UNIVERSITI TUN HUSSEIN ONN MALAYSIAFaculty of Mechanical and Manufacturing Engineering

MECHANICS OF MACHINE LABORATORY

LABORATORY REPORT

ENGINEERING LABORATORY IV (BDA 2721)

TOPIC : BALANCING OF ROTATING MASSES

1.OBJECTIVE

The objective of this experiment is to study the different balancing of the following

body

1.1 Body at single plane and multiple-plane.1.2 Body at static and dynamics state for the multiple plane.2.LEARNING OUTCOME

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

2.1Understand the concept of balancing for the single and multiple planes. 2.2Implement and analyze the required data collectively within member of group.

2.3Produce good technical report according to the required standard.3.THEORY

The balancing of rotating bodies is important to avoid vibrations. In heavy industrial machines such as steam turbines and electric generators, vibration could cause catastrophic failure. Vibrations are noisy and uncomfortable and when a car wheel is out of balance, the ride is quite unpleasant. In this case of a simple wheel, balancing simply involves moving the centre of gravity to the centre of the rotation but as we shall see, for longer and more complex bodies, there is no more to it. For a body to be completely balanced it must have two things:a) Static Balance This occurs where there is no resultant centrifugal force and the centre of gravity is on the axis of rotation.

b) Dynamics Balance This occurs when there is no resulting turning moment along the axis.This experiment is to prove on the basic principle of balancing. Before implementing the experiment (5.1) and (5.2), sector plate B5 and C5 need to be installed at the internal position of disc by using the short screw (5/8) as shown in Figure 1. 4.EQUIPMENTS

5. CALCULATION & OBSERVATION

5.1 BALANCING IN A SINGLE PLANE OF REVOLUTION

NO.DATAOUTPUT

a)m1 = 30, r1 = 60mm

m1 x r1 = 30 x 60 = 1800 ( not zero)

Imbalance

b)m1 = 30, r1 = 60mm

m2 = 30, r2 = 60mm

m1r1 + m2r2 = ( 30 x 60) + (-30 x 60)

Balance

c)m1 = 30, r1 = 60mm

m2 = 60, r2 = 30mm

m1r1 + m2r2 = ( 30 x 60) + (-60 x 30) = (zero)

Balance

d)m1 = 30, r1 = 60mm

m2 = 30, r2 = 30mm

m2 = 15, r2 = 60mm

m1r1 + m2r2 + m3r3 = ( 30 x 60) + (-30 x 30) +

(-15 x 60)

Balance

e1)m1 = 30, r1 = 60mm

m2 = 30, r2 = 60mm

m2 = 30, r2 = 60mm

1 = 0 , 2 = 120 & 3 = 240

m1r1 + m2r2 + m3r3 =

Balance

e2)m1 = 30, r1 = 60mm

m2 = 30, r2 = 60mm

m2 = 30, r2 = 60mm

1 = 0 , 2 = 120 & 3 = 240

m1r1 + m2r2 + m3r3 =

Imbalance

fm1 = 30, r1 = 60mm

m2 = 40, r2 = 45mm

m2 = 60, r2 = 30mm

1 = 0 , 2 = 120 & 3 = 240

m1r1 + m2r2 + m3r3 =

Balance

5.2 BALANCING IN A SEPARATE PLANE OF REVOLUTION

NODATAOUTPUT

a.m1 = 30, r1 = 60mm on plane B

m2 = 40, r2 = 45mm on plane C

m3 = 60, r2 = 30mm on plane D

where position m3 is in opposite

of the radius m1 and m2

L1 = X, L2 = 2X and L3 = 3X

m1r1 + m2r2 + m3r3

= (30 X 60) + (30 X 60) + (-60 X 60) = 0

Vector equation : (m1r1 + m2r2 + m3r3)

Static

StateDynamic

State

balanceImbalance

m1r1L1 + m2r2L2 + m3r3L3

= (30 x 60 x X) + (30 x 60 x 2X) + (-60 x 60 x 3X)

= - 5400 X ( not zero)

Vector equation : (m1r1L1 + m2r2L2 + m3r3L3)

b.m1 = 30, r1 = 60mm on plane B

m2 = 30, r2 = 60mm on plane C

m3 = 30, r2 = 60mm on plane D

m4 = 30, r2 = 60mm on plane A

where position m3 and m4 is in opposite of the radius m1 and m2

L1 = X, L2 = 2X, L3 = 3X and L4 = 0

m1r1 + m2r2 + m3r3 + m4r4

=(30 X 60) + (30 X 60) +(-30 X 60) + (-30 X 60)

= 0

Vector equation : m1r1 + m2r2 + m3r3 + m4r4

balancebalance

m1r1L1 + m2r2L2 + m3r3L3 + m4r4L4

= (30 X 60 x X) + (30 X 60 x 2X) +(-30 X 60 x 3X) + 0

= 0

Vector equation : m1r1L1 + m2r2L2 + m3r3L3 + m4r4L4

c.m1 = 60, r1 = 60mm on plane B

m2 = 60, r2 = 60mm on plane C

m2 = 20, r2 = 60mm on plane D

m2 = 20, r2 = 60mm on plane A

where position m3 and m4 is in opposite of the radius m1 and m2

L1 = X, L2 = 2X, L3 = 3X and L4 = 0

m1r1 + m2r2 + m3r3 + m4r4

=(60 X 60) + (-60 X 60) +(20 X 60) + (-20 X 60)

= 0

Vector equation : m1r1 + m2r2 + m3r3 + m4r4

balancebalance

m1r1L1 + m2r2L2 + m3r3L3 + m4r4L4

=(60 X 60 x X) + (-60 X 60 x 2X) +(20 X 60 x 3X) + 0

= 0 (zero)

Vector equation : m1r1L1 + m2r2L2 + m3r3L3 + m4r4L4

d.m1 = 60, r1 = 60mm on plane B

m2 = 20, r2 = 60mm on plane D

m3 = 40, r2 = 60mm on plane C

where position m2 and m3 is in opposite of the radius m1

L1 = X, L2 = 3X and L3 = 0

m1r1 + m2r2 + m3r3

= (60 X 60) + (-20 X 60) +(- 40 X 60)

= 0

Vector equation : m1r1 + m2r2 + m3r3

balancebalance

m1r1L1 + m2r2L2 + m3r3L3

= (60 X 60 x X) + (-20 X 60 x 3X) 0

= 0

Vector equation : m1r1L1 + m2r2L2 + m3r3L3 + m4r4L4

7. DISCUSSIONS

7.1- Provide the comments regarding to the results obtained from the balancing of rotating masses from single plane

The result obtained be the same whatever the value of (m1 x r1) and = 0. If a mass of M gram is fastened to a shaft rotating at w rad/s at radius r meter, the centrifugal force, producing out of balance effect acting radically outwards on the shaft will be equal Mw2r Newton. This out of balance in any one of the following two ways:

By introducing single revolving mass in the same transverse. Introduce a second mass B gram, called the balance mass, diametrically opposite to M at radius R rotating with same angular speed of w rad/s fig For complete balance, the centrifugal force of the two masses must be equal and opposite in the plane of rotation.

m1 x r1 = m2 x r2 = 0

If several masses are connected to s shaft at different radii in one plane perpendicular to the shaft and the shaft is made to rotate, each mass will set up out of balance centrifugal force on the shaft. In such a case complete balance can be obtained by placing only one balance mass in the same plane whose magnitude and relative angular position can be determined by means of a force diagram. Since all the masses are connected to the shaft, all will have the same angular velocity, we need not calculate the actual magnitude of centrifugal force of any, but deal only with mass moments.

7.2-Provide the comments regarding to the results obtained from the balancing of

rotating masses from single plane

The technique of tackling this problem is to transfer the centrifugal force acting in each plane to a single parallel plane which is usually termed as reference plane and thereafter the procedure for balancing is almost the same as for different forces acting in the same plane. How force acting in one plane is transferred to another plane

Static balance This occurs when there is no resultant centrifugal force and the centre of gravity is on the axis of rotation

Dynamic balance This occurs when there is no resulting turning moment along the axis

7.3 Give two examples on the application of balancing of rotating masses in real time world

Steam turbines

Electric generator

Fan impeller

Car wheel

8.CONCLUSION

The topic for this week lab is Balancing of Rotating Masses and the objective of this experiment is to study the different of the following body. It is body at single plane and multiple plane also body at static and dynamic state for the multiple plane. From the theory known that the balancing of rotating bodies is important to avoid vibration. For example in cement industry, there are many big impeller is used for the process. During the rotating, the blade of this impeller experience distortion because of friction will make the impeller imbalance and can cause damage to the bearing. If the balancing process is not done to the impeller, not only bearing will damage but also can cause damage to the shaft and motor and would result factory process halted altogether give loss to factory income. This experiment is to prove the principle of balancing. By using the formula (m*r) and (m*r*l) we will know the situation if it is balance or imbalance. For single plane, the calculation must equal to zero to make it balance if not it will became imbalance during the rotation. But for the single plane that have angle, we can know if it balance or not by drawing the force polygon. It can be determine by the line of the polygon must return to 0. Finally, for the balancing in separate plane it is same with the single plane by using (mrl) formulas and it has two situations it is during static state and dynamics state. In conclusion balancing is important in engineering field for dynamics or static state. REFFERNCE 1. http://www.freestudy.co.uk/dynamics/balancing.pdf2. http://modelenginenews.org/etw/etw_bal/p1.html3. http://www.rocw.raifoundation.org/mechanical/BTec-Mec/theoryofmachine-II/lecture-notes/lecture-16.pdf4. http://www.engr.wisc.edu/mpac/pdf/paper05.pdfFigure 1.

C5

B5

Set of Weight

Dynamics Balancing Apparatus

ON/OFF Pedal

Toolbox

Tools

m1,m2 & m3 mass at single plane

r1,r2& r3 distance m1 from centre of gravity and plane of rotation

m1r1

m2r2

m1r1

m2r2

m1r1

m3r3

m2r2

m1r1

240

120

0

240

120

0

m2r2

m3r3

m1r1

240

120

0

m1, m2 & m3 mass at single plane

r1, r2& r3 distance m1 from centre of gravity and plane of rotation

m3r3

m2r2

m1r1

m3r3L3

m1r1L1

m2r2L2

EMBED PBrush

m1r1

m2r2

m4r4

m3r3

m1r1L1

m2r2L2

m3r3L3

EMBED PBrush

m1r1

m2r2

m3r3

m4r4

m2r2L2

m1r1L1

m3r3L3

EMBED PBrush

m1r1

m2r2

m3r3

m2r2

m1r1

m2r2

m1r1

A

m1

m2

m3

D

C

B

x

x

x

m4

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