Transcript Vibrations Tutorial

AE 2403 VIBRATIONS AND
ELEMENTS OF AEROELASTICITY
BY
Mr. G.BALAJI
DEPARTMENT OF AERONAUTICAL ENGINEERING
REC,CHENNAI
Fundamentals of Linear Vibrations
1.
2.
3.
4.
Single Degree-of-Freedom Systems
Two Degree-of-Freedom Systems
Multi-DOF Systems
Continuous Systems
Single Degree-of-Freedom Systems
1. A spring-mass system
General solution for any simple oscillator
General approach
Examples
2. Equivalent springs
Spring in series and in parallel
Examples
3. Energy Methods
Strain energy & kinetic energy
Work-energy statement
Conservation of energy and example
A spring-mass system
Governing equation of motion:
General solution for any simple oscillator:
x(t )  xo cos( nt ) 
mx  kx  0
vo
sin( nt )
n
where:
x o  initial displacement; v o  initial velocity  x o ; t  time (sec.)
ωn 
k
2π
 natural frequency (rads/sec. ) 
; T  period of vibration
m
T
v 
1
ω
fn  frequency (cycles/se c. or Hz) 
 n ; C  amplitude  x 2o   o 
T
2π
 ωn 
2
Any simple oscillator
General approach:
1.
2.
3.
4.
Select coordinate system
Apply small displacement
Draw FBD
Apply Newton’s Laws:
d
F 
( mx )
dt
d
M 
( I)
dt
Simple oscillator – Example 1
I  mass moment of inertia
 I cg  md 2  ml 2
ωn 
K
ml 2
 M  I
 K  I
+
ml   K  0
2
Simple oscillator – Example 2
I  I cg  md 2  ml 2
 M o  I o
k a
 
m l 
 (ka)a  ml 2
ωn 
Note limits: When
a
k
 1, ωn 
l
m
a
As
 0, ωn  0 (unstable)
l
+
2


ml   ka   0
2
Simple oscillator – Example 3
I cg   r 2 dm  2  x 2 Adx
l/2
0
Al
ml


12
12
 I cg  md 2
3
Io
2
 M o  I o
2
2
ml 2
ml 2
 m

 m  
12
3
2
3k  b 
ωn 
 
ml
+
ml 
 (kb)b 

3
2
ml 
  kb 2  0
3
Simple oscillator – Example 4
ma 2
From table : I 
2
TL
 JG 

 
  T
JG
 L 
Equivalent stiffness: K 
 n2 
2GJ
ma 2 L
 M z  I
 T  I
JG
L
+
ma 2  GJ
   0
2
L
Equivalent springs
Springs in series:
same force - flexibilities add
1
1

 P
  1   2   
 k1 k 2 
 ( f1  f 2 ) P  f eq P
f eq  f1  f 2
Springs in parallel:
same displacement - stiffnesses add
P  k1  k 2 
 ( k1  k 2 )   keq 
keq  k1  k 2
Equivalent springs – Example 1
mx  K eq x  0
 12EI 3EI 
mx   3  3  x  0
L2 
 L1
Equivalent springs – Example 2
2
ka
 Wl
2
ωn 
ml 2
n  n ( a )
Consider:
ka2 > Wl
ka2 = Wl
ka2 < Wl
 M o  I o
+
 ( ka )a  Wl  ml 2
2 
2
ml    ( ka  Wl )  0
n2 is positive - vibration is stable
statics - stays in stable equilibrium
unstable - collapses
Equivalent springs – Example 3
We cannot define n
 M o  I o
If  < < 1, sin   :
 Wl sin  ml 2
2 
ml   mgl sin  0
since we have sin term
g


   0
l
g
ωn 
l
+
g


  sin  0
l
Energy methods
Strain energy U:
energy in spring = work done
1 2 1
U  k  P
2
2
Conservation of energy:
work done = energy stored
Kinetic energy T:
1  
T  mr  r
2
 
Increment of work done  F  dr
 
 
1
 (mr)  (rdt)  d( 2 mr  r)  dT
 increment of kinetic energy T
Work-Energy principles
Work done = Change in kinetic energy

r2

r1

 
F  dr 
T2
T
1
dT  T2  T1
Conservation of energy for conservative systems
E = total energy = T + U = constant
Energy methods – Example
Work-energy principles have many
uses, but one of the most useful is
to derive the equations of motion.
Conservation of energy: E = const.
1
U  kx 2
2
1 2
T  mx
2
1 2 1 2
E  U  T  kx  mx
2
2
d
(E)0
dt
kxx  mxx  0
mx  kx  0
Same as vector mechanics
Two Degree-of-Freedom Systems
1. Model problem
Matrix form of governing equation
Special case: Undamped free vibrations
Examples
2. Transformation of coordinates
Inertially & elastically coupled/uncoupled
General approach: Modal equations
Example
3. Response to harmonic forces
Model equation
Special case: Undamped system
Two-DOF model problem
Matrix form of governing equation:
m1 0   x1  (c1  c2 )  c2   x1  (k1  k2 )  k2   x1   P1 
 
  
 0 m  x     c


c2   x2    k2
k2   x2   P2 


2  2 
2
where:
[M] = mass matrix;
[C] = damping matrix;
[K] = stiffness matrix;
{P} = force vector
Note: Matrices have positive diagonals and are symmetric.
Undamped free vibrations
Zero damping matrix [C] and force vector {P}
Assumed general solutions:
Characteristic equation:
 x1   A1 
     cos(t   )
 x2   A2 
(k1  k2  m1 2 )
  A1  0
 k2
 

2 
 k2
(k2  m2 )  A2  0

Characteristic polynomial (for det[ ]=0):
 k1  k 2 k 2  2 k1k 2
 
  

0
m2 
m1m2
 m1
4
Eigenvalues (characteristic values):
1
2

2






1  k  k2 k2
k1  k 2 k 2
4k1k 2



1  12   1







2  m1
m2    m1
m2 
m1m2  
2
2


Undamped free vibrations
Special case when k1=k2=k and m1=m2=m
Eigenvalues and frequencies:
12  0.3819 k
1   2   

2
.
618

2

m
 1
ω1  0.618
T

2π
ω
k
 fundamental frequency
m
 fundamental period
Two mode shapes (relative participation of each mass in the motion):
st
1 mode shape
A2 2k  m 2 1.618


A1
k
1
2nd mode shape
A2
k
 0.618


A1 k  m 2
1
The two eigenvectors are orthogonal:
 A1(1)   1 
Eigenvector (1) =  (1)   

1
.
618
A

 2  
 A1( 2 )   1 
Eigenvector (2) =  ( 2 )   


0
.
618
A

 2  
Undamped free vibrations (UFV)
Single-DOF:
x(t )  C cos(nt   )
For two-DOF:
 A1(1) 
 A1( 2) 
 x1 (t ) 
x     C1  (1)  cos(1t  1 )  C2  ( 2)  cos( 2t  2 )
 x2 (t )
 A2 
 A2 
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
UFV – Example 1
Given:
x  0
 1.0 
and xo    

1
.
618


No phase angle since initial velocity is 0:
 x1 
 1.0 
 1.0 
x     C1 
 cos(1t )  C2 
 cos( 2t )
1.618
 0.618
 x2 
From the initial displacement:
xo    
1.0 
 1.0 
 1.0 

C

C



1
2
1
.
618
1
.
618

0
.
618






C1   ; C2  0;

T1 
2
1
UFV – Example 2
1
Now both modes are involved: x  0 and xo     
2
From the given initial displacement:
1  C1 
1
 1 
 1   1
xo       C1 
  C2 

 C 
2
1
.
618

0
.
618
1
.
618

0
.
618
 



 
 2 
Solve for C1 and C2:
C1 
 0.618  1 1
 1.171 
1



 


  1.618 1  2
C

0
.
171

0
.
618

1
.
618

 


 2
Hence,
 1 
 1 
x  (1.171) 
 cos(1t )  (0.171) 
 cos( 2t )
1.618
 0.618
or
x1 (t )  1.171  (1) cos(1t )  0.171  (1) cos( 2t )
x2 (t )  1.171  (1.618) cos(1t )  0.171  (0.618) cos( 2t )
Note: More contribution from mode 1
Transformation of coordinates
UFV model problem:
“inertially uncoupled”
m1 0   x1  (k1  k2 )  k2   x1  0
  
 0 m  x     k

k2   x2  0
2  2 
2


“elastically coupled”
Introduce a new pair of coordinates that represents spring stretch:
or
z1(t) = x1(t)
= stretch of spring 1
z2(t) = x2(t) - x1(t) = stretch of spring 2
x1(t) = z1(t)
x2(t) = z1(t) + z2(t)
Substituting maintains symmetry:
(m1  m2 ) m2   z1  k1 0   z1  0
 
  
 m


m2  z2   0 k2   z2  0
2

“inertially coupled”
“elastically uncoupled”
Transformation of coordinates
We have found that we can select coordinates so that:
1) Inertially coupled, elastically uncoupled, or
2) Inertially uncoupled, elastically coupled.
Big question: Can we select coordinates so that both are uncoupled?
Notes in natural coordinates:
Eigenvecto rs (modal vectors) :
 A1(1)   1 
u1   (1)   

1
.
618
A

 2  
 A1( 2)   1 
u2    ( 2)   


0
.
618
A

 2  
The eigenvectors are orthogonal w.r.t [M]:
The modal vectors are orthogonal w.r.t [K]:
Algebraic eigenvalue problem:
u1T M  u2   0
u2 T M  u1  0
u1T K  u2   0
u2 T K  u1  0
K u1  1M u1
K u2   2 M u2 
Transformation of coordinates
General approach for solution
Governing equation:
M x  K x  0
Let
x  u1 q1 (t )  u2  q2 (t )
or
 x1 (t )  u11 
u12 

q
(
t
)


   1
 q2 (t )
 x2 (t ) u21 
u22 
(**)
We were calling “A” - Change to u to match Meirovitch
Substitution:
(*)
M  u1 q1 (t )  u2  q2 (t )  K u1 q1 (t )  u2  q2 (t )  0
Modal equations:
u1T (*)
u2 T (*)
 q1 (t )  12 q1 (t )  0 
Known solutions

 q2 (t )   22 q2 (t )  0
Solve for these using initial conditions then substitute into (**).
Transformation - Example
Model problem with:
xo    
1

2
and
xo   
0

0
1)
Solve eigenvalue problem:
u   1 
u12   1 
1  0.618;  11   
and


1
.
618
;

 

2
u
1
.
618

 21  
u22   0.618
2)
Transformation:
x  u1 q1 (t )  u2  q2 (t )
and
1   1 
 1 
 
 q1 (0)  
 q2 (0)
2
1
.
618

0
.
618
  



 q1 (0)   1.171 



q
(
0
)

0
.
171

 2
 
So
and
q
1 (t )  12 q1 (t )  0

2


q
(
t
)


2 q2 (t )  0
 2
 q1 (t )  q1 (0) cos(1t )

q2 (t )  q2 (0) cos( 2t )
 1 
 1 
x  
 1.171  cos(1t )  
(0.171  ) cos( 2t )
1
.
618

0
.
618




As we had before.
More general procedure: “Modal analysis” – do a bit later.
Response to harmonic forces
 F1  i t
M x  C x  K x  F (t )    e
F2 
Model equation:
[M], [C], and [K] are full but symmetric.
Assume:
{F}
not function of time
 X 1 (i )  i t
x  X (i )  
e
 X 2 (i )
Substituting gives:

M   i C   K X (i )  F 



2
Z (i )  2x2 impedance matrix
Hence:

Z (i )1 Z (i ) X (i )  Z (i )1 F 
 X1 
 z22  z12   F1 
1
X     
 F 
2 
X

z
z
z
z

z
11   2 
 2  11 22 12  12
All zij are function of (i ) :
zij   ω2 mij  iω cij  kij
i, j  1, 2
Special case: Undamped system
Zero damping matrix [C]
Entries of impedance matrix [Z]:
z11 ( )  k11  m1 2 ; z22 ( )  k22  m2 2 ; z12 ( )  k12
Substituting for X1 and X2:
(k 22  m2 2 ) F1  k12 F2
X1 
;
(k11  m1 2 )( k 22  m2 2 )  k122
 k12 F1  (k11  m1 2 ) F2
X2 
(k11  m1 2 )( k 22  m2 2 )  k122
For our model problem (k1=k2=k and m1=m2=m), let F2 =0:
(k  m 2 ) F1
X1  2 2
;
m (  12 ) ( 2   22 )
X2 
k F1
m 2 ( 2  12 ) ( 2   22 )
Notes:
1) Denominator originally (-)(-) = (+).
As it passes through 1, changes sign.
2) The plots give both amplitude
and phase angle (either 0o or 180o)
Multi-DOF Systems
1. Model Equation
Notes on matrices
Undamped free vibration: the eigenvalue problem
Normalization of modal matrix [U]
2. General solution procedure
Initial conditions
Applied harmonic force
Multi-DOF model equation
Multi-DOF systems are so similar to two-DOF.
Model equation:
M x  C x  K x  Q
We derive using:
1)
2)
3)
Vector mechanics (Newton or D’ Alembert)
Hamilton's principles
Lagrange's equations
Notes on matrices:
They are square and symmetric.
Kinetic energy :
xT M x
T
U  12 x K  x
T
1
2
Strain energy in spring :
[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.
UFV: the eigenvalue problem
Equation of motion:
M q  K q  0
Substitution of
leads to
in terms of the generalized D.O.F. qi
q  u f (t )
f (t )  A1ei t  A2e i t
K u   2 M u
Matrix eigenvalue problem
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]
We know that:
ui  M  u j  ui  M  u j  C  ij
So far, we pick our
eigenvectors to look like:
Instead, let us try to pick
so that:
T
δij  Kronecker delta
1  Let the 1st
uk    entry be 1
 
1 
 
uk new   uk    

 
uk Tnew M  uk new   2 uk T M  uk   1
Then:
where :
U T M U   I 
and
This is a common technique
for us to use after we have solved
the eigenvalue problem.
1 if i  j

0 if i  j
Do this a row at a time to form [U].
U T K U   
where :
12 0

0  22

  
.
.

.
 0





.  n2 
.
.
.
0
.
.
General solution procedure
Consider the cases of:
1.
2.
3.
Initial excitation qo  and qo 
Harmonic applied force
Arbitrary applied force
For all 3 problems:
1.
2.
Form [K]{u} = 2 [M]{u}
(nxn system)
Solve for all 2 and {u}  [U].
Normalize the eigenvectors w.r.t. mass matrix (optional).
Initial conditions
General solution for any D.O.F.:
q(t )  u1C1 cos(1t  1 )  u2 C2 cos(2t  2 )    un Cn cos(nt  n )
2n constants that we need to determine by 2n conditions on qo i and qo i
Alternative: modal analysis
Displacement vectors:
UFV model equation:
q  U 
q(t )  u11 (t )  u2 2 (t )    un n (t )
M q  K q  0
U T M U η  U T K U η  0
    0
n modal equations: 1  121  0 

2   22 2  0 
  Need initial conditions on ,


not q.
2

n   n n  0 
Initial conditions - Modal analysis
Using displacement vectors:
As a result, initial conditions:
q  U 
U T M  q  U T M U η
η  U T M  q
or
T

ηo   U  M  qo 

T






M  qo 
η

U
 o
   2  0 is:
Since the solution of 
hence we can easily solve for
And then solve
q  U η
C cos( t   ) or

 (t )  o cos( t )  o sin(  t )

( )
1 (t )  (o )1 cos(1t )  o 1 sin( 1t )
1

( )
 n (t )  (o ) n cos( nt )  o n sin(  nt )
n
Applied harmonic force
Driving force {Q} = {Qo}cos(t)
Equation of motion:
Substitution of
M q  K q  Q
q  U η
U  known
η unknown
leads to
and
Q  Qo cos( t )
ω  driving frequency
U T M U η  U T K U η  U T Qo cos( t )  N 
Hence,
T

u1  Qo 
1  2
cos( t )
2
1  
T

u2  Qo 
2  2
cos( t )
2
2  
etc.
then
q  U η
Continuous Systems
1. The axial bar
Displacement field
Energy approach
Equation of motion
2. Examples
General solution - Free vibration
Initial conditions
Applied force
Motion of the base
3. Ritz method – Free vibration
Approximate solution
One-term Ritz approximation
Two-term Ritz approximation
The axial bar
Main objectives:
1.
2.
Use Hamilton’s Principle to derive the equations of motion.
Use HP to construct variational methods of solution.
A = cross-sectional area = uniform
E = modulus of elasticity (MOE)
u = axial displacement
 = mass per volume
Displacement field:
u(x, y, z) = u(x, t)
v(x, y, z) = 0
w(x, y, z) = 0
Energy approach
For the axial bar:
1  u  u E  u 
Uo  strain energy density  12 σ ε  12 (E ε x )ε x   E
   
2  x  x 2  x 
V  potential energy  strain energy  U   U o dV
2
V
T  kinetic energy  mu 
1
2
2
1
2
ρ (Adx)u
2
Hamilton’s principle:
t2
0   (T  V ) dt
t1
t2  L 
 
 u  
0      Adx u u  A E   u  dx dt
t1
0
 x  x
 
 
L
L
 L 
 
u  
u
0       Adx u u 
AE
u  dx  A E u dt    A u u tt12 dx


t1
0
0
x 
x  
x
0
  t
L
t2  L 

 
u  
u
0      A u    AE   u dx  AE
u dt
t1
0

t

x

x

x
0


 
t2
Axial bar - Equation of motion

Hamilton’s principle leads to:
If area A = constant 

 A u     AE u   0
t
x 
x 
2
 2u
2  u

t 2
x 2
where :  
2
E

 F L2 
M 
 L3 
Since x and t are independent, must have both sides equal to a constant.
Separation of variables:

d
2
2
u ( x, t )  X ( x) T (t )
 
X    p   X  0
X  C cos p x    D sin  p x  
T  p 2 T  0
T  A cos( pt )  B sin( pt )
Hence
u ( x, t ) 

X dx 2
d 2T dt 2

 contant  - p 2
X
T
2

 A cos( p t )  B sin( p t ) C cos p x    D sin  p x  
i 1
i
i
i
i
i
i
i
i
Fixed-free bar – General solution
Free vibration:
General solution:
u ( x, t ) 
EBC 
u (0, t ) 
NBC 
xL

u (0)  0
NBC:
AE
xL

 A cos( p t )  B sin( p t ) C
i 1

i
i
i
i
i
0


i 1
i
Di pi

i
i
i
xL

Ci  0
cos pi L   Ai cos( pi t )  Bi sin( pi t )  0
pi L



2
or
3
2
or
(i  1, 3, 5, )
 i x  
 i  t 
 i  t 
sin
A
cos

B
sin







i
 i
2
L
2
L
2
L






i  1, 3, 5, 


0
cos pi x    Di sin  pi x  
i
 p L
Either Di  0 (trivial solution) or cos i   0
  
i 
pi 
2L
For any time dependent problem:
u ( x, t ) 
u
x

 C A cos( p t )  B sin( p t )  0
i 1
u
x
EBC:
u
x
E
= wave speed


5

2
Fixed-free bar – Free vibration
For free vibration:
2
 2u
2  u

2
t
x 2
General solution:
Hence
i E
n 
2L 
 i x 
sin 

2
L


u( x, t )  A( x) cos( n t )
are the frequencies (eigenvalues)
(i  1, 3, 5, )
are the eigenfunctions
Fixed-free bar – Initial conditions
Give entire bar an initial stretch.
Release and compute u(x, t).
Initial conditions:
Initial velocity:
u
t
t 0

0
t 0
0

L
L
 Lo  L 
 i x 
 i x   i x 
x
sin
dx

A
sin
sin
dx

A









i
i
0
2
 2L 
 2L   2L 
 L 
i  1, 3, 
2( Lo  L)
or Ai 
L2
Hence
u
t

 Lo  L 
 i x 

 x   Ai sin 

 2L 
 L 
i  1, 3, 
Initial displacement:
L
 L  L
u ( x, 0)   o
 x and
L



i 
 i x 
 
Bi sin 
 Bi  0
0
 2L 
i  1, 3,  2 L
u ( x, t ) 

L
0
( i  1)
8( Lo  L)
 i x 
x sin 
( 1) 2
 dx 
2 2
i
 2L 
8( Lo  L)
2


i  1, 3, 
( 1)
( i  1)
2
(i  1, 3, )
1
 i x 
 i  t 
sin
cos




i2
 2L 
 2L 
Fixed-free bar – Applied force
Now, B.C’s:
From
we assume:
u (0, t )  0

u

A
E
x  L  Fo sin(  t )


x

2
 2u
2  u

2
t
x 2
u( x, t )  X ( x) sin( n t )
Substituting:

 x 
  x 
u ( x, t )   A1 cos
  A2 sin 
 sin  t 







B.C. at x = 0:
u (0, t )  0
B.C. at x = L:
AE
or
Hence
u
x
xL

 AE
A2 
u ( x, t ) 
A1  0

L 
A 2 cos
 sin(  t )  Fo sin(  t )

L


Fo 
L 
sec

AE



Fo 
L   x 
sec
 sin 
 sin  t 
AE



 

Fixed-free bar – Motion of the base
From
2
 2u
2  u

t 2
x 2
Using our approach from before:

 x 
  x 
u ( x, t )   A1 cos
  A2 sin 
 sin  t 







B.C. at x = 0:
u (0, t )  A1 sin(  t )  U o sin(  t )
B.C. at x = L:
AE
u
x
xL
u
x
xL

0
L 

 A 2  U o tan 
  
 U
  L  A2 
  L 
  o sin 

cos


 sin  t   0

  
  
 
Hence
Resonance at:
A1  U o
  x 
  L    x 
u ( x, t )  U o cos
  tan 
 sin 
 sin  t 





 

 
 L  3
 ,
,

2 2
or

  3 
2L
,
2L
, etc.
Ritz method – Free vibration
Start with Hamilton’s principle after I.B.P. in time:
0
t2
t1
 L 
 
 u  

0   A u u  AE   u  dx dt
 x  x
 
  t
Seek an approximate solution to u(x, t):
In time:
harmonic function

cos(t)
( = n)
In space: X(x) = a11(x)
where: a1
= constant to be determined
1(x) = known function of position
1(x) must satisfy the following:
1.
2.
Satisfy the homogeneous form of the EBC.
u(0) = 0 in this case.
Be sufficiently differentiable as required by HP.
One-term Ritz approximation 1
1 ( x)  x  u ( x, t )  a1 1 ( x) cos( t )  a1 x cos( t )
Also approximate :
u  1 ( x) cos( t )  x cos( t )
Pick :


 L

0     a1  2 A ( x)( x)  AE (1)(1) dx  cos 2 ( t ) dt
t1
 0

L
L
 2   A x 2 dx  a1   A E dx a1
in matrix form:  2 M  a  K a
 0

 0

3
3 E 3
2 AL

 AE L   2  2    2  2
3
L  L
1RITZ  x
Substituting:
t2

Hence
 RITZ 
 EXACT 

3

  1.732
L
L

2L
  1.571
 x 

2
L


1EXACT  sin 

L
Ritz estimate is higher than the exact
Only get one frequency
If we pick a different basis/trial/approximation function 1,
we would get a different result.
One-term Ritz approximation 2
 x 

 2L 
1 ( x)  sin 
What if we pick :

 x 
d1 


cos 
dx 2 L
2
L


u ( x, t )  a1 1 ( x) cos( t )  a1 sin  x 2 L  cos( t )
Also approximate :
Substituting:
t2  L 
 

 u  
0      A u u  AE   u  dx dt
t1
0
 x  x
 
  t
0
t2
t1
Hence
u  1 ( x) cos( t )  sin  x 2 L  cos( t )
2
 L 
 





x

2
2
2 x 
  AE   cos 
 dx  cos 2 ( t ) dt
0 a1   A sin 

 2L 
 2L 
 2 L  

 RITZ 

E
2L



2L
   EXACT
Both mode shape and natural frequency are exact.
But all other functions we pick will never give us a
frequency lower than the exact.
Two-term Ritz approximation
Let : X ( x)  a1 x  a2 x 2
If approximate u  1  x : 0  
t2
t1
If approximate u  x 2 :
In matrix form:
where:
t2 
0 
t1

 M 11
 
 M 21
2




  A
L
0
  A
L
0
2
2
dX
 a1  2a2 x
dx


(a1 x  a2 x 2 ) x  AE (a1  2a2 x)(1) dx dt



(a1 x  a2 x 2 ) x 2  AE (a1  2a2 x)(2 x) dx dt

M 12   a1  E  K11
 
M 22  a2    K 21
L

L3
 M 11  0 ( x)( x)dx 
3

L

L4
2
 M 12  M 21  0 ( x )( x)dx 
4

L

L5
2
2
 M 22  0 ( x )( x )dx 
5

K12   a1 
 
K 22  a2 

L
K

 11 0 (1)(1)dx  L

L

2
K

K

 12
21
0 (2 x)(1)dx  L

3
 K  L (2 x)( 2 x)dx  4 L
22
0

3

Two-term Ritz approximation (cont.)
Substitution of:
leads to
   2 and  2 
 ( 2 L   L3 3)
 2 2
4
( L   L 4)
E

( 2 L2   L4 4)   a1  0
    
2
3
5
( 4 L 3   L 5) a2  0
Solving characteristic polynomial (for det[ ]=0) yields 2 frequencies:
(1 ) RITZ  1.5767  L and (2 ) RITZ  5.67  L
(1 ) EXACT  1.5708  L and ( 2 ) EXACT  4.7123  L
Let a1 = 1:
Mode 1:
 2 L (0.1713 a1  0.3785 L a2 )  0  a2   0.4526 L 
Mode 2:
 2 L2 ( 7.043 a1  5.10 L a2 )  0  a2   1.38 L 
Mode shape 1 :
X 1 ( x)  x  0.4526 x 2 L
Mode shape 2 :
X 2 ( x)  x  1.38 x 2 L