Torsion distribution in a prismatic shaft fixed against twist at both ends: A 9 m long shaft is restrained at both ends and a torque of 30 tonne·metre is applied at a point 3 m from the nearer end (therefore 6 m from the far end). What is the reactive (balancing) torque at the nearer end?

Introduction / Context:
When a torque is applied at an intermediate section of a uniform circular shaft whose ends are fixed against rotation, equal and opposite reaction torques develop at the ends. These reactions distribute in proportion to the lengths of the shaft segments on either side to satisfy compatibility of twist.

Given Data / Assumptions:

• Total length L = 9 m; distances from torque point to ends: a = 3 m (near end), b = 6 m (far end).
• Applied torque T = 30 tonne·metre.
• Uniform torsional rigidity J G along the shaft; ends fixed (zero rotation).

Concept / Approach:
Compatibility requires the angle of twist of each segment (relative to the torque point) to be equal and opposite so that the net rotation at the load point is consistent. For a uniform shaft, reactions are proportional to the opposite segment lengths: T_near = T * b / L and T_far = T * a / L, with T_near + T_far = T.

Step-by-Step Solution:
Compute T_near: T_near = T * b / L = 30 * 6 / 9 = 20 tonne·metre.Compute T_far: T_far = T * a / L = 30 * 3 / 9 = 10 tonne·metre.Check: T_near + T_far = 20 + 10 = 30 tonne·metre (balances applied torque).

Verification / Alternative check:
Angles of twist: θ_near ∝ T_near * a and θ_far ∝ T_far * b; substituting shows equal and opposite rotations, confirming compatibility.

Why Other Options Are Wrong:
5, 10, 15 tonne·metre do not satisfy both equilibrium and compatibility with given distances; 30 tonne·metre ignores the share at the far end.

Common Pitfalls:
Allocating reactions proportional to the same-side length (should be opposite-side length), or forgetting the sum must equal the applied torque.

Final Answer:
20 tonne·metre.