For a close-coiled helical spring (stiffness calculation), which expression correctly gives the axial stiffness k = W/δ in terms of material and geometric parameters?

Introduction / Context:Close-coiled helical springs under axial load primarily deform by torsion of the wire. The spring stiffness (load per unit deflection) depends on the shear modulus and the spring’s wire/coil geometry. This relationship is essential for designing springs to meet load–deflection requirements.

Given Data / Assumptions:

• Close-coiled spring; helix angle is small.
• Axial load W produces wire torsion dominated deformation.
• Material characterized by shear modulus G (not Young’s modulus E).
• Symbols: d = wire diameter, D = mean coil diameter, n = number of active coils.

Concept / Approach:The classic formula for axial deflection δ of a close-coiled spring is δ = (8 * W * D^3 * n) / (G * d^4). Inverting gives stiffness k = W / δ = (G * d^4) / (8 * n * D^3). Note the strong dependence on d (to the 4th power) and the inverse cubic dependence on D, reflecting torsional compliance of the wire around the coil diameter.

Step-by-Step Solution:Start from δ = (8 * W * D^3 * n) / (G * d^4).Rearrange: k = W / δ = (G * d^4) / (8 * n * D^3).Confirm units: G has units of stress; geometric ratios render k in force/length.

Verification / Alternative check:Compare with design handbooks; expression matches standard spring design equations for close-coiled springs.

Why Other Options Are Wrong:

• Using E instead of G: Spring twist depends on shear modulus G.
• Other exponents or constants (D^4, 16, D^2) do not match the derived relation.

Common Pitfalls:Substituting E for G; forgetting that only active coils count in n; ignoring effects of end conditions on effective coil count.

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