Mechanical Engineering Principles

SIMPLE MACHINES 203

2. A load of 1.7 kN is lifted by a screw-jack
   having a single-start screw of lead 5 mm.
   The effort is applied at the end of an
   arm of effective length 320 mm from the
   centre of the screw. Calculate the effort
   required if the efficiency at this load is
   25%. [16.91 N]

18.5 Gear trains

A simple gear train is used to transmit rotary
motion and can change both the magnitude and the
line of action of a force, hence is a simple machine.
The gear train shown in Figure 18.4 consists ofspur
gearsand has an effort applied to one gear, called
the driver, and a load applied to the other gear, called
thefollower.

Driver Follower

Figure 18.4

In such a system, the teeth on the wheels are so
spaced that they exactly fill the circumference with
a whole number of identical teeth, and the teeth on
the driver and follower mesh without interference.
Under these conditions, the number of teeth on the
driver and follower are in direct proportion to the
circumference of these wheels, i.e.

number of teeth

on driver

number of teeth

on follower

=

circumference

of driver

circumference

of follower

( 18. 5 )

If there are, say, 40 teeth on the driver and 20
teeth on the follower then the follower makes two
revolutions for each revolution of the driver. In
general:

number of revolutions

made by driver

number of revolutions

made by the follower

=

number of teeth

on follower

number of teeth

on driver

( 18. 6 )

It follows from equation (18.6) that the speeds of

the wheels in a gear train are inversely proportional

to the number of teeth. The ratio of the speed of the

driver wheel to that of the follower is the movement

ratio, i.e.

Movement ratio=

speed of driver

speed of follower

=

teeth on follower

teeth on driver

( 18. 7 )

When the same direction of rotation is required on

both the driver and the follower anidler wheelis

used as shown in Figure 18.5.

Driver (A) Idler (B) Follower (C)

Figure 18.5

Let the driver, idler, and follower beA,Band

C, respectively, and letNbe the speed of rotation

andTbe the number of teeth. Then from equation

(18.7),

NB

NA

=

TA

TB

orNA=NB

TB

TA

and

NC

NB

=

TB

TC

orNC=NB

TB

TC

Thus

speed ofA

speed ofC

=

NA

NC

=

NB

TB

TA

NB

TB

TC

=

TB

TA

×

TC

TB

=

TC

TA