 In the last video, we looked at how shear stress can be related to the normal stress in a situation of bending. However, we did not derive any formulas, so let's take a look at that now. In order to do that though, we need to remember where exactly we left off. Well, we looked at a beam subjected to a transverse shear force as shown here. We looked at a small section of that beam cut away at a distance y' from the neutral axis. And we saw that a horizontal shear stress needed to form on that section plane in order to balance the resultant from the two normal stress distributions. And the key there was that there was this d' or this variation in that normal stress distribution that causes this shear stress to occur. Now in this video, we are actually going to look at equilibrium of all of these, of the resultants of all these stresses in order to derive our equation. But before we do that, we need to think about some of the assumptions we're making in this analysis. And the first major assumption is that the flexure formula describes a normal stress distribution in our beam. Basically, this means that all the assumptions that we used for the flexure formula in the previous unit must also apply here, so that plane sections remain plane and the individual lines along the beam will curve in the nature that we saw in the previous unit. We're also going to allow in this analysis the width or, as it's labeled here, t of the section can vary with our vertical position in the beam. Now you have to remember that for the flexure formula that applied only to symmetric beam sections. There has to be symmetry across the y plane. But we will allow the width to vary with vertical distance. The next assumption we need to make is regarding the distribution of the shear stress on this cut plane. If we're going to look at the resultant of it, we have to have some idea of the distribution. So for this, we are going to make the assumption that at this point, the shear stress is uniform. With our assumptions made, we can now look at the horizontal equilibrium. So if we look at the sum of the forces in the axis of the beam or the z direction, they must equal zero on our element. So let's look at the resultants of our three different stress distributions. First, let's look at tau. So tau is uniform as per our assumption on this face, and it acts upon this face that has a depth of dz and a width of t. So this becomes our resultant due to our shear stress distribution. Acting in the same direction is our normal stress on our left face. So we need to define the area that that acts on, which I will call a prime, and it is on the opposite face of the beam, this cross-sectional area. And what we need to do is integrate our normal stress distribution, sigma, times a differential area element, da, over the domain a prime. Now those are acting in the direction to the left, to the right. We have sigma plus d sigma, so we subtract the integral of sigma plus d sigma times da over the domain a prime. What we can then do is recall the flexure formula that sigma is equal to my over i, and we can substitute this in to our equation, and we get the following. What we can also see here is if we expand out this term here, we see the addition of the integral of my over i, da, minus the integral of my over i, da, because you can separate out this dm. So this is cancelled out by that term. And what we are left with is that 1 over it times dm by dz is equal to the integral over the domain a prime of y times da. If we take a closer look at this result, there are some simplifications that we can make. First, we can recall that the variation in internal bending moment with distance along the beam, or dm over dz, is precisely equal to the internal shear force, v. So we can take this and substitute it in. We can also recall that the integral of y, da is the first moment of area. So it will be equal to the area a prime, which is our domain, times the centroid to that area a prime. So we will call that y prime bar, and that's the distance from the neutral axis, which is here, to the centroid of a prime. And for simplicity, we will give this a label q, and that is the first moment of area a prime about the neutral axis. If we take these simplifications and combine them, we get the result that the shear stress tau is equal to vq over it. So now we actually have brought back in our shear force, which we knew would happen from the previous video. Now the only simplification or remaining thing to do is remember, due to complementary shear, that the horizontal shear stress is equal to the vertical shear stress. So tau in the vertical direction is vq over it. Now we can take a look at this again and see that we can take our thickness term and bring it over to the left hand side and do our shear stress tau times our thickness t, and that's equal to vq over i. Now this might actually ring some bells. The shear stress times thickness is actually known as shear flow. We used this concept of shear flow in torsion of thin walled sections. And it will become important again for shear of thin walled beams. So for now I just indicate it to you, but we will revisit this shear flow concept later on. So now that we've derived this formula, what can we say about the nature of its distribution? Well if we look at the section of our element again, showing a distance y prime, what we can see is this term v will be constant for the section. It is the internal shear force acting on this entire section. Similarly i will be constant for the section. The moment of inertia of that cross section is constant. It's not a function of y prime. Now the area a prime is, and that all comes into play in q, but v and i will be constant for the cross section. T can also vary, but for now let's just bring t over here and equate this to shear flow just to remove t from the equation. So v is constant, i is constant, so the distribution is dependent upon q. And if you recall what q is, q is the distance from the neutral axis to the centroid of the area a prime times the area a prime. Now if we look at the area a prime, it's going to be a linear function of y prime. It's going to decrease with increasing y prime in a linear fashion. You can formulate that and we'll do it in some later examples, but it should be pretty clear that that area will decrease linearly with distance from the neutral axis. Similarly with y prime, it is linear function of y. So that distance to the centroid will vary linearly with our location of our section. So that means that q is proportional to y prime squared, so it should be a parabolic distribution. And because v is constant and i is constant, our shear flow should have a parabolic distribution. So that's one thing we can already say. And if we think about the fact that the shear stress reacts the imbalance between the two normal stress distributions, that also gives us an idea of where it will be maximum and minimum. If you had y prime equal to zero, you have all of the tensile stress acting on both sides, and so you create the maximum difference in resultant normal force. Whereas if y prime brought you all the way up to the top and we included the entire cross section, well we know the normal stress distribution over the entire cross section has zero resultant. So you would have zero force or zero shear stress acting at that top surface. So if we then plotted from our neutral axis a distance of y, and this was the periphery of our beam, it has to be zero at the outer portions because the entire normal stress distribution will produce zero resultant, and it's going to vary parabolically to a maximum at our neutral axis. So that is a little bit about the nature of our shear stress distribution or shear flow distribution. And we're going to talk about this a little bit more in later videos where we look at shear of thin walled structures because we're going to have a few more directions and a few more intricacies introduced.