As I said today, we will review the materials of ME 218 and ME 219. And rather than I going through something that I think you should hear, I would like you to tell me what you would like to be reviewed, what is it that’s mirky, what was it that you didn’t quite understand? Obviously in, what, about less than an hour and a half, we can’t go over the two courses in very much detail, but we can pick a few subjects 2, 3, 4 maybe at maximum 4, in the previous class I did three. So we can pick a subject like that and then review those subjects, but I want it to be the subjects that you want reviewed rather than those that I would like to review. So tell me what subjects you want reviewed. Finding stresses from strain values. In other words the stress-strain diagram, is that what you’re talking about? Very simple, we’ll do that. Okay, what else? Mohr’s circle. Mohr’s circle, okay let’s write that, that’s always one of the choices. What else? So if you don’t want any other subject reviewed, if I give you the quiz right now on combined loading, it’s okay, right? I doubt that. So again, don’t be bashful, you have to answer my question, you don’t have to really, I mean you can just sit there and look at me, but remember if you don’t ask any questions, I would, I don’t know what you don’t know, you have to tell me what you don’t know or what you’ve got questions on and we’ll answer those questions. So, anybody else want to be brave enough to say something? Combined loading. I’m sorry. Combined loading. Combined loading, okay. Anything else? Castigliano’s theorem. Castigliano’s theorem we will do in this class, independent of what you studied. In fact, the emphasis will be on Castigliano’s theorem. Anything else? , okay. Okay. Let’s take a look at these and then if we have more time you can come up with other subjects as well. Take these one by, actually I’ll answer your question first because that’s simple enough. If you’re asking how we find stresses from strains, and if I have misunderstood your question, please let me know. That’s a stress-strain diagram for a ductile material, a part of that diagram is linear and elastic. In other words, when you load the material in this section to some value, there will be some strain associated with it, and if you unload it, the unloading will be exactly on that very same line. So that when you completely unload the material, there will be no effect on the material properties. It is as though you never loaded the material, it’s as good as it was before. If you go beyond a point. Actually there are three points, one is called proportional limit, the next one is called elastic limit, and then we define this third point as the yield point and it is in a stress-strain diagram, such as the one that I have on the board, it is conventional, it’s arbitrary, well not arbitrary, but conventional. In that we say that the yield strength of the material is a stress for which the strain is two-tenths of a percent, 0.002, it’s called two-tenths of a percent offset yield strength, and that’s what we use. The stress-strain diagram and the material may look like this as well. If you’re in this, you’re just past that point, you have definitely passed the 0.002 mark. So we say this is the yield point of the material, and there’s a little upper yield point and lower yield point. Those of you who have taken materials, ME 315, probably saw that in your studies. If this, if the stress or strain is in the linear elastic region, then you can use this equation where E is the slope of that line. That’s where stress and strain are proportional. So if you have strain, then use that equation and find stress and vice versa. Alternatively, if you know the value of the strain, you can put it in here, come up and rate the value of stress. Either of the two will give you the answer. This one probably with some approximation because you’re going to have to read a curve. That one, I don’t want to say it’s more accurate, but you’ll get a number that matches the equation. If, however, the strain is not in the elastic region, you have strain the material past the elastic limit and the yield point and you’re out here somewhere, that equation is no longer valid, you cannot use that equation any more. The only way to do it if you have the stress-strain diagram is go up here, come in here, and read the value of, let’s say, sigma one epsilon one. Is that your question? Are you sure? Okay, alright, but that’s about it, just make sure you don’t use that equation in an inelastic region. You have to use the graph or sometimes you find actual equations for that part, then you can use that equation, of course it will not be linear, it will be a nonlinear equation, but you can use that equation, okay? So Mohr’s circle. Where does Mohr’s circle come from? The stress transformation. Exactly, exactly. Stress transformation. What do we mean by stress transformation? Transform stress into what? Along different, a different plane. That’s correct, yes, that’s correct. So let me, and both answers are very correct, very good. Let me show you something, you’re familiar with this. Regular stress element, right? You’ve had that in 218 and 219 and all of that, and that’s a starting point for drawing Mohr’s circle. These are stresses, where? At the point you solve for. Point, that’s, at the point where you want to solve the problem, whatever critical point if you will. These are stresses at one point in the material, this is not an overall stress, one point in the material. So this square here shows you a point, for example, similar to this. Let’s say, you have all of these forces acting on this object whatever it is, we picked this point and we represent that point by this little square. The stresses on the sides of this little square are shown here, these dimensions are very, very small, this is a point. So apparently having this, we’ve taken our X and Y axes like this, agreed? Horizontal X direction, vertical Y direction. But what if instead of drawing a square like that, at that very same point I draw the square like this, I don’t want to draw it right on top of that point, but this is for the same point. What if instead of that, I draw a square like that? And here are my axes, X prime, Y prime. That’s at this, that very same point. I’ll give you a simple example that you’re all familiar with. If this is 10 KSI stress, and I make a cut here and I take a look at this cut section, what’s the stress here? 10 KSI? Because assuming that the dimensions have not changed, I know I didn’t draw it to the same size, but assuming that the dimensions have not changed, it’s the same force being transmitted from here to here. And that creates the same stress if the areas are all equal, F over A. But what if rather than making this cut, I make that cut. Now I’m looking at this. I have these two stresses, this is my X prime and Y prime direction that I have shown there. This stress, the normal stress, here call it sigma prime, tau prime. The normal stress here is no longer equal to 10 KSI, it’s something else. And as I change this angle of cut, this angle theta , that stress changes. Therefore the normal stress and the shearing stress are functions of this angle. If we can go from this set of stresses to that set of stresses, in other words something like this. Sigma X prime, sigma Y prime, and tau XY prime, and the same on these sides of course, if I can come up with a set of equations that will take me from this set of stresses to that set of stresses, I will have found the transformation equations for stresses. So these are all stresses at the same point at various angles, and the transformation from one set to the other set is called stress transformation. How do you that? Well, pretty simple, all you do is you take that element and you say, what is this angle here, theta with the vertical, so there’s the angle theta. And we take a look at this little triangle. Have sigma X, tau XY, sigma Y, sigma X prime, tau XY prime. This is all at one point. Once we look at it like that in these direction, once we look at the like that in those direction, but it’s all at the same point. To get to the transformation equations, all you have to do is use equilibrium equations. This tiny body here, which is at a point, must be in equilibrium. And remember that equilibrium means that the summation of forces are zero, not stresses. Don’t add stresses, you must add forces. Meaning that if you want to use equilibrium equations you have to multiply each stress by its respective area on which it acts. So for example, to find the force that acts on this area, you multiply sigma X by this area, remember that this is an area into the board. So you say, well what are those areas? Well, it really doesn’t matter that much, I can call this area A, then that area would be A cosine theta, and that area would be A sine theta. The thickness is the same in all cases. You multiply them by their respective areas, write an equilibrium equation in the X prime direction, write an equilibrium equation in the Y prime direction, some of the forces in the X prime and Y prime, in each of those two equations there will be one unknown. In the summation of forces in the X prime direction, you’ll have sigma X prime as the only unknown and that unknown stress will have no component in that direction. Summation of forces in the Y prime direction, we’ll only have that as the unknown and this will not have a component in that direction. So when you write these equilibrium equations and you square them and add them together and simplify them trigonometrically, what you will find is the equation of a circle. That circle is called Mohr’s circle, and all that it does, it represents the results of these equations, called transformation equations, graphically and makes their solution of the problems a little bit simpler. Rather than using those equations, we use Mohr’s circle. So, any questions so far? Please do ask your questions if you do have them, don’t let them go unanswered. So if we have a set of stresses like that and we want to find these, this is what we do. Take a set of axes, sigma and tau, and we plot the stresses at this point on the vertical plane and on the horizontal plane. On the vertical plane there’s a normal stress sigma X and a shear stress tau XY. Normal stresses are positive if their tension, they pull on the material, a negative if they’re compression. Shear stresses, and remember once again this is all at one point. So there has to be only one shear stress at this point, there can’t be very many, two or three, just one shear stress, but for the construction of Mohr’s circle we have a special convention whereby we think of these two shear stresses as two forces and say if they produce, these are both on vertical planes, remember. If they produce about the center of this a counterclockwise moment, then they’re negative, if they produce a clockwise moment, then they’re positive. By that convention this is a negative shear stress and that’s a positive shear stress. And that really doesn’t make sense because at one point you should only have one shear stress, positive or negative, period. But in order to draw Mohr’s circle we use this convention and it works. The sine of shear stresses as well as normal stresses, I’ll talk about that probably next week sometime, there is a way to find the sine of shearing and normal stresses using the same approach. And you don’t have to worry about clockwise and counterclockwise, and if you do use that approach, both of these shearing stresses will be either positive or negative as the case may be. But for Mohr’s circle, we use this convention, okay. So let’s just give these some numbers. Say this is equal to 4 KSI, that’s equal to 3 KSI, and my shear stress is equal to two KSI. Let me erase the top elements of these will not be so clogged, and this too. So let’s draw Mohr’s circle for that, we’ll draw Mohr’s circle and take a break. On the vertical plane we have a normal stress of 4KSI, and a shearing stress, and according to our convention, it’s negative counterclockwise. Oh sorry, negative. This point, which is a point on Mohr’s circle, represents the stresses on the vertical plane, It is sigma X, yes, but sigma X is acting on a vertical plane, on a plane parallel to the y-axis. So we’ll call this the vertical plane. Similarly, on the horizontal plane, we have 3 KSI and 2 KSI. This is going to be tight, but that’s okay. So 3 KSI, not positive 2, so 1, 2. Stresses at these points, 3 and 2, they represent the horizontal plane. These two points are the endpoints of a diameter of Mohr’s circle So if I connect these two and if I draw a circle like that, that’s Mohr’s circle, just like these two points represent the stressors on the horizontal and vertical planes, all other points on Mohr’s circle represent the stresses on all other planes passing through that point. And therefore if we know the orientation of those arbitrary planes, we can find the stresses on them by using Mohr’s circle. Let’s, let me do the maximum and minimum here and then we’ll take a break. You can see that if I calculate the radius of Mohr’s circle, and it’s not very difficult to calculate, it’s the hypotenuse of this little right triangle. The distance from the center of Mohr’s circle to the tau axis is 3.5 KSI. That stress is for this plane, correct? Correct, I’ll get to that, I’ll get to that. This is plane stress. Or I mean – You mean three-dimensional? Yes we’ll, I’ll talk about that today. I’ll extend this Mohr’s circle to three dimensions, that’s why I have parentheses here, these parentheses are for in-plane, now that you asked the question, we’ll put them in. We have stresses on other plans as well. The sigma max and sigma mean are called principal stresses. So what is the definition of principal stress? We have no shear stress. They act on planes of no shear. Not that they should be either maximum or minimum. Although that’s partially true, maximum and minimum stresses are included in the principal stresses because as you can see, the shear stress here is zero. The shear stress there is also zero, but look at this three-dimensionally, this is our sigma X, tau XY, and sigma Y, right? What’s the shear stress on this plane, the plane that I have put a circle in? At that point? This is one point, remember this, this whole cube is one point. I thought you meant a principal, at the angle of the principal stress. No I’m not looking at the angle of the principal, not yet anyway, not looking at that. What I’m looking at are the principal stresses themselves, what are they? They’re stresses that occur on planes of zero shear, correct? Okay. Then my question to you is, and they’re oriented at some angle here. So if I were to draw them on this, I’m not drawing right on it, but probably they would look like that. This on top of that, you see like so. Those would be the planes on which the principal stresses will act, but this is that, this is 4, that’s 2, that’s three, and so on. What’s the shear stress on this plane? Zero. Zero, you don’t see any load? What’s the normal stress at that point, on that plane? Also zero. Also zero, you don’t see any normal loads either, you don’t see any loads perpendicular to that plane, you don’t see any loads parallel to that plane, there is no normal stress on that plane, there is no shear stress on that plane, therefore zero is a principal stress because it acts on a plane of no shear. So coming back here, that is our real minimum principal stress, not that. And if you take a look at this cube, this element, three-dimensional element, from the three sides, for each side you get a Mohr’s circle. Before I draw those, one more thing. the maximum shear stress, I’ll rewrite the values for these after we come back from the break, but the maximum shear stress is that, isn’t it? , in-plane shear stress. So if I have these principal stresses I will have the ends of Mohr’s circles, these two have given me one, these two would give me another one. And these two will give me another one. And the radius of that large green circle is the real maximum shear stress, not the in-plane one. So tau max is the radius of this big circle. Tau max therefore is sigma max minus sigma mean, not in-plane, just sigma mean, sigma max minus sigma mean divided by two. The radius of the largest Mohr’s circle. Okay, any questions? Okay, let’s take a break. Oh, way past the time. Now, any questions so far? The location of the principal stress, meaning the angle at which the plane of the principal stress acts, can be found if we find either of these two angles, either this one or that one, of course they’re both exactly the same. So how does that tell us what the principal stress is? Let’s assume, and I’m not going to go through finding that angle, that’s simple enough in a right triangle, you can do that. Let’s say that angle is, I don’t want to call it theta because I called that one theta, let’s call it alpha. Then, this point representing the vertical plane or this radius representing the vertical plane stresses, if it is rotated to this angle, call it 2 alpha, you recall that the angles on the element are doubled on the Mohr’s circle, therefore I’m calling that 2 alpha. If I rotate the vertical plane radius counterclockwise 2 alpha, I will get to the plane on which the maximum principal stress acts. So if I am to draw that here, and notice that I took the vertical plane counterclockwise 2 alpha. Take the vertical plane counterclockwise alpha. Counterclockwise alpha means like this. This is the plane on which the maximum principal stress act, notice that the maximum principal stress and the minimum in-plane principal stress are 180 degrees apart in the Mohr’s circle, therefore they’ll be 90 degrees apart in the actual element. In other words, if you were to draw an element, it would look like that. Sometimes they include the maximum shear stress in the sketch as well because the maximum shear stress is located 90 degrees from either one of these, that’s the in-plane maximum shear stress, 90 degrees from each one of these, so 45 degrees. So if I want the maximum shear stress to be shown here as well 45, 45, sigma max, sigma mean, in-plane, and the maximum shear stress. So the maximum shear stress depends on whether you rotate clockwise or counterclockwise and come up with the direction, and there’s a normal stress which acts on that plane of maximum shear stress whose value is this, in our case, 3 and a half. The magnitude of the principal stresses can be found, again, by looking at that in-plane Mohr’s circle. For example, sigma max is 3 and a half plus our radius. Sigma mean is 3 and a half minus a radius. So sigma max, in our case, and of course the maximum in-plane shear stress is just equal to the radius. Okay, any questions? Whoever asked for Mohr’s circle, I forgot. Have I answered all of the questions or at least the vast majority that there may have been? If not, if there’s something regarding Mohr’s circle that is still not clear, tell me. Nothing comes to mind, huh? If it comes mind later, ask later, not a big deal. Next subject, combined loading. Whoever asked this question, do you have anything in particular in mind? A problem, for example, in combined loading. Who asked the question? You did. Do you have anything in mind? No Nothing in mind. Okay, no big deal. Let me then give you a problem. We did this in the previous section as well. Say you have a shaft like this and you have pulleys. You have to live with my artwork, it’s not the best the world for sure. But starting next time I’ll be using the projector and things will look much better. Okay. Let’s say it’s simply supported here and here, and you have loads on the two sides of these pulleys, just give them some numbers, let’s say. And let us assume that the radius of this pulley is equal to 6 inches and the radius of that pulley equals 4 inches. So that’s 200 times 4 is 800, let’s make it 8 here so that I can do this. Forces in pounds. And distances, I’ll put them in here like this, 2 feet, 3 feet, 2 feet. Just to have taken some number, is that clear? Especially the last part where I’m showing the length along the member. This is the location of the two pulleys, those are the distances in-between them. One of the things that you have to be sure of, and if given in a problem for example something like this, and this load is not given. F, the question is, the first question is what’s F? Even though this is a shaft and it rotates, it must be in equilibrium. You do know that equilibrium does not mean the state of the material being stationary, not moving that is not what it means, it can be but it doesn’t have to be. Any motion has to be with constant speed, whether rotational or linear. So in this case, if it’s constant speed then that means that the torque that’s being applied on the two sides by these two pulleys must be equal. So if these are my axes, and if I call this pulley one and that pulley two, the forces on pulley one create a torque equal to 200 times 2 or 4 inches, I say radius 4 inches, that’s okay. 200 times 4 inches, that’s 800 inch pounds. And if I were to write this in vectorial notation, what would be the unit vector? Minus I. Minus I, exactly. Does everybody understand? Okay, good. Pound inch and pounds all the same, as a, as a vector. The torque generated by pulley number 2, by necessity, has to be 800 pound inch, otherwise there will be no equilibrium. The two must be equal. So, if only this force is given, you can find that other force. So this is 200 times 8, 1600, you have to subtract 8 from it, so that must be 100. Does that make sense? Is that okay with everybody? If you take a look at that, it will also give you 200, or sorry, 800 pound inch, except that is in the positive X direction.. Positive I. Okay, and let us assume that the question is this. What is the location of the maximum stress in this shaft and what are the stresses at that location? By location, we mean cross-section and a point which is usually on the outside of the shaft because that’s where both the shear and the normal stresses are generally maximum. That’s not the case if there’s stress concentration of course, but not in this case. So if that is the case, what part of the shaft should I be looking at in order to find the location of the maximum stress or stresses? First, what about the shear stress? Where does it act? Maximum shear stress, I’m not talking Mohr’s circle, just talking your regular old shear stress from the TC over J equation. Where does is it act? The center . Remember the stress distribution? At the outside of the shaft, but we’re in the shaft here, here, or here, or any other point in between or section in between. Where does the maximum shear stress occur? I hope you’re just bashful, because that you should be able to answer. What’s the torque here in this section? Zero, there is no torque. That section, zero. In between. So the maximum shear stress occurs from here to here. So it occurs on the left-hand side, immediately on the left-hand side of pulley 2 or immediately on the right-hand side of pulley 1 and every place in between those two. Okay. However, the shear stress is not the only stress that acts on this shaft, there is also a bending moment which creates a normal stress. In order to be able to answer them, the question, what is most critical section of the shaft? Not only do we need the maximum shear stress location, we also need the maximum normal stress location. To do that, you have to draw shear moment diagram. So this would be a good exercise in that as well. We draw the shaft like this. And then we have 400 pounds here and 300 pounds there and this is 2 feet, 3 feet, 2 feet, okay. You can draw shear moment diagrams, which is really not that big of a deal, you just, for the shear diagram, follow the load. Go wherever the load goes. First of course, you have to find, let’s see, let’s see, let’s see, that’s 4, so, ah good, good, good. It’ll work for one, but not the other. Want to change those numbers, I don’t remember the numbers from the previous section. I want to change those numbers so that we get round numbers for the reactions. I make that 4, then I get 3 and 1 and if I make that 4, I may get 3 and 1, it may turn out to be okay, let’s see. I don’t want to write equilibrium equations just yet. So I’ll show you where I get these numbers. That’s 6, 2, so one-third and two-thirds, so this is equal to two-thirds of 400, and that’s equal to one-third of 400. Then this will be equal to two-thirds of 300, one-third of 300, so you add those two, those should be the reactions. So it’s 100 plus 800 over 3. And this is equal to 400 over 3 plus 200 pounds, we’ll figure out what they are. Now that we’re on the subject, let me, let me show you how you get that. If you have a simply supported beam, which that is, if this is a force F, this is A, and that’s B. Yes you can write equilibrium equations and find the reactions. But if you remember this, the easy way is to say that this is divided between these two supports in the ratio of these two distances, and it’s clearly, the reaction due to that force is larger at the left side than it is on the right side. Therefore, and this will be AF divided by A plus B, that’s what I’ve used to come up with that. So I looked at each one of these like that, and for each, I wrote its own ratio, then I added them up. But again, if you’re more comfortable with equilibrium equation, that’s fine. Furthermore, the maximum moment which occurs here, this is the moment diagram. Maximum moment is that. All you got to do is cut a section here and the moment here just next to F is equal to A times that, or on the other side, B times that. So those are your moments, there’s maximum moment and also the reactions for a simple support. This is something good to have in the back of your mind, saves you a lot of time. So what is this? This is 800 over 3, so it is about 267 approximately, plus 1367. And then from here to here nothing. So we drop down to 33 and then nothing from here to here, 1 following the load and 333 and the sum of these better be 333. It is, right? Sir. Oh sorry, come again please. Isn’t the distance 6 over 8 when you do that shortcut? No no no no no no no no. Oh what? Let me, let’s see, let’s see, it’s two-eighths, yeah, you’re right. I divided it in the ratio of these two. So here’s what I did, this has to be divided into these two parts, right? Okay. This one, this one is A over B times that one, ratio of those two, A and B, correct? So coming here, the ratio is 2 to 6, 1 to 3, okay, 2 to 6. And then you do the moment diagram. Which now it’s actually not difficult to calculate. How do I go from this point to whatever, this area? The change in the, actually let me write these for you here, I’ll keep that there. Change in moment equals area under shear diagram, right? That’s from the shear moment diagram. So in the we established that there’s shear force due to the torque, right? Correct. But there’s also shear force due to just. Yes there, yes there is, yes there is. Our shear diagram doesn’t care about the torque shear, right? They’re two independent parts that produce shear stresses, both of them produce shear stresses. Torque may not be there, but the shear stress from the bending diagram is there. On the other hand, that one may not be there, but the torque is there so you get, but in general, you’re right. Correct. The both of them are there so you, they’re either additive or subtractive depending on whether they both act in the same direction or they act in opposite directions. There may be additive or subtract, that’s number 1. Number 2, the transfer shear, which I have down there, what is the value of the transfer shear at the surface? Zero. Zero. Maximum at the neutral axis, right? What’s the maximum value of the shear stress due to torque? Where does it occur? Surface, zero at the neutral axis. Yes there are these two shear stresses and in some locations they add to one another, but at the point of interest that we have, on the surface of this, only the torque has a value. The transfer shear is zero. Remember it looks like this, the variation of the shear stress, I have to continue writing that. If this is your cross-sectional area, that’s the transfer shear, agreed? So it’s zero on the surface. So we get that torque. Okay, here we have 4, 3, 7 this is M in pound foot. And then we subtract from that 300 times 4, that’s 1200, something is not right. It may be what you were talking about, the 8. Did I do that right? Yeah that’s fine. 433, 734. Alright, let’s check the reactions. I believe what you said is correct. Is 2, it’s not two-thirds. This one is 2 over 8 times 400, you were right after all, and from the other one, 6 times 300 over 8. A divided by A plus B and B, yeah. In reducing the ratios, I made a mistake. So let’s check this one more time, so that’s 800, that 100. And what is that? Three-quarters, that’s 900 over 4. And 900 over 4 is 225. Add those two, and rather than 367, we will have 325. Okay 325, that still won’t give us the right answer. Isn’t it 16 times 400 and ? 8 times 2 divided by 8, or sorry, 400 times 2 divided by 8? It’s like in reverse, isn’t it 6 over 8, 6 times 400 over 8 and. 6 times 400, that’s right, okay. One more time. We get 2, 3, 300, right? Is that right? Yeah, that’s correct. So 300 and then this one. I know. I switched those two. 6 times 400 divided by 8 is correct for that one. 2 times 300 over 8 is good for this one. 600 over 8. So what do you get if you add those two? Calculator? Anyone? Divided by 4 is 150, 75. Check that. Correct? So 375. On this side, 400, 2 over 8. And 300, 6 over 8. And what do we get here? That’s 50, that’s 100, that’s 100 plus 1800 divided by 8 is what? Quickly, 2, that’s 900 divided 4, so 225. Is that correct? Okay, now hopefully things will come out right. So 375, 25, 325. And the maximum moment here is this times two, which will give us 0570, and the maximum moment here is 325 times 2650. Does everybody know where I got those numbers from? This times that, that times that. This times that gives me 325 times 2. 375 times 2 gives me 750. So how do I go between these two points? Straight line or a curved line? Straight. Why? Because the shear force is constant. Because the shear force is constant so you go here, we go there, and then we go there. Like that. So continuing on this, change in shear equals area under load diagram. Slope of moment diagram equals value of shear force and slope of shear diagram equals value of load. So the slope of the moment, the slope of the shear diagram here is zero because there is no load here. This is free of load. Slope of the moment diagram is constant because shear is constant. So now that you can see, this is our maximum moment, we said, the maximum shear stress occurred between here and here. Since the maximum moment occurs here, then we should be looking at this section right there on the right-hand side of pulley number 1. And sigma max will be equal to, in this case, 750 times C divided by I. C and I, I don’t have because I didn’t give you the diameter of the shaft. Once you have these two and tau of course, we can calculate that. Tau max, that means the tau on the surface is equal to 800 C over J. Once you have these two, you’re back to Mohr’s circle with your sigma X, sigma Y, and tau XY. Except that in this case, there is no sigma Y or sigma Z. Whenever you have bending of shafts or bending in general, it doesn’t have to be a shaft, it could be a square cross-section, a rectangular, I-beam. If you have bending, the only stress is along the axis, normal stress is along the axis of the shaft in bending. It doesn’t matter whether you bend it like this or like that. Whether you bend it like this or you bend it like that, normal stress is always along the axis of the shaft. They will always be along the axis of the shaft. So, when you have shafts like this, and this is your XYZ, your only normal stress is in the X direction along the axis of the shaft. There is no sigma Y, there is no sigma Z, so don’t come up with those. Your Mohr’s circle in this case will look like this, oh sorry, your element will be one normal stress and one shear stress, and this is sigma X in our case. Any questions? Now what if rather than these loads being like that, they were like this. Still simply supported, and you have a load here like so and a load here like so. The two loads are not in the same plane. One of them and is in the X direction and the other one is in the, sorry, one of them is in the Y direction, the other one is in the Z direction. So this is your x-axis, that’s the load in the Y direction, load in the Z direction. You guys did that in 219, right? Yes? Sure you did. So what you do in this case is draw two moment diagrams, one for each of these, and you end up with something like this. Say moment in the Y direction, our moment in the X direction of course is torque. Moment in the Z direction. The maximum moment occurs either here or here. And the way you find it, come down here, call this MY prime, and then you combine, you say and MY max, square root of MY squared plus MY prime squared. Same thing here, MZ prime, MZ max equals this, I shouldn’t actually call them, I guess it’s okay, I’ll call them MZ max, equal the square root of MZ squared plus MZ prime squared. You compare these two, whichever is larger, that’s your maximum moment, and wherever that occurs, that’s where the maximum normal stress occurs. That’s what you use in this equation in that case, rather than 750, whatever that comes out to be. That ring a bell, of course, from 219. Any questions? I have three minutes, let me talk very quickly about transfer shear. And I forgo the preliminaries and got to do this. Oh I had a picture drawn of transfers shear, I guess it’s not here. This is your cross-section. Shear stress will occur in beams loaded like that. I have various sections, wherever you have a shear force, there will be a shear stress. The shear stress acts both on a vertical plane and on a horizontal plane because shear stresses must occur in pairs and on two perpendicular planes. The magnitude of the distribution of the shear stresses like this. Its maximum at the neutral axis. Its value is found using this equation. V is a shear force, I is the moment of inertia of the cross-section about the neutral axis, T is the thickness of the cross-section wherever you want to find the stress. Now if the thickness varies here, of course this shape will change too. Students have most problem with the value of Q. This is first moment of area from the point of interest to the outer fiber, any outer fiber, doesn’t have to be the closest, about the neutral axis. So for example, if you want to find this shear stress, this tau, Q will be, we’ll call that A times that distance, 1. Moment of the area from the point of interest to the outer fiber, either this outer fiber or this outer fiber, it doesn’t matter, you can go up and down, exactly the same thing, about the neutral axis. Therefore in this case, Q is equal to AY, whatever the value of Y is, which you can find just using simple geometry. Okay? Any questions? I think we’re done. I’ll see you on Tuesday and we will have a quiz on the review stuff.

ifeanyi nwachukuwPost authorI can see the board too much light

Avinash ChaudharyPost authorGREAT LECTURE SIR. FINALLY MY CONCEPT GOT CLEARED BY THIS

Lemuel MajuruPost authorI am following…..from Zimbabwe. Great Lecture indeed

V SharmaPost authorI wish this professor was teaching me back in university days !

anurag7058Post authorWhich book is followed in this lecture series ?

qualifying 2Post author26:22 where did the 3.5 come from?

Abdul QadirPost authorwonderful lectures i am extremely thankful to u sir,,sir i would to request u to , please we want to learn strength of material from u so gifts many more thanks once again

HHemidaPost authorcould you please rise the handout of this course? thanks in advance

vivek koulPost authorOf which university are these lectures?

Lindo LoloPost authorGreat lecture …following from South Africa

Joel VarghesePost authorwhy is the video so bright?

mustafa hilal ateea al zamely al zamelyPost authorNot clear video