Question

A helical spring has coil diameter 50 mm and is subject to axial load W. Another spring has coil diameter 25 mm, but otherwise identical to the first. The deflection of second spring will be ________ of that of first spring.
1. \(\frac{1}{2}\)
2. \(\frac{1}{4}\)
3. \(\frac{1}{8}\)
4. None of these

Answer 1

Correct Answer - Option 3 : \(\frac{1}{8}\)

Concept:

For closed coil helical spring,

\(Deflection\;under\;load,\;δ = \frac{{8W{D^3}n}}{{G{d^4}}}\)

where, W = Load, D = Mean diameter of coil spring, n = Number of turns, G = Modulus of elasticity and d = Wire diameter

From the above formula, it is clear that keeping all other parameters same,

Deflection, δ ∝ D3

Calculation:

Given:

Coil diameter for first spring (D₁) = 50 mm, Coil diameter for second spring (DA₂) = 25 mm, 

∵ Deflection, δ ∝ D³

∴ \(\frac{{{δ _2}}}{{{δ _1}}} = (\frac{{{D_2}}}{{{D_1}}})^3\)

\(\frac{{{δ _2}}}{{{δ _1}}} = (\frac{25}{50})^3\)

\(\frac{{{δ _2}}}{{{δ _1}}} = (\frac{1}{2})^3\)

\(\frac{{{δ _2}}}{{{δ _1}}} = \frac{1}{8}\)

Hence the deflection in the second spring will be \(\frac{1}{8}\) times of the deflection of first spring.