Chapter 3.1, Problem 30P

Fig. P3.30

3.30 (a) For a given allowable shearing stress, determine the ratio T/w of the maximum allowable torque T and the weight per unit length w for the hollow shaft shown, (b) Denoting by (T/w)₀ the value of this ratio for a solid shaft of the same radius express the ratio T/w for the hollow shaft in terms of (T/w)₀ and c₁/c₂ ■

To determine

The ratio $T/w$ of the hollow shaft.

Answer to Problem 30P

The ratio $T/w$ of the hollow shaft is $\underset{\_}{\frac{\left(c_2^2+c_1^2\right){\tau }_{\text{all}}}{2\rho g{c}_2}}$.

Explanation of Solution

Given information:

The maximum allowable torque is T.

The weight per unit length of the hollow shaft is w.

Calculation:

The torsion formula for allowable shear stress in the hollow shaft $\left({\tau }_{\text{all}}\right)$ is expressed as shown below:

${\tau }_{\text{all}}=\frac{T{c}_2}{J}$ (1)

Here, $T$ is the allowable internal torque, J is the polar moment of inertia of the shaft, and $c_2$ is the outer radius of the shaft.

The polar moment of inertia for a hollow shaft $\left(J\right)$ of outer radius $c_2$ and inner radius $c_1$ is $J=\frac{\pi }{2}\left(c_2^4-c_1^4\right)$.

The area of the hollow shaft is $A=\pi \left(c_2^2-c_1^2\right)$.

Substitute $\frac{\pi }{2}\left(c_2^4-c_1^4\right)$ for J in Equation (1).

$\begin{array}{c}{\tau }_{\text{all}}=\frac{T{c}_2}{\frac{\pi }{2}\left(c_2^4-c_1^4\right)}\\ =\frac{2T{c}_2}{\pi \left(c_2^4-c_1^4\right)}\\ =\frac{2T{c}_2}{\pi \left(c_2^2-c_1^2\right)\left(c_2^2+c_1^2\right)}\\ T=\frac{\pi }{2}\frac{\left(c_2^2-c_1^2\right)\left(c_2^2+c_1^2\right)}{c_2}{\tau }_{\text{all}}\end{array}$

Let the specific weight of the shaft be $\rho g$, the total weight of the shaft be W, and the length of the shaft be L.

Total weight of the shaft is $W=\rho gLA$.

The weight per unit length of the hollow shaft is expressed as follows:

$\begin{array}{c}w=\frac{W}{L}\\ =\frac{\rho gLA}{L}\\ =\rho gA\\ =\rho g\pi \left(c_2^2-c_1^2\right)\end{array}$

Find the ratio of $T/w$ as shown below:

$\begin{array}{c}\frac{T}{w}=\frac{\frac{\pi }{2}\frac{\left(c_2^2-c_1^2\right)\left(c_2^2+c_1^2\right)}{c_2}{\tau }_{\text{all}}}{\rho g\pi \left(c_2^2-c_1^2\right)}\\ =\frac{\left(c_2^2+c_1^2\right){\tau }_{\text{all}}}{2\rho g{c}_2}\end{array}$

Therefore, the ratio $T/w$ of the hollow shaft is $\underset{\_}{\frac{\left(c_2^2+c_1^2\right){\tau }_{\text{all}}}{2\rho g{c}_2}}$.

To determine

The ratio $T/w$ of the hollow shaft in terms of ${\left(T/w\right)}_0$ and $c_1/c_2$.

Answer to Problem 30P

The ratio $T/w$ of the hollow shaft in terms of ${\left(T/w\right)}_0$ and $c_1/c_2$ is $\underset{\_}{{\left(\frac{T}{w}\right)}_0\left(1+\frac{c_1^2}{c_2^2}\right)}$.

Explanation of Solution

Given information:

The maximum allowable torque is T.

The weight per unit length of the hollow shaft is w.

Calculation:

The polar moment of inertia for a solid shaft $\left(J\right)$ of radius $c_2$ is $J=\frac{\pi }{2}{c}_2^4$.

The area of the solid shaft is $A=\pi c_2^2$.

Substitute $\frac{\pi }{2}{c}_2^4$ for J in Equation (1).

$\begin{array}{c}{\tau }_{\text{all}}=\frac{T{c}_2}{\frac{\pi }{2}{c}_2^4}\\ =\frac{2T{c}_2}{\pi c_2^4}\\ =\frac{2T}{\pi c_2^3}\\ T=\frac{\pi }{2}{c}_2^3{\tau }_{\text{all}}\end{array}$

The weight per unit length of the solid shaft is expressed as follows:

$\begin{array}{c}w=\frac{W}{L}\\ =\frac{\rho gLA}{L}\\ =\rho gA\\ =\rho g\pi c_2^2\end{array}$

Find the ratio of solid shaft ${\left(T/w\right)}_0$ as shown below:

$\begin{array}{c}{\left(\frac{T}{w}\right)}_0=\frac{\frac{\pi }{2}{c}_2^3{\tau }_{\text{all}}}{\rho g\pi c_2^2}\\ =\frac{c_2{\tau }_{\text{all}}}{2\rho g}\end{array}$

Refer part (a).

The ratio $T/w$ of the hollow shaft is $\frac{\left(c_2^2+c_1^2\right){\tau }_{\text{all}}}{2\rho g{c}_2}$.

$\begin{array}{c}\frac{T}{w}=\frac{\left(c_2^2+c_1^2\right){\tau }_{\text{all}}}{2\rho g{c}_2}\\ =\frac{c_2^2\left(1+c_1^2/c_2^2\right){\tau }_{\text{all}}}{2\rho g{c}_2}\\ =\frac{c_2^{}\left(1+c_1^2/c_2^2\right){\tau }_{\text{all}}}{2\rho g}\end{array}$ (2)

Substitute ${\left(\frac{T}{w}\right)}_0$ for $\frac{c_2{\tau }_{\text{all}}}{2\rho g}$ in Equation (2).

$\frac{T}{w}={\left(\frac{T}{w}\right)}_0\left(1+\frac{c_1^2}{c_2^2}\right)$

Therefore, the ratio $T/w$ of the hollow shaft in terms of ${\left(T/w\right)}_0$ and $c_1/c_2$ is $\underset{\_}{{\left(\frac{T}{w}\right)}_0\left(1+\frac{c_1^2}{c_2^2}\right)}$.