Constitutive Relations Andri Andriyana, Ph.D. 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 equations General elasticity problem Concluding remarks 2 General form of stress-strain relations Linear elastic materials
Review of field equations Review of field equations
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)
Review of field equations General elasticity problem Strain-displacement relations E = 1 2 ( H + H T) + 1 2 HHT = ǫ + 1 2 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
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
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
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)
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
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 12 +... + C 1133 ǫ 33 σ 23 = C 2311 ǫ 11 + C 2312 ǫ 12 +... + C 2333 ǫ 33 σ 33 = C 3311 ǫ 11 + C 3312 ǫ 12 +... + C 3333 ǫ 33 Elasticity tensor C has 81 components
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
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
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 0 0 0 1 E y υyz E y 0 0 0 1 E z 0 0 0 1 G yz 0 0 1 (sym) G xz 0 1 G xy Compliance matrix [C] has 9 independent constants
Linear elastic materials Special case 2 : Transversely isotropic materials Having one plane which is isotropic C 11 C 12 C 13 0 0 0 C 22 C 23 0 0 0 [C] = [D] 1 = C 33 0 0 0 C 44 0 0 (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)
Linear elastic materials Special case 3 : Isotropic materials Material properties independent of directions 1 υ υ 0 0 0 [C] = [D] 1 = 1 1 υ 0 0 0 1 0 0 0 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)