Torsion of a Circular Bar (Statically Indeterminate) – Using Superposition

Example: Bar ABCD shown below is clamped at both ends and subjected to applied torques
T and 2T. The bar has cross-section with polar moment of inertia, J, and shearing modulus of

elasticity, G. Find the restraint torques at each end and the maximum shear stress in the bar.

Strategy: Use a free body diagram to identify the support torques at A and at D (shown
in the top figure below.) Next replace the original bar with three bars the first loaded by T,

the second loaded by 2T, and the third loaded by T_D (as shown in the bottom three figures.)

Then apply the torque-rotation relation given by ∆_i = T_iL_i/J_iG_i for each of the three separate

bars. The geometrical relation is that the net rotation is zero. Solve for the unknown torques.

Finally pass sections to determine the torque in each section of the bar. Use it to calculate

the shear stress in each section.

For equilibrium,

→ΣT_x = 0 – T_A – T + 2T + T_D = 0 and for the data: T_A – T_D = T (1)

Let ∆ denote rotation. Now the total rotation sums to zero.

∆₁ + ∆₂ + ∆₃ = 0

So – T(L/2)/JG + (2T)(3L/4)/JG + (T_D)(L)/JG = 0

which yields T_D = – T (as before) and from (1) T_A = 0 (as before)

The shear stresses in each section is the same as before. Use T_i c/ J where c is the radius

of the bar and T_i is the torque in each section.

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