Welcome to Module 21 of Mechanics of Materials part 4. We're moving along and getting close to the end of the course. We've done deflections, we've now completed buckling, and today we're going to look at combined static loading. This is a combination of all my mechanics and materials courses, the parts one through four. So let me warn you in advance that this is going to be a rather long module, but you can go through it, stop it, take it a step at a time, and be able to digest and understand how to do this type of analysis. So our learning outcome is to solve a combined static loading problem. This is the problem I'm going to look at. I've got an engineering structure. I want to find the principal stresses and the max sheer stress at point A, which is a point at the top of the beam right in the center. I'm going to assume that the material in the structure remains in the linear elastic region for all the loading conditions that is given. That's important because when we do combined static loading, this linear elasticity will allow us to analyze each type of loading individually and then add them together by superposition. So how might we start? What you would say is, to start, we're going to have to make a cut and find out what the internal forces in moments are at that cut that goes through point A. So do that cut, and then draw the free body diagram. If you draw the free body diagram, this is what you get. You've got your external forces and then we have, since this is at when we cut it, it's a completely fixed condition between the two parts. So we have three force reactions RX, RY and RZ, and we have three moment reactions that prohibit rotation MRX, M_RY, and M_RZ. So let's go ahead now and find the forced reactions and do that on your own, come on back, and let's do it together. So to find the force reactions, we're going to add the cut, sum forces vectorally, and set them equal to 0. So I'm going to have R_x in the i direction. We're going to use vector form because that's the easiest way to do it with a 3D problem, plus R_y in the j direction, plus R_z in the k direction, plus now we've got 2,000 Newtons in the i direction. I'm going to leave the units off when you put the units in at the end. Then we have plus 4,000 Newtons, and that's in the k direction. All of that has to equal 0. So I get R_x, if I match components, is equal to minus 2,000 Newtons. R_z is equal to 0, and R_z is equal to minus 4,000 Newtons. So those are the force reactions. Now do the moment reactions on your own and come on back and we'll do them together. For the moment reactions, I'm going to sum moments vectorally. So I've got sum of the moments vectorally equal 0. At this point where I've done the cut, I have my moment reactions. The force reactions at that point are not going to cause any moment because their line of action goes through the point in the center of our engineering member. But we do have M_R_x in the i direction plus the moment reaction in the y direction in the j direction, and plus the moment reaction about the Z-axis in the k direction. Plus then, we have the two moment. If I'm looking at the outer side, if I'm looking at the right-hand side here of my cut, I have the moments due to these two forces. Which we're going to use as R cross F. R is the distance from the point about which we're rotating out to the line of action of the forces. So R will be, in this case, we're going out 800 millimeters in the z direction, or the k direction, and then we're going to go up 200 millimeters in the j direction. We're going to cross that with the forces acting at that point which are 2,000 in the i direction plus 4,000 in the k direction. All that has to equal zero. If you do that math and match components, you get M moment reaction about the x-axis, plus 800,000 in the i direction, plus moment reaction about the y-axis plus 1,600,000 all in the j direction, plus moment reaction about z minus 400,000, all in the k direction equals 0. I match components, I get M_R_x equals minus 800,000 Newton millimeters, M_R_y equals minus 1,600,000 Newton millimeters, and finally M_R_z equals 400,000 Newton millimeters. So here are the total results. Those are my force reactions, and those are my moment reactions. Now with those force reactions and moment reactions, we're going to get various loading conditions at point A. So let's start with looking at torsion. So for torsion, if I take this cut and this is difficult, so you may have to try to take a piece of pool noodle or whatever to look at this, but we're looking in this direction that this cut, we see that the moment reaction in the z direction is 400,000 Newton millimeters. So by the right-hand rule, that's going to be this direction, and so if I draw a cross section looking back out in this direction, I'm going to have my moment. Reaction about the z-axis is shown there in that direction. Here is my point A, right at the top, and we have our axis up is Y. If I'm looking back in this direction, then I'm going to get this axis is X. So let's first calculate the stresses due to torsion out at point A and try to do that on your own and then come on back. We know that the shear stress due to torsion is going to be, if you review back to my Mechanics and Material Part 2 course, the torque times rho which is the radius from the center out to the point we're looking at and that's point A which is at the outside which is going to be 50 millimeters divided by J, where J is the polar moment of inertia. So I've got the torque which in this case for about z-axis is 400,000 Newton millimeters times rho again that's at the outer surface at A. So that's 50 millimeters divided by J. If you go back and look at my Part 2 course in mechanics materials we would calculate J for the circular cross section is pi over 2 times the radius which is 50 millimeters to the fourth. So we find that the shear stress then is equal to 2.037 Newtons per millimeter squared which is the same as megapascals. So I can draw a stress block now. So let's draw a stress block here at point A. Let's put our axis on here first. We have our axis to the right which is Z and our axis up which is X. With the stress block and our coordinates, if we look at this inner edge here of point A, the stress due to torsion will be in the negative X direction. So it's going to be down like this which means it's up over here, this direction here, this direction here. The stress due to torsion is pure shear in this case and the Tau is equal to 2.037 megapascals or Newtons per millimeter squared. Let's now do bending and let's look at our cross section. Again, we're looking back from the cut in this direction to the right. My axis then will be X to the left and Y up. I've got the torsion about the z-axis or the moment about the z-axis, the torque about the z-axis was what caused the torsion at point A, the moment about the X and the y-axis is what's going to potentially cause bending at point A, here again is point A at the top. So I've got my bending moment about the x-axis is in the negative X direction. So it's going to be equal to 800,000 in this direction. The bending about the y-axis is going to be down and it's down in the negative Y direction. It is going to be 1,600,000 Newton millimeters. You can see now for this bending due to this moment about the y-axis, since point A lies on the neutral axis which is right in the center of the cross section, there is not going to be a bending stress or flexural stress due to that torque or moment. So we've got Sigma due to M_RY is equal to 0. However, we can see due to this bending moment or torque about the x-axis, that we are going to get flexural stresses in the Z direction. In this case, it's going to be equal to, by my third course in Mechanics and Materials Part 3, Sigma about the z-axis because it's bending about the x-axis, so it's going to be either coming in or out, we'll talk about whether it's intention or compression in a second. So that's going to be M_RX times Y over I, which is the area moment of inertia. Y is from the center of rotation out to the point of interest which is point A, so that's going to be the radius which is 50 millimeters. So I've got 800,000 Newton millimeters for M_RX, Y is from the neutral axis here out to the point of interest which is up at the top. So that's 50 millimeters, since the diameter is 100 millimeters. We're going to divide that by I, I for the cross section, it's a circular cross section, is pi over 4 times 50 millimeters_4. I guess I left off my units up here, Newtons per millimeter. That's going to give us Newtons per millimeter squared or MPA. So that value is 8.149 MPA, megapascals, or Newtons per millimeter squared. Let's look physically here and see if that's going to be in the Z direction, if that's going to be in tension or compression. So this is the Z direction. You see it's coming from this 4,000 Newton force in the Z direction. That's what caused this bending about the X-direction and so it's going to pull, so you can see physically that at the top of the beam, this is going to be in tension. So this is tension. I can draw a stress block again. Here's my stress block with my coordinates X and I'm looking down on point A, so it's X and Z. So this is X, this is Z, and we see that we're going to get a tensile flexual stress which is equal to 8.149 megapascals or Newtons per millimeter squared and it's going to be on both sides. So that's the stress due to bending. So now we've taken care of torsion, we've taken care of bending, is there any other stresses that are going to be caused by this loading that we need to account for? What you should say is, yes. The next one we're going to look at is transverse shear. Now, realize with transverse shear that we're violating the assumptions that we made in my Mechanics and Material Part 3 course about calculating transverse shear stress formula that we came up with. Because the edges of the cross section we said to use the shear formula must be parallel to the y-axis and so it wouldn't be good for circles or triangles or semi-circles. We also said that there had to be uniform shear stress across the width of the cross section, that's not the case here. But you're going to find that the transverse shear stress contribution to this overall combined loading is quite small and rather negligible. So we'll use the shear stress formula even though it's by rights not exactly what we should apply. So for transverse shear, again, let's look at the cross section. So here is my cross section. Looking back out towards the positive z-axis again at the cut, we have our coordinates Y up and X to the left, and the transverse shear stress will be caused by this 2,000 Newton force that's going to be in the negative X direction. So let's draw that on here. We have 2,000 Newtons and it's in the negative X direction. I can now inappropriately but it's close enough because it's a small effect use the shear stress formula, Tau equals VQ over It. V in this case is the 2,000 Newton shear force, Q will be the first moment of the outer area. Let's look at the outer area here. If I look at my cross section, I'm just looking at the shear stress. For the outer area, we're talking about from point Y here, it'll go out and to the centroid of that outer area if you look at a reference out to the centroid, this distance will be four-thirds r over pi and the area is equal to pi r squared over 2. So if I put those values in, I'm going to have 4r over 3pi which is the Y out to the centroid of the outer area times the outer area itself which is pi r squared over 2. Then, we're going to divide by I, I the area moment of inertia for this circular engineering element will be pi over 4r_4 and finally t is the width. In this case, the width is 2 times the radius or 2 width or thickness, 2r. If you put in those values, you will find out that the Tau due to the transverse shear is pretty small. It's 0.00679 Newtons per millimeter squared. Again, let me write down where r equals 50 millimeters which was the radius. Finally, let's draw our stress block. At A, my x Direction will be in this direction if I'm looking down. On top of point A, my Z will be to the right, and the shear stress caused by this transverse shear force in the y direction 2,000 will cause a shear stress as shown here, and it's equal to 0.00679 Newtons per millimeter squared or megapascals. Okay. Any other stress is due to this combined loading. The last stress that we need to worry about is the axial stress, which we studied way back in mechanics and materials part one. That's going to be due to this 4,000 Newton force pulling in the Z direction. So in this case we've got the axial stress sigma in the Z direction due to this 4,000 Newton force is P over A or R sub Z over A. So sigma sub Z is equal to 4,000 Newtons divided by the cross sectional area pi r squared or pi 50 millimeters squared, which ends up being 0.510 Newtons per millimeter squared or megapascals. Again if we do the cut, we have the 4,000 Newton force pulling to the right. That means it's going to cause tension at point A as far as my axial loading is concerned. So I can draw my stress block one last time, for this last stress, and we've got our coordinates x looking down on the element A at point A, to the right is Z, and we get this tensile axial force which is equal to 0.510 megapascals. All right, so we took care of torsion stress for this loading, we took care of bending stress for this loading, we looked at transfer shear stress due to this loading and we've looked at axial stress. We assume that the elements stayed in the linear elastic region, so we can combine them all by superposition, and here they are. Torsion, we found here looking at our element A, again we're looking down on our element A. So X is in this direction, Z is in this direction. That was our torsion stress. This was our bending stress, because of the loading, the bending flexural stress was tension. This was our transverse shear stress. We used the formula that wasn't quite right for a circular cross-section. However, you can see how small the effect was. Then finally we also had an axial stress due to the 4,000 Newton force pulling in the Z direction, and it also caused the tensile stress. So I can now add all those up together by superposition and this is the result. So here's the result shown again. Remember, we want to find the principal stresses and the maximum shear stress at point A. Again, this is a combination of everything we've been doing in all of our mechanics and materials course. For all of this external loading with the different contributions, this was the overall stress that we're seeing at this point A. But that's not the principle of maximum shear stress. To draw or to find the principal maximum, principle stress in the maximum shear stress, the easiest way will be to go ahead and use Mohr's circle. Let's go ahead and plot. First divide the horizontal, and this is our horizontal by the Mohr's circle. Clockwise is positive so it's going to be plus 2.044 for the tow direction. So this is no normal stress, and 2.044 for shear stress, that's the horizontal face. For the vertical face, we're going to have positive 8.659 for the normal stress and since this is counterclockwise, it's going to be negative 2.044 for the shear stress. So we're going to go over here 8.659 down to minus 8.659 and minus 2.044. We have two points. I can go ahead and now draw a line between them. That's our diameter of our Mohr's circle. I'm not a great circle drawer here but we'll go ahead and draw Mohr's circle. When we do that, we can find the radius, the center. Let's start by finding the center. That's going to be 8.659 divided by two or 4.33 and zero. Which means my radius now is equal to the square root of 4.33 squared plus 2.044 squared, while my radius of my Mohr's circle equals 4.788. So that means, Tau max is going to be up here, and it's going to be equal to the radius which is 4.79 rounding to three significant digits, Newtons per millimeter squared or 4.79 megapascals. So that's one of our answers. Then finally, sigma one, will be our principal stress out here, and it's equal to the value of sigma at the center, which is 4.33 plus the radius which is 4.79 or a tensile stress, 9.12 megapascals tension for our first principal stress. For our second principal stress, it'll be the center minus the radius. So I've got sigma sub two equals 4.33 minus 4.79 or minus 0.46 which means it's 0.46 megapascals in compression. Okay. In fact I can double underline these and these are the answers. So as a recap, we have a very complex combined loading. We've had stresses at point a caused by torsion, bending, transfer shear and axial. We combined all those together, we found the state of stress at point A, and then we used Mohr's circle to find the principal stresses and the maximum shear stress at point A. So that's about as comprehensive as you can get as far as mechanics and materials are concerned. So long module, lots of stuff, probably have to stop and start a few times to get your head wrapped around it. But if you can understand this, you've got a really good handle on mechanics and materials.