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Engineering books and lectures typically make two things clear:

Maximum bending stresses (normal) develop at the beam surfaces, that is, at the greatest distance from the neutral axis.

Shear stresses on the other hand have their maximum at the neutral axis.

Recently we were given a beam with a simple loading condition:

enter image description here

Here you could find the shear stresses at the surface using the flexure formula, and the shear formula for the centroid axis.

This would yield 4.x in as the required diameter with the bending stresses as the limiting factor, which was also the most common answer among students.

However, if you plot the bending stresses at the surface in Mohr circle and rotate the stress element 45 degrees, you're left with a shearing stress at the surface that requires 12.88 in diameter in order to not exceed the requirements.

This seems the most logical to me, and I don't see any reason to differentiate between the shear stresses developed at the centroidal axis and the surface shear due to the rotated stress element.

Am I correct in this assumption and calculation? If / if not, why?

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    $\begingroup$ ΙΜΗΟ, I think you are right. The issue is that the maximum shear stress meaning can have two meanings depending on the context. The one is the maximum shear stress which is related to the principal stresses, while the other has to do with the transverse forces. However from the problem statement (the context) I think that it refers on the latter. The first context (Mohr's circle) is covered by the dual constraints (allowable normal stresses and allowable shear stresses). $\endgroup$
    – NMech
    Nov 14, 2021 at 8:32
  • $\begingroup$ @NMech that seams reasonable, I agree with you interpretation $\endgroup$
    – Erik
    Nov 14, 2021 at 9:03

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The graph below provides clear information on stresses variation along a cross-section. Pay attention to the segment "C", which always has the maximum stresses act on it. The magnitude of the maximum shear stress on the Mohr's Circle shall agree with the result from the shear formula, otherwise, it is meaningless.

enter image description here

Note: To draw a Mohr's Circle, the minimum information required is either the pair of normal stresses $\sigma_1$ (on major axis) & $\sigma_2$ (on minor axis - usually the confining stress in the beam); the maximum shear stress $\tau_{max}$, or a pair of $\sigma_x$ and $\tau_x$ at any point within the beam, from which the radius of the circle can be determined.

On the graph above, at/near the extreme fiber, $\sigma_1 = \sigma_{x,max} = My_{max}/I$, $\sigma_2 = 0$, the resulting stresses $\sigma_1$, $\sigma_2$ & $\tau$ indicate that at that location, both the diagonal stresses and the shear stress are high, and at which, the failure usually starts, the failure plane is at an angle to the longitudinal beam axis (x-axis).

However, for an element at the neutral axis, we know the normal stress $\sigma = 0$, so how to start the circle without additional information? Shear formula $\tau_{max} = VQ/Ib$ provides the key.

IMO, Mohr's Circle analysis is only valuable in providing information on the state of stresses acting on a finite element at a specific location of the beam. It does not provide meaningful data to be used in structural design, as the state of stresses varies from point to point.

However, people have tried to make sense out of it, thus the "Principal Stress Trajectory Diagram" as shown below.

enter image description here

ADD: State of stresses of an element on the extreme surface of the beam and an element on the neutral axis of the beam.

enter image description here

ADD: Verification through a numerical example:

https://www.ecourses.ou.edu/cgi-bin/ebook.cgi?topic=me&chap_sec=07.2&page=case_sol

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  • $\begingroup$ what is the purpose of principal stress trajectories? $\endgroup$ Nov 14, 2021 at 16:20
  • $\begingroup$ @RameezUlHaq For information/visualization on the variation of the principal stress along the beam. $\endgroup$
    – r13
    Nov 14, 2021 at 16:44
  • $\begingroup$ I thought these trajectories show how the stresses would actually travel through the beam when subjected to loadings. $\endgroup$ Nov 14, 2021 at 16:48
  • $\begingroup$ @RameezUlHaq It would, concentrated load produces a diagram with straight lines as the change in stresses is linear. If you are a practicing engineer, you will notice that at the old time for RC fixed end beams, the tension reinforcement always followed the paths of the principal stress with the 45-degree bent bar at the transition in anticipation of the maximum shear stress. For other practical reasons, this practice was outdated but the concept remains intact. $\endgroup$
    – r13
    Nov 14, 2021 at 17:08
  • $\begingroup$ I think you should modify your answer to include why do we need to have principal stress trajectories. The direction of the principal trajectory basically tells us the angle for each and every point on that line of trajectory, at which the principal stresses would occur. $\endgroup$ Nov 14, 2021 at 17:55

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