1.0 LEARNING OUTCOME:

At the end of this experiment, the student should be able to,

1) to find natural frequency for the system

Wn = 2πf

= 2π(0.1639)

= 1.0298 rps

2) to identify the properties of spring used in the system

(wn)2 = k1 + k2 + k3 + k4 + / m1 + m2 + m3 + m4

(1.0298rps)2 = 4k / (1.0+1.0+1.4+1.1)

K = 1.193 N/m

2.0 INTRODUCTION

Balancing of rotating equipment is a very important aspect in the design of any

mechanical system that involves a rotating shaft. Rotating systems are rarely

perfectly balanced ; the degree of balance required depends upon size and location

of the unbalances and the speed of operation. Unbalance is generally caused by an

unbalanced mass, located at some eccentricity, spinning about the center of rotation

at a constant frequency

3.0 SAFETY

No hand touching the machine while still in operation

4.0 THEORY:

The total force unbalanced is the sum of all of unbalance forces. The total moment

unbalanced is the sum of all of the unbalance forces operating at different locations along

the axis rotation. Static balancing refers to a procedure that adds or subtracts mass at

some eccentricity to balance the vector forces. Dynamic balancing refers to a procedure

that adds or subtracts mass at some eccentricity and location along the axis of rotation to

balance the unbalance moments.

Free vibration is initiated by disturbing the system from its static equilibrium position by

imparting the mass some displacement u (0) and velocity u’(0) at time t=0.

Un damped free vibration:

The governing equation for undamped free vibration is

mu” + ku = 0

Where,

m= mass , u” = acceleration, K= stiffness, u = displacement.

Model Math:

Fm + FL = 0

mx’ + (K1 + K2 + K3 + K4)x = 0

5.0 APPARATUS

Balancing Apparatus and steel ruler

Balancing machines consist of horizontal square frame that was hanging at their own

positions using a spring. This machine also has rotating system which is operating by

motor and belt. The rotation system consist one spindle that is supported by bearing at

end of shaft. It is has four discs such as discs A, B, C and D. Where the distance between

each discs is a same (6 inches) and has same radius position which is 3, 4, 5, and 6 inches

respectively.

6.0 PROCEDURE:

6.1 Observed the system weather it is balanced or otherwise without load and mass once the

motor is switched on.

6.2 The motor is switched off

6.3 The load 1.0kg weight on 4 inches radius at 00 is fixed on discs B.

6.4 The load 1.4kg weight on 4 inches radius at 1200 is fixed on discs C

6.5 Fit the load 1.0kg and 1.1 kg weights at 2200 and 2700 on discs A and D. Start the both

side of discs with 6 inch.

6.6 Change the mass position of both discs. Change the both discs position untill the dynamic

balancing is formed

6.7 Record the mass position of the mass in the result table.

6.8 Use the Mr polygon diagram to calculate the “X” value. And the load 1.1 kg is placed at

“X” distance.

7.0 RESULT:

The position of the load radius for load 1.0kg weight discs A is 3” radius

The position of the load radius for load 1.1kg weight at disc D is 4” radius

8.0 DISCUSSION AND CONCLUSION:

Select plane A as a reference plane.

ROTATION

PLANE

MASS , M

(KG)

RADIUS, r

(INCH)

Mr (KG

INCH)

DISTANCE,

L (INCH)

MRL (KG

INCH2)

A 1.0 3” 3.0 6 18

B 1.0 5” 5.0 6 30

C 1.4 4” 5.6 6 33.6

D 1.1 4” 4.4 0 0

Draw the Mr polygon diagram. Compare the experiment results with the result from

drawing method

Period of oscillation = 6.1 second

F = 1/6.1 = 0.1639Hz

Natural frequency, Wn = 2π(0.1639)

= 1.0298 rps