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[MUSIC]

Â This is module 17 of Mechanics and Materials part two and

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

Â As a review, we looked at critical buckling loads for

Â different end conditions.

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

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

Â We called the denominator, the effective length squared.

Â And it's define there, and

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

Â 0:36

And now, let's do a worksheet.

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

Â The normal yield stress for this steel is 36 ksi.

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

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

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

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

Â then see which case will govern.

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

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

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

Â This goes all the way back to my second course,

Â Applications in Engineering Mechanics.

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

Â this should be a quite straight forward problem.

Â With the techniques that we learned in that course.

Â So, how should we begin?

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

Â Do that on your own, come on back.

Â We see we have pinned roller connections, and so

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

Â And then, what?

Â 1:55

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

Â that will give us the equation for solving for By.

Â We find By, then I can continue on solve for

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

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

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

Â members AD and AE.

Â All of these forces and

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

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

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

Â 2:41

And so there is the free body diagram,

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

Â And this is the result, we get.

Â 2:52

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

Â if we put a mirror in the center,

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

Â And so, since we've solved for FAD,

Â that also is the same as FCB, or BC.

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

Â Some forces in the x direction.

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

Â We can substitute that in.

Â We get FAE.

Â And that's our, FAE is this number.

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

Â 3:35

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

Â this member, and this member.

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

Â those?

Â 3:47

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

Â And so lets go ahead and do a section cut,

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

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

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

Â We can solve on our section cut here for FDE.

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

Â so we have solved for the internal forces for

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

Â 4:32

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

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

Â which says that failure will occur when the shear stress in

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

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

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

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

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

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

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

Â So the shear failure stress is 18 ksi and

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

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

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

Â 5:38

Okay, so we have our axial load failure.

Â For shear and we have to now include the factor or

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

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

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

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

Â I show now the actual or allowed shear stress.

Â We see the actual or allowed normal stress.

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

Â Here's our truss.

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

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

Â I know what the cross sectional area is,

Â because I'm given that in the problem statement.

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

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

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

Â where we have a compressive force of 0.75 F.

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

Â than 0.75 F.

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

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

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

Â equal to 3,142 pounds There are 3.142 kips

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

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

Â 7:34

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

Â these are all pinned-pinned conditions.

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

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

Â so we put in our values for our Critical Load and

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

Â but that's equal to 0.75 F critical,

Â the F critical that the trust can hold, and so

Â we see that the critical force that can be held as

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

Â And so, the actual allowed stress for

Â yielding was 3.142 kips.

Â So the lesser.

Â Those are obviously the buckling condition.

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

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

Â And so, that's the limiting condition.

Â And so, that's the case that governs.

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

Â