[MUSIC] Welcome Module 45 of Mechanic of Materials Part I. Today's Learning Outcome is to solve an engineering problem now when thermal effects are present, which we talked about last module. And so, here's our example. Again, this is going to be a little bit longer module than usual, because I want to go through the entire problems, so that there is continuity. Now I have a bar BC, which is aluminum. Cross section area of 2000, modulus of elasticity is 70 gigapascals and the yield stress, again, of 0.28 gigapascals. But now I'm given a coefficient of thermal expansion for aluminum as shown, 22.5 times 10 to the -6, per degree Celsius. The bar DE is going to be made out of brass. It's got a cross sectional area of 1,300 and a modulus of elasticity of 100 GPa, and a yield stress of 0.1 GPa. And I give you, again, the coefficient of thermal expansion for brass, which is 17.6 times 10 to the -6 degrees Celsius. And what we're going to, once again, allow this brass and aluminum bar to deform. But we're going to assume that bar ABDF is rigid. The weight of the bars will be neglected, because they're much less than the forces that the bars are supporting. And I want to determine the axial stress in the aluminum and brass bars when the temperature decreases by 30 degrees. And we'll, again, work in these units. Kilonewtons per millimeter squared and Gigapascals. Okay. So, what do we do now to start the problem? And what you should say is, let's go ahead and look at static equilibrium. How do I do that? What's the first step? At first episode is to draw free body diagram. Go ahead and draw the free body diagram, come on back, and see how you did. And here's the free body diagram. You'll notice that I gotta be a little careful. I've assumed that bars B, C and D, E are in tension. You have to assume them to be in one of the other. If you're consistent and they come out to be negative in the end. If my bar forces come out to be negative, then I'll know they're in compression instead of tension. But you should make that assumption up front. So, go ahead and write the best equilibrium equation to start to solve the problem. What you should say is that summing the moments about F, because that allows us to find, we're going to want to find the axial stretches in B and D. We're not interested in the reaction forces of F at this point in this problem. So, here is my equilibrium equation. And it boils down to this. We want to find the stresses, so lets go ahead and put the stresses in for B and D. The stress in bar B is going to be the, the force in bar B is going to be equal to the stress in bar BC times its cross-sectional area, which is 2,000 times sigma BC. Do the same thing for D, and what you say is that the force in D is equal to sigma DE times its cross-sectional area in 1,300. So now, I have an equation with two stresses in it. And this is shown here. We'll call that equation star. And so, we have one equation and two unknowns. Those are my unknowns. That's what I want to solve for. So my question is, what do we do now? And what you should say is we're going to need an additional equation. What additional equation can we write? Well, we can write a deformation equation or a compatibility equation again. And so, we're going to assume, as we did before, small deformations and small angles. This is my bar in it's original position. This is in the deflected shape. This is the deflection total at D. This is the deflection, okay. So for the deflection total at D, I assume that this member was. That means it shrank quite a bit due to the temperature change. So much so, that I had to pull it back with a normal force to have it end up being at the total deformation at D shown here. I'd like you to do the same thing for the deformation at bar B. And in this case, I've got deformation in bar B. I've got, again, a cooling. So this is going to pull in this bar BC. And I'm going to have to have to pull it back out, to meet the compatibility with the equilibrium and for the deformation to end up at this point. So that's a good deformation equation or compatibility equation. I can use similar triangles, again, to find out the relationship between the total deformation at B and the total deformation at D. That's the relationship I get. So there is my equilibrium equation, a star or asterisk. There is my deformation equation. And let's go ahead and, now, put in the parts of the deformation due to the temperature change and due to the normal force. And so, the force thermal displacement relationship is 1.5 times for DT total. That's the deformation due to the temperature change, minus the deformation due to the normal force at D. On the other side, we had the deformation of B total is equal to, minus the deformation due to temperature, plus the deformation due to the normal force. And I can substitute now in my values for my deformation for temperature chain, which is the strain due to temperature change times the length. And the deformation due to the normal force, which PLL over AE. On this side I have, for the bar, DE. On the right hand side I have, for the bar, BC. Again, I can reduce this. I see that P over A for bar DD is the stress in bar DE. Similarly, the P over A for bar BC is the stress in bar BC. And so, I can calculate this out and I get another equation now, which we're going to call equation double star or double asterisk that is a expression for the axio stresses in the aluminum and the brass bars. And so, we can solve those simultaneously. We find out that the stress in bar BC, the aluminum bar, is .0989 in tension gigapascal. That's less than our yield stress. So, we are still in the linear elastic region. We had made the assumption that delta equals Pl over AE and it holds. As I show here, and so we're okay with that assumption. What's the stress going to be in bar DE? And if you do that calculation, this is what you should have come up with. Again, we find out that it's less than the yield stress for brass. And so, our assumptions hold and this is the stress in the aluminum, excuse me, in the brass bar. And so we've gone ahead and completed a problem. Again, a very good typical problem having to do with coefficients of thermal expansion and what the thermal effects are on a system like this. And, we'll go ahead and wrap up the course next module. [MUSIC]