[MUSIC] This is module 33 of Mechanics of Materials part one. And today's learning outcome is now given the plane strain conditions a point, we're going to determine the principal strains, the principal planes, and the maximum sheer strain, using Mohr's Circle. And so here's where we left off last time. We had our condition of plain strain, we found out that the strain transformation equations came out very similar mathematically analogous to plane stress, and so we could use Mohr's Circle to depict them. So here is the Mohr's Circle outline. The sign convention is the same, tension is positive, compression is negative. On my Mohr Circle the sign convention I'll use for the shear strain, the Mohr Circle sign convention I'm going to use is that clockwise is positive. Counter clockwise is negative. And so here's a problem, we have the measured strain components, and again, we're going to learn how those are measured very shortly in a couple of modules here. And we're going to use now Mohr's Circle to find the principal strains and the maximum shear strain at this point, and find the orientation of that principal plane. So here's our Mohr's Circle equations. Now we have to plot. Instead of plotting gamma xy, we're going to plot gamma over 2. So we're going to plot gamma xy divided by 2, which is equal to 300 mu millimeters per millimeter. And mu, just as a review, is times 10 to the minus 6th. And so let's go ahead and start by plotting the vertical and horizontal face. And so, for the vertical face I've got compressions which is minus 400. Clockwise rotation for the shear strain so that's going to be plus 300. So, I got minus 400 plus 300 and that'll be just about here. And so I'll give it v. And this is minus 400 mu and then 300 mu. And I do the same thing for the horizontal face. Now I have positive 1,000 and counterclockwise so negative 300. So I go out 1000 and down 300, and about there. This is the horizontal face. And that's going to be 1000 mu. And 300, -300 mu. I can now draw a line between those two points and I can draw my Mohr's Circle. Let's see how well I can do a circle here. Not so good but you get the idea. Okay, I can find the center now, that's epsilon x, epsilon x is -400 plus epsilon y, which is plus 1,000, divided by 2, and those are with you I may forget, on occasion, to write the mu, but it's always times 10 to the minus sixth and zero. So that's going to be equal to 600 divided by two is 300 mu and zero. So here is our center at, now let's do this in blue, center at 300 new and 0. We can now find the radius as well. So I've got the square root of epsilon xi minus 400 plus our, excuse me, minus 1,000 divided by 2 and that's all squared. Plus gamma x y over 2 is 300 squared. And if you do that math, you'll find that the radius is equal to 716 mu. So now if I come up from my center, the radius and 716 mu this is the max of gamma over 2 which is 716 mu. And so that means that gamma max and let's again write this in black, gamma max. Is 2 times that, or 1432 mu radians, radians being dimensionless. And so that's one of our answers, A N S. Okay, so that takes care of the max in plane shear strength, okay? We've gotta be careful and make sure that we say in plane sheer strain. Now the principal strains are going to be here and here. So I'm going to add the radius to 300. So that's 300 plus 716 or 1016 for, epsilon 1. So epsilon 1 is equal to 300 mu plus 716 mu or 1016 mu. Again, I'll say millimeters per millimeters, but that's actually dimensionless. And it's tension, since it's positive. So that takes care of one of our principle strains. And the other one is back here, and so that's going to be 300 minus 716 epsilon 2 is equal to 300 minus 716, and again mu, mu or minus 416 mu which is 416 mu millimeters per millimeter. And since it's negative it's going to be in compression. That's the second principle strain. And the last thing we want to do is to draw this on a properly oriented stress block. From the plain that we know. And so we've got here is going to be our rotation on Mohr's Circle, 2 theta sub P. So in finding this 2 theta sub P, we use the trig function sign, our opposite side here is going to be a value of 300. The hypotenuse is 716. So we've got sign of 2 thetas of p equals 300 over 716. So if you do that 2 theta sub p ends up being 24.8 degrees on. Moore's circle that means it's going to be half of that or 12.4 degrees on the stress block. And we're moving from the horizontal face in a counterclockwise direction. And so now if I draw my properly oriented stress block. This angle is going to be 12.4 degrees. And when we move 12.4 degrees on this horizontal face, we're going to have a tensile strain. 1016 mu and no sheer strain there. Go another 90 degrees on the stress block or 180 degrees on Mohr's Circle. We've got -416 or 416 inwards mu for our other principal strain and so you'd have the same on the opposite sides. This would be 416 mu and this would be tan 16 U. And so that's our example. In fact, let's just label this as the horizontal face and we move from there. And so that's a good example of finding the principle strains and max in plane sheer strain, and we'll come back and do part B next module. [SOUND]