Question

A compression spring is made of a music wire of 2 mm diameter having a shear strength and shear modulus of 800 MPa and 80 GPa respectively. The mean coil diameter is 20 mm, the free length is 40 mm and the number of active coils is 10. If the mean coil diameter is reduced to 10 mm, the stiffness of the spring is approximately
1. Decreased by 8 times
2. Decreased by 2 times
3. Increased by 2 times
4. Increased by 8 times

Answer 1

Correct Answer - Option 4 : Increased by 8 times

Concept:

Spring constant / Stiffness of a spring (k): Force required to produce unit deflection.

\(k = \frac{P}{\delta }\)

where P = Load & δ = Deflection.

Deflection of spring in terms of active turns:

\(\delta = \frac{{8P{D^3}N}}{{G{d^4}}}\)

\(∴ k = \frac{P}{{\frac{{8P{D^3}N}}{{G{d^4}}}}} ⇒ \frac{{G{d^4}}}{{8{D^3}N}}\)

where

D = Diameter of the spring coil, d = Diameter of spring wire, N = No. of active turns, and G = Modulus of Rigidity.

Calculation:

Given:

d = 2 mm, τ = 800 MPa, G = 80 GPa, D₁ = 20 mm, N = 10, and D₂ = 10 mm.

\(k = \frac{{G{d^4}}}{{8{D^3}N}}\)

\(k \propto \frac{{{1}}}{{D^3}}\)

\(\frac{{{k_2}}}{{{k_1}}} = {\left( {\frac{{{D_1}}}{{{D_2}}}} \right)^3}\)

\(\frac{{{k_2}}}{{{k_1}}} = {\left( {\frac{{20}}{{10}}} \right)^3} = 8\)

∴ stiffness of spring is increased by 8 times.