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](http://en.wikipedia.org/wiki/Torsion_%28mechanics%29),

$$
\theta = \frac{l\tau}{IG},
$$

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](http://en.wikipedia.org/wiki/List_of_area_moments_of_inertia) of the cross-section of the coil and $G$ the [shear modulus](http://en.wikipedia.org/wiki/Shear_modulus) of the material the coil is made of.

I assume that the coil is circular rod, such that $I$ would be equal to,

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

where $r$ is the radius of the rod.

However the spring constant of such a spring does not relate $\tau$ and $\theta$, but the elongation and (linear) force. The entire spring has length $L$ and consists of $N$ turns with radius $R$. The pitch, $\psi$, of the coil, defined as the the angle the coil makes with the plane normal to the length axis of the spring, will be equal to,

$$
\psi = tan^{-1}\left(\frac{L}{2\pi NR}\right)
$$

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

$$
\Delta\psi = tan^{-1}\left(\frac{L+\Delta L}{2\pi NR}\right) - tan^{-1}\left(\frac{L}{2\pi NR}\right) \approx \frac{2\pi NR}{L^2 + 4\pi^2 N^2R^2}\Delta L.
$$

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

$$
\theta = 4N\Delta\psi \approx \frac{8\pi N^2R}{L^2 + 4\pi^2 N^2R^2}\Delta L.
$$

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

$$
F = \frac{\cos(\psi)}{R} \tau = \frac{2\pi N\tau}{\sqrt{L^2 + 4\pi^2 N^2R^2}}.
$$

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

$$
l = \sqrt{L^2 + 4\pi^2 N^2R^2}.
$$

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

$$
k = \frac{F}{\Delta L} \approx \frac{8\pi^3N^3r^4RG}{\left(L^2+4\pi^2N^2R^2\right)^2} = \frac{8\pi^3N^3r^4RG}{l^4}.
$$

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

![enter image description here][1]

One end of the spring has four tightly packed windings, which are measured to have a height of 1.5 mm, thus the radius of the the rod to be equal to 0.1875 mm. The radius of the spring is measured to be 2 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](http://www.engineeringtoolbox.com/modulus-rigidity-d_946.html) of about 64 GPa. The estimation for the spring constant would then be equal to roughly 170 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 roughly the value for the spring constant.

PS: Since you said that you where unable to find a source of information about this on the internet, I thought you might also be interested in [this document](http://elearning.vtu.ac.in/12/enotes/Des_Mac-Ele2/Unit3-RK.pdf).

  [1]: https://i.sstatic.net/cPlnf.jpg