Finite Elements Part 2 (solid mechanics)

Draft DRAFT Lecture Notes in: FINITE ELEMENT II Solid Mechanics CVEN 6525 c VICTOR E. SAOUMA, SPRING 2001 Dept. ...

Author: Victor E Sauoma

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Draft DRAFT

Lecture Notes in:

FINITE ELEMENT II Solid Mechanics CVEN 6525

c VICTOR

E. SAOUMA,

SPRING 2001

Dept. of Civil Environmental and Architectural Engineering University of Colorado, Boulder, CO 80309-0428

Draft 0–2

Victor Saouma

Finite Elements II; Solid Mechanics

Draft Contents 1 PREREQUISITE 1.1 Variational Formulations . . . . . . . . . . . . . . . . . . . . . 1.2 Finite Element Formulation . . . . . . . . . . . . . . . . . . . 1.2.1 Strain Displacement Relations . . . . . . . . . . . . . 1.2.1.1 Axial Members . . . . . . . . . . . . . . . . . 1.2.1.2 Flexural Members . . . . . . . . . . . . . . . 1.2.2 Virtual Displacement and Strains . . . . . . . . . . . . 1.2.3 Element Stiﬀness Matrix Formulation . . . . . . . . . 1.2.3.1 Stress Recovery . . . . . . . . . . . . . . . . 1.3 Direct Stiﬀness Method . . . . . . . . . . . . . . . . . . . . . 1.3.1 Global Stiﬀness Matrix . . . . . . . . . . . . . . . . . 1.3.1.1 Structural Stiﬀness Matrix . . . . . . . . . . 1.3.1.2 Augmented Stiﬀness Matrix . . . . . . . . . 1.3.2 Logistics . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.2.1 Boundary Conditions, [ID] Matrix . . . . . . 1.3.2.2 LM Vector . . . . . . . . . . . . . . . . . . . 1.3.2.3 Assembly of Global Stiﬀness Matrix . . . . . E 1-1 Assembly of the Global Stiﬀness Matrix . . . . . . . . 1.3.2.4 Algorithm . . . . . . . . . . . . . . . . . . . E 1-2 Direct Stiﬀness Analysis of a Truss . . . . . . . . . . . E 1-3 Analysis of a Frame with MATLAB . . . . . . . . . . E 1-4 Analysis of a simple Beam with Initial Displacements 2 INTRODUCTION 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . 2.2 Elliptic, Parabolic and Hyperbolic Equations . . . E 2-1 Seepage Problem;(Bathe 1996) . . . . . . . E 2-2 Diﬀusion Problem; (Bathe 1996) . . . . . . E 2-3 Wave Equation, (Bathe 1996) . . . . . . . . 2.3 Solution of Discrete-System Mathematical models . 2.3.1 Steady State Problems . . . . . . . . . . . . 2.3.1.1 Elastic Spring . . . . . . . . . . . 2.3.1.2 Heat Transfer . . . . . . . . . . . 2.3.1.3 Hydraulic Network . . . . . . . . . 2.3.1.4 DC Network . . . . . . . . . . . . 2.3.2 Equivalent “Truss”/Direct Stiﬀness Models 2.3.2.1 Nonlinear Elastic Spring . . . . .

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1–1 . 1–1 . 1–1 . 1–1 . 1–1 . 1–4 . 1–5 . 1–5 . 1–6 . 1–6 . 1–6 . 1–7 . 1–7 . 1–8 . 1–8 . 1–10 . 1–10 . 1–10 . 1–12 . 1–13 . 1–18 . 1–20

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2–1 . 2–1 . 2–1 . 2–3 . 2–5 . 2–6 . 2–8 . 2–8 . 2–9 . 2–10 . 2–11 . 2–12 . 2–13 . 2–15

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2.3.3

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. 2–15 . 2–16 . 2–16 . 2–17 . 2–17 . 2–18 . 2–20 . 2–21 . 2–21 . 2–23 . 2–23

3 FUNDAMENTAL RELATIONS 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1.1 Indicial Notation . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1.2 Tensor Notation . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1.3 Voigt Notation . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.2 Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Vector Fields; Solid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Kinetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1.1 Force, Traction and Stress Vectors . . . . . . . . . . . . . . . 3.2.1.2 Traction on an Arbitrary Plane; Cauchy’s Stress Tensor . . . E 3-1 Stress Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 Kinematic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.3 Fundamental Laws of Continuum Mechanics . . . . . . . . . . . . . . 3.2.3.1 Conservation of Mass; Continuity Equation . . . . . . . . . . 3.2.3.2 Linear Momentum Principle; Equation of Motion . . . . . . 3.2.3.3 Conservation of Energy; First Principle of Thermodynamics 3.2.4 Constitutive Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.4.1 General 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.4.2 Transversly Isotropic Case . . . . . . . . . . . . . . . . . . . 3.2.4.3 Special 2D Cases . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.4.3.1 Plane Strain . . . . . . . . . . . . . . . . . . . . . . 3.2.4.3.2 Axisymmetry . . . . . . . . . . . . . . . . . . . . . . 3.2.4.3.3 Plane Stress . . . . . . . . . . . . . . . . . . . . . . 3.2.4.4 Pore Pressures . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.5 † Field Equations for Thermo- and Poro Elasticity . . . . . . . . . . . 3.3 Scalar Field: Diﬀusion Equation . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Derivation of the Diﬀusion Problem . . . . . . . . . . . . . . . . . . . 3.3.2.1 Simple 2D Derivation . . . . . . . . . . . . . . . . . . . . . . 3.3.2.2 †Generalized Derivation . . . . . . . . . . . . . . . . . . . . . 3.3.2.3 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . 3.4 Summary and Tonti Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . .

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3–1 . 3–1 . 3–1 . 3–1 . 3–3 . 3–3 . 3–3 . 3–5 . 3–5 . 3–5 . 3–6 . 3–7 . 3–7 . 3–10 . 3–12 . 3–12 . 3–13 . 3–14 . 3–14 . 3–15 . 3–16 . 3–16 . 3–16 . 3–16 . 3–17 . 3–17 . 3–19 . 3–20 . 3–21 . 3–21 . 3–22 . 3–24 . 3–24

2.4

2.5 2.6

Propagation Problems . . . . . . . 2.3.3.1 Dynamic Elastic System 2.3.3.2 Transient Heat Flow . . . 2.3.4 Eigenvalue Problems . . . . . . . . 2.3.4.1 Free Vibration . . . . . . 2.3.4.2 Column Buckling . . . . Solution Strategies . . . . . . . . . . . . . 2.4.1 Euler Equation . . . . . . . . . . . E 2-4 Flexure of a Beam . . . . . . . . . Computer Programs . . . . . . . . . . . . Examples of applications . . . . . . . . . .

CONTENTS

Victor Saouma

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Finite Elements II; Solid Mechanics

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4 MESH GENERATION 4.1 Introduction . . . . . . . . . . . . 4.2 Triangulation . . . . . . . . . . . 4.2.1 Voronoi Polygon . . . . . 4.2.2 Delaunay Triangulation . 4.2.3 MATLAB Code . . . . . . 4.3 Finite Element Mesh Generation 4.3.1 Boundary Deﬁnition . . . 4.3.2 Interior Node Generation 4.3.3 Final Triangularization .

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5 VARIATIONAL and RAYLEIGH-RITZ METHODS 5.1 Multiﬁeld Variational Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Total Potential Energy Principle . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Static; Euler . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2 Dynamic; Euler/Lagrange . . . . . . . . . . . . . . . . . . . . . . . . . . E 5-1 Hamilton’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 General Hu-Washizu Variational Principle . . . . . . . . . . . . . . . . . . . . . 5.4 Rayleigh Ritz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E 5-2 Uniformly Loaded Simply Supported Beam; Polynomial Approximation E 5-3 Heat Conduction; (Bathe 1996) . . . . . . . . . . . . . . . . . . . . . . .

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4–1 4–1 4–2 4–2 4–3 4–3 4–3 4–3 4–4 4–6

5–1 . 5–1 . 5–2 . 5–2 . 5–4 . 5–5 . 5–9 . 5–13 . 5–15 . 5–15 . 5–16

6 INTERPOLATION FUNCTIONS; NATURAL COORDINATE SYSTEMS 6–1 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–1 6.2 Cartesian Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–1 6.2.1 C 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–2 6.2.1.1 Truss element . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–2 6.2.1.2 Generalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–3 6.2.1.3 Constant Strain Triangle Element . . . . . . . . . . . . . . . . . 6–3 6.2.1.4 Further Generalization: Lagrangian Interpolation Functions . . . 6–5 6.2.1.5 Rectangular Bilinear Element . . . . . . . . . . . . . . . . . . . . 6–6 6.2.1.6 Solid Rectangular Trilinear Element . . . . . . . . . . . . . . . . 6–7 6.2.2 C 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–8 6.2.2.1 Flexural . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–8 6.2.2.2 C 1 : Hermitian Interpolation Functions . . . . . . . . . . . . . . 6–9 6.2.3 Characteristics of Shape Functions . . . . . . . . . . . . . . . . . . . . . . 6–10 6.3 Natural Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–10 6.3.1 Straight Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–11 6.3.2 Triangular Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–12 6.3.3 Volume Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–14 6.3.4 Interpolation Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–15 6.4 Pascal’s Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–16 7 FINITE ELEMENT DISCRETIZATION and REQUIREMENTS 7–1 7.1 Discretization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–1 7.1.1 Discretization of the Variational Statement for the General TPE Variational Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–1 Victor Saouma

Finite Elements II; Solid Mechanics

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7.2 7.3 7.4 7.5

CONTENTS

7.1.2 Discretization of the Variational Statement for the HW Variational Principle7–3 General Element Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–7 Discretization Error and Convergence Rate . . . . . . . . . . . . . . . . . . . . . 7–8 Lower Bound Character of Minimum Potential Energy Based Solutions . . . . . 7–10 Equilibrium and Compatibiliy in the Solution . . . . . . . . . . . . . . . . . . . . 7–10

8 STRAIGHT SIDED ELEMENTS; 1st GENERATION 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Rod Elements . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.1 Truss Element . . . . . . . . . . . . . . . . . . . . 8.2.2 Beam Element . . . . . . . . . . . . . . . . . . . . 8.3 Triangular Elements . . . . . . . . . . . . . . . . . . . . . 8.3.1 Cartesian Coordinate System (CST) . . . . . . . . 8.3.2 Natural Coordinate System . . . . . . . . . . . . . 8.3.2.1 Linear, T3 . . . . . . . . . . . . . . . . . 8.3.2.2 Quadratic Element (T6) . . . . . . . . . . 8.4 Bilinear Rectangular Element . . . . . . . . . . . . . . . . 8.5 Element Assessment . . . . . . . . . . . . . . . . . . . . . 8.5.1 CST . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.2 BiLinear Rectangular . . . . . . . . . . . . . . . .

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9 ISOPARAMETRIC ELEMENTS; 2nd GENERATION 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Element Formulation . . . . . . . . . . . . . . . . . . . . . . . . 9.2.1 Bar Element . . . . . . . . . . . . . . . . . . . . . . . . 9.2.2 Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . 9.2.2.1 Linear Element (Q4) . . . . . . . . . . . . . . . 9.2.2.1.1 Example: Jacobian Operators, (Bathe 9.2.2.2 Quadratic Element . . . . . . . . . . . . . . . . 9.2.2.2.1 Serendipity Element (Q8) . . . . . . . 9.2.2.2.2 Lagrangian element (Q9) . . . . . . . 9.2.2.2.3 Variable (Hierarchical) Element . . . 9.2.3 Triangular Elements . . . . . . . . . . . . . . . . . . . . 9.3 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . 9.3.1 Newton-Cotes . . . . . . . . . . . . . . . . . . . . . . . . 9.3.2 Gauss-Legendre Quadrature . . . . . . . . . . . . . . . . 9.3.2.1 † Legendre Polynomial . . . . . . . . . . . . . 9.3.2.2 Gauss-Legendre Quadrature for n = 2 . . . . . 9.3.3 Rectangular and Prism Regions . . . . . . . . . . . . . . 9.3.4 Triangular Regions . . . . . . . . . . . . . . . . . . . . . 9.4 Stress Recovery; Nodal Stresses . . . . . . . . . . . . . . . . . . 9.5 Nodal Equivalent Loads . . . . . . . . . . . . . . . . . . . . . . 9.5.1 Gravity Load . . . . . . . . . . . . . . . . . . . . . . . . 9.5.2 Traction Load . . . . . . . . . . . . . . . . . . . . . . . . 9.5.3 Initial Strains/Stresses; Thermal Load . . . . . . . . . . 9.6 Computer Implementation . . . . . . . . . . . . . . . . . . . . . 9.6.1 Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6.2 MATLAB Code . . . . . . . . . . . . . . . . . . . . . . . Victor Saouma

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8–1 . 8–1 . 8–1 . 8–1 . 8–2 . 8–3 . 8–3 . 8–5 . 8–5 . 8–6 . 8–7 . 8–9 . 8–9 . 8–10

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9–1 . 9–1 . 9–3 . 9–3 . 9–5 . 9–5 . 9–10 . 9–11 . 9–11 . 9–13 . 9–15 . 9–16 . 9–17 . 9–18 . 9–20 . 9–21 . 9–22 . 9–22 . 9–22 . 9–24 . 9–25 . 9–25 . 9–26 . 9–28 . 9–29 . 9–29 . 9–30

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9.6.3

0–5 9.6.2.1 9.6.2.2 9.6.2.3 9.6.2.4 9.6.2.5 Plott of

stiff.m . . . . . dmat.m . . . . . sfr.m . . . . . . jacob.m . . . . . bmatps.m . . . . Shape Functions

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. 9–30 . 9–32 . 9–33 . 9–35 . 9–35 . 9–36

10 ELEMENT FORMULATION and STRAIN RECOVERY in HW FORMULATION 10–1 10.1 Element Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–1 10.2 Strain Recovery . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–2 10.2.1 C-lumping. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–3 10.2.2 Strain smoothing. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–3 10.2.3 C-splitting. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–4 10.3 Uniqueness and Existence of a Solution . . . . . . . . . . . . . . . . . . . . . . . 10–5 11 WEIGHTED RESIDUAL METHODS 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 General Formulation . . . . . . . . . . . . . . . . . . . . . . . . 11.2.1 Diﬀerential Operators . . . . . . . . . . . . . . . . . . . 11.2.1.1 Application to 1D Axial Member . . . . . . . . 11.2.2 Residual Formulation . . . . . . . . . . . . . . . . . . . 11.3 Weighted Residual Methods . . . . . . . . . . . . . . . . . . . . 11.3.1 † Point Collocation Method . . . . . . . . . . . . . . . . 11.3.2 † Subdomain Collocation Method . . . . . . . . . . . . . 11.3.3 † Least-Squares Method . . . . . . . . . . . . . . . . . . 11.3.4 Galerkin Method . . . . . . . . . . . . . . . . . . . . . . E 11-1 String Vibration . . . . . . . . . . . . . . . . . . . . . . 11.4 Applications of the Galerkin Method to 3D Elasticity Problems 11.4.1 Derivation of the Weak Form . . . . . . . . . . . . . . . 11.4.2 FE Discretization . . . . . . . . . . . . . . . . . . . . . .

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11–1 . 11–1 . 11–1 . 11–1 . 11–2 . 11–2 . 11–4 . 11–4 . 11–4 . 11–5 . 11–5 . 11–5 . 11–7 . 11–7 . 11–9

12 FINITE ELEMENT DISCRETIZATION OF THE FIELD 12.1 Derivation of the Weak Form . . . . . . . . . . . . . . . . . . 12.2 FE Discretization . . . . . . . . . . . . . . . . . . . . . . . . . 12.2.1 No Convection . . . . . . . . . . . . . . . . . . . . . . 12.2.2 Convection . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Illustrative Examples . . . . . . . . . . . . . . . . . . . . . . . E 12-1 Composite Wall . . . . . . . . . . . . . . . . . . . . . . E 12-2 Heat Transfer across a Fin . . . . . . . . . . . . . . . . 12.4 Comparison Between Vector and Scalar Formulations . . . . .

EQUATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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12–1 . 12–1 . 12–2 . 12–3 . 12–4 . 12–4 . 12–4 . 12–7 . 12–12

13 TOPICS in STRUCTURAL MECHANICS 13.1 Condensation/Substructuring . . . . . . . . . 13.2 Element Evaluation . . . . . . . . . . . . . . . 13.2.1 Patch Test . . . . . . . . . . . . . . . 13.2.2 Eigenvalue Test . . . . . . . . . . . . . 13.2.3 Order of Integration . . . . . . . . . .

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13–1 . 13–1 . 13–2 . 13–2 . 13–2 . 13–4

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13.2.3.1 Full Integration . . . . . . . . 13.2.3.2 Reduced Integration . . . . . . 13.2.3.3 Selective Reduced Integration 13.3 Parasitic Shear/Incompatible Elements . . . . . 13.3.1 Q4, The Problem . . . . . . . . . . . . . 13.3.2 Q6, The Solution . . . . . . . . . . . . . 13.3.3 QM6, Further Enhancements . . . . . . 13.4 Rotational D.O.F. . . . . . . . . . . . . . . . . 13.5 Constraints . . . . . . . . . . . . . . . . . . . .

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. 13–4 . 13–5 . 13–6 . 13–7 . 13–7 . 13–8 . 13–8 . 13–9 . 13–10

14 GEOMETRIC NONLINEARITY 14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 14.1.1 Strong Form . . . . . . . . . . . . . . . . . . . . 14.1.1.1 Lower Order Diﬀerential Equation . . . 14.1.1.2 Higher Order Diﬀerential Equation . . 14.1.2 Weak Form . . . . . . . . . . . . . . . . . . . . . 14.1.2.1 Strain Energy . . . . . . . . . . . . . . 14.1.2.2 Euler Equation . . . . . . . . . . . . . . 14.2 Finite Element Discretization . . . . . . . . . . . . . . . 14.3 Elastic Instability; Bifurcation Analysis . . . . . . . . . E 14-1 Column Stability . . . . . . . . . . . . . . . . . . E 14-2 Frame Stability . . . . . . . . . . . . . . . . . . . 14.4 Second-Order Elastic Analysis; Geometric Non-Linearity E 14-3 Eﬀect of Axial Load on Flexural Deformation . . E 14-4 Bifurcation . . . . . . . . . . . . . . . . . . . . . 14.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . .

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14–1 . 14–1 . 14–2 . 14–2 . 14–3 . 14–6 . 14–6 . 14–7 . 14–8 . 14–9 . 14–10 . 14–13 . 14–15 . 14–16 . 14–19 . 14–22

15 PLATES 15.1 Fundamental Relations . . . . . . . . . . . . . . . . . 15.1.1 Equilibrium . . . . . . . . . . . . . . . . . . . 15.1.2 Kinematic Relations . . . . . . . . . . . . . . 15.1.3 Constitutive Relations . . . . . . . . . . . . . 15.2 Plate Theories . . . . . . . . . . . . . . . . . . . . . 15.2.1 Reissner-Mindlin . . . . . . . . . . . . . . . . 15.2.1.1 Fundamental Relations . . . . . . . 15.2.1.2 Diﬀerential Equation . . . . . . . . 15.2.1.3 † Variational Formulation . . . . . . 15.2.2 Kirchhoﬀ . . . . . . . . . . . . . . . . . . . . 15.2.2.1 Fundamental Relations . . . . . . . 15.2.2.2 Diﬀerential Equation . . . . . . . . 15.2.2.3 Stresses . . . . . . . . . . . . . . . . 15.2.2.4 Variational Formulation . . . . . . . 15.2.3 Summary . . . . . . . . . . . . . . . . . . . . 15.3 Finite Element Formulations . . . . . . . . . . . . . 15.3.1 Rectangular Element . . . . . . . . . . . . . . 15.3.1.1 Formulation . . . . . . . . . . . . . 15.3.1.2 Shear Locking . . . . . . . . . . . . 15.3.2 Nonconforming Kirchhoﬀ Triangular Element

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15–1 . 15–1 . 15–1 . 15–4 . 15–5 . 15–6 . 15–6 . 15–6 . 15–8 . 15–8 . 15–9 . 15–9 . 15–10 . 15–10 . 15–11 . 15–11 . 15–13 . 15–13 . 15–13 . 15–16 . 15–17

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15.3.2.1 Formulation . . . 15.3.2.2 Nonconformity . . 15.3.3 Discrete Kirchhoﬀ Triangle 15.4 Summary . . . . . . . . . . . . . .

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. 15–17 . 15–19 . 15–20 . 15–24

16 MATERIAL NONLINEARITIES 16.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1.1 Linearization . . . . . . . . . . . . . . . . . . . . . . . 16.1.2 Solution Strategies . . . . . . . . . . . . . . . . . . . . 16.2 Load Control . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2.1 Newton-Raphson . . . . . . . . . . . . . . . . . . . . . 16.2.1.1 Newton-Raphson/Tangent Stiﬀness Method . 16.2.1.2 Modiﬁed Newton-Raphson . . . . . . . . . . 16.2.1.3 Secant Newton . . . . . . . . . . . . . . . . . 16.2.2 Acceleration of Convergence, Line Search Method . . 16.2.3 Convergence Criteria . . . . . . . . . . . . . . . . . . . 16.3 Direct Displacement Control . . . . . . . . . . . . . . . . . . 16.4 Indirect Displacement Control . . . . . . . . . . . . . . . . . . 16.4.1 Partitioning of the Displacement Corrections . . . . . 16.4.2 Arc-Length . . . . . . . . . . . . . . . . . . . . . . . . 16.4.3 Relative Displacement Criterion . . . . . . . . . . . . 16.4.4 IDC Methods with Approximate Line Searches . . . .

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16–1 . 16–1 . 16–1 . 16–2 . 16–4 . 16–4 . 16–4 . 16–6 . 16–6 . 16–7 . 16–10 . 16–10 . 16–13 . 16–13 . 16–15 . 16–17 . 16–18

A VECTOR OPERATIONS A.1 Vector Diﬀerentiation . . . . . . . . . . . . . . . . . . . . . A.1.1 Derivative WRT to a Scalar . . . . . . . . . . . . . . E A-1 Tangent to a Curve . . . . . . . . . . . . . . . . . . A.1.2 Divergence . . . . . . . . . . . . . . . . . . . . . . . E A-2 Divergence . . . . . . . . . . . . . . . . . . . . . . . A.1.3 Gradient . . . . . . . . . . . . . . . . . . . . . . . . . A.1.4 Scalar . . . . . . . . . . . . . . . . . . . . . . . . . . E A-3 Gradient of a Scalar . . . . . . . . . . . . . . . . . . E A-4 Stress Vector normal to the Tangent of a Cylinder . A.2 Vector Integrals . . . . . . . . . . . . . . . . . . . . . . . . . A.2.1 Integral of a Vector . . . . . . . . . . . . . . . . . . . A.2.2 Line Integral . . . . . . . . . . . . . . . . . . . . . . A.2.3 Integration by Parts . . . . . . . . . . . . . . . . . . A.2.4 Gauss; Divergence Theorem . . . . . . . . . . . . . . A.2.5 Stoke’s Theorem . . . . . . . . . . . . . . . . . . . . A.2.6 Green; Gradient Theorem . . . . . . . . . . . . . . . E A-5 Physical Interpretation of the Divergence Theorem

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A–1 . A–1 . A–1 . A–2 . A–3 . A–5 . A–5 . A–5 . A–6 . A–6 . A–9 . A–9 . A–9 . A–9 . A–10 . A–10 . A–10 . A–10

B CASE-STUDY: FRACTURING of A DAM DUE TO B.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . B.1.1 Elastic and Thermal Properties . . . . . . . . . . B.1.2 Loads . . . . . . . . . . . . . . . . . . . . . . . . B.2 ANALYSIS II; “Thermal Shock” . . . . . . . . . . . . . B.2.1 Thermal Analysis . . . . . . . . . . . . . . . . . . Victor Saouma

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THERMAL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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B.2.2 Stress Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B–7 B.2.3 Data Files . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B–9 B.3 CONCLUSIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B–9 C MISC. C–1 C.1 Units & Conversion Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C–1 C.2 Metric Preﬁxes and Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . C–2

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Draft List of Figures 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

Summary of Variational Methods . . . . . . . . . Duality of Variational Principles . . . . . . . . . Frame Example . . . . . . . . . . . . . . . . . . . Example for [ID] Matrix Determination . . . . . Simple Frame Analyzed with the MATLAB Code . . . . . . . . . . . . . . . . . . . . . . . . . . . Simple Frame Analyzed with the MATLAB Code Stiﬀness Analysis of one Element Structure . . .

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Finite Element Process, (Bathe 1996) . . . . . . . . . . . . . . . . . . . . . . . . 2–2 Seepage Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–3 One Dimensional Heat Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . 2–5 Rod subjected to Step Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–6 System of Rigid Carts Interconnected by Linear Springs, (Bathe 1996) . . . . . . 2–9 Slab Subjected to Temperature Boundary Conditions, (Bathe 1996) . . . . . . . 2–10 Pipe Network, (Bathe 1996) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–11 DC Network, (Bathe 1996) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–12 Equivalent Trusses/Direct Stiﬀness . . . . . . . . . . . . . . . . . . . . . . . . . . 2–13 Heat Transfer Idealization in an Electron Tube, (Bathe 1996) . . . . . . . . . . . 2–17 Stability of a Two Rigid Bars System . . . . . . . . . . . . . . . . . . . . . . . . 2–18 Idealization, Discretization and Solution of a Numerical Simulation, (Felippa 2000)2–20

3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9

Stresses as Tensor Components . . . . . . . . . . . . Cauchy’s Tetrahedron . . . . . . . . . . . . . . . . . Flux Through Area dS . . . . . . . . . . . . . . . . . Equilibrium of Stresses, Cartesian Coordinates . . . Flux vector . . . . . . . . . . . . . . . . . . . . . . . Flux Through Sides of Diﬀerential Element . . . . . *Flow through a surface Γ . . . . . . . . . . . . . . . Components of Tonti’s Diagram, (Felippa 2000) . . . Fundamental Equations of Solid Mechanics and Heat

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4.1 4.2 4.3 4.4

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5.3

Uniformly Loaded Simply Supported Beam Analysed by the Rayleigh-Ritz Method5–15

6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9

Axial Finite Element . . . . . . . . . . . . . . . . . . . . Constant Strain Triangle Element . . . . . . . . . . . . . Rectangular Bilinear Element . . . . . . . . . . . . . . . Solid Trilinear Rectangular Element . . . . . . . . . . . Flexural Finite Element . . . . . . . . . . . . . . . . . . Shape Functions for Flexure of Uniform Beam Element. Natural Coordinate System Along a Straight Line . . . Natural Coordinate System for a Triangle . . . . . . . . Integration over a Triangle . . . . . . . . . . . . . . . . .

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Completness and Compatibility . h, p and r Convergence . . . . . Interelement Continuity of Stress Interelement Continuity of Strain

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Rectangular Bilinear Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–8 Linear Triangular Element Subjected to Pure Bending . . . . . . . . . . . . . . . 8–10

9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11 9.12 9.13 9.14 9.15 9.16 9.17 9.18 9.19 9.20 9.21

Two-Dimensional Mapping of Some Elements . . . . . . . . . . . . . . . . . . . . 9–1 Actual and Parent Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–2 Iso, Super, and Sub Parametric Elements . . . . . . . . . . . . . . . . . . . . . . 9–3 Three-Noded Quadratic Bar Element . . . . . . . . . . . . . . . . . . . . . . . . . 9–3 Four Noded Isoparametric Element . . . . . . . . . . . . . . . . . . . . . . . . . . 9–5 Diﬀerential Element in Curvilinear Coordinate System . . . . . . . . . . . . . . . 9–8 Cross Product of Two Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–8 *Elements with Possible Singular Jacobians . . . . . . . . . . . . . . . . . . . . . 9–9 Serendipity and Lagrangian Quadratic Quadrilaterals . . . . . . . . . . . . . . . 9–11 Pascal Triangles for Quadrilateral and Triangle Elements . . . . . . . . . . . . . . 9–12 Serendipity Isoparametric Quadratic Finite Element: Global and Parent Element 9–13 Shape Functions for 8 Noded Quadrilateral Element . . . . . . . . . . . . . . . . 9–14 Shape Functions for 9 Noded Quadrilateral Element . . . . . . . . . . . . . . . . 9–15 Nodal Numbering for Isoparameteric Elements . . . . . . . . . . . . . . . . . . . 9–16 Newton-Cotes Numerical integration . . . . . . . . . . . . . . . . . . . . . . . . . 9–18 Gauss-Legendre Integration Over a Surface . . . . . . . . . . . . . . . . . . . . . 9–23 Numerical Integration Over a Triangle . . . . . . . . . . . . . . . . . . . . . . . . 9–23 Extrapolation from 4-Node Quad Gauss Points (Felippa 1999) . . . . . . . . . . . 9–24 Gravity Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–26 Traction Load in Isoparametric Elements . . . . . . . . . . . . . . . . . . . . . . 9–27 Traction Load in Contiguous Isoparametric Elements . . . . . . . . . . . . . . . . 9–28

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10.1 Patch test. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–6 12.1 Heat Flow in a Thin Rectangular Fin

. . . . . . . . . . . . . . . . . . . . . . . . 12–9

13.1 Patch Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–2 13.2 Eigenvectors Corresponding to a) Non-Zero and b) Zero Eigenvalues for a Square Bilinear element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–4

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

LIST OF FIGURES

0–3

13.3 Independent Displacement Modes for a Bilinear Element . . . . . . . . . . 13.4 Hourglas Modes in Under-Integrated Quadratic Element . . . . . . . . . . 13.5 Rectangular Bilinear Element Subjected to Bending; Bilinear Element and rect Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6 Displacements Associated with Incompatible Modes for the Q6 Element . 13.7 Side Displacements Induced by Drilling d.o.f. . . . . . . . . . . . . . . . .

. . . . . . Cor. . . . . . . . .

14.1 14.2 14.3 14.4

. 13–6 . 13–6 . 13–7 . 13–8 . 13–9

Level of Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Euler Column . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Simply Supported Beam Column; Diﬀerential Segment; Eﬀect of Axial Force P Solution of the Tanscendental Equation for the Buckling Load of a Fixed-Hinged Column . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5 Summary of Stability Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .

. 14–5 . 14–23

15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 15.9

. . . . . . . . .

. 15–1 . 15–2 . 15–3 . 15–4 . 15–6 . 15–13 . 15–17 . 15–20 . 15–21

16.1 Test Controls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Newton-Raphson Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3 Modiﬁed Newton-Raphson Method, Initial Tangent in Increment . . . . . . . . 16.4 Modiﬁed Newton-Raphson Method, Initial Problem Tangent . . . . . . . . . . 16.5 Incremental Secant, Quasi-Newton Method . . . . . . . . . . . . . . . . . . . . 16.6 Schematic of Line Search, (Reich 1993) . . . . . . . . . . . . . . . . . . . . . . . 16.7 Flowchart for Line Search Algorithm, (Reich 1993) . . . . . . . . . . . . . . . . 16.8 Divergence of Load-Controled Algorithms . . . . . . . . . . . . . . . . . . . . . 16.9 Hydrostatically Loaded Gravity Dam . . . . . . . . . . . . . . . . . . . . . . . . 16.10Load-Displacement Diagrams with Snapback . . . . . . . . . . . . . . . . . . . 16.11Flowchart for an incremental nonlinear ﬁnite element program with indirect displacement control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.12Two points on the load-displacement curve satisfying the arc-length constraint 16.13Flow chart for line search with IDC methods . . . . . . . . . . . . . . . . . . .

. 16–3 . 16–5 . 16–6 . 16–7 . 16–8 . 16–9 . 16–9 . 16–11 . 16–13 . 16–13 . 16–16 . 16–17 . 16–20

A.1 A.2 A.3 A.4 A.5 A.6 A.7 A.8

. A–2 . A–2 . A–3 . A–4 . A–4 . A–7 . A–8 . A–11

Finite Element Formulation . . . . . . . . . . . . . . . . Stresses in a Plate . . . . . . . . . . . . . . . . . . . . . Free Body Diagram of an Inﬁnitesimal Plate Element . Displacements in a Plate . . . . . . . . . . . . . . . . . . Positive Moments and Rotations . . . . . . . . . . . . . Rectangular Plate Element . . . . . . . . . . . . . . . . Triangular Plate Element in Natural Coordinate System Edges of Adjacent Triangular Elements . . . . . . . . . Discrete Kirchhoﬀ Triangular Element . . . . . . . . . .

. . . . . . . . .

Diﬀerentiation of position vector p . . . . . . . . . . . . . Curvature of a Curve . . . . . . . . . . . . . . . . . . . . . Vector Field Crossing a Solid Region . . . . . . . . . . . . Flux Through Area dA . . . . . . . . . . . . . . . . . . . . Inﬁnitesimal Element for the Evaluation of the Divergence Radial Stress vector in a Cylinder . . . . . . . . . . . . . . Gradient of a Vector . . . . . . . . . . . . . . . . . . . . . Physical Interpretation of the Divergence Theorem . . . .

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. 14–1 . 14–2 . 14–4

B.1 Boundary Description of Dam for Transient Thermal Analysis . . . . . . . . . . . B–4 Victor Saouma

Finite Elements II; Solid Mechanics

Draft 0–4

LIST OF FIGURES

B.2 Heat of Hydration Interpolations . . . . . . . . . . . . . . . . . . . . . . . . . . . B–6 B.3 Temperature Distribution in the Transient Thermal Analysis at Day 8 . . . . . . B–7 B.4 Maximum Principal Stresses and Deformed Mesh at Day 8 . . . . . . . . . . . . B–8

Victor Saouma

Finite Elements II; Solid Mechanics

Draft List of Tables 1.1

Summary of Variational Terms Associated with One Dimensional Elements . . . 1–4

3.1 3.2 3.3

Selected Examples of Diﬀusion Problems . . . . . . . . . . . . . . . . . . . . . . . 3–19 Comparison of Scalar and Vector Field Problems . . . . . . . . . . . . . . . . . . 3–26 Classiﬁcation of various Physical Problems, (Kardestuncer 1987) . . . . . . . . . 3–27

5.1 5.2

Functionals in Linear Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–1 Comparison Between Total Potential Energy and Hu-Washizu Formulations . . . 5–12

6.1 6.2 6.3

Characteristics of Beam Element Shape Functions . . . . . . . . . . . . . . . . . 6–9 Interpretation of Shape Functions in Terms of Polynomial Series (1D & 2D) . . . 6–16 Polynomial Terms in Various Element Formulations (1D & 2D) . . . . . . . . . . 6–16

8.1

Shape Functions and Derivatives for T6 Element . . . . . . . . . . . . . . . . . . 8–6

9.1 9.2 9.3 9.4

Shape Functions, and Natural Derivatives for Q8 Element . . . . . . . . . . . . Shape Functions for Variable Node Elements . . . . . . . . . . . . . . . . . . . Weights for Newton-Cotes Quadrature Formulas . . . . . . . . . . . . . . . . . Integration Points and Weights for Gauss-Quadrature Formulaes Over the Interval [−1, 1] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Coordinates and Weights for Numerical Integration over a Triangle . . . . . . . Natural Coordinates of Bilinear Quadrilateral Nodes . . . . . . . . . . . . . . .

9.5 9.6

. 9–13 . 9–16 . 9–19 . 9–21 . 9–23 . 9–25

10.1 Polynomial orders of the shape functions. . . . . . . . . . . . . . . . . . . . . . . 10–1 10.2 Table of α coeﬃcients and spectral radii for CS technique. . . . . . . . . . . . . . 10–5 12.1 Comparison of Scalar and Vector Field Problems, Revisited . . . . . . . . . . . . 12–12 13.1 Full and Reduced Numerical Integrations for Quadrilateral Elements . . . . . . . 13–5 13.2 Bilinsear and Exact Displacements/Strains . . . . . . . . . . . . . . . . . . . . . 13–7 15.1 Comparison of Governing Equations in Elasticity and Plate Bending . . . . . . . 15–12 15.2 Integration Rules for Mindlin Plate Elements . . . . . . . . . . . . . . . . . . . . 15–16 A.1 Similarities Between Multiplication and Diﬀerentiation Operators . . . . . . . . . A–1 B.1 B.2 B.3 B.4 B.5

Concrete Material Properties . . . . . Thermal Properties of the concrete . . Interface Element Material Properties Loads applied on the Dam . . . . . . . Heat of Hydration From the Literature

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. B–2 . B–2 . B–2 . B–3 . B–5

Draft 0–2

LIST OF TABLES

B.6 Heat of Hydration Adopted in the Simulation; Days and J/Kg/Day . . . . . . . . B–5 B.7 Sresses Along the Interface Element; m] and [Pa] . . . . . . . . . . . . . . . . . . B–8 B.8 Crack Opening and Sliding Displacements; [m] . . . . . . . . . . . . . . . . . . . B–9

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

LIST OF TABLES

0–3

NOTATION A c E h I J L Q t T u, v, w U0 U U0∗ U∗ W Π α µ ν ρ θ δM δP δθ δu δφ δU δW

SCALARS Area Speciﬁc heat Elastic Modulus Film coeﬃcient for convection heat transfer Moment of inertia St Venant’s torsional constant Length Rate of internal heat generation per unit volume Time Temperature Translational displacements along the x, y, and z directions Strain energy density Strain energy Complementary strain energy density Complementary strain energy Work Potential energy Coeﬃcient of thermal expansion Shear modulus Poisson’s ratio mass density Rotational displacement Virtual moment Virtual force Virtual rotation Virtual displacement Virtual curvature Virtual internal strain energy Virtual external work TENSORS order 1

a b c F p N ˜ N p P q R R t Victor Saouma

Vector of coeﬃcients in assumed displacement ﬁeld Body force Nodal coordinates Unknown element forces and unknown support reactions Matrix of coeﬃcients of a polynomial series Displacement shape functions Coordinate shape functions Element nodal forces = F Structure nodal forces Flux per unit area Structure reactions Residuals Traction vector Finite Elements II; Solid Mechanics

Draft 0–4

t u (x) u u ue u u V δ δε δσ σ σ0

LIST OF TABLES Speciﬁed tractions along Γt Displacement vector Speciﬁed displacements along Γu Displacement vector Nodal element displacements Nodal displacements in a continuous system Structure nodal displacements Shear forces in a plate Vx , Vy Element nodal displacements Virtual strain vector Virtual stress vector Stress vector Initial stress vector TENSORS order 2

d I k kg K Kg lij M N γ Γ ε ε0 k κ

Element ﬂexibility matrix Idendity matrix Element stiﬀness matrix Geometric element stiﬀness matrix Structure stiﬀness matrix Structure’s geometric stiﬀness matrix Direction cosine of rotated axis i with respect to original axis j Moments in a plate Mxx , Mxy , Myx , Myy Membrane forces Nxx , Nxy , Nyx , Nyy Shear deformations Transformation matrix Strain vector Initial strain vector Conductivity Curvature TENSORS order 4

D

Constitutive matrix CONTOURS, SURFACES, VOLUMES

Γ Γt Γu ΓT Γc Γq Ω

Surface Boundary along Boundary along Boundary along Boundary along Boundary along Volume of body

which which which which which

surface tractions, t are speciﬁed displacements, u are speciﬁed temperatures, T are speciﬁed convection ﬂux, qc are speciﬁed ﬂux, qn are speciﬁed

FUNCTIONS, OPERATORS

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

LIST OF TABLES

u ˜ δ B L ∇φ ∇.u = div .u v 2 . . . ∞

0–5

Neighbour function to u(x) Variational operator Discrete strain-displacement operator Linear diﬀerential operator relating displacement to strains ∂φ ∂φ T Divergence, (gradient operator) on scalar ∂φ ∂x ∂y ∂z ∂u y ∂ux z Divergence, (gradient operator) on vector ∂x + ∂y + ∂u ∂z Euclidian norm. Inﬁnity norm. PROGRAM ARRAYS

ID LM

Victor Saouma

Matrix relating nodal dof to structure dof structure dof of nodes connected to a given element

Finite Elements II; Solid Mechanics

Draft 0–6

Victor Saouma

LIST OF TABLES

Finite Elements II; Solid Mechanics

Draft

LIST OF TABLES 1 2 3 4 5 6

7 8 9 10

12 13 14 15 16

Jan. 16 18 23 25 30 Feb. 1 6 8 13 15 20 22 27 29 Mar. 5 7 12 14 19 21 Apr. 2 4 9 11 16 18 23 25 30 May 2

Victor Saouma

0–7

Introduction; Course objective; Overview; Notation. Mathematical Formulations. Elasticity Direct Method; Field Eq. Variational Methods Mesh Generators; Laboratory Variational Methods, Mechanics; Laboratory FE Discretization and Requirements; C0 Elements. Isoparametric Elements, Bar Element Isoparamteric Element, Bilinear Element Isoparameteric Element; Quadratic, Hierarchical Elements; Numerical integration Isoparameteric Element; Numerical integration Laboratory Weighted Residuals Galerkin; 3D Elasticity; Field Equation Field Equation, Theory, application Field Equation Exam I Lab (Field Equations) Error Analysis SPRING BREAK Topics (Condensation, Transformation, Integration, Test) Order of Integration, Eignevalu tests Plate Bending Plate Bending Plate Bending Geometric Non Linearity Geometric Nonlinearity Material Nonlinearity Dynamics Review

6.1-6.14

Ch. 16

9.1-9.6 11.1-11.5 Ch. 14. Ch. 17 Ch. 13

Finite Elements II; Solid Mechanics

Draft 0–8

Victor Saouma

LIST OF TABLES

Finite Elements II; Solid Mechanics

Draft Chapter 1

PREREQUISITE In the ﬁrst course (CVEN4525/5525, Finite Element I; Framed Structures), the direct stiﬀness method was ﬁrst introduced (element stiﬀness matrix, transformation matrix, global stiﬀness matrix assembly, internal force recovery). As an interlude we then covered the ﬂexibility method and stiﬀness-ﬂexibility relationship. The second part of the course began with a thorough coverage of variational method (duality between extremization of a functional and a corresponding euler diﬀerential equation) followed by a rigorous introduction/derivation of the various energy methods. 1

1.1

Variational Formulations

2 A summary of the various methods introduced in Finite Element I; Framed Structures is shown in Fig. 1.1, Fig. 1.2, and Table 1.1.

1.2 1.2.1

Finite Element Formulation Strain Displacement Relations

The displacement ∆ at any point inside an element can be written in terms of the shape functions N and the nodal displacements {∆} 3

∆ = N{∆}

(1.1)

ε = [B]{∆}

(1.2)

The strain is then deﬁned as: where [B] is the matrix which relates joint displacements to strain ﬁeld. 1.2.1.1

Axial Members

u = (1 −

x L)

N

x L

u1 u2 {∆}

(1.3-a)

Draft 1–2

PREREQUISITE

❄

❄

div σ + ρb = 0 t − t = 0 Γt

✲

def

U0 =

ε 0

✻

Natural B.C. Essential B.C.

δε − Dδu = 0 δu = 0 Γu

def

0

εdσ

Gauss

❄

δεT σdΩ −

σ

δuT bdΩ − Γt δuT tdΓ = 0 δWi − δWe = 0 Ω

❄

✻

❄

Principle of Stationary Potential Energy δΠ = 0 def Π = U − We Π = Ω U0 dΩ − ( Ω ui bi dΩ + Γt ui ti dΓ) ❄

❄

Principle of Complementary Stationary Potential Energy δΠ∗ = 0 ∗ def ∗ ∗ Π = Wi + We ∗ Π = Ω U0 dΩ + Γu ui δti dΓ ❄

Castigliano’s First Theorem

Castigliano’s Second Theorem

∂Wi k =P ∂∆k

∂Wi∗ k =∆ ∂Pk

Rayleigh-Ritz n

cji φji + φj0 uj ≈ ∂cji

❄

Principle of Complementary Virtual Work ε δσ i δti dΓ = 0 ij ij dΩ − Γu u Ω ∗ ∗ δWi − δWe = 0

✻

∂Π

δσij,j = 0 δti = 0 Γt

✻

Gauss

❄

Ω

❄

εij − (ui,j + uj,i ) = 0 Ω = 0 Γu Γ ✲ ui − u

U0∗ =

σdε

Principle of Virtual Work

❄ 1 2

❄

i=1

=0

i = 1, 2, · · · , n;

j = 1, 2, 3

Figure 1.1: Summary of Variational Methods

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

1.2 Finite Element Formulation

1–3

Kinematically Admissible Displacements Displacements satisfy the kinematic equations and the the kinematic boundary conditions ✻

❄

Principle of Stationary Complementary Energy

Principle of Virtual Work

Principle of Complementary Virtual Work

Principle of Stationary Potential Energy

✻

❄

Statically Admissible Stresses Stresses satisfy the equilibrium conditions and the static boundary conditions

Figure 1.2: Duality of Variational Principles

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 1–4

PREREQUISITE U

L

P2 dx 0 AE

1 2

Axial

Shear

Virtual Displacement δU General Linear L L du d(δu) σδεdx E Adx dx dx 0 0 dΩ σ

...

M2 dx 0 EIz

1 2

Torsion

L 0

L

M δφdx 0

T2 dx GJ

0

L

L 0

L

w

w(x)v(x)dx

0L

dθx d(δθx ) GJ dx dx dx

0

L 0

L

L

δT θdx 0

δε

0

M δM dx EIz δσ ε

T δT dx GJ δσ

Virtual Force δW ∗ Σi δPi ∆i Σ δM θ L i i i δw(x)v(x)dx

Virtual Displacement δW Σi Pi δ∆i Σ M δθ L i i i w(x)δv(x)dx

0

...

δM φdx

δε

σ

P M

δV γxy dx

d2 v d2 (δv) EIz 2 dx 2 dx dx σ

T δθdx 0

W 1 Σi 2 Pi ∆i Σi 12 Mi θi

ε

L

...

0L

L

1 2

Flexure

V δγxy dx

δε

L

Virtual Force δU ∗ General Linear L L P δσεdx δP dx AE 0 0 δσ

0

0

Table 1.1: Summary of Variational Terms Associated with One Dimensional Elements   1 1 du   −L L =  ε = εx = dx ∂N1 ∂N2 ∂x

∂x

[B]

1.2.1.2

    u1  u2 {∆}

(1.3-b)

Flexural Members

Using the shape functions for ﬂexural elements previously derived in Eq. 6.41 we have: d2 v y =y 2 ρ dx M = EI d2 v = y 2 dx   6 2 2 6   L2 (2ξ − 1) − L (3ξ − 2) L2 (−2ξ + 1) − L (3ξ − 1) = y

ε =

(1.4-a)

1 ρ

(1.4-b)

∂ 2 N1 ∂x2

∂ 2 N2 ∂x2

[B]

Victor Saouma

∂ 2 N3 ∂x2

∂ 2 N4 ∂x2

        

(1.4-c)  v1    θ1 (1.4-d) v2    θ2 {∆}

Finite Elements II; Solid Mechanics

ε

Draft

1.2 Finite Element Formulation

1.2.2

1–5

Virtual Displacement and Strains δ∆ = [N]{δ∆}

(1.5-a)

δε = [B]{δ∆}

(1.5-b) (1.5-c)

1.2.3

Element Stiﬀness Matrix Formulation

Let us consider the most general case, or element with: Initial strain: (temperature eﬀect, support settlement, or other) such that: εx =

σx E

+

due to load

εix initial strain

(1.6)

thus: σx = Eεx − Eεix

(1.7)

{σ} = [D]{ε} − [D]{εi }

(1.8)

or in matrix form: where [D] is the constitutive matrix which relates stress and strain. Load: q(x) along it. Let us apply the principle of virtual work. = δW δε{σ}dvol δU = vol {σ} = [D]{ε} − [D]{εi }

(1.9-b)

{δε} = [B]{δ∆}

(1.9-d)

{ε} = [B]{∆}

(1.9-e)

δU

δε = δ∆[B]

T

(1.9-a)

(1.9-c)

(1.9-f) (1.9-g)

Combining Eqns. 1.9-a, 1.9-b, 1.9-c, 1.9-f, and 1.9-e, the internal virtual strain energy is given by: T δ∆[B]T [D]{εi }dvol δ∆[B] [D][B]{∆} dvol − δU = vol vol [B]T [D][B] dvol{∆} − δ∆ [B]T [D]{εi }dvol (1.10-a) = δ∆ vol vol the virtual external work in turn is given by: δW = Victor Saouma

{F} + δ∆q(x)dx δ∆ (1.11) l Virt. Nodal Displ. Nodal Force Finite Elements II; Solid Mechanics

Draft 1–6

PREREQUISITE

combining this equation with: yields:

{δ∆} = [N]{δ∆}

(1.12)

l δW = δ∆{F} + δ∆ [N]T q(x) dx

(1.13)

0

Equating the internal strain energy Eqn. 1.10-a with the external work Eqn. 1.13, we obtain: T [B] [D][B] dvol{∆} − δ∆ [B]T [D]{εi }dvol = δ∆ vol vol [K] init {F } l [N]T q(x) dx (1.14-a) δ∆{F} + δ∆ 0 {Fe }

where: The element stiﬀness matrix: [K] =

[B]T [D][B]dvol

(1.15)

[B]T [D]{εi }dvol

(1.16)

vol Element initial force vector: {Fi }

= vol

Element equivalent load vector: {Fe }

l

=

[N] q(x) dx

(1.17)

{σ} = [D] · [N]{∆}

(1.18)

0

1.2.3.1

Stress Recovery

Recall that we have:

1.3 1.3.1

{σ} = [D]{ε} {ε} = [B]{∆}

Direct Stiﬀness Method Global Stiﬀness Matrix

4 The physical interpretation of the global stiﬀness matrix K is analogous to the one of the element, i.e. If all degrees of freedom are restrained, then Kij corresponds to the force along global degree of freedom i due to a unit positive displacement (or rotation) along global degree of freedom j. 5

For instance, with reference to Fig. 1.3, we have three global degrees of freedom, ∆1 , ∆2 , and

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

1.3 Direct Stiﬀness Method

1–7

P

1A 0 0 1

M

11 00 00 11

P/2

B

EI w

∆2 1 0

A B

w C 111 000 000 111

1 K211 0

B

H

K22

B

P

1

C 00 L/2 0000 L/211 0000 1111 1111 00 11

A 1 0 0 1

θ3

1 0 11A 00 0 1 00 111 ∆ 11 000 0 1 00 11

K31

1 0 0 1 1111 0000 0 K11 1

A 11 00 00 11

C 11 00 00 11

C

11 00 K23 0 1

K32

0 1 0 111K121 000 0 1 01 1 B 0 1 0 1 0 1

A 1 0 0 1

111 000 C 000 111

B

K 33

1 0 0 1 1111 0000 0 K13 1 1

C 111 000 000 111

Figure 1.3: Frame Example θ3 . and the global (restrained or structure’s) stiﬀness matrix is   K11 K12 K13 K =  K21 K22 K23  K31 K32 K33

(1.19)

and the ﬁrst column corresponds to all the internal forces in the unrestrained d.o.f. when a unit displacement along global d.o.f. 1 is applied. 1.3.1.1

Structural Stiﬀness Matrix

6 The structural stiﬀness matrix is assembled only for those active degrees of freedom which are active (i.e unrestrained). It is the one which will be inverted (or rather decomposed) to determine the nodal displacements.

1.3.1.2

Augmented Stiﬀness Matrix

7 The augmented stiﬀness matrix is expressed in terms of all the dof. However, it is partitioned into two groups with respective subscript ‘u’ where the displacements are known (zero otherwise), and t where the loads are known.

√ Ktt Ktu Pt ∆t ? √ = Ru ? Kut Kuu ∆u

(1.20)

We note that Ktt corresponds to the structural stiﬀness matrix. Victor Saouma

Finite Elements II; Solid Mechanics

Draft 1–8 8

PREREQUISITE

The ﬁrst equation enables the calculation of the unknown displacements. ∆t = K−1 tt (Pt − Ktu ∆u )

9

(1.21)

The second equation enables the calculation of the reactions Ru = Kut ∆t + Kuu ∆u

(1.22)

10 For internal book-keeping purpose, since we are assembling the augmented stiﬀness matrix, we proceed in two stages:

1. First number all the global unrestrained degrees of freedom 2. Then number separately all the global restrained degrees of freedom (i.e. those with known displacements, zero or otherwise) starting with -1 this will enable us later on to distinguish the restrained from unrestrained dof. The element internal forces (axial and shear forces, and moment at each end of the member) are determined from 11

(e)

pint = k(e) δ (e)

(1.23)

(e)

at the element level where pint is the six by six array of internal forces, k(e) the element stiﬀness matrix in local coordinate systems, and δ (e) is the vector of nodal displacements in local coordinate system. Note that this last array is obtained by ﬁrst identifying the displacements in global coordinate system, and then premultiplying it by the transformation matrix to obtain the displacements in local coordinate system.

1.3.2 1.3.2.1

Logistics Boundary Conditions, [ID] Matrix

12 Because of the boundary condition restraints, the total structure number of active degrees of freedom (i.e unconstrained) will be less than the number of nodes times the number of degrees of freedom per node. 13 To obtain the global degree of freedom for a given node, we need to deﬁne an [ID] matrix such that:

ID has dimensions l × k where l is the number of degree of freedom per node, and k is the number of nodes). ID matrix is initialized to zero. 1. At input stage read ID(idof,inod) of each degree of freedom for every node such that: 0 if unrestrained d.o.f. ID(idof, inod) = (1.24) 1 if restrained d.o.f. Victor Saouma

Finite Elements II; Solid Mechanics

Draft

1.3 Direct Stiﬀness Method

1–9

2. After all the node boundary conditions have been read, assign incrementally equation numbers (a) First to all the active dof (b) Then to the other (restrained) dof, starting with -1. Note that the total number of dof will be equal to the number of nodes times the number of dof/node NEQA. 3. The largest positive global degree of freedom number will be equal to NEQ (Number Of Equations), which is the size of the square matrix which will have to be decomposed. 14

For example, for the frame shown in Fig. 1.4:

Figure 1.4: Example for [ID] Matrix Determination 1. The input data ﬁle may contain: Node No. 1 2 3 4

[ID]T 000 110 000 100

2. At this stage, the [ID] matrix is equal to: 

 0 1 0 1 ID =  0 1 0 0  0 0 0 0

3. After we determined the equation numbers, we  1 −1 ID =  2 −2 3 4

Victor Saouma

would have:  5 −3 6 8  7 9

(1.25)

(1.26)

Finite Elements II; Solid Mechanics

Draft 1–10

1.3.2.2

PREREQUISITE

LM Vector

15 The LM vector of a given element gives the global degree of freedom of each one of the element degree of freedom’s. For the structure shown in Fig. 1.4, we would have:

LM = −1 −2 4 5 6 7 element 1 (2 → 3) LM = 5 6 7 1 2 3 element 2 (3 → 1) LM = 1 2 3 −3 8 9 element 3 (1 → 4) 1.3.2.3

Assembly of Global Stiﬀness Matrix

16 As for the element stiﬀness matrix, the global stiﬀness matrix [K] is such that Kij is the force in degree of freedom i caused by a unit displacement at degree of freedom j.

Whereas this relationship was derived from basic analysis at the element level, at the structure level, this term can be obtained from the contribution of the element stiﬀness matrices [K(e) ] (written in global coordinate system). 17

For each Kij term, we shall add the contribution of all the elements which can connect degree of freedom i to degree of freedom j, assuming that those forces are readily available from the individual element stiﬀness matrices written in global coordinate system. 18

19

Kij is non-zero if degree of freedom i and degree of freedom j 1. Are connected by an element. 2. Share a node. 3. Are connected by an element and the corresponding value in the element stiﬀness matrix in the global coordinate system is zero.

There are usually more than one element connected to a dof. Hence, individual element stiﬀness matrices terms must be added up. 20

21 Because each term of all the element stiﬀness matrices must ﬁnd its position inside the global stiﬀness matrix [K], it is found computationally most eﬀective to initialize the global stiﬀness matrix [KS ](N EQA×N EQA ) to zero, and then loop through all the elements, and then through

(e)

each entry of the respective element stiﬀness matrix Kij . (e)

The assignment of the element stiﬀness matrix term Kij (note that e, i, and j are all known since we are looping on e from 1 to the number of elements, and then looping on the rows and S is made through columns of the element stiﬀness matrix i, j) into the global stiﬀness matrix Kkl the LM vector (note that it is k and l which must be determined). 22

Since the global stiﬀness matrix is also symmetric, we would need to only assemble one side of it, usually the upper one. 23

24

Contrarily to the previous method, we will assemble the full augmented stiﬀness matrix. Example 1-1: Assembly of the Global Stiﬀness Matrix

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

1.3 Direct Stiﬀness Method

1–11 50kN

4 kN/m 2

8m

00 11 11 00 00 11 00 11

2

3

3m 1 000 111 111 000 000 111 000 111 0001 111

7.416 m

8m

Figure 1.5: Simple Frame Analyzed with the MATLAB Code As an example, let us consider the frame shown in Fig. 1.5. The ID matrix is initially set to:   1 0 1 [ID] =  1 0 1  1 0 1 We then modify it to generate the global degrees  −4 [ID] =  −5 −6

(1.27)

of freedom of each node:  1 −7 2 −8  3 −9

(1.28)

Finally the LM vectors for the two elements (assuming that Element 1 is deﬁned from node 1 to node 2, and element 2 from node 2 to node 3): 2 3 −4 −5 −6 1 (1.29) [LM ] = 1 2 3 −7 −8 −9 Let us simplify the operation by designating the element stiﬀness matrices in global coordinates as follows:

K (1) =

−4  −4 A11 −5  A21 −6  A31 1   A41 2  A51 3 A61

−5 A12 A22 A32 A42 A52 A62

−6 A13 A23 A33 A43 A53 A63

1 A14 A24 A34 A44 A54 A64

2 A15 A25 A35 A45 A55 A65

3  A16 A26   A36   A46   A56  A66

1 1 B11 2   B21 3   B31 −7  B41 −8 B51 −9 B61

2 B12 B22 B32 B42 B52 B62

3 B13 B23 B33 B43 B53 B63

−7 B14 B24 B34 B44 B54 B64

−8 B15 B25 B35 B45 B55 B65

−9  B16 B26   B36   B46   B56 



K (2) =

(1.30-a)

(1.30-b)

B66

We note that for each element we have shown the corresponding LM vector. Victor Saouma

Finite Elements II; Solid Mechanics

Draft 1–12

PREREQUISITE

Now, we assemble the global stiﬀness matrix        K=      

A44 + B11 A45 + B12 A46 + B13 A54 + B21 A55 + B22 A56 + B23 A64 + B31 A65 + B32 A66 + B33 A14 A15 A16 A25 A26 A24 A35 A36 A34 B42 B43 B41 B51 B52 B53 B61 B62 B63

A41 A51 A61 A11 A21 A31 0 0 0

A42 A52 A62 A12 A22 A32 0 0 0

A43 A53 A63 A13 A23 A33 0 0 0

B14 B24 B34 0 0 0 B44 B54 B64

B15 B25 B35 0 0 0 B45 B55 B65

B16 B26 B36 0 0 0 B46 B56 B66

             

(1.31)

We note that some terms are equal to zero because we do not have a connection between the corresponding degrees of freedom (i.e. node 1 is not connected to node 3). 1.3.2.4 25

Algorithm

The direct stiﬀness method can be summarized as follows:

Preliminaries: First we shall 1. Identify type of structure (beam, truss, grid or frame) and determine the (a) Number of spatial coordinates (1D, 2D, or 3D) (b) Number of degree of freedom per node (local and global) (c) Number of cross-sectional and material properties 2. Determine the global unrestrained and restrained degree of freedom equation numbers for each node, Update the [ID] matrix (which included only 0’s and 1’s in the input data ﬁle). Analysis : 1. For each element, determine (a) (b) (c) (d) (e)

Vector LM relating local to global degree of freedoms. Element stiﬀness matrix [k(e) ] Angle α between the local and global x axes. Rotation matrix [Γ(e) ] Element stiﬀness matrix in global coordinates [K(e) ] = [Γ(e) ]T [k(e) ][Γ(e) ]

2. Assemble the augmented stiﬀness matrix [K(S) ] of unconstrained and constrained degree of freedom’s. 3. Extract [Ktt ] from [K(S) ] and invert (or decompose into into [Ktt ] = [L][L]T where [L] is a lower triangle matrix. 4. Assemble load vector {P} in terms of nodal load and ﬁxed end actions. 5. Backsubstitute and obtain nodal displacements in global coordinate system. 6. Solve for the reactions. Victor Saouma

Finite Elements II; Solid Mechanics

Draft

1.3 Direct Stiﬀness Method

1–13

7. For each element, transform its nodal displacement from global to local coordinates {δ} = [Γ(e) ]{∆}, and determine the internal forces [p] = [k]{δ}. 26

Some of the prescribed steps are further discussed in the next sections. Example 1-2: Direct Stiﬀness Analysis of a Truss Using the direct stiﬀness method, analyze the truss shown in Fig. 1.6. 4

4

5

5 1

1

6 8

3

2

50k

7

2

12’

3

100k

16’

16’

Figure 1.6: Solution: 1. Determine the structure ID matrix Node # 1 2 3 4 5 ID =

=

Bound. X 0 0 1 0 0

Cond. Y 1 0 1 0 0

0 0 1 0 0 1 0 1 0 0

N ode 1 2 3 4 1 2 −2 4 −1 3 −3 5

(1.32-a) 5 6 7

(1.32-b)

2. The LM vector of each element is evaluated next

Victor Saouma

LM 1 = 1 −1 4 5

(1.33-a)

LM 2 = 1 −1 2 3

(1.33-b)

LM 3 = 2 3 4 5

(1.33-c)

Finite Elements II; Solid Mechanics

Draft 1–14

PREREQUISITE LM 4 = 4 5 6 7

(1.33-d)

LM 5 = −2 −3 4 5

(1.33-e)

LM 6 = 2 3 6 7

(1.33-f)

LM 7 = 2 3 −2 −3

(1.33-g)

LM 8 = −2 −3 6 7

(1.33-h)

3. Determine the element stiﬀness matrix of each element in the global coordinate system noting that for a 2D truss element we have [K (e) ] = [Γ(e) ]T [k(e) ][Γ(e) ]  2  c cs −c2 −cs EA  s2 −cs −s2   cs  = 2  −c −cs c2 cs  L s2 −cs −s2 cs x2 −x1 L ;

where c = cos α =

Element 1 L = 20 , c =

12−0 20

= 0.8, s =

Element 2 L = 16 , c = 1 , s = 0 ,

EA L

1 1 18, 750 −1 0   2 −18, 750 3 0

Element 3 L = 12 , c = 0 , s = 1 ,

[K3 ] =

2  2 0 3 0 4 0 5 0

Element 4 L = 16 , c = 1 , s = 0 ,

[K4 ] =

Victor Saouma

=

(30,000

4 −9, 600 −7, 200 9, 600 7, 200

ksi)(10 in2 ) 20

5  −7, 200 −5, 400   7, 200  5, 400

= 15, 000 k/ft.

(1.35)

= 18, 750 k/ft.



[K2 ] =

EA L

= 0.6,

−1 7200 5, 400 −7, 200 −5, 400

1 1 9, 600 −1 7,  200 4  −9, 600 5 −7, 200 

[K1 ] =

(1.34-b)

Y2 −Y1 L

s = sin α =

16−0 20

(1.34-a)

EA L

−1 2 0 −18, 750 0 0 0 18, 750 0 0

(1.36)

= 25, 000 k/ft.

3 0 25, 000 0 −25, 000 EA L

3  0 0  0 0

4 5  0 0 0 −25, 000    0 0 0 25, 000

(1.37)

= 18, 750 k/ft.

4  4 18, 750 5 0   6 −18, 750 7 0

5 6 0 −18, 750 0 0 0 18, 750 0 0

7  0 0  0 0

(1.38)

Finite Elements II; Solid Mechanics

Draft

1.3 Direct Stiﬀness Method

Element 5 L = 20 , c =

1–15

−16−0 20

[K5 ] =

= − 0.8 , s = 0.6 ,

−2  −2 9, 600 −3  −7, 200 4  −9, 600 5 7, 200

Element 6 L = 20 , c = 0.8 , s = 0.6 ,

[K6 ] =

2  2 9, 600 3  7, 200 6 −9, 600 7 −7, 200

Element 7 L = 16 , c = 1 , s = 0 ,

[K7 ] =

EA L

 −2 −3  6  7

EA L

= 15, 000 k/ft.

4 −9, 600 7, 200 9, 600 −7, 200

5  7, 200 −5, 400   −7, 200  5, 400

(1.39)

= 15, 000 k/ft.

3 7, 200 5, 400 −7, 200 −5, 400

6 −9, 600 −7, 200 9, 600 7, 200

7  −7, 200 −5, 400   7, 200  5, 400

(1.40)

= 18, 750 k/ft.

2  2 18, 750 3  0  −2 −18, 750 −3 0

Element 8 L = 12 , c = 0 , s = 1 ,

[K8 ] =

EA L

−3 −7, 200 5, 400 7, 200 −5, 400

EA L

3 −2 0 −18, 750 0 0 0 18, 750 0 0

−3  0 0   0  0

(1.41)

= 25, 000 k/ft.

−2 −3 0 0 0 25, 000 0 0 0 −25, 000

6 7  0 0 0 −25, 000    0 0 0 25, 000

(1.42)

4. Assemble the global stiﬀness matrix in k/ft Note that we are not assembling the augmented stiﬀness matrix, but rather its submatrix [Ktt ]. 8 > > > > > > > < > > > > > > > :

0 0 −100k 0 0 50k 0

9 2 > > > 6 > > > > 6 = 6 6 =6 6 > > 6 > > > 4 > > ;

9, 600 + 18, 750

−18, 750 9, 600 + (2) 18, 750

0 7, 200 5, 400 + 25, 000

−9, 600 0 0 18, 750 + (2)9, 600

SYMMETRIC

−7, 200 0 −25, 000 7, 200 − 7, 200 25, 000 + 5, 400(2)

0 −9, 600 −7, 200 −18, 750 0 18, 750 + 9, 600

0 −7, 200 −5, 400 0 0 7, 200 25, 000 + 5, 400 (1.43)

5. Convert to k/in and simplify    2, 362.5 −1, 562.5 0 −800 −600 0 0 0          3, 925.0 600 0 0 −800 −600 0         2, 533.33 0 −2, 083.33 −600 −450  −100     = 3, 162.5 0 −1, 562.5 0 0     SYMMETRIC   2, 983.33 0 0  0           2, 362.5 600 50      2, 533.33 0 Pt

Ktt

(1.44) Victor Saouma

Finite Elements II; Solid Mechanics

                   

38 > > > 7 > > 7 > > 7 < 7 7 7 > > 7 > > 5> > > :

U1 U2 V3 U4 V5 U6 V7 ut

u1 u2 v3 u4 v5 u6 v7

                  

9 > > > > > > > = > > > > > > > ;

Draft 1–16

PREREQUISITE

6. Invert stiﬀness matrix and solve for displacements    −0.0223 in U1             U 0.00433 in   2            V3   −0.116 in = U4 −0.0102 in       V −0.0856 in    5            −0.00919 in    U6     V7 −0.0174 in

                  

(1.45)

7. Solve for member internal forces (in this case axial forces) in local coordinate systems   U   1     V1 c s −c −s u1 = (1.46) u2 U  −c −s c s    2   V2 Element 1

p1 p2

1

  −0.0223       1 ft 0.8 0.6 −0.8 −0.6 0 = (15, 000 k/ft)( ) (1.47-a) −0.8 −0.6 0.8 0.6  −0.0102  12 in     −0.0856 52.1 k Compression (1.47-b) = −52.1 k

Element 2

p1 p2

2 = 18, 750 k/ft( =

−43.2 k 43.2 k

1 ft ) 12 in

  −0.0233       1 0 −1 0 0 −1 0 1 0  0.00433      −0.116

Tension

(1.48-a)

(1.48-b)

Element 3

p1 p2

3 = 25, 000 k/ft( =

Victor Saouma

−63.3 k 63.3 k

1 ft ) 12 in

  0.00433       0 1 0 −1 −0.116 0 −1 0 1   −0.0102     −0.0856

Tension

(1.49-a)

(1.49-b)

Finite Elements II; Solid Mechanics

Draft

1.3 Direct Stiﬀness Method

1–17

Element 4

  −0.0102       1 ft 1 0 −1 0 −0.0856 = 18, 750 k/ft( ) (1.50-a) −1 0 1 0  −0.00919  12 in     −0.0174 −1.58 k Tension (1.50-b) = 1.58 k

4

p1 p2

Element 5

p1 p2

  −0.0102       1 ft −0.8 0.6 0.8 −0.6 −0.0856 = 15, 000 k/ft( ) (1.51-a) 0.8 −0.6 −0.8 0.6  0  12 in     0 54.0 k Compression (1.51-b) = −54.0 k

5

Element 6

p1 p2

6 = 15, 000 k/ft( =

−60.43 k 60.43 k

1 ft ) 12 in

  0.00433       0.8 0.6 −0.8 −0.6 −0.116 (1.52-a) −0.8 −0.6 0.8 0.6  −0.00919      −0.0174

Tension

(1.52-b)

Element 7

p1 p2

7 = 18, 750 k/ft( =

6.72 k −6.72 k

1 ft ) 12 in

  0.00433       1 0 −1 0 −0.116 −1 0 1 0  0      0

Compression

(1.53-a)

(1.53-b)

Element 8

p1 p2

8 = 25, 000 k/ft( =

Victor Saouma

36.3 k −36.3 k

1 ft ) 12 in

0 1 0 −1 0 −1 0 1

Compression

   

0 0 −0.00919    −0.0174

      

(1.54-a)

(1.54-b)

Finite Elements II; Solid Mechanics

Draft 1–18

PREREQUISITE

8. Determine the structure’s MAXA vector    1 3 9 14       2 5 8 13 19 25         4 7 12 18 24     MAXA = [K] =  6 11 17 23      10 16 22           15 21   20

1 2 4 6 10 15 20

                  

(1.55)

Thus, 25 terms would have to be stored.

Example 1-3: Analysis of a Frame with MATLAB The simple frame shown in Fig. 1.7 is to be analyzed by the direct stiﬀness method. Assume: E = 200, 000 MPa, A = 6, 000 mm2 , and I = 200 × 106 mm4 . The complete MATLAB solution is shown below along with the results. 50kN

4 kN/m 2

8m

2

0 1 1 0 0 1 0 1 0 1

3

3m 1 111 000 000 111 000 111 0001 111 000 111

7.416 m

8m

Figure 1.7: Simple Frame Analyzed with the MATLAB Code

% zero the matrices k=zeros(6,6,2); K=zeros(6,6,2); Gamma=zeros(6,6,2); % Structural properties units: mm^2, mm^4, and MPa(10^6 N/m) A=6000;II=200*10^6;EE=200000; % Convert units to meter and kN A=A/10^6;II=II/10^12;EE=EE*1000; % Element 1 i=[0,0];j=[7.416,3]; [k(:,:,1),K(:,:,1),Gamma(:,:,1)]=stiff(EE,II,A,i,j); % Element 2 i=j;j=[15.416,3]; [k(:,:,2),K(:,:,2),Gamma(:,:,2)]=stiff(EE,II,A,i,j); % Define ID matrix ID=[ -4 1 -7; -5 2 -8; -6 3 -9]; % Determine the LM matrix LM=[ Victor Saouma

Finite Elements II; Solid Mechanics

Draft

1.3 Direct Stiﬀness Method

1–19

-4 -5 -6 1 2 3; 1 2 3 -7 -8 -9]; % Assemble augmented stiffness matrix Kaug=zeros(9); for elem=1:2 for r=1:6 lr=abs(LM(elem,r)); for c=1:6 lc=abs(LM(elem,c)); Kaug(lr,lc)=Kaug(lr,lc)+K(r,c,elem); end end end % Extract the structures Stiffness Matrix Ktt=Kaug(1:3,1:3); % Determine the fixed end actions in local coordinate system fea(1:6,1)=0; fea(1:6,2)=[0,8*4/2,4*8^2/12,0,8*4/2,-4*8^2/12]’; % Determine the fixed end actions in global coordinate system FEA(1:6,1)=Gamma(:,:,1)’*fea(1:6,1); FEA(1:6,2)=Gamma(:,:,2)’*fea(1:6,2); % FEA_Rest for all the restrained nodes FEA_Rest=[0,0,0,FEA(4:6,2)’]; % Assemble the load vector for the unrestrained node P(1)=50*3/8;P(2)=-50*7.416/8-FEA(2,2);P(3)=-FEA(3,2); % Solve for the Displacements in meters and radians Displacements=inv(Ktt)*P’ % Extract Kut Kut=Kaug(4:9,1:3); % Compute the Reactions and do not forget to add fixed end actions Reactions=Kut*Displacements+FEA_Rest’ % Solve for the internal forces and do not forget to include the fixed end actions dis_global(:,:,1)=[0,0,0,Displacements(1:3)’]; dis_global(:,:,2)=[Displacements(1:3)’,0,0,0]; for elem=1:2 dis_local=Gamma(:,:,elem)*dis_global(:,:,elem)’; int_forces=k(:,:,elem)*dis_local+fea(1:6,elem) end function [k,K,Gamma]=stiff(EE,II,A,i,j) % Determine the length L=sqrt((j(2)-i(2))^2+(j(1)-i(1))^2); % Compute the angle theta (careful with vertical members!) if(j(1)-i(1))~=0 alpha=atan((j(2)-i(2))/(j(1)-i(1))); else alpha=-pi/2; end % form rotation matrix Gamma Gamma=[ cos(alpha) sin(alpha) 0 0 0 0; -sin(alpha) cos(alpha) 0 0 0 0; 0 0 1 0 0; 0 0 0 cos(alpha) sin(alpha) 0; 0 0 Victor Saouma

0

Finite Elements II; Solid Mechanics

Draft 1–20

PREREQUISITE

0 -sin(alpha) cos(alpha) 0; 0 0 0 0 1]; % form element stiffness matrix in local coordinate system EI=EE*II; EA=EE*A; k=[EA/L, 0, 0, -EA/L, 0, 12*EI/L^3, 6*EI/L^2, 0, -12*EI/L^3, 6*EI/L^2; 0, 6*EI/L^2, 4*EI/L, 0, -6*EI/L^2, 2*EI/L; -EA/L, 0, 0, EA/L, 0, 0; 0, -12*EI/L^3, -6*EI/L^2, 0, 12*EI/L^3, -6*EI/L^2; 0, 6*EI/L^2, 2*EI/L, 0, -6*EI/L^2, 4*EI/L]; % Element stiffness matrix in global coordinate system K=Gamma’*k*Gamma;

0

0,

0;

This simple proigram will produce the following results: Displacements = 0.0010 -0.0050 -0.0005 Reactions = 130.4973 55.6766 13.3742 -149.2473 22.6734 -45.3557

int_forces =

int_forces =

141.8530 2.6758 13.3742 -141.8530 -2.6758 8.0315

149.2473 9.3266 -8.0315 -149.2473 22.6734 -45.3557

We note that the internal forces are consistent with the reactions (specially for the second node of element 2), and amongst themselves, i.e. the moment at node 2 is the same for both elements (8.0315).

Example 1-4: Analysis of a simple Beam with Initial Displacements

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

1.3 Direct Stiﬀness Method

1–21

The full stiﬀness matrix of a beam element is given by v1 V1 12EI/L3  M1 6EI/L2 V2  −12EI/L3 M2 6EI/L2

θ1 6EI/L2 4EI/L −6EI/L2 2EI/L



[ke ] =

v2 −12EI/L3 −6EI/L2 12EI/L3 −6EI/L2

θ2  6EI/L2 2EI/L   −6EI/L2  4EI/L

(1.56)

This matrix is singular, it has a rank 2 and order 4 (as it embodies also 2 rigid body motions). 27

We shall consider 3 diﬀerent cases, Fig. 1.8 -3

1 0 0 1 0 1 0 1 -2

P

-4

1 1 0 0 1 0 1

-3

2

1 0 0 1 0 1

1 0 0 1 0 1

1

-4

2

M

-3

1 0 0 1 0 1 0 1

-4

1 0 0 1 0 1

1 ∆

Figure 1.8: Stiﬀness Analysis of one Element Structure Cantilivered Beam/Point Load 1. The element stiﬀness matrix is −4 6EI/L2 4EI/L −6EI/L2 2EI/L

1 −12EI/L3 −6EI/L2 12EI/L3 −6EI/L2

2 3 6EI/L2 2EI/L 7 7 −6EI/L2 5 4EI/L

2. The structure stiﬀness matrix is assembled 1 2 2 1 12EI/L2 −6EI/L2 2 2 6 4EI/L 6 −6EI/L K= −34 −12EI/L3 6EI/L2 −4 −6EI/L2 2EI/L

−3 −12EI/L3 6EI/L2 12EI/L3 6EI/L2

−4 3 −6EI/L2 2EI/L 7 7 6EI/L2 5 4EI/L

k=

−3 2 −3 12EI/L3 2 −46 6 6EI/L 1 4 −12EI/L3 2 6EI/L2

3. The global matrix can be rewritten as 8 √ 9 −P > > > < 0√ > =

2

12EI/L2 6 −6EI/L2 =6 4 −12EI/L3 R3 ? > > > > : ; −6EI/L2 R4 ?

−6EI/L2 4EI/L 6EI/L2 2EI/L

−12EI/L3 6EI/L2 12EI/L3 6EI/L2

38

−6EI/L2 > ∆1 ? > < θ ? 2EI/L 7 2 7 √ 6EI/L2 5 > ∆ > : 3√ 4EI/L θ4

9 > > = > > ;

4. Ktt is inverted (or actually decomposed) and stored in the same global matrix 2

L3 /3EI

6 6 6 L2 /2EI 6 4 −12EI/L3 −6EI/L2

Victor Saouma

L2 /2EI

−12EI/L3

−6EI/L2

L/EI

6EI/L2

2EI/L

6EI/L2 2EI/L

12EI/L3 6EI/L2

6EI/L2 4EI/L

3 7 7 7 7 5

Finite Elements II; Solid Mechanics

Draft 1–22

PREREQUISITE

5. Next we compute the equivalent load, P t = Pt − Ktu ∆u , and overwrite Pt by P t Pt − Ktu ∆u

8 9 > −P > > > > > < =

2

L3 /3EI

6 6 0 L2 /2EI −6 6 > > 3 0 > > > > : ; 4 −12EI/L2 0 −6EI/L 8 9 > > −P > > > > < =

=

=

0 0 0

> > > :

L2 /2EI

−12EI/L3

−6EI/L2

L/EI

6EI/L2

2EI/L

2

6EI/L 2EI/L

3

6EI/L2 4EI/L

12EI/L 6EI/L2

3

9 8 −P > > 7> > < 7 0 = 7 7> 0 > > 5> ; : 0

> > > ;

6. Now we solve for the displacement ∆t = K−1 tt Pt , and overwrite Pt by ∆t

8 > ∆1 > > < θ2 0 0

> > > :

2

9 > > > =

6 6 6 L2 /2EI 6 4 −12EI/L3

=

> > > ;

L3 /3EI

2 8 −6EI/L > −P L3 /3EI > > > < 2

=

> > > > :

−P L /2EI 0 0

L2 /2EI

−12EI/L3

L/EI

6EI/L2

2EI/L

12EI/L3 6EI/L2

6EI/L2 4EI/L

2

6EI/L 2EI/L 9

−6EI/L2

38 9 > > −P > > 7> > 7< 0 = 7 7> 0 > > 5> > : 0 > ;

> > > > = > > > > ;

7. Finally, we solve for the reactions, Ru = Kut ∆tt + Kuu ∆u , and overwrite ∆u by Ru 8 −P L3 /3EI > > > < −P L2 /2EI R3 > > > : R 4

2

9 > > > = > > > ;

L3 /3EI L2 /2EI

6 6 6 6 −12EI/L3 4

=

−6EI/L

2

8 −P L3 /3EI > > > 2 <

L2 /2EI L/EI

−12EI/L3 6EI/L2

6EI/L2

12EI/L3 2

2EI/L

6EI/L

1 6EI/L2 4EI/L −6EI/L2 2EI/L

−4 −12EI/L3 −6EI/L2 12EI/L3 −6EI/L2

9 > > > =

38

−6EI/L2 > > −P L3 /3EI > > > 2EI/L 7 < 7 −P L2 /2EI 7 6EI/L2 7 > 5> > 0 > > : 4EI/L 0

−P L /2EI P > > > > > > : ; PL

=

Simply Supported Beam/End Moment 1. The element stiﬀness matrix is −3 2 −3 12EI/L3 2 1 6 6 6EI/L −44 −12EI/L3 2 6EI/L2

k=

2 3 6EI/L2 2EI/L 7 7 −6EI/L2 5 4EI/L

2. The structure stiﬀness matrix is assembled 1 2 1 4EI/L 2 6 6 2EI/L −34 6EI/L2 −4 −6EI/L2

K=

2 2EI/L 4EI/L 6EI/L2 −6EI/L2

−3 6EI/L2 6EI/L2 12EI/L3 −12EI/L3

−4 3 −6EI/L2 2 7 −6EI/L 7 −12EI/L3 5 12EI/L3

3. The global stiﬀness matrix can be rewritten as 8 √ 9 0 > > > √ > > > > > < M = > > > > :

Victor Saouma

2

4EI/L 6 2EI/L 6 = > 4 6EI/L2 R3 ? > > > ; −6EI/L2 R4 ?

2EI/L 4EI/L 6EI/L2 −6EI/L2

6EI/L2 6EI/L2 12EI/L3 −12EI/L3

38

−6EI/L2 θ1 ? > > < θ ? −6EI/L2 7 2 7 √ −12EI/L3 5 > ∆3 > √ : 3 12EI/L ∆4

9 > > = > > ;

Finite Elements II; Solid Mechanics

9 > > > > > = > > > > > ;

Draft

1.3 Direct Stiﬀness Method 4. Ktt is inverted

2

1–23

6 6 −L/6EI 6 6 4 6EI/L2 −6EI/L

2

3

−L/6EI

6EI/L2

−6EI/L2

L/3EI

6EI/L2

6EI/L2 −6EI/L2

12EI/L3 −12EI/L3

−6EI/L2 7 7 −12EI/L3 5 12EI/L3

L3 /3EI

7 7

5. We compute the equivalent load, P t = Pt − Ktu ∆u , and overwrite Pt by P t Pt − Ktu ∆u

8 > > <

0 M > > : 0 0

=

8 > > <

0 M > > : 0 0

=

9 > > =

2

6 6 −L/6EI −6 6 > > 4 ; 6EI/L2

−6EI/L2

9 > > =

3

−L/6EI

6EI/L2

−6EI/L2

L/3EI

6EI/L2

−6EI/L2

6EI/L2 −6EI/L2

12EI/L3 −12EI/L3

−12EI/L3 12EI/L3

L3 /3EI

9 8 0 > > 7> > < 7 M = 7 7> 0 > > 5> ; : 0

> > ;

6. Solve for the displacements, ∆t = K−1 tt Pt , and overwrite Pt by ∆t

8 > θ1 > > < > > > :

2

9 > > > =

θ2 > 0 > > ; 0

=

L3 /3EI

6 6 −L/6EI 6 6 4 6EI/L2 8 −6EI/L > −M L/6EI > > > < 2

=

> > > > :

−L/6EI

6EI/L2

L/3EI

6EI/L2

2

6EI/L 2 −6EI/L 9

M L/3EI 0 0

> > > > =

12EI/L3 −12EI/L3

3

8 9 0 > 7> > > < = 7 M −6EI/L2 7 7 > 0 > > −12EI/L3 5 > : ; −6EI/L2

12EI/L3

0

> > > > ;

7. Solve for the reactions, Rt = Kut ∆tt + Kuu ∆u , and overwrite ∆u by Ru 8 −M L/6EI > > > < M L/3EI > > > :

R1 R2

2

9 > > > = > > > ;

=

=

6 6 6 6 4

L3 /3EI −L/6EI

−L/6EI L/3EI

6EI/L2 6EI/L2

−6EI/L2 −6EI/L2

6EI/L2

6EI/L2

12EI/L3

−12EI/L3

−6EI/L2

−6EI/L2

−12EI/L3

12EI/L3

8 −M L/6EI > > > > < M L/3EI

9 > > > > =

> > > > :

> > > > ;

M/L −M/L

38 > > −M L/6EI 7> > 7 < M L/3EI 7 7> 5> 0 > > : 0

Cantilivered Beam/Initial Displacement and Concentrated Moment 1. The element stiﬀness matrix is

k=

−2 2 −2 12EI/L3 2 −36 6 6EI/L 3 4 −4 −12EI/L 1 6EI/L2

−3 6EI/L2 4EI/L −6EI/L2 2EI/L

−4 −12EI/L3 −6EI/L2 12EI/L3 −6EI/L2

1 3 6EI/L2 2EI/L 7 7 −6EI/L2 5 4EI/L

−3 2EI/L 6EI/L2 4EI/L −6EI/L2

−4 3 −6EI/L2 −12EI/L3 7 7 −6EI/L2 5 3 12EI/L

2. The structure stiﬀness matrix is assembled

K=

Victor Saouma

1 2 1 4EI/L 2 −26 6 6EI/L 4 −3 2EI/L −4 −6EI/L2

−2 6EI/L2 12EI/L3 6EI/L2 −12EI/L3

Finite Elements II; Solid Mechanics

9 > > > > = > > > > ;

Draft 1–24

PREREQUISITE

3. The global matrix can be rewritten as 8 √ 9 M > > > > < =

2

6EI/L2 12EI/L3 6EI/L2 −12EI/L3

4EI/L 6 6EI/L2 R2 ? =6 R ? > 4 2EI/L > > : 3 > ; −6EI/L2 R4 ?

38

−6EI/L2 θ1 ? > > √ < −12EI/L3 7 ∆ 2 7 √ 2 5 −6EI/L θ > > : 3√ ∆4 12EI/L3

2EI/L 6EI/L2 4EI/L −6EI/L2

9 > > = > > ;

4. Ktt is inverted (or actually decomposed) and stored in the same global matrix 2

−6EI/L2

L/4EI

6EI/L2

2EI/L

−6EI/L2

12EI/L3 6EI/L2 −12EI/L3

6EI/L2 4EI/L −6EI/L2

6 6 6EI/L2 6 4 2EI/L

3 7

−12EI/L3 7 7 −6EI/L2 5 3 12EI/L

5. Next we compute the equivalent load, P t = Pt − Ktu ∆u , and overwrite Pt by P t 2

8 9 M > > > > < =

Pt − Ktu ∆u

6

2 6 − 6 6EI/L > 4 2EI/L > ; 2 −6EI/L 9

0 0 0

=

> > : 8 > M + 6EI∆0 /L2 > > > > > < =

=

0 0 0

> > > :

6EI/L2

2EI/L

−6EI/L2

12EI/L3 6EI/L2 −12EI/L3

6EI/L2 4EI/L −6EI/L2

−12EI/L3 −6EI/L2 12EI/L3

L/4EI

38 > > 7> < 7 7 5> > > :

2

0 > > > : 00 > ; ∆

=

=

L/4EI

6EI/L2

2EI/L

−6EI/L2

12EI/L3 6EI/L2 −12EI/L3

6EI/L2 4EI/L −6EI/L2

6 6 6EI/L2 6 4 2EI/L

∆0

9 > > > = > > > ;

> > > ;

6. Now we solve for the displacements, ∆t = K−1 tt Pt , and overwrite Pt by ∆t

8 9 θ1 > > > > < =

θ1 0 0

8 > M L/4EI + 3∆0 /2L > > <

9 > > > =

> > > :

> > > ;

0 0 ∆0

38 > > M + 6EI∆0 /L2 7> < 3 7 −12EI/L 7 0 −6EI/L2 5 > > 0 > : 3 0 −6EI/L2

12EI/L

∆

9 > > > = > > > ;

7. Finally, we solve for the reactions, Rt = Kut ∆tt + Kuu ∆u , and overwrite ∆u by Ru 8 M L/4EI + 3∆0 /2L > > > <

9 > > > =

> > > :

> > > ;

R2 R3 R4

2

=

=

Victor Saouma

6 6 6 6 6 4

3

L/4EI

6EI/L2

2EI/L

−6EI/L2

6EI/L2

12EI/L3

6EI/L2

−12EI/L3 7

2EI/L

6EI/L2

4EI/L

−6EI/L2

−6EI/L2

−12EI/L3

−6EI/L2

12EI/L3

8 > M L/4EI + 3∆0 /2L > > > <

9 > > > > =

> > > > : 8 > > > > > <

> > > > ;

0 0

0

∆ M L/4EI + 3∆0 /2L 3M/2L − 3EI∆0 /L3

> M/2 − 3EI∆0 /L2 > > > > : −3M/2L + 3EI∆0 /L3

7 7 7 7 5

9 > > > > > = > > > > > ;

Finite Elements II; Solid Mechanics

Draft Chapter 2

INTRODUCTION 2.1

Introduction

Whereas the ﬁrst course focused exclusively on one dimensional ”rod” elements, this course will greatly expand our horizons by considering introducing a methodology to solve partial diﬀerential equations, with special emphasis on solid mechanics. 1

2

The ﬁeld of mechanics, can itself be subdivided into four major disciplines:

Theoretical which deals with the fundamental laws and principles of mechanics. A Continuum Mechanics course is a must. Applied mechanics seeks to apply the theoretical knowledge to engineering applications. Elasticity or Fracture Mechanics solutions are such an example of applied mechanics. Computational mechanics combines mathematical models with numerical methods to solve problems on a digital computer. Experimental mechanics is conducted exclusively in a laboratory through physical measurements. 3 Any problem characterized by a PDE can be analyzed by the ﬁnite element method. The process of ﬁnite element analysis is illustrated by Fig. 2.12.

2.2

Elliptic, Parabolic and Hyperbolic Equations

4 Since the ﬁnite element method is a numerical scheme to solve (partial) diﬀerential equations, let us closely examine some of the major PDE which can be solved.

The general form of a partial diﬀerential equation is (note that we adopt the tensor notation where u,x = du ): dx 5

F (x, y, z, · · · , u, u,x , u,y , u,z , · · · , u,xx , u,yy , · · · , u,xy , u,xz , · · ·) = 0 and the order of the PDE is deﬁned by the order of the highest partial derivatives appearing in the equation. For instance α1 u,xx + α2 u,xy + α3 u,yy + α4 = 0

Draft 2–2

INTRODUCTION

Physical Problem

Change Physical Problem

Mathematical Model

FEM Solution of Mathematical Model

Governed by Differential Equations Assumptions on Geometry Kinematics Material Loading Boundary Conditions Etc.

Improve Mathematical Model

Finite Element Solution Choice of: Finite Elements Mesh Density Solution Paramters Representation of: Loading Boundary Conditions Etc.

Refine mesh, solution parameters

Assessment of accracy of FEM solution of mathematical model

Interpretation of Results

Refine Analysis Design Improvements Structural Optimization

Figure 2.1: Finite Element Process, (Bathe 1996)

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

2.2 Elliptic, Parabolic and Hyperbolic Equations

2–3

is quadratic. It would be linear if αi is a function of (x, y) only, otherwise it is nonlinear. Many signiﬁcant physical systems can be described by second order linear partial diﬀerential equations. The most general form is ∂2u ∂2u ∂u ∂u ∂2u + C(x, y) 2 = φ x, y, u, , (2.1) A(x, y) 2 + 2B(x, y) ∂x ∂x∂y ∂y ∂x ∂y

6

Where u is the unknown state variable. 7

This equation is classiﬁed into three types:

G(x, y) Poisson Equation 0 Laplace Equation 2 u,xx = u,t Heat Equation B − AC = 0 Parabolic B 2 − AC > 0 Hyperbolic u,xx − u,tt = 0 Wave Equation B 2 − AC < 0 Elliptic

8

u,xx + u,yy =

Note: 1. The Laplace equation (∇2 u = 0) is a special case of Poisson’s equation, where the right hand side is zero. Laplace associated with the equilibrium problem. 2. The Heat equation (Hu,t − K 2 ∇2 u = 0) corresponds to exponential decay. Also referred to as Diﬀusion equation (ﬂuid ﬂow through porous media, irrotational ﬂuid ﬂow, Saint Venant torsion of elastic bars...). 3. The Wave equation (ρu,tt − K 2 ∇2 u = 0) corresponds to harmonic motion

9 This classiﬁcation is established when solving Eq. 2.1 using the method of characteristics because it is then observed that the character of the solutions is distinctly diﬀerent for the three categories of equations.

Example 2-1: Seepage Problem;(Bathe 1996) The idealized dam shown in Fig. 2.2 stands on permeable soil. Formulate the diﬀerential

h L

q(y+dy)

h2

dy

h1

111 000 000 111 000 111 000 111 Impermeable wall 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111

q(x)

q(x+dx)

Seepage q(y) y Permeable soil

dx

x

0000000000000000000000000 1111111111111111111111111 1111111111111111111111111 0000000000000000000000000 Impermeable rock

Figure 2.2: Seepage Problem Victor Saouma

Finite Elements II; Solid Mechanics

Draft 2–4

INTRODUCTION

equation governing the steady-state seepage of water through the soil and give the corresponding boundary conditions. Solution: 1. For a typical element of widths dx and dy (and unit thickness), the total ﬂow into the element must be equal to the total ﬂow out of the element. Hence we have ∂qy ∂qx dx + qx − qx + dy = 0 (2.2-a) qy − qy + ∂y ∂x ∂qy ∂qx − dydx − dxdy = 0 (2.2-b) ∂y ∂x

2. Using Darcy’s law, the ﬂow is given in terms of the total potential φ (water elevation), qx = −k

∂φ ∂x

qy = −k

∂φ ∂y

(2.3)

where we assume a uniform permeability k. Substituting from Eq. 2.3 into Eq. 2.2-b, we obtain the Laplace equation k

∂2φ ∂2φ + 2 ∂x2 ∂y

=0

(2.4)

3. It may be noted that this same equation is also obtained in heat transfer analysis and in the solution of electrostatic potential and other ﬁeld problems. 4. The boundary conditions are no-ﬂow boundary conditions in the soil at x = −∞ and x = +∞, ∂φ ∂φ = 0; =0 (2.5) ∂x x=−∞ ∂x x=+∞ at the rock-soil interface,

∂φ =0 ∂y y=0

(2.6)

∂φ =0 ∂y − h ≤x≤+ h ,y=L

(2.7)

and at the dam-soil interface,

2

10

2

In addition, the total potential is prescribed at the water-soil interface, φ(x, L)|x<−(h/2) = h1 and φ(x, L)|x>−(h/2) = h2

(2.8)

The diﬀerential equation in Eq. 2.4 and the boundary conditions in Eq. 2.5 to Eq. 2.8 deﬁne the seepage ﬂow steady-state response.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

2.2 Elliptic, Parabolic and Hyperbolic Equations

111 000 000 111 000 111

x

1.0

0

dz=1.0

z q (t) t>0

111 000 000 111 000 111

1 0 0 1 0 1 0 1 0q(x) 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

11111111 00000000 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00000000 11111111 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11

q(x+dx)

dy =1 .0

y 111 000 000 111 000 000111 111 000 111 000 111

2–5

dx

L

Figure 2.3: One Dimensional Heat Conduction

Example 2-2: Diﬀusion Problem; (Bathe 1996) The very long slab shown in Fig. 2.3 is at a constant initial temperature θi when the surface at x = 0 is suddenly subjected to a constant uniform heat ﬂow input. The surface at x = L of the slab is kept at the temperature θi , and the surfaces parallel to the x, z plane are insulated. Assuming one-dimensional heat ﬂow conditions, show that the problem-governing diﬀerential equation is the heat conduction equation k

∂θ ∂2θ = ρc 2 ∂x ∂t

(2.9)

where ρ is the mass density, c is the heat capacity per unit mass (amount of heat needed to raise a unit mass by one degree), k is the conductivity, and the temperature θ is the state variable. State also the boundary and initial conditions. Solution: 1. We consider a typical diﬀerential element of the slab. The element equilibrium requirement is that the net heat ﬂow input to the element must equal the rate of heat stored in the element. Thus ∂θ ∂q dx A = ρAc (2.10) qA − q + ∂x ∂t 2. The constitutive relation is given by Fourier’s law of heat conduction q = −k

∂θ ∂x

(2.11)

3. Substituting from Eq. 2.11 into Eq. 2.10 we obtain k

Victor Saouma

∂2θ ∂θ = ρc 2 ∂x ∂t

(2.12)

Finite Elements II; Solid Mechanics

Draft 2–6

INTRODUCTION

1 0 0 1 x u(x,t) 0 1 0 11 1 00 00 11 11 00 00 11 00 11

σA

x

dx

F(t)

F0 F0

σA + d σ d x x+dx dx

t

Figure 2.4: Rod subjected to Step Load In this case the element interconnectivity requirements are contained in the assumption that the temperature θ be a continuous function of x and no additional compatibility conditions are applicable. 4. The boundary conditions are (for t > 0) ∂θ ∂x

= −

(0,t)

q0 (t) k

(2.13-a)

θ|(L,t) = θi

(2.13-b)

θ|(x,0) = θi

(2.14)

and the initial condition is

Example 2-3: Wave Equation, (Bathe 1996) The rod shown in Fig. 2.4 is initially at rest when a load F (t) is suddenly applied at its free end. Show that the problem-governing diﬀerential equation is the wave equation 1 ∂2u E ∂2u (2.15) = 2 2 ; and c = 2 ∂x c ∂t ρ where E is the Young’s modulus, ρ the mass density, and A the cross sectional area, c corresponds to the velocity of sound in the elastic medium, and u is the state variable. Also state the boundary and initial conditions. Solution: 1. The element force equilibrium requirements of a typical diﬀerential element are, using d’Alembert’s principle 1 which states that with inertia forces included, a system is in 1

Thinking in terms of equilibrium of forces, it is more appealing to invoke D’Alembert’s principle of dynamic equilibrium rather than Newton’s second law of motion. This principle is based on the notion of a ﬁctitious inertia force, equal to the product of mass times acceleration and acting in a direction opposite to the acceleration.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

2.2 Elliptic, Parabolic and Hyperbolic Equations

2–7

equilibrium at each time instant. σA + A

∂2u ∂σ dx − σA = ρA 2 dx ∂x ∂t

2. The constitutive relation is σ=E

∂u ∂x

(2.16)

(2.17)

3. Combining Eq. 2.16 and Eq. 2.17 we obtain 1 ∂2u ∂2u = ∂x2 c2 ∂t2

(2.18)

4. The element interconnectivity requirements are satisﬁed because we assume that the displacement u is continuous, and no additional compatibility conditions are applicable. 5. The boundary conditions are for t > 0 u|(0,t) = 0 ∂u = F0 EA ∂x

(2.19-a) (2.19-b)

(L,t)

(2.19-c) and the initial conditions are u(x,0) = 0 ∂u = 0 ∂t

(2.20-a) (2.20-b)

(x,0)

With the above two sets of conditions, the formulation of the problem is complete, and Eq. 2.18 can be solved for the displacement response of the rod.

10

Observations

Example 2-1 From the ﬁrst example we observe 1. The unknown state variables (or their normal derivatives) are given on the boundary. These problems are for this reason also called boundary value problems, where we should note that the solution at a general interior point depends on the data at every point of the boundary. 2. A change in only one boundary value aﬀects the complete solution; for instance, the complete solution for φ depends on the actual value of h1 . 3. Elliptic diﬀerential equations generally govern the steady-state response of systems. Example 2-2, 2-3

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 2–8

INTRODUCTION

1. Comparing the governing diﬀerential equations given in the three examples we note that in contrast to the elliptic equation, the parabolic and hyperbolic equations Eq. 2.12 and Eq. 2.18 include time as an independent variable and thus deﬁne propagation problems. These problems are also called initial value problems because the solution depends on the initial conditions. 2. We may note that analogous to the derivation of the dynamic equilibrium equations of lumped-parameter models, the governing diﬀerential equations of propagation problems are obtained from the steady-state equations by including the “resistance to change” (inertia) of the diﬀerential elements. 3. The parabolic and hyperbolic diﬀerential equations Eq. 2.12 and Eq. 2.18 would become elliptic equations if the time-dependent terms were neglected. In this way the initial value problems would be converted to boundary value problems with steady-state solutions. 4. The solution of a boundary value problem depends on the data at all points of the boundary. However, in propagation problem, the solution at an interior point may depend only on the boundary conditions of part of the boundary and the initial conditions over part of the interior domain.

2.3

Solution of Discrete-System Mathematical models

Section adapted from (Bathe 1996) 11

The essence of the solution of discrete-system is 1. System idealization: the actual system is idealized as an assemblage of elements. 2. Element equilibrium: the ”equilibrium” requirements of each element are established in terms of state variables (displacement, temperature, pressure, etc...). 3. Element assemblage: element interconnection requirements are invoked to establish a set of simultaneous equations in terms of the unknown state variables. 4. Calculation of state variables: The set of linear equations is solved to determine the state variables at each discretized points. 5. Calculation of ﬂux variable: or derived variables.

12 In the following sections, we shall illustrate, through a number of diﬀerent physical problems, the solution of discrete-systems. This preliminary exposure is a “snap-shot” of the type of problems which can be addressed by the ﬁnite element method.

2.3.1

Steady State Problems

13 In this ﬁrst class of problem, we shall focus on “equilibrium” problems, that is problems where the solution does not change with time.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

2.3 Solution of Discrete-System Mathematical models 2.3.1.1

2–9

Elastic Spring

14 We seek to determine the displacements of each of the rigid carts connected by linear elastic springs, as well as the force in each spring, Fig. 2.5. Conceptually, this can be viewed as a one dimensional truss, with the spring stiﬀness k corresponding to AE/L

u 1, R

11 00 00 11 11 00

k1

1

1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 1 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

k4

u 2, R

u 3, R

2

0000000 k1111111 3 0000000 1111111

k2

1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 2 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

k5

3

1111111 0000000 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 3 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111

Figure 2.5: System of Rigid Carts Interconnected by Linear Springs, (Bathe 1996) 1. Considering the ﬁrst two springs, we can write the following equations of equilibrium (1)

k1 u1 = F1

(2.21-a)

(2) F1 (2) F2

(2.21-b)

k2 (u1 − u2 ) = k2 (u2 − u1 ) =

2. By extension, we can write the equation of equilibrium for each spring in terms of the displacement state variables (2) 1 −1 u1 F1 (2.22-a) = k 2 (2) u2 −1 1 F2 (4) 1 −1 u1 F1 (2.22-b) = k4 (4) u3 −1 1 F3 (3) 1 −1 u1 F1 (2.22-c) = k3 (3) u2 −1 1 F2 (5) 1 −1 u2 F2 (2.22-d) = k5 (5) u3 −1 1 F3 3. The global equations of equilibrium being  (1) (2) (3) (4)  = R1  F1 + F1 + F1 + F1 (2) (3) (5) F2 + F2 + F2 = R2   (4) (5) F3 + F3 = R3

(2.23)

4. We next substitute the equilibrium equations of each element into the three global equations of equilibrium      −(k2 + k3 ) −k4 (k1 + k2 + k3 + k4 )  u1   R1   (k2 + k3 + k5 ) −k5 u R −(k2 + k3 ) = (2.24) K=  2   2  −k4 −k5 (k4 + k5 ) u3 R3 Victor Saouma

Finite Elements II; Solid Mechanics

Draft 2–10

INTRODUCTION

5. The state variables (nodal displacements) ui can be solved through the inversion of K. 6. The forces in each spring can be determined from equilibrium of each individual spring. 2.3.1.2

Heat Transfer

15 A wall is constructed of two homogeneous slabs in contact. In steady-state conditions the temperatures in the wall is characterized by the external surface temperature θ1 and θ3 and the interface temperature θ2 . Establish the equilibrium equations of the problem in terms of these temperatures when the ambient temperatures θ0 and θ4 are known.

θ0

θ1

θ2

θ3 θ 4

111 000 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000000000 111111111 0 1 0 1 000000000111111111111111111 111111111 000000000000000000 000000000 111111111 0 1 0 1 000000000 111111111 000000000000000000 111111111111111111 000000000 111111111 0 1 0 1 000000000 111111111 000000000000000000 111111111111111111 000000000 111111111 0 1 0 1 000000000 111111111 000000000000000000 111111111111111111 2k 3k 00 000000000 111111111 00 11 0 1 0 1 000000000 111111111 000000000000000000 111111111111111111 11 000000000 111111111 00 11 0 1 0 1 000000000 111111111 000000000000000000 111111111111111111 000000000 111111111 0 1 0 1 000000000000000000 111111111111111111 000000000 111111111 000000000 111111111 0 1 0 1 000000000 111111111 000000000000000000 111111111111111111 000000000 111111111 0 1 0 1 000000000 111111111 000000000000000000 111111111111111111 000000000 111111111 0 1 0 1 000000000 111111111 000000000000000000 111111111111111111 000000000 111111111 0 1 0 1 000000000 111111111 000000000000000000 111111111111111111 000000000 111111111 A 111111111 B C D 0 1 0 1 000000000000000000 111111111111111111 000000000 000000000 111111111 0 1 0 1 000000000000000000 111111111111111111 000000000 111111111 000000000 111111111 0 1 0 1 000000000000000000 111111111111111111 000000000 111111111 3k 000000000 111111111 2k 000000000 111111111 0 1 000000000000000000 111111111111111111 0 1 000000000 111111111 0 1 000000000000000000 111111111111111111 0 1 000000000 111111111 000000000 111111111 0 1 000000000000000000 0 1 111111111111111111

Conductance Figure 2.6: Slab Subjected to Temperature Boundary Conditions, (Bathe 1996)

1. The governing equation is the heat conduction law q = Ak∆θ

(2.25)

where q is the total heat ﬂow, A the area, ∆θ the temperature drop in the direction of heat ﬂow, and k the conductance or surface coeﬃcient. 2. The state variables are θ1 , θ2 and θ3 . 3. We then apply the “equilibrium” equation for each interface  q1 = 3k(θ0 − θ1 ) Convection A → B    q2 = 2k(θ1 − θ2 ) Conduction B q = 3k(θ2 − θ3 ) Conduction C    3 q4 = 2k(θ3 − θ4 ) Convection C → D

(2.26)

4. Heat ﬂow equilibrium must be satisﬁed q1 = q2 = q3 = q4 Victor Saouma

(2.27)

Finite Elements II; Solid Mechanics

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2.3 Solution of Discrete-System Mathematical models

2–11

thus 3k(θ0 − θ1 ) = 2k(θ1 − θ2 )

(2.28-a)

2k(θ1 − θ2 ) = 3k(θ2 − θ3 )

(2.28-b)

3k(θ2 − θ3 ) = 2k(θ3 − θ4 )

(2.28-c)

5. These equations can be rewritten as      5k −2k 0  θ1   3kθ0   −2k 5k −3k  θ 0 =  2    0 −3k 5k θ3 2kθ4 2.3.1.3

(2.29)

Hydraulic Network

In this example, we seek to establish the equations that govern the steady-state pressure and ﬂow distribution in the hydraulic network shown in Fig. 2.7. We assume the ﬂuid to be incompressible and the pressure drop in a branch to be proportional to the ﬂow q through that branch, (Darcy’s law) πd4 ∆p ⇒ ∆p = Rq (2.30) q= 128µL 16

where d is the pipe diameter, µ the ﬂuid kinematic viscosity, L the pipe length, R is the branch eﬀective resistance coeﬃcient. E q

R=10b

1

A

B

Q

R=

5b

q

R=2b

q

5 R=

2

C

R=3b q

Q

b

3

4

q

2

D

Figure 2.7: Pipe Network, (Bathe 1996)

1. We consider each branch of the pipe to constitute an element, and we select as state variables pA , pC , and pD the pressures at A, C, and D respectively. 2. Thus from “equilibrium” of each element  q1     q3  q2 |AC   q |    2 DB q4

Victor Saouma

= = = = =

pA 10b pC −pD 2b pA −pC 5b pD 5b pC −pD 3b

(2.31)

Finite Elements II; Solid Mechanics

Draft 2–12

INTRODUCTION

3. The conservation mass requirement (or continuity of ﬂow) requires that  Q = q1 + q2  q2 |AC = q3 + q4  q2 |DB = q3 + q4

(2.32)

Substituting, we get     3 −2 0  pA   10bQ   −6 31 −25  p = 0   C   pD 0 −1 1 1

(2.33)

    9 −6 0  pA   30bQ   −6 31 −25  p = 0   C   pD 0 0 −25 31

(2.34)





or

4. The pressures are determined through the inversion of the matrix, and the ﬂow can in turn be determined from Eq. 2.31 2.3.1.4

DC Network

Considering the network shown in Fig. 2.8, determine the steady state current distribution in the network. 17

I3 I3

I3

6R 2R

I - I3 1

4

I - I3 2

4R

2 1 I

1

3

2R

I2 B

A

2E

E

Figure 2.8: DC Network, (Bathe 1996)

1. The state variables will be the currents I1 , I2 , and I3 . Ohm’s law will be applied ∆E = RI

(2.35)

where ∆E is the voltage drop across the resistor. 2. The “equilibrium” equation to be satisﬁed across each element interconnection will be Kirchhoﬀ’s law  = 2E 2RI1 + 2R(I1 − I3 )  (2.36) = E 4R(I2 − I3 )  6RI3 + 4R(I3 − I2 ) + 2R(I3 − I1 ) = 0

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

2.3 Solution of Discrete-System Mathematical models

2–13

3. Rewriting this equation in matrix form      4R 0 −2R  I1   2E   0 I = E 4R −4R    2   I3 0 −2R −4R 12R

2.3.2

(2.37)

Equivalent “Truss”/Direct Stiﬀness Models

Each of the preceding problems can be discretized by an “equivalent truss” framework, and the direct stiﬀness method applied to assemble the global “stiﬀness” matrix, Fig. 2.9.

u1 , R1

01 1010 1010

u2 , R2

u3 , R3

1

2

3

4

3K

2K

3K

2K

1 0 0 1 0 1 0 1 0 1

1

0110 100 11 1010 1010

2 5b

110 20 1010

10 b 3 2b

3b

u2 , R2

u1 , R1

1 0 0 31 0 1 0 1

5 5b

4

1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1

u3 , R3 u2 , R2

01 1010 10

1 2R

1 0 0 21 0 1 0 1 0 11 0 1 0 1 0 1

3 4R

2 2R

u1 , R1

1 0 0 1 0 1 0 31 0 1 0 1 0 1 0 1 0 1

4 6R

1 0 0 1 0 1 0 1

u3 , R3

Figure 2.9: Equivalent Trusses/Direct Stiﬀness

Heat Transfer: where the “displacement” correspond to the temperature θ.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 2–14

INTRODUCTION

1. Element “stiﬀness” matrices (1)

Spring 1:

Spring 2: 2K Spring 3: 3K

1 −1 −1 1 1 −1 −1 1

Spring 4:

3Ku1 = F 1 (1) u1 F1 = (2) u2 F2 (3) F2 u2 = (3) u3 F3 (4) 2Ku3 = F3

(2.38)

2. Assemble active degrees of freedom    (2) F1 2K −2K 0 u   1   (2) (3)  −2K 2K + 3K −3K  = u F +F  2 (3) 2  2   u3 0 −3K 3K F3 

   (2.39)

 

3. Equilbrium at each node  (1) (2) (2) (1)  = R1  F1 + F1 = R1 − F1 = 3Kθ0 − 3Ku1 F1 (2) (3) R = 0 ⇒ F2 + F2 = R2 2  (3) (4)  (3) (4) = R3 − F3 = 2Kθ4 − 2Ku3 F F3 + F3 = R3 3

(2.40)

4. Rearrange state variables        5K −2K 0  u1   R1   3Kθ0   −2K 5K −3K  u R 0 = = =   2   2   u3 R3 2Kθ4 0 −3K 5K

(2.41)

Pipe Network: where the “displacement” correspond to the pressure p 1. Element “stiﬀness” matrices Spring 1:

Spring 2:

1 5b

Spring 3:

1 2b

Spring 4:

1 3b

1 −1 −1 1 1 −1 −1 1 1 −1 −1 1

1 10b

u1 u2 u2 u3 u2 u3

1 5b u3

Spring 5:

(1)

= F 1 (2) F1 = (2) F2 (3) F3 = (3) F4 (4) F3 = (4) F4

(2.42)

(5)

= F4

2. Assemble active degrees of freedom  

1 5b 1 − 5b

0

Victor Saouma

1 − 5b 1 1 1 5b + 2b + 3b 1 −f rac12b − 3b

   (2) F1 0 u  1    (2) (3) (4) 1 1  − 2b − 3b u = F3 + F3 + F3  2   1 1  (3) (4) u3 F4 + F4 2b + 3b

    

(2.43)

Finite Elements II; Solid Mechanics

Draft

2.3 Solution of Discrete-System Mathematical models

2–15

3. Equilbrium at each node  (1) (2) (2) (1) 1 = R1  F1 + F1 = R1 − F1 = Q − 10b u1 F1  (2) (3) (4) R2 = 0 ⇒ F3 + F3 + F3 = R2  (3) (4) (5)  (3) (4) (5) 1 = R3 − F4 = 0 − 5b u3 F4 + F4 F4 + F4 + F4 = R3 (2.44) 4. Rearrange state variables and multiply by 30      9 −6 0  u1   30bQ   −6 31 −25  u = 0   2   u3 0 0 −25 31

(2.45)

DC Network: where the “displacement” u correspond to the current intensity I 1. Element “stiﬀness” matrices Spring 1:

(1)

Spring 2: 2R Spring 3: 4R

1 −1 −1 1 1 −1 −1 1

Spring 4:

2Ru1 = F 1 (2) u1 F1 = (2) u2 F3 (3) F2 u2 = (3) u3 F3 (4) 6Ru3 = F3

(2.46)

2. Assemble active degrees of freedom    (2) F1 2R 0 −2R   u1   (3)  0 u = 4R −4R  F2  2    (2) (3) u3 −2R −4R 2R + 4R F3 + F3 

    

(2.47)

3. Equilbrium at each node  (2) (2) (1) + F1 = R1  = R1 − F1 = 2E − 2Ru1 F1  (3) (3) ⇒ F2 = R2 F2 = R2 = E   (2) (3) (4) (2) (3) (4) F3 + F3 + F3 = R3 = R3 − F3 = −6Ru3 F3 + F3 (2.48) (1)

F1

4. Rearrange state variables        4R 0 −2R  u1   2E   3Kθ0   0 u2 0 = E = = 4R −4R        u3 2Kθ4 0 −2R −4R 12R 2.3.2.1

2.3.3

(2.49)

Nonlinear Elastic Spring

Propagation Problems

18 The main characteristic of a propagation dynamic problem is that the response of the system changes with time. In principle, we may apply the same analysis procedure as in steady-state problems, however in this case the state variables and the the equilibrium relations depend on time.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 2–16

2.3.3.1

INTRODUCTION

Dynamic Elastic System

Let us reconsider the system of rigid carts previously analysed.We assume the springs to be massless, but the carts have masses mi . 1. Invoking d’Alembert’s principle, the interelement  (1) (2) (3) (4)   F1 + F1 + F1 + F1 (2) (3) (5) F2 + F2 + F2   (4) (5) F3 + F3 where u ¨i =

d2 ui dt2

for

interconnectivity requirement yields = R1 (t) − m1 u ¨1 = R2 (t) − m2 u ¨2 = R3 (t) − m3 u ¨3

i = 1, 2, 3

(2.50)

(2.51)

2. Thus we obtain the following system of governing equilibrium equations M¨ u + Ku = R(t)

(2.52)

 0 m1 0 M =  0 m2 0  0 0 m3

(2.53)



where

3. The equilibrium equation represents a system of ordinary diﬀerential equations of the second order in time. For the solution of these equations it is also necessary to specify the initial conditions for u and u˙ at time t = 0 (u0 and u˙ 0 respectively). 2.3.3.2

Transient Heat Flow

Fig. 2.10 illustrates an idealization of the heat ﬂow inside an electron tube. A ﬁlament is heated to a temperature θf by an electric current; heat is convected from the ﬁlament to the surrounding gas and is radiated to the wall which also receives heat by convection of the gas. The wall itself convects heat to the surrounding atmosphere, which is at temperature θa . It is required to formulate the system-governing heat ﬂow equilibrium equations. 1. The state variables are the temperature of the gas, θ1 , and the temperature of the wall θ2 . 2. The governing equations for heat transfer are Gas C1 dθ1 = k1 (θf − θ1 ) − k2 (θ1 − θ2 ) dt ! " d θ2 4 4 = kr (θf ) − (θ2 ) + k2 (θ1 − θ2 ) − k3 (θ2 − θa ) Wall C2 dt

(2.54)

Note that the ﬁrst equation is Newton’s law of cooling, and the second is the StefanBoltzman law of radiation. 3. The two equations can be written in matrix form as Cθ˙ + Kθ = Q C1 0 C = 0 C2 Victor Saouma

(2.55-a) (2.55-b)

Finite Elements II; Solid Mechanics

Draft

2.3 Solution of Discrete-System Mathematical models

2–17

Wall

Co n

ve

c ti

on

k

1

Gas θ 1

k2

Atmosphere

Gas

1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

n tio ec nv Co

1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

Filament θ f

Radiation kr

Wall θ 2

Convection k3

Atmosphere θ a

Filament

Figure 2.10: Heat Transfer Idealization in an Electron Tube, (Bathe 1996) −k2 (k1 + k2 ) −k2 (k2 + k3 ) θ1 θ = θ2 k1 θf " ! Q = kr (θf )4 − (θ2 )4 + k3 θa )

K =

(2.55-c) (2.55-d) (2.55-e)

4. We note that because of the radiation boundary conditions, the heat ﬂow equilibrium equations are nonlinear in θ.

2.3.4 2.3.4.1 19

Eigenvalue Problems Free Vibration

The governing equation for the free (undamped) vibration of a structure is M¨ u + Ku = 0

(2.56)

where the motion is referred to being free, since there are no applied loads. 20

By assuming a harmonic motion u = φ sin ωt

(2.57)

the natural frequencies ω and the corresponding mode shapes φ can be determined from the generalized eigenvalue problem (2.58) ω 2 Mφ = Kφ or Since φ is nontrivial

Victor Saouma

(K − ω 2 M)φ = 0

(2.59)

|K − ω 2 M| = 0

(2.60)

Finite Elements II; Solid Mechanics

Draft 2–18

INTRODUCTION

or with ω 2 = λ |K − λM| = 0

(2.61)

which is the characteristic equation, and λ is called the eigenvalue of the equation, and the structure is said to respond in the mode corresponding to a particular frequency. For computational purposes, if we premultiply each side of the preceding equation by M−1 , then 21

(M−1 K − λI)φ = 0

(2.62)

It should be noted that a zero eigenvalue is obtained for each possible rigid body motion of a structure that is not completely supported. 22 Depending on which mass matrix is adopted, slightly diﬀerent results are obtained. In general, lumped mass matrices approach the exact value (consistent mass matrix) from below. 23 The mode shapes φ are “shapes” , anfd give a relative magnitude of the DOF, not the absolute values (since they are the solution to a set of homogeneous equations). 24 The natural frequencies and mode shapes provide a fundamental description of the vibrating structure.

2.3.4.2

Column Buckling

25 Next we consider the two rigid bar problem illustrated in Fig. 2.11. For this problem, we must consider equilibrium of the deformed state (rather than undeformed), because of the large deformation in presence:

P P C

L B k

L

Ak

0000000 1111111 0000 C 1111 1111111 0000000 0000 1111 θ1 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 θ2 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 B 0000000 1111111 0000 1111 L 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 θ1 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 0000000 1111111 C 1 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 L 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 θ 0000000 1111111 0000000 1111111 2 0P 1 0000000 1111111 0 1 0000000 1111111 0 1 0000000 1111111 0 1 0000000 1111111 0 1 0000000 1111111 0 1 0000000 1111111 0 1 0000000 1111111 0 1 0000000 1111111 0 1 0000000 1111111 B 0000B 1111 0000 1111 0000 1111 0000k(θ 1111 k(θ2 - θ)1 1111 0000 2 0000 1111 0000 1111 0000 1111 0000 1111 P 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 A 1111 kθ1 0000 1111 0 1 0 1 0 1 0 1 0 1 0 1 P 0 1 0 1 0 1

0 P1 0 1 0 1

P

1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

θ)1

1 0 0 1 0 1 0 1 0 1 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 1.618 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 1 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

0 P 1 1 0 0 1 0 1 0 1 0 1 0 1 0 1 000 111 000 111 000 111 000 111 000 111 000 111 000 111 .618 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 1 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

Figure 2.11: Stability of a Two Rigid Bars System ΣMB = P Lθ2 − k(θ2 − θ1 ) = 0 PL θ2 ⇒ −θ1 + θ2 = k Victor Saouma

(2.63-a) (2.63-b)

Finite Elements II; Solid Mechanics

Draft

2.3 Solution of Discrete-System Mathematical models

26

ΣMA = P Lθ1 + k(θ2 − θ1 ) − kθ1 = 0 PL θ1 ⇒ 2θ1 − θ2 = k

Those two equations can be cast in matrix form θ1 2 −1 θ1 =λ θ2 θ2 −1 1

where λ = P L/k, this is an eigenvalue formulation and can be rewritten as 0 2 − λ −1 θ1 = λ θ2 0 −1 1 − λ 2 − λ −1 −1 1 − λ = 0 2 − λ − 2λ + λ2 − 1 = 0

λ1,2 =

27

(2.63-d)

(2.64)

(2.65-a) (2.65-b)

√ 3± 5 3± 9−4 = 2 2 √

(2.65-d) (2.65-e)

Hence we now have two critical loads: √ k 3− 5k = 0.382 2√ L L k 3+ 5k = 2.618 2 L L

Pcr1 = Pcr2 =

28

(2.63-c)

(2.65-c)

λ − 3λ + 1 = 0 2

2–19

(2.66) (2.67)

We now seek to determine the deformed shape for each of the ﬁrst critical loads √ 3− 5 λ1 = 2 $ # √ 3− 5 −1 √ 0 θ1 2− 2 = 3− 5 θ 0 2 −1 1− 2 $ # √ 1+ 5 −1 √ 0 θ1 2 = −1+ 5 θ 0 2 −1 1 − 2 0 1.618 −1 θ1 = θ2 0 −1 0.618

(2.68-a) (2.68-b) (2.68-c) (2.68-d)

we now arbitrarily set θ1 = 1, then θ2 = 1/0.618 = 1.618, thus the ﬁrst eigenmode is

Victor Saouma

θ1 θ2

=

1 1.618

(2.69)

Finite Elements II; Solid Mechanics

Draft 2–20 29

INTRODUCTION

Finally, we examine the second mode shape loads

#

λ2 = −1 √ θ1 2 − 3+2 5 = θ2 −1 1 − 3+2 5 # $ √ 1− 5 −1 θ1 2 √ = θ2 −1 1 − −1−2 5 −0.618 −1 θ1 = θ2 −1 −1.618 $

√

√ 3+ 5 2 0 0 0 0 0 0

(2.70-a) (2.70-b) (2.70-c) (2.70-d)

we now arbitrarily set θ1 = 1, then θ2 = −1/1.618 = −0.618, thus the second eigenmode is

2.4

θ1 θ2

=

1 −0.618

(2.71)

Solution Strategies

Within the context computer based simulation of a partial diﬀerential equation, we have three alternatives for a numerical solution 30

Strong Form

Weak Form FE Variational Form Discretization Mathematical Model

Discrete Model

Solution

Physical System

Idealization

FD

Discrete Solution

Solution error Discretization + solution error Modeling + discretization+ solution error

Figure 2.12: Idealization, Discretization and Solution of a Numerical Simulation, (Felippa 2000) 31 Two of the three alternative forms for discretization have already been introduced in Finite Element I and are:

Strong Form/Diﬀerential Form, SF/DF which are the Euler Equations corresponding to the variational formulation. The strong form is represented by a system of ordinary or partial diﬀerential equations in space and/or time complemented by appropriate boundary conditions. The strong form is numerically solved by the ﬁnite diﬀerence FDM method. Victor Saouma

Finite Elements II; Solid Mechanics

Draft

2.4 Solution Strategies

2–21

Variational Form, VF: is a functional which must be stationary (maximum or minimum), and which can lead to either the strong form (corresponding Euler equations), or to the weak form. The variational form is solved by the Rayleigh-Ritz method which leads to a ﬁnite element formulation (FEM). Weak Form, WF: is a weighted integral equation which ”relaxes” the strong form to be satisﬁed on an average sense inside a ﬁnite element through the Weighted Residual method. 32

Note that: 1. The variational method provides a relatively easy way to construct the system of governing equations. This ease stems from the fact that in the variational formulation scalar quantities (energies, potentials and so on) are considered rather than vector quantities (forces, displacements, etc.), 2. Not all problems can be solved by the VF, whereas most can be solved by the WF. 3. In complex problems, a combination of techniques is used • Fluid-structure interaction: FEM for the structure, FDM for the ﬂuid. • Structural dynamics: FEM in space, and FDM in time.

2.4.1 33

Euler Equation

Given a functional

Π(u) =

b

F (x, u, u )dx

(2.72)

a

it can be shown that its corresponding Euler equation is given by d ∂F ∂F − = 0 in a < x < b ∂u dx ∂u

(2.73)

This diﬀerential equation is called the Euler equation associated with Π and is a necessary condition for u(x) to extremize Π. 34 Generalizing for a functional Π which depends on two ﬁeld variables, u = u(x, y) and v = v(x, y) (2.74) Π= F (x, y, u, v, u,x , u,y , v,x , v,y , · · · , v,yy )dxdy

There would be as many Euler equations as dependent ﬁeld variables ∂F ∂F ∂ ∂F ∂ ∂F ∂ 2 ∂F ∂2 ∂ 2 ∂F ∂u − ∂x ∂u,x − ∂y ∂u,y + ∂x2 ∂u,xx + ∂x∂y ∂u,xy + ∂y 2 ∂u,yy ∂F ∂v

−

∂ ∂F ∂x ∂v,x

−

∂ ∂F ∂y ∂v,y

+

∂ 2 ∂F ∂x2 ∂v,xx

+

∂F ∂2 ∂x∂y ∂v,xy

+

∂ 2 ∂F ∂y 2 ∂v,yy

= 0 = 0

(2.75)

Example 2-4: Flexure of a Beam

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 2–22

INTRODUCTION

The total potential energy of a beam supporting a uniform load p is given by L L 1 1 M κ − pw dx = (EIw )w − pw dx Π= 2 2 0 0

(2.76)

F

Solution: 1. From Eq. 2.75

& % = 12 (EIw )w − pw ∂F ∂w ' ( = −p ∂F = (EIw ) ∂w,xx F

∂2 ∂x2

    

⇒ (EIw ) = −p Euler Equation

2. Alternatively, integrating by part twice L L ∂F ∂F δw dx δF dx = δw + δΠ = ∂w ∂w 0 0 L (EIw δw − pδw)dx = 0 L " ! L (EIw ) δw − pδw dx = (EIw δw ) 0 − 0 L " ! L L (EIw ) + p δwdx = 0 = (EIw δw ) 0 − [(EIw ) δw]|0 + 0 Nat. Ess. Nat. Ess. Euler Eq. BC Or

(EIw ) = −p

(2.77)

(2.78-a) (2.78-b) (2.78-c) (2.78-d)

for all x

which is the governing diﬀerential equation of beams and Essential δw = 0 δw = 0

or or

Natural EIw = −M = 0 (EIw ) = −V = 0

at x = 0 and x = L 3. The variational form is thus given by L " ! (EIw ) + p δwdx = 0 0 Euler Eq. 4. The weak form will be given by L 0

Victor Saouma

" (EIw ) + p r(x)dx = 0 Euler Eq. !

(2.79)

(2.80)

Finite Elements II; Solid Mechanics

Draft

2.5 Computer Programs

2.5 35

2–23

Computer Programs

Modern programs are composed of three modules 1. Mesh generator 2. Analysis 3. Post-Pocessor

36

Major ﬁnite element codes

NASTRAN Originally developed by NASA, primarily used by NASA and its contractors. Original version public domain, later version (McNeal Schwindler) commercial. SAP Originally developed by Ed. Wilson at Berkeley. First version public domain, later ones available only for PC (SAP90). NONSAP is the nonlinear version of SAP ANSYS Commercial program mostly used in the Nuclear industry. ABAQUS Probably the most modern and widely used commercial ﬁnite element. (Available in Bechtel Lab). ALGOR Very widely used PC/based code, mechanical/civil applications. FEAP Public domain code listed in Zienkiewicz et al., often used in academia as a base for extension. MERLIN our very own!.

2.6

Examples of applications

1. Aircraft, automobile, submarine 2. Dam, buildings, bridges 3. Mechanical design and optimization 4. Heat transfer 5. Biomechanics (hip joints) 6. Electrical (design of rotors) 7. Coastal engineering 8. Fluid mechanics 9. Coupled problems

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 2–24

Victor Saouma

INTRODUCTION

Finite Elements II; Solid Mechanics

Draft Chapter 3

FUNDAMENTAL RELATIONS 3.1

Introduction

Whereas, ideally, a course in Continuum Mechanics should be taken prior to a ﬁnite element course, this is seldom the case. Most often, students have had a graduate course in Advanced Strength of Materials, which can only provide limited background to a solid ﬁnite element course. 1

2 Accordingly, this preliminary chapter (mostly extracted from the author’s lecture notes in Continuum Mechanics) will partially remedy for occasional deﬃciencies and will be often referenced. 3 It should be noted that most, but not all, of the material in this chapter will be subsequently referenced.

3.1.1 4

Notation

Diﬀerent set of notations are commonly used in Engineering:

Matrix: Finite Element [A], [σ], {F } Indicial: Mechanics cartesian, Fx , σij , Cijkl Tensorial: Mechanics cartesian/curvilinear, F, σ, C Engineering: (Timoshenko/Voigt) Elasticity, σx , γxy T Ax = xi Aij xj = x x.A.x tensor matrix indicial

(3.1)

In the following sections, we shall brieﬂy explain the last two. 3.1.1.1

Indicial Notation

5 Whereas the Engineering notation may be the simplest and most intuitive one, it often leads to long and repetitive equations. Alternatively, the tensor will lead to shorter and more compact forms.

Draft 3–2

FUNDAMENTAL RELATIONS

6 While working on general relativity, Einstein got tired of writing the summation symbol with ) its range of summation below and above (such as n=3 i=1 aij bi ) and noted that most of the time the upper range (n) was equal to the dimension of space (3 for us, 4 for him), and that when the summation involved a product of two terms, the summation was over a repeated index ) (i in our example). Hence, he decided that there is no need to include the summation sign if there was repeated indices (i), and thus any repeated index is a dummy index and is summed over the range 1 to 3. An index that is not repeated is called free index and assumed to take a value from 1 to 3. 7

Hence, this so called indicial notation is also referred to Einstein’s notation.

8

The following rules deﬁne indicial notation: 1. If there is one letter index, that index goes from i to n (range of the tensor). For instance:    a1  a i = 1, 3 (3.2) ai = ai = a1 a2 a3 =  2  a3 assuming that n = 3. 2. A repeated index will take on all the values of its range, and the resulting tensors summed. For instance: a1i xi = a11 x1 + a12 x2 + a13 x3 (3.3) 3. Tensor’s order: • First order tensor (such as force) has only one free index: ai = ai = a1 a2 a3

(3.4)

other ﬁrst order tensors aij bj , Fikk , εijk uj vk • Second order tensor (such as stress or strain) will have two free indeces.   D11 D22 D13 Dij =  D21 D22 D23  D31 D32 D33

(3.5)

other examples Aijip , δij uk vk . • A fourth order tensor (such as Elastic constants) will have four free indeces. 4. Derivatives of tensor with respect to xi is written as , i. For example: ∂Φ ∂xi

= Φ,i

∂vi ∂xi

= vi,i

∂vi ∂xj

= vi,j

∂Ti,j ∂xk

= Ti,j,k

(3.6)

Usefulness of the indicial notation is in presenting systems of equations in compact form. For instance:   x1 = c11 z1 + c12 z2 + c13 z3 x = c21 z1 + c22 z2 + c23 z3 (3.7) xi = cij zj ≡  2 x3 = c31 z1 + c32 z2 + c33 z3 9

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

3.1 Introduction

3–3

Similarly:

Aij = Bip Cjq Dpq

3.1.1.2

 A11    A12 ≡ A    21 A22

= = = =

B11 C11 D11 + B11 C12 D12 + B12 C11 D21 + B12 C12 D22 B11 C11 D11 + B11 C12 D12 + B12 C11 D21 + B12 C12 D22 B21 C11 D11 + B21 C12 D12 + B22 C11 D21 + B22 C12 D22 B21 C21 D11 + B21 C22 D12 + B22 C21 D21 + B22 C22 D22

(3.8)

Tensor Notation

10 In tensor notation the indices are not shown. While cartesian indicial equations apply only to Cartesian coordinates, expressions in tensor notation are independent of the coordinate system and apply to other coordinate systems such as cylindrical, curvilinear, etc. 11 In tensor notation, we indicate tensors of order one or greater in boldface letters. It is usually recommended to use lower case letters for tensors of order one, and upper case letters for tensors of order 2 and above. 12 We distinguish between tensor notation and matrix notation by using dots and colons between terms, as in a.b ≡ ai bi , (the ‘.’ indicates a contraction of a pair of repeated indices which appear in the same order), and A : B ≡ Aij Bij . or σij = Cijkl εkl is equivalent to σ = C : ε.

3.1.1.3

Voigt Notation

In ﬁnite element, symmetric second-order tensors are often written as column matrices. This conversion, and the one of other higher-order tensors into column matrices are called Voigt notation. 13

The Voigt rule depends on whether the kinematic one (such as strain).   σ11 σ12 → σ≡ σ21 σ22    ε11 ε12 → ε≡ ε21 ε22  14

3.1.2

tensor is a kinetic quantity (such as stress) or a    σ11   σ1  σ22 σ = = {σ}   2  σ12 σ3    ε11   ε1  ε22 ε = = {ε}   2  2ε12 ε3

(3.9-a)

(3.9-b)

Tensors

We generalize the concept of a vector by introducing the tensor (T), which essentially exists to operate on vectors v to produce other vectors (or on tensors to produce other tensors!). We designate this operation by T·v or simply Tv. 15

16

We hereby adopt the tensor (or dyadic) notation for tensors as linear vector operators u = T·v or ui = Tij vj u = v·S where S = T

Victor Saouma

(3.10-a) T

(3.10-b)

Finite Elements II; Solid Mechanics

Draft 3–4

FUNDAMENTAL RELATIONS

17 Whereas a tensor is essentially an operator on vectors (or other tensors), it is also a physical quantity, independent of any particular coordinate system yet speciﬁed most conveniently by referring to an appropriate system of coordinates. 18 Tensors frequently arise as physical entities whose components are the coeﬃcients of a linear relationship between vectors. 19 A tensor is classiﬁed by the rank or order. A Tensor of order zero is speciﬁed in any coordinate system by one coordinate and is a scalar. A tensor of order one has three coordinate components in space, hence it is a vector. In general 3-D space the number of components of a tensor is 3n where n is the order of the tensor. 20

A force and a stress are tensors of order 1 and 2 respectively.

21

The sum of two (second order) tensors is simply deﬁned as: Sij = Tij + Uij

22

(3.11)

The multiplication of a (second order) tensor by a scalar is deﬁned by: Sij = λTij

(3.12)

In a contraction, we make two of the indices equal (or in a mixed tensor, we make a subscript equal to the superscript), thus producing a tensor of order two less than that to which it is applied. For example: 2 → 0 Tij → Tii ; 2 → 0 ui vj → ui vi ; mr r (3.13) 4 → 2 Amr ..sn → A..sm = B.s ; 3 → 1 Eij ak → Eij ai = cj ; mp → Ampr Ampr qs qr = Bq ; 5 → 3 23

The outer product of two tensors (not necessarily of the same type or order) is a set of tensor components obtained simply by writing the components of the two tensors beside each other with no repeated indices (that is by multiplying each component of one of the tensors by every component of the other). For example

24

ai bj i

A

Bj.k

= Tij = C

(3.14-a)

i.k

.j

(3.14-b)

vi Tjk = Sijk

(3.14-c)

The inner product is obtained from an outer product by contraction involving one index from each tensor. For example

25

→ ai bi

(3.15-a)

ai Ejk → ai Eik = fk

(3.15-b)

ai bj

Eij Fkm → Eij Fjm = Gim i

A Victor Saouma

Bi.k

→ A

i

Bi.k

=D

k

(3.15-c) (3.15-d)

Finite Elements II; Solid Mechanics

Draft

3.2 Vector Fields; Solid Mechanics

26

3–5

The scalar product of two tensors is deﬁned as T : U = Tij Uij

(3.16)

in any rectangular system.

3.2 3.2.1 3.2.1.1 27

Vector Fields; Solid Mechanics Kinetics Force, Traction and Stress Vectors

There are two kinds of forces in continuum mechanics

body forces: act on the elements of volume or mass inside the body, e.g. gravity, electromagnetic ﬁelds. dF = ρbdV ol. surface forces: are contact forces acting on the free body at its bounding surface. Those will be deﬁned in terms of force per unit area. 28 The surface force per unit area acting on an element dS is called traction or more accurately stress vector. tdS = i tx dS + j ty dS + k tz dS (3.17)

S

S

S

S

Most authors limit the term traction to an actual bounding surface of a body, and use the term stress vector for an imaginary interior surface (even though the state of stress is a tensor and not a vector). The traction vectors on planes perpendicular to the coordinate axes are particularly useful. When the vectors acting at a point on three such mutually perpendicular planes is given, the stress vector at that point on any other arbitrarily inclined plane can be expressed in terms of the ﬁrst set of tractions. 29

30 A stress, Fig 3.1 is a second order cartesian tensor, σij where the 1st subscript (i) refers to the direction of outward facing normal, and the second one (j) to the direction of component force.     σ11 σ12 σ13  t1  t (3.18) σ = σij =  σ21 σ22 σ23  =  2  σ31 σ32 σ33 t3

In fact the nine rectangular components σij of σ turn out to be the three sets of three vector components (σ11 , σ12 , σ13 ), (σ21 , σ22 , σ23 ), (σ31 , σ32 , σ33 ) which correspond to the three tractions t1 , t2 and t3 which are acting on the x1 , x2 and x3 faces (It should be noted that those tractions are not necesarily normal to the faces, and they can be decomposed into a normal and shear traction if need be). In other words, stresses are nothing else than the components of tractions (stress vector), Fig. 3.1. 31

32 The state of stress at a point cannot be speciﬁed entirely by a single vector with three components; it requires the second-order tensor with all nine components.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 3–6

FUNDAMENTAL RELATIONS X3 X3

V3

σ33

σ32

t3

σ23

σ31

t2

σ13

σ

V2 X2

σ22

V1 X1

21

X2

σ

(Components of a vector are scalars)

12

t1

σ 11

V

X 1 Stresses as components of a traction vector (Components of a tensor of order 2 are vectors)

Figure 3.1: Stresses as Tensor Components 3.2.1.2

Traction on an Arbitrary Plane; Cauchy’s Stress Tensor

33 Let us now consider the problem of determining the traction acting on the surface of an oblique plane (characterized by its normal n) in terms of the known tractions normal to the three principal axis, t1 , t2 and t3 . This will be done through the so-called Cauchy’s tetrahedron shown in Fig. 3.2, and will be obtained without any assumption of equilibrium and it will apply

X2

-t 1

B

*

-t

*

∆

S

∆

t*n ∆ S

1

O

h N

S3

3

n A

X1

C

∆V *

ρb

*

-t 2 ∆ S2 *

X3

Figure 3.2: Cauchy’s Tetrahedron in ﬂuid dynamics as well as in solid mechanics. 34

This equation is a vector equation, and the corresponding algebraic equations for the com-

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

3.2 Vector Fields; Solid Mechanics

3–7

ponents of tn are tn1 tn2 tn3 Indicial notation tni Tensor notation tn

= = = = =

σ11 n1 + σ21 n2 + σ31 n3 σ12 n1 + σ22 n2 + σ32 n3 σ13 n1 + σ23 n2 + σ33 n3 σji nj n·σ = σT ·n

(3.19)

We have thus established that the nine components σij are components of the second order tensor, Cauchy’s stress tensor. 35

Example 3-1: Stress Vectors if the stress tensor at point P is given by     7 −5 0  t1  t σ =  −5 3 1  =  2  t3 0 1 2

(3.20)

We seek to determine the traction (or stress vector) t passing through P and parallel to the plane ABC where A(4, 0, 0), B(0, 2, 0) and C(0, 0, 6). Solution: The vector normal to the plane can be found by taking the cross products of vectors AB and AC: e 1 e2 e3 (3.21-a) N = AB×AC = −4 2 0 −4 0 6 (3.21-b) = 12e1 + 24e2 + 8e3 The unit normal of N is given by n=

6 2 3 e1 + e2 + e3 7 7 7

Hence the stress vector (traction) will be   7 −5 0 37 76 27  −5 3 1  = − 97 0 1 2 and thus t = − 97 e1 + 57 e2 +

3.2.2 36

(3.22)

5 7

10 7

(3.23)

10 7 e3

Kinematic

In continuum mechanics, we diﬀerentiate between

Material coordinates (X1 , X2 , X3 ) deﬁned in the undeformed original coordinate system which gives rise to the Lagrangian coordinate system. Victor Saouma

Finite Elements II; Solid Mechanics

Draft 3–8

FUNDAMENTAL RELATIONS

Spatial Coordinates (x1 , x2 , x3 ) deﬁned in the deformed coordinate system. This gigives rise to the Eulerian coordinate system 1. If both the displacement gradients and the displacements themselves are small, then ∂ui ∂ui ∂Xj ≈ ∂xj and thus the Eulerian and the Lagrangian inﬁnitesimal strain tensors may be ∗. taken as equal Eij = Eij 2. If the displacement gradients are small, but the displacements are large, we should use the Eulerian inﬁnitesimal representation. 3. If the displacements gradients are large, but the displacements are small, use the Lagrangian ﬁnite strain representation. 4. If both the displacement gradients and the displacements are large, use the Eulerian ﬁnite strain representation. 37

The Lagrangian ﬁnite strain tensor can be written as 1 Eij = 2

or:

38

∂uj ∂ui ∂uk ∂uk + + ∂Xj ∂Xi ∂Xi ∂Xj

or E =

1 (u∇X + ∇X u + ∇X u·u∇X ) 2 J+Jc J c ·J

# $ ∂u1 2 ∂u1 1 ∂u2 2 ∂u3 2 + + + E11 = ∂X1 2 ∂X1 ∂X1 ∂X1 1 ∂u1 ∂u2 ∂u2 ∂u2 ∂u3 ∂u3 1 ∂u1 ∂u1 + + + + E12 = 2 ∂X2 ∂X1 2 ∂X1 ∂X2 ∂X1 ∂X2 ∂X1 ∂X2 ··· = ···

(3.24)

(3.25-a) (3.25-b) (3.25-c)

If large deformation is accounted for (such as in buckling), the Eulerian ﬁnite strains are: εxx =

∂u 1 + ∂x 2

εyy =

∂v 1 + ∂y 2

εzz = εxy = εxz = εyz =

Victor Saouma

# #

∂u ∂x

2

+

∂v ∂x

2

+

∂w ∂x

2 $

$ 2 ∂u 2 ∂v ∂w 2 + + ∂y ∂y ∂y # $ ∂u 2 ∂w 1 ∂v 2 ∂w 2 + + + ∂z 2 ∂z ∂z ∂z ∂u ∂u ∂u ∂v ∂v ∂w ∂w 1 ∂v + + + + 2 ∂x ∂y ∂x ∂y ∂x ∂y ∂x ∂y 1 ∂w ∂u ∂u ∂u ∂v ∂v ∂w ∂w + + + + 2 ∂x ∂z ∂x ∂z ∂x ∂z ∂x ∂z 1 ∂w ∂v ∂u ∂u ∂v ∂v ∂w ∂w + + + + 2 ∂y ∂z ∂y ∂z ∂y ∂z ∂y ∂z

(3.26) (3.27) (3.28) (3.29) (3.30) (3.31)

Finite Elements II; Solid Mechanics

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3.2 Vector Fields; Solid Mechanics

3–9

or εij =

1 (ui,j + uj,i + uk,i uk,j ) 2

(3.32)

From this equation, we note that: 1. We deﬁne the engineering shear strain as γij = 2εij

(i = j)

(3.33)

2. If the strains are given, then these strain-displacements provide a system of (6) nonlinear partial diﬀerential equation in terms of the unknown displacements (3). 3. εik is the Green-Lagrange strain tensor. 4. The strains have been expressed in terms of the coordinates x, y, z in the undeformed state, i.e. in the Lagrangian coordinate which is the preferred one in structural mechanics. 5. Alternatively we could have expressed ds 2 − ds2 in terms of coordinates in the deformed state, i.e. Eulerian coordinates x , y , z , and the resulting strains are referred to as the Almansi strain which is the preferred one in ﬂuid mechanics. 6. In most cases the deformations are small enough for the quadratic term to be dropped, the resulting equations reduce to εxx = εyy = εzz = γxy = γxz = γyz =

∂u ∂x ∂v ∂y ∂w ∂z ∂u ∂v + ∂x ∂y ∂w ∂u + ∂x ∂z ∂w ∂v + ∂y ∂z

(3.34) (3.35) (3.36) (3.37) (3.38) (3.39)

or εij =

1 (ui,k + uk,i ) 2

(3.40)

which is called the Cauchy strain 39

In ﬁnite element, the strain is often expressed through the linear operator L ε = Lu

Victor Saouma

(3.41) Finite Elements II; Solid Mechanics

Draft 3–10 or

3.2.3

FUNDAMENTAL RELATIONS  εxx     εyy    εzz εxy      ε   xz εyz ε

       



∂ ∂x

 0   0  = ∂  ∂y    ∂     ∂z   0

0 ∂ ∂y

0 ∂ ∂x

0 ∂ ∂z

 0 0   ∂  ∂z   0  ∂   ∂x ∂ ∂y

 ux  uy  uz

(3.42)

u

L

Fundamental Laws of Continuum Mechanics

In this section, we will derive diﬀerential equations governing the way stress and deformation vary at a point and with time. They will apply to any continuous medium, and yet we will not have enough equations to determine unknown tensor ﬁeld. For that we need to wait for the next section where constitututive laws relating stress and strain will be introduced. Only with constitutive equations and boundary and initial conditions would we be able to obtain a well deﬁned mathematical problem to solve for the stress and deformation distribution or the displacement or velocity ﬁelds. 40

41 In this section we shall summarize the diﬀerential equations which express locally the conservation of mass, momentum and energy.

These diﬀerential equations of balance will be derived from integral forms of the equation of balance expressing the fundamental postulates of continuum mechanics. 42

43 Conservation laws constitute a fundamental component of classical physics. A conservation law establishes a balance of a scalar or tensorial quantity in voulme V bounded by a surface S. In its most general form, such a law may be expressed as d AdV + αdS = AdV (3.43) dt V S V Rate of variation

Exchange by Diﬀusion

Source

where A is the volumetric density of the quantity of interest (mass, linear momentum, energy, ...) a, A is the rate of volumetric density of what is provided from the outside, and α is the rate of surface density of what is lost through the surface S of V and will be a function of the normal to the surface n. 44 Hence, we read the previous equation as: The input quantity (provided by the right hand side) is equal to what is lost across the boundary, and to modify A which is the quantity of interest. The dimensions of various quantities are given by

dim(a) = dim(AL−3 )

(3.44-a)

dim(α) = dim(AL

−2 −1

t

)

(3.44-b)

dim(A) = dim(AL

−3 −1

)

(3.44-c)

t

Hence this section will apply the previous conservation law to mass, momentum, and energy. The resulting diﬀerential equations will provide additional interesting relation with regard to 45

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

3.2 Vector Fields; Solid Mechanics

3–11

the imcompressibiltiy of solids (important in classical hydrodynamics and plasticity theories), equilibrium and symmetry of the stress tensor, and the ﬁrst law of thermodynamics. 46 The enunciation of the preceding three conservation laws plus the second law of thermodynamics, constitute what is commonly known as the fundamental laws of continuum mechanics. 47 Prior to the enunciation of the ﬁrst conservation law, we need to deﬁne the concept of ﬂux across a bounding surface.

The ﬂux across a surface can be graphically deﬁned through the consideration of an imaginary surface ﬁxed in space with continuous “medium” ﬂowing through it. If we assign a positive side to the surface, and take n in the positive sense, then the volume of “material” ﬂowing through the inﬁnitesimal surface area dS in time dt is equal to the volume of the cylinder with base dS and slant height vdt parallel to the velocity vector v, Fig. 3.3 (If v·n is negative,

48

v

vn dt

n

vdt

dS Figure 3.3: Flux Through Area dS then the ﬂow is in the negative direction). Hence, we deﬁne the volume ﬂux as

Volume Flux =

v·ndS = S

vj nj dS

(3.45)

S

where the last form is for rectangular cartesian components. 49

We can generalize this deﬁnition and deﬁne the following ﬂuxes per unit area through dS: Mass Flux = ρv·ndS = ρvj nj dS (3.46) S S ρv(v·n)dS = ρvk vj nj dS (3.47) Momentum Flux = S S 1 2 1 ρv (v·n)dS = ρvi vi vj nj dS (3.48) Kinetic Energy Flux = 2 S2 S q·ndS = qj nj dS (3.49) Heat ﬂux = S S J·ndS = Jj nj dS (3.50) Electric ﬂux = S

Victor Saouma

S

Finite Elements II; Solid Mechanics

Draft 3–12

3.2.3.1

FUNDAMENTAL RELATIONS

Conservation of Mass; Continuity Equation ∂vi dρ dρ +ρ + ρ∇·v = 0 = 0 or dt ∂xi dt

(3.51)

The vector form is independent of any choice of coordinates. This equation shows that the divergence of the velocity vector ﬁeld equals (−1/ρ)(dρ/dt) and measures the rate of ﬂow of material away from the particle and is equal to the unit rate of decrease of density ρ in the neighborhood of the particle. 50 If the material is incompressible, so that the density in the neighborhood of each material particle remains constant as it moves, then the continuity equation takes the simpler form

∂vi = 0 or ∇·v = 0 ∂xi

(3.52)

this is the condition of incompressibility 3.2.3.2

Linear Momentum Principle; Equation of Motion

51 The momentum principle states that the time rate of change of the total momentum of a given set of particles equals the vector sum of all external forces acting on the particles of the set, provided Newton’s Third Law applies. The continuum form of this principle is a basic postulate of continuum mechanics. d tdS + ρbdV = ρvdV (3.53) dt V S V

Then we substitute ti = Tij nj and apply the divergence theorm (Appendix A, (Schey 1973)) to obtain dvi ∂Tij dV (3.54-a) + ρbi dV = ρ ∂x j V V dt dvi ∂Tij dV = 0 (3.54-b) + ρbi − ρ ∂xj dt V or for an arbitrary volume ∂Tij dv dvi or ∇T + ρb = ρ + ρbi = ρ ∂xj dt dt

(3.55)

which is Cauchy’s (ﬁrst) equation of motion, or the linear momentum principle, or more simply equilibrium equation. 52

When expanded in 3D, this equation yields: ∂T11 ∂T12 ∂T13 + + + ρb1 = 0 ∂x1 ∂x2 ∂x3 ∂T21 ∂T22 ∂T23 + + + ρb2 = 0 ∂x1 ∂x2 ∂x3 ∂T31 ∂T32 ∂T33 + + + ρb3 = 0 ∂x1 ∂x2 ∂x3

Victor Saouma

(3.56-a)

Finite Elements II; Solid Mechanics

Draft

3.2 Vector Fields; Solid Mechanics

3–13

We note that these equations could also have been derived from the free body diagram shown in Fig. 3.4 with the assumption of equilibrium (via Newton’s second law) considering an inﬁnitesimal element of dimensions dx1 × dx2 × dx3 . Writing the summation of forces, will yield 53

(3.57)

Tij,j + ρbi = 0 where ρ is the density, bi is the body force (including inertia). σ σyy+δ yy d y δy

dy

τyx+

δ τ yx y d δy σxx+

σ xx τ xy

τxy+

δ σxx d x δx

δ τ xy x d δx

τ yx σyy dx

Figure 3.4: Equilibrium of Stresses, Cartesian Coordinates

3.2.3.3

Conservation of Energy; First Principle of Thermodynamics

The ﬁrst principle of thermodynamics relates the work done on a (closed) system and the heat transfer into the system to the change in energy of the system. We shall assume that the only energy transfers to the system are by mechanical work done on the system by surface traction and body forces, by heat transfer through the boundary. 54

If mechanical quantities only are considered, the principle of conservation of energy for the continuum may be derived directly from the equation of motion given by Eq. 3.55. This is accomplished by taking the integral over the volume V of the scalar product between Eq. 3.55 and the velocity vi . dvi dV (3.58) vi Tji,j dV + ρbi vi dV = ρvi dt V V V

55

Applying the divergence theorem, dU dW dK + = +Q dt dt dt Victor Saouma

(3.59)

Finite Elements II; Solid Mechanics

Draft 3–14

FUNDAMENTAL RELATIONS

this equation relates the time rate of change of total mechanical energy of the continuum on the left side to the rate of work done by the surface and body forces on the right hand side. 56 If both mechanical and non mechanical energies are to be considered, the ﬁrst principle states that the time rate of change of the kinetic plus the internal energy is equal to the sum of the rate of work plus all other energies supplied to, or removed from the continuum per unit time (heat, chemical, electromagnetic, etc.). 57 For a thermomechanical continuum, it is customary to express the time rate of change of internal energy by the integral expression d dU ρudV (3.60) = dt dt V

where u is the internal energy per unit mass or speciﬁc internal energy. We note that U appears only as a diﬀerential in the ﬁrst principle, hence if we really need to evaluate this quantity, we need to have a reference value for which U will be null. The dimension of U is one of energy dim U = M L2 T −2 , and the SI unit is the Joule, similarly dim u = L2 T −2 with the SI unit of Joule/Kg.

3.2.4

Constitutive Equations

ceiinosssttuu Hooke, 1676 Ut tensio sic vis Hooke, 1678 3.2.4.1 58

General 3D

The Generalized Hooke’s Law can be written as: σij = Dijkl εkl

(3.61)

i, j, k, l = 1, 2, 3

The (fourth order) tensor of elastic constants Dijkl'has 81((34 ) components however, due to distinct elastic terms. the symmetry of both σ and ε, there are at most 36 9(9−1) 2 59

For the purpose of writing Hooke’s Law, the double indexed system is often replaced by a simple indexed system with a range of six: 60

62 =36

σk = Dkm εm 61

k, m = 1, 2, 3, 4, 5, 6

(3.62)

In terms of Lame’s constants, Hooke’s Law for an isotropic body is written as Tij = λδij Ekk + 2µEij 1 λ Tij − δij Tkk Eij = 2µ 3λ + 2µ

Victor Saouma

or or

T = λIE + 2µE 1 −λ IT + T E= 2µ(3λ + 2µ) 2µ

(3.63) (3.64)

Finite Elements II; Solid Mechanics

Draft

3.2 Vector Fields; Solid Mechanics 62

In terms of engineering constants: 1 E

=

λ =

63

3–15

λ+µ λ ;ν = µ(3λ + 2µ) 2(λ + µ) νE E ;µ = G = (1 + ν)(1 − 2ν) 2(1 + ν)

(3.65) (3.66)

Similarly in the case of pure shear in the x1 x3 and x2 x3 planes, we have σ21 = σ12 = τ all other σij = 0 τ 2ε12 = G

(3.67-a) (3.67-b)

and the µ is equal to the shear modulus G. 64

65

66

Hooke’s law for isotropic material in terms of engineering constants becomes ν E ν E εij + δij εkk ε+ Iε σij = or σ = (3.68) 1+ν 1 − 2ν 1+ν 1 − 2ν ν 1+ν ν 1+ν σij − δij σkk or ε = σ − Iσ (3.69) εij = E E E E When the strain equation is expanded in 3D cartesian coordinates it would yield:      1 −ν −ν 0 0 0 σ ε   xx xx            −ν 1 −ν     0 0 0 ε σ     yy yy             1  −ν −ν 1 0 0 0  εzz σzz =  0 0 1+ν 0 0  γxy (2εxy )  τxy   E 0                    (2ε ) τ 0 0 0 0 1 + ν 0 γ yz yz yz         τzx γzx (2εzx ) 0 0 0 0 0 1+ν If we invert this equation, we obtain       σxx  1−ν ν ν εxx         E        σyy  εyy ν 1−ν ν 0  (1+ν)(1−2ν)          σzz εzz ν ν 1 − ν    =   τ γ 1 0 0    xy  xy (2εxy )                τ  γ (2ε ) 0 G 0 1 0    yz    yz yz τzx γzx (2εzx ) 0 0 1

(3.70)

               (3.71)

3.2.4.2

67

Transversly Isotropic Case

For transversely isotropic, we can express the stress-strain relation in tems of

Victor Saouma

εxx εyy εzz γxy γyz γxz

= = = = = =

a11 σxx + a12 σyy + a13 σzz a12 σxx + a11 σyy + a13 σzz a13 (σxx + σyy ) + a33 σzz (3.72) 2(a11 − a12 )τxy a44 τxy Finite Elements II; Solid Mechanics a44 τxz

Draft 3–16 and a11 =

FUNDAMENTAL RELATIONS

1 ; E

a12 = −

ν ; E

a13 = −

ν ; E

a33 = −

1 ; E

a44 = −

1 µ

(3.73)

where E is the Young’s modulus in the plane of isotropy and E the one in the plane normal to it. ν corresponds to the transverse contraction in the plane of isotropy when tension is applied in the plane; ν corresponding to the transverse contraction in the plane of isotropy when tension is applied normal to the plane; µ corresponding to the shear moduli for the plane of isotropy and any plane normal to it, and µ is shear moduli for the plane of isotropy. 3.2.4.3 68

Special 2D Cases

Often times one can make simplifying assumptions to reduce a 3D problem into a 2D one.

3.2.4.3.1 Plane Strain 69 For problems involving a long body in the z direction with no variation in load or geometry, then εzz = γyz = γxz = τxz = τyz = 0. Thus, replacing into Eq. 3.71 we obtain       σ (1 − ν) ν 0   xx   εxx       E ν (1 − ν) 0  σyy  (3.74) =    εyy  ν ν 0 σ   (1 + ν)(1 − 2ν) zz   γxy   1−2ν τxy 0 0 2 3.2.4.3.2 and

Axisymmetry

70

In solids of revolution, we can use a polar coordinate sytem

εrr = εθθ = εzz = εrz =

71

∂u ∂r u r ∂w ∂z ∂u ∂w + ∂z ∂r

The constitutive relation is again analogous to 3D/plane strain    1−ν ν ν 0 σrr      ν 1 − ν ν 0    E σzz  ν ν 1−ν 0 = σθθ   (1 + ν)(1 − 2ν)     ν ν 1−ν 0   τrz 1−2ν 0 0 0 2

(3.75-a) (3.75-b) (3.75-c) (3.75-d)



   εrr   εzz    εθθ   γrz

      

(3.76)

3.2.4.3.3 Plane Stress 72 If the longitudinal dimension in z direction is much smaller than in the x and y directions, then τyz = τxz = σzz = γxz = γyz = 0 throughout the thickness.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

3.2 Vector Fields; Solid Mechanics

3–17

Again, substituting into Eq. 3.71 we obtain:      1 ν 0 εxx    σxx  1  = σyy ε ν 1 0    yy   1 − ν2 1−ν τxy γ 0 0 xy 2 1 εzz = − ν(εxx + εyy ) 1−ν

3.2.4.4

(3.77-a) (3.77-b)

Pore Pressures

73 In porous material, the water pressure is transmitted to the structure as a body force of magnitude

bx = −

∂p ∂x

by = −

∂p ∂y

(3.78)

where p is the pore pressure. The eﬀective stresses are the forces transmitted between the solid particles and are deﬁned in terms of the total stresses σ and pore pressure p

74

σij = σij + mij p

mT = [ −1, −1, 0 ]

(3.79)

i.e simply removing the hydrostatic pressure component from the total stress.

3.2.5

† Field Equations for Thermo- and Poro Elasticity

Adapted from (Reich 1993) 75

In the absence of thermal/initial stresses and pore pressures, the ﬁeld equations are written

as

+ bi = 0 Equilibrium σij,j σij nj − tˆi = 0 Natural B.C.

(3.80)

where bi and tˆi are the body forces and surface tractions respectively. These equations will form the basis of the variational formulation of the ﬁnite element method. 76

77 To account for the eﬀect of thermal/initial stresses and pore pressures we seek to modify Eq. 3.80, in such a way that bi and tˆi are replaced by b i and tˆ i .

Thermal strains are caused by a change in temperature with respect to the stress-free condition. 78

79 In Thermo- or poro-elasticity problems, the thermal strains and pore pressures are treated as initial strains and stresses, respectively.

In porous media, the total stress σ is equal to the sum of σ and p. This last term is the pore pressures and acts only in the voids of the material, and the eﬀective stresses σ act only on the skeleton of the material, (Terzaghi and Peck 1967). 80

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 3–18

FUNDAMENTAL RELATIONS

81 It must be noted that the pore pressures p being considered in this discussion and throughout the remainder of this course are the steady state pore pressures; excess pore pressures resulting from dilatant behavior in the skeleton of the material are not considered. 82

Step by step generalization of the ﬁeld equations:

Initial thermal strain is caused by a change in temperature ε0ij = α ∆T δij

(3.81)

where ∆T is the temperature (change), α the coeﬃcient of thermal expansion, δij is the Kronecker delta, = D 0 Thermal stress is simply given by σij ijkl εkl or

Net strain ε¯ij = εij − ε0ij or Net eﬀective stress

= α ∆T Dijkl δkl σij

(3.82)

ε¯ij = εkl − α ∆T δkl

(3.83)

= Dijkl (εkl − ε0kl ) σ ¯ij

(3.84)

bi = p,i

(3.85)

0 = −p δij σij

(3.86)

Pore pressure/body force relation

Pore pressure p the pore pressure deﬁned using the compression positive sign convention (the minus sign corrects the discrepancy in the sign conventions)1 Eﬀective body force b i = bi − σij,j

b i = bi − p,i − α Dijkl ∆T,i δkl

(3.87)

tˆ i = tˆi + p ni + α ∆T Dijkl ni δkl

(3.88)

Eﬀective surface traction:

83

The eﬀective stress principle for the case of combined thermo- and poro-elasticity is 0 σij = σij − σij + σij

(3.89)

σ ¯ij

and in the absence of both initial strains and ¯ij Clearly, in the absence of initial stresses σij = σ stresses σij = σij . 1

Pore pressures are typically deﬁned using the sign convention for soil mechanics in which compression is positive, but in the sign convention for standard solid mechanics tension is considered to be positive.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

3.3 Scalar Field: Diﬀusion Equation

3–19

84 The general stress-strain relationship obtained by substituting the constutive law into the eﬀective stress principle is

0 σij = Dijkl (εkl − ε0kl ) + σij

(3.90)

85 When thermal strains and pore pressures are considered in combination the constitutive law is deﬁned as a simple combination of Equations ?? and ??

σij = Dijkl (εkl − α ∆T δkl ) − p δij

(3.91)

86 The equilibrium equation and natural boundary conditions, respectively, can be rewritten in terms of the eﬀective stresses σij

where

3.3

Ω: σij,j + b i = 0 Equilibrium Γt : σij nj − tˆ i = 0 Natural B.C.

(3.92)

b i = bi − p,i − α Dijkl ∆T,i δkl tˆ i = tˆi + p ni + α ∆T Dijkl ni δkl

(3.93)

Scalar Field: Diﬀusion Equation

87 Scalar ﬁeld problems are encountered in almost all branches of engineering and physics. Most of them can be viewed as special forms of the general Helmholtz equation given by ∂ ∂ ∂φ ∂φ ∂φ ∂φ ∂ kx + ky + kz + Q = cρ (3.94) ∂x ∂x ∂y ∂y ∂z ∂z ∂t

where φ(x, y, z) is the ﬁeld variable to be solved. 88

Table 3.1 illustrates selected examples of the diﬀusion equation. Equation Heat Flow Fluid ﬂow through porous media Diﬀusion Saint-Venant Torsion

φ Temperature T Piezometric head h ion concentration Prandtl’s stress function

div(Drφ) + Q = 0 D Q Thermal Heat conductivity supply/sink Permeability Fluid Coeﬃcients supply Permeability Ion coeﬃcients supply 1 Rate of µ twist θ2

q Heat ﬂux Volume ﬂux Ion ﬂux

Constitutive law Fourrier q = −DrT Darcy q = −Drφ Fick q = −Drφ Hooke

Table 3.1: Selected Examples of Diﬀusion Problems

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 3–20

3.3.1 89

FUNDAMENTAL RELATIONS

Heat Transfer

There are three fundamental modes of heat transfer:

Conduction: takes place when a temperature gradient exists within a material and is governed by Fourier’s Law, Fig. 3.5 on Γq :

Figure 3.5: Flux vector ∂T ∂x ∂T = −ky ∂y

qx = −kx

(3.95)

qy

(3.96)

where T = T (x, y) is the temperature ﬁeld in the medium, qx and qy are the componenets of the heat ﬂux (W/m2 or Btu/h-ft2 ), k is the thermal conductivity (W/m.o C or Btu/h∂T ft-o F) and ∂T ∂x , ∂y are the temperature gradients along the x and y respectively. Note that heat ﬂows from “hot” to “cool” zones, hence the negative sign. Convection: heat transfer takes place when a material is exposed to a moving ﬂuid which is at diﬀerent temperature. It is governed by the Newton’s Law of Cooling q = h(T − T∞ ) on Γc

(3.97)

where q is the convective heat ﬂux, h is the convection heat transfer coeﬃcient or ﬁlm coeﬃcient (W/m2 .o C or Btu/h-ft2 .o F). It depends on various factors, such as whether convection is natural or forced, laminar or turbulent ﬂow, type of ﬂuid, and geometry of the body; T and T∞ are the surface and ﬂuid temperature, respectively. This mode is considered as part of the boundary condition. Radiation: is the energy transferred between two separated bodies at diﬀerent temperatures by means of electromagnetic waves. The fundamental law is the Stefan-Boltman’s Law of Thermal Radiation for black bodies in which the ﬂux is proportional to the fourth power of the absolute temperature, which causes the problem to be nonlinear. This mode will not be covered.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

3.3 Scalar Field: Diﬀusion Equation

3.3.2

3–21

Derivation of the Diﬀusion Problem

Consider a solid through which there is a ﬂow q of some quantity, mass, heat, chemical, etc... The rate of transfer per unit area is q which is nothing else than the ﬂux.

90

The direction of ﬂow (i.e. ﬂux) is in the direction of maximum “potential” (such as temperature, piezometric head, or ion concentration) decrease (Fourrier, Darcy, Fick...). Hence the ﬂux is equal to the gradient of the scalar quantity.     ∂φ   q  ∂x   x  ∂φ qy = −D = −D∇φ (3.98) q= ∂y     ∂φ   qz 91

∂z

D is a three by three (symmetric) constitutive/conductivity matrix The conductivity can be either Isotropic

Anisotropic

Orthotropic



 1 0 0 D = k 0 1 0  0 0 1

(3.99)

 kxx kxy kxz D =  kyx kyy kyz  kzx kzy kzz

(3.100)

 0 kxx 0 D =  0 kyy 0  0 0 kzz

(3.101)





Note that for ﬂow through porous media, Darcy’s equation is only valid for laminar ﬂow. 3.3.2.1 92

Simple 2D Derivation

If we consider a unit thickness, 2D diﬀerential body of dimensions dx by dy, Fig. 3.6 then 1. Rate of heat generation/sink is I2 = Qdxdy

(3.102)

2. Heat ﬂux across the boundary of the element is shown in Fig. ?? (note similarity with equilibrium equation) ∂qy ∂qx ∂qy ∂qx dx − qx dx dy + qy + dy − qy dy dx = dxdy + dydx qx + I1 = ∂x ∂y ∂x ∂y (3.103) 3. Change in stored energy is dφ .dxdy (3.104) dt where we deﬁne the speciﬁc heat c as the amount of heat required to raise a unit mass by one degree. I3 = cρ

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 3–22

FUNDAMENTAL RELATIONS

✻q + ∂qy dy y ∂y

qx

✲

qx +

Q

✻

∂qx ∂x dx ✲

dy

❄ ✻ ✛

qy

✲

dx

Figure 3.6: Flux Through Sides of Diﬀerential Element Figure 3.7: *Flow through a surface Γ 93 From the ﬁrst law of thermodynamics, energy produced I2 plus the net energy across the boundary I1 must be equal to the energy absorbed I3 , thus

I1 + I2 − I3 = 0 ∂qy dφ ∂qx dxdy + dydx + Qdxdy − cρ dxdy = 0 ∂x ∂y dt I2

94

(3.105-b)

I3

I1

3.3.2.2

(3.105-a)

†Generalized Derivation

The amount of ﬂow per unit time into an element of volume Ω and surface Γ is q(−n)dΓ = D∇φ.ndΓ I1 = Γ

(3.106)

Γ

where n is the unit exterior normal to Γ, Fig. 3.7 95

Using the divergence theorem

v·ndΓ = Γ

Eq. 3.106 transforms into

div vdΩ

(3.107)

div (D∇φ)dΩ

(3.108)

Ω

I1 =

Ω

Furthermore, if the instantaneous volumetric rate of “heat” generation or removal at a point x, y, z inside Ω is Q(x, y, z, t), then the total amount of heat/ﬂow produced per unit time is Q(x, y, z, t)dΩ (3.109) I2 = 96

Ω

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

3.3 Scalar Field: Diﬀusion Equation

3–23

Finally, we deﬁne the speciﬁc heat of a solid c as the amount of heat required to raise a unit mass by one degree. Thus if ∆φ is a temperature change which occurs in a mass m over a time ∆t, then the corresponding amount of heat that was added must have been cm∆φ, or ρc∆φdΩ (3.110) I3 =

97

Ω

where ρ is the density, Note that another expression of I3 is ∆t(I1 + I2 ). 98 The balance equation, or conservation law states that the energy produced I2 plus the net energy across the boundary I1 must be equal to the energy absorbed I3 , thus

Ω

I1 + I2 − I3 = 0 ∆φ dΩ = 0 div (D∇φ) + Q − ρc ∆t

(3.111-a) (3.111-b)

but since t and Ω are both arbitrary, then ∂φ =0 ∂t

(3.112)

∂φ ∂t

(3.113)

∂φ ∂qx ∂qy + + Q = ρc ∂x ∂y ∂t

(3.114)

div (D∇φ) + Q − ρc or

div (D∇φ) + Q = ρc This equation can be rewritten as

1. Note the similarity between this last equation, and the equation of equilibrium ∂ 2 ux ∂σxx ∂σxy + + ρbx = ρm 2 ∂x ∂y ∂t ∂σxy ∂ 2 uy ∂σyy + + ρby = ρm 2 ∂y ∂x ∂t

(3.115-a) (3.115-b)

2. For steady state problems, the previous equation does not depend on t, and for 2D problems, it reduces to ∂ ∂φ ∂φ ∂ kx + ky +Q=0 (3.116) ∂x ∂x ∂y ∂y 3. For steady state isotropic problems, Q ∂2φ ∂2φ ∂2φ + 2 + 2 =− 2 ∂x ∂y ∂z k

(3.117)

which is Poisson’s equation in 3D.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 3–24

FUNDAMENTAL RELATIONS

4. If the heat input Q = 0, then the previous equation reduces to ∂2φ ∂2φ ∂2φ + 2 + 2 =0 ∂x2 ∂y ∂z

(3.118)

which is an Elliptic (or Laplace) equation. Solutions of Laplace equations are termed harmonic functions (right hand side is zero) which is why Eq. 3.116 is refered to as the quasi-harmonic equation. 5. If the function depends only on x and t, then we obtain ∂ ∂φ ∂φ = kx +Q ρc ∂t ∂x ∂x

(3.119)

which is a parabolic (or Heat) equation. 3.3.2.3

Boundary Conditions

The boundary conditions, for thermal problems, are mainly of three kinds: Essential: Temperature prespeciﬁed on ΓT Natural: Flux Speciﬁed Flux prescribed on Γq , qn = cst Convection Flux prescribed on Γc , qc = h(T − T∞ ). Note this type of boundary condition is analogous to the one in structural mechanics where we have an inclined support on rollers. similar boundary conditions can be written for ﬂuid ﬂow.

3.4 99

Summary and Tonti Diagrams

The analogy between scalar and vector problems is shown in Table 3.2.

To graphically illustrate the inter-relationship between ﬁeld equations and variables, Tonti diagrams, Fig. 3.8 have been used. 100

Specified Primary Variable

Essential B.C. (EBC)

Primary Variable

Kinematic Equation (KE) Intermediate Variable

Field Equations

Constitutive Equations (CE)

Source Function

Balance Equations (BE) Flux Variable

Natural B.C. (NBC)

Specified Flux Variable

Figure 3.8: Components of Tonti’s Diagram, (Felippa 2000)

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

3.4 Summary and Tonti Diagrams

3–25

Essential B.C. ui : Γu

❄

Body Forces

State variable

bi

ui

❄

❄

Balance

Kinematics

LT + ρb = 0

" = Lu

❄ Flux variable σij

❄ ✲

Constitutive Rel. σij = Dijkl εkl

✛

Interm. variable εij

✻

Natural B.C. t i : Γt Heat Supply

Essential B.C.

Q

T : ΓT

❄

❄

Balance r.q

State variable

= −Q

T

❄ Flux variable q

❄ ✲

Constitutive Rel. q = −Dg

✛

Interm. variable g=rT

✻

Natural B.C. qn : Γq ∪ Γc

Figure 3.9: Fundamental Equations of Solid Mechanics and Heat Flow

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 3–26

FUNDAMENTAL RELATIONS Scalar Conduction T g q div q + Q = 0 g = ∇T q = −Dg Fourrier

Vector Solid Mechanics Variables u ε σ Field Equations LT σ + b = 0 ε = Lu σ =Dε

State Intermediate Flux Balance kinematic Constitutive

Hooke Boundary Conditions T on ΓT u on Γu Essential BC qn = qT n on Γq ∪ Γc t = σn on Γt Natural BC (Flux) Table 3.2: Comparison of Scalar and Vector Field Problems 101 However, we here expand on the standard Tonti diagram to graphically show the fundamental equations of solid mechanics and heat ﬂow in Fig. 3.9.

Similarly, many other physical problems can be solved by the ﬁnite element method, Table 3.3. 102

Victor Saouma

Finite Elements II; Solid Mechanics

Victor Saouma

Magnetostatic

Flow through porous media Electrostatics

Fluid ﬂow

Heat transfer

Electric potential Magnetic potential

Conservation of electric ﬂux Conservation of magnetic potential

Magnetic ﬂux

Electric ﬂux

Flow rate

Shear stress

Magnetic permeability

Permitivity

Permeability

Thermal conductivity Viscosity

Material Constants Young’s Modulus, Poisson’s ratio Electric conductivity Shear

Current

Charge

Fluid sources

External electric charge -2× angle of twist Internal or external heat Body forces

Body force or traction

Source

Table 3.3: Classiﬁcation of various Physical Problems, (Kardestuncer 1987)

Hydraulic head

Velocity

Heat ﬂux

Rate of twist

Electric ﬂux

Stress or strain

Displacement or forces Voltage or intensity Stress or warping function Temperature

Flux

State Vriable

Conservation of momentum Conservation of mass

Equilibrium of currents Conservation of potential energy Conservation of energy

Electric work Torsion

net-

Conservation Principle Equilibrium of forces

Physical Problem Deformation of an elastic body

Maxwell’s law

Coulomb’s law

Darcy’s law

Stoke’s law

Fourrier’s law

Hooke’s law

Kirchhoﬀ’s law

Constitutive Equation Hooke’s law

Draft 3.4 Summary and Tonti Diagrams

3–27

Finite Elements II; Solid Mechanics

Draft 3–28

Victor Saouma

FUNDAMENTAL RELATIONS

Finite Elements II; Solid Mechanics

Draft Chapter 4

MESH GENERATION Requires further editing

4.1

Introduction

1 Finite element mesh generation is now an integral part of a ﬁnite element analysis. With the increased computational capabilities, increasingly more complex structures are being analysed. Those structures must be discretized. 2 The task is one of developing a mathematical model (discretization or tessalation) of a continuum model. This is not only necessary in ﬁnite elment analysis, but in computer graphics/rendering also. 3 In computer graphics, we focus on the boundary representation, and assign colors and shades on the basis of light source and outward normal direction of the polygon.

Hence, in the most general case, meshing can be deﬁned as the process of breaking up a physical domain into smaller sub-domains (elements) in order to facilitate the numerical solution of a partial diﬀerential equation. Surface domains may be subdivided into triangle or quadrilateral shapes, while volumes may be subdivided primarily into tetrahedra or hexahedra shapes. The shape and distribution of the elements is ideally deﬁned by automatic meshing algorithms. 4

1. Point placement, followed by triangularization (discussed below). 2. Sub-domain removal. Elements are gradually removed from the domain, one ata time, until the whole domain is decomposed int ﬁnite elements. 3. Recursive subdivision. The domain is broken into simpler parts until the individual parts are single elements or simple regions, that can be meshed directly, for instance by the conformal mapping algorithm. 4. Hierarchical decomposition. The basic principle of a quadtree (or hierarchical decomposition) is to cover a planar region of interest by a square, then recursively partition squares into smaller squares until each square contains a suitably uniform subset of the input.

Draft 4–2

4.2

MESH GENERATION

Triangulation

5 The concept of Voronoi diagrams ﬁrst appeared in works of Descartes as early as 1644. Descartes used Voronoi-like diagrams to show the disposition of matter in the solar system and its environs. 6 The ﬁrst man who studied the Voronoi diagram as a concept was a German mathematician G. L. Dirichlet. He studied the two- and three dimensional case and that is why this concept is also known as Dirichlet tessellation. However it is much better known as a Voronoi diagram because another German mathematician M. G. Voronoi in 1908 studied the concept and deﬁned it for a more general n-dimensional case. 7 Very soon after it was deﬁned by Voronoi it was developed independently in other areas like meteorology and crystalography. Thiessen developed it in meteorology in 1911 as an aid to computing more accurate estimates of regional rainfall averages. In the ﬁeld of crystalography German researchers dominated and Niggli in 1927 introduced the term Wirkungsbereich (area of inﬂuence) as a reference to a Voronoi diagram.

During the years this concept kept being rediscovered over and over again in diﬀerent ﬁelds of science and today it is extensively used in about 15 diﬀerent ﬁelds of sciences. Some of them being mathematics, computer science, biology, cartography, physiology and many others. 8

Vb

Va ρ

a

b

c

Vc 1

2

3

4

4f

5 6

6f

Voronoi Delaunay

Figure 4.1: Voronoi and Delaunay Tessellation

4.2.1

Voronoi Polygon

9 Given a ﬁnite set of poits in the plane, the idea is to assign to each point a region of inﬂuence in such a way that the regions decompose the plane, (Voronoi 1907).

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

4.3 Finite Element Mesh Generation

4–3

To describe a speciﬁc way to do that, let S be a subset of R 2 (S ⊆ R 2 ). We deﬁne the Voronoi region of p ∈ S as the set of points x ∈ R that are at least as close to p as to any other points in S: * + Vp = x ∈ R2 | ||x − p|| ≤ x − 1 , ∀q ∈ S (4.1) 10

Each point x ∈ R 2 has at least one nearest point in S, so it lies in at least one Voronoi region. Two Voronoi regions lie on opposite sides of the perpendicular bisector separating the two generating points.

4.2.2

Delaunay Triangulation

The dual of the Voronoi diagram is obtained by drawing straight Delaunay edges connecting points p, q ∈ S if and only if their Voronoi regions intersect along a common line segment. Thus in general, the Delaunay edges decompose the convex hull of S into triangular regions which are referred to as Delaunay triangles, (Delaunay 1934). 11

12 Using Euler’s relation, it can be shown that a planar graph with n ≥ 3 vertices has at most 3n − 6 edges and at most 2n − 4 faces. THe same bounds hold for the number of Delaunay edges and triangles.

Each Voronoi vertex u = Va ∩ Vb ∩ Vc is the center of a circle with radius ρ = u − a = u − b = u − c . The circle is empty because it encloses no point of S. 13

14 Additional detailed information on Voronoi tesselation, (Okabe, Boots, Sugihara and Chiu 2000) is an excellent reference.

4.2.3

MATLAB Code

rand(’state’,4); x = rand(1,3); y = rand(1,3); TRI = delaunay(x,y); subplot(1,2,1),... trimesh(TRI,x,y,zeros(size(x))); view(2),... axis([0 1 0 1]); hold on; plot(x,y,’o’); set(gca,’box’,’on’); [vx, vy] = voronoi(x,y,TRI); subplot(1,2,2),... plot(x,y,’r+’,vx,vy,’b-’),... axis([0 1 0 1])

4.3 4.3.1

Finite Element Mesh Generation Boundary Deﬁnition

15 In order to discretize the continuum into a ﬁnite element mesh, ﬁrst key geometrical information of the must be speciﬁed hierarchically:

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 4–4

MESH GENERATION

Vertices: with nodal coordinates, and approximate desired element size in the immediate vicinity (thus describing the mesh density). Edges: which connect vertices. Those can be either linear segments, polylines, or curves. Surfaces: composed of edges, deﬁned counterclockwise. Volumes: (3D only) composed of surfaces. 16

Associated with surfaces (2D0, or volumes (3D) are diﬀerent material properties.

17

Examples of the hierarchical boundary deﬁnition is shown in Fig. 4.2 and 4.3.

Figure 4.2: Control Point for a 2D Mesh

4.3.2

Interior Node Generation

18 Once the boundary has been deﬁned, we need to insert internal nodes at a spacing which respect the required mesh density. There are numerous techniques to insert those internal nodes. We present one approach, FIg. 4.4

1. Decompose the region into a disjoint ensemble of subregions with equal mesh density. 2. Shrink the mesh to avoid elements near the boundary with very acute interior angles. 3. Starting with the ﬁrst zone, circumscribe it by the smallest possible rectangle. 4. Superimpose a square rectangular grid over the circumscribing rectangle. Victor Saouma

Finite Elements II; Solid Mechanics

Draft

4.3 Finite Element Mesh Generation

4–5

Figure 4.3: Control Point for a 3D Mesh

Boundary discretization

Zone II, r2

Zone III, r3

Zone I, r1

Region to be meshed

Zonal decomposition with nodal density

Boundary shrinking by ε

r Generation of internal nodes in zone I within shrunk boundary

Disk r

Figure 4.4: A Two Dimensional Triangularization AlgorithmControl Point for a 3D Mesh

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 4–6

MESH GENERATION

5. Use a random number generator to randomly generate one interior node in each square. A disk of radius r centered at each node is used to test that no other surrounding nodes are enclosed in the disk. If so, the node in question is regenerated. 19

ˇ An alternative approach consists in, (Cervenka, J. 1994) 1. Generating a triangularization compatible with the initial nodes. 2. Check lengths of the edges. If an edge does not satisfy the prescribed size r, a new node is inserted in the center of the edge. The prescribed size is interpolated between those of the vertices at each end of the edge. 3. Repeat this operation until convergence. 4. Smoothen the elemnts to assure appropriate aspect ratios.

4.3.3 20

Final Triangularization

With boundary and interior nodes generated: 1. Determine the Voronoi polygons 2. Perform a Delaunay triangularization 3. Smoothen the mesh to ensure that all generated elements have a satisfactory aspect ratio.

21 It should be noted that recent algorithms, which can generate quadrilateral elements out of the Delaunay triangularization have recently emerged.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft Chapter 5

VARIATIONAL and RAYLEIGH-RITZ METHODS Adapted from (Reich 1993)

5.1

Multiﬁeld Variational Principles

A Multiﬁeld variational principle is one that has more than one master ﬁeld (or state variable), that is more than one unknown ﬁeld is subject to independent variations. 1

2 In linear elastostatics, we can have displacement, u, strains ε, or stress σ as potential candidates for master ﬁelds. Hence seven combinations are possible, (Felippa 2000), Table 5.1.

3 In this course, we shall focus on only the Total Potential Energy, and the Hu-Washizu variational principles.

u

ε

σ

Y Y Y Y Y

Y

Y Y Y

Y

Y

Y

Name Single Field Total Potential Energy Total Complementary Potential Energy No name Two Fields Hellinger-Reissner de Veubeke No name Three Fields Hu-Washizu

Table 5.1: Functionals in Linear Elasticity

Draft 5–2

5.2

Total Potential Energy Principle

5.2.1 4

VARIATIONAL and RAYLEIGH-RITZ METHODS

Static; Euler

The expression for the total potential energy (TPE) is given by Π We U

def

=

def

=

def

=

U − We uT bdΩ + uT ˆtdΓ + uP Ω Γt 1 T T D dΩ − D 0 dΩ + T σ 0 dΩ 2 Ω Ω Ω

(5.1-a) (5.1-b) (5.1-c)

the functional for the general form of the potential energy variational principle is obtained 1 Π= 2

D dΩ − T

Ω

D 0 dΩ + Ω T

U

σ 0 dΩ − Ω T

u bdΩ − uT ˆtdΓ − uP Ω Γt T

(5.2)

−We

5 A variational statement is obtained by taking the ﬁrst variation of Π and setting this scalar quantity equal to zero.

The variational statement for the general form of the potential energy functional (i.e. Equation 5.2) is T T T T δ D dΩ − δ D 0 dΩ + δ σ 0 dΩ − δu bdΩ − δuT ˆtdΓ = 0 (5.3) δΠ = 6

Ω

Ω

Ω

Ω

Γt

which is the Principle of Virtual Work. 7 We rewrite the strain-displacement relations in terms of a linear diﬀerential operator L (Eq. 3.41) = Lu (5.4)

where L is a linear diﬀerential operator and u is the  ∂ 0 ∂x ∂  0  ∂y  0 0  L= ∂ ∂  ∂y ∂x  ∂  ∂z 0 ∂ 0 ∂z

displacement vector deﬁned in Eq. 3.42:  0 0   ∂  ∂z  (5.5)  0   ∂  ∂x ∂ ∂y

8 Since the diﬀerential operator L is linear, the variation of the strains δ can be expressed in terms of the variation of the displacements δu

δ = δ(Lu) = Lδu 9

(5.6)

We consider two forms of the variational statement

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

5.2 Total Potential Energy Principle

5–3

1. In terms of strains: which is more suitable for the derivation of the corresponding Euler equations,

δ(Lu) D dΩ − T

δΠ =

T

δ(Lu) D 0 dΩ + Ω

Ω δU − δuT bdΩ − δuT ˆtdΓ = 0 Ω Γt

δ(Lu)T σ 0 dΩ Ω (5.7)

−δWe

2. In terms of displacements: (using Equation 5.7) which is more suitable for the subsequent discretization.

δ(Lu) D(Lu)dΩ − T

δΠ =

Ω

Ω

−

δuT bdΩ −

δ(Lu) D 0 dΩ +

Ω

δ(Lu)T σ 0 dΩ

δU

Ω

T

Γt

−δWe

δuT ˆtdΓ = 0

(5.8)

10 To obtain the Euler equations for the general form of the potential energy variational principle the volume integrals deﬁning the virtual strain energy δU in Equation 5.7 must be integrated by parts in order to convert the variation of the strains δ(Lu) into a variation of the displacements δu. 11

Integration by parts of these integrals using Green’s theorem (Eq. 1.46) , ∂S ∂R SdΩ + RSnx dΓ R dΩ = − ∂x Ω Ω ∂x Γ

(5.9)

yields ,

T

δ(Lu) D dΩ =

Ω

δu G(D )dΓ − T

,Γ

δuT LT (D )dΩ

δ(Lu)T D 0 dΩ = δuT G(D 0 )dΓ − δuT LT (D 0 )dΩ Ω ,Γ Ω T T δ(Lu) σ 0 dΩ = δu Gσ 0 dΓ − δuT LT σ0 dΩ Ω

Γ

(5.10-a)

Ω (5.10-b) (5.10-c)

Ω

where G is a transformation matrix containing the direction cosines for a unit normal vector such that the surface tractions t are deﬁned as t = Gσ and the surface integrals are over the entire surface of the body Γ. 12

Substituting Equation 5.10-a into Equation 5.7, the variational statement becomes δΠ = − δuT {LT [D( − 0 ) + σ 0 ] +b}dΩ Ω σ

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 5–4

VARIATIONAL and RAYLEIGH-RITZ METHODS

ε

u

u

u

σ

b

^ t

^u

ˇ Figure 5.1: Tonti Diagram for the Total Potential Energy, (Cervenka, J. 1994) +

δuT {G [D( − 0 ) + σ 0 ] −ˆt}dΓ = 0 Γ σ

(5.11)

13 Since δu is arbitrary the expressions in the integrands within the braces must both be equal to zero for δΠ to be equal to zero. Recognizing that the stress-strain relationship appears in both the volume and surface integrals, the Euler equations are

(BE): Equilibrium (NBC): Natural B.C.

LT σ + b = 0 on Ω Gσ − ˆt = 0 on Γt

(5.12) (5.13)

where the ﬁrst Euler equation is the equilibrium equation and the second Euler equation deﬁnes the natural boundary conditions. The natural boundary conditions are deﬁned on Γt rather than Γ because both the applied surface tractions ˆt and the matrix-vector product Gσ are identically zero outside Γt . In general, only certain forms of diﬀerential equations are Euler equations of a variational functional. 14

15 For the elastostatic problem, it is possible to start from the Euler equations, and then derive the total potential energy functional by performing the operations just presented in reverse order. 16 The Tonti diagram for the TPE is shown in Fig. 5.2. In this diagram, strong connections are shown by solid lines, weak connections by spring-like symbols, boxes with solid lines denote the primary unknown ﬁeld, variables inside dashed boxes are internally derived ﬁelds, and shaded boxes indicate prescribed ﬁelds.

5.2.2

Dynamic; Euler/Lagrange

In dynamics, we deﬁne the Lagrangian function as the diﬀerence between the kinetic and the potential energies L≡K −Π (5.14) 17

and Hamilton’s principle states that The motion of a particle acted on by conservative forces between two arbitrary instants of time t1 and t2 is such that the line integral over the Lagrangian function is an extremum for the path motion. t2 Ldt (5.15) I≡ t1

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

5.2 Total Potential Energy Principle

5–5

is an extremum. If the path can be deﬁned in terms of generalized coordinates qi (i = 1, 2, 3), then it can be shown that t2 L(q1 , q2 , q3 , q˙1 , q˙2 , q˙3 )dt = 0 (5.16-a) δI = δ 18

t1

t2

=

3

∂L

t1 i=1

∂qi

−

d dt

∂L ∂ q˙i

δqi dt

(5.16-b)

If all qi are linearly independent (i.e. no constraints among them), the variations δqi are independent of t, except δqi = 0 at t1 and t2 . Therefore, the coeﬃcients of δq1 , δq2 , and δq3 vanish separately yielding the Lagrangian equations of motion, also known as the EulerLagrange equations. d ∂L ∂L (5.17) − = 0, i = 1, 2, 3 ∂qi dt ∂ q˙i 19

Hamilton’s principle can be expressed as: t2 (K − Π)dt + ΠH = δΠH

t1 t2

=

δ(K − Π)dt +

t1

t2

Wnc dt

(5.18)

t1 t2

δWnc dt = 0

(5.19)

t1

where K is the kinetic energy, Π the potential energy, and Wnc the work done by nonconservative forces acting on the system (including damping). It should be noted that TPE is a special case of Hamilton’s principle in the absence of kinetic energy. The Kinetic energy is given by ρ ∂u ∂u . dΩ (5.20) K= Ω 2 ∂t ∂t A conservative force is one for which the sum of the potential and kinetic energies is conserved).

Example 5-1: Hamilton’s Principle For a uniform cross section bar of length L, cross sectional area A, Young’s modulus E, and mass density ρ ﬁxed at one end, and connected to a rigid support at the other by a spring with stiﬀness k

k A, E L Victor Saouma

Finite Elements II; Solid Mechanics

Draft 5–6

VARIATIONAL and RAYLEIGH-RITZ METHODS

1. Show that the kinetic energy and the strain energy are given by ∂u 2 dx K = ∂t L k EA ∂u 2 dΩ + [u(L)]2 U = ∂x 2 0 2

L

ρA 0 2

(5.21) (5.22)

2. For Wnc = 0 derive an expression for the ﬁrst variation of the Hamilton’s functional δΠH 3. If we are interested in determining the periodic motion, which has the form u(x, t) = u0 (x)eiωt

(5.23)

where ω is the frequency of natural vibration, and u0 (x) is the amplitude. Show that # 2 $ L du 1 k 0 ρAω 2 (u0 )2 − EA dx − [u0 (L)]2 = 0 (5.24) δ dx 2 0 2 4. Show that the Euler equations of the preceding functional are du0 d EA + ρAω 2 u0 = 0 for 0 < x < L dx dx du0 + ku0 = 0 for x = L EA dx 5. Rewrite Eq. 5.24 in terms of x =

x L,

u=

u0 L,

α=

kL EA ,

λ=

(5.25) (5.26)

ω 2 ρL2 E

6. Show that if this problem was to be solved by the Rayleigh-Ritz method with u = c1 x + c2 x2

(5.27)

then a nontrivial solution c1 = 0 and c2 = 0 exists if an only if 15λ2 − 640λ + 2400 = 0

(5.28)

7. Solve for ω1 and ω2 , and compare with the exact solution given by λ + tan λ = 0

(5.29)

Solution: 1. Kinetic energy L ∂u 2 ρA ∂u 2 dΩ = dx K = ∂t ∂t 0 2 L k E ∂u 2 EA ∂u 2 dΩ = dx + [u(L)]2 U = 2 ∂x 2 ∂x 2 0 Ω

Victor Saouma

ρ Ω2

(5.30) (5.31)

Finite Elements II; Solid Mechanics

Draft

5.2 Total Potential Energy Principle

5–7

2. For Wnc = 0 δΠH

=

t2

δ(K − Π)dt 2 $ t2 L # 2 ∂u 1 ∂u ρA dx − k[u(L)]2 δ − EA 2 ∂t ∂x 0 t1

(5.32)

t1

=

dt

(5.33)

3. If we adopt u(x, t) = u0 (x)eiωt

(5.34)

where ω is the frequency of natural vibration, and u0 (x) is the amplitude. Then substituting, we obtain # $ L k ∂u0 2 1 2 2 dx − [u0 (L)]2 = 0 (5.35) δ ρAω [u0 (x)] − EA 2 ∂x 2 0

t2

where

exp2iωt dt being nonzero is factored out.

t1

4. To determine the corresponding Euler Equation # $ L du0 2 1 k 2 2 ρAω (u0 ) − EA dx − [u0 (L)]2 = δ 2 dx 2 0 L ∂u0 ∂u0 δ dx − ku0 (L)δu0 (L) = 0 ρAω 2 u0 δu0 − EA ∂x ∂x 0

(5.36) (5.37)

but L L ∂u0 ∂u0 ∂u0 EA δ dx = EA δu0 − ∂x ∂x ∂x 0 v v

u

Substituting

u

L 0

∂u0 ∂ EA δu0 dx ∂x ∂x v

(5.38)

u

0

∂u0 ∂u0 δ dx − ku0 (L)δu0 (L) (5.39) ρAω [u0 (x)] − EA ∂x ∂x 0 L L ∂u0 du0 d EA − EA δu0 − ku0 (L)δu0 (L) = 0(5.40) ρAω[u0 (x)] + = dx ∂x dx 0 0 L

2

0

or d dx

Victor Saouma

du0 EA + ρAω 2 u0 = 0 for 0 < x < L dx du0 + ku0 = 0 for x = L EA dx

(5.41) (5.42)

Finite Elements II; Solid Mechanics

Draft 5–8

VARIATIONAL and RAYLEIGH-RITZ METHODS 2

2

kL dx = Ldx, u = uL0 , and α = EA , λ = ω EρL we have 2 $ L # 1 du k 0 ρAω 2 (u0 )2 − EA dx − [u0 (L)]2 dx 2 0 2 2 $ 1 # du0 k 1 λEA (Lu)2 − EA Ldx − [Lu0 (1)]2 = 2 L dx 2 02 # 2 $ 1 du0 kL2 1 EAL λ(u)2 − [u0 (1)]2 dx − = 2 dx 2 0 # $ 1 λ 2 1 du0 2 KL2 dx − [u0 (1)]2 = (u) − 2 α α dx 0 # # $ $ 1 KL2 λ 2 1 du0 2 2 (u) − ⇒ δ dx − [u0 (1)] =0 2 α dx 0 α # $ du0 2 α 1 1 2 λ(u) − dx − [u0 (1)]2 = 0 δ 2 0 dx 2

5. Substituting x =

x L,

(5.43) (5.44) (5.45) (5.46) (5.47) (5.48)

6. Using the Rayleigh-Ritz method with u = c1 x + c2 x2

(5.49)

and rewriting the previous equation in terms of u # 2 $ du α 1 1 λ(u)2 − dx − [u(1)]2 = 0 δ 2 0 dx 2 1 " ! λ(c1 x + c2 x2 )(δc1 x + δc2 x2 ) − (c1 + 2c2 x)(δc1 + 2δc2 x) dx

(5.50)

0

−(c1 + c2 )(δc1 + δc − 2) = 0

(5.51)

Collecting the coeﬃcients of δc1 and δc2 and setting them to zero separately, we have 1 " ! (5.52) λ(c1 x + c2 x2 )x − (c1 + 2c2 x) dx − (c1 + c2 ) = 0 δc1 : 0 1 " ! (5.53) λ(c1 x + c2 x2 )x2 − (c1 + 2c2 x)2x dx − (c1 + c2 ) = 0 δc2 : 0

Integrating, reducing and compacting 2 2 −λ 2 73

1 3 1 4

A nontrivial solution exists if and only if 2− λ 3 2− λ 4

1 4 1 5

c1 c2

2 − λ4 7 λ =0 3 − 5

=

0 0

(5.54)

(5.55)

or 15λ2 − 640λ + 2400 = 0 Victor Saouma

(5.56)

Finite Elements II; Solid Mechanics

Draft

5.3 General Hu-Washizu Variational Principle

5–9

This quadratic equation has two roots λ1 = 4.1545

λ2 = 38.512

(5.57)

which correspond to 2.038 ω1 = L

E ρ

6.206 ω2 = L

E ρ

(5.58)

7. The exact solution is given by given by λ + tan λ = 0

(5.59)

whose ﬁrst two roots are 2.02875 ω1 = L

5.3

E ρ

4.91318 ω2 = L

E ρ

(5.60)

General Hu-Washizu Variational Principle

20 The Hu-Washizu (HW) variational principle is a three-ﬁeld variational principle in which the displacements, strains, and stresses are treated as independent ﬁelds (as opposed to only the displacement in the total potential energy principle). 21 Naturally, the two additional ﬁeld variables, with respect to the TPE variational principle, appear not only in the functional, but also in the discretized system of equations. Consequently, for a domain with a given discretization the discrete system of equations derived from the HW variational principle will be much larger than the discrete system of equations derived from the TPE variational principle. 22 With the increased number of equations, signiﬁcant improvements in accuracy can be observed for the solution obtained from the discrete form of the HW variational principle compared to the solution obtained from the discrete form of the TPE variational principle for the same discretization. This means that coarse discretizations can be used with the discrete form of the HW variational principle to obtain the same degree of accuracy that would be observed with much ﬁner discretizations using the TPE variational principle. 23 The functional for the HW variational principle is derived from the functional for the TPE variational principle by imposing the strain-displacement equation as a ﬁnite subsidiary condition using the method of Lagrange multipliers. 24

The ﬁnite subsidiary condition or constraint is written in residual form as Lu − = 0

(5.61)

and enforced in an average sense over the entire body Ω.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 5–10

VARIATIONAL and RAYLEIGH-RITZ METHODS

By imposing the strain-displacement equation as a constraint C 0 continuous strains and stresses are obtained in the discrete form of the varitional statements, as opposed to the discontiuous strains and stresses obtained in the discrete form of the varitional statement for the TPE variational principle. 25

26

The constrained functional is written as

ΠHW = ΠTPE +

λT (L u − ) dΩ

(5.62)

Ω

Where λ is the Lagrange multiplier and to be consistent with the integrals in the TPE functional (i.e. Equation 5.2) the Lagrange multiplier must have the units of stress. Since this is the case, σ will be used for the Lagrange multiplier instead of the more typical λ such that the physical meaning of the Lagrange multiplier is more apparent. 27

The functional for the HW variational principle thus becomes 1 T T D dΩ − D 0 dΩ + T σ 0 dΩ ΠHW = 2 Ω Ω Ω U − uT b dΩ − uT ˆt dΓ + σT (L u − ) dΩ Γt Ω Ω −We Constraint

(5.63)

28 A variational statement is obtained by taking the ﬁrst variation of the functional and setting this scalar quantity equal to zero.

The ﬁrst variation of the HW functional, with terms arranged according to which ﬁeld variable is varied, is T T δ(L u) σ dΩ − δu b dΩ − δuT ˆt dΓ δΠHW = Ω Ω Γ t (5.64) δ T D dΩ − δ T D 0 dΩ + δ T σ 0 dΩ − δ T σ dΩ + Ω Ω Ω Ω δσ T (L u − ) dΩ = 0 + 29

Ω

Note that the 4th and 7th term were added and cancell each others, and that we are not using Eq. 5.6 in this formulation. Since u, , and σ are independent ﬁeld variables, terms involving δu, δ , and δσ must add up to zero individually and are, therefore grouped together to form three separate variational statements (analogous to the method of separation of variables in the solution of partial diﬀerential equations) δ(Lu)T σ dΩ − δuT b dΩ − δuT ˆt dΓ = 0 (5.65) Ω Ω Γt δ T [D( − 0 ) + σ0 − σ] dΩ = 0 (5.66) Ω δσ T (L u − ) dΩ = 0 (5.67) 30

Ω

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

5.3 General Hu-Washizu Variational Principle

5–11

31 To obtain the corresponding Euler equations for the general form of the HW variational principle the volume integral in Equation 5.65 containing the variation of the strains δ(Lu) deﬁned in terms of the displacements u must be integrated by parts using Green’s theorem in order to obtain a form of the variational statement in terms of the variation of the displacements δu. 32

Integration by parts (Eq. 1.39) of this integral yields , T T δ(L u) σ dΩ = δu G σ dΓ − δuT LT σ dΩ Ω

Γ

(5.68)

Ω

where G is a transformation matrix containing the direction cosines for a unit normal vector such that the surface tractions t are deﬁned as t = G σ and the surface integral is over the entire surface of the body Γ. 33

Substituting Equation 5.68 into Equation 5.65, the ﬁrst variational statement becomes T T δuT (G σ − ˆt) dΓ = 0 (5.69) − δu (L σ + b) dΩ + Ω

Γt

Since δu is arbitrary the expressions in the integrands within the parentheses must both be equal to zero for the sum of the integrals to be equal to zero. 34 Likewise, δ and δσ are also arbitrary and the expressions within the braces in the second variational statement (i.e. Equation 5.66) and within the parentheses in the third variational statement (i.e. Equation 5.67) must both be equal to zero for the integral to be equal to zero. 35

The Euler equations for the HW functional are (BE): Equilibrium (CE): Stress-Strain (KE): Strain-Displacement (NBC): Natural B.C.

LT σ + b = 0

on Ω

(5.70)

D ( − 0 ) + σ0 − σ = 0 on Ω

(5.71)

Lu − = 0 G σ − ˆt = 0

on Ω

(5.72)

on Γt

(5.73)

where the ﬁrst Euler equation is the equilibrium equation; the second Euler equation is the stress-strain relationship; the third Euler equation is the strain-displacement equation; and the fourth Euler equation deﬁnes the natural boundary conditions. The natural boundary conditions are deﬁned on Γt rather than Γ because both the applied surface tractions ˆt and the matrix-vector product G σ are identically zero outside Γt . Starting from the Euler equations, it is possible to derive the HW functional by performing the operations just presented in reverse order. 36

37 This last set of four Euler equations, should be compared with the two (Eq. 5.12 and 5.13) obtained from the original TPE. The additional two equations bring into play stress-strain and strain displacement. Also, whereas the original formulation (Eq. 5.12 and 5.13) was in terms of the displacement only (u), the Hu-Washizu formulation is in terms of three independent variables (u, σ and ), Table 5.2. 38

The Tonti diagram for the HW is shown in Fig. ??.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 5–12

VARIATIONAL and RAYLEIGH-RITZ METHODS

Equilibrium Stress-Strain Strain-Displacement Natural B.C.

Euler Equations LT σ + b = 0 D ( − 0 ) + σ 0 − σ = 0 Lu − = 0 G σ − ˆt = 0 Variables Displacement Strain Stress

TPE

HW

Ω Ω Ω Γt

Y N N Y

Y Y Y Y

u σ

Y N N

Y Y Y

Table 5.2: Comparison Between Total Potential Energy and Hu-Washizu Formulations

u

ε

^ u

ε

u

σ

ε

u

σ

σ

σ

ε

b

^ t

ˇ Figure 5.2: Tonti Diagram for Hu-Washizu, (Cervenka, J. 1994)

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

5.4 Rayleigh Ritz

5.4

5–13

Rayleigh Ritz

39 In the principle of virtual displacement (or minimum total potential energy), the Euler equations are the equilibrium equations. The Euler equations are usually in the form of diﬀerential equations that are not always solvable by exact methods of solutions.

Indeed we do have available a number of approximate methods such as the ﬁnite diﬀerence but an alternative approach is to bypass entirely the Euler equations and start directly from a variational statement of the problem to the solution of the Euler equation.

40

41 Such a direct method was ﬁrst proposed by Lord Rayleigh, and then independently generalized by Ritz.

Hence, this section will brieﬂy review the Rayleigh-Ritz method (which solves the variational problem, but which does not require a topological discretization (i.e. ﬁnite element). The similarity between those two approaches should by now be apparent. In both cases we are dealing with an approximate method. 42

43 Application of the principle of total potential energy (or virtual displacement) requires an assumed displacement ﬁeld. This displacement ﬁeld can be approximated by interpolation functions written in terms of:

1. Unknown polynomial coeﬃcients most appropriate for continuous systems, and the Rayleigh-Ritz method (5.74) y = a1 + a2 x + a3 x2 + a4 x3 A major drawback of this approach, is that the coeﬃcients have no physical meaning. 2. Unknown nodal deformations most appropriate for discrete systems and Potential Energy based formulations y = u = N1 u1 + N2 u2 + . . . + Nn un

(5.75)

44 This chapter will focus on the ﬁrst type of approximation, whereas subsequent ones will adopt an approximation based on the nodal deformations. 45

In general, we may adopt an approximate displacement ﬁeld by a function u1 ≈ u2 ≈ u3 ≈

n

i=1 n

i=1 n

c1i φ1i + φ10

(5.76-a)

c2i φ2i + φ20

(5.76-b)

c3i φ3i + φ30

(5.76-c)

i=1

where cji denote undetermined parameters, and φ are appropriate functions of positions. 46

φ should satisfy three conditions 1. Be continous.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 5–14

VARIATIONAL and RAYLEIGH-RITZ METHODS

2. Must be admissible, i.e. satisfy the essential boundary conditions (the natural boundary conditions are included already in the variational statement. However, if φ also satisfy them, then better results are achieved). 3. Must be independent and complete (which means that the exact displacement and their derivatives that appear in Π can be arbitrary matched if enough terms are used. Furthermore, lowest order terms must also be included). In general φ is a polynomial or trigonometric function. We determine the parameters cji by requiring that the principle of virtual work for arbitrary variations δcji . or 47

δΠ(u1 , u2 , u3 ) =

n

∂Π i=1

δc1 ∂c1i i

∂Π ∂Π + 2 δc2i + 3 δc3i ∂ci ∂ci

=0

(5.77)

for arbitrary and independent variations of δc1i , δc2i , and δc3i , thus it follows that ∂Π ∂cji

=0

i = 1, 2, · · · , n; j = 1, 2, 3

(5.78)

Thus we obtain a total of 3n linearly independent simultaneous equations. From these displacements, we can then determine strains and stresses (or internal forces). Hence we have replaced a problem with an inﬁnite number of d.o.f by one with a ﬁnite number. 48

Some general observations 1. cji can either be a set of coeﬃcients with no physical meanings, or variables associated with nodal generalized displacements (such as deﬂection or displacement). 2. If the coordinate functions φ satisfy the above requirements, then the solution converges to the exact one if n increases. 3. For increasing values of n, the previously computed coeﬃcients remain unchanged. 4. Since the strains are computed from the approximate displacements, strains and stresses are generally less accurate than the displacements. 5. The equilibrium equations of the problem are satisﬁed only in the energy sense δΠ = 0 and not in the diﬀerential equation sense (i.e. in the weak form but not in the strong one). Therefore the displacements obtained from the approximation generaly do not satisfy the equations of equilibrium. 6. Since the continuous system is approximated by a ﬁnite number of coordinates (or d.o.f.), then the approximate system is stiﬀer than the actual one, and the displacements obtained from the Ritz method converge to the exact ones from below. 7. If the functional Π is quadratic, then ∂Π ≡ Kc + f = 0 ∂c

(5.79)

and K will always be symmetric Victor Saouma

Finite Elements II; Solid Mechanics

Draft

5.4 Rayleigh Ritz

5–15 y

w x

111 000

111 000 L

Figure 5.3: Uniformly Loaded Simply Supported Beam Analysed by the Rayleigh-Ritz Method

5.4.1

Examples

Example 5-2: Uniformly Loaded Simply Supported Beam; Polynomial Approximation For the uniformly loaded beam shown in Fig. 5.3 let us assume a solution given by the following inﬁnite series: v = a1 x(L − x) + a2 x2 (L − x)2 + . . .

(5.80)

for this particular solution, let us retain only the ﬁrst term: v = a1 x(L − x)

(5.81)

We observe that: 1. The essential B.C. are immediately satisﬁed at both x = 0 and x = L. ∂Π = 0 (If we had left v in terms of a1 and a2 we 2. We can keep v in terms of a1 and take ∂a 1 ∂Π ∂Π should then take both ∂a1 = 0, and ∂a2 = 0 ).

3. Or we can solve for a1 in terms of vmax (@x = Π= U −W =

L o

Recalling that:

M EIz

Victor Saouma

L 2)

M2 dx − 2EIz

and take

∂Π ∂vmax

= 0.

L

wv(x)dx

(5.82)

0

2 = d v2 , the above simpliﬁes to: dx $ 2 L# EIz d2 v − wv(x) dx Π = 2 dx2 0 L EIz 2 (−2a1 ) − a1 wx(L − x) dx = 2 0 L3 L3 EIz 2 4a1 L − a1 w + a1 w = 2 2 3 3 a wL 1 = 2a21 EIz L − 6

(5.83)

(5.84)

Finite Elements II; Solid Mechanics

Draft 5–16

If we now take

VARIATIONAL and RAYLEIGH-RITZ METHODS ∂Π ∂a1

= 0, we would obtain: 4a1 EIz l −

wL3 6

= 0 wL2 24EIz

a1 =

(5.85)

Having solved the displacement ﬁeld in terms of a1 , we now determine vmax at wL4 v = 24EI z

x x2 − 2 L L

L 2:

a1

=

wL4 96EIz

(5.86) 4

4

wL exact = 5 wL = This is to be compared with the exact value of vmax 384 EIz 76.8EIz which constitutes ≈ 17% error. wL2 w & a2 = 24EI Note: If two terms were retained, then we would have obtained: a1 = 24EI z z exact . (Why?) and vmax would be equal to vmax

Example 5-3: Heat Conduction; (Bathe 1996) Considering the slab in Example 2-2, and assuming θ(t) = θ1 (t) + θ2 (t)x + θ3 (t)x2

(5.87)

where θ1 (t), θ2 (t), and θ3 (t) are the undetermined parameters. Use the Rayleigh-Ritz method procedure to generate the governing heat transfer equilibrium equations, using the following functional 2 L L ∂θ 1 k dx − θq B dx − θ|x=0 q0 (5.88) Π= 2 ∂x 0 0 with the essential boundary condition θ|x=L = θi Solution: 1. Substituting L L " ! " 1 ! k (θ2 )2 + 4θ2 θ3 x + 4(θ3 )2 x2 dx − θ1 + θ2 x + θ3 x2 q B dx − θ1 q0 (5.89) Π= 0 2 0 2. Invoking ∂Π = 0; ∂θ1 we obtain



0 0 k 0 L 0 L2 Victor Saouma

∂Π = 0; ∂θ2

∂Π =0 ∂θ3

  L B  0  θ1    0 q dx + q0 L B θ2 = L2  0 xq dx    4 3  L 2 B θ3 3L 0 x q dx

(5.90)     

(5.91)

Finite Elements II; Solid Mechanics

Draft

5.4 Rayleigh Ritz

5–17

3. In this problem q0 varies with time, hence heat capacity must be accounted for q B = −ρc 4. Substituting  0 0  k 0 L 0 L2

   0 L  θ1  1 2 2   L θ + ρc 2 L  2  4 3 1 3 L θ 3 3 3L

∂θ ∂t

1 2 2L 1 3 3L 1 4 4L

(5.92)

1 3 3L 1 4 4L 1 5 5L

   

   θ˙1   q0  0 = θ˙2    0 θ˙3

(5.93)

5. The ﬁnal equilibrium equation are now obtained by imposingon the last equation the condition that θ|x=L = θi , i.e. θ1 (t) + θ2 (t)L + θ3 (t)L2 = θi

(5.94)

which can be achieved by expressing θ1 in terms of θ2 , θ3 and θi

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 5–18

Victor Saouma

VARIATIONAL and RAYLEIGH-RITZ METHODS

Finite Elements II; Solid Mechanics

Draft Chapter 6

INTERPOLATION FUNCTIONS; NATURAL COORDINATE SYSTEMS 6.1

Introduction

In the Rayleigh Ritz method we solved the variational problem using a functional approximation for the displacement ﬁeld (Chapter ??). This powerful method has its limitation in terms of the complexity of solvable problems. 1

As stated above, an alternate approach consists in adopting an approximate displacement ﬁeld in terms of the nodal displacements via interpolation functions (or shape functions). 2

6.2

Cartesian Coordinate System

3 For an element (ﬁnite or otherwise), we can write an expression for the generalized displacement (translation/rotation), u at any point in terms of all its nodal ones, u.

u=

n

Ni (X)∆i = N(x){u}

(6.1)

i=1

where: 1. ui is the (generalized) nodal displacement corresponding to d.o.f i 2. Ni is an interpolation function, or shape function which has the following characteristics: (a) Ni = 1 at ui (b) Ni = 0 at uj where i = j. 3. N can be derived on the bases of: (a) Assumed deformation state deﬁned in terms of polynomial series. (b) Interpolation function (Lagrangian or Hermitian).

Draft 6–2

INTERPOLATION FUNCTIONS; NATURAL COORDINATE SYSTEMS

Figure 6.1: Axial Finite Element 4

We shall distinguish between two classes of problems:

C 0 where only displacement continuity must be ensured across elements (truss, torsion, plane stress/strain, 3D Elasticity). C 1 where we must ensure continuity of both displacements and their derivatives (such as beams, plates, and shells).

6.2.1 6.2.1.1 5

6

C0 Truss element

With reference to Fig. 6.1 we start with: u = N1 u1 + N2 u2

(6.2)

θx = N1 θx1 + N2 θx2

(6.3)

Since we have 2 d.o.f’s, we will assume a linear deformation state u = a1 x + a2

(6.4)

where u can be either u or θ, and the B.C.’s are given by: u = u1 at x = 0, and u = u2 at x = L. Thus we have:

7

8

u1 = a2

(6.5)

u2 = a1 L + a2

(6.6)

Solving for a1 and a2 in terms of u1 and u2 we obtain: u2 u1 − L L = u1

a1 =

(6.7)

a2

(6.8)

Substituting and rearranging those expressions into Eq. 6.4 we obtain u2 u1 − )x + u1 L L x x u2 = (1 − ) u1 + L L

u = (

N1

Victor Saouma

(6.9) (6.10)

N2

Finite Elements II; Solid Mechanics

Draft

6.2 Cartesian Coordinate System

6–3

or: N1 = 1 − N2 = Lx 6.2.1.2 9

x L

(6.11)

Generalization

The previous derivation can be generalized by writing: a1 u = a1 x + a2 = x 1 a2 [p]

(6.12)

{a}

where [p] corresponds to the polynomial approximation, and {a} is the coeﬃcient vector. 10

We next apply the boundary conditions: 0 1 u1 a1 = u2 a2 L 1 {u}

{a}

[L]

following inversion of [L], this leads to 1 −1 1 u1 a1 = L 0 a2 u2 L {a}

11

(6.13)

[L]−1

(6.14)

{u}

Substituting this last equation into Eq. 6.12, we obtain: u1 x x u = (1 − L ) L u2 −1 [p][L] {u}

(6.15)

[N]

12

Hence, the shape functions [N] can be directly obtained from [N] = [p][L]−1

(6.16)

Note that in some cases L−1 is not always possible to obtain, and that in others there may be considerable algebraic diﬃculties for arbitrary geometries. Hence, we shall introduce later on Lagrangian and Hermitian interpolation functions. 13

6.2.1.3 14

Constant Strain Triangle Element

Next we consider a triangular element, Fig. 6.2 with bi-linear displacement ﬁeld (in both x

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 6–4

INTERPOLATION FUNCTIONS; NATURAL COORDINATE SYSTEMS

Figure 6.2: Constant Strain Triangle Element and y): u = a1 + a2 x + a3 y v = a4 + a5 x + a6 y   u = 1 x y  [p]

(6.17)  a1  a2  a3

(6.18) (6.19)

{a}

15 As before, we ﬁrst seek the shape functions, and hence we apply the boundary conditions at the nodes for the u displacements ﬁrst:      1 0 0  a1   u1  u =  1 x2 0  a (6.20)  2   2  u3 a3 1 x3 y 3

{u}

16

[L]

{a}

We then multiply the inverse of [L] in Eq. 6.20 by [p] and obtain: u = N1 u1 + N2 u2 + N3 u3

(6.21)

where N1 = N2 = N3 =

1 (x2 y3 − xy3 − x2 y + x3 y) x2 y3 1 (xy3 − x3 y) x2 y3 y y3

(6.22)

We observe that each of the three shape functions is equal to 1 at the corresponding node, and equal to 0 at the other two. 17

The same shape functions can be derived for v: v = N1 v 1 + N2 v 2 + N3 v 3

Victor Saouma

(6.23)

Finite Elements II; Solid Mechanics

Draft

6.2 Cartesian Coordinate System 18

6–5

Hence, the displacement ﬁeld will be given by:

u v

=

N1 0 N2 0 N3 0 N1 0 N2 0

  u1       v1         0 u2 N3  v2         u 3      v3

(6.24)

The element is refereed to as Constant Strain Triangle (CST) because it has a linear displacement ﬁeld, and hence a constant strain. 19

6.2.1.4 20

Further Generalization: Lagrangian Interpolation Functions

In our earlier approach, the shape functions were obtained by: 1. Assumption of a polynomial function: u = p{a} 2. Application of the boundary conditions {u} = [L]{a} 3. Inversion of [L] 4. And ﬁnally [N] = [p][L]−1

By following these operations, we have in eﬀect deﬁned the Lagrangian Interpolation Functions for problems with C 0 interelement continuity (i.e continuity of displacement only). 21

22 The Lagrangian interpolation deﬁnes the coeﬃcients ([N] in our case) of a polynomial series representation of a function in terms of values deﬁned at discrete points (nodes in our case). For points along a line this would yield:

Ni =

23

m+1 j=1,j=i (x−xj ) m+1 j=1,j=i (xi −xj )

(6.25)

If expanded, the preceding equation would yield: N1 = N2 = Nm+1 =

24

Q Q

(x − x2 )(x − x3 ) · · · (x − xm+1 ) (x1 − x2 )(x1 − x3 ) · · · (x1 − xm+1 ) (x − x1 )(x − x3 ) · · · (x − xm+1 ) (x2 − x1 )(x2 − x3 ) · · · (x2 − xm+1 ) (x − x1 )(x − x2 ) · · · (x − xm ) (xm+1 − x1 )(xm+1 − x2 ) · · · (xm+1 − xm )

(6.26)

For the axial member, m = 1, x1 = 0, and x2 = L, the above equations will result in: u=

x x x (x − L) u1 + u2 = (1 − ) u1 + u2 −L L L L N1

(6.27)

N2

which is identical to Eq. 6.10. Victor Saouma

Finite Elements II; Solid Mechanics

Draft 6–6

INTERPOLATION FUNCTIONS; NATURAL COORDINATE SYSTEMS

Figure 6.3: Rectangular Bilinear Element 6.2.1.5

Rectangular Bilinear Element

Next we consider a quadrilateral element, Fig. 6.3 with bi-linear displacement ﬁeld (in both x and y). 25

26 Using the Lagrangian interpolation function of Eq. 6.25, and starting with the u displacement, we perform two interpolations: the ﬁrst one along the bottom edge (1-2) and along the top one (4-3). 27

From Eq. 6.25 with m = 1 we obtain: u12 = =

28

x2 − x x1 − x u1 + u2 x2 − x1 x1 − x2 a−x x+a u1 + u2 2a 2a

(6.28)

Similarly u43 = =

x2 − x x1 − x u4 + u3 x2 − x1 x1 − x2 a−x x+a u4 + u3 2a 2a

(6.29)

Next, we interpolate in the y direction along 1-4 and 2-3 between u12 and u43 . Again, we use Eq. 6.25 however this time we replace x by y: 29

y1 − y y2 − y u12 + u43 (6.30) y2 − y1 y1 − y2 b−yx+a y+ba−x y+bx+a b−ya−x u1 + u2 + u4 + u3 = 2b 2a 2b 2a 2b 2a 2b 2a (a − x)(b − y) (a + x)(b − y) (a + x)(b + y) (a − x)(b + y) = u1 + u2 + u3 + u4 4ab 4ab 4ab 4ab

u =

N1

Victor Saouma

N2

N3

N1

Finite Elements II; Solid Mechanics

Draft

6.2 Cartesian Coordinate System

6–7

30 One can easily check that at each node i the corresponding Ni is equal to 1, and all others to zero, and that at any point N1 + N2 + N3 + N4 = 1. Hence, the displacement ﬁeld will be given by:   u1         v  1        u   2     u N1 0 N2 0 N3 0 N4 0 v2 = (6.31) v u3  0 N1 0 N2 0 N3 0 N4         v3         u 4      v4

6.2.1.6

Solid Rectangular Trilinear Element

31 By extension to the previous derivation, the shape functions of a solid rectangular trilinear solid element, Fig. 6.4 will be given by:

Figure 6.4: Solid Trilinear Rectangular Element

   0 N2 0 0 N3 0 0 N4 0 N1 0  u  0 N2 0 0 N3 0 0 N4 v =  0 N1 0   0 N2 0 0 N3 0 0 0 0 N1 0 w

where Ni =

Victor Saouma

(a ± x)(b ± y)(c ± z) 8abc

                    0   0  N4                   

u1 v1 w1 u2 v2 w2 u3 v3 w3 u4 v4 w4

                                        

(6.32)

(6.33)

Finite Elements II; Solid Mechanics

Draft 6–8

INTERPOLATION FUNCTIONS; NATURAL COORDINATE SYSTEMS

Figure 6.5: Flexural Finite Element

6.2.2 6.2.2.1

C1 Flexural

With reference to Fig. 6.5. We have 4 d.o.f.’s, {u}4×1 : and hence will need 4 shape functions, N1 to N4 , and those will be obtained through 4 boundary conditions. Therefore we need to assume a polynomial approximation for displacements of degree 3. 32

v = a1 x3 + a2 x2 + a3 x + a4 dv = 3a1 x2 + 2a2 x + a3 θ = dx 33

(6.34) (6.35)

Note that v can be rewritten as:

v = x3

  a1       a2 2 x x 1  a3      [p] a4

(6.36)

{a}

34

We now apply the boundary conditions: 1. v = v 1 at x = 0 2. v = v 2 at x = L dv at x = 0 3. θ = θ1 = d x 4. θ = θ2 = dv at x = L dx

or:

   v1  0 0       0 0 θ1 =  L3 L2 v   2     3L2 2L θ2 {u}

Victor Saouma

[L]

0 1 L 1

 1  a1    0  a2 1  a   3 0 a4

{a}

      

(6.37)

Finite Elements II; Solid Mechanics

Draft

6.2 Cartesian Coordinate System

N1 N2 N3 N4

6–9 ξ=0 Ni Ni,x 1 0 0 1 0 0 0 0

Function = (1 + 2ξ 3 − 3ξ 2 ) = x(1 − ξ)2 = (3ξ 2 − 2ξ 3 ) = x(ξ 2 − ξ)

ξ=1 Ni Ni,x 0 0 0 0 1 0 0 1

Table 6.1: Characteristics of Beam Element Shape Functions 35

which when inverted yields:   a   1     1 a2 = 3 a   L    3  a4 {a}

36

 2 L −2 L   v1  −3L −2L2 3L −L2   θ1    0 L3 0 0  v   2 3 L 0 0 0 θ2 

[L]−1

{u}

   

Combining Eq. 6.38 with Eq. 6.36, we obtain:   2 L −2 L   v1 2 3L −L2   θ  1 −3L −2L 1  u = x3 x2 x 1 3  L3 0 0  v L  0   2 [p] 0 0 0 L3 θ2 [L]−1

(6.38)

  

{u}

      

  v1     (1 + 2ξ 3 − 3ξ 2 ) x(1 − ξ)2 (3ξ 2 − 2ξ 3 ) (ξ 2 − ξ)  θ1  =  N1 N2 N3 N4  v2    θ  2 [p][L]−1 {u}

(6.39)

(6.40)

[N]

where ξ = xl . 37

Hence, the shape functions for the ﬂexural element are given by: N1 = (1 + 2ξ 3 − 3ξ 2 ) N2 = x(1 − ξ)2 N3 = (3ξ 2 − 2ξ 3 ) N4 = x(ξ 2 − ξ)

(6.41)

and are shown in Fig 6.6. 38

Table 6.1 illustrates the characteristics of those shape functions

6.2.2.2

C 1 : Hermitian Interpolation Functions

For problems involving the ﬁrst derivative of the shape function, that is with C 1 interelement continuity (i.e continuity of ﬁrst derivative or slope) such as for ﬂexure, Hermitian interpolation functions rather than Lagrangian ones should be used. 39

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 6–10

INTERPOLATION FUNCTIONS; NATURAL COORDINATE SYSTEMS

Shape Functions for Flexure (v1; θ1; v2; θ2)

1.0

0.8

N1 N3 N2 N4

N

0.6

0.4

0.2

0.0

−0.2 0.0

0.2

0.4

0.6

0.8

1.0

ξ(x/L)

Figure 6.6: Shape Functions for Flexure of Uniform Beam Element. 40 Hermitian interpolation functions are piecewise cubic functions which satisfy the conditions of displacement and slope (C 0 , C 1 ) continuities. They are extensively used in CAD as Bezier curves.

6.2.3

Characteristics of Shape Functions

1. The basis of derivation of shape functions could be: (a) A polynomial relation i. Exact ii. Approximation (b) Or other i. Logarithmic ii. Trigonometric 2. Shape functions should (a) be continuous, of the type required by the variational principle. (b) exhibit rigid body motion (i.e. v = a1 + . . .) (c) exhibit constant strain. 3. Shape functions should be complete, and meet the same requirements as the coeﬃcients of the Rayleigh Ritz method.

6.3

Natural Coordinate System

Natural coordinates are dimensionless and deﬁned with respect to element length/area/volume rather than the global coordinate system. 41

42 They are often used in element formulation of simplex elements (where in an n dimensional space, there are n + 1 vertices and n + 1 surfaces of dimensionality n − 1).

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

6.3 Natural Coordinate System

6–11

L L2

O

ξ =

L1

1

P x1

x1 L

ξ = 2

x2 L

1 ξ=ξ x + ξ x 1 1

2 2

x =x+L 2

1

Figure 6.7: Natural Coordinate System Along a Straight Line

6.3.1 43

Straight Line

We deﬁne two natural coordinates ξ1 and ξ2 , Fig. 6.7 ξ1 =

L1 L

and ξ2 =

L2 L

(6.42)

since L1 + L2 = L, we have ξ1 + ξ2 = 1 44

We note that ξi = 1 at node i and zero at all other nodes.

45

An arbitrary point P with coordinate x has x = ξ1 x1 + ξ2 x2

(6.43)

46 Thus, for every point x, corresponds a set of (nonindependent) natural coordinates ξ1 , ξ2 . Those equations can be stated as 1 x2 −1 1 ξ1 ξ1 1 1 1 (6.44) and = = ξ2 ξ2 1 x −x x x1 x2 L 1

47

Interpolation can be done in natural coordinates φ1 φ = ξ1 ξ2 φ2

(6.45)

N

48 Integration of polynomials in ξ1 and ξ2 can be done through (a rigorous proof is developed below through Eq. 6.60) k!l! (6.46) ξ1k ξ2l dL = L (1 + k + l)! L

Furthermore

Victor Saouma

xdL = L (ξ1 x1 + ξ2 x2 )dL = L2 2 ξ ξ dL = L = L 1 2 4!

L 2 (x1 L 12

+ x2 )

(6.47)

Finite Elements II; Solid Mechanics

Draft 6–12

INTERPOLATION FUNCTIONS; NATURAL COORDINATE SYSTEMS

L =1 3

A P 1

A2 A

1

L =0 2

1

Side

2

e Sid

L =0

L =1

3

2

Side 3

L =0

L =1

3

1

Figure 6.8: Natural Coordinate System for a Triangle

6.3.2

Triangular Coordinates

49 For a triangular surface, any point P divides the triangle 1-2-3 into three subareas A1 , A2 , A3 , Fig. 6.8 thus we deﬁne area coordinates as ratios of areas

L1 =

A1 , A

L2 =

A21 , A

L3 =

A3 A

(6.48)

where A is the area of the triangle 1-2-3. 50

Since A = A1 + A2 + A3 then L1 + L2 + L3 = 1

51

52

The centroid is at L1 = L2 = L3 =

(6.49)

1 3

Again the constraints equations are      1 1 1  L1   1  L2 x =  x1 x2 x3      y1 y2 y3 L3 y

(6.50)

[A]

inverting      x2 y3 − x3 y2 y2 − y3 x3 − x2  1   L1  L =  x3 y1 − x1 y3 y3 − y1 x1 − x3  x    2  L3 x1 y2 − x2 y1 y1 − y2 x2 − x1 y

=

[A]−1

  2A23 y23 x32  1  1  2A31 y31 x13  x   2A 2A12 y12 x21 y 

(6.51-a)

(6.51-b)

where xij = xi − xj Victor Saouma

Finite Elements II; Solid Mechanics

Draft

6.3 Natural Coordinate System 53

6–13

We observe that the area is equal to the determinant over two. 2A = |A| = x2 − x1 y3 − y1 − x31 y21

(6.52)

and if the labels are reversed such that 1-2-3 is clockwise, then the determinant is negative. 54

A function φ may be expressed in terms of area coordinates φ = φ(L1 , L2 , L3 ) ∂φ ∂φ ∂L1 ∂φ ∂L2 ∂φ ∂L3 = + + ∂x ∂L1 ∂x ∂L2 ∂x ∂L3 ∂x ∂φ ∂L1 ∂φ ∂L2 ∂φ ∂L3 ∂φ = + + ∂y ∂L1 ∂y ∂L2 ∂y ∂L3 ∂y and

∂L1 ∂x ∂L1 ∂y

y2 −y3 2A ; x3 −x2 2A ;

= =

∂L2 ∂x ∂L2 ∂y

y3 −y1 2A ; x1 −x3 2A ;

= =

∂L3 ∂x ∂L3 ∂y

= =

y1 −y2 2A ; x2 −x1 2A ;

(6.53-a) (6.53-b) (6.53-c)

(6.54)

55 Integration for the stiﬀness matrix will require integration in area coordinate of expressions of the form A Lk1 Ll2 Lm 3 dA. To accomplish this operation, and with reference to Fig. 6.9, (Eisenberg and Malvern 1973):

h2

3 θ3

L =0 2

s

ds

1

1

s

1

ds 1

2

2

11 00 00 11 00 11 00 11 00 11 00 11 00 11

L =0

2 L =0 3

h1

Figure 6.9: Integration over a Triangle dA = (ds1 ) csc θ3 (ds2 )

Victor Saouma

(6.55-a)

= (h1 dL1 ) csc θ3 (h2 dL2 )

(6.55-b)

= 2AdL1 dL2

(6.55-c) Finite Elements II; Solid Mechanics

Draft 6–14

Thus

INTERPOLATION FUNCTIONS; NATURAL COORDINATE SYSTEMS A

Lk1 Ll2 Lm 3 dA

= 2A

1 1−L1 0

0

Lk1 Ll2 (1

− L1 − L2 ) dL2 dL1 m

(6.56)

we next substitute L2 = t(1 − L1 ) and dL2 = (1 − L1 )dt which yields A

Lk1 Ll2 Lm 3 dA

1 1 k l+m+1 = 2A L1 (1 − L1 ) dL1 tl (1 − t)m dt 0

(6.57)

0

each of the integrals on the right-hand side is of the form of the beta function (Abramowitz and Stegun 1970) 1 Γ(z)Γ(w) (6.58) tz−1 (1 − t)w−1 dt = B(z, w) = Γ(z + w) 0 where Γ denotes the gamma function which satisﬁes Γ(n + 1) = n! for integers n ≥ 0. Thus Γ(k + 1)Γ(l + 1)Γ(m + 1) (6.59) Lk1 Ll2 Lm 3 dA = 2A Γ(k + l + m + 3) A Since k, l, and m are nonnegative integers, A

Lk1 Ll2 Lm 3 dA = 2A

k!l!m! (k + l + m + 2)!

(6.60)

56 It should be noted that the above relation is used only if we have straight sided elements and the element formulation can thus be analytically derived. Alternatively, if we have a curvilinear side, this formula will not be used, and the element will be numerically integrated within the context of an isoparametric formulation.

6.3.3 57

Volume Coordinates

Volumes coordinates are a direct extension of triangular coordinates.

An internal point P subdivide the tetrahedron into four subtetrahedra, and thus Li = for i = 1, 4. 58

59

The constraint equation is    1        x =  y       z

 L1 1 1 1 1    L x1 x2 x3 x4  2  y1 y2 y3 y4   L   3 z1 z2 z3 z4 L4

Vi V ,

      

(6.61)

[A]

60

Integration is also carried in a similar way, n Lk1 Ll2 Lm 3 L4 dA = 6V V

Victor Saouma

k!l!m!n! (3 + k + l + m + n)!

(6.62)

Finite Elements II; Solid Mechanics

Draft

6.3 Natural Coordinate System

6.3.4

6–15

Interpolation Functions

61 Triangular elements allow a complete polynomial in Cartesian coordinate to be used for the ﬁeld (e.g. temperature, displacement) quantity. Thus, all terms of a truncated Pascal triangle are used in the shape functions. (This will not be the case for the bilinear quadrilateral element). ) 62 We seek to determine φ = Ni φi where φi are nodal d.o.f. and Ni = Ni (L1 , L2 , L3 ). )n q r s 63 We start with φ = i=1 ai L1 , L2 , L3 ) where q, r and s range over the n possible combinations for which q + r + s = p and where p is the order of the polynomial. For example for

Linear element: φ = a1 L1 + a2 L2 + a3 L3 = a 1 + a 2 x + a 3 y

(6.63)

Quadratic element: φ = a1 L21 + a2 L22 + a3 L23 + a4 L1 L2 + a5 L2 L3 + a6 L3 L1 = a 1 + a 2 x+ a 3 y + a 4 x2 + a 5 xy + a 6 y 2 (6.64) 64

The shape functions for the Linear Triangle are simply the area coordinates, N1 = L1

N2 = L2

N3 = L3

(6.65)

and each one is equal to unity at one node, zero at the others, and varies linearly. 65 The shape functions for other elements can be obtained from the Lagrangian interpolation function used earlier Eq. 6.25 -n i=0,i =k (L − Li ) n (6.66-a) lk = -n i=0,i =k (Lk − Li )

=

(L − L0 )(L − L1 ) · · · (L − Lk−1 )(L − Lk+1 ) · · · (L − Ln ) (Lk − L0 )(Lk − L1 ) · · · (Lk − Lk−1 )(Lk − Lk+1 ) · · · (Lk − Ln )

(6.66-b)

66 Hence, denoting a typical node i by three numbers q, r and s corresponding to the position of coordinates L1i , L2i and L3i , we can write the shape functions in terms of three Lagrangian interpolation functions (6.67) Ni = lqq (L1 )lrr (L2 )lss (L3 )

67

Using this formula, the shape functions for higher order elements are

Linear triangle Ni = Li

(6.68)

Quadratic triangle Ni = Li (2Li − 1) Corner nodes Midside nodes, i, j, k are on the same side Nj = 4Li Lk

Victor Saouma

(6.69)

Finite Elements II; Solid Mechanics

Draft 6–16

INTERPOLATION FUNCTIONS; NATURAL COORDINATE SYSTEMS

Cubic triangle Ni = 12 Li (3Li − 1)(3Li − 2) Corner nodes Midside nodes, i, j, k are on the same side Nj = 92 Li Lk (3Li − 1) Internal node N10 = 27L1 L2 L3

6.4

(6.70)

Pascal’s Triangle

68 A schematic interpretation of shape functions in terms of polynomial series terms is given by Pascal’s triangle which is shown in Table 6.2.

Constant Linear Quadratic Cubic Quartic a11 x4

a1 3

a7 x

a4 x2 a12 x3 y

a2 x 2

a8 x y

a3 y a5 xy a13 x2 y 2

a9 xy

2

a6 y 2 a14 xy 3

a10 y 3

(6.71) a15 x4

Table 6.2: Interpretation of Shape Functions in Terms of Polynomial Series (1D & 2D) 69

Polynomial terms present in various element formulations is shown in Table 6.3 Element

Terms

Linear Quadratic Bi-Linear (triangle) Bi-Linear (quadrilateral) Bi-Quadratic (Serendipity) Bi-Quadratic (Lagrangian)

a1 , a1 , a1 , a1 , a1 , a1 ,

a2 a2 , a2 , a2 , a2 , a2 ,

a4 a3 , a3 , a5 a3 , a4 , a5 , a6 , a8 , a9 a3 , a4 , a5 , a6 , a8 , a9 , a13

# of Nodes (terms) 2 3 3 4 8 9

Table 6.3: Polynomial Terms in Various Element Formulations (1D & 2D) 70

A complete polynomial contains all the terms above a certain line in the Pascal triangle.

71 The more terms are included, the higher the accuracy. For instance u = a1 +a2 x+a3 y +a5 xy is more accurate than u = a1 +a2 x+a3 y, however the rate of convergence is unchanged (because the second approximation is not a complete quadratic one). 72 Terms in the approximation which do not improve the rate of convergence are called parasitic terms.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft Chapter 7

FINITE ELEMENT DISCRETIZATION and REQUIREMENTS 7.1

Discretization

This section is mostly extracted from (Reich 1993)

7.1.1

Discretization of the Variational Statement for the General TPE Variational Principle

The discretization of Equation 5.8 will be performed on an element domain Ωe using the procedures described in Chapter 2 of (Zienkiewicz and Taylor 1989); 1

2 The surface of the element subjected to surface tractions Γt comprises one or more surfaces of the element boundary Γ. For the present time this discussion will be kept on a very general level with no mention of the dimensionality of the elements; the number of nodes deﬁning the elements; or the nature of the constitutive law.

The ﬁrst step in the discretization process is to deﬁne the displacements u at a point inside the element in terms of the shape functions N and the nodal displacements ue for the element 3

(7.1)

u = Nue

4 The virtual displacements δu at a point inside the element can also be deﬁned in terms of the shape functions N and the nodal virtual displacements δue for the element

(7.2)

δu = Nδue 5

In order to discretize the volume integral in Equation 5.8 δΠ = δ(Lu)T D(Lu)dΩ − δ(Lu)T D 0 dΩ + δ(Lu)T σ0 dΩ − δuT bdΩ − Ω

Ω

Ω

Ω

δuT ˆtdΓ = 0 Γt

(7.3)

Draft 7–2

FINITE ELEMENT DISCRETIZATION and REQUIREMENTS

deﬁning the virtual strain energy for the element due to the nodal displacements u, the strains at a point inside the element are expressed in terms of the nodal displacements ue using Equation 7.1 = Lu = LNue (7.4) and the virtual strains δ at a point inside the element are expressed in terms of the nodal virtual displacements δue using Equation 7.2 δ = δ(Lu) = LNδue 6

Deﬁning the discrete strain-displacement operator B as def

(7.6)

B = LN the virtual strain energy for an element is written as T T δ(Lu) D(Lu)dΩ = δue Ωe

7

(7.5)

BT DBdΩue

(7.7)

Ωe

Deﬁning the element stiﬀness matrix Ke as BT DBdΩ

Ke =

(7.8)

Ωe

Equation 7.7 can be rewritten as δ(Lu)T D(Lu)dΩ = δuTe Ke ue

(7.9)

Ωe 8 In order to discretize the volume integrals in Equation 7.3 deﬁning the virtual strain energy for the element due to the initial strains 0 and stresses σ 0 , Equations 7.5 and 7.6, which deﬁne the virtual strains δ at a point inside the element in terms of the nodal virtual displacements δue , are substituted into the integrands T T δ(Lu) D 0 dΩ = δue BT D 0 dΩ (7.10)

Ωe

Ωe 9

δ(Lu)T σ 0 dΩ = δuTe

Ωe

Ωe

BT σ 0 dΩ

(7.11)

Deﬁning the initial force vector f0e as f0e =

B D 0 dΩ − T

Ωe

Ωt

BT σ0 dΩ

the strain energy due to the initial strains and stresses is δ(Lu)T D 0 dΩ − δ(Lu)T σ 0 dΩ = δuTe f0e Ωe

Victor Saouma

(7.12)

(7.13)

Ωe

Finite Elements II; Solid Mechanics

Draft

7.1 Discretization

7–3

10 In order to discretize the volume integral deﬁning the work done by the body forces and the surface integral deﬁning the work done by the surface tractions in Equation 7.3, Equation 7.2 is substituted into the integrands δuT bdΩ = δuTe NT bdΩ (7.14)

Ωe

Ωe

δuT ˆtdΓ = δuTe Γt 11

NT ˆtdΓ

(7.15)

NT ˆtdΓ

(7.16)

Γt

Deﬁning the applied force vector fe as

NT bdΩ +

fe = Ωe

Γt

the sum of the internal and external virtual work due to body forces and surface tractions is T δu bdΩ + δuT ˆtdΓ = δuTe fe (7.17) Ωe

Γt

12 Having obtained the discretization of the various integrals deﬁning the variational statement for the TPE variational principle, it is now possible to deﬁne the discrete system of equations. Substituting Equations 7.7, 7.13, and 7.17 into Equation 7.3 and rearranging terms, the discretized Principle of Virtual Work is

δuTe Ke ue = δuTe fe + δuTe f0e

(7.18)

Since δuTe is an arbitrary (i.e. non-zero) vector appearing on both sides of Equation 7.18, the discrete system of equations can be simpliﬁed into 13

Ke ue = fe + f0e + Pu

(7.19)

as the discrete system of equations for an element.

7.1.2

Discretization of the Variational Statement for the HW Variational Principle

14 The discretization of the three variational statements deﬁned in Equation 5.65, 5.66, and 5.67 will be performed on an element domain Ωe using the procedures described in Chapter 2 of (Zienkiewicz and Taylor 1989) assembly of the discrete element equations into a discrete global system of equations is straightforward and will be omitted from this discussion.

The surface of the element subjected to surface tractions Γt comprises one or more surfaces of the element boundary Γe . For the present time this discussion will be kept on a very general level with no mention of the dimensionality of the elements; the number of nodes deﬁning the elements; or the nature of constitutive law. 15

The ﬁrst step in the discretization process is to deﬁne the displacements u, strains , and stresses σ at a point inside the element in terms of the shape functions Nu , N= , and Nσ , respectively, and the element nodal displacements ue , strains e , and stresses σ e 16

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 7–4

FINITE ELEMENT DISCRETIZATION and REQUIREMENTS u = Nu ue

(7.20)

= N = e

(7.21)

σ = Nσ σ e

(7.22)

We note that contrarily to the previous case (Eq. 7.1) we now have three discretizations (instead of just one). 17 The virtual displacements δu, virtual strains δ , and virtual stresses δσ at a point inside the element can also be deﬁned in terms of the shape functions Nu , N= , and Nσ , respectively, and the nodal virtual displacements δue , virtual strains δ e , and virtual stresses δσ e for the element

18

δu = Nu δue

(7.23-a)

δ = N= δ e

(7.23-b)

δσ = Nσ δσ e

(7.23-c)

We now need to discretize each one of the corresponding Euler T T δ(Lu) σ dΩ − δu b dΩ − δuT ˆt dΓ Ω Ω Γt δ T [D( − 0 ) + σ 0 − σ] dΩ Ω δσ T (L u − ) dΩ

equations: = 0

(7.24-a)

= 0

(7.24-b)

= 0

(7.24-c)

Ω

In order to discretize the volume integral in the ﬁrst variational statement (i.e. Equ. 7.24-a) deﬁning the virtual strain energy for the element, Equation 7.23-a is substituted into the virtual strain-displacement relationship (i.e. Equation 5.6) to deﬁne the virtual strains δ at a point inside the element in terms of the nodal virtual displacements δue 19

δ(L u) = L δu = L Nu δue 20

(7.25)

Deﬁning the discrete strain-displacement operator Bu as Bu = L Nu

(7.26)

and substituting Equation 7.22 into the integrand, the virtual strain energy for an element is written as T T δ(Lu) σ dΩ = δue BTu Nσ dΩ σe (7.27) Ωe 21

Ωe

Deﬁning an element operator matrix Fe as T Fe =

BTu Nσ dΩ

(7.28)

Ωe

Equation 7.27 can be rewritten as δ(Lu)T σ dΩ = δuTe FTe σe

(7.29)

Ωe

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

7.1 Discretization

7–5

22 In order to discretize the volume integral deﬁning the work done by the body forces and the surface integral deﬁning the work done by the surface tractions in the ﬁrst variational statement (i.e. the ﬁrst equation in Equation ??), Equation 7.23-a is substituted into the integrands T T δu b dΩ = δue NTu b dΩ (7.30)

Ωe

Ωe

δuT ˆt dΓ = δuTe

NTu ˆt dΓ

Γt

(7.31)

Γt

Deﬁning the applied force vector fe as fe =

NTu b dΩ +

NTu ˆt dΓ

Ωe

(7.32)

Γt

the sum of the internal and external virtual work is T δu b dΩ + δuT ˆt dΓ = δuTe fe Ωe

(7.33)

Γt

23 Having deﬁned the discretization of the various integrals in the ﬁrst variational statement for the HW variational principle (i.e. Equ. 7.24-a), it is now possible to deﬁne the discrete system of equations. Substituting Equations 7.29 and 7.33 into the variational statement and rearranging terms, the discretized Principle of Virtual Work is

δuTe FTe σ e = δuTe fe

(7.34)

where the left-hand side is the virtual strain energy and the right-hand side is the internal and external virtual work. Since δue is an arbitrary (i.e. non-zero) vector appearing on both sides of Equation 7.34, the discrete system of equations can be simpliﬁed into FTe σ e = fe

(7.35)

as the discrete system of equations for an element. 24 In order to discretize the second variational statement (i.e. Equ. 7.24-b), Equations 7.21, 7.22, and 7.23-b are substituted into the integrand T T T T δ [D( − 0 ) + σ0 − σ] dΩ = δ e N= D N= dΩ e − δ e NT= D 0 dΩ

Ω

+ δ Te Ωe 25

Ωe

Ωe

NT= σ 0 dΩ − δ Te

NT= Nσ dΩ σ e = 0

(7.36)

Ωe

Deﬁning a pair of element operator matrices Ae and Ce as NT= D N= dΩ Ae =

(7.37)

Ωe

NT= Nσ dΩ

Ce =

(7.38)

Ωe

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 7–6

FINITE ELEMENT DISCRETIZATION and REQUIREMENTS

and the initial strain/stress vector ge as NT= D 0 dΩ − ge = Ωe

Ωe

NT= σ 0 dΩ

Equation 7.36 can be rewritten as δ T [D( − 0 ) + σ0 − σ] dΩ = δ Te Ae e − δ Te ge − δ Te Ce σ e = 0

(7.39)

(7.40)

Ω

Since the nodal virtual strains δ are arbitrary they can be eliminated from Equation 7.40 yielding Ae e − Ce σ e = ge

(7.41)

as the discretized form of the second variational statement. 26 In order to discretize the third variational statement (i.e. Eq. 7.24-c), Equations 7.20, 7.21, and 7.23-c are substituted into the integrand T T T T δσ (L u − ) dΩ = δσ e Nσ Bu dΩ ue − δσ e NTσ N= dΩ e = 0 (7.42)

Ωe

Ωe

Recognizing that

Ωe

NTσ Bu dΩ = Fe

(7.43)

NTσ N= dΩ = CTe

(7.44)

Ωe

Ωe

Equation 7.42 can be rewritten as δσ T (L u − ) dΩ = δσ Te Fe ue − δσ Te CTe e = 0

(7.45)

Ωe

Since the nodal virtual stresses δσ e are arbitrary they can be eliminated from Equation 7.45 yielding Fe ue − CTe e = 0

(7.46)

as the discretized form of the third variational statement. 27 Having deﬁned the discretized form of all three variational statements, it is now possible to deﬁne the discrete mixed system of equations for an element. Assembling Equations 7.35, 7.41, and 7.46 in matrix form adopting the classic arrangement for a constrained system of equations      Ae −Ce 0  e   ge   −CTe 0 Fe  σe 0 (7.47) =     T 0 Fe 0 ue fe

yields a symmetric system of equations. Although is technically an intermediate variable in the ﬁeld equations indirectly relating σ to u, e is the primary variable and σe to ue are constraint variables in Equation 7.47. Victor Saouma

Finite Elements II; Solid Mechanics

Draft

7.2 General Element Requirements

7–7

28 Since it would be computationally expensive to solve the system of equations in Eq. 7.47 using direct method, an indirect or iterative procedure (i.e. Gauss-Seidel instead of GaussJordan) is often selected, (Zienkiewicz and Taylor 1989).

Step Step Step Step

1: 2: 3: 4:

uk+1 n εk+1 n σk+1 n rk+1 n

= = = =

ukn + K−1 rkn C−T Fuk+1 n C−1 Aεk+1 n f − FT σk+1 n

(7.48)

for k = 0, 1, 2, · · ·, where k is an iteration index and rk+1 is the residual force vector. It should n be noted that this procedure is solved on the structural level, meaning that steps 1 to 3 require a solution of a system of linear equations. Step 1, K corresponds to the classical standard displacement stiﬀness matrix, and this step is used as a pre-conditioner. This implies that at the beginning of the ﬁrst iteration, when u0n = 0 and r0n =, step corresponds to the standard displacement-based formulation of the ﬁnite element method. Steps 1, 2, and 3 above require the solution of simultaneous linear equations. Step 3, however, may be reduced by nodal quadrature and assuming same interpolation functions for strains and stresses to (7.49) σ i = D i In this equation, σ i and i are the stresses at node i, respectively, and D is the stress-strain constitutive matrix. Then, Step 3 is nothing else but direct computation of nodal stresses from nodal strains using the constitutive matrix D. Finally, the uniqueness and the existence of a solution has been addressed by the so-called Babuˇska-Brezzi (BB) condition (Babuˇska 1973, Brezzi 1974). 29

30

Details of the algotithmic implementation will be covered in a later chapter.

7.2

General Element Requirements

A ﬁnite element (just a an approximate displacement ﬁeld in the Rayleigh-Ritz formulation) must satisfy two basic requirements 31

Completeness: The FE discretization must at least accommodate constant displacement and constant strain (or temperature and temperature gradient). This is accomplished by including in two dimensional problems the following φ = a1 + a2 x + a3 y + possibly additional terms

(7.50)

Compatibility or Conformity: The approximation of the ﬁeld over element boundaries must be continuous (C0 or C1 continuity). Most ﬁnite elements are conforming, but some are not. 32 For instance, with respect to Fig. 7.1, element A must be capable of undergoing rigid body motion without internal strains/stresses, and at node B we should have continuity of displacement (but not slope for this element). 33 If those two requirements are satisﬁed, then convergence is assured. In the FE method approximate solution are obtained, and the more elements we use, the more accurate is the

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 7–8

FINITE ELEMENT DISCRETIZATION and REQUIREMENTS

111 000 000 111 000 111 000 111 000 111 111 000 000 111 000 111 000 111 000 111

P

B A

Figure 7.1: Completness and Compatibility approximate solution. In the limit, for inﬁnitely small elements, we require the solution to be also inﬁnitely close to the exact one. Hence, convergence is ensured if completeness and compatibility requirements are satisﬁed. 34

35

Lax equivalence theorem: Convergence criterion=completeness+ compatibility requirements Two essential requirements:

Patch test: Completeness can be assessed through the patch test which will be discussed later. Zero Strain Energy: For structural problems, there should be zero strain energy when the element is subjected to a rigid body motion. Recall that in the stiﬀness matrix formulation, the matrix is singular as it embodies not only the force displacement relations, but the equilibrium equations also. To each of those equations, corresponds a rigid body mode which can be detected by an eigenvalue analysis (more about this later).

7.3

Discretization Error and Convergence Rate

Approximation will yield an exact solution in the limit as the size h of element approaches zero. 36

37 In some cases the exact solution is obtained with a ﬁnite number of elements (or even with only one) if the polynomial expansion used in that element can ﬁt exactly the correct solution. (e.g. Truss, beam elements, plane stress element used in a plate under pure axial load). 38

The exact solution can always be expanded in the vicinity of any node i as a polynomial ∂u ∂u (x − xi ) + (y − yi ) + ... (7.51) u = ui + ∂x i ∂y i

If within an element of size h a polynomial of degree p is used, then the Taylor expansion up to that degree can be accommodated and the error is of the order O(hp+1 ).

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

7.3 Discretization Error and Convergence Rate

7–9

Thus for a linear element, the convergence rate is of order O(h2 ), i.e., the error in displacement is reduced to 1/4 of the original error by halving the mesh. 39

40 By a similar argument, it can be shown that the strains (or stresses) which are given by the mth derivatives of the displacement should converge with an error of O(hp+1−m ), and the strain energy (which is given by the square of the stresses) error will be O(h2(p+1−m) ) 41 From the above arguments, knowledge of the order of convergence may help in extrapolating the solutions to the “correct” one. Hence, for instance, if the displacement converges at O(h2 ), and we have two approximate solutions u1 and u2 obtained with meshes of sizes h and h/2, then we can write O(h2 ) u1 − u = =4 (7.52) u2 − u O(h/2)2

where u is the extrapolated value. 42

Convergence to the exact solution can be accelerated by, Fig. 7.2. p h

r

Figure 7.2: h, p and r Convergence

h reducing the size of the elements (or mesh reﬁnement). p increasing the order of the polynomial (same number of elements but higher number of nodes/dof). A third form of acceleration is the so called r reﬁnement in which the same number of nodes/elements is retained but the mesh is shifted around to increase its density in zones of high stress gradient. 43 The above convergence procedures can be accelerated within the context of a program which can accommodate adaptive remeshing techniques. 44

Finally, additional errors may come from round-oﬀ within the computer.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 7–10

7.4

FINITE ELEMENT DISCRETIZATION and REQUIREMENTS

Lower Bound Character of Minimum Potential Energy Based Solutions

A numerical solution that is derived from the principle of minimum total potential energy is a lower bound solution, because the strain energy is smaller than the exact one (i.e. obtained from an inﬁnite number of elements). 45

This can be readily shown if we consider the displacement ui caused by a load Pi which is increased from zero to its stipulated value. The work done is Pi2ui and must be equal to the internal strain energy U . Alternatively, the potential of the applied load is Pi ui , and the exact potential energy is Pi uexact Pi uexact i i − Pi uexact (7.53) = − Πexact = i 2 2 similarly, the approximate value of the potential energy is 46

Πapprox

Pi uapprox i =− 2

(7.54)

We know that the approximation of Π is algebraically higher than the exact value (since the exact value is a minimum), hence −

Pi uiapprox Pi uiexact <− 2 2 Πexact

(7.55)

Πapprox

or uapprox < uexact i i

(7.56)

or alternatively, the solution is too stiﬀ. Note the similarity with the Rayleigh-Ritz method, the lack of enough terms in our polynomial approximation did also result in a stiﬀ solution. 47

7.5

Equilibrium and Compatibiliy in the Solution

48

In an exact solution over each diﬀerential element equilibrium and compatibility prevails.

49

In a ﬁnite element discretization, this is not necessarily the case:

Equilibrium of Nodal Forces: Those are automatically satisﬁed by deﬁnition (ke u − f = 0) at the nodes. Compatibility at Nodes: Elements connected to one another have the same displacements (along corresponding dof) at the connecting node. Interelement Continuity of Stress: is usually not satisﬁed, that is equilibrium across the elements does not necessarily prevail. For example consider Fig. 7.3 if node 4 is the only one displaced (all others remaining ﬁxed), then there will be a discontinuity of σxx across element boundary 2-3. Victor Saouma

Finite Elements II; Solid Mechanics

Draft

7.5 Equilibrium and Compatibiliy in the Solution

7–11

3

1

1

σx2

4

2

u 2

4

Figure 7.3: Interelement Continuity of Stress Interelement Compatibility of Displacements: Most elements have this requirement satisﬁed. If an element does not then it is labeled as incompatible or nonconforming, Fig. 7.4

6 2

4

v

3

v

y,v

x,u 1

5 Figure 7.4: Interelement Continuity of Strain Equilibrium Inside Elements: is not always satisﬁed. In general satisfaction of equilibrium at every point demands relations between dof which do not necessarily result from ke u − f =0 Displacement Compatibility Inside an Element: is satisﬁed as we only require that the element displacement be continuous and single-valued.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 7–12

Victor Saouma

FINITE ELEMENT DISCRETIZATION and REQUIREMENTS

Finite Elements II; Solid Mechanics

Draft Chapter 8

STRAIGHT SIDED ELEMENTS; 1st GENERATION 8.1

Introduction

1 Having ﬁrst introduced the method of virtual displacements, than the interpolation (or shape) functions [N], which relate internal to external nodal displacements, and ﬁnally having applied the virtual displacement method to ﬁnite element in chapter 7, we derive the stiﬀness matrices of some simple elements.

8.2

Rod Elements

8.2.1

Truss Element

2

The shape functions of the truss element were derived in Eq. 6.11: x N1 = 1 − L x N2 = L

3

The corresponding strain displacement relation [B] is given by: du dx = [ dN1 dN2 ] dx dx = [ − L1 L1 ]

εxx =

(8.1)

[B]

4

For the truss element, the constitutive matrix [D] reduces to the scalar E; Hence, substituting [B]T [D][B]dΩ and with dΩ = Adx for element with constant cross

into Eq. 1.15 [ke ] = Ω

Draft

STRAIGHT SIDED ELEMENTS; 1st GENERATION

8–2

sectional area we obtain: [k ] = A

L

e

0

5

=

AE L2

=

AE L

L

− L1

· E · − L1

1 L

1 −1 −1 1 0 1 −1 −1 1

1 L

dx

(8.2-a)

dx (8.2-b)

We observe that this stiﬀness matrix is identical to the one earlier derived in Eq. ??.

8.2.2

Beam Element

6 For a beam element, for which we have previously derived the shape functions in Eq. 6.41 and the [B] matrix in Eq. 1.4-d, substituting in Eq. 1.15:

l e

[B]T [D][B] y 2 dA dx

[k ] = 0

(8.3)

A

y 2 dA = Iz Eq. 1.15 reduces to

and noting that A

l

e

[B]T [D][B]Iz dx

[k ] =

(8.4)

0

7

For this simple case, we have: [D] = E, thus:

l

[ke ] = EIz

[B]T [B] dx

(8.5)

0 8 Using the shape function for the beam element from Eq. 6.41, and noting the change of integration variable from dx to dξ, we obtain

[ke ] = EIz

   1 0

  

6 L2 (2ξ − 1) − L2 (3ξ − 2) 6 L2 (−2ξ + 1) − L2 (3ξ − 1)

   

.

  

6 L2 (2ξ

− 1) − L2 (3ξ − 2)

6 L2 (−2ξ

+ 1) − L2 (3ξ − 1)

/

Ldξ (8.6) dx

or 

[ke ] =

v1

z V1 12EI L3  6EI z M 1 L2  12EIz V2 − L3  6EI z M 2 2

L

9

θ1

6EIz L2 4EIz L z − 6EI L2 2EIz L

v2

z − 12EI L3 6EIz − L2

12EIz L3 z − 6EI L2

θ2

6EIz L2 2EIz L z − 6EI L2 4EIz L

       

(8.7)

The generation of this stiﬀness matrix with Mathematica is given by:

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

8.3 Triangular Elements

8–3

Beam shape functions; N1[x_,L_]:=1-3 (x/L)^2+2 (x/L)^3; N2[x_,L_]:=x (1-(x/L))^2; N3[x_,L_]:=3 (x/L)^2 -2 (x/L)^3; N4[x_,L_]:=(x-L) (x/L)^2; Visualize the shape functions; Plot[{N1[x,1],N3[x,1]},{x,0,1}]; Plot[{N2[x,1],N4[x,1]},{x,0,1}]; Define the B matrix; BMat= {D[N1[x,L], {x,2}],D[N2[x,L], {x,2}],D[N3[x,L], {x,2}],D[N4[x,L], {x,2}]} // Simplify Determine the element stiffness matrix in various forms and shapes; Integrate[EI Outer[Times,BMat,BMat], {x, 0, L}] % // MatrixForm % / (EI/L^3) // Simplify % // MatrixForm

8.3 8.3.1

Triangular Elements Cartesian Coordinate System (CST)

Having retrieved the stiﬀness matrices of simple one dimensional elements using the principle of virtual displacement, we next consider two dimensional continuum elements starting with the triangular element of constant thickness t made out of isotropic linear elastic material. The element will have two d.o.f’s at each node: 10

{u} = u1 u2 u3 v1 v2 v3 T 11

The strain displacement relations is required to determine [B]

12

For the 2D plane elasticity problem, the strain vector {ε} is given by: {ε} = εxx εyy γxy T

(8.8)

(8.9)

hence we can rewrite the strains in terms of the derivatives of the shape functions through the matrix [B]:    ∂N  0  εxx  u  ∂x ∂N  0 (8.10) ε = ∂y  v  yy  ∂N ∂N γxy ∂y ∂x [B] 13 We note that because we have 3 u and 3 v displacements, the size of [B] and [u] are 9(3×3)×6 and 6 × 1 respectively.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

STRAIGHT SIDED ELEMENTS; 1st GENERATION

8–4 14

Diﬀerentiating the shape functions from Eq. 6.22 we obtain:   1 1   1 − 0 0 0 0 εxx       x x2  1    2   ∂N3    ε yy     ∂x ∂N1 ∂N2   u1       1     ∂x ∂x γxy          u −x x − x 1       2 3 2 3 2     ε 0 0 0 −    xx   u   3 x y x y y 2 2 3 2 3 3 εyy =  v1     2  ∂N1 ∂N2 ∂N3    γxy           ∂y ∂y ∂y       v 2     3       −x3 1 x3 − x2 1 1 εxx       v − − 0 3   3   εyy  x2 y3 x2 y3 y3 x x2     ∂N3  2    γ3  {u} ∂x ∂N1 ∂N2 xy ∂N1 ∂N1 ∂N3 ∂x ∂x ∂y ∂y ∂x {ε}

(8.11)

[B]

15 With the constitutive matrix [D] given by Eq. ??, the strain-displacement relation [B] by Eq. 8.11, we can substitute those two quantities into the general equation for stiﬀness matrix, Eq. 7.8:

[ke ]

Z

[B]T [D][B]dΩ

= Ω

2

6 6 6 6 6 Ω6 6 4

Z =

− x12

| 2 =

6 6 6 γ6 6 6 4

where α =

x3 −x2 x2 y 3 3 − x−x 2 y3 1 y3 − x12 1 x2

0 0 0

1 x2

0 0 0 0

x3 −x2 x2 y 3 3 − x−x 2 y3 1 y3

0

{z

[B]T

y32 + αx23−2 −y32 − αx3 x3−2 αx2 x3−2 −βy3 x3−2 νx3 y3 + αy3 x3−2 −νx2 y3

1−ν 2 ,

β=

1+ν 2 ,

3 7 2 7 1 ν 7 E 7 4 ν 1 7 7 1 − ν2 0 0 7 5| {z

1−ν ) 2

[D]

32 − x12 56 0 4 }|

x3 −x2 x2 y 3

1 x2

0

0 0

3 − x−x 2 y3

1 y3

ET , 2(1−ν 2 )x2 y3

{z

0

0

x3 −x2 x2 y 3 − x12

3 − x−x 2 y3

1 y3

1 x2

0

αx2 x3−2 −αx2 x3 αx22 −αx2 y3 αx2 y3 0

−βy3 x3−2 νy3 x3−2 + αx3 y3 −αx2 y3 αy32 + x23−2 −αy32 − x3 x3−2 x2 x3−2

νx3 y3 + αy3 x3−2 −βx3 y3 αx2 y3 −αy32 − x3 x3−2 αy32 + x23 −x2 x3

−νx2 y3 νx2 y3 0 x2 x3−2 −x2 x3 x22

x3−2 = x3 − x2 , and y3−2 = y3 − y2 .

16 For the evaluation of the internal stresses {σ} = [D][B]{u} hence for this particular element we will have:   u1             u 1 1   2   0 0 0 0 − x2 1 ν 0    σx  x2 E  u   x −x −x 3 1 3 2 3  0 − 0 0 σy = ν 1 0 x 2 y3 x 2 y3 y3  v1     1 − ν2 x3 −x2 −x3 1 1 1   τxy ) 0 0 1−ν   − − 0   2 x 2 y3 x 2 y3 y3 x2 x2  v2      {σ } [D] v3 [B]

{u}



−y3 y3 0 νx3−2 = κ  −νy3 νy3 0 x3−2 αx3−2 −αx3 αx2 −αy3 Victor Saouma

3

0

  u1        u2      −νx3 νx2   u 3 −x3 x2  (8.12) v1     αy3 0      v    2   v3 II; Solid Mechanics Finite Elements

7 tdxdy 5 | {z } } dV ol

[B]

}

−y32 − αx3 x3−2 y32 + αx23 −αx2 x3 νy3 x3−2 + αx3 y3 −βx3 y3 νx2 y3

γ=

0 0

3 7 7 7 7 7 7 5

Draft

8.3 Triangular Elements where κ =

8.3.2

8–5

E (1−ν 2 )x2 y3

Natural Coordinate System

From Eq. 8.12 we note that the element must be positioned in such a way that neither x2 nor y3 is equal to zero. 17

18

A palliative to this problem is to use a formulation based on natural coordinate systems.

8.3.2.1

Linear, T3

19

The 3 noded triangular element has a linear displacemenent ﬁeld.

20

For the linear triangle, the shape functions are given by Eq. 6.68, thus u =

3

Li ui =

3

i=1

i=1

3

3

Ni ui (8.13)

v

=

Li v i

=

i=1

Ni v i

i=1

or Ni = Li . 21

The displacement shape functions are given by Equation 6.51-b:     2A23 y23 x32  L1  1  L2 2A31 y31 x13  =  2A  L3 2A12 y12 x21

(8.14)

where xij = xi − xj . 22

Derivatives required for strain-displacement relationships are obtained from Eq. 6.54: ∂L1 ∂x ∂L1 ∂y

23

24

= =

y2 −y3 2A ; x3 −x2 2A ;

∂L2 ∂x ∂L2 ∂y

= =

Then the ith part of the B matrix becomes:   Ni,x 0 Ni,y  Bi =  0 Ni,y Ni,x

y3 −y1 2A ; x1 −x3 2A ;

∂L3 ∂x ∂L3 ∂y

= =

(i = 1, 2, 3)

The stiﬀness matrix is then obtained from: BT D BdΩ K =

y1 −y2 2A ; x2 −x1 2A ;

(8.15)

(8.16)

(8.17)

Ω 25 This expression is then analytically integrated using Eq. 6.60. In general an isoparametric numerical integration is preferred (Section 9.2.3).

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

STRAIGHT SIDED ELEMENTS; 1st GENERATION

8–6

8.3.2.2 26

Quadratic Element (T6)

Quadratic displacement shape functions may be written in natural coordinates as: u =

6

Ni ui

v =

i=1

6

Ni vi

(8.18)

i=1

where the shape functions were obtained in Eq. 6.69 N1 = (2L1 − 1)L1

N4 = 4L1 L2

N2 = (2L2 − 1)L2

N5 = 4L2 L3

N3 = (2L3 − 1)L3

N6 = 4L3 L1

(8.19)

27 It should be noted that alternatively, we could have obtained those shape functions by assuming (8.20) u(L1 , L2 ) = a1 + a2 L1 + a3 L2 + a4 L21 + a5 L1 L2 + a6 L22

and then determine the coeﬃcients through satisfaction of the appropriate boundary conditions. The shape functions and their derivatives with respect to the natural coordinates L1 and L2 are given in Table 8.1 (in terms of L1 , L2 , and L3 ). 28

i 1 2 3 4 5 6

Ni (2L1 − 1)L1 (2L2 − 1)L2 (2L3 − 1)L3 4L1 L2 4L2 L3 4L1 L3

Ni,L1 4L1 − 1 0 0 4L2 0 4L3

Ni,L2 0 4L2 − 1 0 4L1 4L3 0

Ni,L3 0 0 4L3 − 1 0 4L2 4L1

Table 8.1: Shape Functions and Derivatives for T6 Element 29

The B matrix is obtained from:   ∂  0  εxx   ∂x ∂ ε =  0 ∂y  yy  ∂ ∂ εxy ∂y ∂x L

  

u v

(8.21)

or      εxx (L1 , L2 , L3 )   εxx  ε ε (L , L , L ) =  yy 1 2 3   yy  εxy εxy (L1 , L2 , L3 )

Victor Saouma

(8.22-a)

Finite Elements II; Solid Mechanics

Draft

8.4 Bilinear Rectangular Element  6

 =  i=1

 1   2A

=

where

8–7

∂Ni (L1 ,L2 ,L3 ) ∂x

0 ∂Ni (L1 ,L2 ,L3 ) ∂y

∂Ni (L1 ,L2 ,L3 ) ∂y ∂Ni (L1 ,L2 ,L3 ) ∂x

∂L1 ∂x

B=LN ∂L2 ∂L3 ∂x ∂x

∂L1 ∂y

∂L2 ∂y

∂L3 ∂y

∂N1 ∂L1 ∂N1 ∂L2 ∂N1 ∂L3

∂N2 ∂L1 ∂N2 ∂L2 ∂N2 ∂L3

∂N3 ∂L1 ∂N3 ∂L2 ∂N3 ∂L3

0 

 [NL ] = 



0

 

ui vi

(8.22-b)

[NL ] [NL ]

∂N4 ∂L1 ∂N4 ∂L2 ∂N4 ∂L3



0 ∂L1 ∂y ∂L1 ∂x

∂N5 ∂L1 ∂N5 ∂L2 ∂N5 ∂L3

∂L2 ∂y ∂L2 ∂x

∂N6 ∂L1 ∂N6 ∂L2 ∂N6 ∂L3

∂L3 ∂y ∂L3 ∂x

[NL ]   [NL ]

  

(8.23-a)

 0 0 4L2 0 4L3 4L1 − 1 0 4L2 − 1 0 4L1 4L3 0  =  0 0 4L3 − 1 0 4L2 4L1 

30

The preceding equation can be further simpliﬁed     Nx 0  εxx  u   εyy 0 Ny = v   εxy Ny Nx

where

 !

NTx |NTy

"

=

   1   2A    

∂L1 ∂x (4L1 − 1) ∂L2 ∂x (4L2 − 1) ∂L3 ∂x (4L3 − 1) ∂L2 1 4 ∂L ∂x L2 + 4 ∂x L1 ∂L3 2 4 ∂L ∂x L3 + 4 ∂x L2 ∂L3 1 4 ∂x L1 + 4 ∂L ∂x L3

u (8.22-c) v

∂L1 ∂y (4L1 − 1) ∂L2 ∂y (4L2 − 1) ∂L3 ∂y (4L3 − 1) ∂L2 1 4 ∂L ∂y L2 + 4 ∂y L1 ∂L3 2 4 ∂L ∂y L3 + 4 ∂y L2 ∂L3 1 4 ∂y L1 + 4 ∂L ∂y L3

(8.23-b)

(8.24)

         

u v = u1 u2 u3 u4 u5 u6 v 1 v 2 v 3 v 4 v 5 v 6

(8.25-a)

(8.25-b)

31 This expression is then analytically integrated using Eq. 6.60. In general an isoparametric numerical integration is preferred (Section 9.2.3).

8.4

Bilinear Rectangular Element

The structure stiﬀness matrix of the bilinear rectangular shown in Fig. 8.1 is described by the Mathematica code shown below. 32

Define shape functions; N1[x_,y_,a_,b_] := (1/4)(1-(x/a))(1-(y/b)); N2[x_,y_,a_,b_] := (1/4)(1+(x/a))(1-(y/b)); Victor Saouma

Finite Elements II; Solid Mechanics

Draft 8–8

STRAIGHT SIDED ELEMENTS; 1st GENERATION

Figure 8.1: Rectangular Bilinear Element N3[x_,y_,a_,b_] := (1/4)(1+(x/a))(1+(y/b)); N4[x_,y_,a_,b_] := (1/4)(1-(x/a))(1+(y/b)); Visualize the shape functions; Plot3D[N1[x,y,1,2], {x, -1, 1}, {y, -2, 2}]; Plot3D[N3[x,y,1,2], {x, -1, 1}, {y, -2, 2}]; Define the differential operators; diffx[f_] := D[f,x]; diffy[f_] := D[f,y]; Shape Function Matrix Nmat = { {N1[x,y,a,b],0,N2[x,y,a,b],0,N3[x,y,a,b],0,N4[x,y,a,b],0}, {0,N1[x,y,a,b],0,N2[x,y,a,b],0,N3[x,y,a,b],0,N4[x,y,a,b]} } B Matrix Bmat = {Map[diffx,Nmat[[1]]],Map[diffy,Nmat[[2]]], Map[diffy,Nmat[[1]]]+Map[diffx,Nmat[[2]]]} Define the constitutive matrix Dmat = E/(1-nu^2) * {{1,nu,0},{nu,1,0},{0,0,(1-nu)/2}} Determine the stiffness matrix Intmat = Transpose[Bmat] . Dmat . Bmat; Kmat =Simplify[ t * Integrate[ Integrate[Intmat, {y, -b, b}], {x, -a, a}] ] TeXForm[Kmat] 33

The resulting stiﬀness matrix is ke1−8;1−4 =

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

8.5 Element Assessment                    

e (−a2 −2 b2 +a2 ν ) t 6 a b (−1+ν) (1+ν) −(e t) 8 (−1+ν) e (−a2 +4 b2 +a2 ν ) t 12 a b (−1+ν) (1+ν) −(e (−1+3 ν) t) 8 (−1+ν) (1+ν) −(e (−a2 −2 b2 +a2 ν ) t) 12 a b (−1+ν) (1+ν) et 8 (−1+ν) −(e (−a2 +b2 +a2 ν ) t) 6 a b (−1+ν) (1+ν) e (−1+3 ν) t 8 (−1+ν) (1+ν)

ke1−8;5−8                    

8–9 −(e t) 8 (−1+ν) −(e (2 a2 +b2 −b2 ν ) t) 6 a b (−1+ν) (1+ν) e (−1+3 ν) t 8 (−1+ν) (1+ν) −(e (a2 −b2 +b2 ν ) t) 6 a b (−1+ν) (1+ν) et 8 (−1+ν) e (2 a2 +b2 −b2 ν ) t 12 a b (−1+ν) (1+ν) −(e (−1+3 ν) t) 8 (−1+ν) (1+ν) e (4 a2 −b2 +b2 ν ) t 12 a b (−1+ν) (1+ν)

e (−a2 +4 b2 +a2 ν ) t 12 a b (−1+ν) (1+ν) e (−1+3 ν) t 8 (−1+ν) (1+ν) e (−a2 −2 b2 +a2 ν ) t 6 a b (−1+ν) (1+ν) et 8 (−1+ν) −(e (−a2 +b2 +a2 ν ) t) 6 a b (−1+ν) (1+ν) −(e (−1+3 ν) t) 8 (−1+ν) (1+ν) −(e (−a2 −2 b2 +a2 ν ) t) 12 a b (−1+ν) (1+ν) −(e t) 8 (−1+ν)



−(e (−1+3 ν) t) 8 (−1+ν) (1+ν) −(e (a2 −b2 +b2 ν ) t) 6 a b (−1+ν) (1+ν)

     et  8 (−1+ν)  −(e (2 a2 +b2 −b2 ν ) t)   6 a b (−1+ν) (1+ν)   e (−1+3 ν) t  8 (−1+ν) (1+ν)   e (4 a2 −b2 +b2 ν ) t  12 a b (−1+ν) (1+ν)   −(e t)  8 (−1+ν)  e (2 a2 +b2 −b2 ν ) t 12 a b (−1+ν) (1+ν)

=

−(e (−a2 −2 b2 +a2 ν ) t) 12 a b (−1+ν) (1+ν) et 8 (−1+ν) −(e (−a2 +b2 +a2 ν ) t) 6 a b (−1+ν) (1+ν) e (−1+3 ν) t 8 (−1+ν) (1+ν) e (−a2 −2 b2 +a2 ν ) t 6 a b (−1+ν) (1+ν) −(e t) 8 (−1+ν) e (−a2 +4 b2 +a2 ν ) t 12 a b (−1+ν) (1+ν) −(e (−1+3 ν) t) 8 (−1+ν) (1+ν)

et 8 (−1+ν) e (2 a2 +b2 −b2 ν ) t 12 a b (−1+ν) (1+ν) −(e (−1+3 ν) t) 8 (−1+ν) (1+ν) e (4 a2 −b2 +b2 ν ) t 12 a b (−1+ν) (1+ν) −(e t) 8 (−1+ν) −(e (2 a2 +b2 −b2 ν ) t) 6 a b (−1+ν) (1+ν) e (−1+3 ν) t 8 (−1+ν) (1+ν) −(e (a2 −b2 +b2 ν ) t) 6 a b (−1+ν) (1+ν)

−(e (−a2 +b2 +a2 ν ) t) 6 a b (−1+ν) (1+ν) −(e (−1+3 ν) t) 8 (−1+ν) (1+ν) −(e (−a2 −2 b2 +a2 ν ) t) 12 a b (−1+ν) (1+ν) −(e t) 8 (−1+ν) e (−a2 +4 b2 +a2 ν ) t 12 a b (−1+ν) (1+ν) e (−1+3 ν) t 8 (−1+ν) (1+ν) e (−a2 −2 b2 +a2 ν ) t 6 a b (−1+ν) (1+ν) et 8 (−1+ν)



e (−1+3 ν) t 8 (−1+ν) (1+ν) e (4 a2 −b2 +b2 ν ) t 12 a b (−1+ν) (1+ν) −(e t) 8 (−1+ν) e (2 a2 +b2 −b2 ν ) t 12 a b (−1+ν) (1+ν) −(e (−1+3 ν) t) 8 (−1+ν) (1+ν) −(e (a2 −b2 +b2 ν ) t) 6 a b (−1+ν) (1+ν) et 8 (−1+ν) −(e (2 a2 +b2 −b2 ν ) t) 6 a b (−1+ν) (1+ν)

                  

(8.26-a)

8.5 8.5.1

Element Assessment CST

34 The CST gives good results in a region of the FE model where there is little strain gradient. Otherwise, the CST will not work well. 35 A natural extension would be to go linear, i.e. LST. This element would have midside nodes in addition ot vertex nodes. 36 We should note that for this element the stress is independent of x and y because a linear displacement relation was assumed resulting in a constant strain and stress (for linear elastic material). 37 If we consider the element shown in Fig. 8.2 we apply a moment M such that: 1) origin does not move, and 2) the displacement is u = u at x = l on top of the beam, then from the exact solution (Timoshenko and Goodier 1970), the correct displacements are

u= Victor Saouma

u xy cl

v=

u (−x2 − νy 2 ) 2cl

(8.27)

Finite Elements II; Solid Mechanics

Draft 8–10

STRAIGHT SIDED ELEMENTS; 1st GENERATION

Figure 8.2: Linear Triangular Element Subjected to Pure Bending diﬀerentiating, we recover the anticipated strain ﬁeld εxx =

u y cl

εyy = −ν

u y cl

γxy = 0

(8.28)

38 Imposing nodal displacements consistent with theory (u1 = −u, u2 = u, and u3 = 0) and similar descriptions for v, and determining the strains from ε = Bu for the CST element, we obtain c 1 −ν 2 u εyy = 0 γxy = (8.29) εxx = 0 c 4l

which are clearly wrong. Hence, we conclude the CST element can not properly perform in bending problems (As is expected from a constant strain element behavior when used in a linear strain problem). This deﬁciency will later be addressed by including an additional internal “drilling” degree of freedom. 39

40

Zero energy mode can be veriﬁed by considering for example the strain εxx which is given by εxx =

1 (−y3 u1 + y3 u2 ) x2 y3

(8.30)

since u1 = u2 for a rigid body motion, we see that εxx = 0, similar conditions can be found for εyy and εxy

8.5.2 41

BiLinear Rectangular

Under ﬂexure, the top and bottom sides remain straight giving rise to erroneous strains.

42 From Pascals’s triangle (Sect. 6.4), the displacement ﬁeld for a bilinear rectangular element is given by (8.31) u = a1 + a2 x + a3 y + a5 xy

which is consistent with the element shape functions derived earlier (Eq. 6.31) 43 We note that whereas there is a quadratic term (a5 xy) convergence is governed by the full linear expansion, and that this last term may cause parasitic eﬀects.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

8.5 Element Assessment

8–11

44 For an element with an inclined boundary given by y = ax + b, insertion of this expression into Eq. 8.31 yields (8.32) u = a1 + a3 b + (a2 + a3 a + a5 b)x + a5 ax2

hence, u varies quadratically with x along the inclined boundary. However, this boundary is uniquely deﬁned by only two nodal displacements u1 and u2 , thus the quadratic expression is not uniquely deﬁned and we may have diﬀerent variation along the side from one element to the adjacent one. 45

Continuity across the element not being satisﬁed, the element is non-conforming.

We conclude that the bilinear quadrilateral element must always have its four edges parallel to the coordinate system. 46

47 This severe restriction will be alleviated in the next lecture by Isoparametric element formulations.

Reﬁnement of this element include the addition of two degrees of freedom to allow a state of constant curvature 48

u = Σni=1 Ni ui + (1 − ξ 2 )g1 + (1 − η 2 )g2

(8.33-a)

+ (1 − ξ )g3 + (1 − η )g4

(8.33-b)

v =

Σni=1 Ni v i

2

2

Other reﬁnements (mostly in shells) include the inclusion of additional rotational degrees of freedom termed drilling dof. 49

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 8–12

Victor Saouma

STRAIGHT SIDED ELEMENTS; 1st GENERATION

Finite Elements II; Solid Mechanics

Draft Chapter 9

ISOPARAMETRIC ELEMENTS; 2nd GENERATION 9.1

Introduction

1 We have previously examined simple ﬁnite elements, in this lecture we shall distort those simpler elements into others of more arbitrary shape.

Correspondingly, the natural coordinates will be distorted into new curvilinear sets when plotted in a cartesian x, y, z space, Fig. 9.1. 2

η (-1,1)

η (1,1)

ξ

ξ (-1,-1)

(1,-1)

3 L 1 =0

L 2 =0

1

L 3 =0

2

Figure 9.1: Two-Dimensional Mapping of Some Elements In the isoparametric formulation, displacements are expressed in terms of natural coordinates, however they must be diﬀerentiated with respect to cartesian coordinates x, y, z. This is accomplished through a transformation matrix J, and integration can no longer be performed analytically but must be done numerically. 3

4

Natural coordinates range from -1 to +1, Fig. 9.2

5

Nodal displacements at any point inside the element can be written in terms of the nodal

Draft

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

9–2

y,v

η Mapping

x,u

(1,1)

(-1,1)

ξ

(-1,-1)

(1,-1)

Parent Element

Actual Element

Figure 9.2: Actual and Parent Elements known displacements and the shape functions

  u   1   u 2 = Nue u = N1 u1 + N2 u2 + · · · = N    ..   .     v1  v2 = Nve v = N1 v 1 + N2 v 2 + · · · = N    ..   .     w1  w2 = Nwe w = N1 w 1 + N2 w2 + · · · = N    ..  .

(9.1)

or u = u v w T = [N]ue

(9.2)

When elements are also distorted, the coordinates of any point can also be expressed in terms of nodal coordinates   x   1   ˜ ˜ ˜ x ˜ 2 = Nx x = N1 x1 + N2 x2 + · · · = N    ..   .     y1  ˜ ˜ y2 ˜2 y 2 + · · · = N ˜1 y 1 + N (9.3) = Ny y = N    ..   .     z1  ˜ ˜ ˜ z2 ˜ = Nz z = N1 z 1 + N2 z 2 + · · · = N    ..  .

6

or ˜ c = x y z T = [N]c

(9.4)

˜ are functions of ξ, η, and ζ. Hence, three possible cases Shape function matrices [N] and [N] arise, Fig. 9.3 7

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

9.2 Element Formulation

9–3

Geometric Nodal Config.

Displacement Nodal Config.

Isoparametric

Subparametric

Superparametric

Figure 9.3: Iso, Super, and Sub Parametric Elements ˜ are identical Isoparamteric: [N] and [N] ˜ Subparamteric: [N] is of higher degree than [N] ˜ Superparamteric: [N] is of lower degree than [N] 8

Sub and Super parametric elements are very seldom used.

9.2 9.2.1

Element Formulation Bar Element

9 The simplest introduction to isoparamteric elements is through a straight three noded quadratic elements, Fig. 9.4.

ξ= −1

ξ= 0

ξ= +1

1

3

2

x,u L

Figure 9.4: Three-Noded Quadratic Bar Element 10 The shape functions for the element can be obtained from the Lagrangian interpolation function used earlier, Eq. 6.25, and in which we substitute x by ξ. The kth term in a polynomial

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

9–4

of order n − 1 would be Nkn

-n

=

(ξ − ξi ) -ni=1,i =k i=1,i =k (ξk − ξi )

(9.5-a)

=

(ξ − ξ1 )(ξ − ξ2 ) · · · (ξ − ξk−1 )(ξ − ξk+1 ) · · · (ξ − ξn ) (ξk − ξ1 )(ξk − ξ2 ) · · · (ξk − ξk−1 )(ξk − ξk+1 ) · · · (ξk − ξn )

(9.5-b)

For a three noded quadratic element ξ1 = −1, ξ2 = +1, and ξ3 = 0. Substituting, we obtain the three shape functions N1 (ξ) = N2 (ξ) = N3 (ξ) =

(ξ−ξ2 )(ξ−ξ3 ) (ξ1 −ξ2 )(ξ1 −ξ3 ) (ξ−ξ1 )(ξ−ξ3 ) (ξ2 −ξ1 )(ξ2 −ξ3 ) (ξ−ξ1 )(ξ−ξ2 ) (ξ3 −ξ1 )(ξ3 −ξ2 )

= = =

(ξ−1)(ξ−0) (−1−1)(−1−0) (ξ+1)(ξ−0) (1+1)(1−0) (ξ+1)(ξ−1) (0+1)(0−1)

= =

1 2 2 (ξ 1 2 2 (ξ

− ξ) + ξ)

(9.6)

= 1 − ξ2

Hence, x(ξ) = N x1 x2 x3 T

and

u(ξ) = N u1 u2 u3 T

(9.7)

where N =

1 2 2 (ξ

− ξ)

1 2 2 (ξ

+ ξ) 1 − ξ 2

(9.8)

The strain displacement relation is given by Eq. 7.4, ε = Lu = LNue = Bue , and the diﬀerential operator L is equal to d . For this one dimensional case, this reduces to dx   u1   d du = N u (9.9) εx =  2  dx dx u3 L 11

B

12 We invoke the chain rule since the shape functions are expressed in terms of natural coordinates: dN dξ dN = (9.10) B= dx dξ dx

The ﬁrst term may be readily available from the shape functions, Eq. 9.8, however the second one is not. dξ is not available, we may determine its inverse dx , from Eq. 9.7, which we shall 13 Whereas, dx dξ denote by J or Jacobian. The Jacobian operator J is a scale factor which relates cartesian to natural coordinates dx = Jdξ.      x1   x1  d dx x2 x2 = N (9.11) = 1 (2ξ − 1) 12 (2ξ + 1) −2ξ J(ξ) =   2   dξ dξ x3 x3 dN dξ dx dξ 14

We can rewrite Eq. 9.10 as B=

Victor Saouma

dξ dN dN = (9.12) dx dx dξ J −1 Finite Elements II; Solid Mechanics

Draft

9.2 Element Formulation

9–5

and the B matrix is thus obtained by substituting into Eq. 9.9 B(ξ) =

15

1 1 d N = J dξ J

1 2 (2ξ

− 1)

1 2 (2ξ

+ 1) −2ξ

The diﬀerential area is dΩ = Adx = AJdξ

16

(9.13)

(9.14)

Substituting, the element stiﬀness matrix is ﬁnally obtained from Eq. 7.8

L T

Ke (ξ) =

+1

B (ξ)AEB(ξ)dx =

BT (ξ)AEB(ξ)J(ξ)dξ

(9.15)

−1

0

17 We observe that B, in general, contains ξ terms in both the numerator and denominator, and hence the expression can not be analytically inverted. Furthermore, the limits of integration are now from -1 to +1, and we shall see later on how to numerically integrate it. 18 A simple Mathematica code to generate the stiﬀness matrix of three noded (quadratic) element:

x1=0;x2=L;x3=L/2; N1[x_,L_]:=(x-x2) * (x-x3)/( (x1-x2) * (x1-x3)); N2[x_,L_]:=(x-x1) * (x-x3)/( (x2-x1) * (x2-x3)); N3[x_,L_]:=(x-x2) * (x-x1)/( (x3-x2) * (x3-x1)); BMat={D[N1[x,L],x], D[N2[x,L],x], D[N3[x,L],x]}; Integrate[EA Outer[Times, BMat, BMat], {x,0,L}]; MatrixForm[%]

9.2.2 9.2.2.1

Quadrilaterals Linear Element (Q4)

19 We have previously derived the stiﬀness matrix of a rectangular element (aligned with the coordinate axis), this formulation will generalize it to an arbitrary quadrilateral shape. 20

Fig. 9.5 illustrates the element η

ξ= 1 ξ= 1/2

3

η= 1

η

ξ= −1/2

η= 1/2

ξ= −1 4

ξ

1 4

1 3 1 ξ

η= −1/2

y,v

1

2

η= −1

1 1

2

x,u

Figure 9.5: Four Noded Isoparametric Element

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

9–6 21

For the two-dimensional case u(ξ, η) =

Nij uk =

n

m

Nj (ξ)Ni (η)uk

(9.16)

i=1 j=1

where k = (i − 1)m + j. For a bilinear element, n = m = 2, this can be rewritten as u1 u3 N1 (η) u(ξ, η) = N1 (ξ) N2 (ξ) = NTξ uNη u2 u4 N2 (η) = N1 (ξ)N1 (η)u1 + N2 (ξ)N1 (η)u2 + N1 (ξ)N2 (η)u4 + N2 (ξ)N2 (η)u3 = N1 (ξ, η)u1 + N2 (ξ, η)u2 + N3 (ξ, η)u3 + N4 (ξ, η)u4 4

Ni ui =

(9.17-a) (9.17-b)

i=1

22

Applying the Lagrangian interpolation equation, Eq. 9.5-a we obtain N1 (ξ) = N2 (ξ) = N1 (η) = N2 (η) =

(ξ − 1) 1 (ξ − ξ2 ) = = (1 − ξ) (ξ1 − ξ2 ) (−1 − 1) 2 (ξ + 1) 1 (ξ − ξ1 ) = = (1 + ξ) (ξ2 − ξ1 ) (1 + 1) 2 (η − 1) 1 (η − η2 ) = = (1 − η) (η1 − η2 ) (−1 − 1) 2 (η + 1) 1 (η − η1 ) = = (1 + η) (η2 − η1 ) (1 + 1) 2

(9.18-a) (9.18-b) (9.18-c) (9.18-d)

Substituting into Eq. 9.17-a N1 (ξ, η) = N3 (ξ, η) =

1 4 (1 1 4 (1

− ξ)(1 − η); N2 (ξ, η) = + ξ)(1 + η); N4 (ξ, η) =

1 4 (1 1 4 (1

+ ξ)(1 − η); − ξ)(1 + η);

(9.19)

It should be noted that for this simple case, the shape functions could have been determined by mere inspction. 23

24

Coordinates and displacements are given by ) ) x = ) Ni (ξ, η)xi ; y = ) Ni (ξ, η)y i u = Ni (ξ, η)ui ; v = Ni (ξ, η)v i The strain displacement relation is given by Eq. 7.4   ∂  0 ε  xx   ∂x ∂ 0 εyy = {ε} = ∂y   ∂ ∂ γxy ∂y ∂x L

  

(9.20)

u v

(9.21)

u

However the displacements can be obtained from Eq. 9.20

u v

| {z } u

=

|

Victor Saouma

N1 (ξ, η) 0

0 N1 (ξ, η)

N2 (ξ, η) 0

8 u 9 1 > > > > > v1 > > > > > > > > u2 > > > > < v > = 0 N3 (ξ, η) 0 N4 (ξ, η) 0 2 (9.22) N2 (ξ, η) 0 N3 (ξ, η) 0 N4 (ξ, η) u > > 3 > > > > {z }> v3 > > > > > > > > N > > : u4 > ; v4 Finite Elements II; Solid Mechanics | {z } u

Draft

9.2 Element Formulation 25

9–7

Combining Eq. 9.21 and 9.22 yields ε = LNu = Bu  ∂Ni (ξ,η)   4  εxx (ξ, η)  ∂x

 0 εyy (ξ, η) =    ∂N i (ξ,η) i=1 γxy (ξ, η) ∂y



0

 

∂Ni (ξ,η) ∂y ∂Ni (ξ,η) ∂x

B=LN

 = 

N2,x 0 N2,y

0 N2,y N2,x

ui vi

(9.23-a)

u

N3,x 0 N3,y

0 N3,y N3,x

N4,x 0 N4,y

0 N4,y N4,x

B

 =

0 N1,y N1,x

N1,x 0 N1,y

4

 

0

0

∂Ni (ξ,η) ∂η ∂Ni (ξ,η) ∂ξ

∂Ni (ξ,η) ∂η

i=1



∂Ni (ξ,η) ∂ξ

 

#

∂ξ ∂x ∂ξ ∂x

∂η ∂x ∂η ∂y

                       

u1 v1 u2 v2 u3 v3 u4 v4

                      

$ (9.23-b)

[J]−1

26 Considering the local set of coordinates ξ, η and the corresponding global one x, y, the chain rules would give # $

∂Ni ∂ξ ∂Ni ∂η

∂Ni ∂x ∂Ni ∂y

∂x ∂ξ ∂x ∂η

=

J

−1

= [J]

∂y ∂ξ ∂y ∂η

∂Ni ∂x ∂Ni ∂y

(9.24-a)

∂Ni ∂ξ ∂Ni ∂η

(9.24-b)

This last equation is the key to get all the components which will go inside the B matrix. 27

Expanding the deﬁnition of the Jacobian # $ ∂N (ξ,η) i

∂ξ ∂Ni (ξ,η) ∂η

= # =

=

1 4

∂x ∂ξ ∂x ∂η

J

∂N1 ∂ξ ∂N1 ∂η

∂y ∂ξ ∂y ∂η

∂N2 ∂ξ ∂N2 ∂η

∂Ni ∂x ∂Ni ∂y

∂N3 ∂ξ ∂N3 ∂η

=

4

i=1

∂N4 ∂ξ ∂N4 ∂η

$



#

∂Ni ∂ξ xi ∂Ni ∂η xi

x1  x2   x3 x4

 y1 y2   y3  y4

J

∂Ni ∂x ∂Ni ∂y

(9.25-a)

∂Ni ∂x ∂Ni ∂y

−(1 − η) (1 − η) (1 + η) −(1 + η) −(1 − ξ) −(1 + ξ) (1 + ξ) (1 − ξ)

$

∂Ni ∂ξ y i ∂Ni ∂η y i

(9.25-b) 

x1  x2   x3 x4

 y1 y2   y3  y4

∂Ni ∂x ∂Ni ∂y

(9.25-c)

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

9–8

28

Back to the Jacobian # $ ∂ξ ∂x ∂ξ ∂y

[J]−1 =

def

29

∂η ∂x ∂η ∂y

1 = detJ

#

∂y ∂η − ∂x ∂η

− ∂y ∂ξ

$

∂x ∂ξ

1

= detJ 4

#

i=1

∂Ni ∂ξ y i i − ∂N ∂η xi

i − ∂N ∂η y i ∂Ni ∂ξ xi

$ (9.26)

From calculus, if ξ and η are some arbitrary curvilinear coordinates, Fig. 9.6, then y ✻

η ∂y ∂η dη

✄

ds ✄ ∂y ∂ξ dξ

✄ ✄

✄

✄

B

✄✄✗

C

✄

dΩe

✿ ✘ A✘✘ ✘✘ ✘ ✘ dr✘ ✄ ✘✘✘ ✄ ✘ ✘

O

∂x ∂η dη

ξ ✲

x

∂x ∂ξ dξ

Figure 9.6: Diﬀerential Element in Curvilinear Coordinate System dr =

∂x ∂ξ ∂y ∂ξ

dξ

and

ds =

∂x ∂η ∂y ∂η

dη

(9.27)

are vectors directed tangentially to ξ = constant, and η = constant respectively. 30

From vector algebra, the cross product of two vectors lying in the x-y plane, Fig. 9.7 is C=A x B

B

|B| sinθ θ A

Figure 9.7: Cross Product of Two Vectors C = A×B = |A||B| sin θk Victor Saouma

(9.28-a) (9.28-b) Finite Elements II; Solid Mechanics

Draft

9.2 Element Formulation

9–9

i j k = Ax Ay 0 = (Ax By − Bx Ay ) k Bx By 0 Area |C| = |A||B| sin θ

(9.28-c) (9.28-d)

hence, the diﬀerential area dxdy is then equal to the length of the vector resulting from the cross product of drds and is equal to # d(area) = dxdy = det

31

∂x ∂ξ ∂y ∂ξ

$

∂x ∂η ∂y ∂η

dξdη

J

(9.29)

Finally determine the element stiﬀness matrix from [k]8×8 =

[B]T8×3 [D]3x3 [B]3×8 tdxdy

1

1

= [k] = −1

[B]T [D][B]t|J|dξdη

(9.30)

−1

The evaluation of the element stiﬀness matrix involves dA. If we consider an inﬁnitesimal element, of length dr and ds, at the vertex of an element, it has the boundaries of the element as its sides. Then, from Eq. 9.28-d 32

dA = dx.dy. sin θ

(9.31)

however, from Eq. 9.29 we have dA = detJdξ.dη, thus detJ =

dx.dy sin θ dξ.dη

(9.32)

Thus we observe that if θ is small or close to 180o , then det J will be very small, if the angle is greater than 180 o , the determinant is negative (implying a negative stiﬀness which will usually trigger an error/stop in a FE analysis). 33

In general it is recommended that 30o < θ < 150o .

34 The inverse of the jacobian exists as long as the element is not much distorted or folds back upon itself, Fig. ?? in those cases there is no unique relation between the coordinates.

Figure 9.8: *Elements with Possible Singular Jacobians 35 It can be easily shown that for parallelograms, the Jacobian is constant, whereas for nonparallelograms it is not.

In general J is an indicator of the amount of element distorsion with respect to a 2x2 square one. Some times it is constant, others it varies within the element. 36

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

9–10

9.2.2.1.1 Example: Jacobian Operators, (Bathe 1996) Determine the Jacobian operators J for the following 2 dimensional elements. (Bathe 1982).

y 2

y

1

1 1 x

x

4 3

1

3/4 3

4

6

2

4 2

y

1

2

x 1

60

3

6

4

The coordinates are given by Eq. 9.4, the shape functions by Eq. 9.19, and the Jacobian by Eq. 9.24-b. Element 1: x =

=

y =

=

[J] =

1 1 (1 − ξ)(1 − η)x3 + (1 + ξ)(1 − η)x4 4 4 1 1 + (1 + ξ)(1 + η)x1 + (1 − ξ)(1 + η)x2 4 4 1 1 (1 − ξ)(1 − η)(−3) + (1 + ξ)(1 − η)(3) 4 4 1 1 + (1 + ξ)(1 + η)(3) + (1 − ξ)(1 + η)(−3) 4 4 1 1 (1 − ξ)(1 − η)y 3 + (1 + ξ)(1 − η)y 4 4 4 1 1 + (1 + ξ)(1 + η)y 1 + (1 − ξ)(1 + η)y 2 4 4 1 1 (1 − ξ)(1 − η)(−2) + (1 + ξ)(1 − η)(−2) 4 4 1 1 + (1 + ξ)(1 + η)(2) + (1 − ξ)(1 + η)(2) 4 4 3 0 0 2

(9.33-a)

(9.33-b)

(9.33-c)

(9.33-d) (9.33-e)

We note that A = 24 = det[J](2 × 2) = 6 × 4 Element 2: √ √ 1 1 (1 − ξ)(1 − η)(−(3 + 1/(2 3))) + (1 + ξ)(1 − η)(3 − 1/2 3)) 4 4 √ √ 1 1 + (1 + ξ)(1 + η)(3 + 1/2 3)) + (1 − ξ)(1 + η)(−(3 − 1/2 3))) 4 4 1 1 (1 − ξ)(1 − η)(−2) + (1 + ξ)(1 − η)(−2) y = 4 4

x =

Victor Saouma

(9.34-a)

Finite Elements II; Solid Mechanics

Draft

9.2 Element Formulation

[J] =

9–11

1 1 + (1 + ξ)(1 + η)(2) + (1 − ξ)(1 + η)(2) 4 $ #4 3 0 1 √ 2 3

(9.34-b) (9.34-c)

1 2

Element 3: 1 1 (1 − ξ)(1 − η)(−1) + (1 + ξ)(1) 4 4 1 1 + (1 + ξ)(1 + η)(1) + (1 − ξ)(−1) 4 4 1 1 y = (1 − ξ)(1 − η)(−3/4) + (1 + ξ)(1 − η)(−3/4) 4 4 1 1 + (1 + ξ)(1 + η)(5/4) + (1 − ξ)(1 + η)(1/4) 4 4 1 4 (1 + η) [J] = 4 0 (3 + ξ) x =

9.2.2.2 37

(9.35-a)

(9.35-b) (9.35-c)

Quadratic Element

For a quadratic quadrilateral element, there are two possibilities, Fig. 9.9. η 4

η 4

7

7

3

3

8

8

6

9 6

ξ 1

ξ

1 5 2

5 2

Figure 9.9: Serendipity and Lagrangian Quadratic Quadrilaterals 38 The Pascal triangle, Fig. 9.10, will be later used to justify the choice of terms in the displacement ﬁeld of isoparameteric elements.

9.2.2.2.1

39

Serendipity Element (Q8)

Based on Pascal’s triangle, the displacement ﬁeld is given by u = a1 + a2 x + a3 y + a4 x2 + a5 xy + a6 y 2 + a8 x2 y + a9 xy 2 0 1 2 3

(9.36)

In this formulation, the x2 y 2 term is missing and the 8 terms in the assumed polynomial expansion correspond the 8 nodes (4 corner and 4 midside). 40

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

9–12

1

Constant term

η

ξ

Linear terms ξ2

Quadratic terms ξ3

Cubic terms

η2

ξη ξ2η

ξη2

ξ3η

Quartic terms

Linear element Quadratic element η3

Cubic element

ξη3

Serendipity Quadrilateral Elements 1

Constant term

ξ2

Quadratic terms Cubic terms

η

ξ

Linear terms

ξ3 ξ3η

Quartic terms

ξη2 ξ 2 η2

ξ 3 η2

Quintic terms

η2

ξη ξ2η

Linear element Quadratic element η3

Cubic element

ξη 3 ξ 2 η3

ξ 3 η3

Sixtic terms

Lagrangian Quadrilateral Elements 1 Constant term

ξ2

Quadratic terms Cubic term

η

ξ

Linear terms

ξ3

η2

ξη ξ2η

Linear elements

ξη2

Quadratic element η3

Cubic element

Triangle Elements Figure 9.10: Pascal Triangles for Quadrilateral and Triangle Elements

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

9.2 Element Formulation

9–13

y,v

η 4

7

4

3

7

3

1 6

8

6

8

ξ

1 5

1

2

1

5 1

x,u

2 1

Figure 9.11: Serendipity Isoparametric Quadratic Finite Element: Global and Parent Element 41

The shape functions may be obtained by mere inspection (i.e. serependitiously), Ni = Ni = Ni =

1 4 1 2 1 2

(1 % + ξξ2i&) (1 + ηηi ) (ξξi + ηηi − 1) i = 1, 2, 3, 4 i = 5, 7 1 − ξ (1 % + ηηi&) (1 + ξξi ) 1 + η 2 i = 6, 8

(9.37)

and are tabulated in Table 9.1. i

Ni

1 2 3 4 5 6 7 8

1 4 (1 − ξ)(1 − η)(−ξ − η − 1) 1 4 (1 + ξ)(1 − η)(ξ − η − 1) 1 4 (1 + ξ)(1 + η)(ξ + η − 1) 1 4 (1 − ξ)(1 + η)(−ξ − η − 1) 1 2 2 (1 − ξ )(1 − η) 1 2 2 (1 + ξ)(1 − η ) 1 2 2 (1 − ξ )(1 + η) 1 2 2 (1 − ξ)(1 − η )

Ni,ξ + η)(1 − η) − η)(1 − η) + η)(1 + η) − η)(1 + η) −ξ(1 − η) 1 2 2 (1 − η ) −ξ(1 + η) − 12 (1 − η 2 )

1 4 (2ξ 1 4 (2ξ 1 4 (2ξ 1 4 (2ξ

Ni,η

− ξ)(2η + ξ) + ξ)(2η − ξ) + ξ)(2η + ξ) − ξ)(2η − ξ) − 12 (1 − ξ 2 ) −(1 + ξ)η 1 2 2 (1 − ξ ) −(1 − ξ)η

1 4 (1 1 4 (1 1 4 (1 1 4 (1

Table 9.1: Shape Functions, and Natural Derivatives for Q8 Element

42

The shape functions for the corner and midisde nodes are shown in Fig. 9.12.

9.2.2.2.2

Lagrangian element (Q9)

43 If we were to follow a similar procedure to the one adopted to extract the bilinear shape fuctions, we would obtain 9 shape functions, which must in turn correspond to 9 (rather than

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

9–14

1

1 0.8

0.5

0.6 0 0.4 −0.5

0.2

−1 1

0 1 1

0.5

1

0.5

0.5

0 0

−0.5

0

−0.5

−0.5 −1

0.5

0 −0.5 −1

−1 0

1

1

−1

0.1 0.2

0.8

0.8

0.6 0.6

0.6

−0.4

0.4

0.4

0.2

0.2

0

0 −0.2

−0.2

0.4

0.7

−0.2

0.3

0.8

−0.4

−0.4 −0.6

−0.6

0.2 0.4

−0.8 0.8 −1 −1

0.5 0.9

−0.8

0.6 −0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

−1 −1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Figure 9.12: Shape Functions for 8 Noded Quadrilateral Element 8) nodes. 44

In this element, the displacement ﬁeld is given by u = a1 + a2 x + a3 y + a4 x2 + a5 xy + a6 y 2 + a8 x2 y + a9 xy 2 + a12 x2 y 2 0 1 2 3 4

(9.38)

Serendipity 45 All the quadratic terms are present, hence there are 9 terms in the polynomial expansion, and the 9th node will correspond to an internal node. 46 The shape functions in this case can be directly obtained from the Lagrangian interpolation functions, yielding

Ni Ni Ni N9

= 14 (1 % + ξξi&) (1 + ηηi ) (ξξi + ηηi − 1) = 12 1 − ξ 2 (1 % + ηη2i&) 1 = 2 (1 + ξξi ) 1 − η = (1 − ξ 2 )(1 − η 2 )

i = 1, 2, 3, 4 i = 5, 7 i = 6, 8 i=9

47

The last shape function is often called bubble function

48

Those shape function diﬀer slightly from those of the serendipity element.

Victor Saouma

(9.39)

Finite Elements II; Solid Mechanics

Draft

9.2 Element Formulation

9–15

49

Q9 elements perform much better than the Q8 if edges are not parallel or slightly curved.

50

The shape functions for the corner and midisde nodes are shown in Fig. 9.13.

1 1

0.5

0.8 0.6

0 0.4 0.2

−0.5

0

−1 1

−0.2 1

1

0.5

0

−0.5

−0.5 −1

0.5

0

0

−0.5

−0.5 −1

−1

1

1

0.8

0.8

0.6

0.6

0.4

−1

0.6

0.4 −0.6

0.2

1

0.5

0.5

0

0.2

−0.2

0

0 −0.4

−0.2

−0.2 0.8

−0.4

−0.4 0.4

−0.6

0

−0.8 0.8 −1 −1

−0.8

−0.8

0.2

0.6

−0.6 0.2

0.4 −0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

−1 −1

1

0 −0.8

−0.6

−0.4

−0.2

1

0

0.2

0.4

0.6

0.8

1

0.1

0.8

0.4

1

0.6

0.6 0.2

0.8

0.4 0.8

0.6

0.2

0.4

0

0.2

−0.2

0.9

−0.4

0 1

0.7

1

0.5 0.5

0 0

−0.5

−0.5 −1

−1

0.5

−0.6 0.3

−0.8 −1 −1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Figure 9.13: Shape Functions for 9 Noded Quadrilateral Element

9.2.2.2.3 Variable (Hierarchical) Element 51 Based on the above, we can generalize the formulation to one of a quadrilateral element with variable number of nodes. 52 This element may have diﬀerent order of variation along diﬀerent edges, and is quite useful to facilitate the grading of a ﬁnite element mesh. 53

In its simplest formulation, it has four nodes, and has a linear variation along all sides, and

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

9–16

in the most general case it is a full quadratic element. The shape functions may than be obtained from Table 9.2. Note that these shape functions are for the hierarchical element in which the corner nodes are numbered ﬁrst, and midside ones after, Fig. 9.14. 54

4

7

3

9

8

1

6

5

2

Figure 9.14: Nodal Numbering for Isoparameteric Elements

N1 N2 N3 N4 N5 N6 N7 N8 N9

1 4 (1 − ξ)(1 − η) 1 4 (1 + ξ)(1 − η) 1 4 (1 + ξ)(1 + η) 1 4 (1 − ξ)(1 + η) 1 2 2 (1 − ξ )(1 − η) 1 2 2 (1 + ξ)(1 − η ) 1 2 2 (1 − ξ )(1 + η) 1 2 2 (1 − ξ)(1 − η ) (1 − ξ 2 )(1 − η 2 )

i=5 − 12 N5 − 12 N5

Only if node i is present i=6 i=7 i=8 − 12 N8 1 − 2 N6 − 12 N6 − 12 N7 − 12 N7 − 12 N8

i=9 1 4 N9 1 4 N9 1 4 N9 1 4 N9 − 12 N9 − 12 N9 − 12 N9 − 12 N9

Table 9.2: Shape Functions for Variable Node Elements

9.2.3

Triangular Elements

For the six noded triangle, the partial derivatives of a variable φ with respect to x and y can be expressed as, (Felippa 2000)   # $  )6 φi ∂Ni   )i=1 ∂L1  ∂φ ∂L2 ∂L3 ∂L1 6 ∂Ni ∂x ∂x ∂x ∂x = ∂L (9.40) ∂L ∂L ∂φ i=1 φi ∂L2  1 2 3 )  6 ∂y ∂y ∂y  ∂Ni  ∂y φ i=1 i ∂L3 55

56

Transposing both sides 

)6

∂Ni i=1 φi ∂L1

Victor Saouma

)6

∂Ni i=1 φi ∂L2

)6

∂Ni i=1 φi ∂L3

 

∂L1 ∂x ∂L2 ∂x ∂L3 ∂x

∂L1 ∂y ∂L2 ∂y ∂L3 ∂y

  =

∂φ ∂x

∂φ ∂y

(9.41)

Finite Elements II; Solid Mechanics

Draft

9.3 Numerical Integration 57

9–17

We now make φ ≡ 1, x, and y:  )6 )6 ∂Ni ∂Ni i=1 i=1 ∂L2 ∂L 1  )6 ∂Ni )6 i x xi ∂N  i ∂L1 ∂L2 )i=1 )i=1 6 6 ∂Ni ∂Ni i=1 yi ∂L1 i=1 yi ∂L2

)6 ∂Ni   3 )6 i=1 ∂L ∂Ni   x i ∂L3   )i=1 6 ∂Ni i=1 yi ∂L3

∂L1 ∂x ∂L2 ∂x ∂L3 ∂x

∂L1 ∂y ∂L2 ∂y ∂L3 ∂y





  =

∂1 ∂x ∂x ∂x ∂y ∂x

∂1 ∂y ∂x ∂y ∂y ∂y

  

(9.42)

∂y ∂y ∂1 ∂1 ∂x But ∂x ∂x = ∂y = 1, and ∂x = ∂y = ∂y = ∂x = 0 since x and y are independent coordinates. Furthermore all entries in the ﬁrst row are equal to a constant (3 for the T6 element), and since the corresponding right hand side, this row can be normalized, yielding the Jacobian matrix for this element   ∂L     ∂L1 1 1 1 1 0 0 ∂x ∂y ) ) )  6 6 6 ∂N ∂N ∂N   ∂L ∂L  (9.43)  )i=1 xi ∂L1i )i=1 xi ∂L2i )i=1 xi ∂L3i   ∂x2 ∂y2  =  1 0  6 6 6 ∂Ni ∂Ni ∂Ni ∂L3 ∂L3 0 1 y y y i=1 i ∂L1 i=1 i ∂L2 i=1 i ∂L3 ∂x ∂y 58

J

59

Substituting from Table 8.1 we obtain 2

1 4 x1 (4L1 − 1) + 4x4 L2 + 4x6 L3 y1 (4L1 − 1) + 4y4 L2 + 4y6 L3

|

1 x2 (4L2 − 1) + 4x5 L3 + 4x4 L1 y2 (4L2 − 1) + 4y5 L3 + 4y4 L1

2 6 4

∂L1 ∂x ∂L2 ∂x ∂L3 ∂x

∂L1 ∂y ∂L2 ∂y ∂L3 ∂y

{z J 3

2

0 7=4 1 5 0

3

1 x3 (4L3 − 1) + 4x6 L1 + 4x5 L2 5 y3 (4L3 − 1) + 4y6 L1 + 4y5 L2

}

3

(9.44)

0 0 5 1

60 Next we invert the matrix and solve for the six triangular coordinates partials and substitute in Eq. 9.40 which in turn will enable us to determine the B matrix in Eq. 9.23-a  ∂Ni (L1 ,L2 ,L3 )    0 6  εxx (L1 , L2 , L3 ) 

∂x ui  ∂Ni (L1 ,L2 ,L3 )  0 εyy (L1 , L2 , L3 ) = (9.45)   ∂y vi   ∂N ∂N (L ,L ,L ) (L ,L ,L ) 1 2 3 1 2 3 i i i=1 γxy (L1 , L2 , L3 ) ∂y ∂x u

B=LN

9.3

Numerical Integration

Understanding numerical integration is not only essential for a proper integration of the isoparameteric family of elements, but also helpful in understanding the Weighted Residual methods (Chapter 11), 61

A crucial aspect of isoparametric element formulation is the numerical integration which can be expressed as 62

F(ξ)dξ 63

or

F(ξ, η)dξdη

(9.46)

In practice we perform the integration numerically using

Wi F(ξi ) + Rn or F(ξ, η)dξdη = Wij F(ξi , ηj ) + Rn F(ξ)dξ = i

Victor Saouma

i

(9.47)

j

Finite Elements II; Solid Mechanics

Draft

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

9–18

where the summations extend over all i and j, and Wi , Wij are weighting factors, and F(ξi ) and F(ξi , ηj ) are the matrices evaluated at the points speciﬁed in the arguments. The matrices Rn are error matrices, which are in general not computed.

64

As shown above, F = BT .D.B for ﬁnite element stiﬀness matrix evaluation, and each element is integrated individually. b 66 The integration of is essentially based on passing a polynomial P (ξ) through given a F (ξ)dξ b values of F (ξ) and then use a P (ξ)dξ as an approximation. 65

b

b

F (ξ)dξ ≈

a

(9.48)

P (ξ)dξ a

Using P (ξ) = F (ξ) at n points, and recalling the properties of Lagrangian interpolation functions, we obtain 67

P (ξ) = l1 (ξ)F (ξ1 ) + l2 (ξ)F (ξ2 ) + · · · + ln (ξ)F (ξn ) n

li (ξ)F (ξi ) =

(9.49-a) (9.49-b)

i=1

and note that at ξ = ξi li = 1, while all other li = 0.

9.3.1

Newton-Cotes

68 In Newton-Cotes integration, it is assumed that the sampling points are equally spaced, Fig. 9.15, thus we deﬁne

P( ξ ) F( ξ )

-1

0

1

ξ

Figure 9.15: Newton-Cotes Numerical integration

b

P (ξ)dξ =

b

n

a

or Approximation Weights

b a

b

li (ξ)dξF (ξi )

(9.50)

)n (n) P (ξ)dξ = Wi F (ξi ) i=1 b (n) (n) Wi = a li (ξ)dξ = (b − a)Ci

(9.51)

a i=1

li (ξ)F (ξi )dξ =

n

i=1

a

(n)

where Ci are the “weights” of the Newton-Cotes quadrature for numerical integration with n equally spaced sampling points.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

9.3 Numerical Integration n 2 3 4 5

(n)

C0

1 2 1 6 1 8 7 90

9–19 (n)

C1

1 2 4 6 3 8 32 90

(n)

C2

1 6 3 8 12 90

(n)

C3

1 8 32 90

(n)

C4

7 90

Error 10−1 (b − a)3 F II (ξ) 10−3 (b − a)5 F IV (ξ) 10−3 (b − a)5 F IV (ξ) 10−6 (b − a)7 F V I (ξ)

Table 9.3: Weights for Newton-Cotes Quadrature Formulas 69

Newton-Cotes constants, and corresponding reminder are shown in Table 9.3, (Bathe 1982).

70 It can be shown that this method permits exact integration of polynomial of order n − 1, and that if n is odd, then we can exactly integrate polynomials of order n. Hence we use in general odd values of n,

711 For n = 2 over [−1, 1], we select equally spaced points at ξ1 = −1 and ξ2 = 1 to evaluate −1 P (ξ)dξ P (ξ) =

2

li (ξ)F (ξi )

(9.52-a)

i=1

l1 (ξ) = l2 (ξ) = (2)

=

(2)

=

W1 W2

1 −1

F (ξ)dξ ≈

ξ − ξ2 1 = (1 − ξ) ξ1 − ξ2 2 ξ − ξ1 1 = (1 + ξ) ξ2 − ξ1 2 1 1 1 l1 (ξ)dξ = (1 − ξ)dξ = 1 2 −1 −1 1 1 1 l2 (ξ)dξ = (1 + ξ)dξ = 1 2 −1 −1 1 2

(2) P (ξ)dξ = Wi F (ξi ) = F (−1) + F (1) −1

(9.52-b) (9.52-c) (9.52-d) (9.52-e) (9.52-f)

i=1

which is the trapezoidal rule For n = 3 over [−1, 1], we select equally spaced points at ξ1 = −1 ξ2 = 0, and ξ3 = 1, to 1 evaluate −1 P (ξ)dξ 72

P (ξ) =

3

li (ξ)F (ξi )

(9.53-a)

i=1

1 (ξ − ξ2 )(ξ − ξ3 ) = ξ(ξ − 1) (ξ1 − ξ2 )(ξ1 − ξ3 ) 2 (ξ − ξ1 )(ξ − ξ3 ) = −(1 + ξ)(ξ − 1) l2 (ξ) = (ξ2 − ξ1 )(ξ2 − ξ3 ) 1 (ξ − ξ1 )(ξ − ξ2 ) = ξ(1 + ξ) l3 (ξ) = (ξ3 − ξ1 )(ξ3 − ξ2 ) 2 1 1 1 1 (3) = l1 (ξ)dξ = ξ(ξ − 1)dξ = W1 2 −1 3 −1 l1 (ξ) =

Victor Saouma

(9.53-b) (9.53-c) (9.53-d) (9.53-e)

Finite Elements II; Solid Mechanics

Draft

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

9–20

(3)

W2

1 −1

−1 1

(3)

=

W3

1

=

−1

l2 (ξ)dξ =

−1

1 2

l3 (ξ)dξ =

P (ξ)dξ = −1

ξ(1 + ξ)dξ = −1

i=1

4 3

−(1 + ξ)(ξ − 1)dξ = 1

3

1

F (ξ)dξ ≈

1

(9.53-f)

1 3

(9.53-g)

1 (3) Wi F (ξi ) = [F (−1) + 4F (0) + F (1)] (9.53-h) 3

which is Simpson’s rule

9.3.2

Gauss-Legendre Quadrature

73 In Gauss-Legendre quadrature, the points are not ﬁxed and equally spaced, but are selected to achieve best accuracy. 74

Again we start with

1

−1

F (ξ)dξ ≈

n

1

P (ξ)dξ = −1

(n)

(n)

(9.54)

Wi F (ξi )

i=1

and ξi are unknowns to be yet determined. Thus, we

however in this formulation both Wi have a total of 2n unknowns.

At the integration points P (ξi ) = F (ξi ), however at intermediary points the diﬀerence can be expressed as (9.55) F (ξ) = P (ξ) + χ(ξ)(β0 + β1 ξ + β2 ξ 2 + · · ·) 0 at ξ=ξi ;i=1,2,···,n 75

since we want the left side to be exactly equal to P (ξ) at the integration points, we deﬁne χ(ξ) = (ξ − ξ1 )(ξ − ξ2 )...(ξ − ξn )

(9.56)

as a polynomial of order n, to be equal to zero at the integration points ξi . 76 βi should be appropriately selected in orer to eliminate the gap between F (ξ) and P (ξ) at intermediary points. 77

Integrating

1

1

F (ξ)dξ =

P (ξ) +

−1

−1

∞

βj

j=0

1

χ(ξ)ξ j dξ

(9.57)

−1

We split the last term ∞

j=0

78

βj

1

j

χ(ξ)ξ dξ = −1

n−1

βj

j=0

1

j

χ(ξ)ξ dξ + −1

∞

j=n

βj

1

χ(ξ)ξ j dξ

(9.58)

−1

Truncating the last terms of the expansion

1 −1

Victor Saouma

F (ξ)dξ ≈

1

P (ξ) + −1

n−1

j=0

βj

1

χ(ξ)ξ j dξ

(9.59)

−1

Finite Elements II; Solid Mechanics

Draft

9.3 Numerical Integration

9–21

we observe that the ﬁrst integral on the right-hand side involves a polynomial of order n − 1, and the second integral a polynomial of order 2n − 1. Thus we set 1 χ(ξ)ξ j dξ = 0 j = 0, 1, · · · , n − 1 (9.60) −1

which would result in a set of n simultaneous equations of order n in terms of the unknowns ξi , i = 0, 1, · · · , n − 1. 79

Back to Eq. 9.59

1 −1

F (ξ)dξ ≈

1

P (ξ) = −1

Approximation Weights Gauss Points

F (ξi )

i=1

or

80

n

1

−1 F (ξ)dξ (n) Wi 1 j −1 χ(ξ)ξ dξ

−1

li (ξ)dξ =

(n)

Wi F (ξi )

(9.61)

i=1

)n (n) W F (ξi ) 1i=1 i = −1 li (ξ)dξ = 0 j = 0, 1, · · · , n − 1 (n)

(9.62)

are given in Table 9.4, (Bathe 1982). (n)

ξi √0 √ 3 −1/ 1/ 0 03 − 3/5 0 3/5

n

≈

Integration points ξi and weight coeﬃcients Wi n 1 2 3

1

Wi 2 1 1 5/9 8/9 5/9

Error 1 (2) (ξ) 3F 1 (4) (ξ) 135 F 1 (6) (ξ) F 15,750

Table 9.4: Integration Points and Weights for Gauss-Quadrature Formulaes Over the Interval [−1, 1]

9.3.2.1

† Legendre Polynomial

81 The solutions (Gauss integration points) are equal to the roots of a Legendre polynomial deﬁned by L0 (ξ) = 1 L1 (ξ) = ξ (9.63) 2k−1 k−1 2≤k≤n Lk (ξ) = k ξLk−1 (ξ) − k ξLk−2 (ξ)

and the n Gauss integration points are determined by solving Ln (ξ) = 0 for its roots ξi , i = 0, 1, · · · , n − 1. 82

The weighting functions are then given by (n)

Wi

Victor Saouma

=

2(1 − ξi2 ) [nLn−1 (ξi )]2

(9.64)

Finite Elements II; Solid Mechanics

Draft

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

9–22

9.3.2.2

Gauss-Legendre Quadrature for n = 2

83 First we seek the integration points for n = 2, χ(ξ) = (ξ − ξ1 )(ξ − ξ2 ), and the resulting equations are 1 (ξ − ξ1 )(ξ − ξ2 )ξ 0 dξ = 0 (9.65-a)

−1 1

−1

(ξ − ξ1 )(ξ − ξ2 )ξ 1 dξ = 0

(9.65-b) (9.65-c)

Upon integration, we obtain ξ1 ξ2 = − hence

1 3

1 ξ1 = − √ 3

and ξ1 + ξ2 = 0

(9.66)

1 and ξ2 = √ 3

(9.67)

The weight coeﬃcients are =

(2)

=

W1

(2)

W2

9.3.3

(n) Wi

1

−1 1 −1 1

= −1

li (ξ)dξ

(9.68-a)

−2ξ2 ξ − ξ2 dξ = = 1.0 ξ1 − ξ2 ξ1 − ξ2

(9.68-b)

−2ξ1 ξ − ξ1 dξ = = 1.0 ξ2 − ξ1 ξ1 − ξ2

(9.68-c)

Rectangular and Prism Regions

84 Numerical integration of F (ξ, η) over a rectangular region −1 ≤ ξ ≤ 1, and −1 ≤ η ≤ 1, is accomplished by selecting m and n (not to be confused with the order of the polynomial) integration points in the ξ and η directions

1 −1

1 −1

F (ξ.η)dξdη ≈

1

1

P (ξ, η)dξdη = −1

−1

n

m

(m)

Wi

(n)

Wj F (ξi , ηj )

(9.69)

i=1 j=1

and the total number of integration points will thus be m × n, Fig. 9.16.

9.3.4

Triangular Regions

85 For the numerical integration over a triangle, the Gauss points are shown in Fig. 9.17, and the corresponding triangular coordinates are given by Table 9.5.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

9.3 Numerical Integration

η ξ= −1/ 3

9–23

η ξ= +1/ 3

4

ξ= + 0.6

ξ= − 0.6 η= +1/ 3

2

η= + 0.6

9

6 3 5

2

ξ

1

1

3

8

ξ

4 7

η= −1/ 3

η= − 0.6

Figure 9.16: Gauss-Legendre Integration Over a Surface

b a b

a

a c d

c

Quadratic

Linear

Cubic

Figure 9.17: Numerical Integration Over a Triangle

Order Linear Quadratic

Error R = O(h2 ) R = O(h3 )

Cubic R = O(h4 )

Points a a b c a b c d

Triang. Coord. 1 1 1 3, 3, 3 1 1 2, 2, 0 0, 12 , 12 , 1 1 2 , 0, 2 1 1 1 3, 3, 3 0.6, 0.2, 0.2 0.2, 0.6, 0.2 0.2, 0.2, 0.6

Weights 1 1 3 1 3 1 3 − 27 48 25 48 25 48 25 48

Table 9.5: Coordinates and Weights for Numerical Integration over a Triangle

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 9–24

9.4 86

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

Stress Recovery; Nodal Stresses

Stresses are evaluated from σ = DBu

(9.70)

in general, it is desirable to have them evaluated at the elements nodal points. However, it should be kept in mind that stresses computed at a given nodes from diﬀerent elements need not be the same (since stresses are not required to be continuous in displacement based ﬁnite element formulations). 87

Hence some sort of stress averaging at nodal points may be desirable.

88 In the isoparametric formulation, nodal stresses are very poor, and best results are obtained at the Gauss points. 89

To evaluate nodal stresses, two approaches: 1. Evaluate σ directly at the nodes (ξ = η = ±1) 2. Evaluate the stresses at the Gauss points and then extrapolate.

The second approach yields far better results. Extrapolation from Gauss points will be illustrated through Fig. 9.18 for the 4 noded isoparametric quadrilateral. We specify an “internal” element with its own nodes and natural coordi90

Figure 9.18: Extrapolation from 4-Node Quad Gauss Points (Felippa 1999) nates ξ and η which are related to ξ and η through Table 9.6 or √ √ η = 3η ξ = 3ξ

(9.71)

hence any scalar quantity σ (such as σxx ) whose values σi is known at the Gauss element corners can be interpolated through the usual bilinear shape functions now expressed in terms of ξ and η   σ     1   σ 2 σ(ξ , η ) = N1e N2e N3e N4e (9.72)   σ3     σ4 Victor Saouma

Finite Elements II; Solid Mechanics

Draft

9.5 Nodal Equivalent Loads Corner Node 1 2 3 4

9–25

ξ

η

ξ

η

−1 +1 +1 −1

−1 −1 +1 +1

√ −√3 +√3 +√3 − 3

√ −√3 −√3 +√3 + 3

Gauss Node 1’ 2’ 3’ 4’

ξ

η

ξ

η

√ −1/√3 +1/√3 +1/√3 −1/ 3

√ −1/√3 −1/√3 +1/√3 +1/ 3

−1 +1 +1 −1

−1 −1 +1 +1

Table 9.6: Natural Coordinates of Bilinear Quadrilateral Nodes where N1e

N2e

N3e

N4e

= = = =

1 (1 − ξ )(1 − η ) 4 1 (1 + ξ )(1 − η ) 4 1 (1 + ξ )(1 + η ) 4 1 (1 − ξ )(1 + η ) 4

(9.73-a) (9.73-b) (9.73-c) (9.73-d)

Similarly, we can extrapolate σ to the corners√of the element. For corner 1, for instance, we replace ξ and η in the preceding equations by − 3. Doing that for the four corners, we obtain √ √      − 12√ 1 − 12 3 − 12√ 1 + 12 3 σ  σ1        1      1 1 1 1  σ2 − 2√ 1 + 2 3 − 2√ 1 − 2 3  σ2 = (9.74)  1− 1 3 σ3  σ3   − 12√ 1 + 12 3 − 12√       2     σ4 σ4 − 12 1 − 12 3 − 12 1 + 12 3 As expected, the sum of each row is equal to one, and for stresses we replace σ by σxx , σyy , and τxy As we know, diﬀerent nodal stresses will be obtained from adjacent elements. To obtain a single value we can either take 91

1. Unweighted average of all the nodal stresses. 2. Weighted average of nodal stresses based on the relative sizes of the elements as determined from their area through det(J).

9.5

Nodal Equivalent Loads

In the ﬁnite element formulation, all loads must be replaced by an “energy equivalent” nodal load. 92

93

We shall consider the following cases: nodal load, gravity, tractions, and thermal.

9.5.1

Gravity Load

94 Gravity forces are equivalent to a body force/unit volume acting within the solid in the direction of the gravity axis, Fig. 9.19,

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

9–26

Volume dΩ

6 5

7 η ξ y

4

8

θ x

2

1

3

Direction of Gravity

Figure 9.19: Gravity Loading (which need not be coincident with either of the coordinate axis). dGx = ρgdΩ sin θ

(9.75-a)

= −ρgdΩ cos θ

dGy

(9.75-b)

where g is the gravitational acceleration and ρ is the mass density. 95

Recalling from Eq. 7.16 that

NT bdΩ

fe =

(9.76)

Ωe

we obtain

Pxi Pyi

Ni ρg

= Ωe

sin θ − cos θ

dΩ

(9.77)

or the energy equivalent nodal forces for node i are

96

Pxi Pyi

=

ngaus

ngaus

j=1

k=1

ρgt

sin θ − cos θ

Ni (ξj ηk )Wj Wk detJ(ξj , ηk )

(9.78)

The angle θ is to be measured counter-clockwise from the positive y axis.

9.5.2

Traction Load

Any element edge can have a distributed load per unit length in a normal and tangential direction prescribed. 97

98 The variation of the distributed load is polynomial and its order can not exceed the order of the element. 99

For the sake of consistency, loaded nodes are listed also counterclockwise.

100 First we determine the components of the distributed loads in the x and y directions by considering the forces acting on an incremental length dS of the loaded edge, Fig. 9.20:

dPx = (pt dS cos θ − pn dS sin θ) = (pt dx − pn dy) dPy = (pn dS cos θ − pt dS sin θ) = (pn dx − pt dy) Victor Saouma

(9.79)

Finite Elements II; Solid Mechanics

Draft

9.5 Nodal Equivalent Loads

9–27 6

5

7 ν y

ξ

8

P

t

4

dy α

2

x

Pt

dx

3

1

Pn

Pn

Figure 9.20: Traction Load in Isoparametric Elements 101

But since integration is to be carried on in terms of natural coordinates: dx =

∂x dξ ∂ξ

dy =

∂y dξ ∂ξ

(9.80)

Substituting ∂x ∂y − pn dξ ∂ξ ∂ξ ∂x ∂y + pt dξ = pn ∂ξ ∂ξ

dPx = dPy

(9.81-a)

pt

(9.81-b) (9.81-c)

102

From Eq. 7.16 we have

NT ˆtdΓ

fe =

(9.82)

Γt

or

∂x ∂y − pn dξ = Ni pt ∂ξ ∂ξ Γt ∂x ∂y + pt dξ = Ni pn ∂ξ ∂ξ Γt

Pxi Pyi

(9.83-a) (9.83-b)

The integration is again carried on numerically (along the edge) and the energy equivalent nodal forces for node i are Pxi =

Pyi =

ngaus

j=1 ngaus

j=1

Victor Saouma

Ni Ni

∂x ∂y − pn pt ∂ξ ∂ξ ∂x ∂y + pt pn ∂ξ ∂ξ

Wj

(9.84)

Wj

(9.85)

Finite Elements II; Solid Mechanics

Draft 9–28

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

where the integration is carried on numerically along the edge. and note that taken straight out of the Jacobian matrix. 103

Note that since integration is to be carried along the edge, we have used ξ.

104

For adjacent elements, Fig. 9.21.

∂x ∂ξ

and

∂y ∂ξ

are

Figure 9.21: Traction Load in Contiguous Isoparametric Elements

9.5.3 105

Initial Strains/Stresses; Thermal Load

We distinguish between two cases:

Plane Stress which is the simplest ε0xx = α∆T

(9.86-a)

ε0yy 0 γxy

= α∆T

(9.86-b)

= 0

(9.86-c)

Plane Strain we have 0 νσzz + α∆T E νσ 0 = − zz + α∆T E = 0

ε0xx = −

(9.87-a)

ε0yy

(9.87-b)

0 γxy

ε0zz =

0 σzz

E

+ α∆T = 0

(9.87-c) (9.87-d)

0 , we obtain Using the last expression to eliminate σzz

Victor Saouma

ε0xx = (1 + ν)α∆T

(9.88-a)

ε0yy 0 γxy 0 σzz

= (1 + ν)α∆T

(9.88-b)

= 0

(9.88-c)

= −Eα∆T

(9.88-d)

Finite Elements II; Solid Mechanics

Draft

9.6 Computer Implementation 106

9–29

Those expressions are then substituted into Eq. 7.12 T B D 0 dΩ − BT σ0 dΩ f0e = Ωe

(9.89)

Ωt

and integrated numerically.

9.6 9.6.1

Computer Implementation Algorithm

107 The computer implementation of a numerically integrated isoparametric element is summarized as follows. But ﬁrst, it is assumed that this operation is to be performed in a function called stiff and it takes as input arguments elcod, young, poiss, type, ndime, ndofn, ngaus. In turn it will compute the stiﬀness matrix KELEM of element ielem.

1. Retrieve element geometry and material properties for the current element 2. Zero the stiﬀness matrix 3. Call function dmat to set the constitutive matrix De of the element 4. Enter (nested) loop covering all integration points (a) Look up the sampling position of the current point (ξp , ηq ) (s, t) and their weights (weigp) (b) Call shape function routine sfr given ξp , ηq which will return the shape function Nie ∂N e ∂N e (sfr) and the derivatives ∂ξi and ∂ηi (deriv) at the point ξp , ηq ∂Nie ∂Nie ∂ξ and ∂η at point ξp , ηq will return ∂Nie ∂Nie e ∂x and ∂y (cartd), the Jacobian matrix J determinant det Je (djac) and the x and y

(c) Call another subroutine (jacob), given Nie , cartesian shape function derivatives (jacm), its inverse Je−1 (jaci), its coordinates all at the point ξp , ηq

(d) Call strain matrix (bmatps) routine, given Nie , strain matrix Bei (bmat)

∂Nie ∂Nie ∂x , ∂y ,

at ξp , ηq will return the

(e) Call a routine (dbmat) to evaluate De Bei (dbmat) (f) Evaluate Bei T De Bej detJe × integration weights and assemble them into the element stiﬀness matrix Keij (g) Assemble De Be (smat into a stress array for later evaluation of stresses from the nodal displacements. 5. Write Stiﬀness matrix Suggested list of variables:

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

9–30

idime, ndime idofn,ndofn ielem,nelem igaus,jgaus,ngaus inode, nnode kgasp,ngasp type poiss young elcod(ndime,nnode)

s t gpcod(ndime,ngasp)

posgp(mgaus) weigp(mgaus) shape(nnode)

deriv(ndime,nnode)

cartd(ndime,nnode)

djacb jacm(ndime,ndime) jaci(ndime,ndime) bmat(nstre,nevab)

Index, Number of dimensions (2 for 2D) Index, Number of degree of freedom per node Index, number of elements Index, Index, Number of Gauss rule adopted Index, number of nodes per element Kounter, number of Gauss points used 1 for plane stress; 2 for plane strain Poisson’s ratio Young’s modulus Local array of the element currently under consid of nodal cartesian coordinates x(ξ1 , η1 ) · · · x(ξ8 , η8 ) eration y(ξ1 , η1 ) · · · y(ξ8 , η8 ) ξ coordinate of sampling point η coordinate of sampling point Local array of cartesian coordinates of the Gauss pointsfor element currently x(ξG1 , ηG1 · · · x(ξG5 , ηG5 ) · · · under consideration y(ξG1 , ηG1 · · · y(ξG5 , ηG5 ) · · · ξ coordinates of Gauss point Weight factor for Gauss point Shape function associated with each node of current element at (ξp , ηp ) " # N1 (ξp , ηp ) .. .N8 (ξp , ηp ) Shape at sampling point (ξp , ηp ) within the element " ∂N function derivative # ∂N8 1 (ξ , η ) · · · (ξ , η ) p p p p ∂ξ ∂ξ ∂N1 8 (ξp , ηp ) · · · ∂N (ξp , ηp ) ∂η ∂η Cartesian shape function derivatives associated with the nodes of the element sampled at any point (ξp , ηp ) within the element current ∂N1 8 (ξ , ηp ) · · · ∂N (ξp , ηp ) p ∂x ∂x ∂N1 8 (ξp , ηp ) · · · ∂N (ξp , ηp ) ∂y ∂y Determinant of the Jacobian matrix sampled at any point (ξp , ηp ) within the element Jacobian matrix at sampling point Inverse of Jacobian matrix at sampling point Element strain matrix at any point within the element B = [ B1 B2 · · · B8 ] 2 ∂N 3 i 0 ∂x 6 ∂Ni 7 where Bi = 4 0 ∂y 5 ∂Ni ∂y

dbmat(istre,ievab)

9.6.2 9.6.2.1

∂Ni ∂x

stores DB

MATLAB Code stiff.m

function KELEM = stiff(ngaus,posgp,weigp,type,nelem,lnods,coord,nnode) %----------------------------------------------------------------------------------% The purpose of this function is calculate element stiffness matricies for % bilinear and biquadratic isoparemtric elements using Gaussian integration. % Functions called by this function are: dmat,sfr,jacob,bmatps %----------------------------------------------------------------------------------% VARIABLES %----------------------------------------------------------------------------------% nelem Global variable NELEM % nnode Global variable NNODES % posgp Global variable POSGP % weigp Global variable WEIGP % ngaus Global variable NGAUS % type Global variable TYPE

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

9.6 Computer Implementation

9–31

% lnods Global variable LNODS % coord Global variable COORD % % stifsize Number of columns in the element stiffness matrix % nrowcount Position indicator for element stiffness matrix % elmt Current element for formation of stiffness matrix % young Modulus of elasticity for current element % poiss Poisson’s ration for current element % D Constitutive matix % elcod Matrix of element coordinates % row Counter % kelem Element stiffness matrix % KELEM Element stiffness matricies for all elements returned by function % s Current integration position % t Current integration position % shape Shape function at current point % deriv Derivative of shape function at current point % cartd Cartesian shape function derivatives % jacm Jacobian matrix % jaci Jacobian matrix inverse % djac Determinant of Jacobian matrix % xy x and y coordinates at the current point in the element % bmat Strain matrix [B] % dbmat Strain matrix * constitutive matrix [B]*[D] %----------------------------------------------------------------------------------tic fprintf(’Calculating ELEMENT STIFFNESS matricies\n’) stifsize = 2*nnode; nrowcount = stifsize-1; for ielem = 1:nelem %----------------------------------------% Extract material constants from lnods %----------------------------------------elmt = lnods(ielem,1); young = lnods(ielem,2); poiss = lnods(ielem,3); %----------------------------------------% Extract element coordinates % % elcod = [ X1 X2 X3 . . . Xn] % [ Y1 Y2 Y3 . . . Yn] %----------------------------------------elcod = zeros(2,nnode); for inode = 1:nnode row = find(coord(:,1:1)==lnods(ielem,inode+3)); elcod(:,inode:inode) = coord(row:row,2:3)’; end %----------------------------------------% Constitutive matrix %----------------------------------------D = dmat(young,poiss,type); %----------------------------------------% Element stiffness matrix - element ielem %----------------------------------------kelem = zeros(stifsize); for igaus = 1:ngaus

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

9–32

for jgaus = 1:ngaus s = posgp(igaus); t = posgp(jgaus); W = weigp(igaus)*weigp(jgaus); [shape,deriv] = sfr(s,t,nnode); [cartd,jacm,jaci,djac,xy] = jacob(shape,deriv,elcod); [bmat,dbmat] = bmatps(shape,cartd,D); kelem = kelem + bmat’*dbmat*djac; end end %---------------------------------------------------------------------% Store element stiffness matricies in as a stack in a single matrix : % kelem(1) % KELEM = : % kelem(nelem) %---------------------------------------------------------------------startrow = stifsize*ielem - nrowcount; endrow = ielem*stifsize; KELEM(startrow:endrow,:) = kelem; end t = toc; fprintf(1,’Elapsed time for this operation =%3.4fsec\n\n’,t)

9.6.2.2

dmat.m

function D = dmat(young,poiss,type) %----------------------------------------------------------------------------------% This function calculates the constitutive matrix for an element %----------------------------------------------------------------------------------% VARIABLES %----------------------------------------------------------------------------------% young Young’s modulus/Modulus of elasticity % poiss Poisson’s ratio % E Modulus of elasticity % type type of problem - plain stress = 1, plain strain = 2 % D Constitutive matrix returned by function %----------------------------------------------------------------------------------tic %fprintf(’Calculating element constitutive matrix\n’) E = young; v = poiss; %-------------------------% Plain stress %-------------------------if type == 1.0 D = (E/(1-v^2))*[

1 v 0 %-------------------------% Plain strain %-------------------------else

Victor Saouma

v 1 0

0; 0; (1-v)/2];

Finite Elements II; Solid Mechanics

Draft

9.6 Computer Implementation D = E/((1+v)*(1-2*v))*[ 1-v v 0

9–33 v 1-v 0

0 ; 0 ; (1-2*v)/2];

end t = toc; %fprintf(1,’Elapsed time for this operation =%3.4fsec\n\n’,t)

9.6.2.3

sfr.m

function [shape,deriv] = sfr(s,t,nnode) %----------------------------------------------------------------------------------% This function calculates the shape function and derivative for the current node %----------------------------------------------------------------------------------% VARIABLES %----------------------------------------------------------------------------------% shape Shape function returned by function % deriv Derivative of shape function returned by function % nnode Number of nodes per element % s Natural coordinate (xi) of sampling point - horizontal % t Natural coordinate (eta) of sampling point - vertical %----------------------------------------------------------------------------------tic %fprintf(’Calculating shape functions and derivatives\n’) %------------------% Q9 Element %------------------if nnode == 9 N9 = (1-s^2)*(1-t^2); N8 = .5*(1-s)*(1-t^2) -.5*N9; N7 = .5*(1-s^2)*(1+t) -.5*N9; N6 = .5*(1+s)*(1-t^2) -.5*N9; N5 = .5*(1-s^2)*(1-t) -.5*N9; N4 = .25*(1-s)*(1+t) -.5*N7 - .5*N8 - .25*N9; N3 = .25*(1+s)*(1+t) -.5*N6 - .5*N7 - .25*N9; N2 = .25*(1+s)*(1-t) -.5*N5 - .5*N6 - .25*N9; N1 = .25*(1-s)*(1-t) -.5*N5 - .5*N8 - .25*N9; shape = [N1 N2 N3 N4 N5 N6 N7 N8 N9]’; dN9ds dN9dt dN8ds dN8dt dN7ds dN7dt dN6ds dN6dt dN5ds dN5dt dN4ds dN4dt dN3ds dN3dt dN2ds dN2dt dN1ds dN1dt

= = = = = = = = = = = = = = = = = =

-2*s*(1-t^2); -2*t*(1-s^2); -.5*(1-t^2) -.5*dN9ds; -t*(1-s) -.5*dN9dt; -s*(1+t) -.5*dN9ds; .5*(1-s^2) -.5*dN9dt; .5*(1-t^2) -.5*dN9ds; -t*(1+s) -.5*dN9dt; -s*(1-t) -.5*dN9ds; -.5*(1-s^2) -.5*dN9dt; -.25*(1+t) -.5*dN7ds .25*(1-s) -.5*dN7dt .25*(1+t) -.5*dN6ds .25*(1+s) -.5*dN6dt .25*(1-t) -.5*dN5ds -.25*(1+s) -.5*dN5dt -.25*(1-t) -.5*dN5ds -.25*(1-s) -.5*dN5dt -

Victor Saouma

.5*dN8ds .5*dN8dt .5*dN7ds .5*dN7dt .5*dN6ds .5*dN6dt .5*dN8ds .5*dN8dt

-

.25*dN9ds; .25*dN9dt; .25*dN9ds; .25*dN9dt; .25*dN9ds; .25*dN9dt; .25*dN9ds; .25*dN9dt;

Finite Elements II; Solid Mechanics

Draft 9–34

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

deriv = [dN1ds dN2ds dN3ds dN4ds dN5ds dN6ds dN7ds dN8ds dN9ds; dN1dt dN2dt dN3dt dN4dt dN5dt dN6dt dN7dt dN8dt dN9dt]; %------------------% Q8 Element %------------------elseif nnode == 8 N8 = .5*(1-s)*(1-t^2); N7 = .5*(1-s^2)*(1+t); N6 = .5*(1+s)*(1-t^2); N5 = .5*(1-s^2)*(1-t); N4 = .25*(1-s)*(1+t) -.5*N7 N3 = .25*(1+s)*(1+t) -.5*N6 N2 = .25*(1+s)*(1-t) -.5*N5 N1 = .25*(1-s)*(1-t) -.5*N5

-

.5*N8; .5*N7; .5*N6; .5*N8;

shape = [N1 N2 N3 N4 N5 N6 N7 N8]’; dN8ds dN8dt dN7ds dN7dt dN6ds dN6dt dN5ds dN5dt dN4ds dN4dt dN3ds dN3dt dN2ds dN2dt dN1ds dN1dt

= = = = = = = = = = = = = = = =

-.5*(1-t^2); -t*(1-s); -s*(1+t); .5*(1-s^2); .5*(1-t^2); -t*(1+s); -s*(1-t); -.5*(1-s^2); -.25*(1+t) -.5*dN7ds .25*(1-s) -.5*dN7dt .25*(1+t) -.5*dN6ds .25*(1+s) -.5*dN6dt .25*(1-t) -.5*dN5ds -.25*(1+s) -.5*dN5dt -.25*(1-t) -.5*dN5ds -.25*(1-s) -.5*dN5dt

-

.5*dN8ds; .5*dN8dt; .5*dN7ds; .5*dN7dt; .5*dN6ds; .5*dN6dt; .5*dN8ds; .5*dN8dt;

deriv = [dN1ds dN2ds dN3ds dN4ds dN5ds dN6ds dN7ds dN8ds; dN1dt dN2dt dN3dt dN4dt dN5dt dN6dt dN7dt dN8dt]; %------------------% Q4 Element %------------------else N4 = .25*(1-s)*(1+t); N3 = .25*(1+s)*(1+t); N2 = .25*(1+s)*(1-t); N1 = .25*(1-s)*(1-t); shape = [N1 N2 N3 N4]’; dN4ds dN4dt dN3ds dN3dt dN2ds dN2dt dN1ds dN1dt

= = = = = = = =

-.25*(1+t); .25*(1-s); .25*(1+t); .25*(1+s); .25*(1-t); -.25*(1+s); -.25*(1-t); -.25*(1-s);

deriv = [dN1ds dN2ds dN3ds dN4ds;

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

9.6 Computer Implementation

9–35

dN1dt dN2dt dN3dt dN4dt]; end t = toc; %fprintf(1,’Elapsed time for this operation =%3.4fsec\n\n’,t)

9.6.2.4

jacob.m

function [cartd,jacm,jaci,djac,xy] = jacob(shape,deriv,elcod) %----------------------------------------------------------------------------------% This function calculates the cartesian shape function derivatives % the Jacobian matrix, Jacobian inverse and Jacobian determinant %----------------------------------------------------------------------------------% VARIABLES %----------------------------------------------------------------------------------% shape Shape function at current point % deriv Derivative of shape function at current point % cartd Cartesian shape function derivatives returned by function % jacm Jacobian matrix returned by function % jaci Jacobian matrix inverse returned by function % djac Determinant of Jacobian matrix returned by function % xy x and y coordinates at the current point in the element %----------------------------------------------------------------------------------tic %fprintf(’Calculating Jacobian matrix\n’) %---------------------------------------------------------------% The cartesian shape function derivatives {cartd} are given by: % % {dN/dx} -1 {dN/ds} % {cartd} = { } = [J] { } % {dN/dy} {dN/dt} % Start by calculating Jacobian [J] = jacm: % T T % [dX/ds dY/ds] [{dN/ds}*{x} {dN/ds}*{y}] % [J] = [ ] = [ T T ] % [dX/dt dY/dt] [{dN/dt}*{x} {dN/dt}*{y}] %---------------------------------------------------------------jacm = deriv*elcod’; jaci = inv(jacm); djac = det(jacm); cartd = jaci*deriv; xy = elcod*shape; t = toc; %fprintf(1,’Elapsed time for this operation =%3.4fsec\n\n’,t)

9.6.2.5

bmatps.m

function [bmat,dbmat] = bmatps(shape,cartd,D) %----------------------------------------------------------------------------------% This function calculates the strain matrix B %----------------------------------------------------------------------------------% VARIABLES %----------------------------------------------------------------------------------% shape Shape function at current point % cartd Cartesian shape function derivatives % bmat Strain matrix returned by function % dbmat Strain matrix * constitutive matrix D

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 9–36

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

%----------------------------------------------------------------------------------tic %fprintf(’Calculating strain matrix [B]\n’) numcols = 2*length(cartd); bmat = zeros(3,numcols); cartdcol = 0; for ibmatcol = 1:2:numcols cartdcol = cartdcol+1; bmat(:,ibmatcol:ibmatcol+1) = [cartd(1,cartdcol) 0 ; 0 cartd(2,cartdcol); cartd(2,cartdcol) cartd(1,cartdcol)]; end dbmat = D*bmat; t = toc; %fprintf(1,’Elapsed time for this operation =%3.4fsec\n\n’,t)

9.6.3 108

Plott of Shape Functions

Matlab code to generate shape function plots

% Shape FUnctions X=-1:1/20:1; Y=X; YT=Y’; XT=X’; N9=(1-YT.*YT)*(1-X.*X); N8=0.5*(1-YT.*YT)*(1-X); N7=0.5*(1-XT.*XT)*(1+Y); N6=0.5*(1-YT.*YT)*(1+X); N5=0.5*(1-XT.*XT)*(1-Y); N4=0.25*(1-XT)*(1+Y)-0.5*(N7+N8); N3=0.25*(1+XT)*(1+Y)-0.5*(N6+N7); N2=0.25*(1+XT)*(1-Y)-0.5*(N5+N6); N1=0.25*(1-XT)*(1-Y)-0.5*(N8+N5); meshc(X,Y,N1) print -deps2 shap8-1.eps c=contour(X,Y,N1); clabel(c) print -deps2 shap8-1-c.eps meshc(X,Y,N8) print -deps2 shap8-8.eps c=contour(X,Y,N8); clabel(c) print -deps2 shap8-8-c.eps N1=N1-0.25*N9; meshc(X,Y,N1) print -deps2 shap9-1.eps c=contour(X,Y,N1); clabel(c) print -deps2 shap9-1-c.eps N8=N8-0.5*N9; meshc(X,Y,N8) print -deps2 shap9-8.eps c=contour(X,Y,N8);

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

9.6 Computer Implementation

9–37

clabel(c) print -deps2 shap9-8-c.eps meshc(X,Y,N9) print -deps2 shap9-9.eps c=contour(X,Y,N9); clabel(c) print -deps2 shap9-9-c.eps

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 9–38

Victor Saouma

ISOPARAMETRIC ELEMENTS; 2nd GENERATION

Finite Elements II; Solid Mechanics

Draft Chapter 10

ELEMENT FORMULATION and STRAIN RECOVERY in HW FORMULATION Taken from (ˇ Cervenka, J. 1994)

10.1

Element Formulation

ˇervenka, J. 1994) Taken from (C It is necessary to select appropriate interpolation functions for all three elastic ﬁelds (i.e. u, and σ). The choice of these shape functions must be such that the BB condition is satisﬁed (Appendix ??). In this work, the same interpolation functions are used for all three ﬁelds (i.e. displacements, strains and stresses), which implies that there is a full number of unknowns in each node. dim(un ) = N × dim − R,

dim( n ) = dim(σ n ) = N × dim(σ)

(10.1)

where N denotes the number of nodes, dim is the problem dimension and R is the number of rigid body modes. In Section 10.3, it will be shown that this formulation guarantees the satisfaction of the BB condition. The polynomial orders of the ﬁeld approximations are given in Table 10.1. Table 10.1: Polynomial orders of the shape ﬁeld 2D T3 T6 T4 displacement u linear quadratic linear strain linear quadratic linear stress σ linear quadratic linear

functions. 3D T10 quadratic quadratic quadratic

In general case a variable x is interpolated over a ﬁnite element using the expression: x=

Nen

i

Φi xi

(10.2)

Draft 10–2

ELEMENT FORMULATION and STRAIN RECOVERY in HW FORMULATION

where Nen is the number of element nodes, Φi is an interpolation function associated with node i, and xi is the value of variable x at element node i. For the linear triangular element (T3) the interpolation functions are: Φi = li ,

i = 1, 2, 3

(10.3)

and for the six noded triangular element (T6) with three corner nodes and three mid-side nodes the interpolation functions are: Φi = li (2li − 1)

i = 1, 2, 3,

Φ4 = 4l1 l2 ,

Φ5 = 4l2 l3 ,

Φ6 = 4l3 l1 ,

(10.4)

where element nodes 1 to 3 indicate the element corner nodes and 4 to 6 are the mid-side nodes. Symbols li denote the natural area coordinates of the element, which are related to the element natural coordinates ξ and η by relations: l1 = ξ,

l3 = 1 − ξ − η

l2 = η,

(10.5)

For three-dimensional ﬁnite elements, the interpolation functions are similar. For the linear tetrahedron T4 element they are: Φi = li ,

i = 1, 2, 3, 4

(10.6)

and for the T10 element with four corner nodes and six mid-edge nodes they are deﬁned analogically to the six noded triangular T6 element as: i = 1, 2, 3, 4, Φ5 = 4l1 l2 , Φ6 = 4l2 l3 , Φi = li (2li − 1), Φ8 = 4l1 l4 , Φ9 = 4l2 l4 , Φ10 = 4l3 l4 Φ7 = 4l3 l1 ,

(10.7)

Similarly to the two-dimensional elements, symbols li denotes the volumetric natural coordinates, which are again related to the element natural coordinates by relations: l1 = ξ,

10.2

l2 = η,

l3 = ζ,

l4 = 1 − ξ − η − ζ

(10.8)

Strain Recovery

ˇervenka, J. 1994) Taken from (C In Equation ??, Step 2 is essentially an expression for the computation of nodal strains from nodal displacements. It involves the inversion of a symmetric matrix C T , or in other words the solution of a system of linear simultaneous equations. Three strain recovery techniques, described below, represent an attempt to avoid the direct solution of a large linear system of equations in this step, as its assembly and factorization is computationally expensive. Three algorithms are discussed and compared: (1) C-lumping (CL), (2) Strain smoothing (SS) and (3) C-splitting (CS). The CL as the simplest algorithm, in which a lumped form of matrix C is constructed, and the inversion of the resulting diagonal matrix is trivial. The other two algorithms are diﬀerent iterative techniques to solve the system of equations. Between them, the CS method is tailored for the fastest convergence for linear elements. In the sequel, the MIM iteration index k is omitted as it focuses on faster solution techniques for the Step 2 only.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

10.2 Strain Recovery

10.2.1

10–3

C-lumping.

The inversion of C for the C-lumping (CL) technique is simpliﬁed by forming a diagonalized C matrix. This lumped C matrix 1 is evaluated by the following expression: (I Φ)dV (10.9) CL = V

where I is the identity matrix and Φ is the shape function matrix. Since identical shape functions are used for all three primary ﬁelds, the subscript at Φ is no longer necessary. Replacing C by C L Step 2 of Equation ?? reduces to: n = C −1 L Eun At the element level, the lumped C matrix  5  0 C eL =   0 0

(10.10)

for the four node linear tetrahedron (T4) is:  0 0 0 5 0 0  ψ (10.11) 0 5 0  0 0 5

where ψ is a constant based on the element volume. CL is by far the simplest strain recovery method since no iterations are required to compute the nodal strains. However, the numerical experiments reported in Section ?? indicate that the displacement solution converges to an erroneous value. Hence, the C-lumping technique is kinematically inconsistent.

10.2.2

Strain smoothing.

Strain Smoothing (Zienkiewicz, Vilotte, Toyoshima and Nakazawa 1985) (SS) is an indirect procedure within Step 2 that avoids the direct decomposition of the C matrix. Nodal strains are iteratively evaluated until the ratio of the Euclidean norms of strain correction to total strains satisﬁes a prescribed limit. This technique exploits the diagonal matrix C L previously described and the consistent matrix C deﬁned below. Iteratively the nodal strains are evaluated by: j = jn + C −1 j+1 n L (Eun − C n ).

(10.12)

where j = 0, 1, 2 . . . is the strain-iteration count. Note that this represents an internal iteration, not to be confused with the MIM iteration of (??). The iteration process involves the nodal strains in the whole mesh since Step 2 is equivalent to the least square ﬁt of the nodal based strain ﬁeld to the strain ﬁeld derived from the displacement ﬁeld (Zienkiewicz and Taylor 1989). For a four noded linear tetrahedral element (T4), the consistent matrix C is given at the element level by:   2 1 1 1  1 2 1 1  e  ΦΦT dV =  (10.13) C =  1 1 2 1 ψ V e 1 Note: The diagonalized matrix C L and the consistent matrix in Section 10.2.2 and 10.2.3 are 1 1 C 1described 2 determined in an analogous way as the standard lumped and consistent mass matrices in dynamics. The only diﬀerence is the exclusion of the weight density of the material which is replaced by unity.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 10–4

ELEMENT FORMULATION and STRAIN RECOVERY in HW FORMULATION

where ψ is again a constant based on the element volume. The correction of nodal strain ∆ jn during one iteration is: − jn = A(ρ)∆ nj−1 , ∆ jn = j+1 n where:

(j ≥ 1)

A(ρ) = I − C −1 L C

(10.14) (10.15)

where A(ρ) is a ﬁxed ampliﬁcation matrix having a spectral radius ρ = 45 . The spectral radius ρ is deﬁned as the largest eigenvalue of ampliﬁcation matrix A(ρ). Since Equation 10.15 involves a product of C and inverse of C L , the constants ψ are cancelled out. By Banach’s ﬁxed point theorem (Haser and Sullivan 1991) it is necessary for the spectral radius ρ to be less than 1 to ensure convergence of the iterative process given by Equation 10.12. Thus, this value of the spectral radius indicates an error decay of 15 .

10.2.3

C-splitting.

ˇ A new iterative process was recently developed by (Cervenka, Keating and Felippa 1993) to solve Step 2. This new technique guarantees faster convergence for linear triangular and tetrahedral elements (T3 and T4). This technique is referred to as C-splitting (CS). This method “splits” the consistent matrix C of Equation 10.13 into two matrices. One matrix is diagonalized and the second is formed such that their algebraic sum is equivalent to the original C matrix: C = CD + CR

(10.16)

where: C D = α diag(C)

CR = C − CD

(10.17)

α is a “splitting” coeﬃcient controlling the splitting of the matrix C. Using this method Step 2 in Equation ?? is modiﬁed to: j = C −1 j+1 n D (Eun − C R n ).

(10.18)

For the C-splitting method, the per-iteration strain correction is: − jn = A(ρ)∆ nj−1 , ∆ jn = j+1 n

(j ≥ 1)

(10.19)

where the ampliﬁcation matrix A(ρ) is given by: A(ρ) = −C −1 D C R,

(10.20)

It is possible to select the coeﬃcient α such that the spectral radius of the ampliﬁcation matrix is minimal. For four-node tetrahedron elements using α = 32 , which as shown below minimizes the spectral radius of the ampliﬁcation matrix, C splits at the element level into the following matrices:   3 0 0 0  0 3 0 0   (10.21) C eD =   0 0 3 0 ψ 0 0 0 3 

and C eR Victor Saouma

 −1 1 1 1  1 −1 1 1  ψ =  (10.22)  1 1 −1 1  1 1 1 −1 Finite Elements II; Solid Mechanics

Draft

10.3 Uniqueness and Existence of a Solution

10–5

Table 10.2: Table of α coeﬃcients and spectral radii for CS technique. Element type splitting coef. spectral radius α ρ three node linear triangle T3 5/4 3/5 four node quadrilateral Q4 5/4 4/5 four node tetrahedral element T4 3/2 2/3 eight node brick element B8 7/4 0.929

Then the spectral radius ρ is equal to 23 . Thus, CS has an error decay rate of 13 allowing for a faster convergence than the SS method. For example, 10 steps of CS can be expected to 10 reduce the initial strain errors by ( 23 ) ≈ 0.0173 whereas 10 steps of SS would reduce those 10 errors by only ( 45 ) ≈ 0.1074. This technique was investigated also for other low order element types. The best splitting coeﬃcients α and resulting spectral radii of the operator A(ρ) (Eq. 10.20) for other element types are summarized in Table 10.2. The coeﬃcients α for four node quadrilateral and eight node brick element are however valid only for elements with parallel or almost parallel sides. Therefore, the (SS) technique would be probably more reliable for these element types.

10.3

Uniqueness and Existence of a Solution

The BB condition for uniqueness and existence of a solution of the three-ﬁeld variational principle is stated in Appendix ??. This condition was derived by (Babuˇska 1971), (Babuˇska 1973) and (Brezzi 1974). Xue and Atluri (1985) extended the condition to a general three-ﬁeld problem, and derived its discrete form. The continuous and discrete forms of the BB conditions are again described in Appendix ??, and it is shown that they are equivalent to the following three conditions: rank(E) = nu ≤ nσ (10.23) rank(C) = nσ ≤ n= + nu A is positive deﬁnite where matrices E, C and A are derived in Appendix ?? and are given by the following integrals: T ΦBdV, C = ΦΦ dV, A = ΦDΦT dV (10.24) E= V

V

V

The third condition is satisﬁed as long as the material does not exhibit softening. This is always guaranteed in the discrete crack approach, since softening is modeled only along the interface elements which are not included in the mixed iterative solution. If identical shape functions are used for all three ﬁelds, then the number of unknowns for each ﬁeld is given by Equation 10.1, and the inequalities in the ﬁrst and second condition of Equation 10.23 are always satisﬁed. nσ = N × dim(σ) nu = N × dim − R ≤ nσ = N × dim(σ) ≤ n= + nu = N × dim(σ) + N × dim − R

Victor Saouma

(10.25)

Finite Elements II; Solid Mechanics

Draft 10–6

ELEMENT FORMULATION and STRAIN RECOVERY in HW FORMULATION

We note that the rank condition of matrix C is also satisﬁed, as it is analogous to the consistent mass matrix of isoparametric elements, which has always full rank, as can be seen from Equation 10.13. More complex is the veriﬁcation of the rank condition of matrix E. In Equation 10.24, E is given by the integral: E=

ΦB dV

(10.26)

V

where matrix B is the matrix relating strains at a certain point to the nodal displacements and Φ is the matrix of shape functions relating strains or stresses at a certain point to their nodal counterparts. Matrix E will have a rank equal to nu if the following two conditions are satisﬁed. ∃x ∈ V : B(x)un = 0 ∀un = 0, (10.27) T ∃ (x) = Φ (x) n : (x) is unique ∀ n , The ﬁrst condition is equivalent to the requirement that a nonzero vector of nodal displacements must cause non-zero strain ﬁeld. It should be noted that the rigid body modes are excluded from vector un . They would be the only modes allowed to produce a zero strain ﬁeld. In this case, matrix B corresponds to that of a standard isoparametric triangular or tetrahedral element, and will therefore satisfy this condition. The second condition is also satisﬁed, since the shape functions in matrix Φ are those of a standard isoparametric element, and the uniqueness of the interpolation is guaranteed. For higher order triangular and tetrahedral elements (i.e. 6 noded triangle and 10 noded tetrahedron), it would seem preferable to select interpolation functions for strains and stresses, which are one order lower than those for the displacements. This would correspond to the mathematical relation between strains and displacements, since the strains are determined by diﬀerentiation of the displacement ﬁeld. For this formulation, there would be unknown displacements, strains and stresses at each element corner node, but only displacement unknowns at the midside element nodes. The strains and stresses would be interpolated using linear shape functions and displacement using quadratic shape functions. Now, it is possible to show that this formulation would not guarantee the satisfaction of the inequality in the ﬁrst condition of Equation 10.23. We consider a patch of two six-noded triangular element as shown in Figure 10.1. From F

h

h

5

2F

5 20 5

h

h

5

F 8 and 9 noded isoparametric quadrilaterals; Pure Axial and Flexure 8 noded Isoparametric constant axial stress

Figure 10.1: Patch test.

the previous assumptions of linear stress and strain interpolation and quadratic displacement interpolation, the unknown stresses and strains are only at the element corner nodes, while all nodes have unknown displacements. The total number of stress and displacement unknowns is

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

10.3 Uniqueness and Existence of a Solution

10–7

then given by: nu = 9 × 2 − 3 = 15 > nσ = 4 × 3 = 12

(10.28)

and clearly the important inequality nu ≤ nσ is not satisﬁed and the existence and uniqueness of a solution cannot be guaranteed. This should be contrasted by the previous formulation, in which the same interpolation functions are used for all three ﬁelds, and the satisfaction of the BB is guaranteed by Equation 10.25 and 10.27.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 10–8

Victor Saouma

ELEMENT FORMULATION and STRAIN RECOVERY in HW FORMULATION

Finite Elements II; Solid Mechanics

Draft Chapter 11

WEIGHTED RESIDUAL METHODS Adapted from (Ottosen and Petersson 1992)

11.1

Introduction

1 Thus far, the ﬁnite element method was derived from a variational principle through the Rayleigh-Ritz method. 2 In Elasto-Statics, the variational principle used is the minimization of the total potential energy, and the resulting weak formulation is equivalent to the equation of equilibrium (strong form). 3 In many branches of physical sciences, a variational principle may not be available, and all what is available is a diﬀerential equation and its associated boundary conditions.

It should be noted that a variational formulation is not possible when the order of the diﬀerential equation is odd. 4

5

It is for those problems that the weighted residual methods are employed.

6

The resulting formulation is an integral one, and is thus “weak”.

11.2

General Formulation

11.2.1

Diﬀerential Operators

7

Considering a general diﬀerential equation Governing Diﬀerential Equation Lφ + g = 0; Ω Boundary conditions Bi (φ) = qi ; Γi

(11.1)

where L is a diﬀerential operator, (assuming a 1D problem) φ = φ(x) is an unknown function which represents the state variable to be determined (i.e. displacement, temperature, head, etc...), g(x) is a known forcing function.

Draft 11–2 8

WEIGHTED RESIDUAL METHODS

We assume that the diﬀerential operator to be symmetric and positive deﬁnite, (L[u])vdΩ = (L[v])udΩ Symmetric Ω Ω (L[u])udΩ > 0 Positive Deﬁnite

(11.2)

Ω

where u and v are any functions which satisfy essential and natural boundary conditions. 11.2.1.1

Application to 1D Axial Member

9 As an example, let us consider the steady state response of an axial bar. The governing diﬀerential equation and boundary conditions are

Governing Diﬀerential Equation Essential Boundary conditions Natural Boundary conditions Hence

2

−EA ∂∂xu2 = 0 u|x=0 = 0 EA ∂u ∂x x=l = P

(11.3)

2

∂ φ = u; g = 0; L[] = −EA ∂x 2; ∂ q1 = 0; B2 = EA ∂x ; q2 = P ; B1 = 1;

(11.4)

To determine whether the operator L is symmetric and positive deﬁnite, we consider P = 0, and integrating by part twice1 we obtain 10

l

0

l ∂2u ∂u l ∂u ∂v dx −EA 2 vdx = − EA v + EA ∂x ∂x 0 ∂x ∂x 0 l ∂v l ∂2v ∂u l EA 2 udx = − EA v + EAu − ∂x ∂x ∂x 0

(11.5-a) (11.5-b)

0

0

But since the boundary conditions are u = v = 0 at x = 0 and ∂u/∂x = ∂v/∂x = 0 at x = l, then we only have l l ∂2u ∂2v −EA 2 vdx = −EA 2 udx (11.6) ∂x ∂x 0 0 hence the operator is symmetric. We can also conclude that the operator is positive deﬁnite from Eq. 11.5-a

l 0

11.2.2 11

∂2u −EA 2 udx = ∂x

l

EA 0

∂u ∂x

2 dx

(11.7)

Residual Formulation

We now alter Eq. 11.1, and for a 1D problem rewrite it as

b

w(Lφ + g)dx = 0 Z 1

b

From Eq.1.39: a

Victor Saouma

a

u(x)v (x)dx = u(x)v(x)|ba −

Z

b

(11.8)

v(x)u (x)dx

a

Finite Elements II; Solid Mechanics

Draft

11.2 General Formulation

11–3

where w is an arbitrary weight function2 . We note that through this equation, the governing diﬀerential equation is satisﬁed in a weak sense. 12 We now seek an approximate numerical solution for the unknown function φ(x), hence, in general we may assume

φ(x)app = a1 ψ1 (x) + a2 ψ2 (x) + · · · + an ψn (x) = aψ

(11.9-a)

where ai are the unknown parameters, and ψ(x) are prespeciﬁed trial functions3 of x. 13

Substituting the previous equation into Eq. 11.8

b

w(Lφapp + g)dx = 0

(11.10)

a

Since φapp will in general not satisfy the diﬀerential equation exactly, then Lφapp + g = R b wR = 0

(11.11) (11.12)

a

where R(x) is a measure of the error, and is called the Residual; We thus impose the requirement that weights and residuals be orthogonal functions4 . 14 The residual clearly depends on the coeﬃcients ai , thus our objective is to select the weight w(x) in such a way that the weighted residuals w(x)R(x) over a − b is zero. 15

Arbitrary weight functions are selected ﬁrst w = c1 W1 (x) + c2 W2 (x) + · · · + cn Wn (x)  c1    c2 = W1 (x) W2 (x) · · · Wn (x)  ···   cn

      

(11.13-a)

and Wi (x) are predeﬁned functions of x, ci are parameters. Hence, we can rewrite Eq. 11.8 as

b

T

WT Rdx = 0

c

(11.14)

a

But since c does not depend on the coordinates, then

b

WT Rdx = 0

(11.15)

a

2 Note similarity with the derivation of the principle of virtual work from the equilibrium equation and boundary condition. 3 Later on in the Galerkin Method, which is a special form of Weighted Residuals, φ will correspond to the to the shape functions N, and a to the nodal displacements u. displacement, 4 T R Two vectors a and b are orthogonals if a b = 0, similarly, two functions u(x) and v(x) are orthogonals if v(x)u(x) = 0.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 11–4

WEIGHTED RESIDUAL METHODS

This equation really represents a system of n equations. 16

Substituting Eq. 11.8, and since a does not depend on the coordinates (11.16)

R = L(ψa) + g = L(ψ)a + g Hence, taking the weighted average

b

W L(ψ)dx a = − a T

or

Ke

b

Wgdx

(11.17)

a

f

 b   W1 L(ψ1 )dx W1 L(ψ2 )dx · · · W1 L(ψn )dx  W1 gdx     a a a    ab  b b b        W L(ψ )dx W L(ψ )dx · · · W L(ψ )dx W2 gdx 2 1 2 2 2 n   Ke =  a ;f = a a a    .. .. .. .. ..     . . . . .     b b b     b    Wn L(ψ1 )dx Wn L(ψ2 )dx · · · Wn L(ψn )dx Wn gdx 

b

a

b

a

b

a



                    

a

(11.18) Ke is thus an n by n matrix, and from the preceeding equation we can solve for thee unknown coeﬃcients a. 17

11.3

Weighted Residual Methods

18 The previous formulation was very general, and diﬀerent techniques will vary according to the selected weights, i.e. our choice for W.

11.3.1

† Point Collocation Method

In this method, the weight function w is based on Dirac’s delta function δ(x − xi ) and thus the residual is set to be exactly equal to zero at n points. This is achieved by adopting for the weight function w the Dirac’s delta function δ(x − xi ) where ∞ if x = xi (11.19-a) δ(x − xi ) = 0 otherwise ∞ δ(x − xi )dx = 1 (11.19-b) 19

−∞

11.3.2

† Subdomain Collocation Method

20 We subdivide the region of interest into n subregions, and within each one of them the integral of the residual is set to zero.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

11.3 Weighted Residual Methods

11.3.3 21

† Least-Squares Method

The integral of the square of the residuals are minimized with respect to ai . I = R2 (x, y, z, a1 , · · · , an )dx ∂I ∂ai

11.3.4 22

11–5

= 0;

(11.20-a)

i = 1, 2, · · · , n

(11.20-b)

Galerkin Method

In the Galerkin method, we select W=ψ

(11.21)

i.e. the weight functions are equal to the trial functions. Since the trial functions are orthogonal to the residuals (Eq. 11.8), then Eq. 11.15 becomes b ψ T Rdx = 0 (11.22) a

and

 b   ψ1 L(ψ1 )dx ψ1 L(ψ2 )dx · · · ψ1 L(ψn )dx  ψ1 gdx     a a a    ab  b b b        ψ L(ψ )dx ψ L(ψ )dx · · · ψ L(ψ )dx ψ2 gdx 2 1 2 2 2 n   Ke =  a ;f = a a a    .. .. .. .. ..     . . . . .     b b b     b    ψn L(ψ1 )dx ψn L(ψ2 )dx · · · ψn L(ψn )dx ψn gdx 

b

a

b

a

b

a



                    

a

(11.23)

Example 11-1: String Vibration Formulate the ﬁnite element matrices by the Galerkin method for an an element of a string subjected to a tension T (x), and to an external transverse force per unit length be p(x, t). The mass per unit length is denoted by ρ. Assume small displacements. Solve for the fundamental frequency of vibration. Solution:

1. We ﬁrst write Newton’s equation of motion in the transverse direction (sigf y = ma) for an element of the string ∂w(x, t) ∂ 2 w(x, t) ∂ 2 w(x, t) ∂w(x, t) ∂T (x) dx + = ρdx dx + p(x, t)dx − T (x) T (x) + 2 ∂x ∂x ∂x ∂x ∂t2 (11.24) Victor Saouma

Finite Elements II; Solid Mechanics

Draft 11–6

WEIGHTED RESIDUAL METHODS ∂w(x, t) ∂ 2 w(x, t) ∂ T (x) + p(x, t = ρ ∂x ∂x ∂t2

which reduces to

(11.25)

w T(x)+

p(x,t) δw(x,t) δx

δ T(x) dx δx

2

δw(x,t) + δw(x,t) dx δx δ x2

w(x,t)

T(x)

x

x+dx

x

2. In the case of free vibration p(x, t) = 0, and for constant tension T =cst, we obtain T w,xx − ρw,tt = 0

(11.26)

which is the wave equation. 3. We next apply the Galerkin method L NT [T w,xx − ρw] ¨ dx = 0

(11.27)

0

where w = Nw 4. Integrate by parts −

L

NT,x T w,x dx

−

0

L 0

! "L NT ρwdx ¨ + NT T w,x 0 = 0

(11.28)

The last term is equal to zero because the shape functions at the corresponding nodes are equal to zero. By analogy to the virtual displacement method, N corresponds to the virtual displacement which must be equal to zero at the supports in order to satisfy the essential boundary conditions. 5. Substituting w = Nw and w,x = N,x w we obtain L L T ¨ =0 N,x T N,x dx w + NT ρNdx w 0 0 m

k

where N=

L−x L

x L

6. Substituting k = m =

Victor Saouma

(11.29)

and N,x =

1 −1 1 L

T 1 −1 L −1 1 ρL 2 −1 1 2 6

(11.30)

(11.31) (11.32)

Finite Elements II; Solid Mechanics

Draft

11.4 Applications of the Galerkin Method to 3D Elasticity Problems

11–7

7. If we consider a two elements assemblage, and suppress the ﬁxed d.o.f (1 and 3), then ρL T 2 w 2 + 4 w¨2 = 0 L 6

(11.33)

8. For free vibration w 2 = A sin ωt w¨2 = −ω 2 A sin ωt 9. Substituting

2

ρL T T − 4 ω 2 A sin ωt = 0 ⇒ ω 2 = 3 ρL 2 L 6

Note that the exact answer is w2 =

11.4

23

(11.34) (11.35)

(11.36)

π2 T 4ρL2

Applications of the Galerkin Method to 3D Elasticity Problems

We will apply the Galerkin method to two classes of problems.

We start with elasticity problems, to show that the Galerkin Method would result in exactly the same formulation as the one obtained from the variational principle. 24

25 Following this “validation” of the method, we shall consider (in the next chapter) a class of problems which does not have a variational principle, and thus the Galerkin method is the only applicable technique. The second application will be the heat equation.

11.4.1

Derivation of the Weak Form

26 We will now apply the Galerkin method to the equation of elasticity and show that we will retrieve the principle of virtual work which was derived from a variational principle. Hence, in this particular case the two methods are indeed (as anticipated) equivalent. 27

Starting with the equilibrium equation (Eq. ??) LT σ + b = 0

where ε = Lu and 

∂ ∂x

 LT =  0 0

Victor Saouma

0 ∂ ∂y

0

0 0 ∂ ∂z

∂ ∂y ∂ ∂x

0

∂ ∂z

0 ∂ ∂x

 σxx      σyy   0  σzz  ∂ ∂z  ; σ =  σ xy ∂    ∂y  σ   xz σyz

(11.37)        

   bx  b ; ; b=   y   b  z    

(11.38)

Finite Elements II; Solid Mechanics

Draft 11–8 28

WEIGHTED RESIDUAL METHODS

Expanding the equations of equilibrium we have ∂σxx ∂x ∂σyx ∂x ∂σzx ∂x

+ + +

∂σxy ∂y ∂σyy ∂y ∂σzy ∂y

+ ∂σ∂zxz + bx = 0 ∂σ + ∂zyz + by = 0 +

∂σzz ∂z

(11.39)

+ bz = 0

We multiply the ﬁrst equation in Eq. 11.39 by the arbitrary function wx and we integrate over the volume5 Ω: ∂σxy ∂σxx ∂σxz dΩ + dΩ + dΩ + wx wx wx wx bx dΩ = 0 (11.40) ∂x ∂y ∂z Ω Ω Ω Ω 29

30

31

We now apply Green-Gauss theorem (a form of the divergence theorem6 ) ∂wx ∂wx σxx dΩ + σxy dΩ wx σxx nx dΓ − wx σxy ny dΓ − ∂x ∂y Γ Γ Ω Ω ∂wx σxz dΩ + wx σxz nz dΓ − wx bx dΩ = 0 + Γ Ω ∂z Ω Recalling that the tractions are given by Eq. ?? tx = σxx nx + σxy ny + σxz nz ty = σyx nx + σyy ny + σyz nz tz = σzx nx + σzy ny + σzz nz

then the preceding equation reduces to ∂wx ∂wx ∂wx σxx + σxy + σxz dΩ + wx tx dΓ − ∂x ∂y ∂z Γ Ω 32

(11.41-a)

(11.42)

wx bx dΩ = 0

(11.43)

Ω

Similarly, we obtain for the other two equations of equilibrium ∂wy ∂wy ∂wy σyx + σyy + σyz dΩ + wy ty dΓ − ∂x ∂y ∂z Γ Ω ∂wz ∂wz ∂wz σzx + σzy + σzz dΩ + wz tz dΓ − ∂x ∂y ∂z Γ Ω

wy by dΩ = 0 (11.44-a) Ω

wz bz dΩ = 0 (11.44-b) Ω

Summing those three equations (wx tx + wy ty + wz tz )dΓ + (wx bx + wy by + wz bz )dΩ Γ Ω WT t wT b (11.45) ∂wy ∂wz ∂wx σxx + σyy + σzz + − 5 ∂x ∂y ∂z Note similarity the derivationRof the principle Ω with R ∂w R of virtualwork. 6 T ∂w ∂wIfy we ∂w x y vdΩ = x ndΓ∂w z (rΦ)T vdΩ. z vT = [ Ψ 0 0 ], we obtain From Eq. ∂w 1.43, Φ r Φv − set Ω σ R R + Ω R ∂Φ σxy + Γ + + σyz dΩ = 0 xz + Φ ∂Ψ dΩ = Γ∂y ΦΨnx dΓ∂x − Ω ∂x ΨdΩ ∂z ∂x ∂z ∂y ∂x Ω 33

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

11.4 Applications of the Galerkin Method to 3D Elasticity Problems

11–9

We next focus on the third term in the preceding equation. For w = wx wy wz T , then

34

(Lw)T =

∂wy ∂y ;

∂wx ∂x ;

∂wz ∂z ;

∂wx ∂y

+

∂wy ∂x ;

∂wx ∂z

+

∂wz ∂x ;

∂wy ∂z

+

∂wz ∂y ;

(11.46)

and (Lw)T σ =

∂wx ∂x' σxx x + ∂w ∂y

+ +

∂wy z + ∂w ∂y σ ∂z σzz (yy % ∂wy x σxy + ∂w ∂x ∂z

+

∂wz ∂x

&

' σxz +

Hence, substituting Eq. 11.47 into Eq. 11.45 we obtain T T (Lw) σdΩ = w tdΓ +

35

Ω

Γ

∂wy ∂z

+

wT bdΩ

∂wz ∂y

( σyz

(11.47)

(11.48)

Ω

But since the L operator is symmetric, from Eq. 11.2, the above equation can also be written

36

as

w (L σ)dΩ = T

T

Ω

T

wT bdΩ

w tdΓ + Γ

(11.49)

Ω

which is the weak form of the diﬀerential equation of equilibrium subjected to the traction (natural) boundary conditions. 37 The weight vector w is completely arbitrary. In the principle of virtual displacement, it would correspond to the virtual displacement, compare this last equation with Eq. 5.11. (11.50) + δuT {G [D( − 0 ) + σ 0 ] −ˆt}dΓ = 0 Γ σ

11.4.2 38

FE Discretization

Given that the displacement ﬁeld is written as (11.51)

u = Nu

where N are the shape functions, and u are the nodal known displacements. Note that this is the application of Eq. 11.9-a 39 The Galerkin method uses the shape functions for the weight (functions Wi ) Hence, Eq. 11.13-a will be

wT = cT NT

w = Nc;

(11.52-a)

and Lw = LNc = Bc

(11.53)

40 Inserting the last two equations into the weak form, and recalling that the L operator is symmetric, Eq. 11.49, yields T T T T B σdΩ − N tdΓ − N bdΩ = 0 (11.54) c

Ω

Victor Saouma

Γ

Ω

Finite Elements II; Solid Mechanics

Draft 11–10

WEIGHTED RESIDUAL METHODS

(Note similarity with Eq. 11.14). Since c is an arbitrary matrix, we conclude that BT σdΩ = NT tdΓ + NT bdΩ Ω

Γ

(11.55)

Ω

This equation is applicable to both linear and nonlinear systems. The right hand side represent the eﬀect of the load. 41

Next we introduce a constitutive model, Eq. 3.90 which is rewritten in vector form σ = D(ε − ε0 )

42

43

(11.56)

The stress and strain are given by ε = Lu = LNu = Bu

(11.57-a)

σ = DBu − Dε0

(11.57-b)

Inserting those two equations into Eq. 11.55, yields

T

Ω

B DBdΩ u = Ke

T

Γ

N tdΓ + f1

T

Ω

N bdΩ + fb

Ω

BT Dε0 dΩ

(11.58)

f0

Thus, using the Galerkin approach, we have recovered the same stiﬀness formulation as the one previously obtained from a Variational Principle. 44

45 Whereas, there may not be any clearer advantage in using the Galerkin approach when a variational principle is available, this may be the only formulation possible in the absence of the second.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft Chapter 12

FINITE ELEMENT DISCRETIZATION OF THE FIELD EQUATION Adapted from Ottosen (1992) 1 Having “proven” the validity of the Galerkin method by its ability to recover the variational based formulation, we now turn our attention to a problem which (strictly speaking) could only be handled by such an approach. 2 As a vehicle for such an application, we will consider the ﬁeld equation which was discussed in Chapter 3.3.

12.1 3

Derivation of the Weak Form

Our starting point will be the ﬁeld equation, Eq. 3.113 −div (D∇φ) − Q = ρc

∂φ ∂t

(12.1)

Recalling (Eq. 3.98) that q = −D∇φ, and that for steady state problems the right hand side reduces to zero, and ﬁnally (for the mere sake of clarity) substituting φ by T , this equation reduces to (12.2) div q − Q = 0 4

Lφ+g 5

The boundary conditions are given by Essential, Temperature ΓT T = g Natural, Flux Γq qn = qt n = qf Natural, Convection Flux Γc qc = h(T − T∞ )

6

(12.3)

Note that in a ﬁn, the convection heat loss can be considered as a negative heat source.

Draft 12–2 7

FINITE ELEMENT DISCRETIZATION OF THE FIELD EQUATION

We now apply Eq. 11.8

w(div q − Q)dΩ = 0

(12.4)

Ω

for 2D problems, we replace dΩ by tdA wdiv qtdA −

wQtdA = 0

A

8

Applying Green-Gauss Theorem1 , Eq. ??, the ﬁrst term becomes , wdiv qtdA = wtqT ndΓ − (∇w)T qtdA A

9

(12.5)

A

Γ

(12.6)

A

Thus Eq. 12.5 becomes

,

wtq ndΓ −

T

T

(∇w) qtdA = A

Γ

wQtdA

(12.7)

A

which is the weak form. 10

Applying the boundary conditions, Γ = ΓT + Γq

(∇w)T qtdA = A

wqn tdΓ −

wqf tdΓ + Γq

ΓT

wQtdA

(12.8)

A

where along Γq the ﬂux (which may include convection) is known but the temperature is not, and along ΓT , the temperature is known but ﬂux is not. This equation is valid for any constitutive relation and is analogous to Eq. 11.49. 11

Applying Fourrier Law, we obtain

(∇w) D∇T tdA = −

wqf tdΓ −

T

A

wqn tdΓ +

Γq

wQtdA

ΓT

(12.9)

A

This is the weak form of our problem. 12

By extension, the weak form in 3D would be ,

, wqf dS −

(∇w)T D∇T dV = V

12.2 13

Sq

wqn dS +

ST

wQdV

(12.10)

V

FE Discretization

The temperature ﬁeld is expressed in terms of the nodal ones (12.11)

T = NT where N are the shape functions, and T are the nodal known temperatures. 1

RR

A

φdiv qdA =

Victor Saouma

H

Γ

φqT ndΓ −

RR

A

(rφ)T qdA

Finite Elements II; Solid Mechanics

Draft

12.2 FE Discretization 14

12–3

The ﬂux is given by q = −D∇T where ∇T =

T

∂T ∂y

∂T ∂x

(12.12)

furthermore, the ﬂux qn at the boundary is given by qn = qT n. 15

Substituting ∇T = ∇NT = BT #

where B=

(12.13)

$

∂Ni ∂x ∂Ni ∂y

(12.14)

Note, this deﬁnition of the B matrix for scalar ﬁeld problems, should not be confused with the one for vector problems, Eq. ??   ∂N   ∂N i i 0 0 ∂ξ  ∂x ∂Ni  ∂Ni  −1  (12.15) B= 0  0 ∂y  = J ∂η  16

∂Ni ∂y

12.2.1 17

∂Ni ∂η

∂Ni ∂ξ

No Convection

Substituting the previous set of equations, Eq. 12.9 reduces to T (∇w) DBtdA T = − wqf tdΓ − wqn tdΓ + A

18

∂Ni ∂x

Γq

ΓT

wQtdA

(12.16)

A

In accordance with the Galerkin method, the weight function w is w = Nc

(12.17)

Since w is arbitrary, the matrix c is also arbitrary, thus ∇w = ∇Nc = Bc

(12.18)

w = cT NT

(12.19)

and since w is a scalar

19

Inserting those equations into Eq. 12.16 # T T B DBtdA T + NT qf tdΓ + c A

20

Γq

$

N qn tdΓ − T

ΓT

T

N QtdA = 0

(12.20)

A

But since this expression should hold for any arbitrary cT matrix, we conclude that Ke T Ke fbe e fQ

e = fbe + fQ = BT DBtdA A T = − N qf tdΓ − Γq = − NT QtdA

(12.21) (12.22) NT qn tdΓ

(12.23)

ΓT

(12.24)

A

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 12–4

12.2.2 21

FINITE ELEMENT DISCRETIZATION OF THE FIELD EQUATION

Convection

Next we consider the eﬀet of the convection term, and thus Γ = Γq + ΓT + Γc

(12.25)

The presence of the convection term aﬀects only fb in Eq. 12.23 which now becomes fb = − NT qf tdΓ − NT qn tdΓ − NT qc dΓ Γq

ΓT

(12.26)

Γc

and based on Newton’s convection boundary condition (Eq. 3.97) qc = h(T − T∞ )

(12.27)

qc = hNT − hT∞

(12.28)

or

22

Substituting into Eq. 12.26 yields NT qf tdΓ − NT qn tdΓ − fb = − Γq

23

ΓT

Γc

hNT NdΓ T + T∞

NT hdΓ

(12.29)

Γc

Using this relation, Eq. 12.23 becomes (Ke + Kec )T Ke Kec fbe e fQ

e = fbe + fQ = BT DBtdA A = hNT NdΓ Γc T = − N qf dΓ − Γq = − NT QtdA

(12.30) (12.31) (12.32) T

ΓT

N qn dΓ + T∞

NT hdΓ

(12.33)

Γc

(12.34)

A

It should be noted that convection occurs around the perimeter of an element, and conduction occurs inside its volume. 24 Note the analogy between this “load case” and P-∆ eﬀects in structural engineering. As will be seen in chapter 14, the geometric stiﬀness matrix (KG ), which accounts for large deformation, will be very similar to Kc .

12.3

Illustrative Examples

Example 12-1: Composite Wall

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

12.3 Illustrative Examples

12–5

Let us consider a composite wall made up of three materials, (k1 = 20 W/mo C, k2 = 30 W/mo C, k3 = 50 W/mo C),

o

h

T0 = 20 c

K1

K2

K3

0.3 m

0.15 m 0.15 m

T

inf

1

2

1

2

3

3

4 20

q

o

1 0 0 1 0 1 0 1 0 1 0 1 0 1

c

Natural B.C.

11 00 00 11 00 11 00 11 00 11 00 11 00 11

Essential B.C.

The outer temperature is T0 = 20o C, convection heat transfer takes place on the inner surface of the wall with the ﬂuid temperature T∞ = 800o C, the ﬁlm coeﬃcient (or convection term) h = 25 W/m2 .o C. We seek the temperature distribution across the wall. Solution: 1. Selecting linear elements, we have T (ξ) = NT =

1−ξ 1+ξ T1 + T2 2 2 N1

ξ = dξ = dx = dT dx

= =

B =

(12.35-a)

N2

2 2 (x − x1 ) − 1 = (x − x1 ) − 1 x2 − x1 L 2 dx L L dξ 2 dT dξ 2 dN = .T dξ dx L dξ 2! 1 1 " − 2 2 T = BT L " 1! −1, 1 L

(12.35-b) (12.35-c) (12.35-d)

(12.35-e) (12.35-f)

2. The conductivity matrix for the one dimensional element is D=k

(12.36)

3. Substituting into Eq. 12.34, we obtain Victor Saouma

Finite Elements II; Solid Mechanics

Draft 12–6

FINITE ELEMENT DISCRETIZATION OF THE FIELD EQUATION

Element Conductivity Matrix: L e T B DBtdA = BT DBdx K = 0 A kL 1 T B Bdξ = 2 −1 " kL 1 2 1 −1 ! −1, 1 dξ = 1 2 L −1 k 1 −1 = L −1 1 K1 =

20 0.3

K2 =

30 0.15

K3 =

50 0.15

1 2 1 1 −1 66.67 = −1 1 −66.67

(12.37-b)

3 −200.00 200.00

3 4 3 1 −1 333.34 = −1 1 −333.34

4 −333.34 333.34

1 66.67  −66.67  

2 −66.67 66.67 + 200.00 −200.00

1 66.67  −66.67   0. 0.

2 −66.67 266.67 −200.00 0.

 =

2 −66.67 66.67

2 3 2 1 −1 200.00 = −1 1 −200.00

 K =

(12.37-a)

(12.37-c)

(12.37-d)

3

4

−200.00 200.00 + 333.34 −333.34

3 0. −200.00 533.34 −333.34



  (12.37-e) −333.34  333.34

4  0.  0.  −333.34  333.34

(12.37-f)

Element Convection Matrix: Since convection occurs only at one point, 1, we just add the convection term h = 25 to K1,1 1 66.67 + 25  −66.67   0. 0.  K =

1 91.67  −66.67   0. 0. 

=

Victor Saouma

2 −66.67 266.67 −200.00 0.

2 −66.67 266.67 −200.00 0.

3 0. −200.00 533.34 −333.34

3 0. −200.00 533.34 −333.34

4  0.  0.  −333.34  333.34

4  0.  0.  −333.34  333.34

(12.38-a)

Finite Elements II; Solid Mechanics

Draft

12.3 Illustrative Examples

12–7

Element heat rate vector: Since there is no heat generation Q, fQ = 0, and we do NT qf dΓ − NT qn dΓ + not have a distributed convection (hence fb = − Γ Γ q T T N hdΓ = 0) and we only have a point convection load T∞ Γc

  √   (800)(25)  P1        √     0 P2 √ = 0 P            3  P4 ? P4 ?

(12.39)

where P4 corresponds to the point heat ﬂux transmitted at node 4. Solve  0. 91.67 −66.67 0.   T1 ?   T2 ?  −66.67 266.67 −200.00 0.    0. −200.00 533.34 −333.34    T3 ?  T4 = 20 0. 0. −333.34 333.34 

   

  (800)(25)       0 = 0          P4 ? (12.40)

In this case, one of the essential boundary conditions is non-zero, hence following a procedure similar to the one outlined in Eq. this reduces to        0 81.67 −66.67 0.  20, 000.     T1 ?   −66.67 266.67 −200.00  = 0 − 20. (12.41-a) 0 T2 ?       0 −333.34 T3 ? 0. −200.00 533.34    20, 000.  = 0 (12.41-b)   6, 666.80      T1   304.6  o T2 119.0 C =     T3 57.1

and the solution is

(12.42)

Veriﬁcation We can check the numerical solution by verifying that q=

T1 − T2 x1 −x2 k1

=q=

T2 − T3 x2 −x3 k2

=q=

T3 − T4 x3 −x4 k3

(12.43)

Example 12-2: Heat Transfer across a Fin A metallic ﬁn with thermal conductivity k = 360 W/m.o C, 0.1 cm thick and 10 cm long extends from a plane whose temperature is 235 o C.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 12–8

FINITE ELEMENT DISCRETIZATION OF THE FIELD EQUATION

Tinf= 20o C 2 h = 9 W/m oc

0.1 cm

10 cm 1

Tinf= 20o C

2

3

4

We seek to determine the temperature distribution and amount of heat transfered from the ﬁn to the air at 20o C where the ﬁlm coeﬃcient h=9 W/mo C. Using a three element discretization, and assuming the tip of the ﬁn to be isolated. Solution: 25

The solution described below, considers a unit volume of the ﬁn. Element conductivity matrix: Since we are considering a unit volume K

e

= = = =

K1 =

Victor Saouma

L

BT DBdx 0 kL 1 T B Bdξ 2 −1 " kL 1 −1 ! −1, 1 dξ 1 2 −1 k 1 −1 L −1 1 360 0.033

K2 =

360 0.033

K3 =

360 0.033

1 2 1 −1 −1 1 2 3 1 −1 −1 1 3 4 1 −1 −1 1

(12.44-a)

(12.44-b)

(12.44-c)

(12.44-d)

Finite Elements II; Solid Mechanics

Draft

12.3 Illustrative Examples

12–9 1 2 1 −1 360   −1 1 + 1 −1 0.033 

3



K =

4



  −1  1

−1 1+1 −1

1 2 3 4  1 −1  360   −1 2 −1   −1 2 −1  0.033 −1 1

(12.44-e)



=

(12.44-f)

Convection Matrix The shape functions for the linear element are given by Eq. 12.35-a. 1−ξ 1+ξ N = 2 2 N1 1−ξ 2 1+ξ 2

NT N =

 ' = 

1

T

N N = −1

1 3

N2

1−ξ 2

1−ξ 2

(2

1+ξ 1−ξ 2 2

2 1 1 2

(12.45-a)

1+ξ 2

1+ξ 1−ξ 2 2 ' (2 1+ξ 2

(12.45-b)

 

(12.45-c)

(12.45-d)

With respect to Fig. 12.1 heat transfer by convection occurs across the perimeter of the

co nv

ec

tio

n

he

at

lo ss

B

t

x

q dx

L

Figure 12.1: Heat Flow in a Thin Rectangular Fin ﬁn (P = 2(B + t)); however, as noted above, we are considering a unit volume of the ﬁn

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 12–10

FINITE ELEMENT DISCRETIZATION OF THE FIELD EQUATION

(AC = Bt), hence we rewrite Eq. 12.32 as Kec

= "

but, from Eq. 12.35-d dx =

L 2 dξ,

P L hNT Ndx Ac 0 2h L T N Ndx t 0

(12.46-a) (12.46-b)

thus

hL Kc = t

1

NT Ndξ

(12.47)

−1

Inserting the term from Eq. 12.45-d we ﬁnally obtain hL 2 1 Kc = 3t 1 2

(12.48)

As for the conduction stiﬀness matrix, the assembled convection stiﬀness matrix will

26

be 1 2 −2  (9)(3.33 × 10 )  1 Kc =  3(10−3 ) 

2 3 1 4 1 1 4 1

4

  1 2

Element heat rate vector: Can be determined from T T N qf tdΓ − N qn tdΓ + T∞ fb = − Γq



ΓT

NT htdΓ

(12.49)

(12.50)

Γc

however in this problem, only the third term is non-zero. Following a similar procedure as above, e fb = T∞

= T∞ h = =

NT htdΓ

(12.51-a)

NT dΓ

(12.51-b)

Γc

Γc

2T∞ hL 2t Γc T∞ hL 1 1 t

1−ξ 2 1+ξ 2

Assembling the load vector for this problem would yield    1 (20)(9)(3.33 × 10−2 )  2 fb = 2  10−3   1

Victor Saouma

dξ

(12.51-c) (12.51-d)

      

(12.52)

Finite Elements II; Solid Mechanics

Draft

12.3 Illustrative Examples Solution of the system  1   1   360  −1     0.033   

12–11

becomes 2 3 4  −1  (9)(3.33 × 10−2 ) 2 −1 + −1 2 −1  3(10−3 ) −1 1

    1   (20)(9)(3.33 × 10−2 )  2  2   10−3     1  1 2 3   10, 800. −10, 800. 0.      −10, 800. 21, 600. −10, 800.   0. −10, 800. 21, 600.   0. 0. −10, 800. 

1  2 1  

 2 3 4   1     4 1   1 4 1     1 2 

√ T1 T2 ? T3 ? T4 ?

      

=

(12.53-a)

4  0.  0. + −10, 800.  10, 800.

 P1 ?    12, 000. 12, 000.       6, 000.  0. 0. 11, 000. −10, 700.  −10, 700. 22, 000. −10, 700. 0.   −10, 700. 22, 000. −10, 700. 0. 0. −10, 700 11, 000. 0.

1  200.  100.   0. 0.

2 100. 400. 100. 0.

3 0. 100. 400. 100.

 4   0.  T1 = 235    T2 ? 0.    T 100.   3?   T4 ? 200. 

      

=

   

(12.53-b)     235. P ?     1       T2 ?   12, 000.     T3 ?  =  12, 000.          T4 ? 6, 000.

As in the previous example, we rewrite this last equation as        22, 000. −10, 700. 0.  12, 000.   −10, 700   T2 ?   −10, 700. 22, 000. −10, 700.  = 12, 000. − 235. 0. T ?      3  6, 000. 0. T4 ? 0. −10, 700. 11, 000.    2.514 × 106  12, 000. (12.54-a) =   6, 000. Solving, we obtain

     T2   208.77  o T C (12.55) = 194.25   3   T4 189.50 ) Heat loss in the ﬁn can be computed from H = e H e where H e = h(Tavg − T∞ )As sand the surface area of each element is As = 2(1 × 0.0333)m2

Victor Saouma

Finite Elements II; Solid Mechanics

(12.53-c)

Draft 12–12

FINITE ELEMENT DISCRETIZATION OF THE FIELD EQUATION

12.4

Comparison Between Vector and Scalar Formulations

25 We now complement Table 3.2, by contrasting major governing equations in both scalar and vector problems, Table 12.1.

Scalar

Vector

Conduction

Solid Mechanics

I Diﬀerential Equation div q − Q

qn q T qn

Z

= T rT q = = on on

LT + b

0

"

qT n −DrT ΓT Γq ∪ Γ c

Z

T

w (L )dΩ = Ω

t u t

B

=

(Ke + Kec )T

=

e

K

=

Kec

=

fbe

=

− −

e fQ

=

−

NT bdΩ

III Discretized Solution #

T

B DBtdA

ZΓc

Discretized Weak Form

Ω

∂Ni ∂x ∂Ni ∂y

A

ZΓ

NT tdΓ

+

e fbe + fQ +P Z

Z

n D" Γu Γt

=

wT bdΩ

NT BT

"

Balance/Equilibrium State variable “kinematic” Flux vector Flux at boundary Constitutive law Essential BC Natural BC

Lu

Ω

Ω

= =

BT dΩ

w tdΓ

+ T ∇T

0

IIZ Weak Formulation Z

T

ZΓ

= u = = = on on

u "

= =

Nu Bu 2

B

=

6 4

Ke ue

=

Ke

=

∂Ni ∂x

0

∂Ni ∂y ∂Ni ∂x

0 ∂Ni ∂y

3 7 5

fe + f0e + P

R

T

Ωe

B DBdΩ

hNT NtdΓ NT qf tdΓ

Z

Γq

Z

ΓT A

State variable gradient Balance/Equilibrium Stiﬀness matrix Convection Stiﬀness Matrix

fe

=

fe

=

NT qn tdΓ NT QtdA

State variables Flux

R Γt

NT ˆtdΓ

Z Z

NT bdΩ

Boundary forces

Body forces

Ωe

+

Z

BT D0 dΩ

Initial strain

BT 0 dΩ

Initial stress

Ωe

− Ωt

Table 12.1: Comparison of Scalar and Vector Field Problems, Revisited

Victor Saouma

Finite Elements II; Solid Mechanics

Draft Chapter 13

TOPICS in STRUCTURAL MECHANICS 13.1

Condensation/Substructuring

1 Condensation is the process of reducing the number of d.o.f. by substitution. At an element level, we may want to condense an internal node (such as in the Lagrangian element), at the global level, this is closely associated with substructuring. 2

Let us consider kd = p and let d be partitioned into dr {d} = dc

(13.1)

where the subscripts r and c refer to retained and condensed d.o.f. respectively. 3

For substructuring, subscripts r and c refer to interface and interior d.o.f. Hence, dr pr krr krc = kcr kcc dc pc

or

k dr = p

(13.2)

(13.3)

where k = krr − krc k−1 cc kcr

p

p r

=

−

krc k−1 cc pc

(13.4-a) (13.4-b)

4 Noting the common form of the above two equations, we deduce that condensation may also be viewed as a transformation from a d, p, k system to a dr , p , k through the following transformation

k = ΓTc kΓc

p

=

ΓTc p

(13.5-a) (13.5-b)

Draft 13–2

TOPICS in STRUCTURAL MECHANICS

where Γc =

−k−1 cc kcr I

(13.6)

k is a r by r matrix, whereae k was r+c by r+c (yet we still have to separately decompose kcc which is c by c). 5

13.2

Element Evaluation

13.2.1

Patch Test

6 The patch test is a check which ascertains whether a patch of inﬁnitesimally small elements subjected to constant strain reproduces exacly the constitutive behavior of the material through correct stresses. 7 It has been argued that an element which passes the patch test satisﬁes the two essential conditions for convergence.

First we assemble a small number of elements into a “patch” in such a way that there is at least one internal node shared by two or more elements, and that one or more interelement boundaries exist. Enough support must be provided to prevent rigid body motion, Fig. 13.1. 8

F

h

h

5

2F

5 20 5

h

h

5

F 8 and 9 noded isoparametric quadrilaterals; Pure Axial and Flexure 8 noded Isoparametric constant axial stress

Figure 13.1: Patch Tests Boundary nodes are then loaded by a consistently derived nodal loads appropriate to a state of constant stress. This can be achieved by applying nodal load or nodal displacement. We then verify that a state of constant stress is obtained inside the element. This test must be repeated for each of the appropriate strains (εxx , εyy and γxy ). 9

10

The patch test may pass by an individual element, but fail by an assemblage of such elements.

11 Failure of the patch test is also an indication of lack of stability. That is we have zero-energy deformation (as detected by the eigenvalue test) is present.

13.2.2

Eigenvalue Test

12 The stiﬀness matrix is, by deﬁnition, singular due to the fact that equilibrium relations are embedded in its formulation, or alternatively, the assumed displacement ﬁeld on which it is based is supposed to provide for rigid body motions (translations and/or rotations).

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

13.2 Element Evaluation 13

13–3

The augmented stiﬀness matrix may be expressed as: [d]−1 [B]T [d]−1 [k] = [B][d]−1 [B][d]−1 [B]T

(13.7)

where B is the statics (or equilibrium) matrix, relating external nodal forces to internal forces; d is a ﬂexibility matrix, and d−1 is its inverse or reduced stiﬀness matrix. We note that the stiﬀness matrix is obviously singular, since the second “row” is linearly dependent on the ﬁrst one (through [B]) and thus, its determinent is equal to zero. 14

The reduced stiﬀness matrix, which is the inverse of a ﬂexibility matrix, is not.

15

The internal strain energy stored in an element can be determined from U

= =

1 u {F} 2 1 u[K] {u} 2

(13.8)

16 If we consider a system where the load {F} applied to each node is proportional to the element nodal displacement {u} by a factor λ, we have:

[K] {u} = λ {u}

(13.9)

([K] − λ[I]) {u} = 0

(13.10)

or

17 This is by deﬁnition an eigenproblem. There will be as many eigenvalues λi as there are degrees of freedom (or rows in [K]). To each eigenvalue λi corresponds an eigenvector {u}i . 18

If the eigenvectors are normalized such that: ui {u}i = 1

(13.11)

ui [K] {u}i = λi

(13.12)

then Eq. 13.9 becomes: and this last equation, when compared to Eq. 13.8, shows that the eigenvalue λi is equal to twice the internal strain energy stored in an element undergoing a (normalized) deformation deﬁned by {u}i . 19

Hence, we observe that: 1. In a rigid body motion, all nodes displace by the same amount, and there are no internal strains. Hence in a rigid body motion the strain energy U (and thus corresponding λ) must be equal to zero. 2. There should be as many zero eigenvalues as there are possible independent rigid body motions (i.e. number of equations of equilibrium). For a two dimensional Lagrangian element, there should be three zero eigenvalues, corresponding to two translations and one rotation. 3. Too few zero eigenvalues is an indication of an element lacking the capability of rigid body motion without strain.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 13–4

TOPICS in STRUCTURAL MECHANICS

4. Too many zero eigenvalues is an indication of undesirable mechanism (or failure). 5. Eigenvalues should not change when the element is rotated. 6. Similar modes (such as ﬂexure in two orthogonal directions) will have identical eigenvalues (for isotropic material). 7. When comparing the stiﬀness matrices of two identical elements but based on diﬀerent formulations, the one with the lowest strain energy (tr[K] = Σλi ) is best. 20

Hence, the element stiﬀness matrix will have:

Order: corresponds to the number of degrees of freedom (i.e size of the matrix). Rank: corresponds to the total number of linearly independent equations which is equal to the order minus the number of rigid body motions. Rank Deﬁciency: would be equal to the total number of zero eigenvalues minus the rank. 21

For a square bilinear element the non-zero eigenvectors are shown in Fig. 13.2.

1

2

3

4

5

a

7

6

8

b

Figure 13.2: Eigenvectors Corresponding to a) Non-Zero and b) Zero Eigenvalues for a Square Bilinear element

13.2.3 13.2.3.1

Order of Integration Full Integration

For numerically integrated elements, a question of paramount importance is the order of the numerical integration. If the order of integration is too

22

High: it would certainly result in an exact formulation, which however may be computationally expensive. Low: Then we may have rank deﬁciency. Hence, we ought to properly select the order of numerical quadrature. 23

We recall that

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

13.2 Element Evaluation

13–5

1. We seek to integrate F = BT DBdetJ. 2. The Jacobian is constant only for parallelograms. 3. In Gaussian Quadrature, integration with n points correctly integrates a polynomial of order 2n − 1 (whereas for Newton-Cotes it was n − 1). Hence, for a bilinear parallelogram element, F is a polynomial function, of order 2, in terms of the natural coordinates ξ and η which can be exactly integrated by n = 2 points (i.e 2 × 2). 24 By similar arguments, a quadratic quadrilateral should be integrated by three points (F is now of order 4, and thus 2 × 3 − 1 = 5).

Those orders of approximations are only valid for parallelograms, as the edge becomes inclined, then the error increases. 25

13.2.3.2

Reduced Integration

26 Reduced integration, i.e. using fewer Gauss points than required for an exact integration, will cause the element to soften because higher order polynomial terms may vanish at the points of lower order rule. 27 In an elastic body, the FE solution permits only those displacements which can be represented by a linear combination of the shape functions. In general, the exact displacement ﬁeld will diﬀer from the assumed displacement, in other words the shape functions prevent the structure from deforming the way it wants1 .

Given that the stiﬀness formulation is too stiﬀ, (Ref. Sect. ??), the softening introduced by a reduced numerical integration is beneﬁcial. 28

29

Recommended orders of integration for quadrilaterla elements are shown in Table 13.1. Element 4-node 4-node distorted 8-node 8-node distorted 9-node 9-node distorted

Reliable Integration 2x2 2x2 3x3 3x3 3x3 3x3

Reduced Integration 2x2 2x2 2x2 2x2

Table 13.1: Full and Reduced Numerical Integrations for Quadrilateral Elements . 30 If the element is artiﬁcially softened, then there will be spurious zero energy modes, and the stiﬀness matrix will be rank deﬁcient, i.e. its rank is less than the number of element d.o.f. minus the number of rigid body modes. 31

The independent displacement modes of a bilinear element are shown in Fig. 13.3. First 1

This is analogous to trucating a Fourrier series, which otherwise could have been an exact substitute to a set of physical observations.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 13–6

TOPICS in STRUCTURAL MECHANICS 4

3 2

1

3

4

2

1

5

6

7

8

Figure 13.3: Independent Displacement Modes for a Bilinear Element three modes are rigid body motion, the next three are constant strains, and the last two are bending modes. 32 Since the element does not provide any resistance to all spurious zero-energy modes (which can nevertheless be used in linear combination with the other modes to describe the displacement ﬁeld), then the eﬀect of reduced integration is to soften the stiﬀness of the FE model. 33 Should we adopt a reduced integration scheme, we will substantially reduce the computational eﬀort, however we will forfeit some very special properties of the displacement method, namely: a) Boundedness and b) monotonic convergence toward an exact solution. 34 An illustration of problems arising with reduced integration is illustrated for the following elements

Q4 For a single point integration, the Gauss point is at the center. At the center of an element under bending εxx = εyy = γxy = 0, accordingly U = 0 for the corresponding modes 4 and 5 in Fig. 13.2, those spurious modes will disappear if n = 2. Q9 Displacement is given by u = 3ξ 2 η 2 − ξ 2 − η 2 and v = 0. If the element is integrated by 2x2 points, at ξ = η = ± √13 it can be shown that in those points the strains are zero and the corresponding mechanisms, or hourglass are shown in Fig. 13.4.

η ξ

Eight- or nine-node elements Nine-node element only

Nine-node element only Eight- or nine-node elements

Figure 13.4: Hourglas Modes in Under-Integrated Quadratic Element Q8 13.2.3.3

Selective Reduced Integration

35 In selective integration, we integrate the diﬀerent strain terms with diﬀerent orders of integration.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

13.3 Parasitic Shear/Incompatible Elements

13–7

Figure 13.5: Rectangular Bilinear Element Subjected to Bending; Bilinear Element and Correct Geometry 36

More about this technique in subsequent chapter.

13.3

Parasitic Shear/Incompatible Elements

13.3.1

Q4, The Problem

37

The shape functions for the bilinear element are given by Eq. 9.19 N1 (ξ, η) = N3 (ξ, η) =

1 4 (1 1 4 (1

− ξ)(1 − η); N2 (ξ, η) = + ξ)(1 + η); N4 (ξ, η) =

1 4 (1 1 4 (1

+ ξ)(1 − η); − ξ)(1 + η);

(13.13)

where ξ = x/a and η = y/b. 38 Imposing the displacement ﬁeld ﬁeld of Fig. 13.5 the bilinear element and correct displacements and strains are given by Table 13.2.

u v εxx εyy γxy

Bilinear Element ξηu 0 η ua 0 ξ ub

Exact Formulation ξηu au bu 2 (1 − ξ ) 2b + (1 − η 2 )ν 2a η ua −νη ua 0

Table 13.2: Bilinsear and Exact Displacements/Strains From this table we note that parasitic shear contributes to the strain energy stored in the element (which is equal to PT .d). Hence for a given imposed displacement, the forces would be larger and M1 > M2 . 39

40

Computing the ratio of strain energies of the two elements, we obtain 1 1 ' a (2 1 M1 + = M2 1+ν 1+ν 2 b

(13.14)

hence, as a/b increases, the element “locks” (a term which will be explained in later chapters). 41

Hence, the unwanted shear strain that produces M1 > M2 is called parasitic shear.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 13–8

TOPICS in STRUCTURAL MECHANICS

Figure 13.6: Displacements Associated with Incompatible Modes for the Q6 Element

13.3.2

Q6, The Solution

42 Comparing the displacement ﬁelds in Table 13.2 we observe that the bilinear element errs from the exact solution by omitting the displacement mode associated with (1−ξ 2 ) and (1−η 2 ). 43

We can remedy to this situation by adding the missing modes as internal nodes

u = Ni ui + (1 − ξ 2 )a1 + (1 − η 2 )a2

v = Ni v i + (1 − ξ 2 )a3 + (1 − η 2 )a4

(13.15-a) (13.15-b)

where the Ni are given by Eq. 13.13, and ai are nodeless d.o.f. The element now has 6 degrees of freedom, and hence the element is referred to as Q6. The B matrix would have to be correspondingly adjusted and following its formulation, the additional d.o.f. ai are eliminated through condensation (Sect. 13.1). 44

45 The Q6 element is incompatible (or nonconforming) due to the square distribution of u or v along the edge which is deﬁned by only two nodes, Fig. 13.6.

The softening in this case counters the inherent overstiﬀness of the standard assumed displacement formulation. 46

13.3.3 47

QM6, Further Enhancements

It can be shown that the Q6 element passes the patch test only if it is rectangular.

To remedy to this deﬃciency, we consider that portion of the B matrix associated with the nodeless d.o.f. and label it Ba , and then the corresponding consistent nodal load vector is e e BTa σ 0 dΩ (13.16) pa = − 48

Ω

where σ 0 represent the initial stresses caused by the standard element formulation associated with dr on the nodal d.o.f. 49 Because the standard isoparametric element passes the patch test, when σ 0 is constant, without this additional load associated with the ai , then pea should also be zero for constant σ 0

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

13.4 Rotational D.O.F.

13–9

Figure 13.7: Side Displacements Induced by Drilling d.o.f. or

1 −1

1 −1

BTa tdetJdξdη = 0

(13.17)

For general (non-rectangular elements) this equation is not satisﬁed and the patch test fails. 50 This can be artiﬁcaily remedied in forming Ba and integrating by using J0 where the jacobian is evaluated at ξ = η = 0 rather than at the Gauss quadrature point, and we use the corresponding t0 for the thickness. 51 When such an operation is performed, the resulting element is called the QM6, and it works almost as well as the quadratic element if rectangular, and is considerably more accurate than the bilinear element. 52 Stresses are computed with all the d.o.f., including the ai . Better results are achieved if element nodal loads pe that are associated with incompatible modes are set to zero during recovery of the associated d.o.f. ai (following condensation).

13.4

Rotational D.O.F.

53 Drilling d.o.f. are rotations at corner nodes. In the context of plane elements, they are associated with parabolic displaced shapes of the edges, Fig. 13.7. They are of importance in shells deﬁned as an assemblage of ﬂat elements. Hence, the element can model ﬂexure and membrane action.

Assuming rotation ω i and ω j at nodes i and j of an element side of length L, the end moments of an immaginary beam will be 54

Mi =

Victor Saouma

2EI (2ω i + ω j ) L

(13.18-a)

Finite Elements II; Solid Mechanics

Draft 13–10

TOPICS in STRUCTURAL MECHANICS

Mj

=

2EI (ω i + 2ω j ) L

(13.18-b)

Hence, from M = Mi + (Mj − Mi ) Lx and the vitual force method we can solve for the midside edge-normal displacement L (13.19) δ = (ω j − ω i ) 8 55

Hence the edge displacement is caused by two parts: translation and parabolic distorsion: s L−s L − s ui cos α uj u s(ωj − ωi ) (13.20) + + = sin α vi vj v L L 2L

and similar expressions can be written for other edges.

13.5

Constraints

Victor Saouma

Finite Elements II; Solid Mechanics

Draft Chapter 14

GEOMETRIC NONLINEARITY 14.1 1

Introduction

With refernce to Fig. 14.1, we distinguish diﬀerent levels of analysis:

Load

First order elastic analysis

Stiffening

Bifurcation

Elastic Critical Load Elastic Stability Limit Bifurcation Softening

Bifurcation

Second order elastic analysis

Inelastic Critical Load Plastic Limit Load

First order inelastic analysis

Inelastic Stability Limit Second order inelastic analysis

Displacement

Figure 14.1: Level of Analysis

First Level elastic which excludes anly nonlinearities. This is usually acceptable for service loads. Elastic Critical load is usually determined from an eigenvalue analysis resulting in the buckling load. Secon-order elastic accounts for the eﬀects of ﬁnite deformation and displacements, equilibrium equations are written in terms of the geometry of the deformed shape, does not account for material non-linearilties, may be able to detect bifurcation and or increased stiﬀness (when a member is subjected to a tensile axial load).

Draft 14–2

GEOMETRIC NONLINEARITY

First-order inelastic equilibrium equations written in terms of the geometry of the undeformed structure, accounts for material non-linearity. Second-order inelastic equations of equilibrium written in terms of the geometry of the deformed shape, can account for both geometric and material nonlinearities. Most suitable to determine failure or ultimate loads. 2 This chapter will focus on elastic critical load determination. We will begin by reviewing the derivation of some of the fundamental equations in stability analysis. We will examine both the strong form, and the weak formulation.

14.1.1 3

Strong Form

Column buckling theory originated with Leonhard Euler in 1744.

An initially straight member is concentrically loaded, and all ﬁbers remain elastic until buckling occur. For buckling to occur, it must be assumed that the column is slightly bent as shown in Fig. 14.2. Note, in reality no column is either perfectly straight, and in all cases a minor 4

P

P

x

x and y are principal axes

x

Slightly bent position L

y

Figure 14.2: Euler Column imperfection is present. 14.1.1.1

Lower Order Diﬀerential Equation

5 At any location x along the column, the imperfection in the column compounded by the concentric load P , gives rise to a moment

Mz = −P y

(14.1)

Note that the value of y is irrelevant. 6

Recalling that

Mz d2 y = dx2 EI upon substitution, we obtain the following diﬀerential equation P d2 y y=0 − 2 dx EI

Victor Saouma

(14.2)

(14.3)

Finite Elements II; Solid Mechanics

Draft

14.1 Introduction 7

Letting k2 =

P EI ,

14–3

the solution to this second-order linear diﬀerential equation is y = −A sin kx − B cos kx

8

(14.4)

The two constants are determined by applying the boundary conditions 1. y = 0 at x = 0, thus B = 0 2. y = 0 at x = L, thus A sin kL = 0

(14.5)

This last equation can be satisﬁed if: 1) A = 0, that is there is no deﬂection; 2) kL = 0, that is no applied load; or 3) kL = nπ (14.6) % & 2 P = nπ or Thus buckling will occur if EI L 9

P =

n2 π 2 EI L2

10 The fundamental buckling mode, i.e. a single curvature deﬂection, will occur for n = 1; Thus Euler critical load for a pinned column is

π 2 EI L2

(14.7)

π2 E σcr = % &2 L

(14.8)

Pcr =

11

The corresponding critical stress is

r

where I = Ar 2 . 12

Note that buckling will take place with respect to the weakest of the two axis.

14.1.1.2

Higher Order Diﬀerential Equation

In the preceding approach, the buckling loads were obtained for a column with speciﬁed boundary conditons. A second order diﬀerential equation, valid speciﬁcally for the member being analyzed was used. 13

In the next approach, we derive a single fourth order equation which will be applicable to any column regardelss of the boundary conditions. 14

15 Considering a beam-column subjected to axial and shear forces as well as a moment, Fig. dv between the 14.3, taking the moment about i for the beam segment and assuming the angle dx axis of the beam and the horizontal axis is small, leads to (dx)2 dV dv dM dx + w + V + dx − P dx = 0 (14.9) M− M+ dx 2 dx dx

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 14–4

GEOMETRIC NONLINEARITY w w(x)

1111 0000 0000 1111 0 1 0 1 0 1 0 y, u 1

1111 0000 1111 0000 dx

w

M

P

P

x

V+ δ V dx δx

i

P δv δx

P dx

P M+δ M dx δx

θi

i j

θj

P

Figure 14.3: Simply Supported Beam Column; Diﬀerential Segment; Eﬀect of Axial Force P Neglecting the terms in dx2 which are small, and then diﬀerentiating each term with respect to x, we obtain d2 v dV d2 M − P − =0 (14.10) dx2 dx dx2 16

17

However, considering equilibrium in the y direction gives dV = −w dx

18

(14.11)

From beam theory, neglecting axial and shear deformations, we have M = −EI

d2 v dx2

(14.12)

19 Substituting Eq. 14.11 and 14.12 into 14.10, and assuming a beam of uniform cross section, we obtain d4 v d2 v (14.13) EI 4 − P 2 = w dx dx

P Introdcing k2 = EI , the general solution of this fourth order diﬀerential equation to any set of boundary conditions is 20

v = C1 sin kx + C2 cos kx + C3 x + C4 21

(14.14)

If we consider again the stability of a hinged-hinged column, the boundary conditions are v = 0, v,xx = 0 at x = 0 v = 0, v,xx = 0 at x = L

(14.15)

substitution of the two conditions at x = 0 leads to C2 = C4 = 0. From the remaining conditions, we obtain C1 sin kL + C3 L = 0

(14.16-a)

−C1 k sin kl = 0

(14.16-b)

2

these relations are satisﬁed either if C1 = C3 = 0 or if sin kl = C3 = 0. The ﬁrst alternative leads to the trivial solution of equilibrium at all loads, and the second to kL = nπ for n = 1, 2, 3 · · ·. For n = 1, the critical load is Pcr = Victor Saouma

π 2 EI L2

(14.17)

Finite Elements II; Solid Mechanics

Draft

14.1 Introduction

14–5

which was derived earlier using the lower order diﬀerential equation. Next we consider a column with one end ﬁxed (at x = 0), and one end hinged (at x = L). The boundary conditions are 22

v = 0, v,xx = 0 at x = 0 v = 0, v,x = 0 at x = L

(14.18)

These boundary conditions will yield C2 = C4 = 0, and sin kL − kL cos kL = 0

(14.19)

But since cos kL can not possibly be equal to zero, the preceding equation can be reduced to tan kL = kL

(14.20)

which is a transcendental algebraic equation and can only be solved numerically. We are essentially looking at the intersection of y = x and y = tan x, Fig. 14.4 and the smallest 10.0 8.0 6.0 4.0 2.0 0.0 -2.0 -4.0 -6.0 -8.0 -10.0 0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

Figure 14.4: Solution of the Tanscendental Equation for the Buckling Load of a Fixed-Hinged Column positive root is kL = 4.4934, since k2 =

Pcr =

P EI ,

the smallest critical load is

(4.4934)2 π2 EI = EI 2 L (0.699L)2

(14.21)

Note that if we were to solve for x such that v,xx = 0 (i.e. an inﬂection point), then x = 0.699L. We observe that in using the higher order diﬀerential equation, we can account for both natural and essential boundary conditions. 23

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 14–6

14.1.2 14.1.2.1

GEOMETRIC NONLINEARITY

Weak Form Strain Energy

24 Considering a uniform section prismatic element, Fig. ??, subjected to axial and ﬂexural deformation (no shear), the Lagrangian ﬁnite strain-displacement relation is given by 3.26

1 2 2 εxx = u,x + (u2,x + v,x + w,x ) 2

(14.22)

thus, the total strain would be εxx =

2 1 dv 2 d v du + −y dx dx2 2 dx Axial Flexure Large Deformation

(14.23)

25 We note that the ﬁrst and second terms are the familiar components of axial and ﬂexural strains respectively, and the third one (which is nonlinear) is obtained from large-deﬂection strain-displacement. 26

27

28

The Strain energy of the element is given by 1 e Eε2 dΩ U = 2 Ω xx

(14.24)

Substituting Eq. 14.23 into U e we obtain 2 2 2 # 2 d v 1 1 dv 4 du du e 2 d v +y + − 2y U = 2 L A dx dx2 4 dx dx dx2 2 2 2 $ dv du dv d v + EdAdx −y dx2 dx dx dx

Noting that

dA = A;

y 2 dA = I

ydA = 0;

A

A

A 4

% dv &4 dx

(14.26)

A

for y measured from the centroid, U e reduces to # 4 2 2 2 $ 2 dv dv 1 du d A du v E A +I + +A dx Ue = 2 2 L dx dx 4 dx dx dx We discard the highest order term linear instability formulation.

(14.25-a)

(14.27)

in order to transform the above equation into a

29 Under the assumption of an independent prebuckling analysis for axial loading, the axial load Px is du (14.28) Px = EA dx

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

14.1 Introduction

14–7

Thus Eq. 14.27 reduces to 1 U = 2

#

e

EA

L

du dx

2

+ EI

d2 v dx2

2

+ Px

dv dx

2 $ dx

(14.29)

We can thus decouple the strain energy into two components, one associated with axial and the other with ﬂexural deformations 30

U e = Uae + Ufe 2 1 du e EA dx Ua = 2 L dx # 2 $ 2 2 dv d 1 v EI dx Ufe = + Px 2 2 L dx dx

14.1.2.2

(14.30-a) (14.30-b) (14.30-c)

Euler Equation

Recall, from Eq. 2.74 that for a functional in terms of two ﬁeld variables (u and v) with higher order derivatives of the form (14.31) Π= F (x, y, u, v, u,x , u,y , v,x , v,y , · · · , v,yy )dxdy 31

there would be as many Euler equations as dependent ﬁeld variables, Eq. 2.75  ∂F ∂ ∂F ∂ ∂F ∂ 2 ∂F ∂2 ∂ 2 ∂F  ∂F ∂u − ∂x ∂u,x − ∂y ∂u,y + ∂x2 ∂u,xx + ∂x∂y ∂u,xy + ∂y 2 ∂u,yy = 0  32

∂F ∂v

∂ ∂F − ∂x ∂v,x −

∂ ∂F ∂y ∂v,y

2

∂ ∂F + ∂x 2 ∂v,xx +

∂F ∂2 ∂x∂y ∂v,xy

+

∂ 2 ∂F ∂y 2 ∂v,yy

(14.32)

= 0

For the problem at hand, those two equations reduce to " 1! e 2 2 EIv,xx + Px v,x dx Uf = L 2

(14.33)

F

and the corresponding Euler equation will be −

∂ 2 ∂F ∂ ∂F + 2 =0 ∂x ∂v,x ∂x ∂v,xx

(14.34)

The terms of the Euler Equation are given by ∂F ∂v,x ∂F ∂v,xx

= Px v,x

(14.35-a)

= EIv,xx

(14.35-b)

Substituting into the Euler equation, and assuming constant Px , and EI, we obtain EI

d4 v d2 v − P =0 x dx4 dx2

(14.36)

which is identical to Eq. 14.13 Victor Saouma

Finite Elements II; Solid Mechanics

Draft 14–8

14.2

GEOMETRIC NONLINEARITY

Finite Element Discretization

33 Assuming a functional representation of the transverse displacements in terms of the four joint displacements

v = Nu dv = N,x u dx d2 v = N,xx u dx2

34

(14.37-a) (14.37-b) (14.37-c)

Substituting this last equation into Eq. 14.30-c, the element potential energy is given by Πe = Ufe + W e 1 1 ue [ke ] {ue } + ue [kg ] {ue } − u {P} = 2 2

where

(14.38-a) (14.38-b)

EI {N,xx } N,xx dx

[ke ] =

(14.39)

L

and

[kg ] = P

{N,x } N,x dx

(14.40)

L

where [ke ] is the conventional element ﬂexural stiﬀness matrix. [kg ] introduces the considerations related to elastic instability. We note that its terms solely depend on geometric parameters (length), therefore this matrix is often referred to as the geometric stiﬀness matrix. 35

Using the shape functions for ﬂexural elements, Eq. 6.41, and substituting into Eq. 14.39 and Eq. 14.40 we obtain 36



u1 EA L

 0   0  ke =  EA − L   0 0

v1 0

θ1 0

12EI L3 6EI L2

6EI L2 4EI L

0

0

− 12EI L3

− 6EI L2

6EI L2

2EI L

u2 − EA L 0 0 EA L

0 0

v2 0 − 12EI L3 − 6EI L2 0 12EI L3 − 6EI L2

θ2  0 6EI  L2  2EI  L   0    − 6EI L2

(14.41)

4EI L

which is the same element stiﬀness matrix derived earlier in Eq. 8.7. 37

The geometric stiﬀness matrix is given by u1 0  0  P   0 kg =  L  0   0 0 

Victor Saouma

v1 0

θ1 0

6 5 L 10

L 10 2 2 15 L

0 − 65 L 10

0 L − 10 2 − L30

u2 v2 θ2  0 0 0 L  0 − 65 10  2  L 0 − 10 − L30  (14.42)  0 0 0  6 L  0 − 10  5 Elements Finite II; Solid Mechanics 2 2 L 0 − 10 15 L

Draft

14.3 Elastic Instability; Bifurcation Analysis 38

14–9

The equilibrium relation is thus ku = P

(14.43)

where the element stiﬀness matrix is expressed in terms of both the elastic and geometric components)

39

k = ke + kg

(14.44)

K = Ke + Kg

(14.45)

In a global formulation,we would have

we note that the structure becomes stiﬀer for tensile load P applied through Kg , and weaker in compression. We assume that conservative loading is applied, that is the direction of the load does not “follow” the deﬂected direction of the member upon which it acts.

40

14.3

Elastic Instability; Bifurcation Analysis

41 In elastic instability, the intensity of the axial load system to cause buckling is yet unknown, the incremental stiﬀness matrix must ﬁrst be numerically evaluated using an arbitrary chosen load intensity (since Kg is itself a function of P ). 42 For buckling to occur, the intensity of the axial load system must be λ times the initially ∗ arbitrarily chosen intensity of the force. Note that for a structure, the initial distribution of P must be obtained from a linear elastic analysis. Hence, the buckling load, P is given by

P = λP

∗

(14.46)

43 Since the geometric stiﬀness matrix is proportional to the internal forces at the start, it follows that (14.47) Kg = λK∗g

where K∗g corresponds to the geometric stiﬀness matrix for unit values of the applied loading (λ = 1). 44

The elastic stiﬀness matrix Ke remains a constant, hence we can write ∗

(Ke + λK∗g )u − λP = 0 45

The displacements are in turn given by ∗

u = (Ke + λK∗g )−1 λP

and for the displacements to tend toward inﬁnity (i.e buckling/bifurcation/instability), then |Ke + λK∗g | = 0 Victor Saouma

(14.48)

Finite Elements II; Solid Mechanics

Draft 14–10

GEOMETRIC NONLINEARITY

which can also be expressed as |K−1 g Ke + λI| = 0

(14.49)

46 Alternatively, it can simply be argued that there is no unique solution (bifurcation condition) to u.

The lowest value of λ, λcrit will give the buckling load for the structure and the buckling loads will be given by 47

Pcrit = λcrit P

48

∗

(14.50)

The corresponding deformed shape is directly obtained from the corresponding eigenvector.

Example 14-1: Column Stability Determine the buckling load of the following column. P

011 1010 2 10 1010 1010 10 1010 6 1 0 5 100 4 1 10 1010 1010 010 8 9 1 10 000 111 10107

1 0 0 1 0 1

3

11 00

0110 10 1010 1010 1010 1010 10 1010 1010 1010

L

L

Solution: The following elastic stiﬀness matrices are obtained 

k1e =

Victor Saouma

1 EA L

 0   0   EA − L   0 0

2 0

3 0

12EI L3 6EI L2

6EI L2 4EI L

0

0

− 12EI L3

− 6EI L2

6EI L2

2EI L

4 − EA L 0 0 EA L

0 0

5 0

6 0

− 12EI L3 − 6EI L2 0

6EI L2 2EI L

12EI L3 − 6EI L2

− 6EI L2

0

        

(14.51-a)

4EI L

Finite Elements II; Solid Mechanics

Draft

14.3 Elastic Instability; Bifurcation Analysis 

k2e =

4 EA L

 0   0   EA − L   0 0

5 0

6 0

12EI L3 6EI L2

6EI L2 4EI L

0

0

− 12EI L3

− 6EI L2

6EI L2

14–11 7

2EI L

− EA L 0 0 EA L

0 0

8 0

9 0

− 12EI L3 6EI − L2 0

6EI L2 2EI L

12EI L3 − 6EI L2

− 6EI L2

0

        

(14.51-b)

4EI L

Similarly, the geometric stiﬀness matrices are given by 1 2 0 0 0 6  5 L −P  0  10  L 0 0   0 − 65 L 0 10 

k1g =

4 5 0 65  L 0 10 −P   0 0 L   0 − 65 L 0 10 

k2g =

3 0 L 10 2 2 15 L

0 L − 10 2 − L30 6

L 10 2 2 15 L

0 L − 10 2 − L30

4 5 0 0 0 − 65 L 0 − 10 0 0 6 0 5 L 0 − 10 7 8 0 − 65 L 0 − 10 0 0 6 0 5 L 0 − 10

6  0 L  10  2  − L30   0  L  − 10  2 2 15 L 9 L 10 2 − L30

(14.52-a)



   0  L  − 10  2 2 15 L

(14.52-b)

The structure’s stiﬀness matrices Ke and Kg can now be assembled from the element stiﬀnesses. Eliminating rows and columns 2, 7, 8, 9 corresponding to zero displacements in the column, we obtain 1 4 3 5 6  AL2  AL2 − I 0 0 0 I  − AL2 2 AL2 0 0 0   I I EI  2  (14.53) Ke = 3  0 −6L 2L2  0 4L  L  0 0 −6L 24 0  0 8L2 0 0 2L2 and 1 0 0 −P  0 Kg = L  0 0 

Victor Saouma

4 0 0 0 0 0

3 0 0 2 2 15 L −L 10 −L2 30

5 0 0

−L 10 12 5

0

6  0 0  −L2   30  0 

(14.54)

4 2 15 L

Finite Elements II; Solid Mechanics

Draft 14–12

GEOMETRIC NONLINEARITY

noting that in this case K∗g = Kg for P = 1, the determinant |Ke + λK∗g | = 0 leads to 1 AL2 1 I 2 4 − AL I 3 0 5 0 6 0

introducing φ =

AL2 I

4 2 − AL I 2 2 AL I

0 0 0

3 0 0 2 λL4 2 4L − 15 EI 1 λL3 −6L + 10 EI 1 λL4 2L2 + 30 EI

and µ =

λL2 EI ,

5 0 0 1 λL3 −6L + 10 EI λL2 24 − 12 5 EI 0

6 0 0 1 λL4 2 2L + 30 EI 0 4 λL4 8L2 − 15 EI

=0

(14.55)

the determinant becomes

1 4 3 5 6 1 φ −φ 0 0 0 4 −φ 2φ 0 0 0 % µ& µ µ −6L + 10& 3 0 0 2 2 − 15 2 + 30 = 0 % µ 5 0 0 −6L + 10 12 2 − µ5 % 0 µ & µ 0 4 2 − 15 6 0 0 2 + 30

(14.56)

Expanding the determinant, we obtain the cubic equation in µ 3µ3 − 220µ2 + 3, 840µ − 14, 400 = 0

(14.57)

and the lowest root of this equation is µ = 5.1772 . We note that from Eq. 14.21, the exact solution for a column of length L was Pcr =

(4.4934)2 (4.4934)2 EI = EI = 5.0477 EI L2 l2 (2L)2

(14.58)

and thus, the numerical value is about 2.6 percent higher than the exact one. The mathematica code for this operation is: (* Define elastic stiffness matrices ke[e_,a_,l_,i_]:={ {e a/l , 0 , 0 , {0 , 12 e i/l^3 , 6 e i/l^2 , {0 , 6 e i/l^2 , 4 e i/l , {-e a/l , 0 , 0 , { 0 , -12 e i/l^3 , -6 e i/l^2 , { 0 , 6e i/l^2 , 2 e i/l , } ke1=N[ke[e,a,l,i]] ke2=N[ke[e,a,l,i]]

*) -e a/l 0 0 e a/l 0 0

, , , , , ,

0 -12 e i/l^3 -6 e i/l^2 0 12 e i/l^3 -6 e i/l^2

, , , , , ,

0 6 e i/l^2 2 e i/l 0 -6 e i/l^2 4 e i/l

}, }, }, }, }, }

(* Assemble structure elastic stiffness matrices *) ke={ {ke1[[3,3]], ke1[[3,5]] , ke1[[3,6]] }, { ke1[[5,3]], ke1[[5,5]]+ke2[[2,2]], ke1[[5,6]]+ke2[[2,3]]}, { ke1[[6,3]], ke1[[6,5]]+ke2[[3,2]], ke1[[6,6]]+ke2[[3,3]]} } WriteString["mat.out",MatrixForm[ke1]] WriteString["mat.out",MatrixForm[ke2]] WriteString["mat.out",MatrixForm[ke]] (* Define geometric stiffness matrices *) kg[p_,l_]:=p/l{

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

14.3 Elastic Instability; Bifurcation Analysis

14–13

{0 , 0 , 0 , 0 , 0 , 0 }, {0 , 6/5 , l/10 , 0 , - 6/5 , l/10 }, {0 , l/10 , 2 l^2/15 , 0 , - l/10 , - l^2/30 }, {0 , 0 , 0 , 0 , 0 , 0 }, {0 , -6/5 , - l/10 , 0 , 6/5 , - l/10 }, {0 , l/10 , - l^2/30 , 0 , - l/10 , 2 l^2/15 } } kg1=kg[p,l] kg2=kg[p,l] (* Assemble structure geometric stiffness matrices *) kg={ {kg1[[3,3]], kg1[[3,5]] , kg1[[3,6]] }, { kg1[[5,3]], kg1[[5,5]]+kg2[[2,2]], kg1[[5,6]]+kg2[[2,3]]}, { kg1[[6,3]], kg1[[6,5]]+kg2[[3,2]], kg1[[6,6]]+kg2[[3,3]]} } (* Determine critical loads in terms of p (note p=1) *) p=1 keigen= l^2 (Inverse[kg] . ke)/( e i) pcrit=N[Eigenvalues[keigen]] (* Alternatively*) knew =ke - x kg pcrit2=NSolve[Det[knew]==0,x]

Example 14-2: Frame Stability Determine the buckling load for the following frame. Neglect axial deformation. θ2

P

P

1 0 0 1 0 1 0 1

11 00 u1

θ3 u1

I=200 I=50

6’

10’ I=100

1111 0000

0000 1111 15’ 11111111111111111111 00000000000000000000 Solution: The element stiﬀness matrices are given by u1 20  1, 208   ··· ···  k1e =

Victor Saouma

θ2 1, 208 96, 667 ··· ···

0 ··· ··· ··· ···

0  ··· ···  ··· ···

(14.59-a)

Finite Elements II; Solid Mechanics

Draft 14–14

GEOMETRIC NONLINEARITY u1 0.01  0.10 = −P   ··· ··· 

k1g

0 ··· ···  ··· ··· 

k2e =

θ2 ··· 128, 890 ··· 64, 440

u1 47  1, 678   ··· ··· 

k3e =

0 ··· ··· ··· ··· 0 ··· ··· ··· ···

θ3 1, 678 80, 556 ··· ···

u1 0.01667  0.1 = −P   ··· ··· 

k3g

θ2 0.10 16.00 ··· ···

(14.59-b)

θ3  ··· 64, 440   ···  128, 890

0 ··· ··· ··· ···

θ3 0.1 9.6 ··· ···

0  ··· ···  ··· ···

0 ··· ··· ··· ···

(14.59-c)

0  ··· ···  ··· ···

(14.59-d)

0  ··· ···  ··· ···

(14.59-e)

The global equilibrium relation can now be written as (Ke − P Kg ) δ = 0 u1 (66.75) − P (0.026666) (1, 208.33) − P (0.1) (1, 678.24) − P (0.1)

θ2 (1, 208.33) − P (0.1) (225, 556.) − P (16.) (64, 444.) − P (0)

(14.60)

θ3 (1, 678.24) − P (0.1) (64, 444.4) − P (0) = 0 (209, 444.) − P (9.6)

(14.61)

The smallest buckling load ampliﬁcation factor λ is thus equal to 2, 017 kips. (* Initialize constants *) a1=0 a2=0 a3=0 i1=100 i2=200 i3=50 l1=10 12 l2=15 12 l3=6 12 e1=29000 e2=e1 e3=e1 (* Define elastic stiffness matrices ke[e_,a_,l_,i_]:={ {e a/l , 0 , 0 , {0 , 12 e i/l^3 , 6 e i/l^2 , {0 , 6 e i/l^2 , 4 e i/l , {-e a/l , 0 , 0 , { 0 , -12 e i/l^3 , -6 e i/l^2 ,

Victor Saouma

*) -e a/l 0 0 e a/l 0

, , , , ,

0 -12 e i/l^3 -6 e i/l^2 0 12 e i/l^3

, , , , ,

0 6 e i/l^2 2 e i/l 0 -6 e i/l^2

}, }, }, }, },

Finite Elements II; Solid Mechanics

Draft

14.4 Second-Order Elastic Analysis; Geometric Non-Linearity

14–15

{ 0 , 6e i/l^2 , 2 e i/l , 0 , -6 e i/l^2 , 4 e i/l } } ke1=ke[e1,a1,l1,i1] ke2=ke[e2,a2,l2,i2] ke3=ke[e3,a3,l3,i3] (* Define geometric stiffness matrices *) kg[l_,p_]:=p/l{ {0 , 0 , 0 , 0 , 0 , 0 }, {0 , 6/5 , l/10 , 0 , - 6/5 , l/10 }, {0 , l/10 , 2 l^2/15 , 0 , - l/10 , - l^2/30 }, {0 , 0 , 0 , 0 , 0 , 0 }, {0 , -6/5 , - l/10 , 0 , 6/5 , - l/10 }, {0 , l/10 , - l^2/30 , 0 , - l/10 , 2 l^2/15 } } kg1=kg[l1,1] kg3=kg[l3,1] (* Assemble structure elastic and geometric stiffness matrices *) ke={ { ke1[[2,2]]+ke3[[2,2]] , ke1[[2,3]] , ke3[[2,3]] }, { ke1[[3,2]] , ke1[[3,3]]+ke2[[3,3]] , ke2[[3,6]] }, { ke3[[3,2]] , ke2[[6,3]] , ke2[[6,6]]+ke3[[3,3]] } } kg={ { kg1[[2,2]]+kg3[[2,2]] , kg1[[2,3]] , kg3[[2,3]] }, { kg1[[3,2]] , kg1[[3,3]] , 0 }, { kg3[[3,2]] , 0 , kg3[[3,3]] } } (* Determine critical loads in terms of p (note p=1) *) p=1 keigen=Inverse[kg] . ke pcrit=N[Eigenvalues[keigen]] modshap=N[Eigensystems[keigen]]

14.4

Second-Order Elastic Analysis; Geometric Non-Linearity

49 From Eq. 14.44 it is evident that since kg depends on the magnitude of Px , which itself may be an unknown in a framework, then we do have a geometrically non-linear problem. 50

We rewrite Eq. 14.45 as [Ke + Kg ] u = P

(14.62)

but since Kg depends on the axial laod P , the preoblem is nonlinear. 51 A simple way to solve this nonlinear equation is to use a step-by-step incremental procedure. The linearized incremental formulation can be obtained by applying an incremental operator ∆ * + P i = [Ke + Kg ]i−1 {u}i (14.63) * + {u}i = [Ke + Kg ]−1 i−1 P i (14.64)

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 14–16

GEOMETRIC NONLINEARITY

Example 14-3: Eﬀect of Axial Load on Flexural Deformation Determine the midspan displacement and member end forces for the beam-column shown below in terms of Px ; The concentrated force is 50 kN applied at midspan, E=2 × 109 kN/m2 and I=2 × 10−3 m4 . 50 kN

1 0 0 1 0 1 0 1 0 1

0 80,000 kN1

0 1 1 0 0 1 0 1 0 1 0000000000000000 1 0 6 m 6 m 111111111111111000000000000000 1111111111111111

80,000 kN

Solution: Using two elements for the beam column, the only degrees of freedom are the deﬂection and rotation at midspan (we neglect the axial deformation). The element stiﬀness and geometric matrices are given by 0 0 0 0  0 222, 222.   0 666, 666. 1 [Ke ] =  0 0   0 −222, 222. 0 666, 666.

0 0 666, 666. 2, 666, 666 0 −666, 666. 1, 333, 333

0 0 0 0. 0 0 0.

v1 0 −222, 222. −666, 666. 0 222, 222. −666, 666.

θ2  0 666, 666.   1, 333, 333    0  −666, 666.  2, 666, 666

0 v1 0 0  0 222, 222.   0 666, 666. 2 [Ke ] =  0 0   0 −222, 222. 0 666, 666.

θ2 0 666, 666. 2, 666, 666 0 −666, 666. 1, 333, 333

0 0 0 0. 0 0 0.

0 0 −222, 222. −666, 666. 0 222, 222. −666, 666.

0  0 666, 666.   1, 333, 333    0  −666, 666. 





(14.65)

(14.66)

2, 666, 666

0 0 0 0  0 −16, 000   0 −8, 000 1 [Kg ] =  0 0   0 16, 000 0 −8, 000

0 0 −8, 000 −64, 000 0 8, 000 16, 000

0 v1 0 0 0 16, 000 0 8, 000 0 0 0 −16, 000 0. 8, 000

0 v1 0 0  0 −16, 000   0 −8, 000 2 [Kg ] =  0 0   0 16, 000 0 −8, 000 Victor Saouma

θ2 0 −8, 000 −64, 000 0 8, 000 16, 000

0 0 0  0 0 0 0 16, 000 −8, 000   0 8, 000 16, 000   (14.68)  0 0 0  0 −16, 000 8, 000  0. 8, 000 −64, 000 Finite Elements II; Solid Mechanics





θ2  0 −8, 000   16, 000    0  8, 000  −64, 000

(14.67)

Draft

14.4 Second-Order Elastic Analysis; Geometric Non-Linearity

14–17

Assembling the stiﬀness and geometric matrices we get [K] =

v1 412, 444. 0.

θ2 0. 5, 205, 330

(14.69)

and the displacements would be

and the  Plf t     Vlf t    Mlf t Prgt      Vrgt   Mrgt

v1 θ2

=

−0.00012123 0

(14.70)

member end forces for element 1 are given by    ulf t             v    lf t       " ! 1 θlf t 1 = [Ke ] + [Kg ] urgt               vrgt        θrgt 0 0 0 0  0 206, 222.   0 658, 667. =  0 0   0 −206, 222 0 658, 667.   0         25.       79.8491 = 0.           −25.     79.8491 

0 0 658, 667. 260, 2670 0 −658, 667. 1, 349, 330

0 v1 0 0 0 −206, 222. 0 −658, 667. 0 0 0 206, 222. 0 −658, 667.

θ2   ,0 0         658, 667.  0        1, 349, 330  0  0 0          −658, 667.  −0.00012123     2, 602, 670 0

(14.71-a)

Note that had we not accounted for the axial forces, then −0.0001125 v1 = θ2 0     0  Plf t                V 25.     lf t         Mlf t 75. = Prgt  0.                   Vrgt  −25.         Mrgt 75.

(14.72-a)

(14.72-b)

Alternatively, if instead of having a compressive force, we had a tensile force, then −0.000104944 v1 = (14.73-a) θ2 0

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 14–18

GEOMETRIC NONLINEARITY  Plf t     Vlf t    Mlf t Prgt      V   rgt Mrgt

              

        =

      

0 25. 70.8022 0. −25. 70.8022

        (14.73-b)

      

We observe that the compressive force increased the displacements and the end moments, whereas a tensile one stiﬀens the structure by reducing them. The Mathematica to solve this problem follows (* Initialize constants *) OpenWrite["mat.out"] a1=0 a2=0 e=2 10^9 i=2 10^(-3) i1=i i2=i1 l=6 l1=l l2=6 p=-80000 (* negative compression *) load={-50,0} (* Define elastic stiffness matrices *) ke[e_,a_,l_,i_]:={ {e a/l , 0 , 0 , -e a/l , 0 {0 , 12 e i/l^3 , 6 e i/l^2 , 0 , -12 e i/l^3 {0 , 6 e i/l^2 , 4 e i/l , 0 , -6 e i/l^2 {-e a/l , 0 , 0 , e a/l , 0 { 0 , -12 e i/l^3 , -6 e i/l^2 , 0 , 12 e i/l^3 { 0 , 6e i/l^2 , 2 e i/l , 0 , -6 e i/l^2 } ke1=N[ke[e,a1,l1,i1]] ke2=N[ke[e,a2,l2,i2]] (* Assemble structure elastic stiffness matrices *) ke=N[{ { ke1[[5,5]]+ke2[[2,2]], ke1[[5,6]]+ke2[[2,3]]}, { ke1[[6,5]]+ke2[[3,2]], ke1[[6,6]]+ke2[[3,3]]} }] WriteString["mat.out",MatrixForm[ke1]] WriteString["mat.out",MatrixForm[ke2]] WriteString["mat.out",MatrixForm[ke]] (* Define geometric stiffness matrices *) kg[p_,l_]:=p/l { {0 , 0 , 0 , 0 , 0 , 0 }, {0 , 6/5 , l/10 , 0 , - 6/5 , l/10 }, {0 , l/10 , 2 l^2/15 , 0 , - l/10 , - l^2/30 }, {0 , 0 , 0 , 0 , 0 , 0 }, {0 , -6/5 , - l/10 , 0 , 6/5 , - l/10 }, {0 , l/10 , - l^2/30 , 0 , - l/10 , 2 l^2/15 } } kg1=N[kg[p,l1]] kg2=N[kg[p,l2]] (* Assemble structure geometric stiffness matrices *) kg=N[{ { kg1[[5,5]]+kg2[[2,2]], kg1[[5,6]]+kg2[[2,3]]}, { kg1[[6,5]]+kg2[[3,2]], kg1[[6,6]]+kg2[[3,3]]}

Victor Saouma

, , , , , ,

0 6 e i/l^2 2 e i/l 0 -6 e i/l^2 4 e i/l

}, }, }, }, }, }

Finite Elements II; Solid Mechanics

Draft

14.4 Second-Order Elastic Analysis; Geometric Non-Linearity

14–19

}] (* Determine critical loads and normalize wrt p *) keigen=Inverse[kg] . ke pcrit=N[Eigenvalues[keigen] p] (* Note that this gives lowest pcrit=1.11 10^6, exact value is 1.095 10^6 *) (* Add elastic to geometric structure stiffness matrices *) k=ke+kg (* Invert stiffness matrix and solve for displacements *) km1=Inverse[k] dis=N[km1 . load] (* Displacements of element 1*) dis1={0, 0, 0, 0, dis[[1]], dis[[2]]} k1=ke1+kg1 (* Member end forces for element 1 with axial forces *) endfrc1=N[k1 . dis1] (* Member end forces for element 1 without axial forces knopm1=Inverse[ke] disnop=N[knopm1 . load] disnop1={0, 0, 0, 0, disnop[[1]], disnop[[2]]} (* Displacements of element 1*) endfrcnop1=N[ke1 . disnop1]

*)

Example 14-4: Bifurcation Analyse the stability of the following structure. Compare the axial force caused by the coupled membrane/ﬂexural eﬀects with the case where there is no interaction.

1,000

1 0 0 1 0 1 0000000000 1111111111 12 0000000000 1111111111 0 1 0000000000 1111111111 0000000000 1111111111 12 0 1 0000000000 1111111111 0000000000 1111111111 0000000000 00000000001111111111 1111111111 π/8 π/8 0 1 0 1 0 1 0 1 Solution: In the following solution, we will ﬁrst determine the axial forces based on the elastic stiﬀness matrix only. Then, on the basis of those axial forces, we shall determine the geometric stiﬀness matrix, and solve for the displacements. Because of the non-linearity of the problem, we may have to iterate in order to reach convergence. Following each analysis, we shall recompute the geometric stiﬀness matrix on the basis of the axial loads detemined from the previous iteration. Note that convergence will be reached only for stable problems. If the method fails to converge, it implies possible biﬀurcation which could be caused by elastic displacements approaching L sin θ, due to either θ being too small, or E being too small (i.e not stiﬀ enough). Victor Saouma

Finite Elements II; Solid Mechanics

Draft 14–20

GEOMETRIC NONLINEARITY

NEEDS SOME CORRECTION (* Initialize constants *) a1 = 1 a2 = 1 i1 = 1 1^3/12 i2 = i1 l1 = 12 l2 = 12 e1 = 200000 e2 = e1 e3 = e1 theta1 =N[Pi/8] theta2 = Pi-theta1 load ={0, -1000, 0} normold = 0 epsilon = 0.01 puncpl = load[[2]] / (Sin[theta1] 2) (* Define elastic stiffness matrices *) ke[e_,a_,l_,i_] := { {e a/l , 0 , 0 , -e a/l , 0 {0 , 12 e i/l^3 , 6 e i/l^2 , 0 , -12 e i/l^3 {0 , 6 e i/l^2 , 4 e i/l , 0 , -6 e i/l^2 {-e a/l , 0 , 0 , e a/l , 0 { 0 , -12 e i/l^3 , -6 e i/l^2 , 0 , 12 e i/l^3 { 0 , 6e i/l^2 , 2 e i/l , 0 , -6 e i/l^2 } (* Define geometric stiffness matrix *) kg[l_,p_] := p/l { {0 , 0 , 0 , 0 , 0 , 0 }, {0 , 6/5 , l/10 , 0 , - 6/5 , l/10 }, {0 , l/10 , 2 l^2/15 , 0 , - l/10 , - l^2/30 }, {0 , 0 , 0 , 0 , 0 , 0 }, {0 , -6/5 , - l/10 , 0 , 6/5 , - l/10 }, {0 , l/10 , - l^2/30 , 0 , - l/10 , 2 l^2/15 } } (* Define Transformation matrix and its transpose *) gam[theta_] := { { Cos[theta] , Sin[theta], 0 , 0 , 0 , { -Sin[theta], Cos[theta], 0 , 0 , 0 , { 0 , 0 , 1 , 0 , 0 , { 0 , 0 , 0 , Cos[theta] , Sin[theta] , { 0 , 0 , 0 , -Sin[theta] , Cos[theta] , { 0 , 0 , 0 , 0 , 0 , } gamt[theta_] := { { Cos[theta] , -Sin[theta], 0 , 0 , 0 { Sin[theta] , Cos[theta] , 0 , 0 , 0 { 0 , 0 , 1 , 0 , 0 { 0 , 0 , 0 , Cos[theta] , -Sin[theta] { 0 , 0 , 0 , Sin[theta] , Cos[theta] { 0 , 0 , 0 , 0 , 0 }

Victor Saouma

, , , , , ,

0 6 e i/l^2 2 e i/l 0 -6 e i/l^2 4 e i/l

0 0 0 0 0 1

}, }, }, }, }, }

, , , , , ,

0 0 0 0 0 1

}, }, }, }, }, }

}, }, }, }, }, }

Finite Elements II; Solid Mechanics

Draft

14.4 Second-Order Elastic Analysis; Geometric Non-Linearity (* Define functions for local displacments and loads *) u[theta_,v1_,v2_] := Cos[theta] v1 + Sin[theta] v2 (* Transformation and transpose matrices *) gam1 = gam[theta1] gam2 = gam[theta2] gam1t = gamt[theta1] gam2t = gamt[theta2] (* Element elastic stiffness matrices *) ke1 = ke[e1, a1, l1, i1] ke2 = ke[e2, a2, l2, i2] Ke1 = gam1t . ke1 . gam1 Ke2 = gam2t . ke2 . gam2 (* Structure’s global stiffness matrix *) Ke={ { Ke1[[4,4]] + Ke2[[1,1]] , Ke1[[4,5]] + Ke2[[1,2]] , Ke1[[4,6]] + { Ke1[[5,4]] + Ke2[[2,1]] , Ke1[[5,5]] + Ke2[[2,2]] , Ke1[[5,6]] + { Ke1[[6,4]] + Ke2[[3,1]] , Ke1[[6,5]] + Ke2[[3,2]] , Ke1[[6,6]] + } (* ======= uncoupled analysis ========== *) dise=Inverse[Ke].load u[theta_,diseg1_,diseg2_] := Cos[theta] diseg1 + Sin[theta] diseg2 uu1 = u[ theta1, dise[[1]], dise[[2]] ] uu2 = u[ theta2, dise[[1]], dise[[2]] ] up1 = a1 e1 uu1/l1 up2 = a2 e2 uu2/l2 (* ========== Coupled Nonlinear Analysis ============== Start Iteration *) diseg = N[dise] For[ iter = 1 , iter <= 100, ++iter, (* displacements in local coordinates *) disloc={ 0,0,0, u[ theta1, diseg[[1]], diseg[[2]] ], u[ theta2, diseg[[1]], diseg[[2]] ], 0}; (* local force *) ploc = ke1 . disloc; p1 = ploc[[4]]; p2 = p1; kg1 = kg[ l1 , p1 ]; kg2 = kg[ l2 , p2 ]; Kg1 = gam1t . kg1 . gam1; Kg2 = gam2t . kg2 . gam2; Kg={ { Kg1[[4,4]] + Kg2[[1,1]] , Kg1[[4,5]] + Kg2[[1,2]] , Kg1[[4,6]] + { Kg1[[5,4]] + Kg2[[2,1]] , Kg1[[5,5]] + Kg2[[2,2]] , Kg1[[5,6]] + { Kg1[[6,4]] + Kg2[[3,1]] , Kg1[[6,5]] + Kg2[[3,2]] , Kg1[[6,6]] + };

Victor Saouma

14–21

Ke2[[1,3]] }, Ke2[[2,3]] }, Ke2[[3,3]] }

Kg2[[1,3]] }, Kg2[[2,3]] }, Kg2[[3,3]] }

Finite Elements II; Solid Mechanics

Draft 14–22

GEOMETRIC NONLINEARITY

(* Solve *) Ks = Ke + Kg; diseg = Inverse[Ks] . load; normnew = Sqrt[ diseg . diseg ]; ratio = ( normnew-normold ) / normnew; Print["Iteration ",N[iter],"; u1 ",N[u1],"; p1 ",N[p1]," ratio ",N[ratio]]; normold = normnew; If[ Abs[ ratio ] < epsilon, Break[] ] ] Print[" p1 ",N[p1]," up1 ",N[up1]," p1/up1 ",N[p1/up1]," ratio ",N[ratio]]

14.5

Summary

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

14.5 Summary

14–23

STRONG FORM

❄

WEAK FORM

❄

2nd Order D.E. 2 B.C.

4th Order D.E. 4 B.C.

εx =

du dx

−y

❄ d2 y dx2

− =0 v = −A sin kx − B cos kx P EI

U=

1 2

'❄ ( 2 d v dx2

❄ d4 v EI dx 4

d2 v dx2

−P =w v = C1 sin kx + C2 cos kx + C3 x + C4

+

1 2

% dv &2 dx

❄ 2 Ω Eε dΩ

❄

K=

Kg

Ke

❄

P = (Ke + λKg )u

❄

|Ke + λKg | = 0

Figure 14.5: Summary of Stability Solutions

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 14–24

Victor Saouma

GEOMETRIC NONLINEARITY

Finite Elements II; Solid Mechanics

Draft Chapter 15

PLATES Adapted from Pilkey & Wunderlich

Draft

1 This chapter will cover the transverse deformation of plates. The approach followed will be consistent with the ﬁnite element formulation, Fig. 15.1.

Fundamental Relations

Kinematics

Constitutive

Equilibrium

?

Dierential Equation

?

Variational Formulation

?

Finite Element Discretization

Figure 15.1: Finite Element Formulation

15.1

Fundamental Relations

15.1.1

Equilibrium

2

Considering an arbitrary plate, the stress are given by, Fig. 15.2, resultants per unit width

Draft 15–2

PLATES

Figure 15.2: Stresses in a Plate are given by

    Nxx      t  2 σdz Nyy Membrane Force N =  − 2t         Nxy     Mxx      t  2 σzdz Bending Moments M = Myy  − 2t         Mxy

Transverse Shear Forces

V =

t 2

− 2t

τ dz

= = = = = =

     Vx =     Vy =

t 2

t 2

− 2t t 2

− 2t t 2

− 2t

σxx dz σyy dz σxy dz

t 2

− 2t t 2

− 2t t 2

− 2t

− 2t t 2

− 2t

σxx zdz (15.1-a) σyy zdz σxy zdz

τxz dz τyz dz

Note that in plate theory, we ignore the eﬀect of the membrane forces, those in turn will be accounted for in shells. 3

4 The equation of equilibrium is derived by considering an inﬁnitesimal element tdx dy subjected to an applied transverse load pz . We would have to consider three equations of equilibrium, Fig. 15.3:

Summation of Forces in the z direction

or

Victor Saouma

∂Vy ∂Vx dxdy + dxdy + pz dxdy = 0 ∂x ∂y

(15.2)

∂Vx ∂Vy + + pz = 0 ∂x ∂y

(15.3)

Finite Elements II; Solid Mechanics

Draft Draft

15.1 Fundamental Relations

y

15–3

x -

Vy ; ; 6 Mxy ; ; Myy ; ; z ?V -; x 6 Mxx pz ;; ; -? ; Mxy Mxx + @M@x dx ; Mxy + @M@x dx ? ; Myy + @M@y dy ;; Vx + @V@x dx ?Vy + @V@y dy ; Mxy + @M@y dy xx

yy

xy

x

y

xy

Figure 15.3: Free Body Diagram of an Inﬁnitesimal Plate Element Summation of Moments about the x axis

or

∂Myy ∂Mxy dxdy + dxdy − Vy dxdy = 0 ∂x ∂y

(15.4)

∂Myy ∂Mxy + − Vy = 0 ∂x ∂y

(15.5)

Summation of Moments about the y axis ∂Myx ∂Mxx + − Vx = 0 ∂y ∂x

(15.6)

Since Mxy = Myx , those equations can be expressed in matrix form as 

∂ ∂x

  0 0

0 ∂ ∂y

0

    ∂   −1 0  ∂y  ∂ 0 −1  ∂x  ∂ ∂   0 ∂x  ∂y  LT

1

Mxx Myy Mxy Vx Vy M

     

     0   0  + 0 = 0       p 0  z  

(15.7)

Note that the left matrix corresponds to LT where the 1 term has been substituted by −1

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 15–4

15.1.2

PLATES

Kinematic Relations

5 From Fig. 15.4 we have ﬁve displacements u, v, w, θxx and θyy . However, two of the three displacements, u, v can be expressed in terms of the θxx and θyy which are the rotations of the

Figure 15.4: Displacements in a Plate plate middle surface. The third displacement being the transverse one w. We will use a notation consistent with the traditional one adopted for plate bending, rather than one consistent with the coordinate directions as used in ﬁnite element. 6

7 The middle surface will be assumed to remain unstrained, and that plane sections remain plane. When shear deformations will later be taken into account, then the plane will not necessarily remain normal to the middle surface. 8

9

Based on the plane section remaining plane    z  u  = 0 v   w 0 However, from Eq. ?? we have ε = Lu or   ∂  ε ∂x xx      0     ε   yy      0  εzz  = ∂ γ  ∂y   xy    ∂      γ   ∂z    xz  γyz 0 ε

assumption, we have,   0 0  θxx  θyy z 0    w 0 1

0 ∂ ∂y

0 ∂ ∂x

0 ∂ ∂z

L

 0 0   ∂  ∂z   0  ∂   ∂x ∂ ∂y

 u  v  w

(15.8)

(15.9)

u

10 Since w corresponds to the transverse deﬂection of the middle surface, and does not vary with z, then εzz = 0.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

15.1 Fundamental Relations 11

15–5

Substituting Eq. 15.8, we obtain    zθxx,x εxx             ε zθyy,y   yy        εzz 0 = γ + θyy,x ) z(θ    xy xx,y        (θxx + w,x )  γxz           γyz (θyy + w,y )

       

 zκxx     zκyy    0 = 2zκ   xy         γxz     γyz

        (15.10)

      

Note that this equation assumes that the displacement w is small compared to the thickness of the plate, and the rotation is small. Because the rotation is small, its square is negligible with respect to unity, and hence the curvature κ=

(1

∂θ ∂x + θ 2 )3/2

≈

∂θ ∂x

(15.11)

is equal to the rate of change of rotation. 12

Hence, the kinematic relation for the transverse displacement of a plate is     ∂ 0 0 κ ∂x xx       ∂    0  0   κyy      θxx  ∂y  ∂ ∂ 0  2κxy θyy =  ∂y  ∂x        ∂    γ w   1 0 ∂x   xz   ∂ γyz 0 1 ∂y u κ L

15.1.3 13

(15.12)

Constitutive Relations

For three dimensional continuum, the strain-stress relation is    1 −ν −ν ε xx          εyy  0   −ν 1 −ν     1  εzz −ν −ν 1 =  γxy  2(1 + ν) 0 0  E           γ 0 2(1 + ν) 0 0 xz     γyz 0 0 2(1 + ν)

But From Eq. 15.10 εzz = 0, and we can neglect σzz which is stresses. Hence, inverting the previous equation yields    1 ν 0 σxx       σ   ν 1  0 0  yy  E   0 0 1−ν τxy = 2   1 − ν2  1−ν      τ 0 xz   2   0 1−ν τyz 0 2 σ D Note that the shear modulus is µ = Victor Saouma

               

σxx σyy σzz τxy τxz τyz

              

(15.13)

much smaller than the other            

εxx εyy γxy γxz γyz ε

          

(15.14)

E 2(1+ν) .

Finite Elements II; Solid Mechanics

Draft 15–6

PLATES

Figure 15.5: Positive Moments and Rotations We now seek to write moments in terms of the curvatures, Fig. 15.5, introducing the stresses from Eq. 15.14 into Eq. 15.1-a, and using Eq. 15.10 we derive the ﬁrst term 14

Mxx = =

t 2 E E σxx zdz = (εxx + νεyy )zdz = (zκxx + νzκyy )zdz t t 1 − ν2 t 1 − ν2 −2 −2 −2 t 2 Et3 E 2 (κxx + νκyy ) (κ + νκ ) z dz = (15.15-a) xx yy 1 − ν2 12(1 − ν 2 ) −t t 2

t 2

2

15 Following a similar procedure for the other two curvature relation    1  Mxx  Et3  ν Myy =   12(1 − ν 2 ) Mxy 0

M

D

terms, we obtain the following momentν 1 0

0 0 1−ν 2

   

 κxx  κyy  2κxy κ

(15.16)

3

Et The 12(1−ν 2 ) term is referred to as the ﬂexural rigidity and is analogous to the ﬂexural stiﬀness EI of a beam (if the plate has unit width, and ν = 0, then EI = Et3 /12). 16

15.2

Plate Theories

15.2.1

Reissner-Mindlin

This theory, is primarily applicable to thick plates in which shear deformations can not be neglected.

17

15.2.1.1

Fundamental Relations

In this formulation, shear deformation is accounted for and is the extension of the Timoshenko’s beam theory. 18

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

15.2 Plate Theories

15–7

Kinematic: The kinematic relation of Eq. 15.12 still holds and is written in slightly modiﬁed form     ∂ 0  κxx   ∂x ∂  θxx (15.17-a) κyy =  0 ∂y  θyy   ∂ ∂ 2κxy ∂y ∂x θ κ Lf    w  ∂ γxz ∂x 1 0 θ = (15.17-b) ∂ γyz  xx  ∂y 0 1 θ yy γ Ls u

Equilibrium: The equilibrium equation, Eq. 15.7 also  Mxx    ∂  ∂   0 −1 0  Myy ∂x ∂y   ∂ ∂ Mxy  0 ∂y ∂x 0 −1    ∂ ∂  0 0 0 ∂x ∂y  Vx  Vy LT = LT LTs f M

still holds            0   0  + 0 = 0       p 0  z  

Constitutive: the constitutive relation of Eq. 15.16 must be augmented to account for the relationships between γxz and γyz (which are no longer negligible in this case) with Vx and Vy respectively. This is obtained by equating the internal work with the external one τxz γxz dzdA = Vx γxz dA (15.18) A

z

A

τxz µ ,

and from Eq. 15.40, we obtain Recalling that γxz = t/2 ' z (2 2 2 τxz 1 3 Vx dzdA = 1− 2 τxz γxz dzdA = dzdA 2 t t A z A z µ A −t/2 µ 8 3 Vx 2 t 6 Vx2 6 Vx dA = dA = dA(15.19-a) Vx = 15 2 t µ 5 µt 5 µt A A A Hence, comparing with Eq. 15.18 matrix becomes     1 Mxx        K ν   Myy    0 Mxy =         Vx     Vy M

where K = 19

Et3 12(1−ν 2 )

we conclude that γxz = ν 1 0 0



      0   1−ν  2  1 0     ζ 0 1 0 0

D

and the ﬁnal constitutive κxx κyy 2κxy γxz γyz κ

          

(15.20)

and ζ = 56 µt

The constitutive matrix can be symbolically written as Df 0 D= 0 Ds

Victor Saouma

6 Vx 5 µt ,

(15.21)

Finite Elements II; Solid Mechanics

Draft 15–8

15.2.1.2 19

Diﬀerential Equation

The diﬀerential equation for the Mindlin plate will not be derived in this course. † Variational Formulation

15.2.1.3 20

PLATES

The internal virtual work is expressed through two distinct terms T −δWi = δκ MdA + δγ T VdA A A T δ(Lf θ) Df Lf θdA + δ(Ls u)T Ds Ls udA = A A T T δθ Lf Df Lf θdA + δuT LTs Ds Ls udA = A A

Substituting

"

Z δW

=

−K

A

δ θxx

|

{z

δ

θyy

}

T

|

∂ ∂ ∂ 1−ν ∂ + ∂y ∂x ∂x 2 ∂y ∂ ∂ 1−ν ∂ ∂ ν + ∂y ∂x ∂x 2 ∂y

| 2

Z δ w A

|

θxx

{z

θyy 4

}

δuT

{z

∂ ∂ 1−ν ∂ ν ∂ + ∂y ∂x ∂y 2 ∂x ∂ ∂ ∂ 1−ν ∂ + ∂y ∂y ∂x 2 ∂x

Ke f

{z

5µt − 6

Internal, Bending ∂ ∂ ∂x ∂x

|

|

+ ∂ ∂y ∂ ∂x

∂ ∂ ∂y ∂y

{z

∂ ∂y

1 0

Ke s

{z

38 < 0 5 : 1 }|

∂ ∂x

δ w

+ A

|

θxx

{z

δuT

θyy

8 9 < pz = 0

}: 0 ; | {z } pv

|

(15.22-c)

δ w Γ

|

θxx

{z

δuT

}|

θxx θyy

{z

dA

} }

9

{z u

Z dA +

#

w = θxx dA ; θyy

Internal, Shear

Z

(15.22-b)

Kes

kef

21

(15.22-a)

θyy

}

}

8 9 < ps = 0

dA

p

}

}: 0 ; | {z }

{z

(15.23-a)

External

After substitution, and accounting for the external work, we obtain 6ks (1 − ν) T e δW = −K δθ T Kef θdA + δu K udA s t2 A A δuT pV dA + δuT pdΓ = 0 + A

(15.24)

Γ

Kef

= LTf Df Lf

(15.25)

Kes

LTs Ds Ls

(15.26)

=

where ks = 5/6 is the shear correction factor Z

δW =

2 6 4

δ w θxx A ∂ 5µt ∂ ∂ 5µt ∂ + ∂y ∂x 6 ∂x 6 ∂y ∂ 5µt ∂y 6 5µt ∂ 6 ∂x

Victor Saouma

θyy ∂ 5µt ∂y 6 ∂ ∂ ∂ K(1−ν) ∂ K ∂x + ∂y + 5µt ∂x 2 ∂y 6 ∂ ∂ ∂ K(1−ν) ∂ Kν ∂x + ∂x ∂y 2 ∂y

5µt ∂ 6 ∂x ∂ ∂ ∂ K(1−ν) ∂ Kν ∂y + ∂y ∂x 2 ∂x ∂ ∂ ∂ K(1−ν) ∂ K ∂y + ∂x + 5µt ∂y 2 ∂x 6

38 9 < w = 7 θ 5 : xx ; dA θyy

Finite Elements II; Solid Mechanics

Draft

15.2 Plate Theories

15–9

Z

δ w A

15.2.2

θxx

θyy

8 9 < pz = :

Z

0 dA + ; 0

δ w

θyy

θxx

Γ

8 9 < ps = :

0 dA = 0 ; 0

(15.27-a)

Kirchhoﬀ

This theory, is primarily applicable to thin plates in which shear deformations can be neglected.

22

15.2.2.1

Fundamental Relations

In this formulation, shear deformations are neglected, and formulation is analogous to the conventional Euler-Bernouilli beam theory. 23

Kinematic Relations Since shear deformations are neglected, γxz = γyz = 0 and thus the last two relation of Eq. 15.10 reduce to θxx = −w,x

and θyy = −w,y

(15.28)

and the ﬁrst three strains become Oxx = −z or

∂2w ; ∂x2   κxx κ  yy 2κxy κ

Oyy = −z  

∂2w ; ∂y 2

  

2

∂ − ∂x 2 ∂2 − ∂y2 =    −2 ∂ ∂ ∂x ∂y L

Constitutive Relation The constitutive relation of    1  Mxx  Et3  ν Myy =   12(1 − ν 2 ) Mxy 0 M

D

γxy = −2z

∂2w ; ∂x∂y

(15.29)

  *

+ w  

(15.30)

u

Eq. 15.16 still applies   ν 0  κxx  1 0  κ  yy  1−ν 2κxy 0 2 κ

(15.31)

Equilibrium The equilibrium equation, as expressed by Eq. 15.7 still holds. If we were to substitute the second and third relations into the ﬁrst one, then we would obtain the following equilibrium relation in terms of the moments ∂ 2 Myy ∂ 2 Mxy ∂ 2 Mxx + + 2 + pz = 0 ∂x2 ∂x∂y ∂y 2 (Note the similarity with the corresponding equations for beam ﬂexure

Victor Saouma

(15.32) d2 M dx2

− Vx = 0)

Finite Elements II; Solid Mechanics

Draft 15–10

15.2.2.2

PLATES

Diﬀerential Equation

24 If we combine the kinematic with the constitutive relation, 15.30 and 15.31 respectively we obtain  ∂2    ∂2 + ν ∂y 2  Mxx  ∂x2 3 * + Et  ∂2 ∂2  Myy =− (15.33)  ∂y2 + ν ∂x  w 2 2   12(1 − ν ) ∂ ∂ Mxy (1 − ν)

∂x ∂y

Finally, we substitute the equilibrium equation, 15.32, into the previous one,

25

∂4w ∂4w ∂4w +2 2 2 + = 4 ∂x ∂x ∂y ∂y 4

pz Et3 12(1−ν 2 )

(15.34)

or ∇4 w =

pz Et3 12(1−ν 2 )

(15.35)

Note the similarity with the corresponding equation for beams ∂2w ∂4 pz ∂2 EI = p or = z 2 2 4 ∂x ∂x ∂x EI 15.2.2.3

(15.36)

Stresses

26 Combining the stress-strain relation of Eq. 15.14, with Eq. 15.29, and 15.33, the stresses could be expressed in terms of the moments

σxx =

Mxx t3 12

z;

σyy =

Myy t3 12

z;

τxy =

Mxy t3 12

again we note the analogy with the ﬂexural stress expression in beams σ = dimensional equilibrium equation, Eq. 3.57: ∂τxz ∂σxx ∂τxy + + ∂x ∂y ∂z ∂τyz ∂τyx ∂σyy + + ∂x ∂y ∂z ∂σzz ∂τzx ∂τzy + + ∂x ∂y ∂z integrating the ﬁrst equation yields z ∂σxx ∂τxy + dz = τxz = − ∂x ∂y t/2 12 t/2 zVx dz = 3 t z Finally τxz Victor Saouma

t/2 z

(15.37)

z My I .

Using the three

= 0 = 0

(15.38-a)

= 0

12z t3

∂Mxx ∂Mxy + ∂x ∂y

dz (15.39-a)

# # 2 $ 2 $ 2z 2z 3Vx 3Vy 1− 1− = and τyz = 2t t 2t t

(15.40)

Finite Elements II; Solid Mechanics

Draft

15.2 Plate Theories

15–11

Thus, the shear stress distribution across the plate thickness is parabolic (though they tend to be very small compared to τxy , and the peak shear stresses occur at the middle surface (z = 0) where 3 Vx 3 Vy τxz |max = and τyz |max = (15.41) 2 t 2 t Finally, it can be shown that σzz varies cubically. 15.2.2.4

Variational Formulation

27 Prior to the ﬁnite element discretization, we seek to obtain from the previously derived relations a variational formulation of the problem. 28

The internal virtual work is given by δWi = − δεT σdA = − A

δκT MdA

(15.42)

A

where M = Dκ is obtained from Eq. 15.31, and κ = Lu is obtained from Eq. 15.30. 29

Accounting for the external virtual work, we obtain T T T δW = δWi + δWe = − δu (L DL)udA + δu pv dA + A A

uT pdΓ = 0 Γp

(15.43)

where pv contains the applied transverse loading pz , and p the edge loading ps . 30

Substituting for the LT DL (Eq. 15.30) terms we obtain 

2

∂ Ke = LT DL = − ∂x 2

= K

∂2 ∂x2

2

∂ − ∂y 2

∂2 ∂2 +ν 2 2 ∂x ∂y

∂ ∂ −2 ∂x ∂y

1 ν  ν 1 12(1 − ν 2 ) 0 0

∂ ∂ + 2(1 − ν) ∂x ∂y

0 0

Et3

∂ ∂ ∂x ∂y

∂2 + 2 ∂y

1−ν 2

 ∂2   − ∂x22 ∂  − ∂y 2   −2 ∂ ∂ ∂x ∂y

∂2 ∂2 +ν 2 2 ∂y ∂x

    

(15.44-a)

and ﬁnally, noting that u = w, the virtual work becomes

δW = −K

δw (Lwpz ) dA + A

Γp δwps dγ = 0

(15.45)

Γ

where K = Et3 /12(1 − ν 2 ).

15.2.3

Summary

31 Table 15.1 summarizes some of the major equations governing plate bending, and contrasts them with the equivalent elasticity ones.

Victor Saouma

Finite Elements II; Solid Mechanics

Victor Saouma

|

2 6 6 4

0

0

∂ ∂x

R

{z

σxx σyy σzz σxy σyz σzx

0

1 β β 0 0 0 D

{z

β β 1 0 0 0

∂ ∂z ∂ ∂y

0

L

{z

∂ ∂z

0 0 0 γ 0 0

8 3> > > > > < 7 7 5> > > > > }: |

0 0 0 0 γ 0

∂ ∂x ∂ ∂y

0

∂ ∂z

0 0

{z

0 0 0 0 0 γ

σxx σyy σzz σxy σxz σyz

}

R

R

"

38 ε xx > > > 7 > 7> < εεyy 7 zz 7 εxy 7 > > 5> > > : εyz εzx } | {z > > > > > ; }

9 > > > > > =

8 9 < bx = by +ρ > : bz ; = 0 > > > > ; | {z } ρb }

9 > > > > > =

u

3 7 8 9 7 7 < ux = 7 7 u y 7 : uz ; 7 7 5 | {z }

ν ; γ = 1−2ν α = (1+ν)(1−2ν) ; β = 1−ν 2(1−ν)

β 1 β 0 0 0

∂ ∂x

0

∂ ∂z

0

∂ ∂x

∂ ∂y ∂ ∂z

0

0

E(1−ν)

2 6 6 6 = α6 6 > > > 4 > > ; } |

9 > > > > > =

LT

{z

∂ ∂y ∂ ∂x

0 ∂ ∂y

0

0

∂ ∂x

Elasticity

{z

2 6 6 4

κxx κyy 2κxy

0

0

∂ ∂x

0

M

{z

Mxx Myy Mxy Vx Vy

δW = −K

Z A

0

∂ ∂y

0

0

∂ ∂y

0 0

{z

M

{z

e

Z

Z ν)

D

{z

1−ν 2

0 0

Et3 ; 12(1−ν 2 )

0

ν 1 0

D

{z

ν 1 0

38 > > > < 7 7 5> > > : }|

38 > > > < 7 7 5> > > : }|

}|

;

{z

3 5

A

e

|

"

1 0

0

38 < 5 : }|

Z Γ

Z

0 1

A

0 1

9 = ; }

T

δu pV dA +

{z

0 0 0

0 0 0

Z Γ

T

δu pdΓ = 0

9 > > > = > > > ; }

9 = ;

9 = ;

u

w θxx θyy

9 = : ; } | {z }

#8 <

κxx κyy 2κxy γxz γyz

9 8 = < ;=:

9 8 = < ;=:

38 > > 7 > < 7 7 7 > 5> > : }|

{z

Ls

1 0

Γp δwps dγ = 0

{z

0 0 pz

0 0 pz

∂ ∂x ∂ ∂y

κxx κyy 2κxy

9 > 8 > > = < + : > > > ; }

δu Ks udA +

T

=

9 8 > > > = < + > : > > ; }

}

9 > > > = w > > | {z > ; u } }

ζ = 5 µt 6

ζ

1−ν 2

0 0

M

{z

Mxx Myy Mxy Vx Vy

M

Mxx Myy Mxy Vx Vy

{z

γxz γyz

L

{z

2 − ∂2 ∂y ∂ ∂ −2 ∂x ∂y

2 − ∂2 ∂x

δw (KLwpz ) dA +

K =

1 ν 0

9 2 = =K4 ; } |

1 ν 0

∂ ∂y

∂ ∂x

−1

0

∂ ∂y

−1

0

{z

0

−1

∂ ∂x

0

A 6ks (1 − t2

δW = −

|

−1

}

θxx θyy

8 9 > > = > < ;=> > > } : |

3 7 5

LT s

2 K4

Mxx Myy Mxy

LT f

0

{z

∂ ∂y ∂ ∂x

LT

9 2 > > > = 6 6 =6 6 > > > ; 4 } |

8 < : |

0

{z

κxx κyy 2κxy

∂ ∂y ∂ ∂x

∂ ∂y ∂ ∂x

{z

Lf

8 < : |

Plate Theory (Kirchhoﬀ/Mindlin)

δ Kf dA + T

Virtual Work

8 > > > < > > > : |

Constitutive

|

2 6 6 4

|

0

∂ ∂x

∂ ∂y

0

∂ ∂x

LT =

9 2 = 6 ;=4 } |

Equilibrium

8 < : |

Kinematic

15–12

Table 15.1: Comparison of Governing Equations in Elasticity and Plate Bending

tdΓ = 0 δΠ = Ω δ(Lu)T DdΩ − Ω δuT bdΩ − Γ δuT ˆ t

> > > > > : |

8 > > > > > <

∂ ∂z

0

∂ ∂y

0

0

0

> > > > > : |

8 > > > > > <

2 9 εxx > 6 > > 6 εyy > > = 6 6 εzz =6 6 εxy > > 6 > > εxz > ; 6 4 εyz {z } | "

Draft PLATES

Finite Elements II; Solid Mechanics

Draft

15.3 Finite Element Formulations

15.3

15–13

Finite Element Formulations

32 Numerous elements have been proposed for plate bending. An ideal plate element should have the following properties:

1. Formulation should be based on continuum mechanics and plate theory, the nodal d.o.f. are the transverse displacement w, and sectional rotations θxx and θyy . 2. The element should be “numerically correct”, and convergent. The element stiﬀness matrix must contain the three rigid-body modes and no rank deﬁciency (spurious zeroenergy modes). 3. The element should not “lock” in thin plate analyses. 4. The predictive capability of the element should be insensitive to element geometric distortions. Since plate structure is a special case of shell structures, and plates under large displacement exhibit shell characteristics (through membrane actions), similar formulations can be employed for both plates and shells. 33

15.3.1 15.3.1.1

Rectangular Element Formulation

The stiﬀness matrix, including shear deformation eﬀects, for a rectangular element, Fig. ?? will be developed.

34

Figure 15.6: Rectangular Plate Element The selected degrees of freedom are w, θxx and θyy , thus    0 N1 N2 N3 N4  w   θxx 0 N1 N2 N3 N4 =   θyy 0 0 w Nw 0 = 0 Nθ θ

35

0 0 N1 N2 N3 N4

   w   θ xx   θ yy

where N1 = (1 − ξ)(1 − η); Victor Saouma

N2 = ξ(1 − η);

N3 = ξη;

N4 = η(1 − ξ)

(15.47)

Finite Elements II; Solid Mechanics

Draft 15–14

PLATES

Note that the natural coordinate system selected has its origin at node 1, and 0 ≤ ξ, η ≤ 1. Thus through these shape functions, we will have a ﬁrst order (C0 ) element, for which displacement and rotations will be continuous along the boundaries (since θxx and θyy are also variables). 36 The stiﬀness matrix, for a Mindlin plate, will be derived on the basis of the principle of virtual work from Eq. 15.24  

  −δWi = K  

δθ A

T

6ks (1 − Kef θdA + 2

Flexure

ν)

t

  δuT Kes udA  A Shear

(15.48)

where Kef = LTf Df Lf , and Kes = LTs Ds Ls . The ﬁrst integral will lead to the stiﬀness matrix corresponding to bending, Kf , and the second one to the stiﬀness matrix corresponding to shear deformations Ks . 37

38

We recall that LN = B, and that x = aξ and y = bη, hence 1 ∂ ∂ 1 ∂ ∂ = and = ∂x a ∂ξ ∂y b ∂η

(15.49)

thus substituting into Eq. 15.17-a 2 6 4 |

1 ∂ a ∂ξ

0 1 ∂ b ∂η

3

0

{z

Lf

1 b 1 a

∂ ∂η ∂ ∂ξ

7 5 }

|

(1 − ξ)(1 − η) 0

ξ(1 − η) 0

ξη 0

η(1 − ξ) 0

{z

0 (1 − ξ)(1 − η)

0 ξ(1 − η)

0 ξη

0 η(1 − ξ)

N



− 1−η a = 0 − 1−ξ b

1−η a

η a

0 − ξb

0

− ηa 0

ξ b

1−ξ b

0

0 − ξb

− 1−ξ b − 1−η a

1−η a

0

0

ξ b η a

1−ξ b − ηa

B

}

(15.50)

 

(15.51)

2

t We note that Lf = Lu (used in elasticity) for in plane deformation, and Df = 12 D, thus the stiﬀness matrix for bending can be formed directly from the stiﬀness matrix for in-plane deformation. 39

40

Upon substitution and integration we obtain Kef

=

t2 Et 24(1−ν 2 ) 12

2 w1 w2 w3 w4 θx1 6 0 0 6 0 6 0 6 Aαβ 6 6 6 6 6 6 6 6 6 6 4

θx2

θx3

θx4

θy1

θy2

0 Cαβ Aαβ

−Aαβ /2 Bαβ Aαβ

Bαβ −Aαβ /2 Cαβ Aαβ

ν2 ν3 −ν2 −ν3 Aβα

−ν3 −ν2 ν3 ν2 Bβα Aβα

θy3

θy4

3

7 7 7 7 −ν2 ν3 7 7 −ν3 ν2 7 ν2 −ν3 7 ν3 −ν2 7 7 −Aβα /2 Cβα 7 Cβα −Aβα /2 7 7 Aβα Bβα 7 5 A βα

(15.52)

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

15.3 Finite Element Formulations

15–15

where α = b/a, β = a/b, ν1 = 1 − ν, ν2 = 3(1 + ν), ν3 = 3(1 − 3ν), Aαβ = 8α + 4βν1 , Bαβ = −4α − 2βν1 , Cαβ = −8α + 2βν1 , and Aβα , Bβα , Cβα are obtained by interchanging α and β in the expressions of Aαβ , Bαβ , Cαβ respectively. 41

Considering the second integral in Eq. 15.48, and substituting from Eq. 15.21 Kes = LTs Ds Ls 1 0 Ds = ζ 0 1 5 µt ζ = 6 we obtain

(15.53-a) (15.53-b) (15.53-c)

T

δu

Kζ

Kes udA

= Kζ

A

δ(LTs u)T ILs udA

(15.54-a)

(LTs Nδu)T Ls NudA

(15.54-b)

BTs Bs dAu

(15.54-c)

A

= Kζ

A

= KζδuT A

Recalling that Bs = Ls N, and substituting from Eq. 15.17-b for Ls we obtain BTs Bs dA Ks = B11 B12 0 Bs = 0 B23 B21

(15.55-a) (15.55-b)

Note that the ﬁrst row and second row in the second equations correspond to γxz and γyz respectively. 42

Upon substitution, 1 B11 = B12 = B21 = B12 =

! 1 !

− (1−η) a

(1−η) a

η a

− ηa

2

(1 − η)(1 − ξ) ξ(1 − η) ξη 1 − η 2 (1−ξ) ξ ξ − − (1−ξ) b b b b

(15.56-a)

"

(1 − ξ)(1 − η) ξ(1 − η) ξη η(1 − η)

(15.56-b) "

(15.56-c) (15.56-d)

43 In order to facilitate the integration of Ks , a one point Gauss integration ξ = η = 0.5 is used. Upon substitution 1 1 1 1 1 1 1 1 − 2a 0 2a 2a − 2a 4 4 4 4 s = (15.57) B 1 1 1 1 1 1 1 1 − 2b 0 − 2b 2b 2b 4 4 4 4

Finally Ks is obtained from

TB B s s dA

Ks =

(15.58)

A

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 15–16

PLATES

which gives Ks

=

1 24

2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4

w1 6γ

w2 −6δ 6γ

w3 −6γ 6δ 6γ

w4 6δ −6γ −6δ 6γ

θx1 −3b 3b 3b −3b 1.5ab

θx2 −3b 3b 3b −3b 1.5ab 1.5ab

θx3 −3b 3b 3b −3b 1.5ab 1.5ab 1.5ab

θx4 −3b 3b 3b −3b 1.5ab 1.5ab 1.5ab 1.5ab

θy1 −3a −3a 3a 3a

θy2 −3a −3a 3a 3a

θy3 −3a −3a 3a 3a

θy4 −3a −3a 3a 3a

3

7 7 7 7 7 7 7 7 7 7 7 7 7 7 1.5ab 1.5ab 1.5ab 1.5ab 7 7 1.5ab 1.5ab 1.5ab 7 7 1.5ab 1.5ab 5 1.5ab (15.59-a)

where α = ab , β = ab , γ = α + β, and δ = α − β. 44

The ﬁnal element stiﬀness matrix is ! " Ke = K Kef + ζKes

(15.60)

and in terms of the global stiﬀness matrix we will have [Kf + ζKs ] u =

15.3.1.2

P K

(15.61)

Shear Locking

Through inspection of the previous equation, and noting that ζ = 6ks (1−ν) , we observe that t2 for very thin plates ζ is very large, hence as t → 0, ζ → ∞. Hence, unrealistic u = 0 can be obtained independently of the load. This phenomena is called locking (of the displacements). 45

46

Locking occurs also in thin beam elements when shear deformation eﬀects are accounted for.

To alleviate the locking problems is to make Ks singular so that ζKs is ﬁnite. This can be accomplished through reduced numerical integration for Ks , Table 15.2, which in turn will reduce the rank of the matrix. 47

Bilinear Quadratic Serendipity

4 nodes, 12 d.o.f. 9 nodes, 27 d.o.f. 8 nodes, 24 d.o.f.

Integration Rule Full Reduced Selective e e Kf Ks Kef Kes Kef Kes 2x2 2x2 1x1 1x1 2x2 1x1 3x3 3x3 2x2 2x2 3x3 2x2 3x3 3x3 2x2 2x2 3x3 2x2

Table 15.2: Integration Rules for Mindlin Plate Elements

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

15.3 Finite Element Formulations

15–17

48 The same phenomena occurs for beams. This is illustrated by evaluating the deﬂection of a cantilever beam, due to both ﬂexural and shear deformation, under a point load PL P L3 3EI P L3 + = 1+ (15.62) w= 3EI ks µA 3EI ks µAL2

or w=

P L3 3EI

1+

3 α

;

α=

ks µAL2 µ = ks EI E

2 2 L L =O r h2

(15.63)

Hence, as the slenderness ratio increases, shear deformation decreases. Furthermore, Euler theory is recovered if the shear stiﬀness α → ∞, however shear locking can occur for small α. 49

Shear locking eﬀects are aggravated by large element distortions.

15.3.2 15.3.2.1

Nonconforming Kirchhoﬀ Triangular Element Formulation

Figure 15.7: Triangular Plate Element in Natural Coordinate System 50 Since rectangular elements can hardly accommodate general boundary conditions, and since quadrilateral elements can more readily be obtained through an assemblage of triangular ele-

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 15–18

PLATES

ments rather than through the distortions of a rectangular element, we consider next the three noded, nine d.o.f. Kirchhoﬀ element. 51 The element is shown in Fig. 15.7, has the following 3 dof at each node: displacement w, and ∂w slopes θxx = − ∂w ∂x and θyy = − ∂y . Because the complete cubic polynomial expansion contains 10 terms (1, ξ, η, ξ 2 , ξη, η2 , ξ 3 , ξ 2 η, ξη2 , η 3 ), yet we have only nine dof’s, thus we opt instead to a natural coordinate system and we express the displacement as

w = w1 + w2

(15.64)

where w1 corresponds to the rigid body part, and w2 the displacements incurred when the triangle is simply supported at the nodes. 52

w1 has three rigid body motions, Fig. 15.7

1. Translation: in the z direction with w1 = cst. 2. Rotation about side 1: of the triangle. This results in a rigid body translation in the z direction of line ab which is parallel to side 1 is w1 = h1 α. If we consider a triangle with vertex b, than this point would have L1 = A1 /A = h1 /h, thus w1 = h1 α = hαL1 or w1 = (cst)L1 . 3. Rotation about side 2: where w1 = (cst)L2 . Thus for rigid body motions we can write: 1 L1 + w 2 L2 + w 3 L3 w1 = w

(15.65)

53 Since the element has 9 dof, a cubic polynomial will be used for shape functions; i.e. the shape function can be a linear combination of the cubic terms

L21 L2 , L22 L3 , L23 L1 , L22 L1 , L23 L2 , L21 L3 , L1 L2 L3 Note that if we were to substitute into      x2 y3 −x3 y2 y23 x32  1   L1  L x =  x3 y1 −x1 y3 y31 x13     2  L3 x1 y2 −x2 y1 y12 x21 y

(15.66)

(15.67)

where xij = xi − xj , then we will have a cubic polynomial in terms of x and y. 54

55

Since w1 took already care of the rigid body movements, the cubic terms can be used for w2 . Zienkiewicz found it convenient to form the terms of w2 as follows 1 1 2 2 4 L3 L1 + L1 L2 L3 + w 5 L3 L2 + L1 L2 L3 w2 = w 2 2 1 1 2 2 7 L1 L2 + L1 L2 L3 +w 6 L1 L3 + L1 L2 L3 + w 2 2 1 1 2 2 9 L2 L1 + L1 L2 L3 w 8 L2 L3 + L1 L2 L3 + w 2 2

Victor Saouma

(15.68-a)

Finite Elements II; Solid Mechanics

Draft

15.3 Finite Element Formulations 56

15–19

Hence w = w1 + w2 = Nu w

(15.69)

are deﬁned in Eq. 15.65 and 15.68-a. where Nu and w 57 The unknown quantities w i can in turn be expressed in terms of the nodal unknown dis∂w placements by substituting w, θxx = − ∂w ∂x and θyy = − ∂y in Eq. 15.69. 58

The nodal displacement are hence expressed as uw u=N

(15.70)

−1 u = Gu and =N where u = w1 θx1 θy1 · · · w3 θx3 θy3 thus w u w = Nu Gu = Nu

(15.71)

with w = N1 u1 + N2 u2 + N3 u3 where u1 = wi θxxi θyyi T 59

Upon substitution, it can be shown that T  g 21 L2 + L21 L3 − L1 L2 − L1 L23   L1 + L −b3 (L21 L g 2 + 12 L1 L2 L3 ) + b2 (L21 L g 3 + 12 L1 L2 L3 ) N1 =   g 2 + 12 L1 L2 L3 ) − c2 (L21 L g 3 + 12 L1 L2 L3 ) c3 (L21 L

(15.72)

where b2 = x1 − x3 , b3 = x2 − x1 , c2 = y3 − y1 , and c3 = y1 − y2 . The other two shape functions can be obtained by cyclic permutation of suﬃx 1-2-3. 60

Finally, the stiﬀness matrix is derived from Eq. 15.45 e δw (K wpz ) dA + Γp δwps dγ = 0 δW = − A

(15.73)

S

where Ke = LT DL

(15.74)

Thus Ke =

NT LT DLNdA (15.75)

e

p

A

NT pz dA

=

(15.76)

A

These matrices are evaluated using Lk1 Ll2 Lm 3 dA = 2A A

15.3.2.2

k!l!m! (2 + k + l + m)!

(15.77)

Nonconformity

Considering Fig. 15.8 it can be shown that the displacement w along sides 2-3 (where L1 = 0) are cubic and is the same for element 1 as for element 2 (since they share the same nodal d.o.f.). The four terms of the cubic polynomial can be uniquely determined from w2 , w3 , θxx2 and θxx3 at nodes 2 and 3. Hence, both w and θxx = − ∂w ∂x are continuous along side 2-3 for both elements. 61

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 15–20

PLATES 1

y

1 2 x

3

2

3 2

1

Figure 15.8: Edges of Adjacent Triangular Elements However, for θyy = − ∂w ∂y things are quite diﬀerent. If we set L1 = 0 and take the derivative of w in Eq. 15.69, then we have 62

−θyy =

∂w = α0 + α1 x + α2 x2 ∂y

(15.78)

but since we only have two d.o.f. (θyy at nodes 2 and 3), then the parameters can not be uniquely expressed in terms of these two d.o.f. Hence, we will have discontinuity of normal slope, or kinks. 63

Elements with element boundary discontinuity are referred to as nonconforming elements.

64 This nonconforming element, nevertheless is a viable one as the 3 rigid body modes, and constant strains are present. In addition, it passes the patch test.

Despite its nonconformity, this element is quite appealing because the deformation is expressed only in terms of the middle surface deﬂection w. 65

15.3.3

Discrete Kirchhoﬀ Triangle

66 From the previous formulation, it is apparent that it is diﬃcult to formulate a compatible triangular element with 9 d.o.f. using a single polynomial approximation for w.

A viable approach consists in starting with the thick plate, or Reissner-Mindlin plate theory, by separating the d.o.f. for translation and rotations and making them independent one from the other. These in turn are made continuous at the inter-element boundaries, thus ensuring C0 continuity. Since the plate is very thin, the term corresponding to shear deformation in the virtual work will be neglected. The shape functions will be designed to ensure compatibility, thus the element will be conforming. Finally, the Kirchhoﬀ plate theory assumptions will be only introduced at discrete points along the element boundaries. 67

68

The starting point will be Eq. 15.22-c, Fig. 15.9 δθ T LTf Df Lf θdA + −δWi = A

δuT LTs Ds Ls udA

(15.79)

A

where the ﬁrst term corresponds to ﬂexure, and the second one to the eﬀect of shear which will be neglected. Victor Saouma

Finite Elements II; Solid Mechanics

Draft

15.3 Finite Element Formulations

15–21

3 6 5 y

1 4 2 x

Figure 15.9: Discrete Kirchhoﬀ Triangular Element 69 Since for bending, only the rotations appear in the formulation, the rotations can be approximated by 6 6

Ni θ xi ; θyy = Ni θyi (15.80) θxx =

i=1 70

i=1

The shape functions Ni for the natural coordinates with quadratic expansions are N1 = L1 (2L1 − 1) = (1 − L2 − L3 )(1 − 2L2 − 2L3 ) N2 = L2 (2L2 − 1) N3 = L3 (2L3 − 1) N4 = 4L2 L3 N5 = 4L1 L3 = 4(1 − L2 − L3 )L3 N6 = 4L1 L2 = 4(1 − L2 − L3 )L2

(15.81-a)

For this element, so far we have a total of 12 dof, and we seek to formulate it in terms of only i i 9 dof, 3 at each corner node corresponding to wi , θ xx and θyy thus we need to impose certain constraints for this reduction. 71

72

Hence, Kirchhoﬀ assumptions will be selectively applied:

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 15–22

PLATES

1. Zero transverse shear strains at the corners: or from Eq. 15.17-b    w  ∂ 1 0 −w,x 0 θxx γxz ∂x θ = = = ⇒ ∂ γyz θyy −w,y 0  xx  ∂y 0 1 θyy

(15.82)

2. Zero transverse shear strain at the midside nodes: or k θtk = −w,s

(15.83)

where k is a superscript denoting the midside nodes only, θtk is a rotation about the normal to the boundary at node k, and wsk is the derivative of w with respect to the s direction along the boundary. 3. Linear variation of the tangential slope: , i.e the rotation about the s direction at the middle nodes is the average of the rotations at the end nodes 1 θnk = (θni + θnj ) 2

(15.84)

where the corner nodes i − j are 2-3, 3-1, and 1-2. k corresponds to the midside nodes. 73

We let the displacement of w along the boundary of length lij to be cubic w = a0 + a1 s + a2 s2 + a3 s3 ;

w = a1 + 2a2 s + 3a3 s

(15.85)

and thus w = w,s varies quadratically and the three coeﬃcients can be obtained by matching w,s = −θn at the three points along the edge. Applying the boundary conditions at s = 0 and s = lij in terms of both w and w,j , we have 4 equations to determine the four coeﬃcients, wi = 0 wi wj = a0 + wj

2 a1 lij + a2 lij

+

= a1 + a2 lij +

ats = 0 node i

(15.86-a)

= a1

ats = 0

(15.86-b)

3 a3 lij 2 a3 lij

ats = lij node j

(15.86-c)

ats = lij

(15.86-d)

yielding k =− w,s

3 i 1 i 3 j 1 j w − w,s + w − w,s 2lij 4 2lij 4

(15.87)

74 Since the ﬁrst term of Eq. 15.22-c, does not involve w, there is no need to deﬁne an interpolation function for w inside the element. 75

Relations between θxx , θyy and nodal variables θn and θt are given by cos γij − sin γij θn − cos γij sin γij w,n θxx = = θyy sin γij cos γij θt − sin γij cos γij w,s

and

Victor Saouma

w,n w,s

=

− cos γij − sin γij sin γij − cos γij

θxx θyy

(15.88)

(15.89)

Finite Elements II; Solid Mechanics

Draft

15.3 Finite Element Formulations

15–23

76 Substituting the last three equation into Eq. 15.80 at each node to obtain θ xi and θ yi to ﬁnd the shape functions results in

θ xx = Nx vi ;

θyy = Ny vi

or

(15.90)                 

i θ xx Nx1 Nx2 Nx3 Nx4 Nx5 Nx6 Nx7 Nx8 Nx9 = i Ny1 Ny2 Ny3 Ny4 Ny5 Ny6 Ny7 Ny8 Ny9  θ yy      N   θ        

where Nx1 Nx2 Nx3 Nx4 Nx5 Nx6 Nx1 Nx2 Nx3 x

and ak = − l2ij , bk = ij

= = = = = = = = =

1.5(a6 N6 − a5 N5 ), N1 − b5 N5 − b6 N6 , −c5 N5 − c6 N6 , 1.5(a6 N4 − a5 N6 ), N2 − b5 N6 − b6 N4 , −c6 N6 − c4 N4 , 1.5(a6 N5 − a5 N4 ), N3 − b5 N4 − b5 N5 , −c4 N4 − c5 N5 ,

1 2 x − 1 y2 4 ij 2 ij 2 lij

, ck =

Ny1 Ny2 Ny3 Ny4 Ny5 Ny6 Ny1 Ny2 Ny3

3 xij yij , 2 4 lij

= = = = = = = = =

w1 1 θx 1 θy w2 2 θx 2 θy w3 3 θx 3 θy v

                                

(15.91)

1.5(d6 N6 − d5 N5 ) Nx3 N1 − e5 N5 − e6 N6 1.5(d4 N4 − d6 N6 ) Ny3 N2 − e5 N6 − e4 N4 1.5(d5 N5 − d4 N4 ) Nx3 N3 − e4 N4 − e5 N5

(15.92)

1 2 y − 1 x2 4 ij 2 ij 2 lij

and k = 4, 5, 6

y

dk = − l2ij , ek = ij

corresponds to sides ij=23, 31, and 12 respectively. With the shape functions thus deﬁned, we can ﬁnally determine the stiﬀness matrix from the principle of virtual work (Eq. 15.22-c). Since shear deformations are neglected, we use only the ﬁrst part. δθ T LTf Df Lf θdA (15.93-a) δWi = A 77

= δvi = δvi

T

T

A

Kef

NT LTf Df Lf NdAvi

BT Df BdA vi A

(15.93-b) (15.93-c)

Ke

(15.93-d) Thus the stiﬀness matrix can be evaluated

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 15–24

15.4 78

PLATES

Summary

Major conclusions 1. Two theories: thin plates (Kirchhoﬀ/Euler-Bernouilli), and thick plates (Mindlin-Reissner/Timoshenko), without and with shear deformations. 2. Strong analogy between plates and beams. 3. If membrane actions are accounted for, lead to formulation of shell elements. 4. σ ⇒ M , ε ⇒ κ. 5. If thick plate theory is applied to thin plates, will have shear locking problems. 6. Shear locking is alleviated by reduced integration of the shear components. 7. For thin plate theory, will have non-conforming elements if only 3 d.o.f. are adopted at each node. 8. The DKT element alleviated this problem. 9. Reduced integration will alleviate problems arising from mesh distortions. 10. Shear locking is primarily an issue for 3 node and 4 node elements. Higher order expansions perform much better, whereby Lagrangian 6 and 16 node elements perform signiﬁcantly better than their Serendipity counterparts of 8 node and 12 node elements. 11. The usual 3 DOF/node Kirchhoﬀ elements do not maintain normal slope continuity.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft Chapter 16

MATERIAL NONLINEARITIES 16.1

Introduction

16.1.1

Linearization

1

We deﬁne a constitutive operator as ˜ σ = σ(ε)

(16.1)

˜ denotes the constitutive operator (analogous to the L). where σ 2

3

¯, the corresponding stress will be σ ¯ = σ(˜ ˜ ε). Given a strain state ε ˜ can be expanded into a Taylor series with respect to ε ¯ The constitutive operator σ ˜ ∂σ ˜ ˜ + δε) = σ(ε) δε + · · · + σ(ε ∂ε ε=ε

(16.2)

Neglecting quadratic and higher order terms leads to a linearized constitutive law ˜ + D(ε)δε σ ≈ σ(ε)

(16.3)

which approximates Eq. 16.1 for strains in the neighborhood of ε, and D≡

˜ ∂σ ∂ε

(16.4)

is the tangent stiﬀness matrix which is a function of the current strain. 4

We rewrite Eq. 7.8, 7.12 and 7.16 in terms of the newly deﬁned constitutive operator T T T T ˜ B σ(Bu)dΩ = B D 0 dΩ − B σ 0 dΩ + N bdΩ + NT ˆtdΓ (16.5) Ωe Ωe Ωt Ωe Γt f

int

f 0e

f

ext

fe

Draft 16–2

MATERIAL NONLINEARITIES

or f

int

(u) = f

ext

(16.6)

5 We now develop a linearized expression for the internal forces. Given u as nodal displacements ˜ ˜ = σ(Bu). Then, the Taylor expansion yielding strain ﬁeld ε = Bu, and the stress ﬁeld σ = σ(ε) of the internal forces around u yields int ∂f int int f (u + ∆u) = f (u) + ∆u + · · · (16.7) ∂u

u=u

We again neglect the quadratic and higher order terms, leading to f

int

(u + ∆u) ≈ f

int

(u) + KT (u)∆u

(16.8)

where int

∂f KT ≡ ∂u

(16.9)

is the tangent stiﬀness matrix of the structure 6

Diﬀerentiating Eq. 16.5 int

∂ ∂f = KT = ∂u ∂u

˜ B σdΩ = T

Ω

˜ ∂σ dΩ = B ∂u Ω T

˜ ∂ε ∂σ dΩ = B ∂ε ∂u Ω

T

BT DBdΩ

(16.10)

Ω

which is the well known formula for the stiﬀness matrix, however Del is now replaced by the tangent moduli D

16.1.2

Solution Strategies

7 Before we discuss solution strategies, it may be helpful to point out the parallelism which exists between (numerical) solution strategies, and (experimental) testing methods. Modern testing equipment can be programmed to apply a pre-determined rate of load (as measured by a load cell), of displacement (as measured by an internal displacement transducer), or of strain (or relative displacement such as crack mouth opening displacement) measured by a strain/clip gage or other instruments, Fig. 16.1

Load Control: the cross-head applies an increasing load irrespective of the specimen deformation or response. For all materials, when the tensile strength is reached, there is a sudden and abrupt brittle failure. The strain energy accumulated in the specimen is suddenly released once the ultimate load of the specimen is reached, thus the sudden failure can be explosive. Displacement/Stroke Control: the cross-head applies an increasing displacement to the specimen. For softening material there will be a post-peak response with a gradual decrease in stress accompanying an increase in displacement. In this case, there is a gradual release of strain energy which is then transferred to surface energy during crack formation.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

16.1 Introduction

16–3 P 11 00

11 00 11 00

11 00

∆, 11 00

111 000

11 00 11 00

111 000

P, ∆, CMOD

11 00

t 11 00

11 00

11 00

111 000

CMOD

Actual Programmed

Figure 16.1: Test Controls Strain Control: is analogous to displacement control, except that the feedback is provided by (“strategically positioned”) strain gage or a clip gage or an arbitrary specimen deformation (not necessarily corresponding to the loading direction). To accomplish this test a clip gage or a strain gage has to provide the feedback signal to the testing equipment in order to accordingly adjust the stroke. Similarly, the objective of a nonlinear ﬁnite element analysis is to trace the (nonlinear) response of a structure subjected a given load history. This is best done in an incrementaliterative procedure where the load (or the displacement) is applied through several increments, and within each increment we seek to satisfy equilibrium through an iterative procedure (caused by the nonlinearity of the problem).

8

9

The incremental analysis can be performed under 1. Load control; Load is incrementally applied on the structure. 2. Direct displacement control; An imposed displacement is applied. 3. Indirect displacement control (such as relative displacements between two degrees of freedom) 4. Arc-Length control

10

Alternatively, iterative techniques include 1. Newton-Raphson 2. Modiﬁed Newton-Raphson 3. Initial Stiﬀness 4. Secant Newton

11

Finally, an essential ingredient of an incremental-iterative solution strategy are 1. Convergence Criteria 2. Convergence Accelerators (such as line-search or step-size adjustments).

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 16–4

MATERIAL NONLINEARITIES

16.2

Load Control

16.2.1

Newton-Raphson

For the sake of discussion, we will assume in the following sections that the incremental ext analysis is under load control, with increments of loads ∆f . 12

13 At the end of each load increment, internal forces must be in equilibrium with the external ones. Hence, we deﬁne the vector of residual forces R as

Rn+1 ≡ R(un+1 ) = f where f

int

int

(un+1 ) − f

ext

=0

(16.11)

is the vector of internal forces, also commonly known as reaction vector. int

For equilibrium to be satisﬁed, the vector of reactions internal forces f must be equal to ext the one of external ones f . This is automatically satisﬁed in linear elastic analysis, but not necessarily so in nonlinear analyses. 14

15

We start the analysis from an equilibrium conﬁguration, at the end of increment n such that u = un ,

Rn = 0

(16.12)

ext

and apply an increment of load ∆f n such that ext

ext

ext

f n+1 = f n + ∆f n

(16.13)

and we seek to determine the corresponding change in displacement un+1 = un + ∆un

(16.14)

ext

We will keep ∆f n reasonably small to capture the full nonlinear response. 16.2.1.1

Newton-Raphson/Tangent Stiﬀness Method

This is the most rapidly convergent process (albeit computationally expensive) of non-linear problems. 16

At the beginning of each step n + 1, we start from the displacement un that were computed int ext in the previous step through equilibrium Rn ≈ 0 or f n ≈ f n . The external forces are ext ext ext ext now increased from f n to f n+1 = f n+1 + ∆f , and we seek to determine the corresponding 17

int

ext

displacements un+1 through equilibrium Rn+1 ≈ 0 or f n+1 ≈ f n+1 . 18 Within the current step (identiﬁed through the subscript n), we will be iterating (through superscript k) in order to achieve equilibrium.

As initial guess for u0n+1 we take it to be un and based on the linearization around this initial state we have ext f int (u0n+1 ) + KT (u0n+1 )∆u1n+1 = f n+1 (16.15) 19

where ∆u1n+1 is the ﬁrst approximation for the unknown displacement increment ∆un+1 = un+1 − un . Victor Saouma

Finite Elements II; Solid Mechanics

Draft

16.2 Load Control

16–5

f ext 2 Rn+1

f ext n+1

3 Rn+1

n+1

1 Rn+1

∆fnext

n+1 f int, 2

1 T

n+1 f int, 1

K

n

f ext n

δun2

δun1 ∆ un1

un

∆ un2 1 un+1

2 un+1

3 un+1

u

Figure 16.2: Newton-Raphson Method 20

Alternatively, we begin from a linearization of Eq. 16.11, Fig. 16.2 i ∂R i+1 i R(un+1 ) ≈ R(un+1 ) + δui = 0 ∂u n+1 n

(16.16)

where i is a counter starting from u1n+1 = un . 21

Observing that int

∂f ∂R = = KT ∂u ∂u assuming that f

ext

(16.17)

is constant, and KT is the tangent stiﬀness matrix. Thus, Eq. 16.16 yields i

KiT δuin = −Rn+1

(16.18)

δuin = −(KiT )−1 Rn+1

(16.19)

or i

22

Thus, a series of successive approximations yields i i i ui+1 n+1 = un + ∆un = un+1 + δun

with ∆uin =

δukn

(16.20) (16.21)

k≤i

very rapidly. 23

It should be noted that each iteration involves three computationally expensive steps:

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 16–6

MATERIAL NONLINEARITIES

1. Evaluation of internal forces f

int

(or reactions)

2. Evaluation of the global tangent stiﬀness matrix KT 3. Solution of a system of linear equations 16.2.1.2

Modiﬁed Newton-Raphson

This method is essentially the same as the Newton-Raphson however in Eq. 16.23 (KiT ) is replaced by KT which is the tangent stiﬀness matrix of the ﬁrst iteration of either 1) the ﬁrst increment KT = K1T,0 , Fig. 16.4, or 2) current increment, Fig. 16.3 KT = K1T,n Fig. 16.3 24

f ext 2 Rn+1

f ext n+1

3 4 Rn+1 Rn+1

n+1

1 T

K

n+1 f int, 2

1 Rn+1

∆fnext

n+1 f int, 3

1 K T

n

f ext n ∆ un2

n+1 f int, 1

δun1

δun2

δun3

∆ un1 ∆ un3 un

1 un+1

2 un+1

3 un+1

u

Figure 16.3: Modiﬁed Newton-Raphson Method, Initial Tangent in Increment

δuin = −(KT )−1 Rn+1 i

(16.22)

In general the cpu time required for the extra iterations required by this method is less than the one saved by the assembly and decomposition of the stiﬀness matrix for each iteration. 25

26 It should be mentioned that the tangent stiﬀness matrix does not necessarily have to be the true tangent stiﬀness matrix; an approximation of the true tangent stiﬀness matrix or even the initial stiﬀness matrix will generally produce satisfactory results, albeit at the cost of additional iterations.

16.2.1.3

Secant Newton

27 This method is a compromise between the ﬁrst two. First we seek two displacements by two cycles of modiﬁed Newton-Raphson, then a secant to the curve is established between those

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

16.2 Load Control

16–7

f ext 2 Rn+1

f ext n+1

3 4 Rn+1 Rn+1

n+1

0 T n+1 f int, 3

0 T

K

1 Rn+1

∆fnext

K

n+1 f int, 2 n+1 f int, 1

n

f ext n ∆ un2

δun2

δun1

δun3

∆ un1 ∆ un3 un

1 un+1

2 un+1

3 un+1

u

Figure 16.4: Modiﬁed Newton-Raphson Method, Initial Problem Tangent two points, and a step taken along it, Fig. 16.5. δuin = −(KT )−1 Rn+1 i

(16.23)

28 Subsequently, each step will be taken along a secant connecting the previous two points. Hence, starting with 1 (16.24) δu1n = −K−1 T Rn+1

the secant slope can be determined (K2S )−1 = − and then

δu1n 1

(16.25)

2

(Rn+1 − Rn+1 )

δu2n = −(K2S )−1 Rn+1 2

(16.26)

δuin = −(KiS )−1 Rn+1 i

29

This process can be generalized to (KiS )−1 = −

16.2.2

δuin i−1

i

(Rn+1 − Rn+1 )

(16.27) (16.28)

Acceleration of Convergence, Line Search Method

Adapted from (Reich 1993) 30 The line search is an iterative technique for automatically under- or over-relaxing the displacement corrections δuj so as to accelerate the convergence of nonlinear solution algorithms.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 16–8

MATERIAL NONLINEARITIES f ext

2 Rn+1

f ext n+1

3 Rn+1

4 Rn+1

n+1

n+1 f int, 3

K0

1 Rn+1

∆fnext

n+1 f int, 2 T

n+1 f int, 1

n

f ext n ∆ un2

δun2

δun1

δun3

∆ un1 ∆ un3 un

3 un+1

2 un+1

1 un+1

u

Figure 16.5: Incremental Secant, Quasi-Newton Method 31 The amount of under- or over-relaxation is determined by enforcing an orthogonality condij+1 tion between the displacement corrections δuj and the residual loads R , which amounts to forcing the iterative change in energy to be zero.

The displacement corrections are multiplied by a scalar value sk deﬁning the amount of under- or over-relaxation such that the total displacements uj+1,k are deﬁned as 32

uj+1,k = uj + sk δuj

(16.29)

For k = 0 and k = 1, the values of sk are 0.0 and 1.0, respectively. Therefore, uj+1,0 = uj and uj+1,1 = uj+1 . The orthogonality condition is quantiﬁed by a scalar value gk representing the iterative change in energy, which is deﬁned as 33

j+1,k

gk = δuj · R where

j+1,k

R

=f

ext

−f

int

(16.30)

(uj+1,k )

(16.31)

are the residual loads at the end of solution iteration j and line search iteration k. gk can be expressed as a function of sk (see Figure 16.6) and the object of the line search is to ﬁnd sk such that gk is zero. An estimate of sk+1 such that gk+1 is zero can be computed using a simple extrapolation procedure based on similar triangles 34

sk sk+1 = g0 g0 − gk On rearranging terms, sk+1 is deﬁned as s Victor Saouma

k+1

= s

k

g0 g0 − gk

(16.32)

(16.33)

Finite Elements II; Solid Mechanics

Draft

16.2 Load Control

16–9

As a preventative measure, sk+1 is assigned a value of 5.0 for all sk+1 > 5.0 so that unreint strained over-relaxation is inhibited. Once sk+1 is estimated, uj+1,k+1 , f j+1,k+1 , and Rj+1,k+1 are computed for the next line search iteration, Fig. 16.6. 35

g g0

g1

s1

s

s2

Figure 16.6: Schematic of Line Search, (Reich 1993) 36

The line search terminates after three iterations or when | g0 | ≤ 0.8 | gk |

(16.34)

and g0 gk ≤ 0.001 | g0 |. Smaller tolerances may be used to determine if the line search has converged, Zienkiewicz and Taylor (1989) prefer to use 0.6, but Crisﬁeld, M.A. (1979) concluded that there was little advantange to be gained by doing such. 37

The ﬂowchart illustrating the Line Search algorithm is shown in Fig. 16.7.

Initialize s1

0

Compute R

DO 30 k=1,3 Do 10 j=0, Niter Compute gk and sk+1 j

j+1 0

Computeδ u,∆ u,

g

j+1

j+1

Compute∆ε and∆σ j+1

j+1

Compute fint and R

No Compute∆ u

j+1,k+1

j+1,k+1

Compute∆ε Converged No

Yes

Converged

j+1,k+1

and ∆σ

Yes j+1,k+1

Compute fint

and R

j+1,k+1

10 30

Figure 16.7: Flowchart for Line Search Algorithm, (Reich 1993)

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 16–10

16.2.3

MATERIAL NONLINEARITIES

Convergence Criteria

38 In all preceding methods, iterations are performed until one or all of a variety of convergence criteria are satisﬁed. Relative convergence criteria are optionally enforced on the displacements, loads, and/or incremental energy to deﬁne the termination conditions.

The relative displacement criteria is deﬁned in terms of the displacement corrections δuj and the updated incremental displacements δuj+1 as 39

Ou =

δuj 2 ∆uj+1 2

(16.35)

where . . . 2 is the Euclidean norm. 40

The Euclidean norm, which is also known as the L2 norm, of a vector v is deﬁned as #N

v 2 =

$1/2 vi2

(16.36)

i=1

where N is the size of v. j+1

The relative load criteria is deﬁned in terms of the updated residual loads R int reactions f j+1 as either 41

Or =

or Or =

j+1

2

int f j+1

2

R

j+1

∞

int f j+1

∞

R

where . . . ∞ is the inﬁnity norm. The inﬁnity norm of a vector v is deﬁned as 3N 4

|vi | v ∞ = max

and the

(16.37)

(16.38)

(16.39)

i=1

where N is the size of v. The relative incremental energy criteria is deﬁned in terms of displacement corrections δuj , j+1 the updated residual loads R , the updated incremental displacements ∆uj+1 , and the upint dated reactions f j+1 as 42

j+1

OW =

δuj · R

int

∆uj+1 · f j+1

(16.40)

where the numerator is the change in the incremental energy for iteration j and denominator is the incremental energy.

16.3

Direct Displacement Control

Adapted from (Jirasek and Baˇ zant 2001)

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

16.3 Direct Displacement Control

16–11

43 Independently of the choice of iterative algorithm, any solution strategy using load control fails if the prescribed external loads cannot be maintained in equilibrium by the internal forces. This would typically occur if the load is monotonically increased until the load-carrying capacity of the structure is exhausted, Fig. 16.8

f ext

u

Figure 16.8: Divergence of Load-Controled Algorithms In most engineering analyses, it is simply required to determine the maximum load carrying capacity, and the corresponding displacements. As such, divergence of the iterative process is often taken as an indicator of structural failure, and the last converged step provides information on the state prior to collapse. 44

45 However, ﬁnite element simulations of complex engineering problems can diverge for a number of other reasons, many of which are purely numerical and have nothing to do with the real structural failure. 46 If the load-displacement diagram is to be followed beyond the peak, i.e post-peak response is required, then alternative solution strategy to the load-control one must be devised.

Post-peak response may be of interest not only in problems in structures with imposed displacements (such as initial settlements), but also to assess the ductility of the structure (specially when cracks are present). 47

To outline the displacement controlled algorithm, we divide the displacements into two groups: one with unknown displacements at nodes that are left “free”, and the second with prescribed displacements at nodes that are controlled. Accordingly, we partition the displacement vector into {uf , up }T and the internal and external force vectors into {f int,f , f int,p }T and {f ext,f , f ext,p }T , respectively. External forces f ext,f (corresponding to the unknown displacements uf ) are prescribed, and for simplicity we will assume that they are equal to zero. All external forces acting on the structure are represented by reactions f ext,p at the supports with prescribed displacements up . 48

49

Hence, the equilibrium equations are partitioned as f int,f (uf , up ) = 0

(16.41-a)

f int,p (uf , up ) = f ext,p

(16.41-b)

50 For given up , the unknown displacements uf can be computed by solving Eq. 16.41-a. After that, the reactions f ext,p are obtained by simple evaluation of the left-hand side in (16.41-b).

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 16–12

51

MATERIAL NONLINEARITIES (n−1)

In a typical incremental step number n, we start from the converged displacements uf (n−1)

and up

from the previous step, and we replace Eq. 16.41-a by the linearized equations (n−1)

ff where K11 ≡

∂f int,f ∂uf

(n−1)

+ K11

and K12 ≡ ∂

∂f int,f ∂up

(n,1)

∆uf

(n−1)

+ K12

∆u(n,1) =0 p

(16.42)

are blocks of the global tangent stiﬀness matrix



 ∂f int,f K K ∂up  11 12 = K21 K22 ∂f int,p  ∂up

∂f int,f  ∂f int  ∂uf = K≡ ∂f int,p ∂u ∂uf

(16.43)

The increment of the prescribed displacements up is known in advance, and so we set (n,1) (n) (n) (n−1) = ∆up = up − up and rewrite (16.42) as ∆up 52

(n−1)

K11 (n,1)

Having solved for ∆uf (n,1) up

=

(n−1) up

+

(n,1) ∆up

(n,1)

∆uf

(n−1)

(n−1)

= −f int,f − K12

∆u(n) p (n,1)

, we construct the ﬁrst approximation uf

=

(n−1)

= uf

(n,1)

+ ∆uf

and

(n) up . (n,1)

(n,1)

Equations (16.41-a) are then linearized around (uf , up ), corrections of displacements uf are computed, and the procedure is repeated until the convergence criteria are satisﬁed. 53

54

The iterative process can be described by recursive formulas (n,i−1)

K11

(n,i−1)

(n,i)

= −f int

(n,i)

= uf

δuf uf

(n,i−1)

(n,i−1)

− K12

(n,i)

δup

(n,i)

+ δuf

i = 1, 2, 3, . . .

where (n,0)

(n−1)

(16.44-a)

u(n,0) = u(n−1) p p

(16.44-b)

uf

= uf

(n−1) = u(n) δu(n,1) p p − up

(16.44-c)

δu(n,i) p

(16.44-d)

= 0

for i = 2, 3, . . .

55 Note that, starting from the second iteration, the correction δup is zero, and so the term with K12 on the right-hand side of (16.3) vanishes. This term is present only in the ﬁrst iteration. It (n,0) (n) (n,0) (n−1) = up instead of up = up , and might seem that one could start immediately from up then the correction δup would be zero already in the ﬁrst iteration and the matrix K12 would never have to be evaluated. However, this is in general not a good idea because such an initial approximation would be too far from the equilibrium path and the process might diverge.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

16.4 Indirect Displacement Control

16.4

16–13

Indirect Displacement Control

56 Direct displacement control can be applied only on structures loaded only at one point, or when the load is transmitted by a stiﬀ platen so that all points on the loaded surface exhibit the same displacements. 57 However, this is not always the case. As an example, consider a dam loaded by hydrostatic pressure due to reservoir overﬂow; see Fig. 16.9. Here, the load is applied along a large portion of

Figure 16.9: Hydrostatically Loaded Gravity Dam the boundary, and the shape of the corresponding displacement proﬁle is not known in advance. Another case in which direct displacement control fails is very brittle failure characterized by a load-displacement diagram with a snapback, Fig. 16.10. µ

µ

∆l

µ

∆l ∆l

u Load Control

u Displacement Control

u Arc-Length Control

Figure 16.10: Load-Displacement Diagrams with Snapback 58 Advanced incrementation control techniques abandon the assumption that the values of external loads and/or displacements at supports after each incremental step are prescribed in advance. Instead, the loading program is parameterized by a scalar load multiplier.

16.4.1

Partitioning of the Displacement Corrections

Adapted from (Reich 1993)

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 16–14

MATERIAL NONLINEARITIES

Restricting the applied loading to be proportional, a scalar load parameter β can be used to ext scale an arbitrary set of applied loads f . 59

ext

The applied loads at the start of increment i are deﬁned as the scalar-vector product βi f , where βi is the load parameter at the start of increment i. βi is zero at the start of the ﬁrst increment. 60

The applied incremental loads for increment i are deﬁned as the scalar-vector product ext ∆βi f , where ∆βi is the incremental load parameter for increment i. The updated load parameter βi+1 at the end of increment i is 61

βi+1 = βi + ∆βi

(16.45)

62 The incremental displacements due to the applied incremental loads are obtained using the standard modiﬁed-Newton algorithm, as described in Zienkiewicz & Taylor (1991). The incremental displacements ∆uj+1 at the end of iteration j for a generic increment are deﬁned as (16.46) Duj+1 = ∆uj + δuj

where ∆uj are the incremental displacements at the start of iteration j and δuj are the incremental displacement corrections for iteration j. The incremental load parameter ∆β j+1 at the end of iteration j is deﬁned in an analogous manner as (16.47) ∆β j+1 = ∆β j + δβ j 63

where ∆β j is the incremental load parameter at the start of iteration j and δβ j is the incremental load parameter correction for iteration j. At the start of the ﬁrst iteration ∆uj and ∆β j are identically zero. 64

Incremental displacement corrections are determined by solving K δuj = (βf

ext

+ ∆β j f

ext

+ δβ j f

ext

j

− f int )

(16.48)

where K is the global stiﬀness matrix and j f int

=

N

elem e=1

BT D ( + ∆ j ) δΩ

(16.49)

Ωe

are the reactions for the state of stress at the start of iteration j. j

65

Deﬁning the residual forces R at the start of iteration j as j

R = βf

ext

+ ∆β j f

ext

j

− f int

(16.50)

Equation 16.48 can be written more simply as δuj = K−1 (δβ j f

ext

j

+ R )

(16.51)

ext

The matrix-vector product K−1 f is invariant for the increment and, therefore, can be treated as a vector constant δuT , which Crisﬁeld (1981) referred to as the tangent displacements 66

δuT = K−1 f Victor Saouma

ext

(16.52)

Finite Elements II; Solid Mechanics

Draft

16.4 Indirect Displacement Control

16–15

j

The matrix-vector product K−1 R deﬁnes the displacement corrections δujr due to the residual forces j (16.53) δujr = K−1 R 67

but they are obviously not invariant for the increment. The displacement corrections for iteration j are then deﬁned as δuj = δβ j δuT + δujr (16.54) 68 Figure 16.11 shows a ﬂowchart for an incremental nonlinear ﬁnite element program based on the modiﬁed-Newton algorithm with indirect displacement control capabilities. The numbers in the boxes in Figure 16.11 correspond to those appearing in Figure ??.

16.4.2

Arc-Length

Adapted from (Jirasek and Baˇ zant 2001) 69 The basic idea of a ﬂexible incrementation control technique is that the step size is speciﬁed by a constraint equation that involves the unknown displacements as well as the load multiplier. The original motivation was provided by the requirement that the size of the step measured as the geometric distance between the initial and ﬁnal state in the load-displacement space should be equal to a prescribed constant, Fig. 16.10. 70 Despite the apparent simplicity of the condition of a constant arc length, it must be used with caution. First of all, it is important to realize that forces and displacements have completely diﬀerent units, and so the purely geometrical measure of length in the load-displacement space does not make a good sense. It is necessary to introduce at least one scaling factor, denoted as c, that multiplies the load parameter and converts it into a quantity with the physical dimension of displacement. The length of a step during which the load parameter changes by ∆µ and the displacements change by ∆u is then deﬁned as 5 (16.55) ∆l = ∆uT ∆u + (c ∆µ)2

By adjusting the scaling factor we can amplify or suppress the relative contribution of loads and displacements. One reasonable choice is derived from the condition that the0 contributions should be equal as long as the response remains linear elastic, which leads to c = uTe ue where ue is the solution of Ke ue = f . 71

In some cases, e.g., for frame, plate, and shell models that use both translational and rotational degrees of freedom, the components of the generalized displacement vector u do not have the same physical dimension. It is then necessary to apply scaling also to the vector ∆u. 72

Consider an incremental solution process controled by the arc-length method. In a typical step number n, we start with displacements u(n−1) and load parameter µ(n−1) computed in the previous step, and we search for displacements u(n) and load parameter µ(n) . The state at the end of the step must satisfy the equations of equilibrium between the internal forces f int (u(n) ) and external forces f ext (µ(n) ). Compared to the load control or direct displacement control, the load parameter is an additional unknown. The corresponding additional equation is provided by the constraint that ﬁxes the size of the step. For example, we can require that ¯ We the length of the step evaluated from formula (16.55) be equal to a prescribed value, ∆l. could treat the problem as a system of Ndf + 1 nonlinear equations, where Ndf is the number 73

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 16–16

MATERIAL NONLINEARITIES

Figure 16.11: Flowchart for an incremental nonlinear ﬁnite element program with indirect displacement control

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

16.4 Indirect Displacement Control

16–17

Figure 16.12: Two points on the load-displacement curve satisfying the arc-length constraint of unknown displacement components (degrees of freedom), and solve it by Newton-Raphson iteration. However, a more elegant and computationally more eﬃcient procedure treats the equilibrium equations and the constraint equation to a certain extent separately. Assume for simplicity that the loading program is described by (??). The linearized equations of equilibrium in the i-th iteration read (n,i−1)

K(n,i−1) δu(n,i) = f 0 + µ(n,i−1) f − f int

+ δµ(n,i) f

(16.56)

where δu(n,i) is the unknown displacement correction, and δµ(n,i) is the unknown correction of the load parameter. The ﬁrst three terms on the right-hand side are known, and the last term is an unknown scalar multiple of a given vector f . We can therefore separately solve equations (n,i−1)

K(n,i−1) δu0 = f 0 + µ(n,i−1) f − f int K

(n,i−1)

δuf

= f

(16.57-a) (16.57-b)

and then express the displacement correction as δu(n,i) = δu0 + δµ(n,i) δuf

(16.58)

When this expression is substituted into the constraint condition, ¯ 2 (∆u(n,i−1) + δu(n,i) )T (∆u(n,i−1) + δu(n,i) ) + c2 (∆µ(n,i−1) + δµ(n,i) )2 = (∆l)

(16.59)

we obtain a quadratic equation for a single unknown, δµ(n,i) . This equation usually has two real roots, corresponding to the two points of the equilibrium path that have the prescribed distance from point (u(n−1) , µ(n−1) ); see Fig. 16.12. The correct root is selected depending on the sense in which we march on the equilibrium path (Crisﬁeld, M.A. 1981), and the displacement correction is determined from (16.58). After standard updates of the displacement vector and the load parameter, the iteration cycle is repeated until the convergence criteria are satisﬁed.

16.4.3

Relative Displacement Criterion

Adapted from (Reich 1993) 74 The standard arc-length control performs well if the entire structure or its large portion participates in the failure mechanism. In cases when the failure pattern is highly localized, robustness of the technique may deteriorate. The remedy is to adapt the constraint equation to the particular problem and control the incrementation process by a few carefully selected displacement components. 75 Motivation is again provided by the physical background. If the load-displacement diagram of a brittle structure exhibits snapback, direct displacement control applied in an experiment leads to sudden catastrophic failure. When the displacement imposed by the loading device reaches a critical value, fracture starts propagating even though the imposed displacement at the load point is kept ﬁxed. However, opening of the crack monotonically increases during the entire failure process, and so it can be used as a control variable. If the experimental setup is arranged such that the applied force is continuously adjusted depending on the currently measured value of the crack opening, the response can be traced in a stable manner even after

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 16–18

MATERIAL NONLINEARITIES

the point at which the load-displacement diagram snaps back. The same idea can be exploited by a numerical simulation. It suﬃces to select a suitable linear combination of displacement components that increases monotonically during the entire failure process, and to use this combination as the control variable. de Borst (1985,1986) concluded that arc-length methods (Riks 1979, Ramm 1981, Crisﬁeld 1981), which were the original IDC methods, were not satisfactory for analyses involving cracking accompanied by softening. 76

The main problem with the arc-length methods, when used in this context, was that the constraint involved all displacement components equally when, in fact, only a few displacement components were dominant. The dominant displacement components were typically those for nodes at or near the crack mouth. This being the case, de Borst proposed using a transformed relative displacement component between two nodes as the constraint. 77

78 The transformed relative displacement component can deﬁne the crack mouth opening displacement (CM OD), crack mouth sliding displacement (CM SD), or some arbitrary displacement ∆u between two points on a structure. The arbitrary displacement ∆u may correspond to a relative displacement measured during an experiment such as the relative vertical displacement between a point on the neutral axis of a 3-point bend beam over a support and the bottom of the beam at mid-span.

As it is the most general case, the relative displacement criterion will be described in terms of the arbitrary relative displacement ∆u. A pair of nodes, m and n, are selected to deﬁne ∆u, with their total displacements being (u)m and (u)n , respectively. The direction associated with ∆u is deﬁned by a unit vector v. ∆u is thereby deﬁned as 79

∆u = vT [(u)n − (u)m ]

(16.60)

If m and n are nodes on opposite sides of a discrete crack ∆u ≡ CM OD if v is normal to the crack surface and ∆u ≡ CM SD if v is tangent to the crack surface. The value for ∆u is prescribed for an increment and the applied loads are scaled such that the total displacements at the end of each iteration reﬂect that value. 80

81

Recalling that the total displacements uj+1 at the end of iteration j are deﬁned as uj+1 = uj + δβ j δuT + δujr

(16.61)

the load parameter correction δβ j for iteration j is δβ j =

16.4.4

1 2 ! " ∆u − vT (uj )n − (uj )m − vT (δujr )n − (δujr )m vT [(δuT )n − (δuT )m ]

(16.62)

IDC Methods with Approximate Line Searches

82 Employing a procedure proposed by Crisﬁeld (1983) for use with the arc-length method, the convergence of the solution alogrithm can be accelerated by performing approximate line searches; approximate line searches under ﬁxed (i.e. non-scalable) loads are described in Section ??.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

16.4 Indirect Displacement Control

16–19

83 This procedure requires an extra iterative loop at the beginning of the line search loop in which a combination of δβ j and sk+1 satisifying the constraint conditions (i.e. Equations ?? and 16.60) is computed. As δβ j is initially computed for s1 = 1.0, any change in sk requires a corresponding change in δβ j for the IDC constraint to remain satisﬁed. Consequently, an iterative loop, in which δβ j is recomputed based on the estimated value of sk+1 , is required to obtain a compatible combination of δβ j and sk+1 .

After recomputing δβ j , the values of g0 and gk are also recomputed using Equation 16.30 j+1,k j+1,k caused by the new value of δβ j . f int is not to reﬂect the change in the residual loads R j+1,k , which is strictly not correct, updated to reﬂect the changes in sk+1 when recomputing R but it does signiﬁcantly reduce the number of computations without causing any diﬃculties (Crisﬁeld 1983). 84

Finally, from the new values of g0 and gk , sk+1 is re-estimated using Equation 16.33. The loop is terminated when k+1 | | sk+1 new − s ≤ 0.05 (16.63) | sk+1 new | 85

which generally requires only a few iterations. A ﬂow chart of this procedure is shown in Figure 16.13. 86

Since the total displacements uj+1,k are now deﬁned as uj+1,k = uj + sk (δujr + δβ j δuT )

(16.64)

reﬂecting the introduction of the relaxation parameter sk , the IDC constraint equations must be modiﬁed accordingly. δβ j for the stress criterion is now deﬁned as 1 2    ft − (λj )n + sk (δλjr )n (n)n  (16.65) δβ j = min   sk (δλT )n (n)n and δβ j for the relative displacement criterion is now deﬁned as 1 2 ! " ∆u − vT (uj )n − (uj )m − sk vT (δujr )n − (δujr )m δβ j = sk vT [(δuT )n − (δuT )m ]

(16.66)

It is these general forms of the constraint equations that are implemented in MERLIN.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft 16–20

MATERIAL NONLINEARITIES

Figure 16.13: Flow chart for line search with IDC methods

Victor Saouma

Finite Elements II; Solid Mechanics

Draft Bibliography Abramowitz, M. and Stegun, I.: 1970, Handbook of mathematical functions, Technical report, National Bureau of Standard. Applied Mathematics Series, No. 55. Babuˇska, I.: 1971, Error-bounds for ﬁnite element methods, Numer. Math 20(3), 179–192. Babuˇska, I.: 1973, The ﬁnite element method with lagrange multipliers, Numer. Math. 20(3), 179–182. Bathe, K.: 1996, Finite Element Procedures, Prentice-Hall Inc. Brezzi, F.: 1974, On the existence uniqueness and approximation of saddle-point problems arising from lagrangian multipliers, RAIRO 8-R2 pp. 129–151. ˇ Cervenka, V., Keating, S. and Felippa, C.: 1993, A comparison of strain recovery techniques for the mixed iterative method, Communications in Applied Numerical Methods 9, 925–932. Crisﬁeld, M.A.: 1979, A faster modiﬁed newton-raphson iteration, Computer Methods in Applied Mechanics and Engineering 20(3), 267–278. Crisﬁeld, M.A.: 1981, A fast incremental/iterative solution procedure that handles ‘snap through’, Computers and Structures 13(1-3), 55–62. Delaunay, B.: 1934, Sur la Sph`ere Vide, Izv. Akad. Nauk SSSR, Otdelenie Matematicheskii i Estestvennyka Nauk 7, 793–800. Eisenberg, M. A. and Malvern, L.: 1973, On ﬁnite element integration in natural co-ordinates, International Journal for Numerical Methods in Engineering 7, 574–575. Felippa, C.: 1999, Finite element lecture notes, Technical report, Dept. of Aerospace Engineering, University of Colorado, Boulder. Felippa, C.: 2000, Lecture notes in advanced ﬁnite element methods (asen 5367), Technical report, Dept. of Aerospace Engineering, University of Colorado, Boulder. http://caswww.colorado.edu/courses.d/AFEM.d/Home.html. Haser, N. and Sullivan, J.: 1991, Real Analysis, Dover Publications, New-York. Jirasek, M. and Baˇzant, Z.: 2001, Inelastic Analysis of Structures, John Wiley, Chichester. Kardestuncer, H. (ed.): 1987, Finite Element Handbook, McGraw-Hill. Okabe, A., Boots, B., Sugihara, K. and Chiu, S.: 2000, Spatial Tessellations; Concepts and Applications of Voronoi Diagrams, John Wiley & Sons.

Draft –2

BIBLIOGRAPHY

Ottosen, N. and Petersson, H.: 1992, Introduction to the Finite Element Method, Prentice-Hall. Reich, R.: 1993, On the Marriage of Mixed Finite Element Methods and Fracture Mechanics: An Application to Concrete Dams, PhD thesis, University of Colorado, Boulder. Schey, H.: 1973, Div Grad Curl and all That; An Informal Text on Vector Calculus, W.W. Norton. Terzaghi, K. and Peck, R.: 1967, Soil Mechanics in Engineering Practice, 2nd edition, John Wiley & Sons, New York, NY. Timoshenko, S. and Goodier, J.: 1970, Theory of Elasticity, McGraw Hill. ˇ Cervenka, J.: 1994, Discrete Crack Modeling in Concrete Structures, PhD thesis, University of Colorado, Boulder. Voronoi, G.: 1907, Nouvelles Applications des Param`etres Continus `a la Th´eorie des Formes Quadratiques, J. Reine Angew. Math 133, 97–178. Xue, W. and Atluri, N.: 1985, Existance and stability, and discrete bb and rank conditions for general mixed-hybrid ﬁnite elements in elasticity, Hybrid and Mixed Finite Element Methods pp. 91–112. Zienkiewicz, O. C. and Taylor, R. L.: 1989, The Finite Element Method, Vol. 1, Basic Formulation and Linear Problems, 4th ed., McGraw-Hill, London. Zienkiewicz, O. C., Vilotte, J. P., Toyoshima, S. and Nakazawa, S.: 1985, Iterative method for constrained and mixed approximations. an inexpensive improvement of fem performance, Computer Methods in Applied Mechanics and Engineering 51(1–3), 3–29.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft Appendix A

VECTOR OPERATIONS 1 This appendix covers some elements of Vector Calculus essential to understand some of the derivations. An excellent reference is (Schey 1973).

A.1

Vector Diﬀerentiation

A ﬁeld is a function deﬁned over a continuous region. This includes, Scalar Field g(x), Vector Field v(x), or Tensor Field T(x). 2

3

We ﬁrst introduce the diﬀerential vector operator “Nabla” denoted by ∇ ∇≡

∂ ∂ ∂ i+ j+ k ∂x ∂y ∂z

(A.1)

4 We also note that there are as many ways to diﬀerentiate a vector ﬁeld as there are ways of multiplying vectors, the analogy being given by Table A.1.

Multiplication u·v dot u×v cross u ⊗ v tensor

Diﬀerentiation ∇·v divergence ∇×v curl ∇v gradient

Tensor Order ❄ ✲ ✻

Table A.1: Similarities Between Multiplication and Diﬀerentiation Operators

A.1.1 5

Derivative WRT to a Scalar

The derivative of a vector p(u) with respect to a scalar u, Fig. A.1 is deﬁned by p(u + ∆u) − p(u) dp ≡ lim du ∆u→0 ∆u

6

(A.2)

If p(u) is a position vector p(u) = x(u)i + y(u)j + z(u)k, then dx dy dz dp = i+ j+ k du du du du

(A.3)

Draft A–2

VECTOR OPERATIONS ∆ p=p (u+∆ u)- p(u) C

u) +∆ u ( p

p (u)

Figure A.1: Diﬀerentiation of position vector p is a vector along the tangent to the curve. 7

If u is the time t, then

dp dt

is the velocity

In diﬀerential geometry, if we consider a curve C deﬁned by the function p(u) then dp du is a vector tangent ot C, and if u is the curvilinear coordinate s measured from any point along the curve, then dp ds is a unit tangent vector to C T, Fig. A.2. and we have the following relations 8

N

T

C

B

Figure A.2: Curvature of a Curve dp = T ds dT = κN ds B = T×N

(A.4) (A.5) (A.6)

κ

we also note that p· dp ds Victor Saouma

curvature (A.7) 1 Radius of Curvature (A.8) ρ = κ = 0 if dp ds = 0. Finite Elements II; Solid Mechanics

Draft

A.1 Vector Diﬀerentiation

A–3

Example A-1: Tangent to a Curve Determine the unit vector tangent to the curve: x = t2 + 1, y = 4t − 3, z = 2t2 − 6t for t = 2. Solution:

" dp d ! 2 = (t + 1)i + (4t − 3)j + (2t2 − 6t)k = 2ti + 4j + (4t − 6)k dt dt 0 dp = (2t)2 + (4)2 + (4t − 6)2 dt T = =

A.1.2

0

2ti + 4j + (4t − 6)k

(2t)2 + (4)2 + (4t − 6)2 2 1 2 4i + 4j + 2k 0 = i + j + k for t = 2 3 3 3 (4)2 + (4)2 + (2)2

(1.9-a) (1.9-b) (1.9-c) (1.9-d)

Divergence

The divergence of a vector ﬁeld of a body B with boundary Ω, Fig. A.3 is deﬁned by considering that each point of the surface has a normal n, and that the body is surrounded by a vector ﬁeld v(x). The volume of the body is v(B). 9

v(x)

n Ω B

Figure A.3: Vector Field Crossing a Solid Region 10

The divergence of the vector ﬁeld is thus deﬁned as 1 div v(x) ≡ lim v(B)→0 v(B)

v·ndA

(A.10)

Ω

where v.n is often referred as the ﬂux and represents the total volume of “ﬂuid” that passes through dA in unit time, Fig. A.4 This volume is then equal to the base of the cylinder dA times the height of the cylinder v·n. We note that the streamlines which are tangent to the boundary do not let any ﬂuid out, while those normal to it let it out most eﬃciently. 11

The divergence thus measure the rate of change of a vector ﬁeld.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft A–4

VECTOR OPERATIONS

n dA v v.n

Ω

Figure A.4: Flux Through Area dA The deﬁnition is clearly independent of the shape of the solid region, however we can gain an insight into the divergence by considering a rectangular parallelepiped with sides ∆x1 , ∆x2 , and ∆x3 , and with normal vectors pointing in the directions of the coordinate axies, Fig. A.5. If we also consider the corner closest to the origin as located at x, then the contribution (from 12

x3 -e

e3 -e

∆ x3 e2

2

e1

1

x2 -e

∆ x2

∆ x1

3

x1 Figure A.5: Inﬁnitesimal Element for the Evaluation of the Divergence Eq. A.10) of the two surfaces with normal vectors e1 and −e1 is 1 [v(x + ∆x1 e1 )·e1 + v(x)·(−e1 )]dx2 dx3 lim ∆x1 ,∆x2 ,∆x3 →0 ∆x1 ∆x2 ∆x3 ∆x2 ∆x3 or 1 ∆x1 ,∆x2 ,∆x3 →0 ∆x2 ∆x3 lim

v(x + ∆x1 e1 ) − v(x) ·e1 dx2 dx3 = ∆x1 ∆x2 ∆x3

(A.11)

∆v ·e (1.12-a) 1 ∆x1 →0 ∆x1 ∂v ·e1 (1.12-b) = ∂x1 lim

hence, we can generalize div v(x) =

13

∂v(x) ·ei ∂xi

(1.13)

or alternatively

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

A.1 Vector Diﬀerentiation

A–5

∂ ∂ ∂ e1 + e2 + e3 )·(v1 e1 + v2 e2 + v3 e3 ) (1.14) ∂x1 ∂x2 ∂x3 ∂v2 ∂v3 ∂vi ∂v1 + + = = ∂i vi = vi,i (1.15) ∂x1 ∂x2 ∂x3 ∂xi

div v = ∇·v = ( =

14

The divergence of a vector is a scalar.

15

We note that the Laplacian Operator is deﬁned as ∇2 F ≡ ∇∇F = F,ii

(1.16)

Example A-2: Divergence Determine the divergence of the vector A = x2 zi − 2y 3 z 2 j + xy 2 zk at point (1, −1, 1). Solution: ∂ ∂ ∂ i+ j+ k ·(x2 zi − 2y 3 z 2 j + xy 2 zk) ∇·v = ∂x ∂y ∂z ∂x2 z ∂ − 2y 3 z 2 ∂xy 2 z + + = ∂x ∂y ∂z = 2xz − 6y 2 z 2 + xy 2

= 2(1)(1) − 6(−1) (1) + (1)(−1) = −3 at (1, −1, 1) 2

16

2

2

(1.17-a) (1.17-b) (1.17-c) (1.17-d)

By analogy to Eq. A.10, the divergence of a second-order tensor ﬁeld T is 1 ∇·T = div T(x) ≡ lim v(B)→0 v(B)

T·ndA

(1.18)

Ω

which is the vector ﬁeld ∇·T =

A.1.3

Gradient

A.1.4

Scalar

∂Tpq eq ∂xp

(1.19)

The gradient of a scalar ﬁeld g(x) is a vector ﬁeld ∇g(x) such that for any unit vector v, the directional derivative dg/ds in the direction of v is given by 17

dg = ∇g·v ds Victor Saouma

(1.20) Finite Elements II; Solid Mechanics

Draft A–6

VECTOR OPERATIONS

where v = dp ds We note that the deﬁnition made no reference to any coordinate system. The gradient is thus a vector invariant. To ﬁnd the components in any rectangular Cartesian coordinate system we use

18

v = dg ds

=

dxi dp = ei ds ds ∂g dxi ∂xi ds

(1.21-a) (1.21-b)

which can be substituted and will yield ∇g = or

∂g ei ∂xi

∂ ∂ ∂ i+ j+ k φ ∂x ∂y ∂z ∂φ ∂φ ∂φ i+ j+ k ∂x ∂y ∂z

(1.22)

∇φ ≡ =

(1.23-a) (1.23-b)

and note that it deﬁnes a vector ﬁeld. 19 The physical signiﬁcance of the gradient of a scalar ﬁeld is that it points in the direction in which the ﬁeld is changing most rapidly (for a three dimensional surface, the gradient is pointing along the normal to the plane tangent to the surface). The length of the vector ||∇g(x)|| is perpendicular to the contour lines. 20

∇g(x)·n gives the rate of change of the scalar ﬁeld in the direction of n.

Example A-3: Gradient of a Scalar Determine the gradient of φ = x2 yz+4xz 2 at point (1, −2, −1) along the direction 2i−j−2k. Solution: ∇φ = ∇(x2 yz + 4xz 2 ) = (2xyz + 4z 2 )i + (x2 zj + (x2 y + 8xz)k = 8i − j − 10k at (1, −2, −1) 1 2 2 2i − j − 2k = i− j− k n = 0 2 2 2 3 3 3 (2) + (−1) + (−2) 1 2 16 1 20 37 2 i− j− k = + + = ∇φ·n = (8i − j − 10k)· 3 3 3 3 3 3 3

(1.24-a) (1.24-b) (1.24-c) (1.24-d)

Since this last value is positive, φ increases along that direction.

Example A-4: Stress Vector normal to the Tangent of a Cylinder Victor Saouma

Finite Elements II; Solid Mechanics

Draft

A.1 Vector Diﬀerentiation

A–7

The stress tensor throughout a continuum is  3x1 x2 σ =  5x22 0

given with respect to Cartesian axes as  5x22 0 (1.25) 0 2x23  2x3 0 √ Determine the stress vector (or traction) at the point P (2, 1, 3) of the plane that is tangent to the cylindrical surface x22 + x23 = 4 at P , Fig. A.6. x3 n

x2 P

2

3 1

x

1

Figure A.6: Radial Stress vector in a Cylinder Solution: At point P , the stress tensor is given by   6 5 0 √ 0 2 3  σ= 5 √ 0 0 2 3

(1.26)

The unit normal to the surface at P is given from

At point P ,

∇(x22 + x23 − 4) = 2x2 22 + 2x3 e3

(1.27)

√ ∇(x22 + x23 − 4) = 222 + 2 3e3

(1.28)

√ 3 1 e3 n = e1 + 2 2 determined from     5 0  0   5/2  √ 1/2 3 0 2 3  = √  √   √  3/2 3 0 2 3

and thus the unit normal at P is

Thus the traction vector will be  6 σ= 5 0 √ or tn = 52 e1 + 3e2 + 3e3 Victor Saouma

(1.29)

(1.30)

Finite Elements II; Solid Mechanics

Draft A–8 21

VECTOR OPERATIONS

We can also deﬁne the gradient of a vector ﬁeld as  ∂v  [∇x v] =    [v∇x ] = 

∂vx ∂x ∂vx ∂y ∂vx ∂z ∂vx ∂x ∂vy ∂x ∂vz ∂x

y

∂x ∂vy ∂y ∂vy ∂z ∂vx ∂y ∂vy ∂y ∂vz ∂y

∂vz ∂x ∂vz ∂y ∂vz ∂z ∂vx ∂z ∂vy ∂z ∂vz ∂z

  

(1.31)

  

(1.32)

that is [∇v]ij gives the rate of change of the ith component of v with respect to the jth coordinate axis. Note the diference between v∇x and ∇x v. In matrix representation, one is the transpose of the other. 22

23

The gradient of a vector is a tensor of order 2.

24 We can interpret the gradient of a vector geometrically, Fig. A.7. If we consider two points a and b that are near to each other (i.e ∆s is very small), and let the unit vector m points in the direction from a to b. The value of the vector ﬁeld at a is v(x) and the value of the vector ﬁeld at b is v(x + ∆sm). Since the vector ﬁeld changes with position in the domain, those two vectors are diﬀerent both in length and orientation. If we now transport a copy of v(x) and place it at b, then we compare the diﬀerences between those two vectors. The vector connecting the heads of v(x) and v(x + ∆sm) is v(x + ∆sm) − v(x), the change in vector. Thus, if we divide this change by ∆s, then we get the rate of change as we move in the speciﬁed direction. Finally, taking the limit as ∆s goes to zero, we obtain

lim

∆s→0

v(x + ∆sm) − v(x) ≡ Dv(x)·m ∆s

(1.33)

v(x+∆ s m ) -v(x) v(x+∆ s m ) x3

v(x) a ∆ sm b x2

x1 Figure A.7: Gradient of a Vector The quantity Dv(x)·m is called the directional derivative because it gives the rate of change of the vector ﬁeld as we move in the direction m.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

A.2 Vector Integrals

A.2 A.2.1 25

A–9

Vector Integrals Integral of a Vector

The integral of a vector R(u) = R1 (u)e1 + R2 (u)e2 + R3 (u)e3 is deﬁned as R(u)du = e1 R1 (u)du + e2 R2 (u)du + e3 R3 (u)du

d (S(u)), then if a vector S(u) exists such that R(u) = du d (S(u)) du = S(u) + c R(u)du = du

A.2.2

(1.34)

(1.35)

Line Integral

Given r(u) = x(u)e1 + y(u)e2 + z(u)e3 where r(u) is a position vector deﬁning a curve C connecting point P1 to P2 where u = u1 and u = u2 respectively, anf given A(x, y, z) = A1 e1 + A2 e2 + A3 e3 being a vectorial function deﬁned and continuous along C, then the integral of the tangential component of A along C from P1 to P2 is given by P2 A·dr = A·dr = A1 dx + A2 dy + A3 dz (1.36) 26

C

P1

C

If A were a force, then this integral would represent the corresponding work. 27

If the contour is closed, then we deﬁne the contour integral as , A·dr = A1 dx + A2 dy + A3 dz C

28

(1.37)

C

It can be shown that if A = ∇φ then

P2

A·dr P1

,

A·dr = 0

is independent of the path C connecting P1 to P2

(1.38-a)

along a closed contour line

(1.38-b)

C

A.2.3 29

Integration by Parts

The integration by part formula is a

Victor Saouma

b

u(x)v (x)dx = u(x)v(x)|ba −

b

v(x)u (x)dx

(1.39)

a

Finite Elements II; Solid Mechanics

Draft A–10

A.2.4 30

VECTOR OPERATIONS

Gauss; Divergence Theorem

In the most general case we have

δF =

F

Ω

(1.40)

δΩ

31 The divergence theorem (also known as Ostrogradski’s Theorem) comes repeatedly in solid mechanics and can be stated as follows: (1.41) ∇·vdΩ = v.ndΓ or vi,i dΩ = vi ni dΓ

Ω

Γ

Ω

Γ

That is the vector divergence over a volume is equal to the vector ﬂux over a surface. 32

For 2D-1D transformations, we have ,

∇·qdA = A

qT nds

(1.42)

s

33

This theorem is sometime refered to as Green’s theorem in space.

34

Green-Gauss theorem

ΦvT ndΓ −

Φ∇·vdΩ = Ω

Γ

(∇Φ)T vdΩ

If we select vT = [ Ψ 0 0 ], we obtain ∂Ψ ∂Φ dΩ = ΨdΩ Φ ΦΨnxdΓ − ∂x Ω Γ Ω ∂x

A.2.5 35

(1.43)

Ω

(1.44)

Stoke’s Theorem

Stoke’s theorem states that , A·dr = (∇×A)·ndS = (∇×A)·dS C

(1.45)

S

S

where S is an open surface with two faces conﬁned by C

A.2.6 36

Green; Gradient Theorem

Green’s theorem in plane is a special case of Stoke’s theorem.

, (Rdx + Sdy) =

Γ

Example A-5: Victor Saouma

∂S ∂R − ∂x ∂y

dxdy

(1.46)

Physical Interpretation of the Divergence Theorem Finite Elements II; Solid Mechanics

Draft

A.2 Vector Integrals

A–11

Provide a physical interpretation of the Divergence Theorem. Solution: A ﬂuid has a velocity ﬁeld v(x, y, z) and we ﬁrst seek to determine the net inﬂow per unit time per unit volume in a parallelepiped centered at P (x, y, z) with dimensions ∆x, ∆y, ∆z, Fig. A.8-a. Z

D

E

C V

P(X,Y,Z) H

A

V

∆Z

V

F

V

Y

∆X B

∆Y

G

a) S

X

dV=dxdydz

n V∆t

n

dS

dS

b) c)

Figure A.8: Physical Interpretation of the Divergence Theorem

vx |x,y,z vx x−∆x/2,y,z

≈ vx

(1.47-a)

1 ∂vx ∆x 2 ∂x 1 ∂vx ∆x vx x+∆x/2,y,z ≈ vx + 2 ∂x The net inﬂow per unit time across the x planes is 1 ∂vx ∆x ∆y∆z − vx − vx + ∆Vx = 2 ∂x ∂vx ∆x∆y∆z = ∂x Similarly ≈ vx −

AFED GHCB 1 ∂vx ∆x ∆y∆z 2 ∂x

∂vy ∆x∆y∆z ∂y ∂vz ∆x∆y∆z ∆Vz = ∂z Hence, the total increase per unit volume and unit time will be given by ( ' ∂vy ∂vx ∂vz + + ∆x∆y∆z ∂x ∂y ∂z = div v = ∇·v ∆x∆y∆z ∆Vy =

Victor Saouma

(1.47-b) (1.47-c)

(1.48-a) (1.48-b)

(1.49-a) (1.49-b)

(1.50)

Finite Elements II; Solid Mechanics

Draft A–12

VECTOR OPERATIONS

Furthermore, if we consider the total of ﬂuid crossing dS during ∆t, Fig. A.8-b, it will be given by (v∆t)·ndS = v·ndS∆t or the volume of ﬂuid crossing dS per unit time is v·ndS. Thus for an arbitrary volume, Fig. A.8-c, the total amount of ﬂuid crossing a closed surface ∇·vdV (Eq. 1.50), thus

v·ndS. But this is equal to

S per unit time is S

V

∇·vdV

v·ndS = S

(1.51)

V

which is the divergence theorem.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft Appendix B

CASE-STUDY: FRACTURING of A DAM DUE TO THERMAL LOAD B.1

INTRODUCTION

The dam is a ˜60 m high arch gravity dam built in the early ﬁfties. In its original conﬁguration, the crest supported a 5 m wide road bridge. There is no indication of any malfunction until the ﬂood of 1981 occurred. At that time, the bridge abutment were clogered by trees and debris, preventing the eﬀective overtopping of the water. Following that incident, it was decided to dismantle the bridge from the dam crest, and a new one, parallel to the crest chord, was built. Soon after, a crack was observed inside the entire upper gallery, on the downstream face. Crack mouth opening displacement were recorded, and crack maps drawn. Unfortunately, readings were not taken systematically at the same pool elevation (which varies by +/- 10 m), and no clear trend can be observed. However, it should be noted that the crack opening is stable and there is no indication of increased values. Finally, no observation were conducted on the downstream face to detect the presence of any “daylighting” crack.

B.1.1

Elastic and Thermal Properties

Units adopted: Mass Length Temperature Time

Kg m oK day

Elastic properties of the concrete are given by Table B.1. For the thermal properties, we must account for the diﬀerent mixes used. The core and lower part of the dam had a mix with 180 Kg/m3 of cement, while the dam face and upper part had a mix of 280 Kg/m3 of cement. The respective thermal properties are shown in Table B.2. A key component of the Merlin program (used in this investigation) is the interface element placed along the crack in a non-linear analysis. The formulation of the interface element is

Draft B–2

CASE-STUDY: FRACTURING of A DAM DUE TO THERMAL LOAD

h ρ E ν α hair hwat

1 2,400 36 × 109 0.2 1 × 10−5 34 100

m Kg/m3 Pa m/m/o C W/m2 o C W/m2 o C

Thickness Mass density Young’s modulus Poisson’s ratio Coeﬃcient of thermal expansion Coeﬃcient for heat transfer by convection, air Coeﬃcient for heat transfer by convection, water

Table B.1: Concrete Material Properties Cement Content mz [Kg/m3 ] 180 280

Speciﬁc heat cb [J/Kg.K] 1,000 1,040

Thermal conductivity k [J/sec.m.K] J/day.m.K 2.7 233,280 2.7 233,280

Table B.2: Thermal Properties of the concrete brieﬂy described in the Appendix, and the selected properties are shown in Table B.3. fc ft Kt Kn ΦF ΦD GIF GIIF γ uDmax s1 w1 c1 cw1

25 × 106 3.75 × 106 360 × 109 360 × 109 40o 20o 400 4,000 0.3 1 × 10−2 0.94 × 106 ×10−5 1.21 × 106 6.2 × 10−4

Pa Pa Pa Pa

From Report From Whittman for ’Dam Concrete’ 10 times E 10 times E

N/m N/m

From Whittman for ’Dam Concrete’ 10 times GIF

m Pa m Pa m

ft /4 0.75 GIF /ft c/4 0.75 GII F /c

Table B.3: Interface Element Material Properties

B.1.2

Loads

Loads adopted in the analysis are shown in Table B.4.

B.2

ANALYSIS II; “Thermal Shock”

In this second analysis, we explore a diﬀerent potential cracking scenario: Thermal Shock during construction. That is we speculate that during construction, possibly in winter, the heat

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

B.2 ANALYSIS II; “Thermal Shock” Pool Level Gravity Bridge

B–3

786 m Dam only, not on bridge abutments 62,640 N/m2 (estimated weight of(concrete slab '

Thermal (After Construction) Thermal (During Construction)

(6)m(5.8)m(.3)m(2400)Kg/m3 (10)N/Kg (0.8)m(5)m Water 5o C, Dam 20o C, Air 20o C

Gallery and outside: 0 o C

Table B.4: Loads applied on the Dam of hydration caused by the 280 Kg/m3 concrete caused a thermal gradient large enough in the downstream face to induce cracking. Hence, our ﬁrst step was to create a boundary ﬁle which accounts for the diﬀerent concrete mixes used in in the dam, Fig. B.1

B.2.1

Thermal Analysis

Then, we need to undertake a transient thermal analysis, and the heat of hydration must be known. Ideally, a staged construction simulation should be undertaken, as this would result in a good estimate for the heat of hydration. In this analysis, and as a ﬁrst order approximation, we made a number of simplifying assumptions, the ﬁrst one is to consider the heat of hydration for Portland Cement type I (ASTM) at only 20o C (whereas in actuality, it is well known that this quantity is a function of the temperature, and such a model would be clearly nonlinear). In the literature (Neville), heat of hydration is often tabulated or graphed for few time increments. We adopt such set of values in Table B.5. Bold faced values constitute our starting point, and we deduce all other quantities (i.e Heat/mass/time in J/Kg/day). In our incremental analysis, we will be focusing on the ﬁrst 50-60 days, thus we must interpolate a set of heat of hydration from these selected points. This is accomplished through interpolation in Figure B.2, and extracted values are shown in Table B.6. We note that these set of values were later multiplied by 2.400 Kg/m3 as Merlin needs the heat per volume as opposed to heat per mass. Clearly, a number of simplifying assumptions were made so far, and in addtion to further simplify the analysis we assume: 1. Consider only zones 6, 7, 8 and 9. 2. Assume that at the interfaces 6-3 and 8-5 the temperature is 25o C, whereas 7-4 it is 20o C 3. Assume that the outside temperature, and the one of the gallery is equal to 0o C 4. Do not perform an incremental analysis (i.e. simulating staged construction) 5. Perform a transient analysis with time increment equal to one day. Heat of hydration values are taken from day 10 to 60. 6. Assume uniform heat of hydration within the considered zones. 7. Diﬀerentiate between the two concrete mixes.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft B–4

CASE-STUDY: FRACTURING of A DAM DUE TO THERMAL LOAD

35 34 33 56 32 58 57 55

59 36 60 37

31 54

9

61 19

51 18

52

30

24

2953 30 49 34

35 6 31 29 36 46 15 28 14 18 3 17 23 19 13

33

7

50 48 8

32 22

47 16 26

4

21

16

20

14 12

13

15

17 25 27

40 39 26

5

24 11 12

2

38

27 28

10

41 25 24 42 23 43 22

37

44

11 10

20 45

7

8

21

7 8 6 9

1

9 1

6

2

1

5

2 2

3

3

4

4

5

Figure B.1: Boundary Description of Dam for Transient Thermal Analysis

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

B.2 ANALYSIS II; “Thermal Shock”

Age [days]

Hc [J/g]

0 1.5 3 5 7 17.5 28 59 90 227.5 365 1,368.5 2,372

0

B–5

Hc [J/g.day]

Hb = Hc mz /ρb [J/Kg.day] mz 180 280

85.000

6,375

9,917

20.000

1,500

2,333

3.09524

232.14

361.11

0.48387

36.29

56.45

0.10909

8.18

12.73

0.01495

1.12

1.74

255 335 400 430 460 490

Table B.5: Heat of Hydration From the Literature

Days 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

M-180 510 480 440 410 380 340 310 280 250 220 200 180 165 155 140 130 120 119 110 105

M-280 800 745 690 640 580 530 480 440 380 340 300 275 245 230 220 200 190 180 178 165

Days 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

M-180 100 97 94 90 87 85 80 77 75 72 70 67 65 63 61 59 57 55 53 52

M-280 160 154 147 142 135 128 122 115 110 105 100 95 92 87 85 82 80 77 75 73

Days 50 51 52 53 54 55 56 57 58 59

M-180 50 48 47 45 45 43 42 40 38 37

M-280 70 68 66 65 63 62 60 59 59 57

Table B.6: Heat of Hydration Adopted in the Simulation; Days and J/Kg/Day

Victor Saouma

Finite Elements II; Solid Mechanics

Draft B–6

CASE-STUDY: FRACTURING of A DAM DUE TO THERMAL LOAD

Figure B.2: Heat of Hydration Interpolations

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

B.2 ANALYSIS II; “Thermal Shock”

B–7

Contour Plot, Temperatures, Temperatures 3.00e+1

2.70e+1

2.40e+1

2.10e+1

1.80e+1

1.50e+1

1.20e+1

9.00e+0

6.00e+0

3.00e+0 Y X

-2.25e-3

Z

Figure B.3: Temperature Distribution in the Transient Thermal Analysis at Day 8 When such an analysis was performed, it was observed that the concrete temperatures were very small (about 1-2o C). This was attributed to the lack of incremental analysis, of having one single large ”lift” being investigated, and with an initial temperature of zero, not enough heat was released to warm the temperature. Clearly this is erroneous. To qualitatively alleviate for this limitation (caused by lack of incremental analysis), and keeping in mind that we are primarily interested at this point in determining qualitatively a mechanism which may cause cracking, the speciﬁc heat was reduced by a factor of 10 (no other reduction factors were attempted). With such a set of values, the largest temperature gradient was observed to occur at day 8 with the temperature distribution shown in Fig. B.3. We observe the thermal gradient along the crack line.

B.2.2

Stress Analysis

With the nodal temperature extracted from the previous analysis, a stress analysis was next undertaken. In this analysis, the only loads considered were thermal and gravity (together in one increment). We observe from Fig. B.4 that in this case, the crack did open and the magnitude of the maximum crack opening displacement is about 0.8 mm. The crest displacements are: -4.1 mm (horizontal) and 1.6 mm (vertical) Normal and shear stresses, as well as crack opening and sliding displacements are shown in Table B.7 and B.8.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft B–8

CASE-STUDY: FRACTURING of A DAM DUE TO THERMAL LOAD

Contour Plot, Principal Stresses, Maximum 1.19e+7

1.05e+7

9.03e+6

7.58e+6

6.14e+6

4.69e+6

3.24e+6

1.80e+6

3.53e+5

-1.09e+6 Y

-2.54e+6

X Z

Figure B.4: Maximum Principal Stresses and Deformed Mesh at Day 8

Crack Lig. 0.000E+00 4.168E-01 6.947E-01 1.228E+00 1.317E+00 1.690E+00 1.873E+00 2.205E+00 2.537E+00 2.664E+00 2.996E+00 3.059E+00 3.455E+00 3.455E+00 3.959E+00 4.296E+00 4.633E+00 4.969E+00

σn -1.136E+06 -8.677E+05 -9.974E+05 -1.112E+06 -1.123E+06 -7.825E+05 1.907E+05 5.690E+05 8.226E+05 7.725E+05 7.024E+05 6.063E+05 5.663E+05 4.615E+05 3.259E+05 1.699E+05 3.613E+04 3.750E-04

σt -1.341E+06 -1.264E+06 -9.699E+05 -7.180E+05 -7.853E+05 -1.284E+06 -6.107E+05 -5.777E+05 -5.414E+05 -4.123E+05 -3.491E+05 -2.992E+05 -2.918E+05 -2.169E+05 -1.364E+05 -6.244E+04 -1.220E+04 -1.098E-04

Table B.7: Sresses Along the Interface Element; m] and [Pa]

Victor Saouma

Finite Elements II; Solid Mechanics

Draft

B.3 CONCLUSIONS Crack Lig. 0.00000E+00 2.77864E-01 5.55728E-01 1.08911E+00 1.17780E+00 1.55140E+00 1.73377E+00 2.06576E+00 2.39775E+00 2.52472E+00 2.85671E+00 2.92020E+00 3.31568E+00 3.31568E+00 3.65225E+00 3.98882E+00 4.32540E+00 4.66197E+00

B–9 Crack Opening -3.73773E-06 -1.99149E-06 -2.75825E-06 -3.21438E-06 -3.30575E-06 -2.63906E-06 4.83795E-06 3.44441E-05 1.05330E-04 1.39448E-04 2.40532E-04 2.63296E-04 3.22496E-04 3.95863E-04 4.98598E-04 6.05647E-04 7.18990E-04 8.19540E-04

Crack Sliding -3.78857E-06 -3.59965E-06 -2.88216E-06 -1.65068E-06 -1.47816E-06 -4.04809E-06 -1.74152E-05 -3.95564E-05 -6.29998E-05 -7.66532E-05 -1.16057E-04 -1.25226E-04 -1.71896E-04 -1.67985E-04 -2.02377E-04 -2.19765E-04 -2.29582E-04 -2.30053E-04

Table B.8: Crack Opening and Sliding Displacements; [m]

B.2.3

Data Files

The data ﬁles t10.bd t12.inp t13.dat t14.inp

B.3

used in this analysis are given below. Master ﬁle contains all the various options Thermal analysis File containing the nodal temperature at day 8 Stress analysis

CONCLUSIONS

The following conclusions may be drawn from this preliminary analysis 1. There are two potential reasons for which cracking might have occurred: (a) Thermal gradient between cold water and warm downstream air temperature aggravated by the removal of the bridge which was acting as a corbel on a cathedral buttress (i.e. its dead weight was oﬀsetting ﬂexural tensile stresses). The analysis conducted to investigate this scenario was quite reliable quantitatively. (b) “Thermal shock” during construction. The analysis conducted toward such a cause is mostly qualitative. A more rigorous incremental transient thermal analysis would be needed to further support this theory. However, this analysis may be quite complex and may be outside the scope of this investigation. 2. Diﬀerential settlement was ruled out as there was no indication of foundation measurable excessive deformation.

Victor Saouma

Finite Elements II; Solid Mechanics

Draft B–10

CASE-STUDY: FRACTURING of A DAM DUE TO THERMAL LOAD

3. A three dimensional analysis has been initiated (hydrostatic load), but due to limitation of the author’s PC, it was not extended to the thermal analysis. It would be interesting to contrast results of such an analysis with its 2D counterpart. 4. Through these investigations, some of the possibilities oﬀered by “modern” computational tools which address cracking were exercised. 5. It would be interesting to determine if indeed upstream cracks are present in the ﬁeld (as this investigation predicts).

Victor Saouma

Finite Elements II; Solid Mechanics

Draft Appendix C

MISC. C.1

Units & Conversion Factors length, m (meter) Force, N (Newton) Mass, Kg (kilogram) Density, Kg/m3 Temperature, T Acceleration, m/s2 Stiﬀness, N/m Stress, Pa = N/m2 Work, energy, N-m=Joule Power, J/s=W Convection coeﬃcient, h Heat, J Heat Source/Sink, Q Heat ﬂux (q) Speciﬁc heat, c Thermal conductivity, k

1 inch = 0.0254 m; 1 m = 39.37 inch 1 lb = 4.4482 N; 1 N = 0.22481 lb 1 lbm = 0.45359 Kg; 1 Kg=2.2046 lb 1 lbm/ft3 = 16.018 Kg/m3 ;1 Kg/m3 =0.062428 lbm/ft3 T o F=[(9/5)To C+32] 1 in/s2 = 0.0254 m/s2 ; 1 lb/in = 175.1 N/m 1 psi = 6,894.8 Pa; 1 MPa = 145.04 psi 1 ft-lbf= 1.3558 J; 1 J = 0.73756 ft- lbf Heat Transfer 1 Btu/h.ft2 .o F = 5.6783 W/m2 .o C 1 Btu=1055.06 J; 1 Btu = 778.17 ft-lb W/m3 = 1 Btu/h.ft2 = 3.1546 W/m2 1 Btu/o F = 1,899.108 J/o C 1 Btu/h.ft.o F = 1.7307 W/m.o C Seepage Flow

permeability, k Stress intensity factor, K Fracture energy GF

Fracture Mechanics √ √ 1 MPa m=1.099 ksi in 1 lb/in =.0057 N/m;

Draft C–2

C.2

MISC.

Metric Preﬁxes and Multipliers Preﬁx tera giga mega kilo hecto deca deci centi milli micro nano pico

Victor Saouma

Abbreviation T G M k h da d c m µ n p

Multiplier 1012 109 106 103 102 10 10−1 10−2 10−3 10−6 10−9 10−12

Finite Elements II; Solid Mechanics