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
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
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.
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
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.
(3) (4)
F' L1
F" L2
L
平衡基面 F"2
Ⅱ
平衡基面 F'2
F2
Ⅰm1F1m r2 r1 F'1m3
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'
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
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.
§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.
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
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
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
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)
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)。
w2
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
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
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.