An open-coiled helical spring of mean diameter D, number of coils N and wire diameter d is subjected to an axial force P. The wire of the spring will be subjected to:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 01/03/2017 (Morning)
Detailed Explanation
In an open-coiled spring the coils are steeply inclined (large helix angle α), so the moment P·R is no longer about the wire axis alone.
\[ T = PR\cos\alpha \ (\text{twisting}), \qquad M = PR\sin\alpha \ (\text{bending}), \qquad + \text{direct shear } P \]
Why “combined shear, bending and twisting” Is Correct
The wire therefore carries combined shear, bending and twisting. (A close-coiled spring, with α ≈ 0, has only twisting plus direct shear.)
| 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
- A. direct shear only: Direct shear is the smallest effect, not the only one.
- B. combined shear and bending only: Twisting (PR cos α) is still the main action.
- D. combined shear and twisting only: That is the close-coiled case; the open coil adds bending.
Key Point
Open coil: T = PR cos α, M = PR sin α, plus shear. Close coil: torsion (+ small shear).
Key Concepts for Students
- Open-coiled springs also rotate their free end under axial load.
- Helix angle decides open vs close coil.
- Deflection includes both twist and bending terms.
Practice more: this is question 6 in Spring MCQs – Part 1. Try the whole set and check every answer instantly.