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
Institute of Structural Engineering Identification Methods for Structural Systems 2
Fundamentals
Overview
Multiple DOF Systems
State-space Formulation
Eigenvalue Analysis
The Mode Superposition Method
The effect of Damping on Structural Response
Institute of Structural Engineering Identification Methods for Structural Systems 2
Fundamentals
Overview
Multiple DOF Systems
State-space Formulation
Eigenvalue Analysis
The Mode Superposition Method
The effect of Damping on Structural Response
Institute of Structural Engineering Identification Methods for Structural Systems 2
Fundamentals
Overview
Multiple DOF Systems
State-space Formulation
Eigenvalue Analysis
The Mode Superposition Method
The effect of Damping on Structural Response
Institute of Structural Engineering Identification Methods for Structural Systems 2
Fundamentals
Overview
Multiple DOF Systems
State-space Formulation
Eigenvalue Analysis
The Mode Superposition Method
The effect of Damping on Structural Response
Institute of Structural Engineering Identification Methods for Structural Systems 2
Multiple DOF Systems
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)
Institute of Structural Engineering Identification Methods for Structural Systems 3
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
]
Institute of Structural Engineering Identification Methods for Structural Systems 4
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
Institute of Structural Engineering Identification Methods for Structural Systems 5
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?
Institute of Structural Engineering Identification Methods for Structural Systems 6
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)
Institute of Structural Engineering Identification Methods for Structural Systems 7
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)
Institute of Structural Engineering Identification Methods for Structural Systems 8
Multiple DOF Systems
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.
Institute of Structural Engineering Identification Methods for Structural Systems 9
Multiple DOF Systems
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.
Institute of Structural Engineering Identification Methods for Structural Systems 9
Multiple DOF Systems
The Eigenvalue Problem
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
Institute of Structural Engineering Identification Methods for Structural Systems 10
Multiple DOF Systems
The Eigenvalue Problem
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
]
Institute of Structural Engineering Identification Methods for Structural Systems 11
Multiple DOF Systems
The Eigenvalue Problem
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
Institute of Structural Engineering Identification Methods for Structural Systems 12
Multiple DOF Systems
The Eigenvalue Problem
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)
Institute of Structural Engineering Identification Methods for Structural Systems 13
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
Institute of Structural Engineering Identification Methods for Structural Systems 14
Properties of Eigenvectors
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
Institute of Structural Engineering Identification Methods for Structural Systems 15
Properties of Eigenvectors
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,
Institute of Structural Engineering Identification Methods for Structural Systems 16
Properties of Eigenvectors
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
Institute of Structural Engineering Identification Methods for Structural Systems 17
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
Institute of Structural Engineering Identification Methods for Structural Systems 18
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
Institute of Structural Engineering Identification Methods for Structural Systems 19
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
Institute of Structural Engineering Identification Methods for Structural Systems 20
Dynamic Response with Damping Neglected
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
Institute of Structural Engineering Identification Methods for Structural Systems 21
Dynamic Response with Damping Neglected
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.
Institute of Structural Engineering Identification Methods for Structural Systems 22
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.
Institute of Structural Engineering Identification Methods for Structural Systems 23
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
Institute of Structural Engineering Identification Methods for Structural Systems 24
Example - 2 DOF system
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
Example - 2 DOF system
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
]
Institute of Structural Engineering Identification Methods for Structural Systems 26
Example - 2 DOF system
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)
Institute of Structural Engineering Identification Methods for Structural Systems 27
Example - 2 DOF system
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).
Institute of Structural Engineering Identification Methods for Structural Systems 28
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.
Institute of Structural Engineering Identification Methods for Structural Systems 29
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)
Institute of Structural Engineering Identification Methods for Structural Systems 30
Analysis Including Damping
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
Institute of Structural Engineering Identification Methods for Structural Systems 31
Analysis Including Damping
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.
Institute of Structural Engineering Identification Methods for Structural Systems 32
Example - Damping for an MDOF system
Assume that the approximate damping to be specified for a multiple degree of
freedom system is as follows:
Institute of Structural Engineering Identification Methods for Structural Systems 33
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
Institute of Structural Engineering Identification Methods for Structural Systems 34