Topic: Strength of Materials
A rod has coefficient of thermal expansion $31.4\times10^{-6}/^{\circ}C$ and Young's modulus 200 GPa. It is fitted snugly between the supports as shown in the figure and heated. The stress induced in it due to $20^{\circ}C$ rise in temperature will be
- A. 0.1 MPa
- B. - 0.1 MPa
- C. 0.2 MPa
- D. 0.2 MPa
Answer: D. 0.2 MPa
Explanation: Rod length $L = 0.5~m$, dia $d = 10~mm$, spring stiffness $k = 50~kN/m = 50000~N/m$.
Free thermal expansion $\Delta L_{th} = L \alpha \Delta T = 0.5 \times 31.4 \times 10^{-6} \times 20 = 314 \times 10^{-6}~m$.
When the rod expands against the spring, it undergoes a net expansion $x$. The spring reaction force is $F = kx$.
The rod also shortens elastically due to this force $F$: $\Delta L_{el} = \frac{FL}{AE}$.
Compatibility equation: $\Delta L_{th} = x + \Delta L_{el} \Rightarrow L \alpha \Delta T = \frac{F}{k} + \frac{FL}{AE} = F \left( \frac{1}{k} + \frac{L}{AE} \right)$.
Area $A = \frac{\pi}{4} (0.01)^2 = 78.54 \times 10^{-6}~m^2$.
$\frac{1}{k} = \frac{1}{50000} = 2 \times 10^{-5}~m/N$.
$\frac{L}{AE} = \frac{0.5}{78.54 \times 10^{-6} \times 200 \times 10^9} = 3.18 \times 10^{-8}~m/N$.
Since $\frac{L}{AE}$ is very small compared to $\frac{1}{k}$, $F \approx \frac{L \alpha \Delta T}{1/k} = k L \alpha \Delta T = 50000 \times 314 \times 10^{-6} = 15.7~N$.
Stress $\sigma = \frac{F}{A} = \frac{15.7}{78.54 \times 10^{-6}} = 200,000~Pa = 0.2~MPa$ (Compressive).