This is module 15 of Mechanics of Materials part four. Today's learning outcome is to derive the critical buckling load or what we call the Euler buckling load for a column with pinned-pinned end conditions. Let me say at the beginning here that this module will be slightly longer than some of my modules because it's going to involve, I wanted to flow through and it's going to involve some background mathematical techniques and information that you may have to review on your own to be able to go through. The nice thing is even though I'm going to go quite quickly through the problem, you can stop the video and start it again and go ahead and research on your own to make sure you understand the mathematical steps. So here's where we left off last time. We went ahead and we derived the differential equation for the column buckling and it's shown here. So we'll start at that point and we want to find what is the minimum axial compressive load that will cause buckling. So this is a second order homogeneous ordinary differential equation. Second order because we have d squared y dx squared. It's non-homogeneous because we have a right-hand side and it's an ordinary differential equation because it does not involve partial differentials. So we'll let, just to simplify it a little bit, we'll let D squared equals P over EI and so this just makes it a little bit simpler. We're going to want to go ahead and solve for y. So we're going to use something mathematically we call it the method of undetermined coefficients, and if you review the Method of Undetermined Coefficients it says that solving for y as a function of x is the complimentary solution which is also called the homogeneous solution when the right-hand side is equal to zero and the particular solution for the right-hand side or what's also called the non-homogeneous part of the solution. So let's begin by looking at the complementary solution where we set the right-hand side of the equation equal to zero, we have a function for y that involves the function y itself and its second derivative. So we're going to want to have a function of y that comes back. We're going have to add it together and have it equal to zero. So we're going to want a function of y that when I take derivatives, particularly a second derivative and it needs to come back with the same form so that we can add it together and set it equal to zero. So what form of an assumed solution of y comes back with the same form as you're taking derivative after derivative, after derivative? What you should say is okay, that assumed solution should be an exponential because if I take an assume that y is in the form of an exponential, every time I take a derivative it's still in the form of an exponential so I can say, okay, we're going to assume that solution, if that solution is going to work we're going to have to substitute it into our equation. So if we assume the solution substituted into the equation, this is the result that we get. We know that a e to the yx cannot be equal to zero because if that's the case that's just what we call the trivial solution and y sub c is just equal to zero. So we're going to have to set what's in parentheses equal to zero and if we do that, we come up with these solutions. However, we can throw out this bottom solution because for our problem physically we know that D squared has to be greater than zero because these values P, E, and I are positive and so they're always going to be greater than zero. So we only end up with this value for Lambda is equal to plus or minus i_ D where i is an imaginary number. So given that, we can now say that y is a linear combination of two terms which have both Lambdas one of plus i_D, id, and minus i_D with different coefficients in front of each one. So I show that solution here and again, I'm going really rapidly. However, you should go back and review on your own look at maybe some other videos or the internet how to do this method of undetermined coefficients and how to solve differential equations. It so rather advanced mathematical topic, but there's a lot of resources out there to go through this. Once I had this in complex exponential form, you can also show mathematically that it can be changed with different coefficients into a combination of sines and cosine and so I've written it like that here. So now I've put back in my value for D which is the square root of P over EI. So we now have our complimentary solution. We now need to go back and find the particular solution and for the particular solution we're going to say, okay, let that be y_P we call the complimentary solution y_C. For the particular solution, we're going to say, okay, since the right-hand side is a constant D squared times Delta we'll say it's going to take the form of a constant. We'll just call that an unknown constant C1. If that's going to be the particular solution, it needs to solve the differential equations. So again, we'll substitute it back in. When I do that, since y_P is a constant, when I take the second derivative that zeros out and I end up with what's shown here. I can cancel D squared, I know that y_P is equal to C1. So therefore, I find that C1 the constant is equal to Delta. So now I know that my particular solution is equal to the constant Delta. Now that I have both the complimentary solution and the particular solution, I can add them together and I have the total solution and it's shown here. So there's our total solution and a diagram of the situation. We want to now solve for y and we're using the differential equation that describes this pin-pin connection situation. So we're going to have to use the boundary conditions for the pin-pin connection which will allow us to solve for the undetermined coefficients here of b1 and b2. So the first condition that I'm going to use, boundary condition I'm going to use, is that y at zero is equal to zero for my origin here at the center of the beam, you can see that we haven't yet deflected when we get out to y equals x is equal to Delta. So y is equal to zero, we put in cosine of zero b2 times sine of zero plus Delta now the particular solution. We know that the sine of zero is equal to zero. We know that the cosine of zero is equal to one and so we get b1 is equal to minus Delta. So we've solved for one of these coefficients. Now let's use another boundary condition to solve for the other coefficient. The boundary condition we're going to use in this case has to do with slope. So I'm going to take my solution, I'm going to find the slope or the dy dx equation by taking the derivative. This is the derivative of this deflection equation. Here is the slope equation. Where do I know anything about the slope? What you should say is, okay, here at the origin, we know that the slope is going to equal to zero. If you can't see that turn it 90 degrees in your mind like a beam and for this column you can see that dy dx the slope at this origin is equal to zero. So that's we have x equals zero is equal to zero, we substitute that in now we have sine of zero here. This times cosine of zero here, cosine of zero again is equal to one, sine of zero is equal to zero and so what we end up with is b2 must be equal to zero. So I now have both the constants for b1 and for b2 and I have my solution for y as a function of x or the deflection as a function of x. I just rearranged it here mathematically. We know now also that another boundary condition at x equals L over two y has to be equal to Delta. If we substitute that in, we've got y at L over two is equal to Delta that's equal to the right-hand side and we find out that if this has to be true, what's in the parentheses here has to be equal to zero, one minus cosine times the square root of P over EI times L over two. So this is true when cosine of this argument is equal to one. Well, cosine of an argument is equal to one when the argument is pi over two radians or 90 degrees, cosine is equal to one when it's three pi over two radians, five pi over two radians etc. So when these values are true, this equation is satisfied. So only the first value has physical significance since it determines the minimum value of P for a non-trivial solution. So just the part where this value on the left-hand side is equal to pi over two and so if we solve then we call that the critical P value, that's the minimum axial compressive load that's going to cause buckling since this differential equation we solve was for column buckling. We now solve for P, this is the value we get for the critical buckling load, we call that the Euler buckling load. This was for pinned-pinned end conditions. So again, going back, column buckling is a stability type phenomenon. So the stages of column buckling are as we first compressed to column, there is a range where it's stable and I show stability by this little ball in this trough. So if I unload the column it would go back to its original shape, if I push the ball deform it or perturb it away from its current spot, it'll go back to its original spot. At the verge of buckling where we have the p critical load, we call that neutrally equilibrium. That's right on the verge of stability. So if we perturb it now beyond this, we're going to have a problem and we call that unstable equilibrium and once it goes past the unstable equilibrium or once it gets into the unstable equilibrium condition, if we perturb the column even a little bit more laterally, our situation will just go completely unstable just like we saw for this ball. So that's it, long module but very important. We now know the Euler critical, the Euler column loading or critical buckling loading for the case of a column with pinned-pinned end conditions. I'll see you next time.