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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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


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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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Draft CONTENTS

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

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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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

LOAD . . . . . . . . . . . . . . . . . . . . . . . . .

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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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2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12

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

Voronoi and Delaunay Tessellation . . . . . . . . . . . . . . . . . . . . . . . . Control Point for a 2D Mesh . . . . . . . . . . . . . . . . . . . . . . . . . . . Control Point for a 3D Mesh . . . . . . . . . . . . . . . . . . . . . . . . . . . A Two Dimensional Triangularization AlgorithmControl Point for a 3D Mesh

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LIST OF FIGURES

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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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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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

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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


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


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


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


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


Draft 0–6

Victor Saouma

LIST OF TABLES


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


Draft 0–8

Victor Saouma

LIST OF TABLES


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


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


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 {∆}


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


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


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


Draft


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)


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


Draft


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


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


Draft


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)


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)


Draft


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


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

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)


Draft


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)


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


Draft


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


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


Draft


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


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
> > =

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 > > = > > ;


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 0

=

8 > >
> : 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


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


−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


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 > >
> > =

> > > :

> > > ;

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 > > >
> > =

> > > :

> > > ;

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 > > >
> > > =

> > > > : 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 > > > > > = > > > > > ;


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


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


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) = 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


Draft


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)


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


Draft


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


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


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


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)


Draft

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)


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


Draft


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


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)


Draft


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


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 ï =

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)


Draft


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)


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)


Draft


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)


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


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


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)


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


Draft 2–24

Victor Saouma

INTRODUCTION


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


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)


Draft 3–4


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)


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


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


Draft


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


Draft 3–8


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)


Draft


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


Draft


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


Draft 3–12

3.2.3.1


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)


Draft


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)


Draft 3–14


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)


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 =


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.

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Draft


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 − tî = 0 Natural B.C.

(3.80)

where bi and tî 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 tî 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

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Draft 3–18


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 = tî + 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.

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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 = tî + 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

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Draft 3–20

3.3.1 89


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


Draft


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


Draft 3–22


✻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


Draft


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


Draft 3–24


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


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


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


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


Draft 3–28

Victor Saouma



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


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


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


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


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


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


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


Draft 5–4


ε

u

u

u

σ

b

^ t

û

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


Draft


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


Draft 5–6


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)


Draft


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)


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)


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


Draft 5–10


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


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


Draft 5–12


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


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


Draft 5–14


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


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)


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)


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


Draft 5–18

Victor Saouma



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


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


Draft 6–4


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)


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


Draft 6–6


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


Draft


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)


Draft 6–8


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)


Draft


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


Draft 6–10


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


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)


Draft 6–12


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


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)


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)


Draft 6–16


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


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


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


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


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


Draft 7–6


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


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


Draft 7–8


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


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


Draft 7–10

7.4


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 > > > > < 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
> 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 > > > > >
> > > > : |

8 > > > > >
6 > > 6 εyy > > = 6 6 εzz =6 6 εxy > > 6 > > εxz > ; 6 4 εyz {z } | "

Draft PLATES


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)


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


Draft


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


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


Draft


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


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)


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


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


Draft


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


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)


Draft


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


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


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


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


Draft 16–4


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


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


Draft 16–6


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


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


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)


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


Draft 16–10

16.2.3


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


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


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


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


Draft 16–14


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)


Draft


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


Draft 16–16


Figure 16.11: Flowchart for an incremental nonlinear finite element program with indirect displacement control

Victor Saouma


Draft


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


Draft 16–18


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


Draft


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


Draft 16–20


Figure 16.13: Flow chart for line search with IDC methods

Victor Saouma


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 Sphère 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 Paramètres Continus à la Théorie 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


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


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


Draft


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


Draft


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)


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


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


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)


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


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


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


Draft B–4


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


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


Draft B–6


Figure B.2: Heat of Hydration Interpolations

Victor Saouma


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


Draft B–8


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


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


Draft B–10


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


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


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