Computational Biomechanics Lecture 2: Basic Mechanics 2. Ulli Simon, Frank Niemeyer, Martin Pietsch

Transcription

1 Computational Biomechanics 016 Lecture : Basic Mechanics Ulli Simon, Frank Niemeyer, Martin Pietsch Scientific Computing Centre Ulm, UZWR Ulm University

2 Contents

3 .7 Static Equilibrium Important: Free-body diagram (FBD) first, then equilibrium! 3 equations of equilibrium for each FBD in D: F F 10 N Free-body diagram (FBD) Sum of all forces in x - direction : F 1, x F, x!... 0, Sum of all forces in y - direction : F 1, y F, y!... 0, Sum of all moments w.resp.to P : M P 1, z M P, z! Force EEs can be substituted by moment EEs 3 moment reference points should not lie on one line

4 6 equlibrium equations for one FBD in 3D: Summe Summe Summe Summe Summe Summe aller Kräftein x - Richtung : aller Kräftein y - Richtung : aller Kräftein z - Richtung : 0, 0, 0, aller Momenteum x - Achse bezüglich Punkt P : aller Momenteum y - Achse bezüglich Punkt Q : aller Momenteum z - Achse bezüglich Punkt R : i i i F F F ix iy iz!!! i i i M M M P ix Q iy R iz! 0.! 0.! 0. Force EEs can be substituted by moment EEs Max. moment axis parallel to each other Determinant of coef. matrix not zero

5 .8 Recipe for Solving Problems in Statics Step 1: Model building. Generate a simplified replacement model (diagram with geometry, forces, constraints). Step : Cutting, Free-body diagram. Cut system and develop free-body diagrams. Include forces and moments at cut, as well as weight. Step 3: Equilibrium equations. Write the force- and moment equilibrium equations (only for free-body diagrams). Step 4: Solve the equations. One can only solve for as many unknowns as equations, at most. Step 5: Display results, explain, confirm with experimental comparisons. Are the results reasonable?

6 .9 Classical Example: Biceps Force From: De Motu Animalium G.A. BORELLI ( )

7 Step 1: Model building Rigid fixation Rope (fixed length) Beam (rigid, massless) 10 kg g Joint (frictionless)

8 Schritt : Schneiden und Freikörperbilder 3 B S 3 S 3 C S S 1 10 kg S 1 S N A 4

9 More to Step : Cutting and Free-Body Diagrams A B C S S 1 S 3 h 100 N S h 1 S 1 = 100 N G Step 3 and 4: Equilibrium and Solving the Equations Sum of all forces in vertical direction = N ( S S N! ) 0 Sum of all forces in rope direction = 0 S! ( S3) 0 S 3 S Sum of all moments with respect to Point G = 0 S N 35cm S S h 1 S h 35cm 100 N 5cm!! 5cm N

10 ELASTOSTATICS 3.1 Stresses to account for the loading of the material! 500 N Fotos: Lutz Dürselen

11 Note to Remember: Stress = smeared force Stress = Force per Area or = F/A (Analogy: Nutella bread teast )

12 Units of Stress Pascal: 1 Pa = 1 N/m Mega-Pascal: 1 MPa = 1 N/mm F A 1 3. Example: Tensile stress in Muscles 1 F A 1 F A 700 N 7000 mm 700 N 70mm 10 N 0,1 mm N mm 0,1M Pa 10MPa F A

13 3.3 Normal and Shear Stresses 1 P F P 1 Tensile bar Cut 1: Normal stress 1 Cut : Normal stress Shear stress

14 Note to Remember: First, you must choose a point and a cut through the point, then you can specify (type of) stresses at this point in the body. Normal stresses (tensile and compressive stress) are oriented perpendicular to the cut-surface. Shear stresses lie tangential to the cut-surface.

15 General (3D) Stress State: Stress Tensor... in one point of the body: How much numbers do we need? 3 stress components in one cut (normal str., x shear str.) times 3 cuts (e.g. frontal, sagittal, transversal) results in 9 stress components for the full stress state in the point. But only 6 components are linear independent ( equality of shear stresses ) σ zz τ zy y Stress Tensor xx yx zx xy yy zy xz yz zz x τ yx τ zx σ yy τ yz τ xz τ xy σ xx z

16 General 3D Stress State 6 Components 6 Pictures

17 Problem: How to produce nice Pictures? Which component should I use? Do I need 6 pictures at the same time? So called Invariants are smart mixtures of the components Mises xx yy zz xx yy xx zz yy zz 3 xy 3 xz 3 yz

18 3.4 Strains Finite element model of the fracture callus Global, (external) strains : Change in length Original length L L 0 Local, (internal) strains Units of Strain without a unit 1 1/100 = % 1/ = με (micro strain) = 0,1 % Gap Distortional Strain in a fracture callus

19 3D Local Strain State: Strain Tensor y Displacements [Verschiebungen] xy y+δy z x xy +δ xy undeformed x+δx deformed x y xx, yy, x y xy 1 xy, xz 1 zz z z xz, yz 1 yz Strain tensor xx xy xz xy yy yz xz yz zz

20 Note to Remember: Strain is relative change in length (and shape) Strain = Change in length / Original length

21 Material Laws for Biological Tissues

22 Characterizing Mechanical Properties Structural Properties (Organ Properties) Material Properties (Tissue Properties) L 0 3-Point Bending of a bone specimen Tensile Test of a standardized Specimen

23 Ultimate Elastic Mechanical Properties Organ Properties Tissue Properties Different stiffnesses: load/deformation Young s Modulus Poisson s Ratio Shear Modulus Compressive Modulus Ultimate forces, moments, displacements: tension, compression, bending, torsion Ultimate stresses, strains: tensile, compressive, shear at fracture at end of elastic region, at start of yielding

24 Example: Fracture Callus Haematoma Granulation tissue Fibrous tissue Cartilage Woven bone Cortical bone 0, MPa 0,4 500 MPa MPa GPa Yamada 1970; Hori & Lewis 198; Davy & Connolly 198; Proctor et al. 1989; Mente & Lewis 1994; Rho et al. 1995; Augat et al. 1998; Tanck et al. 1999

25 Overview Simplest Material Law: Linear-elastic, isotropic (Hook s Law) More complex Laws: Nonlinear Hyperelastic (nonlinear elastic + large deformations) Viscoelastic Non-elastic, plastic, yielding, hardening Anisotropic Fatigue, recovery Remodeling, healing

26 Overview In principal: Biological Tissues know all of these bad things. They often combine the bad things

27 Stress-strain curve L 0 B Ultimate stress S Yield point B S E Ultimate strain Strain at yoild point Young s modulus

28 Linear Elastic Law Linear stress-strain relation Stress σ linear E E (81) ij Eijkl kl Strain ε Equality of shear stresses (Boltzmann Continua) and strains (36) Reciprocity Theorem from Maxwell fully anisotropic (1) Orthotropic (9) Transverse Isotropic (5) Isotropy ()

29 E - Young s modulus - Poisson s ratio ( ) G - Shear modulus K - Bulk modulus μ, λ - Lame Constants E zx yz xy zz yy xx zx yz xy zz yy xx v v v v v v v v v v v E ) (1 sym 0 ) (1 0 0 ) ( ) ( ) ( ) (1 ) (1 ) (1 Linear Elastic Law of these

30 Anisotropic Properties Bone tissue Corticale bone Transverse isotropic (5) Like Wood Trabecular bone Orthotropic (9) 3D Grid

31 Anisotropic Properties Material Spongy bone Vertebra E Moduli in MPa 60 (male) 35 (female) Strength in MPa 4,6 (male),7 (female) Fracture strain in % prox. Femur Tibia Bovine ,7...3,8 Ovine Cortical bone Longitudinal ,5 Transversal

32 Nonlinear Elastic and Hyperelastic Law soft, fibrous tissues Stress σ progressive linear degressive Strain ε Hyperelastic Laws: Nonlinear + large deformations SED-stress relation, (Mooney-Rivlin, Neo-Hookean,...)

33 Properties for Ligaments Properties Tensile strength Shear strenght Rupture strain Rupture force Stiffness E Modulus, tension 0 50 MPa Value 1 kpa (fast loading 0.04mm/h) 740 kpa (slow loading 40mm/h) m/m (Shear) 0,1 0,5 m/m (Tension) 1500 N (Finger) N (Ankle) 100 N (ACL) N/mm (ACL) MPa

34 Non-elastic = plastic Bone strength Stress σ loading unloading ISO bone screw Strain ε Local Damage

35 Viscoelastic Laws (Time Dependent) Cartilage, Fibrous Tissue, Bone Stress σ loading unloading Strain ε

36 Measured Applied Viscoelastic Laws (Time Dependent) Retardation (Creep) Relaxation Stress step Strain step Strain history Stress history

37 Viscoelastic Laws (Time Dependent) b 1 b 1 k b k 0 b k 0 k 1 k 1 Typ: Inner damping Typ: Memory effect

38 Poroelastic Solid structure with pores: Cartilage, Fibrous Tissue, Bone Hyaloron

39 Properties for Articular Cartilage Properties Values Density 1300 kg/m 3 Water content 75% Compressive strength 5 MPa Tensile strenght 0 MPa E Modulus, eq. 0,8 MPa E Modulus, initial 3 MPa Shear Modulus 3,5 MPa Friction 0,005 Rupture strain 6...8% Permeability 0, m 4 /Ns

40 When is simplification allowed? Non linear: Plastic: Small strains linear Approximation: Ideal elastic plastic Viscoelastic: Quasi-static Deformations elastic Anisotropic: Worst-case scenario

41 4 Simple Load Cases

42 1 Tension, Compression Tensile bar (tensional rigidity EA) Device stiffness L 0 d 0 d 0 -d F EA L, L 0 k EA L 0 A Axial Stress F A L F Strain ΔL ε unloaded loaded cut ε q L 0 Δd d 0

43 Shear Shear bolt (shear rigidity GA) Device stiffness L F w τ A F GA w, L Shear stress F A k GA L Shear strain 1 xy

44 3 Bending (Cantilever Beam) Beam (bending rigidity EI a, Length L) Cut z x F M w Tension Neutral axis Compression Shear stress Cantilever formula: w L EI 3 L EI a a F F L EI L EI a a M, M.

45 3 Torsion Torsional rod (torsional rigidity GI T ) M GIT, L c GI L T L 0 R M τ Length L, radius r Cut Note to Remember: A hollow bone reaches a high stiffness and strength against bending and torsion with relatively little material.

46 Axial nd Moment of Area I a [Axiales Flächenmoment. Grades] Cross-Sections: h D D d b I a b 1 3 h I a 64 D I a ( D d ) 64

47 Second Moment of Area I (SMA) [Flächenmoment zweiten Grades] h D D d b Axial moment of area (bending) I a b 1 3 h I a 64 D I a ( D d ) 64 Polar (torsional) moment of area (torsion) I 3 D T 3 I P IT I p ( D d )