ELEMEN MESIN(RI.1232) LECTURE II Tegangan Konsep Tegangan Tegangan Tarik dan Tekan Tegangan lentur Tegangan geser dan puntir Dosen: Fahmi Mubarok, ST., MSc. Metallurgy Laboratory Mechanical Engineering ITS- Surabaya 2008 http://www.its.ac.id/personal/material.php?id=fahmi Definition Fahmi Mubarok V Start with internal system of forces as shown below to get proper signs for V, N and M. - Tegangan (stress) intensitas gaya persatuan luas - Regangan (strain) deformasi (perubahan bentuk) akibat tegangan yang bekerja σ= P Ao ∆l ε= lo Jenis-jenis tegangan 1. Tegangan tarik dan tekan (Tensile dan compression stress). 2. Tegangan Geser (Shears stress) disini termasuk tegangan puntir (Torsional Stress ). 3. Tegangan Bending / lengkung ( Bending stress ). 4. Tegangan kombinasi ( Combination stress ). Mech. Eng. Dept. ITS Surabaya I 2 Various of Average Normal Stress Fahmi Mubarok Mech. Eng. Dept. ITS Surabaya I 3 Tegangan Tarik dan Tekan Fahmi Mubarok Untuk membandingkan spesimen dengan berbagai ukuran, maka digunakan konsep tegangan teknik F = beban yang diberikan tegak lurus terhadap penampang spesimen Ao = luas penampang awal sebelum beban diberikan Tegangan dan regangan akan memberikan nilai positif pada kondisi tegangan tarik sedang pada kondisi tegangan tekan akan memberikan nilai negatif ∆l = perpanjangan lo = panjang awal sebelum beban diberikan Mech. Eng. Dept. ITS Surabaya I 4 Normal Stress due to Bending Moment Fahmi Mubarok Key Points: 1. Internal bending moment causes beam to deform. 2. For this case, top fibers in compression, bottom in tension. Mech. Eng. Dept. ITS Surabaya I 5 Normal Stress due to Bending Moment Fahmi Mubarok Internal bending moment, lb-in Bending stress, psi Distance from NA to point of interest, in My σ = I Moment of inertia, in4 I 6 Mech. Eng. Dept. ITS Surabaya Design of Beams Fahmi Mubarok Mech. Eng. Dept. ITS Surabaya I 7 Tegangan Geser dan Puntir Fahmi Mubarok Mech. Eng. Dept. ITS Surabaya I 8 Average Shear Stress Fahmi Mubarok • Forces P and P’ are applied transversely to the member AB. • Corresponding internal forces act in the plane of section C and are called shearing forces. • The resultant of the internal shear force distribution is defined as the shear of the section and is equal to the load P. • The corresponding average shear stress is, τ ave = P A • Shear stress distribution varies from zero at the member surfaces to maximum values that may be much larger than the average value. • The shear stress distribution cannot be assumed to be uniform. Mech. Eng. Dept. ITS Surabaya I 9 Average Shear Stress Fahmi Mubarok Single Shear Double Shear P F τ ave = = A A Mech. Eng. Dept. ITS Surabaya P F τ ave = = A 2A I 10 Torsion Fahmi Mubarok Torque is a moment that tends to twist a member about its longitudinal axis. 4 J=1 π c 2 4 4 J=1 π c − c 2 1 2 Mech. Eng. Dept. ITS Surabaya ( ) I 11 Shear and Bending Moment Diagrams Fahmi Mubarok • Variation of shear and bending moment along beam may be plotted. • Determine reactions at supports. • Cut beam at C and consider member AC, V = + P 2 M = + Px 2 • Cut beam at E and consider member EB, V = − P 2 M = + P (L − x ) 2 • For a beam subjected to concentrated loads, shear is constant between loading points and moment varies linearly. Mech. Eng. Dept. ITS Surabaya I 12 Sample Problem Bending and Shear SOLUTION: Fahmi Mubarok • Treating the entire beam as a rigid body, determine the reaction forces • Section the beam at points near supports and load application points. Apply equilibrium analyses on resulting free-bodies to determine internal shear forces and bending couples • Identify the maximum shear and bending-moment from plots of their distributions. • Apply the elastic flexure formulas to determine the corresponding maximum normal stress. Mech. Eng. Dept. ITS Surabaya I 13 For the timber beam and loading shown, draw the shear and bend-moment diagrams and determine the maximum normal stress due to bending. Sample Problem 5.1 SOLUTION: Fahmi Mubarok • Treating the entire beam as a rigid body, determine the reaction forces from ∑F y = 0 = ∑MB : R B = 46 kN R D = 14 kN • Section the beam and apply equilibrium analyses on resulting free-bodies ∑ Fy = 0 ∑ M1 = 0 ∑ Fy = 0 ∑M2 = 0 − 20 kN − V1 = 0 V1 = − 20 kN M1 = 0 V2 = − 20 kN =0 M 2 = − 50 kN ⋅ m (20 kN )(0 m ) + M 1 = 0 − 20 kN − V2 = 0 (20 kN )(2.5 m ) + M 2 V3 = + 26 kN V4 = + 26 kN V5 = − 14 kN V6 = − 14 kN M 3 = − 50 kN ⋅ m M 4 = + 28 kN ⋅ m M 5 = + 28 kN ⋅ m M6 = 0 I 14 Mech. Eng. Dept. ITS Surabaya Sample Problem 5.1 Fahmi Mubarok • Identify the maximum shear and bendingmoment from plots of their distributions. Vm = 26 kN M m = M B = 50 kN ⋅ m • Apply the elastic flexure formulas to determine the corresponding maximum normal stress. 2 1 (0 .080 m )(0 .250 m )2 S =1 b h = 6 6 = 833 .33 × 10 − 6 m 3 MB 50 × 10 3 N ⋅ m σm = = S 833 .33 × 10 − 6 m 3 σ m = 60 .0 × 10 6 Pa Mech. Eng. Dept. ITS Surabaya I 15 ELEMEN MESIN (RI.1232) LECTURE III Sambungan -Sambungan Keling Dosen: Fahmi Mubarok, ST., MSc. Metallurgy Laboratory Mechanical Engineering ITS- Surabaya 2008