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
[tex]\frac {16T}{\pi d^{3}}[/tex] 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= [tex]\frac {16*30}{\pi *3^{3}}\approx 5.66 ksi[/tex]
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 [tex]\frac {5.66}{6000}\approx 0.000943[/tex]
The maximum tensile strain will be the above divided by 2 whereas the maximum compressive strain will be negative of tensile strain hence [tex]\frac {0.000943}{2}=0.0004715[/tex]
Maximum compressive strain will be [tex]\frac {-0.000943}{2}=-0.0004715[/tex]
