Question

When a shaft is subjected to torsion, the shear stress induced in the shaft varies from ---- at the centre and ---- at the circumference.

• A. minimum, maximum
• B. maximum, minimum
• C. zero, maximum
• D. maximum, zero

Hint:

Shear stress in torsion varies linearly from the center to the outer surface of the shaft.

Answer 1

The Correct Option is C

Step 1: In torsion of a circular shaft, plane cross-sections remain plane and simply rotate relative to one another, so the shear strain at radius \( r \) is directly proportional to \( r \): \( \gamma(r) \propto r \).
Step 2: Since shear stress is proportional to shear strain (\( \tau = G\gamma \)), stress also varies linearly with \( r \): \( \tau(r) \propto r \).
Step 3: Evaluating at the two extreme radii: at \( r = 0 \) (the centre), \( \tau = 0 \); at \( r = R \) (the outer surface), \( \tau \) takes its largest value:
\[ \tau(0) = 0, \qquad \tau(R) = \tau_{\max} \]

Answer 2

A helpful analogy is the way bending stress in a beam varies linearly from zero at the neutral axis to a maximum at the outer fibres. Torsion behaves the same way but with the shaft's central axis playing the role of the neutral axis, since it is the one line in the cross-section that experiences no shear during twisting. Testing the options with this analogy:

1. Minimum, maximum: Just as bending stress is exactly zero (not merely small) at the neutral axis, torsional shear stress is exactly zero at the shaft's axis, so calling it a "minimum" understates this special condition.
2. Maximum, minimum: This would be equivalent to claiming a beam's bending stress is largest at the neutral axis and smallest at the outer fibres, which contradicts the well-known linear stress distribution in both bending and torsion.
3. Zero, maximum: This mirrors the beam analogy exactly: stress is zero along the central axis (where there is no relative sliding) and grows linearly to a maximum at the farthest material from that axis, the outer circumference.
4. Maximum, zero: This would require the outer surface to be the "neutral" location with no shear, which is physically backwards since the outer fibres undergo the largest twist-induced displacement.

The bending-stress analogy confirms shear stress in torsion is zero at the centre and maximum at the circumference.

Therefore, the correct answer is zero, maximum.