exercise transmission forces/couples dense wheels

[Taipan95]

Good morning to all, I am a new user of the forum! I am a student at the polytechnic of torino, I attend the magisterium. I wanted to propose an exercise on dense wheels with adjoining doubts about the transmission of forces/couples: place the scheme with the data and illustrate my procedure.
Basically, by means of a gear system, a driving power is transmitted to two different users. the doubts concern only a part of the exercise, i.e. the one that requires to identify the most stressed wheel and calculate the safety coefficient for the static verification of lewis on that wheel.

I started by calculating the input torque, placing the input power as sum of those on the two outputs: c1 = 267.5 nm.
from here c2 = c1*z2/z1 = 445.8 nm, as well as omega2 = r1/r2 * omega1 = 94.2 rad/s = omega3 since they are cast on the same tree (idem for the pair).

continuing with these formulas I came to conclusion that omega5 = 136.1 rad/s and omega4 = 49 rad/s.

Now I ask the question: the output pair, for example on wheel 4, should be calculated as pu/omega4 or as (pin - pu)/omega4? I ask this because following the first road I find two pairs on wheels 4 and 5 pairs to 612.2 and 88.2 nm, instead in the second case 244.9 and 240.2 nm respectively.

I had this doubt because I can't understand why there is no balance of couples, in the sense that (I believe) that the outgoing pairs in 4 and 5 should balance the incoming couple in 3, so the sum of couples 4 and 5 should give 445.8 nm. using the "second street" I find a result of 465 nm (however incorrect, but much more similar to what I would find using the first alternative!). in both cases the two forces (obtained as pairs / respective rays) would be 2800 and 6997 n (obviously reversed... in the first case 2800 would be of one wheel, in the second case it would be of the other) and therefore (2800+6997)*r3 would give me 445.8 nm confirming the balance. what I can't understand is whether the balance must take place for "pure" couples or for couples intended as "complete force*rage".

I apologize in advance to be extended, I thank in advance anyone who wants to contribute for patience!

 

[PiE81]

the sum of the powers (pair for angular speed) incoming must be equal to the sum of the output powers.
based on their respective reduction reports you will have specific couples related to the previous report.
in the understanding of the powers must be considered the performance of the various rotisms and that therefore part of the incoming power will be dissipated by friction.


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[Taipan95]

Thank you for the answer!

 

[meccanicamg]

it could work the sum of couples but from applied and rational mechanics, as well as from dynamics you should know that the balance of powers is made, most of all in the mtu.
However if you can the power of the unknown user, it will be the power that undergoes z4. In fact the transmission you have to see without the fan and you will see a double train of gears with reduction ratio greater than 1, then the maximum torque will be on z4 and the speed the minimum.
seeing instead without u but only with fan you have to calculate with the pv the pair on z5 that has a last ratio of multiplied and not of reduction, so it should be the stressed less than z4....contti to the hand verify this.
the poor z3 takes a double load because it stops the total power and therefore will have a low duration because oziosa....hertz with two contacts.
z2 takes all the power and z1 also.
a nice excel and determine everything.

 

[meccanicamg]

made calculations very quickly in 5 minutes I'll coach you couples.

 

[PiE81]

made calculations very quickly in 5 minutes I'll coach you couples.

I think the couple on c5 is about 88nm.
at least this is the result considering a pv of 12 kw and the reduction ratio between the wheel 3 and wheel 5.

 

[Taipan95]

I agree, I avoided writing it because it was a blatant mistake dictated by the rush and making the professor I don't really want to

 

[PiE81]

far from me do the professor, moreover with the always accurate and expert mechanicalmg. We are not at university to judge But in order to avoid anyone being confused by reading this post, even in the future, I felt it appropriate to state.

 

[meccanicamg]

You did well to notice the mistake. I actually used n4 to calculate c5 and did not take into account the transmission ratio of z45.
I attach new version with the two updated bold values.

 

[mod29b]

I have these results. . .
calculation of the total transmission ratiothe transmission ratio of the gearbox is the product of the transmission ratio of the individual gears.
i12 = z2 / z1 = 35/21 = 1,67
i34 = z4 / z3 = 50/26 = 1,92
i35 = z5 / z3 = 18/26 = 0,69
i14 = i12 · i34 = 1,67 · 1,92 = 3,21
i15 = i12 · i35 = 1,67 · 0,69 = 1,15
calculation of the number of laps of the intermediate shaft and the output shaftfor calculation the relationship between the transmission ratio and the number of turns is exploited.
i12 = n1 / n2 → n2 = n1 / i12 = 1500 r/min / 1,67 = 900 r/min = nint
N3 = n2 = 900 r/min
i34 = n3 / n4 → n4 = n3 / i34 = 900 r/min / 1,92 = 468 r/min = nu
i35 = n3 / n5 → n5 = n3 / i35 = 900 r/min / 0,69 = 1300 r/min = nv
calculation of torque moments on the three treesthe calculation is carried out considering balance of input/output powers and neglecting yields, as it is in ideal conditions.
ω1 = 2 · π · n1 / 60 = 2 · π · 1500 r/min / 60 = 157,08 rad/s
ping = pu + pv = 30 kw + 12 kw = 42 kw = 42000 w
mt1 = pin / ω1 = 42000 w / 157,08 rad/s = 267,38 nm
mt2 = mt3 = mt1 · i12 = 287,87 nm · 1,67 = 445,63 nm
ω2 = ω3 = z1 / z2 · ω1 = 21 / 35 · 157,08 rad/s = 94,25 rad/s
ω4 = z3 / z4 · ω3 = 26 / 50 · 94,25 rad/s = 49,01 rad/s
mt4 = (ping – pv) / ω4 = (42000 w – 12000 w) / 49,01 rad/s = 612,13 nm
ω5 = z3 / z5 · ω3 = 26 / 18 · 94,25 rad/s = 136,14 rad/s
mt5 = (ping – pu) / ω5 = (42000 w – 30000 w) / 136,14 rad/s = 88,15 nm
calculation of the primitive diameters of the four toothed wheelsthe dimensions of the toothed wheels can be obtained from the knowledge of the module and the number of teeth, taking into account the fact that between two toothed wheels it is possible to engrave only for equal modules. so that the pinion 1 and wheel 2 have the same normal module as happens between wheel 3 and 4 and between wheel 3 and 5.
dp1 = m12 · z1 = 3.5 mm · 21 = 74 mm
dp2 = m12 · z2 = 3.5 mm · 35 = 123 mm
dp3 = m34 · z3 = 3.5 mm · 26 = 91 mm
dp4 = m34 · z4 = 3.5 mm · 50 = 175 m
dp5 = m34 · z5 = 3.5 mm · 18 = 63 m
calculation of interassistancei12 = (dp1 + dp2) / 2 = (74 + 123) / 2 = 98 mm
in34 = (dp3 + dp4) / 2 = (91 + 175) / 2 = 133 mm
in35 = (dp3 + dp5) / 2 = (91 + 63) / 2 = 77 mm