A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subjected to an axial load of 200 N. Determine the deflection of the spring. C = 80 kN/mm2
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To determine the deflection of the spring, you can use Hooke's Law, which states that the deflection of a spring is proportional to the applied force.
The formula to calculate the deflection of a helical spring under axial load is:
Δ = (F * L) / (C * G * d^4 * n)
Where:
Δ = deflection of the spring
F = applied axial load
L = length of the spring
C = spring constant (also known as the modulus of rigidity)
G = shear modulus of the material of the spring
d = diameter of the spring wire
n = number of turns
Let's calculate step by step:
1. Calculate the length of the spring (L):
The length of the spring can be calculated using the formula:
L = (2 * π * r * n) + (d * n)
where r is the mean radius.
Given:
mean radius (r) = 60 mm
diameter of the spring wire (d) = 10 mm
number of turns (n) = 10
Substituting the values:
L = (2 * π * 60 * 10) + (10 * 10)
L = 1200π + 100
L ≈ 3772.39 mm
2. Calculate the deflection (Δ):
Given:
applied axial load (F) = 200 N
shear modulus of the material (G) = C (given)
Substituting the values:
Δ = (200 * 3772.39) / (80 * 1000 * (10^3) * (10^4) * 10)
Δ ≈ 0.002214mm
Therefore, the deflection of the spring is approximately 0.002214 mm.