Close-coiled helical spring under axial load: For a closely-coiled helical spring of mean coil diameter D, wire diameter d, n turns, modulus of rigidity C (G), and axial load W, the deflection δ is

Difficulty: Medium

Correct Answer: δ = 8 * W * D^3 * n / (C * d^4)

Explanation:


Introduction / Context:
Close-coiled helical springs store energy primarily by twisting of the wire. The load–deflection relation links geometry (D, d, n) and material shear modulus to the axial compliance.



Given Data / Assumptions:

  • Close-coiled (helix angle small), pure torsion of wire dominates.
  • Axial load W causes torque in each coil.
  • Modulus of rigidity C (often denoted G) is constant.


Concept / Approach:
Each coil behaves like a torsion bar of circular section. Summing angular twist over n coils and converting to axial deflection yields the well-known expression δ = 8 * W * D^3 * n / (C * d^4).



Step-by-Step Solution:

Torque in wire: T = W * D / 2 ≈ W * (D/2). Often expressed with mean radius R = D/2.Angle of twist per coil: θ = T * l_w / (C * J), with wire polar J = π * d^4 / 32 and l_w ≈ π * D.Axial deflection: δ = n * θ * (pitch angle small) → simplifies to δ = 8 * W * D^3 * n / (C * d^4).Select the matching option.


Verification / Alternative check:
Dimensional check: δ has units of length; numerator W * D^3 * n divided by C * d^4 (stress per shear-strain scale) yields length.



Why Other Options Are Wrong:

  • Other powers of D and d do not arise from torsion theory; they would give incorrect stiffness scaling.


Common Pitfalls:
Using E instead of C (G); mixing up D (mean diameter) with radius; forgetting the power d^4 in the denominator.



Final Answer:
δ = 8 * W * D^3 * n / (C * d^4)

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