Mechanical Vibration - APS
1. Basic Concept
1.1 Definition
- Mechanical Vibration: system moving forward and backward between the Balanced position
- system
- input
- response
1.2 Basic question
- vibration analysis[this] : system,input → response
- system identification : i. r→s
- Load identification : s,r → i
- Steps for vibration analysis
- Mechanical model
- Mathematical model : Motion differential equation
- solve Motion differential equation
- get engineering information
- test for verification
1.3 Elements
- Inertial component : mass,m,J
- Elastic component : stiffness, k
- Damping element : Damping coefficient,c
1.4 Classification
Component
Discrete component: Discrete system
- 1 dof : f(x,˙x,¨x,t)=0
- 2 dof
- multiple dof
Linear Component : Linear System
m : F=m¨x
k : F=k(x1−x2)
c : F=c(˙x1−˙x2)
Time-varying system : parameter change with time
- Time invariant system
- Constant coefficient differential equation
Force
- free vibration
- Homogeneous equation
- forced vibration
- Nonhomogeneous equation
- free vibration
1.5 Simple harmonic vibration
x=xusin(wt±θ)=Im[xuej(wt+θ)]
2. 1 DOF
Generalized equation : at any time
m¨x+c˙x+kx=f(t)
- Generalized coordinates
- Initial conditions
2.1 Free vibration
m¨x+kx=0
- set x=xusin(wt+θ), generalized equation
- wn=√km : Natural frequency
2.2 Damping vibration
m¨x+c˙x+kx=0
Characteristic equation : mλ2+cλ+k=0
Characteristic root : λ1,2=−c2m±√(c2m2)−km\=(−ζ±√ζ2−1)wn
=0, c=2√mk=ce , Critical damping
ζ=cce=c2√mk,Damping ratio
Underdamped system : ζ<1
- Real solution
- x(t)=Ae−ζwntsin(wd+θ)
- wd=√1−ζ2wn
- Overdamped system : ζ>1
- Critical damping system : ζ=1
- x(t)=(c1+c2t)e−wnt
2.3 Forced damping vibration
m¨x+c˙x+kx=f(t) ⇓ [m(jw)2+cjw+k]X=PDownarrow Z(w)X=PDownarrow\X=H(w)P
H(w)=huejθ=1Z(w)=1(k−w2m)+jwc
- hu=1√(k−w2m)2+(wc)2
- h(w) : Amplitude-frequency characteristic
- θ(w) : Phase frequency characteristic
β=xuxst=huk=1√(1−λ2)2+(2ζλ)2
- λ=wwc
- Re[β] :real number frequency characteristics
- Im[β] : Imaginary number frequency characteristics
- Resonance : ζ=0.707
3. Non-harmonic forced vibration
- Frequency domain analysis
3.1 Fourier
Fourier expanation
x(t)=12a0+∞∑1(akcoswkt+bksinwkt)
- Fourier transform
3.2 Steady state response
superposition of linear system
3.3 Transient response
Unit pulse function
Duhamel’s integral
x(t)=xfree(t)+xd(t)
4.Application for 1dof system
4.1 Vibration isolation
double direction effect
Passive : stop the transmission
- Active : stop the base
Vibration isolation coefficient
RTR=|Hf,fT(w)|=|k+jwc||k−w2m+jwc|
Active and passive are same with same H(w)
4.2 Vibration absorber
- absorb at some particular frequency
4.3 Inertial force excitation
u(t)=esinwt
5. 2 DOF
- Mass matrix : [m]
- Damping matrix : [c]
- Stiffness matrix : [k]
5.1 No damping vibration
Impedance matrix : [Z]
Frequency equation : no zero in generalized function
2dof : 2 natural frequency
- first : lower , basic frequency
- second : higher
Mode shape
mode shape matrix
solution
EXAMPLE
5.2 Steady state response
- Frequency response matrix
6.Multiple DOF
Lagrange equation
Generalized coordinates
Generalized force : Virtual displacement method, virtual work,pi=δWiδqi
kinetic energy : Ek=12˙qT[m]˙q
Potential energy : Ep=12qT[k]q
Dissipative function : ED=12˙qT[c]˙q
example
6.1 No damping system
[m]¨x+[k]x=0Downarrowx=usin(wnt+θ) [k]u=w2n[m]u
- Generalized eigenvalue problem : [A]u=λ[B]u
- Frequency function : Δ(wn)=det([Z(wn)])
- Impedance function : [Z(wn)]=−w2n[m]+[k]
Feature vector : ${u}r,[Z(w{nr})]{u}_r={0}$
- wnr : natural frequency
- Modal matrix : [u]=[ui]
General solution : x(t)=[u]arcoswnrt+brsinwnrt
example
Initial conditions : a=[u]−1x0
6.2 Primary coordinate
- Master mass matrix
- Master stiffness matrix
Primary coordinate : y=[u]x
Regular vibration vector : μr=1Mrur
Regular coordinate : x=[μ]η
example
6.3 Multiple DOF forced steady state response
Modal method