# 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:

The maximum shearing stress is the sum of the direct shearing stress τ_{1} = *P*/*A* and the torsional shearing stress τ_{2} = *Tr*/*J*, with *T* = *PR*.

$\tau = \tau_1 + \tau_2$

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

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

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

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

The elongation of the bar is

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

**Springs in Series**

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

where *k*_{1}, *k*_{2},... are the spring constants for different springs.

**Springs in Parallel**

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

- Add new comment
- 85478 reads