机械原理第六章 动平衡

Solution: 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=2.980kg at the chosen radius of Rb=0.806m
F
F Ⅰ
F Ⅱ
4. Theory
平衡基面 F"2
Ⅱ
平衡基面 F'2
F2
Ⅰ
m1
F1
m2 r2 r1 F'1
m3
F3
r3 F"3
F"1
F'3
l1
l2
l3
l
Convert each centrifugal forces to the correction plane Ⅰ and Ⅱ. That is F1,F2 and F3 can be replaced by F1Ⅰ,F2 Ⅰ , F3 Ⅰ and F1Ⅱ,F2Ⅱ , F3Ⅱ
Conclusion：
Requirement for static balance: ∑m1 r1=0
Add
Remove
Conclusion:A balance can be achieved by adding or removing a balance mass in the same plane.
§6-3 Calculation for the dynamic balancing of a rigid rotor 刚性回转体的动平衡
1. geometric condition
F2 m2 r2 r1 m1 F1 m3 r3 F3
B/D > 1/5
Mass maybe unevenly distributed both rotationally around their axis and also longitudinally along their axis.
1. Phenomena of static imbalance If mass center of rotor doesn’t 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.
3.On-spot balancing现场平衡
§ 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。
(1) make use of balancing mechanism to balance (2) make use of balancing mass to balance (3) make use of spring to balance.
本章结束
在平衡基面上分别对两个分力
-F'
L1 L L2
-F"
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。
F1 将力F平行分解到两个平衡基面 上，得F1和F2 :
Ⅰ
F
F2
Ⅱ
F = F1 + F2 F1 L1 = F2L2 L2 F1 = F L L1 F2 = F L
(1) (2)
F2 Ⅰ
F2 F2Ⅱ
Ⅱ
Ⅰ
m r1 1 F1 Ⅰ
m2 r2
m3 r3
F3 F3 Ⅱ
F1Ⅱ
F1 L2 L3 L
F3 Ⅰ
L1
§ 6-4 Balancing experiment of rigid rotor
1. Static Balancing Experiment 静平衡实验
静平衡试验台
2.Dynamic Balancing Experiment动平衡实验
F2 m2 m1 r2 r1 m3 r3 mP FP F3 F1
w2
F1+F2 +F3 +FP = 0 m1 r1w2 + m2 r2w2 + m3 r3w2 + mP rPw2 = 0 m1 r1 + m2 r2 + m3 r3 + mP rP =0
miri ----mass-radius product (质径积)
1.Entire balancing of mechanism (1) make use of symmetrical structure to balance
(2) make use of balancing mass to balance
2.Partial balancing of mechanism
2. Geometric condition
B/D ≤ 1/5
m1
m2
m1
m2
D
m3
m3
B
3. Theory of static balancing centrifugal forces(离心力）of the unbalanced masses（偏心质量）： F1 = m1 r1
F2 m2 m1 r2 r1 m3 r3 mP FP F3 F1
(3) (4)
F' L1
F" L2
L
平衡基面 F"2
Ⅱ
平衡基面 F'2
F2
Ⅰ
m1
F1
m2 r2 r1 F'1
m3
F3
r3 F"3
F"1
F'3
l1
l2
l3
l
F'1 = F1
F"1 = F1
F'3 = F3 F"3 = F3
L- L1 L L1 L L- L3 L L3 L
L- L2 F'2 = F2 L L2 F"2 = F2 L F2 F'
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F2 = m2 r2w2
F3 = m3 r3w2 If , F1+F2 +F3 ≠ 0 Then , imbalance
To balance: Some counterweight质量点(mp) can be added to the rotor to balance its centrifugal force . Fp = mp rp
F2Ⅱ
Ⅱ
spatial force system
F2 Ⅰ
F2
Ⅰ
m r1 1 F1 Ⅰ
m2 r2
m3 r3
F3 F3 Ⅱ
F1Ⅱ
2 planar force systems
F3 Ⅰ
F1 L2 L3 L
L1
平衡原理： 将集中质量点所产生的离心力F
F1
Ⅰ
F
F2
Ⅱ
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；
2
F"2
Ⅱ
Ⅰ F'
m' r' F'1 F'3 L1 L2 m1
m2 r2 r1 m3 r3 F3
r"m" F"1
F"3
F"
F1
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 "。
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)。
Solution: 1. Resolve the position vectors into xy components : R1=1.135m θ1=113.4°; R1x=-0.451,R1y=-1.042 R2=0.822m θ2=48.8 °; R2x=+0.541,R2y=0.618 2. Solve equations mbRbx=-m1R1x-m2R2x=-(1.2)(-0.451)-(1.8)(0.541)=-0.433 mbRby=-m1R1y-m2R2y=-(1.21.042)-(1.8)(0.618-2.363 3. Solve equations
第六章 机械的平衡
Chapter 6. Balancing of Machinery
§6-1 Introduction 1. Purposes
Inertia force (torque)
Compelled oscillation
Dynamic press in kinematic pair efficiency and lifespan
resolution： A. Graphical method miri Scale(比例尺)：W = (kgm/mm) Wi Wi = miri m1 r1w2 + m2 r2w2 + m3 r3w2 + mP rPw2 = 0
F2
m2 m1 r2 r1
F1
m3
r3
mP
FP
F3
W3
W2
WP
W1
2.Force-balance-condition ∑Fi = 0 centrifugal forces ∑Mi = 0 centrifugal moment
3. Correction planes To 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.
F2 B. Analytical method mx1 rx1 + mx2 rx2 + mx3 rx3 + mxP rxP = 0 my1 ry1 + my2 ry2 + my3 ry3 + myP ryP = 0
m2 m1 r2 r1
F1
m3
r3
mP
FP
F3
EXAMPLE Given: The system shown in FIG has the following data: m1=1.2kg R1=1.135m θ1=113.4° m2=1.8kg R2=0.822m θ2=48.8 ° Find: The mass-radius production and its angular location needed to statically balance the system.
2. Contents
The balance of component rotating about a fixed axis
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
rigid rotor
flexible rotor
(avi)
mechanism
§ 6-2 Calculation for static balancing of a rigid rotor
Friction and inner stress in link
The purpose of mechanical balancing is to clear up or decrease the bad effect by balancing the component’s unbalanced inertia.