# Comparison of Different Methods for Stress Analysis

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Year: 2013/14Answer (2): Given data:d=0.2mL=1mT=10πStress at point A,σx=σy=σZ=τxy=τxz=τyz=0Stress at the point (B) Fx=0Vx20π=0Mx=0;Mx20π(1)=0Mx=20πFy=0Vy=0Fz=0VZ=0Stress at point A,σx=σy=σZ=τxy=τxz=τyz=0Principal stress at point A, Since the point A is the origin of the cylinder rod and there are no external applied load, so the principal stress as following, σ1=0(Tensile,compressive)τ=01 | Page
Shear stress at point B, τB=VxQItI=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=0(σx+σy2)2+(τxy)2Principal stress at point B, σ1,2=(σx+σy2)±(σx+σy2)2+(τxy)2¿80,000+02±(80000+02)2+(2666)2σ1,2=40000±40088.74σ1,2=80088.74(Tensile)kN/m2σ1,2=88.74(Compressive)2 | Page
Answer (c)Yielding not occur at point “A”.τmax=±(σx+σy2)2+τxy2¿(80000+02)2+26662¿40088.74kNm2¿40.088MpaAs 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 “B” Answer (4): (A)The bar element separated by three nodal point. At the fixed point of bar, middle point of bar and at the end point of bar. A=1mm2The following equation obtain in order to find the displacement and stress.u(x)=(1xl)u1+(xl)u2u=[N]{d}Where N=[(1xl)xl]{d}=[u1u2]εx=dudx¿(1l)u1+(1l)u2[1l1l]{u1u2}3 | Page

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