[SOUND] Hi, this is module 28 of Mechanics of Materials, part I. Today's learning outcome is to now use the theory that we developed last time, and calculate the maximum stress at a discontinuity in a structural member. And so we had an example, a generic example of an axially loaded bar with a hole last time. Now I've given you some dimensions of our bar, and I've given you some actual average stress on that bar, axially loaded. And,we have our graph that will allow us to find our stress concentration factor. And so, let's start off, we're going to need the gross area, we're going to need how much the normal force is, and we're going to need the net area. So let's start off with the gross area. And the gross area is across the entire cross section without the discontinuity, and so that area equals the width, which is 10 millimeters, times the thickness, which is 2 millimeters, or 20 millimeters squared. Now we use that with our average normal stress calculation to find the normal force. So, the normal force equals sigma times this area, or 0.2 kilo-Newtons per millimeter squared, times that gross area, which is 20 millimeters squared, and so we get four kilonewtons. Finally, let's look at the net area at the discontinuity. And so that will be, we'll call it A sub t, that's going to be, our thickness is the same so that's going to be t times now the entire width minus the diameter of our discontinuity or the hole. And so that's going to be 2 millimeters times the entire width, is 10 millimeters minus the diameter of the hole which is 2 millimeters. And so, the net area ends up being 10 minus 2 is 8 times 2 is 16 millimeter square. And we'll use that later on along with the gross area and the normal force. Next, let's use our graph to find the stress concentration factor, that we'll use in this problem. And so, we're going to need the ratio of D to W, so D to W. The diameter is 2 millimetres, the width is 10 millimetres, and so this is going to be 0.2. And so if I go down here and I mark 0.2 on my graph and I've run a vertical line up and across. I can see that my stress concentration is roughly about 2.5. You can be the more accurate you are, the more accurate answer you'll get, but we'll just use 2.5. So here we go, we have our normal force, we have our K, which is 2.5. That's K sub T, for the K associated with the net area. Here is our net area that we calculated. And here is our gross area, which we also calculated. And so here's our stress concentration factor multiplied by our average stress to come up with the maximum stress. So let's use the net-area method to start. So I've got sigma-max = kt(N/At), the net area. So that's going to be 2.5 times the normal force is 4 kilonewtons, divided by, the net area is 16 millimeter squared. And so sigma max at this continuity here, right close to the hole is going to equal, if you multiply that out, 0.625 KN per millimetre squared. It's going to be in tension, as we can see. And that ends up being 625 MPa in tension since one kilo-Newton per millimeter squared is equal to a 1,000 MPa. So that's the answer for the maximum stress using the net area method. Let's show that the maximum stress is the same using the gross area method. And so with the Gross Area Method I have to convert my K sub t for the net area method to the K sub G, the stress concentration for the gross area and so that is going to equal now K sub t 2.5 w was 10. And w- d was 10- 2 for the diameter, and so K sub G ends up being 3.125. Now I use this equation now except I'm going to use gross area and the k stress concentration area, stress concentration factor for gross area. And so I get sigma max equals 3.125 times N. N is still 4 kilonewtons over now the gross area is 20 millimeters squared. And if you multiply that out, you'll see that again, that equals 0.625 kilo-Newtons per millimeter squared (tension) or sigma max equals 625 MPa in tension. You get the same result whichever method you use. So here's an actual example of a member with a discontinuity, a hole. This is a bracket and you'll notice that it's failed here, and it's failed right at the point of highest stress, at the stress concentration. And so that's a great practical example of the theory we just went through. So, as a wrap up, stress concentrations. We did a simple example for a hole on an axially loaded bar. The same concept works for other types of loading, torsion, bending, which we're going to talk about in my next courses. You can find the graphs and these different examples in mechanics and material references. Stress concentrations are important in design, but they're not as significant for the static loading of ductile material like steel because that material yields around the region of high stress and the stress is redistributed and the equilibrium is established. However, for brittle materials something like glass for example, the stress concentrations may actually cause fracture and so you have to be very careful about that. And so that's it for the topic of stress concentrations and we'll pick up next time. [MUSIC]