Helical Springs

When close-coiled helical spring, composed of a wire of round rod of diameter d wound into a helix of mean radius R with n number of turns, is subjected to an axial load P produces the following stresses and elongation:

Close-coiled helical spring


The maximum shearing stress is the sum of the direct shearing stress τ1 = P/A and the torsional shearing stress τ2 = Tr/J, with T = PR.

$\tau = \tau_1 + \tau_2$

$\tau = \dfrac{P}{\pi d^2 / 4} + \dfrac{16PR}{\pi d^3}$

$\tau = \dfrac{16PR}{\pi d^3} \left( 1 + \dfrac{d}{4R} \right)$


This formula neglects the curvature of the spring. This is used for light spring where the ratio d/4R is small.

For heavy springs and considering the curvature of the spring, A.M. Wahl formula a more precise, it is given by:

$\tau = \dfrac{16PR}{\pi d^3} \left( \dfrac{4m - 1}{4m - 4} + \dfrac{0.615}{m} \right)$

where m is called the spring index and (4m - 1)/(4m - 4) is the Wahl Factor.

The elongation of the bar is

$\delta = \dfrac{64PR^3n}{Gd^4}$


Notice that the deformation δ is directly proportional to the applied load P. The ratio of P to δ is called the spring constant k and is equal to

$k = \dfrac{P}{\delta} = \dfrac{Gd^4}{64R^3n} \, \text{ N/mm}$


Springs in Series
For two or more springs with spring laid in series, the resulting spring constant k is given by



$1/k = 1/k_1 + 1/k_2 + \dots$


where k1, k2,... are the spring constants for different springs.

Springs in Parallel
For two or more springs in parallel, the resulting spring constant is



$k = k_1 + k_2 + \dots$