 In this video I want to walk through an example problem of doing a design using the more rigorous analysis for gears. So in the problem we have a setup such that there are two gears, a gear and a pinion coming together, and I've kind of already prescribed out some of the setup stuff here. The pinion has 20 teeth. It's rotating at 1100 rpm. The gear has 40 teeth. We want to design for a safety factor of 1.5 with 99% reliability. The gears are made out of steel with a Bernal hardness of 350. There's accurate mounting, a dimensional pitch of 8. The gears are 1 inches wide with a 20 degree pressure angle and they're manufactured by a top quality hobbing operation. So then the intended outcome is to determine a horsepower rating for this gear set if we want it to last five years running 60 hours a week for 50 weeks each year. Okay so that's all of our setup. So we're going to start by saying we have our base equation that looks like this and we have to start filling in the gaps. P we have, B we have, J the geometry factor we can look up, Ft we don't have, but if we think about what the question is actually asking it's asking for horsepower rating and remember before we talked about how power can be determined by the the force transmitted. So this is actually what we're trying to solve for so keep that in mind and then we need each of these factors. So let's go ahead and start with the factors. The first is the velocity factor Kv but in order to know the velocity factor we need to get the pitch line velocity. Pitch line velocity is equal to pi dn over 12 and we actually weren't given the diameter but we can go ahead and specify that or to calculate that because we know the pitch and we know the n value. So if we rearranged our pitch equation we'd see that diameter equals n over p so we can actually plug that in so we have n of 20 over p of 8 and we're working with the pinion in this case usually we start we actually analyze the pinion because it's the gear of the two in a pair that's most likely to fail so usually we'll start with that. So we have this plugged in for diameter 1100 rpm over 12 and this gives us 720 and if we carried this out carried out our units this would be in feet per minute. So now we need to go ahead and come up with Kv. If we go to our charts in the book we can actually find Kv here this is figure 15.24 and we said we had a top quality hobbing operation so there's a Hobbs operation here and we'll assume we're on the best line for that which would be the top quality so line c and just below the figure there's an equation given for for line c which is 50 plus square root of the pitch line velocity over 50 so we'll go ahead and use that then 50 plus square root v over 50 and that comes out to be 1.54 for this particular setup. All right next we need or we can get a value for j since we had that figure up so if I go back here I'm looking at a geometry factor let's see we said so there's two charts here a and b the first one is for a 20 degree and the second one is for a 25 degree we have a 20 degree in this case and again we're going to use the the pinion as our matching option the number of teeth we said was 20 so I have a x-axis here of 20 I'm going to go up my mating pair had 40 so the lines all get really close together here and it gets kind of hard to read but what we could do is if we're going to assume worst-case scenario we might say that there's no load sharing which is kind of a you know less likely but also just a worst-case scenario situation so if we do that we could pick this line here which is 0.24 approximately it looks like just before 0.25 so we'll go ahead and say 0.24 and yeah we can move on from there all right next might work on mounting factor mounting factor we can pull from a table in the textbook first on face width we said it was one inch wide so that falls into this first column and then depending on what side of style of mountings it says accurate mountings so probably somewhere between here and here arbitrarily we might pick a middle this this middle value which gives us a little bit more slot built into it kind of more conservative I guess as a as a choice so that's all kind of personal opinion I guess on that and overload factor ko we don't have anything in the problem that specifies whether we should expect any shock loading so we might go ahead and pick a km value of one or excuse me a ko value of one so putting this all together we get ft times 8 over 1 times 0.24 times 1.54 times 1 times 1.6 and this gives us that the stress in our part is 82.1 times the tangential force so again we're trying to find a horsepower rating and this tells us the stress in terms of the force transmitted and force transmitted remember is related to that horsepower that we're passing through our gear set so in that case we don't actually know the numerical solution here instead we need to find something that we can compare this against and our comparison point is you know failure what what the stress would be at failure so we need to go ahead and apply a fatigue analysis to this in order to determine that failure point so separating this out a little bit remember we have the equation that looks something like this let's see if I can get all my correct factors in here something like that so we need to start plugging things in um sn prime we were given that this is uh steel with a 350 hardness and if we look back at our fatigue chapter we'd actually see there's an equation for sn prime based on that which is 250 times that hardness which comes out to be in psi so we can make use of that load factor we'll pick a load factor of actually I'm just going to write it in of one because the the dimensional pitch is greater than five that means that our cg is also one cs surface factor I would read this off of figure 8.13 and we'd get 0.66 reliability factor from table 15.3 99 reliability gives us 0.814 and we don't have anything about high temperature so that gives us a temperature correction of one and this is not given as an idler gear which gives us an kms of 1.4 great so we've got this uh spelled out and we can go ahead and calculate that through so if we do this I think it comes out to be 65812 psi and this we're going to go ahead and set equal to now our sigma value from up there so we have 82.1 ft is equal to 65812 great and what this is doing is it's basically saying here's the stress in our gear here's the stress limit based on fatigue what do we expect to happen now the one thing that I haven't included yet is we said we wanted a safety factor of 1.5 um two ways we can think about this we can think about it as reducing the fatigue limit or increasing the stress in order to factor in that 1.5 so I would go ahead and just pull this out and say I need to multiply this times 1.5 to build in my safety factor there and if I do this then I calculate that ft with a safety factor of 1.5 would be 534.4 pounds so that's the load that I can transmit at the gear tooth and presumably not shear that tooth off under this fatigue loading condition so I can from that calculate my power which is equal to ftv over 33 000 and I get 534.4 times my pitch line velocity that we calculated before and 33 000 and this comes out to be 11.66 horsepower so what this is saying is I can transmit up to 11.66 horsepower through my gear train gear my gear set I should say um and presumably not fail in fatigue and I have built in a safety factor of uh 1.5 into this now you may ask one thing uh that you might notice is I didn't really take into account uh where did it go I didn't really take into account this life here and I probably should have written it down but basically what the only thing I'm really having to check here is how many cycles this would be so if I took five years multiplied it by 50 weeks per year multiplied it by 60 hours per week multiplied that by 60 minutes per hour and multiplied that by 1100 rpm I'd get 9.9 times 10 to the eighth cycles so where this comes into play is that this value is larger than one times 10 to the sixth which is what we usually consider to be um that infinite life condition so I really already did factor that in down here where I said that the sn prime was equal to 250 times 350 because that's an infinite life condition um specification um and therefore I've kind of already built that in I should have made that more clear but I didn't realize till the end that I had skipped over it so we've we're expecting it to last a whole a whole bunch of cycles and to do that with a 1.5 safety factor we have to limit our power to 11.6 horsepower all right thank you