Chapter 3, Problem 151RP

A steel pipe of 12-in. outer diameter is fabricated from ¼-in.-thick plate by welding along a helix that forms an angle of 45° with a plane parallel to the axis of the pipe. Knowing that the maximum allowable tensile stress in the weld is 12 ksi, determine the largest torque that can be applied to the pipe.

Fig. P3.151

To determine

Find the largest torque that can be applied to the pipe.

Answer to Problem 151RP

The largest torque that can be applied to the pipe is $\underset{\_}{637\text{kip}\cdot \text{in}}$.

Explanation of Solution

Given information:

The outer diameter of the steel pipe $\left(d_o\right)$ is $12\text{in}\text{.}$

The thickness of the steel pipe (t) is $\frac{1}{4}\text{in}\text{.}$

The angle of the helix is $45°$ with a plane perpendicular to the axis of the pipe.

The maximum allowable tensile stress in the weld $\left({\tau }_{\max }\right)$ is $12\text{ksi}$.

Calculation:

Calculate the inner radius of the steel pipe $\left(c_1\right)$ using the calculation.

$c_1=\frac{d_o}{2}-t$

Substitute $12\text{in}\text{.}$ for $d_o$ and $\frac{1}{4}\text{in}\text{.}$ for t.

$\begin{array}{c}{c}_1=\frac{12}{2}-\frac{1}{4}\\ =5.75\text{in}\text{.}\end{array}$

Calculate the polar moment of inertia of steel pipe (J) as follows:

$J=\frac{\pi }{2}\left(c_2^4-c_1^4\right)$

Here, $c_2$ is outer radius of the steel pipe.

Substitute $\frac{12}{2}\text{in}\text{.}$ for $c_2$ and $\text{5}\text{.75}\text{in}\text{.}$ for $c_1$.

$\begin{array}{c}J=\frac{\pi }{2}\left[{\left(\frac{12}{2}\right)}^4-{5.75}^4\right]\\ =318.67\text{in}{\text{.}}^4\end{array}$

The torsion formula for maximum shear stress in the solid shaft AB $\left({\tau }_{\max }\right)$ is expressed as shown below:

${\tau }_{\max }=\frac{Tc}{J}$

Here, T is the applied internal torque.

Substitute $12\text{ksi}$ for T, $6\text{in}\text{.}$ for c, and $318.67\text{in}{\text{.}}^4$ for J.

$\begin{array}{c}12\text{ksi}=\frac{T\times 6\text{in}\text{.}}{318.67\text{in}{\text{.}}^4}\\ T=637\text{kip}\cdot \text{in}\end{array}$

Therefore, the largest torque that can be applied to the pipe is $\underset{\_}{637\text{kip}\cdot \text{in}}$.