loads on each rollers 2

[cgelang]

Hi everyone.
I have 9 rollers in semi circle arrangement, receiving 3 tons of total loads.
I want to get the load of the center roller, anybody points me out where can I get this information? or how.
Thank you so much.

Regards,
cgelang

 

[Walterke]

Try this

NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000

 

[MintJulep]

Minimum: 0
Maximum: 3 tons

Generally somewhere in between.

 

[rb1957]

a reasonable assumption is that the vertical component of the roller reactions obeys a cosine distribution.

then sum(cos(theta)^2)*P0 = applied, where theta is the angle (to the vertical) of the roller loads, and P0 is the load in the roller on the CL (the one you're interested in).

another assumption is that the roller loads are equal ... sum(cos(theta))*P0 = applied

Quando Omni Flunkus Moritati

 

[tgrimley]

Do you know the location of the CG? Get a scale and measure each roller?

 

[MintJulep]

a reasonable assumption is that the vertical component of the roller reactions obeys a cosine distribution.

I don't think that's reasonable at all.

Perhaps if all of the rollers are perfectly aligned with each other.

Perhaps if each of the roller supports deflects identically.

Perhaps if the "pallet" (if there even is one) is perfectly flat and very stiff.

 

[rb1957]

"Perhaps if the "pallet" (if there even is one) is perfectly flat" ? ... i took "9 rollers in semi circle arrangement" to mean that the rollers were in a semi-circular trough (or bed) and the load is applied by a cylinder.

alignment and stiffness variations ... sure that an assumption. similar rollers ... i think it's reasonable to assume that the rollers are essentially the same.

Quando Omni Flunkus Moritati

 

[MintJulep]

Ah, we have different pictures of the problem in our minds.

My picture looks like this:

 

[rb1957]

fair enough ! ... another well worded question ... not !

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[dhengr]

The worst possible thing an OP’er. can do is provide a sketch with some loads, sizes and dimensions, which would better explain his/her problem. Sketching used to be a valuable tool for an engineer to explain his problem or concept. Particularly so when the others can’t see it from where they are. This would reduce much of the guessing, but what’s the fun in that, and then most of us would have nothing left to do. Our bosses would see that we had nothing to do on their time and dollar, and fire us. Then we could devote full time to guessing games on Eng-Tips, and not have to worry about the boss getting all pi$$y, thinking we should be doing his work on his computer and dollar.

 

[SnTMan]

For a first pass, how about 3(2000)/9 lbs. Accurate enough?

Regards,

Mike

 

[Tmoose]

My mind picture looks kind of like this, but upside down.

http://www.maintenanceworld.com/Articles/timken/calculatebearing/diagram5.gif

 

[cgelang]

Thank you guys for sharing your ideas to this problem.

Thanks MintJulep, the arrangement of the rollers is like what Tmoose has attached.

@rb1957-I agree with your assumptions, but i modify a little to be more close. I just assume that the force by the vertical vector component on 8 rollers(4 @ left & @right) are the same.(though its not, but fair assumption)
Thus load at center roller = 3tons - sum of 8vertical forces on rollers.

Thanks.

Regards,
cgelang

 

[tbuelna]

cgelang-

If your rollers have an arrangement of constraints similar to a cylindrical roller bearing then you can find an analytical approach in any rolling element bearing text. The actual load distribution at the rollers can be greatly affected by the stiffness of the components, dimensional accuracy of the components, and any installed preload or clearance in the assembly.

Hope that helps.
Terry

 

[rb1957]

it sounds like you want to say each roller is carrying the same load (= 1/3 ton) ... ok, it's your assumption.

if you mean you want a symmeric loading (a reasonable assumption) then my post gave two options for that. note my 2nd loading assumed each roller carried the same load, but as a vector, normal to the contact cylinder; so only the vertical component is effective in reacting the applied load.

btw, Tmoose's link wouldn't open for me.

Quando Omni Flunkus Moritati

 

[Tmoose]

Tmoose's link, now an attachment

 

[Tmoose]

Well downloading the GIF somehow inverted the colors badly.

 

[rb1957]

yeah, that's my original thought, for a cosine distribution.

Quando Omni Flunkus Moritati