Question

Q.3. A hollow steel shaft ACB of outside diameter 50 mm and inside diameter 40 mm is fixed at ends A and B. Horizontal forces P are applied at the ends of a vertical arm that is welded to the shaft at point C. The allowable shear stress in the shaft is tall 45 MPa. Determine the maximum allowable force P 200 mm 200 mm 600 mm 400 mm

Answer 1

T_A = T_0 L_B/L T_B = T + 0L_A/L The larger and hence the layer shear stress, occur in part CB of the shaft T_0 = p(400 mm) L_B = 400 mm L_A = 600 mm L = L_A + L_B = 100 mm d_c = 50 mm, d_1 = 40 mm tau_allow - 45 Mpa T_A = T_OlB/L = p(0.4)(400)/1000 = 0.16 P T_B = T_0L_A/L = p(0.4)(600)/1000 = 0.24 P Therefore the layer torque, and hence the layer shear stress, occurs in part CB of the shaft Therefore T_max = T_B = 0.24 P shear stress in part CB tau_max = T_max(d/2)/I_p, T_max = 2 tau_max I_p/d tau_max = 45 times 10^6: I + p = pi/32(d^4_2 - d^7_1) = 362.26 times 10^- 9 m^4 d = d_2 = 0.05 mm 0.24 p = 2(45 times 10^6)(362.26 times 10^- 9)/0.05 = 652.07 Nm P = 652.07/0.24 = 2717 N P_ = 2720