Topic: Strength of Materials
A composite bar made of aluminum and steel is fixed rigidly at supports P and Q as shown in figure. When the temperature of the bar is increased by 10 °C the reactions at the supports are (given E_al = 70 GPa, E_s = 210 GPa, alpha_al = 11E-6 mm/mm °C and alpha_s = 12E-6 mm/mm °C)
- A. 4060 N
- B. 6780 N
- C. 12240 N
- D. 2412 N
Answer: A. 4060 N
Explanation: Total deflection due to temperature rise is (alpha * t * L)_s + (alpha * t * L)_Al = 0.58 mm. Equating this to mechanical elongation (PL/AE)_s + (PL/AE)_Al gives 0.58 = P/7000, which yields P = 4060 N.