Question

The step shaft is subjected to a torque of 710 lb·in. If the allowable shear stress for the material is τallow = 12 ksi, determine the smallest radius at the junction between the cross sections that can be used to transmit the torque.

Answer 1

Answer:

The smallest radius at the junction between the cross section that can be used to transmit the torque is 0.167 inches.

Explanation:

Torsional shear stress is determined by the following expression:

$\tau = \frac{T\cdot r}{J}$

Where:

$T$ - Torque, measured in $lbf\cdot in$.

$r$ - Radius of the cross section, measured in inches.

$J$ - Torsion module, measured in quartic inches.

$\tau$ - Torsional shear stress, measured in pounds per square inch.

The radius of the cross section and torsion module are, respectively:

$r = \frac{D}{2}$

$J = \frac{\pi}{32}\cdot D^{4}$

Where $D$ is the diameter of the cross section, measured in inches.

Then, the shear stress formula is now expanded and simplified as a function of the cross section diameter:

$\tau = T \cdot \frac{D}{\frac{\pi}{16}\cdot D^{4} }$

$\tau = \frac{16\cdot T}{\pi \cdot D^{3}}$

In addition, diameter is cleared:

$D^{3} = \frac{16\cdot T}{\pi \cdot \tau}$

$D = 2\cdot \sqrt[3] {\frac {2\cdot T}{\pi\cdot \tau}}$

If $T = 710\,lb\cdot in$ and $\tau = 12000\,psi$, then:

$D = \sqrt[3]{\frac{2\cdot (710\,lbf\cdot in)}{\pi \cdot (12000\,psi)} }$

$D \approx 0.335\,in$

$r \approx 0.167\,in$

The smallest radius at the junction between the cross section that can be used to transmit the torque is 0.167 inches.