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Torsional Shear Stresses - Assignment

   

Added on  2019-09-30

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Answer (2) The twisted moment T produces a torsional shear stresses τ=TrIpolar¿2Tπr3¿2x2ππ(0.1)3τ=4000kN/m2The stresses τ1 acts horizontally to the left at point A and vertically downwards at the point B. the bending moment M produce a tensile stress at point A. σbending=MrI¿4Mπr3¿4x4ππ(0.1)3σbending=16000kN/m2There are no stress produces at point B, because B is located on the neutral axis. The shear stress at the top of the bar at point A, the shear stress at the point B is as following: τshear=VQIb¿4V3A¿4x10π3π(0.1)3τshear=13333.33kN/m2σA and τ1 are in point A, while the τ1 and τ2 are acting in point B. Note that the elements is in the plane stress with At point A: σx=σA,σy=0,τxy=τ1σx=NA¿10ππ(0.1)31 | P a g e

σx=10000kN/m2τxy=4000kN/m2Principal stresses: σ1=1200+25002±(σxσy2)2+τxy2σ1=10000+02+(1000002)2+40002¿5000+6403.1242σ1=11403.1243kN/m2σ2=σx+σy2(σxσy2)2+τxy2σ2=10000+02(1000002)2+40002¿50006403.1242¿1403.1242kN/m2τmax=(σxσy2)2+τxy2The stress element in point B is also plane stress and only stresses acting on this elements are shearstress τ1τ2.σx=σy=0¿τxy=(τ1+τ2)¿(4000+13333.33)τxy=17333.33kN/m2Answer (B) &(C) : σn=σ1+σ22+σ1σ22cos2θ¿11403.12431403.12422+11403.1243+1403.12422cos2θ¿50000+6403.12425σn=56403.12425kN/m22 | P a g e

τs=(σ1σ2)2τs=(11403.1243+1403.1242)2τs=6403.12425kN/m2Answer (3): Answer (A):As per given condition, a die press polymeric material in vertical Y-direction (k –direction), therefore as per strain definition the dimension of polymeric material have change in y-direction only. StrainTensor=[εxxεxyεxzεyxεyyεyzεzxεzyεzz]¿[ux12(uy+vx)12(uz+wx)12(uy+vx)vy12(vz+wy)12(uz+wx)12(vz+wy)wz]Assume 1mm pressed after applied vertical force, So the strain in vertical direction = changelenghtoriginallenght=11000=0.001StrainTensorL=[10.000500.00050.0010.000500.00051]3 | P a g e

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