[NOISE] Welcome back to Mechanics of Materials Part II. Well we're nearing the end of the course, we have one more topic we're going to cover and that's statically indeterminate torsional members. And so the learning outcome today is to solve a statically indeterminate torsion problem. If you go back to my Mechanics and Materials Part I course, we saw the statically indeterminate problem for axial loading, and you should probably go back and review that, because this is going to be the same procedure in general an analogous procedure for torsion. I'll also again mention since this is one of the final problems it's rather long and the module may be a little longer than usual, but stick with it and make sure you understand each step. So again let's relate it to a real world problem. We've got a simple model of a torsion bar here. For a tracked vehicle, I've got the model shown below. This time I'm going to model both ends as being fixed, and I don't have a hollow section here. And, I'm applying an eight foot kip torque to the center. One thing I'm going to do is, I'm going to change the G that I give you for steel, instead of 11.6 I'm going to make it 11.0. But you could use 11.6 and redo the problem it doesn't matter. That's the numbers that I'm going to go ahead and work with though. Okay, so, we want to determine the maximum shear stress in each section. And so, how might you start? And what you should say is, hey let's come up with the static equilibrium equations. That's what we did for indeterminate structures with axial loading and so how could I come up with a static equilibrium equations? And what you should do, we've been doing it throughout all of my courses is to draw a free body diagram a very important engineering graphical tool. Draw that free-body diagram on your own, come on back, see how you did. And in this case, I replaced these fixed constraints at the end with a torque in the Monel part of my structure. And a torque in the steel part of my structure. And so now that I have my free-body diagram, what do I do? Well, I'm going to have to come up with the static equilibrium equations. I'm really not concerned forces in the x direction or the y direction, I'm only concerned about the torques or the moments, and so I'm going to have one equilibrium equation. How many unknowns do I have? Okay, I have two unknowns. I have my monel torque and the torque in the steel. So I have these two unknowns. Which means I have an unhappy face and I'm going to have to generate another relationship. So let's first start though, by writing that equilibrium equation to start the solve the problem. The one for the torques or the moment. And if you do that on your own, come on back, see how you did. And so we sum moments about the z direction if I say that the z direction is in this direction, and when I do that I get the monel torque and the steel torque have to balance the eight foot kip applied torque, or that converts to 96 inch kips. I'll work with kips and inches. And so, with that, that'll become equation one, that I'll need to soften the two unknowns. And so, there's my one equation, two unknowns, we're going to need an additional equation. How do you think we'll find an additional equation? Well, if you recall back to the way we approach these indeterminate problems in my first course, in Mechanics and Materials, we can use a deformation equation, or what's also called a compatibility equation. Which, it makes sure that we look at the deformation of the geometry, or the geometry of the deformation of my members and make sure that there's a compatibility between equilibrium and the deformation. And so, here's my structure. Write that deformation or compatibility equation on your own, if you can, and then come on back and let's see how you did. And so here I'm going to have a fixed end on the right-hand side. I'm going to have an angle of twist of C with respect to A. And then I'm going to have an angle of twist, of D with respect to C, that angle of twist is going to be in this direction because of my applied torque. Then I'm going to have to have an angular twist of C with respect to D, excuse me, D with respect to C in the other direction, to bring it back to a total of zero displacement. because I got zero displacement on either end. So I've got the angle of twist of C with respect to A minus the angle of twist of D with respect to A has to equal zero. So here's my equilibrium equation, equation one, here's my deformation equation but they're in terms of different unknowns so I need to have them in terms of the same unknowns. To do that, let's start by saying we're going to use the elastoplastic assumption and assume that the steel and the monel shafts remain in the linear elastic region. And we'll assume that the steel has a torsional yield strength of 18 ksi and the monel has a torsional yield strength of 25 ksi. Here's a plot of an idealized elasto plastic material. And we're saying okay let's assume we're continuing to operate for all these situations in the linear elastic range. Can we find a relationship between the torque and the angle of twist? And the answer is yes. We've done that before. The angle of twist if we assume linear elastic is equal to the torque, the length of this section divided by the module of this rigidity times the polar moment of inertia of this section. And, so let's go ahead and do that for my deformation equation and I've got the angle of twist, the C with respect to A, which is my steel section. So, that's T of Steel, times its length, which is 11 feet, and we're going to convert everything to inches, that's 12 inches per foot, and we're going to divide then by G. I said at the beginning I'm going to change this to, instead of 11.6, I'm going to say 11.0 times 10 to the third or 11,000 Ksi times J, solid circular cylinder. So J is pi over 2 times the radius, which is 3 to the fourth, 3 inches to the fourth. And that's going to equal now. No it's not going to equal. Yes it is going to equal. It's going to equal my phi of D with respect to C. So that's the monel section if I carry down the other side, so I've got T of the monel. Times its length which is four feet convert to inches, 12 inches per foot, divided by G for monel is given is 9.5 times 10 to the third or 9,500 Ksi and J is pi over two times its radius, which is half of 4, or 2 inches, to the fourth. If I boil that down, run all those numbers I end up with the equation 2.13 torque monel is equal to torque steel. And so I have a now, an equation, a second equation, I'll call this EQN 2 for the relationship between Tmonel land Tsteel. So we're in good shape now. Two equations, two unknowns, happy face. All right. So there's my equilibrium equation. There's the deformation equation expressed in terms of T and in torques and assuming that we're working in the linear elastic region, If you solve those simultaneously you get Tmonel is 30.7 in-k. T steel is 65.3 inch kips. So, now we need to calculate the stresses. How would we do that? And as we've done it before if we're in the linear last region we use the elastic torsion formula and that is tao equals the torque times row divided by J. So let's do it for monel. And so I've got sheer stress monel is equal to the torque monel, which is 30.7 inch kips, times Ro. And we've got Ro for the monel section is going to be 2 inches, divided by, J that´s going to be pi over 2 times 2 inches to the fourth. And so, the sheer stress in the monel, ends up equaling 2.443 Ksi. Let's do the same thing for the steel. Tao in the steel. We want the maximum so that's why we're going to the outer radius in both cases. That's going to be equal to the torque. Which is 65.3 inch kips times Rho in this case for the steel section is 3 inches divided by J which is pi over 2 times 3 inches to the fourth or the max sheer stress in the steel section is 1.539 Ksi. Okay, so there's the calculation of the stresses that we came up with but we need to make sure that we stayed in the elastic region because that's the only way we could use this second equation. And so we can see however that, if we check the stresses for the linearly elastic assumption, that 2.44, Ksi, that's rounded to 3 significant figures, is much less than the torsional yield strength for monel which is 25 Ksi and 1.54 Ksi rounded to three significant figures again, is less than, much less than, 18 Ksi. So both sections do indeed remain in the linear elastic range, and so I have my solution. And we've completed the worksheet, and so now you know how to do an indeterminate structure for a torsional problem. And so you're in great shape. We've gone through all the sections of the course, and you should have a really good handle on both thin walled pressure vessels and torsional structures. And we'll wrap up next lesson, or next module. [MUSIC]
