Helical Springs — Effect of Wire Diameter on Stiffness Two closely coiled helical springs 'A' and 'B' are identical in mean coil diameter, number of active turns, material, and overall geometry. However, the wire diameter of spring 'A' is exactly double that of spring 'B'. In terms of linear stiffness (load per unit deflection), the stiffness of spring 'B' will be __________ of the stiffness of spring 'A'.

Introduction:
Helical spring stiffness depends strongly on wire diameter. This question checks whether you recall the correct proportionality and can compare two otherwise identical springs when the wire diameter changes.

Given Data / Assumptions:

• Springs A and B are closely coiled and identical in material, mean coil diameter, and number of active turns.
• Wire diameter of A is double that of B.
• Linear stiffness k is load/deflection in the elastic range.

Concept / Approach:
For a closely coiled helical spring in shear: k = (G * d^4) / (8 * D^3 * n) where G is shear modulus, d is wire diameter, D is mean coil diameter, n is active turns. Holding G, D, n constant, stiffness is proportional to d^4.

Step-by-Step Solution:
Let d_B = d. Then d_A = 2d.k_A / k_B = (d_A^4) / (d_B^4) = (2d)^4 / d^4 = 16.Therefore k_B = k_A / 16.

Verification / Alternative check:
Doubling d massively increases stiffness because polar resistance rises with d^4. A factor of 16 is consistent with this strong dependence.

Why Other Options Are Wrong:

• one-eighth: Implies d^3 dependence, which is incorrect.
• one-fourth: Would correspond to d^2 dependence.
• one-half: Would correspond to linear dependence on d.

Common Pitfalls:
Confusing spring stiffness with beam bending stiffness or misremembering the d^4 proportionality is common.

Final Answer:
one-sixteenth