 Let's warm up with a problem much like the type we were looking at when we finished on Friday. So imagine a tubular member of some kind rigidly fixed on the back wall and inside of it with a much smaller diameter than even the inside diameter of the outer tube is yet another one. So I have two tubes like that, concentric or a tube with a concentric rod in it and there's some kind of rigid cap over those, so I'll draw that as if it's out here, but it's really bonded to the top of those so the cap itself is rigid so there will be no deformation in that. That's sort of always been one of the idealizations we've used and both of the structural member, both parts of that are intimately made and inlay adhered to that rigid cap that's on the outside and then there's some kind of torsion applied to that. So if you need a side view to help, we have this tubular member now cut open in cross-section, another material, remember down the inside and then that cap over the outside to which we've applied some kind of torsion. So if we use our right hand rule, we can also draw that like that then. All right, pretty simple setup, a little more details to it. Let's say that it's 500 millimeters in length, that the inside material is steel and the outside material is aluminum. Maybe there's room for coolant flow between the two or something, for some reason we've got a setup like that. Outside diameter is 76 meters, inside diameter of the rod on the inside is 50 and then the walls of the tube are 8 millimeters. See that well enough? Is that all right? Got the picture, a steel solid shaft inside a tube, all capped off of a rigid cap to which we apply some kind of torsion. All right, some of the material limits with which we must stay below, we have an allowable shear stress on the tube and we need to stay under both of those, 120 megapascals for the steel, 70 for the aluminum, other material properties, modulus of rigidity, 77 megapascals for the steel and 27 for the aluminum. No gigapascals, we struggle with that elastic, I know you'd rather I let you get halfway through before I make those corrections, but let me do them now, okay, I think that's all the pieces we've got. All right, so we need to determine the maximum applied torque, there's one torque that will drive one of the materials or the other to its limit in terms of the shear stress and that then would be the design of maximum on the applied torque. Got the picture, take a second to get the whole drawing down. This is one of those that's statically indeterminate because we certainly know that if we look just at the rigid cap that it's got some applied torque in one direction but because of its immaculate adhesion to the other two materials, they've got some amount of torsion that they're withstanding or applying to the cap and all we know is that total torsion has got equal some load taken by the two inner structures of this machine component of some kind. Static alone is not going to help us, this is statically indeterminate but that's okay, we're not worried because we've got extra tools this term that allows us to finish these problems that we couldn't have finished last fall. That's all the farther we could have gotten is to here but what we need to do is decide what of the stuff from this term is going to be helpful to us. You have to look at the problem to decide what of the tools from this term are going to be of use to us. That's there, not bad. And what are you writing? You're doing something, you want to share it? So two unknowns, how much torsion goes into each, actually this is three unknowns I guess since we don't know any one of those we just know that the applied torques to the individual members have got equal to total torque applied to the entire thing so what other things can we bring to our aid from this term? We're looking at deformation so it's going to have to be something of that. Any ideas? Are you mad at me? Is that what happened? Did I ruin your weekend? You're going to make me pay for it? Chris you're going to say something? Well yeah, but not really much change of length expected here. The angle of twist is going to be the same on the two however much one twist the other is going to twist just as much. That again is on the assumption that whatever this cap is rigid. So we can also use the fact that whatever angle of twist we see in one we'll see in the other. So we started to set that up then depends on a couple things. First the load experienced by that number and the overall length. What else? Which all those last ones independent of the load. G of course is a material. Property and L and GA are properties of the geometry of the problem as set up. So those two must be equal. Trouble is that we're still short because there's three unknowns. This is an independent equation doesn't introduce any more unknowns but we're still short one equation. The equation bank. Now this is one where since we have two materials and we know they're not going to be, well I hope we know, I hope we have a feel that even though they have limits on the stress they're not going to be at the same stress level at the same time and there's going to be one situation that dominates these two. So we have to assume one of them is at the limit and then use that to calculate what the total load is then check that for the other one and if it's over the limit then we know our assumption is wrong and we have to redo it the other way. So we can add that to it by bringing into it the stress limits just for a little bit of help here. I'll give you the two moments, polar moments so you don't have to bother with calculating those the deal is 2.03 x 10 to the minus 6.003 10 to the minus 6 inch millimeters to the fourth that'd be meters meters to the fourth and the aluminum is 0.064 All right, all this stuff here is constants except for the two torsional loads that we're looking at so those can all be calculated and you can get this down to just a simple relationship between the two it's just a proportionality constant between the two that just slimlines things a little bit so then what we need to do is assume one or the other is critical take that through the calculation until we find a load on the other one and just determine whether or not it has exceeded the limit so for example maybe assume the word at the limit on the aluminum assume that's the shear stress we're at the limit on the shear stress of the aluminum remember these are allowable limits so assume we're at the shear stress on the aluminum use that then to find the load on the aluminum so assume one or the other is critical you don't know which, you have to take a guess assume that we're at this critical limit on the aluminum 70 megapascals everything else is known so that we can then find the torsion expected on the aluminum that will make it critical we can then go back and then test that if we're at the limit then on the steel as well because once we find the aluminum torsion we can then find the steel torsion that goes with that we can then check to see if that's at the limit of the steel as well so there's the bare bones plan sir one thing, is it possible you've got the j-aluminum and j-steel switch it's possible j of the aluminum yeah why don't you check these things thanks David trouble was I changed my subscripts from my notes to the board so now they're all messed up alright everybody understand a bit of the game plan then solving this problem Joey make a little sense assume the aluminum stress is critical that's this one put that in here then we can solve for the torsion on the aluminum with that we can then find the torsion on the steel with that in the same way we can find out the shear stress on the steel then and see if we're over this limit if we're over this limit then our assumption is wrong and we have to redo this very same calculation with the opposite material being under the stress limit at the stress limit don't forget that C is the outer radius yeah for this part right here all these values in here are for the aluminum if you're looking we're using the value for the stress limit on the aluminum that works Travis go anywhere that's what you're finding okay now then that gives you a value in the stress and the steel but then you have to also determine then for the steel what the shear stress is in the steel and compare that to this and if we're over that then this assumption was wrong yeah the this assumption for the most for essential purposes is our second equation our third equation we're assuming a value for one of the shear stresses that serves as a second equation a third equation that's the torsion in the aluminum if the aluminum is at the stress limit what are the numbers I have yeah that's what I have yeah check your numbers John show if you're confirming mine well you have 36.9 hang on hang on where's what's 74 not 74, 70 just put it in an actualistic case yeah so you know I appreciate it you're mimicking my style at the board where did I get that from? from Joey it's just way way back there as far away as they can be there's too much flying and chalk dust up here to go close to it alright most people are getting something like 36.90 meters on that with the John and then you can use that to find what it is in the steel then once we've got the torsion in the steel you can find the shear stress on the steel and see if we're over the limit for the steel once you get a torsion in the steel well 4 yeah you've got a decimal point on the of the 76 find a torsion in the steel then you can in the same way find shear stress on the steel now that we've got that compare that number to the limit on the steel find that bar and yeah so the assumption is wrong it's a working bill alright now we've got an estimated load in the steel then the dimensions for the steel and that should give you then a shear stress based upon the assumption that we were already at the limit in the aluminum but if you wasn't wrong by that much we're all on traps and we can find other ones okay yeah and that's passed so good and that tells you the assumption was wrong huh well this would be the steel inside 2 so this number is 25 that would be yours and the j thanks today because we got corrected Bill you got something? not yet? first word you get well that's a little different oh that's that's for the aluminum yeah so now you have to find it for the steel using equation 2 which is just the torsion and the 2 related by a bunch of constants so you have to say this number use it to find this number which is the torsion and the steel then that's the one used in here because it's for the steel yeah did you use j steel to get the 3694 no that's aluminum 2 yeah 2.0 I used the 2.0 I used the right numbers in here and the c was the outside diameter which is half of the 76 so that comes to be about 131 John 131 3 give or take a little bit it doesn't really matter because that's greater than our our allow of 120 which indicates that the assumption is wrong because that's the torsional load in the aluminum you use equation 2 to find the torsional load in the steel and then that should be about this that you use then to find the shear stress on the steel and that we find is over the limit a sine limit of 120 where that comes from we're not saying that's one of the givens in the problem agree David with those numbers coming up over there then you have to assume now that the steel is critical redo essentially the same calculation you did there and we know that it will come below the shear stress for the aluminum because we assume that was critical the first one and went over so we have to come down getting those numbers down Chris that's the torsion in the aluminum we're going to use equation 2 we have the torsion in the aluminum all these others are just geometry things so now we can get this so once you have that which should be that number over there then you can figure out the shear stress in the aluminum you got this okay this conclusion Chris fill you up to it then it's a matter of going through the same numbers you should get then the torsion in the steel you have that Tom? yeah 2950 2.9 if you use kilonewtons kilometers or 2950 doesn't matter, wait as long as you're consistent then back into equation 2 to find the load in the aluminum and then determine the shear stress in the aluminum and we already know it will be below this because we reduced the entire load since we first found the steel was the limiting number of the torsion and the aluminum then is 3375 which we know is less because we have been at 3690 with her original assumption it's not quite the answer the answer was what's the total load and so it's a matter of summing all those up because that's what we've done now is found the division between the two and then adding it together to give us a total Chris you getting those numbers now a little better Joe okay getting those just we had to pick one, we picked the wrong one of course I did that by design so you can see how to recognize that your assumption is wrong so I'm going to go back and calculate the same steps we did here the same steps we did here you don't need to actually calculate the stress on the aluminum because we already know it's below that was these were the stresses at the aluminum limit and we're already below those so we don't need to calculate the stress on the aluminum that doesn't mean you don't want to know that number but we don't need to check it to see if we're alright, we already know we're alright I'm sorry 29 47 3371 oh yeah that's fun that's just round off because if you want three significant figures we agree just fine and don't forget we're not saying where these came from we don't know if there's a factor of safety on those yet we're just going through the calculations in the real design there's more details that need to be brought out yeah so you're right here you know that that's over the limit for the steel so now we go back and then assume this value with the limit value for the steel and essentially we do the population only each of the different pieces swap the roles so now that we know that the steel is the limit calculate the load and the steel using this same equation only for the steel so now we've got that as the limit on the steel j for the steel c for the steel we'll give you the load on the steel then in equation two again we've got the load on the steel we can find the load on the aluminum and we already know that's going to be below the stress limit for the aluminum because we are way below those not way below quite a bit below enough below that we know we're not at the limit of that aluminum Chris okay so you should give us a total load of 663 25 Travis you got that Joe you okay do we have to check that and like we did with the aluminum no because being at the aluminum limit pushes over the steel limit so we drop down on the steel which is automatically going to take all the loads off to a lower level which means exactly below the load on the aluminum anyway so you don't need to check it you might need to know it for some reason but you don't need to check it okay John press those numbers down a little bit later then because we've got a little bit more we've got to hit here you sit and just make sure you're not crossing aluminum values with steel values alright to wrap up our look at torsion typically in drive shafts and the like and there's several problems of doing just that notice that we can also do the very same kind of thing we did with the the surfaces in axial loading that looks kind of terrible let's just draw it on side because we're it's a lot easier to draw a drive shaft on the side we can have these changes in thickness where we get stress concentrations so depending upon what the load is and the geometry of each of these it's no surprise I don't think it depends upon the major and minor diameters of the piece because whatever stress is felt in the large part that same amount of stress is going to have to flow into the smaller one causing then stress concentrations and it also depends upon the radius of the fillet and we define it just like we did just like we did for the axial loading where we have the stress concentration factor some kind of allowable or maximum limit compared to the average values we have been calculating and that's what all of these are remember that assumes that we're not at any limit on the stresses but at a an average if we look at the load levels in the piece the real profile is something like this it's actually quite severe where we've been assuming that we have average stress levels the distribution of the shear stress torsionally loaded shafts was at a maximum of the outside and decreased linearly to the inside but these are all average values so our average distribution there looks something like this it's linearly increasing from the inside to the outside and that's what we've been using whereas the maximum might be there and if that's the value we need to keep below the allowable limit so the stress concentration factor allows us to not have to think that these places of greater concern are possible and the chart we use it looks very, very much like that well it's in your book depending on which edition you have it's 536 or 532 I'm not sure which for some reason the device is not open see if that's necessary is it necessary oh yeah that was nice there we go the table is very, very much like the one we saw for the the axially loaded members our distribution is very much different these tend to be kind of severe so let's try a couple problems so imagine we have a step shaft like that with a major diameter of seven and a half inches minor of 3.75 inches and a fillet radius of 9 sixteenths good old American values the shaft rotates at 900 rpm and the allowable stress that's the one that's on the upper part of the definition of the spring stress concentration factor is 8 ksi so we want to find then the maximum power output so there's a little bit extra in here that we need to work through just to make sure we've got the parts to it the power from a rotating loaded shaft of this kind is 2pi f where f is the frequency of rotation the engineering frequency not the angular frequency so it's essentially this number in per seconds rather than per minute and then whatever the torsional load is this is the very same t we've been using as we go along here for a while so we can take this value then and using the allowable limit and the stress concentration factor find then a maximum load we can put on the piece just to help you out a little bit whether you can do it so that's the engineering frequency the revolution per second or hertz if you will and then that's the number we use in here so we need to find a maximum torsion from the allowable once we find that then we can calculate the maximum allowable power any power load over that is going to over stress the shaft all is calculated on the smaller diameter because that's the part that's under the greatest stress the torsional load is the same the stress is going to be a lot greater and we need to calculate what effect the radius has on the stress we're going to figure out a stress concentration factor use that to find the torsion that keeps us at or below that stress limit that we can find the power and finding K is just like it was before with the other axially loaded members all these are torsionally loaded it just comes right off the chart you can imagine in real design situations there's a lot of these graph spectra probably all embedded in software now so find the K use the allowable limit to find an average stress use that to find the torsional load use that to find the power we need R over D and D over D for this particular problem and then those things can be adjusted as needed as you do with time iteration both in inches of course these little indices values are so we're at about 0.15 and we're on the two line which is this second one here so where those two hit looks like about 1.3 looks pretty close to it easy one to read this is near the end of chapter 5 now we can use that to find the torsion which allow us to find the average load no you put it into horsepower when you go down to A's hardware you're looking for a motor to connect to this shaft you're going to want to know horsepower that's how they sell it down there Earl's a bright guy but he wants horsepower 550 foot-pounds per horsepower intermediate values but you're going to have to move that horsepower that's probably one unit that's going to be with us for a very long time well no European vehicles are sold over here without mention of horsepower what they actually have in their catalog and how they go about the design the torsion the T before you went to the power there should be able to get a T from that and then once you get a T from that you can get a power up there and we're getting 63.7 per pinches do you want to underestimate K or overestimate K overestimate I'm not sure how they come up with those I don't know if it says in the book or not how they actually come up with these charts should get something like 900 horsepower 910 911 is also close there's 550 foot-pounds per horsepower very easily make a change in the radius of the fillet if that's not enough if 900 horsepower isn't enough you can easily change the radius would you go bigger or smaller increase the available power output would you go to a bigger radius or smaller radius on the fillet so size matters you know without even looking at the charts that'll just confirm it but if you have a very very sharp corner in there that's an extreme stress concentration those areas tend to break very easily part of what's good about this too is that when you put these shafts together this fillet might well be just a matter of the welding of the two shafts together well if done well enough conserve that as a fillet just make sure you have the intermediate value 6 times 10 is 6 inch-pounds is what I have to have per second that's a power unit Chris are you alright Joel did you get that you're tired too how do you get J oh back here how do you get J one-half pi radius to the fourth but which radius the smaller one these calculations are all done on the smaller that's going to be under the greatest load this 550 foot-pounds oh foot-pounds per second we have 63.7 kits over here yeah and that goes in here so that becomes units of something like the band where your decimal place is inch-pounds per sec because the frequency is in per seconds but 2 pi here is the revolution but it's actually the radius this is actually the angular flossing fill alright did no same numbers Samantha make sure that all these calculations in here are on the smaller diameter same thing that we did for the actually loaded steps on the quality of the weld and the material used in the weld so yeah it could make it weaker but it could also make it stronger also possible of course the machine needs that one piece material the the science of welding there's you can get a master's degree on welding alone non-destructive testing techniques how to check welds without taking the piece apart which then doesn't help you any because that really weakens it when you saw through the thing me? no I don't want to play with fire alright increasing the fillet radius to just a little bit more to 15-16 raises the power by about 11% so then it's an engineering design decision then if that's worth it drops your k-value yeah if you increase the radius you'll appear the k-value drops and you get more power out if you can put in more torque alright let's do one more and then we're done alright the drive shaft with a diameter change but multiple loads on it so 700 pound inches back here further along so all of these will be in pound inches 300 up here in the opposite direction the only one in the opposite direction 1500 pounds there all of those foot pounds and diameters 1.25 inches fillet radius 0.05 inches so find the maximum expected shear stress that's the same thing as the allowable on the definition of k-factor stress-construction factor we can make that a get out of class question that'll incentivize you so we have a shaft of different loads on it somehow gears are placed there maybe even belts boys what about the length of this we only needed the length when we were looking at the angle of twist other than that the length of any of this doesn't matter how far apart those loads are other than that they're not right over each other we need to find the maximum shear stress maybe label the different regions the shear stress changes in every one of those regions because the load changes and the radius change so you're going to need to look at it for each of these regions A to B, B to C C to D and D to E figure out the maximum stress in each of those determine which is the worst case the design limit you then decide if you have material that can take that or you're going to have to change some of the design specifications each of the sections for example A to B we know that there's a load of 700 pound inches so there must be an internal torsion then of all the very same so you need to do that for each of the sections that the internal torsion loads are and then figure out where the area is to worry about this doesn't seem like too much of a worry because there's other places with greater load and smaller diameter so the suspicion would be A to B is not going to be the limiting region some other region will be that's a pretty low load not a great concern we actually want to take the diameter no you have to look at what's happening in each section that will be a C a little bit longer we've got the 700 and 500 so you know the internal load there is 1200 pound inches and we only do B up to C because that's one diameter there's a transition at C that could be the critical point actually this the 1200 will take you all the way to D we have the 700 there 500 there not quite up at the D load yet so that's not on there so there must be an internal load of 1200 yeah because in the transition there's no increase in load the load doesn't come until down here farther so when we're below D not quite up at D yet remember this section B to D we're going until just until we get the load at D which is out here farther so you have to look at a tau max in here well that wouldn't be a concern because here we're at even smaller diameter but then we also have to take into account the stress concentration then it's probably easier to go from the other end so D to E we have the 1500 there not concerned about the direction we would be if we were looking at the angle and twist on this entire shaft but we're not we're just looking to make sure other than the stress limits so it looks like two areas of big concern this is a large load and a small diameter so A to B is much better off much slower load but a much greater diameter so we're not even concerned with A to B B to C is no real trouble but we are worried about what happens at the transition and then we're worried about what happens at this this region there when we're down in the smaller end of the shaft out here between B and E the areas of concern that area where we have the diameter change and the stress concentration and this area where we have the large T and the small D both of those areas for the areas of concern we know the other areas are not as much of concern the loads are less and the diameters are greater well, you should because it's the largest load near the end of that shaft and it's the smallest diameter so we want to calculate here the expected shear stress with those loads direction doesn't matter the only time direction matters if we're looking at the angle of twist where twist can be one direction and the other this all that matters is this magnitude this is the internal load in that end of the shaft which is the biggest load of any and the smallest diameter so it's going to be this is a critical area that we have to look at then so is this one where you have the area change because that can with the stress concentration factor make things even worse in fact this value will be the average value to use there we'll have that one or we can use that same number just with the diameter in instead of this radius we'll have this one PSI yep that's actually the lower number on this one because remember the calculation actually wouldn't be because we'd have 1200 in here instead we're at the slightly lower load but we have the stress concentration factor so we have to calculate a tau average calculate a from that which is greater than 1 that will be greater concern do you have this intermediate number with 1200 in here using the smaller diameter remember that's that's the average and not taking into account the stress concentration factor which we can squeeze in here and we're done our over D is 0.05 D over D 5 so very small radius on the fillet and there's a 1.25 line it's this one so we've got a pretty high stress concentration factor it looks like 1.58 or so if you take a little bit that gives you a multiply of 1.58 times this average shear stress in the little piece before the 300 foot pounds is added so this comes out to be about 96.60 which is quite a bit worse than the 76.40 so the area of concern is at the step change so I want to investigate increasing the fillet radius if it's machined in that's usually not too difficult a change to make that would increase this number bringing it down to a lower K factor and can significantly drop this number then Phil, okay Travis you got those same numbers Chris David 1232 so I was using problem D to E you got 1232 on here oh yeah at the fillet there's not a load of 1500 there's a load of 1200 at the fillet you wanted us to look at the fillet well you have to look at both this had the greatest load on the smallest diameter so we had to look at it but then we also had to look back at the fillet area even though the load was not as great oh so the area of concern is going to be right at these corners because that's where the stress concentration factor is yes because that's the whole point of the stress concentration factor is these area changes okay that's why I circled that region of its own to compare those two numbers