 All right, so we're working, that's a good start. This is just the type of thing where we might see a stressed element here. So if we imagine a kind of crank arm thing, yeah, that'll do, such that there's some force on it in the Z direction out there at the end of the crank arm. Now maybe this is part of a crankshaft, and that's a piston force that'll be feeling there. Who knows? All kinds of ways we could possibly find something like this. So that load there is 150 pounds. It's at the end of an arm that's 18 inches. This part of the arm is 14 inches. And there's a point right here that is an instance four inches up from the bottom and in the front right there. That also matters as hopefully we've seen. So four inches out on the front facing the Z axis direction is a point where we want to see what kind of stresses are on that element. The force on the cross section there what I'll do is blow that piece up. Remember a lot of times what we do is actually take an imaginary cut through the material, which is up there. That's the Z axis, that's the X axis, the Y axis. We have a little element right there. We want to see what the stresses are on that. We want to determine the principal planes and the principal stresses. Remember that's what we're looking at Friday. This idea that this XY direction, if that's the only direction we look at, we only look at the normal stresses and the shear stresses there, then we're going to miss some of the directions where there are actually more greater shear stresses, greater normal stresses than there would be normally. So we have to be careful in how we look at this. All right. There should be not a great bit of question here what kind of internal forces we're going to see there due to this load. This 150 Z direction force is also going to be seen as a 150 pound shear across the face there. The fact that this 150 pounds is down 18 inches is going to cause this whole thing to twist in that direction. It's going to cause a torsion. We draw that as a vector using the right hand rule. And then the fact that this 150 pounds is a certain amount above our point of interest is going to also cause a moment. Let's see, if we move this to here and then include a moment there, then the right hand rule would put it something like that. So if it would help a moment in two ways, it's got a 350. Does that seem about right? Is that kind of what you'd see with it as well? So if we get a good direction on this thing, which direction did I use? Yeah. Now look at this element. Remember we've been looking at the element, looking down the Z axis. So we have this going to look like that. That's a little view at this element, looking down the Z axis, right at that element. We want to see what are the normal and shear stresses. And then we want to look and see what do we find when we go to the principal plane direction. Remember those are the directions where we see greater shear, greater stress than before. So let's see. We'll look it down here. What x-direction stress, normal stress, are we going to see? Remember normal stress is just simply whatever load there is in that direction divided by whatever area is carrying that load. So what would that stress be? What load is there in the x-direction? One would sound this type of load though that I'm going to put in here. There isn't that. That makes sure what the bending in the light is, but those are all in the other direction. So that happens to be zero. In the y-direction, well in the y-direction, that is where we have the moment. So the way this thing is loaded with that direction moment, we know that an element on the front here is going to be in tension. And an element at the back is going to be in compression because this arm is going to tend to bend like that. The upright part of the arm is going to bend back in the z-direction, which is going to cause tension along the front here, compression along the back there. So we do expect to see tension like that at that front element. The way that's twisting, we can see it's going to... You can even imagine we inscribe a little square there what would happen to it because of that torsion. It's going to drag the top that way. The bottom, imagine it's stuck to the piece so it would stay there. And then so we expect some kind of shear stress, something like that, to find out what those three pieces are. And then we can... This is the type of thing we've been doing for a couple of weeks, so now we put it together with the stuff we did Friday looking what the principal planes are, principal strains and the like. All right, so let's see. We're going to help speed you up a little bit. Let's make sure. This moment is the 150 times the 10-inch moment arm. Yeah, 15 inches. C is, remember, the maximum distance. Oh, I didn't give you a diameter of that thing. 1.2-inch diameter all the way. And what is C? Half of that, the radius, 0.6. Remember what I is for a circular piece. And so we have all those little pieces. All right, everybody agree with each of those pieces? 150 out of this 18-inch moment arm. J is, what, 1.5. Got those pieces yet? 0.8, 0.84. KSI? Does everybody agree that these directions are okay? The easiest way to see is this torsion. You know, it's going to drag the top across there relative to the bottom, so that element's going to be twisted like that so we can fairly easily place these two and then those two just got to balance that because we have the sum of the moments and the forces as well. So is this as drawn seems like the orientations, right? Is that a plus or a minus? Because remember, we need that now for calculating the principal stresses. That's positive. All right, so we can now do a little bit with the rest of it. A couple things now. Remember that we need, now that we've got those, it's useful because it repeats itself in those equations. It's also something we need for the more circle if you want to draw them, which we will. The two together are a big help, I think. Average is the center of the circle. The square root of sigma dip squared plus tau xy squared is the radius of the circle. A lot of the things we need, too. In the direction of the principal plane angles, but it's really easy to check those on the more circle, though it's all distorted because the camera's up on the ceiling. That's not too bad. Better, Jay. And if you want to use a word or some drafting program or something to draw a more circle, it's easy to just draw the circle, figure out what some of the values are, and then put it in the axes and some of the points. Which is nice for a better drawing. It's really hard to start with the points and then draw the circle when you're just trying to get an idea and picture what's going on. All right, so where's the center of the circle? The sigma average? 8.84 divided by 2. So that's at 0.42. And sigma dip, let me write it down. So r, what do you have for r? Right. Now we can start sketching in some of the values and this is 4.42. Remember the x-axis is any normal stress value we have. And now the radius then 9.1 plus the 4.42 gives us 13.52 as the maximum stress and the minimum stress you get is 4.68. So now we know our y-axis. A little bit different than we draw things normally if you would put down the axes but it's just so much easier to draw the circle first and then put in the points you know. How did you know that was a good fist at the point of time? Sorry? How did you know that was a good fist at the point of time? Because there's 4.42 and then it had to be about halfway in between these two. Oh right, it's just the zero point. Yeah. Simple as that. No magic to that. And remember we can plot one point to help us determine when you do this tangent two feet you don't necessarily with the calculator but remember you do get two angles because the tangent repeats itself. So you should get a value for that but then you can check it on more circle if you indeed have the correct one. So one of the two feet come out to be the angle minus 61? Minus 61. We can check that by actually plotting the point sigma x tau xy. So sigma x is zero. Tau xy is 7.96. Well that's got to put us right about here and that's the point, the book called point A from there to the x axis and that looks like it's about 60 degrees or so and it's in our minus direction. That helps us determine everything we need to know about theta B. What else we got? Oh we had R. Remember R now is the maximum shear stress. Oh remember there was also a theta remember that's the angle where we'll see no shear stress maximum and minimum. What was sigma y? The angle of minus 30 degrees P minus 30 degrees, something like that and remember what we see in that direction what do we see in this theta P direction? That's where we see the sigma max which is 13.52. Sigma min we're actually right on this axis which is much smaller and negative so we know it's compression no shear stress since we're right on the x axis. If you remember is just 45 degrees from theta P or on this diagram it's 90 degrees so it's the opposite of the yield on the spoke so that comes out to be what? 15 degrees somewhere real close to 15 degrees what do we see in that element direction? That's where the shear stress is maximum and the normal stress is the same on each phase and is equal to sigma average which we know is positive so we can draw out a tension here to draw the elements that way then you've got all three you want a nice little drawing you see what the angles are everything tends to match up a little bit this works better for me if you see it some other way I know it's wet what's that? Remember theta S and theta P are 45 degrees apart You probably have stated this a lot but when there's maximum stress does that mean you don't ask me the number of degrees? Yes which I work with we're taking these two points and that's at an angle remember this is our operating point the originally loaded point is right here and then an angle from that is when we hit this x axis and when we have the maximum normal stress minimum normal stress and no shear stress because we're right on that axis the equations and the circle together you can pretty much get all things you need in fact you don't even need all of the equations now since it all comes from the two of them once you get sigma average and R you've pretty much got the entire circle then all you need to double check is where point A is and I remember A is sigma x how x, y has applied from the original direction take it the 9.1 is a shears max does that have to be down there because it's positive? is it negative on the top? oh yeah that should be negative but it's plus and minus 9.10 and you can see that the way this was loaded that those shears are going to match that loading in the same way plus don't forget we're only a little bit off of 2AIS 225 sigma dip no, sigma dip and then squared and then sigma average is 4.4 as well oh it's 10 I was putting sigma average in there also so that's fine sigma average is positive well it gets squared anyway so it doesn't matter in this case sigma dip and sigma average are the same right? yeah well opposite signs but when they get squared it's the same are you okay here now? yeah just a little mad thing alright before we take our next step with it possibility of course is that elements are not loaded in only 2 degrees but might be in 3 so what do we do 3 dimensional we can certainly have the possibility here of some kind of x direction stress normal stress y direction and the z direction I happen to draw them all positive but it doesn't matter first the dear stresses to simplify things we'll do problems where there is no z face stresses so that's tau in the x or y direction in this case just the possibility that there are greater stresses than we'd see if we only paid attention to the x-y directions like we just did the problems we've been doing we only looked at the x-y directions we didn't look at the z direction so we have to make another step with it so let me let's put some numbers to it of sigma y 36 mega pascal sigma x is 72 minus so these are drawn in the wrong direction the actual numbers and then we'll let sigma z what we haven't considered before to just be 10 mega pascal positive 10 put some numbers to it alright first thing we do is what we normally do in the x-y direction so we have principal directions and it is very helpful on this to do more circle it doesn't get terribly complicated so it's not too big a deal figure out these things real quick just using the x-y directions then we can plot more circle those values remember the centers are sigma average full planes the then stresses show you what we do in the third direction it's very simple luckily we like simple things because we are simple things average which again for that 54 center of the circle is 54 we can magnify that and write it down 18 so then our radius square root of that squared plus tau xy squared 34 so what's the radius 30 so now we've got our circle this is 84 x-axis 30 on that rate let's put out make sure we've got the right tangent or the right what would you get for then theta p double check that circle by plotting the point A which is sigma x tau xy so sigma x is 72 so what that's about here oh it's minus minus 24 something about there there back to the x-axis is 2 theta p that looks like about 45 degrees half of that minus 45, half of that's about minus 45 that's stuff we've all done before theta z and as we're doing it here we have no shear stress on that face so we don't have to redo the entire thing it's very simple you take theta sorry not theta, sigma z take sigma z and plot it on the same circle so remember sigma is on the x-axis so what's right about there that then another circle sigma min because it's less than what we got with just the x-y directions which was 24 sigma max didn't happen to increase on this one but if sigma z had been greater then we have to plot it out here now we know that we have a greater a greater normal stress in that direction it was inside the circle a little better walk sorry it was inside, wouldn't change anything in that case no it wouldn't then the circle would be on the inside here and you'd still be within the original sigma max and sigma min and then just to make things exciting we have now a third circle too bad that as hard as it gets to draw circles freehand is doing something like this I have to see Hampshire do this that's useful because now we see that there's a greater theta max than we would have had before it's hard to see where it falls because they don't have the same center absolute now shear stress we expect so now we have a loading such that we expect our maximum normal stress to be the 84 megapascals that's the greatest point we saw on the original circle the z direction didn't change that now I have a sigma min that we know to be less than it was before so if we have a material that's weak in a particular direction then we want to make sure that that's our z direction and now we have a new maximum expected shear stress of what 37 megapascals the new min and max have the diameter of the new circle alright then there's a thousand angles other things we can kick off of there it gets really hard I think to visualize this stuff so here's our original direction we have a sigma y we have a negative shear stress so I'll go ahead and draw that in as we'd expected 26 degrees let me redraw here 36 that there 26 degrees we have now an element we've now got remember that was this minimum shear stress but that's not minimum anymore I mean sorry normal stress it's now maybe intermediate 90 degrees to that maximum which was 84 no shear stresses in those directions and then 45 degrees to that remember we had the sigma average all the way around and the maximum shear stress so if I draw that here's just another way to look at it 45 degree so that's a 45 degree isosceles triangle that gives us the theta s direction which was the 4 sigma average and the intermediate maximum shear stress which was absolute I think that's what our book does every book a z part of it take another plane view of it no I do not expect you for the test it gets confusing fast I think there's sigma z same sigma max we're just looking at it in a different direction now that's that new center point which is is what 10 plus is 47 new center that new third circle we have absolute maximum just to picture in your mind we love these 3D problems we always have the best ways to look at those in 3D but that's impossible to draw so that's just an idea of what we do with the third dimension these ones we are taking there to be no shear stress on the c faces I'm not even going to ask if there's any questions but for the way we do it it's just a matter of plotting that next point and then figuring out the geometry pieces to see if you do have greater there's greater shear stress now that we originally would not have gotten we thought it was 30 but in three dimensions now it's 37 degrees so if you're laying down carbon fiber with a particular direction to it you want to pay attention to that the 47, that's the location of the center of the new circle that'll suffice and that wraps it up