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) my 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 my 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 y0w y(t ) e wt y0 cos wD t 0 sin wD t wD Equation (11.216) can also be expressed as: yt K1e wt coswD t where K1 y 2 0 2 v0 y0w wD2 v0 y 0w 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) my 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 mC *i 2 w 2 e iw t cC *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 icw 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 2r ; 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 (2r ) 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 (2r ) 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( ) 2rn 1 rn2 Hence sin 2 ( ) sin 2 ( ) x 2 cos 2 ( ) 1 sin 2 ( ) 1 cos 2 ( ) cos 2 ( ) y 2 2rn (63) (64) 2 1 r 2 2 n Solving Equation (64) for ( x sin( )) and ( y cos( )) , one gets x sin 2rn 1 rn2 2 2rn 2 y cos (65) 1 rn2 1 r 2r 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 2rn cosnw t n 2 2 k 1 rn2 2rn 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 cosnw t 2rn sin nw t 2 2 k 1 rn2 2rn 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 2rn bn 1 rn a n 2rn 1 sin( n w t ) cos( n w t ) 2 2 2 2 2 2 k 1 1 rn 2rn 1 rn 2rn (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 )einw 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 jt (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 nt 1 N 1 F (t n )e Nt 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 jt ) Cn y n (t j ) e 2 k 1 rn i 2rn ~ 2 Cn in w T jt e 2 k 1 r i 2 r n n ~ 2 Cn in Nt jt e 2 k 1 rn i 2rn ~ inj2 / N Cn (74) e 2 k 1 rn i 2rn 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 2rn 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 2rn 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).