 In this video, we are going to examine shear in a beam subjected to transverse loading. Now in previous units, when we introduced a new type of loading and looked at the stress distribution, resultant from that new type of loading, we did it in a particular way. We first looked at the deformation that would occur to that type of loading, then looked at the compatible strain distribution that matched that deformation, and then used Hooke's law to convert that strain distribution into a stress distribution. This isn't quite possible to do with shear stresses in a beam subjected to transverse loading, and we're going to show you why by looking at a simple beam. In the past unit, we looked at beams subjected to pure bending or moment loading, so that when we took a section, saw an internal load equal to that moment, and isolated an element, we had an element subjected to pure moment loading. We could do this and look at the deformation of this element, and use that to derive the normal stress distribution. Now if we look at a beam subjected to transverse loading such as the beam shown here with a point load V applied to the end, and we take a section, we see that we will have an internal shear force and bending moment. And then if we isolate a small element, we see that there will also be an internal shear force and bending moment. So we can actually isolate shear in a beam subjected to transverse loading. You will always have shear and bending. But if we actually take a closer look at this element that we've drawn here, there's a small mistake. If we look, there is a distance here in our element. Even though it's an infinitesimal distance, it does have some distance. And so the two shear forces V will create a moment. So the internal bending moment actually has to vary from one side to the other to maintain equilibrium. Now what this actually means is that there is a variation in the internal moment distribution as a result of a shear force. Hopefully this relationship between internal bending moment and shear force is not too big of a surprise. Earlier in your studies when you learned how to construct shear force and bending moment diagrams, you already established this relationship. You found that the derivative of the bending moment diagram, or the rate of change of the bending moment with position along the beam, was precisely equal to the internal shear force. We can exploit this relationship in order to derive the internal shear stress distribution in terms of bending moments. So we see that there is a relationship between bending moment and shear force. So there should also be a relationship between the bending stresses and the shear stresses. So what do we know about the bending stresses first of all? Well we know that due to that bending moment loading we will have a linear varying normal stress distribution as shown here on the left with a similar distribution on the right which will be slightly different. It will vary by a small amount d sigma due to the variation in our bending moment. Now if we look at our shear stress distribution, we know that there has to be some sort of vertical shear stress distribution acting on those two faces. But we don't actually really know anything about the nature of that distribution. However, since the internal moment and shear forces are related, there must be a relationship between these shear stresses and normal stresses. So let's take a closer look at our normal stress state and see if we can define this relationship. I will redraw our element in a sort of a three-dimensional view so that we can better see what's going on. If we look at our stress distributions it actually does look like this element is not in equilibrium because the magnitudes of the stresses on the left are smaller than the magnitudes of the stresses on the right. However, you need to recall that the normal stress distribution due to bending is such that the resultant force due to the stresses on one face is actually zero. So the compressive stresses on one face create a resultant force that is in equilibrium with the resultant due to the tensile stresses. This is true for the entire section of the beam. However, if we take a section of our section, so we'll cut a line at a distance y' from the neutral axis, we can now create a smaller portion that we remove and now it will no longer be in equilibrium because we've removed the compressive stresses at the top of our beam. We've cut them away and now the resultant tensile force on the left face will be smaller than the resultant tensile force on the right face. As a result, a shear stress has to be formed along our new section plane to maintain equilibrium. So an applied shear force results in variation in bending normal stresses along the length of the beam and this variation necessitates an internal horizontal shear stress within the beam that varies with position y' in the beam. But how does this help? We are interested in the shear stresses acting vertically, not horizontally. Well, this is where the principle of complementary shear stress comes into play. If you recall back from our videos on stress and strain, we derive that every shear stress has an accompanying or complementary shear stress acting perpendicular to the original shear stress. If we apply this principle to the horizontal shear stress we examined, we can see that the vertical shear stress is precisely equal to the horizontal shear stress at every point in the beam. In the next video, we will look at deriving an expression to describe the distribution of shear stress through the thickness of the beam. However, before viewing the next video, please take a moment to think about what you can say about the distribution based upon what you have learned in this video. Where would you expect the maximum shear stress to occur? What about the minimum shear stress? And can we actually say anything about the nature of the distribution? Once you have thought about these answers, you will be ready to go on to the next video where we will derive the expression for shear stress in a beam.