Helical springs – stiffness versus number of turns For a closely-coiled helical spring, the stiffness (k) varies with the number of turns (n) as:

Mechanical Engineering Strength of Materials Difficulty: Easy
Choose an option
  • A
    k is directly proportional to n
  • B
    k is inversely proportional to n
  • C
    k is proportional to n^2
  • D
    k is independent of n
  • E
    k is inversely proportional to n^2

Answer

Correct Answer: k is inversely proportional to n

Explanation

Introduction / Context:Helical compression and tension springs are designed to achieve a target stiffness (spring rate). Understanding how geometry affects stiffness enables quick design adjustments. The number of active turns has a direct, simple influence on k.

Given Data / Assumptions:

  • Closely-coiled helical spring (helix angle small).
  • Wire diameter d, mean coil diameter D, shear modulus G.
  • n = number of active coils.

Concept / Approach:The standard spring rate formula for a closely-coiled helical spring loaded axially is k = (G d^4) / (8 D^3 n). This shows linear inverse dependence on n, holding other parameters constant.

Step-by-Step Solution:Start from k = (G d^4) / (8 D^3 n).Treat G, d, D as constants for a given design iteration.Therefore, k ∝ 1 / n. Increasing the number of active turns reduces stiffness.

Verification / Alternative check:Doubling n halves k; halving n doubles k. This proportionality matches both analytical derivations and empirical spring catalogs.

Why Other Options Are Wrong:Options claiming k ∝ n, n^2, or independent of n contradict the formula; k ∝ 1 / n^2 is also incorrect because the dependence is first power.

Common Pitfalls:Forgetting to count only active coils (end coils may be inactive); changing D or d simultaneously and misattributing the effect to n.

Final Answer:k is inversely proportional to n

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