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Unformatted text preview: and speed ratio or velocity ratio, d1 + d 2
2 ...(i) N1 d2 T2
...(ii)
=
=
N2
d1 T1
From the above equations, we can conveniently find out the values of d1 and d2 (or T 1 and T 2)
and the circular pitch ( pc ). The values of T 1 and T 2, as obtained above, may or may not be whole
numbers. But in a gear since the number of its teeth is always a whole number, therefore a slight
alterations must be made in the values of x , d1 and d2, so that the number of teeth in the two gears may
be a complete number.
Example 13.2. Two parallel shafts, about 600 mm apart are to be connected by spur gears.
One shaft is to run at 360 r.p.m. and the other at 120 r.p.m. Design the gears, if the circular pitch is
to be 25 mm.
Solution. Given : x = 600 mm ; N 1 = 360 r.p.m. ; N 2 = 120 r.p.m. ; pc = 25 mm
Let d1 = Pitch circle diameter of the first gear, and
d2 = Pitch circle diameter of the second gear.
We know that speed ratio, N1 d 2 360
=
=
=3
N2
d1 120 or d2 = 3d1 ...(i) and centre distance between the shafts (x ), 1
(d1 + d2 )
2
From equations (i) and (ii), we find that
600 = or d1 + d2 = 1200 d1 = 300 mm, and d2 = 900 mm
∴ Number of teeth on the first gear, T1 = π d 2 π × 300
=
= 37.7
25
pc ...(ii) 434 l Theory of Machines and number of teeth on the second gear, T2 = π d 2 π × 900
=
= 113.1
25
pc Since the number of teeth on both the gears are to be in complete numbers, therefore let us
make the number of teeth on the first gear as 38. Therefore for a speed ratio of 3, the number of teeth
on the second gear should be 38 × 3 = 114.
Now the exact pitch circle diameter of the first gear,
T1 × pc 38 × 25
=
= 302.36 mm
π
π
and the exact pitch circle diameter of the second gear,
d1′ = T × pc 114 × 25
d 2′ = 2
=
= 907.1 mm
π
π
∴ Exact distance between the two shafts,
d1′ + d 2′ 302.36 + 907.1
=
= 604.73 mm
2
2
Hence the number of teeth on the first and second gear must be 38 and 114 and their pitch
circle diameters must be 302.36 mm and 907.1 mm
respectively. The exac...
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 Fall '08
 CELGHORN

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