2. Variation of τ in the cross-sectional plane

The shear components of traction in the cross sections are τ_yx and τ_zx. As there is an overall shear force V(x) acting on the cross section in y direction, τ_yx must be non-zero. Let us now try to obtain its distribution in the cross-section.

In any beam undergoing loading, shear forces develop at the cross-section. The shear stress components τ_yx (shear stress in y-direction acting on the x face) and τ_zx (shear stress in z-direction) arise from the overall shear force acting on the beam. If the beam is bent, these stresses are important to consider since they affect the structure's integrity. The introduction suggests that we will analyze the distribution of τ_yx across the section to understand how it varies due to the shear force applied.

Think of a loaf of bread. When you press down on it (applying a load), the parts of the loaf near the surface compress more than those further inside. Similarly, when a beam bends and experiences shear forces, different parts of its cross-section experience different amounts of stress due to that force.

The stress component τ_yx in a cross section can be a function of both y and z in general. However, we make a simplifying assumption that it is only a function of y and not of z.

To simplify the calculations, we assume that the shear stress τ_yx only varies with the y-coordinate of the cross-section. This means that at any given height (y), τ_yx is the same across all points that share that height, vertical lines parallel to the z-axis show consistent τ_yx. This is a crucial assumption that allows us to make calculations more straightforward when analyzing stress variations.

Imagine a tall glass of water. If you pour salt into the top, the salt dissolves and you can consider the concentration the same at any horizontal slice of water at the same height, simplifying how you might look at the salt-water mixture. This is similar to how we simplify τ_yx distribution as constant along horizontal lines.

To find the distribution of τ_yx, we cut a small cuboid element from the beam. The bottom surface (-y plane) of this element is at a distance of y from the neutral plane where its free body diagram shows external loads acting on it.

In analysis, we visualize a small cuboidal section of the beam, which we will examine for forces acting upon it. This section helps us to see how the applied loads and moments influence the stress distribution. The forces acting on the surfaces of this cuboid are derived from the external loads and the shear stress acting on the surfaces in relation to the beam's cross-section.

Consider slicing a sponge to examine how water is distributed inside. You observe how much water is held at various depths within the sponge, allowing you to understand stress distribution within that sponge when pressed. The same idea applies here to the cuboid section of a beam experiencing stress.

In order to find τ_yx, we just need to balance the forces on this small cuboidal element in x-direction. The forces due to σ_xx, τ_xy, and τ_zx acting on different faces come into play.

To understand how shear stress varies, we perform a force balance on the cuboid. The total forces acting in the x-direction must equal zero in an equilibrium situation. This includes contributions from the different stress components acting on each face of the cuboid, which add up to explain how τ_yx is determined from the interactions of shear and bending stresses.

Think of balancing scales. If you place weights on both sides, they must balance out; otherwise, one side tips. In our cuboid balance, we ensure all forces counter each other to maintain equilibrium and find the distribution of shear stress.

The final expression for τ_yx becomes τ_yx(x,y) = (V(x)Q(y))/Ib(y) where Q(y) is the first moment of the area and I is the moment of inertia.

This formula gives us a way to calculate the shear stress τ_yx at any point in the beam's cross-section based on the shear force V(x), the first moment of area Q(y) related to y, and the moment of inertia I. It connects how internal stress varies in a solid structure due to applied loading, showing that more shear force results in greater stress on the cross-sections.

Imagine a loaded children’s swing. The heavier the child, the more tension there is in the chains and the greater the stress on the swing’s structure. This formula reflects that relationship within beams, illustrating how loading affects stress distribution.

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