please help me understand shearstresses

This is due to what's called "surface tension" - a molecular phenomenon, responsible to many peculiarities, concerning molecular bonding and reactivity - chemical and mechanical.

Top layers are special in terms molecular structure, and molecular bonding there, is a bit different when compared to the rest of the molecule-lattice underneath.

Molecular bonding is a three-dimensional structure, and every compound has it's typical average mechanical and chemical characteristics, but the most outer layer, being bonded only to the layers underneath, and this creates surface tension which in turn reacts different to an outer contact and interaction, than the rest of the material.

To be able to analyse given characteristics of materials, you have to be very specific in chemical terms, down to the molecular level, which of course is what metallurgy (for instance) is about.

In short: Shear Stress acts slightly different on the surface of the material, than it does inside it, because of surface tension of a specific, given, compound in question

If you load a rectangular beam in pure torsion as implied by the arrows around the little red rectangles, there will be shear stress throughout the section. The maximum shear exists at the middle of the longer face. That would be torsional shear.

If you consider a simple span with gravity load, you can calculate the shear, V and the moment, M at any cross section. For a rectangular section with total shear V, the average shear stress is V/bd, but the stress at any given point varies according to the distance above or below the neutral axis. At the top and bottom the stress is 0. At the neutral axis, it is 1.5V/bd, 50% more than the average. The magnitude of shear stress varies parabolically throughout the height of beam, including the fibers at the edge.

Hope that helps.

I looked at your sketch again. I was mistaken in my last post about what the arrows indicate. The arrows do not show torsion (they would if they were all clockwise or counterclockwise). Each little red rectangle represents a small element of the beam, not an element on the face but an element at right angles to the face (such as the side of the beam).

The arrows represent an element in pure shear. At any point, the horizontal shear stress is precisely equal to the vertical shear stress. They are both 0 at the top and bottom of the beam and maximum at the neutral axis as stated earlier.

Sorry about the confusion.

Yes Bruce that is correct what you say, but you misunderstood my question. I drew the three particles to show that they relate to each other. The reason that one particle has shear stress is because the one over and under has shearstress(this is because of Newtons 3. low and that you need shearstresses in vertical and horizontal direction because of equlibrium). My problem is understanding how the particles right beneth those on the surface can have shearstress, when the ones on top can't.

If your particles have any dimension at all, there are none without shear stress. The particles at the top of the beam have zero shear stress at their upper boundary but no matter how small they are, they feel shear stress at their lower boundary.

Here are a few references which may throw a little light on the subject for you.