We’re going to look at

a basic column, so I’m going to try to make an

image of the column here. It’s a basic column on

of steel and my question is, I want to know if it

can support a certain load. So I am expecting the load on

this column to be 200 kilonewtons, and I want to make sure that this

column can support that much. This column is 5 meters

tall, it’s made of steel, and actually I’m going to make it a

hollow tube instead of a solid tube. So it just has an outer little portion. So let me draw a picture of

that square cross section. So it’s a square cross section,

again, made out of steel. Just to make sure we

know that it’s hollow. I’m going to try some certain

dimensions to start with. So I’ll start with 60 millimeters

for the outer dimension, and I’m going to start with 40

millimeters for the inner dimension, and that will help me calculate

the properties for this column. But this is a column, I’m

designing it, it’s my trial design, and I want to see if it can

carry that 200 kilonewtons. So I’m going to have to go in

and look at a couple things. We decided we wanted

to look at compression. That tends to be the easier one, so

I’m going to start with compression. So if I want to

calculate the stress– So this compressive force is going to

cause an internal stress as we get down the column, and that

equals force over area. Fairly straightforward. So I take that force. We know the force is 200 kilonewtons,

so that’s what I want to try– and I’m going to divide by

the cross sectional area. So let’s go back and calculate

the cross sectional area. So for this cross section, I can

figure out the cross sectional area. It’s a square, so I’m

going to take the outer. So it would be 60 millimeters. I’m going to switch that to meters so

I’m working all in meters, 0.06 meters, and that’ll get squared. That’s the area of the

outer section, and I’ll subtract the area of the

inner section, 0.04 squared. And then I can calculate that

cross sectional area, which is 0.002 meters squared, but

that’s my cross sectional area that I can then use up

here in this equation. And that’ll allow me to

calculate the stress, and that stress is going to be

100,000– fairly high number– kilonewtons per meter squared. So that’s the stress that I get due

to this 200 kilonewtons pushing down on the column with this cross section. I want to make sure

that’s OK, so I’m going to want to look up the

allowable stress for steel. So I want to know how much

stress deal can handle, and that allowable stress is 250,000

kilonewtons per meter squared. So it’s greater than

my stress applied, so I expect my column to be OK if

I’m considering compression. So that’s the first

thing I want to look at. So this column will not

fail by compression. It will not yield, and that’s good news. But now let’s also look at buckling. It’s the other mode of

failure that’s possible. So that mode would be that

I push down the column it would start moving laterally. So I want to figure out

if it’s going to buckle, I need to calculate what’s

called a critical buckling load, and that equals pi

squared EI over L squared. Pi is just a constant. E is the modulus of

elasticity of the material. So we’re talking about steel, and

I can look up for steel figure out what the modulus of

elasticity is for steel. So E is modulus of

elasticity, and for steel, that is 200 times 10 to the sixth

kilonewtons per meter squared, pretty high number, and that’s

just a function of the material. Movement of inertia, so we can

also calculate moment of inertia, and that’s a function

of the cross section. So that’s going to be

based on these dimensions. And that, in this case, equals the

base dimension, this outer dimension. I’m going to start with

that outer dimension. That will my 60 millimeters. That’s going to be to the fourth. Divide that by 12. That’s just an equation we can look

up, and we’ve posted tables of those that you can use to look up. And the inner to the fourth over 12. So in this case, I would

put 60 in and the 40 in, and I can calculate the moment of

inertia, and that moment of inertia ends up being the 8.67 times 10 to

the minus 7 meters to the fourth. Moment of inertia will always

be raised to the fourth. Length to the fourth, so

that can go into my equation. So now back to my equation. I have pi squared times

200, times 10 to the sixth, times this moment of inertia,

8.67, times 10 to the minus 7, and that’s all divided

by my length squared. In this case, it’s 5

meters squared, and that’ll allow me to calculate this P critical,

which ends up being 68 kilonewtons. But I want to apply 200 kilonewtons,

so this is less than my P that I want to apply. So this column is no good. It’s going to fail in

buckling, so I need to go back and reevaluate something. I’m already using steel so I don’t

know if I change the material. Maybe there’s a way I

can reduce the load. Maybe there’s a way I can reduce the

length, either by reducing the length or by adding some type

of brace to it, but I have to reevaluate this column somehow.