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Â This is module 15 of Mechanics of Materials Part IV.

Â And today's learning outcome is to derive the critical buckling load, or what we

Â call the Euler buckling load, for a column with pinned-pinned end conditions.

Â And let me say at the beginning here that this module will be slightly

Â longer than some of my modules because I want it to flow through.

Â And it's going to involve some background mathematical techniques and

Â information that you may have to review on your own to be able to go through.

Â The nice thing is, even though I'm going to go quite quickly

Â through the problem, you can stop the video and start it again.

Â And go ahead and

Â research on your own to make sure you understand the mathematical steps.

Â 0:49

And so here is where we left off last time.

Â We went ahead and

Â we derived the differential equation for the column buckling.

Â And it's shown here, and so we'll start at that point.

Â And we want to find what is the minimum axial compressive load that will cause

Â buckling?

Â And so, this is a second order,

Â homogeneous ordinary differential equation.

Â Okay, it's second order because we have d squared y dx squared.

Â It's nonhomogeneous because we have a right-hand side.

Â And it's an ordinary differential equation because it does not involve

Â partial differentials.

Â And so, we'll let, just to simplify it a little bit,

Â we'll let D squared = P over EI.

Â And so this just makes it a little bit simpler, and

Â we're going to want to go ahead and solve for y.

Â And so we're going to use something called the,

Â mathematically we call it the Method of Undetermined Coefficients.

Â And if you review the Method of Undetermined Coefficients, it says that

Â solving for y as a function of x is the complementary solution, which is

Â 2:07

And so let's begin by looking at the complementary

Â solution where we set the right-hand side of the equation equal to 0.

Â We have a function for y that involves the function y itself and

Â its second derivative.

Â And so we're going to want to have a function of y that comes back.

Â We're going to have to add it together and have it equal to 0.

Â So we're going to want a function of y that when I take derivatives,

Â particularly a second derivative, and it needs to come back with the same form so

Â that we can add it together and set it equal to 0.

Â So what form of an assumed solution of y comes back with the same form

Â after taking derivative after derivative after derivative?

Â And what you should say is okay, that assumed solution should be an exponential.

Â Because if I take and assume that y is in the form of an exponential,

Â every time I take a derivative it's still in the form of an exponential.

Â So I can say, okay, we're going to assume that solution.

Â If that solution's going to work,

Â we're going to have to substitute it into our equation.

Â And so if we assume the solution's substituted into the equation,

Â this is the result that we get.

Â 3:16

We know that a, e to the yx cannot be equal to 0.

Â Because if that's the case, that's just what we call the trivial solution, and

Â y sub c is just equal to 0.

Â So, we're going to have to set what's in parentheses equal to 0, and

Â if we do that, we come up with these solutions.

Â However, we can throw out this bottom solution because for

Â our problem physically we know that D squared has to be greater than 0

Â because these values P, E, and I are positive.

Â And so they're always going to be greater than 0.

Â And so we only end up with this value for lambda as equal to plus or

Â minus i sub D, where i is an imaginary number.

Â And so, given that, we can now say that y is a linear

Â combination of two terms which have both lambdas, one of plus iD,

Â and minus i sub D with different coefficients in front of each one.

Â So I show that solution here, and again, I'm going really rapidly.

Â However, you should go back and review on your own,

Â look at maybe some other videos or the Internet how to do this method of

Â undetermined coefficients and how to solve differential equations.

Â It's a rather advanced mathematical topic,

Â but there's a lot of resources out there to go through this.

Â Once, I have this in complex exponential form,

Â you can also show mathematically that it can be changed with different

Â coefficients into a combination of sines and cosines.

Â And so I've written it like that here.

Â 5:04

And so we now have our complementary solution.

Â We now need to go back and find the particular solution.

Â And for the particular solution, we're going to say okay, let that be y sub p.

Â We called the complimentary solution y sub c.

Â For the particular solution, we're going to say okay,

Â since the right-hand side is a constant,

Â D squared times delta, we'll say it's going to take the form of a constant.

Â We'll just call that an unknown constant c1.

Â 5:30

If that's going to be the particular solution, it needs to solve

Â the differential equation, so again we'll substitute it back in.

Â When I do that,

Â since y sub p is a constant when I take the second derivative, that zeroes out.

Â I end up with what's shown here.

Â I can cancel D squared.

Â I know that y sub p is equal to c1.

Â And so therefore I find that c1, the constant, is equal to delta.

Â And so now I know that my particular solution is equal to the constant delta.

Â Now that I have both the complementary solution and

Â the particular solution, I can add them together.

Â And I have the total solution, and it's shown here.

Â 6:15

And so there's our total solution and a diagram of the situation.

Â And we want to now solve for y, and we're using

Â the differential equation that describes this pinned-pinned connection situation.

Â And so we're going to have to use the boundary conditions for

Â the pinned-pinned connection which will

Â allow us to solve for the undetermined coefficients here of b1 and b2.

Â 6:44

And so the first boundary condition I'm going to use is that

Â y at 0 = 0 for my origin here at the center of the beam.

Â You can see that we haven't yet deflected.

Â When we get out to y equals x, it's equal to delta.

Â So y is equal to 0.

Â We put in cosine of 0, b2 times sine of 0 + delta now, the particular solution.

Â We know that the sine of 0 is equal to 0.

Â We know that the cosine of 0 is equal to 1.

Â And so we get b1 = minus delta, so we've solved for one of these coefficients.

Â Now let's use another boundary condition to solve for the other coefficient,

Â the boundary condition we're going to use in this case has to do with slope.

Â So I'm going to take my solution, I'm going to find the slope, or

Â the dy dx equation by taking the derivative.

Â This is the derivative of this deflection equation.

Â Here is the slope equation.

Â Where do I know anything about the slope?

Â 7:44

And what you should say is, okay,

Â here at the origin, we know that the slope is going to equal to zero.

Â If you can't see that, kind of turn it 90 degrees in your mind, like a beam.

Â 7:54

And for this column you can see that dy dx,

Â the slope at this origin is equal to zero.

Â And so we have dy dx at x equal 0 = 0.

Â We substitute that in, now we have sine of 0 here.

Â This times cosine of 0 here, cosine of 0, again, is equal to 1.

Â Sine of 0 is equal to 0.

Â And so what we end up with is b2 must be equal to 0.

Â And so I now have both the constants for b1 and for b2.

Â And I have my solution for y as a function of x, or

Â the deflection as a function of x.

Â I just rearranged it here mathematically.

Â 8:40

And we know now, also, that another boundary

Â condition at x = L over 2, y has to be equal to delta.

Â And if we substitute that in, we've got y at L over 2 = delta.

Â That's equal to the right-hand side.

Â And we find out that if this is to be true, what's in the parentheses here has

Â to be equal to 0, 1- cosine times the square root of P over EI and L over 2.

Â And so this is true when cosine of

Â this argument is equal to 1.

Â Well cosine of an argument is equal to 1 when the argument is pi over 2 radians,

Â or 90 degrees.

Â Cosine is equal to 1 when it's 3 pi over 2 radians, 5 pi over 2 radians, etc.

Â And so when these values are true, this equation is satisfied.

Â 9:35

And so only the first value has physical significance since it

Â determines the minimum value of P for a nontrivial solution, and so

Â just the part where this value on the left-hand side is equal to pi over 2.

Â And so if we solve then, we call that the critical P value.

Â That's the minimum axial compressive load that's going to cause buckling.

Â Since this differential equation we solved was for

Â column buckling, and we now solve for P.

Â This is the value we get for the critical buckling load.

Â We call that the Euler Buckling Load.

Â This is for pinned-pinned end conditions.

Â And so, again, going back,

Â column buckling is a stability type phenomenon.

Â And so the stages of column buckling are, as we first compress the column,

Â there's a range where it's stable.

Â And I show stability by this little ball in this trough.

Â So if I unload the column, it would go back to its original shape.

Â If I push the ball, deform it,

Â or perturb it away from its current spot, it'll go back to its original spot.

Â At the verge of buckling where we have the P critical load,

Â we called that neutral equilibrium.

Â That's right on the verge of stability.

Â And so if we perturb it now beyond this, we're going to have a problem.

Â And we call that unstable equilibrium.

Â And once it goes past the unstable equilibrium,

Â or once it gets into the unstable equilibrium condition,

Â if we perturb the column even a little bit more laterally.

Â 11:21

Our situation will just go completely unstable, just like we saw for this ball.

Â And so that's it, long module but very important.

Â We now know the Euler column loading or critical buckling loading for

Â the case of a column with pinned-pinned end conditions.

Â And I'll see you next time.

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Â