Continuum Mechanics and the Finite Element Method
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1 Continuum Mechanics and the Finite Element Method 1
2 Assignment 2 Due on March 2 midnight 2
3 Suppose you want to simulate this
4 The familiar mass-spring system l 0 l y i X y i x Spring length before/after l = x - y i l 0 = X - y i Deformation Measure Elastic Energy Forces f int (x) =
5 Mass Spring Systems Can be used to model arbitrary elastic/plastic objects, but Behavior depends on tessellation Find good spring layout Find good spring constants Different types of springs interfere No direct map to measurable material properties
6 Alternative Start from continuum mechanics Discretize with Finite Elements Decompose model into simple elements Setup & solve system of algebraic equations Advantages Accurate and controllable material behavior Largely independent of mesh structure
7 Mass Spring vs Continuum Mechanics Mass spring systems require: 1. Measure of Deformation 2. Material Model 3. Deformation Energy 4. Internal Forces We need to derive the same types of concepts using continuum mechanics principles
8 Continuum Mechanics: 3D Deformations For a deformable body, identify: 3 undeformed state R described by positions 3 deformed state ' R described by positions Displacement field u describes ' u : ' x X u(x ) in terms of X x z u(x) ' X x u(x ) y x
9 Continuum Mechanics: 3D Deformations Consider material points X 1 and X 2 such that d is infinitesimal, where d X Now consider deformed vector x z X 1 d X 2 d' x y 2 x 1 ( I u)d x 1 d' ' x 2 d' 2 X 1 Deformation gradient F= x X u xu xv xw u y y y v w u z v z w z
10 So Displacement field transforms points x X u(x ) Jacobian of displacement field (deformation gradient) transforms differentials (infinitesimal vectors) from undeformed to deformed d' ( I u) d Fd x F= X
11 Displacement Field and Deformation Gradient In general, displacement field is not explicitly described. Nevertheless, toy examples:
12 Displacement Field and Deformation Gradient In general, displacement field is not explicitly described. Nevertheless, toy examples:
13 Displacement Field and Deformation Gradient In general, displacement field is not explicitly described. Nevertheless, toy examples:
14 Displacement Field and Deformation Gradient In general, displacement field is not explicitly described. Nevertheless, toy examples:
15 Displacement Field and Deformation Gradient In general, displacement field is not explicitly described. Nevertheless, toy examples:
16 Measure of deformations Displacement field transforms points x X u(x ) Jacobian of displacement field (deformation gradient) transforms vectors d' ( I u) d Fd x F= X How can we describe deformations?
17 Back to spring deformation Deformation measure (strain): Similar to F Undeformed spring: = 1 Undeformed (infinitesimal) continuum volume: = I?
18 Strain (description of deformation in terms of relative displacement) Deformation measure (strain): Desirable property: if spring is undeformed, strain is 0 (no change in shape) Can we find a similar measure that would work for infinitesimal volumes?
19 3D Nonlinear Strain Idea: to quantify change in shape, measure change in squared length for any arbitrary vector d I F F d d d d d d d ) ( ' ' ' 2 2 T T T T Green strain ) ( 2 1 I F F E T
20 Strain (description of deformation in terms of relative displacement) Deformation measure (strain): Desirable property: if spring is undeformed, strain is 0 (no change in shape) Can we find a similar measure that would work for infinitesimal volumes? T Green strain E ( F F I) 1 2
21 3D Linear Strain Green strain is quadratic in displacements 2 1 w v w u w w v v u v w u v u u z z y z x y z y y x x z x y x ) ( ) ( u u u u I F F E T T T Neglecting quadratic term (small deformation assumption) leads to the linear Cauchy strain (small strain) Written out: Notation ε = 1 2 u + ut = 1 2 F + Ft I
22 3D Linear Strain Linear Cauchy strain Geometric interpretation y,v x,u i : normal strains i : shear strains 2 1 w v w u w w v v u v w u v u u z z y z x y z y y x x z x y x x y xy
23 Cauchy vs. Green strain Nonlinear Green strain is rotation-invariant Apply incremental rotation R to given deformation F to obtain F = RF Then E' 1 2 ( F' T F' I) E Linear Cauchy strain is not rotation-invariant ε = 1 2 F + F t ε artifacts for larger rotations
24 Mass Spring vs Continuum Mechanics Mass spring systems: 1. Measure of Deformation 2. Material Model 3. Deformation Energy 4. Internal Forces Continuum Mechanics: 1. Measure of Deformation: Green or Cauchy strain 2. Material Model
25 Material Model: linear isotropic material Material model links strain to energy (and stress) Linear isotropic material (generalized Hooke s law) Energy density Ψ = 1 2 λtr(ε)2 +μtr(ε 2 ) Lame parameters λ and μ are material constants related to Poisson Ratio and Young s modulus Interpretation tr ε 2 = ε F 2 penalizes all strain components equally tr ε 2 penalizes dilatations, i.e., volume changes
26 Volumetric Strain (dilatation, hydrostatic strain) Consider a cube with side length a For a given deformation ε, the volumetric strain is ΔV/V0 = (a 1 + ε 11 a 1 + ε 22 a 1 + ε 33 a 3 )/a 3 = ε 11 + ε 22 + ε 33 + O ε 2 tr ε a
27 Linear isotropic material Energy density: Ψ = 1 2 λtr(ε)2 +μtr(ε 2 ) Problem: Cauchy strain is not invariant under rotations artifacts for rotations Solutions: Corotational elasticity Nonlinear elasticity
28 Material Model: non-linear isotropic model Replace Cauchy strain with Green strain St. Venant-Kirchhoff material (StVK) Energy density: Ψ StVK = 1 2 λtr(e)2 +μtr(e 2 ) Rotation invariant!
29 Problems with StVK StVK softens under compression Ψ StVK = 1 2 λtr(e)2 +μtr E 2 Energy Deformation
30 Advanced nonlinear materials Green Strain E = 1 2 Ft F I = 1 (C I) 2 Split into deviatoric (i.e. shape changing/distortion) and volumetric (dilation, volume changing) deformations Volumetric: J = det F Deviatoric: C = det F 2/3 C Neo-Hookean material: Ψ NH = μ tr C I μln(j) + λ ln J Energy
31 Different Models St. Venant-Kirchoff Neo-Hookean Linear
32 Mass Spring vs Continuum Mechanics Mass spring systems: 1. Measure of Deformation 2. Material Model 3. Deformation Energy 4. Internal Forces Continuum Mechanics: 1. Measure of Deformation: Green or Cauchy strain 2. Material Model: linear, StVK, Neo-Hookean, etc 3. From Energy Density to Deformation Energy: Finite Element Discretization
33 Finite Element Discretization Divide domain into discrete elements, e.g., tetrahedra Explicitly store displacement values at nodes (x i ). Displacement field everywhere else obtained through interpolation: x X = N i X x i Deformation Gradient: F = x X = N X i x i i X t
34 ҧ ҧ Basis Functions Basis functions (aka shape functions) N i X j : R 3 R Satisfy delta-property: N i X j = δ ij Simplest choice: linear basis functions N i x, ҧ തy, z ҧ = a i x + b i തy + c i z + d i Compute N i (and N i X ) through X 4 ഥV e X 3 X 1 Ω e X2
35 Constant Strain Elements (Linear Basis Functions) Displacement field is continuous in space Deformation Gradient, strain, stress are not Constant Strain per element Deformation Gradient can be computed as F = ee -1 where e and E are matrices whose columns are edge vectors in undeformed and deformed configurations
36 Constant Strain Elements: From energy density to deformation energy Integrate energy density over the entire element: W e = න Ω e Ψ F If basis functions are linear: F is linear in x i F is constant throughout element: W e = න Ω e Ψ F = Ψ F തV e
37 Mass Spring vs Finite Element Method Mass spring systems: 1. Measure of Deformation 2. Material Model 3. Deformation Energy 4. Internal Forces Continuum Mechanics: 1. Measure of Deformation: Green or Cauchy strain 2. Material Model: linear, StVK, Neo-Hookean, etc 3. Deformation Energy: integrate over elements 4. Internal Forces:
38 FEM recipe Discretize into elements (triangles/tetraderons, etc) For each element Compute deformation gradient F = ee -1 Use material model to define energy density Ψ(F) Integrate over elements to compute energy: W Compute nodal forces as:
39 FEM recipe Area/volume of element f = W Ψ F = V x F x First Piola-Kirchhoff stress tensor P
40 FEM recipe Area/volume of element f = W Ψ F = V x F x First Piola-Kirchhoff stress tensor P
41 FEM recipe For a tetrahedron, this works out to: f 1 f 2 f 3 = VPE T ; f 4 = f 1 f 2 f 3 Additional reading:
42 Material Parameters Lame parameters λ and μ are material constants related to the fundamental physical parameters: Poisson s Ratio and Young s modulus (
43 Young s Modulus and Poisson Ratio Lame parameters λ and μ are material constants related to the fundamental physical parameters: Poisson s Ratio and Young s modulus ( 10 Kg 10 Kg Young s modulus (E), measure of stiffness Poisson s ratio (ν), relative transverse to axial deformation
44 What Do These Parameters Mean Stiffness is pretty intuitive 10 Kg OBJECT E= GPa Pascals = force/area 10 Kg OBJECT E=60 KPa
45 What Do These Parameters Mean Poisson s Ratio controls volume preservation OBJECT 1 1 OBJECT
46 What Do These Parameters Mean Poisson s Ratio controls volume preservation OBJECT 2 OBJECT 1/2
47 What Do These Parameters Mean Poisson s Ratio controls volume preservation OBJECT 2 Poisson s Ratio ( ) =? OBJECT 1/2
48 What Do These Parameters Mean Poisson s Ratio controls volume preservation OBJECT PR = 0.5 Poisson s Ratio ( ) = OBJECT PR = 0.5 1/2
49 What Do These Parameters Mean Poisson s ratio is between -1 and 0.5 OBJECT PR = 0.5 Poisson s Ratio ( ) = OBJECT PR =
50 Negative Poisson s Ratio
51 Measurement Where do material parameters come from?
52 Simple Measurement: Stiffness OBJECT
53 Simple Measurement What s the Force (Stress)? What s the Deformation (Strain)? 1kg OBJECT
54 Simple Measurement 1kg OBJECT
55 Simple Measurement 2kg OBJECT
56 Simple Measurement 3kg OBJECT
57 Simple Measurement 7kg OBJECT
58 Simple Measurement How do we get the stiffness? 7kg OBJECT
59 Simple Measurement How do we get the stiffness? 7kg OBJECT
60 Simple Measurement How do we get the stiffness? 7kg Stiffness OBJECT
61 Simple Measurement: Poisson s Ratio OBJECT
62 Simple Measurement: Poisson s Ratio OBJECT
63 Simple Measurement: Poisson s Ratio Compute changes in width and height OBJECT
64 Simple Measurement: Poisson s Ratio Poisson s Ratio OBJECT
65 Measurement Devices
66 Simulating Elastic Materials with CM+FEM You now have all the mathematical tools you need
67 Suppose you want to simulate this
68 Plastic and Elastic Materials Elastic Materials Objects return to their original shape in the absence of other forces Plastic Deformations: Object does not always return to its original shape
69 Example: Crushing a Coke Can Old Reference State New Reference State
70 Example: Crushing a van
71 New Reference State A Simple Model For Plasticity Recall our model for strain: Let s consider how to encode a change of reference shape into this metric Changing undeformed mesh is not easy! We want to exchange F with, a deformation gradient that takes into account the new shape of our object
72 Continuum Mechanics: Deformation deformation gradient maps undeformed vectors (local) to deformed (world) vectors Reference Space World Space
73 Continuum Mechanics: Deformation deformation gradient maps undeformed vectors (reference) to deformed (world) vectors Reference Space World Space
74 Continuum Mechanics: Deformation F transforms a vector from Reference space to World Space Reference Space World Space
75 Continuum Mechanics: Deformation Introduce a new space Reference Space Plastic Space World Space
76 Continuum Mechanics: Deformation Our goal is to use access to but we only have Reference Space Plastic Space World Space
77 Continuum Mechanics: Deformation Our goal is to use access to but we only have Reference Space Plastic Space World Space
78 Continuum Mechanics: Deformation Our goal is to use access to but we only have Keep an estimate of built incrementally per element,
79 How to Compute the Plastic Deformation Gradient We compute the strain/stress for each element during simulation When it gets above a certain threshold store F as
80 How to Compute the Plastic Deformation Gradient We compute the strain/stress for each element during simulation When it gets above a certain threshold store F as
81 How to Compute the Plastic Deformation Gradient Each subsequent simulation step uses
82 So now you too can simulate this
83 Questions?
84 93
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