doubts dynamic analysis shaft press eccentric (texts)

[GunMaker]

save to all, I do my thesis work on a tree (1) of an eccentric molding press, for the purposes of a fem analysis I need the forces that act in the more critical dynamic balance configuration of the shaft. In the press under examination the transmission is constituted by a asynchronous motor of 4 kw, a belt transmission that connects the pulley of the motor directly with the flywheel (2) which is connected to the clutch.
My problem is the following:the press was designed to ensure a nominal force of about 8 tf to 30 ° from the upper dead point (which I thought corresponds to the force necessary to the molding of the sheet) , at a nominal speed of about 20 rad/sec. This means that the tangential force t agent on the crank button is equal to 4 tf (at 30 ° from the pms). the total stroke is 80 mm (wire radius 40 mm and rod length 550 mm). I then know the degree of irregularity and the various inertias of the tree and of the various members calettati on it.

I have reasoned a bit on the question of the force agents on the tree to vary the configuration of the tavellism (see figure below) , not having at the disposal data regarding the course of the acceleration to the change of the angle of the handle (and not having at the disposal the diagrams of the couple motor and the torque itself), I can not accurately determine what is the torque actual agent on the tree to vary the direction of the tanger or is it wrong how reasoning make such approximation?

another thing I can't understand about that system is the following. in the project data appears the nominal force at 30 ° from the pms, but I did not understand if in such position of the crank the mold enters into contact with the sheet to print or is simply a convention that is adopted in the design phase (i.e. if the stamping stroke is larger than the space traveled from the sled in the 30 ° before the pms). because in the first case I have the impression that the tree never chooses but varies only its own angular acceleration that remains always positive (and that doesn't make sense)

for convenience I carry a qualitative diagram of the agent force on the sleigh (to be noted that it is reported to another machine but the course of the diagram is the same,as cinematism is equal) to the variation of the theta angle (valued in anticlockwise from the pms)
in case it is wrong, how is it correct to proceed in these cases?
I hope it was clear that the problem was raised.
thank you very much for the answers:redface:

 

[meccanicamg]

If you're making a fem for a press, I'd say you're in a company that builds presses, so it would be good if you asked your company first.
if they left you all alone if you can talk about it.

 

[GunMaker]

we remain mechanicsmg, for the design of presses and for the calculations rely on external design studies, they cure the part of tolerances/technical drawing and realization of the details, so from this point of view they can not give me a great help (because they produce numerous types of machines not only presses), most likely in the week I will see myself with a person (a firm collaborator) who has shown himself available to give me not a hand

 

[meccanicamg]

beautiful winery company... not even minds have owners.
Let us simplify the concept. the eccentric press is nothing but a biella ordinary centered crank. So once you get or copy the equations of cinematics and excels you have the trend. you have to overlap the effect between vacuum press and working force in the contact period with the lanyard and therefore the processing.
the press provides maximum strength when the crank is in square with the biella, i.e. 90 degrees. but if the race in the mold is smaller, the effort will be carried forward in the corner.
It is also true that if you take the corner later you have more speed and less potential energy....but you should see the mold to understand.
Usually even if you invent or hypothesize some things is ok the same.

 

[GunMaker]

so if I have well understood.. regarding the agent force on the sleigh (supposing the moment constant motor and therefore the tangential force t constant) use the expression:

= t = fsl * sen( θ + β) /cos (β)


where β is the angle that the biella forms with the vertical axis (depending on θ) . made this I find fsl (force agent on the sled to vary of θ). then calculate alternative inertia forces (other diagram).

at this point I should consider the resistant force agent from the moment you have the contact between punch and matrix until you have the perforation of the sheet (i.e. along the entire stamping stroke) right? Can I consider this working force constant and equal to fr=k*s*p (where p is the perimeter to be drawn, the thickness and k the resistance of the material) during the entire contact?

I also don't understand one thing. if by hypothesis at the contact between the sled and the sheet the fsl = fr and such fsl increases to the decrease of θ (as in the diagram on exposed I suppose that's correct.) , means that you will always have an angular acceleration of the tree (although very small give the inertias in play) , as for balance it is worth:

mm-mr'=i dw/dt

and mm will always be greater than m'r (where m'r takes into account the fr and alternative forces of inertia reduced to the axis of the tree)

if it is so how does the tree always rotate at the same speed? Perhaps the contact with the sheet is at an angle θ such that fsl < fr and this involves a deceleration compensated by a subsequent acceleration that occurs when fsl becomes greater than fr ?

I'm obviously wrong or I'm missing something at the conceptual level or I'm not clear about the print dynamics :frown:

 

[meccanicamg]

That's what you're doing. hairstyle and its strength varies to vary penetration even if the first constant approach can go well. but the springs react more and compensate not linearly. the flywheel is needed to stop the car.

 

[GunMaker]

perfect. ..unfortunately now I can't make diagrams with excel (I'm out of the house) but by tonight or in the morning I should post them to check if I understood well and my reasoning square ... apart from the diagrams so it is true that for a certain δθ the sum of the strength resistant that opposes the sheet (we assume the constant) and the reaction force of the springs exceeds the fsl (res from the couple motrice mm and relation to

 

[Sandra_ME30]

Tags 16.12.15 00:19

here are the diagrams that I traced today (on the axis of the ascisses is reported the angle theta in radianti between the pmi and the positive pms in the anticlockwise direction). Supposing the tree moves with an anti-clockwise w:

Annex

As far as the mould is concerned, I have been told that the moulding stroke is around 15 mm (long which the springs supposed for simplicity to linear behaviour react with fk ). From here I have deducted, given the handle radius, the angle from the pms from which you have contact between the punch and the matrix (about 62°=1,09 rad). Moreover I could not measure the fr that opposes the mold I considered it constant from that angle (green line). fsl and fia are the forces recurring respectively from the moment of the engine and the force of alternative inertia , to forces if I have well understood vary with w (while without information on acceleration/deceleration I can not however have a satisfactory quantitative response).

Annex 42645 Annex 42647

However, if the interpretation of the problem is correct, according to your experience:
- is during the δθ between 62° (vertical ascissa) and 30 ° (a meeting point between fr and fsl in the diagram) that the fly returns the energy well?
-both fsl that fia vary to the range of w ,to replace such constant w is correct as approximation being there a very low degree of irregularity?
-What is the strength that I should consider for the purposes of analysis, i.e. t(θ) max ? could I consider t constant and equal to the value it assumes at 30 ° from the pms?

Of course I think the last two points are related as if w is constant also t will be as a driving power=m*w=t*r*w well? (or at least I suppose it is desirable to keep this power transmitted next to the nominal power provided by the engine)

I do not know if I was exhaustive in clarifying my doubts. I apologize if I was confused in exposing the problem.

 

[Sandra_ME30]

Mechanical 16.12.15 02:56

the cinematism biella manovella was born to have variable accelerations of the pms to constant omega of crank. the flywheel tries to go after this bike but will oppose inversion area with maximum down pms and instead will be energy in the ski run. the flywheel also compensates for the arrangement due to the penetration of the actual print.

 

[Sandra_ME30]

Tags 15.08.

ok thanks mechanicmg, I think I have understood the overall operation:

Suppose we are at the top of the sleigh (the shaft rotates counterclockwise). We will find ourselves at a point of operation of the asynchronous engine characteristic (which depends on the frictions though modest involved in the rotation of the group). We are almost synchronism.
at some point the clutch is grafted:
- during the descent of the group it moves at constant speed since there is no resistance such as to alter the operating point.
- At some point you have contact with the sheet to be inserted/printed. At this point you have a deceleration because instantly the value of the mr agent on the handle exceeds the value of mm.tale deceleration is as much less push as the greater is the inertia of the flywheel.
- once the bottom dead end is finished, the sleigh goes up during this phase, the mr is back to be the one preceding the phase of the embutitura/stampaggio. This acceleration depends on the engine used (chosen according to the time available to return the flywheel to the speed of the regime).

So finally the maximum tangential force t depends on fsl (the latter should be considered in the most critical position i.e. when you have the contact with the sheet , which usually happens 90° before the lower dead point, in my case 62° first), in this condition you have the maximum effort of embutitura/printing and therefore you have the maximum strength along the biella .

Is that the correct reasoning? :smile: