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 ?


1 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.

  • $\begingroup$ The formula is how to calculate the rotational stiffness from the screw parameters. But what I'm trying to do is convert the translational stiffness into rotational stiffness. Without knowing G or J. $\endgroup$
    – Crapsy
    Commented Oct 5, 2016 at 20:21
  • $\begingroup$ I improved my answer. Is this what you meant? $\endgroup$ Commented Oct 5, 2016 at 21:23
  • $\begingroup$ There is not any correlation between axial stiffness in a prismatic section and torque or rotational stiffness. J which is mass moment of inertia of the section is dependent on the cross section of the piece and weather it is hollow or solid! $\endgroup$
    – kamran
    Commented Oct 5, 2016 at 21:27
  • $\begingroup$ @kamran there is a correlation between them since I assumed linear elasticity and that the cylinder is solid. $\endgroup$ Commented Oct 5, 2016 at 21:44
  • $\begingroup$ What if the bar is shaped like an "H" or anything other than a cylinder? The OP has not restricted the cross-section to cylinder. $\endgroup$
    – kamran
    Commented Oct 5, 2016 at 22:10

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.