First of all, english is not my first language so if there are some mistakes I made - I'm sorry and I hope everything is understandable.

The example I am trying to solve reads the following: The gear is loaded with torque of M=500Nm it is connected with a clutch with 8 screws made from ST5(S275) steel. The screws are distribiuted on diameter D=320mm. Calculate the diameter of the screws.

enter image description here

So what I think I should do here is to use the definition of torque and calculate the value of the force that is distributed equally between the screws, so it would look like this:

$$F=\frac{M}{D/2}$$ $$F_{\tau} = \frac{F}{8}$$

where 8 is the quantity of the screws/

and then calculate the diameter using shear condition $$\frac{F_{tau}}{A}=<k_t$$


  • $A$ - cross section area of the screw
  • $k_t$ - admissible stresses on shearing

Am I right? Or am I doing a mistake somewhere? If all above is correct is it all I have to do or is there something other to calculate?

Thanks in advance for your help!

  • $\begingroup$ if your problem states that you need to do this with shear analysis, I don't see anything wrong with your approach. My answer, is focused on how I would have approached the design. $\endgroup$
    – NMech
    May 25 '21 at 4:24
  • $\begingroup$ I would add, follow the context of the chapter of the book. Also, it seems odd to find the diameter of the screws so that they fail, but perhaps rather to find the diameter that would see only xx% of yield strength at the specified loading. $\endgroup$
    – Jim Clark
    May 25 '21 at 11:47

The only problem I see with your calculation (some might argue against it) is that you are calculating the screws in shear.

IMHO, the best practice is design the bolt, so that the axial force of the force is such so that the friction force between the two plates will transfer the torque between the machine elements.

So instead of $F_{tau}$, I would calculate the axial force $F_a$:

$$ F_{\tau} = F_a \mu$$


  • $\mu$ is the coefficient of friction.

Therefore: $$ F_{a} = \frac{F_{\tau}}{\mu}$$

And then proceed like before to calculate the minimum required diameter based on the allowed tensile stress $\sigma_{all}$.

$$ F_{a} \le \sigma_{all} A$$ $$ A \ge \frac{F_{a}}{ \sigma_{all}}$$


  • $A = \pi\frac{d^2}{4}$

therefore $$ \pi\frac{d^2}{4} \ge \frac{F_{a}}{ \sigma_{all}}$$

The minimum required bolt diameter (the core) should be more than:

$$ d \ge \sqrt{ \frac{4}{\pi}\frac{F_{a}}{ \sigma_{all}}}$$

If you substitute the rest: $$ d \ge \sqrt{ \frac{4}{\pi}\frac{\frac{F_{\tau}}{\mu}}{ \sigma_{all}}}$$ $$ d \ge \sqrt{ \frac{4}{\pi}\frac{\frac{M}{D/2 n}}{ \mu\sigma_{all}}}$$ $$ d \ge \sqrt{ \frac{4}{\pi}\frac{\frac{2 M}{D n}}{ \mu\sigma_{all}}}$$ $$ d \ge \sqrt{ \frac{8 M}{ \pi\cdot D\cdot n\cdot \mu\cdot \sigma_{all}}}$$


  • $M$: is the transmitted torque
  • $D$: is the diameter of the location of the bolts
  • $n$ : is the number ofbolts (=8)
  • $\sigma_{all}$ the allowable tension of steel
  • $\mu$: the friction coefficient between the connected elements (this is probably the only unknown from the data you presented)
  • $\begingroup$ Shouldn't the bolts also be analyzed as OP initially suggested for shear as well, in case that requires a larger sized fastener? $\endgroup$
    – jko
    May 24 '21 at 15:32
  • 1
    $\begingroup$ @jko Generally, I think that its bad design if the bolts are subjected to shear loads. It might be better if dowel pins or shear pins are used for that purpose. so although, you can do a great many things for the calculation of bolts, I would personally advice against using them under a shear loading to transmit torque. However, since you seem to have a different perspective, and I would genuinely like to listen to your thoughts on the matter. $\endgroup$
    – NMech
    May 24 '21 at 15:43
  • $\begingroup$ Another point, is that although the allowable shear stresses are generally smaller (for isotropic materials about 0.57) than tensile, the coefficient of friction is most of the times even smaller (thus leading to greater diameters). Although the point is that the bolt shouldn't have to be subjected to shear, and the friction due to tightening should be enough. $\endgroup$
    – NMech
    May 24 '21 at 15:45
  • $\begingroup$ I also mentioned it on my post, that the best practice. (Ok maybe I was a bit cocky and I should have put "IMHO the best practice is"). :-) $\endgroup$
    – NMech
    May 24 '21 at 15:47
  • $\begingroup$ @NMech There are many instances where the bolts are precisely sized to break as they are cheaper than other components - these are called shear-bolts for a reason. So it is a good idea to use bolts in a shear situation - often found in agricultural equipment especially when the machine could pick up a stone and cause catastrophic damage. Much easier to spend 5 minutes replacing a shear bolt or two $\endgroup$
    – Solar Mike
    May 24 '21 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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