 Thin walled beam profiles are very common and important in aerospace engineering, thus it's important to be able to evaluate and understand the shear stress distribution in a thin walled beam section. In order to examine this, however, we're going to look at it by looking at a limitation of our shear formula we already developed. And we're going to do that by looking at sort of a common thin walled profile, an eye beam, however we're going to make the flanges very thick. So the shear formula would definitely apply in this thick flange as well as in this web. If I section away a portion of the flange and a portion of the web and applied our shear formula, we would get that there has to be a uniform shear stress acting in the flange and a uniform shear stress over the small area of the web. Now the problem occurs is when these two faces get close together as we approach this sudden transition. At that sudden transition this uniform stress distribution here has to change into this more compact higher magnitude acting over a smaller area. And we know that's not going to happen in some sort of sudden jump. It's not going to just suddenly change from one to the other. Really stresses tend to flow through a structure and that boundary condition will influence the stress distribution and the stresses will begin to flow into that web. This kind of disruption of the stress field that we predict with our equations that assume sort of a well-distributed type of field. We examine this when we introduce the concept of Saint-Finance principle where certain boundary conditions and things can interfere with that sort of well-distributed stress. Well that sudden transition is precisely one of those things. So strictly speaking we should be comfortable with the fact that the shear formula and the stress that it predicts here and here will not be very representative of exactly what happens at that transition. So what if this profile was thin walled? We actually were looking at a similar profile but it had a very thick flange on it. Well if you recall what a thin wall means effectively if we call something thin walled we say that the distribution through the thickness is negligible. We looked at this in torsion where we took the shear stress distribution and made it uniform rather than a linear distribution because we said you could neglect any variation over that very small distance. So that's it precisely what we would do for a shear and a beam however something interesting comes up. If we look at the top surface of our flange it is a free surface and the shear stress has to be zero. If we look at the bottom surface of our flange except in the region where it's connected to the web but just over here then our shear stress is also zero in the vertical direction. That means using this thin walled approximation that the overall shear stress distribution through the thickness of the flange is zero. Now that seems a little bit counterintuitive but it's a result of that assumption that we're making. So all that is actually remaining is that horizontal component that we showed in the previous slide. Shear has to redistribute in the horizontal direction. So what we get is a shear flow that increases from the outer edges, moves its way into the web, flows down through the web and then out again through the bottom flange. And this is the shear flow which we call Q which is just simply thickness times our shear stress. Now if you recall from last video we actually developed an equation for shear and we even rearranged it in terms of shear flow and said that the shear flow is VQ over I. So is it really that simple that we can apply the same formula from the previous videos? Well let's take a closer look at our beam and we'll cut away a section of that flange and we can recall that there will be a bending stress distribution and it'll actually be sigma here. It's a result of M and sigma plus d sigma which is a resultant of the internal moment plus differential moment. Now as a result of the higher magnitude of the stress here we will see that an internal shear flow Q has to develop along this face and that will balance the equilibrium of these two resultants. And because of the principle of complementary shear we will have a complementary shear flow acting in the cross section of our beam as shown here. So yes it really is that simple. We can calculate the shear flow Q with the same formula VQ over I where Q is still equal to Y bar prime times A prime where in this case our neutral axis would be here. This would be A prime and the centroid of this would define Y bar prime. So you have to look at it from a slightly different perspective because now we're doing sections in the vertical direction of the flange here and then when we get here we start doing sections horizontally but the same formula the same mechanics the same principles do apply.