Helical springs – stiffness versus number of turns For a closely-coiled helical spring, the stiffness (k) varies with the number of turns (n) as:
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Ak is directly proportional to n
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Bk is inversely proportional to n
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Ck is proportional to n^2
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Dk is independent of n
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Ek 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