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?
[Physics] n intuitive explanation of transverse shear stress
classical-mechanicsmaterial-sciencenewtonian-mechanics
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It's hard to do this one analytically, but that's why we have computers. I used a commercial FEM package to set up your problem and simulate steel with some dummy loading conditions.
Your intuition is basically correct away from the corner. Indeed, the stress is uniform far from the corner, and it only has interesting behavior close to the feature. However, You're not quite right about how the stress is distributed at the interface. The corner creates a singularity in the stress field that is best described with a picture. The first picture shows contours of the axial component of the stress. Don't worry about the actual values of stress. These are dependent on the loading, and will simply scale with the applied traction.
To get a more quantitative view of how the stress changes at different cross-sections, I created a few plots starting at the top in the thick section and moving down into the thin section. The plots are the axial stress (normalized by the far-field stress in the thick section) vs the x-position. As you can see in the third plot from the bottom, the stress at the corner is much higher than the stress in the bulk. Successively refining the mesh would reveal that this is, in fact, a true singularity, so the stress would go to $\infty$ if the mesh size $\Delta x \rightarrow 0$.
Note, it's not surprising that the small-diameter mean stress is ~4 since this was a 3D calculation of a cylindrical body and the ratio of the $D_{big}/D_{small} = 2$.
Think of it in this way: If there exists a couple on the center of mass by exerting shear stresses on AB and CD, then shouldn't the block rotate? However, we know from experiment that the block will tend to deform (by a very small amount) as shown in Figure A, so there exists no rotation, which means the system is in rotational equilibrium.
The couple created by the shear stresses is opposed by the couple created by the complementary shear stresses, which prevents turning of the object. The object will not deform to the shape as shown in Figure 2 because $ F_{com} = F_{comneg} $ and is actually a reaction couple to the couple produced by the shear stresses, thus causing deformation in only one direction, which is the direction in which the shear stress was applied.
Best Answer
Consider a cantilever beam with a square cross section and an end load pointing downward:
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:
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:
with the maximum value at the center and a value of zero at the top and bottom surfaces.