[SOUND] [MUSIC] Welcome to module seven of Mechanics and Materials Part One. Today's learning outcomes are to review a normal stress, and to define and discuss something called nominal stress or engineering stress, and to define and discuss true stress. We left off last class with the 3-dimensional state of stress at a point. If we let the out of plane stresses go to zero then this became a 2-dimensional or plane stress problem. And it looked like this. Now, if we only look at axial loading, that's a special case of two-dimensional, or plane stress. And so, here's a clip of an axial test being done on a specimen. And you can see it stretches out. And it starts to neck down, and eventually it would fail. Here's another member, and so again, if I were able to put enough force on this member, it would stretch and the area, the cross-sectional area would start to get smaller. You can see a little bit better on some other material here. This is a softer material and so, here I can actually stretch it down and you may be able to see the cross sectional area become a little bit smaller. Here's a rubber band type material. And so if I pull it you can see how the cross section area, it's very thin, its got a width and a length, but you can see how the cross-sectional area will become smaller and that the length stretches. And so, the same thing happens for all these materials if they're isotropic. And we'll talk about isotropic later and homogenous. And if I pull on them, that same sort of effect, it's actually called the pwasons effect. And we'll talk about that more later as well. And so, our L becomes now L plus delta. So, I'm stretching them out. And we can do a cut inside that member to see what's going on. Let's do a transverse cut. It reveals a normal force and a shear force. If we sum forces in the y direction we can see that the shear force for our transverse cut equals zero, and if we sum forces in the x direction we see that the normal force is actually equal to the external force p. And so as we recall, this is a review. We can say that the stress will assume it to be uniformly distributed across the cross section, and so that's the definition of normal stress. Force per unit area perpendicular to the cut surface. And this is the formula for it and this is the sign convention. Positive stress for tension, negative stress for compression. Now, we can also talk about something called engineering stress or nominal stress. As we stretched this thing, we saw that the cross-sectional area got smaller. And so the engineering stress Is based on the initial cross-sectional area of our specimen. True stress however, is based on the actual area, and so as we stretch the member out, the actual area becomes smaller as the specimen gets closer and closer to failure, so the true stress can actually be a larger number. Now in most engineering applications, we use and we'll solve problems using the engineering stress, because the change in the area for most materials and the change of the stresses between nominal stress, engineering stress and true stress are so small that we just go with the initial cross-sectional area. And so let's do a worksheet. I'll let you do the worksheet on your own. I've put the solution in the module hand outs, but we've got a flat alloy bar, it's got a thickness of 10mm and a width of 60 mm, so it's a rectangular cross section. We subject it to a tensile load of 60 kN, and I want you to find the nominal Or engineering axial stress in the bar. And once you've done that, you can check your solution and we'll see you next time. [SOUND] [MUSIC]