This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Step 1: Calculate the allowable working stress. The allowable working stress (_w) is determined by dividing the ultimate stress (_u) by the factor of safety (FOS). _w = (_u)/(FOS) Given _u = 480 \, MPa and FOS = 4: _w = 480 \, MPa4 _w = 120 \, MPa The allowable working stress is 120 MPa. Step 2: Determine the required cross-sectional area of the tube. The tensile load (P) is 400 \, kN, which is 400 × 10^3 \, N. The allowable working stress (_w) is 120 \, MPa, which is 120 × 10^6 \, Pa (N/m^2). The area (A) is calculated using the formula _w = (P)/(A), so A = (P)/(_w). A = 400 × 10^3 \, N120 × 10^6 \, Pa A = (400)/(120000) \, m^2 A = (1)/(300) \, m^2 ≈ 0.003333 \, m^2 Step 3: Calculate the outside diameter of the tube. The tube is hollow with an inside diameter (d_i) of 100 \, mm = 0.1 \, m. The area of a hollow tube is given by: A = ()/(4) (d_o^2 - d_i^2) We need to solve for the outside diameter (d_o). Rearrange the formula: (4A)/() = d_o^2 - d_i^2 d_o^2 = (4A)/() + d_i^2 d_o = sqrt((4A)/() + d_i^2) Substitute the values for A and d_i: d_o = sqrt(4 × (1/300) \, m)^2 + (0.1 \, m)^2 d_o = sqrt((1)/(75) + 0.01) \, m d_o = sqrt(0.00424413 + 0.01) \, m d_o = sqrt(0.01424413) \, m d_o ≈ 0.119348 \, m Convert the outside diameter to millimeters: d_o ≈ 0.119348 × 1000 \, mm d_o ≈ 119.35 \, mm The outside diameter of the tube is 119.35 mm. That's 2 down. 3 left today — send the next one.