1 / 26

第六章 机械的平衡

第六章 机械的平衡. Chapter 6. Balancing of Machinery. §6-1 The purposes( 目的 )and content( 内容 ) Balancing of Machinery. compelled oscillation. 1. purposes. inertia force (torque) 惯性力 ( 惯性力矩 ). dynamic press( 动压力 ) in kinematic pair. Friction( 摩擦力 and inner stress( 内应力 ) in link.

jerry-massey
Download Presentation

第六章 机械的平衡

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 第六章 机械的平衡 Chapter 6. Balancing of Machinery

  2. §6-1 The purposes(目的)and content(内容)Balancing of Machinery compelled oscillation 1. purposes inertia force (torque)惯性力(惯性力矩) dynamic press(动压力)in kinematic pair Friction(摩擦力and inner stress(内应力)in link efficiency and life-span The purpose of mechanical balance is to clear up or decrease the bad effect by balancing the component’s unbalanced inertia.

  3. 2. contents • The balance of component rotating about a fixed axis 回转体的平衡 rotor(转子): Parts constrained to rotate about a fixed axis. • (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 • 机构的平衡

  4. 刚性回转体 挠性回转体 机构 (avi)

  5. §6-2 Calculation for the static balancing of a rigid rotor 刚性转子的静平衡计算 1.The phenomena of static imbalance (静不平衡现象) If the center of mass of the rotor doesn’t coincide with the axis of rotation, their eccentric mass will lead tocentrifugal inertia force (离心力)when rotating, and causes an additional dynamic press (附加动压力)in the linkage.

  6. D m1 m2 m1 m2 B m3 m3 2. geometric condition几何条件 B/D ≤ 1/5

  7. F3 = m3 r3w2 F2 F2 = m2 r2w2 mP m2 F1 m1 F1 = m1 r1w2 r2 r1 FP If , F1+F2+F3 ≠ 0 r3 m3 F3 3. theory of static balancing centrifugal forces(离心力）of the unbalanced masses（偏心质量）： Then , imbalance

  8. m1 r1w2 + m2 r2w2 + m3 r3w2 + mP rPw2 = 0 F2 mP m1 r1+ m2 r2+ m3 r3+ mP rP=0 m2 F1 m1 r2 r1 FP r3 m3 F3 miri ----mass-radius product (质径积) F1+F2+F3 +FP =0 To balance: some counterweight质量点(mp) can be added to the rotor to balance its centrifugal force . Fp = mp rpw2 结论：若欲使回转体处于平衡，则各质量点的质径积(或重径积)的矢量和为零。

  9. Add Remove Conclusion: a balance can be achieved by adding or removing a balance mass in the same plane. 结论：对于静不平衡的转子，不论它有多少个偏心质量，都只需要在同一个平衡面内增加或除去一个平衡质量即可获得平衡

  10. m1 r1w2 + m2 r2w2 + m3 r3w2 + mP rPw2 = 0 F2 mP m2 F1 m1 r2 r1 FP r3 m3 W2 W3 F3 WP W1 miri Scale(比例尺)：μW = (kgm/mm) Wi resolution： A. Graphical method 矢量图解法 其中：Wi = miri

  11. F2 mP m2 F1 m1 r2 r1 FP r3 m3 F3 B. Analytical method解析法 mx1 rx1 + mx2 rx2+ mx3 rx3+ mxP rxP= 0 my1 ry1 + my2 ry2+ my3 ry3+ myP ryP= 0 Discussion: (1) 因为忽略了回转体厚度的影响，故此时，回转体离心惯性为一汇交力系。 (2) 平衡质量点的解包含两个方面：质(重)径积的量和方位。此时，往往已知平衡质量点应在的径向尺寸。 (3) 可分别用矢量方程图解法和坐标轴投影法两种方法求解。

  12. F2 m2 r2 r3 m3 m1 r1 F3 F1 §6-3 Calculation for the dynamic balancing of a rigid rotor刚性回转体的动平衡 1.geometric condition几何条件 B/D > 1/5 即不能忽略力矩的影响。 2.Force-balance-condition 力平衡条件 ∑Fi = 0 centrifugal forces ∑Mi = 0 centrifugal torque

  13. F F Ⅱ F Ⅰ 3.Balance base plane 平衡基面：施加(或减去)平衡质量点所在的回转面。

  14. 平衡基面 F"2 Ⅱ 平衡基面 F2 F'2 m2 Ⅰ F"1 r2 r3 m3 m1 F"3 r1 F3 F1 F'1 F'3 l1 l2 l3 l 4. Theory 平衡原理

  15. F2 Ⅱ Ⅱ F2 F2 Ⅰ m2 Ⅰ F1 Ⅱ r2 r3 m3 m1 F3 Ⅱ r1 F3 F1 F1 Ⅰ L1 L2 F3 Ⅰ L3 L Convert each centrifugal forces to the base plane Ⅰ and Ⅱ. That is F1,F2 and F3 can be replaced by F1Ⅰ,F2 Ⅰ , F3 Ⅰ and F1Ⅱ,F2Ⅱ, F3Ⅱ spatial force system 2 planar force systems

  16. F1 F2 F Ⅰ Ⅱ -F' -F" L2 L1 L 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； 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。

  17. F1 F2 F Ⅰ Ⅱ (3) F' F" (4) L2 F1 =F L L1 F2=F L L2 L1 L 将力F平行分解到两个平衡基面上，得F1和F2 : F = F1 + F2 (1) F1L1 = F2L2 (2)

  18. 平衡基面 F"2 Ⅱ 平衡基面 F2 F'2 m2 Ⅰ F"1 r2 r3 m3 m1 F"3 r1 F3 F1 F'1 F'3 l1 l2 l3 l

  19. L- L2 L- L1 F"2 F'2=F2 F'1=F1 L L Ⅱ L2 L1 F"2=F2 F"1=F1 L L F2 F'2 L- L3 m" m2 F'3=F3 Ⅰ r" F' L F"1 r2 r3 m3 L3 F"3=F3 m1 m' F"3 L r1 F" F3 r' F1 F'1 L1 L2 F'3 L3 L F'1 + F'2 + F'3 +F' = 0 F"1 + F"2 + F"3 +F" = 0 从而求得m'r'和m"r "。

  20. F2 Ⅱ Ⅱ F2 F2 Ⅰ m2 Ⅰ F1 Ⅱ r2 r3 m3 m1 F3 Ⅱ r1 F3 F1 F1 Ⅰ L1 L2 F3 Ⅰ L3 L 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mi ri(或重径积)。

  21. § 6-4 Balancing experiment of rigid rotor 1. Static Balancing Experiment静平衡实验

  22. 2.Dynamic Balancing Experiment动平衡实验 3.On-spot balancing现场平衡

  23. § 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 作用于机构质心的总惯性力和总惯性力偶矩应分别为零。

  24. 1.Entire balancing of mechanism • make use of symmetrical structure to balance • make use of balancing mass to balance

  25. 2.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.

  26. 本章结束

More Related