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Shear Stress and Yield

   

Added on  2019-09-23

8 Pages428 Words190 Views
Answer (2): Given data:d=0.2mL=1mT=20πStress at point (A) ∑Fx=0Vx−20π=0∑Fy=0Vy=−20π∑My=0σyb=NA=200.1π=200π=63.66σy=MyII=πR44=π(0.1)44=0.25πx10−4m4y=4R3πσy=20πx1x4x0.13πx0.25πx10−4¿3.395305x104σy=63.66+33953.05σy=34016.71Find, shear stressτy=VyQItQ=∑A−1y−1¿π(0.1)22(4R3π)1 | P a g e
Shear Stress and Yield_1
¿π(0.1)22(4x0.13π)Q=6.66x10−4I=πR44=π(0.1)44=0.25πx10−4m4τA=20πx6.66x10−40.25πx10−4x0.2τA=2664kN/m2Answer (B)As per given condition if maximum shear stress greater then yield stress then yield occur, but value of maximum shear stress is lower than yield stress therefore yielding not occur at point “A” Principal stress at point A, σ1,2=(σx+σy2)±√(σx+σy2)2+(τxy)2¿34016.71+02±√(34016.71+02)2+(2664)2¿17008.355±34120.327σA=51128.682kNm2(Tensile)σA=−17111.972kNm2(Compressive)Answer (c)τmax=±√(σx+σy2)2+τxy2¿√(34016.71+02)2+26642¿34120.327¿34.120MpaAs per given condition if maximum shear stress greater then yield stress then yield occur, but value of maximum shear stress is lower than yield stress therefore yielding not occur at point “A” Calculation for point: B2 | P a g e
Shear Stress and Yield_2
Answer (A) Stress at the point (B) ∑Fy=0Vy−(−20π)=0∑My=0;Mx−20π(1)=0Mx=−20π∑Fx=0Vx=0∑Fz=0VZ=0Shear stress at point B, τB=VyQItI=πR44=π(0.1)44=0.25πx10−4m4Q=∑A−1y−1¿π(0.1)22(4R3π)¿π(0.1)22(4x0.13π)Q=6.665x10−4τB=20πx6.665x10−40.25πx10−4x0.2¿−2666kN/m2σB=MyI=20πx(1)x(0.1)0.25πx10−4=80,000kN/m2σy=σZ=τxz=τyz=0Answer (B)Principal stress at point B, 3 | P a g e
Shear Stress and Yield_3

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