Constitutive Relations
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1 Constitutive Relations Andri Andriyana, Ph.D. Centre de Mise en Forme des Matériaux, CEMEF UMR CNRS 7635 École des Mines de Paris, Sophia Antipolis, France Spring, 2008
2 Outline Outline 1 Review of field equations General elasticity problem Concluding remarks 2 General form of stress-strain relations Linear elastic materials
3 Review of field equations Review of field equations
4 Review of field equations General elasticity problem General elasticity problem A general elasticity problem involves: Solutions of several partial differential equations i.e. equilibrium equations, strain-displacement relations (kinematic) Solutions must satisfy boundary condition (traction and displacement)
5 Review of field equations General elasticity problem Strain-displacement relations E = 1 2 ( H + H T) HHT = ǫ HHT E : Green-Lagrange strain tensor ǫ : Linearized strain tensor (small strain tensor) H = Grad U : Displacement gradient tensor U : Displacement vector Remarks 6 partial differential equations 9 unknowns : E ij and U i
6 Review of field equations General elasticity problem Equilibrium equations DivP + G = 0 or divσ + g = 0 Remarks P : First Piola-Kirchhoff stress tensor σ : Cauchy stress tensor G,g : Body force vectors (reference and current configurations) 3 partial differential equations 6 unknowns : P ij or σ ij
7 Review of field equations Concluding remarks Concluding remarks Previous equations hold for all continua (solids and fluids) 9 partial differential equations for 15 unknowns : 3 displacements, 6 strains, 6 stresses 6 more equations which characterize (distinguish) material behavior (stress-strain) are needed These are provided by : Constitutive Relations
8
9 General form of stress-strain relations General form of stress-strain relations Lagrangian description : ds = C : 1 2 dc S: Second Piola-Kirchhoff stress tensor C: Right Cauchy-Green strain tensor C = 2 S(C)/ C: elasticity tensor (4th order, 81 components)
10 General form of stress-strain relations Elasticity tensor Properties of C : 1 Minor symmetries, i.e. symmetric in its first and second slots (hold for all elastic materials) : C ijkl = C jikl = C ijlk 2 Major symmetries, i.e. based on strain energy consideration (hold for hyperelastic, linear elastic materials) : C ijkl = C klij
11 Linear elastic materials Linear elastic materials For small strain, stress-strain relations become : σ = C : ǫ σ 11 = C 1111 ǫ 11 + C 1112 ǫ C 1133 ǫ 33 σ 23 = C 2311 ǫ 11 + C 2312 ǫ C 2333 ǫ 33 σ 33 = C 3311 ǫ 11 + C 3312 ǫ C 3333 ǫ 33 Elasticity tensor C has 81 components
12 Linear elastic materials Pseudo-vector representation of anisotropic materials Using symmetries of σ and ǫ (which yields to minor symmetries of C), previous relations can be represented in pseudovector form: σ x = σ 11 D 11 D 12 D 13 D 14 D 15 D 16 ǫ x = ǫ 11 σ y = σ 22 D 21 D 22 D 23 D 24 D 25 D 26 ǫ y = ǫ 22 σ z = σ 33 = D 31 D 32 D 33 D 34 D 35 D 36 ǫ z = ǫ 33 σ yz = σ 23 D 41 D 42 D 43 D 44 D 45 D 46 γ yz = 2ǫ 23 σ xz = σ 13 D 51 D 52 D 53 D 54 D 55 D 56 γ xz = 2ǫ 13 σ xy = σ 12 D 61 D 62 D 63 D 64 D 65 D 66 γ xy = 2ǫ 12 Stiffness matrix [D] has 36 independent components
13 Linear elastic materials Pseudo-vector representation of anisotropic materials Using strain energy consideration (which yields to major symmetries of C), the stiffness matrix [D] is symmetric: D 11 D 12 D 13 D 14 D 15 D 16 D 22 D 23 D 24 D 25 D 26 [D] = D 33 D 34 D 35 D 36 D 44 D 45 D 46 (sym) D 55 D 56 D 66 Stiffness matrix [D] has 21 independent components
14 Linear elastic materials Special case 1 : Orthotropic materials Three mutually perpendicular planes of symmetry Normal stresses do not induce shear strains Shear stresses affect only on corresponding shear strains [C] = [D] 1 = 1 E x υxy E x υxz E x E y υyz E y E z G yz (sym) G xz 0 1 G xy Compliance matrix [C] has 9 independent constants
15 Linear elastic materials Special case 2 : Transversely isotropic materials Having one plane which is isotropic C 11 C 12 C C 22 C [C] = [D] 1 = C C (sym) C 55 0 C 66 Compliance matrix [C] has 5 independent constants If y-z plane is isotropic : E y = E z, υ xy = υ xz, G xy = G xz, G yz = Ey 2(1+υ yz)
16 Linear elastic materials Special case 3 : Isotropic materials Material properties independent of directions 1 υ υ [C] = [D] 1 = 1 1 υ E 2(1 + υ) 0 0 (sym) 2(1 + υ) 0 2(1 + υ) Compliance matrix [C] has 2 independent constants G = E 2(1+υ)
Constitutive Relations
Constitutive Relations Dr. Andri Andriyana Centre de Mise en Forme des Matériaux, CEMEF UMR CNRS 7635 École des Mines de Paris, 06904 Sophia Antipolis, France Spring, 2008 Outline Outline 1 Review of field
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