force distribution

Dear kkrish:

Presumably you have supports at A and B?

You use the fact that the rod is not accelerating rotationally, or linearly. This means the net torques from all the forces is zero, and the sum of the forces is zero. So at support A, write an equation that contains f_B and F1, and the distances from A. Write an equation that has f_A, f_B, and F1 equal to zero.

That should be all you need... then solve the equation-set.

David A. Smith

Thanks for your reply , A and B are supported by bearings .I did not know that the force distribution will change when the rod rotates.It will be of great help if you tell how to proceed when the rod rotates about its axis .Soory for not being clear in my first mail.

N:dlzc D:aol T:com (dlzc) wrote:

Dear kkrish:

Moving your response to the end.

Axial bearings, such that the long axis of the rod and the rotational axis of the bearings is the same?

You didn't do much better here. If the bearings rotate like I said above, there is no change in what I said before.

If the bearings allow motion "in and out", so that the "rod near A" can move towards B (and vice versa), there is no change in what I said before.

If the rod is free to leave contact with either/both A and/or B, good luck to you. ;>) You will need to know the mass of the rod, and its distribution along the rod.

David A. Smith

That's basically a beam with 2 simple supports (or close enough). Statics is enough to solve that. 2 equations will do it.

Now hang on a darn minute. David was considering body rotation of the long axis, not around the long axis. If a round shaft is contrained by two bearings, a cross force is reacted by both bearings.

Saying again: the bearing force at A times its distance from the load equals the force on bearing B times its distance from the load.

If not, the shaft moves like the hand of a clock. Supposing there is no bearing atthe rt end

Brian Whatcott Altus OK