[SOUND] [MUSIC] This is module 44 of Mechanic of Materials part one. And we're at the final topic of the course which is thermal and pre-strain effects. Our learning outcome for the day is to describe the temperature effects on engineering materials. I have a demo here. This is a brake rotor that was made out of aluminum and was subjected to high temperature. And as a result, if you look closely you can see that these outer holes are not circular, not round anymore. They've been deformed, so they've plastically yielded. Which is an adverse effect of temperatures. And so, this had to be redesigned. Let's continue to look at thermal effects. For most engineering materials, they expand when they're heated and they contract when they're cooled. And we define the coefficient of thermal expansion as the strain per one degree temperature change. And so for thermal strain, this is the coefficient of thermal expansion times the change in temperature. And we'll assume, in this course, that alpha, the coefficient of thermal expansion, is constant although it's actually generally higher at higher temperatures. And we're also going to use it for homogeneous, isotropic materials which means that the coefficient of thermal expansion will be the same in all directions. And the thermal stress will exist from this thermal strain if the member is restrained. So let's look at some examples. Let's take a bronze member, we'll have a temperature change of 40 degrees Celsius, and this is our thermal coefficient of expansion. And we have our Young's modulus of 100 GPa. And so let's first look at this specimen in the unrestrained case, which means I'll have it have an initial length of 250. And so it will go ahead and if we heat it up by 40 degrees, it's going to expand. And that's going to be the epsilon times the length is going to give us our deformation. Epsilon is the epsilon due to the temperature change, which is the coefficient of thermal expansion times the temperature change times L. If we calculate that out, we find in this case that the bar will extend 0.169 millimeters. What about the stress in this case? And what you should say is stress is zero because it's unrestrained. It just expands out. Okay, let's now look at a case where it's fully restrained. The bar is again 250 millimeters long, but it's not allowed to expand. And so as a result, it wants to expand that amount that we calculated before, but we're going to have to force it back, it's going to be held back by a normal force, to keep it from expanding, since it's restrained. And so the total deformation has to equal zero. Which means that the temperature change deformation, or the deformation due to temperature must equal the deformation due to the normal force. And so, I substitute in for my deformation due to temperature, which is the epsilon due to temperature times L or the strain times L. And we know that as long as we make the linear elastic assumption that the deformation due to the normal force is going to be PL over AE, and the L's will then cancel. I see that P over A is going to go ahead and give us the sigma that the bar is going to experience, or the stress. And so I come up with the stress, equal to the strain times E, then the strain due to the thermal expansion. And what we have is that the stress in this case for fully restrains ends up being 0.0676 gigapascals in compression, or 67.6 megapascals. Okay, let's look at a final example here. This time, let's look at a case where we have some partially restraints bar. And so the bar and this is not to scale obviously, it's shown so that you can see what's going on. The bar is 250 millimeters long, it can expand a little bit, 0.05 mm, until it hits a restraint. And so we know that if it was unrestrained, it would expand even more than that due to the temperature, but we're going to have to have a normal force that holds it back into place when it hits the restraint. And so in this case, my total deformation is the deformation due to temperature change, minus the deformation due to the normal force, and it has to equal 0.05 millimeters. And so we already calculated what the temperature change, the deformation would be due to the temperature expansion and that was 0.169 if it was unrestrained. But then we've gotta have a normal force that brings it back to a complete deformation of 0.05. And so here's our equation. PL over AE then has to equal 0.119. And P over A, again, is the stress. And so the stress ends up equalling 47.6 megapascals, and again, in compression. And we'll see you next time. [SOUND]