A parallel helical gearset consists of a 19-tooth pinion driving a 57-tooth gear. The pinion has a left-hand helix angle of 30°, a normal pressure angle of 20°, and a normal module of 2.5 mm.
Find:_______.
(a) The normal, transverse, and axial circular pitches
(b) The transverse diametral pitch and the transverse pressure angle
(c) The addendum, dedendum, and pitch diameter of each gear

Respuesta :

Answer:

a)

normal circular pitch = 7.8539 mm

transverse circular pitch = 9.0689 mm

axial circular pitches = 15.7077

b)

transverse diametral pitch is 0.3464 teeth/mm

transverse pressure angle is 22.8°

c)

Addendum = 2.5 mm

dedendum = 3.125 mm

pinion diameter = 54.8482 mm and Gear diameter = 164.5448 mm

Explanation:

Given that;

module m = 2.5 mm

Number of teeth on Gear  nG = 57 TEETH

Number of teeth on Pinion  nP = 19 TEETH

Helix angle W = 30°

Normal Pressure angle β = 20°

finding the circular pitch

Pc = πm

we substitute

Pc = π * 2.5 mm = 7.8539 mm

now the diametral pitch p = π / Pc

= π / 7.8539

= 0.4 teeth/mm

a)

So the normal circular pitch

Pn = π / P

Pn = π / 0.4

Pn = 7.8539 mm

the transverse circular pitch

Pt = Pn / cosW

Pt = 7.8539 / cos30°

Pt = 9.0689 mm

for axial circular pitches

Px = Pt / tanW

Px = 9.0689 / tan30°

Px = 15.7077

b)

The transverse diametral pitch and the transverse pressure angle.

The transverse diametral pitch Pt = PcosW

= 0.4 * cos30°

= 0.3464 teeth/mm

transverse diametral pitch is 0.3464 teeth/mm

transverse pressure angle  β1 = tan^-1 ( tan βn / cos W)

=  tan^-1 ( tan20° / cos 30°)

= tan^-1 ( 0.42027 )

β1 = 22.8°

transverse pressure angle is 22.8°

c)

The addendum, dedendum, and pitch diameter of each gear

Now from table standard Tooth proportions for Helical Gears;

Addendum a = 1/p

= 1 / 0.4

= 2.5 mm

dedendum b = 1.25 / p

= 1.25 / 0.4

= 3.125 mm

now  pinion diameter dP = Np / PcosW

= 19 / 0.4 (cos30°)

= 54.8482 mm

Gear diameter dG = nG / pcosW

= 57 / 0.4 (cos30°)

= 164.5448 mm