[SOUND] Hi, this is module 17 of Mechanics of Materials part one. Today's learning outcome is to derive the equations for stresses on inclined planes for the case of plane stress in general. And so this is where we left off last time. We had our stress block and plane stress, we wanted to look at the stresses on a inclined surface, and so here was the result. We came up with a sign convention for both normal and sheer stresses and the angle. And I said last time I showed a couple of parts in the last module, in general we can find from the external loading condition what the shear and normal stresses are on an x and a y plane. What we want to be able to then transform that into any arbitrary plane to find what the normal stresses and shear stresses are on any arbitrary plane. And so we want to find sigma sub n and tau sub nt. So here again is my inclined plane stress block. This is known, we want to find this. And so what we're going to do is we're going to go ahead and apply equilibrium. But to apply equilibrium, these are stresses, which are forces per unit area and not forces. And so how would I, if I changed these into forces? And what you should have answered is, we need to multiply these stresses by the area that they're acting on. And so I've shown the areas here. We'll call the inclined area dA. And therefore the x surface is dA cosine theta and the y surface is dA sine theta. And so now I've shown my stress block with forces acting on it instead of stresses. So now let's go ahead and apply equilibrium to our stress block, this is our coordinates. And here is my stress block with the forces shown instead of the stresses. And so let's start by summing forces in the normal direction. I'll call up and to the right positive. Set that equal to zero. For purposes of summing these forces I'm going to put the angles in here. By geometry this angle is theta. And this angle is theta. And this angle is theta. And so if I sum forces now in the normal direction I have sigma sub m dA in the positive normal direction. This force, along the parallel to the incline surface, is completely in the tangential direction, so it has no impact on this equation. Let's take this force, which is in the negative normal direction, and it's going to be the cosine component, so we've got minus sigma sub x dA cosine theta. And it's the cosine component. And then we'll take. Well, we've taken care of this force. This force has no effect. We've taken care of this force. Now, let's do the dy dA sine theta, and it's the sine component in the negative normal direction. So we've got minus Sigma sub y, dA sine theta times the sine theta component. So that takes care of that. Now I've got the shear stress. It's going to be the sine component in the negative normal direction. So I've got minus tau xy times dA cosine theta, and it was the sine component. And then, finally, the last force is this shear force, and that's going to be minus tau y x dA sine theta times cosine theta component. That all equals zero. You can see that every term includes a dA. So let's cancel those out. So there's a dA, there's a dA. And let's leave sigma sub n on the left side, carry all the other terms to the right side since they're negative, and we'll end up with sigma sub n equals sigma sub x cosine squared theta. So this is sigma sub x cosine squared theta. Plus, sigma sub y sine squared theta. Plus sigma sub y sine squared theta. We know by equilibrium that tau xy and tau yx are equal so we've got two terms here. So it's going to be plus 2 tao xy sin theta cosine theta. And we're going to use some trig identities here. Cosine squared theta is the same as 1 + cosine 2 theta over 2. Sine squared theta is the same as 1- cosine 2 theta over 2. And you can find these trig identities online or any trig book. And finally sin theta cosine theta is the same as sine of 2 theta over 2. And what we get as a result if we simplify that down, is that sigma n, the normal stress on the incline surface, is equal to sigma sub x plus sigma sub y over two. And then plus sigma sub x minus sigma sub y over 2 times cosine 2 theta and + tau xy sine 2 theta. And as I said, we know the xy terms all on the right hand side, so that allows us to find the normal stress on any arbitrary plane. So that's one result. I would like you to do the same thing for summing forces on your own in the tangential direction, then come on back and see how you did. Okay, so you should have come up with this equation and you could have reduced it down to this. Sine theta, cosine theta again is equal to sine 2 theta over 2. Sine squared was 1 + cosine 2 theta over 2. Sine squared theta was 1- cosine 2 theta over 2. And here is our result. Which is now the sheer stress on the inclined surface. And again we know all the values on the right hand side and so we can find the shear stress on the incline plane. And so here is our overall result. We now have what's called the transformation equations for plane stress. And, we can find for any plane at any angle theta what the normal stress is and what the shear stress is. And so that'll be real important as we analyze structures, analyze engineering parts to find out where's the max, the min, what angle they occur on. And now you have that tool. And so to finish up for today, I'd like you to go ahead and do this worksheet, which is an application of what today's module was about. And when you finish, I've included the solution in the module materials. See you next time. [MUSIC]