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The University of Sydney Slide 1

Mechanics of Biomaterials

Presented byAndrian SueAMME4981/9981Semester 1, 2016

Lecture 7


Mechanics Models


Last Week

– Using motion to find forces and moments in the body (inverse problems)


This Week

– Using the forces and moments to determine the stresses


Elastic Behaviour

Hooke’s Law (Uniaxial) 𝜎 = 𝐸ϵ– Strain is directly proportional to the stress (Young’s modulus)

Hooke’s Law (General) [𝜎] = [𝑆][𝜖]– Stress tensor [𝜎]– Strain tensor [𝜖]– Stiffness tensor [𝑆]

[𝜖] = [𝑆])*[𝜎] = [𝐶][𝜎]– Compliance tensor [𝐶] = [𝑆])*


Stress Calculation

– Undeformed

– Cauchy Stress (True Stress) 𝑇 = -.

– Nominal Stress (Engineering Stress)

𝜖 =Δ𝐿𝐿 =

𝑙 − 𝐿𝐿 =

𝑙𝐿 − 1 → 𝜃 =

𝑙𝐿 = 1 + 𝜖

𝜎 =𝐹𝐴 =

𝐿𝑙𝑎𝐿𝑙𝑎

𝐹𝐴 =

𝑙𝑎𝐿𝐴

𝐿𝑙𝐹𝑎 =

𝑉def𝑉undef

𝐿𝑙𝐹𝑎 =

𝑚/𝜌𝑚/𝜌C

1𝑙/𝐿

𝐹𝑎 =

𝜌C𝜌1𝜃 𝑇

– Deformed


Elastic Constants

Young’s Modulus, E– Relationship between tensile or compressive stress and strain– Only applies for small strains (within the elastic range)

*Assume linear, elastic, isotropic material

Biomaterial E (GPa)*Cancellous bone 0.49Cortical bone 14.7Long bone - Femur 17.2Long bone - Humerus 17.2Long bone - Radius 18.6Long bone - Tibia 18.1Vertebrae - Cervical 0.23Vertebrae - Lumbar 0.16


Elastic Constants

Poisson’s Ratio, ν– Describes the lateral deformation in response to an axial load

𝜈 = −𝜖lateral𝜖axial

aFF

l Density ρ

L ΔL

r R


Elastic Constants

Shear Modulus (or Lame’s second constant), G, μ– Describes the relationship between applied torque and angle of

deformation

𝐺 = 𝜇 =𝜏𝛾 =

ShearStressShearStrain

Bulk Modulus, K– Describes the resistance to uniform compression (hydrostatic pressure)

𝐾 = −Δ𝑃𝑒 = −

Δ𝑃Δ𝑉/𝑉 ≈ −𝑉

𝜕𝑃𝜕𝑉

Lame’s first constant, λ– Used to simplify the stiffness matrix in Hooke’s law


Elastic Constants

– Young’s Modulus, E 𝐸 = X(Z[\]X)[\X = [(*\_)(*)]_)

_ = 2𝐺(1 + 𝜈)

– Poisson’s Ratio, ν 𝜈 = []([\X) =

[(Za)[) =

b]X − 1

– Shear Modulus, G, μ 𝐺 = [(*)]_)]_ = b

](*\_)

– Bulk Modulus, K 𝐾 = bZ(*)]_)

– Lame’s Constant, λ 𝜆 = ]X_*)]_ =

X(b)]X)ZX)b = b_

(*\_)(*)]_)


Hooke’s Law: Tensor Representation

– Hooke’s Law: [𝜖] = [𝐶][𝜎] or [𝜎] = [𝑆][𝜖]

– Stress Tensor: [𝜎] =𝜎dd 𝜎de 𝜎df𝜎ed 𝜎ee 𝜎ef𝜎fd 𝜎fe 𝜎ff

or [𝜎] =𝜎** 𝜎*] 𝜎*Z𝜎]* 𝜎]] 𝜎]Z𝜎Z* 𝜎Z] 𝜎ZZ

– Strain Tensor: [𝜖] =𝜖dd 𝜖de 𝜖df𝜖ed 𝜖ee 𝜖ef𝜖fd 𝜖fe 𝜖ff

or [𝜖] =𝜖** 𝜖*] 𝜖*Z𝜖]* 𝜖]] 𝜖]Z𝜖Z* 𝜖Z] 𝜖ZZ

– In this form, [𝜎] and [𝜖] are 2nd order tensors– In this form, [𝐶] and [𝑆] are 4th order tensors– Too difficult to determine [𝐶] and [𝑆]


Hooke’s Law: Matrix Representation

– Hooke’s Law: {𝜖} = [𝐶]{𝜎}

𝜖 =

𝜖*𝜖]𝜖Z𝜖i𝜖j𝜖k

=

𝜖**𝜖]]𝜖ZZ2𝜖]Z2𝜖*Z2𝜖*]

[𝐶] =

𝐶** 𝐶*] 𝐶*Z𝐶]* 𝐶]] 𝐶]Z𝐶Z* 𝐶Z] 𝐶ZZ𝐶i* 𝐶i] 𝐶iZ𝐶j* 𝐶j] 𝐶jZ𝐶k* 𝐶k] 𝐶kZ

𝐶*i 𝐶*j 𝐶*k𝐶]i 𝐶]j 𝐶]k𝐶Zi 𝐶Zj 𝐶Zk𝐶ii 𝐶ij 𝐶ik𝐶ji 𝐶jj 𝐶jk𝐶kj 𝐶kj 𝐶kk

𝜎 =

𝜎*𝜎]𝜎Z𝜎i𝜎j𝜎k

=

𝜎**𝜎]]𝜎ZZ𝜎]Z𝜎*Z𝜎*]

– In this form, {𝜎} and {𝜖} are 1st order vectors– In this form, [𝐶] is a 2nd order tensor– Much easier to determine [𝐶]– This is called the Voigt notation – reduces the order of the symmetric tensor


Constitutive Material Models


Constitutive Material Models

[𝐶] =


𝐶*i 𝐶*j 𝐶*k𝐶]i 𝐶]j 𝐶]k𝐶Zi 𝐶Zj 𝐶Zk𝐶ii 𝐶ij 𝐶ik𝐶ji 𝐶jj 𝐶jk𝐶kj 𝐶kj 𝐶kk

Constitutive Model Number of Independent Components in [C]

Anisotropy 21Orthotropy 9Transverse Isotropy 5Isotropy 2


Anisotropy

– Most general form of Hooke’s law– 21 independent components– Example: wood

{𝜖} = [𝐶]{𝜎}


=


𝐶*i 𝐶*j 𝐶*k𝐶]i 𝐶]j 𝐶]k𝐶Zi 𝐶Zj 𝐶Zk𝐶ii 𝐶ij 𝐶ik𝐶ji 𝐶jj 𝐶jk𝐶ki 𝐶kj 𝐶kk


– Symmetric matrix: 𝐶*] = 𝐶]*,𝐶*Z = 𝐶Z*,𝑒𝑡𝑐.


Orthotropy

– Material possesses symmetry about three orthogonal planes– 9 independent components

– 3 Young’s moduli: 𝐸*,𝐸],𝐸Z– 3 Poisson’s ratios: 𝜈*] = 𝜈]*,𝜈]Z = 𝜈Z], 𝜈Z* = 𝜈*Z– 3 shear moduli: 𝐺*],𝐺]Z, 𝐺Z*

– Example: cortical bone


=

1𝐸*

−𝜈*]𝐸*

−𝜈*Z𝐸*

0 0 0

−𝜈*]𝐸]

1𝐸]

−𝜈]Z𝐸]

0 0 0

−𝜈*Z𝐸Z

−𝜈]Z𝐸Z

1𝐸Z

0 0 0

0 0 01𝐺]Z

0 0

0 0 0 01𝐺Z*

0

0 0 0 0 01𝐺*]


1

2

3


Orthotropy

– Example: cortical bone

– Large variations in property values are not necessarily (although may possibly be) due to experimental error

Component ValuesE1 6.91–18.1 GPaE2 8.51–19.4 GPaE3 17.0–26.5 GPaG12 2.41–7.22 GPaG12 3.28–8.65 GPaG12 3.28–8.67 GPaνij 0.12–0.62


Transverse Isotropy

– 5 independent components– Young’s moduli: 𝐸* = 𝐸], 𝐸Z– Poisson’s ratios: 𝜈*] = 𝜈]*, 𝜈]Z = 𝜈Z] = 𝜈Z* = 𝜈*Z– Shear modulus: 𝐺]Z = 𝐺Z*, 𝐺*] =

bq](*\_qr)

– Example: skin


=

1𝐸*

−𝜈*]𝐸*

−𝜈*Z𝐸*

0 0 0

−𝜈*]𝐸*

1𝐸*

−𝜈*Z𝐸*

0 0 0

−𝜈*Z𝐸Z

−𝜈*Z𝐸Z

1𝐸Z

0 0 0

0 0 01𝐺Z*

0 0

0 0 0 01𝐺Z*

0

0 0 0 0 02(1 + 𝜈*])

𝐸*


1

23


Isotropy

– 2 independent components

– Young’s modulus: 𝐸 = 𝐸* = 𝐸] = 𝐸Z– Poisson’s ratio: 𝜈 = 𝜈*] = 𝜈]Z = 𝜈Z*, 𝐺 = 𝐺]Z = 𝐺Z* = 𝐺*] =

b](*\_)

– Example: Ti-6Al-4V


=

1𝐸 −

𝜈𝐸 −

𝜈𝐸 0 0 0

−𝜈𝐸

1𝐸 −

𝜈𝐸 0 0 0

−𝜈𝐸 −

𝜈𝐸

1𝐸 0 0 0

0 0 02(1 + 𝜈)

𝐸 0 0

0 0 0 02(1 + 𝜈)

𝐸 0

0 0 0 0 02(1 + 𝜈)

𝐸


1

2

3


Hooke’s Law (Isotropic): Stress-Strain Relationship

𝜎 = 𝑆 𝜖 ⇔ 𝜎tu = 𝜆𝑡𝑟 𝜖 𝛿tu + 2𝜇𝜖tu

𝑡𝑟 𝜖 = 𝜖dd + 𝜖ee + 𝜖ff 𝛿tu = x1𝑖𝑓𝑖 = 𝑗0𝑖𝑓𝑖 ≠ 𝑗

𝜎dd =b

*\_ *)]_[ 1 − 𝜈 𝜖dd + 𝜈 𝜖ee + 𝜖ff ]

𝜎ee =b

*\_ *)]_[ 1 − 𝜈 𝜖ee + 𝜈 𝜖ff + 𝜖dd ]

𝜎ff =b

*\_ *)]_[ 1 − 𝜈 𝜖ff + 𝜈 𝜖dd + 𝜖ee ]

𝜎de =b

*\_𝜖de

𝜎ef =b

*\_𝜖ef

𝜎fd =b

*\_𝜖fd

or

𝜎dd = 𝜆 𝜖dd + 𝜖ee + 𝜖ff + 2𝐺𝜖dd𝜎ee = 𝜆 𝜖dd + 𝜖ee + 𝜖ff + 2𝐺𝜖ee𝜎ff = 𝜆 𝜖dd + 𝜖ee + 𝜖ff + 2𝐺𝜖ff

𝜎de = 2𝐺𝜖de𝜎ef = 2𝐺𝜖ef𝜎fd = 2𝐺𝜖fd


Hooke’s Law (Isotropic): Strain-Stress Relationship

𝜖 = 𝐶 𝜎 ⇔ 𝜖tu =1 + 𝜈𝐸

𝜎tu −𝜈𝐸𝑡𝑟 𝜎 𝛿tu

𝑡𝑟 𝜎 = 𝜎dd + 𝜎ee + 𝜎ff 𝛿tu = x1𝑖𝑓𝑖 = 𝑗0𝑖𝑓𝑖 ≠ 𝑗

𝜖dd =*b[𝜎dd − 𝜈 𝜎ee + 𝜎ff ]

𝜖ee =*b[𝜎ee − 𝜈 𝜎ff + 𝜎dd ]

𝜖ff =*b[𝜎ff − 𝜈 𝜎dd + 𝜎ee ]

𝜖de =*\_b

𝜎de

𝜖ef =*\_b

𝜎ef

𝜖fd =*\_b

𝜎fd

or

𝜖dd =*b[𝜎dd − 𝜈 𝜎ee + 𝜎ff ]

𝜖ee =*b[𝜎ee − 𝜈 𝜎ff + 𝜎dd ]

𝜖ff =*b[𝜎ff − 𝜈 𝜎dd + 𝜎ee ]

𝜖de =*]X𝜎de

𝜖ef =*]X𝜎ef

𝜖fd =*]X𝜎fd


Biomechanics


Biomechanics Methods

There are three methods that can be used to determine the biomechanical responses to loads:

1. Analytical method (Mechanics of Solids 1 and 2)

2. Biomechanical experimentation (testing)

3. Numerical techniques (FEM)


Analytical Method: General Case

𝜖}}𝜖~~𝜖ff2𝜖~f2𝜖f}2𝜖}~

=

1𝐸}

−𝜈~}𝐸~

−𝜈f}𝐸f

0 0 0

−𝜈}~𝐸}

1𝐸~

−𝜈f~𝐸f

0 0 0

−𝜈}f𝐸}

−𝜈~f𝐸~

1𝐸f

0 0 0

0 0 01𝐺~f

0 0

0 0 0 01𝐺f}

0

0 0 0 0 01𝐺}~

𝜎}}𝜎~~𝜎ff𝜎~f𝜎f}𝜎}~

z (3)

y (2) x (1)

ez

et

en


𝜎ff = −-��

𝜖}}𝜖~~𝜖ff2𝜖~f2𝜖f}2𝜖}~

=

1𝐸}

−𝜈~}𝐸~

−𝜈f}𝐸f

0 0 0

−𝜈}~𝐸}

1𝐸~

−𝜈f~𝐸f

0 0 0

−𝜈}f𝐸}

−𝜈~f𝐸~

1𝐸f

0 0 0

0 0 01𝐺~f

0 0

0 0 0 01𝐺f}

0

0 0 0 0 01𝐺}~

00𝜎ff000

=

−𝜈f}𝜎ff𝐸f

−𝜈f~𝜎ff𝐸f𝜎ff𝐸f000

z (3)

y (2) x (1)

ez

et

en

Fz Fz

Analytical Method: Pure Axial Load


Analytical Method: Pure Bending

𝜎ff = ±��e��

𝜎ff = ±��d��

z (3)

y (2) x (1)

ez

et

en

Mxx

z (3)

y (2) x (1)

ez

et

en

Myy


Analytical Method: Eccentric Axial Load

Using the principle of superposition

𝜎ff = −-�� ±

��e��

± ��d��

= 𝐹f −*� ±

e�e��± d�d

��

𝜎 = − -�

𝜎 = ±�e�

z (3)

y (2) x (1)

ez

et

enFz Fz ( )y~,x~

x

y


Example: Analytical Method

Determine the maximum compressive stress on the bone, given F=200N, M=10Nm, the outer diameter of the bone is do=5cm, and the inner diameter of the bone is di=3cm.

Using the principle of superposition:

𝜎 = −�e� −

-� [𝐼 = �

i (𝑟�i − 𝑟ti), 𝐴 = 𝜋 𝑟�] − 𝑟t] ]

𝜎 = − *C×C.C]j��× C.C]j�)C.C*j� −

]CC�× C.C]jr)C.C*jr

𝜎 = −1.095𝑀𝑃𝑎

F FMM


Biomechanical Experimentation: Femoral Testing

Three-pointBending

Four-pointBending

FemoralNeck Test


Numerical Techniques: Bovine Femur Modelling

Bovine Femur Sample CT Scanning ScanIP Modelling

Angela Shi, 2010 (Thesis)


Experimentation & Numerical Techniques: Bovine Femur

Specimen from bovine femur sample

in-vitro experimental setup

ScanCAD model



Experimentation & Numerical Techniques: Bovine Femur

XFEM fracture analysis



Numerical Techniques: Inhomogeneity of Bone

HU

E

pCEHU ρρ =→∝

Material relationAngela Shi, 2010 (Thesis)


Experimentation & Numerical Techniques: Femur Fracture

– In-vitro test of cadaver model – eXtend FEM (XFEM) in Abaqus



Numerical Techniques: Dental Prostheses

– Whole Jaw Model

– Partial Jaw Model

CT Image Segmentation Sectional Curves CAD Model FE Model

PDLMolar


Numerical Techniques: Dental Prostheses

– 3 unit, all ceramic dental bridge

Solid Model Von Mises Stress


Summary

– Mechanics models– Elastic constants

– Constitutive material models– Number of independent components required to describe the material

model– Biomechanics

– Determining the biomechanical response to loads through analytical methods, biomechanical experimentation, and numerical techniques

mechanics of biomaterials - university of...

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