For example, I have an axial force and a torsion force applied on a rock socket installed in bedrock. If I am considering the side friction between socket and rock as a resistance, the socket surface will have two shear stress generated, one in axial direction, and the other in radial direction. If I want to design this, what is the maximum I should use?

Can I combine the two stresses by triangle rule, or simply adding them up, or only take the larger one?

Here is a conceptual sketch:

enter image description here

  • 1
    $\begingroup$ Are you familiar with Von Mise's stress criterion? Essentially you need to look at the combined stress state when these are acting simultaneously. $\endgroup$
    – ShadowMan
    May 3, 2019 at 17:44
  • $\begingroup$ Thank you, ShadowMan. Simply, do you have a simple answer on this. If there are two shear stresses acting on one plane, in perpendicular direction, what is the correct way to combine them? $\endgroup$ May 3, 2019 at 17:52
  • $\begingroup$ Duplicated on Physics see physics.stackexchange.com/q/477678/207455 $\endgroup$
    – Solar Mike
    May 3, 2019 at 18:25
  • $\begingroup$ @Christian777 if you can provide your attempt at looking at the combination so far I can assist you from there. But I think as it stands you have a pretty broad question and you haven't shown any work. You should review Von Mises, attempt to identify which shears you have in his equation. Basically It isn't clear to me what the stresses are on your problem, in your OP it seems different than in your comment. Best to draw them up and name them so you can use Von Mises. $\endgroup$
    – ShadowMan
    May 3, 2019 at 18:27
  • $\begingroup$ Hi ShadowMan, Von Mises is a very theoretical formation, I don't know exactly how to use it. My attempt is that i want to know when I have two shear stresses acting on one plane in perpendicular direction, what is the real stresses on my surface now? Is it the resultant of the two stresses, or sum of the two, or the greater of the two will control, or it is in a way but not a simply like this. I am just not sure what is the final stress level in reality. I initially thought using the resultant, just like two force in two direction, or any vector. But stress seems to be scalar. $\endgroup$ May 3, 2019 at 18:35

1 Answer 1


You are right, the stresses in this case to are at the same plane, at any given point, so simply by triangle method or vector addition just add them, to verify that you do not surpass the allowable stress.

However keep in mind their projection axially and horizontally for later when you need to calculate Say the torque as opposed to pull out force.

If the stresses where not on the same surface then you could calculate maximum stress and its direction using Mohr circle.

  • $\begingroup$ The torque would resolve into an axial stress that is normal to a cross section surface. How can we neglect interaction of stress? $\endgroup$
    – ShadowMan
    May 3, 2019 at 18:56
  • $\begingroup$ torque is orthogonal to axial stress. woul not resolve to axial stress. $\endgroup$
    – kamran
    May 3, 2019 at 19:01
  • $\begingroup$ Hi ShadowMan, see my posted sketch, I think it may help. The torque is generating a side friction on the cylinder, or a rotational stress, tangential to the surface, will not generate a axial stress. $\endgroup$ May 3, 2019 at 19:05
  • $\begingroup$ thank you kamran. I think I got it. i can treat them just as two forces, if I am understanding correctly, and the resultant stress would be the maximum stress on the plane in this case. $\endgroup$ May 3, 2019 at 19:10
  • $\begingroup$ @Christian777, yes stress is actually force/ surface unit, but its a vector and will add like vectors. The reason we don't care about the combination of axial and shear stress in beams is they usually maximize in different zones. Like shear at the supports and moment at the middle. $\endgroup$
    – kamran
    May 3, 2019 at 19:14

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.