 Earlier, we looked at the general state of stress in three dimensions that can be described by this 3x3 matrix shown here. Now luckily for us, a lot of engineering problems can be simplified to a two-dimensional stress state. So this 3x3 matrix here simplifies to a 2x2 matrix. But that's still four whole numbers we have to deal with, and that's a lot for most engineers, so let's see if we can simplify this further. We will start by illustrating this two-dimensional stress state on an infinitesimal element as shown here. The infinitesimal element has a width of dx, a height of dy, and it also has a thickness of dz coming in and out of the screen, although we are drawing a projection view in this case. We're going to look at the interrelationship between the different stress components by examining equilibrium of this infinitesimal element. We will do this first by looking at normal stresses on their own and then at shear stresses. Here we have redrawn our element with only the normal stresses applied. Let's look first at horizontal force equilibrium. If we look at some of the forces in the horizontal direction with the right direction being positive, we know they have to sum to zero. If we look at the right face of this element, we know that the area of it is that height dy times the length dz in and out of the screen. Thus we have a force component equal to sigma x times dy times dz. That has to be balanced by the stress sigma x on the opposite side of the element acting on area dy times dz. So we get exactly this being equal to sigma x times dy times dz, which tells us that sigma x on the right side of the element has to equal sigma x on the left hand side of the element. If we look at vertical force equilibrium, we get a very similar result in the sense that sigma y at the top of the element has to be equal to sigma y at the bottom of the element. Now we have one last equilibrium to look at, and that's moment equilibrium. However, this is trivial for this problem in the sense that we can pick a point A in the center of the element through which all of the resultant forces act through, and if we take some of the moments about this point, then it's automatically equal to zero. Thus we conclude that normal stresses are independent. Let's repeat the same exercise for the general state of shear stress. If we look at horizontal and vertical force equilibrium, we get a very similar result as for normal stresses. If you sum the forces in the horizontal direction, we find out that the shear stress at the top of the element, tau xy at the top side of the element, must equal tau xy at the bottom side of the element. And similarly, for vertical force equilibrium, tau yx at the right side of the element has to equal tau yx at the left side of the element. However, moment equilibrium is slightly different. We can't actually pick a point where the resultant forces all act through. So we will look at some of the moments about the bottom left corner, point A as labeled on your screen. Now the resultant due to tau xy at the bottom and tau yx at the left side pass through this point, so we only have to deal with the other two shear stress components. So if we look at tau xy at the top surface, it acts on area dx times dz and has a moment arm dy. This has to be equal to shear stress tau yx acting on the right of the element on area dy times dz multiplied by moment arm dx. Now what we get out of here is that tau xy has to equal tau yx. The area times moment arm cancels out in the moment equilibrium. Thus we see that shear stresses are not independent. So there we have it. If we apply a shear stress in one plane, it must be accompanied by an equal magnitude shear stress in the orthogonal plane in order to maintain moment equilibrium. We call this the complementary shear stress. And we have shown it for equilibrium in two dimensions, but it also works in three dimensions.