[SOUND] This is module 30 of Mechanics of Materials Part one. Today's learning outcome is to derive the strain transformation equations for the case of plane strain. And we talked about plane strain last time. In general with a plane strain condition, just like with plane stress where we could find sigma x, sigma y, and tau xy. In general, for plane strain, we'll be able to measure and find epsilon x and epsilon y, and gamma xy. In fact, in later modules we'll actually do that using something called strain rosettes. And once we find those, we can then transform to the normal stress and the sheer stress for any angle theta, for normal and tangential coordinates. And so this is very similar to the type of transformation that we did with stresses or plane stress. Now, we're doing the same thing for plane strain. We're gonna have a lot of math in here again, in this module. The good news is, with this video, you can go back and forth slowly to try to make sure you understand every step. I'll go rather quickly as we step through it. So, here is our block in the, our parallel piped in the deformed position. We want to relate the strain in the x and y coordinates to any coordinates n and t. And so here is our unrestrained parallel piped, here it is after it's been deformed. We developed these relationships last time. Now, I'm going to also look at what happens to this normal direction. So, we start off with this dn along this diagonal. And we see now that it goes and becomes stretched out longer. And just like we did with dx and dy, dn prime the new length is equal to dn plus epsilon sub n dn or one plus epsilon n times dn. And remember now, this angle here is going to be equal to pi over 2, minus gamma 1 and gamma 2 or gamma xy. So this angle is pi over 2 minus gamma xy. Okay. So, that's what our deform parallel piped looks like. Here it is shown again, all together. This angle here, pi over two minus gamma xy is this entire angle. And we can now use math. We'll use what's called the law of cosines, which stated here. Let's go ahead and call our a side of this, we're gonna use this triangle here and the law of cosines. So, this will be our side. We'll call this our b side, and this our c side. And we'll call this angle phi. And for that a-b-c triangle, this is the law of cosines that applies. And when I do that, this is the result I get. A is this distance dn prime or we're gonna square it now. So, that's one plus epsilon sub n squared, times d n squared, equals b squared, which is this side squared. Okay. And then c squared, which is this side, or the y side squared. Minus two times b. And times c, which we have here. Times cosine of phi. Okay. Let's recall now our rectangular parallel piped. Here is an X and Y coordinate. Here is a rectangular parallel piped. And you recalled by geometry that this angle, let's just call this generically alpha one and this alpha two. They have to add up to pi because the entire, all of the angles have to add up to two pi. And so, we have alpha one, plus alpha two, equals pi. Let's relate that to our actual parallel piped that we have here. So, alpha one is the same as this angle pi over two, minus gamma xy. So, we have pi over two, minus gamma xy. Plus alpha two, is the same as phi on our actual parallel piping, equals pi. And so I can solve for phi, which is equal to pi minus pi over two, is pi over two, plus gamma xy. And let's go ahead and now put that value down into our equation. And when I do that, that's the result I get. And so, here's the result I came up with. I'm gonna use some more math, Trig identities. This relationship holds true. And so therefore, I can change this last term to minus sin of gamma x y. And so, also note for our original block that adjacent over hypotenuse or dx over dn is equal to cosine theta, or dx equals dn cosine theta and similarly for dy over dn sine theta dy, equals dn sin theta. Let's take all those results now. So, here's the relationship that we have developed along with dx and dy. We're gonna go ahead and substitute those in. And I have dx squared, dy squared, and dx times dy. I can now cancel dn squared out of every term. And I end up with this relationship. Another observation we can make is that the strains are very small. And so, whenever we square the strains, they're going to be much less than the strain itself and so we can neglect those terms. Also we can say for small strains, small shear strains, that sign of gamma is approximately equal to gamma. An so, here's a strain term squared that will neglect, another one will neglect, another one will neglect. We have epsilon x times sine of gamma xy, which is about gamma xy. So, you have epsilon times gamma, very small strain. We'll cancel that term. We'll cancel that term as well. And finally, we'll cancel this term. And when we do that, we also can say that sine of gamma xy is equal to gamma xy. This is the result we arrive at. And so, here's that result shown again. We see that we have a sin. I've rearranged it, so that I have combined terms. I have a sin squared, plus a cosine squared, which is equal to one. So, one cancels with one. I can cancel two in every one of my terms. And I end up with now, finally, the normal strain transformation equation. And it looks like this. We can also use trig identities to express it in a little bit different form that looks like this. And when I do that and put those trig identities in, this is the result. So, I finally now going through the math and the geometry. I have strain transformation equations that take my plane strain known for ex, I know the strains in the xy and the gamma xy shear strain. I can now transform them for any normal and tangential coordinates at any angle theta. And this is the Normal Strain Transformation Equation, completely analogous to the Stress Transformation Equations we came up for the case of plane stress. And this is what they look like. [SOUND]