3. Variation in τ for some representative cross-sections

A typical rectangular cross section is shown in Figure 8 and we want to find the value of τ at a distance of y from the neutral axis.

As τ is independent of z, it will have the same value over lines parallel to the z-axis. For applying equation (13), we need to find Q(y), b(y) and I. As this is a rectangular cross section, width b(y) is constant and equal to b. We have already derived I for a rectangular cross section in one of the previous lectures to be.

We only need to obtain expression for the first moment Q(y) of the area above the y line where τ is to be calculated (shown as the shaded region in Figure 8). The first moment will simply be the y-coordinate of the centroid of the shaded area multiplied by the shaded area. As the height of the shaded area is -y, its centroid will be at half of this distance from the y line and hence at.

Thus, Q(y) becomes

while τ becomes

As V is the total shear force on the cross-section and b is the area of the cross-section, equals average shear stress τ while τ at the neutral axis is.

Likewise, at the periphery of the cross section, τ is.

The variation of y vs τ is shown in Figure 9.

We observe that due to the presence of shear force in the cross section, shear stress is maximum at the centroid and vanishes at the two ends. There is another way to realize the vanishing of shear stress at the ends. The points y = ±h/2 also lie on top and bottom surfaces of the beam, respectively. There is no external traction on the bottom surface whereas on the top surface, the distributed load b(x) acts in the y direction. Thus, τ is zero at both top and bottom surfaces. However, due to τ and τ being equal, shear stress on the cross-sectional plane vanishes at y = ±h/2.

In a rectangular beam, the shear stress τ at a certain distance y from the neutral axis can be calculated using the principles of static equilibrium and previous derivations. The assumption that τ is the same across all points parallel to the z-axis simplifies calculations. The width of the rectangular section remains uniform, and the first moment Q(y) is determined by the area above the y line where τ is being examined. This first moment reflects how the area is distributed concerning the neutral axis. The final shear stress calculation is proportional to the shear force and the geometric properties of the area. Importantly, shear stress reaches its maximum at the neutral axis and reduces to zero at the top and bottom surfaces of the beam due to the lack of external forces at those points.

Imagine holding a wide, flat piece of cardboard (the beam) horizontally with your hands applying gentle pressure downwards at the center (the neutral axis). The middle of the cardboard sags the most (maximum shear stress) whereas the ends stay flat (zero shear stress) since there’s no force acting on them. This illustrates how shear stress behaves in a rectangular beam under uniform loading.

In the derivation for rectangular beams, we had assumed that τ is independent of z coordinate. For a circular beam however, this assumption cannot be used. Consider the beam shown in Figure 10 and analyze one of its cross-sections.

If τ is constant along lines parallel to z-axis, the shear stress will be as shown in Figure 11.

Basically, it is non-zero even at the ends. Let us work with a cylindrical coordinate system and assume a radial distributed load is acting (e.g., pressure load) which could also be zero. In that case τ must be zero on the lateral surface. So, τ must also be zero along the periphery of the cross section, i.e., shear stress cannot have radial component along the periphery of the cross-section in the cross-sectional plane. However, if we look at Figure 11, the assumption of τ being independent of the z coordinate leads to a non-zero radial component which is a contradiction. Thus, we conclude that for circular cross-sections, considering τ independent of z is not a good assumption. Still, this assumption is often used since it gives an approximate distribution of shear stress.

For circular cross-sections, the shear stress τ cannot be assumed to be the same along all parallel planes because of how the geometry influences stress distribution. When considering shear in circular beams, if we assume τ to be constant in z-direction, we may calculate a non-zero shear stress at the ends, which is incorrect. By analyzing the forces acting on a circular beam with a radial load, we see that τ must go to zero on the circular edge due to the absence of lateral forces. This realization is crucial for accurate modeling; however, approximations are still made in practice as constant τ can simplify calculations when the material response is generally understood.

Visualize a water pipe (the circular beam) that is under pressure. You notice that the pressure on the sides of the pipe (the ends) is zero because no water is pushing outward there; it is only in the middle where the force is acting. This shows how shear stress behaves differently in circular structures compared to rectangular ones.

Figure 12 shows the cross-section of an I-beam. The centroid of this section would be at the center because of symmetry. So, the neutral axis passes through the center. There are two different values of width possible in the cross section. To find τ at a distance y from the neutral axis, we can again use equation (13). As the width changes abruptly in this case, the distribution of shear stress will also exhibit a jump corresponding to this abrupt change in width. A plot of y vs. τ is shown in Figure 13 exhibiting this jump.

I-beams are designed with a varying width across their height, creating regions of different shear stress distributions in response to forces applied to them. Given that the centroid is in the center due to symmetry, the neutral axis also lies there. Due to the abrupt change in widths of the sections, the shear stress does not vary uniformly; instead, there’s a noticeable jump in shear stress values as you move between these sections. This means that at certain points, you can have significant changes in stress based on the geometry of the beam. Utilizing the established equations allows for quantifying these variations precisely.

Think of an I-beam as a bridge structure. Depending on where you are on the bridge, the support and weight distribution change; sections with wider tops support more weight and experience different shear stress than narrower sections. This is similar to how an I-beam behaves under load.

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