Final answer:
The initial stress in the tie bar is approximately 50.92 x 10⁶ N/m². When the temperature rises to 50°C, the thermal stress is about 98 x 10⁶ N/m², leading to a resultant stress of approximately 148.92 x 10⁶ N/m². The induced force in the tie bar remains 100 kN, since the bar is rigidly held.
Step-by-step explanation:
To solve for the initial stress in the tie bar, we will apply the formula for stress (σ = F/A), where F is the force applied and A is the cross-sectional area of the bar. The initial cross-sectional area (A) can be calculated using the formula for the area of a circle (A = πd²/4). Given that the diameter (d) is 50 mm, we first convert this to meters (0.050 m) to match the units of the given modulus of elasticity (E = 200 GN/m²).
A = π(0.050)²/4 ≈ 0.001963495 m²
Using F = 100 kN (which is 100,000 N to match unit consistency with E), we get:
σ_initial = 100,000 N / 0.001963495 m² ≈ 50.92 x 10⁶ N/m²
To determine the thermal stress when the temperature rises to 50°C, we use the relationship Δσ = EαΔT, where α is the coefficient of linear expansion and ΔT is the change in temperature.
ΔT = 50°C - 15°C = 35 K
Δσ = (200 x 10⁹ N/m²)(14 x 10⁻⁶ /K)(35 K) = 98 x 10⁶ N/m²
Because the bar is rigidly held, it cannot expand, which means this stress is purely thermal and adds onto the initial mechanical stress. Therefore, the resultant stress will be the sum of the initial stress and the thermal stress:
σ_resultant = σ_initial + Δσ ≈ 50.92 x 10⁶ N/m² + 98 x 10⁶ N/m² = 148.92 x 10⁶ N/m²
Considering the bar is constrained, there will be no additional force apart from the initial force applied; hence, the induced force remains 100 kN.