The Finite Element Method for the Analysis of Non-Linear ... · The Finite Element Method for the Analysis of Non-Linear and Dynamic Systems Prof. Dr. Eleni Chatzi Lecture 6 - 5 November,

The Finite Element Method for the Analysis ofNon-Linear and Dynamic Systems

Prof. Dr. Eleni Chatzi

Lecture 6 - 5 November, 2015

Institute of Structural Engineering Method of Finite Elements II 1

Introduction to Dynamic Analysis

The very basics - Newton’s 2nd law of motion:

Swiss Federal Institute of Technology Page 3

I t d ti t D i A l iIntroduction to Dynamic Analysis

I d i• Introduction

The very basics ☺ k Unstretched positionkΔ

m

p

Δm

kΔNewtons 2‘nd law of motion:

( )mx w k x= − Δ +

⇓ Stretched position(static equilibrium)w mg=

,x xx( ) 0mx w k x

⇓− + Δ + =

⇓ ,x xx

2

( ) 0

0

mx k k x mx kxk

⇓− Δ + Δ + = + =

2 0nkx x x xm

ω+ = + =

⇓

22 2

1 12

nn

n

mk

kfm

πω τ π τ πω

τ π

= ⇒ = =

= =

Method of Finite Elements II

0 0sin cos sin costn n n nx A t B t x x t x tω ω ω ω= + ⇒ = +


I t d ti t D i A l iIntroduction to Dynamic Analysis

I d i• Introduction

The very basics ☺ k Unstretched positionkΔ

m

p

Δm

kΔNewtons 2‘nd law of motion:

( )mx w k x=

⇓ Stretched position(static equilibrium)w mg=

,x xx( ) 0mx w k x

⇓− + Δ + =

⇓ ,x xx

2 0

mx k k x mx kxk

⇓

2 0nkx x x xm

ω+ = + =

⇓

22 2

1 12

nn

n

mk

kfm

πω τ π τ πω

τ π

= ⇒ = =

= =


0 0sin cos sin costn n n nx A t B t x x t x tω ω ω ω= + ⇒ = +

x(t) = Asinωnt+Bcosωnt⇒ x(t) =x0

ωnsinωnt+ x0cosωnt

where ωn =2π

T=

√k

m



We have previously considered the equilibrium equations governingthe linear dynamic response of a system of finite elements:

MU + CU + KU = R

where:

M : Mass Matrix

K : Stiffness Matrix

U : Displacements

U : Velocities

U : Accelerations

FI(t) + FD(t) + FE(t) = R(t)

where:

FI(t) = MU Inertial Force

FD(t) = CU Damping Force

FE(t) = KU Internal Force



When is Dynamic Analysis required?

Whether a dynamic analysis is needed or not is generally up toengineering judgment

requires understanding of the interaction between loading andstructural response!

In general if the loading varies over time with frequencies higherthan the Eigen-frequencies of the structure 7−→ dynamic analysis willbe required



Objective

Solve the dynamic Equation of motion (numerically)

MU + CU + KU = R

In principle the equilibrium equations may be solved by any standardnumerical integration scheme BUT!

Efficiency numerical efforts must be considered and it is worthwhileto look at special techniques of integration which are especiallysuited for the analysis of finite element assemblies.

*The section on Direct Integration Methods is based on

Prof. M. Faber’s notes of the FEM II course - Fall 2009


Direct Integration Methods

These methods rely in discretizing the continuous problem. Theoriginal problem formulation is essentially the dynamic equation ofmotion which is for structural problems is a second order OrdinaryDifferential Equation (ODE).

This involves, after the solution is defined at time zero, the attemptto satisfy dynamic equilibrium at discrete points in time. Mostmethods use equal time intervals at ∆t, 2∆t, 3∆t........N∆t.

However this is not mandatory; in some cases a variable time stepmight be employed. This is most commonly the case for specialclasses of problems such as Impact Problems.



Direct means: The equations are solved in their original form

Two ideas are utilized

1. The equilibrium equations are satisfied only at time steps, i.e. atdiscrete times with intervals ∆t

2. A particular variation of displacements, velocities andaccelerations within each time interval is assumed

The accuracy depends on these assumptions as well as the choice oftime intervals!



Assumptions

The displacements, velocities and accelerations

U0 : Displacement vector at time t

U0 : Velocity vector at time t

U0 : Acceleration vector at time t

are assumed to be known and we aim to establish the solution of theequilibrium equations for the period 0 - T .

For this purpose we sub-divide T into n intervals of length∆t = T/n and establish solutions for the times ∆t, 2∆t, 3∆t,..., T



We distinguish principally between Implicit and Explicit methods

Explicit methods:

Solution is based on the equilibrium equations at time t

1st order Example: The Forward Euler Method

Assuming we want to approximate the solution of the initialvalue problem

y = f(t, y(t)), y(t0) = y0

by using the first two terms of the Taylor expansion of y, whichrepresents the linear approximation around the point (t0, y(t0))we obtain the forward Euler integration rule:

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



Implicit methods:

Solution is based on the equilibrium equations at time t+ ∆t

1st order Example: The Backward Euler Method

Assuming we want to approximate the solution of the sameinitial value problem the backward Euler integration rule isobtained as:

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

Since f(t, y) is, in general, a non-linear function of y, iterationis required to solve this last equation for yn+1.



Explicit methods calculate the state of a system at a later time fromthe state of the system at the current time, while implicit methodsfind a solution by solving an equation involving both the currentstate of the system and the later one.

Note that:

Implicit integration is not necessarily more accurate than explicit. Itcan be less accurate! The major benefit of implicit integration isstability. Many of these methods are able to run with any arbitrarilylarge time step for any input unless we are lying at the limits offloating point math (unconditionally stable). Obviously a large timestep implies throwing away accuracy.

For most real structures, which contain stiff elements, a very smalltime step is required in order to obtain a stable solution. Therefore,all explicit methods are conditionally stable. The condition is the sizeof the selected time step.



Most Commonly used Direct Integration Methods

(for the case of the Dynamic Equation of Motion)

1 The Central Difference Method (CDF)

2 The Houbolt method

3 The Newmark method

4 The Wilson θ method

5 Coupling of integration operators

The difference in items 1-4 lies in the way we choose a discretized

equivalent of the derivatives. The overall setup of the solution is very much

similar for all methods. Additionally depending on the resulting equations

some schemes are explicit (CDF) and others implicit (Houbolt, Newmark,

Wilson θ)




Central Difference Method

Velocity Acceleration

U(t)

t - Δt

t t + Δt

t - Δt

t t + Δt

U(t)

t - Δt

t t + Δt

Reminder: The difference lies in the way we choose a discretizedequivalent of the derivatives.




The Houbolt Method

Displacement Velocity

Acceleration

U(t)

t - 2Δt

t t + Δt t - Δt

Houbolt’s method uses a third-order interpolation of displacements

extending two steps back in time.

These approximations are derived via the displacement approximation.





The Newmark Method Displacement Velocity

The Newmark method uses a second order Taylor expansion for approximating Velocities and

Accelerations:

!Ut+Δt = !Ut + Δt !!UV

Ut+Δt =Ut + Δt !!UV + Δt2

2!!U D

, where !!UV , !!U D are approximations of the Acceleration

t + Δt t

t + Δt t

if if



The Central Difference Method

Approximate the velocity (first derivative) as:

tU =1

2∆t(t+∆tU− t-∆tU) (1)

Approximate the acceleration (second derivative) as:

tU =1

∆t2(t+∆tU− 2tU + t-∆tU) (2)

Substitute Eqns (1),(2) into MU + CU + KU = R:

(1

∆t2M +

1

∆t2C

)t+∆tU = tR−

(K− 2

∆t2M

)tU−

(1

∆t2M− 1

2∆tC

)t-∆tU

⇒ (a0M + a1C) t+∆tU = tR− (K− a2M) tU− (a0M− a1C) t-∆tU



We see that we do not need to factorize the stiffness matrix, i.e,we only rely on info from the current step t to go to the nextone t+ ∆t(explicit method)

We also see that in order to calculate the displacements at time∆t we need to know the displacements at time 0 and ∆t

In general, 0U , 0U , 0U are known and we can use Eqns (1),(2)to obtain −∆tU :

-∆tU = 0U−∆t0U +∆t2

20U



Solution Procedure

A. Initial Calculations



Solution Procedure

B. For each time step


Di t I t ti M th dDirect Integration Methods

Th C l diff h d S l i d• The Central difference method: Solution procedure:

B: For each time step

1) Calculate effective loads at time t:

( ) ( )ˆt t t t t−ΔR R K M U M C U

2) Solve for the displacements U at time t + Δt

( ) ( )2 0 1t t t t ta a a −Δ= − − − −R R K M U M C U

ˆLDL U RT t t t+Δ

3) If required, solve for the corresponding velocities and accelerations

LDL U RT t t t+Δ =

accelerations

0 ( 2 )U U U + Ut t t t t ta −Δ +Δ= −

( )U U + Ut t t t ta −Δ +Δ=


1( )U U + Ua=



The effectiveness of the central difference method depends on theefficiency of the time step solution since generally a smalldiscretization is required.

For this reason the method is usually only applied when a lumped(diagonal) mass matrix can be assumed and when the velocitydependent damping (C) can be neglected, i.e.:



• The Central difference method

The effectiveness of the central difference method depends on theThe effectiveness of the central difference method depends on the efficiency of the time step solution – because we need a lot of themF thi th th d i ll l li d h l dFor this reason the method is usually only applied when a lumped mass matrix can be assumed and when the velocity dependent damping can be neglected, i.e.:

2

1 ˆM U Rt t t

t+Δ =

Δ2 1R R K M U M Ut t t t t−Δ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟

2ˆ ( ), 0t t t

i i iitU R mm

+Δ Δ= >

2 2R R K M U M Ut t t t t

t tΔ⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟Δ Δ⎝ ⎠ ⎝ ⎠

iim

( )( )K U K U F it i t t

i i

= =∑ ∑( ) 1ˆ ( 2 )R R F M U Uit t t t t t−Δ∑


2 ( 2 )R R F M U Ut t t t t t

i tΔ= − − −

Δ∑



Example - 2 DOF system



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 IIFor this system the natural periods are T1 = 4.45, T2 = 2.8








Di t I t ti M th dDirect Integration Methods11

22

2 0 6 2 00 1 2 4 10

UUUU

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

• The Central difference method: Example:

0 1 2 0 32

1 1 1, , 2 ,2

a a a a at t a

= = = =Δ Δ

2

0 12

21 112.8, 1.79,

(0.28) 2 0.281 1

t t a

a a

Δ Δ

= = = =⋅

For Δt = 0.28

2 32

1 12 25.5, 0.0392(0.28) 25.5

a a= ⋅ = = =

0 0 0 00 28 0 0392t−Δ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= − + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥U 0.28 0.03920 0 10 0.0392

2 0 0 0 25.5 0ˆ 12 8 1 79

i = + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥

U

M 12.8 1.790 1 0 0 0 12.8

0 45.0 2 25.5 0ˆt t t t−Δ

= + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + −⎢ ⎥ ⎢ ⎥ ⎢ ⎥

M

R U U


10 2 21.5 0 12.8= +⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦R U U




The equation which must be solved for each time step is:


Di t I t ti M th dDirect Integration Methods11

22

2 0 6 2 00 1 2 4 10

UUUU

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

• The Central difference method: Example:

The equation which must be solved for each time step is:The equation which must be solved for each time step is:

125.5 0 ˆt t

tU+Δ⎡ ⎤⎡ ⎤

=⎢ ⎥⎢ ⎥ R2

0 12.8 t t

U+Δ

=⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

R

0 45 0 2 25 5 0t tt UU−Δ⎡ ⎤⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 11

2 2

0 45.0 2 25.5 0ˆ10 2 21.5 0 12.8 t t

tt

t

UUU U

−Δ

⎡ ⎤⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + − ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦R





The results are:

5

6

CentralDifference_U1

3

4

5CentralDifference_U2

0

1

2

00 0.5 1 1.5 2 2.5 3 3.5 4


The Houbolt Method

Houbolt Method Derivative Approximation



MU + CU + KU = R c

22

1 (2 5 4 )t t t t t t t t t

t+Δ +Δ −Δ − Δ= − −

ΔU U U + U U a

tΔ21 (11 18 9 2 )

6t t t t t t t t t

t+Δ +Δ −Δ − Δ= − −

ΔU U U + U U b

a and b inserted in c

2

2 1265 3 4 3 1 1

t t

t t+Δ⎛ ⎞+ + =⎜ ⎟Δ Δ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

M C K U

22 2 2

5 3 4 3 1 12 3

t t t t t t t

t t t t t t+Δ −Δ − Δ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + − + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟Δ Δ Δ Δ Δ Δ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

R M C U M C U M C U

Method of Finite Elements II Institute of Structural Engineering Method of Finite Elements II 26

The Houbolt Method

We will not consider the Houbolt in more detail however it is notedthat it is necessary to factorize the stiffness matrix

(implicit method)

Furthermore, if the mass and damping terms are neglected, theHoubolt method results in the static analysis equations:

K t+∆tU = t+∆tR

Notice how this is not true for the CDM!


The Newmark Method

In 1959 Newmark presented a family of single-step integrationmethods for the solution of structural dynamic problems for bothblast and seismic loading.



Th N k h d• The Newmark method

This method may be seen as an extension of the Wilson θ method

(1 )t t t t t t tδ δ+Δ +Δ⎡ ⎤= + − + Δ⎣ ⎦U U U U

21( )2

t t t t t t tt tα α+Δ +Δ⎡ ⎤= + Δ + − + Δ⎢ ⎥⎣ ⎦U U U U U

δ and α are parameters which may be adjusted to achieve accuracy and stability

δ =0.5, α = 1/6 corresponds to the linear acceleration method which also correspond to the Wilson θ method with θ = 1

Ne ma k o iginall p oposed δ 0 5 1/4 hich es lts in an


Newmark originally proposed δ =0.5, α = 1/4 which results in an unconditionally stable scheme (the trapetzoidal rule)

The terms essentially result from the use of a Taylor series expansion:

tU = t-∆tU + ∆tt-∆tU +∆t2

2!t-∆tU +

∆t3

3!t-∆t

...U + ...

tU = t-∆tU + ∆tt-∆tU +∆t2

2!t-∆t

...U + ...


The Newmark Method

Notes: δ and a are parameters, effectively acting as weights forcalculating the approximation of the acceleration, and may beadjusted to achieve accuracy and stability

δ = 0.5, a = 1/6 is known as the linear acceleration method,which also correspond to the Wilson θ method with θ = 1

Newmark originally proposed δ=0.5, a = 1/4, which results inan unconditionally stable scheme (the trapezoidal rule)


The Newmark Method

We now solve for the displacements, velocities and accelerations byinserting the above into the dynamic equilibrium equation:

where


The Newmark Method

Solution Procedure

A. For each time step


The Newmark Method

Solution Procedure




• The Newmark method : Solution procedure:

B: For each time stepB: For each time step

1) Calculate effective loads at time t:ˆ ( )t t t t t t t+Δ +ΔR R M U U U0 2 3

1 4 5

( )

( )

t t t t t t t

t t t

a a a

a a a

+Δ +Δ= + + +

+ + +

R R M U U U

C U U U

2) Solve for the displacements U at time t + ΔtˆT t t t t+Δ +Δ=LDL U R

3) Solve for the corresponding velocities and accelerations0 2 3( )t t t t t t ta a a+Δ +Δ= − − −U U U U U

t t t t t tΔ Δ


7 6t t t t t ta a+Δ +Δ= +U U + U U


The Newmark Method

Stability of the Newmark Method

For zero damping the Newmark method is conditionally stable if

δ ≥ 1

2, α ≤ 1

2and ∆t ≤ 1

ωmax

√δ2 − α

where ωmax is the maximum natural frequency.

The Newmark method is unconditionally stable if

2α ≥ δ ≥ 1

2


The Newmark Method

Stability of the Newmark Method

However, if δ ≥ 12 , errors are introduced. These errors are associated

with “numerical damping” and “period elongation”, i.e. a seeminglylarger damping and period of oscillation than in reality.

Because of the unconditional stability of the average accelerationmethod, it is the most robust method to be used for the step-by-stepdynamic analysis of large complex structural systems in which a largenumber of high frequencies, short periods, are present.

The only problem with the method is that the short periods, whichare smaller than the time step, oscillate indefinitely after they areexcited. The higher mode oscillation can however be reduced by theaddition of stiffness proportional (artificial) damping.

source: csiberkeley.com


The Newmark Method

2dof system example ∆t = 0.28s

0 1 2 3 4 5 6 7 8 9 10−2

−1

0

1

2

3

4

5

6

time

Dis

plac

emen

tDiscrete Time Step, Δt=1 sec

1st Floor CMD2nd Floor CMD1st Floor NM2nd Floor NM1st Floor true2nd Floor true


The Newmark Method

2dof system example ∆t = 1s

0 2 4 6 8 10−10

−5

0

5

10

15

20

time

Dis

plac

emen

tDiscrete Time Step, Δt=1 sec

1st Floor CMD2nd Floor CMD1st Floor NM2nd Floor NM1st Floor true2nd Floor true


The Wilson θ Method

APPENDIX The Wilson θ Method In 1973, the general Newmarkmethod was made unconditionally stable by the introduction of a θfactor.

The introduction of this factor is motivated by the observation thatan unstable solution tends to oscillate about the true solution.

Therefore, if the numerical solution is evaluated within the timeincrement the spurious oscillations are minimized. This can beaccomplished by a simple modification to the Newmark method byusing a time step defined by:

∆t′ = θ∆t, θ ≥ 1.0

source: csiberkeley.com

Reminder: For θ = 1 the Wilson θ method is equivalent to theNewmark linear acceleration method with δ = 0.5, a = 1/6.





Th Wil h d• The Wilson θ method

In this method the acceleration is assumed to vary linearly from time t to t + Δt

( )t t t t tτ θτ+ + Δ+U U U U

By integration we obtain

( )t t t t t

tτ θ

θ+ + Δ= + −

ΔU U U U

By integration we obtain

( )2

t t t t t tτ θττ+ + Δ= + + −U U U U U( )2 tθΔ

( )2 31 12 6

t t t t t t t

tτ θτ τ τ

θ+ + Δ= + + + −

ΔU U U U U U


( )2 6 tθΔ





Th Wil h d• The Wilson θ method

Setting τ = θ Δt we get

( )2t t t t t ttθ θθ+ Δ + ΔΔ

= + +U U U U2

( ) ( )21 26

t t t t t t tt tθ θθ θ+ Δ + Δ= + Δ + Δ +U U U U U

from which we can solve

( )6 6 2t t t t t t tθ θ+ Δ + Δ= − − −U U U U U( )

( )

( )

2 2

3 2t t t t t t t

tttθ θ

θθθ+ Δ + Δ

= − − −ΔΔ

Δ= − − −

U U U U U

U U U U U


( ) 2tθΔ



Th Wil h d( )

( )26 6 2

3

t t t t t t t

ttt

θ θ

θθθ

+ Δ + Δ= − − −ΔΔ

Δ

U U U U U

• The Wilson θ method

We now solve for the displacements, velocities and accelerations

( )3 22

t t t t t t ttt

θ θ θθ

+ Δ + Δ Δ= − − −

ΔU U U U U

by inserting into the dynamic equilibrium equation

θ θ θ θΔ Δ Δ Δ

( )

t t t t t t t t

t t t t t t

θ θ θ θ

θ θ

+ Δ + Δ + Δ + Δ

+ Δ +Δ

+ + =

+

M U C U K U R

R R R R( )t t t t t tθ θ= + −R R R R




Solution Procedure

A. Initial Calculations

Implicit Procedure!



Solution Procedure




Th Wil h d S l i d• The Wilson θ method : Solution procedure:

B: For each time step

1) Calculate effective loads at time t+∆t:ˆ ( ) ( 2 )t t t t t t t t tθ θθ+ Δ + ΔR R R R M U U U

2) Solve for the displacements U at time t + Δt

0 2

1 3

( ) ( 2 )

( 2 )

t t t t t t t t t

t t t

a a

a a

θ θθ+ Δ + Δ= + − + + +

+ + +

R R R R M U U U

C U U U2) Solve for the displacements U at time t + Δt

3) Solve for the corresponding velocities and accelerations

ˆT t t t tθ θ+ Δ + Δ=LDL U R

3) Solve for the corresponding velocities and accelerations

4 5 6( )t t t t t t ta a aθ+Δ + Δ= − + +U U U U U

7 ( )t t t t t ta+Δ +Δ= +U U + U U


7 ( )8 ( 2 )t t t t t t tt a+Δ +Δ= Δ + +U U + U U U



Coupling of integration operators

For some problems it may be an advantage to combine the differenttypes of integration schemes e.g. if a structure is subjected todynamic load effect from hydrodynamic loading then the analysis ofthe hydrodynamic forces may be assessed using an explicit schemeand the structural response by using an implicit scheme.

The best choice of strategy will depend on the problem at hand inregard with stability and accuracy!


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The Finite Element Method for the Analysis of Non-Linear ... · The Finite Element Method for the Analysis of Non-Linear and Dynamic Systems Prof. Dr. Eleni Chatzi Lecture 6 - 5 November,