A close-coiled helical spring is subjected to a torque about its axis. The spring wire would experience:
Try it first: pick an option, then read the explanation.
Correct answer: A. bending stress
Asked in SSC JE 01/03/2017; SSC JE 2008
Detailed Explanation
Now the loading is a couple about the spring's own axis (winding or unwinding the coils), not an axial force.
Why “bending stress” Is Correct
In a close-coiled spring the wire lies almost in a plane perpendicular to the spring axis. A couple about the spring axis is therefore a couple about an axis perpendicular to the wire — a pure bending moment M = T, giving σ = 32T/(πd³).
| Close-coiled helical spring | Axial load W | Axial torque (couple) T |
|---|---|---|
| Main action on the wire | Twisting (torsion) T = WR | Bending M = T |
| Maximum stress | τ = 16WR/(πd³) | σ = 32T/(πd³) |
| Deformation | δ = 64WR³n/(Gd⁴) | θ = 64TDn/(Ed⁴) = TL/(EI), L = πDn |
| Stiffness | k = W/δ = Gd⁴/(64R³n) | T/θ = Ed⁴/(64Dn) |
R = mean coil radius, D = 2R mean coil diameter, d = wire diameter, n = number of active coils.
Why the Other Options Are Wrong
- B. direct tensile stress of uniform intensity at its cross-section: There is no axial force in the wire, so no uniform tension.
- C. direct shear stress: No transverse force acts, so there is no direct shear.
- D. torsional shearing stress: Torsion of the wire is caused by an axial load, not an axial torque.
Key Point
Axial load → wire twisted; axial torque → wire bent.
Key Concepts for Students
- Angle of wind-up θ = 64TDn/(Ed⁴) uses E, not G.
- Coil curvature changes under the couple.
- Spiral springs work the same way (bending).
Practice more: this is question 7 in Spring MCQs – Part 1. Try the whole set and check every answer instantly.