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# Calculating loads produced by a jamming event of a rotating shaft

## Calculating loads produced by a jamming event of a rotating shaft

(OP)
Does anyone have any experience in calculating the loads a shaft would see if the tool on the end jammed? An example would be a cutter rotating at 30 RPM which hits an obstruction and stops the rotation.
Thanks
Russ

### RE: Calculating loads produced by a jamming event of a rotating shaft

Wouldn't it be the maximum torque the motor can output? Probably increase it by some sort of impact factor. You might find something in machinery's Handbook.

### RE: Calculating loads produced by a jamming event of a rotating shaft

Hi Russ

It would indeed be the maximum torque of the motor which might be either 2.5 or 3 times the normal running torque, do you have the torque curve for the motor?
With regard to the actual forces seen at the cutter end, that would be the maximum torque divide by cutter radius.

### RE: Calculating loads produced by a jamming event of a rotating shaft

(OP)
Thanks for the replies. I don't have the torque curve for the motor for now. I'll have to dig that out.

### RE: Calculating loads produced by a jamming event of a rotating shaft

Is this an existing system or a new design?
If existing, I would be working backwards from the prime mover and understanding the safety overloads from there and what they are set to. Once you know these limits (I.e. The purposeful weak link in the process) then you can understand the forces required to protect before any real damage is done.
If new, it is back to base principles of design to protect. On a cutter, I would expect numerous devices would come into action before an actual cutter stops. Going from 30rpm to zero when it is not meant to is usually classed as a catastrophe if nothing has stopped a motor or other prime mover beforehand.

### RE: Calculating loads produced by a jamming event of a rotating shaft

Russ 1911,
The load on the shaft would be based on the inertia (Iw^^2) of the driving system and the time that it takes to stop rotation. If the time is very, very short, the forces on the shaft (and the entire driven system) would be very high. This is why these types of systems are usually protected by a clutch or sheer pin, basically a mechanical "fuse". The motor torque capabilities would have very little to do with the shaft loading if the cutter is truly stopped. However, if you analyze all of the drive components, most of the system inertia would be found in the rotor of the motor, as it spins very fast. The mechanical fuse would have to be large enough to handle the peak torque of the motor.
Dave

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