Question

A uniform heavy plank of length L is resting on two rods of identical length and area of cross section, made up of different materials. It is observed that to keep the plank horizontal, one rod is to be at the end of the plank (say rod A) and another rod is to be at separation $\frac{L}{4}$ from centre of plank as shown in figure. Now the rods are kept between two rigid walls such that separation between the walls is equal to the natural length of the rod. Now temperature of both the rods is gradually increased in same manner, it is observed that thermal stress developed in rod A is always thrice the thermal stress developed in another rod. Ratio of coefficient of linear expansion of rod A to coefficient of linear expansion of another rod will be :

Answer 1

6

Answer 2

Let $L$ be the length of the uniform heavy plank and $W$ be its weight. The weight acts at the center of the plank. Let the rods have identical natural length $L_0$ and area of cross section $A$. Let the Young's modulus and coefficient of linear expansion of rod A be $E_A$ and $\alpha_A$, and those of the other rod (rod B) be $E_B$ and $\alpha_B$.

In the first scenario, the plank is resting on the two rods and is horizontal. Rod A is at one end of the plank (say, the left end, $x=0$). The center of the plank is at $x=L/2$. The other rod (rod B) is at a distance $L/4$ from the center. From the figure, rod B is to the right of the center, so its position is $x = L/2 + L/4 = 3L/4$. Let $R_A$ and $R_B$ be the normal reactions exerted by rods A and B on the plank. For vertical equilibrium, $R_A + R_B = W$. For rotational equilibrium about the left end ($x=0$), the torque due to the weight at $x=L/2$ must balance the torque due to the reaction $R_B$ at $x=3L/4$.

$W \times (L/2) = R_B \times (3L/4)$

$R_B = \frac{W(L/2)}{3L/4} = \frac{2W}{3}$.

Then $R_A = W - R_B = W - \frac{2W}{3} = \frac{W}{3}$.

The forces exerted by the plank on the rods are $F_A = R_A = W/3$ and $F_B = R_B = 2W/3$.

Since the plank remains horizontal, the vertical displacement of the top surfaces of the rods must be the same. As the rods have identical natural length $L_0$ and are supported at the bottom on a rigid surface, this means the compression in both rods is the same. Let the compression be $\Delta L$.

The compression in a rod is related to the force by $\Delta L = \frac{FL_0}{EA}$.

For rod A, $\Delta L = \frac{F_A L_0}{E_A A} = \frac{(W/3) L_0}{E_A A}$.

For rod B, $\Delta L = \frac{F_B L_0}{E_B A} = \frac{(2W/3) L_0}{E_B A}$.

Since the compressions are equal, $\frac{(W/3) L_0}{E_A A} = \frac{(2W/3) L_0}{E_B A}$.

$\frac{W/3}{E_A} = \frac{2W/3}{E_B}$.

$\frac{1}{E_A} = \frac{2}{E_B}$, which gives $\frac{E_B}{E_A} = 2$.

In the second scenario, the rods are kept between rigid walls such that the separation is equal to their natural length $L_0$. The temperature of both rods is increased by the same amount $\Delta T$. The thermal expansion is prevented by the rigid walls, leading to thermal stress. The free thermal expansion of a rod would be $\Delta L_{thermal} = L_0 \alpha \Delta T$. The rigid walls exert a compressive force that causes a mechanical compression $\Delta L_{mechanical} = - \Delta L_{thermal} = - L_0 \alpha \Delta T$. The strain is $\epsilon = \frac{\Delta L_{mechanical}}{L_0} = - \alpha \Delta T$. The thermal stress is $\sigma = E \epsilon = E (-\alpha \Delta T)$. The magnitude of the thermal stress is $|\sigma| = E \alpha \Delta T$.

For rod A, the thermal stress is $\sigma_A = E_A \alpha_A \Delta T$.

For rod B, the thermal stress is $\sigma_B = E_B \alpha_B \Delta T$.

We are given that the thermal stress developed in rod A is always thrice the thermal stress developed in another rod: $\sigma_A = 3 \sigma_B$.

$E_A \alpha_A \Delta T = 3 (E_B \alpha_B \Delta T)$.

Since the temperature increase $\Delta T$ is the same and non-zero, we can cancel it out:

$E_A \alpha_A = 3 E_B \alpha_B$.

We want to find the ratio of the coefficient of linear expansion of rod A to that of another rod, which is $\frac{\alpha_A}{\alpha_B}$.

From the equation, $\frac{\alpha_A}{\alpha_B} = 3 \frac{E_B}{E_A}$.

Using the result from the first scenario, $\frac{E_B}{E_A} = 2$.

$\frac{\alpha_A}{\alpha_B} = 3 \times 2 = 6$.

The ratio of the coefficient of linear expansion of rod A to coefficient of linear expansion of another rod is 6.