[SOUND] Hi.

This is Module 20 of Mechanics of Materials Part IV.

Todays learning outcome is to solve the column

problem again to determine if it fails in yielding or in buckling.

And so, I just wanted to give you a couple of real world examples of Buckling.

This shows i-beams with local flange buckling,

which is quite common if you're going to have buckling.

We don't want to have buckling, that's a bad thing, but if you do see buckling,

local flange buckling is Is something that you might see.

Here's another example.

This is the Windsor Tower fire that took place in Madrid.

And you can see that there was some unprotected steel perimeter columns, and

that they buckled.

If we look here that you can see that they actually buckled as the fire took place.

And so those are some real world examples.

Let's go ahead now and do a worksheet.

Let's look at a steel column that is made of a W 305 by 74 wide-flange I beam.

It's pinned at both ends for rotation about the x-axis where the x-axis

is out of the page and so bending in this direction.

And then it's fixed at both ends against rotation about the y-axis and

the y-axis is shown here.

And we're to determine the load P that will cause failure.

And so here's a top view of a 305 by 74 wide-flange I beam.

If you look it up in tables, first of all you can find this,

the yield stress for steel you can find the moduls of elasticity for steel.

And you can find from tables for

this W 305 by 74 I beam that the area is as shown.

The I about the area moment of inertia about the x-axis is 164 times 10 to

the sixth millimeters to the fourth, and

about the y-axis it's 23.4 times 10 to the sixth millimeters to the fourth.

And so the first thing I'm going to do is I'm going to check for yielding or

at the compressive load that would cause yielding and so

we know all the way back to my mechanics of materials course one,

we know that the normal stress and it's going to be compressing this.

Compression on this case is P/A we have

for P yield we're looking for what will cause yielding.

That's equal to sigma yield times the area.

So P yield is equal to

250 mega pascals and

we'll put everything in terms of newtons and millimeters and so

that's the yield stress in steel so we'll say

a Newton per millimeter squared is the same as a mega pascal.

And then we're going to multiply by the area which is 9,485

millimeter squared and if I do that calculation?

I'll find that P yield, P it's going to cause a yielding.