[MUSIC] >> Welcome to module 34 Mechanics of Materials part one. Today's learning outcome, if we're given a set of plane strain conditions we'll be able to determine the strain on any given plane using Mohr's circle. So here is the condition of plane strain and the strain transformation. We had the equations that related to Mohr's circle, and we also had Mohr's circle sign convention, as shown. And last time we did part A, which was to find the principal strains epsilon one and epsilon two, and the maximum in plane shear strain and the orientation to the principal planes. Now we want to do part two, which is the normal and shear strains on plane AA. And so for plane AA we're rotating 50 degrees counter-clockwise on the stress block. So theta sub A = 50 degrees on the stress block. So, 2 theta sub A is going to be 100 degrees on Mohr's circle. And so if I draw that, we're going to go 100 degrees. And so we're going to end up. Somewhere up here. And this is 2 theta sub A. And so, remember from the last time we did this sort of an approach on Mohr's circle for plane stress. We'll call this angle phi. And so, we find phi equals 2 theta sub A -2 theta sub p, or 100- 24.8 degrees, which is equal to 75.2 degrees. Now we want to find, this is going to be point A up here, we want to find what those stresses are. And so, drop a vertical line down here. So, for our triangle now, sine of phi will be this opposite side over the hypotenuse, which is the radius. And so we have sin of phi equals, the opposite side is going to be the shear strain at A, gamma sub A /2 over the radius, which was R, or 716 mu. So we have gamma sub A /2 = R, which is 716 times sin of 75.2 degrees, or gamma sub A over 2. Shear strain /2 at point A will end up being 692 mu radians. And we could do a similar approach to find the normal strain at point A. So we want this distance, and this will be the adjacent side. So it's the cosine. So we have Epsilon A equals 716 times cosine of 75.2 degrees. But that's only this distance here. We also need to add 300 to it to get out to point A. So it's plus 300 mu. And so epsilon A ends up equaling 483 mu, and I'll put in millimeters per millimeter even though that's dimensionless. So now I have my point A. The strain conditions there, the normal strain, is 483 mu, and the shear strain divided by 2 is 692 mu. And the last thing we want to to do is show that, well, let's call that our answer because that's the normal and shear strains on plane AA. Let's show it on a properly oriented stress block. So we are again going from a horizontal face. And so I've got my horizontal face here. I'm turning now to point A on the stress block. I'm turning 50 degrees. And so. This angle is 50 degrees. And we get to that 50 degrees, the strains on that face are going to be 483. It's positive, so it's in tension, mu. So the shear strain is positive, and this is the value of the shear strain divided by 2, 692. So on my stress block it's going to be twice that for the shear strain, and that's going to be 1384 mu. And so here's a properly oriented block. And it's got the strains showed on it, and the proper convention. And you can do that for any plane now at any angle using these Mohr's circle techniques. And so that's where we'll leave off this time. [MUSIC]