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direct example of the current chapter. or TD = 4 TB We know that T D = dD / m = 224 / 4 = 56 Ans. ∴ T B = T D / 4 = 56 / 4 = 14 Ans. ...(i) ...[From equation (i)] Let dB, dC and dD be the pitch circle diameters of sun wheel B , planet wheels C and internally
toothed ring D respectively. Assuming the pitch of all the gears to be same, therefore from the geometry of Fig. 13.13,
dB + 2 dC = dD
Since the number of teeth are proportional to their pitch circle diameters, therefore
TB + 2 TC = TD
∴ or 14 + 2 T C = 56 T C = 21 Ans. Example 13.10. Two shafts A and B are co-axial. A gear C (50 teeth) is rigidly mounted
on shaft A. A compound gear D-E gears with C and an internal gear G. D has 20 teeth and gears
with C and E has 35 teeth and gears with an internal gear G. The gear G is fixed and is concentric with the shaft axis. The compound gear D-E is mounted on a pin which projects from an arm
keyed to the shaft B. Sketch the arrangement and find the number of teeth on internal gear G
assuming that all gears have the same module. If the shaft A rotates at 110 r.p.m., find the speed
of shaft B.
Solution. Given : T C = 50 ; T D = 20 ; T E = 35 ; N A = 110 r.p.m.
The arrangement is shown in Fig. 13.14.
Number of teeth on internal gear G
Let dC , dD , dE and dG be the pitch circle diameters of gears C, D, E and G respectively. From
the geometry of the figure, dG
= C+ D+ E
or dG = dC + dD + dE l Chapter 13 : Gear Trains 449 Let T C , T D , T E and T G be the number of teeth on gears C, D, E and G respectively. Since all
the gears have the same module, therefore number of teeth are proportional to their pitch circle
∴ T G = T C + T D + T E = 50 + 20 + 35 = 105 Ans. Fig. 13.14 Speed of shaft B
The table of motions is given below :
Table 13.9. Table of motions.
Revolutions of elements
No. Conditions of motion Arm Gear C (or
shaft A) Compound
gear D-E TC
TD 1. Arm fixed - gear C rotates through + 1
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- Fall '08
- gear, Gear Trains, Epicyclic gearing, epicyclic gear train