In a reverted gear train, as shown in figure, 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.
Answers
Answer:
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Step-by-step explanation:
Velocity ratio required between A and B = 12
\frac{N_1}{N_2}=12
N
2
N
1
=12
\frac{N_1}{N_2}=\frac{N_2}{N_4}
N
2
N
1
=
N
4
N
2
Also \frac{N_1}{N_2}=\frac{N_3}{N_4}
N
2
N
1
=
N
4
N
3
Take \frac{N_1}{N_2}=\frac{N_3}{N_4}=3.4 \implies 3.4*3.4=11.56
N
2
N
1
=
N
4
N
3
=3.4⟹3.4∗3.4=11.56
Module for gear 1<2 mA =2
Module for gear 3<4 mB =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=90mmd
3
=mZ
3
=4.5∗20=90mm
Z_4=3.4*Z_3=3.4*20=68Z
4
=3.4∗Z
3
=3.4∗20=68
Pitch diameter of gear 4 d_4=mZ_4=4.5*90=306mmd
4
=mZ
4
=4.5∗90=306mm
Center distance c=\frac{d_3+d_4}{2}=\frac{90+306}{2}=198 mmc=
2
d
3
+d
4
=
2
90+306
=198mm
Consider gear 1 and 2
c=\frac{d_1+d_2}{2}=198 mmc=
2
d
1
+d
2
=198mm
\frac{d_2}{d_1}=3.4 \implies d_2 = 3.4 d_1
d
1
d
2
=3.4⟹d
2
=3.4d
1
198=\frac{d_1+3.4d_1}{2}; d_1=90 mm; d_2=3.4*90 =306 mm198=
2
d
1
+3.4d
1
;d
1
=90mm;d
2
=3.4∗90=306mm
Z_1=\frac{d_1}{m_A}=\frac{90}{2}=45Z
1
=
m
A
d
1
=
2
90
=45
Z_2=\frac{d_2}{m_A}=\frac{306}{2}=153Z
2
=
m
A
d
2
=
2
306
=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
2
d
1
+d
2
=
2
d
3
+d
4
=198mm
@ngel..