Energy in Close-Coiled Helical Spring under Axial Load A close-coiled helical spring with mean coil diameter D, wire diameter d, and n active coils carries an axial load W. For material shear modulus G, what is the strain energy stored in the spring?

Introduction / Context:
Close-coiled helical springs deform primarily by torsion of the wire. Under an axial load W, both deflection and stored energy can be expressed using torsion theory and the spring rate. Energy expressions are essential for impact, vibration, and fatigue assessments.

Given Data / Assumptions:

• Mean coil diameter = D, wire diameter = d, active coils = n.
• Axial load = W, shear modulus = G.
• Close-coiled assumption, negligible direct shear component on energy, linear elastic behavior.

Concept / Approach:

The axial deflection of a close-coiled spring is delta = 8 W D^3 n / (G d^4). Strain energy U equals one half times load times deflection, or equivalently W^2 divided by twice the spring stiffness k.

Step-by-Step Solution:

Verification / Alternative check:

Using stiffness k = G d^4 / (8 D^3 n), compute U = W^2 / (2 k) which gives the same value: 4 W^2 D^3 n / (G d^4).

Why Other Options Are Wrong:

W^2 D n / (2 G d^3) has incorrect powers of D and d. W^2 D^3 n / (2 E d^4) uses the wrong modulus and coefficient. W^2 / (2 k) is correct in form but not a final expression in basic variables for direct comparison. 8 W^2 D^3 n / (G d^4) is off by a factor of 2.

Common Pitfalls:

Confusing E with G, forgetting the 1/2 factor in energy, or missing the close-coiled assumption that yields the standard deflection formula.

Final Answer:

4 W^2 D^3 n / (G d^4)