Question

Two steel rods and an aluminium rod of equal length $l_0$ and equal cross-sectional areas are joined rigidly together as shown in the figure below. All the rods are in a state of zero tension at $0{}^{\circ }\text{C}$. 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 $Y_s.$

• A. $[1+\frac{2Y_s{\alpha }_s\theta +Y_a{\alpha }_a\theta }{2Y_s+Y_a}]l_0$
• B. $[1-\frac{2Y_s{\alpha }_s\theta +Y_a{\alpha }_a\theta }{2Y_s+Y_a}]l_0$
• C. $2[1+\frac{2Y_s{\alpha }_s\theta +Y_a{\alpha }_a\theta }{2Y_s+Y_a}]l_0$
• D. $2[1-\frac{2Y_s{\alpha }_s\theta +Y_a{\alpha }_a\theta }{2Y_s+Y_a}]l_0$

Answer 1

The correct option is A $[1+\frac{2Y_s{\alpha }_s\theta +Y_a{\alpha }_a\theta }{2Y_s+Y_a}]l_0$

Given:
Length of each rod $=l_0$
Change in temperature, $\Delta T=\theta -0=\theta$
Coefficient of linear expansion of steel $={\alpha }_s$
Young's modulus of steel $=Y_s$
Coefficient of linear expansion of aluminium $={\alpha }_a$
Young's modulus of aluminium $=Y_a$
To find:
Length of the system, $l=?$


We know that,
Total strain $=$ Total stress $/$ Total Young's modulus
$\Rightarrow \frac{\Delta l}{l_0}=\frac{2\times Y_s\times \frac{\Delta l_s}{l_0}+Y_a\times \frac{\Delta l_a}{l_0}}{2Y_s+Y_a}$
[ from formula of Young's modulus ]

$\Rightarrow \frac{\Delta l}{l_0}=\frac{2\times Y_s\times \frac{l_0{\alpha }_s\theta }{l_0}+Y_a\times \frac{l_0{\alpha }_a\theta }{l_0}}{2Y_s+Y_a}$
[ from formula of linear expansion ]

$\Rightarrow \frac{\Delta l}{l_0}=\frac{2Y_s{\alpha }_s\theta +Y_a{\alpha }_a\theta }{2Y_s+Y_a}$
$\Rightarrow \Delta l=\left[\frac{2Y_s{\alpha }_s\theta +Y_a{\alpha }_a\theta }{2Y_s+Y_a}\right]l_0$
$\therefore l=l_0+\Delta l=[1+\frac{2Y_s{\alpha }_s\theta +Y_a{\alpha }_a\theta }{2Y_s+Y_a}]l_0$