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exercise - drawing of machines (reducer)

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iolupe

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Hello, I am a mechanical engineering student at the 1st year, I would like some advice on the resolution of this design.
 

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a couple of tips never deny!
therefore. . .

1) read the regulation well
2) post precise questions (what do you want us to recommend? We will never give you the task solved. . )
3) start stirring, because 30 is tomorrow!
 
the wording of the text is clear, you ask to draw the reducer in section by adopting cylindrical dense wheels, input shaft with cylindrical roller bearings and cast gear by tab, the out shaft with ball bearings and use the reduction capp. 1:1,4 . having the side pattern you already know how to draw everything. not having any other data from the attached pdf I assume that the sizing of cinematism is at your discretion. . then mbt has already explained all the rest.
 
a minimum of dimensioning takes us. hypothesize power and number of incoming turns, determine the module of toothed wheels, impose z1 and determine z2 with some attached and connected checks. dimensional the trees at least statically with a good safety coefficient, dimensional the bearings at durability, sizing cutting tab and specific pressure.

then start the drawing part.
 
we see how it analyzes and solves the exercise. I prefer to answer here that it is of interest to all users, there are no secrets of state.

- input data for dimensioning: it is necessary to hypothesize power, number of incoming turns. use the transmission ratio 1:1.4 to determine how much greater z2 must be than z1. basically means that if the entrance fee makes 1 round, the output one makes 1.4 turns, then transmission ratio = i = n1/n2 = z2/z1 = 1.4- the shaft 1, the input one with pinion must be obtained from the full "piece pinion", while the shaft ducted has mounted through tongue the wheel conducted. it is necessary to make a static sizing and possible fatigue verification of both trees. from here you can get the ø of the tree and then of the proportions to then simens the pinion (which will have greater diameter than the tree but not very possibly)

- the tab from catalogue the size according to the diameter of the shaft as regards b and h, while the length must be verified to cut cut e Press specifica find in page 20 le formulas

- straight teeth cylindrical wheels: qui e qui and for geometry qui

- dimensioning and bearing verification: see on the skf site that there are formulas and catalogs, as well as the automatic buffer verification program chosen. Of course, it is necessary to realize the suitable locations with the right constraints on the tree and the carcass. the carcass is not espcified, it can be a cube of sheets from 15 / 20 mm (or what you want according to the measurements of the rest) welded internally and externally + cover screwed with subsequent processing to accommodate bearings, seals for lubrication etc.
notes: as I have seen and understood does not need the exact representation of the profile of the evolved but only the simplified unified representation. draw everything as required. I recommend that you are doing engineering that some manuals should be bought, since they are part of the necessary tools for the profession. It is a work that I highly recommend doing partially on paper for the part of accounts to have a calculation strategy, then implement in excel to optimize the search for parameters such as the size of the toothed wheels, rather than the tabs or the trees or the bearings. If you are looking for it has already been designed a reducer here in the forum by your colleague and you can have an idea of what you have to do.

for the rest and for explanations we are here. :biggrin:
 
for the rest and for explanations we are here. :biggrin:
Thank you for your patience.
Let me get this straight. and if you can give me an example starting from 1/1.4(or another report)
I don't understand how to use formulas starting from 1/1.4
I want to understand! !
 
Thank you for your patience.
Let me get this straight. and if you can give me an example starting from 1/1.4(or another report)
I don't understand how to use formulas starting from 1/1.4
I want to understand! !
in the sense that having only the ratio of reduction the nr of teeth of the pinion z1 you establish it and consequently for the z2 calculations multiplying z1*i

as mentioned by mechanicsmg you have to hypothesize power, number of incoming turns to size the input shaft and consequently determine how large the z1 pinion is needed

If by calculations/preliminary size you establish that your pinion z1 has 40 teeth (nb: randomly said value !!!) to determine z2 you just multiply z1 for transmission rapp., ie z2=z1*i = 40*1,4= 56 ie your pinion z2 will have 56 teeth
 
in the sense that having only the ratio of reduction the nr of teeth of the pinion z1 you establish it and consequently for the z2 calculations multiplying z1*i

as mentioned by mechanicsmg you have to hypothesize power, number of incoming turns to size the input shaft and consequently determine how large the z1 pinion is needed

If by calculations/preliminary size you establish that your pinion z1 has 40 teeth (nb: randomly said value !!!) to determine z2 you just multiply z1 for transmission rapp., ie z2=z1*i = 40*1,4= 56 ie your pinion z2 will have 56 teeth
I've done, z1=20 z2=20*1.4=28
the pinion 20 teeth and the other wheel 28.
the wheel with 28 teeth(enter shaft) will spin the pinion with 0.4 extra spin (exit shaft).is it right? ?
 
I've done, z1=20 z2=20*1.4=28
the pinion 20 teeth and the other wheel 28.
the wheel with 28 teeth(enter shaft) will spin the pinion with 0.4 extra spin (exit shaft).is it right? ?
you are correct.
 
you are correct.
now for drawing..pignone:
m= primitive diameter/number of teeth=50 d prim.(choice by case)/20=2.5
Step=2.5*3,141(pgreek)=7.8
addendum=2.5
dedendum...
Axial length(8-12)=11*2.5=27.5
inner diameter=50-2.5*2.5=49.75
outer diameter=50+2*2.5=62.5
height dente=2.25*2.5=5.6

These are the calculations that allow me to draw the pinion.
repeating the same procedure with the 28 tooth wheel.
Is that correct?

primitive diameter=50 is it correct that I chose it?? in the drawing is the diameter that is between the point drawn lines of the toothed wheel? those lines that divide the view of the tooth. .
 
now for drawing..pignone:
m= primitive diameter/number of teeth=50 d prim.(choice by case)/20=2.5
Step=2.5*3,141(pgreek)=7.8
addendum=2.5
dedendum...
Axial length(8-12)=11*2.5=27.5
inner diameter=50-2.5*2.5=49.75
outer diameter=50+2*2.5=62.5
height dente=2.25*2.5=5.6

These are the calculations that allow me to draw the pinion.
repeating the same procedure with the 28 tooth wheel.
Is that correct?

primitive diameter=50 is it correct that I chose it?? in the drawing is the diameter that is between the point drawn lines of the toothed wheel? those lines that divide the view of the tooth. .
calculations are right, allow you to draw the wheel. Usually you do not choose the primitive diameter but the module and everything else turns around it. unified modules are 0.5 - 1 - 1.25 - 1.5 - 2 - 2.5 - 3 - 3.5 - 4 - 5 - 6 - 8 - 10 - 12 and beyond... so if in your case you have to increase or decrease you are appropriate;)

the primitive diameter is that of ingracing the wheels, the ideal/real one depending on the cases, represented with line drawn point.

but have you already sized the bending, cutting and twisting shaft? have you already made the sizing and verification of the forces on the dentate wheels, the static and the fatigued one?
 
calculations are right, allow you to draw the wheel. Usually you do not choose the primitive diameter but the module and everything else turns around it. unified modules are 0.5 - 1 - 1.25 - 1.5 - 2 - 2.5 - 3 - 3.5 - 4 - 5 - 6 - 8 - 10 - 12 and beyond... so if in your case you have to increase or decrease you are appropriate;)

the primitive diameter is that of ingracing the wheels, the ideal/real one depending on the cases, represented with line drawn point.

but have you already sized the bending, cutting and twisting shaft? have you already made the sizing and verification of the forces on the dentate wheels, the static and the fatigued one?
but the module according to what I choose it?

My book doesn't tell me to check the forces of the dentate, static and dynamic wheels.. and what does it mean to dimensional the bending, cutting and twisting shaft? ?
 
but the module according to what I choose it?

My book doesn't tell me to check the forces of the dentate, static and dynamic wheels.. and what does it mean to dimensional the bending, cutting and twisting shaft? ?
What school did you do before? Did you do that?

the shaft on which there is the toothed wheel, bound according to bearings, will have actions/reactions with the other wheel because of the power and number of turns in and with the load applied downstream. then on the tree there will be torque moment, cutting and bending. it is necessary to schematize as beams, to determine the most stressed section, to see with von mises the section and then possibly to correct the diameter of the tree itself. Moreover, the dynamic verification of the tree is used for fatigue testing (cuts, fittings, gorges etc). idem must be done on the dentate wheels that could break immediately (transfer of crash for static resistance failure and then undersized) or to surface wear or bending after medium high cycles (see pitting and tooth bending).

the module or the hypotizzi and do static and dynamic verification, or presize it statically using the static lewis report or according to one 8862 through the synthetic surface factor reported to pressure and with the synthetic factor of the tooth foot reported to bending. then check dynamics to see if it is ok.
 
What school did you do before? Did you do that?

the shaft on which there is the toothed wheel, bound according to bearings, will have actions/reactions with the other wheel because of the power and number of turns in and with the load applied downstream. then on the tree there will be torque moment, cutting and bending. it is necessary to schematize as beams, to determine the most stressed section, to see with von mises the section and then possibly to correct the diameter of the tree itself. Moreover, the dynamic verification of the tree is used for fatigue testing (cuts, fittings, gorges etc). idem must be done on the dentate wheels that could break immediately (transfer of crash for static resistance failure and then undersized) or to surface wear or bending after medium high cycles (see pitting and tooth bending).

the module or the hypotizzi and do static and dynamic verification, or presize it statically using the static lewis report or according to one 8862 through the synthetic surface factor reported to pressure and with the synthetic factor of the tooth foot reported to bending. then check dynamics to see if it is ok.
thank you for the explanation..I did the high school class constructions of machines no...comanywhere in this course the prof did not explain anything about all this.I think I will study it in another course.I have the book technical industrial design, backincasa seems to me that it does not speak of this. .
 
thank you for the explanation..I did the high school class constructions of machines no...comanywhere in this course the prof did not explain anything about all this.I think I will study it in another course.I have the book technical industrial design, backincasa seems to me that it does not speak of this. .
Your decision is very courageous! Good! Since things are like this, you have to try to give a correct proportion to your design. Surely on the return home there will be a parallel axle reducer or his some other book and surely to have an idea of proportions you will find the suitable material. Since a real dimensioning has not yet taught you, try to make the most of the drawing book regarding the data of the toothed wheels, choice of the bearing, tolerances and finishes.

It's a half done job, I know. In fact, I don't understand why you don't do more "suitable" exercises or why you don't put your drawing courses after the design courses. . or because you do not begin to explain something in the drawing courses perhaps because you assume a pre-arranged tacnic institution requirement.
ot:
I feel lucky that in high school we had the mythical peppino that explained to me by thread and by sign all the part of design, and I still printed in the brain the mistakes not to do and why. at the university two drawing courses made a lot to the water of roses, so much so that I did not even purchase books. then natural predisposition to matter I always had the maximum faces. but shouldn't an engineer be the one who can design, draw according to regulations, what has greater and better knowledge? I think there's something to reform the school, but not how they do... otherwise...it's enough to steal and not teach.

Sorry for the oct... but today I am very critical :wink:
 
I've done, z1=20 z2=20*1.4=28
the pinion 20 teeth and the other wheel 28.
the wheel with 28 teeth(enter shaft) will spin the pinion with 0.4 extra spin (exit shaft).is it right? ?
but the motorcycle entry shaft should not be the one to which the smaller wheel is attached, so with the least number of teeth? otherwise it does not reduce.
 
but the motorcycle entry shaft should not be the one to which the smaller wheel is attached, so with the least number of teeth? otherwise it does not reduce.
ratio 1:1.4 is the transmission ratio such as i = z2/z1 = n1/n2. may also not be a reducer but a multiplier of turns. in this case n1 = 1 lap and n2 = 1.4 laps, as I understood it by consulting the vademecum for designers and technicians and other sources, by scruple.
then you have (definition current use):
i = 1:1 direct relationship
i = 1:5 multiplier of turns
i = 5:1 turn reducer
only on a manual of theskf I found that uses the reverse value as a transmission and wikipedia ratio, so we know as before.

it is written that with the use has been reversed the numberer with the denominator. therefore from all this disquisition, according to original definition of transmission ratio we have t = 1/i = n2/n1 then 1:1.4 will be a turn reducer!!!
then you have (definition original use):
i = 1:1 direct relationship
i = 5:1 multiplier of turns
i = 1:5 rpm reducer
 
just a few ideas since I was asked in mp, not to make the same mistakes and the same considerations.

- piece sprocket means tree with cut in single piece the pinion.
- caletta wheel with tongue means tree + wheel + tongue
- reducers have seals etc. this reducer already analyzed and all the recommendations in the development phase

pdf layout of concepts.
 

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just a few ideas since I was asked in mp, not to make the same mistakes and the same considerations.

- piece sprocket means tree with cut in single piece the pinion.
- caletta wheel with tongue means tree + wheel + tongue
- reducers have seals etc. this reducer already analyzed and all the recommendations in the development phase

pdf layout of concepts.
Besides the design of the reducer, in the exercise I have to redesign it and quote it, but some of my colleagues told me that you have to quote only the bigger wheel and not all the reducer and not even the pinion. why?

and then in the registration table I have to write the various calculations of the two wheels or only one?
 
Besides the design of the reducer, in the exercise I have to redesign it and quote it, but some of my colleagues told me that you have to quote only the bigger wheel and not all the reducer and not even the pinion. why?

and then in the registration table I have to write the various calculations of the two wheels or only one?
I would propose to read carefully the text your teacher gave you. I am also normdotated as you, so I can only read the text ... I can give my synthetic interpretation:

insert b: draw and quote in a complete way (quotes, geometric tolerances and position, machining etc.) the cast-wheel. if the pinion is piece and the wheel is the only organs of the exercise... what is there to do?

... because"... because the text says so. If it was a job... you know how many other things were listed and done. . .
 

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