The torsion BAR shown is designed to hold 30000 in ib of tor

The torsion BAR shown is designed to hold 30.000 in ib of torque per degree of twist Given G = 11.4x10^6psg L=12\" determine the rep\'a dimeter of the bar

Solution

Second moment of area of the cross section of cylinder,I = pi r ^4 / 2
but T = I G (theta) / l
80,000in lbs = 1.57*11.4*10^6 psi * r^4 / 12 inch =>
921.69 Nm = 4.04*10^11 r^4
r = 0.69 cm
d = 1.38cm

 The torsion BAR shown is designed to hold 30.000 in ib of torque per degree of twist Given G = 11.4x10^6psg L=12\

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