Dynamics of structures with uncertainties 
S Adhikari 
College of Engineering, Swansea University, Swansea UK 
Email: S.Adhikari@swansea.ac.uk 
http://engweb.swan.ac.uk/ adhikaris/ 
Twitter: @ProfAdhikari 
The University of Campinas, Campinas, Brazil 
November 2014
Swansea University
Swansea University
About me 
Education: 
PhD (Engineering), 2001, University of Cambridge (Trinity College), 
Cambridge, UK. 
MSc (Structural Engineering), 1997, Indian Institute of Science, 
Bangalore, India. 
B. Eng, (Civil Engineering), 1995, Calcutta University, India. 
Work: 
04/2007-Present: Professor of Aerospace Engineering, Swansea 
University (Civil and Computational Engineering Research Centre). 
01/2003-03/2007: Lecturer in dynamics: Department of Aerospace 
Engineering, University of Bristol. 
11/2000-12/2002: Research Associate, Cambridge University 
Engineering Department (Junior Research Fellow, Fitzwilliam College, 
Cambridge) .
Outline of this talk 
1 Introduction 
2 Stochastic single degrees of freedom system 
3 Stochastic multi degree of freedom systems 
Stochastic finite element formulation 
Projection in the modal space 
Properties of the spectral functions 
4 Error minimization 
The Galerkin approach 
Model Reduction 
Computational method 
5 Numerical illustrations 
6 Conclusions
Mathematical models for dynamic systems 
Mathematical Models of Dynamic Systems 
❄ ❄ ❄ ❄ ❄ 
Linear 
Time-invariant 
Elastic 
Non-linear 
Time-variant 
Elasto-plastic 
Viscoelastic 
Continuous 
Discrete 
Deterministic 
Uncertain 
❄ 
❄ ❄ 
Single-degree-of-freedom 
(SDOF) 
✲ Probabilistic 
✲ 
✲ 
Multiple-degree-of-freedom 
(MDOF) 
✲ 
Fuzzy set 
Convex set 
❄ 
Low frequency 
Mid-frequency 
High frequency
A general overview of computational mechanics 
Real System
 
Input
 
( eg , earthquake, 
turbulence ) 
Measured output
 
( eg , velocity, 
acceleration , 
stress)
 
   
System Uncertainty
 
parametric uncertainty
 
model inadequacy
 
model uncertainty
 
calibration uncertainty
 
Physics based model
 
L
(u) = f 
( eg , ODE/ PDE / SDE / SPDE ) 
Input Uncertainty
 
uncertainty in time
 
history
 
uncertainty in
 location
 
Simulated Input
 
(time or frequency
 
domain)
 
Computation
 
( eg , FEM / BEM /Finite 
difference/ SFEM / MCS ) 
uncertain
 
experimental
 
error
 
calibration/updating
 
Computational
 
Uncertainty
 
machine precession,
 
error tolerance
 
‘
h
’ and ‘
p
’ refinements
 
Model output
 
( eg , velocity, 
acceleration , 
stress)
 
verification
 system identification
 
Total Uncertainty =
 
input + system +
 
computational
 
uncertainty
 
model validation
Uncertainty in structural dynamical systems 
Many structural dynamic systems are manufactured in a production line (nom-inally 
identical systems). On the other hand, some models are complex! Com-plex 
models can have ‘errors’ and/or ‘lack of knowledge’ in its formulation.
Model quality 
The quality of a model of a dynamic system depends on the following three 
factors: 
Fidelity to (experimental) data: 
The results obtained from a numerical or mathematical model undergoing 
a given excitation force should be close to the results obtained from the 
vibration testing of the same structure undergoing the same excitation. 
Robustness with respect to (random) errors: 
Errors in estimating the system parameters, boundary conditions and 
dynamic loads are unavoidable in practice. The output of the model 
should not be very sensitive to such errors. 
Predictive capability: 
In general it is not possible to experimentally validate a model over the 
entire domain of its scope of application. The model should predict the 
response well beyond its validation domain.
Sources of uncertainty 
Different sources of uncertainties in the modeling and simulation of dynamic 
systems may be attributed, but not limited, to the following factors: 
Mathematical models: equations (linear, non-linear), geometry, damping 
model (viscous, non-viscous, fractional derivative), boundary 
conditions/initial conditions, input forces. 
Model parameters: Young’s modulus, mass density, Poisson’s ratio, 
damping model parameters (damping coefficient, relaxation modulus, 
fractional derivative order). 
Numerical algorithms: weak formulations, discretisation of displacement 
fields (in finite element method), discretisation of stochastic fields (in 
stochastic finite element method), approximate solution algorithms, 
truncation and roundoff errors, tolerances in the optimization and iterative 
methods, artificial intelligent (AI) method (choice of neural networks). 
Measurements: noise, resolution (number of sensors and actuators), 
experimental hardware, excitation method (nature of shakers and 
hammers), excitation and measurement point, data processing 
(amplification, number of data points, FFT), calibration.
Few general questions 
How does system uncertainty impact the dynamic response? Does it 
matter? 
What is the underlying physics? 
How can we model uncertainty in dynamic systems? Do we ‘know’ the 
uncertainties? 
How can we efficiently quantify uncertainty in the dynamic response for 
large multi degrees of freedom systems? 
What about using ‘black box’ type response surface methods? 
Can we use modal analysis for stochastic systems? Does stochastic 
systems has natural frequencies and mode shapes?
Stochastic SDOF systems 
k
 
 
m
 
u( t ) 
f ( t ) 
f
d
( t ) 
Consider a normalised single degree of freedom system (SDOF): 
u¨(t) + 2ζωn u˙ (t) + ω2 
n u(t) = f (t)/m (1) 
Here ωn = 
p 
k/m is the natural frequency and ξ = c/2√km is the damping 
ratio. 
We are interested in understanding the motion when the natural 
frequency of the system is perturbed in a stochastic manner. 
Stochastic perturbation can represent statistical scatter of measured 
values or a lack of knowledge regarding the natural frequency.
Frequency variability 
4 
3.5 
3 
2.5 
2 
1.5 
1 
0.5 
0 
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 
x 
(x) 
p 
x 
uniform 
normal 
log−normal 
(a) Pdf: a = 0.1 
4 
3.5 
3 
2.5 
2 
1.5 
1 
0.5 
0 
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 
x 
(x) 
p 
x 
uniform 
normal 
log−normal 
(b) Pdf: a = 0.2 
Figure: We assume that the mean of r is 1 and the standard deviation is a. 
Suppose the natural frequency is expressed as ω2 
n = ω2 
n0 
r , where ωn0 is 
deterministic frequency and r is a random variable with a given 
probability distribution function. 
Several probability distribution function can be used. 
We use uniform, normal and lognormal distribution.
Frequency samples 
1000 
900 
800 
700 
600 
500 
400 
300 
200 
100 
0 
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 
Frequency: wn 
Samples 
uniform 
normal 
log−normal 
(a) Frequencies: a = 0.1 
1000 
900 
800 
700 
600 
500 
400 
300 
200 
100 
0 
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 
Frequency: wn 
Samples 
uniform 
normal 
log−normal 
(b) Frequencies: a = 0.2 
Figure: 1000 sample realisations of the frequencies for the three distributions
Response in the time domain 
1 
0.8 
0.6 
0.4 
0.2 
0 
−0.2 
−0.4 
−0.6 
−0.8 
−1 
0 5 10 15 
Normalised time: t/T 
n0 
0 
Normalised amplitude: u/v 
deterministic 
random samples 
mean: uniform 
mean: normal 
mean: log−normal 
(a) Response: a = 0.1 
1 
0.8 
0.6 
0.4 
0.2 
0 
−0.2 
−0.4 
−0.6 
−0.8 
−1 
0 5 10 15 
Normalised time: t/T 
n0 
0 
Normalised amplitude: u/v 
deterministic 
random samples 
mean: uniform 
mean: normal 
mean: log−normal 
(b) Response: a = 0.2 
Figure: Response due to initial velocity v0 with 5% damping
Frequency response function 
120 
100 
80 
60 
40 
20 
0 
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 
Normalised frequency: w/wn0 
|2 
st 
Normalised amplitude: |u/u 
deterministic 
mean: uniform 
mean: normal 
mean: log−normal 
(a) Response: a = 0.1 
120 
100 
80 
60 
40 
20 
0 
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 
Normalised frequency: w/wn0 
|2 
st 
Normalised amplitude: |u/u 
deterministic 
mean: uniform 
mean: normal 
mean: log−normal 
(b) Response: a = 0.2 
Figure: Normalised frequency response function |u/ust |2, where ust = f /k
Key observations 
The mean response is more damped compared to deterministic 
response. 
The higher the randomness, the higher the “effective damping”. 
The qualitative features are almost independent of the distribution the 
random natural frequency. 
We often use averaging to obtain more reliable experimental results - is it 
always true? 
Assuming uniform random variable, we aim to explain some of these 
observations.
Equivalent damping 
Assume that the random natural frequencies are ω2 
n = ω2 
n0(1 + ǫx), where 
x has zero mean and unit standard deviation. 
The normalised harmonic response in the frequency domain 
u(iω) 
f /k 
= 
k/m 
n0 (1 + ǫx)] + 2iξωωn0√1 + ǫx 
[−ω2 + ω2 
(2) 
Considering ωn0 = 
p 
k/m and frequency ratio r = ω/ωn0 we have 
u 
f /k 
= 
1 
[(1 + ǫx) − r 2] + 2iξr√1 + ǫx 
(3)
Equivalent damping 
The squared-amplitude of the normalised dynamic response at ω = ωn0 
(that is r = 1) can be obtained as 
ˆU 
= 
 
|u| 
f /k 
2 
= 
1 
ǫ2x2 + 4ξ2(1 + ǫx) 
(4) 
Since x is zero mean unit standard √√deviation √uniform random variable, its 
pdf is given by px (x) = 1/23,− 
3 ≤ x ≤ 
3 
The mean is therefore 
E 
h 
ˆU 
i 
= 
Z 
1 
ǫ2x2 + 4ξ2(1 + ǫx) 
px (x)dx 
= 
1 
4√3ǫξ 
p 
1 − ξ2 
tan−1 
  √p 
3ǫ 
2ξ 
1 − ξ2 − 
ξ p 
1 − ξ2 
! 
+ 
1 
4√3ǫξ 
p 
1 − ξ2 
tan−1 
  √p 
3ǫ 
2ξ 
1 − ξ2 
+ 
ξ p 
1 − ξ2 
! 
(5)
Equivalent damping 
Note that 
1 
2 
 
tan−1(a + δ) + tan−1(a − δ) 
	 
= tan−1(a) + O(δ2) (6) 
Neglecting terms of the order O(ξ2) we have 
E 
h 
ˆU 
i 
≈ 
1 
2√3ǫξ 
p 
1 − ξ2 
tan−1 
  √p 
3ǫ 
2ξ 
1 − ξ2 
! 
= 
tan−1(√3ǫ/2ξ) 
2√3ǫξ 
(7)
Equivalent damping 
For small damping, the maximum deterministic amplitude at ω = ωn0 is 
1/4ξ2 
e where ξe is the equivalent damping for the mean response 
Therefore, the equivalent damping for the mean response is given by 
(2ξe)2 = 
2√3ǫξ 
tan−1(√3ǫ/2ξ) 
(8) 
For small damping, taking the limit we can obtain 
ξe ≈ 
31/4√ǫ 
√π 
p 
ξ (9) 
The equivalent damping factor of the mean system is proportional to the 
square root of the damping factor of the underlying baseline system
Equivalent frequency response function 
120 
100 
80 
60 
40 
20 
0 
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 
Normalised frequency: w/wn0 
|2 
st 
Normalised amplitude: |u/u 
Deterministic 
MCS Mean 
Equivalent 
(a) Response: a = 0.1 
120 
100 
80 
60 
40 
20 
0 
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 
Normalised frequency: w/wn0 
|2 
st 
Normalised amplitude: |u/u 
Deterministic 
MCS Mean 
Equivalent 
(b) Response: a = 0.2 
Figure: Normalised frequency response function with equivalent damping (e = 0.05 
in the ensembles). For the two cases e = 0.0643 and e = 0.0819 respectively.
Can we extend the ideas based on stochastic SDOF systems to stochastic 
MDOF systems?
Stochastic modal analysis 
Stochastic modal analysis to obtain the dynamic response needs further 
thoughts 
Consider the following 3DOF example: 
k
1
 k
4
 k
5
 k
3
 
m
1
 
m
2
 
m
3
 
k
2
 
k
6
 
Figure: A 3DOF system with parametric uncertainty in mi and ki
Statistical overlap 
2 4 6 8 10 12 
1000 
900 
800 
700 
600 
500 
400 
300 
200 
100 
Samples 
Eigenvalues, lj 
(rad/s)2 
l1 
l2 
l3 
(a) Eigenvalues are seperated 
1 2 3 4 5 6 7 8 
1000 
900 
800 
700 
600 
500 
400 
300 
200 
100 
Samples 
Eigenvalues, lj 
(rad/s)2 
l1 
l2 
l3 
(b) Some eigenvalues are close 
Figure: Scatter of the eigenvalues due to parametric uncertainties 
The SDOF based approach cannot be applied when there is statistical 
overlap in the eigenvalues.
Stochastic partial differential equation 
We consider a stochastic partial differential equation (PDE) for a linear 
dynamic system 
ρ(r, θ) 
∂2U(r, t , θ) 
∂t2 + Lα 
∂U(r, t , θ) 
∂t 
+ LβU(r, t , θ) = p(r, t) (10) 
The stochastic operator Lβ can be 
∂x AE(x, θ) ∂ 
∂x axial deformation of rods 
Lβ ≡ ∂ 
Lβ ≡ ∂2 
∂x2 EI(x, θ) ∂2 
∂x2 bending deformation of beams 
Lα denotes the stochastic damping, which is mostly proportional in nature. 
Here α, β : Rd ×  → R are stationary square integrable random fields, which 
can be viewed as a set of random variables indexed by r ∈ Rd . Based on the 
physical problem the random field a(r, θ) can be used to model different 
physical quantities (e.g., AE(x, θ), EI(x, θ)).
Discretisation of random fields 
The random process a(r, θ) can be expressed in a generalized Fourier 
type of series known as the Karhunen-Lo`eve expansion 
a(r, θ) = a0(r) + 
1X 
i=1 
√νi ξi (θ)ϕi (r) (11) 
Here a0(r) is the mean function, ξi (θ) are uncorrelated standard 
Gaussian random variables, νi and ϕi (r) are eigenvalues and 
eigenfunctions satisfying the integral equation 
Z 
D 
Ca(r1, r2)ϕj (r1)dr1 = νjϕj (r2), ∀ j = 1, 2, · · · (12)
Exponential autocorrelation function 
The autocorrelation function: 
C(x1, x2) = e−|x1−x2|/b (13) 
The underlying random process H(x, θ) can be expanded using the 
Karhunen-Lo`eve (KL) expansion in the interval −a ≤ x ≤ a as 
H(x, θ) = 
1X 
j=1 
p 
λjϕj (x) (14) 
ξj (θ) 
Using the notation c = 1/b, the corresponding eigenvalues and 
eigenfunctions for odd j and even j are given by 
λj = 
2c 
ω2 
j + c2 
, ϕj (x) = 
cos(ωjx) s 
a + 
sin(2ωja) 
2ωj 
, where tan(ωja) = 
c 
ωj 
, 
(15) 
λj = 
2c 
ωj 
2 + c2 , ϕj (x) = 
sin(ωjx) s 
a − 
sin(2ωja) 
2ωj 
, where tan(ωja) = 
ωj 
−c 
. 
(16)
KL expansion 
10−1 
10−2 
10−3 
0 5 10 15 20 25 30 35 
100 
Index, j 
Eigenvalues, lj 
b=L/2, N=10 
b=L/3, N=13 
b=L/4, N=16 
b=L/5, N=19 
b=L/10, N=34 
The eigenvalues of the Karhunen-Lo`eve expansion for different correlation 
lengths, b, and the number of terms, N, required to capture 90% of the infinite 
series. An exponential correlation function with unit domain (i.e., a = 1/2) is 
assumed for the numerical calculations. The values of N are obtained such 
that λN/λ1 = 0.1 for all correlation lengths. Only eigenvalues greater than λN 
are plotted.
Example: A beam with random properties 
The equation of motion of an undamped Euler-Bernoulli beam of length L with 
random bending stiffness and mass distribution: 
∂2 
∂x2 
 
EI(x, θ) 
∂2Y(x, t) 
∂x2 
 
+ ρA(x, θ) 
∂2Y(x, t) 
∂t2 = p(x, t). (17) 
Y(x, t): transverse flexural displacement, EI(x): flexural rigidity, ρA(x): mass 
per unit length, and p(x, t): applied forcing. Consider 
EI(x, θ) = EI0 (1 + ǫ1F1(x, θ)) (18) 
and ρA(x, θ) = ρA0 (1 + ǫ2F2(x, θ)) (19) 
The subscript 0 indicates the mean values, 0  ǫi  1 (i=1,2) are 
deterministic constants and the random fields Fi (x, θ) are taken to have zero 
mean, unit standard deviation and covariance Rij (ξ).
Random beam element 
1 3 
EI(x), m(x), c , c 1 2 
2 4 
l 
y 
x 
Random beam element in the local coordinate.
Realisations of the random field 
4 
3.5 
3 
2.5 
2 
1.5 
1 
0.5 
0 
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 
Length along the beam (m) 
EI (Nm2) 
baseline value 
perturbed values 
Some random realizations of the bending rigidity EI of the beam for 
correlation length b = L/3 and strength parameter ǫ1 = 0.2 (mean 2.0 × 105). 
Thirteen terms have been used in the KL expansion.
Example: A beam with random properties 
We express the shape functions for the finite element analysis of 
Euler-Bernoulli beams as 
N(x) =  s(x) (20) 
where 
 = 
 
 
1 0 −3 
ℓe 
2 
2 
ℓe 
3 
0 1 −2 
ℓe 
2 
1 
ℓe 
2 
0 0 
2 −2 
3 
ℓe 
ℓe 
3 
0 0 −1 
ℓe 
2 
1 
ℓe 
2 
 
 
and s(x) = 
 
1, x, x2, x3 T 
. (21) 
The element stiffness matrix: 
Ke(θ) = 
Z ℓe 
0 
N 
′′ 
(x)EI(x, θ)N 
′′T 
(x)dx = 
Z ℓe 
0 
EI0 (1 + ǫ1F1(x, θ))N 
′′ 
(x)N 
′′T 
(x)dx. 
(22)
Example: A beam with random properties 
Expanding the random field F1(x, θ) in KL expansion 
Ke(θ) = Ke0 +Ke(θ) (23) 
where the deterministic and random parts are 
Ke0 = EI0 
Z ℓe 
0 
′′ 
(x)N 
N 
′′T 
(x) dx and Ke(θ) = ǫ1 
XNK 
j=1 
p 
λKjKej . (24) 
ξKj (θ) 
The constant NK is the number of terms retained in the Karhunen-Lo`eve 
expansion and ξKj (θ) are uncorrelated Gaussian random variables with zero 
mean and unit standard deviation. The constant matrices Kej can be 
expressed as 
Kej = EI0 
Z ℓe 
0 
ϕKj (xe + x)N 
′′T 
(x) dx (25) 
′′ 
(x)N
Example: A beam with random properties 
The mass matrix can be obtained as 
Me(θ) = Me0 +Me(θ) (26) 
The deterministic and random parts is given by 
Me0 = ρA0 
Z ℓe 
0 
N(x)NT (x) dx and Me(θ) = ǫ2 
XNM 
j=1 
p 
λMjMej . (27) 
ξMj (θ) 
The constant NM is the number of terms retained in Karhunen-Lo`eve 
expansion and the constant matrices Mej can be expressed as 
Mej = ρA0 
Z ℓe 
0 
ϕMj (xe + x)N(x)NT (x) dx. (28) 
Both Kej and Mej can be obtained in closed-form.
Example: A beam with random properties 
These element matrices can be assembled to form the global random 
stiffness and mass matrices of the form 
K(θ) = K0 +K(θ) and M(θ) = M0 +M(θ). (29) 
Here the deterministic parts K0 and M0 are the usual global stiffness and 
mass matrices obtained form the conventional finite element method. The 
random parts can be expressed as 
K(θ) = ǫ1 
XNK 
j=1 
p 
λKjKj and M(θ) = ǫ2 
ξKj (θ) 
XNM 
j=1 
p 
λMjMj (30) 
ξMj (θ) 
The element matrices Kej and Mej can be assembled into the global matrices 
Kj and Mj . The total number of random variables depend on the number of 
terms used for the truncation of the infinite series. This in turn depends on the 
respective correlation lengths of the underlying random fields.
Stochastic equation of motion 
The equation for motion for stochastic linear MDOF dynamic systems: 
M(θ)u¨(θ, t) + C(θ)u˙ (θ, t) + K(θ)u(θ, t) = f(t) (31) 
M(θ) = M0 + 
Pp 
i=1 μi (θi )Mi ∈ Rn×n is the random mass matrix, 
K(θ) = K0 + 
Pp 
i=1 νi (θi )Ki ∈ Rn×n is the random stiffness matrix, 
C(θ) ∈ Rn×n as the random damping matrix and f(t) is the forcing vector 
The mass and stiffness matrices have been expressed in terms of their 
deterministic components (M0 and K0) and the corresponding random 
contributions (Mi and Ki ). These can be obtained from discretising 
stochastic fields with a finite number of random variables (μi (θi ) and 
νi (θi )) and their corresponding spatial basis functions. 
Proportional damping model is considered for which 
C(θ) = ζ1M(θ) + ζ2K(θ), where ζ1 and ζ2 are scalars.
Frequency domain representation 
For the harmonic analysis of the structural system, taking the Fourier 
transform  
 
−ω2M(θ) + iωC(θ) + K(θ) 
eu 
(ω, θ) =ef 
(ω) (32) 
whereeu 
(ω, θ) is the complex frequency domain system response 
amplitude,ef 
(ω) is the amplitude of the harmonic force. 
For convenience we group the random variables associated with the 
mass and stiffness matrices as 
ξi (θ) = μi (θ) and ξj+p1 (θ) = νj (θ) for i = 1, 2, . . . , p1 
and j = 1, 2, . . . , p2
Frequency domain representation 
Using M = p1 + p2 which we have 
  
A0(ω) + 
XM 
i=1 
! 
ξi (θ)Ai (ω) 
eu(ω, θ) =ef(ω) (33) 
where A0 and Ai ∈ Cn×n represent the complex deterministic and 
stochastic parts respectively of the mass, the stiffness and the damping 
matrices ensemble. 
For the case of proportional damping the matrices A0 and Ai can be 
written as 
A0(ω) = 
 
−ω2 + iωζ1 
 
M0 + [iωζ2 + 1]K0, (34) 
Ai (ω) = 
 
−ω2 + iωζ1 
 
Mi for i = 1, 2, . . . , p1 (35) 
and Aj+p1 (ω) = [iωζ2 + 1]Kj for j = 1, 2, . . . , p2 .
Time domain representation 
If the time steps are fixed to t , then the equation of motion can be written as 
M(θ) ¨ut+t (θ) + C(θ) ˙ut+t (θ) + K(θ)ut+t (θ) = pt+t . (36) 
Following the Newmark method based on constant average acceleration 
scheme, the above equations can be represented as 
[a0M(θ) + a1C(θ) + K(θ)] ut+t (θ) = peqv 
t+t (θ) (37) 
and, peqv 
t+t (θ) = pt+t + f (ut (θ), ˙ ut (θ), ¨ ut (θ),M(θ),C(θ)) (38) 
where peqv 
t+t (θ) is the equivalent force at time t + t which consists of 
contributions of the system response at the previous time step.
Newmark’s method 
The expressions for the velocities ˙ut+t (θ) and accelerations ¨ut+t (θ) at each 
time step is a linear combination of the values of the system response at 
previous time steps (Newmark method) as 
¨ut+t (θ) = a0 [ut+t (θ) − ut (θ)] − a2 ˙ut (θ) − a3 ¨ut (θ) (39) 
and, ˙ut+t (θ) = ˙ut (θ) + a6 ¨ut (θ) + a7 ¨ut+t (θ) (40) 
where the integration constants ai , i = 1, 2, . . . , 7 are independent of system 
properties and depends only on the chosen time step and some constants: 
a0 = 
1 
αt2 ; a1 = 
δ 
αt 
; a2 = 
1 
αt 
; a3 = 
1 
2α − 1; (41) 
a4 = 
δ 
α − 1; a5 = 
t 
2 
 
δ 
α − 2 
 
; a6 = t(1 − δ); a7 = δt (42)
Newmark’s method 
Following this development, the linear structural system in (37) can be 
expressed as  
A0 + 
XM 
i=1 
ξi (θ)Ai 
# 
| {z } 
A(θ) 
ut+t (θ) = peqv 
t+t (θ). (43) 
where A0 and Ai represent the deterministic and stochastic parts of the 
system matrices respectively. For the case of proportional damping, the 
matrices A0 and Ai can be written similar to the case of frequency domain as 
A0 = [a0 + a1ζ1]M0 + [a1ζ2 + 1]K0 (44) 
and, Ai = [a0 + a1ζ1]Mi for i = 1, 2, . . . , p1 (45) 
= [a1ζ2 + 1]Ki for i = p1 + 1, p1 + 2, . . . , p1 + p2 .
General mathematical representation 
Whether time-domain or frequency domain methods were used, in 
general the main equation which need to be solved can be expressed as 
  
A0 + 
XM 
i=1 
ξi (θ)Ai 
! 
u(θ) = f(θ) (46) 
where A0 and Ai represent the deterministic and stochastic parts of the 
system matrices respectively. These can be real or complex matrices. 
Generic response surface based methods have been used in literature - 
for example the Polynomial Chaos Method
Polynomial Chaos expansion 
After the finite truncation, the polynomial chaos expansion can be written as 
u(θ) = 
XP 
k=1 
Hk ((θ))uk (47) 
where Hk ((θ)) are the polynomial chaoses. We need to solve a nP × nP 
linear equation to obtain all uk ∈ Rn. 
 
 
A0,0 · · · A0,P−1 
A1,0 · · · A1,P−1 
... 
... 
... 
AP−1,0 · · · AP−1,P−1 
 
 
 
 
u0 
u1 
... 
uP−1 
 
 
= 
 
 
f0 
f1 
... 
fP−1 
 
 
(48) 
The number of terms P increases exponentially with M: 
M 2 3 5 10 20 50 100 
2nd order PC 5 9 20 65 230 1325 5150 
3rd order PC 9 19 55 285 1770 23425 176850
Some Observations 
The basis is a function of the pdf of the random variables only. For 
example, Hermite polynomials for Gaussian pdf, Legender’s polynomials 
for uniform pdf. 
The physics of the underlying problem (static, dynamic, heat conduction, 
transients....) cannot be incorporated in the basis. 
For an n-dimensional output vector, the number of terms in the projection 
can be more than n (depends on the number of random variables). This 
implies that many of the vectors uk are linearly dependent. 
The physical interpretation of the coefficient vectors uk is not immediately 
obvious. 
The functional form of the response is a pure polynomial in random 
variables.
Possibilities of solution types 
As an example, consider the frequency domain response vector of the 
stochastic system  u(ω, θ) governed by 
−ω2M((θ)) + iωC((θ)) + K((θ)) 
 
u(ω, θ) = f(ω). Some possibilities are 
u(ω, θ) = 
XP1 
k=1 
Hk ((θ))uk (ω) 
or = 
XP2 
k=1 
k (ω, (θ))k 
or = 
XP3 
k=1 
ak (ω)Hk ((θ))k 
or = 
XP4 
k=1 
ak (ω)Hk ((θ))Uk ((θ)) . . . etc. 
(49)
Deterministic classical modal analysis? 
For a deterministic system, the response vector u(ω) can be expressed as 
u(ω) = 
XP 
k=1 
k (ω)uk 
where k (ω) = 
T 
k f 
−ω2 + 2iζkωkω + ω2 
k 
uk = k and P ≤ n (number of dominantmodes) 
(50) 
Can we extend this idea to stochastic systems?
Projection in the modal space 
There exist a finite set of complex frequency dependent functions k (ω, (θ)) 
and a complete basis k ∈ Rn for k = 1, 2, . . . , n such that the solution of the 
discretized stochastic finite element equation (31) can be expiressed by the 
series 
ˆu(ω, θ) = 
Xn 
k=1 
k (ω, (θ))k (51) 
Outline of the derivation: In the first step a complete basis is generated with 
the eigenvectors k ∈ Rn of the generalized eigenvalue problem 
K0k = λ0kM0k ; k = 1, 2, . . . n (52)
Projection in the modal space 
We define the matrix of eigenvalues and eigenvectors 
0 = diag [λ01 , λ02 , . . . , λ0n ] ∈ Rn×n; = [1,2, . . . ,n] ∈ Rn×n (53) 
Eigenvalues are ordered in the ascending order: λ01  λ02  . . .  λ0n . 
We use the orthogonality property of the modal matrix  as 
TK0 = 0, and TM0 = I (54) 
Using these we have 
TA0 = T  
[−ω2 + iωζ1]M0 + [iωζ2 + 1]K0 
 
 
= 
 
−ω2 + iωζ1 
 
I + (iωζ2 + 1)0 (55) 
This gives TA0 = 0 and A0 = −T0−1, where 
0 = 
 
−ω2 + iωζ1 
 
I + (iωζ2 + 1) 0 and I is the identity matrix.
Projection in the modal space 
Hence, 0 can also be written as 
0 = diag [λ01 , λ02 , . . . , λ0n ] ∈ Cn×n (56) 
where λ0j = 
 
−ω2 + iωζ1 
 
+ (iωζ2 + 1) λj and λj is as defined in 
Eqn. (53). We also introduce the transformations 
eA 
i = TAi ∈ Cn×n; i = 0, 1, 2, . . . ,M. (57) 
Note that eA 
0 = 0 is a diagonal matrix and 
Ai = −TeA 
i−1 ∈ Cn×n; i = 1, 2, . . . ,M. (58)
Projection in the modal space 
Suppose the solution of Eq. (31) is given by 
ˆu(ω, θ) = 
 
A0(ω) + 
XM 
i=1 
#−1 
ξi (θ)Ai (ω) 
f(ω) (59) 
Using Eqs. (53)–(58) and the mass and stiffness orthogonality of  one has 
ˆu(ω, θ) = 
 
−T0(ω)−1 + 
XM 
i=1 
ξi (θ)−TeA 
i (ω)−1 
#−1 
f(ω) 
⇒ ˆu(ω, θ) =  
 
0(ω) + 
XM 
i=1 
ξi (θ)eA 
i (ω) 
#−1 
| {z } 
	(ω,(θ)) 
−T f(ω) 
(60) 
where (θ) = {ξ1(θ), ξ2(θ), . . . , ξM(θ)}T .
Projection in the modal space 
Now we separate the diagonal and off-diagonal terms of the eA 
i matrices as 
eA 
i = i +i , i = 1, 2, . . . ,M (61) 
Here the diagonal matrix 
i = diag 
h 
eA 
i 
= diag [λi1 , λi2 , . . . , λin ] ∈ Rn×n (62) 
and i = eA 
i − i is an off-diagonal only matrix. 
	(ω, (θ)) = 
 
 
0(ω) + 
XM 
i=1 
ξi (θ)i (ω) 
| {z } 
(ω,(θ)) 
+ 
XM 
i=1 
ξi (θ)i (ω) 
| {z } 
(ω,(θ)) 
 
−1 
 
(63) 
where  (ω, (θ)) ∈ Rn×n is a diagonal matrix and (ω, (θ)) is an 
off-diagonal only matrix.
Projection in the modal space 
We rewrite Eq. (63) as 
	(ω, (θ)) = 
h 
(ω, (θ)) 
h 
In + −1 (ω, (θ))(ω, (θ)) 
ii 
−1 
(64) 
The above expression can be represented using a Neumann type of matrix 
series as 
	(ω, (θ)) = 
1X 
s=0 
(−1)s 
h 
−1 (ω, (θ))(ω, (θ)) 
is 
−1 (ω, (θ)) (65)
Projection in the modal space 
Taking an arbitrary r -th element of ˆu(ω, θ), Eq. (60) can be rearranged to have 
ˆur (ω, θ) = 
Xn 
k=1 
rk 
 
 
Xn 
j=1 
	kj (ω, (θ)) 
 
T 
 
j f(ω) 
 
 (66) 
Defining 
k (ω, (θ)) = 
Xn 
j=1 
	kj (ω, (θ)) 
 
T 
 
j f(ω) 
(67) 
and collecting all the elements in Eq. (66) for r = 1, 2, . . . , n one has 
ˆu(ω, θ) = 
Xn 
k=1 
k (ω, (θ)) k (68)
Spectral functions 
Definition 
The functions k (ω, (θ)) , k = 1, 2, . . . n are the frequency-adaptive spectral 
functions as they are expressed in terms of the spectral properties of the 
coefficient matrices at each frequency of the governing discretized equation. 
Each of the spectral functions k (ω, (θ)) contain infinite number of terms 
and they are highly nonlinear functions of the random variables ξi (θ). 
For computational purposes, it is necessary to truncate the series after 
certain number of terms. 
Different order of spectral functions can be obtained by using truncation 
in the expression of k (ω, (θ))
First-order and second order spectral functions 
Definition 
The different order of spectral functions (1) 
k (ω, (θ)), k = 1, 2, . . . , n are 
obtained by retaining as many terms in the series expansion in Eqn. (65). 
Retaining one and two terms in (65) we have 
	(1) (ω, (θ)) = −1 (ω, (θ)) (69) 
	(2) (ω, (θ)) = −1 (ω, (θ)) − −1 (ω, (θ))(ω, (θ)) −1 (ω, (θ)) (70) 
which are the first and second order spectral functions respectively. 
From these we find (1) 
k (ω, (θ)) = 
Pn 
j=1 	(1) 
kj (ω, (θ)) 
 
T 
 
are 
j f(ω) 
non-Gaussian random variables even if ξi (θ) are Gaussian random 
variables.
Nature of the spectral functions 
10−2 
10−3 
10−4 
10−5 
10−6 
10−7 
10−8 
0 100 200 300 400 500 600 
Frequency (Hz) 
Spectral functions of a random sample 
G(4) 
1 
(w,x(q)) 
G(4) 
2 
(w,x(q)) 
G(4) 
3 
(w,x(q)) 
G(4) 
4 
(w,x(q)) 
G(4) 
5 
(w,x(q)) 
G(4) 
6 
(w,x(q)) 
G(4) 
7 
(w,x(q)) 
(a) Spectral functions for a = 0.1. 
10−2 
10−3 
10−4 
10−5 
10−6 
10−7 
10−8 
0 100 200 300 400 500 600 
Frequency (Hz) 
Spectral functions of a random sample 
G(4) 
1 
(w,x(q)) 
G(4) 
2 
(w,x(q)) 
G(4) 
3 
(w,x(q)) 
G(4) 
4 
(w,x(q)) 
G(4) 
5 
(w,x(q)) 
G(4) 
6 
(w,x(q)) 
G(4) 
7 
(w,x(q)) 
(b) Spectral functions for a = 0.2. 
The amplitude of first seven spectral functions of order 4 for a particular 
random sample under applied force. The spectral functions are obtained for 
two different standard deviation levels of the underlying random field: 
σa = {0.10, 0.20}.
Summary of the basis functions (frequency-adaptive spectral functions) 
The basis functions are: 
1 not polynomials in ξi (θ) but ratio of polynomials. 
2 independent of the nature of the random variables (i.e. applicable to 
Gaussian, non-Gaussian or even mixed random variables). 
3 not general but specific to a problem as it utilizes the eigenvalues and 
eigenvectors of the system matrices. 
4 such that truncation error depends on the off-diagonal terms of the matrix 
(ω, (θ)). 
5 showing ‘peaks’ when ω is near to the system natural frequencies 
Next we use these frequency-adaptive spectral functions as trial functions 
within a Galerkin error minimization scheme.
The Galerkin approach 
One can obtain constants ck ∈ C such that the error in the following 
representation 
ˆu(ω, θ) = 
Xn 
k=1 
ck (ω)b 
k (ω, (θ))k (71) 
can be minimised in the least-square sense. It can be shown that the vector 
c = {c1, c2, . . . , cn}T satisfies the n × n complex algebraic equations 
S(ω) c(ω) = b(ω) with 
Sjk = 
XM 
i=0 
ijkDijk ; ∀ j, k = 1, 2, . . . , n;eA 
eA 
ijk = T 
j Aik , (72) 
Dijk = E 
h 
ξi (θ)b 
k (ω, (θ)) 
i 
, bj = E 
h 
T 
i 
. (73) 
j f(ω)
The Galerkin approach 
The error vector can be obtained as 
(ω, θ) = 
  
XM 
i=0 
!  
Ai (ω)ξi (θ) 
Xn 
k=1 
ckb 
k (ω, (θ))k 
! 
− f(ω) ∈ CN×N (74) 
The solution is viewed as a projection where k ∈ Rn are the basis 
functions and ck are the unknown constants to be determined. This is 
done for each frequency step. 
The coefficients ck are evaluated using the Galerkin approach so that the 
error is made orthogonal to the basis functions, that is, mathematically 
(ω, θ)⊥j ⇛ 

 
j , (ω, θ) 
 
= 0 ∀ j = 1, 2, . . . , n (75)
The Galerkin approach 
Imposing the orthogonality condition and using the expression of the 
error one has 
E 
 
T 
j 
  
XM 
i=0 
!  
Ai ξi (θ) 
Xn 
k=1 
ckb 
k ((θ))k 
! 
− T 
j f 
# 
= 0, ∀j (76) 
Interchanging the E [•] and summation operations, this can be simplified 
to 
Xn 
k=1 
  
XM 
i=0 
 
T 
j Aik 
 
E 
h 
ξi (θ)b 
k ((θ)) 
i! 
ck = 
E 
h 
T 
j f 
i 
(77) 
or 
Xn 
k=1 
  
XM 
i=0 
eA 
ijkDijk 
! 
ck = bj (78)
Model Reduction by reduced number of basis 
Suppose the eigenvalues of A0 are arranged in an increasing order such 
that 
λ01  λ02  . . .  λ0n (79) 
From the expression of the spectral functions observe that the 
eigenvalues ( λ0k = ω2 
0k 
) appear in the denominator: 
(1) 
k (ω, (θ)) = 
T 
k f(ω) 
0k (ω) + 
PM 
i=1 ξi (θ)ik (ω) 
(80) 
where 0k (ω) = −ω2 + iω(ζ1 + ζ2ω2 
0k 
) + ω2 
0k 
The series can be truncated based on the magnitude of the eigenvalues 
relative to the frequency of excitation. Hence for the frequency domain 
analysis all the eigenvalues that cover almost twice the frequency range 
under consideration can be chosen.
Computational method 
The mean vector can be obtained as 
¯u = E [ˆu(θ)] = 
Xp 
k=1 
ckE 
h 
b 
k ((θ)) 
i 
k (81) 
The covariance of the solution vector can be expressed as 
u = E 
h 
(ˆu(θ) − ¯u) (ˆu(θ) − ¯u)T 
i 
= 
Xp 
k=1 
Xp 
j=1 
ckcjkjkT 
j (82) 
where the elements of the covariance matrix of the spectral functions are 
given by 
kj = E 
h 
b 
k ((θ)) − E 
h 
b 
k ((θ)) 
i  
b 
j ((θ)) − E 
h 
b 
j ((θ)) 
ii 
(83)
Summary of the computational method 
1 Solve the generalized eigenvalue problem associated with the mean 
mass and stiffness matrices to generate the orthonormal basis vectors: 
K0 = M00 
2 Select a number of samples, say Nsamp. Generate the samples of basic 
random variables ξi (θ), i = 1, 2, . . . ,M. 
3 Calculate the spectral basis functions (for example, first-order): 
k (ω, (θ)) = 
T 
k f(ω) 
0k (ω)+PM 
i=1 ξi (θ)ik 
(ω) 
, for k = 1, · · · p, p  n 
4 Obtain the coefficient vector: c(ω) = S−1(ω)b(ω) ∈ Rn, where 
b(ω) = gf(ω) ⊙ (ω), S(ω) = 0(ω) ⊙ D0(ω) + 
PM 
i=1 
eA 
i (ω) ⊙ Di (ω) and 
Di (ω) = E 
h 
(ω, θ)ξi (θ)T (ω, θ) 
i 
, ∀ i = 0, 1, 2, . . . ,M 
5 Obtain the samples of the response from the spectral series: 
ˆu(ω, θ) = 
Pp 
k=1 ck (ω)k ((ω, θ))k
The Euler-Bernoulli beam example 
An Euler-Bernoulli cantilever beam with stochastic bending modulus for a 
specified value of the correlation length and for different degrees of 
variability of the random field. 
F 
(c) Euler-Bernoulli beam 
6000 
5000 
4000 
3000 
2000 
1000 
0 
0 5 10 15 20 
Natural Frequency (Hz) 
Mode number 
(d) Natural frequency dis-tribution. 
10−1 
10−2 
10−3 
10−4 
0 5 10 15 20 25 30 35 40 
100 
/ lj 
Ratio of Eigenvalues, l1 
Eigenvalue number: j 
(e) Eigenvalue ratio of KL de-composition 
Length : 1.0 m, Cross-section : 39 × 5.93 mm2, Young’s Modulus: 2 × 1011 Pa. 
Load: Unit impulse at t = 0 on the free end of the beam.
Problem details 
The bending modulus of the cantilever beam is taken to be a 
homogeneous stationary Gaussian random field of the form 
EI(x, θ) = EI0(1 + a(x, θ)) (84) 
where x is the coordinate along the length of the beam, EI0 is the 
estimate of the mean bending modulus, a(x, θ) is a zero mean stationary 
random field. 
The covariance kernel associated with this random field is 
ae−(|x1−x2|)/μa (85) 
Ca(x1, x2) = σ2 
where μa is the correlation length and σa is the standard deviation. 
A correlation length of μa = L/5 is considered in the present numerical 
study.
Problem details 
The random field is assumed to be Gaussian. The results are compared with 
the polynomial chaos expansion. 
The number of degrees of freedom of the system is n = 200. 
The K.L. expansion is truncated at a finite number of terms such that 90% 
variability is retained. 
direct MCS have been performed with 10,000 random samples and for 
three different values of standard deviation of the random field, 
σa = 0.05, 0.1, 0.2. 
Constant modal damping is taken with 1% damping factor for all modes. 
Time domain response of the free end of the beam is sought under the 
action of a unit impulse at t = 0 
Upto 4th order spectral functions have been considered in the present 
problem. Comparison have been made with 4th order Polynomial chaos 
results.
Mean of the response 
(f) Mean, a = 0.05. (g) Mean, a = 0.1. (h) Mean, a = 0.2. 
Time domain response of the deflection of the tip of the cantilever for 
three values of standard deviation σa of the underlying random field. 
Spectral functions approach approximates the solution accurately. 
For long time-integration, the discrepancy of the 4th order PC results 
increases.
Standard deviation of the response 
(i) Standard deviation of de-flection, 
a = 0.05. 
(j) Standard deviation of de-flection, 
a = 0.1. 
(k) Standard deviation of de-flection, 
a = 0.2. 
The standard deviation of the tip deflection of the beam. 
Since the standard deviation comprises of higher order products of the 
Hermite polynomials associated with the PC expansion, the higher order 
moments are less accurately replicated and tend to deviate more 
significantly.
Frequency domain response: mean 
10−2 
10−3 
10−4 
10−5 
10−6 
10−7 
0 100 200 300 400 500 600 
Frequency (Hz) 
: 0.1 
damped deflection, sf 
MCS 
2nd order Galerkin 
3rd order Galerkin 
4th order Galerkin 
deterministic 
4th order PC 
(l) Beam deflection for a = 0.1. 
10−1 
10−2 
10−3 
10−4 
10−5 
10−6 
10−7 
0 100 200 300 400 500 600 
Frequency (Hz) 
: 0.2 
damped deflection, sf 
MCS 
2nd order Galerkin 
3rd order Galerkin 
4th order Galerkin 
deterministic 
4th order PC 
(m) Beam deflection for a = 0.2. 
The frequency domain response of the deflection of the tip of the 
Euler-Bernoulli beam under unit amplitude harmonic point load at the free 
end. The response is obtained with 10, 000 sample MCS and for 
σa = {0.10, 0.20}.
Frequency domain response: standard deviation 
10−2 
10−3 
10−4 
10−5 
10−6 
10−7 
0 100 200 300 400 500 600 
Frequency (Hz) 
: 0.1 
Standard Deviation (damped), sf 
MCS 
2nd order Galerkin 
3rd order Galerkin 
4th order Galerkin 
4th order PC 
(n) Standard deviation of the response for 
a = 0.1. 
10−1 
10−2 
10−3 
10−4 
10−5 
10−6 
0 100 200 300 400 500 600 
Frequency (Hz) 
: 0.2 
Standard Deviation (damped), sf 
MCS 
2nd order Galerkin 
3rd order Galerkin 
4th order Galerkin 
4th order PC 
(o) Standard deviation of the response for 
a = 0.2. 
The standard deviation of the tip deflection of the Euler-Bernoulli beam under 
unit amplitude harmonic point load at the free end. The response is obtained 
with 10, 000 sample MCS and for σa = {0.10, 0.20}.
Experimental investigations 
Figure: A cantilever plate with randomly attached oscillators - Probabilistic Engineering Mechanics, 24[4] 
(2009), pp. 473-492
Measured frequency response function 
60 
40 
20 
0 
−20 
−40 
−60 
0 100 200 300 400 500 600 
Frequency (Hz) 
(w) 
(1,1) 
Log amplitude (dB) of H 
Baseline 
Ensemble mean 
5% line 
95% line
Conclusions 
The mean response of a damped stochastic system is more damped 
than the underlying baseline system 
For small damping, ξe ≈ 31/4pǫ pπ √ξ 
Random modal analysis may not be practical or physically intuitive for 
stochastic multiple degrees of freedom systems 
Conventional response surface based methods fails to capture the 
physics of damped dynamic systems 
Proposed spectral function approach uses the undamped modal basis 
and can capture the statistical trend of the dynamic response of 
stochastic damped MDOF systems
Conclusions 
The solution is projected into the modal basis and the associated 
stochastic coefficient functions are obtained at each frequency step (or 
time step). 
The coefficient functions, called as the spectral functions, are expressed 
in terms of the spectral properties (natural frequencies and mode 
shapes) of the system matrices. 
The proposed method takes advantage of the fact that for a given 
maximum frequency only a small number of modes are necessary to 
represent the dynamic response. This modal reduction leads to a 
significantly smaller basis.
Assimilation with experimental measurements 
In the frequency domain, the response can be simplified as 
u(ω, θ) ≈ 
Xnr 
k=1 
T 
k f(ω) 
−ω2 + 2iωζkω0k + ω2 
0k 
+ 
PM 
i=1 ξi (θ)ik (ω) 
k 
Some parts can be obtained from experiments while other parts can come 
from stochastic modelling.

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Dynamics of structures with uncertainties

  • 1. Dynamics of structures with uncertainties S Adhikari College of Engineering, Swansea University, Swansea UK Email: [email protected] http://engweb.swan.ac.uk/ adhikaris/ Twitter: @ProfAdhikari The University of Campinas, Campinas, Brazil November 2014
  • 4. About me Education: PhD (Engineering), 2001, University of Cambridge (Trinity College), Cambridge, UK. MSc (Structural Engineering), 1997, Indian Institute of Science, Bangalore, India. B. Eng, (Civil Engineering), 1995, Calcutta University, India. Work: 04/2007-Present: Professor of Aerospace Engineering, Swansea University (Civil and Computational Engineering Research Centre). 01/2003-03/2007: Lecturer in dynamics: Department of Aerospace Engineering, University of Bristol. 11/2000-12/2002: Research Associate, Cambridge University Engineering Department (Junior Research Fellow, Fitzwilliam College, Cambridge) .
  • 5. Outline of this talk 1 Introduction 2 Stochastic single degrees of freedom system 3 Stochastic multi degree of freedom systems Stochastic finite element formulation Projection in the modal space Properties of the spectral functions 4 Error minimization The Galerkin approach Model Reduction Computational method 5 Numerical illustrations 6 Conclusions
  • 6. Mathematical models for dynamic systems Mathematical Models of Dynamic Systems ❄ ❄ ❄ ❄ ❄ Linear Time-invariant Elastic Non-linear Time-variant Elasto-plastic Viscoelastic Continuous Discrete Deterministic Uncertain ❄ ❄ ❄ Single-degree-of-freedom (SDOF) ✲ Probabilistic ✲ ✲ Multiple-degree-of-freedom (MDOF) ✲ Fuzzy set Convex set ❄ Low frequency Mid-frequency High frequency
  • 7. A general overview of computational mechanics Real System Input ( eg , earthquake, turbulence ) Measured output ( eg , velocity, acceleration , stress) System Uncertainty parametric uncertainty model inadequacy model uncertainty calibration uncertainty Physics based model L (u) = f ( eg , ODE/ PDE / SDE / SPDE ) Input Uncertainty uncertainty in time history uncertainty in location Simulated Input (time or frequency domain) Computation ( eg , FEM / BEM /Finite difference/ SFEM / MCS ) uncertain experimental error calibration/updating Computational Uncertainty machine precession, error tolerance ‘ h ’ and ‘ p ’ refinements Model output ( eg , velocity, acceleration , stress) verification system identification Total Uncertainty = input + system + computational uncertainty model validation
  • 8. Uncertainty in structural dynamical systems Many structural dynamic systems are manufactured in a production line (nom-inally identical systems). On the other hand, some models are complex! Com-plex models can have ‘errors’ and/or ‘lack of knowledge’ in its formulation.
  • 9. Model quality The quality of a model of a dynamic system depends on the following three factors: Fidelity to (experimental) data: The results obtained from a numerical or mathematical model undergoing a given excitation force should be close to the results obtained from the vibration testing of the same structure undergoing the same excitation. Robustness with respect to (random) errors: Errors in estimating the system parameters, boundary conditions and dynamic loads are unavoidable in practice. The output of the model should not be very sensitive to such errors. Predictive capability: In general it is not possible to experimentally validate a model over the entire domain of its scope of application. The model should predict the response well beyond its validation domain.
  • 10. Sources of uncertainty Different sources of uncertainties in the modeling and simulation of dynamic systems may be attributed, but not limited, to the following factors: Mathematical models: equations (linear, non-linear), geometry, damping model (viscous, non-viscous, fractional derivative), boundary conditions/initial conditions, input forces. Model parameters: Young’s modulus, mass density, Poisson’s ratio, damping model parameters (damping coefficient, relaxation modulus, fractional derivative order). Numerical algorithms: weak formulations, discretisation of displacement fields (in finite element method), discretisation of stochastic fields (in stochastic finite element method), approximate solution algorithms, truncation and roundoff errors, tolerances in the optimization and iterative methods, artificial intelligent (AI) method (choice of neural networks). Measurements: noise, resolution (number of sensors and actuators), experimental hardware, excitation method (nature of shakers and hammers), excitation and measurement point, data processing (amplification, number of data points, FFT), calibration.
  • 11. Few general questions How does system uncertainty impact the dynamic response? Does it matter? What is the underlying physics? How can we model uncertainty in dynamic systems? Do we ‘know’ the uncertainties? How can we efficiently quantify uncertainty in the dynamic response for large multi degrees of freedom systems? What about using ‘black box’ type response surface methods? Can we use modal analysis for stochastic systems? Does stochastic systems has natural frequencies and mode shapes?
  • 12. Stochastic SDOF systems k m u( t ) f ( t ) f d ( t ) Consider a normalised single degree of freedom system (SDOF): u¨(t) + 2ζωn u˙ (t) + ω2 n u(t) = f (t)/m (1) Here ωn = p k/m is the natural frequency and ξ = c/2√km is the damping ratio. We are interested in understanding the motion when the natural frequency of the system is perturbed in a stochastic manner. Stochastic perturbation can represent statistical scatter of measured values or a lack of knowledge regarding the natural frequency.
  • 13. Frequency variability 4 3.5 3 2.5 2 1.5 1 0.5 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x (x) p x uniform normal log−normal (a) Pdf: a = 0.1 4 3.5 3 2.5 2 1.5 1 0.5 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x (x) p x uniform normal log−normal (b) Pdf: a = 0.2 Figure: We assume that the mean of r is 1 and the standard deviation is a. Suppose the natural frequency is expressed as ω2 n = ω2 n0 r , where ωn0 is deterministic frequency and r is a random variable with a given probability distribution function. Several probability distribution function can be used. We use uniform, normal and lognormal distribution.
  • 14. Frequency samples 1000 900 800 700 600 500 400 300 200 100 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Frequency: wn Samples uniform normal log−normal (a) Frequencies: a = 0.1 1000 900 800 700 600 500 400 300 200 100 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Frequency: wn Samples uniform normal log−normal (b) Frequencies: a = 0.2 Figure: 1000 sample realisations of the frequencies for the three distributions
  • 15. Response in the time domain 1 0.8 0.6 0.4 0.2 0 −0.2 −0.4 −0.6 −0.8 −1 0 5 10 15 Normalised time: t/T n0 0 Normalised amplitude: u/v deterministic random samples mean: uniform mean: normal mean: log−normal (a) Response: a = 0.1 1 0.8 0.6 0.4 0.2 0 −0.2 −0.4 −0.6 −0.8 −1 0 5 10 15 Normalised time: t/T n0 0 Normalised amplitude: u/v deterministic random samples mean: uniform mean: normal mean: log−normal (b) Response: a = 0.2 Figure: Response due to initial velocity v0 with 5% damping
  • 16. Frequency response function 120 100 80 60 40 20 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Normalised frequency: w/wn0 |2 st Normalised amplitude: |u/u deterministic mean: uniform mean: normal mean: log−normal (a) Response: a = 0.1 120 100 80 60 40 20 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Normalised frequency: w/wn0 |2 st Normalised amplitude: |u/u deterministic mean: uniform mean: normal mean: log−normal (b) Response: a = 0.2 Figure: Normalised frequency response function |u/ust |2, where ust = f /k
  • 17. Key observations The mean response is more damped compared to deterministic response. The higher the randomness, the higher the “effective damping”. The qualitative features are almost independent of the distribution the random natural frequency. We often use averaging to obtain more reliable experimental results - is it always true? Assuming uniform random variable, we aim to explain some of these observations.
  • 18. Equivalent damping Assume that the random natural frequencies are ω2 n = ω2 n0(1 + ǫx), where x has zero mean and unit standard deviation. The normalised harmonic response in the frequency domain u(iω) f /k = k/m n0 (1 + ǫx)] + 2iξωωn0√1 + ǫx [−ω2 + ω2 (2) Considering ωn0 = p k/m and frequency ratio r = ω/ωn0 we have u f /k = 1 [(1 + ǫx) − r 2] + 2iξr√1 + ǫx (3)
  • 19. Equivalent damping The squared-amplitude of the normalised dynamic response at ω = ωn0 (that is r = 1) can be obtained as ˆU = |u| f /k 2 = 1 ǫ2x2 + 4ξ2(1 + ǫx) (4) Since x is zero mean unit standard √√deviation √uniform random variable, its pdf is given by px (x) = 1/23,− 3 ≤ x ≤ 3 The mean is therefore E h ˆU i = Z 1 ǫ2x2 + 4ξ2(1 + ǫx) px (x)dx = 1 4√3ǫξ p 1 − ξ2 tan−1 √p 3ǫ 2ξ 1 − ξ2 − ξ p 1 − ξ2 ! + 1 4√3ǫξ p 1 − ξ2 tan−1 √p 3ǫ 2ξ 1 − ξ2 + ξ p 1 − ξ2 ! (5)
  • 20. Equivalent damping Note that 1 2 tan−1(a + δ) + tan−1(a − δ) = tan−1(a) + O(δ2) (6) Neglecting terms of the order O(ξ2) we have E h ˆU i ≈ 1 2√3ǫξ p 1 − ξ2 tan−1 √p 3ǫ 2ξ 1 − ξ2 ! = tan−1(√3ǫ/2ξ) 2√3ǫξ (7)
  • 21. Equivalent damping For small damping, the maximum deterministic amplitude at ω = ωn0 is 1/4ξ2 e where ξe is the equivalent damping for the mean response Therefore, the equivalent damping for the mean response is given by (2ξe)2 = 2√3ǫξ tan−1(√3ǫ/2ξ) (8) For small damping, taking the limit we can obtain ξe ≈ 31/4√ǫ √π p ξ (9) The equivalent damping factor of the mean system is proportional to the square root of the damping factor of the underlying baseline system
  • 22. Equivalent frequency response function 120 100 80 60 40 20 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Normalised frequency: w/wn0 |2 st Normalised amplitude: |u/u Deterministic MCS Mean Equivalent (a) Response: a = 0.1 120 100 80 60 40 20 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Normalised frequency: w/wn0 |2 st Normalised amplitude: |u/u Deterministic MCS Mean Equivalent (b) Response: a = 0.2 Figure: Normalised frequency response function with equivalent damping (e = 0.05 in the ensembles). For the two cases e = 0.0643 and e = 0.0819 respectively.
  • 23. Can we extend the ideas based on stochastic SDOF systems to stochastic MDOF systems?
  • 24. Stochastic modal analysis Stochastic modal analysis to obtain the dynamic response needs further thoughts Consider the following 3DOF example: k 1 k 4 k 5 k 3 m 1 m 2 m 3 k 2 k 6 Figure: A 3DOF system with parametric uncertainty in mi and ki
  • 25. Statistical overlap 2 4 6 8 10 12 1000 900 800 700 600 500 400 300 200 100 Samples Eigenvalues, lj (rad/s)2 l1 l2 l3 (a) Eigenvalues are seperated 1 2 3 4 5 6 7 8 1000 900 800 700 600 500 400 300 200 100 Samples Eigenvalues, lj (rad/s)2 l1 l2 l3 (b) Some eigenvalues are close Figure: Scatter of the eigenvalues due to parametric uncertainties The SDOF based approach cannot be applied when there is statistical overlap in the eigenvalues.
  • 26. Stochastic partial differential equation We consider a stochastic partial differential equation (PDE) for a linear dynamic system ρ(r, θ) ∂2U(r, t , θ) ∂t2 + Lα ∂U(r, t , θ) ∂t + LβU(r, t , θ) = p(r, t) (10) The stochastic operator Lβ can be ∂x AE(x, θ) ∂ ∂x axial deformation of rods Lβ ≡ ∂ Lβ ≡ ∂2 ∂x2 EI(x, θ) ∂2 ∂x2 bending deformation of beams Lα denotes the stochastic damping, which is mostly proportional in nature. Here α, β : Rd × → R are stationary square integrable random fields, which can be viewed as a set of random variables indexed by r ∈ Rd . Based on the physical problem the random field a(r, θ) can be used to model different physical quantities (e.g., AE(x, θ), EI(x, θ)).
  • 27. Discretisation of random fields The random process a(r, θ) can be expressed in a generalized Fourier type of series known as the Karhunen-Lo`eve expansion a(r, θ) = a0(r) + 1X i=1 √νi ξi (θ)ϕi (r) (11) Here a0(r) is the mean function, ξi (θ) are uncorrelated standard Gaussian random variables, νi and ϕi (r) are eigenvalues and eigenfunctions satisfying the integral equation Z D Ca(r1, r2)ϕj (r1)dr1 = νjϕj (r2), ∀ j = 1, 2, · · · (12)
  • 28. Exponential autocorrelation function The autocorrelation function: C(x1, x2) = e−|x1−x2|/b (13) The underlying random process H(x, θ) can be expanded using the Karhunen-Lo`eve (KL) expansion in the interval −a ≤ x ≤ a as H(x, θ) = 1X j=1 p λjϕj (x) (14) ξj (θ) Using the notation c = 1/b, the corresponding eigenvalues and eigenfunctions for odd j and even j are given by λj = 2c ω2 j + c2 , ϕj (x) = cos(ωjx) s a + sin(2ωja) 2ωj , where tan(ωja) = c ωj , (15) λj = 2c ωj 2 + c2 , ϕj (x) = sin(ωjx) s a − sin(2ωja) 2ωj , where tan(ωja) = ωj −c . (16)
  • 29. KL expansion 10−1 10−2 10−3 0 5 10 15 20 25 30 35 100 Index, j Eigenvalues, lj b=L/2, N=10 b=L/3, N=13 b=L/4, N=16 b=L/5, N=19 b=L/10, N=34 The eigenvalues of the Karhunen-Lo`eve expansion for different correlation lengths, b, and the number of terms, N, required to capture 90% of the infinite series. An exponential correlation function with unit domain (i.e., a = 1/2) is assumed for the numerical calculations. The values of N are obtained such that λN/λ1 = 0.1 for all correlation lengths. Only eigenvalues greater than λN are plotted.
  • 30. Example: A beam with random properties The equation of motion of an undamped Euler-Bernoulli beam of length L with random bending stiffness and mass distribution: ∂2 ∂x2 EI(x, θ) ∂2Y(x, t) ∂x2 + ρA(x, θ) ∂2Y(x, t) ∂t2 = p(x, t). (17) Y(x, t): transverse flexural displacement, EI(x): flexural rigidity, ρA(x): mass per unit length, and p(x, t): applied forcing. Consider EI(x, θ) = EI0 (1 + ǫ1F1(x, θ)) (18) and ρA(x, θ) = ρA0 (1 + ǫ2F2(x, θ)) (19) The subscript 0 indicates the mean values, 0 ǫi 1 (i=1,2) are deterministic constants and the random fields Fi (x, θ) are taken to have zero mean, unit standard deviation and covariance Rij (ξ).
  • 31. Random beam element 1 3 EI(x), m(x), c , c 1 2 2 4 l y x Random beam element in the local coordinate.
  • 32. Realisations of the random field 4 3.5 3 2.5 2 1.5 1 0.5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Length along the beam (m) EI (Nm2) baseline value perturbed values Some random realizations of the bending rigidity EI of the beam for correlation length b = L/3 and strength parameter ǫ1 = 0.2 (mean 2.0 × 105). Thirteen terms have been used in the KL expansion.
  • 33. Example: A beam with random properties We express the shape functions for the finite element analysis of Euler-Bernoulli beams as N(x) = s(x) (20) where =   1 0 −3 ℓe 2 2 ℓe 3 0 1 −2 ℓe 2 1 ℓe 2 0 0 2 −2 3 ℓe ℓe 3 0 0 −1 ℓe 2 1 ℓe 2   and s(x) = 1, x, x2, x3 T . (21) The element stiffness matrix: Ke(θ) = Z ℓe 0 N ′′ (x)EI(x, θ)N ′′T (x)dx = Z ℓe 0 EI0 (1 + ǫ1F1(x, θ))N ′′ (x)N ′′T (x)dx. (22)
  • 34. Example: A beam with random properties Expanding the random field F1(x, θ) in KL expansion Ke(θ) = Ke0 +Ke(θ) (23) where the deterministic and random parts are Ke0 = EI0 Z ℓe 0 ′′ (x)N N ′′T (x) dx and Ke(θ) = ǫ1 XNK j=1 p λKjKej . (24) ξKj (θ) The constant NK is the number of terms retained in the Karhunen-Lo`eve expansion and ξKj (θ) are uncorrelated Gaussian random variables with zero mean and unit standard deviation. The constant matrices Kej can be expressed as Kej = EI0 Z ℓe 0 ϕKj (xe + x)N ′′T (x) dx (25) ′′ (x)N
  • 35. Example: A beam with random properties The mass matrix can be obtained as Me(θ) = Me0 +Me(θ) (26) The deterministic and random parts is given by Me0 = ρA0 Z ℓe 0 N(x)NT (x) dx and Me(θ) = ǫ2 XNM j=1 p λMjMej . (27) ξMj (θ) The constant NM is the number of terms retained in Karhunen-Lo`eve expansion and the constant matrices Mej can be expressed as Mej = ρA0 Z ℓe 0 ϕMj (xe + x)N(x)NT (x) dx. (28) Both Kej and Mej can be obtained in closed-form.
  • 36. Example: A beam with random properties These element matrices can be assembled to form the global random stiffness and mass matrices of the form K(θ) = K0 +K(θ) and M(θ) = M0 +M(θ). (29) Here the deterministic parts K0 and M0 are the usual global stiffness and mass matrices obtained form the conventional finite element method. The random parts can be expressed as K(θ) = ǫ1 XNK j=1 p λKjKj and M(θ) = ǫ2 ξKj (θ) XNM j=1 p λMjMj (30) ξMj (θ) The element matrices Kej and Mej can be assembled into the global matrices Kj and Mj . The total number of random variables depend on the number of terms used for the truncation of the infinite series. This in turn depends on the respective correlation lengths of the underlying random fields.
  • 37. Stochastic equation of motion The equation for motion for stochastic linear MDOF dynamic systems: M(θ)u¨(θ, t) + C(θ)u˙ (θ, t) + K(θ)u(θ, t) = f(t) (31) M(θ) = M0 + Pp i=1 μi (θi )Mi ∈ Rn×n is the random mass matrix, K(θ) = K0 + Pp i=1 νi (θi )Ki ∈ Rn×n is the random stiffness matrix, C(θ) ∈ Rn×n as the random damping matrix and f(t) is the forcing vector The mass and stiffness matrices have been expressed in terms of their deterministic components (M0 and K0) and the corresponding random contributions (Mi and Ki ). These can be obtained from discretising stochastic fields with a finite number of random variables (μi (θi ) and νi (θi )) and their corresponding spatial basis functions. Proportional damping model is considered for which C(θ) = ζ1M(θ) + ζ2K(θ), where ζ1 and ζ2 are scalars.
  • 38. Frequency domain representation For the harmonic analysis of the structural system, taking the Fourier transform −ω2M(θ) + iωC(θ) + K(θ) eu (ω, θ) =ef (ω) (32) whereeu (ω, θ) is the complex frequency domain system response amplitude,ef (ω) is the amplitude of the harmonic force. For convenience we group the random variables associated with the mass and stiffness matrices as ξi (θ) = μi (θ) and ξj+p1 (θ) = νj (θ) for i = 1, 2, . . . , p1 and j = 1, 2, . . . , p2
  • 39. Frequency domain representation Using M = p1 + p2 which we have A0(ω) + XM i=1 ! ξi (θ)Ai (ω) eu(ω, θ) =ef(ω) (33) where A0 and Ai ∈ Cn×n represent the complex deterministic and stochastic parts respectively of the mass, the stiffness and the damping matrices ensemble. For the case of proportional damping the matrices A0 and Ai can be written as A0(ω) = −ω2 + iωζ1 M0 + [iωζ2 + 1]K0, (34) Ai (ω) = −ω2 + iωζ1 Mi for i = 1, 2, . . . , p1 (35) and Aj+p1 (ω) = [iωζ2 + 1]Kj for j = 1, 2, . . . , p2 .
  • 40. Time domain representation If the time steps are fixed to t , then the equation of motion can be written as M(θ) ¨ut+t (θ) + C(θ) ˙ut+t (θ) + K(θ)ut+t (θ) = pt+t . (36) Following the Newmark method based on constant average acceleration scheme, the above equations can be represented as [a0M(θ) + a1C(θ) + K(θ)] ut+t (θ) = peqv t+t (θ) (37) and, peqv t+t (θ) = pt+t + f (ut (θ), ˙ ut (θ), ¨ ut (θ),M(θ),C(θ)) (38) where peqv t+t (θ) is the equivalent force at time t + t which consists of contributions of the system response at the previous time step.
  • 41. Newmark’s method The expressions for the velocities ˙ut+t (θ) and accelerations ¨ut+t (θ) at each time step is a linear combination of the values of the system response at previous time steps (Newmark method) as ¨ut+t (θ) = a0 [ut+t (θ) − ut (θ)] − a2 ˙ut (θ) − a3 ¨ut (θ) (39) and, ˙ut+t (θ) = ˙ut (θ) + a6 ¨ut (θ) + a7 ¨ut+t (θ) (40) where the integration constants ai , i = 1, 2, . . . , 7 are independent of system properties and depends only on the chosen time step and some constants: a0 = 1 αt2 ; a1 = δ αt ; a2 = 1 αt ; a3 = 1 2α − 1; (41) a4 = δ α − 1; a5 = t 2 δ α − 2 ; a6 = t(1 − δ); a7 = δt (42)
  • 42. Newmark’s method Following this development, the linear structural system in (37) can be expressed as A0 + XM i=1 ξi (θ)Ai # | {z } A(θ) ut+t (θ) = peqv t+t (θ). (43) where A0 and Ai represent the deterministic and stochastic parts of the system matrices respectively. For the case of proportional damping, the matrices A0 and Ai can be written similar to the case of frequency domain as A0 = [a0 + a1ζ1]M0 + [a1ζ2 + 1]K0 (44) and, Ai = [a0 + a1ζ1]Mi for i = 1, 2, . . . , p1 (45) = [a1ζ2 + 1]Ki for i = p1 + 1, p1 + 2, . . . , p1 + p2 .
  • 43. General mathematical representation Whether time-domain or frequency domain methods were used, in general the main equation which need to be solved can be expressed as A0 + XM i=1 ξi (θ)Ai ! u(θ) = f(θ) (46) where A0 and Ai represent the deterministic and stochastic parts of the system matrices respectively. These can be real or complex matrices. Generic response surface based methods have been used in literature - for example the Polynomial Chaos Method
  • 44. Polynomial Chaos expansion After the finite truncation, the polynomial chaos expansion can be written as u(θ) = XP k=1 Hk ((θ))uk (47) where Hk ((θ)) are the polynomial chaoses. We need to solve a nP × nP linear equation to obtain all uk ∈ Rn.   A0,0 · · · A0,P−1 A1,0 · · · A1,P−1 ... ... ... AP−1,0 · · · AP−1,P−1     u0 u1 ... uP−1   =   f0 f1 ... fP−1   (48) The number of terms P increases exponentially with M: M 2 3 5 10 20 50 100 2nd order PC 5 9 20 65 230 1325 5150 3rd order PC 9 19 55 285 1770 23425 176850
  • 45. Some Observations The basis is a function of the pdf of the random variables only. For example, Hermite polynomials for Gaussian pdf, Legender’s polynomials for uniform pdf. The physics of the underlying problem (static, dynamic, heat conduction, transients....) cannot be incorporated in the basis. For an n-dimensional output vector, the number of terms in the projection can be more than n (depends on the number of random variables). This implies that many of the vectors uk are linearly dependent. The physical interpretation of the coefficient vectors uk is not immediately obvious. The functional form of the response is a pure polynomial in random variables.
  • 46. Possibilities of solution types As an example, consider the frequency domain response vector of the stochastic system u(ω, θ) governed by −ω2M((θ)) + iωC((θ)) + K((θ)) u(ω, θ) = f(ω). Some possibilities are u(ω, θ) = XP1 k=1 Hk ((θ))uk (ω) or = XP2 k=1 k (ω, (θ))k or = XP3 k=1 ak (ω)Hk ((θ))k or = XP4 k=1 ak (ω)Hk ((θ))Uk ((θ)) . . . etc. (49)
  • 47. Deterministic classical modal analysis? For a deterministic system, the response vector u(ω) can be expressed as u(ω) = XP k=1 k (ω)uk where k (ω) = T k f −ω2 + 2iζkωkω + ω2 k uk = k and P ≤ n (number of dominantmodes) (50) Can we extend this idea to stochastic systems?
  • 48. Projection in the modal space There exist a finite set of complex frequency dependent functions k (ω, (θ)) and a complete basis k ∈ Rn for k = 1, 2, . . . , n such that the solution of the discretized stochastic finite element equation (31) can be expiressed by the series ˆu(ω, θ) = Xn k=1 k (ω, (θ))k (51) Outline of the derivation: In the first step a complete basis is generated with the eigenvectors k ∈ Rn of the generalized eigenvalue problem K0k = λ0kM0k ; k = 1, 2, . . . n (52)
  • 49. Projection in the modal space We define the matrix of eigenvalues and eigenvectors 0 = diag [λ01 , λ02 , . . . , λ0n ] ∈ Rn×n; = [1,2, . . . ,n] ∈ Rn×n (53) Eigenvalues are ordered in the ascending order: λ01 λ02 . . . λ0n . We use the orthogonality property of the modal matrix as TK0 = 0, and TM0 = I (54) Using these we have TA0 = T [−ω2 + iωζ1]M0 + [iωζ2 + 1]K0 = −ω2 + iωζ1 I + (iωζ2 + 1)0 (55) This gives TA0 = 0 and A0 = −T0−1, where 0 = −ω2 + iωζ1 I + (iωζ2 + 1) 0 and I is the identity matrix.
  • 50. Projection in the modal space Hence, 0 can also be written as 0 = diag [λ01 , λ02 , . . . , λ0n ] ∈ Cn×n (56) where λ0j = −ω2 + iωζ1 + (iωζ2 + 1) λj and λj is as defined in Eqn. (53). We also introduce the transformations eA i = TAi ∈ Cn×n; i = 0, 1, 2, . . . ,M. (57) Note that eA 0 = 0 is a diagonal matrix and Ai = −TeA i−1 ∈ Cn×n; i = 1, 2, . . . ,M. (58)
  • 51. Projection in the modal space Suppose the solution of Eq. (31) is given by ˆu(ω, θ) = A0(ω) + XM i=1 #−1 ξi (θ)Ai (ω) f(ω) (59) Using Eqs. (53)–(58) and the mass and stiffness orthogonality of one has ˆu(ω, θ) = −T0(ω)−1 + XM i=1 ξi (θ)−TeA i (ω)−1 #−1 f(ω) ⇒ ˆu(ω, θ) = 0(ω) + XM i=1 ξi (θ)eA i (ω) #−1 | {z } (ω,(θ)) −T f(ω) (60) where (θ) = {ξ1(θ), ξ2(θ), . . . , ξM(θ)}T .
  • 52. Projection in the modal space Now we separate the diagonal and off-diagonal terms of the eA i matrices as eA i = i +i , i = 1, 2, . . . ,M (61) Here the diagonal matrix i = diag h eA i = diag [λi1 , λi2 , . . . , λin ] ∈ Rn×n (62) and i = eA i − i is an off-diagonal only matrix. (ω, (θ)) =   0(ω) + XM i=1 ξi (θ)i (ω) | {z } (ω,(θ)) + XM i=1 ξi (θ)i (ω) | {z } (ω,(θ))  −1  (63) where (ω, (θ)) ∈ Rn×n is a diagonal matrix and (ω, (θ)) is an off-diagonal only matrix.
  • 53. Projection in the modal space We rewrite Eq. (63) as (ω, (θ)) = h (ω, (θ)) h In + −1 (ω, (θ))(ω, (θ)) ii −1 (64) The above expression can be represented using a Neumann type of matrix series as (ω, (θ)) = 1X s=0 (−1)s h −1 (ω, (θ))(ω, (θ)) is −1 (ω, (θ)) (65)
  • 54. Projection in the modal space Taking an arbitrary r -th element of ˆu(ω, θ), Eq. (60) can be rearranged to have ˆur (ω, θ) = Xn k=1 rk   Xn j=1 kj (ω, (θ)) T j f(ω)   (66) Defining k (ω, (θ)) = Xn j=1 kj (ω, (θ)) T j f(ω) (67) and collecting all the elements in Eq. (66) for r = 1, 2, . . . , n one has ˆu(ω, θ) = Xn k=1 k (ω, (θ)) k (68)
  • 55. Spectral functions Definition The functions k (ω, (θ)) , k = 1, 2, . . . n are the frequency-adaptive spectral functions as they are expressed in terms of the spectral properties of the coefficient matrices at each frequency of the governing discretized equation. Each of the spectral functions k (ω, (θ)) contain infinite number of terms and they are highly nonlinear functions of the random variables ξi (θ). For computational purposes, it is necessary to truncate the series after certain number of terms. Different order of spectral functions can be obtained by using truncation in the expression of k (ω, (θ))
  • 56. First-order and second order spectral functions Definition The different order of spectral functions (1) k (ω, (θ)), k = 1, 2, . . . , n are obtained by retaining as many terms in the series expansion in Eqn. (65). Retaining one and two terms in (65) we have (1) (ω, (θ)) = −1 (ω, (θ)) (69) (2) (ω, (θ)) = −1 (ω, (θ)) − −1 (ω, (θ))(ω, (θ)) −1 (ω, (θ)) (70) which are the first and second order spectral functions respectively. From these we find (1) k (ω, (θ)) = Pn j=1 (1) kj (ω, (θ)) T are j f(ω) non-Gaussian random variables even if ξi (θ) are Gaussian random variables.
  • 57. Nature of the spectral functions 10−2 10−3 10−4 10−5 10−6 10−7 10−8 0 100 200 300 400 500 600 Frequency (Hz) Spectral functions of a random sample G(4) 1 (w,x(q)) G(4) 2 (w,x(q)) G(4) 3 (w,x(q)) G(4) 4 (w,x(q)) G(4) 5 (w,x(q)) G(4) 6 (w,x(q)) G(4) 7 (w,x(q)) (a) Spectral functions for a = 0.1. 10−2 10−3 10−4 10−5 10−6 10−7 10−8 0 100 200 300 400 500 600 Frequency (Hz) Spectral functions of a random sample G(4) 1 (w,x(q)) G(4) 2 (w,x(q)) G(4) 3 (w,x(q)) G(4) 4 (w,x(q)) G(4) 5 (w,x(q)) G(4) 6 (w,x(q)) G(4) 7 (w,x(q)) (b) Spectral functions for a = 0.2. The amplitude of first seven spectral functions of order 4 for a particular random sample under applied force. The spectral functions are obtained for two different standard deviation levels of the underlying random field: σa = {0.10, 0.20}.
  • 58. Summary of the basis functions (frequency-adaptive spectral functions) The basis functions are: 1 not polynomials in ξi (θ) but ratio of polynomials. 2 independent of the nature of the random variables (i.e. applicable to Gaussian, non-Gaussian or even mixed random variables). 3 not general but specific to a problem as it utilizes the eigenvalues and eigenvectors of the system matrices. 4 such that truncation error depends on the off-diagonal terms of the matrix (ω, (θ)). 5 showing ‘peaks’ when ω is near to the system natural frequencies Next we use these frequency-adaptive spectral functions as trial functions within a Galerkin error minimization scheme.
  • 59. The Galerkin approach One can obtain constants ck ∈ C such that the error in the following representation ˆu(ω, θ) = Xn k=1 ck (ω)b k (ω, (θ))k (71) can be minimised in the least-square sense. It can be shown that the vector c = {c1, c2, . . . , cn}T satisfies the n × n complex algebraic equations S(ω) c(ω) = b(ω) with Sjk = XM i=0 ijkDijk ; ∀ j, k = 1, 2, . . . , n;eA eA ijk = T j Aik , (72) Dijk = E h ξi (θ)b k (ω, (θ)) i , bj = E h T i . (73) j f(ω)
  • 60. The Galerkin approach The error vector can be obtained as (ω, θ) = XM i=0 ! Ai (ω)ξi (θ) Xn k=1 ckb k (ω, (θ))k ! − f(ω) ∈ CN×N (74) The solution is viewed as a projection where k ∈ Rn are the basis functions and ck are the unknown constants to be determined. This is done for each frequency step. The coefficients ck are evaluated using the Galerkin approach so that the error is made orthogonal to the basis functions, that is, mathematically (ω, θ)⊥j ⇛ j , (ω, θ) = 0 ∀ j = 1, 2, . . . , n (75)
  • 61. The Galerkin approach Imposing the orthogonality condition and using the expression of the error one has E T j XM i=0 ! Ai ξi (θ) Xn k=1 ckb k ((θ))k ! − T j f # = 0, ∀j (76) Interchanging the E [•] and summation operations, this can be simplified to Xn k=1 XM i=0 T j Aik E h ξi (θ)b k ((θ)) i! ck = E h T j f i (77) or Xn k=1 XM i=0 eA ijkDijk ! ck = bj (78)
  • 62. Model Reduction by reduced number of basis Suppose the eigenvalues of A0 are arranged in an increasing order such that λ01 λ02 . . . λ0n (79) From the expression of the spectral functions observe that the eigenvalues ( λ0k = ω2 0k ) appear in the denominator: (1) k (ω, (θ)) = T k f(ω) 0k (ω) + PM i=1 ξi (θ)ik (ω) (80) where 0k (ω) = −ω2 + iω(ζ1 + ζ2ω2 0k ) + ω2 0k The series can be truncated based on the magnitude of the eigenvalues relative to the frequency of excitation. Hence for the frequency domain analysis all the eigenvalues that cover almost twice the frequency range under consideration can be chosen.
  • 63. Computational method The mean vector can be obtained as ¯u = E [ˆu(θ)] = Xp k=1 ckE h b k ((θ)) i k (81) The covariance of the solution vector can be expressed as u = E h (ˆu(θ) − ¯u) (ˆu(θ) − ¯u)T i = Xp k=1 Xp j=1 ckcjkjkT j (82) where the elements of the covariance matrix of the spectral functions are given by kj = E h b k ((θ)) − E h b k ((θ)) i b j ((θ)) − E h b j ((θ)) ii (83)
  • 64. Summary of the computational method 1 Solve the generalized eigenvalue problem associated with the mean mass and stiffness matrices to generate the orthonormal basis vectors: K0 = M00 2 Select a number of samples, say Nsamp. Generate the samples of basic random variables ξi (θ), i = 1, 2, . . . ,M. 3 Calculate the spectral basis functions (for example, first-order): k (ω, (θ)) = T k f(ω) 0k (ω)+PM i=1 ξi (θ)ik (ω) , for k = 1, · · · p, p n 4 Obtain the coefficient vector: c(ω) = S−1(ω)b(ω) ∈ Rn, where b(ω) = gf(ω) ⊙ (ω), S(ω) = 0(ω) ⊙ D0(ω) + PM i=1 eA i (ω) ⊙ Di (ω) and Di (ω) = E h (ω, θ)ξi (θ)T (ω, θ) i , ∀ i = 0, 1, 2, . . . ,M 5 Obtain the samples of the response from the spectral series: ˆu(ω, θ) = Pp k=1 ck (ω)k ((ω, θ))k
  • 65. The Euler-Bernoulli beam example An Euler-Bernoulli cantilever beam with stochastic bending modulus for a specified value of the correlation length and for different degrees of variability of the random field. F (c) Euler-Bernoulli beam 6000 5000 4000 3000 2000 1000 0 0 5 10 15 20 Natural Frequency (Hz) Mode number (d) Natural frequency dis-tribution. 10−1 10−2 10−3 10−4 0 5 10 15 20 25 30 35 40 100 / lj Ratio of Eigenvalues, l1 Eigenvalue number: j (e) Eigenvalue ratio of KL de-composition Length : 1.0 m, Cross-section : 39 × 5.93 mm2, Young’s Modulus: 2 × 1011 Pa. Load: Unit impulse at t = 0 on the free end of the beam.
  • 66. Problem details The bending modulus of the cantilever beam is taken to be a homogeneous stationary Gaussian random field of the form EI(x, θ) = EI0(1 + a(x, θ)) (84) where x is the coordinate along the length of the beam, EI0 is the estimate of the mean bending modulus, a(x, θ) is a zero mean stationary random field. The covariance kernel associated with this random field is ae−(|x1−x2|)/μa (85) Ca(x1, x2) = σ2 where μa is the correlation length and σa is the standard deviation. A correlation length of μa = L/5 is considered in the present numerical study.
  • 67. Problem details The random field is assumed to be Gaussian. The results are compared with the polynomial chaos expansion. The number of degrees of freedom of the system is n = 200. The K.L. expansion is truncated at a finite number of terms such that 90% variability is retained. direct MCS have been performed with 10,000 random samples and for three different values of standard deviation of the random field, σa = 0.05, 0.1, 0.2. Constant modal damping is taken with 1% damping factor for all modes. Time domain response of the free end of the beam is sought under the action of a unit impulse at t = 0 Upto 4th order spectral functions have been considered in the present problem. Comparison have been made with 4th order Polynomial chaos results.
  • 68. Mean of the response (f) Mean, a = 0.05. (g) Mean, a = 0.1. (h) Mean, a = 0.2. Time domain response of the deflection of the tip of the cantilever for three values of standard deviation σa of the underlying random field. Spectral functions approach approximates the solution accurately. For long time-integration, the discrepancy of the 4th order PC results increases.
  • 69. Standard deviation of the response (i) Standard deviation of de-flection, a = 0.05. (j) Standard deviation of de-flection, a = 0.1. (k) Standard deviation of de-flection, a = 0.2. The standard deviation of the tip deflection of the beam. Since the standard deviation comprises of higher order products of the Hermite polynomials associated with the PC expansion, the higher order moments are less accurately replicated and tend to deviate more significantly.
  • 70. Frequency domain response: mean 10−2 10−3 10−4 10−5 10−6 10−7 0 100 200 300 400 500 600 Frequency (Hz) : 0.1 damped deflection, sf MCS 2nd order Galerkin 3rd order Galerkin 4th order Galerkin deterministic 4th order PC (l) Beam deflection for a = 0.1. 10−1 10−2 10−3 10−4 10−5 10−6 10−7 0 100 200 300 400 500 600 Frequency (Hz) : 0.2 damped deflection, sf MCS 2nd order Galerkin 3rd order Galerkin 4th order Galerkin deterministic 4th order PC (m) Beam deflection for a = 0.2. The frequency domain response of the deflection of the tip of the Euler-Bernoulli beam under unit amplitude harmonic point load at the free end. The response is obtained with 10, 000 sample MCS and for σa = {0.10, 0.20}.
  • 71. Frequency domain response: standard deviation 10−2 10−3 10−4 10−5 10−6 10−7 0 100 200 300 400 500 600 Frequency (Hz) : 0.1 Standard Deviation (damped), sf MCS 2nd order Galerkin 3rd order Galerkin 4th order Galerkin 4th order PC (n) Standard deviation of the response for a = 0.1. 10−1 10−2 10−3 10−4 10−5 10−6 0 100 200 300 400 500 600 Frequency (Hz) : 0.2 Standard Deviation (damped), sf MCS 2nd order Galerkin 3rd order Galerkin 4th order Galerkin 4th order PC (o) Standard deviation of the response for a = 0.2. The standard deviation of the tip deflection of the Euler-Bernoulli beam under unit amplitude harmonic point load at the free end. The response is obtained with 10, 000 sample MCS and for σa = {0.10, 0.20}.
  • 72. Experimental investigations Figure: A cantilever plate with randomly attached oscillators - Probabilistic Engineering Mechanics, 24[4] (2009), pp. 473-492
  • 73. Measured frequency response function 60 40 20 0 −20 −40 −60 0 100 200 300 400 500 600 Frequency (Hz) (w) (1,1) Log amplitude (dB) of H Baseline Ensemble mean 5% line 95% line
  • 74. Conclusions The mean response of a damped stochastic system is more damped than the underlying baseline system For small damping, ξe ≈ 31/4pǫ pπ √ξ Random modal analysis may not be practical or physically intuitive for stochastic multiple degrees of freedom systems Conventional response surface based methods fails to capture the physics of damped dynamic systems Proposed spectral function approach uses the undamped modal basis and can capture the statistical trend of the dynamic response of stochastic damped MDOF systems
  • 75. Conclusions The solution is projected into the modal basis and the associated stochastic coefficient functions are obtained at each frequency step (or time step). The coefficient functions, called as the spectral functions, are expressed in terms of the spectral properties (natural frequencies and mode shapes) of the system matrices. The proposed method takes advantage of the fact that for a given maximum frequency only a small number of modes are necessary to represent the dynamic response. This modal reduction leads to a significantly smaller basis.
  • 76. Assimilation with experimental measurements In the frequency domain, the response can be simplified as u(ω, θ) ≈ Xnr k=1 T k f(ω) −ω2 + 2iωζkω0k + ω2 0k + PM i=1 ξi (θ)ik (ω) k Some parts can be obtained from experiments while other parts can come from stochastic modelling.