Elasticity Theory Basics



Similar documents
State of Stress at Point

[1] Diagonal factorization

Physics 235 Chapter 1. Chapter 1 Matrices, Vectors, and Vector Calculus

Linear algebra and the geometry of quadratic equations. Similarity transformations and orthogonal matrices

by the matrix A results in a vector which is a reflection of the given

x = + x 2 + x

Notes on Orthogonal and Symmetric Matrices MENU, Winter 2013

Lecture L3 - Vectors, Matrices and Coordinate Transformations

Scalars, Vectors and Tensors

Electromagnetism - Lecture 2. Electric Fields

When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid.

Unified Lecture # 4 Vectors

Lectures notes on orthogonal matrices (with exercises) Linear Algebra II - Spring 2004 by D. Klain

Chapter 6. Orthogonality

LS.6 Solution Matrices

University of Lille I PC first year list of exercises n 7. Review

Finite Element Formulation for Plates - Handout 3 -

Data Mining: Algorithms and Applications Matrix Math Review

Similarity and Diagonalization. Similar Matrices

The elements used in commercial codes can be classified in two basic categories:

Linear Algebra Notes for Marsden and Tromba Vector Calculus

1 of 79 Erik Eberhardt UBC Geological Engineering EOSC 433

The Basics of FEA Procedure

CBE 6333, R. Levicky 1 Differential Balance Equations

Chapter 7: Polarization

A Primer on Index Notation

MATH 423 Linear Algebra II Lecture 38: Generalized eigenvectors. Jordan canonical form (continued).

Bending Stress in Beams

Introduction to Seismology Spring 2008

Linear Algebra Notes

Introduction to Matrix Algebra

v w is orthogonal to both v and w. the three vectors v, w and v w form a right-handed set of vectors.

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS

9 MATRICES AND TRANSFORMATIONS

Plates and Shells: Theory and Computation - 4D9 - Dr Fehmi Cirak (fc286@) Office: Inglis building mezzanine level (INO 31)

Chapter 22: The Electric Field. Read Chapter 22 Do Ch. 22 Questions 3, 5, 7, 9 Do Ch. 22 Problems 5, 19, 24

3 Concepts of Stress Analysis

Solving Simultaneous Equations and Matrices

Chapter 17. Orthogonal Matrices and Symmetries of Space

is in plane V. However, it may be more convenient to introduce a plane coordinate system in V.

Lecture 8 : Coordinate Geometry. The coordinate plane The points on a line can be referenced if we choose an origin and a unit of 20

8 Hyperbolic Systems of First-Order Equations

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS. + + x 2. x n. a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 a 31 a 32 a 3n b 3. a m1 a m2 a mn b m

Recall that two vectors in are perpendicular or orthogonal provided that their dot

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS

LINEAR ALGEBRA W W L CHEN

Figure 1.1 Vector A and Vector F

6 J - vector electric current density (A/m2 )

x1 x 2 x 3 y 1 y 2 y 3 x 1 y 2 x 2 y 1 0.

CITY UNIVERSITY LONDON. BEng Degree in Computer Systems Engineering Part II BSc Degree in Computer Systems Engineering Part III PART 2 EXAMINATION

Analysis of Stresses and Strains

Lecture 3 Fluid Dynamics and Balance Equa6ons for Reac6ng Flows

The Matrix Elements of a 3 3 Orthogonal Matrix Revisited

Problem Set 5 Due: In class Thursday, Oct. 18 Late papers will be accepted until 1:00 PM Friday.

Numerical Analysis Lecture Notes

The Characteristic Polynomial

Lecture 1: Schur s Unitary Triangularization Theorem

Feature Commercial codes In-house codes

8.2 Elastic Strain Energy

MATH10212 Linear Algebra. Systems of Linear Equations. Definition. An n-dimensional vector is a row or a column of n numbers (or letters): a 1.

CS3220 Lecture Notes: QR factorization and orthogonal transformations

Au = = = 3u. Aw = = = 2w. so the action of A on u and w is very easy to picture: it simply amounts to a stretching by 3 and 2, respectively.

Cross product and determinants (Sect. 12.4) Two main ways to introduce the cross product

Structural Axial, Shear and Bending Moments

Copyright 2011 Casa Software Ltd. Centre of Mass

Orthogonal Projections

Bindel, Spring 2012 Intro to Scientific Computing (CS 3220) Week 3: Wednesday, Feb 8

Linear Algebra Review. Vectors

Stress Analysis, Strain Analysis, and Shearing of Soils

DYNAMIC ANALYSIS OF THICK PLATES SUBJECTED TO EARTQUAKE

arxiv: v2 [physics.acc-ph] 27 Oct 2014

Numerical Analysis Lecture Notes

CONTINUUM MECHANICS. (Lecture Notes) Zdeněk Martinec

Stress Strain Relationships

ASEN Structures. MDOF Dynamic Systems. ASEN 3112 Lecture 1 Slide 1


11.1. Objectives. Component Form of a Vector. Component Form of a Vector. Component Form of a Vector. Vectors and the Geometry of Space

Mechanics lecture 7 Moment of a force, torque, equilibrium of a body

Differential Relations for Fluid Flow. Acceleration field of a fluid. The differential equation of mass conservation

Methods for Finding Bases

Unit 3 (Review of) Language of Stress/Strain Analysis

Lecture 2: Homogeneous Coordinates, Lines and Conics

Let s first see how precession works in quantitative detail. The system is illustrated below: ...

1 Symmetries of regular polyhedra

Math 312 Homework 1 Solutions

Probability and Statistics Prof. Dr. Somesh Kumar Department of Mathematics Indian Institute of Technology, Kharagpur

Vector Calculus: a quick review

13 MATH FACTS a = The elements of a vector have a graphical interpretation, which is particularly easy to see in two or three dimensions.

Geometric Transformations

1 Introduction to Matrices

Plane Stress Transformations

Physics 9e/Cutnell. correlated to the. College Board AP Physics 1 Course Objectives

Mechanical Properties - Stresses & Strains

n 2 + 4n + 3. The answer in decimal form (for the Blitz): 0, 75. Solution. (n + 1)(n + 3) = n lim m 2 1

Chapter 22: Electric Flux and Gauss s Law

Objectives. Experimentally determine the yield strength, tensile strength, and modules of elasticity and ductility of given materials.

Chapter Outline. Mechanical Properties of Metals How do metals respond to external loads?

Algebra 2 Chapter 1 Vocabulary. identity - A statement that equates two equivalent expressions.

1. Fluids Mechanics and Fluid Properties. 1.1 Objectives of this section. 1.2 Fluids

Vector Algebra and Calculus

Transcription:

G22.3033-002: Topics in Computer Graphics: Lecture #7 Geometric Modeling New York University Elasticity Theory Basics Lecture #7: 20 October 2003 Lecturer: Denis Zorin Scribe: Adrian Secord, Yotam Gingold Introduction This lecture is the summary of the basics of the elasticity theory. We will discuss the fundamental concepts of stress and strain, and the equations describing the deformation of an elastic object in equilibrium. We will focus on materials following the linear material law. The linear material law can be thought of as a generalization Hooke s law f = kx to 3D deformable objects. The most straightforward generalization would be to state that stretching or compressing an object in any direction results in an elastic force in the same direction which is proportional to the extension. However, this would match well only a small class of materials. In three dimensions in most cases a stretch in one direction changes the dimensions in the perpendicular directions. The two aspects of deformation are characterized using two constants: the Young s modulus, which corresponds to the constant in Hooke s law, and is denoted E or Y, and the Poisson s ratio, which is denoted ν. The Poisson ratio is the ratio of stretch to perpendicular change. To formulate the equations of balance for elastic objects and define the linear material law more precisely, we start with a discussion of the counterparts of forces and displacements in the equations of elasticity: strain and stress tensors. Stress and Strain There are two kinds of forces acting on a small volume of a body: 1. olume forces f is the volume force density; df vol = fd is the differential volume force, where d is the differential volume element. 2. Surface forces t(n) the vector surface force density, where n is the normal to the differential surface element da; t(n) is also called the stress vector; df surf = t(n)da is the differential surface force, acting on the differential surface element da.

2 G22.3033-002: Lecture #7 Stress tensor As stated, the stress vector t(n) can have any form at all, but it turns out that it can be expressed in the special form t(n) = T n, where T is called the Cauchy stress tensor. An aside on tensors. 2D tensors are matrices which follow certain transformation rules when we change a coordinate system. Typically these rules follow from the definition of the quantity described by the matrix. In the case of the stress tensor, this can be seen as follows. Let A be the matrix representing a coordinate transformation x = Ax. Under an arbitrary linear change of coordinates, normals are not generally transformed as points or displacement vectors. The correct transformation rule for normals can be derived if we keep in mind that if n is the normal to a plane P the transformed normal n should remain perpendicular to the transformed plane P. Then it is easy to derive that n = A T n. Note that in the case of rigid transformations we have A 1 = A T, so A T = A. On the other hand force vectors are transformed as all other vectors. Using these considerations, we have t (n) = T n At(n) = T A T n t(n) = A 1 T A T n which implies that T = A 1 T A T or T = AT A T. So the stress tensor is transformed using T = AT A T. Back to the stress tensor. In general, the stress tensor is T = σ 11 σ 12 σ 13 σ 21 σ 22 σ 23 σ 31 σ 32 σ 33, (1) which means that, for example, T î = σ 11 σ 12 σ 13 σ 21 σ 22 σ 23 σ 31 σ 32 σ 33 1 0 0 = σ 11 σ 21 σ 31 (2) is the force if the normal to the surface area points along the x-axis. We assume that the volume torques will balance, i.e. the volume forces create no net torque. This is generally assumed in elasticity. One can derive from this constraint that T is symmetric and has only six distinct elements.

G22.3033-002: Lecture #7 3 An aside on symmetric matrices. matrix S are As an aside, some properties of a general symmetric The eigenvalues are always real (and as a consequence the eigenvectors have real components); R T SR is diagonal for some (rotation) matrix R, which means that S = R T DR, where D is a diagonal matrix with the eigenvalues of S along its diagonal. This means that we can always find some change of basis that transforms T into a nice diagonal form σ 1 T = σ 2. (3) The σ i s are called the principal stresses. Choosing these coordinates brings us closer to our goal of defining the generalization of Hooke s law for 3D materials. The second ingredient that we need is the measure of the deformation. σ 3 Strain tensor The strain tensor measures the change of distances between close points in the deformed state with respect to the distances in the undeformed state. Let ϕ : Ω R 3 be the mapping from the reference domain Ω to the deformed state. Compare the distance between two points that are close together in the reference frame, (x + dx) x 2, with the distance between them in the deformed frame ϕ(x + dx) ϕ(x) 2. We have in the reference frame and (x + dx) x 2 = dx 2 = dx T dx ϕ(x + dx) ϕ(x) 2 = (ϕ(x + dx) ϕ(x)) T (ϕ(x + dx) ϕ(x) (4) in the deformed frame. For small deformations, we can approximate ϕ(x + dx) with ϕ(x) + ϕdx, so (4) becomes (ϕ(x + dx) ϕ(x)) T (ϕ(x + dx) ϕ(x) ( ϕdx) T ( ϕdx) = dx T ϕ T ϕdx Aside: the gradient of a vector is a matrix. Specifically, if ϕ(z) = (ϕ 1 (z), ϕ 2 (z), ϕ 3 (z)) T, then ϕ = ϕ 1 ϕ 2 ϕ 3.

4 G22.3033-002: Lecture #7 The quantity ϕ T ϕ is called the Cauchy-Green deformation tensor. Finally, if we take the difference between the reference and deformed squared distances, we get dx T ϕ T ϕdx dx T dx = dx T ( ϕ T ϕ I)dx The quantity ϕ T ϕ I is twice the strain tensor. Note that strain is non-dimensional. Note that it is a matrix, and it is easy to derive its transformation rules, which end up the same as for the stress tensor. It follows directly from the definition that this tensor is symmetric: ɛ = ɛ 11 ɛ 12 ɛ 13 ɛ 12 ɛ 22 ɛ 23 ɛ 13 ɛ 23 ɛ 33 We can also choose a coordinate system in which the strain is diagonal: ɛ 1 ɛ = ɛ 2. The diagonal elements are called principal strains. This mathematical fact has important physical meaning: any deformation locally can be regarded as a simple stretch/compression along three perpendicular directions. To get a better idea about the meaning of the principal strains, let s consider the change in length along a principal direction; according to the definition of strain, the new squared length is given by (dx 1 0, 0)(2ɛ + I)(dx 1 0, 0) T, i.e. ɛ 3, ( dx1 0 0 ) 2ε 1 + 1 2ε 2 + 1 2ε 3 + 1 dx 1 0 0 = (2ε 1 + 1)dx 2 dx 2 1 Assuming small strains, 2ɛ 1 + 1 1 + ɛ 1. So dx 1 dx 1 dx 1 the relative change in length in the x 1 direction. = ε 1, where the first quantity is Equations of balance We will focus on elastostatic equations. Elasto-static equations are simpler to understand than elastodynamic; at the same time, it is relatively easy to generalize elastostatic equations. These equations are derived from the equations of the balance of force and the balance of torques: fi = 0 sum of forces on body ri f i = 0 sum of moments of forces on body and can be written either in the integral or differential form. We start with the the integral form. We examine a small region of a deformed object d. d denotes its surface.

G22.3033-002: Lecture #7 5 If f is the volume force density, f d + t(n)ds = d } {{ } integrate over surface which must be zero to stay balanced. Given two volumes in contact along surface element da, f d + d T n ds (5) the two surface forces with which the volumes act on each other are the same. In this way all internal surface forces cancel out. From equation (5), summing over all small volumes into which we have partitioned our object, we get the balance of forces equation: f d + T n ds = 0 Similarly, the torques need to stay balanced: r f d + r T n ds = 0 To get the differential form of these equations, we use the Gauss Integral Theorem: T n ds = div T d, where (div T ) is the divergence of T.

6 G22.3033-002: Lecture #7 An aside on tensor divergence. FOr a vector function f 1 (x) f = f 2 (x), f 3 (x) divergence is scalar: div f = f 1 x 1 + f 2 x 2 + f 3 x 3. For a tensor, divergence is vector (sum over each column). Returning to our two equations, f d + T n ds = 0 (6) r f d + r T n ds = 0 (7) From equation (6), (f + div T ) d = 0 f + div T = 0 The last equation is the differential form of the force balance. Recall that the symmetry of the stress tensor was a consequence of the torque balance, so the torque equation is captured by symmetry of the stress tensor and the boundary conditions. The stress of the boundary is equal to the external stress: T = T ext on (stress continuity). An important fact to mention is that these equations are written with respect to the coordinates in the deformed object. In practice, it is much more convenient to use the coordinates in the undeformed object to write these equations, as typically the deformation is unknown and needs to be computed. Material laws, a.k.a. constitutive equations. As we have mentioned in the beginning, we are primarily interested in materials following a generalization of Hooke s law. In general the elastic material behavior is captured by writing the strain as a function of stress. We fix the coordinate system in which the stress tensor Σ is diagonal. In this coordinate system the forces acting on the elementary volume act along the coordinate axes; for each axis, the forces acting on opposite sides are equal in magnitude and opposite in direction.

G22.3033-002: Lecture #7 7 s 2 s 3 s 1 s 1 s 3 s 2 In this coordinate system, the linear material law has the form ɛ i = 1 E σ i ν E (σ j + σ k ) (8) where E is the Young modulus and ν is the Poisson ratio. The first term in the formula reflects the assumption that the strain in a given direction depend linearly on the stress in the same direction. The second term takes into account the deformation due to the stresses in other directions, taking into account the degree to which the object can compress. For example, for incompressible objects ν = 1/2, and the volume is preserved. While this is a simple an intuitive way to write these equations, this requires using a coordinate system which might change from one point to another. In general, we need the form of these equations in an arbitrarily chosen coordinate system. A simple way to arrive at such a form is to express everything in terms of tensor invariants. An aside on tensor invariants. What are invariants of a symmetric 3D tensor? (In other words, functions of matrix elements that do not change when the coordinates change.) Eigenvalues can be regarded as invariants as we look at the whole set (the order may change as we change the coordinate system). It is more convenient to use symmetric functions of eigenvalues: λ 1 + λ 2 + λ 3 = Tr ɛ = ɛ ii λ 1 λ 2 λ 3 = det ɛ λ 1 λ 2 + λ 2 λ 3 + λ 3 λ 1 = Tr ɛ 2 = Tr λ 2 1 λ 2 2 = 1 2 [ (Tr ɛ) 2 ( Tr ɛ 2)] Intuitively, one can think of the action of the deformation on a small sphere. The result is an ellipsoid with major axes lengths proportional to eigenvalues. Then λ 2 3 λ 1 + λ 2 + λ 3 = 3 average cross section λ 1 λ 2 λ 3 = volume λ 1 λ 2 + λ 2 λ 3 + λ 3 λ 1 = 3 length of diagonal

8 G22.3033-002: Lecture #7 Equation (8) can be rewritten as Since σ i + σ j + σ k = Tr Σ, we can write ɛ i = 1 + ν E σ i ν E (σ i + σ j + σ k ). ɛ = 1 + ν E σ ν (Tr Σ) I (9) E This is the coordinate-independent form of the material linear law. Materials that follow this law are called St. enant-kirchhoff materials. Note that this law is more universal than it might seem: most isotropic elastic materials are quite well described by it. The reason for this is that as long as the strain is a function of stress, in most cases for small strains it is approximated well by the linear term in its Taylor series expansion. But strains tend to be small in most cases, excluding highly deformable objects such as rubber. Lamé constants We can invert equation (9) to write Σ as a function of ɛ. For this, we need to express the trace of stress in terms of the trace of strain, which can be done by taking the trace of both sides of the linear material law: Tr ɛ = 1 + ν 3ν Tr Σ E E Tr Σ Substituting the resulting expressions back, we obtain an expression of the form Σ = λ(tr ɛ)i + 2µɛ. Where the constant λ and µ are called Lamé constants: Summary and f + div T = 0 Σ = λ(tr ɛ)i + 2µɛ The stress tensors in these equations are denoted with different letters for a reason. The first equation is written with respect to the deformed body, and the constitutive law is written with variables specified in the original body coordinates. So these stress tensors are different: the second equation uses what is called the second Piola-Kirchhoff tensor and these two tensors are related by Σ = (det φ) 1 φt φ T

G22.3033-002: Lecture #7 9 Fortunately, the form of the equation of the balance of forces does not change its form when we switch to the undeformed body quantities. The unknowns in these equations are the deformations (recall that the stress is defined using the gradient of the deformation function) and the stress tensor. The deformations are defined by three coordinate functions, and the stresses by six, so we have a total of nine unknown functions. The force balance results in three equations with the constitutive law providing additional six, so now the system of equations is complete. Of course, this simple counting argument does not prove that the solution exists or is unique.