Convert translational stiffness to rotational stiffness of a screw

Question

I have the formula for the translational stiffness which is :

However, what I would like to find is the rotational stiffness. Is there any formulas to convert from one to an other knowing the pitch of the screw and the diameter ?

Answer 1

The translational stiffness is written as $$ k_l = \frac{A\,E}{\ell},$$ where the stiffness $k_l$ is in $\left[\frac{N}{m}\right]$, the area $A$ is in $\left[m^2\right]$, the young's modulus is in $\left[\frac{N}{m^2}\right]$ and the length $\ell$ is in $\left[m\right]$.

$$\,$$

The rotational stiffness is written as $$ k_r = \frac{G\,J}{\ell},$$ where the stiffness $k_r$ is in $\left[\frac{Nm}{rad}\right]$, the second moment of area (or torsion constant) $J$ is in $\left[m^4\right]$, the rigidity modulus is in $\left[\frac{N}{m^2}\right]$ and the length $\ell$ is in $\left[m\right]$.

$$\,$$

Assuming linear elasticity the following relation between the rigidity modulus and the young modulus holds $$ G = \frac{E}{2(1+v)},$$ where $v$ is the poison ratio.

$$\,$$

The second moment of area (or torsion constant) for a circle is $$ \iint\limits_{R} r^2\,dA = \int_0^{2\pi}\int_0^r r^2\left(r\,dr\,d\theta\right) = \int_0^{2\pi}\int_0^r r^3\,dr\,d\theta = \int_0^{2\pi} \frac{r^4}{4}\,d\theta = \frac{\pi}{2}r^4 = \frac{A\,r^2}{2}. $$

$$\,$$

Hence the rotational stiffness of a cylinder can be rewritten as function of $k_l$ $$ k_r = \frac{\frac{E}{2(1+v)} \frac{A\,r^2}{2}}{\ell} = \frac{A\,E}{\ell} \frac{r^2}{4(1+v)} = k_l \, \frac{r^2}{4(1+v)} $$

This equation does not hold for any other shape!

When looking at a screw, there is a thread at the outside which has a significant effect on the rotational stiffness.

For example, leaving half the diameter open in the cylinder, the stiffness will only decrease $0.0625\%$ which is insignificant. Increasing the outer radius will increase the stiffness significant.