Engineering Mechanics

Chapter 34 : Transmission of Power by Gear Trains „„„„„ 713

First of all, prepare the table of motions as given below :

Step No. Conditions of motions

Revolution of

Arm Wheel C Wheel B Wheel A

1. Arm fixed; wheel C 0+ 1– C
   B

T

T

• CB
  BA

T T

TT

×

rotates through + 1

revolution

2. Arm fixed; wheel C 0+ x – C
   B

T

x

T

• C
  A

T

x

T

×

rotates through

+ x revolutios

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


4. Total motion (2 +3) + yx + y –

C

B

T

yx

T

• C
  A

T

yx

T

×

Speed of wheel C

We know that the speed of the arm,

y= 18 r.p.m.

Since the wheel A is fixed, therefore

• C
  A

T

yx

T =0

or

32

18 –

72

x× =0

∴ x=

18 × 72

=40·5 r.p.m.

32

and speed of wheel C, NC=x + y = 40·5 + 18 = 58·5 r.p.m. Ans.

Speed of wheel B

Let dA, dB and dC be the pitch circle diameters of wheels A, B and C respectively. From the
geometry of Fig. 34.14, we find that

2

C

B

d

d + =

2

dA

or 2 dB + dC=dA

Since the no. of teeth are proportional to their diameters, therefore

2 TB + TC=TA

or 2 TB + 32 = 7 2

∴ TB=20

and speed of wheel B, NB=

32

–18–40·5

20

C

B

T

yx

T

=×= – 46·8 r.pm. Ans.