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.


1 Answer 1


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.

coil spring illustration

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.

enter image description here

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.

  • $\begingroup$ 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. $\endgroup$
    – Mark
    Jun 26, 2015 at 17:23
  • $\begingroup$ @Mark if you use that substitution for $l$, shouldn't the equation be, $$ k = \frac{r^4\, G}{2\,\pi\,R^3 N} $$ $\endgroup$
    – fibonatic
    May 10, 2017 at 12:43
  • 1
    $\begingroup$ 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. $\endgroup$
    – JMac
    May 10, 2017 at 12:43
  • $\begingroup$ 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. $\endgroup$ Nov 23, 2020 at 10:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.