Closely coiled helical spring deflection: A closely coiled helical spring of mean coil radius R has n turns and is subjected to an axial load W. If the wire radius is r and the material has shear modulus C, what is the axial deflection of the spring?

Introduction / Context:
Close-coiled helical springs store energy primarily through torsion of the wire. The load–deflection relation connects axial deflection to coil geometry and material stiffness.

Given Data / Assumptions:

• Mean coil radius = R; number of active turns = n.
• Wire radius = r (so wire diameter d = 2 r).
• Shear modulus = C.
• Small deflection; close-coil assumption (helix angle small).

Concept / Approach:
For a close-coiled spring under axial load W, the torque in the wire is T = W R for each turn. The twist per turn is θ_turn = T l_w / (J C), but standard derivation yields the familiar formula δ = 64 W R^3 n / (C d^4). Substituting d = 2 r gives δ = 64 W R^3 n / (C (2 r)^4) = 64 W R^3 n / (C * 16 r^4) = 4 W R^3 n / (C r^4).

Step-by-Step Solution:
Start with δ = 64 W R^3 n / (C d^4).Replace d with 2 r → d^4 = 16 r^4.Hence δ = (64 / 16) * (W R^3 n) / (C r^4) = 4 W R^3 n / (C r^4).

Verification / Alternative check:
Dimensional check: numerator has force * length^3; denominator C r^4 has force/area * length^4 = force * length^2; ratio gives length as required.

Why Other Options Are Wrong:
Options with 8 or 64 numerators misuse diameter vs radius or alter powers erroneously.Expressions involving r^3, or R^2 instead of R^3, are dimensionally inconsistent.

Common Pitfalls:
Forgetting to convert from wire diameter to radius; mixing up shear modulus symbols (G vs C).

Final Answer:
δ = (4 W R^3 n) / (C r^4).