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Continuum Mechanics and Elements of Elasticity Structural Mechanics - Victor E.Saouma

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Lecture notes by Victor E. Saouma (Univ. of Colorado, Boulder), written during a 1997-98 sabbatical at EPFL Lausanne for second-year materials students. The contents cover vectors and tensors, stress and strain, general principles and constitutive relations, elasticity and beam theory, variational methods (virtual work, Rayleigh-Ritz), and theoretical strength of solids. It is a third-party reference kept in Phil's tensor support files.

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Draft DRAFT Lecture Notes Introduction to CONTINUUM MECHANICS and Elements of Elasticity/Structural Mechanics c/circlecopyrtVICTOR E. SAOUMA Dept. of Civil Environmental and Architectural Engineering University of Colorado, Boulder, CO 80309-0428 Draft0–2 Victor Saouma Introduction to Continuum Mechanics Draft0–3 PREFACE Une des questions fondamentales que l’ing´ enieur des Mat´ eriaux se pose est de conna ˆitre le comporte- ment d’un materiel sous l’eﬀet de contraintes et la cause de sa rupture. En d´ eﬁnitive, c’est pr´ ecis´ement la r´eponse `a c/mat es deux questions qui vont guider led´ eveloppement de nouveauxmat´ eriaux, et d´ eterminer leur survie sous diﬀ´ erentes conditions physiques et environnementales. L’ing´enieur en Mat´ eriaux devra donc poss´ eder une connaissance fondamentale de la M´ ecanique sur le plan qualitatif, et ˆ etre capable d’eﬀectuer des simulations num´ eriques (le plus souvent avec les El´ ements Finis) et d’en extraire les r´ esultats quantitatifs pour un probl` eme bien pos´ e. Selon l’humble opinion de l’auteur, ces nobles buts sont id´ ealement atteints en trois ´ etapes. Pour commencer, l’´ el`eve devra ˆ etre confront´ ea u xp r i n c i p e sd eb a s ed el aM ´ ecanique des Milieux Continus. Une pr´esentation d´ etaill´ee des contraintes, d´ eformations, et principes fondamentaux est essentiel. Par la suite une briefe introduction ` a l’Elasticit´ e( a i n s iq u ’ ` al at h ´eorie des poutres) convaincra l’´ el`eve qu’un probl`eme g´en´eral bien pos´ e peut avoir une solution analytique. Par contre, ceci n’est vrai (` a quelques exceptions prˆ ets) que pour des cas avec de nombreuses hypoth` eses qui simpliﬁent le probl` eme (´elasticit´e lin´eaire, petites d´ eformations, contraintes/d´ eformations planes, ou axisymmetrie). Ainsi, la troisi` eme et derni`ere ´etape consiste en une briefe introduction ` al aM ´ecanique des Solides, et plus pr´ ecis´ement au Calcul Variationel. A travers la m´ ethode des Puissances Virtuelles, et celle de Rayleigh-Ritz, l’´ el`eve sera enﬁn prˆ et `a un autre cours d’´ el´ements ﬁnis. Enﬁn, un sujet d’int´ erˆet particulier aux ´ etudiants en Mat´eriaux a ´ et´ea j o u t ´e, `a savoir la R´ esistance Th´ eorique des Mat´ eriaux cristallins. Ce sujet est capital pour une bonne compr´ ehension de la rupture et servira de lien ` au n´eventuel cours sur la M´ ecanique de la Rupture. Ce polycopi´ ea´et´ee n t i `erement pr´ epar´e par l’auteur durant son ann´ ee sabbatique ` a l’Ecole Poly- technique F´ ed´erale de Lausanne, D´ epartement des Mat´ eriaux. Le cours ´ etait donn´ ea u x´etudiants en deuxi`eme ann´ee en Fran¸ cais. Ce polycopi´ ea´et´e´ecrit avec les objectifs suivants. Avant tout il doit ˆ etre complet et rigoureux. A tout moment, l’´ el`eve doit ˆ etre `am ˆeme de retrouver toutes les ´ etapes suivies dans la d´ erivation d’une ´equation. Ensuite, en allant ` a travers toutes les d´ erivations, l’´ el`eve sera ` am ˆeme de bien conna ˆitre les limitations et hypoth` eses derri` ere chaque model. Enﬁn, la rigueur scientiﬁque adopt´ ee, pourra servir d’exemple ` a la solution d’autres probl` emes scientiﬁques que l’´ etudiant pourrait ˆ etre emmen´ e`ar ´esoudre dans le futur. Ce dernier point est souvent n´ eglig´e. Le polycopi´ e est subdivis´ ed ef a ¸con tr`es hi´erarchique. Chaque concept est d´ evelopp´e dans un para- graphe s´epar´e. Ceci devrait faciliter non seulement la compr´ ehension, mais aussi le dialogue entres ´ elev´es eux-mˆemes ainsi qu’avec le Professeur. Quand il a ´ et´ej u g ´en ´ecessaire, un bref rappel math´ ematique est introduit. De nombreux exemples sont pr´esent´es, et enﬁn des exercices solutionn´ es avec Mathematica sont pr´ esent´es dans l’annexe. L’auteur ne se fait point d’illusions quand au complet et ` a l’exactitude de tout le polycopi´ e. Il a ´et´e enti`erement d´ evelopp´e durant une seule ann´ ee acad´emique, et pourrait donc b´ en´eﬁcier d’une r´ evision extensive. A ce titre, corrections et critiques seront les bienvenues. Enﬁn, l’auteur voudrait remercier ses ´ elev´es qui ont diligemment suivis son cours sur la M´ ecanique de Milieux Continus durant l’ann´ ee acad´emique 1997-1998, ainsi que le Professeur Huet qui a ´ et´es o n hˆote au Laboratoire des Mat´ eriaux de Construction de l’EPFL durant son s´ ejour `a Lausanne. Victor Saouma Ecublens, Juin 1998 Victor Saouma Introduction to Continuum Mechanics Draft0–4 PREFACE One of the most fundamental question that a Material Scientist has to ask him/herself is how a material behaves under stress, and when does it break. Ultimately, it its the answer to those twoquestions which would steer the development of new materials, and determine their survival in various environmental and physical conditions. The Material Scientist should then have a thorough understanding of the fundamentals of Mechanics on the qualitative level, and be able to perform numerical simulation (most often by Finite Element Method) and extract quantitative information for a speciﬁc problem. In the humble opinion of the author, this is best achieved in three stages. First, the student should be exposed to the basic principles of Continuum Mechanics. Detailed coverage of Stress, Strain, General Principles, and Constitutive Relations is essential. Then, a brief exposure to Elasticity (along with BeamTheory) would convince the student that a well posed problem can indeed have an analytical solution. However, this is only true for problems problems with numerous simplifying assumptions (such as linear elasticity, small deformation, plane stress/strain or axisymmetry, and resultants of stresses). Hence, the last stage consists in a brief exposure to solid mechanics, and more precisely to Variational Methods. Through an exposure to the Principle of Virtual Work, and the Rayleigh-Ritz Method the student willthen be ready for Finite Elements. Finally, one topic of special interest to Material Science students was added, and that is the Theoretical Strength of Solids. This is essential to properly understand the failure of solids, and would later on lead to a Fracture Mechanics course. These lecture notes were prepared by the author during his sabbatical year at the Swiss Federal Institute of Technology (Lausanne) in the Material Science Department. The course was oﬀered to second year undergraduate students in French, whereas the lecture notes are in English. The notes were developed with the following objectives in mind. First they must be complete and rigorous. At any time, a student should be able to trace back the development of an equation. Furthermore, by going throughall the derivations, the student would understand the limitations and assumptions behind every model. Finally, the rigor adopted in the coverage of the subject should serve as an example to the students of the rigor expected from them in solving other scientiﬁc or engineering problems. This last aspect is oftenforgotten. The notes are broken down into a very hierarchical format. Each concept is broken down into a small section (a byte). This should not only facilitate comprehension, but also dialogue among the students or with the instructor. Whenever necessary, Mathematical preliminaries are introduced to make sure that the student is equipped with the appropriate tools. Illustrative problems are introduced whenever possible, and last but not least problem set using Mathematica is given in the Appendix. The author has no illusion as to the completeness or exactness of all these set of notes. They were entirely developed during a single academic year, and hence could greatly beneﬁt from a thorough review. As such, corrections, criticisms and comments are welcome. Finally, the author would like to thank his students who bravely put up with him and Continuum Mechanics in the AY 1997-1998, and Prof. Huet who was his host at the EPFL. Victor E. Saouma Ecublens, June 1998 Victor Saouma Introduction to Continuum Mechanics Draft Contents I CONTINUUM MECHANICS 0–9 1 MATHEMATICAL PRELIMINARIES; Part I Vectors and Tensors 1–1 1 . 1 V e c t o r s ..............................................1 – 1 1 . 1 . 1 O p e r a t i o n s ........................................1 – 2 1 . 1 . 2 C o o r d i n a t eT r a n s f o r m a t i o n ...............................1 – 4 1.1.2.1†G e n e r a lT e n s o r s ................................1 – 4 1.1.2.1.1 †C o n t r a v a r i a n t T r a n s f o r m a t i o n...................1 – 5 1.1.2.1.2 Covariant Transformation . . . . . . . . . . . . . . . . . . . . . . 1–6 1 . 1 . 2 . 2 C a r t e s i a nC o o r d i n a t eS y s t e m.........................1 – 6 1 . 2 T e n s o r s ..............................................1 – 8 1 . 2 . 1 I n d i c i a lN o t a t i o n.....................................1 – 8 1.2.2 Tensor Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–10 1.2.2.1 Sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–101.2.2.2 Multiplication by a Scalar . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–10 1.2.2.3 Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–10 1.2.2.4 Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–11 1.2.2.4.1 Outer Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–11 1.2.2.4.2 Inner Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–11 1.2.2.4.3 Scalar Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–11 1.2.2.4.4 Tensor Product . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–11 1.2.2.5 Product of Two Second-Order Tensors . . . . . . . . . . . . . . . . . . . . 1–13 1.2.3 Dyads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–13 1.2.4 Rotation of Axes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–13 1.2.5 Trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–141.2.6 Inverse Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–14 1.2.7 Principal Values and Directions of Symmetric Second Order Tensors . . . . . . . . 1–14 1.2.8 Powers of Second Order Tensors; Hamilton-Cayley Equations . . . . . . . . . . . . 1–15 2 KINETICS 2–1 2 . 1 F o r c e ,T r a c t i o na n dS t r e s sV e c t o r s ...............................2 – 12 . 2 T r a c t i o no na nA r b i t r a r yP l a n e ;C a u c h y ’ sS t r e s sT e n s o r ...................2 – 3 E2 - 1 S t r e s sV e c t o r s.......................................2 – 4 2 . 3 S y m m e t r yo fS t r e s sT e n s o r ...................................2 – 5 2 . 3 . 1 C a u c h y ’ sR e c i p r o c a lT h e o r e m..............................2 – 6 2 . 4 P r i n c i p a lS t r e s s e s.........................................2 – 7 2 . 4 . 1 I n v a r i a n t s.........................................2 – 82.4.2 Spherical and Deviatoric Stress Tensors . . . . . . . . . . . . . . . . . . . . . . . . 2–9 2 . 5 S t r e s sT r a n s f o r m a t i o n ......................................2 – 9 E 2-2 Principal Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–10 E 2-3 Stress Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–10 2.5.1 Plane Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–112.5.2 Mohr’s Circle for Plane Stress Conditions . . . . . . . . . . . . . . . . . . . . . . . 2–11 Draft0–2 CONTENTS E 2-4 Mohr’s Circle in Plane Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–13 2.5.3†Mohr’s Stress Representation Plane . . . . . . . . . . . . . . . . . . . . . . . . . . 2–15 2.6 Simpliﬁed Theories; Stress Resultants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–15 2.6.1 Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–16 2.6.2 Plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–19 3 MATHEMATICAL PRELIMINARIES; Part II VECTOR DIFFERENTIATION 3–1 3 . 1 I n t r o d u c t i o n............................................3 – 1 3 . 2 D e r i v a t i v eW R Tt oaS c a l a r...................................3 – 1 E3 - 1 T a n g e n tt oaC u r v e ...................................3 – 3 3 . 3 D i v e r g e n c e ............................................3 – 4 3 . 3 . 1 V e c t o r...........................................3 – 4E3 - 2 D i v e r g e n c e ........................................3 – 6 3 . 3 . 2 S e c o n d - O r d e rT e n s o r...................................3 – 7 3 . 4 G r a d i e n t..............................................3 – 8 3 . 4 . 1 S c a l a r...........................................3 – 8 E3 - 3 G r a d i e n to faS c a l a r ...................................3 – 8E3 - 4 S t r e s sV e c t o rn o r m a lt ot h eT a n g e n to faC y l i n d e r ..................3 – 9 3.4.2 Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–10 E 3-5 Gradient of a Vector Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–113.4.3 Mathematica Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–12 3.5 Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–12 E 3-6 Curl of a vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–13 3.6 Some useful Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–13 4KINEMATIC 4–1 4 . 1 E l e m e n t a r yD e ﬁ n i t i o no fS t r a i n.................................4 – 1 4 . 1 . 1 S m a l la n dF i n i t eS t r a i n si n1 D .............................4 – 1 4 . 1 . 2 S m a l lS t r a i n si n2 D ...................................4 – 2 4 . 2 S t r a i nT e n s o r...........................................4 – 3 4.2.1 Position and Displacement Vectors; ( x,X).......................4 – 3 E 4-1 Displacement Vectors in Material and Spatial Forms . . . . . . . . . . . . . . . . . 4–4 4.2.1.1 Lagrangian and Eulerian Descriptions; x(X,t),X(x,t)...........4 – 5 E 4-2 Lagrangian and Eulerian Descriptions . . . . . . . . . . . . . . . . . . . . . . . . . 4–64 . 2 . 2 G r a d i e n t s.........................................4 – 6 4.2.2.1 Deformation; ( x∇ X,X∇x)..........................4 – 6 4.2.2.1.1 †Change of Area Due to Deformation . . . . . . . . . . . . . . . 4–7 4.2.2.1.2 †Change of Volume Due to Deformation . . . . . . . . . . . . . 4–8 E4 - 3 C h a n g eo fV o l u m ea n dA r e a ...............................4 – 8 4.2.2.2 Displacements; ( u∇X,u∇x) .........................4 – 9 4.2.2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–10 E 4-4 Material Deformation and Displacement Gradients . . . . . . . . . . . . . . . . . . 4–104.2.3 Deformation Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–10 4.2.3.1 Cauchy’s Deformation Tensor; ( dX) 2. . . . . . . . . . . . . . . . . . . . 4–11 4.2.3.2 Green’s Deformation Tensor; ( dx)2. . . . . . . . . . . . . . . . . . . . . . 4–12 E 4-5 Green’s Deformation Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–12 4.2.4 Strains; ( dx)2−(dX)2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–13 4.2.4.1 Finite Strain Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–13 4.2.4.1.1 Lagrangian/Green’s Tensor . . . . . . . . . . . . . . . . . . . . . 4–13 E 4-6 Lagrangian Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–14 4.2.4.1.2 Eulerian/Almansi’s Tensor . . . . . . . . . . . . . . . . . . . . . 4–14 4.2.4.2 Inﬁnitesimal Strain Tensors; Small Deformation Theory . . . . . . . . . . 4–15 4.2.4.2.1 Lagrangian Inﬁnitesimal Strain Tensor . . . . . . . . . . . . . . 4–154.2.4.2.2 Eulerian Inﬁnitesimal Strain Tensor . . . . . . . . . . . . . . . . 4–16 Victor Saouma Introduction to Continuum Mechanics DraftCONTENTS 0–3 4.2.4.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–16 E 4-7 Lagrangian and Eulerian Linear Strain Tensors . . . . . . . . . . . . . . . . . . . . 4–16 4.2.5 Physical Interpretation of the Strain Tensor . . . . . . . . . . . . . . . . . . . . . . 4–17 4.2.5.1 Small Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–17 4.2.5.2 Finite Strain; Stretch Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . 4–19 4.2.6 Linear Strain and Rotation Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . 4–21 4.2.6.1 Small Strains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–21 4.2.6.1.1 Lagrangian Formulation . . . . . . . . . . . . . . . . . . . . . . . 4–214.2.6.1.2 Eulerian Formulation . . . . . . . . . . . . . . . . . . . . . . . . 4–23 4.2.6.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–24 E 4-8 Relative Displacement along a speciﬁed direction . . . . . . . . . . . . . . . . . . . 4–24E 4-9 Linear strain tensor, linear rotation tensor, rotation vector . . . . . . . . . . . . . . 4–24 4.2.6.3 Finite Strain; Polar Decomposition . . . . . . . . . . . . . . . . . . . . . . 4–25 E 4-10 Polar Decomposition I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–26 E 4-11 Polar Decomposition II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–27 E 4-12 Polar Decomposition III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–274.2.7 Summary and Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–29 4.2.8†Explicit Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–29 4.2.9 Compatibility Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–34E 4-13 Strain Compatibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–35 4.3 Lagrangian Stresses; Piola Kirchoﬀ Stress Tensors . . . . . . . . . . . . . . . . . . . . . . 4–36 4.3.1 First . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–36 4.3.2 Second . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–37 E 4-14 Piola-Kirchoﬀ Stress Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–38 4.4 Hydrostatic and Deviatoric Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–38 4.5 Principal Strains, Strain Invariants, Mohr Circle . . . . . . . . . . . . . . . . . . . . . . . 4–38 E 4-15 Strain Invariants & Principal Strains . . . . . . . . . . . . . . . . . . . . . . . . . . 4–40E 4-16 Mohr’s Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–42 4.6 Initial or Thermal Strains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–43 4.7†Experimental Measurement of Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–43 4.7.1 Wheatstone Bridge Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–45 4.7.2 Quarter Bridge Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–45 5 MATHEMATICAL PRELIMINARIES; Part III VECTOR INTEGRALS 5–1 5 . 1 I n t e g r a lo faV e c t o r........................................5 – 1 5 . 2 L i n eI n t e g r a l ...........................................5 – 15 . 3 I n t e g r a t i o nb yP a r t s .......................................5 – 2 5 . 4 G a u s s ;D i v e r g e n c eT h e o r e m...................................5 – 2 5 . 5 S t o k e ’ sT h e o r e m .........................................5 – 2 5 . 6 G r e e n ;G r a d i e n tT h e o r e m....................................5 – 2 E5 - 1 P h y s i c a lI n t e r p r e t a t i o no ft h eD i v e r g e n c eT h e o r e m .................5 – 3 6 FUNDAMENTAL LAWS of CONTINUUM MECHANICS 6–1 6 . 1 I n t r o d u c t i o n............................................6 – 1 6 . 1 . 1 C o n s e r v a t i o nL a w s....................................6 – 16 . 1 . 2 F l u x e s...........................................6 – 2 6 . 2 C o n s e r v a t i o no fM a s s ;C o n t i n u i t yE q u a t i o n ..........................6 – 3 6 . 2 . 1 S p a t i a lF o r m .......................................6 – 36 . 2 . 2 M a t e r i a lF o r m ......................................6 – 4 6 . 3 L i n e a rM o m e n t u mP r i n c i p l e ;E q u a t i o no fM o t i o n.......................6 – 5 6 . 3 . 1 M o m e n t u mP r i n c i p l e...................................6 – 5 E 6-1 Equilibrium Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–6 6 . 3 . 2 M o m e n to fM o m e n t u mP r i n c i p l e ............................6 – 7 6 . 3 . 2 . 1 S y m m e t r yo ft h eS t r e s sT e n s o r........................6 – 7 Victor Saouma Introduction to Continuum Mechanics Draft0–4 CONTENTS 6.4 Conservation of Energy; First Principle of Thermodynamics . . . . . . . . . . . . . . . . . 6–8 6 . 4 . 1 S p a t i a lG r a d i e n to ft h eV e l o c i t y .............................6 – 8 6 . 4 . 2 F i r s tP r i n c i p l e ......................................6 – 8 6.5 Equation of State; Second Principle of Thermodynamics . . . . . . . . . . . . . . . . . . . 6–10 6.5.1 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–11 6.5.1.1 Statistical Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–11 6.5.1.2 Classical Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . 6–11 6.5.2 Clausius-Duhem Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–12 6.6 Balance of Equations and Unknowns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–13 6.7†Elements of Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–14 6.7.1 Simple 2D Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–156.7.2†Generalized Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–16 7 CONSTITUTIVE EQUATIONS; Part I LINEAR 7–1 7.1†T h e r m o d y n a m i cA p p r o a c h ...................................7 – 1 7 . 1 . 1 S t a t eV a r i a b l e s......................................7 – 1 7 . 1 . 2 G i b b sR e l a t i o n......................................7 – 27.1.3 Thermal Equation of State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–3 7 . 1 . 4 T h e r m o d y n a m i cP o t e n t i a l s ...............................7 – 3 7 . 1 . 5 E l a s t i cP o t e n t i a lo rS t r a i nE n e r g yF u n c t i o n......................7 – 4 7 . 2 E x p e r i m e n t a lO b s e r v a t i o n s ...................................7 – 5 7 . 2 . 1 H o o k e ’ sL a w .......................................7 – 6 7 . 2 . 2 B u l kM o d u l u s .......................................7 – 6 7 . 3 S t r e s s - S t r a i nR e l a t i o n si nG e n e r a l i z e dE l a s t i c i t y ........................7 – 7 7 . 3 . 1 A n i s o t r o p i c........................................7 – 77.3.2 Monotropic Material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–8 7 . 3 . 3 O r t h o t r o p i cM a t e r i a l...................................7 – 9 7 . 3 . 4 T r a n s v e r s e l y I s o t r o p i cM a t e r i a l.............................7 – 97.3.5 Isotropic Material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–10 7.3.5.1 Engineering Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–12 7.3.5.1.1 Isotropic Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–12 7.3.5.1.1.1 Young’s Modulus . . . . . . . . . . . . . . . . . . . . . . . 7–12 7.3.5.1.1.2 Bulk’s Modulus; Volumetric and Deviatoric Strains . . . . 7–137.3.5.1.1.3 Restriction Imposed on the Isotropic Elastic Moduli . . . 7–14 7.3.5.1.2 Transversly Isotropic Case . . . . . . . . . . . . . . . . . . . . . 7–15 7.3.5.2 Special 2D Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–15 7.3.5.2.1 Plane Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–15 7.3.5.2.2 Axisymmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–16 7.3.5.2.3 Plane Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–16 7.4 Linear Thermoelasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–16 7.5 Fourrier Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–177.6 Updated Balance of Equations and Unknowns . . . . . . . . . . . . . . . . . . . . . . . . . 7–18 8 INTERMEZZO 8–1 II ELASTICITY/SOLID MECHANICS 8–3 9 BOUNDARY VALUE PROBLEMS in ELASTICITY 9–1 9 . 1 P r e l i m i n a r yC o n s i d e r a t i o n s ...................................9 – 19.2 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–1 9.3 Boundary Value Problem Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–4 9 . 4 C o m p a c t e dF o r m s ........................................9 – 4 9 . 4 . 1 N a v i e r - C a u c h y E q u a t i o n s ................................9 – 5 Victor Saouma Introduction to Continuum Mechanics DraftCONTENTS 0–5 9 . 4 . 2 B e l t r a m i - M i t c h e l lE q u a t i o n s ...............................9 – 5 9.4.3 Ellipticity of Elasticity Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–5 9 . 5 S t r a i nE n e r g ya n dE x t e n a lW o r k................................9 – 59 . 6 U n i q u e n e s so ft h eE l a s t o s t a t i cS t r e s sa n dS t r a i nF i e l d ....................9 – 6 9 . 7 S a i n tV e n a n t ’ sP r i n c i p l e.....................................9 – 6 9 . 8 C y l i n d r i c a lC o o r d i n a t e s .....................................9 – 7 9 . 8 . 1 S t r a i n s ...........................................9 – 8 9.8.2 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–99.8.3 Stress-Strain Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–10 9.8.3.1 Plane Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–11 9.8.3.2 Plane Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–11 10 SOME ELASTICITY PROBLEMS 10–1 10.1 Semi-Inverse Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–1 10.1.1 Example: Torsion of a Circular Cylinder . . . . . . . . . . . . . . . . . . . . . . . . 10–1 10.2 Airy Stress Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–3 10.2.1 Cartesian Coordinates; Plane Strain . . . . . . . . . . . . . . . . . . . . . . . . . . 10–3 10.2.1.1 Example: Cantilever Beam . . . . . . . . . . . . . . . . . . . . . . . . . . 10–6 10.2.2 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–7 10.2.2.1 Plane Strain Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–710.2.2.2 Axially Symmetric Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–8 10.2.2.3 Example: Thick-Walled Cylinder . . . . . . . . . . . . . . . . . . . . . . . 10–9 10.2.2.4 Example: Hollow Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–11 10.2.2.5 Example: Stress Concentration due to a Circular Hole in a Plate . . . . . 10–11 11 THEORETICAL STRENGTH OF PERFECT CRYSTALS 11–1 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–1 11.2 Theoretical Strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–3 11.2.1 Ideal Strength in Terms of Physical Parameters . . . . . . . . . . . . . . . . . . . . 11–311.2.2 Ideal Strength in Terms of Engineering Parameter . . . . . . . . . . . . . . . . . . 11–6 11.3 Size Eﬀect; Griﬃth Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–6 12 BEAM THEORY 12–1 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–1 12.2 Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–2 12.2.1 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–2 12.2.2 Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–3 12.2.3 Equations of Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–412.2.4 Static Determinacy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–4 12.2.5 Geometric Instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–5 12.2.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–5E 12-1 Simply Supported Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–5 12.3 Shear & Moment Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–6 12.3.1 Design Sign Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–6 12.3.2 Load, Shear, Moment Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–7 12.3.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–9E 12-2 Simple Shear and Moment Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . 12–9 12.4 Beam Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–10 12.4.1 Basic Kinematic Assumption; Curvature . . . . . . . . . . . . . . . . . . . . . . . . 12–1012.4.2 Stress-Strain Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–12 12.4.3 Internal Equilibrium; Section Properties . . . . . . . . . . . . . . . . . . . . . . . . 12–12 12.4.3.1 Σ F x= 0; Neutral Axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–12 12.4.3.2 Σ M= 0; Moment of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . 12–13 12.4.4 Beam Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–13 Victor Saouma Introduction to Continuum Mechanics Draft0–6 CONTENTS 12.4.5 Limitations of the Beam Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–14 12.4.6 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–14 E 12-3 Design Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–14 13 VARIATIONAL METHODS 13–1 13.1 Preliminary Deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–1 13.1.1 Internal Strain Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–2 13.1.2 External Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–4 13.1.3 Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–4 13.1.3.1 Internal Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–5 13.1.3.2 External Virtual Work δW. . . . . . . . . . . . . . . . . . . . . . . . . . 13–6 13.1.4 Complementary Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–613.1.5 Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–6 13.2 Principle of Virtual Work and Complementary Virtual Work . . . . . . . . . . . . . . . . 13–6 13.2.1 Principle of Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–7 E 13-1 Tapered Cantiliver Beam, Virtual Displacement . . . . . . . . . . . . . . . . . . . . 13–8 13.2.2 Principle of Complementary Virtual Work . . . . . . . . . . . . . . . . . . . . . . . 13–10E 13-2 Tapered Cantilivered Beam; Virtual Force . . . . . . . . . . . . . . . . . . . . . . . 13–11 13.3 Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–12 13.3.1 Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–1213.3.2 Rayleigh-Ritz Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–14 E 13-3 Uniformly Loaded Simply Supported Beam; Polynomial Approximation . . . . . . 13–16 13.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–17 14INELASTICITY (incomplete) –1 A SHEAR, MOMENT and DEFLECTION DIAGRAMS for BEAMS A–1B SECTION PROPERTIES B–1 C MATHEMATICAL PRELIMINARIES; Part IV VARIATIONAL METHODS C–1 C . 1 E u l e rE q u a t i o n..........................................C – 1 EC - 1E x t e n s i o no faB a r....................................C – 4 EC - 2F l e x u r eo faB e a m ....................................C – 6 D MID TERM EXAM D–1E MATHEMATICA ASSIGNMENT and SOLUTION E–1 Victor Saouma Introduction to Continuum Mechanics Draft List of Figures 1 . 1 D i r e c t i o nC o s i n e s(t ob ec o r r e c t e d )...............................1 – 2 1 . 2 V e c t o rA d d i t i o n..........................................1 – 2 1 . 3 C r o s sP r o d u c to fT w oV e c t o r s..................................1 – 31 . 4 C r o s sP r o d u c to fT w oV e c t o r s..................................1 – 4 1 . 5 C o o r d i n a t eT r a n s f o r m a t i o n ...................................1 – 5 1 . 6 A r b i t r a r y3 DV e c t o rT r a n s f o r m a t i o n..............................1 – 7 1 . 7 R o t a t i o no fO r t h o n o r m a lC o o r d i n a t eS y s t e m .........................1 – 8 2.1 Stress Components on an Inﬁnitesimal Element . . . . . . . . . . . . . . . . . . . . . . . . 2–2 2 . 2 S t r e s s e sa sT e n s o rC o m p o n e n t s.................................2 – 2 2 . 3 C a u c h y ’ sT e t r a h e d r o n ......................................2 – 3 2 . 4 C a u c h y ’ sR e c i p r o c a lT h e o r e m..................................2 – 62 . 5 P r i n c i p a lS t r e s s e s.........................................2 – 7 2.6 Mohr Circle for Plane Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–12 2.7 Plane Stress Mohr’s Circle; Numerical Example . . . . . . . . . . . . . . . . . . . . . . . . 2–14 2.8 Unit Sphere in Physical Body around O . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–15 2.9 Mohr Circle for Stress in 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–162.10 Diﬀerential Shell Element, Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–17 2.11 Diﬀerential Shell Element, Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–17 2.12 Diﬀerential Shell Element, Vectors of Stress Couples . . . . . . . . . . . . . . . . . . . . . 2–182.13 Stresses and Resulting Forces in a Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–19 3 . 1 E x a m p l e so faS c a l a ra n dV e c t o rF i e l d s ............................3 – 2 3.2 Diﬀerentiation of position vector p...............................3 – 2 3 . 3 C u r v a t u r eo faC u r v e.......................................3 – 3 3 . 4 M a t h e m a t i c aS o l u t i o nf o rt h eT a n g e n tt oaC u r v ei n3 D...................3 – 43 . 5 V e c t o rF i e l dC r o s s i n g aS o l i dR e g i o n..............................3 – 5 3.6 Flux Through Area dA......................................3 – 5 3.7 Inﬁnitesimal Element for the Evaluation of the Divergence . . . . . . . . . . . . . . . . . . 3–63.8 Mathematica Solution for the Divergence of a Vector . . . . . . . . . . . . . . . . . . . . . 3–7 3 . 9 R a d i a lS t r e s sv e c t o ri naC y l i n d e r ................................3 – 9 3.10 Gradient of a Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–11 3.11 Mathematica Solution for the Gradients of a Scalar and of a Vector . . . . . . . . . . . . . 3–12 3.12 Mathematica Solution for the Curl of a Vector . . . . . . . . . . . . . . . . . . . . . . . . 3–14 4 . 1 E l o n g a t i o no fa nA x i a lR o d...................................4 – 1 4 . 2 E l e m e n t a r yD e ﬁ n i t i o no fS t r a i n si n2 D.............................4 – 2 4 . 3 P o s i t i o na n dD i s p l a c e m e n tV e c t o r s...............................4 – 34.4 Undeformed and Deformed Conﬁgurations of a Continuum . . . . . . . . . . . . . . . . . 4–11 4.5 Physical Interpretation of the Strain Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . 4–18 4.6 Relative Displacement duofQrelative to P. . . . . . . . . . . . . . . . . . . . . . . . . . 4–21 4.7 Strain Deﬁnition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–31 4.8 Mohr Circle for Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–40 Draft0–2 LIST OF FIGURES 4.9 Bonded Resistance Strain Gage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–43 4.10 Strain Gage Rosette . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–44 4.11 Quarter Wheatstone Bridge Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–454.12 Wheatstone Bridge Conﬁgurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–46 5.1 Physical Interpretation of the Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . 5–3 6.1 Flux Through Area dS......................................6 – 3 6.2 Equilibrium of Stresses, Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . 6–6 6.3 Flux vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–156.4 Flux Through Sides of Diﬀerential Element . . . . . . . . . . . . . . . . . . . . . . . . . . 6–16 6.5 *Flow through a surface Γ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–17 9.1 Boundary Conditions in Elasticity Problems . . . . . . . . . . . . . . . . . . . . . . . . . . 9–2 9.2 Boundary Conditions in Elasticity Problems . . . . . . . . . . . . . . . . . . . . . . . . . . 9–3 9.3 Fundamental Equations in Solid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . 9–4 9 . 4 S t - V e n a n t ’ sP r i n c i p l e.......................................9 – 7 9 . 5 C y l i n d r i c a lC o o r d i n a t e s .....................................9 – 7 9 . 6 P o l a rS t r a i n s ...........................................9 – 89 . 7 S t r e s s e si nP o l a r C o o r d i n a t e s ..................................9 – 9 10.1 Torsion of a Circular Bar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–2 10.2 Pressurized Thick Tube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–1010.3 Pressurized Hollow Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–11 10.4 Circular Hole in an Inﬁnite Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–12 11.1 Elliptical Hole in an Inﬁnite Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–1 11.2 Griﬃth’s Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–2 11.3 Uniformly Stressed Layer of Atoms Separated by a 0. . . . . . . . . . . . . . . . . . . . . 11–3 11.4 Energy and Force Binding Two Adjacent Atoms . . . . . . . . . . . . . . . . . . . . . . . 11–4 11.5 Stress Strain Relation at the Atomic Level . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–5 12.1 Types of Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–3 12.2 Inclined Roller Support . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–4 12.3 Examples of Static Determinate and Indeterminate Structures . . . . . . . . . . . . . . . . 12–5 12.4 Geometric Instability Caused by Concurrent Reactions . . . . . . . . . . . . . . . . . . . . 12–512.5 Shear and Moment Sign Conventions for Design . . . . . . . . . . . . . . . . . . . . . . . . 12–7 12.6 Free Body Diagram of an Inﬁnitesimal Beam Segment . . . . . . . . . . . . . . . . . . . . 12–7 12.7 Deformation of a Beam under Pure Bending . . . . . . . . . . . . . . . . . . . . . . . . . . 12–11 13.1 *Strain Energy and Complementary Strain Energy . . . . . . . . . . . . . . . . . . . . . . 13–2 13.2 Tapered Cantilivered Beam Analysed by the Vitual Displacement Method . . . . . . . . . 13–813.3 Tapered Cantilevered Beam Analysed by the Virtual Force Method . . . . . . . . . . . . . 13–11 13.4 Single DOF Example for Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–13 13.5 Graphical Representation of the Potential Energy . . . . . . . . . . . . . . . . . . . . . . . 13–1413.6 Uniformly Loaded Simply Supported Beam Analyzed by the Rayleigh-Ritz Method . . . . 13–16 13.7 Summary of Variational Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–18 13.8 Duality of Variational Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–19 1 4 . 1t e s t ................................................. – 1 1 4 . 2m o d 1 ................................................ – 2 1 4 . 3v - k v ................................................ – 2 1 4 . 4v i s ﬂ ................................................ – 3 1 4 . 5v i s ﬂ ................................................ – 31 4 . 6c o m p................................................ – 3 Victor Saouma Introduction to Continuum Mechanics DraftLIST OF FIGURES 0–3 1 4 . 7e p p................................................. – 3 1 4 . 8e h s................................................. – 4 C.1 Variational and Diﬀerential Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C–2 Victor Saouma Introduction to Continuum Mechanics Draft0–4 LIST OF FIGURES Victor Saouma Introduction to Continuum Mechanics DraftLIST OF FIGURES 0–5 NOTATION Symbol Deﬁnition Dimension SI Unit SCALARS A Area L2m2 c Speciﬁc heat e Volumetric strain N.D. - E Elastic Modulus L−1MT−2Pa g Specicif free enthalpy L2T−2JKg−1 h Film coeﬃcient for convection heat transfer h Speciﬁc enthalpy L2T−2JKg−1 I Moment of inertia L4m4 J Jacobian K Bulk modulus L−1MT−2Pa K Kinetic Energy L2MT−2J L Length Lm p Pressure L−1MT−2Pa Q Rate of internal heat generation L2MT−3W r Radiant heat constant per unit mass per unit time MT−3L−4Wm−6 s Speciﬁc entropy L2T−2Θ−1JKg−1K−1 S Entropy ML2T−2Θ−1JK−1 t Time Ts T Absolute temperature Θ K u Speciﬁc internal energy L2T−2JKg−1 U Energy L2MT−2J U∗Complementary strain energy L2MT−2J W Work L2MT−2J W P o t e n t i a lo fE x t e r n a lW o r k L2MT−2J Π Potential energy L2MT−2J α Coeﬃcient of thermal expansion Θ−1T−1 µ Shear modulus L−1MT−2Pa ν Poisson’s ratio N.D. - ρ mass density ML−3Kgm−3 γij Shear strains N.D. - 1 2γij Engineering shear strain N.D. - λ Lame’s coeﬃcient L−1MT−2Pa Λ Stretch ratio N.D. -µG Lame’s coeﬃcient L −1MT−2Pa λ Lame’s coeﬃcient L−1MT−2Pa Φ Airy Stress FunctionΨ (Helmholtz) Free energy L 2MT−2J Iσ,IEFirst stress and strain invariants IIσ,IIESecond stress and strain invariants IIIσ,IIIEThird stress and strain invariants Θ Temperature Θ K TENSORS order 1 b Body force per unit mass LT−2NKg−1 b Base transformation q Heat ﬂux per unit area MT−3Wm−2 t Traction vector, Stress vector L−1MT−2Pa hatwidet Speciﬁed tractions along Γ t L−1MT−2Pa u Displacement vector Lm Victor Saouma Introduction to Continuum Mechanics Draft0–6 LIST OF FIGURES hatwideu(x) Speciﬁed displacements along Γ u Lm u Displacement vector Lm x Spatial coordinates Lm X Material coordinates Lm σ0 Initial stress vector L−1MT−2Pa σ(i) Principal stresses L−1MT−2Pa TENSORS order 2 B−1Cauchy’s deformation tensor N.D. - C Green’s deformation tensor; metric tensor, right Cauchy-Green deformation tensor N.D. - D Rate of deformation tensor; Stretching tensor N.D. - E Lagrangian (or Green’s) ﬁnite strain tensor N.D. - E∗Eulerian (or Almansi) ﬁnite strain tensor N.D. - E/primeStrain deviator N.D. - F Material deformation gradient N.D. - H Spatial deformation gradient N.D. - I Idendity matrix N.D. - J Material displacement gradient N.D. - k Thermal conductivity LMT−3Θ−1Wm−1K−1 K Spatial displacement gradient N.D. - L Spatial gradient of the velocity R Orthogonal rotation tensor T0 First Piola-Kirchoﬀ stress tensor, Lagrangian Stress Tensor L−1MT−2Pa ˜T Second Piola-Kirchoﬀ stress tensor L−1MT−2Pa U Right stretch tensor V Left stretch tensor W Spin tensor, vorticity tensor. Linear lagrangian rotation tensor ε0 Initial strain vector k Conductivity κ Curvature σ,T Cauchy stress tensor L−1MT−2Pa T/primeDeviatoric stress tensor L−1MT−2Pa Ω Linear Eulerian rotation tensor ω Linear Eulerian rotation vector TENSORS order 4 D Constitutive matrix L−1MT−2Pa CONTOURS, SURFACES, VOLUMES C Contour line S Surface of a body L2m2 Γ Surface L2m2 Γt Boundary along which surface tractions, tare speciﬁed L2m2 Γu Boundary along which displacements, uare speciﬁed L2m2 ΓT Boundary along which temperatures, Tare speciﬁed L2m2 Γc Boundary along which convection ﬂux, qcare speciﬁed L2m2 Γq Boundary along which ﬂux, qnare speciﬁed L2m2 Ω,V Volume of body L3m3 FUNCTIONS, OPERATORS Victor Saouma Introduction to Continuum Mechanics DraftLIST OF FIGURES 0–7 ˜u Neighbour function to u(x) δ Variational operator L Linear diﬀerential operator relating displacement to strains ∇φ Divergence, (gradient operator) on scalar ⌊∂φ ∂x∂φ ∂y∂φ ∂z⌋T ∇·u Divergence, (gradient operator) on vector (div . u=∂u x ∂x+∂u y ∂y+∂u z ∂z ∇2Laplacian Operator Victor Saouma Introduction to Continuum Mechanics Draft0–8 LIST OF FIGURES Victor Saouma Introduction to Continuum Mechanics Draft Part I CONTINUUM MECHANICS Draft Draft Chapter 1 MATHEMATICAL PRELIMINARIES; Part I Vectors and Tensors 1Physical laws should be independent of the position and orientation of the observer. For this reason, physical laws are vector equations ortensor equations , since both vectors and tensors transform from one coordinate system to another in such a way that if the law holds in one coordinate system, it holds in any other coordinate system. 1.1 Vectors 2A vector is a directed line segment which can denote a variety of quantities, such as position of point with respect to another ( position vector ), a force, or a traction. 3A vector may be deﬁned with respect to a particular coordinate system by specifying the components of the vector in that system. The choice of the coordinate system is arbitrary, but some are more suitable than others (axes corresponding to the major direction of the object being analyzed). 4Therectangular Cartesian coordinate system is the most often used one (others are the cylin- drical, spherical or curvilinear systems). The rectangular system is often represented by three mutually perpendicular axes Oxyz, with corresponding unit vector triad i ,j,k(ore1,e2,e3) such that: i×j=k;j×k=i;k×i=j; (1.1-a) i·i=j·j=k·k= 1 (1.1-b) i·j=j·k=k·i= 0 (1.1-c) Such a set of base vectors constitutes an orthonormal basis . 5An arbitrary vector vmay be expressed by v=vxi+vyj+vzk (1.2) where vx=v·i=vcosα (1.3-a) vy=v·j=vcosβ (1.3-b) vz=v·k=vcosγ (1.3-c) are the projections of vonto the coordinate axes, Fig. 1.1. Draft1–2 MATHEMATICAL PRELIMINARIES; Part I Vectors and Tensors V αβ γXY Z Figure 1.1: Direction Cosines (to be corrected) 6The unit vector in the direction of vis given by ev=v v=c o sαi+c o sβj+c o sγk (1.4) Sincevis arbitrary, it follows that any unit vector will have direction cosines of that vector as its Cartesian components . 7The length or more precisely the magnitude of the vector is denoted by /bardblv/bardbl=radicalbig v2 1+v2 2+v2 3. 8We will denote the contravariant components of a vector by superscripts vk,a n di t scovariant components by subscripts vk(the signiﬁcance of those terms will be clariﬁed in Sect. 1.1.2.1. 1.1.1 Operations Addition: of two vectors a+bis geometrically achieved by connecting the tail of the vector bwith the head ofa, Fig. 1.2. Analytically the sum vector will have components ⌊a1+b1a2+b2a3+b3⌋. v θu u+v Figure 1.2: Vector Addition Scalar multiplication: αawill scale the vector into a new one with components ⌊αa1αa2αa3⌋. Vector Multiplications ofaandbcomes in three varieties: Victor Saouma Introduction to Continuum Mechanics Draft1.1 Vectors 1–3 Dot Product (or scalar product) is a scalar quantity which relates not only to the lengths of the vector, but also to the angle between them. a·b≡/bardbla/bardbl/bardblb/bardblcosθ(a,b)=3summationdisplay i=1aibi (1.5) where cosθ(a,b) is the cosine of the angle between the vectors aandb. The dot product measures the relative orientation between two vectors. The dot product is both commutative a·b=b·a (1.6) anddistributive αa·(βb+γc)=αβ(a·b)+αγ(a·c) (1.7) The dot product of awith a unit vector ngives the projection of ain the direction of n. The dot product of base vectors gives rise to the deﬁnition of the Kronecker delta deﬁned as ei·ej=δij (1.8) where δij=braceleftbigg 1i fi=j 0i fi/negationslash=j (1.9) Cross Product (or vector product) cof two vectors aandbis deﬁned as the vector c=a×b=(a2b3−a3b2)e1+(a3b1−a1b3)e2+(a1b2−a2b1)e3 (1.10) which can be remembered from the determinant expansion of a×b=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglee 1e2e3 a1a2a3 b1b2b3vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle(1.11) and is equal to the area of the parallelogram described by aandb, Fig. 1.3. a x b abA(a,b)=||a x b|| Figure 1.3: Cross Product of Two Vectors A(a,b)=/bardbla×b/bardbl (1.12) Victor Saouma Introduction to Continuum Mechanics Draft1–4MATHEMATICAL PRELIMINARIES; Part I Vectors and Tensors The cross product is not commutative, but satisﬁes the condition of skew symmetry a×b=−b×a (1.13) The cross product is distributive αa×(βb+γc)=αβ(a×b)+αγ(a×c) (1.14) Triple Scalar Product: of three vectors a,b,a n dcis desgnated by ( a×b)·cand it corresponds to the (scalar) volume deﬁned by the three vectors, Fig. 1.4. ||a x b|| abc c.nn=a x b Figure 1.4: Cross Product of Two Vectors V(a,b,c)=(a×b)·c=a·(b×c) (1.15) =vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglea xayaz bxbybz cxcyczvextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle(1.16) The triple scalar product of base vectors represents a fundamental operation (ei×ej)·ek=εijk≡  1i f (i,j,k) are in cyclic order 0i f a n y o f ( i,j,k)a r ee q u a l −1i f (i,j,k) are in acyclic order (1.17) The scalars εijkis thepermutationtensor . A cyclic permutation of 1,2,3 is 1 →2→3→1, an acyclic one would be 1 →3→2→1. Using this notation, we can rewrite c=a×b⇒ci=εijkajbk (1.18) Vector Triple Product is a cross product of two vectors, one of which is itself a cross product. a×(b×c)=(a·c)b−(a·b)c=d (1.19) and the product vector dlies in the plane of bandc. 1.1.2 Coordinate Transformation 1.1.2.1†General Tensors 9Let us consider two bases bj(x1,x2,x3)a n d bj( x1, x2 x3), Fig. 1.5. Each unit vector in one basis must be a linear combination of the vectors of the other basis bj=ap jbpandbk=bk q bq (1.20) Victor Saouma Introduction to Continuum Mechanics Draft1.1 Vectors 1–5 (summed on pandqrespectively) where ap j(subscript new, superscript old) and bk qare the coeﬃcients for the forward and backward changes respectively from btobrespectively. Explicitly   e1 e2 e3  = b1 1b12b13 b21b22b23 b3 1b32b33    e1 e2 e3  and   e1 e2 e3  = a1 1a21a31 a12a22a32 a1 3a23a33   e1 e2 e3  (1.21) XX X32 XX X12 312 cos a-1 1 Figure 1.5: Coordinate Transformation 10The transformation must have the determinant of its Jacobian J=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle∂ x1 ∂x1∂ x1 ∂x2∂ x1 ∂x3 ∂ x2 ∂x1∂ x2 ∂x2∂ x2 ∂x3 ∂ x3 ∂x1∂ x3 ∂x2∂ x3 ∂x3vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle/negationslash=0 (1.22) diﬀerent from zero (the superscript is a label and not an exponent). 11It is important to note that so far, the coordinate systems are completely general and may be Carte- sian, curvilinear, spherical or cylindrical. 1.1.2.1.1 †Contravariant Transformation 12The vector representation in both systems must be the same v= vq bq=vkbk=vk(bq k bq)⇒( vq−vkbq k) bq=0 (1.23) since the base vectors bqare linearly independent, the coeﬃcients of bqmust all be zero hence vq=bq kvkand inversely vp=ap j vj (1.24) showing that the forward change from components vkto vqused the coeﬃcients bq kof the backward change from base bqto the original bk. This is why these components are called contravariant . 13Generalizing, a Contravariant Tensor of order one (recognized by the use of the superscript) transforms a set of quantities rkassociated with point Pinxkthrough a coordinate transformation into Victor Saouma Introduction to Continuum Mechanics Draft1–6 MATHEMATICAL PRELIMINARIES; Part I Vectors and Tensors an e ws e t rqassociated with xq rq=∂ xq ∂xkbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright bq krk (1.25) 14By extension, the Contravariant tensors of order two requires the tensor components to obey the following transformation law rij=∂ xi ∂xr∂ xj ∂xsrrs (1.26) 1.1.2.1.2 Covariant Transformation 15Similarly to Eq. 1.24, a covariant component transformation (recognized by subscript) will be deﬁned as vj=ap jvpand inversely vk=bk q vq (1.27) We note that contrarily to the contravariant transformation, the covariant transformation uses the same transformation coeﬃcients as the ones for the base vectors. 16Finally transformation of tensors of order one and two is accomplished through rq=∂xk ∂ xqrk (1.28) rij=∂xr ∂ xi∂xs ∂ xjrrs(1.29) 1.1.2.2 Cartesian Coordinate System 17If we consider two diﬀerent sets of cartesian orthonormalcoordinate systems {e1,e2,e3}and{ e1, e2, e3}, any vector vcan be expressed in one system or the other v=vjej= vj ej (1.30) 18To determine the relationship between the two sets of components, we consider the dot product of v with one (any) of the base vectors ei·v= vi=vj( ei·ej) (1.31) (since vj( ej· ei)= vjδij= vi) 19We can thus deﬁne the nine scalar values aj i≡ ei·ej=c o s ( xi,xj) (1.32) which arise from the dot products of base vectors as the direction cosines . (Since we have an or- thonormal system, those values are nothing else than the cosines of the angles between the nine pairingof base vectors.) 20Thus, one set of vector components can be expressed in terms of the other through a covariant transformation similar to the one of Eq. 1.27. Victor Saouma Introduction to Continuum Mechanics Draft1.1 Vectors 1–7 vj=ap jvp(1.33) vk=bk q vq(1.34) we note that the free index in the ﬁrst and second equations appear on the upper and lower index respectively. 21Because of the orthogonality of the unit vector we have as pasq=δpqandam ranr=δmn. 22As a further illustration of the above derivation, let us consider the transformation of a vector Vfrom (X,Y,Z)c o o r d i n a t es y s t e mt o( x,y,z), Fig. 1.6: Figure 1.6: Arbitrary 3D Vector Transformation 23Eq. 1.33 would then result in Vx=aX xVX+aY xVY+aZ xVZ (1.35) or   Vx Vy Vz  = aX xaYxaZx aXyaYyaZy aX zaYzaZz   VX VY VZ  (1.36) andaj iis the direction cosine of axis iwith respect to axis j •aj x=(axX,aY x,aZx) direction cosines of xwith respect to X,YandZ •aj y=(ayX,aY y,aZy) direction cosines of ywith respect to X,YandZ •aj z=(azX,aY z,aZz) direction cosines of zwith respect to X,YandZ 24Finally, for the 2D case and from Fig. 1.7, the transformation matrix is written as T=bracketleftbigga1 1a21 a12a22bracketrightbigg =bracketleftbiggcosαcosβ cosγcosαbracketrightbigg (1.37) but sinceγ=π 2+α,a n dβ=π 2−α,t h e nc o sγ=−sinαand cosβ=s i nα, thus the transformation matrix becomes T=bracketleftbigg cosαsinα −sinαcosαbracketrightbigg (1.38) Victor Saouma Introduction to Continuum Mechanics Draft1–8 MATHEMATICAL PRELIMINARIES; Part I Vectors and Tensors XX γXX 1122 αα β Figure 1.7: Rotation of Orthonormal Coordinate System 1.2 Tensors 25We now seek to generalize the concept of a vector by introducing the tensor (T), which essentially exists to operate on vectors vto produce other vectors (or on tensors to produce other tensors!). We designate this operation by T·vor simply Tv. 26We hereby adopt the dyadic notation for tensors as linear vector operators u=T·vorui=Tijvj (1.39-a) u=v·SwhereS=TT(1.39-b) 27†In general the vectors may be represented by either covariant or contravariant components vjorvj. Thus we can have diﬀerent types of linear transformations ui=Tijvj;ui=Tijvj ui=T.j ivj;ui=Ti .jvj (1.40) involving the covariant components Tij,t h econtravariant components Tijand themixed com- ponentsTi .jorT.j i. 28Whereas 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. 29Tensors frequently arise as physical entities whose components are the coeﬃcients of a linear relation- ship between vectors. 30A 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, henceit is a vector. In general 3-D space the number of components of a tensor is 3 nwhere n is the order of the tensor. 31A force and a stress are tensors of order 1 and 2 respectively. 1.2.1 Indicial Notation 32Whereas the Engineering notation may be the simplest and most intuitive one, it often leads to long and repetitive equations. Alternatively, the tensor and the dyadic form will lead to shorter and morecompact forms. Victor Saouma Introduction to Continuum Mechanics Draft1.2 Tensors 1–9 33While working on general relativity, Einstein got tired of writing the summation symbol with its range of summation below and above (such assummationtextn=3 i=1aijbi) 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 ( iin our example). Hence, he decided that there is no need to include the summation signsummationtextif 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 calledfree index and assumed to take a value from 1 to 3. 34Hence, this so called indicial notation is also referred to Einstein’s notation . 35The following rules deﬁne indicial notation: 1. If there is one letter index, that index goes from iton(range of the tensor). For instance: ai=ai=⌊a1a2a3⌋=  a1 a2 a3  i=1,3 (1.41) assuming that n=3 . 2. A repeated index will take on all the values of its range, and the resulting tensors summed. For instance: a1ixi=a11x1+a12x2+a13x3 (1.42) 3. Tensor’s order: •First order tensor (such as force) has only one free index: ai=ai=⌊a1a2a3⌋ (1.43) other ﬁrst order tensors aijbj,Fikk,εijkujvk •Second order tensor (such as stress or strain) will have two free indeces. Dij= D11D22D13 D21D22D23 D31D32D33  (1.44) other examples Aijip,δijukvk. •A fourth order tensor (such as Elastic constants) will have four free indeces. 4. Derivatives of tensor with respect to xiis written as ,i. For example: ∂Φ ∂x i=Φ,i∂v i ∂x i=vi,i∂v i ∂x j=vi,j∂T i,j ∂x k=Ti,j,k (1.45) 36Usefulness of the indicial notation is in presenting systems of equations in compact form. For instance: xi=cijzj (1.46) this simple compacted equation, when expanded would yield: x1=c11z1+c12z2+c13z3 x2=c21z1+c22z2+c23z3 (1.47-a) x3=c31z1+c32z2+c33z3 Similarly: Aij=BipCjqDpq (1.48) Victor Saouma Introduction to Continuum Mechanics Draft1–10 MATHEMATICAL PRELIMINARIES; Part I Vectors and Tensors A11=B11C11D11+B11C12D12+B12C11D21+B12C12D22 A12=B11C11D11+B11C12D12+B12C11D21+B12C12D22 A21=B21C11D11+B21C12D12+B22C11D21+B22C12D22 A22=B21C21D11+B21C22D12+B22C21D21+B22C22D22 (1.49-a) 37Using indicial notation, we may rewrite the deﬁnition of the dot product a·b=aibi (1.50) and of the cross product a×b=εpqraqbrep (1.51) we note that in the second equation, there is one free index pthus there are three equations, there are two repeated (dummy) indices qandr, thus each equation has nine terms. 1.2.2 Tensor Operations 1.2.2.1 Sum 38The sum of two (second order) tensors is simply deﬁned as: Sij=Tij+Uij (1.52) 1.2.2.2 Multiplication by a Scalar 39The multiplication of a (second order) tensor by a scalar is deﬁned by: Sij=λTij (1.53) 1.2.2.3 Contraction 40In a contraction, we make two of the indeces equal (or in a mixed tensor, we make a ubscript equal to the superscript), thus producing a tensor of order two less than that to which it is applied. For example: Tij→Tii;2 →0 uivj→uivi;2 →0 Amr ..sn→Amr ..sm=Br .s;4→2 Eijak→Eijai=cj;3→1 Ampr qs→Ampr qr=Bmp q;5→3(1.54) Victor Saouma Introduction to Continuum Mechanics Draft1.2 Tensors 1–11 1.2.2.4Products 1.2.2.4.1 Outer Product 41The 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 aibj=Tij (1.55-a) AiB.k j=Ci.k.j (1.55-b) viTjk=Sijk (1.55-c) 1.2.2.4.2 Inner Product 42The inner product is obtained from an outer product by contraction involving one index from each tensor. For example aibj→aibi (1.56-a) aiEjk→aiEik=fk (1.56-b) EijFkm→EijFjm=Gim (1.56-c) AiB.k i→AiB.k i=Dk(1.56-d) 1.2.2.4.3 Scalar Product 43The scalar product of two tensors is deﬁned as T:U=TijUij (1.57) in any rectangular system. 44The following inner-product axioms are satisﬁed: T:U=U:T (1.58-a) T:(U+V)=T:U+T:V (1.58-b) α(T:U)=(αT):U=T:(αU) (1.58-c) T:T>0 unlessT=0 (1.58-d) 1.2.2.4.4 Tensor Product 45Since a tensor primary objective is to operate on vectors, the tensor product of two vectors provides a fundamental building block of second-order tensors and will be examined next. Victor Saouma Introduction to Continuum Mechanics Draft1–12 MATHEMATICAL PRELIMINARIES; Part I Vectors and Tensors 46TheTensor Product of two vectors uandvis a second order tensor u⊗vw h i c hi nt u r no p e r a t e s on an arbitrary vector was follows: [u⊗v]w≡(v·w)u (1.59) In other words when the tensor product u⊗voperates on w(left hand side), the result (right hand side) is a vector that points along the direction of u, and has length equal to ( v·w)||u||, or the original length ofutimes the dot (scalar) product of vandw. 47Of particular interest is the tensor product of the base vectors ei⊗ej. With three base vectors, we have a set of nine second order tensors which provide a suitable basis for expressing the components of a tensor. Again, we started with base vectors which themselves provide a basis for expressing any vector, and now the tensor product of base vectors in turn provides a formalism to express the components ofat e n s o r . 48Thesecond order tensor T c a nb ee x p r e s s e di nt e r m so fi t sc o m p o n e n t s Tijrelative to the base tensorsei⊗ejas follows: T=3summationdisplay i=13summationdisplay j=1Tij[ei⊗ej] (1.60-a) Tek=3summationdisplay i=13summationdisplay j=1Tij[ei⊗ej]ek (1.60-b) [ei⊗ej]ek=(ej·ek)ei=δjkei (1.60-c) Tek=3summationdisplay i=1Tikei (1.60-d) ThusTikis theith component of Tek. We can thus deﬁne the tensor component as follows Tij=ei·Tej (1.61) 49Now we can see how the second order tensor Toperates on any vector vby examining the components of the resulting vector Tv: Tv= 3summationdisplay i=13summationdisplay j=1Tij[ei⊗ej] parenleftBigg3summationdisplay k=1vkekparenrightBigg =3summationdisplay i=13summationdisplay j=13summationdisplay k=1Tijvk[ei⊗ej]ek (1.62) which when combined with Eq. 1.60-c yields Tv=3summationdisplay i=13summationdisplay j=1Tijvjei (1.63) which is clearly a vector. The ith component of the vector Tvbeing (Tv)i=3summationdisplay i=1Tijvj (1.64) 50The identity tensor Ileaves the vector unchanged Iv=vand is equal to I≡ei⊗ei (1.65) Victor Saouma Introduction to Continuum Mechanics Draft1.2 Tensors 1–13 51A simple example of a tensor and its operation on vectors is the projection tensorPwhich generates the projection of a vector von the plane characterized by a normal n: P≡I−n⊗n (1.66) the action of PonvgivesPv=v−(v·n)n. To convince ourselves that the vector Pvlies on the plane, its dot product with nmust be zero, accordingly Pv·n=v·n−(v·n)(n·n)=0√. 1.2.2.5 Product of Two Second-Order Tensors 52The product of two tensors is deﬁned as P=T·U;Pij=TikUkj (1.67) in any rectangular system. 53The following axioms hold (T·U)·R=T·(U·R) (1.68-a) T·(R+U)=T·R+t·U (1.68-b) (R+U)·T=R·T+U·T (1.68-c) α(T·U)=(αT)·U=T·(αU) (1.68-d) 1T=T·1=T (1.68-e) Note again that some authors omit the dot. Finally, the operation is not commutative 1.2.3 Dyads 54Theindeterminate vector product ofaandbdeﬁned by writing the two vectors in juxtaposition as abis called a dyad.Adyadic D corresponds to a tensor of order two and is a linear combination of dyads: D=a1b1+a2b2···anbn (1.69) Theconjugate dyadic ofDis written as Dc=b1a1+b2a2···bnan (1.70) 1.2.4 Rotation of Axes 55The rule for changing second order tensor components under rotation of axes goes as follow: ui=aj iuj From Eq. 1.33 =aj iTjqvq From Eq. 1.39-a =aj iTjqaq p vpFrom Eq. 1.33(1.71) But we also have ui= Tip vp(again from Eq. 1.39-a) in the barred system, equating these two expressions we obtain Tip−(aj iaq pTjq) vp= 0 (1.72) hence Tip=aj iaq pTjqin Matrix Form [ T]=[A]T[T][A] (1.73) Tjq=aj iaq p Tipin Matrix Form [ T]=[A][ T][A]T(1.74) Victor Saouma Introduction to Continuum Mechanics Draft1–14MATHEMATICAL PRELIMINARIES; Part I Vectors and Tensors By extension, higher order tensors can be similarly transformed from one coordinate system to another. 56If we consider the 2D case, From Eq. 1.38 A= cosαsinα0 −sinαcosα0 00 1  (1.75-a) T= TxxTxy0 TxyTyy0 00 0  (1.75-b) T=ATTA=  Txx Txy0 Txy Tyy0 00 0  (1.75-c) = cos2αTxx+s i n2αTyy+s i n2αTxy1 2(−sin2αTxx+s i n2αTyy+2c o s2αTxy0 1 2(−sin2αTxx+s i n2αTyy+2c o s2αTxysin2αTxx+c o sα(cosαTyy−2sinαTxy0 00 0  (1.75-d) alternatively, using sin2 α=2s i nαcosαand cos2α=c o s2α−sin2α, this last equation can be rewritten as    Txx Tyy Txy  = cos2θ sin2θ 2sinθcosθ sin2θ cos2θ−2sinθcosθ −sinθcosθcosθsinθcos2θ−sin2θ   Txx Tyy Txy   (1.76) 1.2.5 Trace 57Thetraceof a second-order tensor, denoted tr Tis a scalar invariant function of the tensor and is deﬁned as trT≡Tii (1.77) Thus it is equal to the sum of the diagonal elements in a matrix. 1.2.6 Inverse Tensor 58An inverse tensor is simply deﬁned as follows T−1(Tv)=vandT(T−1v)=v (1.78) alternatively T−1T=TT−1=I,o rT−1 ikTkj=δijandTikT−1 kj=δij 1.2.7 Principal Values and Directions of Symmetric Second Order Tensors 59Since the two fundamental tensors in continuum mechanics are of the second order and symmetric (stress and strain), we examine some important properties of these tensors. 60For every symmetric tensor Tijdeﬁned at some point in space, there is associated with each direction (speciﬁed by unit normal nj) at that point, a vector given by the inner product vi=Tijnj (1.79) Victor Saouma Introduction to Continuum Mechanics Draft1.2 Tensors 1–15 If the direction is one for which viisparallel toni, the inner product may be expressed as Tijnj=λni (1.80) and the direction niis calledprincipal direction ofTij.S i n c eni=δijnj, this can be rewritten as (Tij−λδij)nj= 0 (1.81) which represents a system of three equations for the four unknowns niandλ. (T11−λ)n1+T12n2+T13n3=0 T21n1+(T22−λ)n2+T23n3= 0 (1.82-a) T31n1+T32n2+(T33−λ)n3=0 To have a non-trivial slution ( ni= 0) the determinant of the coeﬃcients must be zero, |Tij−λδij|=0 (1.83) 61Expansion of this determinant leads to the following characteristic equation λ3−ITλ2+IITλ−IIIT=0 (1.84) the roots are called the principal values ofTijand IT=Tij=t rTij (1.85) IIT=1 2(TiiTjj−TijTij) (1.86) IIIT=|Tij|=d e tTij (1.87) are called the ﬁrst, second and third invariants respectively of Tij. 62It is customary to order those roots as λ1>λ 2>λ 3 63For a symmetric tensor with real components, the principal values are also real. If those values are distinct, the three principal directions are mutually orthogonal. 1.2.8 Powers of Second Order Tensors; Hamilton-Cayley Equations 64When expressed in term of the principal axes, the tensor array can be written in matrix form as T= λ(1)00 0λ(2)0 00 λ(3)  (1.88) 65By direct matrix multiplication, the quare of the tensor Tijis given by the inner product TikTkj,t h e cube asTikTkmTmn. Therefore the nth power of Tijcan be written as Tn= λn (1)00 0λn (2)0 00 λn (3)  (1.89) Victor Saouma Introduction to Continuum Mechanics Draft1–16 MATHEMATICAL PRELIMINARIES; Part I Vectors and Tensors Since each of the principal values satisﬁes Eq. 1.84 and because the diagonal matrix form of Tgiven above, then the tensor itself will satisfy Eq. 1.84. T3−ITT2+IITT−IIITI=0 (1.90) whereIis the identity matrix. This equation is called the Hamilton-Cayley equation . Victor Saouma Introduction to Continuum Mechanics Draft Chapter 2 KINETICS Or How Forces are Transmitted 2.1 Force, Traction and Stress Vectors 1There are two kinds of forcesin continuum mechanics body forces: act on the elements of volume or mass inside the body, e.g. gravity, electromagnetic ﬁelds. dF=ρbdVol. surface forces: arecontactforcesactingonthe freebodyatitsboundingsurface. Those will be deﬁned in terms of force per unit area. 2The surface force per unit area acting on an element dSis calledtraction or more accurately stress vector . integraldisplay StdS=iintegraldisplay StxdS+jintegraldisplay StydS+kintegraldisplay StzdS (2.1) Most authors limit the term traction to an actual bounding surface of a body, and use the termstress vector for an imaginary interior surface (even though the state of stress is a tensor and not a vector). 3The traction vectors on planes perpendicular to the coordinate axes are particularly useful. When the vectors acting at a point on three such mutually perpendicular planesis given, the stress vector at that point on any other arbitrarily inclined plane can be expressed in terms of the ﬁrst set of tractions. 4Astress, Fig 2.1 is a second order cartesian tensor, σijwhere the 1st subscript ( i) refers to the direction of outward facing normal, and the second one ( j) to the direction of component force. σ=σij= σ11σ12σ13 σ21σ22σ23 σ31σ32σ33 =  t1 t2 t3  (2.2) 5In fact the nine rectangular components σijofσturn out to be the three sets of three vector components ( σ11,σ12,σ13), (σ21,σ22,σ23), (σ31,σ32,σ33) which correspond to Draft2–2 KINETICS 2X∆X3X 12 X3 σσ 11σσ13 21σ23 σ22σ31σ32σ33 12 ∆X1∆X Figure 2.1: Stress Components on an Inﬁnitesimal Element the three tractions t1,t2andt3which are acting on the x1,x2andx3faces (It should be noted that those tractions are not necesarily normal to the faces, and they can bedecomposed into a normal and shear traction if need be). In other words, stresses arenothing else than the components of tractions (stress vector), Fig. 2.2. 13σ 21σ23 σ22σ31 1σ33σ32 X2X1V1X3 X2 (Components of a vector are scalars)VV V2 X3 (Components of a tensor of order 2 are vectors)X3 11σσ 12σ Stresses as components of a traction vectortt t123 Figure 2.2: Stresses as Tensor Components 6The 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 Introduction to Continuum Mechanics Draft2.2 Traction on an Arbitrary Plane; Cauchy’s Stress Tensor 2–3 2.2 Traction on an Arbitrary Plane; Cauchy’s Stress Tensor 7Let 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,t2andt3. This will be done through the so-called Cauchy’s tetrahedron shown in Fig. 2.3. b*∆ V-t ∆S* 1 1 *ρS XX X 312 OhnB n CA N -t * 2∆S2*-t* 3∆S3 t∆ Figure 2.3: Cauchy’s Tetrahedron 8The components of the unit vector nare the direction cosines of its direction: n1=c o s ( /negationslash AON);n2=c o s ( /negationslash BON);n3=c o s ( /negationslash CON); (2.3) The altitude ON,o fl e n g t h his a leg of the three right triangles ANO,BNOandCNO with hypothenuses OA,OB andOC. Hence h=OAn1=OBn2=OCn3 (2.4) 9The volume of the tetrahedron is one third the base times the altitude ∆V=1 3h∆S=1 3OA∆S1=1 3OB∆S2=1 3OC∆S3 (2.5) which when combined with the preceding equation yields ∆S1=∆Sn1;∆S2=∆Sn2;∆S3=∆Sn3; (2.6) or ∆Si=∆Sni. 10In Fig. 2.3 are also shown the averagevalues of the body force and of the surface tractions (thus the asterix). The negative sign appears because t∗ idenotes the average Victor Saouma Introduction to Continuum Mechanics Draft2–4 KINETICS traction on a surface whose outward normal points in the negative xidirection. We seek to determine t∗ n. 11We invoke the momentum principle of a collection of particles (more about it later on) which is postulated to apply to our idealized continuous medium. This principle states that the vector sum of all external forces acting on the free body is equal to the rate of change of the total momentum1. The total momentum isintegraldisplay ∆mvdm.B y t h e mean-value theorem of the integral calculus, this is equal to v∗∆mwherev∗is average value of the velocity. Since we are considering the momentum of a given collection of particles, ∆ mdoes not change with time and ∆ mdv∗ dt=ρ∗∆Vdv∗ dtwhereρ∗is the average density. Hence, the momentum principle yields t∗ n∆S+ρ∗b∗∆V−t∗ 1∆S1−t∗ 2∆S2−t∗ 3∆S3=ρ∗∆Vdv∗ dt(2.7) Substituting for ∆ V,∆Sifrom above, dividing throughout by ∆ Sand rearanging we obtain t∗ n+1 3hρ∗b∗=t∗ 1n1+t∗ 2n2+t∗ 3n3+1 3hρ∗dv dt(2.8) and now we let h→0a n do b t a i n tn=t1n1+t2n2+t3n3=tini (2.9) We observe that we dropped the asterix as the length of the vectors approached zero. 12It is important to note that this result was obtained without any assumption of equi- librium and that it applies as well in ﬂuid dynamics as in solid mechanics. 13This equation is a vector equation, and the corresponding algebraic equations for the components of tnare tn1=σ11n1+σ21n2+σ31n3 tn2=σ12n1+σ22n2+σ32n3 tn3=σ13n1+σ23n2+σ33n3 Indicial notation tni=σjinj dyadic notation tn=n·σ=σT·n (2.10) 14We have thus established that the nine components σijare components of the second order tensor, Cauchy’s stress tensor . 15Note that this stress tensor is really deﬁned in the deformed space (Eulerian), and this issue will be revisited in Sect. 4.3. Example 2-1: Stress Vectors 1Thisisreally Newton’s second law F =ma=mdv dt Victor Saouma Introduction to Continuum Mechanics Draft2.3 Symmetry of Stress Tensor 2–5 if the stress tensor at point Pis given by σ= 7−50 −531 01 2 =  t1 t2 t3  (2.11) We seek to determine the traction (or stress vector) tpassing through Pand parallel to the plane ABCwhereA(4,0,0),B(0,2,0) andC(0,0,6).Solution: The vector normal to the plane can be found by taking the cross products of vectors AB andAC: N=AB×AC=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglee1e2e3 −420 −406vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle(2.12-a) =1 2e1+24e2+8e3 (2.12-b) The unit normal of Nis given by n=3 7e1+6 7e2+2 7e3 (2.13) Hence the stress vector (traction) will be ⌊3 76 72 7⌋ 7−50 −531 01 2 =⌊−9 75 710 7⌋ (2.14) and thust=−9 7e1+5 7e2+10 7e3 2.3 Symmetry of Stress Tensor 16From Fig. 2.1 the resultant force exerted on the positive X1face is ⌊σ11∆X2∆X3σ12∆X2∆X3σ13∆X2∆X3⌋ (2.15) similarly the resultant forces acting on the positive X2face are ⌊σ21∆X3∆X1σ22∆X3∆X1σ23∆X3∆X1⌋ (2.16) 17We now consider moment equilibrium (M=F×d). The stress is homogeneous, and the normal force on the opposite side is equal opposite and colinear. The moment (∆X2/2)σ31∆X1∆X2is likewise balanced by the moment of an equal component in the opposite face. Finally similar argument holds for σ32. 18The net moment about the X3axis is thus M=∆X1(σ12∆X2∆X3)−∆X2(σ21∆X3∆X1) (2.17) which must be zero, hence σ12=σ21. Victor Saouma Introduction to Continuum Mechanics Draft2–6 KINETICS 19We generalize and conclude that in the absence of distributed body forces, the stress matrix is symmetric, σij=σji (2.18) 20A more rigorous proof of the symmetry of the stress tensor will be given in Sect. 6.3.2.1. 2.3.1 Cauchy’s Reciprocal Theorem 21If we consider t1as the traction vector on a plane with normal n1,a n dt2the stress vector at the same point on a plane with normal n2,t h e n t1=n1·σandt2=n2σ (2.19) or in matrix form as {t1}=⌊n1⌋[σ]a n d{t2}=⌊n2⌋[σ] (2.20) If we postmultiply the ﬁrst equation by n2and the second one by n1, by virtue of the symmetry of [ σ]w eh a v e [n1σ]n2=[n2σ]n1 (2.21) or t1·n2=t2·n1 (2.22) 22In the special case of two opposite faces, this reduces to /0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0 /1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0 /1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1 /0/0/0/0/0/0/0/0/0/0/0/0/0 /1/1/1/1/1/1/1/1/1/1/1/1/1/0/0/0/0/0/0/0/0 /1/1/1/1/1/1/1/1/0/0/0/0/1/1/1/1 /0/0/0/0/0/0/1/1/1/1/1/1/0/0/0/0/0/0/1/1/1/1/1/1 /0/0/0/0/0/0/0 /1/1/1/1/1/1/1/0/0/0/0/0/0/0/0/0/0/0/0/0/0 /1/1/1/1/1/1/1/1/1/1/1/1/1/1 /0/0/0/0/0/0/0/0 /1/1/1/1/1/1/1/1/0/0/0/0/0/0/0/0 /1/1/1/1/1/1/1/1 /0/0/0/0/0/0/0/0/0/0 /1/1/1/1/1/1/1/1/1/1t /0/0/0/0/0/0/0/0/0/0/0/0 /1/1/1/1/1/1/1/1/1/1/1/1Γ /0/0/0/0/0/0/0/0 /1/1/1/1/1/1/1/1/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0/0 /1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/1/0/0/0/0/0/0/0/0/1/1/1/1/1/1/1/1 -nn Ω Γn -nt t Figure 2.4: Cauchy’s Reciprocal Theorem tn=−t−n (2.23) Victor Saouma Introduction to Continuum Mechanics Draft2.4Principal Stresses 2–7 23We should note that this theorem is analogous to Newton’s famous third law of motion To every action there is an equal and opposite reaction . 2.4 Principal Stresses 24Regardless of the state of stress (as long as the stress tensor is symmetric), at a given point, it is always possible to choose a special set of axis through the point so that the shear stress components vanish when the stress components are referred to this systemof axis. these special axes are called principal axes of theprincipal stresses . 25To determine the principal directions at any point, we consider nto be a unit vector in one of the unknown directions. It has components ni.L e tλrepresent the principal-stress component on the plane whose normal is n(note both nandλare yet unknown). Since we know that there is no shear stress component on the plane perpendicular to n, 11σ12 tn tn1n2t tnnσ=σ σ =0 t n1tn2nn n1n2σ11σ12tn == n Arbitrary PlaneInitial (X1) Plane Principal Planetn2 t n1 σnσs s Figure 2.5: Principal Stresses the stress vector on this plane must be parallel to nand tn=λn (2.24) 26From Eq. 2.10 and denoting the stress tensor by σwe get n·σ=λn (2.25) in indicial notation this can be rewritten as nrσrs=λns (2.26) or (σrs−λδrs)nr= 0 (2.27) in matrix notation this corresponds to n([σ]−λ[I]) = 0 (2.28) Victor Saouma Introduction to Continuum Mechanics Draft2–8 KINETICS whereIcorresponds to the identity matrix. We really have here a set of three homoge- neous algebraic equations for the direction cosines ni. 27Since the direction cosines must also satisfy n2 1+n2 2+n2 3= 1 (2.29) they can not all be zero. hence Eq.2.28has solutions which are not zero if and only if the determinant of the coeﬃcients is equal to zero, i.e vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleσ11−λσ12σ13 σ21σ22−λσ23 σ31σ32σ33−λvextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle= 0 (2.30) |σrs−λδrs|= 0 (2.31) |σ−λI|= 0 (2.32) 28For a given set of the nine stress components, the preceding equation constitutes a cubic equation for the three unknown magnitudes of λ. 29Cauchy was ﬁrst to show that since the matrix is symmetric and has real elements, the roots are all real numbers. 30The three lambdas correspond to the three principal stresses σ(1)>σ(2)>σ(3).W h e n any one of them is substituted for λin the three equations in Eq. 2.28those equations reduce to only two independent linear equations, which must be solved together with thequadratic Eq. 2.29 to determine the direction cosines n i rof the normal nito the plane on which σiacts. 31The three directions form a right-handed system and n3=n1×n2 (2.33) 32In 2D, it can be shown that the principal stresses are given by: σ1,2=σx+σy 2±radicalBigg parenleftbiggσx−σy 2parenrightbigg2 +τ2xy (2.34) 2.4.1 Invariants 33The principal stresses are physical quantities, whose values do not depend on the coordinate system in which the components of the stress were initially given. They arethereforeinvariants of the stress state. 34When the determinant in the characteristic Eq. 2.32 is expanded, the cubic equation takes the form λ3−Iσλ2−IIσλ−IIIσ=0 (2.35) where the symbols Iσ,IIσandIIIσdenote the following scalar expressions in the stress components: Victor Saouma Introduction to Continuum Mechanics Draft2.5 Stress Transformation 2–9 Iσ=σ11+σ22+σ33=σii=t rσ (2.36) IIσ=−(σ11σ22+σ22σ33+σ33σ11)+σ2 23+σ2 31+σ2 12(2.37) =1 2(σijσij−σiiσjj)=1 2σijσij−1 2I2 σ (2.38) =1 2(σ:σ−I2 σ) (2.39) IIIσ=d e t σ=1 6eijkepqrσipσjqσkr (2.40) 35In terms of the principal stresses, those invariants can be simpliﬁed into Iσ=σ(1)+σ(2)+σ(3) (2.41) IIσ=−(σ(1)σ(2)+σ(2)σ(3)+σ(3)σ(1)) (2.42) IIIσ=σ(1)σ(2)σ(3) (2.43) 2.4.2 Spherical and Deviatoric Stress Tensors 36If we letσdenote the mean normal stress p σ=−p=1 3(σ11+σ22+σ33)=1 3σii=1 3trσ (2.44) then the stress tensor can be written as the sum of two tensors: Hydrostatic stress in which each normal stress is equal to −pand the shear stresses are zero. The hydrostatic stress produces volume change without change in shape in an isotropic medium. σhyd=−pI= −p00 0−p0 00−p  (2.45) Deviatoric Stress: which causes the change in shape. σdev= σ11−σσ12σ13 σ21σ22−σσ23 σ31σ32σ33−σ  (2.46) 2.5 Stress Transformation 37From Eq. 1.73 and 1.74, the stress transformation for the second order stress tensor is given by σip=aj iaq pσjqin Matrix Form [ σ]=[A]T[σ][A] (2.47) σjq=aj iaq p σipin Matrix Form [ σ]=[A][ σ][A]T(2.48) Victor Saouma Introduction to Continuum Mechanics Draft2–10 KINETICS 38For the 2D plane stress case we rewrite Eq. 1.76    σxx σyy σxy  = cos2α sin2α2sinαcosα sin2α cos2α−2sinαcosα −sinαcosαcosαsinαcos2α−sin2α   σxx σyy σxy   (2.49) Example 2-2: Principal Stresses The stress tensor is given at a point by σ= 311 102 120  (2.50) determine the principal stress values and the corresponding directions. Solution:From Eq.2.32 we have vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle3−λ11 10−λ2 12 0 −λvextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle= 0 (2.51) Or upon expansion (and simpliﬁcation) ( λ+2 ) (λ−4)(λ−1) = 0, thus the roots are σ(1)=4 ,σ(2)=1a n dσ(3)=−2. We also note that those are the three eigenvalues of the stress tensor. If we let x1axis be the one corresponding to the direction of σ(3)andn3 ibe the direction cosines of this axis, then from Eq. 2.28we have   (3+2)n3 1+n3 2+n3 3=0 n3 1+2n3 2+2n3 3=0 n3 1+2n3 2+2n3 3=0⇒n3 1=0 ;n3 2=1 √ 2;n3 3=−1 √ 2(2.52) Similarly If we let x2axis be the one corresponding to the direction of σ(2)andn2 ibe the direction cosines of this axis,   2n2 1+n2 2+n2 3=0 n2 1−n2 2+2n2 3=0 n2 1+2n2 2−n2 3=0⇒n2 1=1 √ 3;n2 2=−1 √ 3;n2 3=−1 √ 3(2.53) Finally, if we let x3axis be the one corresponding to the direction of σ(1)andn1 ibe the direction cosines of this axis,   −n1 1+n1 2+n1 3=0 n1 1−4n1 2+2n1 3=0 n1 1+2n1 2−4n1 3=0⇒n1 1=−2 √ 6;n1 2=−1 √ 6;n1 3=−1 √ 6(2.54) Finally, we can convince ourselves that the two stress tensors have the same invariants Iσ,IIσandIIIσ. Example 2-3: Stress Transformation Victor Saouma Introduction to Continuum Mechanics Draft2.5 Stress Transformation 2–11 Show that the transformation tensor of direction cosines previously determined trans- forms the original stress tensor into the diagonal principal axes stress tensor.Solution: From Eq. 2.47 σ= 01 √ 2−1 √ 21 √ 3−1 √ 3−1 √ 3 −2 √ 6−1 √ 6−1 √ 6  311 102122  01 √ 3−2 √ 61 √ 2−1 √ 3−1 √ 6 −1 √ 2−1 √ 3−1 √ 6 (2.55-a) = −200 01 0 00 4  (2.55-b) 2.5.1 Plane Stress 39Plane stress conditions prevail when σ3i= 0, and thus we have a biaxial stress ﬁeld. 40Plane stress condition prevail in (relatively) thin plates, i.e when one of the dimensions is much smaller than the other two. 2.5.2 Mohr’s Circle for Plane Stress Conditions 41The Mohr circle will provide a graphical mean to contain the transformed state of stress ( σxx, σyy, σxy) at an arbitrary plane (inclined by α) in terms of the original one (σxx,σyy,σxy). 42Substituting cos2α=1+cos2α 2sin2α=1−cos2α 2 cos2α=c o s2α−sin2αsin2α=2 s i nαcosα(2.56) into Eq. 2.49 and after some algebraic manipulation we obtain σxx=1 2(σxx+σyy)+1 2(σxx−σyy)cos2α+σxysin2α (2.57-a) σxy=σxycos2α−1 2(σxx−σyy)sin2α (2.57-b) 43Points (σxx,σxy),(σxx,0),(σyy,0) and [(σxx+σyy)/2,0] are plotted in the stress repre- sentation of Fig. 2.6. Then we observe that 1 2(σxx−σyy)=Rcos2β (2.58-a) σxy=Rsin2β (2.58-b) Victor Saouma Introduction to Continuum Mechanics Draft2–12 KINETICS xy 2β2ασxxτxy X( , ) 21 σσxx yy( + ) 21 σσ1 2( + ) 21 σσ1 2( - )21 σ σxx yy( - )σnσxxτxyσxxτxy τxy τyxσxx σyyσyy σyyτxy τyxτxyτyx σxxσxxy x σxxσyy σyyτyxτyx τxyτxy σxx τα yxτ(a) (b) (c)(d)Oσ2 σ1 Cσ σyy xx DR −2β2αX( , )nτ xαxy αA BQxy xy Figure 2.6: Mohr Circle for Plane Stress Victor Saouma Introduction to Continuum Mechanics Draft2.5 Stress Transformation 2–13 where R=radicalBigg 1 4(σxx−σyy)2+σ2 xy (2.59-a) tan2β=2σxy σxx−σyy(2.59-b) then after substitution and simpliﬁation, Eq. 2.57-a and 2.57-b would result in σxx=1 2(σxx+σyy)+Rcos(2β−2α) (2.60) σxy=Rsin(2β−2α) (2.61) We observe that the form of these equations, indicates that σxxand σxyare on a circle centered at1 2(σxx+σyy) and of radius R. Furthermore, since σxx,σyy,Randβare deﬁnite numbers for a given state of stress, the previous equations provide a graphical solution for the evaluation of the rotated stress σxxand σxyfor various angles α. 44By eliminating the trigonometric terms, the Cartesian equation of the circle is given by [ σxx−1 2(σxx+σyy)]2+ σ2 xy=R2(2.62) 45Finally, the graphical solution for the state of stresses at an inclined plane is summa- rized as follows 1. Plot the points ( σxx,0), (σyy,0),C:[1 2(σxx+σyy),0], andX:(σxx,σxy). 2. Draw the line CX, this will be the reference line corresponding to a plane in the physical body whose normal is the positive xdirection. 3. Draw a circle with center Cand radius R=CX. 4. To determine the point that represents any plane in the physical body with normal making a counterclockwise angle αwith the xdirection, lay oﬀ angle 2 αclockwise fromCX. The terminal side C Xof this angle intersects the circle in point Xwhose coordinates are ( σxx, σxy). 5. To determine σyy, consider the plane whose normal makes an angle α+1 2πwith the positivexaxis in the physical plane. The corresponding angle on the circle is 2 α+π measured clockwise from the reference line CX.T h i sl o c a t e sp o i n t Dwhich is at the opposite end of the diameter through X. The coordinates of Dare ( σyy,− σxy) Example 2-4: Mohr’s Circle in Plane Stress An element in plane stress is subjected to stresses σxx= 15,σyy=5a n d τxy=4 . Using the Mohr’s circle determine: a) the stresses acting on an element rotated through an angle θ=+ 4 0o(counterclockwise); b) the principal stresses; and c) the maximum shear stresses. Show all results on sketches of properly oriented elements.Solution: With reference to Fig. 2.7: Victor Saouma Introduction to Continuum Mechanics Draft2–14 KINETICS 15445 15 4 54 4014.81 4.235.19 o 10.006.4010.00 25.7o41.34 16.43.6 19.3oσn θ=40onτ 10 544 6.4X(15,4) 580o 15ooθ=0o oθ=90θ=19.3o θ=64.3θ=−25.7 oo θ=109.3o 38.66 Figure 2.7: Plane Stress Mohr’s Circle; Numerical Example 1. The center of the circle is located at 1 2(σxx+σyy)=1 2(15+5) = 10 . (2.63) 2. The radius and the angle 2 βare given by R=radicalBigg 1 4(15−5)2+42=6.403 (2.64-a) tan2β=2(4) 15−5=0.8⇒2β=3 8.66o;β=1 9.33o(2.64-b) 3. The stresses acting on a plane at θ=+ 4 0oare given by the point making an angle of−80o(clockwise) with respect to point X(15,4) or−80o+38.66o=−41.34owith respect to the axis. 4. Thus, by inspection the stresses on the xface are σxx=1 0 + 6 .403cos−41.34o= 14.81 (2.65-a) τxy=6.403sin−41.34o= −4.23 (2.65-b) 5. Similarly, the stresses at the face yare given by σyy=1 0 + 6 .403cos(180o−41.34o)= 5.19 (2.66-a) τxy=6.403sin(180o−41.34o)= 4.23 (2.66-b) Victor Saouma Introduction to Continuum Mechanics Draft2.6 Simpliﬁed Theories; Stress Resultants 2–15 6. The principal stresses are simply given by σ(1)=1 0 + 6 .4= 16.4 (2.67-a) σ(2)=1 0−6.4= 3.6 (2.67-b) σ(1)acts on a plane deﬁned by the angle of +19 .3oclockwise from the xaxis, and σ(2)acts at an angle of38.66o+180o 2= 109.3o with respect to the xaxis. 7. The maximum and minimum shear stresses are equal to the radius of the circle, i.e 6.4a ta na n g l eo f 90o−38.66o 2= 25.70 (2.68) 2.5.3†Mohr’s Stress Representation Plane 46Therecanbeaninﬁnitenumberofplanespassingthroughapoint O,eachcharacterized by their own normal vector along ON, Fig. 2.8. To each plane will correspond a set of σnandτn. AGB H F DON CY Zσ σII IIIγαβ JE Figure 2.8: Unit Sphere in Physical Body around O 47It can be shown that all possible sets of σnandτnwhich can act on the point Oare within the shaded area of Fig. 2.9. 2.6 Simpliﬁed Theories; Stress Resultants 48For many applications of continuum mechanics the problem of determining the three- dimensionalstressdistributionistoodiﬃculttosolve. However, inmany(civil/mechanical)applications, Victor Saouma Introduction to Continuum Mechanics Draft2–16 KINETICS 1( 2σσ ) Ι+ ΙΙΙ1( 2σσ ) ΙΙΙΙ- σI σ IIIσII n 1( 2σσ )+ ΙΙΙΙΙ1( 2σσ ) ΙΙΙ-1( 2σσ ) ΙΙΙ- ΙΙ O C C CIII IIIτ σn Figure 2.9: Mohr Circle for Stress in 3D one or more dimensions is/are small compared to the others and possess certain symme- tries of geometrical shape and load distribution. 49In those cases, we may apply “ engineering theories ” for shells, plates or beams. In those problems, instead of solving for the stress components throughout the body,we solve for certain stress resultants (normal, shear forces, and Moments and torsions) resulting from an integration over the body. We consider separately two of those three cases. 50Alternatively, if a continuum solution is desired, and engineering theories prove to be either too restrictive or inapplicable, we can use numerical techniques (such as the Finite Element Method ) to solve the problem. 2.6.1 Arch 51Fig. 2.10 illustrates the stresses acting on a diﬀerential element of a shell structure. The resulting forces in turn are shown in Fig. 2.11 and for simpliﬁcation those actingper unit length of the middle surface are shown in Fig. 2.12. The net resultant forces Victor Saouma Introduction to Continuum Mechanics Draft2.6 Simpliﬁed Theories; Stress Resultants 2–17 Figure 2.10: Diﬀerential Shell Element, Stresses Figure 2.11: Diﬀerential Shell Element, Forces Victor Saouma Introduction to Continuum Mechanics Draft2–18 KINETICS Figure 2.12: Diﬀerential Shell Element, Vectors of Stress Couples are given by: Membrane Force N=integraldisplay+h 2 −h 2σparenleftbigg 1−z rparenrightbigg dz  Nxx=integraldisplay+h 2 −h 2σxxparenleftBigg 1−z ryparenrightBigg dz Nyy=integraldisplay+h 2 −h 2σyyparenleftbigg 1−z rxparenrightbigg dz Nxy=integraldisplay+h 2 −h 2σxyparenleftBigg 1−z ryparenrightBigg dz Nyx=integraldisplay+h 2 −h 2σxyparenleftbigg 1−z rxparenrightbigg dz Bending Moments M=integraldisplay+h 2 −h 2σzparenleftbigg 1−z rparenrightbigg dz  Mxx=integraldisplay+h 2 −h 2σxxzparenleftBigg 1−z ryparenrightBigg dz Myy=integraldisplay+h 2 −h 2σyyzparenleftbigg 1−z rxparenrightbigg dz Mxy=−integraldisplay+h 2 −h 2σxyzparenleftBigg 1−z ryparenrightBigg dz Myx=integraldisplay+h 2 −h 2σxyzparenleftbigg 1−z rxparenrightbigg dz Transverse Shear Forces Q=integraldisplay+h 2 −h 2τparenleftbigg 1−z rparenrightbigg dz  Qx=integraldisplay+h 2 −h 2τxzparenleftBigg 1−z ryparenrightBigg dz Qy=integraldisplay+h 2 −h 2τyzparenleftbigg 1−z rxparenrightbigg dz(2.69) Victor Saouma Introduction to Continuum Mechanics Draft2.6 Simpliﬁed Theories; Stress Resultants 2–19 2.6.2 Plates 52Considering an arbitrary plate, the stresses and resulting forces are shown in Fig. 2.13, and resultants per unit width are given by Figure 2.13: Stresses and Resulting Forces in a Plate Membrane Force N =integraldisplayt 2 −t 2σdz  Nxx=integraldisplayt 2 −t 2σxxdz Nyy=integraldisplayt 2 −t 2σyydz Nxy=integraldisplayt 2 −t 2σxydz Bending Moments M =integraldisplayt 2 −t 2σzdz  Mxx=integraldisplayt 2 −t 2σxxzdz Myy=integraldisplayt 2 −t 2σyyzdz Mxy=integraldisplayt 2 −t 2σxyzdz Transverse Shear Forces V =integraldisplayt 2 −t 2τdz  Vx=integraldisplayt 2 −t 2τxzdz Vy=integraldisplayt 2 −t 2τyzdz(2.70-a) 53Note that in plate theory, we ignore the eﬀect of the membrane forces, those in turn will be accounted for in shells. Victor Saouma Introduction to Continuum Mechanics Draft2–20 KINETICS Victor Saouma Introduction to Continuum Mechanics Draft Chapter 3 MATHEMATICAL PRELIMINARIES; Part II VECTOR DIFFERENTIATION 3.1 Introduction 1Aﬁeldis a function deﬁned over a continuous region. This includes, Scalar Field g(x),Vector Field v (x), Fig. 3.1 or Tensor Field T (x). 2We ﬁrst introduce the diﬀerential vector operator “Nabla” denoted by ∇ ∇≡∂ ∂xi+∂ ∂yj+∂ ∂zk (3.1) 3We 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 3.1. Multiplication Diﬀerentiation Tensor Order u·vdot ∇·vdivergence ❄ u×vcross ∇×vcurl ✲ u⊗vtensor ∇vgradient ✻ Table 3.1: Similarities Between Multiplication and Diﬀerentiation Operators 3.2 Derivative WRT to a Scalar 4The derivative of a vector p(u) with respect to a scalar u, Fig. 3.2 is deﬁned by dp du≡lim ∆u→0p(u+∆u)−p(u) ∆u(3.2) Draft3–2 MATHEMATICAL PRELIMINARIES; Part II VECTOR DIFFERENTIATION ‡Scalar and Vector Fields ContourPlot @Exp@−Hx^2 +y^2LD,8x,−2, 2 <,8y,−2, 2 <, ContourShading −>False D -2 -1 0 1 2-2-1012 Ö ContourGraphics Ö Plot3D @Exp@−Hx^2 +y^2LD,8x,−2, 2 <,8y,−2, 2 <, FaceGrids −>AllD -2 -1 0 1 2-2-1012 00.250.50.751 -2 -1 0 1 Ö SurfaceGraphics Öm−fields.nb 1 Figure 3.1: Examples of a Scalar and Vector Fields (u+ u)∆C (u)p pp(u+ u)- (u)= ∆ p p ∆ Figure 3.2: Diﬀerentiation of position vector p Victor Saouma Introduction to Continuum Mechanics Draft3.2 Derivative WRT to a Scalar 3–3 5Ifp(u)i saposition vector p (u)=x(u)i+y(u)j+z(u)k,t h e n dp du=dx dui+dy duj+dz duk (3.3) is a vector along the tangent to the curve. 6Ifuis the time t,t h e ndp dtis the velocity 7Indiﬀerential geometry ,i fw ec o n s i d e rac u r v e Cdeﬁned by the function p(u)t h e n dp duis a vector tangent ot C,a n di fuis the curvilinear coordinate smeasured from any point along the curve, thendp dsis a unit tangent vector to CT, Fig. 3.3. and we have the CTN B Figure 3.3: Curvature of a Curve following relations dp ds=T (3.4) dT ds=κN (3.5) B=T×N (3.6) κcurvature (3.7) ρ=1 κRadius of Curvature (3.8) we also note that p·dp ds=0i fvextendsinglevextendsinglevextendsingledp dsvextendsinglevextendsinglevextendsingle/negationslash=0 . Example 3-1: Tangent to a Curve Determine the unit vector tangent to the curve: x=t2+1,y=4t−3,z=2t2−6t fort=2 . Solution: Victor Saouma Introduction to Continuum Mechanics Draft3–4MATHEMATICAL PRELIMINARIES; Part II VECTOR DIFFERENTIATION dp dt=d dtbracketleftBig (t2+1)i+(4t−3)j+(2t2−6t)kbracketrightBig =2ti+4j+(4t−6)k(3.9-a) vextendsinglevextendsinglevextendsinglevextendsinglevextendsingledp dtvextendsinglevextendsinglevextendsinglevextendsinglevextendsingle=radicalBig (2t)2+(4)2+(4t−6)2 (3.9-b) T=2ti+4j+(4t−6)k radicalBig (2t)2+(4)2+(4t−6)2(3.9-c) =4i+4j+2k radicalBig (4)2+(4)2+(2)2=2 3i+2 3j+1 3kfort=2 ( 3 . 9 - d ) Mathematica solution is shown in Fig. 3.4 ‡Parametric Plot in 3D ParametricPlot3D @8t^2 +1, 4 t −3, 2 t^2 −6t<,8t, 0, 4 <D 0 5 10 150510 05 0 5 10 150510 Ö Graphics3D Öm−par3d.nb 1 Figure 3.4: Mathematica Solution for the Tangent to a Curve in 3D 3.3 Divergence 3.3.1 Vector 8Thedivergence of a vector ﬁeld of a body Bwith boundary Ω, Fig. 3.5 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). Victor Saouma Introduction to Continuum Mechanics Draft3.3 Divergence 3–5 v(x) BΩn Figure 3.5: Vector Field Crossing a Solid Region 9The divergence of the vector ﬁeld is thus deﬁned as divv(x)≡lim v(B)→01 v(B)integraldisplay Ωv·ndA (3.10) wherev.nis often referred as the ﬂuxand represents the total volume of “ﬂuid” that passes through dAin unit time, Fig. 3.6 This volume is then equal to the base of the v Ωv.ndAn Figure 3.6: Flux Through Area dA cylinderdAtimes 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. 10The divergence thus measure the rate of change of a vector ﬁeld. 11The 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,a n d∆x3, and with normal vectors pointing in the directions of the coordinate axies, Fig. 3.7. If we also consider the corner closest to the origin as located at x,t h e n the contribution (from Eq. 3.10) of the two surfaces with normal vectors e1and−e1is lim ∆x1,∆x2,∆x3→01 ∆x1∆x2∆x3integraldisplay ∆x2∆x3[v(x+∆x1e1)·e1+v(x)·(−e1)]dx2dx3(3.11) or lim ∆x1,∆x2,∆x3→01 ∆x2∆x3integraldisplay ∆x2∆x3v(x+∆x1e1)−v(x) ∆x1·e1dx2dx3= lim ∆x1→0∆v ∆x1·e1(3.12-a) Victor Saouma Introduction to Continuum Mechanics Draft3–6 MATHEMATICAL PRELIMINARIES; Part II VECTOR DIFFERENTIATION x∆x∆x∆ 2133e-e e -e e -exx x2 13 11 22 3 Figure 3.7: Inﬁnitesimal Element for the Evaluation of the Divergence =∂v ∂x1·e1(3.12-b) hence, we can generalize divv(x)=∂v(x) ∂xi·ei (3.13) 12or alternatively divv=∇·v=(∂ ∂x1e1+∂ ∂x2e2+∂ ∂x3e3)·(v1e1+v2e2+v3e3) (3.14) =∂v1 ∂x1+∂v2 ∂x2+∂v3 ∂x3=∂vi ∂xi=∂ivi=vi,i (3.15) 13The divergence of a vector is a scalar. 14We note that the Laplacian Operator is deﬁned as ∇2F≡∇∇F=F,ii (3.16) Example 3-2: Divergence Determine the divergence of the vector A=x2zi−2y3z2j+xy2zkat point (1 ,−1,1). Solution: ∇·v=parenleftBigg∂ ∂xi+∂ ∂yj+∂ ∂zkparenrightBigg ·(x2zi−2y3z2j+xy2zk) (3.17-a) =∂x2z ∂x+∂−2y3z2 ∂y+∂xy2z ∂z(3.17-b) Victor Saouma Introduction to Continuum Mechanics Draft3.3 Divergence 3–7 =2xz−6y2z2+xy2(3.17-c) = 2(1)(1) −6(−1)2(1)2+(1)(−1)2=−3a t( 1,−1,1) (3.17-d) Mathematica solution is shown in Fig. 3.8 ‡Divergence of a Vector <<Calculus‘VectorAnalysis‘ V=8x^2z, −2y^3z^2, xy^2z <; Div@V, Cartesian @x, y, z DD -6z2y2+xy2+2xz <<Graphics‘PlotField3D‘ PlotVectorField3D @8x^2 z, −2y^3z^2, x y^2 z <,8x,−10, 10 <,8y,−10, 10 <,8z,−10, 10 <, Axes −>Automatic, AxesLabel −>8"X", "Y", "Z" <D -10 -5 0 5 10X-10-50510Y -10-50510 Z -10 -5 0 5X10-505Y Ö Graphics3D Ö Div@Curl @V, Cartesian @x, y, z DD, Cartesian @x, y, z DD 0m−diver.nb 1 Figure 3.8: Mathematica Solution for the Divergence of a Vector 3.3.2 Second-Order Tensor 15By analogy to Eq. 3.10, the divergence of a second-order tensor ﬁeld Tis ∇·T=d i vT(x)≡lim v(B)→01 v(B)integraldisplay ΩT·ndA (3.18) which is the vector ﬁeld ∇·T=∂Tpq ∂xpeq (3.19) Victor Saouma Introduction to Continuum Mechanics Draft3–8 MATHEMATICAL PRELIMINARIES; Part II VECTOR DIFFERENTIATION 3.4 Gradient 3.4.1 Scalar 16Thegradient of a scalar ﬁeld g(x) is a vector ﬁeld ∇g(x) such that for any unit vector v, the directional derivative dg/dsin the direction of vis given by dg ds=∇g·v (3.20) wherev=dp dsWe note that the deﬁnition made no reference to any coordinate system. T h eg r a d i e n ti st h u sa vector invariant . 17To ﬁnd the components in any rectangular Cartesian coordinate system we use v=dp ds=dxi dsei (3.21-a) dg ds=∂g ∂xidxi ds(3.21-b) which can be substituted and will yield ∇g=∂g ∂xiei (3.22) or ∇φ≡parenleftBigg∂ ∂xi+∂ ∂yj+∂ ∂zkparenrightBigg φ (3.23-a) =∂φ ∂xi+∂φ ∂yj+∂φ ∂zk (3.23-b) and note that it deﬁnes a vector ﬁeld . 18The physical signiﬁcance ofthegradientofascalarﬁeld isthatitpointsinthedirection in which the ﬁeld is changing most rapidly (for a three dimensional surface, the gradientis pointing along the normal to the plane tangent to the surface). The length of thevector||∇g(x)||is perpendicular to the contour lines. 19∇g(x)·ngives the rate of change of the scalar ﬁeld in the direction of n. Example 3-3: Gradient of a Scalar Determine the gradient of φ=x2yz+4xz2at point (1 ,−2,−1) along the direction 2i−j−2k. Solution: ∇φ=∇(x2yz+4xz2)=( 2xyz+4z2)i+(x2zj+(x2y+8xz)k(3.24-a) Victor Saouma Introduction to Continuum Mechanics Draft3.4Gradient 3–9 =8i−j−10kat (1,−2,−1) (3.24-b) n=2i−j−2k radicalBig (2)2+(−1)2+(−2)2=2 3i−1 3j−2 3k (3.24-c) ∇φ·n=( 8i−j−10k)·parenleftbigg2 3i−1 3j−2 3kparenrightbigg =16 3+1 3+20 3=37 3(3.24-d) Since this last value is positive, φincreases along that direction. Example 3-4: Stress Vector normal to the Tangent of a Cylinder The stress tensor throughout a continuum is given with respect to Cartesian axes as σ= 3x1x25x2 20 5x2 202x2 3 02x30  (3.25) Determine the stress vector (or traction) at the point P(2,1,√ 3) of the plane that is tangent to the cylindrical surface x2 2+x2 3=4a tP, Fig. 3.9. n 12 3 xxx 123 P Figure 3.9: Radial Stress vector in a Cylinder Solution: At point P, the stress tensor is given by σ= 65 0 502√ 3 02√ 30  (3.26) Victor Saouma Introduction to Continuum Mechanics Draft3–10 MATHEMATICAL PRELIMINARIES; Part II VECTOR DIFFERENTIATION The unit normal to the surface at Pis given from ∇(x2 2+x2 3−4) = 2x222+2x3e3 (3.27) At point P, ∇(x2 2+x2 3−4) = 222+2√ 3e3 (3.28) and thus the unit normal at Pis n=1 2e1+√ 3 2e3 (3.29) Thus the traction vector will be determined from σ= 65 0 502√ 3 02√ 30   0 1/2√ 3/2  =  5/2 3√ 3  (3.30) ortn=5 2e1+3e2+√ 3e3 3.4.2 Vector 20We can also deﬁne the gradient of a vector ﬁeld. If we consider a solid domain Bwith boundary Ω, Fig. 3.5, then the gradient of the vector ﬁeld v(x) is a second order tensor deﬁned by ∇xv(x)≡lim v(B)→01 v(B)integraldisplay Ωv⊗ndA (3.31) and with a construction similar to the one used for the divergence, it can be shown that ∇xv(x)=∂vi(x) ∂xj[ei⊗ej] (3.32) where summation is implied for both iandj. 21The components of ∇xvare simply the various partial derivatives of the component functions with respect to the coordinates: [∇xv]= ∂vx ∂x∂vy ∂x∂vz ∂x ∂vx ∂y∂vy ∂y∂vz ∂y ∂vx ∂z∂vy ∂z∂vz ∂z (3.33) [v∇x]= ∂vx ∂x∂vx ∂y∂vx ∂z ∂vy ∂x∂vy ∂y∂vy ∂z ∂vz ∂x∂vz ∂y∂vz ∂z (3.34) that is [ ∇v]ijgives the rate of change of the ith component of vwith respect to the jth coordinate axis. 22Note the diference between v∇xand∇xv. In matrix representation, one is the trans- pose of the other. Victor Saouma Introduction to Continuum Mechanics Draft3.4Gradient 3–11 23The gradient of a vector is a tensor of order 2. 24We can interpret the gradient of a vector geometrically, Fig. 3.10. If we consider two pointsaandbthat are near to each other (i.e ∆ sis very small), and let the unit vector mpoints in the direction from atob. The value of the vector ﬁeld at aisv(x)a n d the value of the vector ﬁeld at bisv(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)a n dv(x+∆sm)i s v(x+∆sm)−v(x), the change in vector. Thus, if we divide this change by ∆ s,t h e nw e get the rate of change as we move in the speciﬁed direction. Finally, taking the limit as∆sgoes to zero, we obtain lim ∆s→0v(x+∆sm)−v(x) ∆s≡Dv(x)·m (3.35) sm∆sm v(x+ )∆sm v(x+ ) ∆-v(x) xxx 123 abv(x) Figure 3.10: Gradient of a Vector The quantity Dv(x)·mis called the directional derivative because it gives the rate of change of the vector ﬁeld as we move in the direction m. Example 3-5: Gradient of a Vector Field Determine the gradient of the following vector ﬁeld v(x)=x1x2x3(x1e1+x2e2+x3e3). Solution: ∇xv(x)=2x1x2x3[e1⊗e1]+x2 1x3[e1⊗e2]+x2 1x2[e1⊗e3] +x2 2x3[e2⊗e1]+2x1x2x3[e2⊗e2]+x1x2 2[e2⊗e3] (3.36-a) +x2x2 3[e3⊗e1]+x1x2 3[e3⊗e2]+2x1x2x3[e3⊗e3] =x1x2x3 2x1/x2x1/x3 x2/x12x2/x3 x3/x1x3/x22  (3.36-b) Victor Saouma Introduction to Continuum Mechanics Draft3–12 MATHEMATICAL PRELIMINARIES; Part II VECTOR DIFFERENTIATION 3.4.3 Mathematica Solution 25Mathematica solution of the two preceding examples is shown in Fig. 3.11. Gradient Scalar f=x^2yz+4xz ^2; Gradf=Grad@f, Cartesian@x, y, zDD 84z2+2xyz,x2z,yx2+8zx< <<Graphics‘PlotField3D‘ PlotGradientField3D@f, 8x, 0, 2<, 8y, -3, -1<, 8z, -2, 0<D Graphics3D x=1;y=-2;z=-1; vect=82, -1, -2< Sqrt@ 4+1+4D 92 3,-1 3,-2 3= Gradf.vect 37 3 Gradient of a Vector vecfield=x1x2x3 8x1, x2, x3< 8x12x2 x3, x1 x22x3, x1 x2 x32<m−grad.nb 1PlotVectorField3D@vecfield, 8x1, -10, 10<, 8x2, -10, 10<, 8x3, -10, 10<, Axes->Automatic, AxesLabel->8"x1", "x2", "x3"<D -10 0 10x1-10010x2 -10010 x3 -10 0 10x1-10010x2 Graphics3D MatrixForm@Grad@vecfield, Cartesian@x1, x2, x3DDD i kjjjjjjjjjjj2x 1x 2x 3 x 1 2x3 x12x2 x22x 3 2x 1x 2x 3 x 1x 22 x2 x32x1 x322x 1x 2x 3y {zzzzzzzzzzzm−grad.nb 2 Figure 3.11: Mathematica Solution for the Gradients of a Scalar and of a Vector 3.5 Curl 26When the vector operator ∇operates in a manner analogous to vector multiplication, the result is a vector, curl vcalled the curl of the vector ﬁeld v(sometimes called the rotation). Victor Saouma Introduction to Continuum Mechanics Draft3.6 Some useful Relations 3–13 curlv=∇×v=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglee1e2e3 ∂ ∂x1∂ ∂x2∂ ∂x3 v1v2v3vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle(3.37) =parenleftBigg∂v3 ∂x2−∂v2 ∂x3parenrightBigg e1+parenleftBigg∂v1 ∂x3−∂v3 ∂x1parenrightBigg e2+parenleftBigg∂v2 ∂x1−∂v1 ∂x2parenrightBigg e3(3.38) =eijk∂jvk (3.39) Example 3-6: Curl of a vector Determine the curl of the following vector A=xz3i−2x2yzj+2yz4kat (1,−1,1). Solution: ∇×A=parenleftBigg∂ ∂xi+∂ ∂yj+∂ ∂zkparenrightBigg ×(xz3i−2x2yzj+2yz4k) (3.40-a) =vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleijk ∂ ∂x∂ ∂y∂ ∂z xz3−2x2yz2yz4vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle(3.40-b) =parenleftBigg∂2yz4 ∂y−∂−2x2yz ∂zparenrightBigg i+parenleftBigg∂xz3 ∂z−∂2yz4 ∂xparenrightBigg j+parenleftBigg∂−2x2yz ∂x−∂xz3 ∂yparenrightBigg k(3.40-c) =( 2z4+2x2y)i+3xz2j−4xyzk (3.40-d) =3j+4kat (1,−1,1) (3.40-e) Mathematica solution is shown in Fig. 3.12. 3.6 Some useful Relations 27Some useful relations d(A·B)=A·dB+dA·B (3.41-a) d(A×B)=A×dB+dA×B (3.41-b) ∇(φ+ξ)= ∇φ+∇ξ (3.41-c) ∇×(A+B)= ∇×A+∇×B (3.41-d) ∇·v/negationslash=v∇ (3.41-e) ∇·(φA)=( ∇φ)·A+φ(∇×A) (3.41-f) ∇·(A×B)=B·(∇×A)−A·(∇×B) (3.41-g) ∇(A·B)=(B·∇)A+(A·∇)B+B×(∇×A)+A×(∇×B) (3.41-h) ∇·(∇φ)≡∇2φ≡∂2φ ∂x2+∂2φ ∂y2+∂2φ ∂z2Laplacian Operator (3.41-i) ∇·(∇×A) = 0 (3.41-j) ∇×(∇φ)=0 (3.41-k) Victor Saouma Introduction to Continuum Mechanics Draft3–14MATHEMATICAL PRELIMINARIES; Part II VECTOR DIFFERENTIATION ‡Curl <<Calculus ‘VectorAnalysis ‘ A=8xz^3, −2x^2yz, 2yz^4 <; CurlOfA =Curl @A, Cartesian @x, y, z DD 82z4+2x2y,3xz2,-4xyz < <<Graphics ‘PlotField3D ‘ PlotVectorField3D @CurlOfA, 8x, 0, 2 <,8y,−2, 0<,8z, 0, 2 <, Axes −>Automatic, AxesLabel −>8"x", "y", "z" <D 0 0.5 1 1.5 2x-2-1.5-1-0.50y 00.511.52 z 0 0.5 1 1.5 2x2-1.5-1-0.50y Ö Graphics3D Ö Div@CurlOfA, Cartesian @x, y, z DD 0 x=1; y =−1;z =1; CurlOfA 80, 3, 4 <m−curl.nb 1 Figure 3.12: Mathematica Solution for the Curl of a Vector Victor Saouma Introduction to Continuum Mechanics Draft3.6 Some useful Relations 3–15 Victor Saouma Introduction to Continuum Mechanics Draft3–16 MATHEMATICAL PRELIMINARIES; Part II VECTOR DIFFERENTIATION Victor Saouma Introduction to Continuum Mechanics Draft Chapter 4 KINEMATIC Or on How Bodies Deform 4.1 Elementary Deﬁnition of Strain 20We begin our detailed coverage of strain by a simpliﬁed and elementary set of deﬁni- tions for the 1D and 2D cases. Following this a mathematically rigorous derivation ofthe various expressions for strain will follow. 4.1.1 Small and Finite Strains in 1D 21We begin by considering an elementary case, an axial rod with initial lenght l0,a n d subjected to a deformation ∆ linto a ﬁnal deformed length of l, Fig. 4.1. l0l∆ l Figure 4.1: Elongation of an Axial Rod 22We seek to quantify the deformation of the rod and even though we only have 2 variables ( l0andl), there are diﬀerent possibilities to introduce the notion of strain.W e ﬁrst deﬁne the stretch of the rod as λ≡l l0(4.1) This stretch is one in the undeformed case, and greater than one when the rod is elon- gated. Draft4–2 KINEMATIC 23Usingl0,landλwe next introduce four possible deﬁnitions of the strain in 1D: Engineering Strain ε≡l−l0 l0=λ−1 Natural Strain η=l−l0 l=1−1 λ Lagrangian Strain E≡1 2parenleftBigl2−l2 0 l2 0parenrightBig =1 2(λ2−1) Eulerian Strain E∗≡1 2parenleftBigl2−l2 0 l2parenrightBig =1 2parenleftBig 1−1 λ2parenrightBig (4.2) we note the strong analogy between the Lagrangian and the engineering strain on the one hand, and the Eulerian and the natural strain on the other. 24The choice of which strain deﬁnition to use is related to the stress-strain relation (or constitutive law) that we will later adopt. 4.1.2 Small Strains in 2D 25The elementary deﬁnition of strains in 2D is illustrated by Fig. 4.2 and are given by ∆ux∆uy ∆ux ∆uy2 ∆∆Y XUniaxial Extension Pure Shear Without Rotation ∆∆ XY θθ ψ 1 Figure 4.2: Elementary Deﬁnition of Strains in 2D εxx≈∆ux ∆X(4.3-a) εyy≈∆uy ∆Y(4.3-b) γxy=π 2−ψ=θ2+θ1 (4.3-c) εxy=1 2γxy≈1 2parenleftbigg∆ux ∆Y+∆uy ∆Xparenrightbigg (4.3-d) In the limit as both ∆ Xand ∆Yapproach zero, then εxx=∂ux ∂X;εyy=∂uy ∂Y;εxy=1 2γxy=1 2parenleftBigg∂ux ∂Y+∂uy ∂XparenrightBigg (4.4) We note that in the expression of the shear strain, we used tan θ≈θwhich is applicable as long as θis small compared to one radian. 26We have used capital letters to represent the coordinates in the initial state, and lower case letters for the ﬁnal or current position coordinates ( x=X+ux). This corresponds to the Lagrangian strain representation. Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–3 4.2 Strain Tensor 27Following the simpliﬁed (and restrictive) introduction to strain, we now turn our at- tention to a rigorous presentation of this important deformation tensor. 28The presentation will proceed as follow. First, with reference to Fig. 4.3 we will derive expressions for the position and displacement vectors of a single point Pfrom the undeformed to the deformed state. Then, we will use some of the expressions in the introduction of the strain between two points PandQ. 4.2.1 Position and Displacement Vectors; (x,X) 29We consider in Fig. 4.3 the undeformed conﬁguration of a material continuum at time t= 0 together with the deformed conﬁguration at coordinates for each conﬁguration. IIIiiiu Spatialb XXX xxx PP 123 1231 12 233 0t=0t=t Xx OoU Material Figure 4.3: Position and Displacement Vectors 30In the initial conﬁguration P0has theposition vector X=X1I1+X2I2+X3I3 (4.5) which is here expressed in terms of the material coordinates (X1,X2,X3). 31In the deformed conﬁguration, the particle P0has now moved to the new position P and has the following position vector x=x1e1+x2e2+x3e3 (4.6) which is expressed in terms of the spatial coordinates . 32Therelativeorientationofthematerialaxes( OX1X2X3)andthespatialaxes( ox1x2x3) is speciﬁed through the direction cosines aX x. Victor Saouma Introduction to Continuum Mechanics Draft4–4 KINEMATIC 33The displacement vector uconnecting P0annPis thedisplacement vector which can be expressed in both the material or spatial coordinates U=UkIk (4.7-a) u=ukik (4.7-b) againUkandukare interrelated through the direction cosines ik=aK kIK. Substituting above we obtain u=uk(aK kIK)=UKIK=U⇒UK=aK kuk (4.8) 34The vector brelates the two origins u=b+x−Xor if the origins are the same (superimposed axis) uk=xk−Xk (4.9) Example 4-1: Displacement Vectors in Material and Spatial Forms With respect to superposed material axis Xiand spatial axes xi, the displacement ﬁeld of a continuum body is given by: x1=X1,x2=X2+AX3,a n dx3=AX2+X3 whereAis constant. 1. Determine the displacement vector components in both the material and spatial form. 2. Determine the displaced location of material particles which originally comprises the plane circular surface X1=0 ,X2 2+X2 3=1/(1−A2)i fA=1/2. Solution: 1. From Eq. 4.9 the displacement ﬁeld can be written in material coordinates as u1=x1−X1= 0 (4.10-a) u2=x2−X2=AX3 (4.10-b) u3=x3−X3=AX2 (4.10-c) 2. The displacement ﬁeld can be written in matrix form as   x1 x2 x3  = 100 01A 0A1   X1 X2 X3  (4.11) or upon inversion   X1 X2 X3  =1 1−A2 1−A200 01 −A 0−A1   x1 x2 x3  (4.12) that isX1=x1,X2=(x2−Ax3)/(1−A2), andX3=(x3−Ax2)/(1−A2). Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–5 3. The displacement ﬁeld can be written now in spatial coordinates as u1=x1−X1= 0 (4.13-a) u2=x2−X2=A(x3−Ax2) 1−A2(4.13-b) u3=x3−X3=A(x2−Ax3) a−A2(4.13-c) 4. For the circular surface, and by direct substitution of X2=(x2−Ax3)/(1−A2), and X3=(x3−Ax2)/(1−A2)i nX2 2+X2 3=1/(1−A2), the circular surface becomes the elliptical surface (1+ A2)x2 2−4Ax2x3+(1+A2)x2 3=( 1−A2)o rf o rA=1/2, 5x2 2−8x2x3+5x2 3=3 . 4.2.1.1 Lagrangian and Eulerian Descriptions; x (X,t),X(x,t) 35When the continuum undergoes deformation (or ﬂow), the particles in the continuum move along various paths which can be expressed in either the material coordinates orin the spatial coordinates system giving rise to two diﬀerent formulations: Lagrangian Formulation: gives the present location x iof the particle that occupied the point ( X1X2X3)a tt i m e t= 0, and is a mapping of the initial conﬁguration into the current one. xi=xi(X1,X2,X3,t)o rx=x(X,t) (4.14) Eulerian Formulation: provides a tracing of its original position of the particle that now occupies the location ( x1,x2,x3)a tt i m e t, and is a mapping of the current conﬁguration into the initial one. Xi=Xi(x1,x2,x3,t)o rX=X(x,t) (4.15) and the independent variables are the coordinates xiandt. 36(X,t)a n d(x,t) are the Lagrangian and Eulerian variables respectivly. 37IfX(x,t)islinear, thenthedeformationissaidtobe homogeneous andplanesections remain plane. 38For both formulation to constitute a one-to-one mapping, with continuous partial derivatives, they must be the unique inverses of one another. A necessary and uniquecondition for the inverse functions to exist is that the determinant of the Jacobian should not vanish |J|=vextendsinglevextendsinglevextendsinglevextendsinglevextendsingle∂xi ∂Xivextendsinglevextendsinglevextendsinglevextendsinglevextendsingle/negationslash= 0 (4.16) Victor Saouma Introduction to Continuum Mechanics Draft4–6 KINEMATIC For example, the Lagrangian description given by x1=X1+X2(et−1);x2=X1(e−t−1)+X2;x3=X3 (4.17) has the inverse Eulerian description given by X1=−x1+x2(et−1) 1−et−e−t;X2=x1(e−t−1)−x2 1−et−e−t;X3=x3 (4.18) Example 4-2: Lagrangian and Eulerian Descriptions The Lagrangian description of a deformation is given by x1=X1+X3(e2−1), x2=X2+X3(e2−e−2), andx3=e2X3whereeis a constant. Show that the jacobian does not vanish and determine the Eulerian equations describing the motion.Solution: The Jacobian is given by vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle10 (e2−1) 01(e2−e−2) 00 e2vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle =e2/negationslash= 0 (4.19) Inverting the equation  10 (e2−1) 01(e2−e−2) 00 e2 −1 = 10(e−2−1) 01(e−4−1) 00 e−2 ⇒  X1=x1+(e−2−1)x3 X2=x2+(e−4−1)x3 X3=e−2x3(4.20) 4.2.2 Gradients 4.2.2.1 Deformation; (x∇X,X∇x) 39Partial diﬀerentiation of Eq. 4.14 with respect to Xjproduces the tensor ∂xi/∂Xj which is the material deformation gradient . In symbolic notation ∂xi/∂Xjis repre- sented by the dyadic F≡x∇X=∂x ∂X1e1+∂x ∂X2e2+∂x ∂X3e3=∂xi ∂Xj (4.21) The matrix form of Fis F=  x1 x2 x3  ⌊∂ ∂X1∂ ∂X2∂ ∂X3⌋= ∂x1 ∂X1∂x1 ∂X2∂x1 ∂X3∂x2 ∂X1∂x2 ∂X2∂x2 ∂X3∂x3 ∂X1∂x3 ∂X2∂x3 ∂X3 =bracketleftBigg∂xi ∂XjbracketrightBigg (4.22) Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–7 40Similarly, diﬀerentiation of Eq. 4.15 with respect to xjproduces the spatial defor- mation gradient H=X∇x≡∂X ∂x1e1+∂X ∂x2e2+∂X ∂x3e3=∂Xi ∂xj (4.23) The matrix form of His H=  X1 X2 X3  ⌊∂ ∂x1∂ ∂x2∂ ∂x3⌋= ∂X1 ∂x1∂X1 ∂x2∂X1 ∂x3∂X2 ∂x1∂X2 ∂x2∂X2 ∂x3∂X3 ∂x1∂X3 ∂x2∂X3 ∂x3 =bracketleftBigg∂Xi ∂xjbracketrightBigg (4.24) 41The material and spatial deformation tensors are interrelated through the chain rule ∂xi ∂Xj∂Xj ∂xk=∂Xi ∂xj∂xj ∂Xk=δik (4.25) and thusF−1=Hor H=F−1 (4.26) 4.2.2.1.1 †Change of Area Due to Deformation 42In order to facilitate the derivation of thePiola-Kirchoﬀ stress tensor later on, we need to derive an expression for the change in area due to deformation. 43If we consider two material element dX(1)=dX1e1anddX(2)=dX2e2emanating fromX, the rectangular area formed by them at the reference time t0is dA0=dX(1)×dX(2)=dX1dX2e3=dA0e3 (4.27) 44At timet,dX(1)deforms into dx(1)=FdX(1)anddX(2)intodx(2)=FdX(2),a n dt h e new area is dA=FdX(1)×FdX(2)=dX1dX2Fe1×Fe2=dA0Fe1×Fe2(4.28-a) =dAn (4.28-b) where the orientation of the deformed area is normal to Fe1andFe2which is denoted by the unit vector n.T h u s , Fe1·dAn=Fe2·dAn= 0 (4.29) and recalling that a·b×cis equal to the determinant whose rows are components of a, b,a n dc, Fe3·dA=dA0(Fe3·Fe1×Fe2)bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright det(F)(4.30) or e3·FTn=dA0 dAdet(F) (4.31) Victor Saouma Introduction to Continuum Mechanics Draft4–8 KINEMATIC andFTnis in the direction of e3so that FTn=dA0 dAdetFe3⇒dAn=dA0det(F)(F−1)Te3 (4.32) which implies that the deformed area has a normal in the direction of ( F−1)Te3.A generalization of the preceding equation would yield dAn=dA0det(F)(F−1)Tn0 (4.33) 4.2.2.1.2 †Change of Volume Due to Deformation 45If we consider an inﬁnitesimal element it has the following volume in material coordinate system: dΩ0=(dX1e1×dX2e2)·dX3e3=dX1dX2dX3 (4.34) in spatial cordiantes: dΩ=(dx1e1×dx2e2)·dx3e3 (4.35) If we deﬁne Fi=∂xi ∂Xjei (4.36) then the deformed volume will be dΩ=(F1dX1×F2dX2)·F3dX3=(F1×F2·F3)dX1dX2dX3 (4.37) or dΩ=d e tFdΩ0 (4.38) andJis called the Jacobian and is the determinant of the deformation gradient F J=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle∂x1 ∂X1∂x1 ∂X2∂x1 ∂X3∂x2 ∂X1∂x2 ∂X2∂x2 ∂X3∂x3 ∂X1∂x3 ∂X2∂x3 ∂X3vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle (4.39) and thus the Jacobian is a measure of deformation. 46We observe that if a material is incompressible than detF=1 . Example 4-3: Change of Volume and Area For the following deformation: x1=λ1X1,x2=−λ3X3,a n dx3=λ2X2, ﬁnd the deformed volume for a unit cube and the deformed area of the unit square in the X1−X2 plane. Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–9 Solution: [F]= λ100 00−λ3 0λ20  (4.40-a) detF=λ1λ2λ3 (4.40-b) ∆V=λ1λ2λ3 (4.40-c) ∆A0= 1 (4.40-d) n0=−e3 (4.40-e) ∆An= (1)(det F)(F−1)T(4.40-f) =λ1λ2λ3 1 λ100 00−1 λ3 01 λ20   0 0 −1  =  0 λ1λ2 0  (4.40-g) ∆An=λ1λ2e2 (4.40-h) 4.2.2.2 Displacements; (u∇X,u∇x) 47We now turn our attention to the displacement vector uias given by Eq. 4.9. Partial diﬀerentiation of Eq. 4.9 with respect to Xjproduces the material displacement gradient ∂ui ∂Xj=∂xi ∂Xj−δijorJ≡u∇X=F−I (4.41) The matrix form of Jis J=  u1 u2 u3  ⌊∂ ∂X1∂ ∂X2∂ ∂X3⌋= ∂u1 ∂X1∂u1 ∂X2∂u1 ∂X3∂u2 ∂X1∂u2 ∂X2∂u2 ∂X3∂u3 ∂X1∂u3 ∂X2∂u3 ∂X3 =bracketleftBigg∂ui ∂XjbracketrightBigg (4.42) 48Similarly, diﬀerentiation of Eq. 4.9 with respect to xjproduces the spatial displace- ment gradient ∂ui ∂xj=δij−∂Xi ∂xjorK≡u∇x=I−H (4.43) The matrix form of Kis K=  u1 u2 u3  ⌊∂ ∂x1∂ ∂x2∂ ∂x3⌋= ∂u1 ∂x1∂u1 ∂x2∂u1 ∂x3∂u2 ∂x1∂u2 ∂x2∂u2 ∂x3∂u3 ∂x1∂u3 ∂x2∂u3 ∂x3 =bracketleftBigg∂ui ∂xjbracketrightBigg (4.44) Victor Saouma Introduction to Continuum Mechanics Draft4–10 KINEMATIC 4.2.2.3 Examples Example 4-4: Material Deformation and Displacement Gradients A displacement ﬁeld is given by u=X1X2 3e1+X2 1X2e2+X2 2X3e3, determine the material deformation gradient Fand the material displacement gradient J,a n dv e r i f y thatJ=F−I. Solution:The material deformation gradient is: ∂u i ∂Xj=J=u∇x== ∂uX1 ∂X1∂uX1 ∂X2∂uX1 ∂X3∂uX2 ∂X1∂uX2 ∂X2∂uX2 ∂X3∂uX3 ∂X1∂uX3 ∂X2∂uX3 ∂X3  (4.45-a) = X2 302X1X3 2X1X2X2 10 02X2X3X2 2  (4.45-b) Sincex=u+X, the displacement ﬁeld is also given by x=X1(1+X2 3)bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright x1e1+X2(1+X2 1)bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright x2e2+X3(1+X2 2)bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright x3e3 (4.46) and thus F=x∇X≡∂x ∂X1e1+∂x ∂X2e2+∂x ∂X3e3=∂xi ∂Xj(4.47-a) = ∂x1 ∂X1∂x1 ∂X2∂x1 ∂X3∂x2 ∂X1∂x2 ∂X2∂x2 ∂X3∂x3 ∂X1∂x3 ∂X2∂x3 ∂X3  (4.47-b) = 1+X2 302X1X3 2X1X21+X2 10 02X2X31+X2 2  (4.47-c) We observe that the two second order tensors are related by J=F−I. 4.2.3 Deformation Tensors 49Having derived expressions for∂xi ∂Xjand∂Xi ∂xjwe now seek to determine dx2anddX2 wheredXanddxcorrespond to the distance between points PandQin the undeformed and deformed cases respectively. 50We consider next the initial (undeformed) and ﬁnal (deformed) conﬁguration of a continuum in which the material OX1,X2,X3and spatial coordinates ox1x2x3are super- imposed. Neighboring particles P0andQ0in the initial conﬁgurations moved to Pand Qrespectively in the ﬁnal one, Fig. 4.4. Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–11 2,X3x3, X0 2x X1O x1,u xt=0 +dX XX+d ddxt=t Xu u Q 0Q P P Figure 4.4: Undeformed and Deformed Conﬁgurations of a Continuum 4.2.3.1 Cauchy’s Deformation Tensor; (dX)2 51The Cauchy deformation tensor, introduced by Cauchy in 1827, B−1(alternatively denoted as c) gives the initial square length ( dX)2of an element dxin the deformed conﬁguration. 52This tensor is the inverse of the tensor Bwhich will not be introduced until Sect. 4.2.6.3. 53The square of the diﬀerential element connecting PoandQ0is (dX)2=dX·dX=dXidXi (4.48) however from Eq. 4.15 the distance diﬀerential dXiis dXi=∂Xi ∂xjdxjordX=H·dx (4.49) thus the squared length ( dX)2in Eq. 4.48may be rewritten as (dX)2=∂Xk ∂xi∂Xk ∂xjdxidxj=B−1 ijdxidxj (4.50-a) =dx·B−1·dx (4.50-b) in which the second order tensor B−1 ij=∂Xk ∂xi∂Xk ∂xjorB−1=∇xX·X∇xbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Hc·H (4.51) isCauchy’s deformation tensor . Victor Saouma Introduction to Continuum Mechanics Draft4–12 KINEMATIC 4.2.3.2 Green’s Deformation Tensor; (dx)2 54The Green deformation tensor, introduced by Green in 1841, C(alternatively denoted asB−1), referred to in the undeformed conﬁguration, gives the new square length ( dx)2 of the element dXis deformed. 55The square of the diﬀerential element connecting PoandQ0is now evaluated in terms of the spatial coordinates (dx)2=dx·dx=dxidxi (4.52) however from Eq. 4.14 the distance diﬀerential dxiis dxi=∂xi ∂XjdXjordx=F·dX (4.53) thus the squared length ( dx)2in Eq. 4.52 may be rewritten as (dx)2=∂xk ∂Xi∂xk ∂XjdXidXj=CijdXidXj (4.54-a) =dX·C·dX (4.54-b) in which the second order tensor Cij=∂xk ∂Xi∂xk ∂XjorC=∇Xx·x∇Xbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Fc·F (4.55) isGreen’sdeformationtensor alsoknownas metrictensor ,o rdeformationtensor orright Cauchy-Green deformation tensor . 56Inspection of Eq. 4.51 and Eq. 4.55 yields C−1=B−1orB−1=(F−1)T·F−1 (4.56) Example 4-5: Green’s Deformation Tensor A continuum body undergoes the deformation x1=X1,x2=X2+AX3,a n dx3= X3+AX2whereAis a constant. Determine the deformation tensor C. Solution: From Eq. 4.55 C=Fc·FwhereFwas deﬁned in Eq. 4.21 as F=∂xi ∂Xj(4.57-a) = 100 01A 0A1  (4.57-b) Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–13 and thus C=Fc·F (4.58-a) = 100 01A 0A1 T 100 01A 0A1 = 10 0 01 +A22A 02A1+A2 (4.58-b) 4.2.4 Strains; (dx)2−(dX)2 57With (dx)2and (dX)2deﬁned we can now ﬁnally introduce the concept of strain through ( dx)2−(dX)2. 4.2.4.1 Finite Strain Tensors 58We start with the most general case of ﬁnite strains where no constraints are imposed on the deformation (small). 4.2.4.1.1 Lagrangian/Green’s Tensor 59The diﬀerence ( dx)2−(dX)2for two neighboring particles in a continuum is used as themeasure of deformation . Using Eqs. 4.54-a and 4.48this diﬀerence is expressed as (dx)2−(dX)2=parenleftBigg∂xk ∂Xi∂xk ∂Xj−δijparenrightBigg dXidXj=2EijdXidXj(4.59-a) =dX·(Fc·F−I)·dX=2dX·E·dX (4.59-b) in which the second order tensor Eij=1 2parenleftBigg∂xk ∂Xi∂xk ∂Xj−δijparenrightBigg orE=1 2(∇Xx·x∇Xbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Fc·F=C−I) (4.60) is called the Lagrangian (or Green’s) ﬁnite strain tensor which was introduced by Green in 1841 and St-Venant in 1844. 60To express the Lagrangiantensor in terms of the displacements, we substitute Eq. 4.41 in the preceding equation, and aftersome simple algebraic manipulations, the Lagrangianﬁnite strain tensor can be rewritten as Eij=1 2parenleftBigg∂ui ∂Xj+∂uj ∂Xi+∂uk ∂Xi∂uk ∂XjparenrightBigg orE=1 2(u∇X+∇Xubracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright J+Jc+∇Xu·u∇Xbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Jc·J) (4.61) Victor Saouma Introduction to Continuum Mechanics Draft4–14 KINEMATIC or: E11=∂u1 ∂X1+1 2 parenleftBigg∂u1 ∂X1parenrightBigg2 +parenleftBigg∂u2 ∂X1parenrightBigg2 +parenleftBigg∂u3 ∂X1parenrightBigg2  (4.62-a) E12=1 2parenleftBigg∂u1 ∂X2+∂u2 ∂X1parenrightBigg +1 2bracketleftBigg∂u1 ∂X1∂u1 ∂X2+∂u2 ∂X1∂u2 ∂X2+∂u3 ∂X1∂u3 ∂X2bracketrightBigg (4.62-b) ···=··· (4.62-c) Example 4-6: Lagrangian Tensor DeterminetheLagrangianﬁnitestraintensor Eforthedeformationofexample 4.2.3.2. Solution: C= 10 0 01 +A22A 02A1+A2  (4.63-a) E=1 2(C−I) (4.63-b) =1 2 00 0 0A22A 02AA2  (4.63-c) Note that the matrix is symmetric. 4.2.4.1.2 Eulerian/Almansi’s Tensor 61Alternatively, the diﬀerence ( dx)2−(dX)2for the two neighboring particles in the continuum can be expressed in terms of Eqs. 4.52 and 4.50-b this same diﬀerence is now equal to (dx)2−(dX)2=parenleftBigg δij−∂Xk ∂xi∂Xk ∂xjparenrightBigg dxidxj=2E∗ ijdxidxj(4.64-a) =dx·(I−Hc·H)·dx=2dx·E∗·dx (4.64-b) in which the second order tensor E∗ ij=1 2parenleftBigg δij−∂Xk ∂xi∂Xk ∂xjparenrightBigg orE∗=1 2(I−∇xX·X∇x)bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Hc·H=B−1 (4.65) is called the Eulerian (or Almansi) ﬁnite strain tensor . Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–15 62For inﬁnitesimal strain it was introduced by Cauchy in 1827, and for ﬁnite strain by Almansi in 1911. 63To express the Eulerian tensor in terms of the displacements, we substitute 4.43 in the preceding equation, and after some simple algebraic manipulations, the Eulerian ﬁnite strain tensor can be rewritten as E∗ ij=1 2parenleftBigg∂ui ∂xj+∂uj ∂xi−∂uk ∂xi∂uk ∂xjparenrightBigg orE∗=1 2(u∇x+∇xubracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright K+Kc−∇xu·u∇xbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Kc·K) (4.66) 64Expanding E∗ 11=∂u1 ∂x1−1 2 parenleftBigg∂u1 ∂x1parenrightBigg2 +parenleftBigg∂u2 ∂x1parenrightBigg2 +parenleftBigg∂u3 ∂x1parenrightBigg2  (4.67-a) E∗ 12=1 2parenleftBigg∂u1 ∂x2+∂u2 ∂x1parenrightBigg −1 2bracketleftBigg∂u1 ∂x1∂u1 ∂x2+∂u2 ∂x1∂u2 ∂x2+∂u3 ∂x1∂u3 ∂x2bracketrightBigg (4.67-b) ···=··· (4.67-c) 4.2.4.2 Inﬁnitesimal Strain Tensors; Small Deformation Theory 65Thesmall deformation theory of continuum mechanics has as basic condition the requirement that the displacement gradients be small compared to unity. The funda-mental measure of deformation is the diﬀerence ( dx) 2−(dX)2, which may be expressed in terms of the displacement gradients by inserting Eq. 4.61 and 4.66 into 4.59-b and 4.64-brespectively. Ifthedisplacement gradients aresmall, theﬁnite strain tensors inEq.4.59-b and 4.64-b reduce to inﬁnitesimal strain tensors and the resulting equations represent small deformations . 66For instance, if we were to evaluate sepsilonv+sepsilonv2,f o rsepsilonv=1 0−3and 10−1, then we would obtain 0.001001≈0.001 and 0 .11 respectively. In the ﬁrst case sepsilonv2is “negligible” compared to sepsilonv, in the other it is not. 4.2.4.2.1 Lagrangian Inﬁnitesimal Strain Tensor 67In Eq. 4.61 if the displacement gradient components∂ui ∂Xjare each small compared to unity, then the third term are negligible and may be dropped. The resulting tensor is theLagrangian inﬁnitesimal strain tensor denoted by Eij=1 2parenleftBigg∂ui ∂Xj+∂uj ∂XiparenrightBigg orE=1 2(u∇X+∇Xubracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright J+Jc) (4.68) or: E11=∂u1 ∂X1(4.69-a) Victor Saouma Introduction to Continuum Mechanics Draft4–16 KINEMATIC E12=1 2parenleftBigg∂u1 ∂X2+∂u2 ∂X1parenrightBigg (4.69-b) ···=··· (4.69-c) Note the similarity with Eq. 4.4. 4.2.4.2.2 Eulerian Inﬁnitesimal Strain Tensor 68Similarly, inn Eq. 4.66 if the displacement gradient components∂ui ∂xjare each small compared to unity, then the third term are negligible and may be dropped. The resulting tensor is the Eulerian inﬁnitesimal strain tensor denoted by E∗ ij=1 2parenleftBigg∂ui ∂xj+∂uj ∂xiparenrightBigg orE∗=1 2(u∇x+∇xubracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright K+Kc) (4.70) 69Expanding E∗ 11=∂u1 ∂x1(4.71-a) E∗ 12=1 2parenleftBigg∂u1 ∂x2+∂u2 ∂x1parenrightBigg (4.71-b) ···=··· (4.71-c) 4.2.4.3 Examples Example 4-7: Lagrangian and Eulerian Linear Strain Tensors A displacement ﬁeld is given by x1=X1+AX2,x2=X2+AX3,x3=X3+AX1 whereAis constant. Calculate the Lagrangian and the Eulerian linear strain tensors, and compare them for the case where Ais very small. Solution:The displacements are obtained from Eq. 4.9 u k=xk−Xkor u1=x1−X1=X1+AX2−X1=AX2 (4.72-a) u2=x2−X2=X2+AX3−X2=AX3 (4.72-b) u3=x3−X3=X3+AX1−X3=AX1 (4.72-c) then from Eq. 4.41 J≡u∇X= 0A0 00A A00  (4.73) Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–17 From Eq. 4.68: 2E=(J+Jc)= 0A0 00A A00 + 00A A00 0A0  (4.74-a) = 0AA A0A AA0  (4.74-b) To determine the Eulerian tensor, we need the displacement uin terms of x,t h u s inverting the displacement ﬁeld given above:   x1 x2 x3  = 1A0 01A A01   X1 X2 X3  ⇒  X1 X2 X3  =1 1+A3 1−AA2 A21−A −AA21   x1 x2 x3   (4.75) thus from Eq. 4.9 uk=xk−Xkwe obtain u1=x1−X1=x1−1 1+A3(x1−Ax2+A2x3)=A(A2x1+x2−Ax3) 1+A3(4.76-a) u2=x2−X2=x2−1 1+A3(A2x1+x2−Ax3)=A(−Ax1+A2x2+x3) 1+A3(4.76-b) u3=x3−X3=x3−1 1+A3(−Ax1+A2x2+x3)=A(x1−Ax2+A2x3) 1+A3(4.76-c) From Eq. 4.43 K≡u∇x=A 1+A3 A21−A −AA21 1−AA2  (4.77) Finally, from Eq. 4.66 2E∗=K+Kc (4.78-a) =A 1+A3 A21−A −AA21 1−AA2 +A 1+A3 A2−A1 1A2−A −A1A2 (4.78-b) =A 1+A3 2A21−A1−A 1−A2A21−A 1−A1−A2A2  (4.78-c) asAis very small, A2andhigher power may be neglected with the results, then E∗→E. 4.2.5 Physical Interpretation of the Strain Tensor 4.2.5.1 Small Strain 70We ﬁnally show that the linear lagrangian tensor in small deformation Eijis nothing else than the strain as was deﬁned earlier in Eq.4.4. Victor Saouma Introduction to Continuum Mechanics Draft4–18 KINEMATIC 71We rewrite Eq. 4.59-b as (dx)2−(dX)2=(dx−dX)(dx+dX)=2EijdXidXj (4.79-a) or (dx)2−(dX)2=(dx−dX)(dx+dX)=dX·2E·dX (4.79-b) but since dx≈dXunder current assumption of small deformation, then the previous equation can be rewritten as dubracehtipdownleft bracehtipuprightbracehtipupleft bracehtipdownright dx−dX dX=EijdXi dXdXj dX=Eijξiξj=ξ·E·ξ (4.80) 72We recognize that the left hand side is nothing else than the change in length per unit original length, and is called the normal strain for the line element having direction cosinesdXi dX. 73With reference to Fig. 4.5 we consider two cases: normal and shear strain. 0 P0 dX2 XXX 2Q3P 10MuXXX dX 123 dX3Normal ShearP 0Q 0 2 1xxx M Q ee 12e3 3 2n nθ3 Figure 4.5: Physical Interpretation of the Strain Tensor Normal Strain: When Eq. 4.80 is applied to the diﬀerential element P0Q0which lies along the X2axis, the result will be the normal strain because sincedX1 dX=dX3 dX=0 anddX2 dX= 1. Therefore, Eq. 4.80 becomes (with ui=xi−Xi): dx−dX dX=E22=∂u2 ∂X2 (4.81) Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–19 Likewise for the other 2 directions. Hence the diagonal terms of the linear strain tensor represent normal strains in the coordinate system. Shear Strain: For the diagonal terms Eijwe consider the two line elements originally located along the X2and theX3axes before deformation. After deformation, the original right angle between the lines becomes the angle θ. From Eq. 4.96 ( dui=parenleftBig ∂ui ∂XjparenrightBig P0dXj) a ﬁrst order approximation gives the unit vector at Pin the direction ofQ,a n dMas: n2=∂u1 ∂X2e1+e2+∂u3 ∂X2e3 (4.82-a) n3=∂u1 ∂X3e1+∂u2 ∂X3e2+e3 (4.82-b) and from the deﬁnition of the dot product: cosθ=n2·n3=∂u1 ∂X2∂u1 ∂X3+∂u2 ∂X3+∂u3 ∂X2(4.83) or neglecting the higher order term cosθ=∂u2 ∂X3+∂u3 ∂X2=2E23 (4.84) 74Finally taking the change in right angle between the elements as γ23=π/2−θ, and recalling that for small strain theory γ23is very small it follows that γ23≈sinγ23=s i n (π/2−θ)=c o sθ=2E23. (4.85) Therefore the oﬀ diagonal terms of the linear strain tensor represent one half of the angle change between two line elements originally at right angles to one another.These components are called the shear strains . 74TheEngineering shear strain is deﬁned as one half the tensorial shear strain, and the resulting tensor is written as Eij= ε111 2γ121 2γ13 1 2γ12ε221 2γ23 1 2γ131 2γ23ε33  (4.86) 75We note that a similar development paralleling the one just presented can be made for the linear Eulerian strain tensor (where the straight lines and right angle will be in thedeformed state). 4.2.5.2 Finite Strain; Stretch Ratio 76The simplest and most useful measure of the extensional strain of an inﬁnitesimal element is the stretch orstretch ratio asdx dXwhich may be deﬁned at point P0in the Victor Saouma Introduction to Continuum Mechanics Draft4–20 KINEMATIC undeformed conﬁguration or at Pin the deformed one (Refer to the original deﬁnition given by Eq, 4.1). 77Hence, from Eq. 4.54-a, and Eq. 4.60 the squared stretch at P0for the line element along the unit vector m=dX dXis given by Λ2 m≡parenleftBiggdx dXparenrightBigg2 P0=CijdXi dXdXj dXor Λ2 m=m·C·m (4.87) Thus for an element originally along X2, Fig. 4.5, m=e2and therefore dX1/dX= dX3/dX=0a n ddX2/dX= 1, thus Eq. 4.87 (with Eq. ??) yields Λ2 e2=C22=1+2E22 (4.88) and similar results can be obtained for Λ2 e1and Λ2e 3. 78Similarly from Eq. 4.50-b, the reciprocal of the squared stretch for the line element at Palong the unit vector n=dx dxis given by 1 λ2 n≡parenleftBiggdX dxparenrightBigg2 P=B−1 ijdxi dxdxj dxor1 λ2 n=n·B−1·n (4.89) Again for an element originally along X2, Fig. 4.5, we obtain 1 λ2e2=1−2E∗ 22 (4.90) 79we note that in general Λ e2/negationslash=λe2since the element originally along the X2axis will not be along the x2after deformation. Furthermore Eq. 4.87 and 4.89 show that in the matrices of rectangular cartesian components the diagonal elements of both CandB−1 must be positive, while the elements of Emust be greater than −1 2and those of E∗must be greater than +1 2. 80The unit extension of the element is dx−dX dX=dx dX−1=Λ m−1 (4.91) and for the element P0Q0along the X2axis, theunit extension is dx−dX dX=E(2)=Λe2−1=radicalBig 1+2E22−1 (4.92) for small deformation theory E22<<1, and dx−dX dX=E(2)=( 1+2E22)1 2−1/similarequal1+1 22E22−1/similarequalE22 (4.93) which is identical to Eq. 4.81. 81For the two diﬀerential line elements of Fig. 4.5, the change in angle γ23=π 2−θis given in terms of both Λ e2and Λ e3by sinγ23=2E23 Λe2Λe3=2E23 √ 1+2E22√ 1+2E33(4.94) Again, when deformations are small, this equation reduces to Eq. 4.85. Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–21 4.2.6 Linear Strain and Rotation Tensors 82Strain components are quantitative measures of certain type of relative displacement between neighboring parts of the material. A solid material will resist such relative displacement giving rise to internal stresses. 83Not all kinds of relative motion give rise to strain (and stresses). If a body moves as a rigid body , the rotational part of its motion produces relative displacement. Thus the general problem is to express the strain in terms of the displacements by separating oﬀ that part of the displacement distribution which does not contribute to the strain. 4.2.6.1 Small Strains 84From Fig. 4.6 the displacements of two neighboring particles are represented by the vectorsuP0anduQ0and the vector dui=uQ0 i−uP0 iordu=uQ0−uP0(4.95) is called the relative displacement vector of the particle originally at Q0with respect to the one originally at P0. 0 QQ 0 0pQ dXdu dxu uP0 P Figure 4.6: Relative Displacement duofQrelative to P 4.2.6.1.1 Lagrangian Formulation 85Neglecting higher order terms, and through a Taylor expansion dui=parenleftBigg∂ui ∂XjparenrightBigg P0dXjordu=(u∇X)P0dX (4.96) Victor Saouma Introduction to Continuum Mechanics Draft4–22 KINEMATIC 86We also deﬁne a unit relative displacement vector dui/dXwheredXis the mag- nitude of the diﬀerential distance dXi,o rdXi=ξidX,t h e n dui dX=∂ui ∂XjdXj dX=∂ui ∂Xjξjordu dX=u∇X·ξ=J·ξ (4.97) 87The material displacement gradient∂ui ∂Xjcan be decomposed uniquely into a symmetric and an antisymetric part, we rewrite the previous equation as dui= 1 2parenleftBigg∂ui ∂Xj+∂uj ∂XiparenrightBigg bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Eij+1 2parenleftBigg∂ui ∂Xj−∂uj ∂XiparenrightBigg bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Wij dXj (4.98-a) or du= 1 2(u∇X+∇Xu) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright E+1 2(u∇X−∇Xu) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright W ·dX (4.98-b) or E= ∂u1 ∂X11 2parenleftBig ∂u1 ∂X2+∂u2 ∂X1parenrightBig 1 2parenleftBig ∂u1 ∂X3+∂u3 ∂X1parenrightBig 1 2parenleftBig ∂u1 ∂X2+∂u2 ∂X1parenrightBig ∂u2 ∂X21 2parenleftBig ∂u2 ∂X3+∂u3 ∂X2parenrightBig 1 2parenleftBig ∂u1 ∂X3+∂u3 ∂X1parenrightBig 1 2parenleftBig ∂u2 ∂X3+∂u3 ∂X2parenrightBig ∂u3 ∂X3 (4.99) We thus introduce the linear lagrangian rotation tensor Wij=1 2parenleftBigg∂ui ∂Xj−∂uj ∂XiparenrightBigg orW=1 2(u∇X−∇Xu) (4.100) in matrix form: W= 01 2parenleftBig ∂u1 ∂X2−∂u2 ∂X1parenrightBig 1 2parenleftBig ∂u1 ∂X3−∂u3 ∂X1parenrightBig −1 2parenleftBig ∂u1 ∂X2−∂u2 ∂X1parenrightBig 01 2parenleftBig ∂u2 ∂X3−∂u3 ∂X2parenrightBig −1 2parenleftBig ∂u1 ∂X3−∂u3 ∂X1parenrightBig −1 2parenleftBig ∂u2 ∂X3−∂u3 ∂X2parenrightBig 0 (4.101) 88In a displacement for which Eijis zero in the vicinity of a point P0, the relative displacement at that point will be an inﬁnitesimal rigid body rotation .I t c a n b e shown that this rotation is given by the linear Lagrangian rotation vector wi=1 2sepsilonvijkWkjorw=1 2∇X×u (4.102) or w=−W23e1−W31e2−W12e3 (4.103) Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–23 4.2.6.1.2 Eulerian Formulation 89The derivation inanEulerian formulationparallels theone forLagrangianformulation. Hence, dui=∂ui ∂xjdxjordu=K·dx (4.104) 90Theunit relative displacement vector will be dui=∂ui ∂xjdxj dx=∂ui ∂xjηjordu dx=u∇x·η=K·β (4.105) 91The decomposition of the Eulerian displacement gradient∂ui ∂xjresults in dui= 1 2parenleftBigg∂ui ∂xj+∂uj ∂xiparenrightBigg bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright E∗ ij+1 2parenleftBigg∂ui ∂xj−∂uj ∂xiparenrightBigg bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Ωij dxj (4.106-a) or du= 1 2(u∇x+∇xu) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright E∗+1 2(u∇x−∇xu) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Ω ·dx (4.106-b) or E= ∂u1 ∂x11 2parenleftBig ∂u1 ∂x2+∂u2 ∂x1parenrightBig 1 2parenleftBig ∂u1 ∂x3+∂u3 ∂x1parenrightBig 1 2parenleftBig ∂u1 ∂x2+∂u2 ∂x1parenrightBig ∂u2 ∂x21 2parenleftBig ∂u2 ∂x3+∂u3 ∂x2parenrightBig 1 2parenleftBig ∂u1 ∂x3+∂u3 ∂x1parenrightBig 1 2parenleftBig ∂u2 ∂x3+∂u3 ∂x2parenrightBig ∂u3 ∂x3  (4.107) 92We thus introduced the linear Eulerian rotation tensor wij=1 2parenleftBigg∂ui ∂xj−∂uj ∂xiparenrightBigg orΩ=1 2(u∇x−∇xu) (4.108) in matrix form: W= 01 2parenleftBig ∂u1 ∂x2−∂u2 ∂x1parenrightBig 1 2parenleftBig ∂u1 ∂x3−∂u3 ∂x1parenrightBig −1 2parenleftBig ∂u1 ∂x2−∂u2 ∂x1parenrightBig 01 2parenleftBig ∂u2 ∂x3−∂u3 ∂x2parenrightBig −1 2parenleftBig ∂u1 ∂x3−∂u3 ∂x1parenrightBig −1 2parenleftBig ∂u2 ∂x3−∂u3 ∂x2parenrightBig 0  (4.109) and thelinear Eulerian rotation vector will be ωi=1 2sepsilonvijkωkjorω=1 2∇x×u (4.110) Victor Saouma Introduction to Continuum Mechanics Draft4–24 KINEMATIC 4.2.6.2 Examples Example 4-8: Relative Displacement along a speciﬁed direction A displacement ﬁeld is speciﬁed by u=X2 1X2e1+(X2−X2 3)e2+X2 2X3e3. Determine the relative displacement vector duin the direction of the −X2axis atP(1,2,−1). Deter- mine the relative displacements uQi−uPforQ1(1,1,−1),Q2(1,3/2,−1),Q3(1,7/4,−1) andQ4(1,15/8,−1) and compute their directions with the direction of du. Solution:From Eq. 4.41, J=u∇ Xor ∂ui ∂Xj= 2X1X2X2 10 01 −2X3 02X2X3X2 2  (4.111) thus from Eq. 4.96 du=(u∇X)PdXin the direction of −X2or {du}= 410 012 0−44   0 −1 0  =  −1 −1 4  (4.112) By direct calculation from uwe have uP=2e1+e2−4e3 (4.113-a) uQ1=e1−e3 (4.113-b) thus uQ1−uP=−e1−e2+3e3 (4.114-a) uQ2−uP=1 2(−e1−e2+3.5e3) (4.114-b) uQ3−uP=1 4(−e1−e2+3.75e3) (4.114-c) uQ4−uP=1 8(−e1−e2+3.875e3) (4.114-d) and it is clear that as Qiapproaches P, the direction of the relative displacements of the two particles approaches the limiting direction of du. Example 4-9: Linear strain tensor, linear rotation tensor, rotation vector Under the restriction of small deformation theory E=E∗, a displacement ﬁeld is given byu=(x1−x3)2e1+(x2+x3)2e2−x1x2e3. Determine the linear strain tensor, the linear rotation tensor and the rotation vector at point P(0,2,−1). Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–25 Solution: the matrix form of the displacement gradient is [∂ui ∂xj]= 2(x1−x3)0 −2(x1−x3) 02 ( x2+x3)2 (x2+x3) −x2−x1 0  (4.115-a) bracketleftBigg∂ui ∂xjbracketrightBigg P= 20−2 022 −20 0  (4.115-b) Decomposing this matrix into symmetric and antisymmetric components give: [Eij]+[wij]= 20−2 021 −21 0 + 000 001 0−10  (4.116) and from Eq. Eq. 4.103 w=−W23e1−W31e2−W12e3=−1e1 (4.117) 4.2.6.3 Finite Strain; Polar Decomposition 93When the displacement gradients are ﬁnite, then we no longer can decompose∂ui ∂Xj(Eq. 4.96) or∂ui ∂xj(Eq. 4.104) into a unique sum of symmetric and skew parts (pure strain and pure rotation). 94Thus in this case, rather than having an additive decomposition, we will have a multiplicative decomposition. 95wecallthisa polardecomposition anditshoulddecomposethedeformationgradient in the product of two tensors, one of which represents a rigid-body rotation, while the other is a symmetric positive-deﬁnite tensor. 96We apply this decomposition to the deformation gradient F: Fij≡∂xi ∂Xj=RikUkj=VikRkjorF=R·U=V·R (4.118) whereRis theorthogonal rotation tensor ,a n dUandVare positive symmetric tensors known as the right stretch tensor and theleft stretch tensor respectively. 97The interpretation of the above equation is obtained by inserting the above equation intodxi=∂xi ∂XjdXj dxi=RikUkjdXj=VikRkjdXjordx=R·U·dX=V·R·dX (4.119) and we observe that in the ﬁrst form the deformation consists of a sequential stretching (byU) and rotation ( R) to be followed by a rigid body displacement to x. In the second case, the orders are reversed, we have ﬁrst a rigid body translation to x, followed by a rotation (R) and ﬁnally a stretching (by V). Victor Saouma Introduction to Continuum Mechanics Draft4–26 KINEMATIC 98To determine the stretch tensor from the deformation gradient FTF=(RU)T(RU)=UTRTRU=UTU (4.120) Recalling that Ris an orthonormal matrix, and thus RT=R−1then we can compute the various tensors from U=√ FTF(4.121) R=FU−1(4.122) V=FRT(4.123) 99It can be shown that U=C1/2andV=B1/2 (4.124) Example 4-10: Polar Decomposition I Givenx1=X1,x2=−3X3,x3=2X2, ﬁnd the deformation gradient F,t h er i g h t stretch tensor U, the rotation tensor R, and the left stretch tensor V. Solution: From Eq. 4.22 F= ∂x1 ∂X1∂x1 ∂X2∂x1 ∂X3∂x2 ∂X1∂x2 ∂X2∂x2 ∂X3∂x3 ∂X1∂x3 ∂X2∂x3 ∂X3 = 10 0 00−3 02 0  (4.125) From Eq. 4.121 U2=FTF= 100 0020−30  10 0 00−3 02 0 = 100 040009  (4.126) thus U= 100 020003  (4.127) From Eq. 4.122 R=FU−1= 10 0 00−3 02 0  100 01 20 001 3 = 10 0 00−1 01 0  (4.128) Finally, from Eq. 4.123 V=FRT= 10 0 00−3 02 0  100 0010−10 = 100 030002  (4.129) Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–27 Example 4-11: Polar Decomposition II For the following deformation: x1=λ1X1,x2=−λ3X3,a n dx3=λ2X2, ﬁnd the rotation tensor.Solution: [F]= λ100 00−λ3 0λ20  (4.130) [U]2=[F]T[F] (4.131) = λ100 00 λ2 0−λ30  λ100 00−λ3 0λ20 = λ2 100 0λ2 20 00λ2 3  (4.132) [U]= λ100 0λ20 00λ3  (4.133) [R]=[F][U]−1= λ100 00−λ3 0λ20  1 λ100 01 λ20 001 λ3 = 10 0 00−1 01 0 (4.134) Thus we note that Rcorresponds to a 90orotation about the e1axis. Example 4-12: Polar Decomposition III Victor Saouma Introduction to Continuum Mechanics Draft4–28 KINEMATIC Polar Decomposition Using Mathematica Given x1=X1+2X2, x2=X2, x3=X3, a) Obtain C, b) the principal values of C and the corresponding directions, c) the matrix U and U-1 with respect to the principal directions, d) Obtain the matrix U and U-1 with respect to the ei bas obtain the matrix R with respect to the ei basis. Determine the F matrix In[1]:=F=881, 2, 0 <,80, 1, 0 <,80, 0, 1 << Out[1]=i kjjjjjjj120 010001y {zzzzzzz Solve for C In[2]:=CST=Transpose @FD.F Out[2]=i kjjjjjjj120 250 001y {zzzzzzz Determine Eigenvalues and Eigenvectors In[3]:=N@Eigenvalues @CSTDD Out[3]= 81., 0.171573, 5.82843 <In[4]:= 8v1, v2, v3 <=N@Eigenvectors @CSTD,4D Out[4]=i kjjjjjjj00 1 . -2.414 1. 0 0.4142 1. 0y {zzzzzzz In[5]:= <<LinearAlgebra ‘Orthogonalization ‘ In[6]:=vnormalized =GramSchmidt @8v3,−v2, v1 <D Out[6]=i kjjjjjjj0.382683 0.92388 0 0.92388 -0.382683 0 00 1 .y {zzzzzzz In[7]:=CSTeigen =Chop @[email protected], 4 DD Out[7]=i kjjjjjjj5.828 0 0 0 0.1716 0 00 1 .y {zzzzzzz Determine U with respect to the principal directions In[8]:=Ueigen =N@Sqrt @CSTeigen D,4D Out[8]=i kjjjjjjj2.414 0 0 0 0.4142 000 1 .y {zzzzzzz In[9]:=Ueigenminus1 =Inverse @Ueigen D Out[9]=i kjjjjjjj0.414214 0. 0. 0. 2.41421 0.0. 0. 1.y {zzzzzzz2 m− Determine U and U-1with respect to the ei basis In[10]:= [email protected], 3 D Out[10]=i kjjjjjjj0.707 0.707 0. 0.707 2.12 0. 0. 0. 1.y {zzzzzzz In[11]:= U_einverse =N@Inverse @%D,3D Out[11]=i kjjjjjjj2.12 -0.707 0. -0.707 0.707 0. 0. 0. 1.y {zzzzzzz Determine R with respect to the ei basis In[12]:= R=N@F.%,3 D Out[12]=i kjjjjjjj0.707 0.707 0. -0.707 0.707 0. 0. 0. 1.y {zzzzzzzm−polar.nb Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–29 4.2.7 Summary and Discussion 100From the above, we deduce the following observations: 1. If both the displacement gradients and the displacements themselves are small, then ∂ui ∂Xj≈∂ui ∂xjand thus the Eulerian and the Lagrangianinﬁnitesimal strain tensors may be taken as equal Eij=E∗ ij. 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. 4.2.8†Explicit Derivation 101If the derivations in the preceding section was perceived as too complex through a ﬁrst reading, this section will present a “gentler” approach to essentially the same results albeit in a less “elegant” mannser. The previous derivation was carried out using indicialnotation, in this section we repeat the derivation using explicitly. 102Similarities between the two approaches is facilitated by Table 4.2. 103Considering two points AandBin a 3D solid, the distance between them is ds ds2=dx2+dy2+dz2(4.135) As a result of deformation, point Amoves to A/prime,a n dBtoB/primethe distance between the two points is ds/prime, Fig. 12.7. ds/prime2=dx/prime2+dy/prime2+dz/prime2(4.136) 104The displacement of point AtoA/primeis given by u=x/prime−x⇒dx/prime=du+dx (4.137-a) v=y/prime−y⇒dy/prime=dv+dy (4.137-b) w=z/prime−z⇒dz/prime=dw+dz (4.137-c) 105Substituting these equations into Eq. 4.136, we obtain ds/prime2=dx2+dy2+dz2 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright ds2+2dudx+2dvdy+2dwdz+du2+dv2+dw2(4.138) Victor Saouma Introduction to Continuum Mechanics Draft4–30 KINEMATIC IIIiiiu Spatialb XXX xxx PP 123 1231 12 233 0t=0t=t Xx OoU Material2,X3x3, X0 2x X1O x1,u xt=0 +dX XX+d ddxt=t Xu u Q 0Q P P LAGRANGIAN EULERIAN Material Spatial Position Vector x=x(X,t) X=X(x,t) GRADIENTS Deformation F=x∇X≡∂x i ∂X j H=X∇x≡∂X i ∂x j H=F−1 Displacement ∂u i ∂X j=∂x i ∂X j−δijor ∂u i ∂x j=δij−∂X i ∂x jor J=u∇X=F−I K≡u∇x=I−H TENSOR dX2=dx·B−1·dx dx2=dX·C·dX Cauchy Green Deformation B−1 ij=∂X k ∂x i∂X k ∂x jor Cij=∂x k ∂X i∂x k ∂X jor B−1=∇xX·X∇x=Hc·H C=∇Xx·x∇X=Fc·F C−1=B−1 STRAINS Lagrangian Eulerian/Almansi dx2−dX2=dX·2E·dX dx2−dX2=dx·2E∗·dx Finite Strain Eij=1 2parenleftBig ∂x k ∂X i∂x k ∂X j−δijparenrightBig or E∗ ij=1 2parenleftBig δij−∂X k ∂x i∂X k ∂x jparenrightBig or E=1 2(∇Xx·x∇Xbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Fc·F−I) E∗=1 2(I−∇xX·X∇xbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Hc·H) Eij=1 2parenleftBig ∂u i ∂X j+∂u j ∂X i+∂u k ∂X i∂u k ∂X jparenrightBig or E∗ ij=1 2parenleftBig ∂u i ∂x j+∂u j ∂x i−∂u k ∂x i∂u k ∂x jparenrightBig or E=1 2(u∇X+∇Xu+∇Xu·u∇X)bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright J+Jc+Jc·J E∗=1 2(u∇x+∇xu−∇xu·u∇x)bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright K+Kc−Kc·K Small Eij=1 2parenleftBig ∂u i ∂X j+∂u j ∂X iparenrightBig E∗ ij=1 2parenleftBig ∂u i ∂x j+∂u j ∂x iparenrightBig Deformation E=1 2(u∇X+∇Xu)=1 2(J+Jc) E∗=1 2(u∇x+∇xu)=1 2(K+Kc) ROTATION TENSORS Small [1 2parenleftBig ∂u i ∂X j+∂u j ∂X iparenrightBig +1 2parenleftBig ∂u i ∂X j−∂u j ∂X iparenrightBig ]dXj bracketleftBig 1 2parenleftBig ∂u i ∂x j+∂u j ∂x iparenrightBig +1 2parenleftBig ∂u i ∂x j−∂u j ∂x iparenrightBigbracketrightBig dxj deformation [1 2(u∇X+∇Xu) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright E+1 2(u∇X−∇Xu) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright W]·dX [1 2(u∇x+∇xu) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright E∗+1 2(u∇x−∇xu) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Ω]·dx Finite Strain F=R·U=V·R STRESS TENSORS Piola-Kirchoﬀ Cauchy First T0=(d e tF)Tparenleftbig F−1parenrightbigT Second ˜T=(d e tF)parenleftbig F−1parenrightbig Tparenleftbig F−1parenrightbigT Table 4.1: Summary of Major Equations Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–31 Tensorial Explicit X1,X2,X3,dX x,y,z,ds x1,x2,x3,dx x/prime,y/prime,z/prime,ds/prime u1,u2,u3 u,v,w Eij εij Table 4.2: Tensorial vsExplicit Notation Figure 4.7: Strain Deﬁnition Victor Saouma Introduction to Continuum Mechanics Draft4–32 KINEMATIC 106From the chain rule of diﬀerrentiation du=∂u ∂xdx+∂u ∂ydy+∂u ∂zdz (4.139-a) dv=∂v ∂xdx+∂v ∂ydy+∂v ∂zdz (4.139-b) dw=∂w ∂xdx+∂w ∂ydy+∂w ∂zdz (4.139-c) 107Substituting this equation into the preceding one yields the ﬁnite strains ds/prime2−ds2=2  ∂u ∂x+1 2 parenleftBigg∂u ∂xparenrightBigg2 +parenleftBigg∂v ∂xparenrightBigg2 +parenleftBigg∂w ∂xparenrightBigg2   dx2 +2  ∂v ∂y+1 2 parenleftBigg∂u ∂yparenrightBigg2 +parenleftBigg∂v ∂yparenrightBigg2 +parenleftBigg∂w ∂yparenrightBigg2   dy2 +2  ∂w ∂z+1 2 parenleftBigg∂u ∂zparenrightBigg2 +parenleftBigg∂v ∂zparenrightBigg2 +parenleftBigg∂w ∂zparenrightBigg2   dz2 +2parenleftBigg∂v ∂x+∂u ∂y+∂u ∂x∂u ∂y+∂v ∂x∂v ∂y+∂w ∂x∂w ∂yparenrightBigg dxdy +2parenleftBigg∂w ∂x+∂u ∂z+∂u ∂x∂u ∂z+∂v ∂x∂v ∂z+∂w ∂x∂w ∂zparenrightBigg dxdz +2parenleftBigg∂w ∂y+∂v ∂z+∂u ∂y∂u ∂z+∂v ∂y∂v ∂z+∂w ∂y∂w ∂zparenrightBigg dydz (4.140-a) 108We observe that ds/prime2−ds2is zero if there is no relative displacement between Aand B(i.e. rigid body motion), otherwise the solid is strained. Hence ds/prime2−ds2can be selected as an appropriate measure of the deformation of the solid, and we deﬁne thestrain components as ds /prime2−ds2=2εxxdx2+2εyydy2+2εzzdz2+4εxydxdy+4εxzdxdz+4εyzdydz(4.141) where Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–33 εxx=∂u ∂x+1 2 parenleftBigg∂u ∂xparenrightBigg2 +parenleftBigg∂v ∂xparenrightBigg2 +parenleftBigg∂w ∂xparenrightBigg2 (4.142) εyy=∂v ∂y+1 2 parenleftBigg∂u ∂yparenrightBigg2 +parenleftBigg∂v ∂yparenrightBigg2 +parenleftBigg∂w ∂yparenrightBigg2 (4.143) εzz=∂w ∂z+1 2 parenleftBigg∂u ∂zparenrightBigg2 +parenleftBigg∂v ∂zparenrightBigg2 +parenleftBigg∂w ∂zparenrightBigg2 (4.144) εxy=1 2parenleftBigg∂v ∂x+∂u ∂y+∂u ∂x∂u ∂y+∂v ∂x∂v ∂y+∂w ∂x∂w ∂yparenrightBigg (4.145) εxz=1 2parenleftBigg∂w ∂x+∂u ∂z+∂u ∂x∂u ∂z+∂v ∂x∂v ∂z+∂w ∂x∂w ∂zparenrightBigg (4.146) εyz=1 2parenleftBigg∂w ∂y+∂v ∂z+∂u ∂y∂u ∂z+∂v ∂y∂v ∂z+∂w ∂y∂w ∂zparenrightBigg (4.147) or εij=1 2(ui,j+uj,i+uk,iuk,j) (4.148) From this equation, we note that: 1. We deﬁne the engineering shear strain as γij=2εij(i/negationslash=j) (4.149) 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.εikis theGreen-Lagrange strain tensor . 4. The strains have been expressed interms of the coordinates x,y,zin the undeformed state, i.e. in the Lagrangian coordinate which is the preferred one in structural mechanics. 5. Alternatively we could have expressed ds/prime2−ds2in terms of coordinates in the deformed state, i.e. Eulerian coordinates x/prime,y/prime,z/prime, 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 Victor Saouma Introduction to Continuum Mechanics Draft4–34 KINEMATIC εxx=∂u ∂x(4.150) εyy=∂v ∂y(4.151) εzz=∂w ∂z(4.152) γxy=∂v ∂x+∂u ∂y(4.153) γxz=∂w ∂x+∂u ∂z(4.154) γyz=∂w ∂y+∂v ∂z(4.155) or εij=1 2(ui,k+uk,i) (4.156) which is called the Cauchy strain 109In ﬁnite element, the strain is often expressed through the linear operator L ε=Lu (4.157) or   εxx εyy εzz εxy εxz εyz   bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright ε= ∂ ∂x00 0∂ ∂y0 00∂ ∂z∂ ∂y∂ ∂x0 ∂ ∂z0∂ ∂x 0∂ ∂z∂ ∂y  bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright L  ux uy uz   bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright u (4.158) 4.2.9 Compatibility Equation 110Ifεij=1 2(ui,j+uj,i) then we have six diﬀerential equations (in 3D the strain ten- sor has a total of 9 terms, but due to symmetry, there are 6 independent ones) for determining (upon integration) three unknowns displacements ui. Hence the system is overdetermined, and there must be some linear relations between the strains. 111It can be shown (through appropriate successive diﬀerentiation of the strain expres- sion) that the compatibility relation for strain reduces to: ∂2εik ∂xj∂xj+∂2εjj ∂xi∂xk−∂2εjk ∂xi∂xj−∂2εij ∂xj∂xk=0.or∇x×L×∇ x=0 (4.159) Victor Saouma Introduction to Continuum Mechanics Draft4.2 Strain Tensor 4–35 There are 81 equations in all, but only six are distinct ∂2ε11 ∂x2 2+∂2ε22 ∂x2 1=2∂2ε12 ∂x1∂x2(4.160-a) ∂2ε22 ∂x2 3+∂2ε33 ∂x2 2=2∂2ε23 ∂x2∂x3(4.160-b) ∂2ε33 ∂x2 1+∂2ε11 ∂x2 3=2∂2ε31 ∂x3∂x1(4.160-c) ∂ ∂x1parenleftBigg −∂ε23 ∂x1+∂ε31 ∂x2+∂ε12 ∂x3parenrightBigg =∂2ε11 ∂x2∂x3(4.160-d) ∂ ∂x2parenleftBigg∂ε23 ∂x1−∂ε31 ∂x2+∂ε12 ∂x3parenrightBigg =∂2ε22 ∂x3∂x1(4.160-e) ∂ ∂x3parenleftBigg∂ε23 ∂x1+∂ε31 ∂x2−∂ε12 ∂x3parenrightBigg =∂2ε33 ∂x1∂x2(4.160-f) In 2D, this results in (by setting i=2 ,j=1a n dl=2 ) : ∂2ε11 ∂x2 2+∂2ε22 ∂x2 1=∂2γ12 ∂x1∂x2 (4.161) (recall that 2 ε12=γ12.) 112When he compatibility equation is written in term of the stresses, it yields: ∂2σ11 ∂x2 2−ν∂σ222 ∂x2 2+∂2σ22 ∂x2 1−ν∂2σ11 ∂x2 1=2( 1+ν)∂2σ21 ∂x1∂x2(4.162) Example 4-13: Strain Compatibility For the following strain ﬁeld  −X2 X2 1+X2 2X1 2(X2 1+X2 2)0 X1 2(X2 1+X2 2)00 00 0  (4.163) does there exist a single-valued continuous displacement ﬁeld? Solution: ∂E11 ∂X2=−(X2 1+X2 2)−X2(2X2) (X2 1+X2 2)2=X2 2−X2 1 (X2 1+X2 2)2(4.164-a) 2∂E12 ∂X1=(X2 1+X2 2)−X1(2X1) (X2 1+X2 2)2=X2 2−X2 1 (X2 1+X2 2)2(4.164-b) ∂E22 ∂X2 1= 0 (4.164-c) Victor Saouma Introduction to Continuum Mechanics Draft4–36 KINEMATIC ⇒∂2E11 ∂X2 2+∂2E22 ∂X2 1=2∂2E12 ∂X1∂X2√(4.164-d) Actually, it can be easily veriﬁed that the unique displacement ﬁeld is given by u1= arctanX2 X1;u2=0 ;u3= 0 (4.165) to which we could add the rigid body displacement ﬁeld (if any). 4.3 Lagrangian Stresses; Piola Kirchoﬀ Stress Tensors 113In Sect. 2.2 the discussion of stress applied to the deformed conﬁguration dA(us- ing spatial coordiantes x), that is the one where equilibrium must hold. The deformed conﬁguration being the natural one in which to characterize stress. Hence we had df=tdA (4.166-a) t=Tn (4.166-b) (note the use of Tinstead of σ). Hence the Cauchy stress tensor was really deﬁned in the Eulerian space. 114However, there are certain advantages in referring all quantities back to the unde- formed conﬁguration (Lagrangian) of the body because often that conﬁguration has ge-ometric features and symmetries that are lost through the deformation. 115Hence, if we were to deﬁne the strain in material coordinates (in terms of X), we need also to express the stress as a function of the material point Xin material coordinates. 4.3.1 First 116The ﬁrst Piola-Kirchoﬀ stress tensor T0is deﬁned in the undeformed geometry in such a way that it results in the same total force as the traction in the deformed conﬁguration (where Cauchy’s stress tensor was deﬁned). Thus, we deﬁne df≡t0dA0 (4.167) wheret0is apseudo-stress vector in that being based on the undeformed area, it does not describe the actual intensity of the force, however it has the same direction asCauchy’s stress vector t. 117The ﬁrst Piola-Kirchoﬀ stress tensor (also known as Lagrangian Stress Tensor )i s thus the linear transformation T0such that t0=T0n0 (4.168) and for which df=t0dA0=tdA⇒t0=dA dA0t (4.169) Victor Saouma Introduction to Continuum Mechanics Draft4.3 Lagrangian Stresses; Piola Kirchoﬀ Stress Tensors 4–37 using Eq. 4.166-b and 4.168the preceding equation becomes T0n0=dA dA0Tn=TdAn dA0(4.170) and using Eq. 4.33 dAn=dA0(detF)(F−1)Tn0we obtain T0n0=T(detF)parenleftBig F−1parenrightBigTn0 (4.171) the above equation is true for all n0, therefore T0=( d e tF)TparenleftBig F−1parenrightBigT(4.172) T=1 (detF)T0FTorTij=1 (detF)(T0)imFjm(4.173) and we note that this ﬁrst Piola-Kirchoﬀ stress tensor is not symmetric in general. 118To determine the corresponding stress vector, we solve for T0ﬁrst, then for dA0and n0fromdA0n0=1 detFFTn(assuming unit area dA), and ﬁnally t0=T0n0. 4.3.2 Second 119The second Piola-Kirchoﬀ stress tensor, ˜Tis formulated diﬀerently. Instead of the actual force dfondA, it gives the force d˜frelated to the force dfin the same way that a material vector dXatXis related by the deformation to the corresponding spatial vectordxatx. Thus, if we let d˜f=˜tdA0 (4.174-a) and df=Fd˜f (4.174-b) whered˜fis the pseudo diﬀerential force which transforms, under the deformation gradientF, the (actual) diﬀerential force dfat the deformed position (note similarity withdx=FdX). Thus, the pseudo vector tis in general in a diﬀernt direction than that of the Cauchy stress vector t. 120The second Piola-Kirchoﬀ stress tensor is a linear transformation ˜Tsuch that ˜t=˜Tn0 (4.175) thus the preceding equations can be combined to yield df=F˜Tn0dA0 (4.176) we also have from Eq. 4.167 and 4.168 df=t0dA0=T0n0dA0 (4.177) Victor Saouma Introduction to Continuum Mechanics Draft4–38 KINEMATIC and comparing the last two equations we note that ˜T=F−1T0 (4.178) which gives the relationship between the ﬁrst Piola-Kirchoﬀ stress tensor T0and the second Piola-Kirchoﬀ stress tensor ˜T. 121Finally the relation between the second Piola-Kirchoﬀ stress tensor and the Cauchy stress tensor can be obtained from the preceding equation and Eq. 4.172 ˜T=( d e tF)parenleftBig F−1parenrightBig TparenleftBig F−1parenrightBigT (4.179) and we note that this second Piola-Kirchoﬀ stress tensor is always symmetric (if the Cauchy stress tensor is symmetric). 122To determine the corresponding stress vector, we solve for ˜Tﬁrst, then for dA0and n0fromdA0n0=1 detFFTn(assuming unit area dA), and ﬁnally ˜t=˜Tn0. Example 4-14: Piola-Kirchoﬀ Stress Tensors 4.4 Hydrostatic and Deviatoric Strain 93The lagrangian and Eulerian linearstrain tensors can each be split into spherical anddeviator tensor as was the case for the stresses. Hence, if we deﬁne 1 3e=1 3trE (4.180) then the components of the strain deviator E/primeare given by E/prime ij=Eij−1 3eδijorE/prime=E−1 3e1 (4.181) We note that E/primemeasures the change in shape of an element, while the spherical or hydrostatic strain1 3e1represents the volume change. 4.5 Principal Strains, Strain Invariants, Mohr Circle 94Determination of the principal strains ( E(3)<E(2)<E(1), strain invariants and the Mohr circle for strain parallel the one for stresses (Sect. 2.4) and will not be repeated Victor Saouma Introduction to Continuum Mechanics Draft4.5 Principal Strains, Strain Invariants, Mohr Circle 4–39 Piola−Kirchoff Stress Tensors The deformed configuration of a body is described by x1=X1ê2, x2=−X2/2, x3=4X3; If the Cauchy stress tensor is given byi kjjjjjjjj100 00 0 00 0 0 0 y {zzzzzzzzMPa; What are the corresponding first and second Piola−Kirchoff stress tensors, and calculate the respective stress tensors on the e 3plane in the deformed state. ‡F tensor CST=880, 0, 0 <,80, 0, 0 <,80, 0, 100 << 880, 0, 0 <,80, 0, 0 <,80, 0, 100 << F=881ê2, 0, 0 <,80, 0, −1ê2<,80, 4, 0 << 991ÄÄÄÄÄ2,0 ,0 =,90, 0, -1ÄÄÄÄÄ2=,80, 4, 0 <= Finverse =Inverse @FD 982, 0, 0 <,90, 0,1ÄÄÄÄÄ4=,80,-2, 0<= ‡First Piola−Kirchoff Stress Tensor Tfirst [email protected] @Finverse D 880, 0, 0 <,80, 0, 0 <,80, 25, 0 << MatrixForm @%D i kjjjjjj000 00002 50y {zzzzzz‡Second Piola−Kirchoff Stress Tensor Tsecond =Inverse @FD.Tfirst 980, 0, 0 <,90,25ÄÄÄÄÄÄÄÄ4,0=,80, 0, 0 <= MatrixForm @%D i kjjjjjjjj000 0 25ÄÄÄÄÄÄ40 000y {zzzzzzzz ‡Cuchy stress vector Can be obtained from t=CST n tcauchy =MatrixForm @CST. 80, 0, 1 <D i kjjjjjj0 0 100y {zzzzzz ‡Pseudo−Stress vector associated with the First Piola−Kirchoff stress tensor For a unit area in the deformed state in the e3direction, its undeformed area d A0n0is given by d A0n0=FTnÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄdetF detF =Det@FD 1 n=80, 0, 1 < 80, 0, 1 <2 m−piola.nb MatrixForm @Transpose @FD.nêdetF D i kjjjjjj0 40y {zzzzzz Thus n0=e2and using t0=T0 n0 we obtain t01st =MatrixForm @Tfirst. 80, 1, 0 <D i kjjjjjj0 0 25y {zzzzzz We note that this vector is in the same direction as the Cauchy stress vector, its magnitude is one fourth of that of the Cauchy stress vector, because the undeformed area is 4 times that of the deformed area ‡Pseudo−Stress vector associated with the Second Piola−Kirchoff stress tensor t0second =MatrixForm @Tsecond. 80, 1, 0 <D i kjjjjjjjj0 25ÄÄÄÄÄÄ4 0y {zzzzzzzz We see that this pseudo stress vector is in a different direction from that of the Cauchy stress vector (and we note that the tensor F transforms e2into e3).m−piola.nb 3 Victor Saouma Introduction to Continuum Mechanics Draft4–40 KINEMATIC ε εIε εγ 2 III II Figure 4.8: Mohr Circle for Strain here. λ3−IEλ2−IIEλ−IIIE=0 (4.182) where the symbols IE,IIEandIIIEdenote the following scalar expressions in the strain components: IE=E11+E22+E33=Eii=t rE (4.183) IIE=−(E11E22+E22E33+E33E11)+E2 23+E2 31+E2 12(4.184) =1 2(EijEij−EiiEjj)=1 2EijEij−1 2I2 E (4.185) =1 2(E:E−I2 E) (4.186) IIIE=d e tE=1 6eijkepqrEipEjqEkr (4.187) 95In terms of the principal strains, those invariants can be simpliﬁed into IE=E(1)+E(2)+E(3) (4.188) IIE=−(E(1)E(2)+E(2)E(3)+E(3)E(1)) (4.189) IIIE=E(1)E(2)E(3) (4.190) 96T h eM o h rc i r c l eu s e st h e Engineering shear strain deﬁnition of Eq. 4.86, Fig. 4.8 Example 4-15: Strain Invariants & Principal Strains Victor Saouma Introduction to Continuum Mechanics Draft4.5 Principal Strains, Strain Invariants, Mohr Circle 4–41 Determine the planes of principal strains for the following strain tensor  1√ 30√ 300 00 1  (4.191) Solution: The strain invariants are given by IE=Eii= 2 (4.192-a) IIE=1 2(EijEij−EiiEjj)=−1+3 = +2 (4.192-b) IIIE=|Eij|=−3 (4.192-c) The principal strains by Eij−λδij= 1−λ√ 30√ 3−λ0 00 1 −λ  (4.193-a) =( 1−λ)parenleftBigg λ−1+√ 13 2parenrightBiggparenleftBigg λ−1−√ 13 2parenrightBigg (4.193-b) E(1)=λ(1)=1+√ 13 2=2.3 (4.193-c) E(2)=λ(2)= 1 (4.193-d) E(3)=λ(3)=1−√ 13 2=−1.3 (4.193-e) The eigenvectors for E(1)=1+√ 13 2give the principal directions n(1):  1−1+√ 13 2√ 30√ 3−1+√ 13 20 00 1 −1+√ 13 2   n(1) 1 n(1) 2 n(1) 3  =  parenleftBig 1−1+√ 13 2parenrightBig n(1) 1+√ 3n(1) 2√ 3n(1) 1−parenleftBig 1+√ 13 2parenrightBig n(1) 2parenleftBig 1−1+√ 13 2parenrightBig n(1) 3  =  0 00   (4.194) which gives n(1) 1=1+√ 13 2√ 3n(1) 2 (4.195-a) n(1) 3= 0 (4.195-b) n(1)·n(1)=parenleftBigg1+2√ 13+13 12+1parenrightBiggparenleftBig n(1) 2parenrightBig2=1⇒n1 2=0.8; (4.195-c) ⇒n(1)=⌊0.80.60⌋ (4.195-d) For the second eigenvector λ(2)=1 :  1−1√ 30√ 3−10 00 1 −1   n(2) 1 n(2) 2 n(2) 3  =  √ 3n(2) 2√ 3n(2) 1−n(2) 2 0  =  0 00  (4.196) Victor Saouma Introduction to Continuum Mechanics Draft4–42 KINEMATIC which gives (with the requirement that n(2)·n(2)=1 ) n(2)=⌊001⌋ (4.197) Finally, the third eigenvector can be obrained by the same manner, but more easily from n(3)=n(1)×n(2)=d e tvextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglee1e2e3 0.80.60 001vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle=0.6e1−0.8e2 (4.198) Therefore aj i=  n(1) n(2) n(3)  = 0.80.60 00 1 0.6−0.80  (4.199) and this results can be checked via [a][E][a]T= 0.80.60 00 1 0.6−0.80  1√ 30√ 300 00 1  0.800.6 0.60−0.8 01 0 = 2.30 0 01 000−1.3  (4.200) Example 4-16: Mohr’s Circle Construct the Mohr’s circle for the following plane strain case:  00 0 05√ 3 0√ 33  (4.201) Solution: 1 21 345660o2B D EF εεs n23 Victor Saouma Introduction to Continuum Mechanics Draft4.6 Initial or Thermal Strains 4–43 We note that since E(1)= 0 is a principal value for plane strain, ttwo of the circles are drawn as shown. 4.6 Initial or Thermal Strains 97Initial (or thermal strain) in 2D: εij=bracketleftBigg α∆T0 0α∆TbracketrightBigg bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Plane Stress=( 1+ν)bracketleftBigg α∆T0 0α∆TbracketrightBigg bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Plane Strain(4.202) note there is no shear strains caused by thermal expansion. 4.7†Experimental Measurement of Strain 98Typically, the transducer to measure strains in a material is the strain gage. The most common type of strain gage used today for stress analysis is the bonded resistance strain gage shown in Figure 4.9. Figure 4.9: Bonded Resistance Strain Gage 99These gages use a grid of ﬁne wire or a metal foil grid encapsulated in a thin resin backing. The gage is glued to the carefully prepared test specimen by a thin layer of epoxy. The epoxy acts as the carrier matrix to transfer the strain in the specimen to the strain gage. As the gage changes in length, the tiny wires either contract or elongatedepending upon a tensile or compressive state of stress in the specimen. The crosssectional area will increase for compression and decrease in tension. Because the wirehas an electrical resistance that is proportional to the inverse of the cross sectional area, Rα 1 A, a measure of the change in resistance can be converted to arrive at the strain in the material. 100Bonded resistance strain gages are produced in a variety of sizes, patterns, and resis- tance. One type of gage that allows for the complete state of strain at a point in a plane to be determined is a strain gage rosette. It contains three gages aligned radially from a common point at diﬀerent angles from each other, as shown in Figure 4.10. The straintransformation equations to convert from the three strains a t any angle to the strain atap o i n ti nap l a n ea r e : sepsilonv a=sepsilonvxcos2θa+sepsilonvysin2θa+γxysinθacosθa (4.203) Victor Saouma Introduction to Continuum Mechanics Draft4–44 KINEMATIC sepsilonvb=sepsilonvxcos2θb+sepsilonvysin2θb+γxysinθbcosθb (4.204) sepsilonvc=sepsilonvxcos2θc+sepsilonvysin2θc+γxysinθccosθc (4.205) Figure 4.10: Strain Gage Rosette 101When the measured strains sepsilonva,sepsilonvb,a n dsepsilonvc, are measured at their corresponding angles from the reference axis and substituted into the above equations the state of strain at apoint may be solved, namely, sepsilonv x,sepsilonvy,a n dγxy. In addition the principal strains may then be computed by Mohr’s circle or the principal strain equations. 102Dueto the wide variety ofstyles ofgages, many factors must be considered in choosing the right gage for a particular application. Operating temperature, state of strain, andstability of installation all inﬂuence gage selection. Bonded resistance strain gages are well suited for making accurate and practical strain measurements because of their high sensitivity to strains, low cost, and simple operation. 103The measure of the change in electrical resistance when the strain gage is strained is known as the gage factor. The gage factor is deﬁned as the fractional change in resistance divided by the fractional change in length along the axis of the gage. GF=∆R R ∆L LCommon gage factors are in the range of 1.5-2 for most resistive strain gages. 104Common strain gages utilize a grid pattern as opposed to a straight length of wire in order to reduce the gage length. This grid pattern causes the gage to be sensitive to deformations transverse to the gage length. Therefore, corrections for transverse strains should be computed and applied to the strain data. Some gages come with the tranversecorrection calculated into the gage factor. The transverse sensitivity factor, K t, is deﬁned as the transverse gage factor divided by the longitudinal gage factor. Kt=GFtransverse GFlongitudinal These sensitivity values are expressed as a percentage and vary from zero to ten percent. 105A ﬁnal consideration for maintaining accurate strain measurement is temperature compensation. The resistance of the gage and the gage factor will change due to thevariation of resistivity and strain sensitivity with temperature. Strain gages are produced with diﬀerent temperature expansion coeﬃcients. In order to avoid this problem, the expansion coeﬃcient of the strain gage should match that of the specimen. If no largetemperature change is expected this may be neglected. 106The change in resistance of bonded resistance strain gages for most strain measure- ments is very small. From a simple calculation, for a strain of 1 µsepsilonv(µ=1 0−6)w i t h Victor Saouma Introduction to Continuum Mechanics Draft4.7†Experimental Measurement of Strain 4–45 a 120 Ω gage and a gage factor of 2, the change in resistance produced by the gage is ∆R=1×10−6×120×2 = 240×10−6Ω. Furthermore, it is the fractional change in resistance that is important and the number to be measured will be in the order of acouple of µohms. For large strains a simple multi-meter may suﬃce, but in order to acquire sensitive measurements in the µΩ range a Wheatstone bridge circuit is necessary to amplify this resistance. The Wheatstone bridge is described next. 4.7.1 Wheatstone Bridge Circuits 107Duetotheiroutstandingsensitivity, Wheatstone bridgecircuits arevery advantageous for the measurement of resistance, inductance, and capacitance. Wheatstone bridges arewidely used for strain measurements. A Wheatstone bridge is shown in Figure 4.11. It consists of 4 resistors arranged in a diamond orientation. An input DC voltage, or excitation voltage, is applied between the top and bottom of the diamond and the outputvoltage is measured across the middle. When the output voltage is zero, the bridge issaid to be balanced. One or more of the legs of the bridge may be a resistive transducer, such as a strain gage. The other legs of the bridge are simply completion resistors with resistance equal to that of the strain gage(s). As the resistance of one of the legs changes,by a change in strain from a resistive strain gage for example, the previously balancedbridge is now unbalanced. This unbalance causes a voltage to appear across the middle of the bridge. This induced voltage may be measured with a voltmeter or the resistor in the opposite leg may be adjusted to re-balance the bridge. In either case the changein resistance that caused the induced voltage may be measured and converted to obtainthe engineering units of strain. Figure 4.11: Quarter Wheatstone Bridge Circuit 4.7.2 Quarter Bridge Circuits 108If a strain gage is oriented in one leg of the circuit and the other legs contain ﬁxed resistors as shown in Figure 4.11, the circuit is known as a quarter bridge circuit. Thecircuit is balanced when R1 R2=Rgage R3. When the circuit is unbalanced Vout=Vin(R1 R1+R2− Rgage Rgage+R3). 109Wheatstone bridges may also be formed with two or four legs of the bridge being composed of resistive transducers and are called a half bridge and full bridge respectively. Victor Saouma Introduction to Continuum Mechanics Draft4–46 KINEMATIC Depending upon the type of application and desired results, the equations for these circuits will vary as shown in Figure 4.12. Here E0is the output voltage in mVolts, E is the excitation voltage in Volts, sepsilonvis strain and νis Poisson’s ratio. 110In order to illustrate how to compute a calibration factor for a particular experiment, suppose a single active gage in uniaxial compression is used. This will correspond to the upper Wheatstone bridge conﬁguration of Figure 4.12. The formula then is Figure 4.12: Wheatstone Bridge Conﬁgurations Victor Saouma Introduction to Continuum Mechanics Draft4.7†Experimental Measurement of Strain 4–47 E0 E=Fsepsilonv(10−3) 4+2Fsepsilonv(10−6)(4.206) 111The extra term in the denominator 2 Fsepsilonv(10−6) is a correction factor for non-linearity. Because this term is quite small compared to the other term in the denominator it willbe ignored. For most measurements a gain is necessary to increase the output voltagefrom the Wheatstone bridge. The gain relation for the output voltage may be written as V=GE 0(103), where V is now in Volts. so Equation 4.206 becomes V EG(103)=Fsepsilonv(10−3) 4 sepsilonv V=4 FEG(4.207) 112Here, Equation 4.207 is the calibration factor in units of strain per volt. For common values where F=2.07,G= 1000,E= 5, the calibration factor is simply4 (2.07)(1000)(5)or 386.47 microstrain per volt. Victor Saouma Introduction to Continuum Mechanics Draft4–48 KINEMATIC Victor Saouma Introduction to Continuum Mechanics Draft Chapter 5 MATHEMATICAL PRELIMINARIES; Part III VECTOR INTEGRALS 5.1 Integral of a Vector 20The integral of a vector R(u)=R1(u)e1+R2(u)e2+R3(u)e3is deﬁned as integraldisplay R(u)du=e1integraldisplay R1(u)du+e2integraldisplay R2(u)du+e3integraldisplay R3(u)du (5.1) if a vector S(u) exists such that R(u)=d du(S(u)), then integraldisplay R(u)du=integraldisplayd du(S(u))du=S(u)+c (5.2) 5.2 Line Integral 21Givenr(u)=x(u)e1+y(u)e2+z(u)e3wherer(u) is a position vector deﬁning a curveCconnecting point P1toP2whereu=u1andu=u2respectively, anf given A(x,y,z)=A1e1+A2e2+A3e3being a vectorial function deﬁned and continuous along C, then the integral of the tangential component of AalongCfromP1toP2is given by integraldisplayP2 P1A·dr=integraldisplay CA·dr=integraldisplay CA1dx+A2dy+A3dz (5.3) IfAwere a force, then this integral would represent the corresponding work. 22If the contour is closed, then we deﬁne the contour integral as contintegraldisplay CA·dr=integraldisplay CA1dx+A2dy+A3dz (5.4) 23It can be shown that if A=∇φthen integraldisplayP2 P1A·dris independent of the path Cconnecting P1toP2(5.5-a) contintegraldisplay CA·dr= 0 along a closed contour line (5.5-b) Draft5–2 MATHEMATICAL PRELIMINARIES; Part III VECTOR INTEGRALS 5.3 Integration by Parts 24The integration by part formula is integraldisplayb au(x)v/prime(x)dx=u(x)v(x)|b a−integraldisplayb av(x)u/prime(x)dx (5.6) 5.4 Gauss; Divergence Theorem 25The divergence theorem (also known as Ostrogradski’s Theorem) comes repeatedly in solid mechanics and can be stated as follows: integraldisplay Ω∇·vdΩ=integraldisplay Γv.ndΓo rintegraldisplay Ωvi,idΩ=integraldisplay ΓvinidΓ (5.7) That is the integral of the outer normal component of a vector over a closed surface (which is the volume ﬂux ) is equal to the integral of the divergence of the vector over the volume bounded by the closed surface. 26For 2D-1D transformations, we have integraldisplay A∇·qdA=contintegraldisplay sqTnds (5.8) 27This theorem is sometime refered to as Green’s theorem in space. 5.5 Stoke’s Theorem 28Stoke’s theorem states that contintegraldisplay CA·dr=integraldisplayintegraldisplay S(∇×A)·ndS=integraldisplayintegraldisplay S(∇×A)·dS (5.9) whereSis an open surface with two faces conﬁned by C 5.6 Green; Gradient Theorem 29Green’s theorem in plane is a special case of Stoke’s theorem. contintegraldisplay (Rdx+Sdy)=integraldisplay ΓparenleftBigg∂S ∂x−∂R ∂yparenrightBigg dxdy (5.10) Victor Saouma Introduction to Continuum Mechanics Draft5.6 Green; Gradient Theorem 5–3 Example 5-1: Physical Interpretation of the Divergence Theorem 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. 5.1-a. V∆tn dSdV=dxdydzS dSn c)b)Y BCDE F G ∆∆ XZ YV VV VZ XP(X,Y,Z)HA ∆ a) Figure 5.1: Physical Interpretation of the Divergence Theorem vx|x,y,z≈vx (5.11-a) vxvextendsinglevextendsinglevextendsingle x−∆x/2,y,z≈vx−1 2∂vx ∂x∆xAFED (5.11-b) vxvextendsinglevextendsinglevextendsingle x+∆x/2,y,z≈vx+1 2∂vx ∂x∆xGHCB (5.11-c) The net inﬂow per unit time across the xplanes is ∆Vx=parenleftBigg vx+1 2∂vx ∂x∆xparenrightBigg ∆y∆z−parenleftBigg vx−1 2∂vx ∂x∆xparenrightBigg ∆y∆z(5.12-a) =∂vx ∂x∆x∆y∆z (5.12-b) Similarly ∆Vy=∂vy ∂y∆x∆y∆z (5.13-a) Victor Saouma Introduction to Continuum Mechanics Draft5–4MATHEMATICAL PRELIMINARIES; Part III VECTOR INTEGRALS ∆Vz=∂vz ∂z∆x∆y∆z (5.13-b) Hence, the total increase per unit volume and unit time will be given by parenleftBig ∂vx ∂x+∂vy ∂y+∂vz ∂zparenrightBig ∆x∆y∆z ∆x∆y∆z=d i vv=∇·v (5.14) Furthermore, if we consider the total of ﬂuid crossing dSduring ∆ t, Fig. 5.1-b, it will be given by ( v∆t)·ndS=v·ndS∆tor the volume of ﬂuid crossing dSper unit time is v·ndS. Thus for an arbitrary volume, Fig. 5.1-c, the total amount of ﬂuid crossing a closed surfaceSper unit time isintegraldisplay Sv·ndS. But this is equal tointegraldisplay V∇·vdV(Eq. 5.14), thus integraldisplay Sv·ndS=integraldisplay V∇·vdV (5.15) which is the divergence theorem. Victor Saouma Introduction to Continuum Mechanics Draft Chapter 6 FUNDAMENTAL LAWS of CONTINUUM MECHANICS 6.1 Introduction 20We have thus far studied the stress tensors (Cauchy, Piola Kirchoﬀ), and several other tensors which describe strain at a point. In general, those tensors will vary from pointto point and represent a tensor ﬁeld . 21We have also obtained only one diﬀerential equation, that was the compatibility equa- tion. 22In this chapter, we will derive additional 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 chapter where constitututive laws relating stressand strain will be introduced. Only with constitutive equations and boundary and initialconditions would we be able to obtain a well deﬁned mathematical problem to solve forthe stress and deformation distribution or the displacement or velocity ﬁelds. 23In this chapter we shall derive diﬀerential equations expressing locally the conservation of mass, momentum and energy. These diﬀerential equations of balance will be derivedfrom integral forms of the equation of balance expressing the fundamental postulates ofcontinuum mechanics. 6.1.1 Conservation Laws 24Conservation laws constitute a fundamental component of classical physics. A conser- vation law establishes a balance of a scalar or tensorial quantity in voulme Vbounded by a surface S. In its most general form, such a law may be expressed as d dtintegraldisplay VAdV bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Rate of variation+integraldisplay SαdS bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Exchange by Diffusion=integraldisplay VAdV bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Source(6.1) Draft6–2 FUNDAMENTAL LAWS of CONTINUUM MECHANICS whereAis the volumetric density of the quantity of interest (mass, linear momentum, energy, ...) a,Ais the rate of volumetric density of what is provided from the outside, andαis the rate of surface density of what is lost through the surface SofVand will be a function of the normal to the surface n. 25Hence, 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 Awhich is the quantity of interest. The dimensions of various quantities are given by dim(a)=d i m ( AL−3) (6.2-a) dim(α)=d i m ( AL−2t−1) (6.2-b) dim(A)=d i m ( AL−3t−1) (6.2-c) 26Hence this chapter will apply the previous conservation law to mass, momentum, and energy. the resulting diﬀerential equations will provide additional interesting relation with regard to the imcompressibiltiy of solids (important in classical hydrodynamics and plasticity theories), equilibrium and symmetry of the stress tensor, and the ﬁrst law ofthermodynamics. 27Theenunciationofthepreceding threeconservationlawsplusthesecondlawofthermo- dynamics, constitutewhatiscommonlyknown asthe fundamental lawsofcontinuum mechanics . 6.1.2 Fluxes 28Prior to the enunciation of the ﬁrst conservation law, we need to deﬁne the concept of ﬂux across a bounding surface. 29Theﬂuxacross 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 nin the positive sense, then the volume of “material” ﬂowing through the inﬁnitesimal surface area dSin timedtis equal to the volume of the cylinder with base dSand slant height vdtparallel to the velocity vector v, Fig. 6.1 (If v·nis negative, then the ﬂow is in the negative direction). Hence, we deﬁne the volume ﬂux as Volume Flux =integraldisplay Sv·ndS=integraldisplay SvjnjdS (6.3) where the last form is for rectangular cartesian components. 30We can generalize this deﬁnition and deﬁne the following ﬂuxes per unit area through dS: Victor Saouma Introduction to Continuum Mechanics Draft6.2 Conservation of Mass; Continuity Equation 6–3 v n dSvdtv dtn Figure 6.1: Flux Through Area dS Mass Flux =integraldisplay Sρv·ndS=integraldisplay SρvjnjdS (6.4) Momentum Flux =integraldisplay Sρv(v·n)dS=integraldisplay SρvkvjnjdS (6.5) Kinetic Energy Flux =integraldisplay S1 2ρv2(v·n)dS=integraldisplay S1 2ρvivivjnjdS(6.6) Heat ﬂux =integraldisplay Sq·ndS=integraldisplay SqjnjdS (6.7) Electric ﬂux =integraldisplay SJ·ndS=integraldisplay SJjnjdS (6.8) 6.2 Conservation of Mass; Continuity Equation 6.2.1 Spatial Form 31If we consider an arbitrary volume V, ﬁxed in space, and bounded by a surface S.I f a continuous medium of density ρﬁlls the volume at time t, then the total mass in Vis M=integraldisplay Vρ(x,t)dV (6.9) whereρ(x,t) is a continuous function called the mass density . We note that this spatial form in terms of xis most common in ﬂuid mechanics. 32The rate of increase of the total mass in the volume is ∂M ∂t=integraldisplay V∂ρ ∂tdV (6.10) 33TheL a wo fc o n s e r v a t i o no fm a s s requires that the mass of a speciﬁc portion of the continuum remains constant. Hence, if no mass is created or destroyed inside V,t h e n the preceding equation must eqaul the inﬂow of mass (ofﬂux) through the surface. The outﬂow is equal to v·n, thus the inﬂow will be equal to −v·n. integraldisplay S(−ρvn)dS=−integraldisplay Sρv·ndS=−integraldisplay V∇·(ρv)dV (6.11) Victor Saouma Introduction to Continuum Mechanics Draft6–4FUNDAMENTAL LAWS of CONTINUUM MECHANICS must be equal to∂M ∂t.T h u s integraldisplay VbracketleftBigg∂ρ ∂t+∇·(ρv)bracketrightBigg dV= 0 (6.12) since the integral must hold for any arbitrary choice of dV, then we obtain ∂ρ ∂t+∇·(ρv)o r∂ρ ∂t+∂(ρvi) ∂xi= 0 (6.13) 34The chain rule will in turn give ∂(ρvi) ∂xi=ρ∂vi ∂xi+vi∂ρ ∂xi(6.14) 35It can be shown that the rate of change of the density in the neighborhood of a particle instantaneously at xby dρ dt=∂ρ ∂t+v·∇ρ=∂ρ ∂t+vi∂ρ ∂xi(6.15) where the ﬁrst term gives the local rate of change of the density in the neighborhood of the place of x, while the second term gives the convective rate of change of the density in the neighborhood of a particle as it moves to a place having a diﬀerent density.The ﬁrst term vanishes in a steady ﬂow, while the second term vanishes in a uniformﬂow. 36Upon substitution in the last three equations, we obtain the continuity equation dρ dt+ρ∂vi ∂xi=0 o rdρ dt+ρ∇·v=0 (6.16) 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 densityρin the neighborhood of the particle. 37If 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 thesimpler form ∂vi ∂xi=0 o r ∇·v=0 (6.17) this is the condition of incompressibility 6.2.2 Material Form 38If material coordinates Xare used, the conservation of mass, and using Eq. 4.38 (dV=|J|dV0), implies integraldisplay V0ρ(X,t0)dV0=integraldisplay Vρ(x,t)dV=integraldisplay V0ρ(x,t)|J|dV0 (6.18) Victor Saouma Introduction to Continuum Mechanics Draft6.3 Linear Momentum Principle; Equation of Motion 6–5 or integraldisplay V0[ρ0−ρ|J|]dV0= 0 (6.19) and for an arbitrary volume dV0, the integrand must vanish. If we also suppose that the initial density ρ0is everywhere positive in V0(no empty spaces), and at time t=t0, J=1 ,t h e nw ec a nw r i t e ρJ=ρ0 (6.20) or d dt(ρJ)=0 (6.21) which is the continuity equation due to Euler ,o rt h eLagrangian diﬀerential formof the continuity equation. 39We note that this is the same equation as Eq. 6.16 which was expressed in spatial form. Those two equations can be derived one from the other. 40The more commonly used form if the continuity equation is Eq. 6.16. 6.3 Linear Momentum Principle; Equation of Motion 6.3.1 Momentum Principle 41The 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 ab a s i cpostulate of continuum mechanics. integraldisplay StdS+integraldisplay VρbdV=d dtintegraldisplay VρvdV (6.22) Then we substitute ti=Tijnjand apply the divergence theorm to obtain integraldisplay VparenleftBigg∂Tij ∂xj+ρbiparenrightBigg dV=integraldisplay Vρdvi dtdV (6.23-a) integraldisplay VbracketleftBigg∂Tij ∂xj+ρbi−ρdvi dtbracketrightBigg dV= 0 (6.23-b) or for an arbitrary volume ∂Tij ∂xj+ρbi=ρdvi dtor∇T+ρb=ρdv dt (6.24) which isCauchy’s (ﬁrst) equation of motion ,o rthe linear momentum principle , or more simply equilibrium equation . Victor Saouma Introduction to Continuum Mechanics Draft6–6 FUNDAMENTAL LAWS of CONTINUUM MECHANICS 42When expanded in 3D, this equation yields: ∂T11 ∂x1+∂T12 ∂x2+∂T13 ∂x3+ρb1=0 ∂T21 ∂x1+∂T22 ∂x2+∂T23 ∂x3+ρb2= 0 (6.25-a) ∂T31 ∂x1+∂T32 ∂x2+∂T33 ∂x3+ρb3=0 43We note that these equations could also have been derived from the free body diagram shown in Fig. 6.2 with the assumption of equilibrium (via Newton’s second law) con- sidering an inﬁnitesimal element of dimensions dx1×dx2×dx3. Writing the summation of forces, will yield Tij,j+ρbi=0 (6.26) whereρis the density, biis the body force (including inertia). σ+ δyyδσyy ydy τxyσσ σ+δ xxdy yyxxσδxx xdx τ+ δxyτδ xyd τyxτ+ δτδ ydyyxyx xx dx Figure 6.2: Equilibrium of Stresses, Cartesian Coordinates Example 6-1: Equilibrium Equation In the absence of body forces, does the following stress distribution  x2 2+ν(x2 1−x2 x)−2νx1x2 0 −2νx1x2x2 1+ν(x2 2−x2 1)0 00 ν(x2 1+x2 2)  (6.27) whereνis a constant, satisfy equilibrium? Victor Saouma Introduction to Continuum Mechanics Draft6.3 Linear Momentum Principle; Equation of Motion 6–7 Solution: ∂T1j ∂xj=∂T11 ∂x1+∂T12 ∂x2+∂T13 ∂x3=2νx1−2νx1=0√(6.28-a) ∂T2j ∂xj=∂T21 ∂x1+∂T22 ∂x2+∂T23 ∂x3=−2νx2+2νx2=0√(6.28-b) ∂T3j ∂xj=∂T31 ∂x1+∂T32 ∂x2+∂T33 ∂x3=0√(6.28-c) Therefore, equilibrium is satisﬁed. 6.3.2 Moment of Momentum Principle 44The moment of momentum principle states that the time rate of change of the total moment of momentum of a given set of particles equals the vector sum of the momentsof all external forces acting on the particles of the set . 45Thus, in the absence of distributed couples (this theory of Cosserat will not be covered in this course) we postulate the same principle for a continuum as integraldisplay S(r×t)dS+integraldisplay V(r×ρb)dV=d dtintegraldisplay V(r×ρv)dV (6.29) 6.3.2.1 Symmetry of the Stress Tensor 46We observe that the preceding equation does not furnish any new diﬀerential equation of motion. If we substitute tn=Tnand the symmetry of the tensor is assumed, then the linear momentum principle (Eq. 6.24) is satisﬁed. 47Alternatively, we may start by using Eq. 1.18( ci=εijkajbk) to express the cross product in indicial form and substitute above: integraldisplay S(εrmnxmtn)dS+integraldisplay V(εrmnxmbnρ)dV=d dtintegraldisplay V(εrmnxmρvn)dV (6.30) we then substitute tn=Tjnnj, and apply Gauss theorem to obtain integraldisplay VεrmnbracketleftBigg∂xmTjn ∂xj+xmρbnbracketrightBigg dV=integraldisplay Vεrmnd dt(xmvn)ρdV (6.31) but since dxm/dt=vm, this becomes integraldisplay VεrmnbracketleftBigg xmparenleftBigg∂Tjn ∂xj+ρbnparenrightBigg +δmjTjnbracketrightBigg dV=integraldisplay VεrmnparenleftBigg vmvn+xmdvn dtparenrightBigg ρdV(6.32) Victor Saouma Introduction to Continuum Mechanics Draft6–8 FUNDAMENTAL LAWS of CONTINUUM MECHANICS butεrmnvmvn=0s i n c e vmvnis symmetric in the indeces mnwhileεrmnis antisymmetric, and the last term on the right cancels with the ﬁrst term on the left, and ﬁnally withδ mjTjn=Tmnwe are left withintegraldisplay VεrmnTmndV= 0 (6.33) or for an arbitrary volume V, εrmnTmn=0 (6.34) at each point, and this yields forr=1T23−T32=0 forr=2T31−T13=0 forr=3T12−T21=0 (6.35) establishing the symmetry of the stress matrix without any assumption of equilibrium or of uniformity of stress distribution as was done in Sect. 2.3. 48The symmetry of the stress matrix is Cauchy’s second law of motion (1827). 6.4 Conservation of Energy; First Principle of Thermodynam- ics 49The 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 assumethat the only energy transfers to the system are by mechanical work done on the systemby surface traction and body forces, by heat transfer through the boundary. 6.4.1 Spatial Gradient of the Velocity 50We deﬁne Las thespatial gradient of the velocity a n di nt u r nt h i sg r a d i e n tc a nbe decomposed into a symmetric rate of deformation tensor D (orstretching tensor ) and a skew-symmeteric tensor Wcalled the spin tensor orvorticity tensor1. Lij=vi,jorL=v∇x (6.36) L=D+W (6.37) D=1 2(v∇x+∇xv)a n dW=1 2(v∇x−∇xv) (6.38) this term will be used in the derivation of the ﬁrst principle. 6.4.2 First Principle 51If mechanical quantities only are considered, the principle of conservation of en- ergyfor the continuum may be derived directly from the equation of motion given by 1NotesimilaritywithEq. 4.106-b. Victor Saouma Introduction to Continuum Mechanics Draft6.4Conservation of Energy; First Principle of Thermodynamics 6–9 Eq. 6.24. This is accomplished by taking the integral over the volume Vof the scalar product between Eq. 6.24 and the velocity vi. integraldisplay VviTji,jdV+integraldisplay VρbividV=integraldisplay Vρvidvi dtdV (6.39) If we consider the right hand side integraldisplay Vρvidvi dtdV=d dtintegraldisplay V1 2ρvividV=d dtintegraldisplay V1 2ρv2dV=dK dt(6.40) which represents the time rate of change of the kinetic energy Kin the continuum. 52Also we have viTji,j=(viTji),j−vi,jTjiand from Eq. 6.37 we have vi,j=Lij+Wij. It can be shown that since Wijis skew-symmetric, and Tis symmetric, that TijWij=0 , and thus TijLij=TijDij.T¨Dis called the stress power . 53If we consider thermal processes, the rate of increase of total heat into the continuum is given by Q=−integraldisplay SqinidS+integraldisplay VρrdV (6.41) Qhas the dimension of power, that is ML2T−3, and the SI unit is the Watt (W). qis the heat ﬂux per unit area by conduction, its dimension is MT−3and the corresponding SI unit is Wm−2. Finally, ris theradiant heat constant per unit mass, its dimension isMT−3L−4and the corresponding SI unit is Wm−6. 54We thus have dK dt+integraldisplay VDijTijdV=integraldisplay V(viTji),jdV+integraldisplay VρvibidV+Q (6.42) 55We next convert the ﬁrst integral on the right hand side to a surface integral by the divergence theorem (integraltext V∇·vdV=integraltext Sv.ndS) and since ti=Tijnjwe obtain dK dt+integraldisplay VDijTijdV=integraldisplay SvitidS+integraldisplay VρvibidV+Q(6.43) dK dt+dU dt=dW dt+Q (6.44) 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 handside. 56If 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 sumoftherateofworkplusallotherenergiessuppliedto, orremovedfromthecontinuumper unit time (heat, chemical, electromagnetic, etc.). 57For a thermomechanical continuum, it is customary to express the time rate of change of internal energy by the integral expression dU dt=d dtintegraldisplay VρudV (6.45) Victor Saouma Introduction to Continuum Mechanics Draft6–10 FUNDAMENTAL LAWS of CONTINUUM MECHANICS whereuis the internal energy per unit mass or speciﬁc internal energy .W en o t et h a t Uappears 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 Uwill be null. The dimension ofUis one of energy dim U=ML2T−2, and the SI unit is the Joule, similarly dim u=L2T−2with the SI unit of Joule/Kg. 58In terms of energy integrals, the ﬁrst principle can be rewritten as Rate of increaebracehtipdownleft bracehtipuprightbracehtipupleft bracehtipdownright d dtintegraldisplay V1 2ρvividV bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright dK dt+d dtintegraldisplay VρudV bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright dU dt=Exchangebracehtipdownleft bracehtipuprightbracehtipupleft bracehtipdownrightintegraldisplay StividS+Sourcebracehtipdownleft bracehtipuprightbracehtipupleft bracehtipdownrightintegraldisplay VρvibidV bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright dW dt+Sourcebracehtipdownleft bracehtipuprightbracehtipupleft bracehtipdownrightintegraldisplay VρrdV−Exchangebracehtipdownleft bracehtipuprightbracehtipupleft bracehtipdownrightintegraldisplay SqinidS bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Q(6.46) we apply Gauss theorem to convert the surface integral, collect terms and use the fact thatdVis arbitrary to obtain ρdu dt=T:D+ρr−∇·q(6.47) or ρdu dt=TijDij+ρr−∂qj ∂xj(6.48) 59This equation expresses the rate of change of internal energy as the sum of the stress power plus theheatadded to the continuum. 60In ideal elasticity, heat transfer is considered insigniﬁcant, and all of the input work is assumed converted into internal energy in the form of recoverable stored elastic strainenergy, which can be recovered as work when the body is unloaded. 61In general, however, the major part of the input work into a deforming material is not recoverably stored, but dissipated by the deformation process causing an increase in the body’s temperature and eventually being conducted away as heat. 6.5 Equation of State; Second Principle of Thermodynamics 62The complete characterization of a thermodynamic system is said to describe the stateof a system (here a continuum). This description is speciﬁed, in general, by several thermodynamic andkinematic statevariables . Achange intime ofthose statevariables constitutes a thermodynamic process . Usually state variables are not all independent, and functional relationships exist among them through equations of state .A n ys t a t e variablewhichmaybeexpressed asasinglevaluedfunctionofasetofotherstatevariables is known as a state function . 63The ﬁrst principle of thermodynamics can be regarded as an expression of the inter- convertibility of heat and work, maintaining an energy balance. It places no restrictionon the direction of the process. In classical mechanics, kinetic and potential energy canbe easily transformed from one to the other in the absence of friction or other dissipative mechanism. Victor Saouma Introduction to Continuum Mechanics Draft6.5 Equation of State; Second Principle of Thermodynamics 6–11 64The ﬁrst principle leaves unanswered the question of the extent to which conversion process is reversible orirreversible . If thermal processes are involved (friction) dis- sipative processes are irreversible processes, and it will be up to the second principle ofthermodynamics to put limits on the direction of such processes. 6.5.1 Entropy 65The basic criterion for irreversibility is given by the second principle of thermo- dynamics through the statement on the limitation of entropy production .T h i sl a w postulates the existence of two distinct state functions: θtheabsolute temperature andStheentropy with the following properties: 1.θis a positive quantity. 2. Entropy is an extensive property, i.e. the total entropy is in a system is the sum of the entropies of its parts. 66Thus we can write ds=ds(e)+ds(i)(6.49) whereds(e)is the increase due to interaction with the exterior, and ds(i)is the internal increase, and ds(e)>0 irreversible process (6.50-a) ds(i)= 0 reversible process (6.50-b) 67Entropy expresses a variation of energy associated with a variation in the temperature. 6.5.1.1 Statistical Mechanics 68In statistical mechanics, entropy is related to the probability of the occurrence of that state among all the possible states that could occur. It is found that changes of states are more likely to occur in the direction of greater disorder when a system is left to itself. Thus increased entropy means increased disorder. 69Hence Boltzman’s principle postulates that entropy of a state is proportional to the logarithm of its probability, and for a gas this would give S=kN[lnV+3 2lnθ]+C (6.51) whereSis the total entropy, Vis volume, θis absolute temperature, kis Boltzman’s constant, and Cis a constant and Nis the number of molecules. 6.5.1.2 Classical Thermodynamics 70In a reversible process (more about that later), the change in speciﬁc entropy sis given by ds=parenleftBiggdq θparenrightBigg rev(6.52) Victor Saouma Introduction to Continuum Mechanics Draft6–12 FUNDAMENTAL LAWS of CONTINUUM MECHANICS 71If we consider an ideal gas governed by pv=Rθ (6.53) whereRis the gas constant, and assuming that the speciﬁc energy uis only a function of temperature θ, then the ﬁrst principle takes the form du=dq−pdv (6.54) and for constant volume this gives du=dq=cvdθ (6.55) whercvis the speciﬁc heat at constant volume. The assumption that u=u(θ) implies thatcvis a function of θonly and that du=cv(θ)dθ (6.56) 72Hence we rewrite the ﬁrst principle as dq=cv(θ)dθ+Rθdv v(6.57) or division by θyields s−s0=integraldisplayp,v p0,v0dq θ=integraldisplayθ θ0cv(θ)dθ θ+Rlnv v0 (6.58) which gives the change in entropy for any reversible process in an ideal gas. In this case, entropy is a state function which returns to its initial value whenever the temperaturereturns to its initial value that is pandvreturn to their initial values. 6.5.2 Clausius-Duhem Inequality 73We restate the deﬁnition of entropy as heat divided by temperature, and write the second principle d dtintegraldisplay Vρs bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Rate of Entropy Increase=integraldisplay Vρr θdV bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Sources−integraldisplay Sq θ·ndS bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Exchange+Γbracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright Internal production;Γ≥0 (6.59) dS dt=Q θ+Γ; Γ≥0 (6.60) Γ = 0 for reversible processes, and Γ >0 in irreversible ones. The dimension of S=integraldisplay vρsdVis one of energy divided by temperature or L2MT−2θ−1, and the SI unit for entropy is Joule/Kelvin. 74The second principle postulates that the time rate of change of total entropy Sin a continuum occupying a volume Vis always greater or equal than the sum of the entropy inﬂuxthrough thecontinuum surfaceplustheentropyproducedinternallybybodysources . Victor Saouma Introduction to Continuum Mechanics Draft6.6 Balance of Equations and Unknowns 6–13 75The previous inequality holds for any arbitrary volume, thus after transformation of the surface integral into a volume integral, we obtain the following local version of theClausius-Duhem inequality which must holds at every point ρds dtbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Rate of Entropy Increase≥ρr θbracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright Sources−∇·q θbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Exchange (6.61) 76We next seek to express the Clausius-Duhem inequality in terms of the stress tensor, ∇·q θ=1 θ∇·q−q·∇1 θ=1 θ∇·q−1 θ2q·∇θ (6.62) thus ρds dt≥−1 θ∇·q+1 θ2q·∇θ+ρr θ(6.63) but since θis always positive, ρθds dt≥−∇·q+ρr+1 θq·∇θ (6.64) where−∇·q+ρris the heat input into Vand appeared in the ﬁrst principle Eq. 6.47 ρdu dt=T:D+ρr−∇·q (6.65) hence, substituting, we obtain T:D−ρparenleftBiggdu dt−θds dtparenrightBigg −1 θq·∇θ≥0 (6.66) 6.6 Balance of Equations and Unknowns 77In the preceding sections several equations and unknowns were introduced. Let us count them. for both the coupled and uncoupled cases. Coupled Uncoupled dρ dt+ρ∂vi ∂xi=0 Continuity Equation 1 1 ∂Tij ∂xj+ρbi=ρdvi dt Equation of motion 3 3 ρdu dt=TijDij+ρr−∂qj ∂xj Energy equation 1 Total number of equations 5 4 78Assuming that the body forces biand distributed heat sources rare prescribed, then we have the following unknowns: Victor Saouma Introduction to Continuum Mechanics Draft6–14FUNDAMENTAL LAWS of CONTINUUM MECHANICS Coupled Uncoupled Density ρ 1 1 Velocity (or displacement) vi(ui) 3 3 Stress components Tij 6 6 Heat ﬂux components qi 3 - Speciﬁc internal energy u 1 - Entropy density s 1 - Absolute temperature θ 1 - Total number of unknowns 16 10 and in addition the Clausius-Duhem inequalityds dt≥r θ−1 ρdivq θwhich governs entropy production must hold. 79We thus need an additional 16 −5 = 11 additional equations to make the system determinate. These will be later on supplied by: 6 constitutive equations 3 temperature heat conduction 2 thermodynamic equations of state 11 Total number of additional equations 80The next chapter will thus discuss constitutive relations, and a subsequent one will separately discuss thermodynamic equations of state. 81We note that for the uncoupled case 1. The energy equation is essentially the integral of the equation of motion. 2. The 6 missing equations will be entirely supplied by the constitutive equations. 3. Thetemperatureﬁeldisregardedasknown, oratmost, theheat-conductionproblem must be solved separately and independently from the mechanical problem. 6.7†Elements of Heat Transfer 82One of the relations which we will need is the one which relates temperature to heat ﬂux. This constitutive realtion will be discussed in the next chapter under Fourrier’s law. 83However to place the reader in the right frame of reference to understand Fourrier’s law, this section will provide some elementary concepts of heat transfer. 84There 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. 6.3 on Γ q: qx=−kx∂T ∂x(6.67) qy=−ky∂T ∂y(6.68) Victor Saouma Introduction to Continuum Mechanics Draft6.7†Elements of Heat Transfer 6–15 Figure 6.3: Flux vector whereT=T(x,y) is the temperature ﬁeld in the medium, qxandqyare the componenets of the heat ﬂux (W/m2or Btu/h-ft2),kis the thermal conductiv- ity (W/m.oC or Btu/h-ft-oF) and∂T ∂x,∂T ∂yare the temperature gradients along the xandyrespectively. 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∞)o nΓc (6.69) whereqis the convective heat ﬂux, his the convection heat transfer coeﬃcient or ﬁlm coeﬃcient (W/m2.oCo rB t u / h - f t2.oF). It depends on various factors, such as whether convection is natural or forced, laminar or turbulent ﬂow, type of ﬂuid, andgeometry of the body; TandT ∞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 tem- peratures 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 propor- tional to the fourth power of the absolute temperature., which causes the problem to be nonlinear. This mode will not be covered. 6.7.1 Simple 2D Derivation 85If we consider a unit thickness, 2D diﬀerential body of dimensions dxbydy, Fig. 6.4 then 1. Rate of heat generation/sink is I2=Qdxdy (6.70) 2. Heat ﬂux across the boundary of the element is shown in Fig. ??(note similarity Victor Saouma Introduction to Continuum Mechanics Draft6–16 FUNDAMENTAL LAWS of CONTINUUM MECHANICS ✲qx ✲qx+∂q x ∂xdx ✻ qy ✻qy+∂q y ∂ydy ✲ ✛ dx ✻ ❄dy Q Figure 6.4: Flux Through Sides of Diﬀerential Element with equilibrium equation) I1=bracketleftBiggparenleftBigg qx+∂qx ∂xdxparenrightBigg −qxdxbracketrightBigg dy+bracketleftBiggparenleftBigg qy+∂qy ∂ydyparenrightBigg −qydybracketrightBigg dx=∂qx ∂xdxdy+∂qy ∂ydydx (6.71) 3. Change in stored energy is I3=cρdφ dt.dxdy (6.72) w h e r ew ed e ﬁ n et h e speciﬁc heat cas the amount of heat required to raise a unit mass by one degree. 86From the ﬁrst law of thermodaynamics, energy produced I2plus the net energy across the boundary I1must be equal to the energy absorbed I3,t h u s I1+I2−I3= 0 (6.73-a) ∂qx ∂xdxdy+∂qy ∂ydydx bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright I1+Qdxdybracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright I2−cρdφ dtdxdy bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright I3= 0 (6.73-b) 6.7.2†Generalized Derivation 87The amount of ﬂow per unit time into an element of volume Ω and surface Γ is I1=integraldisplay Γq(−n)dΓ=integraldisplay ΓD∇φ.ndΓ (6.74) wherenis the unit exterior normal to Γ, Fig. 6.5 Victor Saouma Introduction to Continuum Mechanics Draft6.7†Elements of Heat Transfer 6–17 Figure 6.5: *Flow through a surface Γ 88Using the divergence theorem integraldisplay ΓvndΓ=integraldisplay ΩdivvdΩ (6.75) Eq. 6.74 transforms into I1=integraldisplay Ωdiv (D∇φ)dΩ (6.76) 89Furthermore, if the instantaneous volumetric rate of “heat” generation or removal at ap o i n tx,y,zinside Ω is Q(x,y,z,t), then the total amount of heat/ﬂow produced per unit time is I2=integraldisplay ΩQ(x,y,z,t)dΩ (6.77) 90Finally, we deﬁne the speciﬁc heat of a solid cas the amount of heat required to raise a unit mass by one degree. Thus if ∆ φis 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,o r I3=integraldisplay Ωρc∆φdΩ (6.78) whereρis the density, Note that another expression of I3is ∆t(I1+I2). 91The balance equation, or conservation law states that the energy produced I2plus the net energy across the boundary I1must be equal to the energy absorbed I3,t h u s I1+I2−I3= 0 (6.79-a) integraldisplay ΩparenleftBigg div (D∇φ)+Q−ρc∆φ ∆tparenrightBigg dΩ = 0 (6.79-b) but since tand Ω are both arbitrary, then div (D∇φ)+Q−ρc∂φ ∂t=0 ( 6 . 80 ) or div (D∇φ)+Q=ρc∂φ ∂t (6.81) This equation can be rewritten as ∂qx ∂x+∂qy ∂y+Q=ρc∂φ ∂t (6.82) 1. Note the similarity between this last equation, and the equation of equilibrium ∂σxx ∂x+∂σxy ∂y+ρbx=ρm∂2ux ∂t2(6.83-a) ∂σyy ∂y+∂σxy ∂x+ρby=ρm∂2uy ∂t2(6.83-b) Victor Saouma Introduction to Continuum Mechanics Draft6–18 FUNDAMENTAL LAWS of CONTINUUM MECHANICS 2. For steady state problems, the previous equation does not depend on t,a n df o r2 D problems, it reduces to bracketleftBigg∂ ∂xparenleftBigg kx∂φ ∂xparenrightBigg +∂ ∂yparenleftBigg ky∂φ ∂yparenrightBiggbracketrightBigg +Q=0 ( 6 . 84 ) 3. For steady state isotropic problems, ∂2φ ∂x2+∂2φ ∂y2+∂2φ ∂z2+Q k=0 ( 6 . 85 ) which isPoisson’s equation in 3D. 4. If the heat input Q= 0, then the previous equation reduces to ∂2φ ∂x2+∂2φ ∂y2+∂2φ ∂z2=0 ( 6 . 86 ) which is an Elliptic (orLaplace) equation. Solutions of Laplace equations are termedharmonic functions (right hand side is zero) which is why Eq. 6.84 is refered to as thequasi-harmonic equation. 5. If the function depends only on xandt, then we obtain ρc∂φ ∂t=∂ ∂xparenleftBigg kx∂φ ∂xparenrightBigg +Q (6.87) which is a parabolic (or Heat) equation. Victor Saouma Introduction to Continuum Mechanics Draft Chapter 7 CONSTITUTIVE EQUATIONS; Part I LINEAR ceiinosssttuu Hooke, 1676 Ut tensio sic vis Hooke, 1678 7.1†Thermodynamic Approach 7.1.1 State Variables 20The method of local state postulates that the thermodynamic state of a continuum at a given point and instant is completely deﬁned by several state variables (also known as thermodynamic or independent variables ). A change in time of those state variables constitutes a thermodynamic process . Usually state variables are not all independent, and functional relationships exist among them through equations of state. Any state variable which may be expressed as a single valued function of a set of other state variables is known as a state function . 21The time derivatives of these variables are not involved in the deﬁnition of the state, this postulate implies that any evolution can be considered as a succession of equilibriumstates (therefore ultra rapid phenomena are excluded). 22Thethermodynamic state is speciﬁed by n+1v ariables ν1,ν2,···,νnandswhere νiare thethermodynamic substate variables andsthe speciﬁc entropy. The former have mechanical (or electromagnetic) dimensions, but are otherwise left arbitrary in thegeneral formulation. In ideal elasticity we have nine substate variables the componentsof the strain or deformation tensors. 23Thebasic assumption of thermodynamics is that in addition to the nsubstate variables, just one additional dimensionally independent scalar paramer suﬃces to deter-mine the speciﬁc internal energy u. This assumes that there exists a caloric equation Draft7–2 CONSTITUTIVE EQUATIONS; Part I LINEAR of state u=u(s,ν,X) (7.1) 24In general the internal energy ucan not be experimentally measured but rather its derivative. 25For instance we can deﬁne the thermodynamic temperature θand thethermo- dynamic “tension” τjas θ≡parenleftBigg∂u ∂sparenrightBigg ν;τj≡parenleftBigg∂u ∂νjparenrightBigg s,νi(i/negationslash=j);j=1,2,···,n (7.2) where the subscript outside the parenthesis indicates that the variables are held constant. 26By extension Ai=−ρτiwould be the thermodynamic “force” and its dimension depends on the one of νi. 7.1.2 Gibbs Relation 27From the chain rule we can express du dt=parenleftBigg∂u ∂sparenrightBigg νds dt+τpdνp dt(7.3) 28substituting into Clausius-Duhem inequality of Eq. 6.66 T:D−ρparenleftBiggdu dt−θds dtparenrightBigg −1 θq·∇θ≥0 (7.4) we obtain T:D+ρds dtbracketleftBigg θ−parenleftBigg∂u ∂sparenrightBigg νbracketrightBigg +Apdνp dt−1 θq·∇θ≥0 (7.5) but the second principle must be satisﬁed for all possible evolution and in particular the one for which D=0,dνp dt=0a n d ∇θ=0for any value ofds dtthus the coeﬃcient ofds dt is zero or θ=parenleftBigg∂u ∂sparenrightBigg ν(7.6) thus T:D+Apdνp dt−1 θq·∇θ≥0 (7.7) and Eq. 7.3 can be rewritten as du dt=θds dt+τpdνp dt(7.8) Victor Saouma Introduction to Continuum Mechanics Draft7.1†Thermodynamic Approach 7–3 and if we adopt the diﬀerential notation, we obtain Gibbs relation du=θds+τpdνp (7.9) 29For ﬂuid, the Gibbs relation takes the form du=θds−pdv;a n dθ≡parenleftBigg∂u ∂sparenrightBigg v;−p≡parenleftBigg∂u ∂vparenrightBigg s(7.10) wherepis the thermodynamic pressure; and the thermodynamic tension conjugate to the speciﬁc volume vis−p,j u s ta sθis conjugate to s. 7.1.3 Thermal Equation of State 30From the caloric equation of state, Eq. 7.1, and the the deﬁnitions of Eq. 7.2 it follows that the temperature and the thermodynamic tensions are functions of the thermody- namic state: θ=θ(s,ν);τj=τj(s,ν) (7.11) we assume the ﬁrst one to be invertible s=s(θ,ν) (7.12) and substitute this into Eq. 7.1 to obtain an alternative form of the caloric equation of statewithcorresponding thermalequationsofstate (obtainedbysimplesubstitution). u=u(θ,ν,bX)←(7.13) τi=τi(θ,ν,X) (7.14) νi=νi(θ,θ,X) (7.15) 31The thermal equations of state resemble stress-strain relations, but some caution is necessary in interpreting the tesnisons as stresses and the νjas strains. 7.1.4 Thermodynamic Potentials 32Based on the assumed existence of a caloric equation of state, four thermodynamic potentialsareintroduced, Table7.1. Thosepotentialsarederived throughthe Legendre- Potential Relation to u Independent Variables Internal energy u u s,νj Helmholtz free energy Ψ Ψ=u−sθ θ,νj← Enthalpy h h=u−τjνj s,τj Free enthalpy g g=u−sθ−τjνj θ,τj Table 7.1: Thermodynamic Potentials Victor Saouma Introduction to Continuum Mechanics Draft7–4CONSTITUTIVE EQUATIONS; Part I LINEAR Fenchel transformation on the basis of selected state variables best suited for a given problem. 33By means of the preceding equations, any one of the potentials can be expressed in terms of any of the four choices of state variables listed in Table 7.1. 34In any actual or hypothetical change obeying the equations of state, we have du=θds+τjdνj (7.16-a) dΨ=−sdθ+τjdνj← (7.16-b) dh=θds−νjdτj (7.16-c) dg=−sdθ−νjdτj (7.16-d) and from these diﬀerentials we obtain the following partial derivative expressions θ=parenleftBigg∂u ∂sparenrightBigg ν;τj=parenleftBigg∂u ∂νjparenrightBigg s,νi(i/negationslash=j)(7.17-a) s=−parenleftBigg∂Ψ ∂θparenrightBigg ν;τj=parenleftBigg∂Ψ ∂νjparenrightBigg θ← (7.17-b) θ=parenleftBigg∂h ∂sparenrightBigg τ;νj=−parenleftBigg∂h ∂τjparenrightBigg s,νi(i/negationslash=j)(7.17-c) =−parenleftBigg∂g ∂θparenrightBigg τ;νj=−parenleftBigg∂g ∂τjparenrightBigg θ(7.17-d) where the free energy Ψ is the portion of the internal energy available for doing work at constant temperature, the enthalpy h(as deﬁned here) is the portion of the internal energy that can be released as heat when the thermodynamic tensions are held constant. 7.1.5 Elastic Potential or Strain Energy Function 35Green deﬁned an elastic material as one for which a strain-energy function exists. Such a material is called Green-elastic orhyperelastic if there exists an elastic potential function Worstrain energy function , a scalar function of one of the strain or de- formation tensors, whose derivative with respect to a strain component determines thecorresponding stress component. 36Forthefullyrecoverable caseofisothermaldeformationwithreversible heatconduction we have ˜TIJ=ρ0parenleftBigg∂Ψ ∂EIJparenrightBigg θ(7.18) henceW=ρ0Ψ is an elastic potential function for this case, while W=ρ0uis the potential for adiabatic isentropic case ( s= constant). 37Hyperelasticity ignores thermal eﬀects and assumes that the elastic potential function always exists, it is a function of the strains alone and is purely mechanical ˜TIJ=∂W(E) ∂EIJ(7.19) Victor Saouma Introduction to Continuum Mechanics Draft7.2 Experimental Observations 7–5 andW(E)i st h estrain energy per unit undeformed volume . If the displacement gradients are small compared to unity, then we obtain Tij=∂W ∂Eij (7.20) which is written in terms of Cauchy stress Tijand small strain Eij. 38We assume that the elastic potential is represented by a power series expansion in the small-strain components. W=c0+cijEij+1 2cijkmEijEkm+1 3cijkmnpEijEkmEnp+··· (7.21) wherec0is a constant and cij,cijkm,cijkmnpdenote tensorial properties required to main- tain the invariant property of W. Physically, the second term represents the energy due to residual stresses, the third one refers to the strain energy which corresponds to linear elastic deformation, and the fourth one indicates nonlinear behavior. 39Neglecting terms higher than the second degree in the series expansion, then Wis quadratic in terms of the strains W=c0+c1E11+c2E22+c3E33+2c4E23+2c5E31+2c6E12 +1 2c1111E2 11+c1122E11E22+c1133E11E33+2c1123E11E23+2c1131E11E31+2c1112E11E12 +1 2c2222E2 22+c2233E22E33+2c2223E22E23+2c2231E22E31+2c2212E22E12 +1 2c3333E2 33+2c3323E33E23+2c3331E33E31+2c3312E33E12 +2c2323E2 23+4c2331E23E31+4c2312E23E12 +2c3131E2 31+4c3112E31E12 +2c1212E2 12 (7.22) we require that Wvanish in the unstrained state, thus c0=0 . 40We next apply Eq. 7.20 to the quadratic expression of Wand obtain for instance T12=∂W ∂E12=2c6+c1112E11+c2212E22+c3312E33+c1212E12+c1223E23+c1231E31(7.23) if the stress must also be zero in the unstrained state, then c6= 0, and similarly all the coeﬃcients in the ﬁrst row of the quadratic expansion of W. Thus the elastic potential function is a homogeneous quadratic function of the strains and we obtain Hooke’s law 7.2 Experimental Observations 41We shall discuss two experiments which will yield the elastic Young’s modulus ,a n d then thebulk modulus . In the former, the simplicity of the experiment is surrounded by the intriguing character of Hooke, and in the later, the bulk modulus is mathemat-ically related to the Green deformation tensor C, the deformation gradient Fand the Lagrangian strain tensor E. Victor Saouma Introduction to Continuum Mechanics Draft7–6 CONSTITUTIVE EQUATIONS; Part I LINEAR 7.2.1 Hooke’s Law 42Hooke’s Law is determined on the basis of a very simple experiment in which a uniaxial force is applied on a specimen which has one dimension much greater than the other two (such as a rod). The elongation is measured, and then the stress is plotted in terms of the strain (elongation/length). The slope of the line is called Young’s modulus . 43Hooke anticipated some of the most important discoveries and inventions of his time but failed to carry many of them through to completion. He formulated the theory of planetary motion as a problem in mechanics, and grasped, but did not develop mathe- matically, the fundamental theory on which Newton formulated the law of gravitation. His most important contribution was published in 1678in the paper De Potentia Restitutiva . It contained results of his experiments with elastic bodies, and was the ﬁrst paper in which the elastic properties of material was discussed. “Takea wire string of 20, or 30, or 40 ft long, and fasten the upper part thereof to a nail, and to the other end fasten a Scale to receive the weights: Then with a pair of compasses take the distance of the bottom of the scale from the ground or ﬂoor underneath, and set down the said distance, then put inweights intothe said scale and measure the several stretchings of the said string, and setthem down. Then compare the several stretchings of the said string, and you will ﬁnd that they will always bear the same proportions one to the other that the weights do that made them” . This became Hooke’s Law σ=Eε (7.24) 44Because he was concerned about patent rights to his invention, he did not publish his law when ﬁrst discovered it in 1660. Instead he published it in the form of an anagram“ceiinosssttuu ” in 1676 and the solution was given in 1678. Ut tensio sic vis (at the time the two symbols uandvwere employed interchangeably to denote either the vowel uor the consonant v), i.e.extension varies directly with force . 7.2.2 Bulk Modulus 45If, instead of subjecting a material to a uniaxial state of stress, we now subject it to a hydrostatic pressure pand measure the change in volume ∆ V. 46From the summary of Table 4.1 we know that: V=( d e tF)V0 (7.25-a) detF=√ detC=radicalBig det[I+2E] (7.25-b) therefore, V+∆V V=radicalBig det[I+2E] (7.26) Victor Saouma Introduction to Continuum Mechanics Draft7.3 Stress-Strain Relations in Generalized Elasticity 7–7 we can expand the determinant of the tensor det[ I+2E] to ﬁnd det[I+2E]=1+2IE+4IIE+8IIIE (7.27) but for small strains, IE/greatermuchIIE/greatermuchIIIEsince the ﬁrst term is linear in E, the second is quadratic, and the third is cubic. Therefore, we can approximate det[ I+2E]≈1+2IE, hence we deﬁne the volumetric dilatation as ∆V V≡e≈IE=t rE (7.28) this quantity is readily measurable in an experiment. 7.3 Stress-Strain Relations in Generalized Elasticity 7.3.1 Anisotropic 47FromEq. 7.22and7.23weobtainthestress-strainrelationforhomogeneousanisotropic material   T11 T22 T33 T12 T23 T31   bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Tij= c1111c1112c1133c1112c1123c1131 c2222c2233c2212c2223c2231 c3333c3312c3323c3331 c1212c1223c1231 SYM. c2323c2331 c3131  bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright cijkm  E11 E22 E33 2E12(γ12) 2E23(γ23) 2E31(γ31)   bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Ekm (7.29) which isHooke’s law for small strain in linear elasticity. 48We also observe that for symmetric cijwe retrieve Clapeyron formula W=1 2TijEij (7.30) 49In general the elastic moduli cijrelating the cartesian components of stress and strain depend on the orientation of the coordinate system with respect to the body. If the formof elastic potential function Wand the values c ijare independent of the orientation, the material is said to be isotropic , if not it is anisotropic . Victor Saouma Introduction to Continuum Mechanics Draft7–8 CONSTITUTIVE EQUATIONS; Part I LINEAR 50cijkmis a fourth order tensor resulting with 34=81t e r m s .    c1,1,1,1c1,1,1,2c1,1,1,3 c1,1,2,1c1,1,2,2c1,1,2,3 c1,1,3,1c1,1,3,2c1,1,3,3  c1,2,1,1c1,2,1,2c1,2,1,3 c1,2,2,1c1,2,2,2c1,2,2,3 c1,2,3,1c1,2,3,2c1,2,3,3  c1,3,1,1c1,3,1,2c1,3,1,3 c1,3,2,1c1,3,2,2c1,3,2,3 c1,3,3,1c1,3,3,2c1,3,3,3   c2,1,1,1c2,1,1,2c2,1,1,3 c2,1,2,1c2,1,2,2c2,1,2,3 c2,1,3,1c2,1,3,2c2,1,3,3  c2,2,1,1c2,2,1,2c2,2,1,3 c2,2,2,1c2,2,2,2c2,2,2,3 c2,2,3,1c2,2,3,2c2,2,3,3  c2,3,1,1c2,3,1,2c2,3,1,3 c2,3,2,1c2,3,2,2c2,3,2,3 c2,3,3,1c2,3,3,2c2,3,3,3   c3,1,1,1c3,1,1,2c3,1,1,3 c3,1,2,1c3,1,2,2c3,1,2,3 c3,1,3,1c3,1,3,2c3,1,3,3  c3,2,1,1c3,2,1,2c3,2,1,3 c3,2,2,1c3,2,2,2c3,2,2,3 c3,2,3,1c3,2,3,2c3,2,3,3  c3,3,1,1c3,3,1,2c3,3,1,3 c3,3,2,1c3,3,2,2c3,3,2,3 c3,3,3,1c3,3,3,2c3,3,3,3   (7.31) But the matrix must be symmetric thanks to Cauchy’s second law of motion (i.e sym-metry of both the stress and the strain), and thus for anisotropic material we will have a symmetric 6 by 6 matrix with (6)(6+1) 2= 21 independent coeﬃcients. 51By means of coordinate transformation we can relate the material properties in one coordinate system (old) xi, to a new one xi, thus from Eq. 1.27 ( vj=ap jvp)w ec a n rewrite W=1 2crstuErsEtu=1 2crstuar iasjatkaum Eij Ekm=1 2cijkm Eij Ekm (7.32) thus we deduce cijkm=ar iasjatkaumcrstu (7.33) thatisthefourthordertensorofmaterialconstantsinoldcoordinatesmaybetransformed into a new coordinate system through an eighth-order tensor ar iasjatkaum 7.3.2 Monotropic Material 52Aplane of elastic symmetry exists at a point where the elastic constants have the same values for every pair of coordinate systems which are the reﬂected images of oneanother with respect to the plane. The axes of such coordinate systems are referred to as “equivalent elastic directions”. 53If we assume x1=x1, x2=x2and x3=−x3, then the transformation xi=aj ixjis deﬁned through aj i= 10 0 01 0 00−1  (7.34) where the negative sign reﬂects the symmetry of the mirror image with respect to the x3 plane. 54We next substitute in Eq.7.33, and as an example we consider c1123=ar 1as1at2au3crstu= a1 1a11a22a33c1123= (1)(1)(1)( −1)c1123=−c1123, obviously, this is not possible, and the only way the relation can remanin valid is if c1123= 0. We note that all terms in cijklwith the index 3 occurring an odd number of times will be equal to zero. Upon substitution, Victor Saouma Introduction to Continuum Mechanics Draft7.3 Stress-Strain Relations in Generalized Elasticity 7–9 we obtain cijkm= c1111c1122c1133c111200 c2222c2233c221200 c3333c331200 c121200 SYM. c2323c2331 c3131  (7.35) we now have 13 nonzero coeﬃcients. 7.3.3 Orthotropic Material 55Ifthematerialpossesses threemutually perpendicularplanesofelasticsymmetry, (that is symmetric with respect to two planes x2andx3), then the transformation xi=aj ixj is deﬁned through aj i= 10 0 0−10 00−1  (7.36) where the negative sign reﬂects the symmetry of the mirror image with respect to the x3 plane. Upon substitution in Eq.7.33 we now would have cijkm= c1111c1122c1133000 c2222c2233000 c3333000 c121200 SYM. c23230 c3131  (7.37) We note that in here all terms of cijklwith the indices 3 and 2 occuring an odd number of times are again set to zero. 56Woodisusually considered anorthotropicmaterialandwillhave 9nonzerocoeﬃcients. 7.3.4 Transversely Isotropic Material 57A material is transversely isotropic if there is a preferential direction normal to all but one of the three axes. If this axis is x3, then rotation about it will require that aj i= cosθsinθ0 −sinθcosθ0 00 1  (7.38) substituting Eq. 7.33 into Eq. 7.41, using the above transformation matrix, we obtain c1111=( c o s4θ)c1111+(cos2θsin2θ)(2c1122+4c1212)+(sin4θ)c2222 (7.39-a) c1122=( c o s2θsin2θ)c1111+(cos4θ)c1122−4(cos2θsin2θ)c1212+(sin4θ)c2211(7.39-b) +(sin2θcos2θ)c2222 (7.39-c) c1133=( c o s2θ)c1133+(sin2θ)c2233 (7.39-d) Victor Saouma Introduction to Continuum Mechanics Draft7–10 CONSTITUTIVE EQUATIONS; Part I LINEAR c2222=( s i n4θ)c1111+(cos2θsin2θ)(2c1122+4c1212)+(cos4θ)c2222 (7.39-e) c1212=( c o s2θsin2θ)c1111−2(cos2θsin2θ)c1122−2(cos2θsin2θ)c1212+(cos4θ)c1212(7.39-f) +(sin2θcos2θ)c2222+sin4θc1212 (7.39-g) ... But in order to respect our initial assumption about symmetry, these results require that c1111=c2222 (7.40-a) c1133=c2233 (7.40-b) c2323=c3131 (7.40-c) c1212=1 2(c1111−c1122) (7.40-d) yielding cijkm= c1111c1122c1133 00 0 c2222c2233 00 0 c3333 00 0 1 2(c1111−c1122)0 0 SYM. c23230 c3131  (7.41) we now have 5 nonzero coeﬃcients. 58It should be noted that very few natural or man-made materials are truly orthotropic (certain crystals as topaz are), but a number are transversely isotropic (laminates, shist,quartz, roller compacted concrete, etc...). 7.3.5 Isotropic Material 59An isotropic material is symmetric with respect to every plane and every axis, that is the elastic properties are identical in all directions. 60To mathematically characterize an isotropic material, we require coordinate trans- formation with rotation about x2andx1axes in addition to all previous coordinate transformations. This process will enforce symmetry about all planes and all axes. 61The rotation about the x2axis is obtained through aj i= cosθ0−sinθ 01 0 sinθ0c o sθ  (7.42) we follow a similar procedure to the case of transversely isotropic material to obtain c1111=c3333 (7.43-a) c3131=1 2(c1111−c1133) (7.43-b) Victor Saouma Introduction to Continuum Mechanics Draft7.3 Stress-Strain Relations in Generalized Elasticity 7–11 62next we perform a rotation about the x1axis aj i= 10 0 0c o sθsinθ 0−sinθcosθ  (7.44) it follows that c1122=c1133 (7.45-a) c3131=1 2(c3333−c1133) (7.45-b) c2323=1 2(c2222−c2233) (7.45-c) which will ﬁnally give cijkm= c1111c1122c1133000 c2222c2233000 c3333000 a00 SYM. b0 c  (7.46) witha=1 2(c1111−c1122),b=1 2(c2222−c2233), andc=1 2(c3333−c1133). 63If we denote c1122=c1133=c2233=λandc1212=c2323=c3131=µthen from the previous relations we determine that c1111=c2222=c3333=λ+2µ,o r cijkm= λ+2µλ λ 000 λ+2µλ 000 λ+2µ000 µ00 SYM. µ0 µ (7.47) =λδijδkm+µ(δikδjm+δimδkj) (7.48) and we are thus left with only two independent non zero coeﬃcients λandµwhich are calledLame’s constants . 64Substituting the last equation into Eq. 7.29, Tij=[λδijδkm+µ(δikδjm+δimδkj)]Ekm (7.49) Or in terms of λandµ,Hooke’s Law for an isotropic body is written as Tij=λδijEkk+2µEijorT=λIE+2µE (7.50) Eij=1 2µparenleftBigg Tij−λ 3λ+2µδijTkkparenrightBigg orE=−λ 2µ(3λ+2µ)IT+1 2µT(7.51) 65It should be emphasized that Eq. 7.47 is written in terms of the Engineering strains (Eq. 7.29) that is γij=2Eijfori/negationslash=j. On the other hand the preceding equations are written in terms of the tensorial strains Eij Victor Saouma Introduction to Continuum Mechanics Draft7–12 CONSTITUTIVE EQUATIONS; Part I LINEAR 7.3.5.1 Engineering Constants 66The stress-strain relations were expressed in terms ofLame’s parameters which can not bereadily measured experimentally. Assuch, inthefollowing sections we willreformulate those relations in terms of “engineering constants” (Young’s and the bulk’s modulus). This will be done for both the isotropic and transversely isotropic cases. 7.3.5.1.1 Isotropic Case 7.3.5.1.1.1 Young’s Modulus 67In order to avoid certain confusion between the strain Eand the elastic constant E, we adopt the usual engineering notation Tij→σijandEij→εij 68If we consider a simple uniaxial state of stress in the x1direction, then from Eq. 7.51 ε11=λ+µ µ(3λ+2µ)σ (7.52-a) ε22=ε33=−λ 2µ(3λ+2µ)σ (7.52-b) 0=ε12=ε23=ε13 (7.52-c) 69Yet we have the elementary relations in terms engineering constants EYoung’s mod- ulusandνPoisson’s ratio ε11=σ E(7.53-a) ν=−ε22 ε11=−ε33 ε11(7.53-b) then it follows that 1 E=λ+µ µ(3λ+2µ);ν=λ 2(λ+µ)(7.54) λ=νE (1+ν)(1−2ν);µ=G=E 2(1+ν)(7.55) 70Similarly in the case of pure shear in the x1x3andx2x3planes, we have σ21=σ12=τall other σij= 0 (7.56-a) 2ε12=τ G(7.56-b) and theµis equal to the shear modulus G. 71Hooke’s law for isotropic material in terms of engineering constants becomes Victor Saouma Introduction to Continuum Mechanics Draft7.3 Stress-Strain Relations in Generalized Elasticity 7–13 σij=E 1+νparenleftbigg εij+ν 1−2νδijεkkparenrightbigg orσ=E 1+νparenleftbigg ε+ν 1−2νIεparenrightbigg (7.57) εij=1+ν Eσij−ν Eδijσkkorε=1+ν Eσ−ν EIσ (7.58) 72When the strain equation is expanded in 3D cartesian coordinates it would yield:   εxx εyy εzz γxy(2εxy) γyz(2εyz) γzx(2εzx)  =1 E 1−ν−ν000 −ν1−ν000 −ν−ν10 0 0 000 1 + ν00 000 01 + ν0 000 0 01 + ν   σxx σyy σzz τxy τyz τzx   (7.59) 73If we invert this equation, we obtain   σxx σyy σzz τxy τyz τzx  = E (1+ν)(1−2ν) 1−νν ν ν1−νν νν 1−ν  0 0 G 100 010 001    εxx εyy εzz γxy(2εxy) γyz(2εyz) γzx(2εzx)   (7.60) 7.3.5.1.1.2 Bulk’s Modulus; Volumetric and Deviatoric Strains 74We can express the trace of the stress Iσin terms of the volumetric strainIεFrom Eq. 7.50 σii=λδiiεkk+2µεii=( 3λ+2µ)εii≡3Kεii (7.61) or K=λ+2 3µ (7.62) 75We can provide a complement to the volumetric part of the constitutive equations by substracting the trace of the stress from the stress tensor, hence we deﬁne the deviatoric stress and strains as as σ/prime≡σ−1 3(trσ)I(7.63) ε/prime≡ε−1 3(trε)I(7.64) and the corresponding constitutive relation will be σ=KeI+2µε/prime(7.65) ε=p 3KI+1 2µσ/prime(7.66) wherep≡1 3tr (σ) is the pressure, and σ/prime=σ−pIis the stress deviator. Victor Saouma Introduction to Continuum Mechanics Draft7–14CONSTITUTIVE EQUATIONS; Part I LINEAR 7.3.5.1.1.3 Restriction Imposed on the Isotropic Elastic Moduli 76We can rewrite Eq. 7.20 as dW=TijdEij (7.67) but since dWis a scalar invariant (energy), it can be expressed in terms of volumetric (hydrostatic) and deviatoric components as dW=−pde+σ/prime ijdE/prime ij (7.68) substituting p=−Keandσ/prime ij=2GE/prime ij, and integrating, we obtain the following expres- sion for the isotropic strain energy W=1 2Ke2+GE/prime ijE/prime ij (7.69) and since positive work is required to cause any deformation W>0t h u s λ+2 3G≡K>0 (7.70-a) G>0 (7.70-b) ruling out K=G= 0, we are left with E>0;−1<ν<1 2 (7.71) 77The isotropic strain energy function can be alternatively expressed as W=1 2λe2+GEijEij (7.72) 78From Table 7.2, we observe that ν=1 2impliesG=E 3,a n d1 K= 0 or elastic incom- pressibility . λ,µE,ν µ ,ν E,µK,ν λ λνE (1+ν)(1−2ν)2µν 1−2νµ(E−2µ) 3µ−E3Kν 1+ν µ µE 2(1+ν)µµ3K(1−2ν) 2(1+ν) K λ+2 3µE 3(1−2ν)2µ(1+ν) 3(1−2ν)µE 3(3µ−E)K E µ(3λ+2µ) λ+µE 2µ(1+ν)E 3K(1−2ν) ν λ 2(λ+µ)ννE 2µ−1ν Table 7.2: Conversion of Constants for an Isotropic Elastic Material 79The elastic properties of selected materials is shown in Table 7.3. Victor Saouma Introduction to Continuum Mechanics Draft7.3 Stress-Strain Relations in Generalized Elasticity 7–15 Material E(MPa) ν A316 Stainless Steel 196,000 0.3 A5 Aluminum 68,000 0.33 Bronze 61,000 0.34 Plexiglass 2,900 0.4 Rubber 2 →0.5 Concrete 60,000 0.2 Granite 60,000 0.27 Table 7.3: Elastic Properties of Selected Materials at 200c 7.3.5.1.2 Transversly Isotropic Case 80For transversely isotropic, we can express the stress-strain relation in tems of εxx=a11σxx+a12σyy+a13σzz εyy=a12σxx+a11σyy+a13σzz εzz=a13(σxx+σyy)+a33σzz γxy=2 (a11−a12)τxy γyz=a44τxy γxz=a44τxz(7.73) and a11=1 E;a12=−ν E;a13=−ν/prime E/prime;a33=−1 E/prime;a44=−1 µ/prime(7.74) whereEis the Young’s modulus in the plane of isotropy and E/primethe one in the plane normal to it. νcorresponds to the transverse contraction in the plane of isotropy when tension is applied in the plane; ν/primecorresponding to the transverse contraction in the plane of isotropy when tension is applied normal to the plane; µ/primecorresponding to the shear moduli for the plane of isotropy and any plane normal to it, and µis shear moduli for the plane of isotropy. 7.3.5.2 Special 2D Cases 81Often times one can make simplifying assumptions to reduce a 3D problem into a 2D one. 7.3.5.2.1 Plane Strain 82For problems involving a long body in the zdirection with no variation in load or geometry, then εzz=γyz=γxz=τxz=τyz= 0. Thus, replacing into Eq. 5.2 we obtain   σxx σyy σzz τxy  =E (1+ν)(1−2ν) (1−ν)ν0 ν(1−ν)0 νν 0 001−2ν 2   εxx εyy γxy  (7.75) Victor Saouma Introduction to Continuum Mechanics Draft7–16 CONSTITUTIVE EQUATIONS; Part I LINEAR 7.3.5.2.2 Axisymmetry 83In solids of revolution, we can use a polar coordinate sytem and εrr=∂u ∂r(7.76-a) εθθ=u r(7.76-b) εzz=∂w ∂z(7.76-c) εrz=∂u ∂z+∂w ∂r(7.76-d) 84The constitutive relation is again analogous to 3D/plane strain   σrr σzz σθθ τrz  =E (1+ν)(1−2ν) 1−νν ν 0 ν1−νν 0 νν 1−ν0 νν 1−ν0 0001−2ν 2   εrr εzz εθθ γrz  (7.77) 7.3.5.2.3 Plane Stress 85If the longitudinal dimension in zdirection is much smaller than in the xandy directions, then τyz=τxz=σzz=γxz=γyz= 0 throughout the thickness. Again, substituting into Eq. 5.2 we obtain:   σxx σyy τxy  =1 1−ν2 1ν0 ν10 001−ν 2   εxx εyy γxy  (7.78-a) εzz=−1 1−νν(εxx+εyy) (7.78-b) 7.4 Linear Thermoelasticity 86If thermal eﬀects are accounted for, the components of the linear strain tensor Eijmay be considered as the sum of Eij=E(T) ij+E(Θ) ij (7.79) whereE(T) ijis the contribution from the stress ﬁeld, and E(Θ) ijthe contribution from the temperature ﬁeld. 87When abodyis subjected toatemperature change Θ −Θ0with respect to the reference state temperature, the strain componenet of an elementary volume of an unconstrainedisotropic body are given by E (Θ) ij=α(Θ−Θ0)δij (7.80) Victor Saouma Introduction to Continuum Mechanics Draft7.5 Fourrier Law 7–17 whereαis thelinear coeﬃcient of thermal expansion . 88Inserting the preceding two equation into Hooke’s law (Eq. 7.51) yields Eij=1 2µparenleftBigg Tij−λ 3λ+2µδijTkkparenrightBigg +α(Θ−Θ0)δij (7.81) which is known as Duhamel-Neumann relations. 89If we invert this equation, we obtain the thermoelastic constitutive equation : Tij=λδijEkk+2µEij−(3λ+2µ)αδij(Θ−Θ0) (7.82) 90Alternatively, if we were to consider the derivation of the Green-elastic hyperelastic equations, (Sect. 7.1.5), we required the constants c1toc6in Eq. 7.22 to be zero in order that the stress vanish in the unstrained state. If we accounted for the temperature changeΘ−Θ 0with respect to the reference state temperature, we would have ck=−βk(Θ−Θ0) fork= 1 to 6 and would have to add like terms to Eq. 7.22, leading to Tij=−βij(Θ−Θ0)+cijrsErs (7.83) for linear theory, we suppose that βijis independent fromthe strain and cijrsindependent of temperature change with respect to the natural state. Finally, for isotropic cases weobtain T ij=λEkkδij+2µEij−βij(Θ−Θ0)δij (7.84) which is identical to Eq. 7.82 with β=Eα 1−2ν. Hence TΘ ij=Eα 1−2ν (7.85) 91In terms of deviatoric stresses and strains we have T/prime ij=2µE/prime ijandE/prime ij=T/prime ij 2µ (7.86) and in terms of volumetric stress/strain: p=−Ke+β(Θ−Θ0)a n de=p K+3α(Θ−Θ0) (7.87) 7.5 Fourrier Law 92Consider a solid through which there is a ﬂowqof heat (or some other quantity such as mass, chemical, etc...) 93The rate of transfer per unit area is q Victor Saouma Introduction to Continuum Mechanics Draft7–18 CONSTITUTIVE EQUATIONS; Part I LINEAR 94The direction of ﬂow is in the direction of maximum “potential” (temperature in this case, but could be, piezometric head, or ion concentration) decreases (Fourrier, Darcy,Fick...). q=  qx qy qz  =−D  ∂φ ∂x∂φ ∂y ∂φ ∂z  =−D∇φ (7.88) Dis a three by three (symmetric) constitutive/conductivity matrix The conductivity can be either Isotropic D=k 100 010001  (7.89) Anisotropic D= kxxkxykxz kyxkyykyz kzxkzykzz  (7.90) Orthotropic D= kxx00 0kyy0 00 kzz  (7.91) Note that for ﬂow through porous media, Darcy’s equation is only valid for laminar ﬂow. 7.6 Updated Balance of Equations and Unknowns 95In light of the new equations introduced in this chapter, it would be appropriate to revisit our balance of equations and unknowns. Coupled Uncoupled dρ dt+ρ∂vi ∂xi=0 Continuity Equation 1 1 ∂Tij ∂xj+ρbi=ρdvi dt Equation of motion 3 3 ρdu dt=TijDij+ρr−∂qj ∂xj Energy equation 1 T=λIE+2µE Hooke’s Law 6 6 q=−D∇φ Heat Equation (Fourrier) 3 Θ=Θ (s,ν);τj=τj(s,ν) Equations of state 2 Total number of equations 16 10 and we repeat our list of unknowns Victor Saouma Introduction to Continuum Mechanics Draft7.6 Updated Balance of Equations and Unknowns 7–19 Coupled Uncoupled Density ρ 1 1 Velocity (or displacement) vi(ui) 3 3 Stress components Tij 6 6 Heat ﬂux components qi 3 - Speciﬁc internal energy u 1 - Entropy density s 1 - Absolute temperature Θ 1 - Total number of unknowns 16 10 and in addition the Clausius-Duhem inequalityds dt≥r Θ−1 ρdivq Θwhich governs entropy production must hold. 96Hence we now have as many equations as unknowns and are (almost) ready to pose and solve problems in continuum mechanics. Victor Saouma Introduction to Continuum Mechanics Draft7–20 CONSTITUTIVE EQUATIONS; Part I LINEAR Victor Saouma Introduction to Continuum Mechanics Draft Chapter 8 INTERMEZZO Inlight ofthelengthy andrigorousderivation ofthe fundamentalequations ofContinuum Mechanics in the preceding chapter, the reader may be at a loss as to what are the mostimportant ones to remember. Hence, since the complexity of some of the derivation may have eclipsed the ﬁnal results, thishandoutseekstosummarizethemostfundamentalrelationswhichyoushould always remember. 13σ 21σ23 σ22σ31 1σ33σ32 X2X1V1X3 X2 (Components of a vector are scalars)VV V2 X3 (Components of a tensor of order 2 are vectors)X3 11σσ 12σ Stresses as components of a traction vectortt t123 Stress Vector/Tensor ti=Tijnj (8.1-a) Strain Tensor E∗ ij=1 2 ∂ui ∂xj+∂uj ∂xi−∂uk ∂xi∂uk ∂xjbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright  (8.1-b) = ε111 2γ121 2γ13 1 2γ12ε221 2γ23 1 2γ131 2γ23ε33  (8.1-c) Engineering Strain γ23≈sinγ23=s i n (π/2−θ)=c o sθ=2E23(8.1-d) Equilibrium∂Tij ∂xj+ρbi=ρdvi dt(8.1-e) Draft8–2 INTERMEZZO Boundary Conditions Γ=Γu+Γt (8.1-f) Energy Potential Tij=∂W ∂Eij(8.1-g) Hooke’s Law Tij=λδijEkk+2µEij (8.1-h)  εxx εyy εzz γxy γyz γxz  = 1 E−ν E−ν E000 −ν E1 E−ν E000 −ν E−ν E1 E000 0001 G00 000 01 G0 000 0 01 G   σxx σyy σzz τxy τyz τxz  (8.1-i) Plane Stress σzz=0 ;εzz/negationslash=0 ( 8. 1 - j ) Plane Strain εzz=0 ;σzz/negationslash=0 ( 8. 1 - k ) Victor Saouma Introduction to Continuum Mechanics Draft Part II ELASTICITY/SOLID MECHANICS Draft Draft Chapter 9 BOUNDARY VALUE PROBLEMS in ELASTICITY 9.1 Preliminary Considerations 20All problems in elasticity require three basic components: 3 Equations of Motion (Equilibrium): i.e. Equations relating the applied tractions and body forces to the stresses (3) ∂Tij ∂Xj+ρbi=ρ∂2ui ∂t2(9.1) 6 Stress-Strain relations: (Hooke’s Law) T=λIE+2µE (9.2) 6 Geometric (kinematic) equations: i.e. Equations of geometry of deformation re- lating displacement to strain (6) E∗=1 2(u∇x+∇xu) (9.3) 21Those 15 equations are written in terms of 15 unknowns: 3 displacement ui, 6 stress components Tij, and 6 strain components Eij. 22In addition to these equations which describe what is happening inside the body, we must describe what is happening on the surface or boundary of the body. These extraconditions are called boundary conditions . 9.2 Boundary Conditions 23In describing the boundary conditions (B.C.), we must note that: 1. Either we know the displacement but not the traction, or we know the traction and not the corresponding displacement. We can never know both ap r i o r i. Draft9–2 BOUNDARY VALUE PROBLEMS in ELASTICITY 2. Not all boundary conditions speciﬁcations are acceptable. For example we can not apply tractions to the entire surface of the body. Unless those tractions are speciallyprescribed, they may not necessarily satisfy equilibrium. 24Properlyspeciﬁed boundaryconditionsresultin well-posed boundaryvalueproblems, while improperly speciﬁed boundary conditions will result in ill-posed boundary value problem. Only the former can be solved. 25Thus we have two types of boundary conditions in terms of knownquantitites, Fig. 9.1: Ω ΓΤ ut Figure 9.1: Boundary Conditions in Elasticity Problems Displacement boundary conditions along Γuwith the three components of uipre- scribed on the boundary. The displacement is decomposed into its cartesian (or curvilinear) components, i.e. ux,uy Traction boundary conditions along Γtwith the three traction components ti= njTijprescribed at a boundary where the unit normal is n. The traction is de- composed into its normal and shear(s) components, i.e tn,ts. Mixed boundary conditions wheredisplacement boundaryconditionsareprescribed on a part of the bounding surface, while traction boundary conditions are prescribedon the remainder. We note thatat some points, traction may be speciﬁed in one direction, and displacement at another. Displacement and tractions can never be speciﬁed at the same point in thesame direction. 26Various terms have been associated with those boundary conditions in the litterature, those are suumarized in Table 9.1. 27Often time we take advantage of symmetry not only to simplify the problem, but also to properly deﬁne the appropriate boundary conditions, Fig. 9.2. Victor Saouma Introduction to Continuum Mechanics Draft9.2 Boundary Conditions 9–3 u,Γu t,Γt Dirichlet Neuman Field Variable Derivative(s) of Field Variable Essential Non-essential Forced Natural Geometric Static Table 9.1: Boundary Conditions in Elasticity xy ? ABCD Eσ Note: Unknown tractions=ReactionstnyuuΓ xu CDBCAB ? 0 ? DE EAtsΓt 0 ? σ ??? ? 0 ? ?00 0 0 0 0 Figure 9.2: Boundary Conditions in Elasticity Problems Victor Saouma Introduction to Continuum Mechanics Draft9–4BOUNDARY VALUE PROBLEMS in ELASTICITY 9.3 Boundary Value Problem Formulation 28Hence, the boundary value formulation is suumarized by ∂Tij ∂Xj+ρbi=ρ∂2ui ∂t2in Ω (9.4) E∗=1 2(u∇x+∇xu) (9.5) T=λIE+2µEin Ω (9.6) u= uin Γu (9.7) t= tin Γt (9.8) and is illustrated by Fig. 9.3. This is now a well posed problem . Natural B.C. ti:Γt Stresses Tij Equilibrium ∂T ij ∂x j+ρbi=ρdv i dt Body Forces bi Constitutive Rel. T=λIE+2µE Strain Eij Kinematics E∗=1 2(u /D6x+ /D6xu) Displacements ui Essential B.C. ui:Γu ✻ ❄ ❄ ❄ ❄ ❄ ✲ ✛ Figure 9.3: Fundamental Equations in Solid Mechanics 9.4 Compacted Forms 29Solvingaboundaryvalueproblemwith15unknownsthrough15equationsisaformidable task. Hence, there are numerous methods to reformulate the problem in terms of fewer Victor Saouma Introduction to Continuum Mechanics Draft9.5 Strain Energy and Extenal Work 9–5 unknows. 9.4.1 Navier-Cauchy Equations 30One such approach is to substitute the displacement-strain relation into Hooke’s law (resulting in stresses in terms of the gradient of the displacement), and the resultingequation into the equation of motion to obtain three second-order partial diﬀerential equations for the three displacement components known as Navier’s Equation (λ+µ)∂2uk ∂Xi∂Xk+µ∂2ui ∂Xk∂Xk+ρbi=ρ∂2ui ∂t2(9.9) or (λ+µ)∇(∇·u)+µ∇2u+ρb=ρ∂2u ∂t2(9.10) (9.11) 9.4.2 Beltrami-Mitchell Equations 31Whereas Navier-Cauchy equation was expressed in terms of the gradient of the dis- placement, we can follow a similar approach and write a single equation in term of the gradient of the tractions. ∇2Tij+1 1+νTpp,ij=−ν 1−νδij∇·(ρb)−ρ(bi,j+bj,i) (9.12) or Tij,pp+1 1+νTpp,ij=−ν 1−νδijρbp,p−ρ(bi,j+bj,i) (9.13) 9.4.3 Ellipticity of Elasticity Problems 9.5 Strain Energy and Extenal Work 32For the isotropic Hooke’s law, we saw that there always exist a strain energy function Wwhich is positive-deﬁnite, homogeneous quadratic function of the strains such that, Eq. 7.20 Tij=∂W ∂Eij(9.14) hence it follows that W=1 2TijEij (9.15) 33The external work done by a body in equilibrium under body forces biand surface tractiontiis equal tointegraldisplay ΩρbiuidΩ+integraldisplay ΓtiuidΓ. Substituting ti=Tijnjand applying Gauss theorem, the second term becomes integraldisplay ΓTijnjuidΓ=integraldisplay Ω(Tijui),jdΩ=integraldisplay Ω(Tij,jui+Tijui,j)dΩ (9.16) Victor Saouma Introduction to Continuum Mechanics Draft9–6 BOUNDARY VALUE PROBLEMS in ELASTICITY butTijui,j=Tij(Eij+Ωij)=TijEijand from equilibrium Tij,j=−ρbi,t h u s integraldisplay ΩρbiuidΩ+integraldisplay ΓtiuidΓ=integraldisplay ΩρbiuidΩ+integraldisplay Ω(TijEij−ρbiui)dΩ (9.17) or integraldisplay ΩρbiuidΩ+integraldisplay ΓtiuidΓ bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright External Work=2integraldisplay ΩTijEij 2dΩ bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Internal Strain Energy (9.18) that isFor an elastic system, the total strain energy is one half the work done by the external forces acting through their displacements ui. 9.6 Uniqueness of the Elastostatic Stress and Strain Field 34Because the equations of linear elasticity are linear equations, the principles of super- position may be used to obtain additional solutions from those established. Hence, given two sets of solution T(1) ij,u(1) i,a n dT(2) ij,u(2) i,t h e nTij=T(2) ij−T(1) ij,a n dui=u(2) i−u(1) i withbi=b(2) i−b(1) i= 0 must also be a solution. 35Hence for this “diﬀerence” solution, Eq. 9.18would yieldintegraldisplay ΓtiuidΓ=2integraldisplay Ωu∗dΩ but the left hand side is zero because ti=t(2) i−t(1) i=0o nΓ u,a n dui=u(2) i−u(1) i=0o n Γt,t h u sintegraldisplay Ωu∗dΩ=0 . 36Butu∗is positive-deﬁnite and continuous, thus the integral can vanish if and only if u∗= 0 everywhere, and this is only possible if Eij= 0 everywhere so that E(2) ij=E(1) ij⇒T(2) ij=Tij(1) (9.19) hence, there can not be two diﬀerent stress and strain ﬁelds corresponding to the same externally imposed body forces and boundary conditions1and satisfying the linearized elastostatic Eqs 9.1, 9.14 and 9.3. 9.7 Saint Venant’s Principle 37This famous principle of Saint Venant was enunciated in 1855 and is of great im- portance in applied elasticity where it is often invoked to justify certain “simpliﬁed” solutions to complex problem. In elastostatics, if the boundary tractions on a part Γ 1of the boundary Γ are replaced by a statically equivalent traction distribution, the eﬀects on the stress distribution in the body are negligible at points whose distance from Γ 1is large compared to the maximum distance between points of Γ 1. 1ThistheoremisattributedtoKirchoﬀ(1858). Victor Saouma Introduction to Continuum Mechanics Draft9.8 Cylindrical Coordinates 9–7 38For instance the analysis of the problem in Fig. 9.4 can be greatly simpliﬁed if the tractions on Γ 1are replaced by a concentrated statically equivalent force. F=tdxtdx Figure 9.4: St-Venant’s Principle 9.8 Cylindrical Coordinates 39So farallequations have been written ineither vector, indicial, orengineering notation. The last two were so far restricted to an othonormal cartesian coordinate system. 40We now rewrite some of the fundamental relations in cylindrical coordinate system, Fig. 9.5, as this would enable us to analytically solve some simple problems of greatpractical usefulness (torsion, pressurized cylinders, ...). This is most often achieved by reducing the dimensionality of the problem from 3 to 2 or even to 1. z θr Figure 9.5: Cylindrical Coordinates Victor Saouma Introduction to Continuum Mechanics Draft9–8 BOUNDARY VALUE PROBLEMS in ELASTICITY 9.8.1 Strains 41With reference to Fig. 9.6, we consider the displacement of point PtoP∗.t h e uuu * xy rθu θr xy PP θθ Figure 9.6: Polar Strains displacements can be expressed in cartesian coordinates as ux,uy, or in polar coordinates asur,uθ. Hence, ux=urcosθ−uθsinθ (9.20-a) uy=ursinθ+uθcosθ (9.20-b) substituting into the strain deﬁnition for εxx(for small displacements) we obtain εxx=∂ux ∂x=∂ux ∂θ∂θ ∂x+∂ux ∂r∂r ∂x(9.21-a) ∂ux ∂θ=∂ur ∂θcosθ−ursinθ−∂uθ ∂θsinθ−uθcosθ (9.21-b) ∂ux ∂r=∂ur ∂rcosθ−∂uθ ∂rsinθ (9.21-c) ∂θ ∂x=−sinθ r(9.21-d) ∂r ∂x=c o sθ (9.21-e) εxx=parenleftBigg −∂ur ∂θcosθ+ursinθ+∂uθ ∂θsinθ+uθcosθparenrightBiggsinθ r +parenleftBigg∂ur ∂rcosθ−∂uθ ∂rsinθparenrightBigg cosθ (9.21-f) Noting that as θ→0,εxx→εrr,s i nθ→0, and cos θ→1, we obtain εrr=εxx|θ→0=∂ur ∂r(9.22) 42Similarly, if θ→π/2,εxx→εθθ,s i nθ→1, and cos θ→0. Hence, εθθ=εxx|θ→π/2=1 r∂uθ∂θ+ur r(9.23) Victor Saouma Introduction to Continuum Mechanics Draft9.8 Cylindrical Coordinates 9–9 ﬁnally, we may express εxyas a function of ur,uθandθand noting that εxy→εrθas θ→0, we obtain εrθ=1 2bracketleftBigg εxy|θ→0=∂uθ ∂r−uθ r+1 r∂ur ∂θbracketrightBigg (9.24) 43In summary, and with the addition of the zcomponents (not explicitely derived), we obtain εrr=∂ur ∂r(9.25) εθθ=1 r∂uθ ∂θ+ur r(9.26) εzz=∂uz ∂z(9.27) εrθ=1 2bracketleftBigg1 r∂ur ∂θ+∂uθ ∂r−utheta rbracketrightBigg (9.28) εθz=1 2bracketleftBigg∂uθ ∂z+1 r∂uz ∂θbracketrightBigg (9.29) εrz=1 2bracketleftBigg∂uz ∂r+∂ur ∂zbracketrightBigg (9.30) 9.8.2 Equilibrium 44Whereas the equilibrium equation as given In Eq. 6.24 was obtained from the linear momentum principle (without any reference to the notion of equilibrium of forces), its derivation (as mentioned) could have been obtained by equilibrium of forces considera-tions. This is the approach which we will follow for the polar coordinate system withrespect to Fig. 9.7.θθdθd drdr θθrr δ δδ θT r + rr+drθθd rf fr θδ δθθr + +δ δrr r θr Trrrθθr T TTTTTT T +δ θθθθ T Figure 9.7: Stresses in Polar Coordinates Victor Saouma Introduction to Continuum Mechanics Draft9–10 BOUNDARY VALUE PROBLEMS in ELASTICITY 45Summation of forces parallel to the radial direction through the center of the element with unit thickness in the zdirection yields: parenleftBigg Trr+∂Trr ∂rdrparenrightBigg (r+dr)dθ−Trr(rdθ) (9.31-a) −parenleftBigg Tθθ+∂Tθθ ∂θ+TθθparenrightBigg drsindθ 2 +parenleftBigg Tθr+∂Tθr ∂θdθ−TθrparenrightBigg drcosdθ 2+frrdrdθ= 0 (9.31-b) we approximate sin( dθ/2) bydθ/2a n dc o s ( dθ/2) by unity, divide through by rdrdθ, 1 rTrr+∂Trr ∂rparenleftBigg 1+dr rparenrightBigg −Tθθ r−∂Tθθ ∂θdθ dr+1 r∂Tθr ∂θ+fr= 0 (9.32) 46Similarly we can take the summation of forces in the θdirection. In both cases if we were to drop the dr/randdθ/rin the limit, we obtain ∂Trr ∂r+1 r∂Tθr ∂θ+1 r(Trr−Tθθ)+fr= 0 (9.33) ∂Trθ ∂r+1 r∂Tθθ ∂θ+1 r(Trθ−Tθr)+fθ= 0 (9.34) 47It is often necessary to express cartesian stresses in terms of polar stresses and vice versa. This can be done through the following relationships bracketleftBigg TxxTxy TxyTyybracketrightBigg =bracketleftBigg cosθ−sinθ sinθcosθbracketrightBiggbracketleftBigg TrrTrθ TrθTθθbracketrightBiggbracketleftBigg cosθ−sinθ sinθcosθbracketrightBiggT (9.35) yielding Txx=Trrcos2θ+Tθθsin2θ−Trθsin2θ (9.36-a) Tyy=Trrsin2θ+Tθθcos2θ+Trθsin2θ (9.36-b) Txy=(Trr−Tθθ)sinθcosθ+Trθ(cos2θ−sin2θ) (9.36-c) (recalling that sin2θ=1/2sin2θ,a n dc o s2θ=1/2(1+cos2 θ)). 9.8.3 Stress-Strain Relations 48In orthogonal curvilinear coordinates, the physical components of a tensor at a point are merely the Cartesian components in a local coordinate system at the point with itsaxes tangent to the coordinate curves. Hence, Trr=λe+2µεrr(9.37) Tθθ=λe+2µεθθ(9.38) Trθ=2µεrθ (9.39) Tzz=ν(Trr+Tθθ) (9.40) Victor Saouma Introduction to Continuum Mechanics Draft9.8 Cylindrical Coordinates 9–11 withe=εrr+εθθ. alternatively, Err=1 EbracketleftBig (1−ν2)Trr−ν(1+ν)TθθbracketrightBig (9.41) Eθθ=1 EbracketleftBig (1−ν2)Tθθ−ν(1+ν)TrrbracketrightBig (9.42) Erθ=1+ν ETrθ (9.43) Erz=Eθz=Ezz= 0 (9.44) 9.8.3.1 Plane Strain 49For Plane strain problems, from Eq. 7.75:   σrr σθθ σzz τrθ  =E (1+ν)(1−2ν) (1−ν)ν0 ν(1−ν)0 νν 0 001−2ν 2   εrr εθθ γrθ  (9.45) andεzz=γrz=γθz=τrz=τθz=0 . 50Inverting,   εrr εθθ γrθ  =1 E 1−ν2−ν(1+ν)0 −ν(1+ν)1−ν20 νν 0 00 2 ( 1 + ν   σrr σθθ σzz τrθ  (9.46) 9.8.3.2 Plane Stress 51For plane stress problems, from Eq. 7.78-a   σrr σθθ τrθ  =E 1−ν2 1ν0 ν10 001−ν 2   εrr εθθ γrθ  (9.47-a) εzz=−1 1−νν(εrr+εθθ) (9.47-b) andτrz=τθz=σzz=γrz=γθz=0 52Inverting   εrr εθθ γrθ  =1 E 1−ν0 −ν10 00 2 ( 1 + ν)   σrr σθθ τrθ  (9.48-a) Victor Saouma Introduction to Continuum Mechanics Draft9–12 BOUNDARY VALUE PROBLEMS in ELASTICITY Victor Saouma Introduction to Continuum Mechanics Draft Chapter 10 SOME ELASTICITY PROBLEMS 20Practical solutions of two-dimensional boundary-value problem in simply connected regions can be accomplished by numerous techniques. Those include: a) Finite-diﬀerenceapproximation of the diﬀerential equation, b) Complex function method of Muskhelisvili(mostusefulinproblemswithstressconcentration), c)Variationalmethods(whichwillbe covered in subsequent chapters), d) Semi-inverse methods, and e) Airy stress functions. 21Only the last two methods will be discussed in this chapter. 10.1 Semi-Inverse Method 22Often a solution to an elasticity problem may be obtained without seeking simulate- neous solutions to the equations of motion, Hooke’s Law and boundary conditions. Onemay attempt to seek solutions by making certain assumptions or guesses about the com-ponents of strain stress or displacement while leaving enough freedom in these assump- tions so that the equations of elasticity be satisﬁed. 23If the assumptions allow us to satisfy the elasticity equations, then by the uniqueness theorem, we have succeeded in obtaining the solution to the problem. 24This method was employed by Saint-Venant in his treatment of the torsion problem, hence it is often referred to as the Saint-Venant semi-inverse method . 10.1.1 Example: Torsion of a Circular Cylinder 25Let us consider the elastic deformation of a cylindrical bar with circular cross section of radius aand length Ltwisted by equal and opposite end moments M1, Fig. 10.1. 26From symmetry, it is reasonable to assume that the motion of each cross-sectional plane is a rigid body rotation about the x1axis. Hence, for a small rotation angle θ,t h e displacement ﬁeld will be given by: u=(θe1)×r=(θe1)×(x1e1+x2e2+x3e3)=θ(x2e3−x3e2) (10.1) or u1=0 ;u2=−θx3;u3=θx2 (10.2) Draft10–2 SOME ELASTICITY PROBLEMS nXX XLMM 12 3T Taθ n Figure 10.1: Torsion of a Circular Bar whereθ=θ(x1). 27The corresponding strains are given by E11=E22=E33= 0 (10.3-a) E12=−1 2x3∂θ ∂x1(10.3-b) E13=1 2x2∂θ ∂x1(10.3-c) 28The non zero stress components are obtained from Hooke’s law T12=−µx3∂θ ∂x1(10.4-a) T13=µx2∂θ ∂x1(10.4-b) 29We need to check that this state of stress satisﬁes equilibrium ∂Tij/∂xj= 0. The ﬁrst onej= 1 is identically satisﬁed, whereas the other two yield −µx3d2θ dx2 1= 0 (10.5-a) µx2d2θ dx2 1= 0 (10.5-b) thus, dθ dx1≡θ/prime= constant (10.6) Physically, this means that equilibrium is only satisﬁed if the increment in angular rota- tion (twist per unit length) is a constant. Victor Saouma Introduction to Continuum Mechanics Draft10.2 Airy Stress Functions 10–3 30We next determine the corresponding surface tractions. On the lateral surface we have a unit normal vector n=1 a(x2e2+x3e3), therefore the surface traction on the lateral surface is given by {t}=[T]{n}=1 a 0T12T13 T2100 T3100   0 x2 x3  =1 a  x2T12 0 0  (10.7) 31Substituting, t=µ a(−x2x3θ/prime+x2x3θ/prime)e1=0 (10.8) which is in agreement with the fact that the bar is twisted by end moments only, the lateral surface is traction free. 32On the face x1=L, we have a unit normal n=e1and a surface traction t=Te1=T21e2+T31e3 (10.9) this distribution of surface traction on the end face gives rise to the following resultants R1=integraldisplay T11dA= 0 (10.10-a) R2=integraldisplay T21dA=µθ/primeintegraldisplay x3dA= 0 (10.10-b) R3=integraldisplay T31dA=µθ/primeintegraldisplay x2dA= 0 (10.10-c) M1=integraldisplay (x2T31−x3T21)dA=µθ/primeintegraldisplay (x2 2+x2 3)dA=µθ/primeJ(10.10-d) M2=M3= 0 (10.10-e) We note thatintegraltext(x2 2+x3 3)2dAis thepolar moment of inertia of the cross section and is equal to J=πa4/2, and we also note thatintegraltextx2dA=integraltextx3dA= 0 because the area is symmetric with respect to the axes. 33From the last equation we note that θ/prime=M µJ(10.11) which implies that the shear modulus µcan be determined froma simple torsion experi- ment. 34Finally, in terms of the twisting couple M, the stress tensor becomes [T]= 0−Mx3 JMx2 J −Mx3 J00 Mx2 J00  (10.12) 10.2 Airy Stress Functions 10.2.1 Cartesian Coordinates; Plane Strain 35If the deformation of a cylindrical body is such that there is no axial components of the displacement and that the other components do not depend on the axial coordinate, Victor Saouma Introduction to Continuum Mechanics Draft10–4SOME ELASTICITY PROBLEMS then the body is said to be in a state of plane strain. If e3is the direction corresponding to the cylindrical axis, then we have u1=u1(x1,x2),u2=u2(x1,x2),u3= 0 (10.13) and the strain components corresponding to those displacements are E11=∂u1 ∂x1(10.14-a) E22=∂u2 ∂x2(10.14-b) E12=1 2parenleftBigg∂u1 ∂x2+∂u2 ∂x1parenrightBigg (10.14-c) E13=E23=E33= 0 (10.14-d) and the non-zero stress components are T11,T12,T22,T33where T33=ν(T11+T22) (10.15) 36Considering a static stress ﬁeld with no body forces, the equilibrium equations reduce to: ∂T11 ∂x1+∂T12 ∂x2= 0 (10.16-a) ∂T12 ∂x1+∂T22 ∂x2= 0 (10.16-b) ∂T33 ∂x1= 0 (10.16-c) we note that since T33=T33(x1,x2), the last equation is always satisﬁed. 37Hence, it can be easily veriﬁed that for any arbitrary scalar variable Φ, if we compute the stress components from T11=∂2Φ ∂x2 2(10.17) T22=∂2Φ ∂x2 1(10.18) T12=−∂2Φ ∂x1∂x2(10.19) then the ﬁrst two equations of equilibrium are automatically satisﬁed. This function Φ is calledAiry stress function . 38However, if stress components determined this way are statically admissible (i.e. they satisfy equilibrium), they are not necessarily kinematically admissible (i.e. sat- isfy compatibility equations). Victor Saouma Introduction to Continuum Mechanics Draft10.2 Airy Stress Functions 10–5 39To ensure compatibility of the strain components, we obtain the strains components in terms of Φ from Hooke’s law, Eq. 5.1 and Eq. 10.15. E11=1 EbracketleftBig (1−ν2)T11−ν(1+ν)T22bracketrightBig =1 EbracketleftBigg (1−ν2)∂2Φ ∂x2 2−ν(1+ν)∂2Φ ∂x2 1bracketrightBigg (10.20-a) E22=1 EbracketleftBig (1−ν2)T22−ν(1+ν)T11bracketrightBig =1 EbracketleftBigg (1−ν2)∂2Φ ∂x2 1−ν(1+ν)∂2Φ ∂x2 2bracketrightBigg (10.20-b) E12=1 E(1+ν)T12=−1 E(1+ν)∂2Φ ∂x1∂x2(10.20-c) 40For plane strain problems, the only compatibility equation, 4.159, that is not auto- matically satisﬁed is ∂2E11 ∂x2 2+∂2E22 ∂x2 1=2∂2E12 ∂x1∂x2(10.21) thus we obtain the following equation governing the scalar function Φ (1−ν)parenleftBigg∂4Φ ∂x4 1+2∂4Φ ∂x2 1∂x22+∂4Φ ∂x4 1parenrightBigg = 0 (10.22) or ∂4Φ ∂x4 1+2∂4Φ ∂x2 1∂x22+∂4Φ ∂x4 1=0 o r∇4Φ=0 (10.23) Hence, any function which satisﬁes the preceding equation will satisfy bothequilibrium and kinematic and is thus an acceptable elasticity solution. 41We can also obtain from the Hooke’s law, the compatibility equation 10.21, and the equilibrium equations the following parenleftBigg∂2 ∂x2 1+∂2 ∂x2 2parenrightBigg (T11+T22)=0 o r∇2(T11+T22)=0 (10.24) 42Any polynomial of degree three or less in xandysatisﬁes the biharmonic equation (Eq. 10.23). A systematic way of selecting coeﬃcients begins with Φ=∞summationdisplay m=0∞summationdisplay n=0Cmnxmyn(10.25) 43The stresses will be given by Txx=∞summationdisplay m=0∞summationdisplay n=2n(n−1)Cmnxmyn−2(10.26-a) Tyy=∞summationdisplay m=2∞summationdisplay n=0m(m−1)Cmnxm−1yn(10.26-b) Txy=−∞summationdisplay m=1∞summationdisplay n=1mnCmnxm−1yn−1(10.26-c) Victor Saouma Introduction to Continuum Mechanics Draft10–6 SOME ELASTICITY PROBLEMS 44Substituting into Eq. 10.23 and regrouping we obtain ∞summationdisplay m=2∞summationdisplay n=2[(m+2)(m+1)m(m−1)Cm+2,n−2+2m(m−1)n(n−1)Cmn+(n+2)(n+1)n(n−1)Cm−2,n+2]xm−2yn−2= (10.27) but since the equation must be identically satisﬁed for all xandy, the term in bracket must be equal to zero. (m+2)(m+1)m(m−1)Cm+2,n−2+2m(m−1)n(n−1)Cmn+(n+2)(n+1)n(n−1)Cm−2,n+2=0 (10.28) Hence, the recursion relation establishes relationships among groups of three alternate coeﬃcients which can be selected from  00 C02C03 C04 C05C06··· 0C11C12C13C14C15 ··· C20C21 C22 C23C24··· C30C31C32C33 ··· C40 C41C42··· C50C51 ··· C60 ···  (10.29) For example if we consider m=n=2 ,t h e n (4)(3)(2)(1) C40+(2)(2)(1)(2)(1) C22+(4)(3)(2)(1) C04= 0 (10.30) or 3C40+C22+3C04=0 10.2.1.1 Example: Cantilever Beam 45We consider the homogeneous fourth-degree polynomial Φ4=C40x4+C31x3y+C22x2y2+C13xy3+C04y4(10.31) with 3C40+C22+3C04=0 , 46The stresses are obtained from Eq. 10.26-a-10.26-c Txx=2C22x2+6C13xy+12C04y2(10.32-a) Tyy=1 2C40x2+6C31xy+2C22y2(10.32-b) Txy=−3C31x2−4C22xy−3C13y2(10.32-c) These can be used for the end-loaded cantilever beam with width balong the zaxis, depth 2aand length L. 47If all coeﬃcients except C13are taken to be zero, then Txx=6C13xy (10.33-a) Tyy= 0 (10.33-b) Txy=−3C13y2(10.33-c) Victor Saouma Introduction to Continuum Mechanics Draft10.2 Airy Stress Functions 10–7 48This will give a parabolic shear traction on the loaded end (correct), but also a uniform shear traction Txy=−3C13a2on top and bottom. These can be removed by superposing uniform shear stress Txy=+ 3C13a2corresponding to Φ 2=−3C13a2xy.T h u s Txy=3C13(a2−y2) (10.34) note that C20=C02=0 ,a n d C11=−3C13a2. 49The constant C13is determined by requiring that P=bintegraldisplaya −a−Txydy=−3bC13integraldisplaya −a(a2−y2)dy (10.35) hence C13=−P 4a3b(10.36) and the solution is Φ=3P 4abxy−P 4a3bxy3(10.37-a) Txx=−3P 2a3bxy (10.37-b) Txy=−3P 4a3b(a2−y2) (10.37-c) Tyy= 0 (10.37-d) 50We observe that the second moment of area for the rectangular cross section is I= b(2a)3/12 = 2a3b/3, hence this solution agrees with the elementary beam theory solution Φ=C11xy+C13xy3=3P 4abxy−P 4a3bxy3(10.38-a) Txx=−P Ixy=−My I=−M S(10.38-b) Txy=−P 2I(a2−y2) (10.38-c) Tyy= 0 (10.38-d) 10.2.2 Polar Coordinates 10.2.2.1 Plane Strain Formulation 51In polar coordinates, the strain components in plane strain are, Eq. 9.46 Err=1 EbracketleftBig (1−ν2)Trr−ν(1+ν)TθθbracketrightBig (10.39-a) Eθθ=1 EbracketleftBig (1−ν2)Tθθ−ν(1+ν)TrrbracketrightBig (10.39-b) Victor Saouma Introduction to Continuum Mechanics Draft10–8 SOME ELASTICITY PROBLEMS Erθ=1+ν ETrθ (10.39-c) Erz=Eθz=Ezz= 0 (10.39-d) and the equations of equilibrium are 1 r∂Trr ∂r+1 r∂Tθr ∂θ−Tθθ r= 0 (10.40-a) 1 r2∂Trθ ∂r+1 r∂Tθθ ∂θ= 0 (10.40-b) 52Again, itcan beeasily veriﬁed thatthe equations ofequilibrium areidentically satisﬁed if Trr=1 r∂Φ ∂r+1 r2∂2Φ ∂θ2(10.41) Tθθ=∂2Φ ∂r2(10.42) Trθ=−∂ ∂rparenleftBigg1 r∂Φ ∂θparenrightBigg (10.43) 53In order to satisfy the compatibility conditions, the cartesian stress components must also satisfy Eq. 10.24. To derive the equivalent expression in cylindrical coordinates, wenote that T 11+T22is the ﬁrst scalar invariant of the stress tensor, therefore T11+T22=Trr+Tθθ=1 r∂Φ ∂r+1 r2∂2Φ ∂θ2+∂2Φ ∂r2(10.44) 54We also note that in cylindrical coordinates, the Laplacian operator takes the following form ∇2=∂2 ∂r2+1 r∂ ∂r+1 r2∂2 ∂θ2(10.45) 55Thus, the function Φ must satisfy the biharmonic equation parenleftBigg∂2 ∂r2+1 r∂ ∂r+1 r2∂2 ∂θ2parenrightBiggparenleftBigg∂2 ∂r2+1 r∂ ∂r+1 r2∂2 ∂θ2parenrightBigg =0 o r∇4=0 (10.46) 10.2.2.2 Axially Symmetric Case 56If Φ is a function of ronly, we have Trr=1 rdΦ dr;Tθθ=d2Φ dr2;Trθ= 0 (10.47) and d4Φ dr4+2 rd3Φ dr3−1 r2d2Φ dr2+1 r3dΦ dr= 0 (10.48) 57The general solution to this problem; using Mathematica: Victor Saouma Introduction to Continuum Mechanics Draft10.2 Airy Stress Functions 10–9 DSolve[phi’’’’[r]+2 phi’’’[r]/r-phi’’[r]/r^2+phi’[r]/r^3==0,phi[r],r] Φ=Alnr+Br2lnr+Cr2+D (10.49) 58The corresponding stress ﬁeld is Trr=A r2+B(1+2lnr)+2C(10.50) Tθθ=−A r2+B(3+2lnr)+2C(10.51) Trθ= 0 (10.52) and the strain components are (from Sect. 9.8.1) Err=∂ur ∂r=1 EbracketleftBigg(1+ν)A r2+(1−3ν−4ν2)B+2(1−ν−2ν2)Blnr+2(1−ν−2ν2)CbracketrightBigg (10 Eθθ=1 r∂uθ ∂θ+ur r=1 EbracketleftBigg −(1+ν)A r2+(3−ν−4ν2)B+2(1−ν−2ν2)Blnr+2(1−ν−2ν2)CbracketrightBigg (10 Erθ=0 (10 59Finally, the displacement components can be obtained by integrating the above equa- tions ur=1 EbracketleftBigg −(1+ν)A r−(1+ν)Br+2(1−ν−2ν2)rlnrB+2(1−ν−2ν2)rCbracketrightBigg (10.56) uθ=4rθB E(1−ν2) (10.57) 10.2.2.3 Example: Thick-Walled Cylinder 60If we consider a circular cylinder with internal and external radii aandbrespectively, subjected to internal and external pressures piandporespectively, Fig. 10.2, then the boundary conditions for the plane strain problem are Trr=−piatr=a (10.58-a) Trr=−poatr=b (10.58-b) 61These Boundary conditions can be easily shown to be satisﬁed by the following stress ﬁeld Trr=A r2+2C (10.59-a) Tθθ=−A r2+2C (10.59-b) Trθ= 0 (10.59-c) Victor Saouma Introduction to Continuum Mechanics Draft10–10 SOME ELASTICITY PROBLEMS Saint Venant p a bpio Figure 10.2: Pressurized Thick Tube These equations are taken from Eq. 10.50, 10.51 and 10.52 with B= 0 and therefore represent a possible state of stress for the plane strain problem. 62We note that if we take B/negationslash=0 ,t h e n uθ=4rθB E(1−ν2) and this is not acceptable because if we were to start at θ= 0 and trace a curve around the origin and return to t h es a m ep o i n t ,t h a n θ=2πand the displacement would then be diﬀerent. 63Applying the boundary condition we ﬁnd that Trr=−pi(b2/r2)−1 (b2/a2)−1−p01−(a2/r2) 1−(a2/b2)(10.60) Tθθ=pi(b2/r2)+1 (b2/a2)−1−p01+(a2/r2) 1−(a2/b2)(10.61) Trθ= 0 (10.62) 64We note that if only the internal pressure piis acting, then Trris always a compressive stress, and Tθθis always positive. 65If the cylinder is thick, then the strains are given by Eq. 10.53, 10.54 and 10.55. For a very thin cylinder in the axial direction, then the strains will be given by Err=du dr=1 E(Trr−νTθθ) (10.63-a) Eθθ=u r=1 E(Tθθ−νTrr) (10.63-b) Victor Saouma Introduction to Continuum Mechanics Draft10.2 Airy Stress Functions 10–11 Ezz=dw dz=ν E(Trr+Tθθ) (10.63-c) Erθ=(1+ν) ETrθ (10.63-d) 66It should be noted that applying Saint-Venant’s principle the above solution is only valid away from the ends of the cylinder. 10.2.2.4Example: Hollow Sphere 67We consider next a hollow sphere with internal and xternal radii aiandaorespectively, and subjected to internal and external pressures of piandpo, Fig. 10.3. aoipo p ai Figure 10.3: Pressurized Hollow Sphere 68With respect to the spherical ccordinates ( r,θ,φ), it is clear due to the spherical symmetry of the geometry and the loading that each particle of the elastic sphere will expereince only a radial displacement whose magnitude depends on ronly, that is ur=ur(r),uθ=uφ= 0 (10.64) 10.2.2.5 Example: Stress Concentration due to a Circular Hole in a Plate 69Analysing the inﬁnite plate under uniform tension with a circular hole of diameter a, and subjected to a uniform stress σ0, Fig. 10.4. 70The peculiarity of this problem is that the far-ﬁeld boundary conditions are better expressed in cartesian coordinates, whereas the ones around the hole should be writtenin polar coordinate system. 71First we select a stress function which satisﬁes the biharmonic Equation (Eq. 10.23), and the far-ﬁeld boundary conditions. From St Venant principle, away from the hole,the boundary conditions are given by: T xx=σ0;Tyy=Txy= 0 (10.65) Recalling (Eq. 10.19) that Txx=∂2Φ ∂y2, this would would suggest a stress function Φ of the form Φ = σ0y2. Alternatively, the presence of the circular hole would suggest a polar representation of Φ. Thus, substituting y=rsinθwould result in Φ = σ0r2sin2θ. Victor Saouma Introduction to Continuum Mechanics Draft10–12 SOME ELASTICITY PROBLEMS rr rθbrrσ IIIb θθσ τ a aaτrθ σoθb xσrry σo Figure 10.4: Circular Hole in an Inﬁnite Plate 72Since sin2θ=1 2(1−cos2θ), we could simplify the stress function into Φ=f(r)cos2θ (10.66) Substituting this function into the biharmonic equation (Eq. 10.46) yields parenleftBigg∂2 ∂r2+1 r∂ ∂r+1 r2∂2 ∂θ2parenrightBiggparenleftBigg∂2Φ ∂r2+1 r∂Φ ∂r+1 r2∂2Φ ∂θ2parenrightBigg = 0 (10.67-a) parenleftBiggd2 dr2+1 rd dr−4 r2parenrightBiggparenleftBiggd2f dr2+1 rdf dr−4f r2parenrightBigg = 0 (10.67-b) 73The general solution of this ordinary linear fourth order diﬀerential equation is f(r)=Ar2+Br4+C1 r2+D (10.68) thus the stress function becomes Φ=parenleftbigg Ar2+Br4+C1 r2+Dparenrightbigg cos2θ (10.69) Using Eq. 10.41-10.43, the stresses are given by Trr=1 r∂Φ ∂r+1 r2∂2Φ ∂θ2=−parenleftbigg 2A+6C r4+4D r2parenrightbigg cos2θ (10.70-a) Tθθ=∂2Φ ∂r2=parenleftbigg 2A+12Br2+6C r4parenrightbigg cos2θ (10.70-b) Trθ=−∂ ∂rparenleftBigg1 r∂Φ ∂θparenrightBigg =parenleftbigg 2A+6Br2−6C r4−2D r2parenrightbigg sin2θ(10.70-c) 74Next we seek to solve for the four constants of integration by applying the boundary conditions. We will identify two sets of boundary conditions: 1. Outerboundaries: aroundaninﬁnitelylargecircleofradius binsideaplatesubjected to uniform stress σ0, the stresses in polar coordinates are obtained from Eq. 9.35 bracketleftBigg TrrTrθ TrθTθθbracketrightBigg =bracketleftBigg cosθ−sinθ sinθcosθbracketrightBiggbracketleftBigg σ00 00bracketrightBiggbracketleftBigg cosθ−sinθ sinθcosθbracketrightBiggT (10.71) Victor Saouma Introduction to Continuum Mechanics Draft10.2 Airy Stress Functions 10–13 yielding (recalling that sin2θ=1/2sin2θ,a n dc o s2θ=1/2(1+cos2 θ)). (Trr)r=b=σ0cos2θ=1 2σ0(1+cos2 θ) (10.72-a) (Trθ)r=b=1 2σ0sin2θ (10.72-b) (Tθθ)r=b=σ0 2(1−cos2θ) (10.72-c) For reasons which will become apparent later, it is more convenient to decompose the state of stress given by Eq. 10.72-a and 10.72-b, into state I and II: (Trr)I r=b=1 2σ0 (10.73-a) (Trθ)I r=b= 0 (10.73-b) (Trr)II r=b=1 2σ0cos2θ (10.73-c) (Trθ)II r=b=1 2σ0sin2θ (10.73-d) Where state I corresponds to a thick cylinder with external pressure applied on r=band of magnitude σ0/2. This problem has already been previously solved. Hence, only the last two equations will provide us with boundary conditions. 2. Around the hole: the stresses should be equal to zero: (Trr)r=a= 0 (10.74-a) (Trθ)r=a= 0 (10.74-b) 75Upon substitution in Eq. 10.70-a the four boundary conditions (Eq. 10.73-c, 10.73-d, 10.74-a, and 10.74-b) become −parenleftbigg 2A+6C b4+4D b2parenrightbigg =1 2σ0 (10.75-a) parenleftbigg 2A+6Bb2−6C b4−2D b2parenrightbigg =1 2σ0 (10.75-b) −parenleftbigg 2A+6C a4+4D a2parenrightbigg = 0 (10.75-c) parenleftbigg 2A+6Ba2−6C a4−2D a2parenrightbigg = 0 (10.75-d) 76Solving for the four unknowns, and takinga b= 0 (i.e. an inﬁnite plate), we obtain: A=−σ0 4;B=0 ;C=−a4 4σ0;D=a2 2σ0 (10.76) 77To this solution, we must superimpose the one of a thick cylinder subjected to a uniform radial traction σ0/2 on the outer surface, and with bmuch greater than a.T h e s e Victor Saouma Introduction to Continuum Mechanics Draft10–14SOME ELASTICITY PROBLEMS stresses were derived in Eqs. 10.60 and 10.61 yielding for this problem (carefull about the sign) Trr=σ0 2parenleftBigg 1−a2 r2parenrightBigg (10.77-a) Tθθ=σ0 2parenleftBigg 1+a2 r2parenrightBigg (10.77-b) Thus, upon substitution into Eq. 10.70-a, we obtain Trr=σ0 2parenleftBigg 1−a2 r2parenrightBigg +parenleftBigg 1+3a4 r4−4a2 r2parenrightBigg1 2σ0cos2θ (10.78-a) Tθθ=σ0 2parenleftBigg 1+a2 r2parenrightBigg −parenleftBigg 1+3a4 r4parenrightBigg1 2σ0cos2θ (10.78-b) Trθ=−parenleftBigg 1−3a4 r4+2a2 r2parenrightBigg1 2σ0sin2θ (10.78-c) 78We observe that as r→∞,b o t hTrrandTrθare equal to the values given in Eq. 10.72-a and 10.72-b respectively. 79Alternatively, at the edge of the hole when r=awe obtain Trr=Trθ=0a n d (Tθθ)r=a=σ0(1−2cos2θ) (10.79) which for θ=π 2and3π 2gives a stress concentration factor (SCF) of 3. For θ=0a n d θ=π,Tθθ=−σ0. Victor Saouma Introduction to Continuum Mechanics Draft Chapter 11 THEORETICAL STRENGTH OF PERFECT CRYSTALS This chapter (taken from the author’s lecture notes in Fracture Mechanics) is of primary interest to students in Material Science. 11.1 Introduction 20In Eq.??we showed that around a circular hole in an inﬁnite plate under uniform traction, we do have a stress concentration factor of 3. 21Following a similar approach (though with curvilinear coordinates), it can be shown that if we have an elliptical hole, Fig. ??,w ew o u l dh a v e (σββ)β=0,π α=α0=σ0parenleftbigg 1+2a bparenrightbigg (11.1) Weobserve thatfor a=b, we recover thestress concentrationfactorof3ofacircularhole, and that for a degenerated ellipse, i.e a crack there is an inﬁnite stress. Alternatively, x x2b 2a σσ α = α οο2 o 1 Figure 11.1: Elliptical Hole in an Inﬁnite Plate Draft11–2 THEORETICAL STRENGTH OF PERFECT CRYSTALS Theoretical Strength DiameterStrength (P/A) Figure 11.2: Griﬃth’s Experiments the stress can be expressed in terms of ρ, the radius of curvature of the ellipse, (σββ)β=0,π α=α0=σ0parenleftBigg 1+2radicalBigg a ρparenrightBigg (11.2) From this equation, we note that the stress concentration factor is inversely proportional to the radius of curvature of an opening. 22This equation, derived by Inglis, shows that if a=bwe recover the factor of 3, and the stress concentration factor increase as the ratio a/bincreases. In the limit, as b=0 we would have a crack resulting in an inﬁnite stress concentration factor, or a stress singularity . 23Around 1920, Griﬃth was exploring the theoretical strength of solids by performing a series of experiments on glass rods of various diameters. 24He observed that the tensile strength ( σt) of glass decreased with an increase in diam- eter, and that for a diameter φ≈1 10,000in.,σt= 500,000 psi; furthermore, by extrapo- lation to “zero” diameter he obtained a theoretical maximum strength of approximately 1,600,000 psi, and on the other hand for very large diameters the asymptotic values was around 25,000 psi. AreaA1<A2<A3<A4 Failure Load P1<P2<P3>P4 Failure Strength ( P/A)σt 1>σt 2>σt 3>σt 4(11.3) Furthermore, as the diameter was further reduced, the failure strength asymptotically approached a limit which will be shown later to be the theoretical strength of glass, Fig. 11.2. 25Clearly, one would have expected the failure strength to be constant, yet it was not. So Griﬃth was confronted with two questions: 1. What is this apparent theoretical strength, can it be derived? 2. Why is there a size eﬀect for the actual strength? Victor Saouma Introduction to Continuum Mechanics Draft11.2 Theoretical Strength 11–3 Figure 11.3: Uniformly Stressed Layer of Atoms Separated by a0 The answers tothosetwo questions areessential toestablish a linkbetweenMechanics and Materials . 26In the next sections we will show that the theoretical strength is related to the force needed to break a bond linking adjacent atoms, and that the size eﬀect is caused by thesize of imperfections inside a solid. 11.2 Theoretical Strength 27We start, [ ?] by exploring the energy of interaction between two adjacent atoms at equilibrium separated by a distance a0, Fig. 11.3. The total energy which must be supplied to separate atom C from C’ is U0=2γ (11.4) whereγis thesurface energy1, and the factor of 2 is due to the fact that upon sepa- ration, we have two distinct surfaces. 11.2.1 Ideal Strength in Terms of Physical Parameters 28Weshallﬁrstderiveanexpression fortheidealstrengthintermsofphysicalparameters, and in the next section the strength will be expressed in terms of engineering ones. Solution I: Force being the derivative of energy, we have F=dU da,t h u sF=0a ta=a0, Fig. 11.4, and is maximum at the inﬂection point of the U0−acurve. Hence, the slope of the force displacement curve is the stiﬀness of the atomic spring and shouldbe related to E.I fw el e t x=a−a 0, then the strain would be equal to ε=x a0. 1Fromwatchingraindropsandbubblesitisobviousthatliquidwaterhassurfacetension. Whenthesurfaceofaliquid is extended (soap bubble, insect walking on liquid) work is done against this tension, and energy is stored in the new surface. Wheninsectswalkonwateritsinksuntilthesurfaceenergyjustbalancesthedecreaseinitspotentialenergy. For solids,thechemicalbondsarestrongerthanforliquids,hencethesurfaceenergyisstronger. Thereasonwhywedonotnoticeitisthatsolidsaretoorigidtobedistortedbyit. Surfaceenergy γisexpressedin J/m 2andthesurfaceenergies ofwater,mostsolids,anddiamondsareapproximately.077,1.0,and5.14respectively. Victor Saouma Introduction to Continuum Mechanics Draft11–4THEORETICAL STRENGTH OF PERFECT CRYSTALS Distance Interatomic DistanceInteratomic0Energy Repulsion AttractionForcea Figure 11.4: Energy and Force Binding Two Adjacent Atoms Furthermore, if we deﬁne the stress as σ=F a2 0, then the σ−εcurve will be as shown in Fig. 11.5. From this diagram, it would appear that the sine curve would be an adequate approximation to this relationship. Hence, σ=σtheor maxsin2πx λ(11.5) and the maximum stress σtheor maxwould occur at x=λ 4. The energy required to separate two atoms is thus given by the area under the sine curve, and from Eq.11.4, we would have 2γ=U 0=integraldisplayλ 2 0σtheor maxsinparenleftbigg 2πx λparenrightbigg dx (11.6) =λ 2πσtheor max[−cos(2πx λ)]|λ 2 0 (11.7) =λ 2πσtheor max[−−1bracehtipdownleft bracehtipuprightbracehtipupleft bracehtipdownright cos(2πλ 2λ)+1bracehtipdownleft bracehtipuprightbracehtipupleft bracehtipdownright cos(0)] (11.8) ⇒λ=2γπ σtheor max(11.9) Also for very small displacements (small x)s i nx≈x, thus Eq. 11.5 reduces to σ≈σtheor max2πx λ≈Ex a0(11.10) elliminating x, σtheor max≈E a0λ 2π(11.11) Victor Saouma Introduction to Continuum Mechanics Draft11.2 Theoretical Strength 11–5 Figure 11.5: Stress Strain Relation at the Atomic Level Substituting for λfrom Eq. 11.9, we get σtheor max≈radicalBigg Eγ a0 (11.12) Solution II: For two layers of atoms a0apart, the strain energy per unit area due to σ (for linear elastic systems) is U=1 2σεao σ=EεbracerightBigg U=σ2ao 2E(11.13) Ifγis the surface energy of the solid per unit area, then the total surface energy of two new fracture surfaces is 2 γ. For our theoretical strength, U=2γ⇒(σtheor max)2a0 2E=2γorσtheor max=2radicalBig γE a0 Note that here we have assumed that the material obeys Hooke’s Law up to failure, since this is seldom the case, we can simplify this approximation to: σtheor max=radicalBigg Eγ a0 (11.14) which is the same as Equation 11.12 Example: As an example, let us consider steel which has the following properties: γ= 1J m2;E=2×1011N m2;a n da0≈2×10−10m. Thus from Eq. 11.12 we would have: σtheor max≈radicalBigg (2×1011)(1) 2×10−10(11.15) ≈3.16×1010N m2(11.16) ≈E 6(11.17) Thus this would be the ideal theoretical strength of steel. Victor Saouma Introduction to Continuum Mechanics Draft11–6 THEORETICAL STRENGTH OF PERFECT CRYSTALS 11.2.2 Ideal Strength in Terms of Engineering Parameter 29We note that the force to separate two atoms drops to zero when the distance between them isa0+awherea0corresponds to the origin and atoλ 2.T h u s ,i fw et a k e a=λ 2or λ=2a, combined with Eq. 11.11 would yield σtheor max≈E a0a π(11.18) 30Alternatively combining Eq. 11.9 with λ=2agives a≈γπ σtheor max(11.19) Combining those two equations will give γ≈E a0parenleftbigga πparenrightbigg2 (11.20) 31However, since as a ﬁrst order approximation a≈a0then the surface energy will be γ≈Ea0 10(11.21) This equation, combined with Eq. 11.12 will ﬁnally give σtheor max≈E √ 10 (11.22) which is an approximate expression for the theoretical maximum strength in terms of E. 11.3 Size Eﬀect; Griﬃth Theory 32In his quest for an explanation of the size eﬀect, Griﬃth came across Inglis’s paper, and his “strike of genius” was to assume that strength is reduced due to the presence of internal ﬂaws . Griﬃth postulated that the theoretical strength can only be reached at the point of highest stress concentration, and accordingly the far-ﬁeld applied stress willbe much smaller. 33Hence, assuming an elliptical imperfection, and from equation 11.2 σtheor max=σact crparenleftBigg 1+2radicalBigg a ρparenrightBigg (11.23) σis the stress at the tip of the ellipse which is caused by a (lower) far ﬁeld stress σact cr. Asssuming ρ≈a0and since 2radicalBig a a0/greatermuch1, for an ideal plate under tension with only one single elliptical ﬂaw the strength may be obtained from σtheor max=2σact crradicalBigg a a0(11.24) Victor Saouma Introduction to Continuum Mechanics Draft11.3 Size Eﬀect; Griﬃth Theory 11–7 hence, equating with Eq. 11.12, we obtain σtheor max=2σact crradicalBigg a aobracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Macro=radicalBigg Eγ a0bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Micro (11.25) From this very important equation, we observe that 1. The left hand side is based on a linear elastic solution of a macroscopic problem solved by Inglis. 2. The right hand side is based on the theoretical strength derived from the sinu- soidal stress-strain assumption of the interatomic forces, and ﬁnds its roots in micro- physics. Finally, this equation would give (at fracture) σact cr=radicalBigg Eγ 4a (11.26) As an example, let us consider a ﬂaw with a size of 2 a=5,000a0 σact cr=radicalBig Eγ 4a γ=Ea0 10bracerightBigg σact cr=radicalBig E2 40ao aa a0=2,500  σact cr=radicalBig E2 100,000=E 100√ 10(11.27) Thus if we set a ﬂaw size of 2 a=5,000a0inγ≈Ea0 10this is enough to lower the theoretical fracture strength fromE √ 10to a critical value of magnitudeE 100√ 10,o raf a c t o r of 100. As an example σtheor max=2σact crradicalBig a ao a=1 0−6m=1µ ao=1˚A=ρ=1 0−10m  σtheor max=2σact crradicalBigg 10−6 10−10= 200σact cr (11.28) Therefore at failure σact cr=σtheor max 200 σtheor max=E 10bracerightBigg σact cr≈E 2,000(11.29) which can be attained. For instance for steelE 2,000=30,000 2,000=1 5k s i Victor Saouma Introduction to Continuum Mechanics Draft11–8 THEORETICAL STRENGTH OF PERFECT CRYSTALS Victor Saouma Introduction to Continuum Mechanics Draft Chapter 12 BEAM THEORY This chapter is adapted from the Author’s lecture notes in Structural Analysis. 12.1 Introduction 20In the preceding chapters we have focused on the behavior of a continuum ,a n dt h e1 5 equations and 15 variables we introduced, were all derived for an inﬁnitesimal element. 21In practice, few problems can be solved analytically, and even with computer it is quite diﬃcult to view every object as a three dimensional one. That is why we introduced the 2D simpliﬁcation (plane stress/strain), or 1D for axially symmetric problems. In thepreceding chapter we saw a few of those solutions. 22Hence, to widen the scope of application of the fundamental theory developed previ- ously, we could either resort to numerical methods (such as the ﬁnite diﬀerence, ﬁnite element, or boundary elements), or we could further simplify the problem. 23Solid bodies, in general, have certain peculiar geometric features amenable to a reduc- tion from three to fewer dimensions. If one dimension of the structural element1under consideration is much greater or smaller than the other three, than we have a beam, or a plate respectively. If the plate is curved, then we have a shell. 24For those structural elements, it is customary to consider as internal variables the resultant of the stresses as was shown in Sect. ??. 25Hence, this chapter will focusonabriefintroduction tobeam theory. This will however be preceded by an introduction to Statics as the internal forces would also have to be in equilibrium with the external ones. 26Beam theory is perhaps the most successful theory in all of structural mechanics, and it forms the basis of structural analysis which is so dear to Civil and Mechanical engineers. 1Sofarwehaverestrictedourselvestoacontinuum,inthischapterwewillconsiderastructuralelement. Draft12–2 BEAM THEORY 12.2 Statics 12.2.1 Equilibrium 27Any structural element, or part of it, must satisfy equilibrium. 28Summation of forces and moments, in a static system must be equal to zero2. 29In a 3D cartesian coordinate system there are a total of 6 independent equations of equilibrium: ΣFx=ΣFy=ΣFz=0 ΣMx=ΣMy=ΣMz=0(12.1) 30In a 2D cartesian coordinate system there are a total of 3 independent equations of equilibrium: ΣFx=ΣFy=ΣMz=0 (12.2) 31All the externally applied forces on a structure must be in equilibrium. Reactions are accordingly determined. 32For reaction calculations, the externally applied load may be reduced to an equivalent force3. 33Summation of the moments can be taken with respect to anyarbitrary point. 34Whereas forces are represented by a vector, moments are also vectorial quantities and are represented by a curved arrow or a double arrow vector. 35Not all equations are applicable to all structures, Table 12.1 Structure Type Equations Beam, no axial forces ΣFy ΣMz 2D Truss, Frame, Beam ΣFxΣFy ΣMz Grid ΣFzΣMxΣMy 3D Truss, Frame ΣFxΣFyΣFzΣMxΣMyΣMz Alternate Set Beams, no axial Force ΣMA zΣMB z 2 D Truss, Frame, Beam ΣFxΣMA zΣMB z ΣMA zΣMB zΣMC z Table 12.1: Equations of Equilibrium 36The three conventional equations of equilibrium in 2D: Σ Fx,ΣFyand ΣMzcan be replaced by the independent moment equations Σ MA z,ΣMB z,ΣMC zprovided that A, B, and Care not colinear . 2InadynamicsystemΣ F=mawheremisthemassand aistheacceleration. 3Howeverforinternalforces(shearandmoment)wemustusetheactualloaddistribution. Victor Saouma Introduction to Continuum Mechanics Draft12.2 Statics 12–3 37It is always preferable to checkcalculations by another equation of equilibrium. 38Before you write an equation of equilibrium, 1. Arbitrarily decide which is the +vedirection 2. Assume a direction for the unknown quantities3. The right hand side of the equation should be zero If your reaction is negative, then it will be in a direction opposite from the one assumed. 39Summation of external forces is equal and opposite to the internal ones (more about this below). Thus the net force/moment is equal to zero. 40The external forces give rise to the (non-zero) shear and moment diagram. 12.2.2 Reactions 41In the analysis of structures, it is often easier to start by determining the reactions. 42Once the reactions are determined, internal forces (shear and moment) are determined next; ﬁnally, internal stresses and/or deformations (deﬂections and rotations) are deter- mined last. 43Depending onthe type of structures, there can be diﬀerent types of support conditions, Fig. 12.1. Figure 12.1: Types of Supports Roller: provides a restraint in only one direction in a 2D structure, in 3D structures a roller may provide restraint in one or two directions. A roller will allow rotation. Hinge:allows rotation but no displacements. Victor Saouma Introduction to Continuum Mechanics Draft12–4 BEAM THEORY Fixed Support: will prevent rotation and displacements in all directions. 12.2.3 Equations of Conditions 44If a structure has an internal hinge (which may connect two or more substructures), then this will provide an additional equation (Σ M= 0 at the hinge) which can be exploited to determine the reactions. 45Those equations are often exploited in trusses (where each connection is a hinge) to determine reactions. 46In aninclined roller support with SxandSyhorizontal and vertical projection, then the reaction R would have, Fig. 12.2. Rx Ry=Sy Sx (12.3) Figure 12.2: Inclined Roller Support 12.2.4 Static Determinacy 47In statically determinate structures, reactions depend only on the geometry, boundary conditions and loads. 48If the reactions can not be determined simply from the equations of static equilibrium (and equations of conditions if present), then the reactions of the structure are said tobestatically indeterminate . 49Thedegree of static indeterminacy is equal to the diﬀerence between the number of reactions and the number of equations of equilibrium (plus the number of equations of conditions if applicable), Fig. 12.3. 50Failure of one support in a statically determinate system results in the collapse of the structures. Thus a statically indeterminate structure is saferthan a statically determi- nate one. 51For statically indeterminate structures, reactions depend also on the material proper- ties (e.g. Young’s and/or shear modulus) and element cross sections (e.g. length, area,moment of inertia). Victor Saouma Introduction to Continuum Mechanics Draft12.2 Statics 12–5 Figure 12.3: Examples of Static Determinate and Indeterminate Structures 12.2.5 Geometric Instability 52The stability of a structure is determined not only by the number of reactions but also by their arrangement. 53Geometric instability will occur if: 1. Allreactions are parallel and a non-parallel load is applied to the structure. 2. Allreactions are concurrent , Fig. 12.4. Figure 12.4: Geometric Instability Caused by Concurrent Reactions 3. The number of reactions is smaller than the number of equations of equilibrium, that is amechanism is present in the structure. 54Mathematically, this can be shown if the determinant of the equations of equilibrium is equal to zero (or the equations are inter-dependent). 12.2.6 Examples Example 12-1: Simply Supported Beam Victor Saouma Introduction to Continuum Mechanics Draft12–6 BEAM THEORY Determine the reactions of the simply supported beam shown below. Solution: The beam has 3 reactions, we have 3 equations of static equilibrium, hence it is statically determinate. (+ ✲)ΣFx=0 ;⇒Rax−36 k=0 (+ ✻)ΣFy=0 ;⇒Ray+Rdy−60 k−(4) k/ft(12) ft=0 (+ ✁ ✛)ΣMc z=0 ;⇒12Ray−6Rdy−(60)(6) = 0 or through matrix inversion (on your calculator)  10 0 01 101 2−6   Rax Ray Rdy  =  36 108360  ⇒  Rax Ray Rdy  =  36 k 56 k 52 k   Alternatively we could have used another set of equations: (+ ✁ ✛)ΣMa z= 0; (60)(6)+(48)(12) −(Rdy)(18) = 0 ⇒Rdy= 52 k ✻ (+ ✁ ✛)ΣMd z=0 ; (Ray)(18)−(60)(12)−(48)(6) = 0 ⇒Ray= 56 k ✻ Check: (+ ✻)ΣFy=0;;56−52−60−48= 0√ 12.3 Shear & Moment Diagrams 12.3.1 Design Sign Conventions 55Beforewederive theShear-Momentrelations, letus arbitrarily deﬁneasignconvention. 56The sign convention adopted here, is the one commonly used for design purposes4. With reference to Fig. 12.5 LoadPositive along the beam’s local y axis (assuming a right hand side convention), that is positive upward. Axial:tension positive. 4NotethatthissignconventionistheoppositeoftheonecommonlyusedinEurope! Victor Saouma Introduction to Continuum Mechanics Draft12.3 Shear & Moment Diagrams 12–7 Figure 12.5: Shear and Moment Sign Conventions for Design Flexure A positive moment is one which causes tension in the lower ﬁbers, and com- pression in the upper ones. For frame members, a positive moment is one whichcauses tension along the inner side. ShearA positive shear force is one which is “up” on a negative face, or “down” on a positive one. Alternatively, a pair of positive shear forces will cause clockwiserotation. 12.3.2 Load, Shear, Moment Relations 57Let us derive the basic relations between load, shear and moment. Considering an inﬁnitesimal length dxof a beam subjected to a positive load5w(x), Fig. 12.6. The Figure 12.6: Free Body Diagram of an Inﬁnitesimal Beam Segment inﬁnitesimal section must also be in equilibrium. 58There are no axial forces, thus we only have two equations of equilibrium to satisfy ΣFy=0a n dΣ Mz=0 . 59Sincedxis inﬁnitesimally small, the small variation in load along it can be neglected, therefore we assume w(x) to be constant along dx. 60To denote that a small change in shear and moment occurs over the length dxof the element, we add the diﬀerential quantities dVxanddMxtoVxandMxon the right face. 5Inthisderivation,asinallotheronesweshouldassumeallquantitiestobepositive. Victor Saouma Introduction to Continuum Mechanics Draft12–8 BEAM THEORY 61Next considering the ﬁrst equation of equilibrium (+ ✻)ΣFy=0⇒Vx+wxdx−(Vx+dVx)=0 or dV dx=w(x) (12.4) T h es l o p eo ft h es h e a rc u r v ea ta n yp o i n ta l o n gt h ea x i so fam e m b e r is given by the load curve at that point. 62Similarly (+ ✁ ✛)ΣMo=0⇒Mx+Vxdx−wxdxdx 2−(Mx+dMx)=0 Neglecting the dx2term, this simpliﬁes to dM dx=V(x) (12.5) The slope of the moment curve at any point along the axis of a member is given by the shear at that point. 63Alternative forms of the preceding equations can be obtained by integration V=integraldisplay w(x)dx (12.6) ∆V21=Vx2−Vx1=integraldisplayx2 x1w(x)dx(12.7) The change in shear between 1and2,∆V21, is equal to the area under the load between x1andx2. and M=integraldisplay V(x)dx (12.8) ∆M21=M2−M1=integraldisplayx2 x1V(x)dx(12.9) The change in moment between 1and2,∆M21, is equal to the area under the shear curve between x1andx2. 64Note that we still need to have V1andM1in order to obtain V2andM2respectively. 65It can be shown that the equilibrium of forces and of moments equations are nothing else than the three dimensional linear momentum∂Tij ∂xj+ρbi=ρdvi dtand moment of momentumintegraldisplay S(r×t)dS+integraldisplay V(r×ρb)dV=d dtintegraldisplay V(r×ρv)dVequations satisﬁed on the average over the cross section. Victor Saouma Introduction to Continuum Mechanics Draft12.3 Shear & Moment Diagrams 12–9 12.3.3 Examples Example 12-2: Simple Shear and Moment Diagram Draw the shear and moment diagram for the beam shown below Solution: The free body diagram is drawn below Victor Saouma Introduction to Continuum Mechanics Draft12–10 BEAM THEORY Reactions are determined from the equilibrium equations (+ ✛)ΣFx=0 ;⇒−RAx+6=0⇒RAx=6 k (+ ✁ ✛)ΣMA=0 ;⇒(11)(4)+(8)(10)+(4)(2)(14+2) −RFy(18) = 0⇒RFy=1 4 k (+ ✻)ΣFy=0 ;⇒RAy−11−8−(4)(2)+14 = 0 ⇒RAy=1 3 k Shearare determined next. 1. AtAthe shear is equal to the reaction and is positive. 2. AtBthe shear drops (negative load) by 11 kto 2 k. 3. AtCit drops again by 8 kto−6k. 4. It stays constant up to Dand then it decreases (constant negative slope since the load is uniform and negative) by 2 kper linear foot up to −14 k. 5. As a check, −14 kis also the reaction previously determined at F. Moment is determined last: 1. The moment at Ais zero (hinge support). 2. The change in moment between AandBis equal to the area under the corre- sponding shear diagram, or ∆ MB−A= (13)(4) = 52. 3. etc... 12.4 Beam Theory 12.4.1 Basic Kinematic Assumption; Curvature 66Fig.12.7 shows portion of an originally straight beam which has been bent to the radiusρby end couples M. support conditions, Fig. 12.1. It is assumed thatplane cross-sections normal to the length of the unbent beam remain plane afterthe beam is bent . 67Except for the neutral surface all other longitudinal ﬁbers either lengthen or shorten, thereby creating a longitudinal strain εx. Considering a segment EFof length dxat a distanceyfrom the neutral axis, its original length is EF=dx=ρdθ (12.10) and dθ=dx ρ(12.11) 68To evaluate this strain, we consider the deformed length E/primeF/prime E/primeF/prime=(ρ−y)dθ=ρdθ−ydθ=dx−ydx ρ(12.12) Victor Saouma Introduction to Continuum Mechanics Draft12.4Beam Theory 12–11 Neutral Axis dxρ E’ F’ EFO MMdθ XY ZdA+ve Curvature, +ve bending -ve Curvature, -ve Bending Figure 12.7: Deformation of a Beam under Pure Bending The strain is now determined from: εx=E/primeF/prime−EF EF=dx−ydx ρ−dx dx(12.13) or after simpliﬁcation εx=−y ρ (12.14) whereyismeasuredfromtheaxisofrotation(neutralaxis). Thusstrainsareproportional to the distance from the neutral axis. 69ρ(Greek letter rho)i st h eradius of curvature . In some textbook, the curvature κ (Greek letter kappa)i sa l s ou s e dw h e r e κ=1 ρ(12.15) thus, εx=−κy (12.16) 70It should be noted that Galileo (1564-1642) was the ﬁrst one to have made a contri- bution to beam theory, yet he failed to make the right assumption for the planar crosssection. This crucial assumption was made later on by Jacob Bernoulli (1654-1705), who did not make it quite right. Later Leonhard Euler (1707-1783) made signiﬁcant contribu- tions to the theory of beam deﬂection, and ﬁnally it was Navier (1785-1836) who clariﬁedthe issue of the kinematic hypothesis. Victor Saouma Introduction to Continuum Mechanics Draft12–12 BEAM THEORY 12.4.2 Stress-Strain Relations 71So far we considered the kinematic of the beam, yet later on we will need to consider equilibrium in terms of the stresses. Hence we need to relate strain to stress. 72For linear elastic material Hooke’s law states σx=Eεx (12.17) whereEisYoung’s Modulus . 73Combining Eq. with equation 12.16 we obtain σx=−Eκy (12.18) 12.4.3 Internal Equilibrium; Section Properties 74Just as external forces acting on a structure must be in equilibrium, the internal forces must also satisfy the equilibrium equations. 75The internal forces are determined by slicingthe beam. The internal forces on the “cut” section must be in equilibrium with the external forces. 12.4.3.1 ΣFx=0;N e u t r a lA x i s 76The ﬁrst equation we consider is the summation of axial forces. 77Since there are no external axial forces (unlike a column or a beam-column), the internal axial forces must be in equilibrium. ΣFx=0⇒integraldisplay AσxdA= 0 (12.19) whereσxwas given by Eq. 12.18, substituting we obtain integraldisplay AσxdA=−integraldisplay AEκydA= 0 (12.20-a) But since the curvature κand the modulus of elasticity Eare constants, we conclude that integraldisplay AydA=0 (12.21) or the ﬁrst moment of the cross section with respect to the zaxis is zero. Hence we conclude that the neutral axis passes through the centroid of the cross section . Victor Saouma Introduction to Continuum Mechanics Draft12.4Beam Theory 12–13 12.4.3.2 ΣM=0;M o m e n to fI n e r t i a 78The second equation of internal equilibrium which must be satisﬁed is the summation of moments. However contrarily to the summation of axial forces, we now have an external moment to account for, the one from the moment diagram at that particular location where the beam was sliced, hence ΣMz=0 ; ✁ ✛+ve;Mbracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright Ext.=−integraldisplay AσxydA bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Int.(12.22) wheredAis an diﬀerential area a distance yfrom the neutral axis. 79Substituting Eq. 12.18 M=−integraldisplay AσxydA σx=−Eκy  M=κEintegraldisplay Ay2dA (12.23) 80We now pause and deﬁne the section moment of inertia with respect to the zaxis as Idef=integraldisplay Ay2dA (12.24) and section modulus as Sdef=I c (12.25) 12.4.4 Beam Formula 81We now have the ingredients in place to derive one of the most important equations in structures, the beam formula. This formula will be extensively used for designof structural components. 82We merely substitute Eq. 12.24 into 12.23, M=κEintegraldisplay Ay2dA I=integraldisplay ay2dA   M EI=κ=1 ρ (12.26) which shows that the curvature of the longitudinal axis of a beam is proportional to the bending moment Mand inversely proportional to EIwhich we call ﬂexural rigidity . 83Finally, inserting Eq. 12.18above, we obtain σx=−Eκy κ=M EIbracerightBigg σx=−My I (12.27) Hence, for a positive y(above neutral axis), and a positive moment, we will have com- pressive stresses above the neutral axis. Victor Saouma Introduction to Continuum Mechanics Draft12–14 BEAM THEORY 84Alternatively, the maximum ﬁber stresses can be obtained by combining the preceding equation with Equation 12.25 σx=−M S (12.28) 12.4.5 Limitations of the Beam Theory 12.4.6 Example Example 12-3: Design Example A 20 ft long, uniformly loaded, beam is simply supported at one end, and rigidly connected at the other. The beam is composed of a steel tube with thickness t=0.25 in. Select the radius such that σmax≤18 ksi,a n d∆max≤L/360. 20’r 0.25’1 k/ft Solution: 1. Steel has E=2 9,000 ksi, and from above Mmax=wL2 8,∆max=wL4 185EI,a n dI=πr3t. 2. The maximum moment will be Mmax=wL2 8=(1) k/ft(20)2ft2 8=5 0 k.ft (12.29) 3. We next seek a relation between maximum deﬂection and radius ∆max=wL4 185EI I=πr3tbracerightBigg∆=wL4 185Eπr3t =(1) k/ft(20)4ft4(12)3in3/ft3 (185)(29,000) ksi(3.14)r3(0.25) in =65.65 r3(12.30) 4. Similarly for the stress σ=M S S=I r I=πr3t  σ=M πr2t =(50) k.ft(12) in/ft (3.14)r2(0.25) in =764 r2(12.31) 5. We now set those two values equal to their respective maximum ∆max=L 360=(20) ft(12) in/ft 360=0.67 in=65.65 r3⇒r=3radicalBigg 65.65 0.67=4.61 in(12.32-a) σmax=( 1 8) ksi=764 r2⇒r=radicalBigg 764 18= 6.51 in (12.32-b) Victor Saouma Introduction to Continuum Mechanics Draft12.4Beam Theory 12–15 Victor Saouma Introduction to Continuum Mechanics Draft12–16 BEAM THEORY Victor Saouma Introduction to Continuum Mechanics Draft Chapter 13 VARIATIONAL METHODS Abridged section from author’s lecture notes in finite elements. 20Variational methods provide a powerful method to solve complex problems in contin- uum mechanics (and other ﬁelds as well). 21As shown in Appendix C, there is a duality between the strong form ,i nw h i c ha diﬀerential equation (or Euler’s equation) is exactly satisﬁed at every point (such as inFinite Diﬀerences ), and the weak form where the equation is satisﬁed in an averaged sense (as in ﬁnite elements ). 22Since only few problems in continuum mechanics can be solved analytically, we often have to use numerical techniques, Finite Elements being one of the most powerful andﬂexible one. 23At the core ofthe ﬁnite element formulation arethe variationalformulations (or energy based methods) which will be discussed in this chapter. 24For illustrative examples, we shall use beams, but the methods is obviously applicable to 3D continuum. 13.1 Preliminary Deﬁnitions 25Work is deﬁned as the product of a force and displacement Wdef=integraldisplayb aF.ds (13.1-a) dW=Fxdx+Fydy (13.1-b) 26Energy is a quantity representing the ability or capacity to perform work. 27The change in energy is proportional to the amount of work performed. Since only the change of energy is involved, any datum can be used as a basis for measure of energy.Hence energy is neither created nor consumed. 28The ﬁrst principle of thermodynamics (Eq. 6.44), states Draft13–2 VARIATIONAL METHODS U0U0U* 0 U* 0 A AA Aσ σ ε ε Nonlinear Linear Figure 13.1: *Strain Energy and Complementary Strain Energy The time-rate of change of the total energy (i.e., sum of the kinetic energy and the internal energy) is equal to the sum of the rate of work done by the externalforces and the change of heat content per unit time: d dt(K+U)=We+H (13.2) whereKis the kinetic energy, Uthe internal strain energy, Wthe external work, and H the heat input to the system. 29For an adiabatic system (no heat exchange) and if loads are applied in a quasi static manner (no kinetic energy), the above relation simpliﬁes to: We=U (13.3) 13.1.1 Internal Strain Energy 30Thestrain energy density of an arbitrary material is deﬁned as, Fig. 13.1 U0def=integraldisplayε 0σ:dε (13.4) 31Thecomplementary strain energy density is deﬁned U∗ 0def=integraldisplayσ 0ε:dσ (13.5) 32The strain energy itself is equal to Udef=integraldisplay ΩU0dΩ (13.6) U∗def=integraldisplay ΩU∗ 0dΩ (13.7) Victor Saouma Introduction to Continuum Mechanics Draft13.1 Preliminary Deﬁnitions 13–3 33To obtain a general form of the internal strain energy, we ﬁrst deﬁne a stress-strain relationship accounting for both initial strains and stresses σ=D:(ε−ε0)+σ0 (13.8) whereDis the constitutive matrix (Hooke’s Law); /epsilon1is the strain vector due to the displacements u;/epsilon10is the initial strain vector; σ0is the initial stress vector; and σis the stress vector. 34The initial strains and stresses are the result of conditions such as heating or cooling of a system or the presence of pore pressures in a system. 35The strain energy Ufor a linear elastic system is obtained by substituting σ=D:ε (13.9) with Eq. 13.4 and 13.8 U=1 2integraldisplay ΩεT:D:εdΩ−integraldisplay ΩεT:D:ε0dΩ+integraldisplay ΩεT:σ0dΩ (13.10) w h e r eΩi st h ev o l u m eo ft h es y s t e m . 36Considering uniaxial stresses , in the absence of initial strains and stresses, and for linear elastic systems , Eq. 13.10 reduces to U=1 2integraldisplay ΩεEεbracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright σdΩ (13.11) 37When this relation is applied to various one dimensional structural elements it leads to Axial Members: U=integraldisplay Ωεσ 2dΩ σ=P A ε=P AE dΩ=Adx   U=1 2integraldisplayL 0P2 AEdx (13.12) Flexural Members: U=1 2integraldisplay ΩεEεbracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright σ σx=Mzy Iz ε=Mzy EIz dΩ=dAdxintegraldisplay Ay2dA=Iz   U=1 2integraldisplayL 0M2 EIzdx (13.13) Victor Saouma Introduction to Continuum Mechanics Draft13–4VARIATIONAL METHODS 13.1.2 External Work 38External work Wperformed by the applied loads on an arbitrary system is deﬁned as Wedef=integraldisplay ΩuT·bdΩ+integraldisplay ΓtuT·ˆtdΓ (13.14) wherebis the body force vector; ˆtis the applied surface traction vector; and Γ tis that portion of the boundary where ˆtis applied, and uis the displacement. 39For point loads and moments, the external work is We=integraldisplay∆f 0Pd∆+integraldisplayθf 0Mdθ (13.15) 40Forlinear elastic systems ,(P=K∆) we have for point loads P=K∆ We=integraldisplay∆f 0Pd∆  We=Kintegraldisplay∆f 0∆d∆=1 2K∆2 f (13.16) When this last equation is combined with Pf=K∆fwe obtain We=1 2Pf∆f (13.17) whereKis thestiﬀness of the structure. 41Similarly for an applied moment we have We=1 2Mfθf (13.18) 13.1.3 Virtual Work 42Wedeﬁnethe virtual work done by the load on a body during a small, admissible (continuous and satisfying the boundary conditions) change in displacements. Internal Virtual Work δWidef=−integraldisplay Ωσ:δεdΩ (13.19) External Virtual Work δWedef=integraldisplay Γtˆt·δudΓ+integraldisplay Ωb·δudΩ (13.20) where all the terms have been previously deﬁned and bis the body force vector. 43Note that the virtual quantity (displacement or force) is one that we will approxi- mate/guess as long as it meets some admissibility requirements. Victor Saouma Introduction to Continuum Mechanics Draft13.1 Preliminary Deﬁnitions 13–5 13.1.3.1 Internal Virtual Work 44Next we shall derive a displacement based expression of δUfor each type of one di- mensional structural member. It should be noted that the Virtual Force method would yield analogous ones but based on forces rather than displacements. 45Two sets of solutions will be given, the ﬁrst one is independent of the material stress strain relations, and the other assumes a linear elastic stress strain relation. Elastic Systems In this set of formulation, we derive expressions of the virtual strain energies which are independent of the material constitutive laws. Thus δUwill be left in terms of forces and displacements. Axial Members: δU=integraldisplayL 0σδεdΩ dΩ=Adx   δU=AintegraldisplayL 0σδεdx (13.21) Flexural Members: δU=integraldisplay σxδεxdΩ M=integraldisplay AσxydA⇒M y=integraldisplay AσxdA δφ=δε y⇒δφy=δε dΩ=integraldisplayL 0integraldisplay AdAdx   δU=integraldisplayL 0Mδφdx (13.22) Linear Elastic Systems Should we have a linear elastic material ( σ=Eε) then: Axial Members: δU=integraldisplay σδεdΩ σx=Eεx=Edu dx δε=d(δu) dx dΩ=Adx   δU=integraldisplayL 0Edu dxbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright “σ/prime/primed(δu) dxbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright “δε/prime/primeAdxbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright dΩ (13.23) Flexural Members: δU=integraldisplay σxδεxdΩ σx=My Iz M=d2v dx2EIzbracerightBigg σx=d2v dx2bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright κEy δεx=δσx E=d2(δv) dx2y dΩ=dAdx  δU=integraldisplayL 0integraldisplay Ad2v dx2Eyd2(δv) dx2ydAdx (13.24) or: Eq. 13.24integraldisplay Ay2dA=Iz   δU=integraldisplayL 0EIzd2v dx2bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright “σ/prime/primed2(δv) dx2bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright “δε/prime/primedx (13.25) Victor Saouma Introduction to Continuum Mechanics Draft13–6 VARIATIONAL METHODS 13.1.3.2 External Virtual Work δW 46For concentrated forces (and moments): δW=integraldisplay δ∆qdx+summationdisplay i(δ∆i)Pi+summationdisplay i(δθi)Mi (13.26) where:δ∆i= virtual displacement. 13.1.4 Complementary Virtual Work 47We deﬁne the complementary virtual work done by the load on a body during a small, admissible (continuous and satisfying the boundary conditions) change in displacements. Complementary Internal Virtual Work δW∗ idef=−integraldisplay Ωε:δσdΩ (13.27) Complementary External Virtual Work δW∗ edef=integraldisplay Γuˆu·δtdΓ (13.28) 13.1.5 Potential Energy 48The potential of external work Win an arbitrary system is deﬁned as Wedef=integraldisplay ΩuT·bdΩ+integraldisplay ΓtuT·ˆtdΓ+u·P (13.29) whereuare the displacements, bis the body force vector; ˆtis the applied surface traction vector; Γ tis that portion of the boundary where ˆtis applied, and Pare the applied nodal forces. 49Note that the potential of the external work ( W) is diﬀerent from the external work itself (W) 50The potential energy of a system is deﬁned as Πdef=U−We (13.30) =integraldisplay ΩU0dΩ−parenleftbiggintegraldisplay Ωu·bdΩ+integraldisplay Γtu·ˆtdΓ+u·Pparenrightbigg (13.31) 51Note that in the potential the full load is always acting, and through the displacements of its points of application it does work but loses an equivalent amount of potential, this explains the negative sign. 13.2 Principle of Virtual Work and Complementary Virtual Work 52The principles of Virtual Work and Complementary Virtual Work relate forcesystems which satisfy the requirements of equilibrium ,a n ddeformation systems which satisfy the Victor Saouma Introduction to Continuum Mechanics Draft13.2 Principle of Virtual Work and Complementary Virtual Work 13–7 requirement of compatibility : 1. Inanyapplicationtheforcesystem couldeitherbetheactualsetof external loadsdp or somevirtualforce system which happens to satisfy the condition of equilibrium δ p. This set of external forces will induce internal actual forces dσor internal hypothetical forces δ σcompatible with the externally applied load. 2. Similarly the deformation could consist of either the actual joint deﬂections duand compatible internal deformations dεof the structure, or some hypothetical external and internal deformation δ uandδ εwhich satisfy the conditions of compatibility . 53Thus we may have 2 possible combinations, Table 13.1: where: dcorresponds to the Force Deformation Formulation External Internal External Internal 1 δ p δ σ du dε δU∗ 2 dp dσ δ u δ ε δU Table 13.1: Possible Combinations of Real and Hypothetical Formulations actual, and δ(with an overbar) to the hypothetical values. 13.2.1 Principle of Virtual Work 54Derivation of the principle of virtual work starts with the assumption of that forces are in equilibrium and satisfaction of the static boundary conditions. 55The Equation of equilibrium (Eq. 6.26) which is rewritten as ∂σxx ∂x+∂τxy ∂y+bx= 0 (13.32) ∂σyy ∂y+∂τxy ∂x+by= 0 (13.33) wherebrepresenting the body force. In matrix form, this can be rewritten as bracketleftBigg∂ ∂x0∂ ∂y 0∂ ∂y∂ ∂xbracketrightBigg  σxx σyy τxy  +braceleftBigg bx bybracerightBigg = 0 (13.34) or LTσ+b=0 (13.35) Note that this equation can be generalized to 3D. 56The surface Γ of the solid can be decomposed into two parts Γ tand Γuwhere tractions and displacements are respectively speciﬁed. Γ=Γ t+Γu (13.36-a) t=ˆton ΓtNatural B.C. (13.36-b) u=ˆuon ΓuEssential B.C. (13.36-c) Victor Saouma Introduction to Continuum Mechanics Draft13–8 VARIATIONAL METHODS Figure 13.2: Tapered Cantilivered Beam Analysed by the Vitual Displacement Method Equations 13.35 and 13.36-b constitute a statically admissible stress ﬁeld. 57Theprinciple of virtual work (or more speciﬁcally of virtual displacement) can be stated as A deformable system is in equilibrium if the sum of the external virtual work and the internal virtual work is zero for virtual displacements δuwhich are kinematically admissible. The major governing equations are summarized integraldisplay ΩδεT:σdΩ bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright −δWi−integraldisplay ΩδuT·bdΩ−integraldisplay ΓtδuT·ˆtdΓ bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright −δWe= 0 (13.37) δε=L:δuin Ω (13.38) δu=0 o n Γ u(13.39) 58Note that the principle is independent of material properties, and that the primary unknowns are the displacements. Example 13-1: Tapered Cantiliver Beam, Virtual Displacement Analyse the problem shown in Fig. 13.2, by the virtual displacement method. Solution: 1. For this ﬂexural problem, we must apply the expression of the virtual internal strain energy as derived for beams in Eq. 13.25. And the solutions must be expressed in terms of the displacements which in turn must satisfy the essential boundary conditions. Theapproximate solutions proposed to this problem are v=parenleftbigg 1−cosπx 2lparenrightbigg v2 (13.40) v=bracketleftBigg 3parenleftbiggx Lparenrightbigg2 −2parenleftbiggx Lparenrightbigg3bracketrightBigg v2 (13.41) Victor Saouma Introduction to Continuum Mechanics Draft13.2 Principle of Virtual Work and Complementary Virtual Work 13–9 2. These equations do indeed satisfy the essential B.C. (i.e kinematic), but for them to also satisfy equilibrium they must satisfy the principle of virtual work. 3. Using the virtual displacement method we evaluate the displacements v2from three diﬀerent combination of virtual and actual displacement: Solution Total Virtual 1 Eqn. 13.40 Eqn. 13.41 2 Eqn. 13.40 Eqn. 13.40 3 Eqn. 13.41 Eqn. 13.41 Where actual and virtual values for the two assumed displacement ﬁelds are givenbelow. Trigonometric (Eqn. 13.40) Polynomial (Eqn. 13.41) v parenleftBig 1−cosπx 2lparenrightBig v2 bracketleftbigg 3parenleftBig x LparenrightBig2−2parenleftBig x LparenrightBig3bracketrightbigg v2 δv parenleftBig 1−cosπx 2lparenrightBig δv2 bracketleftbigg 3parenleftBig x LparenrightBig2−2parenleftBig x LparenrightBig3bracketrightbigg δv2 v/prime/prime π2 4L2cosπx 2lv2 parenleftBig 6 L2−12x L3parenrightBig v2 δv/prime/prime π2 4L2cosπx 2lδv2 bracketleftBig 6 L2−12x L3bracketrightBig δv2 δU=integraldisplayL 0δv/prime/primeEIzv/prime/primedx (13.42) δW=P2δv2 (13.43) Solution 1: δU=integraldisplayL 0π2 4L2cosparenleftbiggπx 2lparenrightbigg v2parenleftbigg6 L2−12x L3parenrightbigg δv2EI1parenleftbigg 1−x 2Lparenrightbigg dx =3πEI1 2L3bracketleftbigg 1−10 π+16 π2bracketrightbigg v2δv2 =P2δv2 (13.44) which yields: v2=P2L3 2.648EI1(13.45) Solution 2: δU=integraldisplayL 0π4 16L4cos2parenleftbiggπx 2lparenrightbigg v2δv2EI1parenleftbigg 1−x 2lparenrightbigg dx =π4EI1 32L3parenleftbigg3 4+1 π2parenrightbigg v2δv2 =P2δv2 (13.46) which yields: v2=P2L3 2.57EI1(13.47) Victor Saouma Introduction to Continuum Mechanics Draft13–10 VARIATIONAL METHODS Solution 3: δU=integraldisplayL 0parenleftbigg6 L2−12x L3parenrightbigg2parenleftbigg 1−x 2lparenrightbigg EI1δv2v2dx =9EI L3v2δv2 =P2δv2 (13.48) which yields: v2=P2L3 9EI(13.49) 13.2.2 Principle of Complementary Virtual Work 59Derivation of the principle of complementary virtual work starts from the assumption of akinematicaly admissible displacements and satisfaction of the essential boundary conditions. 60Whereas we have previously used the vector notation for the principle of virtual work, we will now use the tensor notation for this derivation. 61The kinematic condition (strain-displacement): εij=1 2(ui,j+uj,i) (13.50) 62The essential boundary conditions are expressed as ui=ˆuon Γu (13.51) 63Theprinciple of virtual complementary work (or more speciﬁcally of virtual force) which can be stated as A deformable system satisﬁes all kinematical requirements if the sum of the external complementary virtual work and the internal complementary virtual work is zero for all statically admissible virtual stresses δσij. The major governing equations are summarized integraldisplay ΩεijδσijdΩ bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright −δW∗ i−integraldisplay ΓuˆuiδtidΓ bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright δW∗e= 0 (13.52) δσij,j= 0 in Ω (13.53) δti=0 o n Γ t(13.54) 64Note that the principle is independent of material properties, and that the primary unknowns are the stresses. Victor Saouma Introduction to Continuum Mechanics Draft13.2 Principle of Virtual Work and Complementary Virtual Work 13–11 Figure 13.3: Tapered Cantilevered Beam Analysed by the Virtual Force Method 65Expressions for the complimentary virtual work in beams are given in Table 13.3 Example 13-2: Tapered Cantilivered Beam; Virtual Force “Exact” solution of previous problem using principle of virtual work with virtual force. integraldisplayL 0δMM EIzdx bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright Internal=δP∆bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright External (13.55) Note: This represents the internal virtual strain energy and external virtual work written in terms of forcesand should be compared with the similar expression derived in Eq. 13.25 written in terms of displacements: δU∗=integraldisplayL 0EIzd2v dx2bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright σd2(δv) dx2bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright δεdx (13.56) Here:δMandδPare the virtual forces, andM EIzand ∆ are the actual displacements. See Fig. 13.3 If δP=1 ,t h e n δM=xandM=P2xor: (1)∆ =integraldisplayL 0xP2x EI1(.5+x L)dx =P2 EI1integraldisplayL 0x2 L+x 2ldx =P22L EI1integraldisplayL 0x2 L+xdx (13.57) FromMathematica we note that: integraldisplay0 0x2 a+bx=1 b3bracketleftbigg1 2(a+bx)2−2a(a+bx)+a2ln(a+bx)bracketrightbigg (13.58) Thus substituting a=Landb= 1 into Eqn. 13.58, we obtain: ∆=2P2L EI1bracketleftbigg1 2(L+x)2−2L(L+x)+L2ln(L+x)bracketrightbigg |L 0 Victor Saouma Introduction to Continuum Mechanics Draft13–12 VARIATIONAL METHODS =2P2L EI1bracketleftBigg 2L2−4L2+L2ln2L−L2 2+2L2+L2logLbracketrightBigg =2P2L EI1bracketleftbigg L2(ln2−1 2)bracketrightbigg =P2L3 2.5887EI1(13.59) Similarly: θ=integraldisplayL 0M(1) EI1parenleftBig .5+x LparenrightBig=2ML EI1integraldisplayL 01 L+x=2ML EI1ln(L+x)|L 0 =2ML EI1(ln2L−lnL)=2ML EI1ln2 =ML .721EI1(13.60) 13.3 Potential Energy 13.3.1 Derivation 66From section ??,i fU0is a potential function, we take its diﬀerential dU0=∂U0 ∂εijdεij (13.61-a) dU∗ 0=∂U0 ∂σijdσij (13.61-b) 67However, from Eq. 13.4 U0=integraldisplayεij 0σijdεij (13.62-a) dU0=σijdεij (13.62-b) thus, ∂U0 ∂εij=σij(13.63) ∂U∗ 0 ∂σij=εij(13.64) 68We now deﬁne the variation of the strain energy density at a point1 δU0=∂U ∂εijδεij=σijδεij (13.65) 69Applying the principle of virtual work, Eq. 13.37, it can be shown that 1Note that the variation of strain energy density is, δU0=σijδεij, and the variation of the strain energy itself is δU=integraltext ΩδU0dΩ. Victor Saouma Introduction to Continuum Mechanics Draft13.3 Potential Energy 13–13 k= 500 lbf/in 100 lbf mg= Figure 13.4: Single DOF Example for Potential Energy δΠ = 0 (13.66) Πdef=U−We (13.67) =integraldisplay ΩU0dΩ−parenleftbiggintegraldisplay Ωu·bdΩ+integraldisplay Γtu·ˆtdΓ+u·Pparenrightbigg (13.68) 70We have thus derived the principle of stationary value of the potential energy: Of all kinematically admissible deformations (displacements satisfying the es- sential boundary conditions), the actual deformations (those which correspondto stresses which satisfy equilibrium) are the ones for which the total potential energy assumes a stationary value. 71For problems involving multiple degrees of freedom, it results from calculus that δΠ=∂Π ∂∆1δ∆1+∂Π ∂∆2δ∆2+...+∂Π ∂∆nδ∆n (13.69) 72It can be shown that the minimum potential energy yields a lower bound prediction of displacements. 73As an illustrative example (adapted from Willam, 1987), let us consider the single dof system shown in Fig. 13.4. The strain energy Uand potential of the external work W are given by U=1 2u(Ku) = 250u2(13.70-a) We=mgu= 100u (13.70-b) Thus the total potential energy is given by Π = 250u2−100u (13.71) Victor Saouma Introduction to Continuum Mechanics Draft13–14VARIATIONAL METHODS 0.00 0.10 0.20 0.30 Displacement [in]−40.0−20.00.020.0Energy [lbf −in]Potential Energy of Single DOF Structure Total Potential Energy Strain Energy External Work Figure 13.5: Graphical Representation of the Potential Energy and will be stationary for ∂Π=dΠ du=0⇒500u−100 = 0⇒ u=0.2in (13.72) Substituting, this would yield U= 250(0 .2)2=1 0 l b f - i n W= 100(0 .2) = 20 lbf-in Π=1 0−20 =−10 lbf-in(13.73) Fig. 13.5 illustrates the two components of the potential energy. 13.3.2 Rayleigh-Ritz Method 74Continuous systems have inﬁnite number of degrees of freedom, those are the dis- placements at every point within the structure. Their behavior can be described by the Euler Equation, or the partial diﬀerential equation of equilibrium. However, only thesimplest problems have an exact solution which (satisﬁes equilibrium, and the boundaryconditions). 75Anapproximate method of solution is the Rayleigh-Ritz method which is based on the principle of virtual displacements. In this method we approximate the displacement ﬁeld by a function u1≈nsummationdisplay i=1c1 iφ1i+φ1 0 (13.74-a) u2≈nsummationdisplay i=1c2 iφ2i+φ2 0 (13.74-b) Victor Saouma Introduction to Continuum Mechanics Draft13.3 Potential Energy 13–15 u3≈nsummationdisplay i=1c3 iφ3i+φ3 0 (13.74-c) wherecj idenote undetermined parameters, and φare appropriate functions of positions. 76φshould satisfy three conditions 1. Be continuous. 2. Must be admissible , i.e. satisfy the essential boundary conditions (the natural boundary conditions are included already in the variational statement. However,ifφalso 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 φis a polynomial or trigonometric function. 77We determine the parameters cj iby requiring that the principle of virtual work for arbitrary variations δcj i.o r δΠ(u1,u2,u3)=nsummationdisplay i=1parenleftBigg∂Π ∂c1 iδc1 i+∂Π ∂c2 iδc2 i+∂Π ∂c3 iδc3 iparenrightBigg = 0 (13.75) for arbitrary and independent variations of δc1 i,δc2 i,a n dδc3 i, thus it follows that ∂Π ∂cj i=0i=1,2,···,n;j=1,2,3 (13.76) Thus we obtain a total of 3 nlinearly 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. 78Some general observations 1.cj icaneitherbeasetofcoeﬃcients withnophysical meanings, orvariablesassociated with nodal generalized displacements (such as deﬂection or displacement). 2. If the coordinate functions φsatisfy the above requirements, then the solution con- verges to the exact one if nincreases. 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 generally do not satisfy the equations of equilibrium. Victor Saouma Introduction to Continuum Mechanics Draft13–16 VARIATIONAL METHODS Figure 13.6: Uniformly Loaded Simply Supported Beam Analyzed by the Rayleigh-Ritz Method 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 thedisplacements obtainedfromtheRitzmethodconverge totheexactones frombelow. Example 13-3: Uniformly Loaded Simply Supported Beam; Polynomial Approximation For the uniformly loaded beam shown in Fig. 13.6let us assume a solution given by the following inﬁnite series: v=a 1x(L−x)+a2x2(L−x)2+... (13.77) for this particular solution, let us retain only the ﬁrst term: v=a1x(L−x) (13.78) We observe that: 1. Contrarily to the previous example problem the geometric B.C. are immediately satisﬁed at both x=0a n dx=L. 2. We can keep vin terms of a1and take∂Π ∂a1=0( I fw eh a dl e f t vin terms of a1and a2we should then take both∂Π ∂a1=0 ,a n d∂Π ∂a2=0) . 3. Or we can solve for a1in terms of vmax(@x=L 2)a n dt a k e∂Π ∂vmax=0 . Π=U−W=integraldisplayL oM2 2EIzdx−integraldisplayL 0wv(x)dx (13.79) Victor Saouma Introduction to Continuum Mechanics Draft13.4Summary 13–17 Recalling that:M EIz=d2v dx2, the above simpliﬁes to: Π=integraldisplayL 0 EIz 2parenleftBiggd2v dx2parenrightBigg2 −wv(x) dx (13.80) =integraldisplayL 0bracketleftbiggEIz 2(−2a1)2−a1wx(L−x)bracketrightbigg dx =EIz 24a2 1L−a1wL3 2+a1wL3 3 =2a2 1EIzL−a1wL3 6(13.81) If we now take∂Π ∂a1= 0, we would obtain: 4a1EIzl−wL3 6=0 a1=wL2 24EIz(13.82) Having solved the displacement ﬁeld in terms of a1, we now determine vmaxatL 2: v=wL4 24EIzbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright a1parenleftBiggx L−x2 L2parenrightBigg =wL4 96EIz(13.83) This is to be compared with the exact value of vexact max=5 384wL4 EIz=wL4 76.8EIzwhich constitutes ≈17% error. Note: If two terms were retained, then we would have obtained: a1=wL2 24EIz&a2= w 24EIzandvmaxwould be equal to vexact max.( W h y ? ) 13.4 Summary 79Summary of Virtual work methods, Table 13.2. Starts with Ends with In terms of virtual Solve for Virtual Work U KAD SAS Displacement/strains Displacement Complimentary Virtual Work U∗ SAS KAD Forces/Stresses Displacement KAD: Kinematically Admissible Dispacements SAS: Statically Admissible Stresses Table 13.2: Comparison of Virtual Work and Complementary Virtual Work 80A summary of the various methods introduced in this chapter is shown in Fig. 13.7. Victor Saouma Introduction to Continuum Mechanics Draft13–18 VARIATIONAL METHODS Ω Γ ∇σ+ρb=0 t−hatwidet=0 Γt U0def=integraltextε 0σ:dε ❄ ✻ Gauss δε−D:δu=0 δu=0 Γu ❄ Principle of Virtual Work integraltext ΩδεT:σdΩ−integraltext ΩδuT·bdΩ−integraltext ΓtδuT·hatwidetdΓ=0 δWi−δWe=0 Principle of Stationary Potential Energy δΠ=0 Πdef=U−We Π=integraltext ΩU0dΩ−(integraltext ΩuibidΩ+integraltext ΓtuihatwidetidΓ) ❄ Rayleigh-Ritz uj≈nsummationdisplay i=1cj iφji+φj 0 ∂Π ∂cj i=0i=1,2,···,n;j=1,2,3 εij−1 2(ui,j+uj,i)=0 ui−hatwideu=0 Γu U∗ 0def=integraltextσ 0ε:σ ❄ ✻ Gauss δσij,j=0 δti=0 Γt ❄ Principle of Complementary Virtual Workintegraltext ΩεijδσijdΩ−integraltext ΓuhatwideuiδtidΓ=0 δW∗ i−δW∗ e=0 ❄Natural B.C. ❄ ❄ Essential B.C. ❄ ❄ Figure 13.7: Summary of Variational Methods Victor Saouma Introduction to Continuum Mechanics Draft13.4Summary 13–19 Kinematically Admissible Displacements Displacements satisfy the kinematic equations and the the kinematic boundary conditions Principle of Stationary Complementary Energy Principle of Complementary Virtual Work Principle of Virtual Work Principle of Stationary Potential Energy Statically Admissible Stresses Stresses satisfy the equilibrium conditions and the static boundary conditions ✻ ✻ ❄ ❄ Figure 13.8: Duality of Variational Principles 81The duality between the two variational principles is highlighted by Fig. 13.8, where beginning with kinematically admissible displacements, the principle of virtual work pro- vides statically admissible solutions. Similarly, for statically admissible stresses, theprinciple of complementary virtual work leads to kinematically admissible solutions. 82Finally, Table 13.3 summarizes some of the major equations associated with one di- mensional rod elements. Victor Saouma Introduction to Continuum Mechanics Draft13–20 VARIATIONAL METHODS U Virtual Displacement δU Virtual Force δU∗ General Linear General Linear Axial 1 2integraldisplayL 0P2 AEdx integraldisplayL 0σδεdx integraldisplayL0 Edu dxbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright σd(δu) dxbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright δεAdxbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright dΩ integraldisplayL 0δσεdx integraldisplayL0 δPbracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright δσP AEbracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright εdx Flexure 1 2integraldisplayL 0M2 EIzdx integraldisplayL0 Mδφdx integraldisplayL0 EIzd2v dx2bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright σd2(δv) dx2bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright δεdx integraldisplayL 0δMφdx integraldisplayL0 δMbracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright δσM EIzbracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright εdx W Virtual Displacement δW Virtual Force δW∗ P Σi1 2Pi∆i ΣiPiδ∆i ΣiδPi∆i M Σi1 2Miθi ΣiMiδθi ΣiδMiθi w integraldisplayL 0w(x)v(x)dx integraldisplayL 0w(x)δv(x)dx integraldisplayL 0δw(x)v(x)dx Table 13.3: Summary of Variational Terms Associated with One Dimensional Elements Victor Saouma Introduction to Continuum Mechanics Draft Chapter 14 INELASTICITY (incomplete) t Relaxation∆F t Creep Figure 14.1: test Draft–2 INELASTICITY (incomplete) σ ε tσε t Elastic Perfectly PlasticPerfectly Elastic ViscoelasticRelaxation Creep CreepStrain Hardening Rigid Perfectly Plasticσ εσ σ ε tσε tσ ε tσε t ε Elastoplastic HardeingCreep RelaxationRelaxation Figure 14.2: mod1 0Ε ησ σ ε Figure 14.3: v-kv Victor Saouma Introduction to Continuum Mechanics Draft–3 0σ E ση ε Figure 14.4: visﬂ E i EEEη ησσ1 1 n nηi Figure 14.5: visﬂ s−σ < σ < σ ss−ε < ε < εσ=Ε ε .σ=ηε ε.1/Nσ=λε.σ ε. sε σ 0σ 0 σ εσ εStress Threshold Strain ThresholdNonlinear ViscosityLinear VisosityLinear Elasticity σ λ σ 0ση 00σ E σ Figure 14.6: comp SEσ 0σ σ Figure 14.7: epp Victor Saouma Introduction to Continuum Mechanics Draft–4INELASTICITY (incomplete) Eεεpi σσSi Sj mEi EjE σ σ 0 Figure 14.8: ehs Victor Saouma Introduction to Continuum Mechanics Draft Appendix A SHEAR, MOMENT and DEFLECTION DIAGRAMS for BEAMS Adapted from [ ?]1) Simple Beam; uniform Load x V VL / 2 L / 2 M max. MomentShearRRw LL R=V Vx=wparenleftbiggL 2−xparenrightbigg at center Mmax=wL2 8 Mx=wx 2(L−x) ∆max=5 384wL4 EI ∆x=wx 24EI(L3−2Lx2+x3) 2) Simple Beam; Unsymmetric Triangular Load DraftA–2 SHEAR, MOMENT and DEFLECTION DIAGRAMS for BEAMS R1=V1=W 3 MaxR2=V2=2W 3 Vx =W 3−Wx2 L2 atx=.577LMmax=.1283WL Mx=Wx 3L2(L2−x2) atx=.5193L∆max=.01304WL3 EI ∆x=Wx3 180EIL2(3x4−10L2x2+7L4) 3) Simple Beam; Symmetric Triangular Load R=V=W 2 forx<L 2Vx=W 2L2(L2−4x2) at center Mmax=WL 6 forx<L 2Mx=WxparenleftBigg1 2−2 3x2 L2parenrightBigg forx<L 2∆x=Wx 480EIL2(5L2−4x2)2 ∆max=WL3 60EI 4) Simple Beam; Uniform Load Partially Distributed Max when a<c R1=V1=wb 2L(2c+b) Max when a>c R2=V2=wb 2L(2a+b) whena<x<a +bVx =R1−w(x−a) whenx<aM x=R1x whena<x<a +bMx=R1x−w 2(x−a)2 whena+b<x M x=R2(L−x) atx=a+R1 wMmax=R1parenleftbigg a+R1 2wparenrightbigg 5) Simple Beam; Concentrated Load at Center Victor Saouma Introduction to Continuum Mechanics DraftA–3 maxR1=V1=wa 2L(2L−a) R=V=2P atx=L 2Mmax=PL 4 whenx<L 2Mx=Px 2 whenx<L 2∆x=Px 48EI(3L2−4x2) atx=L 2∆max=PL3 48EI 6) Simple Beam; Concentrated Load at Any Point max when a<b R1=V1=Pb L max when a>b R2=V2=Pa L atx=aMmax=Pab L whenx<aM x=Pbx L atx=a∆a=Pa2b2 3EIL whenx<a∆x=Pbx 6EIL(L2−b2−x2) atx=radicalBig a(a+2b) 3&a>b∆max=Pab(a+2b)radicalBig 3a(a+2b) 27EIL 7) Simple Beam; Two Equally Concentrated Symmetric Loads R=V=P Mmax=Pa ∆max=Pa 24EI(3L2−4a2) whenx<a∆x=Px 6EI(3La−3a2−x2) whena<x<L −a∆x=Pa 6EI(3Lx−3x2−a2) 8) Simple Beam; Two Equally Concentrated Unsymmetric Loads Victor Saouma Introduction to Continuum Mechanics DraftA–4SHEAR, MOMENT and DEFLECTION DIAGRAMS for BEAMS max when a<b R1=V1=P L(L−a+b) max when b<aR2=V2=P L(L−b+a) whena<x<L −bVx =P L(b−a) max when b<aM1=R1a max when a<b M2=R2b whenx<aM x=R1x whena<x<L −bMx=R1x−P(x−a) 9) Cantilevered Beam, Uniform Load R1=V1=3 8wL R2=V2=5 8wL Vx =R1−wx Mmax=wL2 8 atx=3 8LM1=9 128wL2 Mx=R1x−wx2 2 ∆x=wx 48EI(L3−3Lx+2x3) atx=.4215L∆max=wL4 185EI 10) Propped Cantilever, Concentrated Load at Center R1=V1=5P 16 R2=V2=11P 16 atx=LMmax=3PL 16 whenx<L 2Mx=5Px 16 whenL 2<x M x=PparenleftbiggL 2−11x 16parenrightbigg atx=.4472L∆max=.009317PL3 EI 11) Propped Cantilever; Concentrated Load Victor Saouma Introduction to Continuum Mechanics DraftA–5 R1=V1=Pb2 2L3(a+2L) R2=V2=Pa 2L3(3L2−a2) atx=aM1=R1a atx=LM2=Pab 2L2(a+L) atx=a∆a=Pa2b3 12EIL3(3L+a) whena<.414Latx=LL2+a2 3L2−a2∆max=Pa 3EI(L2−a2)3 (3L2−a2)2 when.414L<aatx=LradicalBig a 2L+a∆max=Pab2 6EIradicalBigg a 2L+a2 12) Beam Fixed at Both Ends, Uniform Load R=V=wL 2 Vx=wparenleftbiggL 2−xparenrightbigg atx=0a n dx=LMmax=wL2 12 atx=L 2M =wL2 24 atx=L 2∆max=wL4 384EI ∆x=wx2 24EI(L−x)2 13) Beam Fixed at Both Ends; Concentrated Load R=V=P 2 atx=L 2Mmax=PL 8 whenx<L 2Mx=P 8(4x−L) atx=L 2∆max=PL3 192EI whenx<L 2∆x=Px2 48EI(3L−4x) 14) Cantilever Beam; Triangular Unsymmetric Load Victor Saouma Introduction to Continuum Mechanics DraftA–6 SHEAR, MOMENT and DEFLECTION DIAGRAMS for BEAMS R=V=8 3W Vx=Wx2 L2 atx=LMmax=WL 3 Mx=Wx2 3L2 ∆x=W 60EIL2(x5−5L2x+4L5) atx=0 ∆ max=WL3 15EI 15) Cantilever Beam; Uniform Load R=V=wL Vx=wx Mx=wx2 2 atx=LMmax=wL2 2 ∆x=w 24EI(x4−4L3x+3L4) atx=0 ∆ max=wL4 8EI 16) Cantilever Beam; Point Load R=V=P atx=LMmax=Pb whena<x M x=P(x−a) atx=0 ∆ max=Pb2 6EI(3L−b) atx=a∆a=Pb3 3EI whenx<a∆x=Pb2 6EI(3L−3x−b) whena<x∆x=P(L−x)2 6EI(3b−L+x) 17) Cantilever Beam; Point Load at Free End Victor Saouma Introduction to Continuum Mechanics DraftA–7 R=V=P atx=LMmax=PL Mx=Px atx=0 ∆ max=PL3 3EI ∆x=P 6EI(2L3−3L2x+x3) 18) Cantilever Beam; Concentrated Force and Moment at Free End R=V=P Mx=PparenleftbiggL 2−xparenrightbigg atx=0a n dx=LMmax=PL 2 atx=0 ∆ max=PL3 12EI ∆x=P(L−x)2 12EI((L+2x) Victor Saouma Introduction to Continuum Mechanics Draft Appendix B SECTION PROPERTIES Section properties for selected sections are shown in Table B.1. DraftB–2 SECTION PROPERTIES byx hY XA=bh x=b 2 y=h 2 Ix=bh3 12 Iy=hb3 12 b’ bh’hx yXY A=bh−b/primeh/prime x=b 2 y=h 2 Ix=bh3−b/primeh/prime3 12 Iy=hb3−h/primeb/prime3 12 h bY X ya A=h(a+b) 2 y=h(2a+b) 3(a+b) Ix=h3(a2+4ab+b2 36(a+b) h bx X yc YA=bh 2 x=b+c 3 y=h 3 Ix=bh3 36 Iy=bh 36(b2−bc+c2) XY r A=πr2=πd2 4 Ix=Iy=πr4 4=πd4 64 XY r tA=2πrt=πdt Ix=Iy=πr3t=πd3t 8 bb aaXY A=πab Ix=πab3 3 Iy=πba3 4 Table B.1: Section Properties Victor Saouma Introduction to Continuum Mechanics Draft Appendix C MATHEMATICAL PRELIMINARIES; Part IV VARIATIONAL METHODS Abridged section from author’s lecture notes in finite elements. C.1 Euler Equation 20The fundamental problem of the calculus of variation1is to ﬁnd a function u(x)s u c h that Π=integraldisplayb aF(x,u,u/prime)dx (3.1) is stationary. Or, δΠ=0 (3.2) whereδindicates the variation 21We deﬁne u(x) to be a function of xin the interval ( a,b), andFto be a known function (such as the energy density). 22We deﬁne the domainofa functionalas the collection ofadmissible functions belonging to a class of functions in function space rather than a region in coordinate space (as is the case for a function). 23We seek the function u(x) which extremizes Π. 24Letting ˜uto be a family of neighbouring paths of the extremizing function u(x)a n d we assume that at the end points x=a,bthey coincide. We deﬁne ˜ uas the sum of the extremizing path and some arbitrary variation, Fig. C.1. ˜u(x,ε)=u(x)+εη(x)=u(x)+δu(x) (3.3) 1Diﬀerentialcalculusinvolvesafunctionofoneormorevariable,whereasvariationalcalculusinvolvesafunctionofa function,orafunctional. DraftC–2 MATHEMATICAL PRELIMINARIES; Part IV VARIATIONAL METHODS u, u xx=b x=c x=aABC dxduu(x)u(x) Figure C.1: Variational and Diﬀerential Operators whereεis a small parameter, and δu(x)i st h evariation ofu(x) δu=˜u(x,ε)−u(x) (3.4-a) =εη(x) (3.4-b) andη(x) is twice diﬀerentiable, has undeﬁned amplitude, and η(a)=η(b)=0 . W e note that ˜ ucoincides with uifε=0 25The variationaloperator δand the diﬀerential calculus operator dhave clearly diﬀerent meanings. duis associated with a neighboring point at a distance dx, however δuis a smallarbitrary change in ufor a given x(there is no associated δx). 26For boundaries where uis speciﬁed, its variation must be zero, and it is arbitrary elsewhere. The variation δuofuis said to undergo a virtualchange. 27To solve the variational problem of extremizing Π, we consider Π(u+εη)=Φ (ε)=integraldisplayb aF(x,u+εη,u/prime+εη/prime)dx (3.5) 28Since ˜u→uasε→0, the necessary condition for Π to be an extremum is dΦ(ε) dεvextendsinglevextendsinglevextendsinglevextendsinglevextendsingle ε=0=0 ( 3 . 6 ) 29From Eq. 3.3 and applying the chain rule with ε=0 ,˜u=u,w eo b t a i n dΦ(ε) dεvextendsinglevextendsinglevextendsinglevextendsinglevextendsingle ε=0=integraldisplayb aparenleftBigg η∂F ∂u+η/prime∂F ∂u/primeparenrightBigg dx=0 ( 3 . 7 ) 30It can be shown (through integration by part and the fundamental lemma of the Victor Saouma Introduction to Continuum Mechanics DraftC.1 Euler Equation C–3 calculus of variation) that this would lead to ∂F ∂u−d dx∂F ∂u/prime=0 (3.8) 31This diﬀerential equation is called the Euler equation associated with Π and is a necessary condition for u(x) to extremize Π. 32Generalizing for a functional Π which depends on two ﬁeld variables, u=u(x,y)a n d v=v(x,y) Π=integraldisplayintegraldisplay F(x,y,u,v,u ,x,u,y,v,x,v,y,···,v,yy)dxdy (3.9) There would be as many Euler equations as dependent ﬁeld variables   ∂F ∂u−∂ ∂x∂F ∂u,x−∂ ∂y∂F ∂u,y+∂2 ∂x2∂F ∂u,xx+∂2 ∂x∂y∂F ∂u,xy+∂2 ∂y2∂F ∂u,yy=0 ∂F ∂v−∂ ∂x∂F ∂v,x−∂ ∂y∂F ∂v,y+∂2 ∂x2∂F ∂v,xx+∂2 ∂x∂y∂F ∂v,xy+∂2 ∂y2∂F ∂v,yy=0(3.10) 33We note that the Functional and the corresponding Euler Equations, Eq. 3.1 and 3.8, or Eq. 3.9 and 3.10 describe the same problem. 34The Euler equations usually correspond to the governing diﬀerential equation and are referred to as the strong form (or classical form). 35The functional is referred to as the weak form (or generalized solution). This clas- siﬁcation stems from the fact that equilibrium is enforced in an average sense over thebody (and the ﬁeld variable is diﬀerentiated mtimes in the weak form, and 2 mtimes in the strong form). 36Euler equations are diﬀerential equations which can not always be solved by exact methods. Analternativemethodconsists inbypassingtheEulerequationsandgodirectlyto the variational statement of the problem to the solution of the Euler equations. 37Finite Element formulation are based on the weak form, whereas the formulation of Finite Diﬀerences are based on the strong form. 38Finally, we still have to deﬁne δΠ δF=∂F ∂uδu+∂F ∂u/primeδu/prime δΠ=integraltextb aδFdxbracerightBigg δΠ=integraldisplayb aparenleftBigg∂F ∂uδu+∂F ∂u/primeδu/primeparenrightBigg dx (3.11) As above, integration by parts of the second term yields δΠ=integraldisplayb aδuparenleftBigg∂F ∂u−d dx∂F ∂u/primeparenrightBigg dx (3.12) 39We have just shown that ﬁnding the stationary value of Π by setting δΠ=0i s equivalent to ﬁnding the extremal value of Π by settingdΦ(ε) dεvextendsinglevextendsinglevextendsingle ε=0equal to zero. 40Similarly, it can be shown that as with second derivatives in calculus, the second vari- ationδ2Π can be used to characterize the extremum as either a minimum or maximum. Victor Saouma Introduction to Continuum Mechanics DraftC–4MATHEMATICAL PRELIMINARIES; Part IV VARIATIONAL METHODS 41Revisiting the integration by parts of the second term in Eq. 3.7, we obtain integraldisplayb aη/prime∂F ∂u/primedx=η∂F ∂u/primevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleb a−integraldisplayb aηd dx∂F ∂u/primedx (3.13) We note that 1. Derivation of the Euler equation required η(a)=η(b) = 0, thus this equation is a statement of the essential (or forced) boundary conditions, where u(a)=u(b)=0 . 2. If we left ηarbitrary, then it would have been necessary to use∂F ∂u/prime=0a tx=aand b. These are the naturalboundary conditions. 42For a problem with, one ﬁeld variable, in which the highest derivative in the governing diﬀerential equation is of order 2 m(or simply min the corresponding functional), then we have Essential (or Forced, or geometric) boundary conditions, involve derivatives of or- der zero (the ﬁeld variable itself) through m-1. Trial displacement functions are explicitely required to satisfy this B.C. Mathematically, this corresponds to Dirich- let boundary-value problems . Nonessential (or Natural, or static) boundary conditions, involve derivatives of or- dermand up. This B.C. is implied by the satisfaction of the variational statement but not explicitly stated in the functional itself. Mathematically, this correspondstoNeuman boundary-value problems . These boundary conditions were already introduced, albeit in a less formal way, in Table 9.1. 43Table C.1 illustrates the boundary conditions associated with some problems Problem Axial Member Flexural Member Distributed load Distributed load Diﬀerential Equation AEd2u dx2+q=0 EId4w dx4−q=0 m 1 2 Essential B.C. [0 ,m−1] u w,dw dx N a t u r a lB . C .[ m,2m−1] du dx d2w dx2andd3w dx3 orσx=Eu,x orM=EIw,xxandV=EIw,xxx Table C.1: Essential and Natural Boundary Conditions Example C-1: Extension of a Bar The total potential energy Π of an axial member of length L, modulus of elasticity E,c r o s ss e c t i o n a la r e a A, ﬁxed at left end and subjected to an axial force Pat the right one is given by Π=integraldisplayL 0EA 2parenleftBiggdu dxparenrightBigg2 dx−Pu(L) (3.14) Victor Saouma Introduction to Continuum Mechanics DraftC.1 Euler Equation C–5 Determine the Euler Equation by requiring that Π be a minimum. Solution: Solution I The ﬁrst variation of Π is given by δΠ=integraldisplayL 0EA 22parenleftBiggdu dxparenrightBigg δparenleftBiggdu dxparenrightBigg dx−Pδu(L) (3.15) Integrating by parts we obtain δΠ=integraldisplayL 0−d dxparenleftBigg EAdu dxparenrightBigg δudx+EAdu dxδuvextendsinglevextendsinglevextendsinglevextendsinglevextendsingleL 0−Pδu(L) (3.16-a) =−integraldisplayL 0δud dxparenleftBigg EAdu dxparenrightBigg bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipuprightdx+ parenleftBigg EAdu dxparenrightBiggvextendsinglevextendsinglevextendsinglevextendsinglevextendsingle x=L−P bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright δu(L) =−parenleftBigg EAdu dxparenrightBigg bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipuprightvextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle x=0δu(0) (3.16-b) The last term is zero because of the speciﬁed essential boundary condition which implies that δu(0) = 0. Recalling that δin an arbitrary operator which can be assigned any value, we set the coeﬃcients of δubetween (0 ,L) and those for δuat x=Lequal to zero separately, and obtain Euler Equation: −d dxparenleftBigg EAdu dxparenrightBigg =0 0<x<L (3.17) Natural Boundary Condition: EAdu dx−P=0 a t x=L (3.18) Solution II We have F(x,u,u/prime)=EA 2parenleftBiggdu dxparenrightBigg2 (3.19) (note that since Pis an applied load at the end of the member, it does not appear as part of F(x,u,u/prime) To evaluate the Euler Equation from Eq. 3.8, we evaluate ∂F ∂u=0 &∂F ∂u/prime=EAu/prime(3.20-a) Thus, substituting, we obtain ∂F ∂u−d dx∂F ∂u/prime= 0 Euler Equation (3.21-a) d dxparenleftBigg EAdu dxparenrightBigg = 0 B.C. (3.21-b) Victor Saouma Introduction to Continuum Mechanics DraftC–6 MATHEMATICAL PRELIMINARIES; Part IV VARIATIONAL METHODS Example C-2: Flexure of a Beam The total potential energy of a beam is given by Π=integraldisplayL 0parenleftbigg1 2Mκ−pwparenrightbigg dx=integraldisplayL 0parenleftbigg1 2(EIw/prime/prime)w/prime/prime−pwparenrightbigg dx (3.22) Derive the ﬁrst variational of Π. Solution:Extending Eq. 3.11, and integrating by part twice δΠ= integraldisplayL 0δFdx=integraldisplayL 0parenleftBigg∂F ∂w/prime/primeδw/prime/prime+∂F ∂wδwparenrightBigg dx (3.23-a) ==integraldisplayL 0(EIw/prime/primeδw/prime/prime−pδw)dx (3.23-b) =(EIw/prime/primeδw/prime)|L 0−integraldisplayL 0[(EIw/prime/prime)/primeδw/prime−pδw]dx (3.23-c) =(EIw/prime/primeδw/prime)|L 0−[(EIw/prime/prime)/primeδw]|L 0+integraldisplayL 0[(EIw/prime/prime)/prime/prime+p]δwdx= 0 (3.23-d) Or (EIw/prime/prime)/prime/prime=−pfor all x which is the governing diﬀerential equation of beams and Essential Natural δw/prime=0 o r EIw/prime/prime=−M=0 δw=0 o r ( EIw/prime/prime)/prime=−V=0 atx=0a n dx=L Victor Saouma Introduction to Continuum Mechanics Draft Appendix D MID TERM EXAM Continuum Mechanics LMC/DMX/EPFL Prof. Saouma Exam I (Closed notes), March 27, 1998 3H o u r s There are 19 problems worth a total of 63 points. Select any problems you want as long as the total number of corresponding points is equal to or larger than 50. 1. (2 pts) Write in matrix form the following 3rd order tensor DijkinR2space.i,j,k range from 1 to 2. 2. (2 pts) Solve for Eijaiin indicial notation. 3. (4 pts) if the stress tensor at point Pis given by σ= 10−20 −241 01 6  determinethetraction(orstressvector) tontheplanepassingthrough Pandparallel to the plane ABCwhereA(6,0,0),B(0,4,0) andC(0,0,2). 4. (5 pts) For a plane stress problem charaterized by the following stress tensor σ=bracketleftBigg 62 24bracketrightBigg use Mohr’s circle to determine the principal stresses, and show on an appropriate ﬁgure the orientation of those principal stresses. 5. (4 pts) The stress tensor throughout a continuum is given with respect to Cartesian axes as σ= 3x1x25x2 20 5x2 202x2 3 02x2 30  (a) Determine the stress vector (or traction) at the point P(2,1,√ 3) of the plane that is tangent to the cylindrical surface x2 2+x2 3=4a tP, DraftD–2 MID TERM EXAM n 12 3 xxx 123 P (b) Are the stresses in equlibrium, explain. 6. (2 pts) A displacement ﬁeld is given by u=X1X2 3e1+X2 1X2e2+X2 2X3e3, determine the material deformation gradient Fand the material displacement gradient J,a n d verify that J=F−I. 7. (4pts)Acontinuumbodyundergoesthedeformation x1=X1+AX2,x2=X2+AX3, andx3=X3+AX1whereAis a constant. Determine: 1) Deformation (or Green) tensorC; and 2) Lagrangian tensor E. 8. (4 pts) Linear and ﬁnite strain tensors can be decomposed into the sum or product of two other tensors. (a) Which strain tensor can be decomposed into a sum, and which other one into a product. (b) Why is such a decomposition performed? 9. (2 pts) Why do we have a condition imposed on the strain ﬁeld (compatibility equa- tion)? 10. (6 pts) Stress tensors: (a) When shall we use the Piola-Kirchoﬀ stress tensors? (b) What is the diﬀerence between Cauchy, ﬁrst and second Piola-Kirchoﬀ stress tensors? (c) In which coordinate system is the Cauchy and Piola-Kirchoﬀ stress tensors ex- pressed? 11. (2pts)Whatisthediﬀerencebetweenthetensorialandengineeringstrain( Eij,γij,i/negationslash= j)? 12. (3 pts) In the absence of body forces, does the following stress distribution  x2 2+ν(x2 1−x2 x)−2νx1x2 0 −2νx1x2x2 1+ν(x2 2−x2 1)0 00 ν(x2 1+x2 2)  whereνis a constant, satisfy equilibrium in the X1direction? 13. (2 pts) From which principle is the symmetry of the stress tensor derived?14. (2 pts) How is the First principle obtained from the equation of motion? 15. (4 pts) What are the 1) 15 Equations; and 2) 15 Unknowns in a thermoelastic formulation. Victor Saouma Introduction to Continuum Mechanics DraftD–3 16. (2 pts) What is free energy Ψ? 17. (2 pts) What is the relationship between strain energy and strain? 18. (5 pts) If a plane of elastic symmetry exists in an anisotropic material,   T11 T22 T33 T12 T23 T31  = c1111c1112c1133c1112c1123c1131 c2222c2233c2212c2223c2231 c3333c3312c3323c3331 c1212c1223c1231 SYM. c2323c2331 c3131   E11 E22 E33 2E12(γ12) 2E23(γ23) 2E31(γ31)   then, aj i= 10 0 01 000−1  show that under these conditions c1131is equal to zero. 19. (6 pts) The state of stress at a point of structural steel is given by T= 620 2−30 000 MPa withE= 207 GPa, µ=80G P a ,a n d ν=0.3. (a) Determine the engineering strain components (b) If a ﬁve centimer cube of structural steel is subjected to this stress tensor, what would be the change in volume? Victor Saouma Introduction to Continuum Mechanics DraftD–4 MID TERM EXAM Victor Saouma Introduction to Continuum Mechanics Draft Appendix E MATHEMATICA ASSIGNMENT and SOLUTION Connect to Mathematica using the following procedure: 1. login on an HP workstation2. Open a shell (window)3. Type xhost+ 4. type rlogin mxsg1 5. Onthe newly opened shell, enter your password ﬁrst, and then type setenv DISPLAY xxx:0.0 where xxxis the workstation name which should appear on a small label on the workstation itself. 6. Type mathematica & and then solve the following problems: 1. The state of stress through a continuum is given with respect to the cartesian axes Ox 1x2x3by Tij= 3x1x25x2 20 5x2 202x3 02x30 MPa Determine the stress vector at point P(1,1,√ 3) of the plane that is tangent to the cylindrical surface x2 2+x2 3=4a tP. 2. For the following stress tensor Tij= 6−30 −360 00 8  (a) Determine directly the three invariants Iσ,IIσandIIIσof the following stress tensor (b) Determine the principal stresses and the principal stress directions. (c) Show that the transformation tensor of direction cosines transforms the original stress tensor into the diagonal principal axes stress tensor. (d) Recompute the three invariants from the principal stresses. DraftE–2 MATHEMATICA ASSIGNMENT and SOLUTION (e) Split the stress tensor into its spherical and deviator parts. (f) Show that the ﬁrst invariant of the deviator is zero. 3. The Lagrangian description of a deformation is given by x1=X1+X3(e2−1), x2=X2+X3(e2−e−2,a n dx3=e2X3whereeis a constant. SHow that the Jacobian Jdoes not vanish and determine the Eulerian equations describing this motion. 4. A displacement ﬁeld is given by u=X1X2 3e1+X2 1X2e2+X2 2X3e3. Determine independently the material deformation gradient Fand the material displacement gradientJand verify that J=F−I. 5. A continuum body undergoes the deformation x1=X1,x2=X2+AX3,x3= X3+AX2whereAis a constant. Compute the deformation tensor Cand use this to determine the Lagrangian ﬁnite strain tensor E. 6. A continuum body undergoes the deformation x1=X1+AX2,x2=X2+AX3, x3=X3+AX2whereAis a constant. (a) Compute the deformation tensor C (b) Use the computed Cto determine the Lagrangian ﬁnite strain tensor E. (c) COmpute the Eulerian strain tensor E∗and compare with Efor very small values ofA. 7. A continuum body undergoes the deformation x1=X1+2X2,x2=X2,x3=X3 (a) Determine the Green’s deformation tensor C (b) Determine the principal values of Cand the corresponding principal directions. (c) Determine the right stretch tensor UandU−1with respect to the principal directions. (d) Determine the right stretch tensor UandU−1with respect to the eibasis. (e) Determine the orthogonal rotation tensor Rwith respect to the eibasis. 8. A continuum body undergoes the deformation x1=4X1,x2=−1 2X2,x3=−1 2X3 and the Cauchy stress tensor for this body is Tij= 100 0 0 00 000 0 MPa (a) Determine the corresponding ﬁrst Piola-Kirchoﬀ stress tensor. (b) Determine the corresponding second Piola-Kirchoﬀ stress tensor. (c) Determine the pseudo stress vector associated with the ﬁrst Piola-Kirchoﬀ stress tensor on the e1plane in the deformed state. (d) Determine the pseudo stress vector associated with the second Piola-Kirchoﬀ stress tensor on the e1plane in the deformed state. 9. Show that in the case of isotropy, the anisotropic stress-strain relation cAniso ijkm= c1111c1112c1133c1112c1123c1131 c2222c2233c2212c2223c2231 c3333c3312c3323c3331 c1212c1223c1231 SYM. c2323c2331 c3131  Victor Saouma Introduction to Continuum Mechanics DraftE–3 reduces to ciso ijkm= c1111c1122c1133000 c2222c2233000 c3333000 a00 SYM. b0 c  witha=1 2(c1111−c1122),b=1 2(c2222−c2233), andc=1 2(c3333−c1133). 10. Determine the stress tensor at a point where the Lagrangian strain tensor is given by Eij= 30 50 20 50 40 0 2 003 0 ×10−6 and the material is steel with λ= 119.2G P aa n d µ=7 9.2G P a . 11. Determine the strain tensor at a point where the Cauchy stress tensor is given by Tij= 100 42 6 42−20 60 1 5 MPa withE= 207 GPa, µ=7 9.2G P a ,a n d ν=0.30 12. Determine the thermally induced stresses in a constrained body for a rise in temer- ature of 50oF,α=5.6×10−6/0F 13. Show that the inverse of   εxx εyy εzz γxy(2εxy) γyz(2εyz) γzx(2εzx)  =1 E 1−ν−ν000 −ν1−ν000 −ν−ν10 0 0 000 1 + ν00 000 01 + ν0 000 0 01 + ν   σxx σyy σzz τxy τyz τzx  (5.1) is  σxx σyy σzz τxy τyz τzx  = E (1+ν)(1−2ν)bracketleftBigg 1−νν ν ν1−νν νν1−νbracketrightBigg 0 0 GbracketleftBigg 100 010001bracketrightBigg   εxx εyy εzz γxy(2εxy) γyz(2εyz) γzx(2εzx)  (5.2) and then derive the relations between stresses in terms of strains, and strains in terms of stress, for plane stress and plane strain. 14. Show that the function Φ = f(r)cos2θsatisﬁes the biharmonic equation ∇(∇Φ) = 0 Note: You must <<Calculus‘VectorAnalysis‘ , deﬁne Φ, and SetCoordinates[Cylindrical[r, θ,z]],a n d ﬁnally use the Laplacian (orBiharmonic ) functions. 15. Solve forbracketleftBigTrrTrθ TrθTθθbracketrightBig =bracketleftBigcosθ−sinθ sinθcosθbracketrightBigbracketleftBigσ00 00bracketrightBigbracketleftBigcosθ−sinθ sinθcosθbracketrightBigT (5.3) 16. If a point load pis applied on a semi-inﬁnite medium Victor Saouma Introduction to Continuum Mechanics DraftE–4MATHEMATICA ASSIGNMENT and SOLUTION θ1p r show that for Φ = −p πrθsinθwe have the following stress tensors: bracketleftbigg −2p πcosθ r0 00bracketrightbigg =bracketleftBigg −2pcos3θ πr−2psinθcos2θ πr −2psinθcos2θ πr−2psin2θcosθ πrbracketrightBigg (5.4) Determine the maximum principal stress at an y arbitrary point, (contour) plot the magnitude of this stress below p.N o t et h a t D[Φ,r], D[ Φ,{θ,2}]would give the ﬁrst and second derivatives of Φ with respect to randθrespectively. Victor Saouma Introduction to Continuum Mechanics