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Gears load 4

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Nashanas

Nashanas

Petroleum
Apr 23, 2021
54
Hello everyone,

I have a very basic question about gear train design.

Immagine_zqgxrm.png


In this design, it can be seen that there is a set of 3 gears with same ratio, which are being used to move the bigger gear. I want to usa only one gear instead of three to move the bigger gear. But I have been told that 3 gears are used to reduce tooth stress. But according to Shigley, the delivered load from one gear tooth to another depends on the pitch, face width and material. If all these variables are same, I dont see any reason for using 3 gears instead of one, as the load delviered will remain the same.
 
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If the table has 4 legs, does the force in each individual leg equal to the weight of the entire table or weight of the entire table divided by 4?

What about if the table has 3 legs? Is the load on the individual leg the same as if it had 4 legs?


the delivered load from one gear tooth to another depends on the pitch, face width and material
Click to expand...
Load doesn't depend on these things. Stress in the tooth depends on gear size (which depends on pitch) and face width.


Make a free body diagram of your gear and see how it turns out.
 
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Theoretically the two gears share the load, each carrying half. Of course that is only true on paper, because microscopic differences will cause one to be in contact when the other is not. They will never share the load equally. But they will share it, and the more they wear the more they will share it.
 
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The gears can be mounted in a way that assures the load is shared equally.
 
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And just like the caveat with respect to the gears never sharing the load equally, so too, a four-legged table will only be distributing the load to three of its legs.

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That three pinion set will be a bugger to set center distances to get them to mesh and to do it equally. Just a little deviation in the distances would transfer the load to one or the other.

Would much rather have a single pinion (or pair, not three) and if tooth load is too high, go with wider teeth.
 
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Nashanas,

Consider spring loading the small gears.

A gear manufacturer has told me that they can cut gear oversized so that they work spring loaded. Normally, a gear is cut so that the optimal geometry works with a slight clearance.

Note that while that arrangement has half the stress, it also has twice the fatigue cycles.

--
JHG
 
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Nashanas,

Another thought.

You can work out the stresses on the gears. What are the stresses on those gears? If you went to a larger pitch and you used one gear only, what would the stresses be?

--
JHG
 
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I see this arrangement as more expensive and and more difficult to manufacture. as pointed out maintaining the center distance. an idler is generally used to change direction of the driven gear with an idler. unless this is necessary a single gear will function properly if it is calculated properly to maintain for wear, tooth bending and ect.
 
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JohnRBaker said:
And just like the caveat with respect to the gears never sharing the load equally, so too, a four-legged table will only be distributing the load to three of its legs.
Click to expand...
Correction - a perfectly rigid four-legged table will only be distributing the load to three of its legs.
The reason is the fact that 3 points define a plane, and you can never make legs perfectly enough so that the tip of the fourth one also lies on the same plane as the other 3.

Table is not a rigid flat plane, though.

An unloaded 4 legged table might wobble on 3 legs. Have 2 people sit on it and it will stand firmly on all 4 legs.
 
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it is well know 3 points not four or more are best to stabilize objects
 
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so long as the load is applied within the footprint of those 3 points ...

another day in paradise, or is paradise one day closer ?
 
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I have found the equation for tangential load on the gear teeth:

Wt = (60000*P)/(pi*d*n)
P = Power (KW)
d = Diameter (mm)
n = rpm
Wt = Tangential load in KN

Since the two idlers will not be transmitting force to there shafts, theoretically, the power of pinion will be divided equally among two idlers because of same diameters, and hence the load on idler teeth will become half if we add another idler. I think in the current arrangement max two idelrs can be used, and both should be of same diameter. Ofcourse this is theoretical, in reality there could be friction losses on idler shafts and other losses because of non-ideal mounting.

Thank you for your replies.
 
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mfgenggear said:
it is well know 3 points not four or more are best to stabilize objects
Click to expand...
Well, this linear axis for a robot has not 3, but 18 points of contact with the ground.
And the structure is incredibly stable and precise.

There are times to use minimum constraint design, and then there are times where MCD is worthless.
 
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"18 points of contact with the ground."

Should read "18 INDIVIDUALLY ADJUSTABLE points of contact with the ground."
 
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Well, office chair has 5 points of contact, none of which are individually adjustable. :)
 
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Office chair 5 points of contact,few of which are equally loaded.

Ted
 
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Depending on the depth of the carpet that would be about 1X 10^30th points of contact.
 
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t is well know 3 points not four or more are best to stabilize objects.
Click to expand...

Disagree, three points of contact is generally a terrible method of supporting anything. Stability is a function of the Cg's relationship to the supports. More supports/points of contact provide a larger/wider "footprint" and thus more stability, hence tables commonly having four legs, office chairs 4-5, etc.
 
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CWB1, don't tell toolmakers and fixturemakers that. They would certainly disagree.

Ted
 
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