Question

The assembly has the diameters and material make-up indicated. It fits securely between its fixed supports when the temperature is T1 = 70?F,determine the average normal stress in each material when the temperature reaches T2 = 108

Answer 1

Concepts and reason

The concepts required to solve the question are stress, strain, and thermal expansion.

First, write the equilibrium condition of the deflection of the bar by using the expression of the deflections due to stress and strain, and thermal expansion to find the force due to fixed ends.

Use the calculated force to find the normal stress in each bar of the material.

Stress: It is defined as the internal force which is offered by a material per unit area.

Strain: It is defined as the change in length of material to the original length of material.

Hooke’s law: Within elastic limit, stress is directly proportional to strain.

Thermal expansion: In thermal expansion, material changes its shape and size due to change in temperature.

Fundamentals

Write the expression of the stress.

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

Here, $\sigma$ is the stress, $P$ is the internal force offered by material and $A$ is the area of material at which force acts.

Write the expression of the strain.

$\varepsilon = \frac{\delta }{L}$

Here, $\varepsilon$ is the strain, $\delta$ is the change in length and $L$ is the original length.

From Hooke’s law, the stress can be written as,

$\begin{array}{c}\\\sigma \propto \varepsilon \\\\\sigma = E\varepsilon \\\\E = \frac{\sigma }{\varepsilon }\\\end{array}$

Here, $E$ is the modulus of elasticity.

Substitute $\frac{P}{A}$ for $\sigma$ and $\frac{\delta }{L}$ for $\varepsilon$ .

$\begin{array}{c}\\E = \frac{{\left( {\frac{P}{A}} \right)}}{{\left( {\frac{\delta }{L}} \right)}}\\\\E = \frac{{PL}}{{\delta A}}\\\\\delta = \frac{{PL}}{{EA}}\\\end{array}$

Write the expression of change in length due to thermal expansion.

${\delta _t} = \alpha L\left( {{T_2} - {T_1}} \right)$

Here, ${\delta _t}$ is the change in length due to thermal expansion, $\alpha$ is the coefficient of thermal expansion, $L$ is the length of material, ${T_1}$ is the initial temperature and ${T_2}$ is the final temperature.

Write the expression of the cross-sectional area of the bar.

$A = \frac{{\pi {d^2}}}{4}$

Here, $A$ is the area of the bar and $d$ is the diameter of the bar.

(A)

Free body diagram of the assembly can be shown as,

From above diagram, the displacement between $A$ to $D$ is zero because both the ends are fixed that can be written as,

$\begin{array}{l}\\{\delta _{AD}} = 0\\\\\left\{ \begin{array}{l}\\\frac{{ - F{L_{AB}}}}{{{A_{AB}}{E_{Al}}}} + {\alpha _{Al}}{L_{AB}}\left( {{T_2} - {T_1}} \right) + \frac{{ - F{L_{BC}}}}{{{A_{BC}}{E_{Br}}}}\\\\ + {\alpha _{Br}}{L_{BC}}\left( {{T_2} - {T_1}} \right) + \frac{{ - F{L_{CD}}}}{{{A_{CD}}{E_{St}}}} + {\alpha _{St}}{L_{CD}}\left( {{T_2} - {T_1}} \right) = 0\\\end{array} \right\}\\\\\left\{ \begin{array}{l}\\ - \frac{{4F{L_{AB}}}}{{\pi d_{AB}^2{E_{Al}}}} + {\alpha _{Al}}{L_{AB}}\left( {{T_2} - {T_1}} \right) - \frac{{4F{L_{BC}}}}{{\pi d_{BC}^2{E_{Br}}}}\\\\ + {\alpha _{Br}}{L_{BC}}\left( {{T_2} - {T_1}} \right) - \frac{{4F{L_{CD}}}}{{\pi d_{CD}^2{E_{St}}}} + {\alpha _{St}}{L_{CD}}\left( {{T_2} - {T_1}} \right) = 0\\\end{array} \right\}\\\end{array}$

Here, $F$ is the force that applied on the material, ${L_{AB}}$ is the length of the aluminum bar, ${A_{AB}}$ is the area of the bar $AB$ , ${E_{Al}}$ is the modulus of elasticity of aluminium bar, ${\alpha _{Al}}$ is the coefficient of the thermal expansion, ${d_{AB}}$ is the diameter of the aluminum bar and similarly for bronze and stainless steel.

From the experimental data, the coefficient of thermal expansion of ${\rm{2014 - T6}}$ aluminium is $12.8 \times {10^{ - 6}}{\rm{/^\circ F}}$ , the coefficient of thermal expansion of ${\rm{C86100}}$ bronze is $9.60 \times {10^{ - 6}}{\rm{/^\circ F}}$ , and the coefficient of thermal expansion of $304$ stainless steel is $9.6 \times {10^{ - 6}}{\rm{/^\circ F}}$ .

From the experimental data, the modulus of elasticity of the aluminum material is $10.6 \times {10^6}{\rm{ psi}}$ , the modulus of elasticity of the bronze material is $15 \times {10^6}{\rm{ psi}}$ , and the modulus of elasticity of stainless steel material is $28 \times {10^6}{\rm{ psi}}$ .

Substitute $\left( {4 \times 12} \right){\rm{ in}}$ for ${L_{AB}}$ , $12{\rm{ in}}$ for ${d_{AB}}$ , $10.6 \times {10^6}{\rm{ psi}}$ for ${E_{Al}}$ , $12.8 \times {10^{ - 6}}{\rm{/^\circ F}}$ for ${\alpha _{Al}}$ , $\left( {6 \times 12} \right){\rm{ in}}$ for ${L_{BC}}$ , ${\rm{8 in}}$ for ${d_{BC}}$ , $15 \times {10^6}{\rm{ psi}}$ for ${E_{Br}}$ , $9.60 \times {10^{ - 6}}{\rm{/^\circ F}}$ for ${\alpha _{Br}}$ , $\left( {3 \times 12} \right){\rm{ in}}$ for ${L_{CD}}$ , ${\rm{4 in}}$ for ${d_{CD}}$ , $28 \times {10^6}{\rm{ psi}}$ for ${E_{St}}$ , $9.60 \times {10^{ - 6}}{\rm{/^\circ F}}$ for ${\alpha _{St}}$ , $70{\rm{^\circ F}}$ for ${T_1}$ and $108{\rm{^\circ F}}$ for ${T_2}$ . $\begin{array}{c}\\\left[ \begin{array}{l}\\ - \frac{{4\left( {\left( {4 \times 12} \right){\rm{ in}}} \right)}}{{\pi {{\left( {12{\rm{ in}}} \right)}^2}\left( {10.6 \times {{10}^6}{\rm{ psi}}} \right)}} + \left( {12.8 \times {{10}^{ - 6}}{\rm{/^\circ F}}} \right)\left( {\left( {4 \times 12} \right){\rm{ in}}} \right)\left( {108{\rm{^\circ F}} - 70{\rm{^\circ F}}} \right)\\\\ - \frac{{4F\left( {\left( {6 \times 12} \right){\rm{ in}}} \right)}}{{\pi {{\left( {{\rm{8 in}}} \right)}^2}\left( {15 \times {{10}^6}{\rm{ psi}}} \right)}} + \left( {9.60 \times {{10}^{ - 6}}{\rm{/^\circ F}}} \right)\left( {\left( {6 \times 12} \right){\rm{ in}}} \right)\left( {108{\rm{^\circ F}} - 70{\rm{^\circ F}}} \right) - \\\\\frac{{4F\left( {\left( {3 \times 12} \right){\rm{ in}}} \right)}}{{\pi {{\left( {{\rm{4 in}}} \right)}^2}\left( {28 \times {{10}^6}{\rm{ psi}}} \right)}} + \left( {9.60 \times {{10}^{ - 6}}{\rm{/^\circ F}}} \right)\left( {\left( {3 \times 12} \right){\rm{ in}}} \right)\left( {108{\rm{^\circ F}} - 70{\rm{^\circ F}}} \right)\\\end{array} \right] = 0\\\\\left[ \begin{array}{l}\\ - 4 \times {10^{ - 8}}F + 23347.2 \times {10^{ - 6}}\\\\ - 9.55 \times {10^{ - 8}}F + 26265.6 \times {10^{ - 6}}\\\\ - 10.23 \times {10^{ - 8}}F + 13132.8 \times {10^{ - 6}}\\\end{array} \right]{\rm{ = 0}}\\\end{array}$

Further, solve,

$\begin{array}{c}\\23.78 \times {10^{ - 8}}F = 62745.6 \times {10^{ - 6}}\\\\F = 263858.70{\rm{ lbf}}\left( {\frac{{1{\rm{ kip}}}}{{1000{\rm{ lbf}}}}} \right)\\\\ = 263.8587{\rm{ kip}}\\\end{array}$

Calculate the average normal stress in the aluminum bar.

${\sigma _{AB}} = \frac{{4F}}{{\pi d_{AB}^2}}$

Substitute $12{\rm{ in}}$ for ${d_{AB}}$ and $263.8587{\rm{ kip}}$ for $F$ .

$\begin{array}{c}\\{\sigma _{AB}} = \frac{{4\left( {263.8587{\rm{ kip}}} \right)}}{{\pi {{\left( {12{\rm{ in}}} \right)}^2}}}\\\\ = 2.34{\rm{ ksi}}\\\end{array}$

(B)

Calculate the average normal stress in the bronze bar.

${\sigma _{BC}} = \frac{{4F}}{{\pi d_{BC}^2}}$

Substitute ${\rm{8 in}}$ for ${d_{BC}}$ and $263.8587{\rm{ kip}}$ for $F$ .

$\begin{array}{c}\\{\sigma _{BC}} = \frac{{4\left( {263.8587{\rm{ kip}}} \right)}}{{\pi {{\left( {{\rm{8 in}}} \right)}^2}}}\\\\ = 5.25{\rm{ ksi}}\\\end{array}$

Ans: Part A

The average normal stress in the aluminum bar is $2.34{\rm{ ksi}}$ .

Part B

The average normal stress in the bronze bar is $5.25{\rm{ ksi}}$ .