To estimate the spring rate, we can use the formula:
k = Gd^4 / (8D^3n)
where:
k = spring rate (N/mm)
G = shear modulus of the material (GPa)
d = wire diameter (mm)
D = mean coil diameter (mm)
n = number of active coils
Assuming a shear modulus of 80 GPa for music wire, we have:
k = (80 GPa * (2.5 mm)^4) / (8 * 31.5 mm^3 * 14) ≈ 15.8 N/mm
To find the force needed to compress the spring to closure, we can use the formula:
F = k * Δx
where:
F = force (N)
Δx = compression distance (mm)
Since the spring has 14 total coils, and assuming no initial compression, the free length (L0) can be calculated as:
L0 = Dn = 31 mm * 14 = 434 mm
If we compress the spring to closure (i.e. until all coils are touching), the compression distance (Δx) can be calculated as:
Δx = L0 - (d/2 * n) = 434 mm - (2.5 mm / 2 * 14) ≈ 400.63 mm
Thus, the force needed to compress the spring to closure is:
F = 15.8 N/mm * 400.63 mm ≈ 6332.3 N
To determine the free length that ensures the torsional stress does not exceed the yield strength, we can use the formula for maximum shear stress in a helical compression spring:
τmax = 16F * D / (πd^3n)
where:
τmax = maximum shear stress (MPa)
F = force (N)
D = mean coil diameter (mm)
d = wire diameter (mm)
n = number of active coils
Assuming a yield strength of 2100 MPa for music wire, we can solve for the free length (L0) that gives a maximum shear stress of 2100 MPa:
L0 = (16F * D) / (πd^3n * τmax)
Taking τmax = 2100 MPa and using the values for D, d, and n given in the problem, we get:
L0 = (16 * 6332.3 N * 31.5 mm) / (π * (2.5 mm)^3 * 14 * 2100 MPa) ≈ 416.7 mm
Therefore, the free length should be at least 416.7 mm to ensure that the torsional stress does not exceed the yield strength when the spring is compressed solid.