Two concentric springs, having same number of turns and free axial length, are made of same material. One spring has mean coil diameter of 12 cm and its wire diameter is 1.0 cm. The other one has mean coil diameter of 8 cm and wire diameter of 0.6 cm. If the set of springs is compressed by an axial load of 2000 N, the loads shared by the springs will be

Concept:

The extension of spring is given as, 

where, P = load, D = diameter of coil, d = diameter of wire, n = no. of turns, G = modulus of rigidity

Here, the concentrated spring is compressed by load W, so the deflection of both spring is in same quantity. i.e. Δ₁ = Δ₂

Calculation:  

Given:  

D₁ = 12 cm, D₂ = 8 cm, d₁ = 1 cm, d₂ = 0.6 cm, W = W1 + W2 = 2000, n₁ = n₂, G₁ = G₂ (material is same)     

Since, for concentric springs, the deflection will be same

∴ Δ₁ = Δ₂

∴ 

 = 2.286   

∴ W₁ = 2.286W₂

⇒ 2.286W₂ + W₂ = 2000

∴ W₂ = 608.64 N and W₁ = 1391.357 N