disk joint problem

[riccardo432532423]

a disc joint connects the shaft of a drive machine that develops the power of 40 kw with the shaft of an operating machine whose resilient moment, at regime, is 700 n*m.
determine

1) dimensional the tree;
2) perform the sizing of bolts connecting joint disks;

Hi, you could explain to me the resolution of this problem from the beginning, where I would be very grateful!

 

[meccanicamg]

if power, torque and rotation regime are the three main variables that describe a mechanical action that performs a mechanical work.

the pair is a force multiplied by its arm of action and therefore measured in [Nm]. the rotation regime would be the omega angle speed that is measured in [rad/s]. the power is the indication of the torque in the speed effect, then:

♪[ P=C•\omega \]where the power is in [W]as the angle speed is more comfortably indicated in [giri/min] you have the compact formula of power equal to:
♪[ P=\frac{C•n}{9549,2} \]where the power is in [kW]after this premise you have to consider that the data you have provided are:
- power (no use at all)
- Couple.
you can make out if you need it, as completion
- rotation system
now you have to dimension the tree that undergoes as the only action described in the text: the torque or twisting moment, which is equivalent to a round force.
so now it has to be sized torsion the section, assuming a material with its characteristics.
we take a carbon steel for general use c45. being a non- fragile ductile steel has as a characteristic of yielding the yield.
I look for an online catalog or a manual and see that through its yield load rs=400mpa.
the permissible voltage torsion octahedral is:
♪[ \tau_{adm}=0,58•\frac{Rs}{K} \]with k the safety coefficient that normally applies 1.5 for normal applications but can also rise to doubles beyond if there are strong shocks and special conditions.
a circular section with a torsion diameter from a torque moment mt generates a torque tension that is worth:
♪[ \tau=\frac{Mt}{Wt} \]where wt is the torsion module of the section under examination that for a full round is worth:
♪[ Wt=\frac{Jp}{r} \]where jp is the moment of polar inertia and r is half diameter, that is the maximum baricentric distance.
♪[ Wt=\frac{\pi•d^3}{16} \]now equaling the admissible voltage and the actual one is obtained as a reverse formula the diameter of the tree
♪[ d≥\sqrt[3]{\frac{16*mt•k}{\pi•0,58•rs}}\]

The theory also finds it qui.

so find about this....30mm of tree:
all clear here?
You're counting and telling me.
then we talk about the joint....

 

[riccardo432532423]

hello by mt what is the formula? I remember the prof telling us that in some cases
we could hold a specific value, like 9.549,000 if I don't remember badly
for the rest thanks again of the attention:d

 

[riccardo432532423]

ah then admissible tens can you also write with the symbol of sigma?

 

[PIETRO2002]

ah then admissible tens can you also write with the symbol of sigma?

no sigma is for logitudinal tensions due to tensile, compression, bending efforts.
tau is for transversal efforts like twisting and cutting.. try to understand the formulas of mechanicsmg .. and I recommend studying mechanics without copying the formulas! :

 

[riccardo432532423]

ah ok thank you, absolutely yes

 

[meccanicamg]

hello by mt what is the formula? I remember the prof telling us that in some cases
we could hold a specific value, like 9.549,000 if I don't remember badly
for the rest thanks again of the attention:d

depends on the units of measurement.... You can find it on your book for sure and then you can get it.
yours is for working in n•mm while normally using n•m in fact there is 1000 difference between formulas.


Then... before you go crazy... 0.58•rs is equal to \[ \frac{1•Rs}{\sqrt{3}} \]

 

[PIETRO2002]

depends on the units of measurement.... You can find it on your book for sure and then you can get it.
yours is for working in n•mm while normally using n•m in fact there is 1000 difference between formulas.


Then... before you go crazy... 0.58•rs is equal to \[ \frac{1•Rs}{\sqrt{3}} \]

this simplified formula makes you get tau knowing sigma.

 

[meccanicamg]

As for the hard disk joint, we have talked about it many times on the forum.

I will train you the formulas for:
- proportional the size of the joint
- calculate the friction joint
- determine the type of screws second a standard that is still taught but has been withdrawn uni cnr 10011
as regards the vines, using the current legislation, i.e. en iso 1993-1-8 which is the eurocode 3 has:

and therefore, for the traction bolt you will have depending on the precarious system that you need to check:

and determine how

and therefore occurs disc and and screws.

right to give you possible solution numbers we are talking about these:
and then select screws that are sufficient to ensure a traction force that keeps the two discs united by friction:understandable?
We want to see your thoughts too.

 

[LupoLucio]

in the first image on the disc joints (it does not want to be a correction, but only to try to understand...) the writing is reported: 'junction for friction between two discs with bolt cut resistance' but shouldn't it be a tensile strength? Or do I misinterpret?

 

[meccanicamg]

in the first image on the disc joints (it does not want to be a correction, but only to try to understand...) the writing is reported: 'junction for friction between two discs with bolt cut resistance' but shouldn't it be a tensile strength? Or do I misinterpret?

What you point out is true. the question is that the two records work attrite.
to generate friction it is necessary to pull the screws. if the screws fail to transmit it will have that the screws will undergo cutting action (any book does all three checks so that the joint is safe to 200%).
Unfortunately that series of notes was part of a series of pages where there were also joints with the pins and therefore with cutting resistance.

however the classic friction joint has the screws under traction.

 

[riccardo432532423]

hello thanks again of attention, a question if I choose a cementing steel
to make the sizing of the tree is fine or better one by reclaiming

also the 18nicrmo5

 

[PIETRO2002]

As a rule, it is better to use a steel that guarantees the appropriate working conditions and has a low cost!
cementing steels are used if you need a high surface hardness and a tenacious heart.
If your tree is mounted on bearings, I think a rectifying steel can go more than well.

 

[meccanicamg]

hello thanks again of attention, a question if I choose a cementing steel
to make the sizing of the tree is fine or better one by reclaiming

also the 18nicrmo5

In your case, given the efforts that are still not low, even properly broken down, it makes suitable a cementing and tempering steel like the 18nicrmo5 or similar, which are able to guarantee superficially excellent hardness but also high tenacity for the bending resistance of the tooth.
However if you put everything in excel you have the possibility to clear the numbers of the mechanical characteristics and see if you are there or not with the checks.
maybe only the inner toothed crown you can make it in steel by transfer it must be tied 42crmo4 with 56hrc.

 

[PIETRO2002]

In your case, given the efforts that are still not low, even properly broken down, it makes suitable a cementing and tempering steel like the 18nicrmo5 or similar, which are able to guarantee superficially excellent hardness but also high tenacity for the bending resistance of the tooth.
However if you put everything in excel you have the possibility to clear the numbers of the mechanical characteristics and see if you are there or not with the checks.
maybe only the inner toothed crown you can make it in steel by transfer it must be tied 42crmo4 with 56hrc.

I didn't know the tree had a tooth.. Did I miss something?

 

[meccanicamg]

I didn't know the tree had a tooth.. Did I miss something?

the solar has few teeth and is of a small diameter, therefore all the integral and for this better than cemented it.

 

[PIETRO2002]

I'm sorry, but aren't you wrong? here we are in the discussion of the joint to disks not to the epicloidal reducer! :

 

[meccanicamg]

I'm sorry, but aren't you wrong? here we are in the discussion of the joint to disks not to the epicloidal reducer! :

You're right. I got a wrong place.
I'm really getting reluctant? ? ? ? ?

 

[meccanicamg]

hello thanks again of attention, a question if I choose a cementing steel
to make the sizing of the tree is fine or better one by reclaiming

also the 18nicrmo5

Forgive me, I have a wrong post to answer.
You who are making a disc joint, since it is small, I would not go to take quality steels, otherwise microscopic joints come out.
Moreover the trend is at normal industrial level fairies in c45 and tied only if integral with the pinions.

 

[riccardo432532423]

hi the project is finished, everything went well but I have to make finishings to expose it to the best,
in the formula of the diameter of the tree what is 0.58?