Computational Biomechanics 18 Lecture : Introduction, Basic Mechanics Ulli Simon, Lucas Engelhardt, Martin Pietsch Scientific Computing Centre Ulm, UZWR Ulm University
Contents
Mechanical Basics
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
Static Equilibrium Important: ree-body diagram (BD) first, then equilibrium! 3 equations of equilibrium for each BD in D: 1 N ree-body diagram (BD) Sum of all forcesin x - direction : 1, x, x!..., Sum of all forcesin y - direction : 1, y, y!..., Sum of all moments w.resp.to P : M P 1, z M P, z!.... orce EEs can be substituted by moment EEs 3 moment reference points should not lie on one line
6 equlibrium equations for one BD in 3D: Summealler Kräfte in x - Richtung: Summealler Kräfte in y - Richtung: Summealler Kräfte in z - Richtung:,,, Summealler Momente um x- Achse bezüglich Punkt P : Summealler Momenteum y - Achsebezüglich Punkt Q : Summealler Momenteum z - Achsebezüglich Punkt R : i i i ix iy iz!!! i i i M M M P ix Q iy R iz!.!.!. orce EEs can be substituted by moment EEs Max. moment axis parallel to each other Determinant of coef. matrix not zero
Classical Example: Biceps orce rom: De Motu Animalium G.A. BORELLI (168-1679)
Step 1: Model building Rigid fixation Rope (fixed length) Beam (rigid, massless) 1 kg g Joint (frictionless)
Schritt : Schneiden und reikörperbilder 3 B S 3 S 3 C S S 1 1 kg S 1 S 1 1 N A 4
More to Step : Cutting and ree-body Diagrams A B C S S 1 S 3 h 1 N S h 1 S 1 = 1 N G Step 3 and 4: Equilibrium and Solving the Equations Sum of all forces in vertical direction = 1 N ( S S 1 1 1N! ) Sum of all forces in rope direction = S! ( S3) S 3 S Sum of all moments with respect to Point G = S 1 1 N 35cm S S h 1 S h! 5cm 35cm 1 N 7 N 5cm!
Units of Stress Pascal: 1 Pa = 1 N/m Mega-Pascal: 1 MPa = 1 N/mm A 1 3. Example: Tensile stress in Muscles 1 A 1 A 7N 7mm 7N 7mm N,1 mm N 1 mm,1mpa 1MPa A
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 ) Stress Tensor xx yx zx xy yy zy xz yz zz
Symmetry of the Stress Tensor Boltzmann Continua: Only volume forces (f x und f y ), no volume moments assumed Equality of corresponding shear stresses [ Gleichheit der einander zugeordneten Schubspannungen ] σ yy y σ xx τ xy f y C τ yx f x τ xy σ xx xx. sym xy. yy xz yz zz τ yx x σ yy
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
Strains inite element model of the fracture callus Global, (external) strains : Change in length Original length L L Local, (internal) strains Units of Strain without a unit 1 1/1 = % 1/1.. = με (micro strain) =,1 % Gap Distortional Strain in a fracture callus
3D Local Strain State: Strain Tensor y Displacements [Verschiebungen] xy z x undeformed deformed
3D Local Strain State: Strain Tensor Definition: Universal Strain Definition: xx xy ij lim x 1 1 x, lim y yy zz x y y z, xz 1,, yz u u, i, j { x, y, } i, j j, i z 1 lim z z xx xy xz xy yy yz xz yz zz Note to Remember: Strain is relative change in length (and shape) Similar to stress tensor 6 independent components Strain Tensor
Displacement vs. Strain x z Displacement u_x Strain, eps_xx
Isotropic vs. Anisotropic (linear elastic) Linear stress-strain relation Stress σ linear E E (81 Param.) E (36 Param.) Strain ε ull 3 4 material properties tensor of 4 th order Equality of shear stresses (Boltzmann Continua) and strains: Reciprocity Theorem from Maxwell fully anisotropic: Orthotropic: Transverse isotropic: ull isotropic: (81 Param.) (36 Param.) (1 Param.) (9 Param.) (5 Param.) ( Param.)
E - Young s modulus - Poisson s ratio (....5) 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 ) (1 ) (1 ) (1 ) (1 ) (1 ) (1 ) (1 Isotropic vs. Anisotropic (linear elastic) of these
our Simple Load Cases
A Tension, Compression Tensile bar (tensional rigidity EA) Global behavior Device stiffness L d d -d A L unloaded loaded cut EA L, L Strain ΔL ε L k Local behavior Axial Stress ε q A Δd d EA L
B Shear Shear bolt (shear rigidity GA) w L τ A Global behavior Device stiffness GA w, L Local behavior Shear stress A Shear strain 1 xy k GA L
C Bending (Cantilever Beam) Beam (bending rigidity EI a, Length L) Cut z x M w Tension Neutral axis Compression Shear stress Global behavior, compliance Cantilever formula [Kragbalkenformel] Local behavior Stresses in transverse cut w 3 L 3EI L EI a a L EI L EI a a M M xx xz ( x, z) ( x, z) M A ( x l) I a z
D Torsion Torsional rod (torsional rigidity GI T ) Global behavior, stiffness GIT M, L c GI L T L R M Length L, radius r Cut τ Local behaviour Stresses in transverse cut M I T DistancefromCenter
Second Moment of Area I (SMA) [lächenmoment zweiten Grades] h D D d Axial (second) moment of area [axiales lächenmoment] I a b b h 1 3 I a 64 D 4 4 4 I a ( D d ) 64 Polar/torsional (second) moment of area [polares lächenmoment] I T 3 D 3 I 4 4 4 P IT I p ( D d ) Note to Remember: A hollow bone reaches a high stiffness and strength against bending and torsion with relatively little material.
Experiment: Alu-Stäbe biegen