Chapter 2.3, Problem 55P

Two steel bars (E_s = 200 GPa and α_s = 11.7 × 10^-6/°C) are used to reinforce a brass bar (E_b =105 GPa, α_b = 20.9 × 10^-6/°C) that is subjected to a load P = 25 kN. When the steel bars were fabricated; the distance between the centers of the holes that were to fit on the pins was made 0.5 mm smaller than the 2 m needed. The steel bars were then placed in an oven to increase their length so that they would just fit on the pins. Following fabrication, the temperature in the steel bars dropped back to room temperature. Determine (a) the increase in temperature that was required to fit the steel bars on the pins, (b) the stress in the brass bar after the load is applied to it.

Fig. P2.55

To determine

The required temperature increment to fit the steel bars on the pins.

Answer to Problem 55P

The required temperature increment to fit the steel bars on the pins is $\underset{\_}{21.4°C}$.

Explanation of Solution

Given information:

The magnitude of load $P$ is $25\text{kN}$.

The coefficient $\left({\alpha }_b\right)$ of thermal expansion of brass is $20.9\times {10}^{-6}/°\text{C}$.

The coefficient $\left({\alpha }_s\right)$ of thermal expansion of steel is $11.7\times {10}^{-6}/°\text{C}$.

The young’s modulus $\left(E_S\right)$ of steel core is $200\text{GPa}$.

The young’s modulus $\left(E_b\right)$ of aluminium shell is $10\text{5GPa}$.

Determine the temperature change require to expand steel bar by this amount.

${\delta }_T={\alpha }_s\left(\Delta T\right)L$ (1)

Here, $L$ is length of bar and ${\delta }_T$ is thermal strain deformation.

Substitute $2\text{m}$ for $L$, $11.7\times {10}^{-6}/°\text{C}$  for ${\alpha }_s$, and $0.5\text{mm}$ for ${\delta }_T$ in Equation (1).

$\begin{array}{l}0.5\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)=2\times 11.7\times {10}^{-6}\left(\Delta T\right)\\ 0.5\times {10}^{-3}=2.34\times {10}^{-5}\left(\Delta T\right)\\ \Delta T=21.368°\text{C}\end{array}$

Thus, the required temperature increment to fit the steel bars on the pins is $\underset{\_}{21.4°C}$.

To determine

The normal stress in the brass bar after load applied.

Answer to Problem 55P

The normal stress due to steel is $\underset{\_}{57.0\text{MPa}}$.

The normal stress due to brass is $\underset{\_}{3.67\text{MPa}}$.

Explanation of Solution

Calculation:

After the assembly of the bars, a compressive force is developed in the brass bars and a tensile force is developed in the steel bars.

Determine the area $\left(A_s\right)$ of elongation steel using the relation:

$A_s=b\times d\times t$ (2)

Here, b is the width of the steel bar , d is the depth of the steel bar, and t is the thickness of the bar.

Substitute $40\text{mm}$ for b, $2\text{m}$ for d, and $5\text{mm}$ for t in Equation (2).

$\begin{array}{c}{A}_s=40\times 2\times 5\\ =400{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =400\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Determine the deformation of steel ${\left({\delta }_P\right)}_s$ using the relation:

${\left({\delta }_P\right)}_s=\frac{FL}{A_s{E}_s}$ (3)

Here, F is the compressive force, L is the length of the bar, and $\left(A_s\right)$ is cross sectional area of steel.

Substitute $400\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_s$, $200\text{GPa}$ for $E_s$, $P$ for F, and $2\text{m}$ for $L$ in Equation (3).

$\begin{array}{c}{\left({\delta }_P\right)}_s=\frac{P\times 2.00}{400\times {10}^{-6}\left(200\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\right)}\\ =\frac{P\times 2.00}{80,000,000}\\ =25\times {10}^{-9}P\end{array}$

Determine the area $\left(A_b\right)$ of elongation brass using the relation:

$A_b=b\times d$ (4)

Here, b is the width of the brass bar, d is the depth of the brass bar.

Substitute $40\text{mm}$ for b and $15{\text{mm}}^2$ for d, in Equation (4).

$\begin{array}{c}{A}_b=40\times 15\\ =600{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =600\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Determine the deformation of brass ${\left({\delta }_P\right)}_b$ using the relation:

${\left({\delta }_P\right)}_b=\frac{FL}{A_b{E}_b}$ (5)

Here, $\left(A_b\right)$ is cross sectional area of brass.

Substitute $600\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_b$, $105\text{GPa}$ for $E_b$, $P$ for F, and $2\text{m}$ for $L$ in Equation (5).

$\begin{array}{c}{\left({\delta }_P\right)}_s=\frac{P\times 2.00}{600\times {10}^{-6}\left(105\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\right)}\\ =\frac{2.00P}{600\times {10}^{-6}\left(105\times {10}^9\right)}\\ =\frac{P\times 2.00}{63,000,000}\\ =31.746\times {10}^{-9}P\end{array}$

The deformation of bar and steel is equal to the initial amount of misfit.

${\left({\delta }_P\right)}_s+{\left({\delta }_P\right)}_b=0.5\text{mm}$ (6)

Here, ${\left({\delta }_P\right)}_b$ is thermal strain of bar and ${\left({\delta }_P\right)}_s$ is thermal strain of steel.

Substitute $31.746\times {10}^{-9}P$ for ${\left({\delta }_P\right)}_s$ and $25\times {10}^{-9}P$ for ${\left({\delta }_P\right)}_b$ in Equation (6).

$\begin{array}{l}31.746\times {10}^{-9}P+25\times {10}^{-9}P=0.5\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\\ 56.746\times {10}^{-9}P=0.5\times {10}^{-3}\\ P=8.8112\times {10}^3\text{N}\end{array}$

Find the stresses due to fabrication for steel using the relation:

${\sigma }_s^{*}=\frac{P}{A_s}$ (7)

Substitute $8.8112\times {10}^3\text{N}$ for P and $400\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_s$ in Equation (7).

$\begin{array}{c}{\sigma }_s^{*}=\frac{8.8112\times {10}^3}{400\times {10}^{-6}}\\ =22.028\times {10}^6\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =22.028\text{MPa}\end{array}$

Find the stresses due to fabrication for brass using the relation:

${\sigma }_b^{*}=-\frac{P}{A_b}$ (8)

Substitute $8.8112\times {10}^3\text{N}$ for P and $600\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_b$ in Equation (8).

$\begin{array}{c}{\sigma }_s^{*}=\frac{8.8112\times {10}^3}{600\times {10}^{-6}}\\ =-14.6853\times {10}^6\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =-14.68\text{MPa}\end{array}$

The stresses should be added the stress due to the $25\text{kN}$.

Consider ${\delta }^{\prime }$ is the additional displacement and $P_s,P_b$ is the additional force developed in the steel and brass.

Find the force developed in steel bar $P_s$ using the relation:

$\begin{array}{c}{\delta }^{\prime }=\frac{P_{s}L}{A_s{E}_s}=\frac{P_{b}L}{A_b{E}_b}\\ P_s=\frac{A_s{E}_s}{L}{\delta }^{\prime }\end{array}$ (9)

Substitute $400\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_s$, $200\text{GPa}$ for $E_s$, $P$ for F, and $2\text{m}$ for $L$ in Equation (9).

$\begin{array}{c}{P}_s=\frac{400\times {10}^{-6}\left(200\times {10}^9\right)}{2.00}{\delta }^{\prime }\\ =40\times {10}^6{\delta }^{\prime }\end{array}$ (10)

Find the force developed in brass bar $P_b$ using the relation:

$P_b=\frac{A_b{E}_b}{L}{\delta }^{\prime }$ (11)

Substitute $600\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_s$, $105\text{GPa}$ for $E_s$, $P$ for F, and $2\text{m}$ for $L$ in Equation (11).

$\begin{array}{c}{P}_b=\frac{600\times {10}^{-6}\left(105\times {10}^9\right)}{2.00}{\delta }^{\prime }\\ =31.5\times {10}^6{\delta }^{\prime }\end{array}$ (12)

Find the force P as follows:

$P=P_s+P_b$ (13)

Here, $P_b$ is force developed in brass bar and $P_s$ is force developed in steel bar.

Substitute $31.5\times {10}^6{\delta }^{\prime }$ for $P_s$ and $40\times {10}^6{\delta }^{\prime }$ for $P_b$, and $25\text{kN}$ for P in Equation (13).

$\begin{array}{c}25\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)=31.5\times {10}^6{\delta }^{\prime }+40\times {10}^6{\delta }^{\prime }\\ 25,000=71,500,000{\delta }^{\prime }\\ {\delta }^{\prime }=349.65\times {10}^{-6}\text{m}\end{array}$

Substitute $349.65\times {10}^{-6}\text{m}$ for ${\delta }^{\prime }$ in Equation (10).

$\begin{array}{c}{P}_s=40\times {10}^6\left(349.65\times {10}^{-6}\right)\\ =13,986\text{N}\end{array}$

Substitute $349.65\times {10}^{-6}\text{m}$ for ${\delta }^{\prime }$ in Equation (12).

$\begin{array}{c}{P}_b=31.5\times {10}^6\left(349.65\times {10}^{-6}\right)\\ =11,013.975\text{N}\end{array}$

Find the normal stress $\left({\sigma }_s\right)$ for steel using the relation:

${\sigma }_s=\frac{P_s}{A_s}$ (14)

Substitute $13,986\text{N}$ for $P_s$ and $400\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_s$ in Equation (14).

$\begin{array}{c}{\sigma }_s=\frac{13,986}{400\times {10}^{-6}}\\ =34.965\times {10}^6\text{N}/{\text{m}}^{\text{2}}\end{array}$

Find the normal stress $\left({\sigma }_b\right)$ for steel using the relation:

${\sigma }_b=\frac{P_b}{A_b}$ (15)

Substitute $11,013.975\text{N}$ for $P_b$ and $600\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_b$ in Equation (15).

$\begin{array}{c}{\sigma }_b=\frac{13.9860\times {10}^3}{600\times {10}^{-6}}\\ =18.3566\times {10}^6\text{N}/{\text{m}}^{\text{2}}\end{array}$

The stress due to fabrication added.

Find the total stress using the relation:

${\sigma }_s={\left({\sigma }_s\right)}_b+{\sigma }_s^{*}$ (16)

Here, ${\left({\sigma }_s\right)}_b$ is normal stress due to fabrication for brass and ${\sigma }_s^{*}$ is normal stress due to fabrication for steel.

Substitute $34.965\times {10}^6\text{N}/{\text{m}}^{\text{2}}$ for ${\left({\sigma }_s\right)}_b$ and $22.028\text{MPa}$ for ${\left({\sigma }_s\right)}_s$ in Equation (16).

$\begin{array}{c}{\sigma }_s=34.965\times {10}^6+22.028\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\\ =34.965\times {10}^6+22.028\times {10}^6\\ =56.991\times {10}^6\text{Pa}\end{array}$

Thus, the normal stress due to steel is $\underset{\_}{57.0\text{MPa}}$.

Find the total stress using the relation:

${\sigma }_b={\left({\sigma }_b\right)}_b-{\sigma }_b^{*}$ (17)

Substitute $18.3566\times {10}^6\text{N}/{\text{m}}^{\text{2}}$ for ${\left({\sigma }_s\right)}_b$ and $14.6853\text{MPa}$ for ${\sigma }_s^{*}$ in Equation (17).

$\begin{array}{c}{\sigma }_s=18.3566\times {10}^6+14.6853\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\\ =18.3566\times {10}^6-14.6853\times {10}^6\\ =3.6713\times {10}^6\text{Pa}\end{array}$

Thus, the normal stress due to brass is $\underset{\_}{3.67\text{MPa}}$.