Engineering Mechanics

(^716) „„„„„ A Textbook of Engineering Mechanics
First of all, prepare the table of motions as below.
Step No. Condition of motion
Revolution of
Arm Wheel A Wheel B Wheel C

1. Arm fixed; wheel A
   0+ 1

A

B

T

T

+ – AB

BC

TT

TT

rotates through + 1 ×

revolutions

• A
  C

T

T

=

2. Arm fixed; wheel A 0+ x A
   B

T

x

T

+ – A

C

T

x

rotates through + x T

revolution

3. Add + y revolutions + y + y + y + y
   to all elements


4. Total motion + yx + y

A

B

T

yx

T

+ – A

C

T

yx

T

Since the speed of the arm is 100 r.p.m. (clockwise), therefore

y= + 100

Moreover, as the speed of the driving shaft is 50 r.p.m. (anticlockwise), Therefore

x + y=– 50

or x= – 50 – y = – 50 – 100 = – 150

∴ Speed of the driving shaft

= Speed of wheel C

40

• 100 150 r.p.m.
  20

A

C

T

yx

T

==+×

= 400 r.p.m. Ans.

EXERCISE 34.2

1. Design the spur wheels, having a velocity ratio of 4 and the pitch of teeth as 7·5 mm. The
   approximate distance between the two shafts is 300 mm.
   [Ans. T 1 = 200; T 2 = 50; d 1 = 477·3 mm; d 2 = 119·3 mm; l = 298·3 mm]


2. Two spur gears A and B of an epicyclic gear train (as shown in Fig. 34.12) have 24 and 30
   teeth respectively. The arm rotates at 100 r.p.m. in the clockwise direction. Find the speed of
   the gear B on its own axis, when the gear A is fixed. [Ans. 180 r.p.m.]


3. If in the above example, the wheel A rotates at 200 r.p.m. in the anticlockwise direction, what
   will be the speed of the gear B.[Ans. 260 r.p.m.]


4. In an epicyclic gear train (as shown in Fig. 34.15) the arm is fixed to the shaft. The sun wheel
   S having 100 teeth rotates freely on the shaft and the wheel A with 150 teeth is separately
   driven. If the arm runs at 200 r.p.m. and the wheel A at 100 r.p.m. in the direction, find
   (i) the number of teeth on the wheel P, and
   (ii) Speed of the sun wheel S. [Ans. 25; 350 r.p.m. in the same direction]