Answer (2). Given data:. Stress at point (A). Internal


Added on  2019-09-23

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Answer (2) Given data: ri=0.5m,di=1mr0=0.52m,d0=1.04mP=100barst=r0ri=0.520.50=0.02mk=rori=0.520.50=1.04Stress at point (A) Internal pressure, Axial tension σx1=pa2t=100x0.520.04=1300Nm2Hoop tension σy=pat=100x0.520.02=2600Nm2Axial compression Axial Normal stress σx2=P(k21)=100(1.0421)=1225.4901Nm2Twisting moment:Maximum out-plane shear stress, τ=Pr2t=100x0.520.04=1300Nm2Answer (a) State of stress: σx=σx1+σx2=1300+1225.5=2525.2Nm2σy=2600Nm2τxy=1300Nm2σZ=τyz=τxz=0The maximum in-plane shear stress: τxy=(σxσy2)2+τxy2=1387.56+1690000=1300.50Nm21 | P a g e

Answer (b) Principal stresses: σ1=σx+σy2+(σxσy2)2+τxy2¿2525.2+26002+¿1300.50¿3863.1N/m2σ2=σx+σy2(σxσy2)2+τxy2¿2525.2+260021300.50σ2=1261.7Nm2σ3=0Answer (c) τmax=(σ1σ3)2=3863.12=1931.55Nm2Yielding not occur because maximum shear stress less than yield stress. Stress at point (B) Internal pressure, Axial tension σx1=pa2t=100x0.500.04=1200Nm2Hoop tension σy=pat=100x0.500.02=2500Nm2Twisting moment:Maximum out-plane shear stress, τ=Pr4t=100x0.500.04=625Nm2Answer (A) State of stress: 2 | P a g e

σx=1200Nm2σy=2500Nm2τxy=625Nm2σZ=τyz=τxz=0The maximum in-plane shear stress: τxy=(σxσy2)2+τxy2=1387.56+1690000=901.73Nm2Answer (B) Principal stresses: σ1=1200+25002+(σxσy2)2+τxy2¿1850+901.73¿2751.73N/m2σ2=σx+σy2(σxσy2)2+τxy2¿1850901.73σ2=948.27Nm2σ3=0Answer(c) τmax=(σ1σ3)2=2751.732=1375.865Nm2Shear stress of given cylinder lower then yield stress, so yielding not occur. 3 | P a g e

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