Problem with the reaction forces in the gearbox

[Amaelle]

Homework Statement

Look at the image

Relevant Equations

Monet equilibrium
Conservation of energy
Gears ratio

Greetings, I have tried to find the reactions of the bearings on the Y direction in the intermediate shaft. but I couldn´t find the numerical results of the solutions

here is my attempt:

in order to find the forces I did the moment equilibruim around the point A and B

With FT1 FT2 and FT3 the tangential forces applied by the gears on the intermediate shaft and RyA and RyB are the reaction forces I´m trying to find.

as you could see my reaction forces are
RAy= 3831,63 N
Rby=-1402,92
the solution says
Ray=-3687N
Rby=278,52N
I don´t know what´s wrong with my approach!

thank you!

 

[Lnewqban]

The last picture is barely visible.
How are you obtaining the value of each tangential force on the gears?

 

[Amaelle]

thank you!

• for Ft1 :we started by calculating the torque of the motor

Cm =power motor/Wmotor=194,26Nm
then Ft1 =Cm/r1
with r1=(z1*m/2) m is the modulus of the gear
we got FT1=6167 N.

• For Ft3:
• we calculated w6 the velocity of the load 2 with the gears ratio we got w6=18,75 rad/s
• C6=Power load1/w6=985N.m
• by the gear ratio we could get C4=518,47N
• FT3=C4/r4=7014N
• For FT7
• w7=34,89 rad/s (by gear ratio)
• C load 2=power load 2/w7 =343,94N
• C8 =C load 2* z5/z6 =3275,6N,m
• FT7=C8/r8=3275,6 N

In order to have the reactions RyA and RyB I just made the torque equilibruim around A and B

 

[Lnewqban]

Amaelle,
Here are your errors:
The calculated values of the tangential forces are all correct.
The equations of moments about bearings A and B are not.

Revise the distances between each force and the bearings.
You need to make a distinction regarding the direction of tangential forces on the intermediate gears and shaft.
Consider the flow of energy, it moves downstream: the input driven gear "feels" different than the output driving gears.
The function of the reaction forces is to keep the shaft in its spatial position regarding the forces on the gears.

It seems that you have a persistent difficult with Free body diagrams and the direction of the forces.
I recommend you to re-visit that chapter, since you seem to be very strong in everything else you have done in this and other threads.

Also, are you supposed to calculate the reaction forces on both planes?
I ask because the solution values that you show seem to be accounting for the contact pressure angle (thrust force on plane z-x), rather than being reasonable for the power force on plane z-y.

 

[Amaelle]

Amaelle,
Here are your errors:
The calculated values of the tangential forces are all correct.
The equations of moments about bearings A and B are not.

Revise the distances between each force and the bearings.
You need to make a distinction regarding the direction of tangential forces on the intermediate gears and shaft.
Consider the flow of energy, it moves downstream: the input driven gear "feels" different than the output driving gears.
The function of the reaction forces is to keep the shaft in its spatial position regarding the forces on the gears.

It seems that you have a persistent difficult with Free body diagrams and the direction of the forces.
I recommend you to re-visit that chapter, since you seem to be very strong in everything else you have done in this and other threads.

Also, are you supposed to calculate the reaction forces on both planes?
I ask because the solution values that you show seem to be accounting for the contact pressure angle (thrust force on plane z-x), rather than being reasonable for the power force on plane z-y.

thanks a million! you mean that my feeybody diagram is wrong? if so this might explain why my results are wrong?
Could you tell what are the wrong directions?
yes the exercice ask about the the reactions on the the other plane by I can solve them once i fix this first!
thanks a million!

 

[Amaelle]

I used an online software to calculate the reactions and I got the same results if my initially supposed directions are correct.
(ignore the fact that I inverted the bearings here as I´m not calculating forces in the x directions right now)