1. Room307,No.3 Teaching Building
The Theory of Machines and Mechanisms Chapter 6. Balancing of Machinery
Sun Zhihong
Room307,No.3 Teaching Building

2. Today’s Topics Purposes of balancing Methods of balancing
Balancing of rigid disk-like rotors
Static balancing
Dynamic balancing

3. Purposes of Balancing
Making machine element’s center of mass coincide with their rotating centers
The center of mass of some machine may not coincide with their rotating centers, the reasons are:
The asymmetry of the structure
Uneven distribution of materials
Errors in machining , casting and forging
Improper boring
By keys
By assembly

4. Method of Balancing
Balancing of rotors
Balancing of mechanisms

5. Balancing of Rotors What is rotor? Rigid rotors
Any link or machine member that is in pure rotation is called rotor.
Rigid rotors
If the rotating frequency of a rotor is less than (0.6~0.75)nC1 ( the rotor’s first resonant共振 frequency)，the rotor is supposed to have no deformation during rotation, it is called rigid rotor.

6. Balancing of Rotors Flexible rotors
If the working rotating frequency of the rotor is higher than (0.6~0.75)nc1,then the rotor will have large elastic deformation due to the imbalance during rotation. The elastic deformation makes the eccentricity larger than the original one. So that a new imbalance factor is added and the balancing problem become more serious and complicated. Such rotor is called flexible rotor.

7. Types of Rigid Rotor Disk-like rigid rotor:
The axial dimension B is smaller compared to the diameter D ( usually B/D≤0.2 )
The mass of such rotors are assumed practically lie in a common transverse plane.
Such as cams, gears, pulleys, flywheels, fans, grinding wheels, impellers, etc.

8. Types of Rigid Rotor
Non-disk like rotor: when B/D>0.2, such as camshafts, spindles in machine tools(机床), axels of electric motors, etc.

9. Balancing of Disk-like Rotor - Static balance/single plane balance
A rigid rotor rotates with a constant angular velocity ω.
Due to some reasons, there are three unbalance masses m1, m2, m3 at radial distance r1, r2, r3.
Centrifugal forces produced by m1, m2, m3 are miriω2.
All centrifugal forces Fi in this disk-like rotor are planar and concurrent.
If ∑Fi=0, then the mass center of the system coincides with the shaft center and the rotor is balanced. Other wise, it is called imbalance. m1 m2 m3 r1 r2 r3 o ω

10. Static balance/single plane balance
Since the imbalance can be shown statically, so it is called static imbalance.
In order to balance the disk-like rotor, a fourth mass mc with rotation radius of rc is added to the system to make:
∑Fi + Fc = 0
Mc is called counterweight. Such balance is called static balance.
Because,
That is, mcRcω2 + ∑ miRiω2=0
miRi - is called mass-radius product(质径积) m1 m2 m3 r1 r2 r3 o ω

11. ω x mcRcy=[(- ∑ miRi cosθi )2+(∑ miRi sinθi )2]1/2 (1) y
Break mcRc , miRi into x and y components:
mcRcx+ (m1R1x+m2R2x+m3R3x)=0
mcRcy+(m1R1y+ m2R2y+m3R3y)=0
So,
mcRccosθc= - (m1R1cosθ1+m2R2cosθ2+m3R3cosθ3)
= - ∑ miRi cosθi
mcRcsinθc= - (m1R1sinθ1+m2R2sinθ2+m3R3sinθ3)
= - ∑ miRi sinθi m1 m2 m3 r1 r2 r3 o ω x y
Mass-radius product is :
mcRcy=[(- ∑ miRi cosθi )2+(∑ miRi sinθi )2]1/ (1)
Angle between rc and x coordinate:
Θc=arctan(∑ miRi sinθi / ∑ miRi sinθi ) (2)
Notice: the proper quadrant（象限） of Θc must be determined by the signs of both the numerator and denominator of equation (2).

12. Static Balance/Single Plane Balance
Usually the counterweight mc are placed at as large a rotating radius as is practicable to minimize the amount of the added mass.
In practice, rotors are often balanced by removing mass in the direction of imbalance (e.g. drilling a hole as the open circle mc’), rather than by adding counterweights to it.
Any number of masses in a disk-like rotor can be balanced by adding a single mass or removing a mass at an appropriate position.
This is also called single-plane balance. m1 m2 m3 r1 r2 r3 o ω x y rc θ2 θ1 θc

13. Static Balancing Machine
The static balancing machine has two hard horizontal parallel rails. The rigid rotor for test is laid on the rails.
If the centre of mass of the system coincides with the axis of rotation, the rotor will not roll regardless of the angular position of the rotor.

14. Static Balancing Machine
The experiment can be conducted as follows:
Roll the disk gently by hand
The disk will rock until it comes to rest . The lowest angular location of the mass centre is indicated.
A trial weight (e.g. a lump of plastic) be attached to the highest point of the disk.
The corrections must be made by trial and error until the disk does not roll from any initial location.
This will determines the amount and the location of unbalance.
The balancing may be achieved by drilling out material or adding mass.

15. Example of Static Balancing
Given : The system shown in the right figure has the following data:
m1=1.2kg
m2=1.8kg
ω=40rad/sec
Find: The mass-radius product and its angular location needed to statically balance the system.

16. Example of Static Balancing
Solution
(1) Resolve the position vectors into x, y components:
R1x=-0.451m, R1y=1.042m
R2x=+0.541m, R2y= 0.618m
(2) So the mass-radius product of the counterweight is:
mbRbx = -(m1R1x+m2R2x) = -[(1.2)(-0.451)-(1.8)(0.541)]
= kg.m
mbRby = -(m1R1y+ m2R2y)= -[(1.2)(1.042)-(1.8)(0.618)]
= kg.m

17. Example of Static Balancing
(3) If a value for Rb=0.806m is desired, the mass required for this counterweight design is:
mb=(2.402kg-)/(0.806m)=2.980kg
at a chosen CG radius of:
Rb=0.806m m1 m2 r1 ω r2 rb mc

18. Balancing of non-disk like rigid rotor dynamic balance/two-plane balance
m1=m2, r1=r2
So, m1r1=m2r2
The disk-like rotor is statically balanced.
But, when it rotates, the centrifugal forces:
F1=m1R1ω2
F2=m2R2ω2
will act outwards. They are not collinear and will produce a resultant couple whose direction changes during rotation.
The couple will act on the frame and tend to produce rotation vibration of the frame.

19. Balancing of non-disk like rigid rotor dynamic balance/two-plane balance
This kind of imbalance can only be detected by means of a dynamic test in which the rotor is spinning.
It is called dynamic imbalance.
The left rotor is statically balanced and dynamically unbalanced.
The criterion for the balancing of a non-disk rigid rotor is:
Both the vector sum of all inertia forces and the vector sum of all moments of inertia forces about any point must be zero, i.e.
∑Fi=0 and ∑Mi=0

20. Dynamic Balance
    图示为一回转体，其上有不平衡质量m1=1kg, m2=2kg,，与转动轴线的距离分别为r1=300mm, r2=150mm, m1r1, m2r2与x轴正向的夹角分别为45°和315°。试求在P, Q 两平衡校正面上应加的平衡质径积(mbrb)P和(mbrb)Q的大小及方位。

21. Dynamic Balance 选择平衡面P,Q 把不平衡力向两平衡基面上分解。 向P面分解： 向Q面分解：
      m1r1=1×300=300kg·mm     m2r2=2×150=300kg·mm
向P面分解：
(m1r1)200=(m1'r1')300,  m1'r1'=300/300×200=200kg·mm
(m2r2)100=(m2'r2')300,         m2'r2'=300/300×100=100kg·mm
向Q面分解：
(m1r1)100=(m1"r1")300,         m1"r1"=300/300×100=100kg·mm
(m2r2)200=(m2"r2")300,         m2"r2"=300/300×200=200kg·mm

22. Dynamic Balance 在P面上有： 与x轴正向的夹角为
θx'=arctan[(mb'rb')y/(mb'rb')x]=198.43°
在Q面上有：
θx''=arctan[(mb''rb'')y/(mb''rb'')x]=161.57
            

23. Homework
思考题：Page 151
6-1, 6-2
Page 152:
6-4, 6-6

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