[MUSIC] Hi, and welcome to module three of Mechanics and Materials part II. Today's learning outcome is to develop an expression for longitudinal stress for a thin walled pressure vessel in terms of the pressure itself and the dimensions of the vessel. So here's where we left off last time where we had an engineering structure, a pressure vessel, thin walled pressure vessel. We had some external loads which came from the pressure due to the liquid or gas inside. And now we're going to look at the internal forces and moments and move onto the stresses. So again lets look at a section cut and so here is my pressure vessel and I'm going to cut through the center of it. And we're going to go ahead and neglect the weight of the contents and the weight of the structure itself. Because this forces are orders of magnitude less than the forces due to the internal pressure and the resulting stresses on the cross section. So why don't you take a minute and try to see if you can draw the stresses on a cross sectional cut. And so here I've drawn them, I've got my pressure pushing inward on the cross section, and it's a force per unit area. And then around the outside I have my what I'm going to call longitudinal stress acting on the material itself. And so again going back to our general analysis approach, we have our engineering structure. Now we've gone ahead and we've applied external loads. And now we're looking at internal forces due to the stresses acting on the cross section. And we're going to go ahead and find those stresses next. Okay so here is my cross section again, and on the right here, I've shown the cross section with the diameters, and I've got an inner radius and an outer radius. And remember that the overall diameter, because the thickness is so small, we're just going to call D. And so now what I want to do is, I want to go ahead and do a balance of equilibrium on this cross section. And so I have, let's go ahead and sum forces in what I'm calling the Z direction here. So I'm going to sum forces in the z direction, I´ll call as shown on my coordinate system, down into the left is being positive. And so, first of all, I've got my longitudinal stresses, which cause a force to the left, or in the z direction so I've got sigma long, remember that's a force per unit area, so we're going to have to multiply by the area. And the area now that the longitudinal stress is going to act on is the outer circle minus the inner circle. So the outer circle is pi r squared. So that's pi times R outer squared then I'm going to subtract out the inner circle. Which is minus phi R inner squared. And so that takes care of the force stood the longitudinal stress now the pressure is in the negative z direction, so that's going to be a minus times the pressure. And it's area we'll say is overall circle is going to be pi D squared over 4, and all that's going to equal 0. And so I can see that I can cancel out my pis now, because pi is in every term. And what I can do now is I can say sigma long, and I'm going to go ahead and factor this r outer, radius outer squared minus radius inner squared, and I'll factor it to be r outer plus r inner times R outer minus R inner. And you can see if you multiply those terms together that's R outer squared minus R inner squared. And that's going to be equal to, on the right hand side if I carry this over, PD squared over 4. And now I can notice that, well what is R outer plus R inner approximately equal to? And so R inner plus R outer is approximately equal to D for the cross section. And R outer minus R inner, what is that going to equal? And what you should see, R outer minus R inner, is going to be equal to t, or the thickness of my pressure vessel wall. I end up with the result of an expression for my longitudinal stress, which is PD over 4t. And so we're moving right along, and let's go ahead and show that again written out, our longitudinal stress, and we'll pick from that location at the next module. [MUSIC]