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A pair of 20 deg full depth involute spur gears having 30 and 50 teeth respectively of module 4 mm are in mesh. The smaller gear rotates at 1000 rpm.

Determine:

  1. sliding velocities at engagement at an disengagement of pair of a teeth, and

  2. Contact ratio.


Subject: Kinematics of Machinery

Topic: Gears and Gear Trains

Difficulty: Medium

1 Answer
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Given ϕ=20; t= 30; T= 50; m=4; N= 1000 rpm; or ω1=2π×(1000/60)=104.7rad/s.

1.Sliding velocities at engagement and at disengagement of pair of a teeth

First of all, let us find the Radius of the addendum circles of the smaller gear and the larger gear.

We know that,

Addendum of the smaller gear,

=mt2[1+Tt(Tt+2)sin2ϕ1]

=4×302[1+5030(5030+2)sin2201]

=60(1.311)=18.6mm

And Addendum of the larger gear,

=mt2[1+tT(tT+2)sin2ϕ1]

=4×502[1+3050(3050+2)sin2201]

=100(1.091)=9 mm

Pitch circle Radius of the smaller gear,

r= mt/2=4(30/2)=60 mm

Radius of the Addendum circle of the smaller gear,

rA=r+Addendum of the smaller gear= 60+18.6=78.6 mm

Pitch circle Radius of the larger gear,

R=mt/2=4(50/2)=100 mm

Radius of addendum circle of the larger gear,

RA=R+Addendum of the larger gear = 100+9=109 mm

We know that the path of approach(i.e path of contact when engagement occurs),

KP=(RA)2R2cos2ϕRsinϕ

=(109)2(100)2cos220100sin20=55.234.2=21mm

and the path of recess(i.e path of contact when disengagement occurs),

PL=(rA)2r2cos2ϕrsinϕ

=(78.6)2(60)2cos22060sin20=54.7620.52=34.24mm

Let ω2 = Angular speed of the larger gear in rad/s

We know that ω1ω2=Tt or ω2=ω1×tT=10.47×3050=62.82rad/s

Sliding velocity at engagement of a pair of teeth

=(ω1+ω2)KP=(104.7+62.82)21=3518 rad/s

=3.518 m/s

And Sliding velocity at Disengagement of a pair of teeth

=(ω1+ω2)PL=(104.7+62.82)34.24=5736 rad/s

=5.736 m/s

2. Contact Ratio: We know that the length of the arc of contact

=Length of the path of contact/ cosϕ

=KP+PLcosϕ=21+34.24cos20=58.78 mm

And circular pitch=π×m=3.142×4=12.568mm

Contact Ratio= Length of arc of contact / circular pitch= 58.78/12.568=4.67 say 5

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