Cam Roller positionning to prevent slip

[Chronik]

Hi,

I am working on a bracket that holds a 150 pounds shaft that can support rolls that weight up to 2500 pounds.

The shaft is going to be rolling on the cam at variable speed throughout the process, as the rolls become smaller and lighter. (init speed ~ 20 RPM, end speed( smaller roll, light weight) ~ 200 RPM.)

My question consists on the positioning of the cam followers. What angle/position should i position the right cam follower so that the shaft doesnt lift up and hit the marked (GAP) area on the drawing? I tried some stuff but i can't come up to any equation that makes sense to me.

FYI, the shaft is rolling clockwise.

Thank you for your help.

 

[zekeman]

Where is the cam? I only see 2 roller supports for the variable diameter rolling mass. What drives this mass?
What is F1?

 

[Chronik]

Hi,

thanks for your reply.

F1 represents a force applied by the belt system. This system consists of an arm that applies pressure to the roll, and the belts on this system generate the rotation of the shaft (by rotating the roll).



R1 is a resulting force as well as R2.

I do believe that as long as the roll is heavy, there will never be a problem. I am worried about when the roll will be lighter, lets say around 50 pounds, the force F will make the roll shift up on and become in contact with only one of the cam followers (the one to the right) and the bracket.


Thank you.

 

[SteveMartin]

Presuming your 'cam followers' are the two round supports shown, a simple answer would be to position the two rollers such that the line of action of F1 (or less conservatively, the resultant of F1 and mg) passes anywhere between the center points of the two rollers. This will drive your large roll into the support rollers, not up & over them.

 

[Chronik]

Hi,

thanks for the reply,

It is impossible to place the two cam rollers so that the F points toward the middle of the two rollers.

We might need to add a third roller, but i would really like to be able to calculate if the "mg" is enough to compensate for the horizontal portion of the F force.

Thanks

 

[zekeman]

Good problem for entry engineering candidate?? LOL

If you can give us an idea of the belt geometry with some dimensions then you can get a good answer.

As a first conservative model, assume the right roller is stalled and the belt with a vector force to the mass attempts to overturn it where the tangent point at the right roller to mass is the pivot.Then you can write the condition of stability

(1) MgRsin@> FxL

which assures that the left roller is in contact.

F= belt vector force
L=vector distance from pivot to belt contact
FxL= F* L* sine angle between vectors
R= Mass radius
@ angle of R vector to the vertical


If (1) is not satisfied, then you can write a less conservative expression involving friction in the right roller.

 

[Chronik]

Hi,

Thanks for your reply,

I am uncertain of the consistency of the resulting FBD of your equation.

I tried to change the axis system (17 degrees, c-clockwise so that the "x" axis is now straight up from center roller 1 to center roller 2.)

i tried to sum forces on the new Y axis, considering the sum is 0. thinking i could find the resulting force R1 and R2, and then projecting them on the x axis to calculate if

R1 cos @ + F sin @ > R2 cos @ + mg sin @ .

if left part was bigger, it would meant that there is the unwanted movement to the right.

Tho, i did not manage to find a way to calculate the resulting R1 and R2 forces... (too many unknowns for the number of equations i manage to formulate).


Feel free to reply, but don't feel bad if you do not have the time to. I am heading back home and i will probably just decide to build the bracket as is and we'll work our way around by adding another roller on top on a pivoting bracket if we find out that the shaft does go hit the bracket.

Thanks again.

 

[tygerdawg]

Without knowing the entire situation, I'd say add a third roller and make it bombproof. Don't "hope" for things to work, force them to work. For a little extra effort & extra money, your name will be forgotten. When the machine fails to work well and becomes a chronic mainteance or downtime problem, you'll be remembered forever.

TygerDawg
Blue Technik LLC
Virtuoso Robotics Engineering

http://www.bluetechnik.com

 

[zekeman]

"R1 cos @ + F sin @ > R2 cos @ + mg sin @ "

You have 2 problems with the problem
1) if you know the vector force F which should include a tangential component, then the exercise in equilibrium to get R1 and R2 is textbook simple. Then, if
R2 <0 it unstable
R2>0stable
2) But, I doubt that you know the vector force, F.

 

[Chronik]

tygerdawg,

i guess you are right. i'll go this way as i have to fix this asap.

zekeman,
you are right, i do not know everything about "F". i could try to calculate it, but that would probably be a waste of time.

Thanks for your help guys.