Let's break down the problem: ## **Given Data** - Length of steel beam, \( L = 5 \) m - Temperature difference, \( \Delta T = 50^\circ C \) - Coefficient of linear expansion, \( \alpha = 11.7 \times 10^{-6} / ^\circ C \) - Young's modulus, \( E = 210 \) MPa = \( 210 \times 10^6 \) Pa --- ## **(a) Stress when both ends are fixed** Thermal stress when expansion is prevented: \[ \sigma = E \alpha \Delta T \] Substitute the values: \[ \sigma = 210 \times 10^6 \times 11.7 \times 10^{-6} \times 50 \] Calculate stepwise: - \( 11.7 \times 50 = 585 \) - \( 585 \times 10^{-6} = .000585 \) - \( 210 \times 10^6 \times .000585 = 122,850 \) Pa = **112.85 MPa** --- ## **(b) Change in length when one end is free** Thermal expansion: \[ \delta = L \alpha \Delta T \] \[ \delta = 5 \times 11.7 \times 10^{-6} \times 50 \] - \( 11.7 \times 50 = 585 \) - \( 5 \times 585 = 2925 \) - \( 2925 \times 10^{-6} = .002925 \) m --- ## **Matching with Options** - (a) \( \sigma = 112.85 \) MPa, (b) \( \delta = .002925 \) m ### **Correct Option:** > **Option 1** > > (a) \( \sigma = 112.85 \) MPa (b) \( \delta = .002925 \) m **(Note: Watch the units: .002925 m = 2.925 mm. The option uses meters, which matches the calculation.)**