Applications of FFT in a Civil Engineering Problem

Transcription

Applications of FFT in a Civil Engineering Problem
Chapter 11.00C
Physical Problem for Fast Fourier Transform
Civil Engineering
Introduction
In this chapter, applications of FFT algorithms [1-5] for solving real-life problems such as
computing the dynamical (displacement) response [6-7] of single degree of freedom (SDOF)
water tower structure will be demonstrated.
Free Vibration Response of Single Degree-Of- Freedom, (SDOF) Systems
Figure 1 SDOF dynamic (water tower structure) system.
11.00B.1
11.00C.2
Chapter 11.00C
Figure 2 Water tower structure subjected to dynamic loads.
a) Water tower structure, Idealized as SDOF system.
b) Impulse blast loading F (t ) , or earthquake ground acceleration g (t ) .
The dynamical equilibrium for a SDOF system (shown in Figure 1) can be given as:
(1)
my  cy  ky  F (t )  F0 sin( w t )
where
m, c and k  mass, damping and spring stiffness, respectively (which are related to
inertia, damping and spring forces, respectively).
y, y , y  displacement, velocity, and acceleration, respectively.
Practical structural models such as the water tower structure subjected to applied blast
loading (or earthquake ground acceleration) etc. can be conveniently modeled and studied as
a simple SDOF system (shown in Figure 2).
For free vibration response, Equation (1) simplifies to
my  cy  ky  F (t )
(2)
0
The solution (displacement response y ) of Equation (2) can be expressed as
(3)
y(t )  Qe pt  displacement
Hence
dy
y  Qpe pt  velocity 
(4)
dt
d2y
y  Qp 2 e pt  acceleration  2
(5)
dt
Substituting Equations (3-5) into Equation (2), one obtains
(6)
mp 2  cp  k  0
The two roots of the above quadratic equation can be obtained as
Physical Problem for FFT: Civil Engineering
11.00B.3
 c  c 2  4(m)( k )
2m
(7)
c
k
 c 

 
 
2m
m
 2m 
(8)
p
2
Critical Damping (Ccr )
In this case, the term under the square root in Equation (8) is set to be zero, hence
2
k
 Ccr 

  0
m
 2m 
(9)
or
Ccr  2 km
(10)
k
m
(11)
since
w
Hence
Ccr  2mw
2k

w
The two identical roots of Equation (8) can be computed as
 Ccr
p1 , p 2 
2m
and the solution y (t ) in Equation (3) can be given as
y(t )  Q1e p1t  Q2 te p2t
 Ccr 

t
 2m 
 (Q1  Q2 t )e
which can be plotted as shown in Figure 3.
(12)
(13)
(14)
(15)
11.00C.4
Chapter 11.00C
Figure 3 Free vibration with critical damping.
Over damping C  Ccr 
In this case, one has
2
k
 C 

  0
 2m  m
The solution of y (t ) from Equation (3) can be given as
(16)
y(t )  Q1e p1t  Q2e p2 t
The response of over damping system is similar to Figure 3.
(17)
Under Damping C  C cr 
In this case, one has
2
k
 C 

  0
 2m  m
and the two “complex” roots from Equation (8) can be given as
(18)
2
C
k  C 
p1 , p2  
i

(19)

2m
m  2m 
Substituting Equation (19), and using Euler’s equation e i  cos( )  i sin(  ) , Equation (3)
or Equation (17) becomes
(20)
y(t )  e c / 2 m t  A cos wD t  B sin wD t 
where


2
wD 
k  C 

 see Equation (19)
m  2m 
(21)
Physical Problem for FFT: Civil Engineering
 w 1 2
C

C cr
C

2 km
Using the initial conditions:
@ t  0; y  y0 ; y  v0
Then, the two constants ( A and B ) can be solved, and Equation (20) becomes


v  y0w
y(t )  e wt  y0 cos wD t  0
 sin wD t 
wD


Equation (11.216) can also be expressed as:
yt   K1e wt  coswD t   
where
K1 
y
2
0
2

v0  y0w

wD2
v0  y 0w
wD y0
Equation (26) can be plotted as shown in Figure 4.
tan   
11.00B.5
(22)
(23)
(24)
(25)
(26)
(27)
(28)
11.00C.6
Chapter 11.00C
Figure 4 Free vibration of SDOF under damped system.
Force Vibration Response of SDOF Systems
For force vibration problem, the right-hand-side (RHS) of Equation (1) F (t )  0, and the
general solution for Equation (1) can be given as
y (t )  y c (t )  y p (t )
(29)
where the complimentary solution yc (t ) can be obtained as (see. Equation (20)) assumed
under-damped (C  Ccr ) case
yc (t )  e  (C / 2 m )t ( A cos wDt  B sin wDt )
e
 C k

t


*

k
2 m m 

(20, repeated)
( A cos wD t  B sin wD t )
 e C ( k / m)t / 2 km ( A cos wD t  B sin wD t )
Using Equations (10) and (11), Equation (30) becomes
yc (t )  e Cwt / Ccr ( A cos wDt  B sin wDt )
Using Equation (23), the above equation becomes
y c (t )  e wt ( A cos wD t  B sin wD t )
(30)
(31)
Physical Problem for FFT: Civil Engineering
11.00B.7
The particular solution y p (t ), associated with the particular sine term forcing function
F (t )  F0 sin( w t ) see Equation (1) can be given as
y p (t )  C1 sin( w t )  C 2 cos( w t )
(32)
The unknown constants C1 and C2 can be found by substituting Equation (32) into Equation
(1), and equating the coefficients of the sine and cosine functions.
Using Euler’s identity, one has
(33)
e iw t  cos( w t )  i sin( w t )
Thus, the RHS of Equation (1) can be expressed as
(34)
my  cy  ky  F0  Imaginary portion of eiw t
Hence, the response will consist of ONLY the imaginary portion of Equation (29).
The particular solution y p (t ), shown in Equation (32), can be more conveniently expressed
as
(35)
y p (t )  C *e iw t
Substituting Equation (35) into Equation (34), one gets
mC *i 2 w 2 e iw t  cC *iw e iw t  k C * e iw t   F0  e iw t
(36)
or
C * k  icw  mw 2   F0
(37)
Hence
F0
(38)
C* 
k  mw 2  icw
Substituting Equation (38) into Equation (35), one obtains
F0

 iw t
(39)
y p (t )  
e
2
 k  mw  icw 
In Equation (39), the “complex” number
(40)
d  k  mw 2  icw 
can be symbolically expressed as
(41)
d  (d R )  i (d I )
or in polar coordinates, one has (see Figure 5)
(42)
d  d e i  d  cos( )  i sin(  )


sin(  )
cos( )
cw

k  mw 2
tan( ) 
(43)
d R  k  mw 2
d I  cw
(44)
(45)
where
d 
d R 2  d I 2
(46)
11.00C.8
Chapter 11.00C


2
 k  mw 2  (cw ) 2
Thus, Equation (39) can be re-written as:
F0 e iw t
y p (t ) 
k  mw 2 2  (cw ) 2  e i

F0 e i ( w t  )
k  mw 
2 2
 (cw )



(48)
(49)
2
Figure 5 Polar coordinates.
The “imaginary” portion of Equation (49) can be given as
F0 sin( w t   )
y p (t ) 
2
k  mw 2  (cw ) 2
Define
F0
= amplitude of the steady state motion
Y
2
k  mw 2  (cw ) 2
F
yst  0 = static deflection of a spring acted by the force F0
k
w
r  = frequency ratio (of applied load/structure)
w
Then, Equations (43) and (50) become

(47)
(50)
(51)
(52)
(53)
Physical Problem for FFT: Civil Engineering
2r
; also refer to Equation (23)
1 r2
y p (t )  Y sin( w t   )
tan(  ) 
11.00B.9
(54)
(55)
y st sin( w t   )

(56)
(1  r 2 ) 2  (2r ) 2
The complimentary (or transient) solution yc (t ) shown in Equation (31), and the particular
solution y p (t ) shown in Equation (56) can be substituted into the general solution (see
Equation (29)) to obtain
y st sin( w t   )
(57)
y(t )  e wt ( A cos wD t  B sin wD t ) 
(1  r 2 ) 2  (2r ) 2
Define
Y
D
(58)
y st
1

(1  r 2 ) 2  (2r ) 2
D = Dynamic Magnification Factor
(59)
Dynamical Response by Fourier Series, DFT and FFT.
The dynamic load F (t ) acting on the SDOF system can also be expressed in Fourier series
as

F (t )  a 0   a n cos( nw t )  bn sin( nw t )
(60)
n 1
where the unknown Fourier coefficients can be computed as
1
a0   
T 
t 0 T
2
an   
T 
 F (t )dt
t0
t 0 T
 F (t ) cos(nw t )dt
(61)
t0
2
bn     F (t ) sin( nw t )dt
T 
If the forcing function contains only sine terms, then the particular (steady state) solution can
be found as (see Equation (56)):
y n  y pn
(62)
b 
 n 
k
sin( nw t   )
1  r   2r  
2 2
n
2
n
11.00C.10
Chapter 11.00C
 b  sin( nw t ) cos( )  sin(  ) cos( nw t )
 n 
2
k 
1  r 2  ( 2r  ) 2

n

n
Recalled Equation (54), one has
sin(  )
tan( ) 
cos( )
2rn

1  rn2
Hence
sin 2 ( ) sin 2 ( )  x 2

cos 2 ( )
1  sin 2 ( )



1  cos 2 ( )
cos 2 ( )  y 2

2rn 
(63)
(64)

2

1  r 
2 2
n
Solving Equation (64) for ( x  sin(  )) and ( y  cos( )) , one gets
x  sin 
2rn

1  rn2 2  2rn 2
y  cos 

(65)
1  rn2
1  r   2r 
2 2
n
2
n
Substituting Equation (65) into Equation (63) to obtain:
y n (t )  y pn




(66)
2
 b  1  rn sin nw t   2rn  cosnw t 
 n 
2
2
k
1  rn2  2rn 
Similarly, if the forcing function contains only the cosine terms, then the particular (steady
state) solution can be found as:
y n (t )  y pn
(67)


2
 an  1  rn cosnw t   2rn sin nw t 
 
2
2
k 
1  rn2  2rn 
Finally, if the forcing function contains both sine and cosine terms, then the total response
can be computed by combining both equations (66) and (67), including the constant forcing
term a0 , as following


Physical Problem for FFT: Civil Engineering
11.00B.11
a 
y (t )   y n t    0 
 k 
2

a n 1  rn2  bn 2rn 
 bn 1  rn  a n 2rn 

1  
   

sin(
n
w
t
)


cos(
n
w
t
)

2
2
2
2
2
2
k 1 

1  rn  2rn 
 1  rn  2rn 









(68)
Remarks
Using Euler’s relationships, the dynamic load F (t ) as shown in Equation (60), can also be
expressed in exponential form as
F (t ) 

C e
n  
inw t
(18, Ch. 11.02)
n
where
1
Cn    F (t )einw t dt
 T 0
For DFT, define
T
t 
; with t 0 , t1 , t 2 ,...........t N 1
N
T
(20, Ch. 11.02)
(69)
where
t j  jt
(70)
Then, the DFT pairs of Equations (21, 1, Ch. 11.04) becomes:

2 
t n

 ik  w0 
~  1  N 1
Ck     F (tn )e  T
 N  n 0
 2 
ik 
 nt
 1  N 1
   F (t n )e  Nt 
 N  n 0
 2 
in
j
 1  N 1
   F (t j )e  N  ; with n  0,1,2,......N  1
 N  j 0
(71)
and
N 1
 2 
~ in  j
F (t j )   C n e  N  ; with j  0,1,2,......N  1
(72)
n 0
Since both Equations 71 and 72 do have similar operations, with the exceptions of the factor
1
  and the sign  or   of the exponential term, both these equations can be handled by
N
the same “general_dft” program given at
http://numericalmethods.eng.usf.edu/simulations/mtl/11fft/fft_civil_engg_example12.m
Introduce the unit amplitude exponential forcing function
F (t )  ( F0  1)  e iwnt
(73)
11.00C.12
Chapter 11.00C
into RHS of Equation (1), the steady state solution can also be obtained as (see Equation 39):

 iwnt
1
e
(39, repeated)
y(t )  y p (t )  
2

k

m
w

ic
w
n
n 

Using the notations defined in Equations (23) and (53), the above equation can be written as,
~
for a harmonic force component of amplitude Cn .
~

 i  wn  nw (t j  jt )
Cn
y n (t j )  
 e
2
 k 1  rn  i 2rn 
~
2 



Cn

 in w  T  jt

 e
2
k
1

r

i
2

r


n
n


~
 2 


Cn

 in Nt  jt


e

2

 k 1  rn  i 2rn 

~

 inj2 / N
Cn

(74)
e
2
 k 1  rn  i 2rn 
and the total (steady state) response due to “ n ” harmonic force components can be calculated
as
~
N 1
Cn einj2 / N
(75)
y (t j )  
2
n  0 k 1  rn  i 2rn






Dynamic Response of “Water Tank Structure” by FFT.
The dynamic response y (t j ) in frequency domain of a general SDOF system (such as the
“water tank structure”) can be obtained by Equation (75), and the required coefficients c~
n
can be computed by Equation (71). Both of these equations can be represented (except for the
sign), by the following general exponential function
N 1
A( j )  factor *  A ( 0 ) (n)W
jn
(76)
n 0
where
W  e sign * i 2 / N
(77)
1
~
If Equation (71) needs be computed for Cn , then one should define factor  , sign = -1,
N
( 0)
and A  F (t j ). However, if Equation (75) needs be computed for y (t j ) , then one should
~
Cn
(0)
define factor  1 , sign = +1, and A 
.
k 1  rn2  i 2rn 
It is important to notice that Equation (76) has the same form as shown in the earlier
Equation (74). However, the definition of W in Equation (77) is different from the one
shown in Equation (4, Ch. 11.05) by a negative sign in the power of W . Therefore, efficient
Physical Problem for FFT: Civil Engineering
11.00B.13
FFT subroutine (with user’s specified SIGN = 1, or -1) can be utilized, as given at
http://numericalmethods.eng.usf.edu/simulations/mtl/11fft/fft_civil_engg_example12.m
References
[1] E.Oran Brigham, The Fast Fourier Transform, Prentice-Hall, Inc. (1974).
[2] S.C. Chapra, and R.P. Canale, Numerical Methods for Engineers, 4th Edition, Mc-Graw
Hill (2002).
[3] W.H . Press, B.P. Flannery, S.A. Tenkolsky, and W.T. Vetterling, Numerical Recipies,
Cambridge University Press (1989), Chapter 12.
[4] M.T. Heath, Scientific Computing, Mc-Graw Hill (1997).
[5] H. Joseph Weaver, Applications of Discrete and Continuous Fourier Analysis, John
Wiley & Sons, Inc. (1983).
[6] Mario Paz, Structural Dynamics: Theory and Computation, 2nd Edition, Van Nostrand
Inc. (1985).
[7] R.W. Clough, and J. Penzien, Dynamics of Structures, Mc-Graw Hill (1975).

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