# Shear Stress Analysis in Structural Components

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Answer (2): Given data:d=0.2mL=1mT=20πStress at point (A) Fx=0Vx20π=0Fy=0Vy=20πMy=0σyb=NA=200.1π=200π=63.66σy=MyII=πR44=π(0.1)44=0.25πx104m4y=4R3πσy=20πx1x4x0.13πx0.25πx104¿3.395305x104σy=63.66+33953.05σy=34016.71Find, shear stressτy=VyQItQ=A1y1¿π(0.1)22(4R3π)1 | P a g e
¿π(0.1)22(4x0.13π)Q=6.66x104I=πR44=π(0.1)44=0.25πx104m4τA=20πx6.66x1040.25πx104x0.2τA=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
Answer (A) Stress at the point (B) Fy=0Vy(20π)=0My=0;Mx20π(1)=0Mx=20πFx=0Vx=0Fz=0VZ=0Shear stress at point B, τB=VyQItI=πR44=π(0.1)44=0.25πx104m4Q=A1y1¿π(0.1)22(4R3π)¿π(0.1)22(4x0.13π)Q=6.665x104τB=20πx6.665x1040.25πx104x0.2¿2666kN/m2σB=MyI=20πx(1)x(0.1)0.25πx104=80,000kN/m2σy=σZ=τxz=τyz=0Answer (B)Principal stress at point B, 3 | P a g e

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