Final answer:
The average normal stress at the midsection of rod BC is calculated to be 43.3 MPa, using the cross-sectional area and the given force applied to the rod.
Step-by-step explanation:
The average normal stress at the midsection of rod BC can be found by first calculating the cross-sectional area of the rod and then using the formula for stress, which is σ = P / A, where σ is stress, P is the applied force, and A is the area. Given that d2 = 50 mm for rod BC, convert this to meters (0.05 m) for consistency in units. The area, A, for a cylinder is π(d/2)^2. Therefore, A = π(0.05/2)^2 square meters. With P = 85 kN (which is 85000 N), we can calculate:
Area A = π(0.025)^2 ≈ 1.9635e-3 m²Stress σ = P / A = 85000 N / 1.9635e-3 m² ≈ 4.33e7 Pa
Converting to megapascals (MPa), the stress is 43.3 MPa. This is the average normal stress at the midsection of rod BC under the given load.