Mechanics of Biomaterials

Transcription

1 Mechanics of Biomaterials Lecture 7 Presented by Andrian Sue AMME498/998 Semester, 206 The University of Sydney Slide

2 Mechanics Models The University of Sydney Slide 2

3 Last Week Using motion to find forces and moments in the body (inverse problems) The University of Sydney Slide 3

4 This Week Using the forces and moments to determine the stresses The University of Sydney Slide 4

5 Elastic Behaviour Hooke s Law (Uniaxial) σ = Eϵ Strain is directly proportional to the stress (Young s modulus) Hooke s Law (General) Stress tensor [σ] Strain tensor [ε] Stiffness tensor [S] Compliance tensor [C] = [S] )* [σ] = [S][ε] [ε] = [S] )* [σ] = [C][σ] The University of Sydney Slide 5

6 Stress Calculation Undeformed Deformed Cauchy Stress (True Stress) T = -. Nominal Stress (Engineering Stress) ε = ΔL l L = = l L L L θ = l L = + ε σ = F A = Lla F Lla A = la L F LA l a = V def L F V undef l a = m/ρ F m/ρ C l/l a = ρ C ρ θ T The University of Sydney Slide 6

7 Elastic Constants Young s Modulus, E Relationship between tensile or compressive stress and strain Only applies for small strains (within the elastic range) Biomaterial Cancellous bone 0.49 Cortical bone 4.7 Long bone - Femur 7.2 Long bone - Humerus 7.2 Long bone - Radius 8.6 Long bone - Tibia 8. Vertebrae - Cervical 0.23 E (GPa)* Vertebrae - Lumbar 0.6 *Assume linear, elastic, isotropic material The University of Sydney Slide 7

8 Elastic Constants Poisson s Ratio, ν Describes the lateral deformation in response to an axial load ν = ε lateral ε axial L ΔL F a r R F l Density ρ The University of Sydney Slide 8

9 Elastic Constants Shear Modulus (or Lame s second constant), G, μ Describes the relationship between applied torque and angle of deformation G = μ = τ Shear Stress = γ Shear Strain Bulk Modulus, K Describes the resistance to uniform compression (hydrostatic pressure) K = ΔP e = ΔP ΔV/V V P V Lame s first constant, λ Used to simplify the stiffness matrix in Hooke s law The University of Sydney Slide 9

10 Elastic Constants Young s Modulus, E E = X(Z[\]X) [\X = [(*\_)(*)]_) _ = 2G( + ν) Poisson s Ratio, ν ν = [ ]([\X) = [ (Za)[) = b ]X Shear Modulus, G, μ G = [(*)]_) ]_ = b ](*\_) Bulk Modulus, K K = b Z(*)]_) Lame s Constant, λ λ = ]X_ *)]_ = X(b)]X) ZX)b = b_ (*\_)(*)]_) The University of Sydney Slide 0

11 Hooke s Law: Tensor Representation Hooke s Law: [ε] = [C][σ] or [σ] = [S][ε] Stress Tensor: [σ] = Strain Tensor: [ε] = σ dd σ de σ df σ ed σ fd σ ee σ fe σ ef σ ff or [σ] = ε dd ε de ε df ε ed ε fd ε ee ε fe ε ef ε ff or [ε] = σ ** σ *] σ *Z σ ]* σ ]] σ ]Z σ Z* σ Z] σ ZZ ε ** ε *] ε *Z ε ]* ε ]] ε ]Z ε Z* ε Z] ε ZZ In this form, [σ] and [ε] are 2 nd order tensors In this form, [C] and [S] are 4 th order tensors Too difficult to determine [C] and [S] The University of Sydney Slide

12 Hooke s Law: Matrix Representation Hooke s Law: {ε} = [C]{σ} ε = ε * ε ] ε Z ε i ε j ε k = ε ** ε ]] ε ZZ 2ε ]Z 2ε *Z 2ε *] [C] = C ** C *] C *Z C ]* C ]] C ]Z C Z* C Z] C ZZ C i* C i] C iz C j* C j] C jz C k* C k] C kz C *i C *j C *k C ]i C ]j C ]k C Zi C Zj C Zk C ii C ij C ik C ji C jj C jk C kj C kj C kk σ = σ * σ ] σ Z σ i σ j σ k = σ ** σ ]] σ ZZ σ ]Z σ *Z σ *] In this form, {σ} and {ε} are st order vectors In this form, [C] is a 2 nd order tensor Much easier to determine [C] This is called the Voigt notation reduces the order of the symmetric tensor The University of Sydney Slide 2

13 Constitutive Material Models The University of Sydney Slide 3

14 Constitutive Material Models [C] = C ** C *] C *Z C ]* C ]] C ]Z C Z* C Z] C ZZ C i* C i] C iz C j* C j] C jz C k* C k] C kz C *i C *j C *k C ]i C ]j C ]k C Zi C Zj C Zk C ii C ij C ik C ji C jj C jk C kj C kj C kk Constitutive Model Anisotropy 2 Orthotropy 9 Transverse Isotropy 5 Isotropy 2 Number of Independent Components in [C] The University of Sydney Slide 4

15 Anisotropy Most general form of Hooke s law 2 independent components Example: wood {ε} = [C]{σ} ε ** ε ]] ε ZZ 2ε ]Z 2ε *Z 2ε *] = C ** C *] C *Z C ]* C ]] C ]Z C Z* C Z] C ZZ C i* C i] C iz C j* C j] C jz C k* C k] C kz C *i C *j C *k C ]i C ]j C ]k C Zi C Zj C Zk C ii C ij C ik C ji C jj C jk C ki C kj C kk σ ** σ ]] σ ZZ σ ]Z σ *Z σ *] Symmetric matrix: C *] = C ]*,C *Z = C Z*,etc. The University of Sydney Slide 5

16 Orthotropy 3 Material possesses symmetry about three orthogonal planes 9 independent components 3 Young s moduli: E *,E ],E Z 3 Poisson s ratios: ν *] = ν ]*,ν ]Z = ν Z], ν Z* = ν *Z 3 shear moduli: G *],G ]Z, G Z* Example: cortical bone ν *] ν *Z E * E * E * ε ** ε ]] ε ZZ 2ε ]Z 2ε *Z 2ε *] = ν *] E ] E ] ν ]Z E ] ν *Z E Z ν ]Z E Z E Z 0 G ]Z G Z* 0 G *] σ ** σ ]] σ ZZ σ ]Z σ *Z σ *] 2 The University of Sydney Slide 6

17 Orthotropy Example: cortical bone Component E E 2 E 3 G 2 G 2 G 2 Values GPa GPa GPa GPa GPa GPa ν ij Large variations in property values are not necessarily (although may possibly be) due to experimental error The University of Sydney Slide 7

18 Transverse Isotropy independent components Young s moduli: E * = E ], E Z Poisson s ratios: ν *] = ν ]*, ν ]Z = ν Z] = ν Z* = ν *Z Shear modulus: G ]Z = G Z*, G *] = b q ](*\_ qr ) Example: skin ε ** ε ]] ε ZZ 2ε ]Z 2ε *Z 2ε *] = E * ν *] E * ν *Z E * ν *] E * E * ν *Z E * ν *Z E Z ν *Z E Z E Z G Z* 0 0 G Z* 0 2( + ν *] ) E * σ ** σ ]] σ ZZ σ ]Z σ *Z σ *] The University of Sydney Slide 8

19 Isotropy 3 2 independent components 2 Young s modulus: E = E * = E ] = E Z Poisson s ratio: ν = ν *] = ν ]Z = ν Z*, G = G ]Z = G Z* = G *] = b Example: Ti-6Al-4V ε ** ε ]] ε ZZ 2ε ]Z 2ε *Z 2ε *] = E ν E ν E ν E E ν E ν E ν E E 2( + ν) E ( + ν) 0 0 E 0 ](*\_) 2( + ν) E σ ** σ ]] σ ZZ σ ]Z σ *Z σ *] The University of Sydney Slide 9

20 Hooke s Law (Isotropic): Stress-Strain Relationship σ = S ε σ tu = λtr ε δ tu + 2με tu tr ε = ε dd + ε ee + ε ff if i = j δ tu = x 0 if i j σ dd = σ ee = σ ff = b *\_ *)]_ [ ν ε dd + ν ε ee + ε ff ] b [ ν ε *\_ *)]_ ee + ν ε ff + ε dd ] b [ ν ε *\_ *)]_ ff + ν ε dd + ε ee ] σ de = σ ef = σ fd = b ε *\_ de b ε *\_ ef b ε *\_ fd or σ dd = λ ε dd + ε ee + ε ff σ ee = λ ε dd + ε ee + ε ff σ ff = λ ε dd + ε ee + ε ff σ de = 2Gε de σ ef = 2Gε ef σ fd = 2Gε fd + 2Gε dd + 2Gε ee + 2Gε ff The University of Sydney Slide 20

21 Hooke s Law (Isotropic): Strain-Stress Relationship ε = C σ ε tu = + ν E σ tu ν E tr σ δ tu tr σ = σ dd + σ ee + σ ff if i = j δ tu = x 0 if i j ε dd = * b [σ dd ν σ ee + σ ff ] ε ee = * b [σ ee ν σ ff + σ dd ] ε ff = * b [σ ff ν σ dd + σ ee ] ε de = *\_ b ε ef = *\_ b ε fd = *\_ b σ de σ ef σ fd or ε dd = * b [σ dd ν σ ee + σ ff ] ε ee = * b [σ ee ν σ ff + σ dd ] ε ff = * b [σ ff ν σ dd + σ ee ] ε de = * ]X σ de ε ef = * ]X σ ef ε fd = * ]X σ fd The University of Sydney Slide 2

22 Biomechanics The University of Sydney Slide 22

23 Biomechanics Methods There are three methods that can be used to determine the biomechanical responses to loads:. Analytical method (Mechanics of Solids and 2) 2. Biomechanical experimentation (testing) 3. Numerical techniques (FEM) The University of Sydney Slide 23

24 Analytical Method: General Case e n y (2) x () e t e z z (3) ε }} ε ~~ ε ff 2ε ~f 2ε f} 2ε }~ = E } ν ~} E ~ ν f} E f ν }~ E } E ~ ν f~ E f ν }f E } ν ~f E ~ E f 0 G ~f G f} 0 G }~ σ }} σ ~~ σ ff σ ~f σ f} σ }~ The University of Sydney Slide 24

25 Analytical Method: Pure Axial Load e n y (2) x () F z e t e z F z z (3) σ ff = - ε }} ε ~~ ε ff 2ε ~f 2ε f} 2ε }~ = E } ν ~} E ~ ν f} E f ν }~ E } E ~ ν f~ E f ν }f E } ν ~f E ~ E f 0 G ~f G f} 0 G }~ 0 0 σ ff = ν f}σ ff E f ν f~σ ff E f σ ff E f The University of Sydney Slide 25

26 Analytical Method: Pure Bending e n y (2) x () e t e z M xx z (3) σ ff = ± ƒƒe ƒƒ e n y (2) x () e t e z M yy z (3) σ ff = ± d The University of Sydney Slide 26

27 Analytical Method: Eccentric Axial Load e n y (2) x () F z F z ( x~, y~ ) e t e z y z (3) x Using the principle of superposition σ = - σ = ± e σ ff = - ± ƒƒe ƒƒ ± d = F f * ± e e ƒƒ ± d d The University of Sydney Slide 27

28 Example: Analytical Method Determine the maximum compressive stress on the bone, given F=200N, M=0Nm, the outer diameter of the bone is d o =5cm, and the inner diameter of the bone is d i =3cm. M F F M Using the principle of superposition: σ = e σ = - *C C.C]j Œ C.C]j )C.C*j σ =.095MPa [I = ˆ i (r i r t i ), A = π r ] r t ] ] ]CC ˆ C.C]j r )C.C*j r The University of Sydney Slide 28

29 Biomechanical Experimentation: Femoral Testing Three-point Bending Four-point Bending Femoral Neck Test The University of Sydney Slide 29

30 Numerical Techniques: Bovine Femur Modelling Bovine Femur Sample CT Scanning ScanIP Modelling Angela Shi, 200 (Thesis) The University of Sydney Slide 30

31 Experimentation & Numerical Techniques: Bovine Femur in-vitro experimental setup Specimen from bovine femur sample ScanCAD model Angela Shi, 200 (Thesis) The University of Sydney Slide 3

32 Experimentation & Numerical Techniques: Bovine Femur XFEM fracture analysis Angela Shi, 200 (Thesis) The University of Sydney Slide 32

33 Numerical Techniques: Inhomogeneity of Bone HU E ρ E = Cρ p HU Material relation Angela Shi, 200 (Thesis) The University of Sydney Slide 33

34 Experimentation & Numerical Techniques: Femur Fracture In-vitro test of cadaver model extend FEM (XFEM) in Abaqus Angela Shi, 200 (Thesis) The University of Sydney Slide 34

35 Numerical Techniques: Dental Prostheses Whole Jaw Model Molar PDL CT Image Segmentation Sectional Curves CAD Model FE Model Partial Jaw Model The University of Sydney Slide 35

36 Numerical Techniques: Dental Prostheses 3 unit, all ceramic dental bridge Solid Model Von Mises Stress The University of Sydney Slide 36

37 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 The University of Sydney Slide 37