Elementary Tutorial
Prepared by Dr. An Tran
in collaboration with Professor P. R. Heyliger
Department of Civil EngineeringColorado State University
Fort Collins, ColoradoJune 2003
Fundamentals of Linear Vibrations
Developed as part of the Research Experiences of Undergraduates Program on “Studies of Vibration and Sound” , sponsored by National Science Foundation
and Army Research Office (Award # EEC-0241979). This support is gratefully
acknowledged.
Fundamentals of Linear Vibrations
1. Single Degree-of-Freedom Systems
2. Two Degree-of-Freedom Systems3. Multi-DOF Systems4. Continuous Systems
Single Degree-of-Freedom Systems
1. A spring-mass systemGeneral solution for any simple oscillatorGeneral approachExamples
2. Equivalent springsSpring in series and in parallelExamples
3. Energy MethodsStrain energy & kinetic energy Work-energy statementConservation of energy and example
A spring-mass system
General solution for any simple oscillator:
Governing equation of motion:
0 kxxm
)sin()cos()( tv
txtx nn
ono
2
n
o2o
nn
n
ooo
ωv
xamplitudeC;2ω
T1
Hz)or c.(cycles/sefrequencyf
vibrationofperiodT;T2
)(rads/sec.frequencynaturalmk
ω
(sec.)timet;xvelocityinitialvnt;displacemeinitialx
π
π
where:
Any simple oscillator
General approach:
1. Select coordinate system2. Apply small displacement3. Draw FBD4. Apply Newton’s Laws:
)(
)(
Idt
dM
xmdt
dF
Simple oscillator – Example 1
22 mlmdI
inertiaofmomentmassI
cg
IK
IM
02 Kml 2ml
Kωn
+
Simple oscillator – Example 2
l
a
m
kω
mlmdII
n
cg22
2)( mlaak
IM oo
022 kaml
+
(unstable)ω,l
aAs
m
kω,
l
aWhenits:limNote
n
n
00
1
Simple oscillator – Example 3
l
b
m
kω
mlmm
ml
mdII
mllA
AdxxdmrI
n
cgo
cg
l
3
3212
1212
2
222
2
23
22 2/
0
3)(
2mlbbk
IM oo
03
22
kbml
+
Simple oscillator – Example 4
Lma
GJL
JGK:stiffnessEquivalent
TL
JG
JG
TL
maI:tableFrom
n 22
2
2
2
IT
IM z
02
2
L
GJma
+
Equivalent springs
Springs in series:same force - flexibilities add
Springs in parallel:same displacement - stiffnesses add
21 kkkeq
eqkkk
kkP
)( 21
21
PfPff
Pkk
eq
)(
11
21
2121
21 fffeq
Equivalent springs – Example 1
0 xKxm eq
0312
32
31
x
L
EI
L
EIxm
Equivalent springs – Example 2
)a(ml
Wlkaω
nn
n
2
22
2mllWa)ak(
IM oo
022 )Wlka(ml
+
Consider:
ka2 > Wl n2 is positive - vibration is stable
ka2 = Wl statics - stays in stable equilibrium
ka2 < Wl unstable - collapses
Equivalent springs – Example 3
02
2
sinmglml
mlsinWl
IM oo
0 sinl
gl
gω
l
g
n
0
+We cannot define n
since we have sin term
If < < 1, sin :
Energy methods
Strain energy U:energy in spring = work done
Kinetic energy T:
Conservation of energy:work done = energy stored
PkU2
1
2
1 2
Tenergy kinetic ofincrement
done work ofIncrement
dT) rrm d(dt) r()r (m
rdF
21
rrmT
2
1
Work-Energy principles
Work done = Change in kinetic energy
Conservation of energy for conservative systems
E = total energy = T + U = constant
122
1
2
1
TTdT rdFT
T
r
r
Energy methods – Example
0
0
xxmxkx
)E(dt
d
0 kxxm 22
2
2
2
1
2
12
12
1
xmkxTUE
xmT
kxU
Same as vector mechanics
Work-energy principles have many uses, but one of the most useful is to derive the equations of motion.
Conservation of energy: E = const.
Two Degree-of-Freedom Systems
1. Model problemMatrix form of governing equation Special case: Undamped free vibrationsExamples
2. Transformation of coordinates
Inertially & elastically coupled/uncoupledGeneral approach: Modal equationsExample
3. Response to harmonic forcesModel equationSpecial case: Undamped system
Two-DOF model problem
Matrix form of governing equation:
2
1
2
1
22
221
2
1
22
221
2
1
2
1 )()(
0
0
P
P
x
x
kk
kkk
x
x
cc
ccc
x
x
m
m
where:[M] = mass matrix; [C] = damping matrix;[K] = stiffness matrix; {P} = force vector
Note: Matrices have positive diagonals and are symmetric.
Undamped free vibrationsZero damping matrix [C] and force vector {P}
)cos(2
1
2
1
tA
A
x
xAssumed general solutions:
Characteristic polynomial (for det[ ]=0):
021
212
2
2
1
214
mm
kk
m
k
m
kk
21
21
21
2
2
2
1
21
2
2
1
212
21
21
4
2
1
mm
kk
m
k
m
kk
m
k
m
kk
Eigenvalues (characteristic values):
Characteristic equation:
0
0
)(
)(
2
1
2222
22
121
A
A
mkk
kmkk
Undamped free vibrationsSpecial case when k1=k2=k and m1=m2=m
Eigenvalues and frequencies:
period lfundamenta
frequency lfundamenta
ω
π T
m
k.ω
2
61801
m
k
618.2
3819.021
21
21
Two mode shapes (relative participation of each mass in the motion):
1
618.12 2
1
2 k
mk
A
A shape mode 1st
1
618.02
1
2
mk
k
A
Ashape mode 2nd
The two eigenvectors are orthogonal:
618.1
1)1(
2
)1(1
A
A
618.0
1)2(
2
)2(1
A
AEigenvector (1) = Eigenvector (2) =
Undamped free vibrations (UFV)
For any set of initial conditions:
We know {A}(1) and {A}(2), 1 and 2
Must find C1, C2, 1, and 2 – Need 4 I.C.’s
)cos()cos()(
)(22)2(
2
)2(1
211)1(2
)1(1
12
1
tA
ACt
A
AC
tx
txx
Single-DOF:
For two-DOF:
)cos()( tCtx n
UFV – Example 1
)cos(618.0
0.1)cos(
618.1
0.12211
2
1 tCtCx
xx
Given:
No phase angle since initial velocity is 0:
618.1
0.10 oxx and
618.0
0.1
618.1
0.1
618.1
0.121 CCxo
From the initial displacement:
11
21
2
;0;
T
CC
UFV – Example 2
)cos(618.0
1)171.0()cos(
618.1
1)171.1( 21 ttx
Now both modes are involved:
Solve for C1 and C2:
2
10 oxx and
2
121 618.0618.1
11
618.0
1
618.1
1
2
1
C
CCCxo
From the given initial displacement:
171.0
171.1
2
1
1618.1
1618.0
618.1618.0
1
2
1 C
C
Hence,
or
Note: More contribution from mode 1
)cos()618.0(171.0)cos()618.1(171.1)(
)cos()1(171.0)cos()1(171.1)(
212
211
tttx
tttx
Transformation of coordinates
Introduce a new pair of coordinates that represents spring stretch:
0
0)(
0
0
2
1
22
221
2
1
2
1
x
x
kk
kkk
x
x
m
m
UFV model problem:
“inertially uncoupled”
“elastically coupled”
z1(t) = x1(t) = stretch of spring 1
z2(t) = x2(t) - x1(t) = stretch of spring 2
or x1(t) = z1(t) x2(t) = z1(t) + z2(t)
Substituting maintains symmetry:
0
0
0
0)(
2
1
2
1
2
1
22
221
z
z
k
k
z
z
mm
mmm
“inertially coupled” “elastically uncoupled”
Transformation of coordinates
We have found that we can select coordinates so that:1) Inertially coupled, elastically uncoupled, or2) Inertially uncoupled, elastically coupled.
Big question: Can we select coordinates so that both are uncoupled?
Notes in natural coordinates:
The eigenvectors are orthogonal w.r.t [M]:
The modal vectors are orthogonal w.r.t [K]:
Algebraic eigenvalue problem:
618.0
1
618.1
1
: vectors)(modal rsEigenvecto
)2(2
)2(1
2)1(2
)1(1
1A
Au
A
Au
0
0
12
21
uMu
uMuT
T
0
0
12
21
uKu
uKuT
T
222111 uMuKuMuK
Transformation of coordinates
Governing equation:
Modal equations:
Solve for these using initial conditions then substitute into (**).
0 xKxM
)()()(
)(
)()(
222
121
21
11
2
1
2211
tqu
utq
u
u
tx
tx
tqutqux
(**)
General approach for solution
We were calling “A” - Change to u to match Meirovitch
0)()((*)
0)()((*)
22222
12111
tqtqu
tqtquT
T
0)()()()( 22112211 tqutquKtqutquM (*)
Substitution:
Let
or
Known solutions
Transformation - Example
)cos()171.0(618.0
1)cos(171.1
618.1
121 ttx
2) Transformation:
618.0
1;618.1
618.1
1;618.0
22
122
21
111 u
u
u
u and
1) Solve eigenvalue problem:
)cos()0()(
)cos()0()(
171.0
171.1
)0(
)0(
)0(618.0
1)0(
618.1
1
2
1
222
111
2
1
21
tqtq
tqtq
q
q
and
So
As we had before.More general procedure: “Modal analysis” – do a bit later.
Model problem with:
0
0
2
1oo xx and
0)()(
0)()()()(
2222
1211
2211tqtq
tqtqtqutqux
and
Response to harmonic forces
Model equation:
[M], [C], and [K] are full but symmetric.
tieF
FtFxKxCxM
2
1)(
{F}
not function of time
Assume: tie
iX
iXiXx
)(
)()(
2
1
Substituting gives: FiXKCiM )(2
matrix impedance 2x2)( iZ
FiZiXiZiZ 11 )()()()(
2
1
1112
1222
21222112
1 1
F
F
zz
zz
zzzX
XX
Hence:
212 ,ji,kciωmωz ijijijij
:)(i of function are z All ij
Special case: Undamped system
Zero damping matrix [C]Entries of impedance matrix [Z]:
For our model problem (k1=k2=k and m1=m2=m), let F2 =0:
212
2222
2111
22
11111222
122
2222
111
21212
2221 ))((
)(;
))((
)(
kmkmk
FmkFkX
kmkmk
FkFmkX
Notes:1) Denominator originally (-)(-) = (+). As it passes through 1, changes
sign.2) The plots give both amplitude and phase angle (either 0o or 180o)
Substituting for X1 and X2:
12122
222222
11111 )(;)(;)( kzmkzmkz
)()(;
)()(
)(22
221
221
222
221
221
2
1
m
FkX
m
FmkX
Multi-DOF Systems
1. Model EquationNotes on matrices Undamped free vibration: the eigenvalue problemNormalization of modal matrix [U]
2. General solution procedureInitial conditionsApplied harmonic force
Multi-DOF model equation
Model equation:
Notes on matrices:
They are square and symmetric.
[M] is positive definite (since T is always positive)[K] is positive semi-definite:
all positive eigenvalues, except for some potentially 0-eigenvalues which occur during a rigid-body motion.
If restrained/tied down positive-definite. All positive.
Q xKxCxM
1) Vector mechanics (Newton or D’ Alembert)2) Hamilton's principles3) Lagrange's equations
We derive using:
Multi-DOF systems are so similar to two-DOF.
xKxU
xMxTT
T
21
21
:spring inenergy Strain
:energy Kinetic
UFV: the eigenvalue problem
Matrix eigenvalue problem
Equation of motion:
titi eAeAtftfuq 21)()(
0 qKqM
Substitution of
in terms of the generalized D.O.F. qi
leads to
uMuK 2
For more than 2x2, we usually solve using computational techniques.
Total motion for any problem is a linear combination of the natural modes contained in {u} (i.e. the eigenvectors).
Normalization of modal matrix [U]
Do this a row at a time to form [U].
This is a common technique for us to use after we have solved the eigenvalue problem.
We know that: ijjT
iji CuMuuMu
1
ku
j i
j i
δij
if
if
deltaKronecker
:where
0
1So far, we pick our eigenvectors to look like:
Instead, let us try to pickso that:
1
knewk uu
12 kT
knewkT
newk uMuuMu
Then: IUMU T UKU Tand
2
22
21
..0
....
..0
0.0
n
:where
Let the 1st entry be 1
General solution procedure
For all 3 problems:
1. Form [K]{u} = 2 [M]{u} (nxn system)Solve for all 2 and {u} [U].
2. Normalize the eigenvectors w.r.t. mass matrix (optional).
Consider the cases of:
1. Initial excitation 2. Harmonic applied force3. Arbitrary applied force
oo qq and
Initial conditions
2n constants that we need to determine by 2n conditions
General solution for any D.O.F.:
Alternative: modal analysis
)cos()cos()cos()( 22221111 nnnn tCutCutCutq
Displacement vectors:
ioio qq and on
)()()()( 2211 tutututq
Uq
nn
UFV model equation:
0
0
0
ηUKUηUMU
qKqMTT
n modal equations:
0
0
0
2
2222
1211
nnn
Need initial conditions on , not q.
Initial conditions - Modal analysis
Using displacement vectors: ηUMUqMU
UqTT
As a result, initial conditions:
Since the solution of
oT
o
oT
o
qMUη
qMUη
)sin()(
)cos()()(
)sin()(
)cos()()( 11
1111
ttt
ttt
nn
nonnon
oo
And then solve
hence we can easily solve for
qMUη T or
02 is:
)sin()cos()(
)cos(
ttt
tC
oo
or
ηUq
Applied harmonic force Driving force {Q} = {Qo}cos(t)
Equation of motion:
unknownη
known U
ηUq
Q qKqM Substitution of
leads to
NtQUηUKUηUMU oTTT )cos(
requency driving fω
tQQ o
)cos(
and
Hence,
.
)cos(
)cos(
222
22
221
11
etc
tQu
tQu
oT
oT
then
ηUq
Continuous Systems
1. The axial barDisplacement field Energy approachEquation of motion
2. ExamplesGeneral solution - Free vibrationInitial conditionsApplied forceMotion of the base
3. Ritz method – Free vibrationApproximate solution One-term Ritz approximationTwo-term Ritz approximation
The axial bar
Main objectives:1. Use Hamilton’s Principle to derive the equations of
motion.2. Use HP to construct variational methods of solution.
A = cross-sectional area = uniformE = modulus of elasticity (MOE)u = axial displacement = mass per volume
Displacement field: u(x, y, z) = u(x, t)v(x, y, z) = 0w(x, y, z) = 0
Energy approach
L tt
t
t
L L
t
t
L
dxuuAdtux
uEAdxu
x
uEA
xuudxA
t
dtdxuxx
uEAuudxA
00 0
0
2
1
2
1
2
1
0
0
221
21
21
2
1
um
x
u
x
uE)εε(Eε σ xx
energy kinetic T
U energy strain energy potentialV
densityenergy strainUo
For the axial bar:
Hamilton’s principle:
dtux
uEAdxu
x
uEA
xuA
t
t
t
L L
2
1 0 0
0
2
1
)(0t
tdtVT
221 u(Adx)ρ
V odVU
2
2
x
uE
Axial bar - Equation of motion
2
22
2
2
x
u
t
u
Hamilton’s principle leads to:
If area A = constant
0
x
uEA
xuA
t
Since x and t are independent, must have both sides equal to a constant.
Separation of variables: )()(),( tTxXtxu
)sin()cos(
02
tpBtpAT
TpT
xpDxpCX
XpX
sincos
02
Hence
1
sincos)sin()cos(),(i
iiiiiiii xpDxpCtpBtpAtxu
3
22
LM
LFE
:where
22222
2 contant p-T
dtTd
X
dxXd
Fixed-free bar – General solution
0cos0
Lp
D ii or solution) (trivial Either
= wave speed
E
For any time dependent problem:
,5,3,1 2sin
2cos
2sin),(
iii L
tiB
L
tiA
L
xitxu
Free vibration:
1
sincos)sin()cos(),(i
iiiiiiii xpDxpCtpBtpAtxu
EBC:
NBC:
0)0( u
00
LxLx x
u
x
uEA
General solution:
EBC
1
0)sin()cos(),0(i
iiiii tpBtpACtu
1
0)sin()cos(cosi
iiiiiii
Lx tpBtpALppD
x
u
0iC
2
5
2
3
2
ororLpi
),5,3,1(2
iL
ipi
NBC
Fixed-free bar – Free vibration
E
L
in 2
are the eigenfunctions
L
xi
2sin
For free vibration:
General solution:
Hence
)cos()(),( txAtxu n
are the frequencies (eigenvalues)
2
22
2
2
x
u
t
u
),5,3,1( i
Fixed-free bar – Initial conditions
or
,3,12
2
)1(
2 2cos
2sin
1)1(
)(8),(
i
io
L
ti
L
xi
i
LLtxu
Give entire bar an initial stretch.Release and compute u(x, t).
0)0,( 0
to
t
ux
L
LLxu and
Initial conditions:
Initial velocity:
Initial displacement:
0
2sin
2,3,10
iit L
xiB
L
i
t
u 0iB
22sin
2sin
2sin
2sin
,3,100
,3,1
LAdx
L
xi
L
xiAdx
L
xix
L
LL
L
xiAx
L
LL
ii
L
i
Lo
ii
o
),3,1()1()(8
2sin
)(2 2
)1(
2202
ii
LLdx
L
xix
L
LLA
io
Lo
i
Hence
Fixed-free bar – Applied force
or
txL
EA
Ftxu o
sinsinsec),(
Now, B.C’s:
)sin(
0),0(
tFx
uEA
tu
oLx
From
B.C. at x = 0:
B.C. at x = L:
0),0( tu 01 A
L
EA
FA o sec2
Hence
2
22
2
2
x
u
t
u
)sin()(),( txXtxu nwe assume:
Substituting:
txA
xAtxu
sinsincos),( 21
)sin()sin(cos2 tFtL
LAEA
x
uEA oLx
Fixed-free bar – Motion of the base
)sin()sin(),0( 1 tUtAtu o
2
22
2
2
x
u
t
u
Using our approach from before:
Resonance at:
txLxUtxu o
sinsintancos),(
oUA 1
L
UA o tan2
Hence
txA
xAtxu
sinsincos),( 21
0sincossin 2 tLALU
x
u oLx
0
Lxx
uEA
From
B.C. at x = 0:
B.C. at x = L:
or,2
3,
2
L.,
2
3,
2etc
LL
Ritz method – Free vibration
Start with Hamilton’s principle after I.B.P. in time:
Seek an approximate solution to u(x, t):In time: harmonic function cos(t) ( = n)
In space: X(x) = a11(x)
where: a1 = constant to be determined
1(x) = known function of position
dtdxuxx
uEAuuA
t
t
t
L
2
1 00
1(x) must satisfy the following:
1. Satisfy the homogeneous form of the EBC. u(0) = 0 in this case.
2. Be sufficiently differentiable as required by HP.
One-term Ritz approximation 1
Ritz estimate is higher than the exactOnly get one frequencyIf we pick a different basis/trial/approximation function 1, we would get a different result.
)cos()cos()(
)cos()cos()(),()(
1
1111
txtxu
txatxatxuxx
:eapproximat Also
:Pick
dttdxEAxxAat
t
L)(cos)1)(1())((0 2
0
21
2
1
Substituting:
222
23
2 33
3
L
E
LLEA
LA
LLRITZ
732.13
LLEXACT
571.12
1010
22 adxEAadxxALL
Hence
aKaM 2:formmatrix in
L
xEXACT
2sin1
xRITZ 1
One-term Ritz approximation 2
Both mode shape and natural frequency are exact.But all other functions we pick will never give us a frequency lower than the exact.
L
xx
2sin)(1
:pick we ifWhat
dttdxL
x
LEA
L
xAa
dtdxuxx
uEAuuA
t
t
t
L
t
t
L
)(cos2
cos22
sin0
0
2
0
2
2
221
0
2
1
2
1
Substituting:
EXACTRITZ L
E
L
22Hence
)cos(2sin)cos()(
)cos(2sin)cos()(),(
1
111
tLxtxu
tLxatxatxu
:eapproximat Also
L
x
Ldx
d
2cos
21
Two-term Ritz approximation
221)( xaxaxX :Let
dtdxxaaEAxxaxaAt
t
L
2
1 0 212
212 )1()2()(0
where:
:1 xu eapproximat If
xaadx
dX21 2
:2xu eapproximat If dtdxxxaaEAxxaxaAt
t
L
2
1 0 2122
212 )2()2()(0
2
1
2221
1211
2
1
2221
12112
a
a
KK
KKE
a
a
MM
MM
5))((
4))((
3))((
5
0
2222
4
0
22112
3
011
LdxxxM
LdxxxMM
LdxxxM
L
L
L
In matrix form:
3
4)2)(2(
)1)(2(
)1)(1(
3
022
2
02112
011
LdxxxK
LdxxKK
LdxK
L
L
L
Two-term Ritz approximation (cont.)
E
22 and
LaaLaL 4526.00)3785.01713.0( 2212
0
0
)534()4(
)4()3(
2
1
532422
42232
a
a
LLLL
LLLL
leads to
Solving characteristic polynomial (for det[ ]=0) yields 2 frequencies:
LL RITZRITZ 67.5)(5767.1)( 21 and
Substitution of:
LL EXACTEXACT 7123.4)(5708.1)( 21 and
Mode 1:
Let a1 = 1:
LxxxX 21 4526.0)(
:1 shape Mode
LaaLaL 38.10)10.5043.7( 22122
Mode 2:
LxxxX 22 38.1)(
:2 shape Mode