This is Module 17 of Mechanics of Materials Part 2, 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 call the denominator the effective length squared, and it's defined there, and then we saw physically the effective lengths as shown with the graphics. 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 the 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 allowable load due to axial loading, and then to find the allowable load due to buckling, and then see which case will gather. Since this is we're going to consider this a perfect truss, we're going to say that we have end end conditions for our members. 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 quite straightforward problem with 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? Well, we're going to go ahead and sum moments about point A here, and that'll give us the equation for solving for B_y, we find B_y. Then I can continue on solve for the forces in the y-direction set them equal to 0 and I'll solve for A_y. Now, let's go ahead and do a joint cut. The joint cut will allow us now to find the actual forces in members AD and AE. All of these forces in all of these members will be in terms of the force F that we're trying to support. So go ahead and do this on your own. Find F_AD and F_EA or AE, and then come on back, and let's see how you did. So there's the free body diagram. If we sum forces in the y-direction, we can solve for F_AD, and this is the result we get. You can see by symmetry now, since this is symmetric about the center of this truss, if we put a mirror in the center of the left-hand side, looks the same as the right hand side. So since we've solved for F_AD, that also is the same as F_CB or BC, and so we've taken care of two of the members. Let's continue on. Sum forces in the x-direction, and when we do that, we know what F_AD is now, we can substitute that in. We get F_AE. F_AE is this member, and so by symmetry it's the same as F_BE. 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? Well, we could continue the joint method, or we can do a section cut, and so let's go ahead and do a section cut. Draw a free body diagram of that section, we'll sum moments about E, and we get this equation, so we can solve for F_CD which is up here. We then go ahead and sum forces in the y-direction, we can solve on our section cut here for F_DE. If we know F_DE by symmetry it's the same as F_CE, and so now we have solved for the internal forces for all of this structural members of our truss in terms of this load F. Now, to find the 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 than any of the members is greater than the failure shear stress. This theory is used because it's good for ductile materials, as we've seen in my past courses and lessons. So ductile material being a steel in this case. For an axial load then, the failure shear stress will be equal to the failure normal stress or yield stress divided by two, or in this case the normal yield stress for steel, is equal to 36 ksi, so the shear failure stress is 18 ksi. This is based on a simple tension test as we've talked about before, if we use Mohr 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. So we have our axial load failure for shear, and we have to now include the factor of safety which we want to be three for yielding. So therefore, we have our failure stress which is 18 over whatever our actual stress is going to be which says okay, our actual or allowed stress can only be 6 ksi or less. So here again, the simple tension test for Mohr Circle, I've shown 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 1.5. 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 Ps in any given situation. So P allowed then, the worst case will be when in member CD where we have a compressive force of 0.75F. All the rest of them are going to be a little bit less than 0.75F. So this is the critical condition. We take the six ksi we multiply by two, we multiply it by the cross-sectional area, and that we find then the actual or allowed F that we can support in our truss has to be less than or equal to 3,142 pounds or 3.142 kips to avoid yielding. So we've done Part A, we found what we can hold as far as yielding is concerned in our truss members. Now let's go ahead and look at the allowable load due to buckling. These are all pinned-pinned conditions. Again, the situation with the worst. The largest load will be in CD. Remember CD it'll 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 0.75F that the truss 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 pounds. So the actual allowed stress for yielding was 3.142 kips, and so the lesser of those are obviously the buckling condition. 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. So that's the case that governs, we've solved our problem, and we'll come back and continue on.