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Forces on Worm gear tooth

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harrai

Aerospace
Aug 4, 2007
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HI All:
I am calculating the forces on worm gear tooth. could anyone suggest me which formulae to use?

Thanks
 
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Any college mechanical design book such as from Shigley or Norton has the formulas. The Machinery's handbook 22nd edition has an extensive treatment of worm gearing. Latest edition dropped the issue.
 
Hi israelkk:

Thanks for ur reply. I did calculate using the following the formulae.But I am getting huge stress on worm gear tooth & huge Tangential force on gear (2174.lbf)which I sould'nt expect because the gear I selected is rated for .23 HP and 1550 lbf*in output torque.In my case power is only .03 hp & output torque is 600 lbf*in(this gear has to lift weight of 200.lbf, worm wheel is connected to pulley (dia=6in, hence torque=600 lbf*in).

Please can any one check these calculation for me & tell why Tangential force on worm is so less & Tangential force on gear is so high& also stress on the worm gear tooth is so high when rated capacity for the chosen gear are high.


T1=255lbf*in, d1=2*in (dia of worm), ? n=14.5 deg, ?=4.4, ?=.04 gear ratio=20:1

d2=3.3*in (dia of worm gear), diametrical pitch=6

Tangential force on worm ( F wt )= axial force on wormwheel
F wt = F ga = 2.T 1 / d 1

F wt=255.lbf(my answer)

Axial force on worm ( F wa ) = Tangential force on gear
F wa = F gt = F wt.[ (cos ? n - ? tan ? ) / (cos ? n . tan ? + ? ) ]

F gt=2174.lbf(my answer)

Modified Lewis equation for stress induced in worm gear teeth
? a = F wt / ( p x. b a. y )(N)

b a=1*in(face width) y=0.1

F wt = Worm gear tangential Force (N)

module m=d2/20=.164*in
Axial circuler pitch px=3.142*m=.51*in

 
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