Worm Torque & tooth Strength

[harrai]

Hi All:

I am working on lifing mechines with handle, worm gear and pulley.

Weight to be lifted at the pulley is 200lb.I am applying 30 lbs on the handle (radius=8.5 in & I am using 20:1 worm gear reduction ratio.Handle is rotating at 30rpm).

This gear is rated for .23 HP and 1550 lbf*in output torque.

I have calculated the torque and the forces on gear tooth. Stress calculated on tooth

is more than the allowable(When both rated input power & output torque of the gear are more than the calculated)

Could anyone through some lights on this & show the mistakes?

T1=255lbf*in, d1=2*in (dia of worm), an=14.5 deg, langle=4.77(lead angle), µ=.04 gear

ratio=20:1

Hp=T1*RPM/63025=.12(input)<.23 HP

Output Torque=200lb*3in=600lbf*in<1550 lbf*in output torque(3in is the drum dia to

which weight is connected.

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=255lbf(my answer)

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

F gt=2040lbf(my answer)

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

ba=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

Stress=274N/mm2(my answer)

The mechanical properties for gear (phosphor bronze)are:
Tensile Strength, 35,000PSI min(241N/mm2)