机械原理第六章机械的平衡

机械原理第六章机械的平衡第六章机械的平衡Chapter6.BalancingofMachinery

1.Purposes§6-1Introduction

DynamicpressinkinematicpairFrictionandinnerstressinlinkefficiencyandlife-spanInertiaforce(torque)

Compelledoscillation

Thepurposeofmechanicalbalancingistoclearupordecreasethebadeffectbybalancingthecomponent’sunbalancedinertia.Thebalanceofcomponentrotatingaboutafixedaxis

Rotor(转子):Partsconstrainedtorotateaboutafixedaxis.2.Contents(1)Balancingofrigidrotor刚性转子的平衡

staticbalancing(静平衡)dynamicbalancing(动平衡)(2)Balancingofflexiblerotor绕性转子的平衡

Balancingofmechanisms

rigidrotorflexiblerotormechanism(avi)1.Phenomenaofstaticimbalance

§

6-2Calculationforstaticbalancingofarigidrotor

Ifmasscenterofrotordoesn’tcoincidewiththeaxisofrotation,theireccentricmasswillleadtocentrifugalinertiaforce(离心力)whenrotating,andcausesanadditionaldynamicpress(附加动压力)inthelinkage.2.Geometriccondition

B/D≤1/5BDm3m1m2m3m1m23.TheoryofstaticbalancingF1F2F3m1m2m3r2r1r3centrifugalforces(离心力）oftheunbalancedmasses（偏心质量）：F1=m1r1w2F2=m2r2w2F3=m3r3w2If,F1+F2

+F3≠0FPmPThen,imbalanceF1F2F3m1m2m3r2r1r3

F1+F2

+F3+FP=

0FPmPSomecounterweight质量点(mp)canbeaddedtotherotortobalanceitscentrifugalforce.Tobalance:Fp=mprpw2m1r1w2+m2r2w2+m3r3w2+mPrPw2=0m1r1

+m2r2

+m3r3

+mPrP

=0miri----mass-radiusproduct(质径积)Conclusion：Requirementforstaticbalance:∑m1r1=0

AddRemoveConclusion:Abalancecanbeachievedbyaddingorremovingabalancemassinthesameplane.resolution：A.GraphicalmethodF1F2F3m1m2m3r2r1r3FPmPWPW1W2W3Scale(比例尺)：μW=(kgm/mm)miriWiWi=mirim1r1w2+m2r2w2+m3r3w2+mPrPw2=0B.Analyticalmethod

mx1rx1+mx2rx2

+mx3rx3

+mxPrxP

=0my1ry1+my2ry2

+my3ry3

+myPryP

=0F1F2F3m1m2m3r2r1r3FPmPEXAMPLEGiven:ThesystemshowninFIGhasthefollowingdata:m1=1.2kg

R1=1.135m

θ1=113.4°m2=1.8kg

R2=0.822m

θ2=48.8°Find:Themass-radiusproductionanditsangularlocationneededtostaticallybalancethesystem.Solution:Resolvethepositionvectorsintoxycomponents:

R1=1.135m

θ1=113.4°;R1x=-0.451,R1y=-1.042R2=0.822m

θ2=48.8°;R2x=+0.541,R2y=0.618SolveequationsmbRbx=-m1R1x-m2R2x=-(1.2)(-0.451)-(1.8)(0.541)=-0.433

mbRby=-m1R1y-m2R2ySolveequationsSolution:Themass-radiusproductcanbeobtainedwithavarietyofshapes.WhenRb=0.806matrequiredangleof259.6°,themassforthiscounterweightdesignisthen:

mb=2.042kg.m/0.806m=2.980kg

atthechosenradiusofRb=0.806m§6-3Calculationforthedynamicbalancingofarigidrotor刚性回转体的动平衡geometriccondition

B/D>1/5

Massmaybeunevenlydistributedbothrotationallyaroundtheiraxisandalsolongitudinallyalongtheiraxis.

F1F2F3r1m1m2r2m3r32.Force-balance-condition∑Fi=0centrifugalforces

∑Mi=0centrifugalmomentF

FⅠ

FⅡ3.CorrectionplanesTocorrectdynamicimbalancerequireseitheraddingorremovingtherightamountofmassattheproperangularlocationsintwocorrectionplanesseperatedbysomedistancealongtheshaft.l1l2l3lF1F2F3r1m1m2r2m3r3F'3F"3F'1F"1F'2F"2Ⅱ

Ⅰ平衡基面平衡基面4.TheoryL1L2L3LF1F2F3r1m1m2r2m3r3Ⅱ

ⅠF2ⅠF1ⅠF3ⅠF2F1F3ⅡⅡⅡConverteachcentrifugalforcestothecorrectionplaneⅠandⅡ.ThatisF1,F2andF3canbereplacedbyF1Ⅰ,F2Ⅰ,F3ⅠandF1Ⅱ,F2Ⅱ

,F3Ⅱ

spatialforcesystem2planarforcesystems平衡原理：将集中质量点所产生的离心力F向两个平衡基面上分解，得到两个分力F1和F2；合力F对系统的影响可以完全有两分力F1、

F2对系统的影响所代替；F-F'-F"L1

L2

LⅠⅡF2F1

在平衡基面上分别对两个分力F1、

F2进行平衡，得平衡力F'和F"，从而完成对集中质量点的平衡。FF'F"

将力F平行分解到两个平衡基面上，得F1和F2:L1

L2

LⅠⅡF2

=

FL1L(4)F1=

FL2L(3)F2F1F=F1+F2(1)F1L1=F2L2(2)l1l2l3lF1F2F3r1m1m2r2m3r3F'3F"3F'1F"1F'2F"2Ⅱ

Ⅰ平衡基面平衡基面L1L2L3LF1F2F3r1m1m2r2m3r3F'2F'1F'3F"3F"1F"2Ⅱ

ⅠF'1

=

F1

L-L1LF"1

=

F1L1LF'2

=

F2

L-L2LF"2

=

F2L2LLF'3

=

F3

L-L3LF"3

=

F3L3m'r'F'F'1+F'2+F'3+F'=0F"1+F"2+F"3+F"=0从而求得m'r'和m"r"。m"F"r"L1L2L3LF1F2F3r1m1m2r2m3r3Ⅱ

ⅠF2ⅠF1ⅠF3ⅠF2F1F3ⅡⅡⅡ步骤：(1)分别将各回转平面上集中质量点mi所产生的惯性力Fi(或质径积、重径积)向两个平衡基面上分解，得到F'i和F"i

。(2)分别在两个平衡基面上用静平衡的方法求解平衡质量点的质径积miri(或重径积)。1.StaticBalancingExperiment