Identification Methods for Structural Systems. Prof. Dr. Eleni Chatzi Lecture March, PDF Free Download

Prof. Dr. Eleni Chatzi Lecture 4-09. March, 2016

Fundamentals Overview Multiple DOF Systems State-space Formulation Eigenvalue Analysis The Mode Superposition Method The effect of Damping on Structural Response

Multiple DOF Systems Let s revisit the simple 2DOF system k 1 F 1 ( t ) k 2 F ( t 2 ) FBD k1x1 F 1 ( t ) k ( x x ) 2 2 1 k2( x2 x1) F ( t 2 ) c 1 m 1 x ( t 1 ) c 1 m 2 x 2 ( t ) cx 1 1 m 1 c2( x 2 x 1) c2( x 2 x 1) (Lumped Mass System) m 2 The equations of motion can be written as m 1 ẍ 1 + (c 1 + c 2 )ẋ 1 c 2 ẋ 2 + (k 1 + k 2 )x 1 k 2 x 2 = F 1 (t) m 2 ẍ 2 + c 2 ẋ 2 c 2 ẋ 1 + k 2 x 2 k 2 x 1 = F 2 (t) The system can be written in matrix form as follows: [ ] [ ] [ ] [ ] [ m1 0 ẍ1 c1 + c 2 c 2 ẋ1 k1 + k 2 k 2 + + 0 m 2 ẍ 2 c 2 c 2 ẋ 2 k 2 k 2 ] [ x1 x 2 ] = [ F1(t) F 2(t) Eq. (1) ]

State Space Equation Formulation for MDOF systems 2dof Mass Spring System or otherwise more compactly: Mx d + Cx d + Kx d = F where x d = [ x 1 x 2 ] T We now introduce the augmented state vector: x = [ x 1 x 2 ẋ 1 ẋ 2 ] T. Then, ẋ = 0 0 1 0 0 0 0 1 [ M 1 K ] [ M 1 C ] x 1 x 2 ẋ 1 ẋ 2 + 0 0 0 0 [ M 1 ] [ F1 F 2 ]

State Space Equation Formulation 2dof Mass Spring System We obtain the following equivalent 1st order ODE system where u = [ F 1 F 2 ] T ẋ = Ax + Bu Assume you would like to monitor (measure) both displacements x 1, x 2. Then the observation vector is: y = 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 x 1 x 2 x 3 x 4 + O 4 2 u(t) y = Cx + Du

State Space Equation Formulation Note Using the state space representation we have converted a 2nd order ODE into an equivalent 1st order ODE system. We can now use any of the aforementioned 1st order ODE integration methods in order to convert the continuous system into a discrete one and obtain an approximate solution For instance MATLAB s ode45, which is a Runge Kutta integration scheme may be used. What are other integration schemes that may be utilized?

Numerical Integration for 1st order ODEs Using these methods a continuous system is brought into an equivalent discrete formulation and an approximative solution is sought. 1st order ODE Integration Methods Assume dy dt = f (t, y(t)), y(t 0) = 0 Forward Euler Method y n+1 = y n + hf (t n, y n ) where h is the integration time step. This explicit expression is obtained from the truncated Taylor Expansion of y(t n + h) Backward Euler Method y n+1 = y n + hf (t n+1, y n+1 ) This implicit expression (since y n+1 is on the right hand side) is obtained from the truncated Taylor Expansion of y(t n+1 h)

Numerical Integration for 1st order ODEs 2nd Order Runge Kutta (RK2) k 1 = hf (t n, y n ), k 2 = hf (t n + 1 2 h, y n + 1 2 k 1) y n+1 = y n + k 2 + O(h 3 ) 4th Order Runge Kutta (RK4) - MATLAB ode45 funcction k 1 = hf (t n, y n ), k 2 = hf (t n + 1 2 h, y n + 1 2 k 1) k 3 = hf (t n + 1 2 h, y n + 1 2 k 2), k 4 = hf (t n + h, y n + k 3 ) y n+1 = y n + 1 6 k 1 + 1 3 k 2 + 1 3 k 3 + 1 6 k 4 + O(h 5 )

Multiple DOF Systems The Eigenvalue Problem The state-space formulation provides an easy way for solving the equations of motion and is a method that is directly implementable in the time domain. An alternative however exists for solving the problem and for additionally extracting significant information on the properties of the system in the frequency domain, meaning its natural frequencies, modal shapes and damping characteristics.

Multiple DOF Systems The Eigenvalue Problem The system of equations (1) is coupled in the sense that each equation involves more than one coordinate (DOF) furthermore, it can be rewritten in matrix - vector notation as: Mẍ + Cẋ + Kx = F(t) k For an SDOF system the natural frequency is obtained as ω n = m. In a similar manner for an n-dof system we consider the following: The undamped - free vibration case can be written as: Mẍ + Kx = 0 n n

Multiple DOF Systems The Eigenvalue Problem As for the SDOF system, the general solution for this 2nd order homogeneous (free response) problem is of the following type (harmonic): [ x1 x 2 ] = [ X1 X 2 ] e iωt We can plug this in the homogeneous differential equation: { [ ] [ ]} [ ] [ ω 2 m1 0 k1 + k + 2 k 2 X1 e iωt 0 = 0 m 2 k 2 k 2 X 2 0 ]

Multiple DOF Systems The Eigenvalue Problem For x 1 = x 2 = 0 we obtain the trivial solution x(t) = 0. So, instead we demand: { [ ] [ ]} [ ] [ ] ω 2 m1 0 k1 + k + 2 k 2 X1 0 = 0 m 2 k 2 k 2 X 2 0 From det(k Mω 2 ) = 0 we obtain the eigenfrequencies (or natural frequencies) ω 1,2. Then for ω = ω 1 we have that: [ ω 2 1m 1 + (k 1 + k 2 )]X 1 k 2 X 2 = 0 k 2 X 1 + [ ω 2 1m 2 + (k 2 + k 3 )]X 2 = 0 The two equations are linearly dependent, hence we solve one to finally obtain: [ ] [ ] X1 1 k 2 = ω1 2m X 1 X 1 2 φ 1 = 1 τ1 1 + (k 1 + k 2 ) X 1 = X 2 1 1 1 1st eigenvector }{{} τ 1

Multiple DOF Systems The Eigenvalue Problem Similarly we solve for ω = ω 2 to obtain the 2nd eigenvector: [ ] τ2 φ 2 = X1 2 1 The total solution occurs through the superposition of the two modes: [ ] x1 (t) = C x 2 (t) 1 φ 1 e iω1t + C 1 φ 2 e iω 2t Example: for m 1 = 2m, m 2 + m, k 1 = k 2 = k 1 st mode 2 nd mode 2 2 k 1 F 1 ( t ) k 2 m 1 1 (out of (in phase) phase) Institute of Structural FBD Engineering F 1 ( t ) Identification Methods for Structural F 2Systems ( t ) 1 c 1 x ( t 1 ) c 1

Properties of Eigenvectors The previously described process is known as: Solution of the Eigenproblem ω 2 Mφ + Kφ = 0 (eigenproblem) The above homogeneous linear system of equations can only have a solution if the determinant is equal to 0: ω 2 M + K = 0 The solution of the above equation will yield: n eigenvalues ωi 2, i = 1,..., n with 0 ω 1...ω n (the eigenfrequencies) and The solution of ω 2 Mφ = Kφ (eigenproblem) will yield: n eigenvectors (or modal vectors) φ i

Properties of Eigenvectors M-Orthonormality From Matrix Properties we know that (AB) T = B T A T, Thus, for two eigenvectors φ n, φ r we obtain: (φ T nkφ r ) T = φ T r K T φ n Since the Stiffness matrix is symmetric, K = K T, hence (φ T nkφ r ) T = φ T r Kφ n The same applies for the Mass matrix, M = M T, yielding: (φ T nmφ r ) T = φ T r Mφ n

Properties of Eigenvectors M-Orthonormality Then eigenproblem for vector r can then be written as: (multiply with φ T n on the left) ωr 2 φ T nmφ r = φ T transpose nkφ r ω 2 r φ T r Mφ n = φ T r Kφ n and the eigenproblem for vector n is: (multiply with φ T r on the left) ω 2 nφ T r Mφ n = φ T r Kφ n By subtracting the two previous formulas we obtain: (ω 2 r ω 2 n)φ T r Mφ n = 0 We conclude that for n r φ T r Mφ n = 0, φ T r Mφ n = 0,

Properties of Eigenvectors M-Orthonormality We choose φ n, such that: Φ T MΦ = I therefore from MΦΩ 2 = KΦ we have that: Φ T KΦ = Ω 2 The principle of M-orthonormality can then be written as: { 1, n = r φ T r Mφ n = 0, n r { ω 2 φ T r Kφ n = n, n = r 0, n r

Mode Superposition Method Matrix Form Defining a matrix Φ whose columns are the eigenvectors φ i and a diagonal matrix Ω 2 which stores the eigenvalues ωi 2 on its diagonal, i.e: ω 1 Φ = [ ] φ 1, φ 2,... φ n ; Ω 2 = ω 2... ; we can write the n solutions to the eigenproblem as: MΦΩ 2 = KΦ ω n

Mode Superposition Method Decoupling the Equation of Motion Assume a regular coupled system where we note the displacement vector as U(t) and therefore we can write the following equation of motion: MÜ + C U + KU = F(t) We can now use Φ as a transformation matrix by defining: U(t) = ΦX(t) Which leads to the transformed equilibrium equation: Ẍ(t) + Φ T CΦẊ(t) + Ω 2 X(t) = Φ T R(t) where using the property of M-orthonormality, the initial conditions will be: 0 X = Φ T M 0 U; 0 Ẋ = Φ T M 0 U

Dynamic Response with Damping Neglected If we neglect the velocity dependent damping effects the equilibrium equation reduces to: Ẍ (t) + Ω 2 X(t) = Φ T R(t) i.e n individual equations of the form (since Ω 2 is diagonal): ẍ i (t) + ωi 2x } i(t) = r i (t) r i (t) = φ T i R(t) i = 1, 2,..., n with 0 x i = φ T i M 0 U; 0ẋ i = φ T i M 0 U

Dynamic Response with Damping Neglected Individual SDOF system response Each equation of the previous system describes a single degree of freedom system with unit mass and stiffness ωi 2. The solution to this equation for a random input excitation can either be obtained by using the SDOF integration methods (like direct integration) or by using the Duhamel Integral: x i (t) = 1 t r i (τ)sinω i (t τ) dτ + α i sinω i t + β i cosω i t ω i 0 where α i, β i are determined from the initial conditions. Therefore, the SDOF response id owed to two contributions A dynamic (steady - state) response obtained by multiplying the static response by a dynamic load factor (this is the particular solution of the governing differential equation), and An additional dynamic response called the transient response

Dynamic Response with Damping Neglected Complete Response The solution of all n SDOF equations are calculated and the finite element nodal point displacements are obtained by superposition of the response in each mode: U(t) = ΦX(t) U(t) = n φ i x i (t) i=1 Therefore the solution scheme is: Solve for the eigenvalues and eigenvectors of problem Solve for the response of the decoupled SDOF equations Use superposition to find the total response.

Superposition Principle An alternative View Superposition A dynamic load can be designed as Fourier series of harmonic sine and cosine contributions The total solution of such a problem is equal to the superposition of solution of the Fourier terms. A dynamic load can be designed as Fourier series The total solution of such a problem is equal to the superposition of solution of the Fourier terms. U(t) = ΦX(t) U(t) = n φ i x i (t) i=1 13.11.2009 Mode Superposition 11

Swiss Federal Institute of Technology Example - 2 DOF system Direct Integration ti Methods Calculate the displacement response of the system Pag k 1 = 4 U, U, U 1 1 1 m = 2 R = 0 1 2 k 2 = 2 1 0 m 2 = 1 R 2 = 10 2 0 U 6 2 1 U1 0 + = 0 1 U 2 2 4 U 2 10 U, U, U 2 2 2 k 3 = 2 Method of Finite Elements II

Example - 2 DOF system Eigenproblem Setup ω 2 nmφ n = Kφ n ω 2 n Eigenvalue Calculation [ 2 0 0 1 ] = [ 6 2 2 4 ] φ n K ω 2 n M 6 2ω 2 = 0 n 2 2 4 1ωn 2 = 2ω4 n 14ωn 2 + 20 = 0 ω1 2 = 14 196 160 4 Eigenvector Calculation = 2, ω 2 2 = 14 + 196 160 4 = 5 ( K ω 2 1 M ) φ 1 = 0 [ 2 2 2 2 ] [ φ11 φ 12 ] = 0 [ φ11 φ 12 ] [ 1 = λ 1 ]

Example - 2 DOF system Use of M - orthonormality φ T 1Mφ 1 = 1 λ 2 (m 1 + m 2 ) = 1 λ = 1 3 Hence, φ 1 = 1 3 1 3 Similarly, φ 2 = 2 2 3 2 3 Transformed Equilibrium Equation [ 2, 0 Ẍ(t) + 0 5 Ẍ(t) + Ω 2 X(t) = Φ T R(t) ] X(t) = 1 1 3 3 2 2 2 3 3 [ 0 10 ]

Example - 2 DOF system Decoupled SDOF equations Initial Conditions ẍ 1 + 2x 1 = 10 3 U(0) = 0 U(0) = 0 and 2 ẍ 2 + 5x 2 = 10 3 x i (0) = φ T i MU(0) ẋ i (0) = φ T i M U(0) x 1 (0) = 0, ẋ 1 (0) = 0, x 2 (0) = 0, ẋ 2 (0) = 0 Then the exact solutions to the ODEs are: x 1 = 5 (1 cos 2 2t) x 2 = 2 3 3 ( 1 + cos 5t) And the total solution is: U(t) = n φ i x i (t) i=1

Example - 2 DOF system Note In this case we dealt with a 2DOF example and used an analytical solution to solve the two decoupled SDOF system equations In practice for multi degree of freedom systems, higher modes are neglected and instead of an analytical solution we use a numerical scheme in order to solve for each SDOF ODE Neglecting higher modes offers the tremendous advantage of obtaining a Reduced Order System (ROM).

The effect of Damping Problems with neglected dampi The presence of damping reduces the dynamic load factor (which then cannot be infinite) and damps out the transient response The response in the modes with ˆω large is ω negligible For ˆω close to zero the ω system follows the loads statically Effectively only the first p modes need to be used p n, in order to obtain a good approximate solution. 13.11.2009 Mode Superposition

Analysis Including Damping Transformed Equilibrium Equation Ẍ(t) + Φ T CΦẊ(t) + Ω 2 X(t) = Φ T R(t) Proportional Damping Assumption φ T i Cφ j = 2ω i ξ i δ ij (1) where ξ i is a modal damping parameter and δ ij is the Kronecker delta (δ ij = 1 for i = j, δ ij = 0 for i j)

Analysis Including Damping Therefore, we still end up with Decoupled SDOF equations for each x i : ẍ i (t) + 2ω i ξ i ẋ i (t) + ω 2 i x i (t) = r i (t) with the Duhamel Integral now being: x i (t) = 1 ω t { r i (τ) e ξ i ω i (t τ) sin ω i (t τ) dτ i 0 } + e ξ i ω i t (α i sin ω i t + β i cos ω i t) where ω i = ω i 1 ξ 2 i

Analysis Including Damping Rayleigh Damping If there are only two different damping ratios ξ i, Damping can be used: i = 1, 2 Rayleigh C = αm + βk (2) Eqns (1), (2) now yield: φ T i (αm + βk)φ i = 2ω i ξ i α + βω 2 i = 2ω i ξ i The 2 2 system can be solved to obtain α, β. In actual analysis it may well be that the damping ratios are known for many more than two frequencies. In that case two average values say ξ 1, ξ 2 are used to evaluate α, β.

Example - Damping for an MDOF system Assume that the approximate damping to be specified for a multiple degree of freedom system is as follows:

Example - Damping for an MDOF system Damping as function of frequency A problem with Rayleigh damping is that higher modes are much more damped than lower modes. However, in general Rayleigh damping provides a good approximation

Identification Methods for Structural Systems. Prof. Dr. Eleni Chatzi Lecture March, 2016