 Let's start today's video off with a little bit of a thought experiment. Consider the uniaxial-loaded tensile member shown here on the left, subjected to a force P. What we would like to do is figure out what is the stress at point C. Well, we've done this many times before in our unit on uniaxial-loaded members, and we would simply section this profile at the point C. We would then see that there is a local coordinate system for that section plane, and we saw that there was a uniform, normal stress distribution acting on that plane, and its magnitude was equal to the uniaxial force P divided by the cross-sectional area. But why did we section this bar as we have shown here? Could we not section it another way? Perhaps on an angle? If we did this, there would also be a local coordinate system for this section plane, and we would have a normal stress acting on that section plane. But what we can see here is that the resultant from this normal stress is not going to be in complete equilibrium with our uniaxial force P. There would be a component that would tend to pull the bar off to the left. Therefore, there would also have to be a shear stress acting on this section plane to maintain equilibrium of the bar. What this little thought experiment shows us is that although we are looking at the same point, point C, in both these cases, the state of stress at a point is dependent upon the orientation that you are looking at for that point. Another way to look at this is by looking at the differences in strain or the resultant deformation in different orientations. Take for instance the book shown here. We have drawn a square element that is two inches by two inches, one oriented with the book on the left, and one oriented 45 degrees to the plane of the book on the right. Now if we apply a shear deformation to this book by sliding the two covers parallel to each other, we would get the following deformation. The element on the left shows a state of pure shear because the lengths of the sides remain the same, but the shape changes, and that is exactly the type of deformation that occurs in pure shear. However, the element on the right actually maintains its shape. It still has approximately 90 degree angle between the sides, but the lengths of the sides change. This is representative of a biaxial normal stress type of loading. So let's take a look if we can derive some equations that can help us to calculate these changes in stress with orientation. We will call these the plane stress transformation equations. I want to emphasize the fact that it says plane stress. If you recall from earlier units, a state of plane stress is one where you have stresses acting only in one two-dimensional plane. So that as the element shown here, we have our sigma x, sigma y, and tau xy. Stresses acting in and out of the screen, the sigma z, are actually zero. So what we're going to do is section this element, which will represent a point. Now we've drawn it as a square, but as we have done in many times, many cases before, but it is representative of a single point. And we'll section it at an arbitrary angle theta, which would give us the element here shown on the right. Now we have to apply our normal stress acting at our new x direction and our shear stress tau x prime y prime acting in our new local y coordinate. What we need to do then is to determine what this new sigma x prime and tau x prime y prime, these are our new normal and shear stress in our transformed coordinate system. And we do this by looking at equilibrium. But it's very important to remember that we can't actually directly do equilibrium of stresses. Equilibrium applies to the resultants of stresses. Therefore we need areas. We need areas for which these stresses act upon. Now this sometimes can cause confusion because we're looking at the stress state of a point. But what happens is when we section it, even though these are infinitesimal areas, there is a difference between the areas that these stresses are acting upon. So for now we will label this cross-sectional area dA. Therefore this lower area becomes dA times sine of theta. And this side becomes dA times cos of theta. So now we're going to look at some of the forces in the local x prime coordinate direction. I've just done this because we want to solve for sigma x prime so it's easier to do equilibrium in this direction. All of these forces have to sum to zero. So what we'll see is we will get a resultant force due to sigma x prime acting over area dA. We will also get a component due to tau xy acting on area sine theta dA. But then it has to be transformed into the x prime coordinate direction through cos of theta. Similarly we get sigma y acting upon this area which is dA sine theta and it has to be transformed into the x prime direction by sine theta. We can do the same for the shear stress and normal stress acting on this phase. And so we have our different areas dA cos theta and our different projection corrections sine theta and cos theta. All of this has to sum to zero for equilibrium. Now we can see right away nicely that our areas dAs cancel out. So although we had to have the relative differences of them, the absolute magnitude of them doesn't actually matter. So if we cancel those out, rearrange and solve, we can actually get that this formula for sigma x prime. Now in the textbook and there is further simplification of this formula using trigonometric identities which are a little bit beyond what I want to get into right now. So I will just show you the final resultant of that that uses those trigonometric identities to get to this final version of this formula. You're welcome to look those up yourself but in fact I don't actually want you to memorize this formula at all. We're going to come up with a graphical method for determining these stresses later on but I want to derive these formulas so that you see where they come from. They come from this equilibrium and we're going to use them to plot a simple case at the end of this video as well. We now need to repeat the same process in the y prime direction so some of the forces in the y prime direction have to equal zero and we get the same lengthy expression due to the shear stress tau x prime y and all these other stress components acting on their area and projected into the y prime direction and we see that the da's cancel out again and if we simplify it apply our trigonometric identities we get the following result. tau x prime y prime is equal to negative sigma x minus sigma y all divided by 2 times sin 2 theta plus tau x y times cos 2 theta. Let's take a look at a simple example of what these two equations will predict for the variation in normal and shear stress with orientation at a point. For this example we will consider a stress state that has sigma x equal to 100 megapascals sigma y equal to 50 megapascals and a shear stress of 45 megapascals. Now what we will do is plot what sigma x prime and tau x prime y prime are for this local coordinate system that is varies as a function of theta. If we plot these two functions out we see that we get a sinusoidal distribution for the normal stress and a sinusoidal distribution for the shear stress. Now there's some interesting properties of these two sinusoidal functions. We can see that the normal stress has a local maximum and minimum value and we call these the principal stresses sorry and they are the maximum and minimum normal stresses predicted by the stress transformation equations. Now there's some interesting properties of the principal stresses that we see here that are true for all cases. First they are out of phase with each other by 90 degrees so it is a normal stress state that acts on two orthogonal planes. Furthermore they occur where the shear stress is zero so the principal stresses will act on a set of orthogonal planes oriented in such a way that there are zero shear stresses. Now if we look at the shear stresses we also get a local minimum and maximum and we see that they are also separated by 90 degrees but they are also separated from the plane of principal stresses by 45 degrees. So hopefully you can now understand the concept of stress transformations and how we get to them through equilibrium.