Question

Consider a 1-in-square steel thin-walled tube loaded in torsion. The tube has a wall thickness t 5 1 in, is 36 in long, and has a maximum allowable shear stress of 12 kpsi. Determine the 16 maximum torque that can be applied and the corresponding angle of twist of the tube.

Answer 1

Answer:

The maximum torque is given by

=2Am*t*r

Were r is the shear stress

Were Am median area =(1-51)^2=2500

Were median line lenght = 4(1-51)=2000

Maximum torque = 2*2500*51*12000=3.06*10^9lbf-in

Corresponding angle of twist©=T*Lm*l/(4*G*Am^2*t)

3.06*10^9*2000*36=2.2*10^12

4*11.5*10^9*2500*10^2*5=1.46*10^19

©=(2.2E12/1.46E19)=1.5E-7rad

Explanation: