《汽车理论》课程教学课件（英文讲稿）第6章 汽车平顺性 Ride Performance 6.3 Simplify of Automobile Vibration System and Vibration of Single Mass System（汽车振动系统的简化及单质量系统的振动）

6.3 Simplify of Automobile Vibration System and Vibration of Single Mass System汽车振动系统的简化及单质量系统的振动1.SimplifyofAutomobileVibrationSystem7freedommodel自由度m模型4wheels车轮，Ix lyZ7mi善V（

7 freedom model自由度 模型 4 wheels车轮，Ix Iy Z 1. Simplify of Automobile Vibration System

x() = y(I) => only vertical vibration"z"and pitch vibration"β" = simplify to plane model1）总质量保持不变m2f + m2r + m2c = m2m2f2）质心位置不变m2rm2ra -m2rb = 0aam2c3）转动惯量保持不变MI, = m2P, = m2ra2 + m,b?mmirp-p-m2 = m2baLm2r=mPm2e=m,4freedoms:2wheels,Iy,Zab

( ) ( ) vertical vibration" " and pitch vibration" " simplify to z plane model x I y I only     1）总质量保持不变 2）质心位置不变 3）转动惯量保持不变                      ab m m bL m m aL m m y y y 2 2c 2 2 2r 2 2 2f 2 1    4 freedoms：2 wheels, Iy,Z m m m m 2 f 2 r 2 c 2    2f 2 r m a m b   0 2 2 2 y y 2 2f 2 r I m m a m b    

sprung mass distribution coefficient悬挂质量分配系数：ε=abs=1→ connection mass联系质量m2。=0= The vertical movement of m2r and m2rare independent. = Two freedoms modelTwo freedoms: one wheel ,Z

Two freedoms：one wheel ,Z 2 sprung mass distribution coefficient y a b  悬挂质量分配系数：  2 2 2 1 connection mass 0 The vertical movement of and are independent. Two freedoms model c f r m m m      ＝ 联系质量

Dualmass vibrationsystem双质量振动系统m2rm21mmf

m1r m2r m2 f m1 f Dual mass vibration system 双质量振动系统

Singlemassvibration system单质量振动系统f < 5Hz = deflection of tyre is smallm2(f, : wheel natural frequency车轮固有频率,f～10～15Hz=Single mass vibration sys.9

( : wheel natural fr 5 deflection of t equency 10 ~ 1 yre is l 5 sma l t t Z f f z H f H    车轮  Single mass vibration 固有频率, ) sys. Single mass vibration system 单质量振动系统

2.FreeVibrationofSingleMassSystemsecond-order eguation with constant coefficients二阶常系数方程m2m,z+C(z-g)+K(z-g)=0K艺+ (z - g) = 0(2-) +一mm9

second-order equation with constant coefficients 二阶常系数方程 2. Free Vibration of Single Mass System 2 2 2 ( ) ( ) 0 ( ) ( ) 0 m z C z q K z q C K z z q z q m m          

K2n =assume :1m.m.の。:system natural circular frequency系统固有圆频率f。=2元2元LC：damping ratio阻尼比，=2/m,K

2 0 2 2 2 , C K n m m assume:    0 2 damping rati 2 o n C m K    ： 阻尼比，   0 0 0 2 : system natural circular frequency 1 2 2 K f m       系统固有圆频率

homogenous differential equation齐次方程z+2nz +のjz = 0 0.25, belong to small damp:. z = Ae-nt sin(Jo? -n?t + α)O, : system natural frequency with damp有阻尼固有频率,,=-n2ne-vibration scope attenuation according to振幅按e-nt衰减

2 2 0 0.25, belong to small damp sin( ) nt z Ae n t          2 2 0 : system natural frequency with damp vibration scope attenuation according to , r r nt nt n e e      有阻尼固有频率,   振幅按 衰减 2 0 homogenous differential equation 2 0 : z nz z     齐次方程

Z一nteA2t2Ot1-Ae-ntT1Attenuation vibration curve衰减振动曲线

Attenuation vibration curve 衰减振动曲线

Attenuation vibration inf luenced by E阻尼比对衰减振动的影响（1）-n2=の。Or =1a个=0,S=1= 0,=0big damp=no vibration大阻尼时系统不振动0 - 0r ~ 3% = 0, ~ 0o~0.25=0o有阻尼后振动频率变化不大

2 2 2 0 0 1 r       n ＝ － Attenuation vibration inf luenced by  0 0 0 0.25 3 r r             ％ 有阻尼后振动频率变化不大        r r 1 0  ＝ ＝ big damp no vibration 大阻尼时系统不振动