CIVIL ENGINEERING 365 ALL ABOUT CIVIL ENGINEERING



Method ExpositionComparing Eq. (11) with the eigenvalue Eq. (13), we obtain the following relationships: (31) (32) The evolution of αi can be studied in two different cases.When |Θi|>1, the second-order Eq. (32) has two real solutions, but only that with the module less than 1 must be taken into account to ensure the convergence of the power series of Eq. (10) (cf. Fig. 8) (33) ∀  Θi>1αi(Θi)=Θi−Θi2−1∀  Θi<−1αi(Θi)=Θi+Θi2−1When |Θi|<1, solutions α1 and α2 are conjugate complex roots with a unitary modulus. In order to choose between the two solutions, it is necessary to introduce a damping term as a damping matrix C in Eq. (8) and choose the complex solution with the modulus less than 1. By passing to the limit by making the matrix C tend toward 0, it can be shown that the solution to be taken into account is that of the upper half circle (cf. Fig. 9) (34) |Θi|<1αi=cosβi+isinβiwithcosβi=ΘiA physical interpretation of this choice will be given in the case of a simple degrees of freedom system by attaching it to propagation waves and the Sommerfield boundary conditions.The general solution Un is the combination of the particular solution defined in Eq. (10) (35) ∀n≥1Un=∑i=1mαinHiYi+Un¯with the unknown Hi, which is the participation of each particular solution; and Un¯ belongs to the kernel of the operator A.According to the PCE, the response becomes (36) U(x)=U0+∑n=1∝(∑i=1mαinHiYi+Un¯)Pn(x)The analytical calculation of the coeffcients Hi and the vectors Un¯ using Eq. (24) is given in the Appendix, and the results are (37) (38) The vector U0 is obtained in the following form: (39) U0=U¯+∑i=1mαi21aiYiYiTAU¯If we consider that the probability density of the random variable x is given in Fig. 1, the orthogonality relationships of Eq. (7) show that U0 corresponds to the statistic mean value of the response E(U).The analytical variance expression of the response for each component Uk of the response vector U is equal to (40) V(Uk)=E((Uk−U0k)(U˜k−U˜0k))where S˜ is the complex conjugate notation of term S.By using Eq. (3), we can write (41) Uk−U0k=∑n=1∝UnkPn(x)By using Eq. (35), the variance of the component Uk is expressed as (42) V(Uk)=(∑n=1∝∑i=1mαinHiYik)(∑l=1∝∑j=1mα˜jlH˜jY˜jk)E(PnPl)By applying the orthogonality relation from Eq. (7), we obtain (43) V(Uk)=∑i=1m∑j=1mHiH˜jY˜jkYik∑n=1∝(αiα˜j)nThe following identity can be used: (44) ∑n=1∝(αiαj)n=∑n=1∝αiα˜j1−αiα˜j(1−limn→+∝(αiα˜j)n)by taking into account the damping of the mechanical system, limn→+∞(αiαj)n=0 because |αiαj|<1.Finally, the variance expression becomes (45) V(Uk)=∑i=1m∑j=1mαiα˜j1−αiα˜j(1+αi2)(1+α˜j2)Y˜jkYikdid˜jwith (46) Interpretation of Results Using Simple ModelsIn this section, the analytical results determined previously will be applied to simple systems to physically interpret the polynomial chaos expansion proposed. The first random system studied is a single DOF (1DOF) system without damping subjected to the harmonic force shown in Fig. 10.The governing equation of the system is (47) (−ω2m+k+xA)U=Fwithu=U exp(iωt)andf=F exp(iωt)In this simple case, the unique solution Θ1 is given by the solution of the spectral problem defined in Eq. (13) (48) Nondimensional terms are introduced: D=(A/k)<1 is the dispersion factor Ustat=(F/k), and Ω=(ω/ω0) is the reduced frequency where ωo=(k/m) is the natural frequency. Eq. (48) then becomes (49) The curve giving DΘ1 as a function of Ω in the event of D=0.2 highlights a frequency band where the αi value is imaginary with a module of 1 while its limit values are Ω1=1−D and Ω2=1+D (cf. Fig. 11). A physical interpretation can link this zone to a propagation band of a discrete semi-infinite periodic system. In this way, the equation system in Eq. (8) in the case of a single DOF system can be illustrated by the mechanical model shown in Fig. 12. In order to obtain a physical model, coefficient A has been replaced by −h, the latter parameter h considered positive is interpreted as the stiffness of a spring. The frequency band [Ω1;Ω2] corresponds to a propagation zone, and the propagation coefficient α1 must have a negative imaginary part in order to comply with the Sommerfeld condition. In Eq. (32), the solution chosen αi must have a positive imaginary part because coefficient D is considered positive, while in the mechanical system, h is positive and A is negative.According to Eq. (39), the mean response U0 is obtained as follows: (50) Note that Y1AY1T is equal to a1, and the terms Y1Y1TA can be switched in this case of 1DOF system; therefore the previous equation becomes (51) U0=U¯(1+α12)withU¯=Fk(1−Ω2)Three zones have to be considered (cf. Fig. 11) •Zone 1: Ω2<Ω12 so Θ1<−1(52) According to Eq. (51) (53) U0=2FkD1(−Θ1+Θ12−1)Using Eq. (49) (54) U0=2Fk(1−Ω2+(1−Ω2)2−D2)In the static case, Ω2=0, therefore Eq. (54) becomes (55) U0=U¯21+1−D2withU¯=Fk=UstatUsing equation Eq. (7), it can be demonstrated that this result is exactly the same as the statistical mean of the static response of a random spring (56) U0=2Fkπ∫−111−u21+Dudu•Zone 2: Ω2>Ω22 so Θ1>1(57) The same calculation as that used previously gives the following result: (58) U0=2Fk(1−Ω2−(1−Ω2)2−D2)•Zone 3: Ω12<Ω2<Ω22 so Θ1∈[−1;1](59) According to Eqs. (49) and (51) (60) U0=2Fk(1−Ω2+iD2−(1−Ω2)2)It is now possible to evaluate the amplitude at the resonance frequency when Ω2=1 (so Θ1=0) (61) This last value can be compared to that viscous damped single degree of freedom system (62) where ξ is the specific damping factor. This analysis gives an equivalence between the structural damping η=2ξ and D/2 the dispersion damping factor. This equivalent damping associated with the uncertainty can be considered as a numerical radiation loss in the fictitious semi-infinite periodic mechanical model shown in Fig. 12. As can be seen in Fig. 13, the mean response modulus |U0| is constant in the third zone of the frequency domain. The real and imaginary complex parts of the mean response are given in Figs. 14 and 15. It can be seen that the imaginary parts are equal to zero in the first and second zones. In the case where we introduce a damping factor c in the 1DOF system, the eigenvalue Θ1 is given by (63) or in nondimensional form (64) Solution α1 is obtained with the complex second-order equation, Eq. (32). The mean response modulus |U0| has been plotted for different values of ξ, as shown in Fig. 16. The expression of the response variance, given in Eq. (45) becomes in this simple case (65) V(U)=|U0|2|α1|21−|α1|2Thus, the normalized standard deviation σ is defined by (66) σ=V(U)|U0|=|α1|1−|α1|2The evolution of σ with respect to frequency Ω for many values of D is shown in Fig. 17. In the case of a 2DOF system, schematized in Fig. 18, the equilibrium equations can be written in the following matrix form: (67) k[−Ω2+2+DΘ−(1+DΘ)−(1+DΘ)−Ω2+1+DΘ][U1U2]=[F0]with ω02=(k/m), Ω2=(ω2/ω02) and D=(A/k).The nullity of the determinant of this matrix gives the characteristic equation (68) Ω4+(−2DΘ−3)Ω2+DΘ+1=0Eq. (68) gives only one value for Θ(69) DΘ1=Ω4−3Ω2+12Ω2−1=−1λ1DThus, there are two solutions for the eigenvalue problem given in Eq. (12), with only one different from zero (70) The first eigenvector is Y1 corresponding to the unique eigenmodes of perturbation (71) where r1 is calculated with respect to the normalization Eq. (14): r1=m1/2(Ω4+(1−Ω2)2)1/2.The mode shape of the first eigenmodes of perturbation is shown in Fig. 19. The curve in Fig. 20 corresponding to D=0.2 shows two propagation bands associated with Θ1 and that Ω=1/2 corresponds to Θ1→∞. The αi are calculated in each zone defined in Fig. 20. For example, in the two propagation bands (72) α1=Ω4−3Ω2+1D(2Ω2−1)+i1−(Ω4−3Ω2+1)2D2(2Ω2−1)2The mean response is calculated by applying Eq. (50) at the two degrees of freedom system (73) U0=(I+α12(2Ω2−1)[Ω2−Ω21−Ω2−1+Ω2])U¯(74) The mean value modulus is given in Fig. 21 for several values of D. If viscous damping is introduced, the eigenvector Y1 and λ1 are complex. In this case of the 2DOF system defined in Fig. 2, the mean value modulus is given in Fig. 22 for several values of D and the damping coefficient ξ equal to 2.10−2.


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