Engineering Mechanics

Chapter 34 : Transmission of Power by Gear Trains „„„„„ 715

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

TA=TS + 2TP = 30 + (2 × 50) = 130

and as the wheel A is fixed, therefore

x + y= 0 or x = – y ...(i)

Moreover, as the sun wheel S rotates at 300 r.p.m. Therefore

300

130 13

–– –

30 3

A

S

T x

yx yx y

T

=×=×=

=

13 13 16

–(–) 1

333

y

yyy

⎛⎞

=+=⎜⎟

⎝⎠

∴ y=

300 3 900

16 16

×

= and

900

16

xy=− =−

Therefore speed of the driven shaft,

N= Speed of arm =

900

= = 56·25 r.p.m.

16

y Ans.

Torque transmitted by the driven shaft

Let T= Torque transmitted by the shaft.

We know that power transmitted by the shaft,

3800 = 2π NT = 2π × 56·25 T = 353·4 T

∴ T=

3800

10·75 N-m

353·4

= Ans.

34.17.EPICYCLIC GEAR TRAIN WITH BEVEL WHEELS

The problems on epicyclic gear train, with bevel wheels, may be solved exactly in the same
manner as in the case of epicyclic gear train with spur gears. The only important point, in such
problems, is that the direction of mostion of the two bevel (i.e. slanting) wheels (connected by an
intermediate spindle) is unlike.

Example 34.9. An epicyclic gear train consists of bevel wheel as shown in Fig. 34.16. The
wheel A, which is keyed to driving shaft X, has 40 teeth
and meshes with the wheel B (50 teeth) which in turn
meshes with the wheel C having 20 teeth. The wheel C is
keyed to driven shaft Y.

The wheel B turns freely on the arm which is rigidly
attached to the hollow sleeve. The hollow sleeve is riding
freely loose on the axis of the shafts X and Y.

If the driving shaft rotates at 50 r.p.m.
anticlockwise and the arm rotates at 100 r.p.m. clock-
wise, determine speed of the driven shaft.

Solution. Given: No. of teeth on wheel A (TA) = 40; No. of teeth on wheel B (TB) = 50; No.
of teeth on wheel C (TC) = 20; Speed of driving shaft = – 50 r.p.m. (anticlockwise); Speed of arm
= 100 r.p.m. (clockwise).

Fig. 34.16.