design of a gearbox with two pairs of straight toothed wheels powered by a four-time diesel engine

[Calender]

At the "qualitative" level you will be led to choosing ball bearings, which still tolerate a slight axial load (without oblique contacts, contact points etc. that usually cost more),the system would not be labile in axial sense. .use the rollers only if the radial loads are very large and you do not have much mounting space, but in that case you have to limit the axial deviation somehow (ragion so I have often seen use at least one ball bearing even in these cases). .but it depends very much on the order of magnitude of the reactions/forces involved, let us know as soon as it arrives to the degrees of stress (binding reactions, bending motion, cutting etc.)

 

[Gabbro01]

quali formule should I use? perché non ho idea...

 

[meccanicamg]

But aren't you the third year of high school?
Did you follow her to the third and fourth year of mechanics?
Trave, you put the forces that generate the gears.... search the mechanical book (but do you have it?) or look on the skf manual how the radial force and tangential is calculated.
take a plan and calculate the sum of the horizontal forces equal to zero....then the vertical ones equal to zero....then the moment regarding a point equal to zero.
Same thing on the other floor.
so find the reactions to the bearings.

Is that what you do?

 

[Gabbro01]

for the balance of forces itself, only that these radial and tangential forces never made. .

 

[meccanicamg]

for the balance of forces itself, only that these radial and tangential forces never made. .

then look for how to calculate on the wheels....from the couple.... primitive diameter....forza

 

[PIETRO2002]

for the balance of forces itself, only that these radial and tangential forces never made. .

due to the pressure angle (20° ) the contact between the teeth is composed in two forces, a tangential and a radial. This applies to straight teeth.

 

[Gabbro01]

then I made the calculations:ft=4972,2 n fr=1810 n
ftotal = 5291,4 n
I ran the calculations compared to the wheel radius, all right?

 

[meccanicamg]

here the scheme of forces on a helical wheel.
If the angle of helix is zero, that is, straight teeth, you will have that axial force is nothing.
the radial force wr is always and however directed towards the center of the wheel in the studio.
the tangential wt force of the wheel in the studio is always the force that undergoes.

 

[PIETRO2002]

then I made the calculations:ft=4972,2 n fr=1810 n
ftotal = 5291,4 n
I ran the calculations compared to the wheel radius, all right?

If you have two gears, on the intermediate shaft you will have two radial forces, the tangential forces transmit torque do not charge the shaft in bending or cutting. .

 

[meccanicamg]

then I made the calculations:ft=4972,2 n fr=1810 n
ftotal = 5291,4 n
I ran the calculations compared to the wheel radius, all right?

Why do I need to search in a thousand posts that data you used? You can't write it all together so let's see if you did it right?
then sooner or later?

 

[meccanicamg]

If you have two gears, on the intermediate shaft you will have two radial forces, the tangential forces transmit torque do not charge the shaft in bending or cutting. .

No. The cut you have on your strength. also the bending because the radial and tangential forces make the trees sting, that if not rigid enough eat out the teeth.

 

[Gabbro01]

 

[Calender]

from the manual skf

radial forces as well as tangential forces are transmitted to the tree and its supports. because the contact point is one for each pair of dense wheels.

 

[Gabbro01]

the number deleted is an 8

 

[meccanicamg]

View attachment 58305

I lost something.
the transmitted torque must be the multiplied nominal 1.5 which is the worst working condition.
Also do not compose now with pitagora the forces because you have to calculate the internal actions otherwise you are not with the plans and angles... but the pitagora you hold it for the calculation of the duration of the bearings passing by the equivalent load.

Are you sure you made the calculation on the pinion of the second couple and on the wheel of the first couple?

 

[Gabbro01]

 

[Gabbro01]

I was wrong about the first couple of wheels
the couple is already multiplied

 

[meccanicamg]

input 30kw-1600rpm-179nm multiplied by ka=1.5 makes 268nm but incoming.... then the multiplication by the reduction ratio of the first stage (i=2).... and becomes 537nm on the wheel conducted....not what you used in the first account.

for straight teeth wheels the formulas of forces are these:

 

[Gabbro01]

 

[Gabbro01]

these are tangential and radial forces