Answer (2):. Given data:. Stress at point (A). Find, sh
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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=4R3πσyb=20πx1x4x0.13πx0.25πx10−4¿3.395305x104σy=63.66+33953.05σy=34016.71Find, shear stressτBy=VyQItQ=∑A−1y−1¿π(0.1)2(4R3π)¿π(0.1)2(4x0.13π)
Q=13.33x10−4I=πR44=π(0.1)44=0.25πx10−4m4τA=20πx13.33x10−40.25πx10−4x0.2τA=5332kN/m2Answer (B)Yielding not occur at point “A”.τmax=±√(σx+σy2)2+τxy2¿√(80000+02)2+53322¿40353.81kNm2¿40.353MpaAs per given condition if maximum shear stress greater then yield stress then yield occur, but valueof maximum shear stress is lower than yield stress therefore yielding not occur at point “B”Principal stress at point A,σ1,2=(σx+σy2)±√(σx+σy2)2+(τxy)2¿34016.71+02±√(34016.71+02)2+(5332)2¿17008.355±34432.060σA=51440.4153kNm2(Tensile)σA=−17423.705kNm2(Compressive)Answer (c)Yielding not occur at point “A”.τmax=±√(σx+σy2)2+τxy2
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