introduction to finite element modeling in biomechanics
DESCRIPTION
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 - PowerPoint PPT PresentationTRANSCRIPT
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