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.
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.