Answer to Question #200264 in Mechanical Engineering for Muhammad Idrees

Question

In a reverted gear train, as shown in Fig. 13.32, two shafts A and B are in the same straight line and are geared together through an intermediate parallel shaft C. The gears connecting the shafts A and C have a module of 2 mm and those connecting the shafts C and B have a

module of 4.5 mm. The speed of shaft A is to be about but greater than 12 times the speed of shaft B, and the ratio at each reduction is same.

Find suitable number of teeth for gears. The number of teeth of each gear is to be a minimum but not less than 16. Also find the exact velocity ratio and the distance of shaft C from A and B.

Answer 1

Velocity ratio required between A and B = 12

$\frac{N_1}{N_2}=12$

$\frac{N_1}{N_2}=\frac{N_2}{N_4}$

Also $\frac{N_1}{N_2}=\frac{N_3}{N_4}$ Take $\frac{N_1}{N_2}=\frac{N_3}{N_4}=3.4 \implies 3.4*3.4=11.56$

Module for gear 1<2 m_A =2

Module for gear 3<4 m_B =4.5

Consider gear that has higher module

Take Z₃ (number of teeth on pinion 3) =20

Diameter (pitch) of gear 3 $d_3=mZ_3=4.5*20=90mm$

$Z_4=3.4*Z_3=3.4*20=68$

Pitch diameter of gear 4 $d_4=mZ_4=4.5*90=306mm$

Center distance $c=\frac{d_3+d_4}{2}=\frac{90+306}{2}=198 mm$

Consider gear 1 and 2

$c=\frac{d_1+d_2}{2}=198 mm$

$\frac{d_2}{d_1}=3.4 \implies d_2 = 3.4 d_1$

$198=\frac{d_1+3.4d_1}{2}; d_1=90 mm; d_2=3.4*90 =306 mm$

$Z_1=\frac{d_1}{m_A}=\frac{90}{2}=45$

$Z_2=\frac{d_2}{m_A}=\frac{306}{2}=153$

Hence gear 1= 45; gear 2= 153; gear 3= 20; gear 4= 68

Center distance , $\frac{d_1+d_2}{2}=\frac{d_3+d_4}{2}=198 mm$