Question

A steel bolt of cross sectional area${A_b} = 5 \times {10^{ - 5}}{m^2}$ is passed through a cylindrical tube made of aluminum. Cross-sectional area of the tube material is${A_t} = {10^{ - 4}}{m^2}$ and its length is$L = 50cm$. The bolt is just taut so that there is no stress in the bolt and temperature of the assembly increases through$\Delta \theta = 10^\circ C$. Given, coefficient of linear thermal expansion of steel, ${\alpha _b} = {10^{ - 5}}\dfrac{1}{{^\circ C}}$.
Young’s modulus of steel, ${Y_b} = 2 \times {10^{11}}\dfrac{N}{{{m^2}}}$.
Young’s modulus of Al,${Y_t} = {10^{11}}\dfrac{N}{{{m^2}}}$, coefficient of linear thermal expansion of Al ${\alpha _t} = 2 \times {10^{ - 5}}\dfrac{1}{{^\circ C}}$.
The tensile strength of bolt is:
A. ${10^4}\dfrac{N}{{{m^2}}}$
B. ${10^7}\dfrac{N}{{{m^2}}}$
C. $2 \times {10^8}\dfrac{N}{{{m^2}}}$
D. ${10^{10}}\dfrac{N}{{{m^2}}}$

Answer 1

Hint:-The stress is the ratio of applied force and area on the surface. Strain is defined as the ratio of change in length to the applied length. Also the Young’s modulus is defined as the ratio of stress and strain.
Formula used: The formula of relationship between stress and strain is given by ${\text{Young's Modulus}} = \dfrac{{{\text{stress}}}}{{{\text{strain}}}}$. The formula of strain is given by${\text{Strain}} = \dfrac{{\Delta l}}{l}$.

Complete step-by-step solution:It is given that a cylindrical tube made of aluminum material of cross sectional area ${A_t} = {10^{ - 4}}{m^2}$ and length $L = 50cm$also the cross section of the steel bolt is${A_b} = 5 \times {10^{ - 5}}{m^2}$. The increase in the temperature is$\Delta \theta = 10^\circ C$. The coefficient of thermal expansion of steel is${\alpha _b} = {10^{ - 5}}\dfrac{1}{{^\circ C}}$.
Let us calculate the actual increase in the length of the aluminum as there is an increase in the temperature of the aluminum by$\Delta \theta = 10^\circ C$.
The change in length of the aluminum is given by,
$ \Rightarrow \Delta L = \alpha {L_{al.}}\Delta T$
Replace the value of length of the aluminum with the thermal expansion of the aluminum and change in temperature of the aluminum.
$L = 0 \cdot 50m$, $\Delta T = 10^\circ C$, ${\alpha _t} = 2 \times {10^{ - 5}}\dfrac{1}{{^\circ C}}$.
$ \Rightarrow \Delta {L_{al.}} = \alpha {L_{al.}}\Delta T$
$ \Rightarrow \Delta {L_{al.}} = \left( {2 \times {{10}^{ - 5}}} \right) \cdot \left( {0 \cdot 5} \right) \cdot \left( {10} \right)$
$ \Rightarrow \Delta {L_{al.}} = {10^{ - 4}}m$
The change in length of the aluminum tube is$\Delta {L_{al.}} = {10^{ - 4}}m$.
As the change in the length of bolt is equal to the change in the length of aluminum.
The strain produced in aluminum tube is given by,
$strain = \dfrac{{\Delta {L_{{\text{aluminum}}}}}}{{{L_{{\text{aluminum}}}}}}$
Replace the value of the change in length and original length of aluminum.
$ \Rightarrow strain = \dfrac{{\Delta {L_{{\text{aluminum}}}}}}{{{L_{{\text{aluminum}}}}}}$
$strain = \dfrac{{{{10}^{ - 4}}}}{{0 \cdot 5}}$
$strain = 5 \times {10^{ - 5}}$
As the strain will be the same in bolt as well as aluminum. Now we can calculate the stress in the bolt.
Since,
$stress = {Y_{steel}} \cdot \left( {strain} \right)$
Replace the value of strain and young’s modulus of steel we get.
$ \Rightarrow stress = {Y_{steel}} \cdot \left( {strain} \right)$
$ \Rightarrow stress = \left( {2 \times {{10}^{11}}} \right) \cdot \left( {5 \times {{10}^{ - 5}}} \right)$
$ \Rightarrow stress = {10^7}\dfrac{N}{{{m^2}}}$
The stress on the bolt is equal to$stress = {10^7}\dfrac{N}{{{m^2}}}$. The correct option is option B.

Note:-
 The aluminum tube material would have expanded more but due to the presence of the steel bolt the aluminum tube cannot expand so the steel bolt applies certain force onto the aluminum material to stop its expansion.