An open-coiled helical spring made of 8 mm diameter wire has 12 coils. If the spring is subjected to an axial load of 600 N, calculate the deflection of the spring.

Answer 1

First, we need to calculate the shear modulus of the material of the wire.

Given:
Diameter of wire (d) = 8mm = 0.008m
Number of coils (N) = 12
Axial load (F) = 600N
Deflection (δ) = ?
Shear modulus (G) = ?

The formula to calculate the shear modulus is:
G = (8*(1+(2*v)) * (d^4)) / (16 * (D)^3) * (N^3)

Where:
v = Poission's ratio (for steel, it is approximately 0.3)

Substitute the given values into the formula:
G = (8*(1+2*0.3) * (0.008^4)) / (16 * (0.008)^3) * (12^3)
G = 67.2 GPa

Next, we need to calculate the spring constant (k) using the formula:
k = G * (D^4) / (8 * (d^3) * N)

Substitute the values into the formula:
k = 67.2 * (0.008^4) / (8 * (0.008^3) * 12)
k = 3520 N/m

Finally, we can calculate the deflection of the spring using Hooke's Law:
δ = F / k
δ = 600 / 3520
δ = 0.17m

Therefore, the deflection of the spring under an axial load of 600 N is 0.17 meters.