This model was designed to help students understand how shear stresses arise in beams. Why those stresses vary from place to place over a beam cross-section, and why they produce specific shear flow patterns. To provide context for the model, you might imagine that it is part of a cantilever beam, like this one. If that were the case, the bending moment acting on the model at section B would be larger than that acting at section A. We represent the tensile bending stresses that these moments would produce using red arrows overlaid on plexiglass blocks, and the compressive ones with blue arrows overlaid on the beam. As you can see, the bending stresses at B are also larger than those at A. As we will discover shortly, differences in bending stresses from one section to another along the length of a beam are the reason that shear stresses arise. Before we begin that discussion, however, notice that the length of each arrow is proportional to the magnitude of the stress that it represents. As a result, the volume of each plexiglass block is proportional to the force generated over its base area. As you can see, large forces are generated in the flanges and smaller ones in the web. Not only that, but the flange forces are relatively far from the neutral axis of the beam, and so they contribute much to the bending moment. The forces on the web are much smaller, and they act at a shorter moment arm. As a result, they contribute relatively little to the moment. To understand how shear stresses arise in the beam, imagine that we used a saw to separate one fiber of the flange from the rest of the beam. As you can see, the force from the bending stresses on end B of the fiber is larger than that produced at end A. In the worked example included at the end of this video, the difference in these forces is 15 units. So, for this fiber to remain in equilibrium, the rest of the beam must exert a force of 15 units on it. We treat the fibers if it were a free body, and therefore draw the arrow on the fiber in the direction of the force that it exerts on it. At the same time that the beam is exerting a force of 15 units on the fiber, the fiber is exerting an equal and opposite force on the rest of the beam, as shown by this arrow. As you can see, the forces between the fiber and the rest of the beam act parallel to the cut, and so we call them shear forces, and we represent them using arrows that have a single-sided head. In contrast, forces that act normal to a surface, like those produced by these bending stresses, are called normal loads, and we represent them using arrows that have symmetrical heads. If we now imagine a cut that removes two fibers of the flange, that pair will be out of balance in the axial direction by twice as much as the single fiber we originally considered. So, to keep this new free body in equilibrium, a shear force of 30 units must act, as shown by these arrows. If we cut off all five of the fibers that make up the top flange from the rest of the beam, the total axial imbalance is 5 times 15 or 75 units, and that is the magnitude of the shear force that must act on the newly cut surface that is between them and the rest of the beam. A fiber closer to the neutral axis of the beam experiences bending stresses that are smaller. The difference between the forces on the two ends of this fiber is only nine force units. Adding 9 to 75 gives a total shear of 84 units. The difference in the end forces for the fiber just above the neutral axis is only three units. Adding that to 84 gives a total of 87 units. Fibers below the neutral axis carry compression rather than tension. As a result, those fibers are out of balance in the opposite axial direction compared to those above it. If we imagine a cut below the neutral axis, some of these fibers are now included, and they reduce the total axial imbalance compared to a beam cut right at the neutral axis. That is why the maximum shear force is always found at the neutral axis. Notice that these shear forces arise because the bending stresses and the moments that cause them vary with position along the length of the beam. It is customary to report the shear stresses on a beam cross section rather than the shear forces. For the sake of simplicity, the web and flanges in this model are assumed to be of unit thickness. Also, the two cross sections are considered to be separated by a unit amount, even though the physical dimensions of the model may suggest otherwise. As a result of these dimensional choices, the shear stress along any of the cuts we have made will be numerically equal to the shear force it carries. Notice too that our cuts always go across the thickness of the web or flanges. That way all of the points in the cut are very near to each other and would generally be expected to carry similar stresses. We never cut the flange parallel to its top surface, for example. Points along such a cut would be relatively distant from each other, and the stresses at various points along the cut would differ substantially. One can plot the magnitude of the shear stress as a function of the distance between the cut and the outer extremity of the flange. As you can see, the graph is linear, a result confirmed by the calculations shown on the right. Similar graphs can be made for other regions of the flanges, and by convention the axes are typically oriented as shown here. As these calculations show, the shear stress in the web takes a parabolic form. Using all of these graphs on a single figure reveals what is called the shear stress distribution for that cross section. In order to understand shear flow, we must transfer the shear stresses we just calculated from their respective longitudinal cutting planes to the beam cross section. We do that using the simple fact that shear in a plane always involves four matched stresses, with arrows that go head to head and tail to tail. If we concentrate on this inside corner of a flange fiber, the shear on the cross section must go tail to tail with a longitudinal stress arrow, and so it must point outwards, and its magnitude must be 15. If we focused on the outside corner of the adjacent flange fiber, we would get exactly the same result. We can show the shear stress on the cross section using a single-sided arrow. At two fibers in, the shear stress is 30, and so that is the stress shown on the cross section at that location. To analyze the web, we note that its shear operates in a vertical plane, unlike that in the flanges which operates in a horizontal plane. Collectively, these shear arrows show how the shear flows over the cross section, and when taken together, they reveal what is known as the shear flow. These arrows represent forces or stresses that must act on the beam for it to remain in equilibrium, so clearly an upwards external force must act on the end of the beam that carries the larger bending stresses. And a downwards force must act on the other end. If you examine the beam from which this model is taken, you can see that those shear directions are indeed correct. Not only that, but if we estimate the total vertical load on the beam cross section by summing the indicated vertical forces, we get 405 units. This value is quite close to the shear of 412 units noted in the worked example that follows the credits. It may seem surprising that the shear forces acting on a beam could be determined solely from the bending stresses acting on two nearby cross sections. However, the differences between the stresses on those two sections are direct result of changes in the bending moment M with axial position X. And that rate of change can be used to calculate the beam shear by using the well-known formula V equals DM DX. This relationship explains mathematically why shear and changes in bending moment are so closely related. We hope this video helped you to understand how shear stresses arise in beams, why they vary over a beam cross section, and how they produce specific shear flow patterns. Thanks for watching.