Identification Methods for Structural Systems · ODE into an equivalent 1st order ODE system. We...

Identification Methods for Structural Systems

Prof. Dr. Eleni Chatzi

Lecture 4 - 09. March, 2016

Institute of Structural Engineering Identification Methods for Structural Systems 1

Fundamentals

Overview

Multiple DOF Systems

State-space Formulation

Eigenvalue Analysis

The Mode Superposition Method

The effect of Damping on Structural Response



Let’s revisit the simple 2DOF system

1m1 1xk

1 1c x

( )1 tF

2m

( )2 2 1x xk −

( )2 2 1x xc −

( )2 tF

( )2 2 1x xk −

( )2 2 1x xc −

FBD

(Lumped Mass System)

1m

1k

1c( )1 tx

( )1 tF

2m

2k

1c( )2 tx

( )2 tF

The equations of motion can be written as

m1x1 + (c1 + c2)x1 − c2x2 + (k1 + k2)x1 − k2x2 = F1(t)

m2x2 + c2x2 − c2x1 + k2x2 − k2x1 = F2(t)

The system can be written in matrix form as follows:[m1 00 m2

] [x1

x2

]+

[c1 + c2 −c2

−c2 c2

] [x1

x2

]+

[k1 + k2 −k2

−k2 k2

] [x1

x2

]=

[F1(t)F2(t)

]Eq. (1)


State Space Equation Formulation for MDOF systems

2dof Mass Spring System

or otherwise more compactly:

Mxd + Cxd + Kxd = F

where xd =[x1 x2

]TWe now introduce the augmented state vector:

x =[x1 x2 x1 x2

]T. Then,

x =

0 0 1 00 0 0 1[−M−1K

] [−M−1C

]

x1

x2

x1

x2

+

0 00 0[M−1

][ F1

F2

]


State Space Equation Formulation

2dof Mass Spring System

We obtain the following equivalent 1st order ODE system

⇒ x = Ax + Bu

where u =[F1 F2

]TAssume you would like to monitor (measure) both displacements x1,x2. Then the “observation vector” is:

y =

1 0 0 00 1 0 00 0 0 00 0 0 0

x1

x2

x3

x4

+O4×2u(t)

⇒ y = Cx + Du


State Space Equation Formulation

Note

Using the state space representation we have converted a 2nd orderODE into an equivalent 1st order ODE system.

We can now use any of the aforementioned 1st order ODEintegration methods in order to convert the continuous system intoa discrete one and obtain an approximate solution

For instance MATLAB’s ode45, which is a Runge Kutta integrationscheme 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 equivalentdiscrete formulation and an approximative solution is sought. 1st orderODE Integration Methods

Assumedy

dt= f (t, y(t)), y(t0) = 0

Forward Euler Method

yn+1 = yn + hf (tn, yn)

where h is the integration time step. This explicit expression isobtained from the truncated Taylor Expansion of y(tn + h)

Backward Euler Method

yn+1 = yn + hf (tn+1,yn+1)

This implicit expression (since yn+1 is on the right hand side) isobtained from the truncated Taylor Expansion of y(tn+1 − h)


Numerical Integration for 1st order ODEs

2nd Order Runge Kutta (RK2)

k1 = hf (tn, yn), k2 = hf (tn +1

2h, yn +

1

2k1)

yn+1 = yn + k2 + O(h3)

4th Order Runge Kutta (RK4) - MATLAB ode45funcction

k1 = hf (tn, yn), k2 = hf (tn +1

2h, yn +

1

2k1)

k3 = hf (tn +1

2h, yn +

1

2k2), k4 = hf (tn + h, yn + k3)

yn+1 = yn +1

6k1 +

1

3k2 +

1

3k3 +

1

6k4 + O(h5)



The Eigenvalue Problem

The state-space formulation provides an easy way for solving theequations of motion and is a method that is directly implementablein the time domain.

An alternative however exists for solving the problem and foradditionally extracting significant information on the properties ofthe system in the frequency domain, meaning its naturalfrequencies, modal shapes and damping characteristics.




The system of equations (1) is coupled in the sense that eachequation involves more than one coordinate (DOF)

furthermore, it can be rewritten in matrix - vector notation as:

Mx + Cx + Kx = F(t)

For an SDOF system the natural frequency is obtained as ωn =

√k

m.

In a similar manner for an n-DOF system we consider the following:The undamped - free vibration case can be written as:

Mx + Kx = 0n×n




As for the SDOF system, the general solution for this 2nd orderhomogeneous (free response) problem is of the following type(harmonic):

[x1

x2

]=

[X1

X2

]e iωt

We can plug this in the homogeneous differential equation:−ω2

[m1 00 m2

]+

[k1 + k2 −k2

−k2 k2

][X1

X2

]e iωt =

[00

]




For x1 = x2 = 0 we obtain the trivial solution x(t) = 0. So, insteadwe demand:

−ω2

[m1 00 m2

]+

[k1 + k2 −k2

−k2 k2

][X1

X2

]=

[00

]From det(K−Mω2) = 0 we obtain the eigenfrequencies (or naturalfrequencies) ω1,2. Then for ω = ω1 we have that:

[−ω21m1 + (k1 + k2)]X1 − k2X2 = 0

− k2X1 + [−ω21m2 + (k2 + k3)]X2 = 0

The two equations are linearly dependent, hence we solve one tofinally obtain:

X 11 =

k2

−ω21m1 + (k1 + k2)︸︷︷︸

τ1

X 12 ⇒ φ1 =

[X 1

1

X 12

]=

[τ1

1

]X 1

1 1st eigenvector




Similarly we solve for ω = ω2 to obtain the 2nd eigenvector:

φ2 =

[τ2

1

]X 2

1

The total solution occurs through the superposition of the twomodes: [

x1(t)x2(t)

]= C1φ1e

iω1t + C1φ2eiω2t

Example: for m1 = 2m, m2 + m, k1 = k2 = k

1m1 1xk

1 1c x

( )1 tF

2m

( )2 2 1x xk −

( )2 2 1x xc −

( )2 tF

( )2 2 1x xk −

( )2 2 1x xc −

FBD

(Lumped Mass System)

1m

1k

1c( )1 tx

( )1 tF

2m

2k

1c( )2 tx

( )2 tF

2

1

2−

1

1st mode 2nd mode

(in phase) (out of phase)


Properties of Eigenvectors

The previously described process is known as:

Solution of the Eigenproblem

−ω2Mφ + Kφ = 0 (eigenproblem)

The above homogeneous linear system of equations can only have asolution if the determinant is equal to 0:∥∥−ω2M + K

∥∥ = 0

The solution of the above equation will yield:n eigenvalues ω2

i , i = 1, ..., nwith 0 ≤ ω1 ≤ ...ωn (the eigenfrequencies) and

The solution of ω2Mφ = Kφ (eigenproblem) will yield:n eigenvectors (or modal vectors) φi



M-Orthonormality

From Matrix Properties we know that (AB)T = BTAT,

Thus, for two eigenvectors φn, φr we obtain:

(φTnKφr )T = φT

r KTφn

Since the Stiffness matrix is symmetric, K = KT, hence

(φTnKφr )T = φT

r Kφn

The same applies for the Mass matrix, M = MT, yielding:

(φTnMφr )T = φT

r Mφn



M-Orthonormality

Then eigenproblem for vector r can then be written as: (multiplywith φT

n on the left)

ω2rφ

TnMφr = φT

nKφrtranspose−→ ω2

rφTr Mφn = φT

r Kφn

and the eigenproblem for vector n is: (multiply with φTr on the left)

ω2nφ

Tr Mφn = φT

r Kφn

By subtracting the two previous formulas we obtain:

(ω2r − ω2

n)φTr Mφn = 0

We conclude that for n 6= r ⇒ φTr Mφn = 0, φT

r Mφn = 0,



M-Orthonormality

We choose φn, such that:

ΦTMΦ = I

therefore from MΦΩ2 = KΦ we have that:

ΦTKΦ = Ω2

The principle of M-orthonormality can then be written as:

φTr Mφn =

1, n = r0, n 6= r

φTr Kφn =

ω2n, n = r

0, n 6= r


Mode Superposition Method

Matrix Form

Defining a matrix Φ whose columns are the eigenvectors φi and adiagonal matrix Ω2 which stores the eigenvalues ω2

i on its diagonal,i.e:

Φ =[φ1, φ2, ... φn

]; Ω2 =

ω1

ω2

...ωn

;

we can write the n solutions to the eigenproblem as:

MΦΩ2 = KΦ


Mode Superposition Method

Decoupling the Equation of Motion

Assume a regular coupled system where we note the displacementvector as U(t) and therefore we can write the following equation ofmotion:

MU + CU + 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:

X(t) + ΦTCΦX(t) + Ω2X(t) = ΦTR(t)

where using the property of M-orthonormality, the initial conditionswill be:

0X = ΦTM0U; 0X = ΦTM0U


Dynamic Response with Damping Neglected

If we neglect the velocity dependent damping effects the equilibriumequation reduces to:

X (t) + Ω2X(t) = ΦTR(t)

i.e n individual equations of the form (since Ω2 is diagonal):

xi (t) + ω2i xi (t) = ri (t)

ri (t) = φTi R(t)

i = 1, 2, ..., n

with

0xi = φTi M0U; 0xi = φT

i M0U



Individual SDOF system response

Each equation of the previous system describes a single degree of freedomsystem with unit mass and stiffness ω2

i . The solution to this equation for arandom input excitation can either be obtained by using the SDOFintegration methods (like direct integration) or by using the DuhamelIntegral:

xi (t) =1

ωi

∫ t

0

ri (τ)sinωi (t − τ) dτ + αi sinωi t + βicosωi t

where αi , βi are determined from the initial conditions. Therefore, theSDOF response id owed to two contributions

A dynamic (steady - state) response obtained by multiplying the staticresponse by a dynamic load factor (this is the particular solution ofthe governing differential equation), and

An additional dynamic response called the transient response



Complete Response

The solution of all n SDOF equations are calculated and the finiteelement nodal point displacements are obtained by superposition ofthe response in each mode:

U(t) = ΦX(t)⇒ U(t) =n∑

i=1

φixi (t)

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

A dynamic load can be designed asFourier series of harmonic sine and cosinecontributionsThe total solution of such a problem isequal to the superposition of solution ofthe Fourier terms.

U(t) = ΦX(t)⇒ U(t) =n∑

i=1

φixi (t)

Superposition

13.11.2009 Mode Superposition 11

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.


Example - 2 DOF system

Calculate the displacement response of the system

Swiss Federal Institute of Technology Page 15

Di t I t ti M th dDirect Integration Methods

1 4 k =

2 0R

112 0 6 2 00 1 2 4 10

UUUU

−⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦⎣ ⎦

2 2 k =

1 2m =

1 1 1, , U U U

1 0 R =

220 1 2 4 10UU −⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦⎣ ⎦

2 1m = 2 10 R =

3 2 k =2 2 2, , U U U

Method of Finite Elements II



Eigenproblem Setup

ω2nMφn = Kφn ⇒ ω2

n

[2 00 1

]=

[6 −2−2 4

]φn

Eigenvalue Calculation

∥∥K− ω2nM∥∥ = 0⇒

∥∥∥∥ 6− 2ω2n −2

−2 4− 1ω2n

∥∥∥∥ = 2ω4n − 14ω2

n + 20 = 0⇒

ω21 =

14−√

196− 160

4= 2, ω2

2 =14 +

√196− 160

4= 5

Eigenvector Calculation

(K− ω2

1M)φ1 = 0⇒

[2 −2−2 2

] [φ11

φ12

]= 0⇒

[φ11

φ12

]= λ ∗

[11

]Institute of Structural Engineering Identification Methods for Structural Systems 25


Use of M - orthonormality

φT1Mφ1 = 1⇒ λ2(m1 + m2) = 1⇒ λ =

1

3

Hence, φ1 =

1√3

1√3

Similarly, φ2 =

√

2

2√

3

−√

2√3

Transformed Equilibrium Equation

X(t) + Ω2X(t) = ΦTR(t)⇒

X(t) +

[2, 00 5

]X(t) =

1√3

1√3√

2

2√

3−√

2√3

[ 010

]



Decoupled SDOF equations

x1 + 2x1 =10√

3x2 + 5x2 = −10

√2

3

Initial Conditions

U(0) = 0

U(0) = 0and

xi (0) = φTi MU(0)

xi (0) = φTi MU(0)

⇒

x1(0) = 0, x1(0) = 0, x2(0) = 0, x2(0) = 0

Then the exact solutions to the ODEs are:

x1 =5√3

(1− cos√

2t) x2 = 2

√2

3(−1 + cos

√5t)

And the total solution is:

U(t) =n∑

i=1

φixi (t)



Note

In this case we dealt with a 2DOF example and used an analyticalsolution to solve the two decoupled SDOF system equations

In practice for multi degree of freedom systems, higher modes areneglected and instead of an analytical solution we use a numericalscheme in order to solve for each SDOF ODE

Neglecting higher modes offers the tremendous advantage ofobtaining a Reduced Order System (ROM).


The effect of Damping

The presence of damping reduces the dynamic load factor (which thencannot be infinite) and damps out the transient response

The response in the

modes withω

ωlarge is

negligible

Forω

ωclose to zero the

system follows the loadsstatically

Problems with neglected damping

13.11.2009 Mode Superposition 12

Effectively only the first p modes need to be used p ≤ n, in order to obtaina good approximate solution.


Analysis Including Damping

Transformed Equilibrium Equation

X(t) + ΦTCΦX(t) + Ω2X(t) = ΦTR(t)

Proportional Damping Assumption

φTi Cφj = 2ωiξiδij (1)

where ξi is a modal damping parameter and δij is the Kronecker delta(δij = 1 for i = j , δij = 0 for i 6= j)



Therefore, we still end up with Decoupled SDOF equations foreach xi :

xi (t) + 2ωiξi xi (t) + ω2i xi (t) = ri (t)

with the Duhamel Integral now being:

xi (t) =1

ωi

∫ t

0

ri (τ) e−ξiωi (t−τ) sinωi (t − τ) dτ

+ e−ξiωi t(αi sinωi t + βicosωi t)

where ωi = ωi

√1− ξ2

i



Rayleigh Damping

If there are only two different damping ratios ξi , i = 1, 2 RayleighDamping can be used:

C = αM + βK (2)

Eqns (1), (2) now yield:

φTi (αM + βK)φi = 2ωiξi ⇒

α + βω2i = 2ωiξi

The 2× 2 system can be solved to obtain α, β.In actual analysis it may well be that the damping ratios are knownfor many more than two frequencies. In that case two average valuessay ξ1, ξ2 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 Rayleighdamping is that highermodes are much moredamped than lower modes.

However, in generalRayleigh damping providesa good approximation


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