Unformatted text preview: therefore
Since the gear A is fixed, therefore from the fourth row of the table,
x =–y =–1
N C and N D = Revolutions of gears C and D respectively.
From the fourth row of the table, the revolutions of gear C,
NC = y + x × A = 1 – 1 ×
and the revolutions of gear D,
ND = y + x × A = 1 –
From above we see that for one revolution of the arm B , the gear C rotates through 1/101
revolutions in the same direction and the gear D rotates through 1/99 revolutions in the opposite
Example 13.14. In the gear drive as shown in Fig.
13.18, the driving shaft A rotates at 300 r.p.m. in the clockwise direction, when seen from left hand. The shaft B is the
driven shaft. The casing C is held stationary. The wheels E
and H are keyed to the central vertical spindle and wheel F
can rotate freely on this spindle. The wheels K and L are
rigidly fixed to each other and rotate together freely on a
pin fitted on the underside of F. The wheel L meshes with
internal teeth on the casing C. The numbers of teeth on the
different wheels are indicated within brackets in Fig. 13.18.
Find the number of teeth on wheel C and the speed
and direction of rotation of shaft B.
Solution. Given : N A = 300 r.p.m. (clockwise) ;
T D = 40 ; T B = 30 ; T F = 50 ; T G = 80 ; T H = 40 ; T K = 20 ; T L = 30 Fig. 13.18 In the arrangement shown in Fig. 13.18, the wheels D and G are auxillary gears and do not
form a part of the epicyclic gear train. 454 l Theory of Machines Speed of wheel E, N E = N A × 40
= 300 ×
= 400 r.p.m. (clockwise)
TE Number of teeth on wheel C
T C = Number of teeth on wheel C.
Assuming the same module for all teeth and since the pitch circle diameter is proportional to
the number of teeth ; therefore from the geometry of Fig.13.18,
T C = T H + T K + T L = 40 + 20 + 30 = 90 Ans.
Speed and direction of rotation of shaft B
The table of motions is given below. The wheel F acts as an arm.
Table 13.13. Table of motions.
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- Fall '08
- gear, Gear Trains, Epicyclic gearing, epicyclic gear train