The solid cylindrical rod BC of length L = 24 in. is attached to the rigid lever AB of len...

Calculate the maximum allowable angular twist.

$$ \phi_{\text {all }}=\frac{s_{\max }}{\mathrm{a}} $$

Here, \(s_{\max }\) is the maximum allowable displacement of point \(A\) and \(a\) is the distance of the line of action of force from the axis of the shaft.

Substitute \(0.1\) in. for \(s_{\max }\) and 15 in. for \(a\).

$$ \begin{aligned} \phi_{\text {all }} &=\frac{1 \mathrm{in} .}{15 \mathrm{in} .} \\ &=0.067 \mathrm{rad} \end{aligned} $$

This angular twist is same for both \(B\) and point \(A\).

Write the equation for the angular of twist in the shaft.

$$ \begin{array}{l} \phi=\frac{T L}{J G} \\ \phi_{\text {all }}=\frac{T L}{\frac{\pi}{32} d^{4} G} \\ d=\left(\frac{32 T L}{\pi G \phi_{\text {all }}}\right)^{\frac{1}{4}} \end{array} $$

Here, \(G\) is the rigidity modulus of the shaft.

Substitute \(11.2 \times 10^{6} \mathrm{lb} / \mathrm{in} .{ }^{2}\) for \(G, 1500 \mathrm{lb} \cdot \mathrm{in} .\) for \(T, 15 \times 10^{3} \mathrm{lb} / \mathrm{in} .{ }^{2}\) for \(\tau_{\mathrm{all}}, 24 \mathrm{in} .\) for \(L\), and

\(0.067 \mathrm{rad}\) for \(\phi_{\text {all }}\).

$$ d=\left[\frac{32(1500 \mathrm{lb} \cdot \mathrm{in} .)(24 \mathrm{in} .)}{\pi\left(11.2 \times 10^{6} \mathrm{lb} / \mathrm{in}^{2}\right)(0.067)}\right]^{1 / 4} $$

\(=0.837\) in.

Consider the two values obtained, the higher value holds the key for the design condition considering the minimum required diameter.

Hence, the minimum diameter of the shaft is \(0.837\) in.