linear thermal expansion coef brass 19e-6 /K ∆L = αL∆T = (19e-6)(1.85)(110) = 0.00387 meter or 3.87 mm
Second part involves linear elasticity. for brass, young's modulus is 15e6 psi or 100 GPa cross-sectional area of rod is π(0.008)² = 0.0002 m² F = EA∆L/L F = (100e9)(0.0002)(0.00387) / (1.85) F = 42000 or 42 kN
