How does the torsional stiffness of a hollow shaft compare to that of a solid shaft of the same length and weight?

• A. Equal to that of the solid shaft
• B. Less than that of the solid shaft
• C. More than that of the solid shaft
• D. Exactly half of that of the solid shaft

Answer: (C)

The torsional stiffness of a hollow shaft is indeed more than that of a solid shaft of the same length and weight. This is due to the geometric properties of the shafts.

Torsional stiffness is a measure of how resistant a shaft is to twisting. For shafts, this resistance is influenced heavily by their cross-sectional area and moment of inertia. A hollow shaft effectively offers more material distribution away from the center axis compared to a solid shaft, which increases its polar moment of inertia. The polar moment of inertia for a hollow shaft is given by the formula ( J = \frac{\pi}{2} (R_o^4 - R_i^4) ), where ( R_o ) is the outer radius and ( R_i ) is the inner radius.

Since both the height and weight constraints are kept the same, a hollow shaft can have a larger moment of inertia while being lighter than the solid shaft. This geometry allows the hollow shaft to be stiffer under torsional loads because the stiffness is directly proportional to the polar moment of inertia. Thus, for equal weight, the hollow shaft can exhibit superior torsional stiffness compared to a solid shaft.

The mechanics of materials behind this principle highlight how material distribution and shape directly impact performance