机械原理第六章动平衡.ppt
第六章 机械的平衡Chapter 6.Balancing of Machinery 2021/9/2311.Purposes6-1 IntroductionDynamic press in kinematic pairFriction and inner stress in linkefficiency and life-spanInertia force(torque)Compelled oscillation The purpose of mechanical balancing is to clear up or decrease the bad effect by balancing the components unbalanced inertia.2021/9/232The balance of component rotating about a fixed axis Rotor(转子转子):Parts constrained to rotate about a fixed axis.2.Contents (1)Balancing of rigid rotor刚性转子的平衡 static balancing(静平衡)dynamic balancing(动平衡)(2)Balancing of flexible rotor绕性转子的平衡 Balancing of mechanisms 2021/9/233rigid rotorflexible rotormechanism(avi)2021/9/2341.Phenomena of static imbalance 6-2 Calculation for static balancing of a rigid rotor If mass center of rotor doesnt coincide with the axis of rotation,their eccentric mass will lead to centrifugal inertia force(离心力)when rotating,and causes an additional dynamic press(附加动压力)in the linkage.2021/9/2352.Geometric condition B/D 1/5BDm3m1m2m3m1m22021/9/2363.Theory of static balancingF1F2F3m1m2m3r2r1r3centrifugal forces(离心力）离心力）of the unbalanced masses（偏心质量）：（偏心质量）：F1=m1 r1w w2F2=m2 r2w w2F3=m3 r3w w2If,F1+F2+F3 0FPmPThen,imbalance2021/9/237F1F2F3m1m2m3r2r1r3 F1+F2+F3+FP =0FPmPSome counterweight质量点质量点(mp)can be added to the rotor to balance its centrifugal force.To balance:Fp=mp rpw w2m1 r1w w2+m2 r2w w2+m3 r3w w2+mP rPw w2=0m1 r1+m2 r2+m3 r3+mP rP=0miri-mass-radius product(质质径径积积)Conclusion：Requirement for static balance:m1 r1=0 2021/9/238AddRemoveConclusion:A balance can be achieved by adding or removing a balance mass in the same plane.2021/9/239resolution：A.Graphical methodF1F2F3m1m2m3r2r1r3FPmPWPW1W2W3Scale(比例尺比例尺)：W=(kgm/mm)miriWiWi=mirim1 r1w w2+m2 r2w w2+m3 r3w w2+mP rPw w2=02021/9/2310B.Analytical method mx1 rx1+mx2 rx2+mx3 rx3+mxP rxP=0 my1 ry1+my2 ry2+my3 ry3+myP ryP=0F1F2F3m1m2m3r2r1r3FPmP2021/9/2311EXAMPLEGiven:The system shown in FIG has the following data:m1=1.2kg R1=1.135m 1=113.4 m2=1.8kg R2=0.822m 2=48.8 Find:The mass-radius production and its angular location needed to statically balance the system.2021/9/2312Solution:1.Resolve the position vectors into xy components:R1=1.135m 1=113.4;R1x=-0.451,R1y=-1.042 R2=0.822m 2=48.8;R2x=+0.541,R2y=0.618Solve equations mbRbx=-m1R1x-m2R2x=-(1.2)(-0.451)-(1.8)(0.541)=-0.433 mbRby=-m1R1y-m2R2y=-(1.21.042)-(1.8)(0.618-2.363Solve equations2021/9/2313Solution:4.The mass-radius product can be obtained with a variety of shapes.When Rb=0.806m at required angle of 259.6,the mass for this counterweight design is then:mb=2.042kg.m/0.806m=2.980kg4.at the chosen radius of Rb=0.806m2021/9/23146-3 Calculation for the dynamic balancing of a rigid rotor刚性回转体的动平衡刚性回转体的动平衡1.geometric condition B/D 1/5 Mass maybe unevenly distributed both rotationally around their axis and also longitudinally along their axis.F1F2F3r1m1m2r2m3r32.Force-balance-condition Fi=0 centrifugal forces Mi=0 centrifugal moment2021/9/2315F F F3.Correction planesTo correct dynamic imbalance requires either adding or removing the right amount of mass at the proper angular locations in two correction planes seperated by some distance along the shaft.2021/9/2316l1l2l3lF1F2F3r1m1m2r2m3r3F3F3F1F1F2F2 平衡基面平衡基面4.Theory2021/9/2317L1L2L3LF1F2F3r1m1m2r2m3r3 F2F1F3F2F1F3Convert each centrifugal forces to the correction plane and.That is F1,F2 and F3 can be replaced by F1,F2 ,F3 and F1,F2,F3 spatial force system 2 planar force systems 2021/9/2318平衡原理：平衡原理：将集中质量点所产生的离心力将集中质量点所产生的离心力F向两个平衡基面上分解，得到两个向两个平衡基面上分解，得到两个分力分力F1和和F2；合力合力F 对系统的影响可以完全对系统的影响可以完全有两分力有两分力F1、F2对系统的影响所代对系统的影响所代替；替；F-F-F L1 L2 L F2F1 在平衡基面上分别对两个分力在平衡基面上分别对两个分力F1、F2进行平衡，得平衡力进行平衡，得平衡力F 和和F，从而完成对集中质量点的平衡。，从而完成对集中质量点的平衡。2021/9/2319F F F 将力将力F平行分解到两个平衡基面平行分解到两个平衡基面上，得上，得F1和和F2:L1 L2 L F2=FL1 L(4)F1=FL2 L(3)F2F1 F=F1+F2 (1)F1 L1=F2L2 (2)2021/9/2320l1l2l3lF1F2F3r1m1m2r2m3r3F3F3F1F1F2F2 平衡基面平衡基面2021/9/2321L1L2L3LF1F2F3r1m1m2r2m3r3F2F1F3F3F1F2 F1=F1 L-L1 LF1=F1L1 LF2=F2 L-L2 LF2=F2L2 LLF3=F3 L-L3 LF3=F3L3 mrFF1+F2+F3+F=0F1+F2+F3+F=0从而求得从而求得mr和和mr。mFr2021/9/2322L1L2L3LF1F2F3r1m1m2r2m3r3 F2F1F3F2F1F3步骤：步骤：(1)分别将各回转平面上集中质量点分别将各回转平面上集中质量点mi所产生的惯性力所产生的惯性力Fi(或或质径积、重径积质径积、重径积)向两个平衡基面上分解，得到向两个平衡基面上分解，得到Fi和和Fi。(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的质径积mi ri(或重径积或重径积)。2021/9/23231.Static Balancing Experiment 静平衡实验静平衡实验 6-4 Balancing experiment of rigid rotor 2021/9/2324静平衡试验台2021/9/23252.Dynamic Balancing Experiment动平衡实验动平衡实验3.On-spot balancing现场平衡现场平衡2021/9/23262021/9/2327 6-5 Balancing of planar mechanisms Condition for balancing of mechanisms:The total inertia force and inertia torque acting on the center of mass are zero.机构平衡的条件机构平衡的条件 作用于机构质心的总惯性力和总惯性力偶矩应分别为零。2021/9/23281.Entire balancing of mechanism(1)make use of symmetrical structure to balance(2)make use of balancing mass to balance2021/9/23292.Partial balancing of mechanism(1)make use of balancing mechanism to balance(2)make use of balancing mass to balance(3)make use of spring to balance.2021/9/2330本章结束本章结束2021/9/2331
