How Does Rolling Resistance Affect Torque in a Sealpress Mechanism?

[ram18]

hi ,

I am going to design sealpress mechanism for rotary system . I need to find out how much torque required to rotate my steel roller which is connected to end of my cylinder. from wikipedia i find out rolling resistance =


The force of rolling resistance, not adjusted for velocity, can be calculated by[3]:
F = Crr x Nf

where

F is the rolling resistance force (shown in figure 1),
Crr is the dimensionless rolling resistance coefficient or coefficient of rolling friction (CRF), and
Nf is the normal force (equal to W shown in figure 1, not just N shown in figure 1).
The coefficient of rolling friction for a slow rigid wheel, not adjusted for velocity, can be calculated by[1][2]

Crr = Sqr of z/d
where

z is the sinkage depth
d is the diameter of the rigid wheel
The force of rolling resistance can also be calculated by[1]:

F= Nf Xb/r

where

F is the rolling resistance force (shown in figure 1),
r is the wheel radius,
b is the rolling resistance coefficient or coefficient of rolling friction with dimension of length, and
Nf is the normal force (equal to W shown in figure 1, not just N shown in figure 1).

In my case I have steel roller and i am going to press buna(rubber) seal by roller .

So if i assume b = 0.4 inch , r = 1.75 , nf =500 pound

F = Nf x b /r = 500x 0.4 / 1.75 = 114.28 pound
so torque = force x radius = 114.28 x 1.75 = 199.99 inch-lb

If i change the radius of roller r = 2.25

F = Nf x b / r = 500x 0.4 / 2.25 = 88.88 pound

Torque = 88.88 x 2.25 = 199.99 inch -lb

so i got same torque that means radius of roller doesn't effect of torque . i am confused about that so please give me some right direction .

Thanks.

 

[Unrest]

hi ,
so i got same torque that means radius of roller doesn't effect of torque . i am confused about that so please give me some right direction

Kind of makes sense that it wouldn't affect torque. But don't forget that a larger roller will need fewer revs to cover the same distance, so it'll use less power at the same sealing speed. I assume you have a motor driving the roller's axle and not just pushing it by the cylinder.

 

[ram18]

yes i put motor to roller's axle.

 

[ram18]

hey my roller width is 18 inch so how can i use that in my equation . i think i need more torque becoz of that can anyone help me on that.

 

[Unrest]

hey my roller width is 18 inch so how can i use that in my equation . i think i need more torque becoz of that can anyone help me on that.

The same equations should still give you the torque. But the normal force would be higher with a wider roller to achieve the same deformation depth. You'd have to do an elasticity calculation to work out how much normal force is required.

 

[ram18]

thanks unrest . which elasticity calculation work here ?

 

[Unrest]

thanks unrest . which elasticity calculation work here ?

Yea somewhere you'll have to work out the relationship between force and deformation. It'll be tricky to do it very accurately because of the non-linear elasticity of rubber.

I'd start by treating the material as a rectangular block with about the area of the projected area of the roller. Have it compressed uniformly into a thinner rectangular block. Then you can just use pressure=stress=elastic modulus * strain.

To do it properly and account for the curved shape you'd probably need to do a non-linear (because it's rubber and because you don't know the area in contact with the roller) finite element analysis.

I guess there are probably books of data and approximate formulas for this type of problem tho, because it must be a very common one.