1. BE 276
Biomechanics
part 1
Roy Kerckhoffs

2. Cardiac Modeling Circulation Ventricles Cross-bridges 3-D myocardium
Image © Research Machines plc
Ventricles
Cross-bridges
3-D myocardium
Myofilament kinetic model
Roff
Ron A1 *
kon f g
Ca2+ kb kn
Boundary conditions: circulatory systems model
Hyperelastic constitutive
equation
3-D finite-element ventricular model

3. Soft Tissue Biomechanics
Conservation of mass, momentum and energy
Large deformations (geometric nonlinearity)
Nonlinear, anisotropic stress-strain relations
Boundary conditions: displacement and traction (e.g. pressure)
Passive Stress (kPa)
1.2
1.1
1.0 10 20 30
Tff
Tcc
Extension Ratio

4. Deformation F ∂xi FiR = ∂XR F = Grad(x) F = RU
Deformation gradient tensor F:
∂xi
FiR =
∂XR
F = Grad(x)
F describes rotation and deformation
F = RU O

5. Deformation Lagrangian strain tensor describes deformation only
E = ½(FTF – I )
E = ½((RU)T(RU) – I )
E = ½(UTRTRU – I )
E = ½(UT I U – I )
E = ½(UT U – I )

6. Equilibrium of a body All stress components acting in a 3D continuum y
Tyy
All stress components acting in a 3D continuum
Tyx
Txy
Tyz
Tzy
Txx x
Txz
Tzz
Tzx z

7. Equilibrium of a body Cauchy Stress tensor T y Tyy Tyx Txy Tyz Tzy Txx
Txz
Tzz
Tzx
The stress tensor defines a relation between the stress vector t which acts on a plane and the normal n of that plane in the deformed state of the body. z

8. Equilibrium of a body
Conservation of momentum: Txy=Tyx Txz=Tzx Tzy=Tyz T=TT
Hence there are only 6 independent stress components y
Tyy
Tyx
Txy
Tyz
Tzy
Txx x
Txz
Tzz
Tzx
The stress tensor defines a relation between the stress vector t which acts on a plane and the normal n of that plane in the deformed state of the body. z

9. Equilibrium and virtual workS
ρ is mass density is gravity S is surface of body is stress vector

10. Equilibrium and virtual work
Apply Green-Gauss or divergence theorem on first term S
Physical interpretation on divergence theorem for fluid flow is that a volume integral of the divergence of the flow field equals the net flow over the boundary

11. Equilibrium and virtual work
Because this must hold for each subpart of the body, we can remove the integrals S

12. Equilibrium and virtual work
Or, in components:
This is Newton’s second law, written in the strong form.

13. Equilibrium and virtual work
To obtain weak form, we apply the notion of virtual work
Now apply integration by parts and Green-Gauss theorem again on first term:
Hence we multiply the strong form equation with a virtual displacement field and integrate over the volume V

14. Equilibrium and virtual work
This leads to the weak form:
Internal virtual work
This is the principle of virtual work, expressed in the current (deformed) configuration
Virtual work by gravity
Virtual work by boundary tractions

15. Equilibrium and virtual work
In the weak form we impose lower order differentiability requirements on the solution
Here, no assumptions have been made yet regarding
material properties
magnitude of strains
Therefore, this equation is valid for both linear and finite elasticity mechanics.
This is the principle of virtual work, expressed in the current (deformed) configuration

16. Equilibrium and virtual work
In a deforming body, volumes, surfaces, mass density, stresses and strains continuously change
At t+Δt the state of the body is unknown
To solve this, refer stresses and strains to a known configuration
Refer to known configuration in finite elasticity, typically the unloaded state. For linear elasticity, one assumes that volumes and surfaces don’t change, so the integration is always performed over the original volume.

17. Lagrangian stress tensors
The (half) Lagrangian Nominal stress tensor or First Piola-Kirchhoff stress tensor S
S = detF.F-1.T  ST
S provides forces in the deformed state measured per unit reference (unloaded) area.
Useful experimentally.

18. Lagrangian stress tensors
The symmetric (fully) Lagrangian Second Piola-Kirchhoff stress tensor P
Useful mathematically but no direct physical interpretation
For small strains differences between T, P, S disappear
To get T from P:
Just like the Lagrangian strain E, the 2nd PK stress tensor P components are invariant to rigid body rotations
This represents the Cauchy stress tensor transformed back to the original geometric configuration (by inverse of F). Perhaps one can think of this as stresses being stretched and rotated, just like the body material.

19. Large Deformation Mechanics

20. Example: uniaxial stress
undeformed length = L
undeformed area = A
mass density is ρ0
volume is V0=LA
deformed length = l
deformed area = a
mass density is ρ
volume is V=la L A a F l
Stretch
Cauchy Stress

21. Example: uniaxial stress
undeformed length = L
undeformed area = A
mass density is ρ0
volume is V0=LA
deformed length = l
deformed area = a
mass density is ρ
volume is V=la L A a F l
Nominal Stress
Second Piola-Kirchhoff Stress

22. Need relation between stress and strain: Constitutive Models
Equate stress in a body to the strain
Metals are elastic for small strains
Hyperelastic
Elastic
Tissues are classified as a hyper-elastic material
Plastic
Exponential relationship used to model soft tissue
Young’s
Modulus
The derivative
In large deformation finite elasticity, relate stress and strain through
a strain energy function W=W(E):

23. 2-D Example: Exponential Strain-Energy Function

24. 2-D Example: Exponential Strain-Energy Function
If b1=b2: tissue is isotropic
If b1≠b2: tissue is anisotropic

25. 3-D Orthotropic Exponential Strain-Energy Function
From: Choung CJ, Fung YC. On residual stress in arteries. J Biomech Eng 1986;108:

26. Strain Energy Functions
Transversely Isotropic (Isotropic + Fiber) Exponential
Transversely Isotropic Exponential
Transversely Isotropic Polynomial
Orthotropic Power Law

27. ( ) Summary of Equations + = Kinematics Constitutive law
Deformation gradient tensor
Strain-displacement relation
Constitutive law
Hyperelastic relation
Lagrangian 2nd Piola-Kirchoff stress
W is the strain energy function
Eulerian Cauchy stress
Conservation of Momentum
Force balance
Moment balance
Conservation of Mass
Lagrangian form
r: mass density =
F = Grad(x)
∂XR
∂xi
FiR
E = ½(FTF – I ) 1 2 ∂W
∂ERS
∂ESR + =
PRS ( )
det 1 T F =
F•P•FT
Div(P•FT) + rb = 0
P = PT
r = r0detF

28. Lagrangian FE Formulation
Dependent Variable
Deformed coordinates of the nodes
Choice of basis functions:
Linear Interpolation
Cubic Hermite interpolation
Boundary Conditions
Essential boundary conditions
Displacement constraints
Natural boundary conditions
Traction (stress) boundary conditions
No nodal boundary condition is equivalent to traction-free

29. Solution Scheme
Non-linear strain-displacement and stress strain relationships
Stiffness matrices are functions of the dependent variables
Need a nonlinear solution scheme
Newton Method
f’(x) is an nxn Jacobian matrix J
Gives us a linear system of equations for x(k+1)

30. Accelerating Newton’s Method
Each step
Requires the solution of the linear system
n2 entries of Jij have to be computed
Modified Newton’s Method
Incremental loading in elasticity
Re-evaluate Jij only occasionally
Evaluate derivatives of J using finite differences
Determine components of J over several processors (parallel solving)

31. Newton’s method in 1D force true mechanical response (nonlinear)
force = K(u)u
displacement

32. Newton’s method in 1D l Apply an external force F, find displacement

33. Newton’s method in 1D
Determine derivative f’ (stiffness K(0)) at unloaded state
force F F l K
displacement

34. Newton’s method in 1D l Calculate displacement u1 = F/K(0) force F F K

35. Newton’s method in 1D
Evaluating force at u1 leads to F1=K(u1)u1 ≠ F
and we have an imbalance of F-F1
force F F1 F l K u1
displacement

36. Newton’s method in 1D
Use stiffness at u1 and calculate next incremental displacement u2
force F K F l u1 u2
displacement

37. Newton’s method in 1D l Again, re-evaluating the force leads to F2
displacement

38. Newton’s method in 1D
Repeating these steps leads to a small enough difference between
F and Fn leading to a total final displacement of u=u1+u2+u3
force F Fn K F l u1 u2 u3
displacement

39. Newton’s method in 1D
Therefore, stop Newton’s loop when either force imbalance is
small enough, or the incremental displacement is small enough
force F Fn K F l u1 u2 u3
displacement

40. Modified Newton’s method in 1D
force
etcetera F Fn F l
Use same K as in
unloaded state u1 u2
…. lots of un’s
displacement

41. Myocardial Infarction

42. Myocardial Infarction FE mesh of left ventricle
32 elements with tricubic basis functions
80 nodes

43. Myocardial Infarction fibers
endo>0
epi<0

44. Myocardial Infarction fibers
Endocardium with fiber direction
Epicardium with fiber direction

45. Myocardial Infarction Constitutive law
C=0.45 kPa for healthy myocardium
C=4.5 kPa for infarcted myocardium

46. Myocardial Infarction Location of infarction
Healthy
Infarct

47. Myocardial Infarction Displacement boundary conditions
coor2
s(2)
coor3
s(1)
coor1
s(3)
Fixed Node
coordinate 1
coordinate 2
coordinate 3
BASE
value, s(1), s(3), s(1)s(3)
YCONS
ZCONS
APEX
s(1), s(2), s(1)s(2), s(2)s(3), s(1)s(3),
s(1)s(2)s(3)
s(1), s(3)

48. Myocardial Infarction Displacement boundary conditions
coor2
s(2)
coor3
s(1)
coor1
s(3)
Fixed Node
coordinate 1
coordinate 2
coordinate 3
BASE
value, s(1), s(3), s(1)s(3)
YCONS
ZCONS
APEX
s(1), s(2), s(1)s(2), s(2)s(3), s(1)s(3),
s(1)s(2)s(3)
s(1), s(3)

49. Myocardial Infarction Displacement boundary conditions
coor2
s(3)
coor3
coor1
Fixed Node
coordinate 1
coordinate 2
coordinate 3
BASE
value, s(1), s(3), s(1)s(3)
YCONS
ZCONS
APEX
s(1), s(2), s(1)s(2), s(2)s(3), s(1)s(3),
s(1)s(2)s(3)
s(1), s(3)

50. Myocardial Infarction Displacement boundary conditions
coor3
coor2
s(2)
coor1
s(1)
s(3)
Fixed Node
coordinate 1
coordinate 2
coordinate 3
BASE
value, s(1), s(3), s(1)s(3)
YCONS
ZCONS
APEX
s(1), s(2), s(1)s(2), s(2)s(3), s(1)s(3),
s(1)s(2)s(3)
s(1), s(3)

51. Myocardial Infarction Pressure boundary conditions
coor2
PLV
Pext=0
coor3
coor1
Passive filling:
Increase pressure incrementally
on endocardial elements
Pressure turned into force
boundary conditions parallel
to normal of element faces