Question

A circular copper bar with diameter d 5 3 in. is subjected to torques T 5 30 kip-in. at its ends. Find the maximum shear, tensile, and compressive stresses in the tube and their corresponding strains. Assume that G 5 6000 ksi.

Answer 1

Answer:

Maximum shear stress= 5.66 ksi

Maximum tensile stress= 5.66 ksi

Maximum compressive stress=-5.66 ksi

Maximum shear strain=0.000943

Maximum tensile strain= 0.0004715

Maximum compressive strain= -0.0004715

Explanation:

For acircular bar, the maximum shear stress will be given by

$\frac {16T}{\pi d^{3}}$ where d is the diameter and T is torque.

By substituting 30 kip-in for torque and 3 in for d then

Maximum shear stress= $\frac {16*30}{\pi *3^{3}}\approx 5.66 ksi$

Also, the maximum tensile and compressive stresses will be 5.66 ksi and -5.66 ksi respectively.

The maximum shear strain will be given by stress divided by modulus of elasticity, in this case 6000 G

Maximum shear strain will be $\frac {5.66}{6000}\approx 0.000943$

The maximum tensile strain will be the above divided by 2 whereas the maximum compressive strain will be negative of tensile strain hence $\frac {0.000943}{2}=0.0004715$

Maximum compressive strain will be $\frac {-0.000943}{2}=-0.0004715$

Answer 2

Answer:

Maximum shear stress= 5.66 ksi

Maximum tensile stress= 5.66 ksi

Maximum compressive stress=-5.66 ksi

Maximum shear strain=0.000943

Maximum tensile strain= 0.0004715

Maximum compressive strain= -0.0004715

Explanation: