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This is module 15 of Mechanics of Materials Part IV.

And 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.

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

longer than some of my modules because I want it 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.

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And so here is where we left off last time.

We went ahead and

we derived the differential equation for the column buckling.

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

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

buckling?

And so, this is a second order,

homogeneous ordinary differential equation.

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

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

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

partial differentials.

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

we'll let D squared = P over EI.

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

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

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

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 complementary solution, which is

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And so let's begin by looking at the complementary

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

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

its second derivative.

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

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

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 0.

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

after taking derivative after derivative after derivative?

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

Because if I take and 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's going to work,

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

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

this is the result that we get.

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

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

y sub c is just equal to 0.

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

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

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

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

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

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

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

and minus i sub 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's a rather advanced mathematical topic,

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

Once, I have this in complex exponential form,

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

coefficients into a combination of sines and cosines.

And so I've written it like that here.

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And so we now have our complementary 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 sub p.

We called the complimentary solution y sub 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.

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

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

When I do that,

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

I end up with what's shown here.

I can cancel D squared.

I know that y sub p is equal to c1.

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

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

Now that I have both the complementary solution and

the particular solution, I can add them together.

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

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

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

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

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

the pinned-pinned connection which will

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

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

y at 0 = 0 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, it's equal to delta.

So y is equal to 0.

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

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

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

And so we get b1 = 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?

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And 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, kind of turn it 90 degrees in your mind, like a beam.

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And for this column you can see that dy dx,

the slope at this origin is equal to zero.

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

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

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

Sine of 0 is equal to 0.

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

And 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.

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And we know now, also, that another boundary

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

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

That's equal to the right-hand side.

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

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

And so this is true when cosine of

this argument is equal to 1.

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

or 90 degrees.

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

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

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And so only the first value has physical significance since it

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

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

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 solved was for

column buckling, and we now solve for P.

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

We call that the Euler Buckling Load.

This is for pinned-pinned end conditions.

And so, again, going back,

column buckling is a stability type phenomenon.

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

there's 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 called that neutral equilibrium.

That's right on the verge of stability.

And 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.

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

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

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

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

And I'll see you next time.

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