Computational Biomechanics Lecture 2: Basic Mechanics 2. Ulli Simon, Martin Pietsch, Lucas Engelhardt

Transcription

1 Computational Biomechanics 2017 Lecture 2: Basic Mechanics 2 Ulli Simon, Martin Pietsch, Lucas Engelhardt Scientific Computing Centre Ulm, UZWR Ulm University

2 Contents

3 Mechanical Basics

4 Temperature 1.3 Variables, Dimensions and Units Standard: ISO 31, DIN 1313 Variable Length L = Number Unit = 2 m = 2 m {Variable} = Number [Variable] = Unit 1 2 Three mechanical SI-Units: m (Meter) kg (Kilogram) s (Seconds) Length L [m] Length L / m Length L in m

5 2.1 orce 2 STATICS O RIGID BODIES We all believe to know what a force is. But, force is an invention not a discovery! it can not be measured directly. Newton s 2 nd Law [Axiom]: orce = Mass times Acceleration or = m a Note to Remember: A force is the cause of acceleration or deformation of a body

6 Representation of orces... with arrows orces are Vectors with Magnitude Direction Sense of Direction Line of action 5 N Screw

7 Units of orce Newton N = kgm/s 2 G = mg = 0,1 kg 9,81 m/s 2 = 0,981 kg m/s 2 1 N 1 N Note to Remember: 1 Newton Weight of a bar of chocolate (100 g)

8 2.2 Method of Sections (Euler) [Schnittprinzip] ree-body Diagramm (BD) [reikörper-bild] Leonhard Euler, kg 10 N reikörperbild Note to Remember: irst, cut the system, then include forces and moments. ree-body diagram = completely isolated part.

9 2.2 Method of Sections Cut through joint (2D) Cut through bone (2D)

10 2.2 Method of Sections Cut through joint (2D) H V V H Q M N φ Cut through N M Q bone (2D)

11 2.3 Combining and Decomposing orces Summation of Magnitudes Vector Addition R 7 N 8 N Subtraction of Magnitudes Decomposition into Components 2 N axial trans

12 2.4 The Moment [Das Moment] Slotted screw with screwdriver blade Blade Screw M = a a orce Couples (, a) Moment M Note to remember: The moment M = a is equivalent to a force couple (, a). A moment is the cause for angular acceleration or angular deformation (Torsion, Bending) of a body.

13 Units for Moment Newton-Meter Nm = kgm 2 /s 2 Rechte-Hand-Regel: Representation of Moments... with rotation arrows or double arrows Moments are Vectors with... Magnitude Direction Sense of Direction Achse Drehpfeil 5 Nm oder 5 Nm Doppelpfeil

14 2.5 Moment of a orce about a Point [Versetzungsmoment] Versetzungsmoment P h P M = h M P P axial quer Note to Remember: Moment = orce times lever-arm

15 2.7 Static Equilibrium Important: ree-body diagram (BD) first, then equilibrium! 10 N ree-body diagram (BD) or 2D Problems max. 3 equations for each BD: The sum of all forces in x-direction equals zero: The sum of all forces in y-direction equals zero: The sum of Moments with respect to P equals zero: 1, x 2, x 1, y 2, y M...!...! 0 0 P P 1, z M 2, z...! 0 (or 3D Problems max. 6 equations for each BD)

16 2.7 Static Equilibrium Important: ree-body diagram (BD) first, then equilibrium! 3 equations of equilibrium for each BD in 2D: 10 N ree-body diagram (BD) Sum of all forces in x - direction : 1, x 2, x!... 0, Sum of all forces in y - direction : 1, y 2, y!... 0, Sum of all moments w.resp.to P : M P 1, z M P 2, z! orce EEs can be substituted by moment EEs 3 moment reference points should not lie on one line

17 6 equlibrium equations for one BD 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 ix iy iz!!! i i i M M M P ix Q iy R iz! 0.! 0.! 0. orce EEs can be substituted by moment EEs Max. 2 moment axis parallel to each other Determinant of coef. matrix not zero

18 2.8 Recipe for Solving Problems in Statics Step 1: Model building. Generate a simplified replacement model (diagram with geometry, forces, constraints). Step 2: Cutting, ree-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?

19 2.9 Classical Example: Biceps orce rom: De Motu Animalium G.A. BORELLI ( )

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

21 Schritt 2: Schneiden und reikörperbilder 3 B S 3 S 3 C S 2 2 S kg S 1 S N A 4

22 More to Step 2: Cutting and ree-body Diagrams A B C S 2 S 1 S 3 h N S 2 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! 2 ( S3) 0 S 3 S 2 Sum of all moments with respect to Point G = 0 S N 35cm S S h 2 1 S 2 h 35cm 100 N 5cm 2! 2! 5cm N

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

24 Note to Remember: Stress = smeared force Stress = orce per Area or = /A (Analogy: Nutella bread teast )

25 Units of Stress Pascal: 1 Pa = 1 N/m 2 Mega-Pascal: 1 MPa = 1 N/mm 2 A Example: Tensile stress in Muscles 1 2 A 1 A N 7000 mm 700 N 70mm N 0,1 mm N mm 2 2 0,1M Pa 10MPa A 2

26 3.3 Normal and Shear Stresses 1 P 2 P 1 Tensile bar Cut 1: Normal stress 1 Cut 2: Normal stress 2 Shear stress 2

27 Note to Remember: irst, 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.

28 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., 2x 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 ) Stress Tensor xx yx zx xy yy zy xz yz zz

29 Symmetry of the Stress Tensor Boltzmann Continua: Only volume forces (f x und f y ), no volume moments assumed Equality of corresponding shear stresses σ yy y σ xx τ xy τ yx f y C τ yx f x τ xy σ xx xx. sym xy. yy xz yz zz x σ yy

30 General 3D Stress State 6 Components 6 Pictures

31 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 2 xx 2 yy 2 zz xx yy xx zz yy zz 3 2 xy 3 2 xz 3 2 yz

32 3.4 Strains inite 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

33 3D Local Strain State: Strain Tensor y Displacements [Verschiebungen] z xy x y+δy xy +δ xy undeformed x+δx deformed

34 3D Local Strain State: Strain Tensor Definition: Universal Strain Definition: xx xy ij lim x x, lim y yy zz 0 x y y z 0, xz 1 2,, yz u u, i, j { x, y, } i, j j, i z 1 2 lim 0 z z 0 xx xy xz xy yy yz xz yz zz Note to Remember: Strain is relative change in length (and shape) Strain Tensor

35 Displacement vs. Strain Displacement u_x Strain, eps_xx

36 Material Laws for Biological Tissues

37 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

38 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

39 Example: racture Callus Haematoma Granulation tissue ibrous tissue Cartilage Woven bone Cortical bone 0, MPa 0,4 500 MPa MPa GPa Yamada 1970; Hori & Lewis 1982; Davy & Connolly 1982; Proctor et al. 1989; Mente & Lewis 1994; Rho et al. 1995; Augat et al. 1998; Tanck et al. 1999

40 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 atigue, recovery Remodeling, healing

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

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

43 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 (21) Orthotropic (9) Transverse Isotropic (5) Isotropy (2) *) also known as Betti's theorem or Maxwell-Betti reciprocal work theorem

44 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 2 ) 2 (1 sym 0 2 ) 2 ( ) 2 ( ) ( ) ( ) (1 ) 2 (1 ) (1 Linear Elastic Law 2 of these

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

46 Anisotropic Properties Material Spongy bone Vertebra E Moduli in MPa 60 (male) 35 (female) Strength in MPa 4,6 (male) 2,7 (female) racture strain in % prox. emur Tibia Bovine ,7...3,8 Ovine Cortical bone Longitudinal ,5 Transversal

47 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,...)

48 Properties for Ligaments Properties Tensile strength Shear strenght Rupture strain Rupture force Stiffness E Modulus, tension MPa Value 1 kpa (fast loading 0.04mm/h) 740 kpa (slow loading 420mm/h) 2 m/m (Shear) 0,1 0,5 m/m (Tension) 1500 N (inger) N (Ankle) 1200 N (ACL) N/mm (ACL) MPa

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

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

51 Measured Applied Effects of Viscoelastic Materials Retardation (Creep) Relaxation Stress steps Strain steps Strain history Stress history

52 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

53 Poroelastic Solid structure with pores: Cartilage, ibrous Tissue, Bone Hyaloron

54 Properties for Articular Cartilage Properties Values Density 1300 kg/m 3 Water content 75% Compressive strength 5 MPa Tensile strenght 20 MPa E Modulus, eq. 0,8 MPa E Modulus, initial 3 MPa Shear Modulus 3,5 MPa riction 0,005 Rupture strain 6...8% Permeability 0, m 4 /Ns

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

56 4 Simple Load Cases

57 1 Tension, Compression Tensile bar (tensional rigidity EA) Device stiffness L 0 d 0 d 0 -d EA L, L 0 k EA L 0 A Axial Stress A L Strain ΔL ε unloaded loaded cut ε q L 0 Δd d 0

58 2 Shear Shear bolt (shear rigidity GA) Device stiffness L w τ A GA w, L Shear stress A k GA L Shear strain 1 xy 2

59 3 Bending (Cantilever Beam) Beam (bending rigidity EI a, Length L) Cut z x M w Tension Neutral axis Compression Shear stress Cantilever formula: w L EI 3 L EI 2 a a L EI 2 L EI a a M, M.

60 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.

61 Second Moment of Area I (SMA) [lächenmoment zweiten Grades] h D D d b Axial moment of area (bending) I a b 12 3 h I a 64 D I a ( D d ) 64 Polar (torsional) moment of area (torsion) I 32 D T 32 I P IT I p ( D d )