Question

A brass rod of length \[50cm\] and diameter \[3.0mm\]is joined to a steel rod with the same length and diameter. What is the change in the length of the combined rod at $250^{\circ}C$, if the original lengths are at $40^{\circ}C$? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (coefficient of linear expansion of brass=$2.0\times 10^{-5}K^{-1},$ steel $=1.2\times 10^{-5} K^{-1})$

Answer 1

Hint: To find the change in total length, we need to find the individual length and add them up. To find the individual length, we can use $\alpha_{L}=\dfrac{dL}{dT}$. Since all the required values, like the coefficient of linear expansion, initial length, initial and final temperature of the material is given, we can substitute the values to find the change in individual length.

Formula used:
$\alpha_{L}=\dfrac{dL}{dT}$

Complete step by step answer:
When materials, like brass and steel, are heated for a long period of time, they undergo thermal expansion. This property is called, Coefficient of linear expansion, which is the measure at which unit length changes with respect to change in temperature.
It is written as $\alpha_{L}=\dfrac{dL}{dT}$, where $\alpha_{L}$ is the coefficient of linear expansion, \[dL\] is the change in unit length and \[dT\]is the unit change in temperature.
Given that, the initial temperature $T_{1}=40^{\circ}C$ and $T_{2}=250^{\circ}C$, then $dT=T_{2}-T_{1}=250^{\circ}-40^{\circ}=210^{\circ}C.$
Also, the length and diameter of the brass rod at $T_{1}$ is , $L_{b}=50cm, D_{b}=3cm$ respectively.
Similarly, the length and diameter of the steel rod at $T_{1}$is , $L_{s}=50cm, D_{s}=3cm$ respectively.
The coefficient of linear expansion of brass $\alpha_{b}=2.0\times 10^{-5}K^{-1}$ and the coefficient of linear expansion of steel$\alpha_{s}=1.2\times 10^{-5} K^{-1}$
Then, $dL=\dfrac{\Delta L}{L}=\alpha_{L}dT$ or $\Delta L=\alpha_{L}\times dT \times L$
Substituting the values we get, change in length of the brass rod as
$\Delta L_{b}=\alpha_{b} \times dT\times L_{b}=2.0\times 10^{-5} \times 210\times 50 =0.22cm$
Similarly, the change in length of the steel rod will be given by:
 $\Delta L_{s}=\alpha_{s} \times dT\times L_{s}= 1.2\times 10^{-5} \times 210\times 50 =0.12cm$
Then the change in total length is given by, $\Delta L=\Delta L_{s}+\Delta L_{b}=0.22+0.12=0.34cm$
Since there is no thermal stress, the rod can expand freely on both sides.

Note:
The diameter of the rods is not needed to solve the question. Also, the change in temperature is the same for both Celsius and Kelvin scales. Also note that $dT=T_{2}-T_{1}$ and $dL=\dfrac{\Delta L}{L}$. Clearly, $\alpha \propto dL$ and $\alpha \propto \dfrac{1}{dT}$. \[dL\] has no units, and thus, $\alpha_{L}$ has units $K^{-1}$.