# Equation for calculating spring costant

I'd like to design some of my own springs in order to obtain some very specific forces for a project. There are plenty of guides on how to make an arbitrary spring, but none I've read explain how to make one in such a way as to obtain a specific spring constant.

Given specifications, like the material (music wire), the material's gauge, number of turns in the spring, diameter of the spring, and space between turns, is there a formula for estimating the spring constant which I can then use with Hooke's law for estimating the force produced by the spring under load?

Googling for formulas or calculators of spring dimensions just gets me a bunch of manufacturer websites selling springs.

A simple model for a coil spring would be that, when the spring is subjected to a force, the entire coil is subjected to a torsion $\tau$. This torque causes the coil the twist by an angle, which can be approximated with,

$$\theta = \frac{l\, \tau}{I\, G},$$

where $\theta$ is the angle of twist in radians, $l$ the length of the coil (not to be confused with the length of the spring), $I$ the second moment of inertia of the cross-section of the coil and $G$ the shear modulus of the material the coil is made of.

Assuming that the coil is circular rod, then $I$ would be equal to,

$$I = \frac{\pi}{2} r^4 = \frac{\pi}{32} d^4,$$

where $r$ and $d$ are the radius and diameter of the rod respectively.

However the spring constant of such a spring does not relate $\tau$ and $\theta$, but the elongation and (linear) force. The entire spring has free length $L$ and consists of $N$ turns with mean diameter $D$. The coil angle, $\alpha$, is defined as the the angle the coil makes with the plane normal to the length axis of the spring. Also see the Figure below.

If you unroll the spring onto a flat plain, the rod will be the diagonal of a rectangle with height $L$ and width $\pi\, N\, D$. Combining this with the fact that $\alpha$ is the angle between the diagonal and horizontal of this rectangle and therefore will be equal to,

$$\alpha= \tan^{-1}\left(\frac{L}{\pi\, N\, D}\right).$$

The relationship between $\theta$ and the elongation, $\Delta L$, can be found by looking at the change in $\alpha$ due to the elongation,

$$\Delta\alpha = \tan^{-1}\left(\frac{L+\Delta L}{\pi\, N\, D}\right) - \tan^{-1}\left(\frac{L}{\pi\, N\, D}\right) = \frac{\pi\, N\, D}{L^2 + \pi^2 N^2 D^2}\Delta L + O(\Delta L^2).$$

Namely the change in $\alpha$ is equal to the twist angle of a quarter of a turn of the spring, thus,

$$\theta = 4\, N\, \Delta\alpha \approx \frac{4\, \pi N^2 D}{L^2 + \pi^2 N^2 D^2} \Delta L.$$

The relationship between $\tau$ and $F$ can be found by looking at the lever of this torque in the spring,

$$F = \frac{2 \cos(\alpha)}{D} \tau = \frac{2\, \pi\, N\, \tau}{\sqrt{L^2 + \pi^2 N^2 D^2}}.$$

By using Pythagorean theorem it can be shown that the length of the coil is equal to,

$$l = \sqrt{L^2 + \pi^2 N^2 D^2}.$$

By substituting $F$ and $\Delta L$ from these equations, the spring constant can be approximated by,

$$k = \frac{F}{\Delta L} \approx \frac{\pi^3 N^3 d^4 D\, G}{4 \left(L^2 + \pi^2 N^2 D^2\right)^2} = \frac{\pi^3 N^3 d^4 D\, G}{4\,l^4}.$$

To test this we can try to calculate the spring constant of a spring from a ballpoint pen.

One end of the spring has four tightly packed windings, which are measured to have a height of 1.5 mm, thus the diameter of the the rod to be equal to 0.375 mm. The mean diameter of the spring is measured to be about 4 mm. The spring counts 8.5 windings. The length of the spring is measured to be 2 cm. The shear modulus is harder to determine, because I am not certain from which metal it is made of, but because it is attracted to a magnet I will assume it is iron, which has a shear modulus of about 64 GPa. The estimation for the spring constant would then be equal to roughly 173 N/m. Since the spring is relatively small I did not had an accurate way of measuring the elongation of the spring while applying a force. I used my mobile phone, which according to the the online specs weighs 162 g, which causes a compression of roughly 0.7 cm, which would yield a spring constant of roughly 227 N/m. So the predicted spring constant is of by 23.8%, which is quite a large relative error, but at least the same order of magnitude. Sources for this error might be: the fact that the compression was quite large, so the linear approximation in $\Delta L$ might not hold; the measured dimensions of the spring might not be totally accurate, especially the values for $d$ and $l$, which are raised to the fourth power could contribute a lot if the of my a little.

• My text has a simpler formula - due to the fact that $l = 2 \pi N R$, for $$k = \frac{r^4 G}{4 R^3 N}$$. Source: Shigley and Mischke, Mechanical Engineering Design, 7th Ed.
– Mark
Commented Jun 26, 2015 at 17:23
• @Mark if you use that substitution for $l$, shouldn't the equation be, $$k = \frac{r^4\, G}{2\,\pi\,R^3 N}$$ Commented May 10, 2017 at 12:43
• It blew my mind when I was exposed to the derivation of the spring constant and the torsion came into play; then they pointed out that tension/compression springs are actually a long bar being purely twisted. Torsion springs are the opposite, they are a long bar being purely bent (essentially pure lateral movement if it was a big bar) which leads to a pure rotation of the ends. The pure torsion in the extension spring leads to a linear movement. I've always found it crazy that the helix shapes seem to transmit the motion from rotation to linear, and vice versa.
– JMac
Commented May 10, 2017 at 12:43
• I would think the assumption that the spring is made of iron might lead to a large inaccuracy, because springs are usually made of steel that has been at least partially turned into "spring steel" by quenching. Ordinary iron is not very elastic. A ball bearing made of iron would not bounce against a hard metal surface as high as one made of spring steel. Commented Nov 23, 2020 at 10:00