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Example 2DOF System
Unforced Case — free responses
Review SDOF: modal free responses of is
where the ai & bi , or c i & " i , depend on and
k 1!=!4
k 2!=!2
2
1
˙q i +" i
2q i = 0
q i (t ) =a i sin" i t +b i cos" i t = c i sin(" i t +# i )
q i (0) !!!!!!! q i (0)
For example, with initial conditions:
q 1 (0) =1= a
1sin0+ b
1cos0 =c
1sin("
1)
q 1 (0) = 0 =a 1# 1cos0 $b 1# 1sin0 = c 1# 1cos(" 1)
q 2(0) = 0 = a 2 sin0+ b 2 cos0 = c 2 sin(" 2)
q 2(0) =1=a 2#
2cos0 $b 2#
2sin0 = c 2#
2cos("
2)
then: a1=0, b1=1, c 1=1, " 1=–! /2 and a 2=1/2, b 2=0, c 2=1/2, " 2=0.
q 1 (t ) = cost
q 2(t ) = 0.5sin2t
More generally, the responses in the original coordinates are:
u(t ) = "i a i sin#
i t +b
i cos#
i t [ ]
i
$ = c i "i sin(#
i t +%
i )
i
$
= ai sin"
i t +bi cos"
i t [ ]i
# = ci sin("
i t +$ i )
i
#
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Example 2DOF System
Example Free Responses
Displaced in shape of: 2nd mode1st mode combination
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c 2!=!0.1
Example 2DOF System
Unforced Case with Damping
Let’s add damping. (What is damping???)
k 1!=!4
k 2!=!2
2
1
k 1!=!4 k 2!=!1m 1!=
!2 m 2
!=
!1
c 1!=!0.2
c 2!=!0.1
c 1!=!0.2
Already know " =
1 1
2 #1
$
% &
'
( ) . Let u ="q and premultiply by " T
M ˙u+Cu+Ku =0
C =
c 1+c
2"c
2
"c 2
c 2
#
$ %
&
' ( =
0.3 "0.1
"0.1 0.1
#
$ %
&
' (
1 2
1 "1
#
$ %
&
' (
2 0
0 1
#
$ %
&
' (
1 1
2 "1
#
$ %
&
' (
˙q 1
˙q 2
) * +
, - . +
1 2
1 "1
#
$ %
&
' (
0.3 "0.1
"0.1 0.1
#
$ %
&
' (
1 1
2 "1
#
$ %
&
' ( q
1
q 2
) * +
, - . +
1 2
1 "1
#
$ %
&
' (
6 "2
"2 2
#
$ %
&
' (
1 1
2 "1
#
$ %
&
' ( q
1
q 2
) * +
, - .
=
6 0
0 3
#
$ %
&
' (
˙q 1
˙q 2
) * +
, - . +
0.3 0
0 0.6
#
$ %
&
' ( q
1
q 2
) * +
, - . +
6 0
0 12
#
$ %
&
' ( q
1
q 2
) * +
, - . =
0
0
) * +
, - .
"TM" ˙q+"TC"q+"TK"q =0
"
˙q 1 +
0.05q 1 +
1q 1 = 0
˙q 2+0.20q
2+ 4q
2= 0
˙q i +2" i # i
˙q i +# i q i = 0
" 1
= 1, # 1=0.025 =2.5%
" 2 =
2, # 2 =
0.050 =5.0%
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c 2!=!0.1
Example 2DOF System
Unforced Case with Damping — free response
So modal damped free responses are
where the ai & bi , or c i & " i , depend on and , and.
k 1!=!4
k 2!=!2
2
1
k 1!=!4 k 2!=!1m 1!=
!2 m 2
!=
!1
c 1!=!0.2
c 2!=!0.1
c 1!=!0.2
q i (t ) =a i e "# i $ i t sin$ i
dt +b i e "# i $ i t cos$ i
dt = c i e "# i $ i t sin($ i
dt +% i )
q i (0) !!!!!!! q i (0)
More generally, the responses in the original coordinates are:
u(t ) = e "# i $ i t %
i a i sin$
i
dt +b
i cos$
i
dt [ ]
i
& = c i %i e "# i $ i t sin($
i
dt +
i
&
= e "# i $ i t a
i sin$
i
dt +b
i cos$
i
dt [ ]
i
& = ci e "# i $ i t sin($
i
dt +'
i
i
&
" i d=" i [1#$ i
2]1/2
USC ViterbiSchool of Engineering
General MDOF Systems
Equation of motion: M˙
u+C˙
u+Ku =Ef
First use the undamped, unforced equation to find natural
frequencies and mode shapes:(M"
2#K)$ =0 or ("
2I#M
#1K)$ =0
The mode shapes have the properties that
"i TM" j =M i # ij =
M i , i = j
0, i $ j
% & '
"i TK" j =K i # ij =
K i , i = j
0, i $ j
% & '
The fact that !i TM! j = 0 and !i
TK! j = 0 for i ! j means the
mode shapes are orthogonal . Note also that K i /M i = ! i 2.
Note: some systems have (M! 2 – K) be less than rank (n – 1), whichimplies there are multiple modes at one or more frequencies.
Note: some systems (e.g., satellites) may have ! = 0 be a solution;
this implies one or more rigid body modes in which the wholestructure moves or rotates, possibly without internal deformation.
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General MDOF Systems
Forced Harmonic Response
Forced response of a linear system is the sum of the freeresponse from any non-zero initial conditions and theeffects of the external forcing.
Let us assume the initial conditions are zero and consider a harmonic excitation.
M ˙u +Cu+Ku =Ef =Ef0sin" t
Then the response is also harmonic. This is most easily
demonstrated using a SDOF example …
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Damping Models
In the previous damped 2DOF example, we used specific
damping coefficients in the model: c 1 = 0.2, c 2 = 0.1.
k 1!=!4
k 2!=!2
2
1c 2!=!0.1
c 1!=!0.2
2 0
0 1
"
# $
%
& '
˙u 1
˙u 2
( ) *
+ , - +
0.3 .0.1
.0.1 0.1
"
# $
%
& ' u 1
u 2
( ) *
+ , - +
4 +2 .2
.2 2
"
# $
%
& ' u 1
u 2
( ) *
+ , - =
0
0
( ) *
+ , -
" TC" =
1 2
1 #1
$
% &
'
( )
0.4 #0.2
#0.2 0.2
$
% &
'
( ) 1 1
2 #1
$
% &
'
( ) =
0.4 #0.2
#0.2 1
$
% &
'
( )
"
˙q 1 +
1
15q 1 #
1
30q 2 +
1q 1
= 0
˙q 2+
1
3q 2#
1
15q 1+ 4q
2= 0
" 1
= 1, # 1=2.5%
" 2=2, #
2=5.0%
We found that the damping #TC# diagonalized.
However, a different choice (c 1 = 0.2, c 2 = 0.2) may not:
0.2
c 2!=!0.1
k 1!=!4 k 2!=!1m 1!=!2 m 2!=!1
c 1!=!0.2 0.2
This is what is called a non-proportionally
damped or non-classically damped system.
The equations
are still coupled!
We will see later how to solve systems like this.(! 1=1.00037, ! 2=1.99926, & 1=3.3313%, & 2=8.3368%)
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Damping Models
Special Cases of Proportional Damping: Modal Damping
We often do not know the damping coefficients (while mass
and stiffness can be computed or easily measured, dampingis much harder to quantify).
Damping ratios & i in individual modes are easier to estimate(e.g., excite the structure in a particular mode, and estimate & from the decay e –&! t ; repeat for other modes).
So, one way to construct the full damping matrix C is byassuming a modal decomposition
" TC" =
! 02# i $ i M i
0 !
%
&
' ' '
(
)
* * *
C =" # T! 0
2$ i % i M i
0 !
&
'
( ( (
)
*
+ + + " #1 =M"
! 02$ i % i
0 !
&
'
( ( (
)
*
+ + + " #1
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Damping Models
Special Cases of Proportional Damping: Modal Damping
So, the equation of motion is: M ˙u+M"!
2# i $ i
!
%
& ' '
(
) * * " +1u+Ku =0
Decouple method 1: let u = #q and premultiply by #T:
" TM" ˙q+" TM"!
2# i $ i
!
%
& ' '
(
) * * " +1"q+" TK"q =0
!
M i
!
%
& ' '
(
) * * ˙q+
!
M i
!
%
& ' '
(
) * *
!
2# i $ i
!
%
& ' '
(
) * * q+
!
K i
!
%
& ' '
(
) * * q =0
!
M i
!
%
& ' '
(
) * * ˙q+
!
M i
!
%
& ' '
(
) * *
!
2# i $ i
!
%
& ' '
(
) * * q+
!
K i
!
%
& ' '
(
) * * q =0
˙q+
!
2# i $ i
!
%
& '
'
(
) *
* ˙q+
!
$ i
2
!
%
& '
'
(
) *
* q=
0
˙q i +2# i $ i q i +$ i 2q i =0, i =1,...,n
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Damping Models
Special Cases of Proportional Damping: Modal Damping
So, the equation of motion is: M ˙u+M"!
2# i $ i
!
%
& ' '
(
) * * " +1u+Ku
Decouple method 2: let u = #q and premultiply by # –1M –1:
" #1M#1M" ˙q+" #1M#1M"!
2$ i % i
!
&
' ( (
)
* + + " #1"q+" #1M#1K"q =
" #1" ˙q+" #1"!
2$ i % i
!
&
' ( (
)
* + + " #1"q+" #1M#1K"q =0
Let " =#$1M
$1K#
% M#" =K# % #T
M#" =#TK#
% ! M i
!
&
' (
(
)
* +
+ " =
!
K i
!
&
' (
(
)
* +
+
% " =
!
K i M i
!
&
' ( (
)
* + + =
!
, i 2
&
' ( (
˙q+
!
2" i # i
!
$
% & &
'
( ) ) q+
!
# i 2
!
$
% & &
'
( ) ) q =0
˙q i +2" i # i q i +# i 2q i = 0, i =1,...,n
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Damping Models
Special Cases of Proportional Damping: Rayleigh Damping
Rayleigh damping (1877) is given by C = ' M + ( K.Since both M and K diagonalize with #, it is easy to see
that #TC# = ' #TM# + ( #TK# is diagonal.
The result: 2& i ! i = ' + (! i 2 or & i = (' /! i + (! i )/2.
Since there are only two parameters, ' and ( , we can
choose the damping of two modes; all other modal
damping ratios are given by the above equation.
If we know (or choose) & r and & s then' = 2! r ! s(& r ! s – & s! r ) / (! s
2 – ! r 2)
( = 2(& s! s – & r ! r ) / (! s2 – ! r
2)& i = (' /! i + (! i )/2
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Damping Models
Conditions for Proportional Damping
It can be shown that the system is classically damped if either of the following are true:• C = M F(M –1K) + K G(K –1M)
• C = F(KM –1)!M + G(MK –1) K for some matrix functions F(") and G(").
Note: should add references here.
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Laplace and Fourier Transforms
Time history p(t ) is related to its
Laplace transform P (s) and Fourier Transform P ( j ! ) as
P (s ) = p (t )e "st dt "#
#
$
p (t ) =1
2"
P (s )e st ds
#$
$
%
P ( j " ) = p (t )e # j " t dt
#$
$
%
p (t ) =1
2" P ( j # )e j # t d #
$%
%
&
Laplace transform P (s) may also be denoted L{ p(t )}
and Fourier Transform P ( j ! ) by F { p(t )}.
Note: L p (t ){ } = p (t )e "st dt "#
#
$ = pe "st "#
#+ p (t )se "st dt
"#
#
$ = s L p (t ){ }
(assuming p(t )$0 as t $#)integration by parts
Similarly: F p (t ){ } = j " F p (t ){ }
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Transfer Functions
SDOF
Consider single degree of freedom system m u +c u +ku =f
ms 2U (s )+csU (s )+kU (s ) =F (s )
(ms 2+cs +k )U (s ) =F (s )
U (s )
F (s )=
1
ms 2+cs +k
Laplace transform both sides:
Or, rearranging:
U ( j " )
F ( j " )=
1
m ( j " )2+cj "
=
1
k #m " 2+cj
Could do similar using Fourier transform:
If f (t ) = F 0 sin! t , then
u (t ) = F 0
1
k "m # 2+cj #
sin(# t +$ )
" = angle1
k #m $ 2+cj $
%
& '
(
) * = tan
#1 c
m $ 2 #k
These are calledtransfer function
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Transfer Functions
SDOF Example
For example:
k !=!8
2c !=!0.2
2˙u +0.2u +8u =f
10−1
100
101
10−2
10−1
100
Frequency [rads/sec]
M a g n i t u d e
10−1
100
101
−150
−100
−50
0
Frequency [rads/sec]
P h a s e [ d
e g r e e s ]
10−1
100
101
−2
0
2
Frequency [rads/sec]
R e a l P a r t
10−1
100
101
−3
−2
−1
0
Frequency [rads/sec]
I m a g i n a r y P a r t
U (s )
F (s )=
1/2
s 2+0.1s + 4 Graph with s = j ! :
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Transfer Functions
SDOF Example — Poles
For example:
k !=!8
2c !=!0.2
2˙u +0.2u +8u =f
U (s )
F (s )=
1/2
s 2+0.1s + 4
Roots of the denominator polynomial are called the poles
of the system:
s poles ="0.1± 0.12 " 4
2= "0.05 ± j 3.9975 # "0.05 ±1.9994
R
Im
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Transfer Functions
SDOF Example — Poles and Zeros
Some TFs have a numerator that is also a polynomial in s.For example, the transfer function from ground accelerationto the absolute acceleration of a SDOF system:
k !=!8
2c !=!0.2
2˙u +0.2u +8u = "2˙v g
L ˙v (t ){ }
L ˙v g(t ){ }=
L ˙u + ˙v g{ }L ˙v g(t ){ }
=
L "0.1u " 4u { }
L ˙v g(t ){ }
= ("0.1s " 4)L u { }
L ˙v g(t ){ }=
0.1s + 4
s 2+0.1s + 4
Roots of the numerator polynomial are
called the zeros of the system:
s root =
"40
Re
Im
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Transfer Functions
MDOF
For a MDOF system, must be careful to handle matricescorrectly in determining the transfer function:
M ˙u+Cu+Ku =Ef
Ms 2+Cs +K[ ]U(s ) =EF(s )
U(s ) = Ms 2+Cs +K[ ]
"1
EF(s )
Now, the transfer function [Ms2 + Cs + K] –1E is a matrix,
each element of which is a scalar transfer function.
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State-Space Formulation
Unforced Response — Mode Shapes
If the damping term does not decouple, then the state-
space approach must be used and
must be solved directly to get the complex eigenvalues and
complex eigenvectors. Once the the eigenvalues ) i ,) i * are
found, then ! i = |) i | and & i = –Re{) i }/(2! i ).
If some eigenvalues are purely real, then no oscillator
mode corresponds to that eigenvalue.
(" 2I+ " M
#1C +M
#1K)$
d=0
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k 3!=!a
1c 3!=!c
k 2!=!a
1c 2!=!c
State-Space Formulation
Visualization of Non-Proportional Damping
Let’s look at an example of a
3DOF structure with classicaldamping (-c = 0). The natural
frequencies & damping are:! 1 = 1 , & 1 = 3.14%;! 2 = 2.80, & 2 = 8.80%;
! 3 = 4.05, & 3 = 12.72%.
A non-classically dampedversion with additional
damping in the first floor (-c = 15c ) has natural
frequencies & damping:! 1 = 1.15, & 1 = 23.6%;! 2 = 2.71, & 2 = 85.9%;
! 3 = 3.64, & 3 = 15.7%.
k 1!=!a
1
c 1!=!c !+!-c
c !=!0.01a
a !=!199.3233(so first natural
frequency is 1!Hz)
"0.328 0.737 "0.591
"0.591 0.328 0.737
"0.737 "0.591 "0.328
0.065± 2.06 j "1.142!12.9 j 1.913±1
0.117± 3.71 j "0.508! 5.8 j "2.385!1
0.146± 4.63 j 0.916±10.4 j 1.061±
#
$
% % % % % % %
"0.233! 0.19 j "0.545! 3.99 j 0.003± 0
"0.557! 0.08 j "0.977! 0.93 j 0.537! 0
"0.737 "0.591 "0.328
"0.966±1.97 j "26.85± 63.2 j 3.749! 0
0.409± 4.04 j 6.213± 22.1 j "3.575!1
1.260± 5.17 j 8.655± 5.2 j 1.174± 7
#
$
% % % % % % %
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State-Space Formulation
Visualization of Non-Proportional Damping
ProportionallyDamped:
Non-Proportionally
Damped:
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State-Space Formulation
Unforced Free Response
The unforced free response of the state-space system
could be found through modal decomposition, but there is aneasier way. To compute the response at time t , break the time
up into r smaller steps. Given the definition of a derivative:
˙" =A"
"(1
r t ) # "(0
r t )+ " 1
r t # "(0)+A"(0) 1
r t = [I+A t
r ]"(0)
The same can be used to approximate ( after each t /r .
"(2r t ) # [I+A t
r ]"(1
r $t ) = [I+A t
r ]2"(0)
!
"(r
r t ) # [I+A t
r ]"(r %1
r $t ) = [I+A t
r ]r "(0)
To eliminate the approximation, let r $#.
"(t ) = [I+ (At ) + 1
2!(At )
2+
1
3!(At )
3+!]"(0)
"(t ) =e At "(0)
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State-Space Formulation
Unforced Free Response — Notes about eAt
eAt is a matrix; its elements are not the exponential of theelements of At . The computation of eAt can be computedthrough the power series formulation, though that is not
computationally efficient.
An efficient computation uses the fact that the eigenvectors of
a power of a matrix are the same as those of the matrix itself.
If A+ = +., where + is the eigenmatrix and . is a diagonalmatrix of eigenvalues, then it can be shown that An+ = +.n.
Thus, e At
= I+ (At ) + 1
2!(At )
2+
1
3!(At )
3+!
="#1" +"#1$"t +1
2!"#1$"2
t 2+
1
3!"#1$3"t
3+!
="#1[I+ ($t ) + 1
2!($t )
2+
1
3!($t )
3+!]"
= "#1
"
e % i t
"
&
'
( ( (
)
*
+ + +
"
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State-Space Formulation
Impulse Response
The state impulse response matrix is given by
H" =AH" +B# (t ), H" (0$) = 0
Hy =CyH" +Dy# (t )
Integrating from t = 0 – to t = 0+, will give the initial conditions
that make a free response equal to the impulse response.
H"(0
+
) #H"(0#) =AH"
avg(0
+
#0#)+B
0 0
H" (0+
) =B
For t > 0, the impulse response is just an unforced free
response (which was solved several slides ago), so
H"(t ) =e
At H
"(0
+
)
H" (t ) =e At B
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State-Space Formulation
Forced Response
The forced response is, then, the combination of the initial
condition free response and the effects of the force
"(t ) =e At "(0)+ e
A(t #$ )Bf($ )d $
0
t
%
Add Example!USC ViterbiSchool of Engineering
Discrete-Time State-Space System
We’ve already seen the free response of the unforced state-
space system . So, ((t + -t ) = eA-t ((t ).
The effect of the forced response of in [t , t + -t ]
is the superposition of the free response with the impulse
response of f during the time step.
˙" =A"
Defining ((k ) = ((k -t ), then the discrete-time state space form
"(k +1) = Ad"(k )+B
df(k )
y(k ) =Cy"(k )+Dyf(k )
˙" =A" +Bf
"(t +
#t )=e
A#t
"(t )+ e
A(t +#t $% )
t
t +#t
& Bf
(%
)d %
In the interval [t , t + -t ), if f($ ) is constant (a zero-order hold )
"(t +#t ) =e A#t
"(t ) + e A(#t $% )
0
#t
& d %
'
( )
*
+ , Bf(t )
Ad
Bd
Ad Bd
0 I
"
# $
%
& ' =e
A
0
"
# $
Note: one can also assume a first-o
(linear) hold on f to determine Bd.
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Distributed Parameter Systems
Euler-Bernoulli Beam Example
Continuous systems (e.g., beams, plates, shells, or more
complex structures) also have modes of vibration (generallyinfinitely many). Let’s look at an Euler-Bernoulli beam;
transverse displacement v ( x ,t ) must satisfy the partialdifferential equation of motion where (")/ = 0/0 x :
(EI " " v " " ) + # A ˙v =f (x ,t )
L
x
v (x ,t )
EI , * A
Substitute v ( x ,t ) = V ( x )T (t ) into
the unforced, undamped system.
(EI " " V " " ) T + # AV T = 0
(EI " " V " " )
# AV = $
˙T
T
LHS is a function of x , the RHS of t , so both must be constant.
=" 2
˙T +"
2T = 0
" " " " V # $ 4V =0
Assuming EI and * A are constant, let ) 4 = [* A! 2/EI ]1/2 so:
V (x ) =c 1sinh" x +c
2cosh" x +c
3sin" x +c
4cos" x
T (t ) = c sin(" t +# )
USC ViterbiSchool of Engineering
Distributed Parameter Systems
Euler-Bernoulli Beam Example — exact simply-supported
The coefficients c i depend on the boundary conditions:
• V = 0 and V / = 0 at a fixed (cantilevered) end
• V = 0 and V // = 0 at a simply supported end
• V // = 0 and V /// = 0 at a free end
1. Simply supported at both ends: c 2 = c 4 = 0
c 1sinh" L + c
3sin" L = 0
" 2c 1sinh" L # "
2c 3 sin" L = 0
$ sinh" L sin" L
" 2sinh" L #"
2sin" L
= 0
The non-trivial solution is ) i L = i ! where n = 1, 2, …; the
resulting eigenfunctions are V i ( x ) = C sin i ! x /L and thenatural frequencies are ! i = (i ! /L)2 [* A/EI ]1/2.
! 3!=!9! 1
! 2!=!4! 1! 1
USC ViterbiSchool of Engineering
Distributed Parameter Systems
Euler-Bernoulli Beam Example — exact cantilevered
cos" L cosh" L +1= 0
which has solutions ) L = 1.8751, 4.6941, 7.8548, 10.996, …;
larger solutions are approximately ! (1"+ 2n)/2.
! 3!=!17.55! 1
! 2!=!6.27! 1
! 1
2. Cantilevered at x = 0 and free at x = L: c 4 = –c 2, c 3 = –c 1 Using the remaining equations leads to
" 1=
3.516
L2
EI
# A
" 2=
22.03
L2
EI
# A
" 3=
61.70
L2
EI
# A
V i (x ) =C cosh"
i x # cos"
i x #
cosh" i L+cos" i L
sinh" i L+sin" i Lsinh"
i x # sin"
i x [ ]{ }
USC ViterbiSchool of Engineering
Distributed Parameter Systems
Euler-Bernoulli Beam Example — assumed modes method
Sometimes solving for the exact mode shapes from thepartial differential equations of motion is difficult.
An alternate approach is to assume a set of mode shapes
V i ( x ) that satisfy the boundary conditions. Then, Lagrange’s
equations can be used to determine the equations of motion.
For example, for an Euler-Bernoulli beam:
V =1
2EI ( " " v )2dx
0
L
# =
1
2v i
2(t ) EI " " V i
2(x )dx
0
L
# i
$
T =1
2% Av
2dx
0
L
# =
1
2v i
2(t ) % AV
i
2(x )dx
0
L
# i
$
f i (t ) = f (x ,t )V i (x )d0
L
"
v (x ,t ) = V i (x )v
i (t
i
"
Which leads to: M˙v +Kv = f
k ij = EI " " V i
(x ) " " V j (x )dx 0
L
#
m ij = " AV i (x )V j (x )dx 0
L
#