The Finite Element Method for the Analysis of Non-Linear and ...

The Finite Element Method for the Analysis of
Non-Linear and Dynamic Systems
Prof. Dr. Eleni Chatzi
Lecture 9 - 27 November, 2012
Institute of Structural Engineering Method of Finite Elements II 1

State Space Equation Formulation
2dof Mass Spring System
k 1
c 1
m 1
( t 1 )
x 1 ( t )
F k
2
m
2
c 1
F ( t 2 )
x 2 ( t )
FBD
k1x1
cx
1 1
m 1
( t 1 )
k
( x2 − x1)
k2( x2 − x1)
2
m
2
F
c2( x2 − x1)
c2( x2 − x1)
(Lumped Mass System)
F ( t 2 )
We introduce the augmented state vector:
x = [ x 1 x 2 ẋ 1 ẋ 2
] T (controllable form equivalent). Then,
ẋ =
⎡
⎢
⎣
⎤
0 0 1 0
0 0 0 1
[
−m −1 k ] [ −m −1 c ] ⎥
⎦
⎡ ⎤ ⎡
x 1
⎢ x 2
⎥
⎣ ẋ 1
⎦ + ⎢
⎣
ẋ 2
⎤
0 0
0 0
[
m
−1 ] ⎥
⎦
[
F1
F 2
]
Institute of Structural Engineering Method of Finite Elements II 2

State Space Equation Formulation
2dof Mass Spring System
where u = [ F 1
F 2
] T
⇒ ẋ = Ax + Bu
Assume you would like to monitor the displacement x 1 , x 2 . Then
the “observation vector” is:
y =
⎡
⎢
⎣
1 0 0 0
0 1 0 0
0 0 0 0
0 0 0 0
⎤ ⎡
⎥ ⎢
⎦ ⎣
⎤
x 1
x 2
x 3
x 4
⎥
⎦ + O 4×2 u(t)
⇒ y = Cx + Du
Institute of Structural Engineering Method of Finite Elements II 3

State Space Equation Formulation
Using the state space representation we have converted a 2nd order ODE
into an equivalent 1st order ODE system. We can now use any 1st order
ODE integration method to convert the continuous system into a
discrete one and obtain an approximate solution:
1st order ODE Integration Methods
Assume dy
dt = f(t, y(t)), y(t 0) = 0
Forward Euler Method
y n+1 = y n + hf(t n , y n )
where h is the integration time step. This explicit expression is
obtained from the truncated Taylor Expansion of y(t n + h)
Backward Euler Method
y n+1 = y n + hf(t n+1, y n+1 )
This implicit expression (since y n+1 is on the right hand side) is
obtained from the truncated Taylor Expansion of y(t n+1 − h)
Institute of Structural Engineering Method of Finite Elements II 4

State Space Equation Formulation
2nd Order Runge Kutta (RK2)
k 1 = hf(t n , y n ), k 2 = hf(t n + 1 2 h, y n + 1 2 k 1)
y n+1 = y n + k 2 + O(h 3 )
4th Order Runge Kutta (RK4)
k 1 = hf(t n , y n ), k 2 = hf(t n + 1 2 h, y n + 1 2 k 1)
k 3 = hf(t n + 1 2 h, y n + 1 2 k 2), k 4 = hf(t n + h, y n + k 3 )
y n+1 = y n + 1 6 k 1 + 1 3 k 2 + 1 3 k 3 + 1 6 k 4 + O(h 5 )
Institute of Structural Engineering Method of Finite Elements II 5

Mode Superposition Method
Direct Integration Methods
Assuming diagonal mass matrix and no damping, implicit methods
require on average 2nm k calculations per time step
where n is the the dimension of the stiffness matrix and m k the
half-bandwidth
The central difference method (explicit scheme) requires significantly
fewer operations per time step however has other drawbacks (requires
T < T cr ⇒ more time steps!)
The total number of operations in a Direct Integration Method are
directly proportional to the number of load steps ⇒ only effective for
short durations (small number of time steps)
Mode Superposition Methods
These methods work by transforming the equilibrium equations into a
form in which step by step solution is less costly
In particular we aim at a reduction of m k which would proportionally
decrease the step by step solution
Institute of Structural Engineering Method of Finite Elements II 6

Mode Superposition Method
Equilibrium Equations Transformation - Change of Basis
The following transformation is used on the n finite element nodal
point displacements U:
U(t) = PX(t)
where P is an n × n square matrix and X(t) is the “generalized
displacement” vector (time dependent)
The dynamic equation of motion then can be written as:
˜MẌ(t) + ˜CẊ(t) + ˜KX(t) = ˜R(t)
where ˜M = P T MP ˜C = P T CP ˜K = P T KP ˜R = P T R
are the transformed matrices corresponding to a smaller bandwidth
than the original system.
Institute of Structural Engineering Method of Finite Elements II 7

Mode Superposition Method
Choice of Transformation Matrix P
P is established using the displacement solution for Undamped, Free
vibrations:
MÜ + KU = 0 (1)
The solution of this 2nd Order ODE is of the form:
U = φsinω(t − t 0 )
where φ is a vector of order n, t is the time variable, t 0 is a time
constant corresponding to the phase and ω is the vibration frequency
of vector φ, then
Ü = −ω 2 φsinω(t − t 0 )
and (1) becomes:
−ω 2 Mφ + Kφ = 0 (eigenproblem)
Institute of Structural Engineering Method of Finite Elements II 8

Mode Superposition Method
Matrix Form
Defining a matrix Φ whose columns are the eigenvectors φ i and a
diagonal matrixΩ 2 which stores the eigenvalues ωi 2 on its diagonal,
i.e:
⎡
⎤
ω 1
Φ = [ ]
φ 1 , φ 2 , ... φ n ; Ω 2 = ⎢ ω 2 ⎥
⎣ ... ⎦ ;
ω n
we can write the n solutions to the eigenproblem as:
MΦΩ 2 = KΦ
Institute of Structural Engineering Method of Finite Elements II 9

Properties of Eigenvectors
Solution of the Eigenproblem
−ω 2 Mφ + Kφ = 0
(eigenproblem)
The above homogeneous linear system of equations can only have a
solution if the determinant is equal to 0:
∥
∥−ω 2 M + K ∥ ∥ = 0
The solution of the above equation will yield:
n eigenvalues ωi 2,
i = 1, ..., n
with 0 ≤ ω 1 ≤ ...ω n (the eigenfrequencies) and
The solution of ω 2 Mφ = Kφ (eigenproblem) will yield:
n eigenvectors (or modal vectors) φ i
Institute of Structural Engineering Method of Finite Elements II 10

Properties of Eigenvectors
M-Orthonormality
From Matrix Properties we know that (AB) T = B T A T ,
Thus, for two eigenvectors φ n , φ r we obtain:
(φ T nKφ r ) T = φ T rK T φ n
Since the Stiffness matrix is symmetric, K = K T , hence
(φ T nKφ r ) T = φ T rKφ n
The same applies for the Mass matrix, M = M T , yielding:
(φ T nMφ r ) T = φ T rMφ n
Institute of Structural Engineering Method of Finite Elements II 11

Properties of Eigenvectors
M-Orthonormality
The eigenproblem for vector n then can be written as:
ωrφ 2 T nMφ r = φ T transpose
nKφ r −→
ω 2 rφ T rMφ n = φ T rKφ n
and the eigenproblem for vector n is: The eigenproblem for vector n
then can be written as:
ω 2 nφ T rMφ n = φ T rKφ n
By subtracting the two previous formulas we obtain:
(ω 2 r − ω 2 n)φ T rMφ n = 0
We conclude that for n ≠ r ⇒ φ T rMφ n = 0, φ T rMφ n = 0,
Institute of Structural Engineering Method of Finite Elements II 12

Properties of Eigenvectors
M-Orthonormality
We choose φ n , such that:
Φ T MΦ = I
therefore from MΦΩ 2 = KΦ we have that:
Φ T KΦ = Ω 2
The principle of M-orthonormality can then be written as:
{ 1, n = r
φ T rMφ n =
0, n ≠ r
{ ω
2
φ T rKφ n = n , n = r
0, n ≠ r
Institute of Structural Engineering Method of Finite Elements II 13

Mode Superposition Method
It is now obvious that we can use Φ as the transformation matrix P.
Using:
U(t) = ΦX(t)
we obtain the transformed equilibrium equation:
Ẍ(t) + Φ T CΦẊ(t) + Ω 2 X(t) = Φ T R(t)
where using the property of M-orthonormality, the initial conditions
will be:
0 X = Φ T M 0 U;
0 Ẋ = Φ T M 0 ˙U
Institute of Structural Engineering Method of Finite Elements II 14

Dynamic Response with Damping Neglected
If we neglect the velocity dependent damping effects the equilibrium
equation reduces to:
Ẍ(t) + Ω 2 X(t) = Φ T R(t)
i.e n individual equations of the form (since Ω 2 is diagonal):
ẍ i (t) + ωi 2x }
i(t) = r i (t)
r i (t) = φ T i R(t) i = 1, 2, ..., n
with
0 x i = φ T i M 0 U;
0ẋ i = φ T i M 0 ˙U
Institute of Structural Engineering Method of Finite Elements II 15

Dynamic Response with Damping Neglected
SDOF Response
Each equation of the previous system describes a single degree of freedom
system with unit mass and stiffness ωi 2 . The solution to this equation for a
random input excitation can either be obtained by using the direct
integration methods or by using the Duhamel Integral:
x i (t) = 1 ∫ t
r i (τ)sinω i (t − τ) dτ + α i sinω i t + β i cosω i t
ω i
0
where α i , β i are determined from the initial conditions. Therefore, the
SDOF response id owed to two contributions
A dynamic (steady - state) response obtained by multiplying the static
response by a dynamic load factor (this is the particular solution of
the governing differential equation), and
An additional dynamic response called the transient response
Institute of Structural Engineering Method of Finite Elements II 16

Dynamic Response with Damping Neglected
Complete Response
The solution of all n SDOF equations are calculated and the finite
element nodal point displacements are obtained by superposition of
the response in each mode:
U(t) = ΦX(t) ⇒ U(t) =
n∑
φ i x i (t)
i=1
Therefore the solution scheme is:
Solve for the eigenvalues and eigenvectors of problem
Solve for the response of the decoupled SDOF equations
Use superposition to find the total response.
Institute of Structural Engineering Method of Finite Elements II 17

Superposition Principle
An alternative View
Superposition
A dynamic load can be designed as
Fourier series of harmonic sine and cosine
contributions
The total solution of such a problem is
equal to the superposition of solution of
the Fourier terms.
A dynamic load can be designed as Fourier series
The total solution of such a problem is equal to
the superposition of solution of the Fourier terms.
U(t) = ΦX(t) ⇒ U(t) =
n∑
φ i x i (t)
i=1
13.11.2009 Mode Superposition 11
Institute of Structural Engineering Method of Finite Elements II 18

The effect of Damping
Problems with neglected dampi
The presence of damping reduces the dynamic load factor (which then
cannot be infinite) and damps out the transient response
The response in the
modes with ˆω large is
ω
negligible
For ˆω close to zero the
ω
system follows the loads
statically
Effectively only the first p modes need to be used p ≤ n, in order to obtain
a good approximate solution ⇒ Advantage over Direct Integration Methods
13.11.2009 Mode Superposition
Institute of Structural Engineering Method of Finite Elements II 19

Swiss Federal Institute of Technology
Example - 2 DOF system
Direct Integration ti Methods
Calculate the displacement response of the system
Pag
k
1
= 4
U , U , U
1 1 1
m = 2 R = 0
1
2
k
2
= 2
1
0
m
2
= 1 R
2
= 10
⎡2 0⎤⎡U
⎤ 6 2
1 ⎡ − ⎤⎡U1
⎤ ⎡0⎤
⎢ ⎥⎢ ⎥+ ⎢ =
⎥⎢ ⎥ ⎢ ⎥
⎣0 1 ⎦
⎣U
2
⎦
⎣−2 4 ⎦ ⎣U
2
⎦
⎣10
⎦
U , U , U
2 2 2
k
3
= 2
Method of Finite Elements II
Institute of Structural Engineering Method of Finite Elements II 20

Example - 2 DOF system
Eigenproblem Setup
ω 2 nMφ n = Kφ n ⇒ ω 2 n
Eigenvalue Calculation
[ 2, 0
0 1
]
=
[ 6, −2
−2 4
]
φ n
∥ K − ω
2
n M ∥ ∥ ∥∥∥ 6 − 2ω 2
= 0 ⇒ n,
−2
−2
4 − 1ωn
2
∥ = 2ω4 n − 14ωn 2 + 20 = 0 ⇒
ω1 2 = 14 − √ 196 − 160
4
Eigenvector Calculation
= 2, ω 2 2 = 14 + √ 196 − 160
4
= 5
(
K − ω
2
1 M ) φ 1 = 0 ⇒
[ 2, −2
−2 2
] [ ]
φ11
= 0 ⇒
φ 12
[
φ11
φ 12
]
[
1
= λ ∗
1
]
Institute of Structural Engineering Method of Finite Elements II 21

Example - 2 DOF system
Use of M - orthonormality
φ T 1Mφ 1 = 1 ⇒ λ 2 (m 1 + m 2 ) = 1 ⇒ λ = 1 3
⎡
1
⎤
⎡ √ ⎤ 2
√
⎢
Hence, φ 1 = ⎣ 3 ⎥
1 ⎦ Similarly, φ 2 = ⎢ 2 √ 3
√ ⎥
⎣
√ 2 ⎦
−√ 3 3
Transformed Equilibrium Equation
[ 2, 0
Ẍ(t) +
0 5
Ẍ(t) + Ω 2 X(t) = Φ T R(t) ⇒
⎡
]
X(t) = ⎢
⎣
⎤
1 1
√ √3 [
√ 3 √ ⎥ 0
2 2
2 √ ⎦ 10
−√ 3 3
]
Institute of Structural Engineering Method of Finite Elements II 22

Example - 2 DOF system
Decoupled SDOF equations
ẍ 1 + 2x 1 = 10 √
3
√
2
ẍ 2 + 5x 2 = −10
3
Initial Conditions
U(0) = 0
˙U(0) = 0
and
x i (0) = φ T i MU(0)
ẋ i (0) = φ T i M ˙U(0)
⇒
x 1 (0) = 0, ẋ 1 (0) = 0, x 2 (0) = 0, ẋ 2 (0) = 0
Then the exact solutions to the ODEs are:
x 1 = 5 √
3
(1 − cos √ 2t) x 2 = 2√
2
3 (−1 + cos√ 5t)
And the total solution is:
n∑
U(t) = φ i x i (t)
i=1
Institute of Structural Engineering Method of Finite Elements II 23

Example - 2 DOF system
Comparison of Direct Integration and Modal Superposition
Methods
Institute of Structural Engineering Method of Finite Elements II 24

Example - 2 DOF system
Note
In this case we dealt with a 2DOF example and used an analytical
solution to solve the two decoupled SDOF system equations
In practice for multi degree of freedom systems, higher modes are
neglected and instead of an analytical solution we again use a Direct
Integration scheme to solve for each SDOF ODE
In essence the finally achieved accuracy might be the same between
the two approaches, however teh mode superposition method usually
requires for less computational cost
Institute of Structural Engineering Method of Finite Elements II 25

Analysis Including Damping
Transformed Equilibrium Equation
Ẍ(t) + Φ T CΦẊ(t) + Ω 2 X(t) = Φ T R(t)
Proportional Damping Assumption
φ T i Cφ j = 2ω i ξ i δ ij (2)
where ξ i is a modal damping parameter and δ ij is the Kronecker delta
(δ ij = 1 for i = j, δ ij = 0 for i ≠ j)
Institute of Structural Engineering Method of Finite Elements II 26

Analysis Including Damping
Therefore, we still end up with Decoupled SDOF equations for
each x i :
ẍ i (t) + 2ω i ξ i ẋ i (t) + ω 2 i x i (t) = r i (t)
with the Duhamel Integral now being:
x i (t) = 1¯ω ∫ t {
r i (τ) e −ξ iω i (t−τ) sin¯ω i (t − τ) dτ
i 0
}
+ e −ξ iω i t (α i sin¯ω i t + β i cos¯ω i t)
where ¯ω i = ω i
√1 − ξ 2 i
Institute of Structural Engineering Method of Finite Elements II 27

Analysis Including Damping
Rayleigh Damping
If there are only two different damping ratios ξ i ,
Damping can be used:
i = 1, 2 Rayleigh
C = αM + βK (3)
Eqns (2), (3) now yield:
φ T i (αM + βK)φ i = 2ω i ξ i ⇒
α + βω 2 i = 2ω i ξ i
The 2 × 2 system can be solved to obtain α, β.
In actual analysis it may well be that the damping ratios are known
for many more than two frequencies. In that case two average values
say ¯ξ 1 , ¯ξ 2 are used to evaluate α, β.
Institute of Structural Engineering Method of Finite Elements II 28

Measuring Damping
Logarithmic decrement method can be used to measure damping in time
domain. In this method, the free vibration displacement amplitude history of a
system to an impulse is recorded. Logarithmic decrement is the natural
logarithmic value of the ratio of two adjacent peak values of displacement in free
decay vibration.
Institute of Structural Engineering Method of Finite Elements II 29

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 Method of Finite Elements II 30

Example - Damping for an MDOF system
Damping as function of frequency
A problem with Rayleigh
damping is that higher
modes are much more
damped than lower modes.
However, in general
Rayleigh damping provides
a good approximation
Institute of Structural Engineering Method of Finite Elements II 31