Force/moment resolution for shaft loading

[ewans]

Hi all,
I'm having some difficulty reconciling the loading for the following scenario and was hoping for some help.

I have a Y-shaped structural member (looks like a tuning fork or motorcycle rear swingarm) with the "legs" of the Y rigidly attached to a rotating shaft. Overall length (or height) of the Y is approximately 48", and the spread of the legs is approximately 12". The Y-shaped member has an applied external load at its end, causing a moment about the axis of the rotating shaft (i.e. with a 48" long moment arm).

Is the moment where each "leg" of the Y attaches to the shaft is simply 1/2 the total moment? I'm thinking about what the moment diagram should look like so I could consider torsional loading on the shaft.

Additionally, if I had an opposing moment applied to one of the legs (for example, via a gear which was mounted rigidly to one leg), what moment load (if any) would exist at the other leg, and would there be torsional loading of the shaft between the two legs?

I hope the above description makes some sense, and thanks in advance for any help.

Regards,
Ewan

 

[rb1957]

just so as understand your problem ... the V of the Y is attached to the axle and the tail of the Y is sticking out radially ? and you have a tangential force applied at the end of the tail ??

the moment you're dealing with is out-of-plane, so 50/50 doesn't sound too bad. the applied force would also divide 50/50, a componet would cause bending in the arm and the other component would cause axial load. note the moment reacted by the arms is less and the difference is taken up in the couple between the arms' axial loads.

 

[ewans]

Hi rb1957,
Thanks for your reply... appreciated.

You described the scenario correctly with the 'V' attached to the shaft and the tangential force applied at the end of the tail.

If I applied a moment load on only ONE of the legs of the V, and this load opposed the one applied at the end of the tail, what's happening at the other leg ?... is it seeing any load, or is it just being stabilized by the shaft?

Thanks again.

 

[Wrenchbender]

I think drawing FBDs of the various components would be the best approach. Also add in rotational effects if appreciable.

 

[rb1957]

if you apply load to one of the arms of the Y, most of the load will be reacted at the base of the arm, and some would be reacted by the other arm ... the arm isn't a cantilever

 

[ewans]

Thanks Bestwrench and rb1957 for your replies.

I've attached a PDF sketch of the scenario just to clarify the situation. I'm wondering if the angled legs of the 'Y' that I described initially was misleading. The legs are actually perpendicular to the axis of the shaft, but I don't suspect that it changes the loading.

With reference to the sketch, the moment about the shaft due to Fc is opposed by Fb*Lb.

rb1957,
I presume the sketch depicts what you already understood. Any chance there's a simple method you could outline for me that would allow approximating the load reacted at A and B? As I mentioned before, I'm trying to determine the torque at positions A and B for the purposes of sizing the shaft for torsion and taperlock hubs for the shaft/leg interfaces.

Thanks again to all.

 

[rb1957]

a picture's worth 1,000 words, and some ...

if Fb is being reacted by Fc, ie if the "Y" is mounted onto the shaft with sliding bushings, then the other leg of the "Y" (away from Fb) is doing just about squat ('cause it can't react torque).

i'd expect Fb to be reacted at B, (if a part of Fb is reacted at A, what reacts the moment induced ?) and Fc would be split between A and B.

like Bestwrench said, start with a free body diagram.

 

[ivymike]

a simple FEA should suffice?

you could force the shaft torsion to be very close to zero by allowing the "fork" to pivot on the shaft.

As it is, unless this fork thing is made of something very flexible compared to the shaft, the torsion in the shaft will be small.

 

[ewans]

Hi rb1957,
The "Y" isn't mounted with sliding bushings to the shaft, it's rigidly connected to the shaft and the whole shebang is supported by outboard pillow block bearings.

This being the case, I presume there's something happening at A and in the shaft between A & B, I just don't know what.

Can I bother you for one last reply given the above clarification?

Thanks again... appreciated.

Ewan

 

[rb1957]

ok then ... Fb isn't balancing Fc, they're just two different shears/torques applied to the Y. if the Y is welded (rigidly attached) at the end of both the arms then i'd share the remote load (Fc? ... I can't open your pic this morning) between the two bearings 50/50, shear and torque, and Fb (to load near the shaft) would all be reacted at the near housing (ptB?)

 

[swearingen]

The attached rudimentary sketches show my interpretation of what's going on. The top sketch is just the FBD to show what goes into the pillow blocks.

The bottom sketch shows a severely deformed Y under static load. If the Y is welded to the rod, then torque does develop between the forks. This torque is dependent on the stiffness of the Y in torsion, as well as what type of torsional restraint (if any) is at point C in the OP's sketch.

Of course, not welding the forks to the shaft would relieve all torque on it, but would stress the forks more.

To get comfortable with the numbers, I think a good FEA model is in order.


If you "heard" it on the internet, it's guilty until proven innocent. - DCS

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[Wrenchbender]

For finding the reactions and moments at the supports, I would treat it as a statics problem. You have force vectors, radius vectors in cartiesian i, j, k, cross product for the moments, etc. I think that might work as an alternative to FEA. (If you like vectors, that is.)

 

[ewans]

Many thanks to all for the replies... much appreciated.

Particular thanks to swearingen for the sketch and rb1957 for sticking with this thread.

To quickly respond to some of the suggestions...

- Unfortunately, I don't have any experience with an FEA package, otherwise I would certainly have started there.

- It's not the reactions and moments at the bearing supports that I'm concerned with... I can resolve these forces using a FBD as suggested. It's the reactions where the legs of the Y are welded to the shaft and, by extension, the torsion generated in the shaft between the legs.

 

[rb1957]

so (now) it looks like the torque from Fb is reacted by the torque from Fc ? so there'd be some (50%) torque between the arms of the Y.

 

[swearingen]

The issue is that it depends on the torsional stiffness of the Y as a whole. Look at the extreme case of the Y being infinitely stiff - there would be no torque in the shaft because the Y would not torsionally deflect as shown in my sketch.

This problem is very difficult to do by hand (accurately), although I believe there is a way to bracket the problem and get close to the answer.

Start by calculating the bending stiffness of the arms perpindicular to the shaft, the torsional stiffness of the two halves of the base of the Y parallel to the shaft, and the stiffness of the shaft between the forks. When this thing is twisted as shown in my sketch, it is these stiffnesses that are doing the primary reacting of the moment vector that is parallel to the shaft.

You'll need to then write equations that relate the rotational deflection of the shaft to the slope angles of the ends of the two forks as your compatibility equation to solve for the indeterminancy.

I'll mull it over this evening and see if I can bang out the relationships...


If you "heard" it on the internet, it's guilty until proven innocent. - DCS

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