Engineering Mechanics

Chapter 22 : Combined Motion of Rotation and Translation „„„„„ 463

Example 22.3. In a reciprocating pump, the lengths of connecting rod and crank is 1125
mm and 250 mm respectively. The crank is rotating at 420 r.p.m. Find the velocity, with which the
piston will move, when the crank has turned through an angle of 40° from inner dead centre.

Solution. Given : Length of connecting rod (l) = 1125 mm = 1.125 m ; Length of crank
(r) = 250 mm = 0.25 m ; Angular rotation of crank (N) = 420 r.p.m and angle through which the
crank has turned (θ) = 40°.

We know that angular velocity of crank,

2 2 420

44 rad/s

60 60

ππ×N

ω= = =

and velocity of B (vB) = ωr = 44 × 0.25 = 11 m/s

Fig. 22.10.
Now let us locate position of the instantaneous centre of the rod BC as shown in Fig. 22.10
as discussed below :

1. Select a suitable point A, and draw a circle with A as centre and 0.25 m radius to some
   suitable scale to represent crank of the pump.


2. Through A, draw a horizontal line meeting at D on the circumference of the crank,
   representing the inner dead centre. Extend the line AD.


3. Draw a line AB, such that angle BAD is equal to 40°.


4. With B cut off BC equal to 1.125 m to the scale on the horizontal line through A, representing
   the connecting rod of the pump.


5. Extend the line AB, and through C draw a line at right angles to AC, meeting the extended
   line AB at O, which is the instantaneous centre of the rod BC.