Free body diagram, bearing forces

  • Thread starter Kalus
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Hi,

I wish to work out the reaction forces on the following diagram:

[PLAIN]http://img227.imageshack.us/img227/7055/imag0211b.jpg [Broken]

The circle is a gear with tangential force acting at Ft and radial force acting at Fr.

A and B are bearings that take the Ft and Fr forces.

I have found the reaction forces opposing the downward Fr in the X-Y plane, which are Ay = Fr*(La/L) and By= Fr*(Lb/L)

In the X-Z plane there is the sideways reaction forces due to Ft, which are Az= Ft*(La/L) and By= Ft*(La/L).

Now, I know that is not all that is to it, because the force Ft acts at half the diameter of the gear wheel, producing a torque.

My question is, which direction does that torque go and why? One of the above reaction forces needs to be modified to take this into account.

Many Thanks,

Kalus
 
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Answers and Replies

  • #2
nvn
Science Advisor
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Kalus: By the way, please do not post wide images directly to the forum page. Just post a text link to wide images.

If r is the gear radius, then your applied torque is T = r*Ft. Assuming the bearings at ends A and B are roughly the same, then applied torque T would be reacted by a torsional moment reaction, Mx, about the x axis, at each end A and B, where Mx is approximately, Mx = -0.5*T. But if the ends cannot generate this much torsional resistance, the gear will rotate.
 

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