[NOISE] [MUSIC] This is module 25 of mechanics of materials part one. And the learning outcome today is to find the stress on any given plane at a point for a given set of plane stress conditions. And so here was our development of our Mohr's circle that we did a couple of modules ago. And then last module we solved this example with actual numbers and we found the principle stresses sigmus of one and sigmus of two. And the maximum sheer stress and we also drew it on a properly oriented stress block. Today we're going to look at the normal stresses and the sheer stresses on plane AA on our stress block and so we'll start by finding out what this angle theta sub A is on our stress block and so we have theta sub A, is equal to, that's going to be the opposite over the adjacent, or the inverse tangent of three over four which ends up being 36.9 degrees on the stress block. But we know on Mohr's circle, that's going to be two theta sub A and if we round off to three significant figures, that ends up being 73.7 degrees on Mohr's Circle and so we're going to go now from our horizontal face counterclockwise 73.7 degrees and this is two theta sub A. And this is going to be our point A and so we now have two theta sub A, we have two theta sub P. We also, let's look at this angle here which is phi. I'll call it phi and so we have phi which is equal to two theta sub A minus two theta sub P or 73.7 degrees minus, if you recall back from last lesson two theta sub P was 22.6 degrees and so, phi ends up being 51.1 degrees and so this angle phi is 51.1 degrees. Now, I can drop a a vertical line here and so, I know what this angle is and now I can find, at point A, my shear stress and my normal stress at point A. So, Sine of phi is equal to the opposite side, which is going to be the shear stress at point A divided by the radius and we know the radius is 130 from last time, it's also written here and so, we've got tau sub A equals 130 times the sine of 51 degrees, which ends up being 101 MPa. And so, that's one of our answers. Similarly, we can go and find the normal stress at point A. This is the adjacent side, we have the hypotenuse. So, sigma sub A is equal to 130, the radius, times cosine of 51.1 degrees and this should have been 0.1 degrees up her but this radius, remember now, we want the normal stress from here to here. So, we're going to have to subtract out this 20 from our center and we find that the normal stress at point A is going to end up being 61.6 or sigma sub a equals and this is positive. So, it's 61.6 megapascals in tension and that's our other answer. So, on our Mohr circle, we can plot it up here. This state of stress or the principle stress condition is 61.6 and the normal stress and 101 for the sheer stress and you can notice now that for this state of plane stress, you could do this for any plane through that stress block. So, you could find the normal sheer stress on any plane. Let's go ahead and draw the stress on a properly-oriented stress block. We're moving from our horizontal face here. So, this is my horizontal face. I'm going to turn my block, in this case I've turned it 36.9 degrees. This is 36.9 degrees and on that face our stresses are 61.6 normal stress mega pascals in tension. And we have 101 and it's positive. So, that means it's going to be in a clockwise direction on our stress block and so our shear stress on that face is 101 megapascals and so that's the solution. And you should now have a pretty good grasp on how to use Mohr's Circle to find the principal stresses, the max shear stress and actually the normal and shear stress condition for any plane for plane stress at a point and I'll see you next time. [MUSIC]
