The maximum shear stress occurs on the outermost fibres of a circular shaft under torsion. In a close-coiled helical spring, the maximum shear stress occurs on the:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 01/03/2017 (Evening)
Detailed Explanation
In a straight shaft the shear stress is the same all round the surface. In a spring wire three effects combine on a section: torsional shear (± across the diameter), direct shear from W (same direction everywhere), and curvature of the coil.
Why “innermost fibres” Is Correct
On the side facing the spring axis the direct shear adds to the torsional shear, and because the coil is curved the fibres on the inside are shorter and strain more. So the maximum shear stress is at the innermost fibres — which is where spring wires usually crack. Wahl's factor K = (4C − 1)/(4C − 4) + 0.615/C (C = D/d) accounts for this.
Why the Other Options Are Wrong
- A. outermost fibres: At the outer fibres the direct shear opposes the torsional shear.
- B. fibres at mean diameter: The fibre at the mean coil diameter is the wire centre, where torsional shear is zero.
- D. none of these: Option (c) is correct.
Key Point
Spring wire: τmax at the inner fibre; τ = K·16WR/πd³ with Wahl's factor K.
Key Concepts for Students
- Spring index C = D/d.
- Smaller C → bigger Wahl correction.
- Fatigue cracks start on the inside of the coil.
Practice more: this is question 12 in Spring MCQs – Part 1. Try the whole set and check every answer instantly.