The Theory of Machines and Mechanisms

Chapter 6. Balancing of Machinery

Sun Zhihong

Zhsun@dhu.edu.cn

62373317

62373319

Room307,No.3 Teaching Building

Today’s Topics

Purposes of balancing

Methods of balancing

Balancing of rigid disk-like rotorsStatic balancingDynamic balancing

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

Method of Balancing

Balancing of rotors

Balancing of mechanisms

Balancing of Rotors

What is rotor?Any link or machine member that is in pure r

otation is called rotor.

Rigid rotorsIf the rotating frequency of a rotor is less tha

n (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.

Balancing of Rotors

Flexible rotorsIf 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.

Types of Rigid Rotor 1. 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.

Types of Rigid Rotor2. Non-disk like rotor: whe

n B/D>0.2, such as camshafts, spindles in machine tools( 机床 ), axels of electric motors, etc.

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

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 t

he system to make:

∑Fi + Fc = 0Mc is called counterweight. Such balance is c

alled static balance.

Because,

∑Fi + Fc = 0That is, mcRcω2 + ∑ miRiω2=0 miRi - is called mass-radius product( 质径积 )

m1

m2

m3

r1 r2

r3

o

ω

Break mcRc , miRi into x and y components: mcRcx+ (m1R1x+m2R2x+m3R3x)=0 mcRcy+(m1R1y+ m2R2y+m3R3y)=0So, mcRccosθc= - (m1R1cosθ1+m2R2cosθ2+m3R3cosθ3)

= - ∑ miRi cosθi mcRcsinθc= - (m1R1sinθ1+m2R2sinθ2+m3R3sinθ3) = - ∑ miRi sinθi

Mass-radius product is :

mcRcy=[(- ∑ miRi cosθi )2+(∑ miRi sinθi )2]1/2 (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).

m1

m2

m3

r1 r2

r3

o

ω

x

y

Static Balance/Single Plane BalanceUsually 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

Static Balancing MachineThe 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.

Static Balancing MachineThe 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.

Example of Static Balancing

Given : The system shown in the right figure has the following data:

m1=1.2kg R1=1.135m @ 113.4°∠ m2=1.8kg R2=0.822m @ 48.8°∠ ω=40rad/sec Find: The mass-radius product and its

angular location needed to statically balance the system.

Solution

(1) Resolve the position vectors into x, y components:

R1=1.135m@ 113.4, R∠ 1x=-0.451m, R1y=1.042m

R2=0.822m@ 48.8, R∠ 2x=+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)]

= -0.433kg.m

mbRby = -(m1R1y+ m2R2y)= -[(1.2)(1.042)-(1.8)(0.618)]

= -2.363kg.m

Example of Static Balancing

mkgRm bb

b

402.2)363.2()433.0(

6.259433.0

363.2arctan

22

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

mc

rb

m1

m2

r1

r2ω

Balancing of non-disk like rigid rotordynamic 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.

m1

m2

r1

r2

F1

F2

s

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

m1

m2

r1

r2

F1

F2

s

Balancing of non-disk like rigid rotordynamic balance/two-plane balance

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, m2

r2 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。

Dynamic Balance选择平衡面 P,Q

把不平衡力向两平衡基面上分解。      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

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：

与 x 轴正向的夹角为θx

''=arctan[(mb''rb

'')y/(mb''rb

'')x]=161.57

Dynamic Balance

Homework

思考题： Page 1516-1, 6-2

Page 152:6-4, 6-6

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