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I know the formulas and the equations etc. But how can I visualize transverse shear stress in lets say a beam with square cross section? And what is a physical explanation of why is transverse shear stress is maximum in the neutral plane despite the fact that neutral plane neither experiences stretch nor compression during bending?

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Consider a cantilever beam with a square cross section and an end load pointing downward:

enter image description here

If you cut the beam at any point, there must be a transverse distributed load that adds up to $F$ or $-F$ (depending on what side of the cut you're looking at) for static equilibrium to apply.

In addition, note that a true state of shear requires traction forces on all four sides to prevent the infinitesimal element from displacing and prevent it from rotating:

enter image description here

But there can be no such traction force on the top surface of the beam because we haven't applied one.

The simplest explanation is that the internal transverse distributed load looks like this:

enter image description here

with the maximum value at the center and a value of zero at the top and bottom surfaces.

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  • $\begingroup$ Can you elaborate this line: "But there can be no such traction force on the top surface of the beam because we haven't applied one." I still don't quite get it physically that why is it minimum at the top and bottom and maximum at the center. Whats make the center different from top and bottom surfaces in this case? and pls don't bring the equation into it cuz i already know the equation...i'm kind of looking for a physical interpretation of that equation. $\endgroup$ – Engineer SAM Nov 14 '17 at 13:41
  • $\begingroup$ The top and bottom are different because no shear forces pointing left or right exist there. (If they existed, we’d have to apply them, because we’re talking about open surfaces.) Because of the nature of shear couples, as illustrated above, if no left or right shear forces exist, then no up or down shear forces can exist. But downward shear forces must exist somewhere in the cross section to balance the downward point load. Therefore, they must develop away from the surfaces. A parabolic distribution fits these requirements. $\endgroup$ – Chemomechanics Nov 14 '17 at 15:16
  • $\begingroup$ So the neutral plane defines the area of maximum shear. Slice the beam in half (lengthwise in top and bottom halves) and the shear forces trying to slide the two parts against each other appear as vertical shear in the cross-section (per the diagram above). To best visualize the shear stress distribution slice the beam in many slices lengthwise and calculate the total shear force needed to balance the forces/moments. $\endgroup$ – ja72 Nov 14 '17 at 20:06
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Consider a cantilever beam with a single loading on the free end

pic

Now recognize that as we move from left to right the internal moment decreases, and this causes the stress distribution (red arrows above) to have less slope across the cross-section.

As a result, when only a small part of the cross-section is considered near the surface you can see that the sum of the forces horizontal won't balance out unless there are shear forces acting on the bottom (blue arrows above).

Now if an element has shear acting on one face, it must have an equal amount on an adjacent face, and an equal and opposite about on the opposite face. The diagram in @Chemomechanics answer shows this clearly.

So now we have a situation where the vertical shear forces are a function of the horizontal force imbalance. For each location along the beam, this imbalance depends on the vertical height of the sub-element.

In the extreme case of the sub-element being super thin and encompassing only the outer fibers of the beam then the imbalance is zero. On the other hand, if the sub-element covers half the cross-section, from the neutral axis to the outer fiber then the imbalance is maximum.

I hope this helps you visualize the source of the shear stress distribution.

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