Contact between skewed cylinders

[Luiz Cavatte]

Hi,

when two cylinders in contact, but not parallet to each other (skewed), are rotating, there is a development of forces between them which tends to push one cylinder along its rotation axis and the another one to the opposite side. In other words, there will be translation movement between them.

Could someone explain me why that happens?

This is the main mechanism to control thrust force in rotary kilns. Rotary kilns are normally supported with riding rings and support rollers on each pier. As the rotary kiln has a small slope to allowd the material fed in one side to go throught it, the thrust force due to the weight is controlled by skewing some support rollers in order to push the rotary kiln upwards and then keep it in balance.

Thank you

 

[lpfr]

I suggest that you make a drawing: Y axis upward. Cylinder A transparent and under it, cylinder B opaque.
Take a cylinder (call it A) whose axe is aligned in the Y direction. This cylinder turns and the near side moves right and the far (lower) side modes left.
Under this cylinder there is another (call it B) whose rotation axe is tilted (say) 30° clockwise.
The rotation of cylinder A exerts a force directed to the left on cylinder B. You can decompose this force y two perpendicular forces: one parallel to the axis of rotation of cylinder B. The force perpendicular to the axis of rotation of B will make B turn and it will serve to overcome frictional forces that oppose rotation.
The force parallel to the axis of B will push the cylinder B downwards. If nothing retains this cylinder in place it will descend (down and left).
But if cylinder B is impeached to move along its rotation axis, the force needed to do this will be transmitted to cylinder A which will "feel" a force directed upwards and rightward (at 30° of the Y axis in this example). To make the cylinder A turn you must provide a torque (and a power) to do this. You must also avoid the cylinder A to move upwards. This can be done tilting the drawing (and cylinders in the drawing) in sort that the upper side of the drawing is higher than the lower side of the cylinder. This is what you described about the inclination of the cylinder kiln.
As the kiln glides over the "misaligned" rollers, there will be a work done by the friction forces and some power lost.
Does this helps you?

 

[Luiz Cavatte]

Hi,

thanks for your explanation!
Actually i didn't understand your cylinders references and in which direction cylinder A rotates.
I attached a picture of a rotary kiln. You can see the kiln shell, the riding ring and a support roller. I put arrows to indicate the riding ring and roller rotation direction. The arrows on the bolts of the roller bearing housing shows one example of how the roller can be skewed. Here the upper bearing is pushed "out" and the lower bearing is pushed "in".

Could you explain how the forces develop in this example?

Many papers say that when the rollers are perfect parallel to the riding ring, there is pure rolling motion but, on the other hand, when the roller is skewed, actually sliding also occur in the ring/roller contact.

Could you explain that?

Thank you