218 Structural elements 4.3.1 Equations of motion projected onto a modal basis Let us consider the forced dynamical problem governed by the linear partial differential equations of the general type: K[  X]+C[ ˙  X]+M[ ¨  X]=  F (e) (r; t) + I.C. and B.C.  [4.35] I.C. stands for the initial conditions and B.C. for the conservative boundary con- ditions. The stiffness, damping and mass operators K[], C[], M[]can depend on the vector position r in the Euclidean space. To the forced problem governed by [4.35], we associate the modal problem [4.36], which satisfies the same boundary conditions: [K(r) −ω 2 M(r)][ϕ]=0 + B.C.  [4.36] solutions of which are defining the following modal quantities: {ϕ n }; {ω n }; {K n }; {M n }; n = 1, 2, 3, [4.37] As in the case of discrete systems, the mode shapes can be used to determine an orthonormal basis with respect to the stiffness and mass operators, in which the solution  X(r, t) of [4.35] can be expanded as the modal series:  X(r, t) = ∞  n=1 q n (t) ϕ n (r) [4.38] where the time functions q n (t) n = 1, 2, , termed modal displacements, are the components of the displacement field  X(r; t) in the modal coordinate system. Formal proof of such a statement is not straightforward and is omitted as already mentioned in the introduction. As the dimension of the functional vec- tor space is infinite, a delicate problem of convergence and space completeness arises. Substitution of [4.38] into the equation of motion [4.35], leads to: ∞  n=1 {K[ϕ n (r)]q n (t) +C[ϕ n (r)]˙q n (t) +M[ϕ n (r)]¨q n (t)}=  F (e) (r; t) + I.C.        [4.39] The system of equations [4.39] is projected on the k-th mode shape ϕ k by using the scalar product [1.43] and the orthogonality properties [4.2]. The transformation Modal analysis methods 219 results into the following ordinary differential equation: K k q k + ∞  n=1 C kn ˙q n + M k ¨q k = Q (e) k (t) where C kn =ϕ k , C[ϕ n ] (V) and Q (e) k (t) =ϕ k ,  F (e)  (V) [4.40] The subscript (V) accounts for the space integration domain involved in the scalar product. Q (e) k (t) k = 1, 2, , termed modal forces, are the components of the external force field  F (e) (r; t), as expressed in the modal coordinate system. As already discussed in [AXI 04], Chapter 7, the result [4.40] can be drastically simplified if coupling through the damping operator is negligible. If it is the case, the mode shapes of the conservative system can be assumed to be orthogonal with respect to the C damping operator and [4.40] reduces to a single equation which governs the forced vibration of an harmonic oscillator, called modal oscillator: M k (ω 2 k q k + 2ω k ς k ˙q k +¨q k ) = Q (e) k (t) where ϕ k , C[ϕ n ] (V) =  c k = 2ω k M k ς k > 0; if n = k 0; otherwise [4.41] By using the Laplace transform, the image of the modal displacement is obtained as: ˜q k (s) = ˜ Q k (s) M k  ω 2 k + 2ω k ς k s + s 2  where ˜ Q k (s) = ˜ Q (e) k (s) +˙q k (0) +(2ω k ς k + s)q k (0) The Laplace transform of the displacement field in the physical coordinates system is finally obtained by using the modal series [4.38]:  ˜ X(r, s) = ∞  k=1 ˜ Q k (s) ϕ k (r) M k  ω 2 k + 2ω k ς k s + s 2  [4.42] Here, the number of terms of the series is infinite, which was obviously not the case for N-DOF systems. Thus, in practice, to compute modal expansions like [4.42], the series must be truncated to a finite number of terms. Truncation criteria which lead to acceptable errors are discussed in the next subsections. 220 Structural elements 4.3.2 Deterministic excitations To study the time response of continuous systems it is necessary to know both the spatial distribution ofthe excitations and their variation with time. The deterministic external forcing functions are described by a vector  F (e) (r; t) which may stand either for an external force field which fluctuates with time, or for a prescribed motion assigned to some degrees of freedom of the mechanical system. Firstly, it is useful to make the distinction between forcing functions in which space and time variables are separated and those where they are not. It is convenient to start with the former case which is more common. 4.3.2.1 Separable space and time excitation The general form of this kind of excitation can be written as:  F (e) (r, t) = F 0 (r)u(r)f(t) [4.43] (r) is a function, or a distribution, which specifies the space distribution of the loading and which complies with the norm condition:  (V) |(r)|dV = 1 [4.44] (V) is the domain in which  is defined. F 0 is the scale factor of the load magnitude. The unit vector u(r) specifies the load direction in the Euclidean space and finally f(t) describes the time evolution, or time-history of the loading.  and f(t)are assumed to be both integrable and square integrable. The resultant of the loading is given by:  R(t) =  (V)  F (e) (r, t)dV = f(t)  (V) (r)u(r)dV [4.45] The point r a at which the resultant is applied is time independent. It is given by the barycenter of the function : r a =  (V) r dV  (V) dV [4.46] The modal forces related to this type of excitation are written as: Q (e) n (t) =  ϕ n ,  F (e) (r; t)  (V) = F 0 f(t) n ; with  n =ϕ n ,  u (V) [4.47] Modal analysis methods 221 Figure 4.22. Example of a travelling load 4.3.2.2 Non-separable space and time excitation When the position vector used to describe the spatial distribution of the load is time dependent it becomes impossible to separate the time and space variables. This is typically the case of the so called travelling loads, which are of practical importance in many applications. Let us consider, for instance, a train running on a flexible bridge at a constant cruising speed V 0 , see Figure 4.22. One is interested in analysing the response of the bridge loaded by the weight of the running train. Let 2λ designates the length of the train. As a first approximation made here for the sake of simplicity, the total weight  P =−M g is assumed to be distributed uniformly along the train. So, the forcing function is written as:  F (e) (x, t) =  P 2λ [U(x − V 0 t +λ) −U (x − V 0 T −λ)] U(x) is the Heaviside step function and the running abscissa V 0 t is taken at the middle of the train. The resultant and the abscissa x a (t) at which it is applied are obtained by using the relationships [4.45] and [4.46]:  R =  P 2λ  +∞ −∞ [U(x − V 0 t +λ) −U (x − V 0 t −λ)]dx =  P 2λ   +∞ V t−λ dx −  +∞ V t+λ dx  =  P x a (t) = 1 2λ  +∞ −∞ x[U(x − V 0 t +λ) −U (x − V 0 t −λ)]dx =  Vt+λ Vt−λ xdx= V 0 t 222 Structural elements Now, if the length of the train remains much smaller than the bridge span denoted L, the load can be reasonably modelled as a concentrated load which leads to the modal forces: Q (e) n =ϕ n ,  Pδ(x − V 0 t)=  P ·ϕ n (V t) [4.48] According to the result [4.48], the whole series of modes is excited as time elapses. If the actual length of the train is accounted for, the modal forces are found to be: Q (e) n = 1 2λ  L 0 (  P ·ϕ n (x))[U(x − V 0 t +λ) −U (x − V 0 T −λ)]dx = 1 2λ  VT+λ VT−λ (  P ·ϕ n (x)) dx [4.49] So, the modal excitation is found to depend on the ratio of the modal wavelength on the train length. Assuming for instance that the bridge deck can be modelled as a pinned-pinned beam, the modal force is expressed as: Q (e) n = PL 2nπλ  cos  nπ(V 0 T −λ) L  − cos  nπ(V 0 T +λ) L  which is found to vanish for all the modes such that λ = mL/n where m is an integer less than n. So, such modes do not contribute to the bridge motion and can be removed from the modal basis. 4.3.3 Truncation of the modal basis 4.3.3.1 Criterion based on the mode shapes The last result of the preceding subsection can be restated as a general rule, according to which all the mode shapes which are orthogonal to the spatial distri- bution of the excitation can be discarded from the modal model. In other words, the only modes which contribute to the response series [4. 42] are those modes which are not orthogonal to the spatial distribution of the excitation: ϕ n , (r)u(r) (V) =  n = 0 [4.50] where for convenience the criterion [4.50] is formulated in the case of a separated variables forcing function. It may be noted that a modal truncation based on the mode shape criterion is similar to the elimination of some components of the physical displacements in a solid body, based on the orthogonality with the loading vector field. Modal analysis methods 223 Figure 4.23. Pinned-pinned beam loaded by a transverse concentrated force example. – Beam loaded by a concentrated transverse force The problem is sketched in Figure 4.23. The equilibrium equation is written as: EI ∂ 4 Z ∂x 4 + C ˙ Z + ρS ¨ Z = f(t)δ(x− x 0 ); Z(0) = Z(L) = 0; ∂ 2 Z ∂x 2     0 = ∂ 2 Z ∂x 2     L = 0 The beam being provided with pinned support conditions at both ends, the modal quantities relevant to the problem are: ϕ n (ξ) = sin(nπ ξ ); M n = ρSL/2 = M b /2; K n = (nπ) 2 EI 2L 3 ; ω 2 n = (nπ) 4 L 4 c 2 , where ξ = x/L and c 2 = EI/ρS The Laplace transform of the response is: Z(ξ, ξ 0 ; s) = 2 ˜ f(s) M b ∞  n=1 sin(nπξ ) sin(nπ ξ 0 ) (c 2 (nπ/L) 4 + 2scς n (nπ/L) 2 ) If ξ 0 = 0.5, the non-vanishing terms of the series are related to the odd modes n = 2k + 1, k = 0, 1, 2, only: Z(0.5, ξ ; s) = 2 ˜ f(s) M b ∞  k=0 (−1) k sin((2k + 1)πξ) c 2 ((2k + 1)π/L) 4 + 2scς 2 n ((2k + 1)π/L) 2 example. – Beam symmetrically or skew symmetrically loaded 224 Structural elements Figure 4.24. Symmetric loading of a beam provided with symmetric supports If the load is symmetrically, or skew symmetrically, distributed with respect to the cross-section at mid-span of the beam and if the boundary conditions are symmetric, the only modes which contribute to the response must verify the same conditions of symmetry as the loading function, see Figure 4.24. Although this rule is correct from the mathematical standpoint, it is still necessary to be careful when using it, because real structures present inevitably mater- ial and geometrical defects which spoil the symmetry of the ideal model. One striking example of the importance of such ‘small’ defects will be outlined in subsection 4.4.3.3. 4.3.3.2 Spectral criterion The spectral considerations made in [AXI 04] Chapter 9, concerning the dynam- ical response of forced N -DOF systems, can be extended to continuous structures to produce a very useful criterion for restricting the modal basis to a finite number of modes. Figure 4.25 is a plot of the power density spectrum of some excitation signal, which in practice extends over a finite bandwidth, limited by a lower cut-off frequency f c1 and by a upper cut-off frequency f c2 ; that is, outside the interval f c1 , f c2 the excitation power density becomes negligible. The dots on the frequency axis mark the sequence of the natural frequencies of the excited structure, which of course extends to infinity. Figure 4.25. Spectral domains of excitation versus structure response properties Modal analysis methods 225 Let us consider first the response of a single mode f n to the excitation signal. The response is found to be quasi-inertial if f n /f c1 ≪ 1, in the resonant range if f c1 <f n <f c2 , and quasi-static if f n /f c2 ≫ 1. Therefore, a finite number of low frequency modes can lie in the quasi-inertial range, and infinitely many other modes lie in the quasi-static range. The contribution to the total response of the modes lying in the quasi-inertial range can be accounted for by neglecting the stiffness and damping terms of the modal oscillators and only a finite number of such modal contributions are to be determined. On the other hand, the contribution to the total response of the modes lying in the quasi-static range can be accounted for by neglecting the damping and inertial terms of the modal oscillators; however there are still infinitely many modal contributions to be accounted for. The method for avoiding the actual calculation of such an infinite series is best described starting from a specific example. Let us consider again a vehicle of mass M, travelling at speed V 0 on a flex- ible bridge. Assuming the bridge deck is modelled as an equivalent straight beam provided with pinned supports at both ends and damping is neglected, the equations of the problem are written as: EI ∂ 4 Z ∂x 4 + ρS ¨ Z =−Mgδ(x − V 0 t) Z(0) = Z(L) = 0; ∂ 2 Z ∂x 2     0 = ∂ 2 Z ∂x 2     L = 0 By projecting this system on the pinned-pinned modal basis, we get the system of uncoupled ordinary differential equations, comprising an infinite number of rows of the type: ω 2 n q n +¨q n =−g 2M M b sin  nπV 0 t L  ; where ω 2 n =  nπ L  4 EI ρS It immediately appears that the dynamic response strongly depends upon the cruising speed V 0 of the load. In particular, the resonant response of the n-th mode occurs if the following condition is fulfilled: nπV n L =  nπ L  2  EI ρS ⇒ V n = nπ L  EI ρS Shifting to the spectral domain, the Fourier transform of the beam deflection is found to be: ⌢ Z (x, ω) =−g M M b ∞  n=1 (δ( n − ω) − δ( n + ω)) i(ω 2 n − ω 2 ) sin  nπx L  226 Structural elements where  n = nπV 0 L . which becomes infinite at the undamped resonances ω n =  n . Nevertheless, it may be also realized that in most cases of practical interest even the smallest cruising speed V 1 needed to excite the first resonance of the beam is likely to be far beyond the realistic speed range of the vehicle. This is because a bridge is designed to withstand large static transverse loads. As a consequence, the dynamical response of the loaded bridge can be determined entirely by using the quasi-static approximation:  nπ L  4 EI ρS q n =−g 2M M b sin  nπV 0 t L  ⇒ q n =− gM K n sin  nπV 0 t L  ; where K n = M b 2  nπ L  4 EI ρS and the bridge deflection is found to be: Z(x; t) =−gM ∞  n=1 1 K n sin  nπV 0 t L  sin  nπx L  To restate the conclusions of this example as a general rule, one-dimensional problems are considered for mathematical convenience. Further extension to the case of two or three-dimensional problems is straightforward as it suffices to deal in the same way with each of the Euclidean dimensions of the problem. As outlined above, it is relevant to discuss the relative importance of the modal expansion terms of the response in relation to the spectral content of the excitation. The Fourier transform of the response is written as: ⌢ Z (x; ω) = F 0 ⌢ f (ω) ∞  n=1 ϕ n (x) n M n  ω 2 n + 2iω n ως n − ω 2  [4.51] The excitation spectrum S ff (ω) = 2 ⌢ f (ω) ⌢ f * (ω) is again assumed to be negli- gible outside the finite interval ω c1 , ω c2 .Ifω n designates the natural frequencies of the structure, ordered as an increasing sequence, the three following distinct cases have to be discussed: 1. Resonant response range: ω n ∈[ω c1 , ω c2 ] It is clearly necessary to retain all the modes whose frequencies lie within the spectral range of the excitation, except if they are orthogonal to the spatial distribution of the excitation, that is if ψ n vanishes. The contribution of such modes to the response is given by the full expression [4.51]. The spectrum Modal analysis methods 227 of the modal response can be conveniently related to the excitation spectrum (cf. Volume 1 Chapter 9) as: S (n) ZZ (x; ω) =  ϕ n (x) n K n  2  1 (1 −(ω/ω n ) 2 ) 2 + 4(ως n /ω n ) 2  S ff (ω) [4.52] Spectral relationships like [4.52] are especially useful to characterize the magnitude of the vibration without entering into the detailed time-history of the response. It is recalled that the mean square value of the modal response can be inferred from the modal response spectrum S (n) ZZ (x; ω) by integration with respect to frequency as detailed in [AXI 04], Chapter 8. 2. Quasi-static responses: ω c2 ≪ ω N In the same way as in the travelling load example, the series [4.51] can be reduced to the quasi-static form: ⌢ Z (x; ω, N) = F 0 ⌢ f (ω) ∞  n=N ϕ n (x) n K n [4.53] The displacement is synchronous with the time evolution of the excitation. Furthermore, the series [4.53] is found to converge, as appropriate from the physical standpoint. Here, convergence can be immediately checked as the modal stiffness coefficients K n are proportional to n 4 for bending modes and to n 2 in the case of torsion and longitudinal modes. Furthermore, the series [4.53] calculated from n = 1 instead of n = N, must converge to the static solution Z s of the forced problem related to the system [4.35]: K(x)Z s = F 0 (x) + B.C.  [4.54] This is because the Hilbert space of the solutions is complete by definition. So, the modal expansion of Z s is found to be: Z s (x) = F 0 ∞  n=1 ϕ n (x)ψ n K n [4.55] As a consequence, the quasi-static part of the dynamical response [4.51] may be conveniently written by using a finite number of terms only: ⌢ Z (x; ω, N) = F 0 ⌢ f (ω)R N (x) [4.56] where R N (x) = Z s (x) F 0 − N  n=1 ϕ n (x)ψ n K n [...]... Taking for instance N = 4, [4 .73 ] is written as follows:      2 2 2 L + ̟1 − ̟ 2 L 2 L 2 L   2 L 2 L 2 L q1 2 2γL + 2 − ̟ 2 2 L 2 L  q2   q3 2 2 L 2 L + ̟3 − ̟ 2 2 L q4 2 2γL 2 L 2 L + ̟4 − ̟ 2 0 0 = 0 0 by solving this system, the following values of the modified natural frequencies are obtained numerically: ̟1 = 1.96 52; 2 = 4. 878 8; ̟3 = 7. 9 571 ; ̟4 = 11. 071 If the projection is restricted... nπ ; an = 0 ⇒ ϕn (ξ ) = cos(nπ ξ ); n = 0, 1, 2 If four modes are retained in the projection, the following homogeneous system to be solved is:      2( γL + ̟ 2 ) 2 L 2 L 2 L q1 0  q2  0  2 L 2 L + π 2 − ̟ 2 2γL 2 L  q = 0  2 2 2 L 2 L 2 L 2 L 2 L + 4π − ̟ 2 L 2 L 2 L + 9π 2 − ̟ 2 3 q4 0 The first row of the modal equation is the projection of the constrained local equation onto the rigid... truncation flexibility is given by: 1 1 1 2 = ′− − 2 γT γ γf π N n =2 2 1 1 = − 2 2 3 n n N n =2 1 n2 where n = 2 is the order of the first non-rigid mode of the free-free beam Using this corrective term, a truncation to N = 5 gives: γe = 0.9465; ̟1 = 0.86 07; 2 = 3. 426 3; ̟3 = 6.4386; ̟4 = 9.5 322 which are very close to the exact values 4.4.1.3 Bending modes of an axially prestressed beam A beam clamped... is plotted versus the modal cut-off index N at ξ = 0 .75 and ξ = 0.1 Incidentally, this kind of calculation is a convenient method to determine the value of a fairly large number of series For instance, here it is found that: ∞ n=0 1 2 = 8 (1 + 2n )2 Modal analysis methods 23 1 Figure 4 . 27 Convergence of modal series of displacement Figure 4 .28 Convergence of modal series of stress To determine the stress... system can be further simplified by using the variable transformation: d(2k+1) = q(2k+1) − p(2k+1) Then the linear system to be solved is written as a modal system, of current row n = 2j + 1: [(2k + 1)4 − ̟ 2 ] (d(2k+1) ) − 2 kj j d2j +1 = 0 The results obtained by solving numerically such a set of equations, projected on the first ten odd modes of the pinned-pinned beams, are shown in Figures 4. 37. .. equation of the constrained system is: −ES d 2X − 2 ρSX = −KL Xδ(x − L); dx 2 with X(0) = 0 [4 .71 ] 23 4 Structural elements Using the dimensionless quantities already defined above, the modal system is written as: ¯ ¯ ¯ −X ′′ − ̟ 2 X = −γL Xδ(ξ − 1); ¯ X(0) = 0 [4. 72 ] The mode shapes φn are expanded as a linear combination of the fixed-free beam modes: φn (ξ ) = ∞ qj ϕj (ξ ) j The projection of [4. 72 ] onto... )) Z + ρIw 4 ∂x ∂x x0 δ ′ (x − x0 ) = 0 [4 .76 ] The modal basis is described by the following quantities: ϕn = sin(nπ ξ ); Mn = Kn = n4 K0 ; ρSL Mb = ; 2 2 EI b π 4 ; 2L3 K0 2 ω0 = Mn K0 = 2 2 ωn = n4 ω0 ; Projecting [4 .76 ] on this basis, a coupled system of equations is obtained, which is written in terms of reduced quantities as: (n4 − ̟ 2 )qn −̟ 2 µR − ̟ 2 µZ ∞ sin j =1 where ̟ = ∞ j =1 j cos nπ x0... linear spring 24 4 Structural elements As is easily understood, only the odd order modes give rise to coupling terms, since the even modes have a node at mid-span Then the system is reduced to : [(2k + 1)4 − ̟ 2 ]q2k+1 = γkj [(2k + 1)4 − ̟ 2 ]p2k+1 = γkj j j (pj − qj ) (qj − pj ) γkj = γ sin((2k + 1)π /2) sin((2j + 1)π /2) = γ (−1)k+j Furthermore, it is noted that only the out -of- phase modes of the beam... ̟n = (1 + 2n) ; ϕn (ξ ) = sin(̟n ξ ); 2 n = 0, 1, 2 (b) Spring stiffness coefficient is infinite: 1/γL = 0 ωn = nπ ; (c) Spring stiffness coefficient has a nonzero finite value, for instance: γL = π/4 The first roots of the transcendental equation x cot(x) − π/4 are found to be: ̟1 = 1.9531; 2 ϕn (ξ ) = sin[ωn ξ ] 2 = 4.8 72 2 ; ̟3 = 7. 9 524 ; ̟4 = 11.0 67 Modal projection of the constrained system The... frequency of an equivalent beam of half length which would be pinned at x = 0 and clamped at x = L /2 ̟∞ = 4 (3. 926 6/π )2 ≃ 1.99π Figure 4. 37 Natural frequency of the first out -of- phase mode versus spring stiffness coefficient Modal analysis methods 24 5 Figure 4.38 Mode shape of the first out -of- phase mode versus spring stiffness coefficient (deflection of only one beam is displayed for sake of clarity) . = 4, [4 .73 ] is written as follows:    2 L + ̟ 2 1 − ̟ 2 2 L 2 L 2 L 2 L 2 L + ̟ 2 2 − ̟ 2 2 L 2 L 2 L 2 L 2 L + ̟ 2 3 − ̟ 2 2 L 2 L 2 L 2 L 2 L + ̟ 2 4 − ̟ 2      q 1 q 2 q 3 q 4   =   0 0 0 0   by. homogeneous system to be solved is:    2( γ L + ̟ 2 ) 2 L 2 L 2 L 2 L 2 L + π 2 − ̟ 2 2γ L 2 L 2 L 2 L 2 L + 4π 2 − ̟ 2 2γ L 2 L 2 L 2 L 2 L + 9π 2 − ̟ 2      q 1 q 2 q 3 q 4   =   0 0 0 0   The. roots of the transcendental equation x cot(x)−π/4 are found to be: ̟ 1 = 1.9531; ̟ 2 = 4.8 72 2 ; ̟ 3 = 7. 9 524 ; ̟ 4 = 11.0 67 2. Modal projection of the constrained system The equilibrium equation of