Equation: Tao = (T*C)/J

i know the equation but it just seems counter-intuitive that the maximum shear stress is at the biggest radius for 3 reasons that I have(thought of).

1- imagine/assume that the applied force propagates (like water waves) in an equidistant manner, or in a circular manner for a cross-sectional view which is in 2D. As the force propagates further away from the center (assuming the torsion is applied at the center), the stress should decrease because the surface area increases as the radius increases (stress is N/M^2)

2- the axle-wheel friction analogy: it is a well known fact that as the axle diameter decreases the axle-wheel friction decreases because the friction's leverage is smaller at smaller axle diameters (thus static & dynamic friction are lesser in magnitude for smaller axle diameters). Thinking of it Intuitively, a smaller axle will try to push the wheel closer to its center which is harder. Similarly, the reaction friction force of the wheel will be less effective on a smaller diameter axle since it is pushing the axle from its edges which are close to the center as well. I don't know whether I can say there is friction/connection between the particles of a solid circular shaft in the same way but I think that the concept of leverage applies here just as it applies for the axle-wheel analogy since torsion is essential bending moment(s). If so, then in a circular cross-section, it should be easier for smaller diameters to shear relative to each other (rather than move with each other; lesser friction/connection force at smaller diameters due to reduced leverage). For the same reasons, larger diameters should feel the opposite. Consequently, this should cause earlier yielding at smaller diameters than at larger diameters for the same applied force(s). (Stress increases as strain increases. So again stress should be greater at smaller diameters) *** (the main contradiction of this part to the literature is that: if this was true, a material at smaller diameters will need SLIGHTLY lesser stress to strain for all points on the curve(since it is easier to strain at smaller leverage) than needed at the larger diameters. This would contradict (only a little bit) the notion that each material has a constant, defined stress-strain curve/behavior.) )

3- based on (2-): as smaller diameters are more prone to elastically deform than larger diameters. There should be some force that goes into motion-deformation at smaller diameters more than at larger diameters (some force is translated into motion-deformation (relieved) instead of stress, thus the force(s) propagating towards later, bigger radii is/are less than the initial force(s), which should further decrease the shear stress at larger diameters than at smaller diameters)

I am not sure whether these reasons are valid/true that's why I was skeptical about mentioning them at first. So can these reasons be valid ?? If so, then are there any other missing factors that make stress maximum at the biggest radii regardless of the 3 factors I mentioned ??

Thanks in advance.

  • $\begingroup$ @Mr.Bean I think you should see it in terms of deformation instead of force (since stress depends on the deformation of the material). You may imagine two hollow shafts (tubes) that have different diameters (one smaller than the other), and you deform both of them applying a torsional force, so both of them are deformed by 90 degrees, and think about which of them has deformed the most link $\endgroup$ – user190081 Sep 10 '18 at 22:38
  • $\begingroup$ @ user190081, " so both of them are deformed by 90 degrees, and think about which one of them has deformed the most" you just said that they both are deformed by 90 degrees so how can one deform more than the other ?? $\endgroup$ – Mr. Bean Sep 10 '18 at 22:50
  • $\begingroup$ @user190081, you don't answer my questions directly. Also what is wrong with seeing it in terms of force (if stress is dependent on deformation, force is the cause of deformation). Are you sure you understood my reasoning and what I am trying to say ?? I know my writing style is quite clumsy so I don't really blame you for not understanding. $\endgroup$ – Mr. Bean Sep 10 '18 at 22:53

I think you may be overcomplicating this. The equation for torsional stress a distance $r$ from the center of circular solid or thick-walled shafts (thickness > 0.1r) is given by $$τ=\frac{Tr}{J}$$ where $J$ is the polar moment of inertia for a circular solid of radius $a$ given by, $$J=\frac {πa^4}{2}$$ So for a given polar moment of inertia, the torsional stress is proportional to the distance $r$ from the center, thereby being maximum at maximum $r$. So the outer portion of the shaft experiences maximum torsional shear stress. One reason why drive shafts are hollow. Saves on material.

Hope this helps

  • $\begingroup$ Thanks but I know this already. However knowing it is one thing, relating it to intuition is something else. As it stands, there is a contradiction for me between what is generally accepted and what I intuitively believe in (my reasoning) $\endgroup$ – Mr. Bean Sep 10 '18 at 23:57
  • $\begingroup$ @Mr.Bean. OK understood. Just wanted to make sure about what you knew. I will look again at your 3 reasons. $\endgroup$ – Bob D Sep 11 '18 at 0:03
  • $\begingroup$ Thanks a lot mr. Bob. Appreciate your help and enthusiasm. $\endgroup$ – Mr. Bean Sep 11 '18 at 0:04

I think deformation is the answer.As pointed out by user190081, the angular deformation may be same but if we consider two circular strips one near the centre and one far from it, the points on the outer strips will be displaced more than the points on the inner strip for the same angular deformation.This might be a reason.

  • 1
    $\begingroup$ That is a simplistic view. You are assuming that there is no relative deformation between the smaller and bigger diameters of a solid circular cross section under torsion. this is probably a wrong assumption. Appreciate your help though, thanks. $\endgroup$ – Mr. Bean Sep 18 '18 at 8:43
  • $\begingroup$ I am arguing that there should be relative deformation in favor of smaller diameters (more deformation at smaller diameters) because of the weaker leverage they have to transmit the stress to larger diameters (if we assume that the torque is applied from the center). THAT'S BRIEFLY WHAT POINT 2 IS ALL ABOUT !!! $\endgroup$ – Mr. Bean Sep 18 '18 at 8:44
  • $\begingroup$ But the problem is stress is not the action, it is the reaction(actually resistence).Deformation or elongation is the action.When you twist the member, the points near the centre deform less and far away from the centre deform more.Then they divide the load in such a way that the points near the centre resist smaller part and those farther resist larger part of the load. $\endgroup$ – Mohan Sep 18 '18 at 9:57
  • $\begingroup$ Deformation/strain is not the cause for stress. Strain does increase stress however as it happens. Regardless, what is your proof or reasoning that the points closer to the center deform less than the farthest points ?? And how do they divide the force between them ?? You must have some reasoning if you are talking like that, so please share it. Thanks. $\endgroup$ – Mr. Bean Sep 18 '18 at 14:19
  • $\begingroup$ Ok.What is the cause of stress? $\endgroup$ – Mohan Sep 18 '18 at 14:22

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.