机械原理第六章动平衡

Fp = mp rpw2
F1+F2 +F3 +FP = 0
m1 r1w2 + m2 r2w2 + m3 r3w2 + mP rPw2 = 0
m2 r2
r1 m1
F1
r3 m3
mP
FP F3
m1 r1 + m2 r2 + m3 r3 + mP rP =0
miri ----mass-radius product (质径积)
机械原理第六章动平衡
§6-3 Calculation for the dynamic balancing of a rigid rotor 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.
r2
r1 m1
F'1
F1
l1
l2 l3
l
m3 r3
F3
F"3
平衡基面 Ⅱ
F"1
机械原理第六章动平衡
F'1
=
F1
L- L1
L
F"1
=
F1
L1 L
F'3=F3Fra bibliotekL- L3
L
F"3
=
F3
L3 L
F'2
=
F2
L-
L
L2
F"2
=
F2
L2 L
F'2
Ⅰ F'
F2 m2
r2
m' r'
r1 m1
F'1
F1
F"2 Ⅱ
m3 r3
mechanism
机械原理第六章动平衡
§ 6-2 Calculation for static balancing of a rigid rotor
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.
F
FⅠ
FⅡ
机械原理第六章动平衡
4. Theory
F"2
F'2 F'3
平衡基面
F2
Ⅰ
m2
r2
r1 m1
F'1
F1
l1
l2 l3
l
m3 r3
F3
F"3
平衡基面 Ⅱ
F"1
机械原理第六章动平衡
Convert each centrifugal forces to the correction plane Ⅰ and Ⅱ.
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 4. at the chosen radius of Rb=0.806m
机械原理第六章动平衡
2. Geometric condition
B/D ≤ 1/5
m1 m2
m1 m2
m3
m3
D B
机械原理第六章动平衡
3. Theory of static balancing
centrifugal forces(离心力）of the unbalanced masses（偏心质量）：
F3
F"3
r"m" F"1 F"
F'3
L1
F'1 + F'2 + F'3 +F' = 0 F"1 + F"2 + F"3 +F" = 0
L2 L3
L
从而求得m'r'和m"r "。
机械原理第六章动平衡
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 。
mbRbx=-m1R1x-m2R2x=-(1.2)(-0.451)-(1.8)(0.541)=-0.433 1. mbRby=-m1R1y-m2R2y=-(1.21.042)-(1.8)(0.618-2.363 • Solve equations
机械原理第六章动平衡
Solution: 4. The mass-radius product can be obtained with a variety of shapes.
That is F1,F2 and F3 can be replaced by F1Ⅰ,F2 Ⅰ , F3 Ⅰ and F1Ⅱ,F2Ⅱ , F3Ⅱ
F2Ⅱ Ⅱ
spatial force system
F2Ⅰ Ⅰ
F2 m2
2 planar force systems
r2
m3 r3
F1Ⅱ
r1m1
F3 F3Ⅱ
efficiency and lifespan
The purpose of mechanical balancing is to clear up or decrease the bad effect by balancing the component’s unbalanced inertia.
机械原理第六章动平衡
机械原理第六章动平衡
Solution: 1. Resolve the position vectors into xy components :
R1=1.135m θ1=113.4°; R1x=-0.451,R1y=-1.042 1. R2=0.822m θ2=48.8 °; R2x=+0.541,R2y=0.618 • Solve equations
F1 = m1 r1w2 F2 = m2 r2w2 F3 = m3 r3w2
F2
m2 r2
r1 m1
F1
r3 m3
mP
FP F3
If , F1+F2 +F3 ≠ 0 Then , imbalance
机械原理第六章动平衡
To balance:
F2
Some counterweight质量点(mp) can be added to the rotor to balance its centrifugal force .
机械原理第六章动平衡
静平衡试验台
机械原理第六章动平衡
2.Dynamic Balancing Experiment动平衡实验
3.On-spot balancing现场平衡
机械原理第六章动平衡
机械原理第六章动平衡
§ 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。
Conclusion： Requirement for static balance: ∑m1 r1=0
机械原理第六章动平衡
Add
Remove
Conclusion:A balance can be achieved by adding or removing a balance mass in the same plane.
-F'
-F"
L1 L2
L
机械原理第六章动平衡
将力F平行分解到两个平衡基面 上，得F1和F2 :
F1
F
F2
Ⅰ
Ⅱ
F = F1 + F2
(1)
F1 L1 = F2L2
(2)
F1 = F
L2 L
(3)
F2 = F
L1 L
(4)
F'
F"
L1 L2 L
机械原理第六章动平衡
F"2
F'2 F'3
平衡基面
F2
Ⅰ
m2
(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)。
F2Ⅱ Ⅱ
F2Ⅰ Ⅰ
F2 m2
r2
m3 r3
F1Ⅱ
r1m1
F3 F3Ⅱ
F1Ⅰ F1
F3Ⅰ
L1 L2 L3 L
机械原理第六章动平衡
§ 6-4 Balancing experiment of rigid rotor 1. Static Balancing Experiment 静平衡实验
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
机械原理第六章动平衡
1.Entire balancing of mechanism (1) make use of symmetrical structure to balance (2) make use of balancing mass to balance
机械原理第六章动平衡
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.
机械原理第六章动平衡
3rew
演讲完毕，谢谢听讲!
机械原理第六章动平衡
2020/11/18
机械原理第六章动平衡
§6-1 Introduction
1. Purposes
Inertia force (torque)
Friction and inner stress in link
Compelled oscillation
Dynamic press in kinematic pair
机械原理第六章动平衡
resolution：
A. Graphical method
Scale(比例尺)： W =mWiiri (kgm/mm)
Wi = miri
F2
m2 r2
r1 m1
F1
r3 m3
mP
FP F3
m1 r1w2 + m2 r2w2 + m3 r3w2 + mP rPw2 = 0
W3
W2
WP
W1
机械原理第六章动平衡
F2
B. Analytical method mx1 rx1 + mx2 rx2 + mx3 rx3 + mxP rxP = 0 my1 ry1 + my2 ry2 + my3 ry3 + myP ryP = 0
m2 r2
r1 m1
F1
r3 m3
mP
FP F3
机械原理第六章动平衡
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.Force-balance-condition ∑Fi = 0 centrifugal forces ∑Mi = 0 centrifugal moment
F2 m2 r2
r1 m1
F1
m3 r3 F3
机械原理第六章动平衡
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.
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
机械原理第六章动平衡
rigid rotor
flexible rotor
(avi)
F1Ⅰ F1
F3Ⅰ
L1 L2 L3 L
机械原理第六章动平衡
平衡原理：
F1
Ⅰ
将集中质量点所产生的离心力F
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；
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。