Helical (Coil) Springs
The figure shows a helical coil spring with ends adapted to support a compressive load. The notation applied are:
P = axial load, N D = mean diameter of coil, mm = (D_{o} + D_{i}) / 2 d = diameter of wire, mm = (D_{o} – D_{i}) / 2 p = pitch of coils, mm δ = deflection of spring, mm n = number of active coils C = spring index = D/d = (D_{o}d)/d = (4 : 12); less than 4; it is difficult to manufacture, more than 12; it is likely to buckle. G = torsional modulus of elasticity, N/mm^{2} τ_{s }= shearing stress, N/mm^{2} 

Stresses on coil spring (t):
The coil spring wire is subjected to the following stresses:
a Torsional stress due to the load P
b Direct shear stress due to load P
c Torsional stress due to wire curvature
Torsional stress due to load P:
_{}
In order to include the effects of both direct shear and wire curvature, a stress factor (K_{w}) had been determined by the use of approximate analytical methods developed by A. M. Wahl:
_{}
which may be used in the above equation to determine the maximum shearing stress in the wire as follows:
_{}
Various lengths associated with a spring:
L_{f} = Free length
L_{a} = Installed length
L_{m} = Operating length
L_{s} = Shut height or Solid length
· Compression springs in which the free length is more than four times the mean diameter of the coils may fail by sidewise buckling.
Spring deflection (d):
_{}
Spring rate (stiffness) (k):
_{}
Spring ends conditions:
Helical springs ends may be either plain, plain ground, squared, or squared and ground as shown in the figure below. This results in a decrease of the number of active coils and affects the free length and solid length of the spring as shown below.
p=(D/3 : D/4), n= ( 3: 15)
Finding spring stress and deflection:
a Using the above equations of t_{max} , and d.
b Using the nomogram
c Using the Excel program given
Explanation of the nomogram:
The nomogram applies to cylindrical helical extension and compression springs made of round steel wire (shear modulus G=81,400 N/mm^{2}). In the case of materials with a different shear modulus G^{’} spring deflection must be multiplied by G/G^{’}.
The nomogram indicates the deflection s^{’} of one coil. Total deflection s (d) is obtained by multiplying (s^{’}) by the number of active coils n:
s (d) = n . s^{’}.
Example:
Find the stress and the deflection of a spring with the following dimensions:
D = 30 mm
d = 4 mm
F = P = 100 N
Solution steps:
1 enter the value of D (point 1), and d (point 2) into the nomogram
2 connect point 1, 2 with the line A and extend the line to points 3 and 3’ on either end.
3 enter the value of F (point 4) into the nomogram
4 connect point 3 to the value of the force F (point 4) with line B, extend the line B to point (5)
5 calculate the value of C and enter its value into the nomogram (point 6)
6 connect point 6 and 5 by line C to obtain point (7)
7 connect point 5 to point 3’ to obtain point (8)
Solution:
 point 5 represents the value of t_{t} (MPa)
 point 7 represents the value of t_{max } (MPa)
 point 8 represents the value of d’ = d/n (mm)
 enter the values of d, D, P, n, G.
The program will give you the values of C, K_{w}, t_{t} (Tau”) , t_{max} (Tau), d/n (Delta”), d (Delta), and K