|May5-07, 10:33 AM||#1|
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.
|May6-07, 09:19 AM||#2|
Start with the simple case when the rollers are crossed at right angles. In that case, it's obvious that if one roller turns, it tries to push the other one along its length.
When the rollers are at an arbitrary angle, the rotation of one roller can be resolved into components "parallel" and "perpendicular" to the other roller. If the rollers are nearly parallel, the "perpendicular" component is small, but it gives the same effect as if the rollers were crossed at right angles.
|May18-07, 12:08 AM||#3|
thanks for your explanation!
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?
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