mws_civ_fft_phy_fourierapplications

mws_civ_fft_phy_fourierapplications - 11.00B.1 Chapter...

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Unformatted text preview: 11.00B.1 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. 11.00C.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 ) ( t F , or earthquake ground acceleration ) ( t g . The dynamical equilibrium for a SDOF system (shown in Figure 1) can be given as: ) sin( ) ( t w F t F ky y c y m = = + + (1) where c m , 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 ) ( t F ky y c y m = + + (2) = The solution (displacement response y ) of Equation (2) can be expressed as nt displaceme Qe t y pt = = ) ( (3) Hence dt dy velocity Qpe y pt = = = (4) 2 2 2 dt y d on accelerati e Qp y pt = = = (5) Substituting Equations (3-5) into Equation (2), one obtains 2 = + + k cp mp (6) The two roots of the above quadratic equation can be obtained as Physical Problem for FFT: Civil Engineering 11.00B.3 m k m c c p 2 ) )( ( 4 2 = (7) m k m c m c = 2 2 2 (8) Critical Damping ) ( cr C In this case, the term under the square root in Equation (8) is set to be zero, hence 2 2 = m k m C cr (9) or km C cr 2 = (10) since m k w = (11) Hence mw C cr 2 = (12) w k 2 = The two identical roots of Equation (8) can be computed as m C p p cr 2 , 2 1 = (13) and the solution ) ( t y in Equation (3) can be given as t p t p te Q e Q t y 2 1 2 1 ) ( + = (14) t m C cr e t Q Q + = 2 2 1 ) ( (15) which can be plotted as shown in Figure 3. 11.00C.4 Chapter 11.00C Figure 3 Free vibration with critical damping. Over damping ( ) cr C C > In this case, one has 2 2 > m k m C (16) The solution of ) ( t y from Equation (3) can be given as t p t p e Q e Q t y 2 1 2 1 ) ( + = (17) The response of over damping system is similar to Figure 3....
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mws_civ_fft_phy_fourierapplications - 11.00B.1 Chapter...

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