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.

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:

mistakes are ok