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[MUSIC]

This is module 17 of Mechanics and Materials part two and

today's outcome is to solve an actual column buckling problem.

As a review, we looked at critical buckling loads for

different end conditions.

This was for pinned-pinned, this was for pin-fixed.

And then, we had fixed-fixed and fixed-free.

We called the denominator, the effective length squared.

And it's define there, and

then we saw physically the effective lengths as shown with the graphics.

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And now, let's do a worksheet.

We've got a truss below composed of half inch diameter structural steel bars.

The normal yield stress for this steel is 36 ksi.

The modulus of elasticity for the steel is 29,000 ksi.

We're going to require a factor of safety of three with respect to yielding,

due to the axial load and we want to find first the act allowable load

due to axial loading, and then to find the allowable load due to buckling and

then see which case will govern.

And since this is considered, we're going to consider this a perfect trust,

we're going to say that we have end to end conditions for our members.

And so, we want to start by solving the axial loads in the truss member.

This goes all the way back to my second course,

Applications in Engineering Mechanics.

If you need to go back and review that, I recommend that you do, but

this should be a quite straight forward problem.

With the techniques that we learned in that course.

So, how should we begin?

Well, we're going to do a free-body diagram of the entire structure.

Do that on your own, come on back.

We see we have pinned roller connections, and so

these are the force reactions that we get as a result.

And then, what?

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Well we're going to go ahead and sum moments about point A here, and

that will give us the equation for solving for By.

We find By, then I can continue on solve for

the forces in the y directions, set them equal to zero and I'll solve for Ay.

And now, let's go ahead and do a joint cut.

And the joint cut will allow us now to find the actual forces in

members AD and AE.

All of these forces and

all of these members will be terms of the force F that we're trying to support.

And so, go ahead and do this on your own.

Find FAD, FEA or AE, and then come on back and let's see how you did.

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And so there is the free body diagram,

if we sum forces in the y direction, we can solve for FAD.

And this is the result, we get.

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You can see by symmetry now, since this is symmetric about the center of this trust,

if we put a mirror in the center,

the left hand side looks the same as the right hand side.

And so, since we've solved for FAD,

that also is the same as FCB, or BC.

And so we've taken care of two of the members, lets continue on.

Some forces in the x direction.

And when we do that, we know what FAB, AD is now.

We can substitute that in.

We get FAE.

And that's our, FAE is this number.

And so by symmetry, it's the same as FBE.

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And so, now we've taken care of this member, this member,

this member, and this member.

So we still need these internal three members, how would we go about solving for

those?

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Well, we could continue the joint method, or we can do a section cut.

And so lets go ahead and do a section cut,

draw a free body diagram of that section, we'll sell moments about e.

And we get this equation, so we can solve for FCD which is up here.

We then, go ahead and sum forces in the Y direction.

We can solve on our section cut here for FDE.

If we know FDE by symmetry, it's the same as FCE, and

so we have solved for the internal forces for

all of this structural members of our truss, in terms of this load F.

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Now, to find part a, the allowable load ,F,

due to axial loading We're going to use the maximum shear stress theory,

which says that failure will occur when the shear stress in

any of the members is greater than the failure shear stress.

And this theory is used because it's good for ductile materials,

as we've seen in my past courses and lessons.

And so, ductile material being a steel, in this case, for

an axial load then The failure sheer stress will be equal to

the failure normal stress or yield stress divided by 2 or in this case,

the yield stress for the normal yield stress for steel is equal to 36 ksi.

So the shear failure stress is 18 ksi and

this is based on a simple tension test, as we've talked about before.

If we use Mohr's Circle and we were doing a simple tension test on any of these

members, we would see that sigma failure is two times the tau failure.

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Okay, so we have our axial load failure.

For shear and we have to now include the factor or

safety which we want to be three for yielding and so, therefore, we have

our failure stress which is 18 over or whatever our actual stress is going to be

would says, okay, our actual or allowed stress can only be six ksi or less.

And so, here again, the Simple Tension Test for Mohr's Circle.

I show now the actual or allowed shear stress.

We see the actual or allowed normal stress.

We know that they're related by a factor of one-half.

Here's our truss.

And so, for our loads in our members since these are axial loads,

the normal stress is equal to the force in those members divided by the area.

I know what the cross sectional area is,

because I'm given that in the problem statement.

I know what the forces are, because I've solved for them.

These equate to the P's in any given situation.

And so, P allowed then the worst case will be when in member CD,

where we have a compressive force of 0.75 F.

All the rest of them are going to be a little bit less than are going to be less

than 0.75 F.

And so, this is the critical condition, we take the 6 ksi, we multiply by two,

we multiply by the cross section area and that we find then the actual or

allowed F that we can support in our trust has to be less than or

equal to 3,142 pounds There are 3.142 kips

to avoid yielding, and so we have done part A, we found

what we could hold as far as yielding is concerned in our trust members.

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Now let's go ahead and look at the allowable load due to buckling, and

these are all pinned-pinned conditions.

Again, the situation with the worst or the largest load will be in CD.

Remember CD, it will be in compression so it can buckle,

so we put in our values for our Critical Load and

we find out that the Critical Load can only be 0.1694 kips,

but that's equal to 0.75 F critical,

the F critical that the trust can hold, and so

we see that the critical force that can be held as

a result of member CD Is going to be 226 lb.

And so, the actual allowed stress for

yielding was 3.142 kips.

So the lesser.

Those are obviously the buckling condition.

And so if we get to a load, a force that's greater than this,

we are going to go ahead and have buckling in member CD.

And so, that's the limiting condition.

And so, that's the case that governs.

We've solved our problem, and we'll come back and continue on.