Difficulty: Easy
Correct Answer: δ = (8 * W * R^3 * n) / (G * d^4)
Explanation:
Introduction / Context:
Close-coiled helical springs are common in mechanical and civil engineering devices (balances, suspensions, bearings). When loaded axially, the spring wire primarily undergoes torsion. This question asks for the standard deflection formula, highlighting how geometry and material properties control compliance.
Given Data / Assumptions:
Concept / Approach:
Under axial load, each coil twists. The total angle of twist over the wire length converts to axial deflection. Using torsion of circular shafts and spring geometry, the standard result for a close-coiled spring is obtained, showing strong sensitivity to wire diameter (d^4) and mean radius (R^3).
Step-by-Step Solution:
Torsion in wire: angle of twist θ = (T * L) / (J * G), with T ≈ W * R.Total wire length L = 2 * π * R * n for close-coiled spring.Polar moment for circular section: J = (π * d^4) / 32.Axial deflection δ relates to total twist by δ = θ * R.Combining and simplifying gives δ = (8 * W * R^3 * n) / (G * d^4).
Verification / Alternative check:
Stiffness k = W / δ = (G * d^4) / (8 * R^3 * n) is a widely cited result; inverting yields the same δ expression.
Why Other Options Are Wrong:
Common Pitfalls:
Final Answer:
δ = (8 * W * R^3 * n) / (G * d^4)
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