[MUSIC] Well can you believe it? We're already on module 18 of Mechanics and Materials Part one. Today's learning outcomes are to derive the angles to the Principal Planes where maximum and minimum normal stresses are going to occur and we're going to define those as principal stresses. And we're going to show that the shear stress is zero on these principal planes. So here's where we left off last time. I applied equilibrium to our stress block. We came up with our stress transformation equations. So if we knew the stresses, the normal stresses and shear stresses on any two planes form the external loading condition, we could find sheer stress and normal stress on an arbitrary inclined plane. And again, these were the transformation equations, and you'll note that these equations, we've derived these solely from equilibrium. And so they're applicable to stresses for any material. Whether it's linear or non linear, elastic or inelastic, it doesn't matter. And so since we're talking about stresses we use the forces on the stress block. But in the future I'm just going to refer from now on to the stresses themselves. So this is the stress block I'll show in the future. And so for any plane at an angle theta we find sigma sub n, sigma sub t. As I mentioned before, I showed some parts and any engineering part, any engineering structure. All structural machine design, we want to know what planes and at what angles the maximum values are for the sheer stress and the normal stress. Because that's going to tell us where our material is or our structure is likely to fail. And we've already done tensions tests to find out what material properties the type of material has. And so, this all starts to come together. And so, let's first find the angle at where the max and minimum normal stresses are going to occur. And my question to you is, how can we do that? How can we find that angle? And so what you should say is well, to find the angle to the maximum, minimum, normal stress, we want to find where sigma sub n, the derivative of sigma sub n with respect to theta goes to zero, because that's the mathematical definition of how to find a max or a min. And so we're going to take d theta sub n with respect to theta and this term doesn't have a theta so that goes to 0. This term we've got a cosine 2 theta. So the derivative of that is going to be minus since we're going to go from cosine theta to sin theta. The 2 comes out, so it cancels with this 2 so we get minus sigma sub x minus sigma sub y sin 2 theta. And on this one we're going to take the derivative of sin 2 theta so the 2's going to come out so we get plus 2 tau xy cosine to theta. And so that we want to set equal to 0 for the maximum or the min. What we get is, if I work this out, is I now have sin of two theta over cosine of two theta which is the same as tangent of 2 theta. And this is going to be the tangent to the principal plane. So I'm going to label that now as theta sub p because this is a very special theta. That's equal to, we get 2 Tau xy over sigma sub x minus sigma sub y. And so that is the result. That's how we can find the angle to the principal planes. And so theta sub p is the angles to what are defined as the principal planes and these are the planes where the maximum and minimum normal stresses are going to occur, and we're going to define those as principal stresses. Okay, so here we go with those results. You'll note that for the principal planes when I took the derivative this is what I found. If I now compare that with my shear equation, you can see if I divide by 2 here, this cancels and I divide by 2, that means that the value on the right hand side here of my shear stress equation is equal to 0. So what that means is on the principal planes, a very important note is that the shear stress is zero on the principal planes. So that completes module 18 and we'll see you next time. [MUSIC]
