Introduction to Finite Element Modeling in

Biomechanics

Dr. N. Fatouraee

Biomedical Engineering Faculty

December, 2004

Overview

• Introduction and Definitions

• Basic finite element methods– 1-D model problem

• Application Examples

Overview• Finite Element Method

– numerical method to solve differential equations

E.g.:

dr

dur

dr

d

r

u

dx

dP )(0 Flow Problem

u(r)

Heat Transfer Problem T(r,t)

The “Continuum” Concept

• biomechanics example: blood flow through aorta– diameter of aorta 25 mm– diameter of red blood cell 8 m (0.008 mm)

– treat blood as homogeneous and ignore cells

The “Continuum” Concept

• biomechanics example: blood flow through capillaries– diameter of capillary can be 7 m– diameter of red blood cell 8 m

– clearly must include individual blood cells in model

Continuous vs. Discrete Solution

• What if the equation had no “analytical solution” (e.g., due to nonlinearities)?

dr

dur

dr

d

r

u

dx

dP )(0

0 0

at 0

constant

rdrdu

aru

dxdP

Continuous vs. Discrete Solution

• What if the equation had no “analytical solution” (e.g., due to nonlinearities)?

• How would you solve an ordinary differential equation on the computer?

• Numerical methods– Runge-Kutta– Euler method

Discretization

0 1

0 1

Discretization

0 1

in general, Euler method is given by:

xyxfyy

yxfx

yyy

dx

dy

yxfy

nnnn

nnnn

,

,

),(

1

1

• Start with initial condition: y(x0)=y0

• Calculate f(x0,y0)• Calculate y1=y0 + f(x0,y0) x• Calculate f(x1,y1) …………..

Euler Example

2 Steps

4 Steps

8 Steps

Exact Solution

x

y

ODE: dy/dx (x,y) = 0.05 yInitial Cond.: y(0)=100

Euler, 2 steps: dy/dt(0,100) = 5 ; x = 20y(20) = y(0) + x*dy/dt(0,100) = 100 + 20*5= 200y(40) = y(20) + x*dy/dt(20,200) = 200 + 20* 10 = 400

Problem:Use Euler with 2 steps: Calculate y(x) between at x=20 and x=40

Discretization

• in general, the process by which a continuous, differential equation is transformed into a set of algebraic equations to be solved on a computer

• various forms of discretization– finite element, finite difference, finite volume

Finite Element Method

• discretization

• steps in finite element method– weak form of differential equation– interpolation functions within elements– solution of resulting algebraic equations

Basic Finite Element Methods:A 1-D Example

],[on 0)( baxxuu

0)(

0)(

bu

ausolve for u(x)

Basic Finite Element Methods:A 1-D Example

0)()(

],[on 0)(

buau

baxxuu

Note that for a=0, b=1:

)1sinh(

)sinh()(

xxxu

Basic Finite Element Method

• seek solution to allied formulation referred to as “weak” statement

Basic Finite Element Method

• seek solution to allied formulation referred to as “weak” statement

)( allfor

0)()( :subject to

0)(

xw

buau

dxxuuxwb

a

Basic Finite Element Method

)( allfor

0)()( :subject to

0)(

xw

buau

dxxuuxwb

a

0)()(

],[on 0)(

buau

baxxuu

The integral form is as valid as the original differential equation.

Basic Finite Element Method

b

a

dxxuuxw 0)(

note that by the chain rule:

uwuwuw

Basic Finite Element Method

b

a

dxxuuxw 0)(

note that by the chain rule:

uwuwuw

0

b

a

dxwxwuuwuw

Basic Finite Element Method

0

b

a

dxwxwuuwuw

0 b

a

b

adxwxwuuwuw

Basic Finite Element Method

0

b

a

dxwxwuuwuw

0 b

a

b

adxwxwuuwuw

recall: w(x) is arbitraryno loss in generality to require w(a)=w(b)=0

i.e., subject w to same boundary conditions as u

Basic Finite Element Method

“weak statement”:

1oH allfor

0

such that find

w

dxwxwuuw

u(x)b

a

the above expression is “continuous”i.e., must be evaluated for all x

Discretization

0 1

0 1

“nodes” “elements”

Discretization

“nodes”

“elements”

1 2 3 4 5 6

1 2 3 4 5

u defined at nodes u1, u2 … = u(x1), u(x2) …

goal solve for ui

Discretization

)(

1

ielements

x

x

b

a

i

i

dxwxwuuwdxwxwuuw

“nodes”

“elements”

1 2 3 4 5 6

1 2 3 4 5

Consider a Typical Element

2

1

x

x

dxwxwuuw

e

x1x2

Interpolation Functions

1)(12

11 xx

xxxN

12

12 )(

xx

xxxN

2211

2

1

)()( uNuNuxNxueln

iii

h

Within the element we interpolate between u1 and u2:

Interpolation Functions

1)(12

11 xx

xxxN

0,1:at 211 NNxx

Interpolation Functions

12

11 1)(

xx

xxxN

12

12 )(

xx

xxxN

0,1:at 211 NNxx

1,0:at 212 NNxx

Interpolation Functions

2211

2

1

)()( uNuNuxNxui

iih

e

x1x2

at x = x1: u = u1

at x = x2: u = u2

x1 < x < x2: interpolation between u1 and u2

u1, u2 unknowns to be solved for i.e., nodal values of u

Approximation Functions

2

1

)()(i

iih uxNxu

2

1

)()(i

iih wxNxw

- referred to as “Galerkin” method

Now we have to choose functions for w:

2

1

)()(i

iih uxNxu

2

1

)()(i

iih wxNxw

2

1

2

1

2

1

2

1j i

x

x

x

x

jjiij

ij dxxNdxNNdx

dN

dx

dNuw

2

1

x

x

dxwxwuuw

We end up with a system of algebraic equations, that canbe solved by the computer

0)(

1

ielements

x

x

b

a

i

i

dxwxwuuwdxwxwuuw

How many elements do we need?

0 1

“nodes”

“elements”

1 2 3 4 5 6

1 2 3 4 5

2 elements 5 elements

10 elements 20 elements

Practical Finite Element Analysis

• many commercial finite element codes exist for different disciplines

– FIDAP, FLUENT: fluid mechanics

– ANSYS, LS-Dyna, Abaqus: solid mechanics

Using a Commercial Code

• choose most appropriate software for problem at hand

– not always trivial

– can the code handle the key physical processes

• e.g., spatially varying material properties, nonlinearities

Steps in Finite Element Method (FEM)• Geometry Creation

– Material properties (e.g. mass density)– Initial Conditions (e.g. temperature)– Boundary Conditions– Loads (e.g. forces)

• Mesh Generation• Solution

– Time discretization (for transient problems)– Adjustment of Loads and Boundary Conditions

• Visualization– Contour plots (on cutting planes)– Iso surfaces/lines– Vector plots– Animations

• Validation

Model Validation

• most important part of the process, but hardest and often not done

• two types of validation

– code validation: are the equations being solved correctly as written (i.e., grid resolution, etc.)

– model validation: is the numerical model representative of the system being simulated (very difficult)

Example 1: Liver Cancer Treatment

Radiofrequency Ablation forLiver Cancer

• Surgical Resection is currently the gold-standard, and offers 5-year survival of around 30%

• Surgical Resection only possible in 10-20% of the cases

• Radiofrequency Ablation heats up tissue by application of electrical current

• Once tumor tissue reaches 50°C, cancer cells die

Effects of RF energy on tissue

• Electrical Current is applied to tissue• Electrical current causes heating by ionic friction• Temperatures above ~50 °C result in cell death (necrosis)

Na+

Na+Cl-

K+

Cl-

Cl-

Electric Field

Clinical procedure

Insertion

Probe Extension

Application of RF power

(~12-25 min)

• Ground pad placed on patients back or thighs

• Patient under local anesthesia and conscious sedation, or light general anesthesia

9-prong probe, 5 cm diameter, (Rita Medical)

Cool-Tip probe, 17-gauge needle, (Radionics / Tyco)

12-prong probe,4 cm diameter,

(Boston Scientific)

200W RF-generator (Radionics / Tyco)

Current RF Devices

RF Lesion Pathology

Coagulation Zone

(= RF lesion, >50 °C)

Hyperemic Zone (increased perfusion)

Finite Element Modeling for Radiofrequency Ablation

• Purpose of Models:– Investigate shortcomings of current devices– Simulate improved devices– Estimate RF lesion dimensions for treatment planning

• Thermo-Electrically Coupled Model:– Solve Electric Field problem (Where is heat generated)– Solve thermal problem (Heat Conduction in Tissue, Perfusion,

Vessels)

Electric Field Problem (Where is heat being generated?)

Laplace’s Equation

P

M

Boundary Conditions

Electric Field

Thermal Problem:Conservation of Energy

rate of change of energy in a body =

+ rate of energy generation

+ rate of energy addition

- rate of energy lost

energy storageby tissue

energy added byelectric current(Power = current*voltage)

energy added due to metabolism

energy transfer to blood flow carrying heat away (“convected”)

energy transferred (“conducted”)back to electrode

energy transfer (“conducted”) to surrounding tissue

Model Geometry

1 cm

2-D axisymmetric model

Animations

Electrical Current Density(Where is heat being generated?)

Temperature

Model Results

1 cm

Temperature at end of ablation

Ex-vivo Validation in Animal Tissue

• Verify Temperature, Impedance and Lesion Diameter

• We applied same power as in computer model

Experimental Setup

Comparison Model Experiment

0

20

40

60

80

100

120

0 100 200 300

t (s)

Z (

Oh

ms)

Z experiment

Z model Impedance

Temperature

Conclusion

• Lesion Diameter:Model: 33 mmExperiment: 29 ± 3 mm

• RF Lesion in model 14% larger

• Information on Electrical Tissue Conductivity vs. Temperature needed

Computer Model Geometry:12-prong probe next to 10mm-vessel (e.g. portal vein)

Flow rate 23 cm/s

• Vessel cooling simulated by estimating convective heat transfer coefficient

Impact of large vessels

Temperatureat end of ablation

50 °C

100 °C

37 °C

Model Results

• Cancer cells next to vessel could survive

Computer 3D-Model Geometry

• Improved configuration heats from both sides, and may create lesions closer to vessel

Improved Configuration

Bipolar

50 °C

100 °C

37 °C

Monopolar

• Improved configuration creates lesion up to vessel

• Next Step: Experimental Validation

Temperatureat End of Ablation

Example 2: Simulation of Artificial Heart Valve

Phantom I

MR Imaging: Bioprosthetic Valve

Comparison between Experiment and Simulation

MRI simulation

Example 3: Artificial Heart Valve II

J. De Hart et al. / Journal of Biomechanics 36 (2003) 699–712 703

Configurations of the fiber-reinforced stentless valve and corresponding velocityvector fields taken at six successive points in time. The leftand right diagram at the bottom of each frame denote the applied

velocityand pressure curves, respectively.

Configurations of the fiber-reinforced stentless valve and corresponding velocityvector fields taken at six successive points in time. The leftand right diagram at the bottom of each frame denote the applied

velocityand pressure curves, respectively.

Maximum principle Cauchystresses in the leaflet matrix material during systole. In all frames the right leaflet is taken from the nonreinforced model for comparison.

MPSr denotes the maximum principle stress ratio of the reinforced and non-reinforced leaflets. The stress scale on the bottom is given in kPa.

Maximum principle Cauchystresses in the leaflet matrix material during systole. In all frames the right leaflet is taken from the nonreinforced model for comparison.

MPSr denotes the maximum principle stress ratio of the reinforced and non-reinforced leaflets. The stress scale on the bottom is given in kPa.

Other Examples in Biomedical Engineering

from Shirazi-Adl et al., J. Biomech. Engr. 123:391 2001

from Miga et al., J. Biomech. Engr. 123:354 2001

Pressure on vertebrae disks

Ene-Iordache et al., 2001

Blood flow in Vessel Aneurism

Blood flow in Vessel Aneurism

Blood flow in Vessel Aneurism

Weiss et al., 2001

Strain in Knee Ligaments

Electric Heart Activity

McLeod et al., 2001