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Chapter 2: Elastic Constitutive Equations of a Laminate

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Chapter 2: Elastic Constitutive Equations of a Laminate 2.0 Introduction • Equations of Motion • Symmetric of Stresses • Tensorial and Engineering Strains • Symmetry of Constitutive Equations 2.1 Three-Dimensional Constitutive Equations • General Anisotropic Materials • Orthotropic Materials • Transversely Isotropic Materials • Isotropic Materials 2.2 Relation Between Mathematical & Engineering Constants • Isotropic Materials • Orthotropic Materials 2.3 Constitutive Equations for an Orthotropic Lamina • Plane Strain Condition • Plane Stress Condition 2.4 Constitutive Equations for an Arbitrarily Oriented Lamina • Coordinate Transformation • Stress Transformation • Strain Transformation • Stiffness and Compliance Matrix Transformation 2.5 Engineering Constants of a Laminate • Lamina • Laminate 2.6 Hygrothermal Coefficients of a Lamina 2.7 Summary 2.0 INTRODUCTION x2 u 2 2.0.1 Equations of Motion of Elastic Solids xP(,xxx123 , ) • Equations of Equilibrium (Kinetics) 2 x1 u1 ∂ u σρ += f i ij, = 123, , ij, j i ∂t2 σ x3 u3 22 x2 u2 σ σ 12 23 σ • Equations of Kinematics σ 12 32 σ σ 11 31 (strain-displacement) σ x u ε =+12(uu) 33 1 1 ij i,, j j i ε 22 x3 u3 x u ε 2 2 21 • Constitutive Equations (stress-strain) ε 23 ε 12 σε= ε ε ij Cijklijkl kl , , , = 123, , 31 ε 11 ε 13 ε 32 x1 u1 33 x3 u3 2.0.2 Symmetry of Stresses Consider a plane 1-2. Equilibrium x σ 2 22 σσσσ∗∗− ∗∗+ ∗∗− ∗∗= σ in x1 1111110tttt 11 21 21 21 σ σσ−∗∗−−∗∗= σσ 12 in x2 ()()22 22110tt 12 12 1 σ σ σσ∗∗− ∗∗= 11 11 Moment about A: 12110tt 21 1 ∴=σσ σ 12 21 12 A σσ= . Similarly we can show, from 2-3 plane 23 32 σ σσ= 21 σ 1-3 plane 13 31 22 x σσ= 1 Therefore, ij ji ij, = 123, , Stress tensor is Symmetric. Tensorial and Contracted Notation Tensorial Contracted σ σ σ11 σ1 σ22 σ2 στ33 = στ3 23 23 = 4 or 4 στ = = στ or στ31= 31 στ5 5 12 12 = 6 or 6 2.0.3 Tensorial and Engineering Strains x2 Tensorial Strains: ε ε =+12(uu) 21 ij i,, j j i ε ε = 12 iiuij i, i = normal strains. 1 ε =+12(uu) ij≠ tensorial shear strain. ij i,, j j i 1 ε Engineering shear strain 12 .A ε γεε=+=(uu +) = Total shear strain 21 ij ij ji i,, j j i x1 ε Engineering Strains ∂ ∂ 22 εγ==u2 +u3 ∂u 44 εε==1 ∂x ∂x x u ε 111∂ 3 2 2 2 21 x1 ε ∂ ∂u ∂u 23 ε u2 εγ==3 +1 12 εε== 55 ε 222∂ ∂x ∂x ε 11 x2 1 3 31 ε 13 ∂ ∂ ∂ ε εε==u3 εγ==u1 +u2 32 333∂ 66∂ ∂ x u x3 x2 x1 ε 1 1 33 x3 u3 Generalized Hooke’s Law (3-D Constitutive Equation) σε Stress-Strain Equation iijj = C i, j = 1,2,3,4,5,6 σ ε 1 CCCCCC11 12 13 14 15 16 1 σ ε 2 CCCCCC21 22 23 24 25 26 2 σ CCCCCC ε 3 = 31 32 33 34 35 36 3 τ γ 4 CCCCCC41 42 43 44 45 46 4 τ γ 5 CCCCCC51 52 53 54 55 56 5 τ γ 6 CCCCCC61 62 63 64 65 66 6 C is called the stiffness matrix. εσ Strain-Stress Equation iijj = S i, j = 1,2,3,4,5,6 ε σ 1 SSSSSS11 12 13 14 15 16 1 ε σ 2 SSSSSS21 22 23 24 25 26 2 ε σ 3 SSSSSS31 32 33 34 35 36 3 γ = τ 4 SSSSSS41 42 43 44 45 46 4 γ τ 5 SSSSSS51 52 53 54 55 56 5 γ τ 6 SSSSSS61 62 63 64 65 66 6 S is called the compliance matrix. 2.0.4 Symmetry of Constitutive Matrix 1 Strain energy density, U = σε - - - -(1) 0 2 ii 1 UC= εε 0 2 ij j i ∂U σ ==0 C ε i ∂ε ij j i ∂ 2U 0 = C ∂ε ∂ε ij ji 1 Eqn.(1) can be written as U = σε 0 2 jj 1 UC= εε 0 2 ji i j ∂U σ ==0 C ε j ∂ε ji i j ∂ 2U 0 = C ∂ε ∂ε ji ij Since the order of differentiating a scalar quantity U0 shouldnot change the result. Therefore, Cij = Cji .Stiffness matrix is symmetric. Similarly, Sij = Sji 2.1 3-D CONSTITUTIVE EQUATIONS (a) General Anisotropic Material (no plane of material symmetry). σ ε 1 1 CCCCCC11 12 13 14 15 16 σ ε 2 2 CCCCCC21 22 23 24 25 26 σ ε 3 3 = CCCCCC31 32 33 34 35 36 τ γ 4 4 CCCCCC41 42 43 44 45 46 τ γ 5 5 CCCCCC51 52 53 54 55 56 τ γ 6 6 CCCCCC61 62 63 64 65 66 • Number of unknowns = 6x 6 = 36 • Because symmetry of Cij, number of unknowns = 6x 7/ 2 = 21 (b) Specially Orthotropic Materials (3 mutually perpendicular planes of material symmetry). Reference coordinate system is parallel to the material coordinate system. σ ε 1 1 C11 σ ε 2 2 C21 C 22 Sym σ ε 3 3 = CCC31 32 33 τ γ 4 4 000C44 τ γ 5 5 0000C55 τ γ 6 6 0000 0C66 Number of unknowns = 9 Features • No interaction between normal stresses (σ1, σ2, σ3) and shear strains (γ4, γ5, γ6 ). Normal stresses acting along principal material directions produce only normal strains. • No interaction between shear stresses (τ4, τ5, τ6) and normal strains (ε1, ε2, ε3). Shear stresses acting on principal material planes produce only shear strains. • No interaction between shear stresses and shear strains on different planes. That is shear stress acting on a principal plane produces a shear strain only on that plane. (c) Transversely Isotropic Material An orthotropic material is called transversely isotropic when one of its principal plane is a plane of isotropy. At every point on this plane, the mechanical properties are the same in all directions. (2-3): Plane of Isotropy σ 1 C ε 11 1 σ 2 CC ε 21 22 2 σ 3 C12 C 23 C 22 Sym ε = − 3 τ CC22 23 4 γ 000 4 τ 2 5 γ 000 0 C55 5 τ 6 γ 000 0 0C55 6 Number of unknowns = 5 (d) Isotropic Material A material having infinite number of planes of material symmetry through a point. σ ε 1 C11 1 σ ε 2 C12 C 11 Sym 2 σ CCC ε 3 = 12 12 11 3 τ γ 4 000C44 4 τ 0000C γ 5 44 5 τ γ 6 0000 0C44 6 CC− where C = 11 12 44 2 Number of unknowns = 2 Summary Material Independent Elastic constants 1. Anisotropic material 36 2. Anisotropic elastic materials 21 3. Orthotropic material 9 4. Orthotropic material with 5 transverse isotropy 5. Isotropic material 2 2.2 Relations Between Mathematical and Engineering Constants (a) Isotropic Materials (E & ν) x2 ενσ=− 31/ E ενσ=− σ 21/ E 1 εσ= 11/ E x1 x3 Definition: σ /ε Elastic Modulus (E) = Stress/Strain = 1 1 ν ε /ε Poisson’s Ratio ( ) = - Transverse strain/Applied strain = - 2 1 x2 ενσ=− 31/ E ενσ=− 21/ E εσ= 11/ E x1 x3 Applied Stresses σ σ σ Normal Strains 1 2 3 in− x σ −νσ −νσ 1 1 / E 2 / E 3 / E − −νσ σ −νσ in x2 1 / E 2 / E 3 / E − −νσ −νσ σ in x3 1 / E 2 / E 3 / E Shear stresses Planes x2-x3, x3 - x1 x1 - x2 Shear Strains τ τ τ 23 31 12 γ τ / G 4 23 γ τ 5 31 / G γ τ / G 6 12 x2 Constitutive Equation σ 1 x1 εσ= iijjS x3 ε 1 −−νν σ 1 EEE000 1 ε −−νν1 σ 2 EE E000 2 ε −−νν1 σ 3 EEE000 3 = εγ= 1 στ= 423 000G 00 423 εγ= 00001 0 στ= 5 31 G 5 31 εγ= 1 στ= 612 00000G 612 −1 {}σε= []{} {}σε= []S {} or C Restrictions of Elastic Constants E Shear modulus G = 21( +ν) for Shear modulus to be positive, ν > - 1 E Bulk modulus K = 31( − 2ν) for Bulk modulus to be positive, ν < 1/2 −1 < ν < 1/2 (b) Orthotropic Materials x 2 E2 ενσ=− 31311/ E σ 1 ενσ=− 21211/ E x E εσ= 1 1 111/ E x3 E3 Definition: σ /ε Elastic Modulus (E1) = Stress/Strain = 1 1 ν ε /ε Poisson’s Ratio ( 12) = - Transverse strain/Applied strain = - 2 1 x2 ενσ=− 31311/ E ενσ=− 21211/ E εσ= 111/ E x1 x3 Applied Stresses Normal Strains σ σ σ 1 2 3 in− x σ −νσ −νσ/ E 1 11/ E 21 2/ E 2 31 3 3 − −νσ σ −νσ/ E in x2 12 1/ E 1 22/ E 32 3 3 − −νσ −νσ σ in x3 13 1/ E 1 23 2/ E 2 33/ E Shear stresses Planes x2-x3, x3 - x1 x1 - x2 Shear Strains τ τ τ 23 31 12 γ τ 4 23/ G 23 γ τ 5 31/ G 31 τ γ 12/ G 12 6 Constitutive Equation {}εσ= []S {} −ν −ν ε 1 21 31 000 σ 1 EEE11 22 33 1 −ν −ν ε 12 1 32 000 σ 2 EE11 22 E33 2 −−νν ε 13 23 1 σ 3 000 3 = EEE11 22 33 εγ= 0001 00στ= 423 G23 423 εγ= 00001 0 στ= 5 31 G31 5 31 εγ= 000001 στ= 612 G12 612 {}σε= −1{} {}σε= []{} []S or C = ≠ from Symmetry of S- matrix: SSij ji when i j = = = SS12 21 SS13 31 SS23 32 νν ν νννν νν ij ji ij E 12 21 13 31 23 = 32 = = i That is = , = , and Therefore or ν EE EE EE EEi j ji Ej 1 2 1 3 2 3 This is the well known Betti’s reciprocal law of orthotropic material properties.
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