Let's consider an example of the relationship between stress and strain. In this case, we have a rod. The rod is made out of aluminum. And I would have to use some sort of materials textbook or the internet to determine a relationship between, to determine the modulus of elasticity of aluminum, which has a value of, we'll say, 10,000 KSI, which is kilopounds per square inch. So, if I look at the aluminum, there's a certain loading on the aluminum here. In this case, 54,000 kilopounds or 54,000 pounds is being applied to a rod that is 3 inches in diameter. 3 inches in diameter. So, let's see what information we can determine from this. In particular, let's figure out how much that rod stretches as a result of this load. We can begin by establishing an initial area of the rod. And we recognize that the rod is drawn as a circle, probably described as a circular rod. We use our geometry. The area is equal to pi d squared over 4, or pi r squared. And we calculate 3.14 times our 3 inch diameter squared divided by 4. When we do so, we get 7.07 square inches for the cross-sectional area of the rod. Again, when we apply stress, we need to apply it. We need to calculate it based on a perpendicular face. So, we cut across the rod and hence we're looking for the cross-sectional area of the rod. Well, with that area and the initial load, we know the relationship between stress and our axial force is P over A. So, we take our 54 kilopounds and we divide it by the 7.07 square inches. When we do that, we get a value of 7.64 KSI, or the KSI stands for kilopounds per square inch. Now that we've established that, we recognize that sigma equals E epsilon. Well, we're going to have to manipulate that a little bit. If we do so, we know that our strain is equal to our stress divided by our modulus of elasticity. In this case, that value is going to be R, 7.64 KSI divided by the 10,000 KSI, both of our kilopounds per square inch cancel, and we get a strain value of 0.00763. It's not strange when working with metals to get very small strain values. Well, what does that information help us with? If we're interested in figuring out how much the metal actually stretches, we need to remember our relationship that strain is equal to the change in length divided by the length, the initial length. So, if we want to determine our change in length, we multiply the strain by the initial length. So, 40 meters times, or length of 40 meters above times our strain gives us a value of 0.031 inches. So, roughly 3.001 inches of an inch. I'm sorry, compared to 40 inches, not 40 meters. So, notice this very, very small amount of deformation is one of the reasons why we assume that there is very little or no deformation when we initially calculate our stresses in something like a truss. Let's consider a second stress and strain example. In this case, we have a bar that's made of steel. We know the dimensions of the bar. It's only 1 meter long. It has a depth of 4 millimeters, and it has a height of 2 millimeters. When we apply an axial force to the bar of 1.6 kilo-neutons, we see a deformation of 1 millimeter. Occasionally, one of the letters to represent deformation is the letter small Greek letter delta, which will sometimes be used for representing deformation as well. In this case, we have a change in the length of 1 millimeter when we apply a force of 1.6 kilo-neutons. Let's see what information we can determine from this. Notice I haven't given the modulus. We can look up the modulus to see what might be reasonable for steel. But let's see if we can find the modulus since we're given all of the information about the bar. First, we recognize that our stress is equal to the force applied divided by the cross-sectional area. That's our 1.6 kilo-neutons applied to an area of 2 millimeters times 4 millimeters. I'll 2 times 4 is 8. The 1.6 divided by 8 gives us 0.2. And then our units are in kilo-neutons per square millimeter. Let's sort of change those units, shall we? 0.2, in this case, kilo is 1,000. Millimeter is per 1,000. So in this case, we take kilo times 1,000 times 1,000 again for each of the millimeters times 1,000 and times 1,000 again. And we can get giga-neutons per meter. Well, a Newton per meter is also a unit of pressure known as a pascal, 0.2 giga-pascal. Now let's calculate the strain. Our strain is our change in length divided by the initial length, which we see is 1 millimeter divided by 1 meter or 1,000 millimeters. The millimeter cancels out, and we get a strain value of 0.001. If we recognize again our relationship, sigma equals e-epsilon, this time let's solve for the Young's modulus, e, which is equal to the stress divided by the strain. In this case, we get 0.2 giga-pascals divided by 0.001. Move that decimal place, three places, by multiplying by 1,000, and we get a value of 200 GPA, giga-pascals, which is a central value for the modulus of elasticity of steel. You