Question

A close helical spring of 100 mm mean diameter is made of 10 mm diameter rod, and has 20 turns. The spring carries an axial load of 200 kN with G = 8.4 × 10⁴ N/mm². The stiffness of the spring is nearly
1. 5.25 N/mm
2. 6.50 N/mm
3. 7.25 N/mm
4. 8.50 N/mm

Answer 1

Correct Answer - Option 1 : 5.25 N/mm

Concept:

Stiffness of helical spring is given by,

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

Calculation:

Mean diameter of helical spring ‘D’ = 100 mm

Diameter of wire or rod ‘d’ = 10 mm

Number of turns ‘N’ = 20

Axial load ‘W’ = 200 kN

Modulus of rigidity ‘G’ = 8.4 × 104 MPa

Stiffness of helical spring,

\(k = \frac{{8.4 \times {{10}^4} \times {{10}^4}}}{{8 \times {{100}^3} \times 20}} = 5.25\;N/mm\)