The Question is from shear stresses in beams

Consider a beam made of circular cross section, in which at any cross section the shear force is V.

enter image description here

The textbook I'm following states that -

When a beam has a circular cross section, we can no longer assume that the shear stresses act parallel to the y axis. For instance, we can easily prove that at point m (on the boundary of the cross section) the shear stress $\tau$ must act tangent to the boundary. This observation follows from the fact that the outer surface of the beam is free of stress, and therefore the shear stress acting on the cross section can have no component in the radial direction.

I'm having trouble with understanding why the shear stress at m, should be tangent to the boundary.

Say I enlarge the element,

enter image description here

and assume that the shear stress acts along y axis. this shear stress can be resolved into tangential and radial components to get $\tau_t$ and $\tau_r$. The book says this $\tau_r$ can't be balanced by any other stress in the radial direction and hence it must be zero. But it could happen that a normal stress is developed on the surface as shown in (b), which balances $\tau_r$

  • $\begingroup$ What is the definition of shear stress? $\endgroup$ Feb 20, 2022 at 20:23
  • $\begingroup$ Can you provide a link to the source material? $\endgroup$
    – r13
    Feb 21, 2022 at 1:30
  • $\begingroup$ To further elaborate upon my question, the quantity $\tau$ is figure (a) is not a shear stress and the quantity $\tau_x$ in Figure (b) is, by definition, a normal stress. The implication is that you don't have a clear idea of what a shear stress is. Perhaps you should return to the part of the book that defines this quantity? $\endgroup$ Feb 21, 2022 at 6:27
  • $\begingroup$ Biswajit, this might clear things up - drive.google.com/file/d/1vdIOKVY4wofJDLtgQSirwrj5BWeOaqQC/… $\endgroup$ Feb 21, 2022 at 7:53
  • 1
    $\begingroup$ I think you misunderstand. I know what the author is talking about and why the author is correct. I'm just pointing out that you have to think a bit about the basic definitions to understand why the author is correct. You probably need to just look at a book on elasticity (or advanced mechanics of materials) to get the ideas clear in your mind. $\endgroup$ Feb 22, 2022 at 21:13

1 Answer 1


Let's take a small element of the circular shaft and apply the shear stress $\tau_1$ in an arbitrary direction on the surface of the cut. The shear stress can be resolved into a radial shear stress $\tau_r$ and tangential shear stress $\tau_t$ as shown below.

enter image description here

We recall the state of shear stresses on a cube element - when shear stress presents on one plane, there must exist a shear stress on the face normal to that plane. In this case, there must be a shear stress $\tau_z$ corresponding to $\tau_r$, however, we know that the stress on the outermost surface of an element is zero, from this realization we can conclude that $\tau_r$ must be zero as well, so only $\tau_t$ remains effective.

Note that this phenomenon is valid for shear stress on the perimeter only.

  • $\begingroup$ Thanks r13, do you happen to know how other shear stresses on the c/s will be oriented wrt the y axis? or if you know about any source? $\endgroup$ Feb 22, 2022 at 18:07
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    $\begingroup$ As noted, other than the surface layer, there will exist a tau_z, thus the tangential orientation of the shear stress is no longer valid. I don't know if this answers your question or not. $\endgroup$
    – r13
    Feb 22, 2022 at 19:20

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