[MUSIC] Hi, and welcome to Module 24 of Mechanics of Materials Part 1. In the last few modules, we've done quite a bit of math to come up with equations for the stress transformation equations in plane stress conditions. And we show that those relationships and equations could be graphically portrayed in something that we call Mohr's Circle. And so today we're going to use Mohr's Circle, to determine the principle stresses, principle plains, and maximum shear stress, for a given set of plane stress conditions. And so, this is a review from last time. This was our equations for the transformation of plane stress. And in the form of the equations for a circle. And we took this, we had a sign convention for the horizontal face and the vertical face. And we went ahead and drew Mohr's Circle with general terms. And so today we're going to apply this to an actual example. So here is our example. I have a stress block that's shown. And we're going to go ahead and use Mohr's Circle today to determine the principal stresses. The max sheer stress and to show the plains on which they act on a port. The principal stresses act on a properly oriented stress block. And so let's go ahead and start. Let's look at the horizontal faces by our sign convention. This is positive 100. Normal stress and minus 50 because it's counter clockwise, sheer stress. And so if I go out 100 and down 50, I'll put this point right here. I'm going to label it as the horizontal face. I'll use H with a circle around it, and that's positive 100 and -50. Now we'll do the same thing for the vertical faces. For the vertical faces we have compressive normal stresses so that's negative. Minus 140 and this is clockwise rotation on the vertical phases so that's going to be positive shear stress and so we're going to go back minus 140 and a positive 50 shear stress. And so this is the vertical phase And we have -140 and 50. Now, I'm going to go ahead and draw a line between those two points, And I'm going to go ahead and draw my Mohr's Circle. And that's about as good a circle as I can draw, pretty close. All right, let's start by labeling the center of the circle, and you recall from the equations for the center that's the average stress. And so the center is going to be the stress on the X face, which is minus 140, and then plus the stress on the, normal stress on the Y face which is plus 100, and we divide that by two so we get the average. And it's got zero sheer stress at that point, and so our center is at minus 20, and 0. Okay, now let's go ahead and do the radius. And we had the equation for the radius from last time. It was the square root of sigma sub x, which is -140 minus sigma sub y, so that's -100 / 2, and that quantity is squared plus the shear stress, which is -50 squared. And what we get for a radius is 130, and so our radius here of our circle is 130. And now if we go out from our radius, straight up, this should be the highest point in the circle. And so this should be the highest point up here, and we find that the talmax is equal to one hundred and thirty megapascals. And so that's one of our answers. Okay, now we want to find the maximum, and minimum principal stresses, and so they're going to occur where the sheer stress equals zero at this point which is furthest to the right. And at this point, which is furthest to the left and so sigma one, our first principal stress is equal to, we go from minus 20 and then we add a radius of 130. So this is minus 20 plus 130 or 110 megapascals, and since that is a positive value that means that principal stress is intentioned. We'll do the same thing for sigma 2, and that's -20 and then -130 for the radius, so that's equal to -150. Mega pascals and that is in compression because it's negative. Okay, so now let's go ahead and find the angle to that principal stress or those principal stresses. And so that angle is going to be this angle here on more circle which is 2 beta sub p, and so 2 theta sub p so we end up with for this right triangle, we have the opposite side is 50. The magnitude of the opposite side is 50. The adjacent side is here, less than 130. The hypotenuse is 130. So I have sine of 2 theta sub P equals opposite 50 over hypotenuse 130. Which means that 2 thetas of p ends up being 22.6 degrees. But remember, the 22.6 degrees is the angle on the more circle, the angle that we're going to move to the principal planes on the stress block is going to be half that and so, theta sub-P equals 11.3 degrees, and we've moved in a counter-clockwise direction. And so if I want to finally show the principal stresses on a properly oriented stress block. We've got the horizontal face. That's this point here. And we're going to rotate 11.3 degrees on the stress block. So here's our stress block. Turn from the position that we see here, this angle is 11.3 degrees, and on this horizontal face, we have a pencil principal stress of 110 mega pascals. And that's on the other horizontal face as well. And if we go another 180 degrees or 90 degrees on our stress block, we have a second principal stress of 150 megapascals in compression. And that's that point there. And so that's the answer to the answer to the stress block. And here again was our principle stress 1 and our principle stress 2. And just as a quick check, remember the stress in variant formula. Stress in variant formula said that sigma 1 plus sigma 2 equals sigma x plus sigma y. That means that the orthogonal stress or normal stress, two orthogonal normal stresses have to be equal or constant all the way around the stress block and so we have here sigma 1 is 110 plus sigma 2 which is minus 150. And does that equal sigma x plus sigma y? Well, sigma x is minus 140 and then we have a plus 100, so we have minus 40 on the left side, minus 40 on the right side. And we're good to go. So, that's our solution. That's a good Mohr's Circle. We'll start back up next time and solve part b. [SOUND]