Question

Two steel rods and an aluminium rod of equal length $l_0$ and equal cross section are joined rigidly at their ends as shown in the figure below. All the rods are in a state of zero tension at $0^{\circ }C$. Find the length of the system when the temperature is raised to $\theta$. Coefficient of linear expansion of aluminium and steel are ${\alpha }_a$ and ${\alpha }_s$ respectively. Young's modulus of aluminium is $Y_a$ and of steel is $Y_s$.

$\begin{array}{c}\text{Steel}\\ \text{Aluminium}\\ \text{Steel}\end{array}$

Answer 1

Let the initial length of the system at $0^{\circ }C=l_0$

When temperature changes by $\theta$

Final length of the a tool at system temperature

So, strain of the system $=\frac{(l-l_0)}{l_0}$

$\begin{array}{c}\text{Steel}\\ \text{Aluminium}\\ \text{Steel}\end{array}$

But the total strain of the system

$=\frac{\text{total stress of system}}{\text{total Young's modulus of system}}$

Now, total stress due to two steel rod + stress due to aluminium

$={\gamma }_s{\alpha }_s\theta +{\gamma }_s{\alpha }_s\theta +{\gamma }_{Al}{\alpha }_{Al}\theta$

$=2{\gamma }_s{\alpha }_s\theta +{\gamma }_{Al}{\alpha }_{Al}\theta$

Now Young's modulus system

$={\gamma }_s+{\gamma }_s+{\gamma }_{Al}=2{\gamma }_s+{\gamma }_{Al}$

$\therefore$ Strain of system $=\frac{2{\gamma }_s{\alpha }_s\theta +{\gamma }_{Al}{\alpha }_{Al}\theta }{2{\gamma }_s+{\gamma }_{Al}}$

$\Rightarrow \frac{l-l}{l_0}=\frac{2{\gamma }_s{\alpha }_s\theta +{\gamma }_{Al}{\alpha }_{Al}\theta }{2{\gamma }_s+{\gamma }_{Al}}$

$\Rightarrow l=l_0[1+\frac{2{\gamma }_s{\alpha }_s\theta +2{\gamma }_{Al}{\alpha }_{Al}\theta }{2{\gamma }_s+{\gamma }_{Al}}]$