Two helical springs of the same material and of equal circular cross-section and length and the number of turns, but having radii 20 mm and 40 mm, kept concentrically (smaller radius spring within the larger radius spring), are compressed between two parallel planes with a load P. The inner spring will carry a load equal to:

Concept:

Axial deflection of closed coil helical spring is given by,

\(Deflection\left( \delta \right) = \frac{{8P{C^3}N}}{{Gd^4}}\)

where P = Applied load, d = Spring wire diameter, D = Mean diameter of the coil, G = Modulus of rigidity, N = Number of turns

\(C = Spring~index = \frac{D}{d}\)

Calculation:

Given:

Radius of inner coil, R₁ = 20 mm or D₁ = 40 mm

Radius of outer coil, R₂ = 40 mm or D₂ = 80 mm

Since both the wires are made up of the same material and have equal cross-section and length. There the deflection of spring varies as,

δ ∝ P × C³ ∝ P × D³

Since both the Springs are kept concentrically and applied by axial compressive load P, Hence the deflection of both spring will be same.

Let the load shared by innerspring be P₁ and the load shared by outer spring be P₂.

\({\delta _1} = {\delta _2}\)

P₁ × D₁³  = P₂ × D₂³

P₁ × 40³ = P₂ × 80³

P­₁ = 8P₂ .........(1)

Also, P₁ + P₂ = P

Put in (1)

P₁ = 8 × (P – P₁)

\({P_1} = \frac{{8P}}{9}\)