Numericals on Gears & Gears Trains
1 Answer

Q.1 Calculate 1) Length of path of contact 2) Arc of contact 3) Contact ratio when a penion having twenty three teeth drive a gear having 57 teeth the profile of gear is involute with stress,Angle=20,module=8 mm,addendum=1 module


$T_2=23$ , $T_1=57$ m=8 mm, $\phi=20$ ,a=8 mm.

$m=\frac{D}{t}$ so, $\frac{D}{2}=\frac{mt}{2}$

$R=\frac{8*57}{2}=228 mm$

Addendum circle padi for pinion $r_a=r+a=92+8=100 mm$

$R_a=R+a=228+8=236 mm$

1)Path of contact

$=\sqrt {R_a^2-R^2cos\phi}+\sqrt {r_a^2-r^2cos^2\phi}-(R+r)sin\phi=39.77 mm$

Arc of contact $\frac{39.77}{cos\phi}=42.32 mm.$

Contact ratio=$\frac{42.33}{T_1m}=1.684$

Q.2 Figure shows epicyclic gear train pinion .A has 15 teeth a bar is rigidly fix to meter shaft .The wheel B has 20 teeth and mushes with gear A and also with annular fix wheel D,pinion C has 15 teeth and integral with B gear C match with annular wheel E which is keyed to the m/c shaft .The arm rotates on about same shaft on which A is fixed and carries the compound wheel B and C is motor and at 1000 rpm find speed of m/c shaft?

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$T_D=15+40=55$ teeth


$T_e=T_A+T_B+T_C=15+20+15=50$ teeth

sl no condition $Arm$ A B/C D E
1 Arm fix 1 rev to A 0 1 -$\frac{T_A}{T_B}$ -$\frac{T_B}{T_D}\times \frac{T_A}{T_B}$ -$\frac{T_C}{T_e}\times \frac{T_A}{T_B}$
2 x 0 x -$\frac{xT_A}{T_B}$ -$\frac{xT_A}{T_D}$ -$\frac{T_C}{T_e}\times \frac{xT_A}{T_B}$
3 y y y y y y
4 Total y x+y $y-\frac{xT_A}{T_B}$ $y-\frac{xT_A}{T_D}$ $y-\frac{T_C}{T_e}\times \frac{xT_A}{T_B}$


$N_E=\frac{T_C}{T_E}\times N_C$

$=-\frac{T_C}{T_E}\times \frac{T_A}{T_B}$

$N_E=-\frac{T_C}{T_E}\times \frac{T_A}{T_B}$

$N_D=N_B\times \frac{T_B}{T_D}$

$=-\frac{T_B}{T_D}\times \frac{T_A}{T_B}$ $=-\frac{T_A}{T_D}$


$x=785.7 \quad y=214.3$

$y-x\times \frac{T_A}{T_D}=0$

$N_E=y-\frac{T_C}{T_E}\times \frac{T_A}{T_B}$

$N_E=37.5$ rpm.

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