Chapter 7, Problem 158RP

A steel pipe of 12-in. outer diameter is fabricated from $\frac{1}{4}$-in.-thick plate by welding along a helix that forms an angle of 22.5° with a plane perpendicular to the axis of the pipe. Knowing that a 40-kip axial force P and an 80-kip-in. torque T, each directed as shown, are applied to the pipe, determine the normal and in-plane shearing stresses in directions, respectively, normal and tangential to the weld.

Fig. P7.158

To determine

Fund the normal and in-plane shearing stresses, normal and tangential stress to the weld.

Answer to Problem 158RP

The normal stress in x-axis is $\underset{\_}{{\sigma }_x=0}$.

The normal stress in y-axis is $\underset{\_}{{\sigma }_y=-4.3344\text{ksi}}$.

The shearing stress in xy-plane is $\underset{\_}{{\tau }_{xy}=1.5063\text{ksi}}$.

The normal stress in the weld is $\underset{\_}{{\sigma }_w=-4.76\text{ksi}}$.

The tangential stress in the weld is $\underset{\_}{{\tau }_w=-0.467\text{ksi}}$.

Explanation of Solution

Given information:

The outer diameter of the steel pipe is $d_o=12\text{in}\text{.}$.

The thickness of the plate is $t=\frac{1}{4}\text{in}\text{.}$.

The plate makes an angle with plane perpendicular to the pipe is $\theta =22.5°$.

The magnitude of the axial force P is 40 kips.

The torque applied in the pipe T is 80 kip-in.

Calculation:

Find the inner diameter of the steel $\left(d_i\right)$ pipe using the relation.

$d_i=d_o-2t$

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

$\begin{array}{c}{d}_i=12-2\left(\frac{1}{4}\right)\\ =11.5\text{in}\text{.}\end{array}$

Find the area of the steel pipe (A) using the equation.

$A=\frac{\pi \left(d_o^2-d_i^2\right)}{4}$

Substitute 12 in. for $d_o$ and 11.5 in. for $d_i$.

$\begin{array}{c}A=\frac{\pi \left({12}^2-{11.5}^2\right)}{4}\\ =9.22843\text{in}{\text{.}}^2\end{array}$

Find the polar moment of inertia (J) of the steel pipe using the equation.

$J=\frac{\pi \left(d_o^4-d_i^4\right)}{32}$

Substitute 12 in. for $d_o$ and 11.5 in. for $d_i$.

$\begin{array}{c}J=\frac{\pi \left({12}^4-{11.5}^4\right)}{32}\\ =318.669\text{in}{\text{.}}^4\end{array}$

Find the normal stress $\left(\sigma \right)$ in the pipe using the relation.

$\sigma =-\frac{P}{A}$

Substitute 40 kips for P and $9.22843\text{in}{\text{.}}^2$ for A.

$\begin{array}{c}\sigma =-\frac{40}{9.22843}\\ =-4.3344\text{ksi}\end{array}$

The normal stress in x-axis is ${\sigma }_x=0$.

The normal stress in y-axis is ${\sigma }_y=-4.3344\text{ksi}$.

Find the shearing stress $\left(\tau \right)$ in the pipe using the relation.

$\tau =\frac{T\left(\frac{d_o}{2}\right)}{J}$

Substitute 80 kip-in. for T, 12 in. for $d_o$, and $318.669\text{in}{\text{.}}^4$ for J.

$\begin{array}{c}\tau =\frac{80\times \left(\frac{12}{2}\right)}{318.669}\\ =1.5063\text{ksi}\end{array}$

The shearing stress in xy-plane is ${\tau }_{xy}=1.5063\text{ksi}$.

Show the stress element for pipe as in Figure 1.

Therefore,

The normal stress in x-axis is $\underset{\_}{{\sigma }_x=0}$.

The normal stress in y-axis is $\underset{\_}{{\sigma }_y=-4.3344\text{ksi}}$.

The shearing stress in xy-plane is $\underset{\_}{{\tau }_{xy}=1.5063\text{ksi}}$.

Consider the axes $x^{\prime }\text{and}{y}^{\prime }$ are tangential and normal to the weld.

The normal stress in weld is ${\sigma }_w={\sigma }_{y^{\prime }}$.

The shearing stress in weld is ${\tau }_w={\tau }_{x^{\prime }{y}^{\prime }}$.

Find the normal stress in the weld using the equation.

${\sigma }_w=\frac{{\sigma }_x+{\sigma }_y}{2}-\frac{{\sigma }_x-{\sigma }_y}{2}\cos 2\theta -{\tau }_{xy}\sin 2\theta$

Substitute 0 for ${\sigma }_x$, –4.3344 ksi for ${\sigma }_y$, $22.5°$ for $\theta$, and 1.5063 ksi for ${\tau }_{xy}$.

$\begin{array}{c}{\sigma }_w=\frac{0+\left(-4.3344\right)}{2}-\frac{0-\left(-4.3344\right)}{2}\cos 2\left(22.5°\right)-1.5063\sin 2\left(22.5°\right)\\ =-2.1672-1.53244-1.06511\\ =-4.76\text{ksi}\end{array}$

Find the tangential stress in the weld using the equation.

${\tau }_w=-\frac{{\sigma }_x-{\sigma }_y}{2}\sin 2\theta +{\tau }_{xy}\cos 2\theta$

Substitute 0 for ${\sigma }_x$, –4.3344 ksi for ${\sigma }_y$, $22.5°$ for $\theta$, and 1.5063 ksi for ${\tau }_{xy}$.

$\begin{array}{c}{\tau }_w=-\frac{0-\left(-4.3344\right)}{2}\sin 2\left(22.5°\right)+1.5063\cos 2\left(22.5°\right)\\ =-1.53244+1.06511\\ =-0.467\text{ksi}\end{array}$

Show the stress element for weld as in Figure 2.

Therefore,

The normal stress in the weld is $\underset{\_}{{\sigma }_w=-4.76\text{ksi}}$.

The tangential stress in the weld is $\underset{\_}{{\tau }_w=-0.467\text{ksi}}$.