[MUSIC] This is module five, the Mechanics of Materials part four and today's learning outcome is to use the double integration method now, to determine the maximum deflection of a simply supported beam with a moment applied at the right end and where that max deflection occurs. And so here's the differential equation for the elastic curve of a beam. We know that if we have this equation, we can integrate. We did that last module and we came up with an expression for the deflection. Now we want to determine the max deflection and where it occurs given this deflection equation. And so, my first question to you is, how do we find the max deflection and where it occurs? How should we proceed? And so mathematically, the max deflection occurs where dy/dx = 0 where y is a function of x or physically where the slope of the beam is equal to 0. So let's go ahead and do that. We have EI, dy is a function of x dx, is equal to 0. Is equal to now M end over 2L times x squared minus M end, times L over 6. And we can cancel the M ends now and we end up with x squared equals L squared over 3 or the max deflection occurs where x equals L over the square root of 3. Okay, now using that result, we found that the max deflection occurs at x equals L over the square root of 3 or 0.577L. Now we want to find the max deflection at that point. And so try to do that on your own, and then come on back and see how you did. So we have EI equals y X equals L over the square root of 3, which is equal to now M sub end over 6L times X cubed, X cubed is going to be L cubed over 3 times the square of 3 minus M end, L over 6 times x which is, L over the square root of 3. And we can then cancel L with L cubed and this becomes L squared and we get now Y at the plate place where it's going to be the max deflection which is, L over the square of 3 is equal to M end over EI, I've divided through by EI now, times L squared over 18 times the square root of 3 minus L squared over 6 times the square root of 3. We'll need a common denominator here, so we've got M sub end over EI times now, L squared over 18 square root of 3 minus 3L squared over 18 square root of 3, which is equal to M over end, Ms of end over EI times, minus 2L squared over 18 square root of 3. So we have y max is equal to, minus M end L squared over 9, square root of three or EI. And it's negative because remember I said that y was always defined as positive up and so we're going to have the max deflection down, here's that written more clearly. The max deflection is minus M end, L squared over square root of 9, times the square root of 3, EI. It occurs at x equals 0.577L and sure if I drew a sketch of that, this is what it would look like. We know that the max deflection is going to be 0.577L, so a little bit further out in the middle. And so we got this moment at the end, it's going to cause as to deflect down, and then come back up. So the max deflection here is Y max and it's value is given here. And so that's the sketch. We've gone ahead and solved for the max deflection and the location where occurs for the specific loading of a simply supported beam with a moment applied at the right end. And that will finish this module. [MUSIC]