Dynamics of Machinery / Balancing of Rotating Masses / Dynamic Balancing / Graphical Method / Problem Solving /

1. Graphical Method
DYNAMIC BALANCING PROBLEM
MOHITH DADU
Assistant Professor
Dept. of Mechanical Engg.
TKMCE, Kollam,Kerala

2. QUESTION
Q. A rotating shaft carries three unbalanced masses of 4 kg, 3 kg and 2.5 kg at radial
distances of 75 mm, 85 mm and 50 mm and at an angular positions of 45o, 135o and 240o
respectively. The shaft length is 0.8 m between bearings. Determine the amount of
balancing masses in planes at 75 mm from the bearings for complete balance of the shaft.
The first balancing mass is in a plane between first mass and the bearing and second
balancing mass is in a plane between third mass and the other bearing as shown in figure.
Plane of first balancing mass and the bearing at that end is 225 mm.
Take the radial distances of first and second balancing masses as 40 mm and 75 mm.
For solving the problem you may visit ;
https://www.youtube.com/watch?v=CoGkpJuV7uY&t=5s
2 Mohith Dadu / Mechanical Engineering

3. m1 =4 kg
Ө1 = 45o
A
m2 =3 kg
Ө2 = 135o
B
m3 =2.5 kg
Ө3 = 240o
C
BP 2
BP 1
m1r1ω2
m2r2ω2
m3r3ω2
3 Mohith Dadu / Mechanical Engineering

4. Planes Mass
(kg)
Radius
(m)
m * r Distance from RP
(l)
(m)
m*r*l Angle
with RA
BP1 (First Balancing mass )
-Reference Plane RP
mB1 rB1 mB1rB1
0 0 ӨB1
A (m1) 4 0.075 0.3 l1 = 0.15 0.045 45o
B (m2) 3 0.085 0.255 l2 = 0.350 0.0893 135o
C (m3) 2.5 0.05 0.125 l3 = 0.525 0.0656 240o
BP2 (Second Balancing
mass )
mB2 rB2 mB2rB2
lBP2 = 0.650
0.65mB2rB2
ӨB2
4 Mohith Dadu / Mechanical Engineering

5. m1r1l1=0.045
m2r2 l2= 0.0893
m3r3 l3= 0.0656
Couple Polygon on RP
Scale : 1 cm = 0.01
Ө1 = 45o
Ө2 = 135o
Ө3 = 240o
ӨB2=329o
0.65mB2rB2 = 7.4 cm = 0.074
Given rB2 = 0.04 m mB2= 0.074/ (.04*0.65) = 2.85 kg
ӨB2=329o
ӨB2=329o
mB2= 2.85
Ө1 = 45o
BP 1 (RP)
A
B
C
Ө2 = 135o
Ө3 = 240o
5 Mohith Dadu / Mechanical Engineering

6. m1r1=0.3
m2r2= 0.255
m3r3= 0.125
mB2rB2 = 0.114
ӨB2=329o
Force Polygon on
RP
Scale : 1 cm = 0.1
Ө1 = 45o
Ө2 = 135o
Ө3 = 240o
ӨB2=329o
ӨB1=253o
mB1rB1 = 2.35 cm = 0.235
Given rB1 = 0.075 m mB1= 0.235/0.075 = 3.13 kg
mB1= 3.13 kg
ӨB1=253o
ӨB1=253o
Ө2 = 135o
Ө3 = 240o
Ө1 = 45o
BP 1 (RP)
A
B
C
6 Mohith Dadu / Mechanical Engineering