Chapter 7.6, Problem 127P

(a)

To determine

Find the tensile stress in the steel ring.

Answer to Problem 127P

The tensile stress in the steel ring is $\underset{\_}{2.28\text{ksi}}$.

Explanation of Solution

Given information:

Steel Ring:

The thickness $\left(t_s\right)$ of the steel ring is $0.25\text{in}\text{.}$

The Young’s modulus $\left(E_s\right)$ of the steel ring is $29\times {10}^6\text{psi}$.

The coefficient of thermal expansion $\left({\alpha }_s\right)$ of the steel ring is $6.5\times {10}^{-6}/°\text{F}$.

Brass Ring:

The thickness $\left(t_b\right)$ of the steel ring is $0.125\text{in}\text{.}$

The Young’s modulus $\left(E_b\right)$ of the steel ring is $15\times {10}^6\text{psi}$.

The coefficient of thermal expansion $\left({\alpha }_b\right)$ of the steel ring is $11.6\times {10}^{-6}/°\text{F}$.

Consider the strain in the steel ring and the brass ring due to tensile stress are denoted by ${\epsilon }_{sp}$ and ${\epsilon }_{bp}$.

Consider the strain in the steel ring and the brass ring due to temperature stress are denoted by ${\epsilon }_{sT}$ and ${\epsilon }_{bT}$.

Consider the total strain in the steel ring and the brass ring are denoted by ${\epsilon }_s$ and ${\epsilon }_b$.

Consider the internal pressure in the steel ring is denoted by p.

Consider the external pressure in the steel ring is same as p.

The initial and final temperature of the ring are $T_i=50°\text{F}$ and $T_f=125°\text{F}$.

Calculation:

Calculate the change in the temperature $\left(\Delta T\right)$ of the ring as follows:

$\begin{array}{c}\Delta T=T_f-T_i\\ =125°\text{F}-50°\text{F}\\ \text{=75}°\text{F}\end{array}$

Calculate the mean radius (r) of the ring as follows:

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

Substitute $5\text{in}\text{.}$ for d.

$\begin{array}{c}r=\frac{5}{2}\\ =2.5\text{in}\text{.}\end{array}$

Consider the tensile stress in the steel ring in the brass ring and the steel ring is denoted by ${\sigma }_b$ and ${\sigma }_s$.

Calculate the total strain in the steel ring using the relation:

$\begin{array}{c}{\epsilon }_s={\epsilon }_{sp}+{\epsilon }_{sT}\\ =\left(\frac{{\sigma }_s}{E_s}\right)+\left({\alpha }_s\Delta T\right)\\ =\left[\frac{\left(\frac{pr}{t_s}\right)}{E_s}\right]+\left({\alpha }_s\Delta T\right)\\ =\frac{pr}{E_s{t}_s}+{\alpha }_s\Delta T\end{array}$

Calculate the change in length of circumference of the steel ring $\left(\Delta L_s\right)$ as follows:

$\begin{array}{c}\Delta L_s=2\pi r{\epsilon }_s\\ =2\pi r\left(\frac{pr}{E_s{t}_s}+{\alpha }_s\Delta T\right)\end{array}$ (1)

Calculate the total strain in the steel ring using the relation:

$\begin{array}{c}{\epsilon }_b={\epsilon }_{bp}+{\epsilon }_{bT}\\ =\left(\frac{{\sigma }_b}{E_b}\right)+\left({\alpha }_b\Delta T\right)\\ =\left[\frac{\left(\frac{-pr}{t_b}\right)}{E_b}\right]+\left({\alpha }_b\Delta T\right)\\ =-\frac{pr}{E_b{t}_b}+{\alpha }_b\Delta T\end{array}$

Calculate the change in length of circumference of the steel ring $\left(\Delta L_b\right)$ as follows:

$\begin{array}{c}\Delta L_b=2\pi r{\epsilon }_b\\ =2\pi r\left(-\frac{pr}{E_b{t}_b}+{\alpha }_b\Delta T\right)\end{array}$ (2)

Equate Equation (1) and (2).

$\begin{array}{c}\Delta L_s=\Delta L_b\\ 2\pi r\left(\frac{pr}{E_s{t}_s}+{\alpha }_s\Delta T\right)=2\pi r\left(-\frac{pr}{E_b{t}_b}+{\alpha }_b\Delta T\right)\\ \left(\frac{pr}{E_s{t}_s}+{\alpha }_s\Delta T\right)=\left(-\frac{pr}{E_b{t}_b}+{\alpha }_b\Delta T\right)\\ p\left(\frac{r}{E_s{t}_s}+\frac{r}{E_b{t}_b}\right)=\left({\alpha }_b-{\alpha }_s\right)\Delta T\end{array}$ (3)

Substitute $2.5\text{in}\text{.}$ for r, $29\times {10}^6\text{psi}$ for $E_s$, $0.25\text{in}\text{.}$ for $t_s$, $15\times {10}^6\text{psi}$ for $E_b$, $0.125\text{in}\text{.}$ for $t_b$, $11.6\times {10}^{-6}/°\text{F}$ for ${\alpha }_b$, $6.5\times {10}^{-6}/°\text{F}$ for ${\alpha }_s$, and $75°\text{F}$ for $\Delta T$ in Equation (3).

$\begin{array}{l}p\left(\frac{2.5}{29\times {10}^6\times 0.25}+\frac{2.5}{15\times {10}^6\times 0.125}\right)=\left(11.6\times {10}^{-6}-6.5\times {10}^{-6}\right)\times 75\\ p\left(0.3448\times {10}^{-6}+1.3333\times {10}^{-6}\right)=\left(5.1\times {10}^{-6}\right)\times 75\\ 1.6781\times {10}^{-6}p=382.5\times {10}^{-6}\\ p=\frac{382.5\times {10}^{-6}}{1.6781\times {10}^{-6}}\end{array}$

$p=227.93\text{psi}$

Calculate the tensile stress in the steel ring using the relation:

${\sigma }_s=\left(\frac{pr}{t_s}\right)$

Substitute $227.93\text{psi}$ for p, $2.5\text{in}\text{.}$ for r, and $0.25\text{in}\text{.}$ for $t_s$.

$\begin{array}{c}{\sigma }_s=\left(\frac{227.93\times 2.5}{0.25}\right)\\ =2,280\text{psi}\times \frac{1\text{ksi}}{1,000\text{psi}}\\ =2.28\text{ksi}\end{array}$

Thus, the tensile stress in the steel ring is $\underset{\_}{2.28\text{ksi}}$.

(b)

To determine

Find the pressure exerted by the brass ring on the steel ring.

Answer to Problem 127P

The pressure exerted by the brass ring on the steel ring is $\underset{\_}{227.93\text{psi}}$.

Explanation of Solution

Refer part (a);

Get the value of the pressure exerted by the brass ring on the steel ring as $p=227.93\text{psi}$.

Thus, the external pressure on the brass ring is $\underset{\_}{227.93\text{psi}}$.