=6m a = 4 m An Above Hopper b=6m a=6m Openning0.5x0.5m 30m 5m 7m φ50cm 3m ENGC6353 Dr. Mohammed Arafa Page 2 Assume angle of response = =25 2 3tan 25 1.4 1.0 3 1 sin 25 0.577 4 1.0 4 4 ' 1.0 4 2 4 6 ' ' 4.8 1.2 4 6 4 s s a b b h m h m k a R m a R m a a R m ρ φ = = ⇒ ≅ = − = = = = = = × × = = = = + Overpressure Factor C 1 d d / 40/10 4 upper H c 1.5 lower 2/3 H c 1.85 Hooper 1.5 d H D c = = = = = d At the bottom of the silos ( ) ' 2 2 2 2 At the bottom of the silos 30-1.0 29.0 1 ' ( 1.0) 4.65 t/m 0.577 4.65 2.7 t/m ( 1.2) 5.53t/m 0.577 5.53 3.2t/m kY R Y m R q e k p kq For short wall R q p kq For long wall R q p kq µ γ µ − = = ( = − ¸ ¸ = = = = = × = = = = = × = Vertical Loads Due to Friction ( ) ( ) ( ) Short Wall 0.8 30 4.65 1.0 19.35ton Long Wall 0.8 30 5.53 1.2 22.16ton Friction V Y q R V V γ = − = × − × = = × − × = Wall tension and bending moment ( ) ( ) , , Short Wall 1.7 1.85 3.2 6 2 30.2ton/m Long Wall 1.7 1.85 2.7 4 2 17.0ton/m a u b u F F = × × = = × × = Frame action analysis using moment distribution Analysis Assume wall thickness h=30cm 25cm ENGC6353 Dr. Mohammed Arafa Page 3 The moment distribution is computed for an idealized rectangular frame 6.3 by 4.4 m Using symmetry a 2 2 Short Wall =0.465I 4.3 2 2 Long Wall =0.317I 6.3 0.465 DF 0.6 0.465 0.317 0.4 a a b b b I I K L I I K L DF = = = = = ≅ + ≅ Short Wall Long Wall DF 0.6 0.4 FEM 4.16 -10.6 Balancing 3.86 2.58 FINAL 8.0 -8.0 Assume fillit (hunch) at the corner =25cm Negative moment will be calculated at the face of the hunch 2 b,-ve 2 a,-ve M 8.0 3.2 0.4 / 2 10.1 0.4 4.2 . M 8.0 2.7 0.4 / 2 5.8 0.4 5.9 . t m t m = + × − × = = + × − × = Check for thickness ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ' 2 , 2 5 3 2 , 2 5 3 2 , 2 6 2 2 350 37.4 / For long Wall 6 4.2 10 1 17 10 11 / 1.7 1.85 30 100 30 100 For short Wall 6 5.9 10 1 30.2 10 15.7 / 1.7 1.85 30 100 30 100 The wall thicknes t b r c t b r t b r T M f f f kg cm bt bt f kg cm f f kg cm f = + ≤ = = = ( × × ( = + = < × ( ¸ ¸ ( × × ( = + = < × ( ¸ ¸ sis oK Design for Reinforcement Long Wall negative moment M Check for small eccentricity -ve 3.2 t/m 2 2 . 7 t / m 2 6.3m 4.3m 7.9 t.m -8.0 -8.0 - 1 . 5 25cm ENGC6353 Dr. Mohammed Arafa Page 4 ( ) 4.2 100 24.7 '' 15 5.7 9.3 17 2 u u M h e d F = = = > − = − = Small eccentricity approach can not be used 3 2 st '' 15 5.7 9.3 2 Direct tension reinforcement 17 10 A 4.5 / 0.9 4200 y h e d T cm m f φ = − = − = × = = = × Bending Moment Reinforcement ( ) ( ) ( )( )( ) ' , 5 2 2 ( ) 2 , 4.2 17 9.3/100 2.6 . d=30-5.7=24.3 2.61 10 2.6 0.85 350 1 1 0.00117 4200 100 24.3 350 0.00117 100 24.3 2.85 4.5 2.85 7.35 / 14@20 u ve ve s ve s total M t m A cm A cm m use cm ρ φ − − − = − × = ( ⋅ ⋅ ( = − − = ⋅ ⋅ ( ¸ ¸ = = = + = Design for Positive Moment at Midspan ( ) ( ) ( )( )( ) ' , 5 2 2 ( ) 2 , 7.9 17 9.3/100 6.32 . d=30-5.7=24.3 2.61 10 6.32 0.85 350 1 1 0.00289 4200 100 24.3 350 0.00289 100 24.3 7.0 4.5 7 11.5 / 16@15 u ve ve s ve s total M t m A cm A cm m use cm ρ φ + + + = − × = ( ⋅ ⋅ ( = − − = ⋅ ⋅ ( ¸ ¸ = = = + = Design for Short Wall negative moment M Check for small eccentricity -ve ( ) 5.9 100 19.5 '' 15 5.7 9.3 30.2 2 u u M h e d F = = = > − = − = Small eccentricity approach can not be used ENGC6353 Dr. Mohammed Arafa Page 5 3 2 st '' 15 5.7 9.3 2 Direct tension reinforcement 30.2 10 A 8.0 / 0.9 4200 y h e d T cm m f φ = − = − = × = = = × Bending Moment Reinforcement ( ) ( ) ( )( )( ) ' , 5 2 2 ( ) 2 , 5.9 30.2 9.3/100 3.0 . d=30-5.7=24.3 2.61 10 3.0 0.85 350 1 1 0.00137 4200 100 24.3 350 0.00137 100 24.3 303 8.0 3.3 11.3 / 16@15 u ve ve s ve s total M t m A cm A cm m use cm ρ φ − − − = − × = ( ⋅ ⋅ ( = − − = ⋅ ⋅ ( ¸ ¸ = = = + = Design at Mid-span Design of the Hopper Walls The pressure changes very little with depth of the hopper, so use the pressure at the top of the hopper with Cd=1.35 2 , 2 , 2 , 2 , 1.35 4.65 6.28t/m 1.35 0.577 4.65 3.6t/m 1.35 5.53 7.47 t/m 1.35 0.577 5.53 4.3t/m a des a des b des b des q p q q = × = = × × = = × = = × × = 1 1 Angleof slopes 3 tan 48 3 0.3 3 tan 60.5 2 0.3 a a α α − − | | = = | − \ . | | = = | − \ . 2 2 2 , 2 2 2 , 3.6sin 48 6.28cos 48 4.8t/m 4.3sin 60.5 7.47cos 60.5 5.0 t/m a des b des q q α α = + = = + = ENGC6353 Dr. Mohammed Arafa Page 6 Horizontal Ultimate tensile forces ( )( ) ( ) ( )( ) ( ) 2 2 1.7 5.0 6/2 sin 48 =19.0t/m 1.7 4.8 4/2 sin 60.5 =14.2t/m tau tbu F F = = The own weight of the Hopper and its contents ( )( )( ) ( )( )( ) 3 4 6 3 0.8 60 3 4 6 3 0.2 2.5 38 L L W ton W ton π π = × = = × × = For simplicity neglect the opening area at the bottom of the hopper. Hopper side A a and A b can be calculated as: ( ) ( ) 2 2 2 1 4 3 6 2 1 6 2 6 2 6 1/ 4 a b a b a b A m A m A A m c c = × = = × = = = = = ( ) ( ) ( )( ) ( ) ( ) ( )( ) , , 1.7 1.4 1.7 0.25 60 6 6.28 1.4 0.25 38 34.6 sin 4sin 48 1.7 1.4 1.7 0.25 60 6 7.47 1.4 0.25 38 22 sin 6sin 60.5 a L a a des b g mau a b L b b des b g mbu b c W A q c W F ton a c W A q c W F ton b α α + + × + × + = = = + + × + × + = = = Hopper wall bending can be computed using Tables for triangular slabs: For Hopper wall A ( ) 2 2 4.3 6.3/ 2 3 4.35 / 1.0 a m c m a c = = + = ≅ From table 16.4 in Appendix At the centre of the top edge n x = -0.209 and n y ( ) ( ) ( ) ( ) 2 2 1.7 0.209 4.8 4.3 / 64 0.493 . 1.7 1.255 4.8 4.3 / 64 2.89 . xau yau M t m M t m = = = = =-1.255 This slab is to be designed for bending and tensile force similarly as shown above. ENGC6353 Dr. Mohammed Arafa Page 7 Design of the edge beam Dowels are provided to transfer the vertical loads from hopper edge beam into the vertical walls 3 2 , sin 34.6sin 48 25.7 / 25.7 10 6.8 / 0.9 4200 mau a st dowels T F ton m A cm m α = = = × = = × Since the edge beam is to be joining the vertical wall using dowels. The upper wall shear and horizontal components of the hopper are assumed to be in equilibrium. Thus no horizontal load is carried by the edge beam. Its only purpose is to simplify construction. Minimum longitudinal steel and shear stirrups are provided Vertical Wall The vertical walls are analyzed as deep girder (strut and tie analysis can be used) to carry vertical the following vertical loads: From dowel 25.7 ton/m Friction 1.7(19.35) = 32.9 ton/m Wall weight, 1.4(2.5)(0.3)(30)= 31.5 ton/m Total = 90 ton/m 8 × 30 − 4.2 p= = kq 0.53) ×1.2 ton/m = = Fb .u 1.16ton Wall tension and bending moment Short Wall Long Wall Fa .2m b 4+6 4 Overpressure Factor C d H / D 40 /10 4 = = upper H1 cd = 1.65 t/m 2 2.u 1.2) q = = 5.53 = t/m 2 Vertical Loads Due to Friction Friction Short Wall Long Wall V = V = V = (γY − q ) R ( 0.0 29.65 = t/m 2 For long wall (R 1.8 R= = 1.0m 4 4 a' R= = 1.0m b 4 2× 4× 6 a' a' = = 4.2 ) × 6 2 30.0) q = = 4.577 × 5.7 p= = kq 0.85 Hooper cd = 1.53t/m 2 3.577 k = 25 = a 4 R a= = = 1.8 × 30 − 5.0m = 3 1 − sin 0.5 25cm At the bottom of the silos At the bottom of the silos Y 30 -1.4m ⇒ hs ≅ 1.7 (1.35 ton 22.85 × 3.85 × 2.65) ×1. Mohammed Arafa Page 2 .Assume angle of response ρ =φ =25 2 hs= 3 tan 25 1.7 ) × 4 2 17.0 = ( 0.0m = = γR −( µ ' kY R ) q = 1 − e µ 'k p = kq For short wall (R 1.5 lower 2/3 H cd = 1.0ton/m = = Frame action analysis using moment distribution Analysis Assume wall thickness h=30cm ENGC6353 Dr.2 = 19.577 × 4.7 (1. 16 3.9 t .86 8.4 = 4.2 t/m2 6.8 × 0.58 -8.m Check for thickness T 6M f t .4 -1.0 -10.3 2.85 ( 30 )(100 ) ( 30 ) (100 ) For short Wall 30.0 -8.4 kg / cm 2 bt bt For long Wall 25cm 17 ×103 6 ( 4.m M a.6 2.3m 7.465 + 0.0 + 3.9 ×105 ) 1 =15.4 m Using symmetry Short Wall Long Wall DFa = = Ka Kb = 2I 2I = =0.The moment distribution is computed for an idealized rectangular frame 6.317 DFb ≅ 0.7 t/m2 4.b = + 2 1.4 = 5.465 ≅ 0.85 ( 30 )(100 ) ( 30 ) (100 ) The wall thickness is oK Design for Reinforcement Long Wall negative moment M -ve Check for small eccentricity ENGC6353 Dr.b = + 2 ≤ f r = 2 f c' = 2 350 = 37.1× 0.6 0.6 4.4 Short Wall Long Wall 0.9 t.0 + 2.42 / 2 − 5.7 × 0.-ve = 8.3 by 4.-ve = 8.317I Lb 6.7 kg / cm 2 < f r f t .3m 3.b = + 11 2 1.2 ×105 ) 1 =kg / cm 2 < f r f t .2 ×103 6 ( 5.0 DF FEM Balancing FINAL 0.7 ×1.0 Assume fillit (hunch) at the corner =25cm Negative moment will be calculated at the face of the hunch M b.2 t .5 -8. Mohammed Arafa Page 3 .3 2I 2I = =0.465I La 4.2 × 0.7 ×1.m 0.42 / 2 − 10. 3 /100 = 2.9 (100 ) h = = 19.+ve = 7.35 A s .total = + 2.7 = 9.61 ⋅105 ( 2.5 7.e= M u 4.−ve = 4.00117 2 4200 100 ⋅ ( 24.85 = cm 2 / m use φ14@ 20cm Design for Positive Moment at Midspan M u' .2 (100 ) h = = 24.7 = 9.3) ⋅ 350 2 = (= 2.3 2 Direct tension reinforcement = A st T 17 ×103 = = 4.m d=30-5.61 ⋅105 ( 6.32 ) 0.9 × 4200 Bending Moment Reinforcement M u' . Mohammed Arafa Page 4 .3 /100 = 6.32 t .5 + 7 = 11.5 > − d '' = 15 − 5.3) 4.0 cm A s ( +ve ) 0.3 30.00117 )(100 )( 24.3) ⋅ 350 2 = (= 7.3 2.total = 4.2 2 Fu Small eccentricity approach can not be used ENGC6353 Dr.3 17 2 Fu Small eccentricity approach can not be used h e = − d '' =15 − 5.00289 )(100 )( 24.7 =9.m d=30-5.7 > − d '' = 15 − 5.6 ) 0.85cm A s ( −ve ) 0.5 cm 2 / m φ f y 0.3) = ρ +ve A s .7=24.85 ⋅ 350 1 − 1 − = 0.5 cm 2 / m use φ16@15cm Design for Short Wall negative moment M -ve Check for small eccentricity e= M u 5.85 ⋅ 350 1 − 1 − = ρ −ve = 0.7=24.6 t .00289 2 4200 100 ⋅ ( 24.2 − 17 × 9.9 − 17 × 9.3 2. des = 4.3 3 = tan −1 α a = 60.35 × 5.00137 2 4200 100 ⋅ ( 24.h e = − d '' =15 − 5.2 ×103 = = 8.577 × 4.3sin 2 60.des = ENGC6353 Dr.6sin 2 48 + 6.0 ) 0.−ve =5.53 = 7.35 × 4.7 =9. Mohammed Arafa Page 5 .des = 1.28cos 2 48 = 4.35 × 0.7=24.0 + 3.3cm 2 / m use φ16@15cm Design at Mid-span Design of the Hopper Walls The pressure changes very little with depth of the hopper.3 3.85 ⋅ 350 1 − 1 − = ρ −ve = 0.3) A s .5 = t/m 2 5.3 = 11.8t/m 2 qα a .00137 )(100 )( 24.35 × 0.m d=30-5.2 × 9.0 cm 2 / m φ f y 0.3 /100 =3.0 t .total = 8.65 = 6.28 t/m 2 p a .9 − 30.3 2 Direct tension reinforcement = A st T 30.9 × 4200 Bending Moment Reinforcement M u' .577 × 5.0 qαb .6 t/m 2 qb .47 t/m 2 qb .53 = 4.35 q a .3) ⋅ 350 2 = (= 303cm A s ( −ve ) 0.des =1.des =1.3 2.3 t/m 2 Angle of slopes 3 = tan −1 α a = 48 3 − 0.47 cos 2 60.65 = 3.5 + 7.5 2 − 0.des = 1.61 ⋅105 ( 3. so use the pressure at the top of the hopper with Cd=1. ENGC6353 Dr.Horizontal Ultimate tensile forces Ftau = 1.8 ) = π 3 60 ton W L = π 3 ( 4 × 6 )( 3)( 0.m = 1.0 )( 6/2 ) sin ( 48 ) =19. Mohammed Arafa Page 6 .5 ) =14.7 ( 0.8 ) 4.4cbW g 1.7 (1.5 Hopper wall bending can be computed using Tables for triangular slabs: For Hopper wall A a = 4.35m a / c ≅ 1.25 × 60 + 6 × 6.7 (cbW L + Ab qb .493 t .25 × 60 + 6 × 7.3m = c ( 6.32 / 64 ) 0.5 )= 38 ton For simplicity neglect the opening area at the bottom of the hopper.8 ) 4.47 ) + 1.7 (c aW L + Aaq a .4 ( 0.0t/m 2 Ftbu = 1.209 ( 4.des ) + 1.8 )( 4/2 ) sin ( 60.25 )( 38 ) = = 34.6 ton a sin α a 4sin 48 1.3 / 2 = ) + 32 2 4.28 ) + 1.255 ( 4.209 and n y =-1.7 ( 0.4 ( 0.4cbW g 1.7 ( 4.32 / 64 ) 2.7 ( 5.2 × 2.0 From table 16.m = This slab is to be designed for bending and tensile force similarly as shown above.255 M xau M yau 1.des ) + 1.4 in Appendix At the centre of the top edge n x = -0. Hopper side A a and A b can be calculated as: 1 ( 4 × 3)= 6 m 2 2 1 Ab = ( 6 × 2 )= 6 m 2 2 = = A a Ab 6 m 2 Aa = c= c= 1/ 4 a b Fmau Fmbu 1.25 )( 38 ) = = 22 ton b sin αb 6sin 60.7 ( 0.2t/m 2 The own weight of the Hopper and its contents W L =( 4 × 6 )( 3)( 0.89 t . The upper wall shear and horizontal components of the hopper are assumed to be in equilibrium.7 ton/m Friction 1. Thus no horizontal load is carried by the edge beam.8 cm 2 / m 0.7(19.4(2.9 ton/m Wall weight.35) = 32.6sin 48 25.5 ton/m 90 ton/m ENGC6353 Dr. Mohammed Arafa Page 7 . 1.7 ×103 = 6.9 × 4200 Since the edge beam is to be joining the vertical wall using dowels.3)(30)= Total = 31.5)(0. Minimum longitudinal steel and shear stirrups are provided Vertical Wall The vertical walls are analyzed as deep girder (strut and tie analysis can be used) to carry vertical the following vertical loads: From dowel 25.dowels 25.7 ton / m T = = = A st . Its only purpose is to simplify construction.Design of the edge beam Dowels are provided to transfer the vertical loads from hopper edge beam into the vertical walls = Fmau sin α a 34.