Question

There is a copper rod of square cross-section, the dimension of each side being 2a, but having a square hole through out the rod, dimension of each side of hole being a. A brass rod exactly fitting inside the hole is inserted in the hole. Now, both sides of the rod are welded together. The temperature is elevated by q. Neglecting any lateral expansion or stresses developed, the composite length of the system at temperature q will be –

• A. L₀$\left\lbrack 1 + \frac{\alpha_{Br}Y_{Br} + \alpha_{Cu}Y_{Cu}}{Y_{Br} + Y_{Cu}}(\theta) \right\rbrack$
• B. L₀$\left\lbrack 1 + \frac{\alpha_{Br}Y_{Br} - \alpha_{Cu}Y_{Cu}}{Y_{Br} - Y_{Cu}}(\theta) \right\rbrack$
• C. L₀$\left\lbrack 1 + \frac{3\alpha_{Br}Y_{Br} + \alpha_{Cu}Y_{Cu}}{3Y_{Br} + Y_{Cu}}(\theta) \right\rbrack$
• D. L₀$\left\lbrack 1 + \frac{\alpha_{Br}Y_{Br} + 3\alpha_{Cu}Y_{Cu}}{Y_{Br} + 3Y_{Cu}}(\theta) \right\rbrack$

Answer 1

Answer: (D)

L₀$\left\lbrack 1 + \frac{\alpha_{Br}Y_{Br} + 3\alpha_{Cu}Y_{Cu}}{Y_{Br} + 3Y_{Cu}}(\theta) \right\rbrack$

Answer 2

F₁ = F₂

$\frac{Y_{Cu}(3a^{2})(x–L_{0}\alpha_{Cu}\theta)}{L_{0}}$

= $\frac{Y_{Br}(a^{2})(L_{0}\alpha_{Br}\theta –x)}{L_{0}}$

Composite length = L₀ + x

= L₀ $\left\lbrack 1 + \frac{\alpha_{Br}Y_{Br} + 3Y_{Cu}\alpha_{Cu}}{3Y_{Cu} + Y_{Br}}(\theta) \right\rbrack$