An 'SS' tube is inserted into an 'Al' tube. They are permanently fixed at one end. The other end is attached to a rigid plate. A torque 'T' is applied to the rigid plate. The circumference of the 'Al' tube at dia 'D' at the plate and with respect to the fixed and rotates by a distance ............... mm due torque 'T'.
- A. DLT / 2(G_{Al}J_{Al} - G_{ss}J_{ss})
- B. DLT / 2G_{Al}J_{Al}
- C. DLT / 2(G_{Al}J_{Al} + G_{ss}J_{ss})
- D. 2DLT / G_{ss}J_{ss}
Answer: C. DLT / 2(G_{Al}J_{Al} + G_{ss}J_{ss})
Explanation: Tubes in parallel. Total Torque T = T_Al + T_ss. Twist \theta is same. T_i = G_i J_i \theta / L. T = (G_Al J_Al + G_ss J_ss) \theta / L. So \theta = TL / (G_Al J_Al + G_ss J_ss). Circumferential displacement d = R\theta = (D/2) \theta = DLT / 2(G_Al J_Al + G_ss J_ss).