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Two involute gears of 20 deg pressure angle are in mesh. The no. of teeth on pinion is 20 and the gear ratio is 2.

If the pitch expressed in module is 5 mm and the pitch line speed is 1.2 m/s, assuming addendum as standard and equal to one module, find:

1.The angle turn through by pinion when one pair of teeth is in mesh; and

2.The maximum velocity of sliding.


Subject: Kinematics of Machinery

Topic: Gears and Gear Trains

Difficulty: High

1 Answer
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Given $\phi=20^{\circ}$; t=20; G=T/t=2; m=5mm; v=1.2m/s; addendum=1 modul=5mm

1. Angle turned through by pinion when one pair of teeth is in mesh

We know that pitch circle radius of pinion,

r=mt/2=5\times 20 / 2 =50 mm

and pitch circle radius of wheel,

R=mt/2=m.G.t/2=5 2 20 /2 =100 mm

Radius of Addendum circle of pinion,

$r_{A}$= r + Addendum=50 + 5 = 55 mm

and radius of Addendum circle of wheel,

$R_{A}$ = R+ Addendum = 100 + 5 = 105 mm

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

$KP=\sqrt{(R_{A})^{2}-R^{2}\cos^{2}\phi}-R\sin\phi$

$=\sqrt{(105)^{2}-100^{2}\cos^{2}20^{\circ}}-100\sin20^{\circ}$

= 46.85 -34.2 = 12.65 mm

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

$PL=\sqrt{(r_{A})^{2}-r^{2}\cos^{2}\phi}-r\sin\phi$

$=\sqrt{(55)^{2}-50^{2}\cos^{2}20^{\circ}}-50\sin20^{\circ}$

= 28.6 - 17.1 = 11.5 mm

So, Length of the path of the contact,

KL = KP+PL = 12.65+11.5 = 24.15 mm

And length of the arc of contact =(length of path of contact)/$\cos\phi$ =24.15/cos 20 =25.7 mm

We know that the angle turn through by pinion

(length of arc of contact ×360$^{\circ}$)/circumference of pinion =(25.7×360$^{\circ}$)/(2$\pi$×50)= 29.45$^{\circ}$

2.Maximum velocity of sliding

Let $\omega_{1}$ =angular speed of pinion and

$\omega_{2}$ =angular speed of wheel

we know that pitch line speed,

$ν = \omega_{1}.r = \omega_{2}.R$

$\omega_{1}$ =120/5 =24 rad/s,

$\omega_{2}$ = 120/10 =12 rad/s

Maximum velocity of sliding

$ν_{S} = (\omega_{1}+ \omega_{2})$.KP

= (24+12)12.65 = 455.4 mm/s

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