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Unformatted text preview: utions of elements
Step
No. Conditions of motion Arm or
wheel F Wheel E Wheel H 1. Arm fixedwheel E
rotated through – 1
revolution ( i . e . 1
revolution clockwise) 0 –1 2. Arm fixedwheel E
rotated through – x
revolutions 0 –x –x + x× 3. Add – y revolutions to
all elements –y –y –y –y 4. Total motion –y –x–y –x–y – 1( 3 E and
H are on the
same shaft) Compound
wheel KL + x× Wheel C TH
TK TH
TK TH
–y
TK + TH TL
×
TK TC + x× TH TL
×
TK TC –y x× TH TL
×
–y
TK TC Since the speed of wheel E is 400 r.p.m. (clockwise), therefore from the fourth row of the table, or – x – y = – 400 or
Also the wheel C is fixed, therefore
T
T
x× H × L – y=0
TK TC
40
30
x×
– y=0
×
20
90
2x
∴
– y=0
3
From equations (i) and (ii),
x = 240 and x + y = 400 ...(i) ...(ii)
y = 160 ∴ Speed of wheel F, N F = y = – 160 r.p.m.
Since the wheel F is in mesh with wheel G, therefore speed of wheel G or speed of shaft B = – NF × 50 TF = – – 160 × = 100 r.p.m.
80 TG ...(3 Wheel G will rotate in opposite direction to that of wheel F.) = 100 r.p.m. anticlockwise i.e. in opposite direction of
shaft A . Ans. Chapter 13 : Gear Trains 455 l Example 13.15. Fig. 13.19 shows a compound epicyclic gear in which the casing C contains
an epicyclic train and this casing is inside the larger casing D.
Determine the velocity ratio of the output shaft B to the input shaft A when the casing D is
held stationary. The number of teeth on various wheels are as follows :
Wheel on A = 80 ; Annular wheel on B = 160 ; Annular wheel on C = 100 ; Annular wheel
on D = 120 ; Small pinion on F = 20 ; Large pinion on F = 66. Fig. 13.19 Solution. Given : T 1 = 80 ; T 8 = 160 ; T 4 = 100; T 3 = 120 ; T 6 = 20 ; T 7 = 66
First of all, let us consider the train of wheel 1 (on A ), wheel 2 (on E), annular wheel 3 (on D)
and the arm i.e. casing C. Since the pitch circle diameters of wheels are proportional to the number of
teeth, therefore from the geometry of Fig. 13.19,
T 1 + 2 T2 = T 3
or 80 + 2 T 2 = 120
∴
T 2 = 20
The tabl...
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 Fall '08
 CELGHORN

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