Helical spring stiffness under axial load For a closely-coiled helical spring of mean coil diameter D subjected to an axial load W, the stiffness k (load per unit deflection) is:

Introduction / Context:
Stiffness of a helical spring tells how much load is required to cause a unit deflection. For closely-coiled springs, the coils are nearly perpendicular to the spring axis, so the wire primarily experiences torsion rather than pure bending. Designers use stiffness to size the wire diameter and count the number of active coils for the target deflection under service load.

Given Data / Assumptions:

• Closely-coiled helical spring, mean coil diameter D.
• Wire diameter d; number of active coils n.
• Material shear modulus G; small deformations; linear elasticity.
• Axial load W causes an axial deflection delta.

Concept / Approach:
The axial deflection of a closely-coiled spring under W is delta = 8 * W * D^3 * n / (G * d^4). Stiffness is k = W / delta, so invert the deflection expression to obtain k in terms of geometry and G.

Step-by-Step Solution:
Start with delta = 8 * W * D^3 * n / (G * d^4).Rearrange for k: k = W / delta.Substitute delta: k = W / [8 * W * D^3 * n / (G * d^4)].Cancel W and invert: k = (G * d^4) / (8 * D^3 * n).Hence stiffness increases strongly with d and decreases with D and n.

Verification / Alternative check:
Dimensional check: G has units of stress, d^4 has length^4, denominator has length^3, giving stress * length = force per deflection, consistent with stiffness units N/m.

Why Other Options Are Wrong:
Option b swaps numerator/denominator (that is the deflection expression inverted incorrectly). Option c incorrectly uses E (Young’s modulus) instead of G (shear modulus). Option d has wrong powers of D and d. Option e is Euler buckling, not spring stiffness.

Common Pitfalls:
Confusing E with G; forgetting that only active coils count; misreading wire vs coil diameters (d vs D).

Final Answer:
k = (G * d^4) / (8 * D^3 * n)