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RE: Shaft Mounted Gearbox Free Body Diagram

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Rusthandle

Mechanical
Jan 17, 2011
3
Hi,

This should be straight forward. Are my free body diagrams correct?


Thx in advance.
 
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The torque reaction arm reacts the torque.

There will be some out-of-plane forces too.
 
i would Strongly suggest that your free body diagram should show forces, not masses. you should change 50kg to 490N, etc.

i think there's a problem with the "torque brkt". as drawn the pin cannot react the torsion applied; ie, what's stopping the gearbox from pivoting about the pin ?

i think there's a problem with the free body of the "torsion brkt". as shown, the loads applied are weight and torque. the pin to the brkt can only (pretty much) react shear (as drawn), but this is the reaction to the weight. if you're going to react the applied torque as a couple, and so the pin to the brkt reacts more than just the weight, you need to show use both halves of the couple.

looking at the pic again, and going through your calcs, i think you're saying that the torque is being reacted as a couple between the shaft and the pin to the brkt (ie 0.14m arm). if this is so then the weight would be reacted by the shaft and the brkt. show All the forces and reaction points and you should be able to see more clearly how the loads are being reacted.

the torque is reacted by a couple of some 1400N (0.14m apart). one reaction is one the pin to the brkt, the other is on the shaft, presumably split between the two ends of the shaft ... so the FBD of the sahft should show this. i guess you could say that the key allows the shaft to react the torque (as shown). but i think you have a problem with how you're reacting the weight, your reaction (at the pin) is offset from the applied load ... where is this torque being reacted ? is there more than 1 loadpath for vertical load ??
 
Hi,

Thanks for your replies. Yes should be Newtons or kgf.

I think the torque arm bracket is correct??., except the 72Nm moment should be clockwise.

My reasoning being that if I remove the torque arm pin the conveyor shaft would remain stationary and the geared motor would rotate around the shaft opposite to the direction of normal shaft rotation. Hence the gearbox applies a force of 200Nm/0.14m = 1430 N to the torque arm bracket.

Because the centre of gravity of the gearbox is 'roughly' over shaft centreline line I've assumed all the gearbox weight i.e. 50kg is carried on the conveyor shaft.

I was mainly interested in whether I've missed anything with respect to any other forces the geared motor applies to the shaft.

For example, if the conveyor was driven by a timing belt. The shaft would have a torque applied to it and a force due to the timing belt tension.

Am I right that with a shaft mounted geared motor. The shaft has a torque applied to it and a force due to the weight of the gearbox?
 
is the shaft supported so it can react vertical load ? (i think the answer is obviouly yes). then i think the weight should be shared between the shaft and the brkt. (sort of like a beam on two supports)

i question that the shaft can react torque (if the shaft is recting the applied torque then it can't rotate). i think the torpque is reacted by a couple betwwen the shaft and the brkt.

if you've got questions about directions of forces, draw the FBDs together (exploded), reactions on one piece and reversed to become loads on the next.
 
Both free body diagrams are incorrect.


1)The shaft should be in equilibrium so you must have out of plane forces and torques to match the values you have. In addition you are missing the force of the couple which is 200nm/.14m=1428n upward on the shaft.

2) The bracket should have a shear force at the mount equal and opposite the vertical force, 1428n

 
Thanks for your comments/help.

rb1957 - Agree that in reality a portion of the gearbox weight would be carried on the torque arm bracket. Because the centre of gravity of the gearbox is close to shaft cl most will be on the shaft.

rb1957/zekeman - Agree I should have a couple upward on the shaft. Shaft diagram is incomplete and doesn't show out of plane reactions & counter torques. I think we agree that the geared motor applies a torque, a force due to the weight of the gearbox & a couple force to the shaft.

zekeman - I am showing an equal & opposite shear force at the mount am I not??



 
please don't show us an incomplete diagram and ask us to comment ... maybe there's an ESL thing going on ?

i think you're heading in the wrong direction ... the torque brkt reacts one force of the couple that reacts the gearbox torque (and a small component due to weight) as your sketch shows (200/0.14) but it is in the other direction ... the torque will be trying to pull up on the brkt, the torque load on the brkt is up.
 
"zekeman - I am showing an equal & opposite shear force at the mount am I not??"

Yes, I missed your shear force.

But as a piece of advice, when you put a free body diagram up for comments, as Rb stated, you must show it in equilibrium and the out of plane forces and moments must be shown in the diagram

Anyway, in summary your free body on the shaft should read.
200nM ccw, 200nm cw ,wihth an asterisk 96*9,8n upwards and 98,6 downwards with an asterisk. The asterisks are for the out-of plane forces.

The bracket diagram is Ok except for the trivial kg to newton conversion.

 
"The bracket diagram is Ok except for the trivial kg to newton conversion." ... except that the forces are in the wrong direction (the load from the gearbox onto the brkt is acting up).

also, i thought the shaft wasn't reacting the torque (as shown)
 
""The bracket diagram is Ok except for the trivial kg to newton conversion." ... except that the forces are in the wrong direction (the load from the gearbox onto the brkt is acting up)."

Yes, the force diagram shown is pretty fuzzy, the up and down vectors weren't clear to me.

"also, i thought the shaft wasn't reacting the torque (as shown)"

Of course it is ; otherwise the shaft will accelerate continuously. There is the implied takeoff torque load = to the gearbox torque satisfying shaft equilibrium.
 
Attached is a JPEG sketch of the free body diagram. Vr and Mr are the resisting shear force and moment at the wall bracket attachment. I used letters to express torque, moment and weights instead of specific numbers. Remember we are looking at one plane. There two other planes requiring further evaluation and consequently there will be combined moments and shear forces at the wall as pointed by others above.
 
 http://files.engineering.com/getfile.aspx?folder=4c12509f-dea7-4018-99da-6d48fb1e2921&file=reply_to_thread_404-289965.jpg
how does the single bolt attaching the brkt react torque ?

is the torque shown correctly ? zeke is posting as though the torque is (sensibly enough) along an axis in-plane though the gearbox ??
 
There should be some form of a locking pin with the bolt to keep the attachment from rotating at the bolt.
 
"how does the single bolt attaching the brkt react torque ?"

Easy, the pure couple ,F*0.140M reacts it, period. F at the bracket pin and the shaft provide the force, F. This certainly puts the gearbox in moment equilibrium normal to the X-Y plane; however, if the CM of the GB is not in the same plane as the couple, then the shaft would have to react with a couple normal to the Y-Z plane.

Why don't we simplify the problem by asserting that the CM of the GB coincides with the shaft CL.

Then it is clear that quite simply

1 The weight of the GB is borne solely by the shaft.
2 The couple , F*0.14M reacts the GB torque with F additional vector force on the shaft and F the sole force on the bracket.
3 Out-of-plane takeoff moments and shaft support forces will be needed to put the shaft in equilibrium.

Maybe time to put this to bed!!

 
sigh,

zeke, i was questioning the FBD showing a moment on the bolt (ie the applied torque wasn't being reacted as a couple).
 
"sigh,

zeke, i was questioning the FBD showing a moment on the bolt (ie the applied torque wasn't being reacted as a couple)."

Rb,

Sorry about that.I should know you know better. I hadn't seen the FBD you mentioned .
 
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