3D and Planar Constitutive Relations

3D and Planar Constitutive Relations A School on Mechanics of Fibre Reinforced Polymer Composites Knowledge Incubation for TEQIP Indian Institute of Technology Kanpur PM Mohite Department of Aerospace Engineering Indian Institute of Technology Kanpur -5 January 017

3D Constitutive Relations

Generalized Hooke s Law: Generalized Hooke s law of the proportionality of stress and strain: Each of the six component of the stress at any point is a linear function of the six components of strain at that point. Concept of initial state Loading under two situations: - Isothermal and Reversible; - Adiabatic and Reversible. Stress components are the partial differential coefficients of a function (W) of the strain-components. 1 W W σ ij = + ε ij ε ji

Generalized Hooke s Law: Form of the Strain Energy Density Function (W): Homogeneous quadratic function of the strain components. 1 W = constant + C ij ε ij + C ijkl ε ij ε kl W is invariant Cij and Cijkl are tensors. W is taken to be zero when body is in the initial state in which Then, constant is zero. ε ij are zero.

Generalized Hooke s Law: For unstrained and unstressed body, Cij are zero. This leads to 1 W = Cijkl ε ij ε kl and σ ij = C ijkl ε kl i, j, k, l = 1,, 3 Cijkl is a fourth order (stiffness) tensor/matrix. (3 ) 4 = 81 independent constants!

Generalized Hooke s Law: Simply, you can view this as if you have a vector of 9 stress components which is related to a vector of 9 strain components through a matrix of 9x9! {σ }9 1 = [C ]9 9 {ε }9 1 C1111 C111 C1311 C111 [C] = C11 C311 C 3111 C311 C3311 C111 C1113 C111 C11 C113 C1131 C113 C1133 C11 C113 C11 C1 C13 C131 C13 C133 C131 C1313 C131 C13 C133 C1331 C133 C1333 C11 C113 C11 C1 C13 C131 C13 C133 C1 C13 C1 C C3 C31 C3 C33 C31 C313 C31 C3 C33 C331 C33 C333 C311 C3113 C311 C31 C313 C3131 C313 C3133 C31 C313 C31 C3 C33 C331 C33 C333 C331 C3313 C331 C33 C333 C3331 C333 C3333

Stress Tensor Symmetry: Stress symmetry: σ ij = σ ji σ ij = C ijkl ε kl σ ji = C jikl ε kl σ ij σ ji = 0 ( Cijkl C jikl ) ε kl = 0 C ijkl = C jikl six independent ways to express when i and j are taken together and still 9 ways to express k and l taken together. 6 9 = 54 independent constants!

Stress Tensor Symmetry: Simply, you can view this as if you have a vector of 6 stress components which is related to a vector of 9 strain components through a matrix of 6x9! {σ }6 1 = [C ]6 9 {ε }9 1 C1111 C 11 C3311 [C] = C 311 C 1311 C111 C111 C1 C331 C31 C1113 C13 C3313 C313 C111 C1 C331 C31 C11 C C33 C3 C113 C3 C333 C33 C1131 C31 C3331 C331 C113 C3 C333 C33 C131 C1313 C131 C13 C133 C1331 C133 C11 C113 C11 C1 C13 C131 C13 C1133 C33 C3333 C333 C1333 C133

Stress and Strain Tensor Symmetry: Strain symmetry: ε ij = ε ji σ ij = C ijkl ε kl σ ij = C ijlk ε lk σ ij σ ij = 0 ( Cijkl Cijlk ) ε kl = 0 C ijkl = C ijlk six independent ways to express when i and j are taken together and 6 ways to express k and l taken together. 6 6 = 36 independent constants!

Stress and Strain Tensor Symmetry: Or simply, you can view this as if you have a vector of 6 stress components which is related to a vector of 6 strain components through a matrix of 6x6! {σ }6 1 = [C ]6 6 {ε }6 1 C1111 C 11 C3311 [C] = C 311 C 1311 C111 C11 C C33 C3 C1133 C33 C3333 C333 C113 C3 C333 C33 C1113 C13 C3313 C31 C13 C1333 C133 C1313 C1 C133 C13 C11 C111 C1 C331 C31 C131 C11

Stress and Strain Tensor Symmetry: In other words, σ11 C1111 σ C 11 σ33 C3311 = σ3 C311 σ13 C1311 σ1 C111 C11 C1133 C113 C1113 C111 ε11 C C33 C3 C13 C1 ε C33 C3333 C333 C3313 C331 ε33 C3 C333 C33 C31 C31 ε3 C13 C1333 C133 C1313 C131 ε13 C1 C133 C13 C11 C11 ε1

Voigt Notation: Using Voigt notation - a way to represent a symmetric tensor by reducing its order For stress components: {σ11 6 5 σ11 σ1 σ13 1 σ σ 3 4 σ33 3 σ σ33 σ3 σ13 σ1} ={σ1 σ σ3 σ4 σ5 σ6} Strain Components: {ε11 ε ε33 ε3 ε13 ε1} ={ε1 ε ε3 ε4 ε5 ε6}

Stress and Strain Tensor Symmetries: Instead of writing C as a fourth order tensor, written as a second order tensor and stress and strains tensors are written as vectors! σ1 C11 σ C 1 σ3 C31 = σ4 C41 σ5 C51 σ6 C61 C1 C13 C14 C15 C16 ε1 C C3 C4 C5 C6 ε C3 C33 C34 C35 C36 ε3 C4 C43 C44 C45 C46 ε4 C51 C53 C54 C55 C56 ε5 C6 C63 C64 C65 C66 ε6

Existence of W: Existence of W : Hyperelastic materials Invariant Positive Definite 1 W = C ij ε j ε i 1 W = C ij ε i ε j ( Cij C ji ) ε iε j = 0 C ij = C ji

Existence of W: σ1 C11 C1 C13 σ C C 3 σ3 C33 = σ4 σ5 σ6 C14 C15 C16 ε1 C4 C5 C6 ε C34 C35 C36 ε3 C44 C45 C46 ε4 C55 C56 ε5 C66 ε6 1 independent constants!

Symmetries: Stress symmetry C ijkl = C jikl Minor Symmetries Strain symmetry C ijkl = C jikl Existence of W: C ijkl = C klij Major Symmetry

Transformations: Transformations Prime denotes the transformed coordinates. aij denotes the components of a transformation matrix

Material Symmetry: Further reduction in constants obtained by material symmetry Symmetry Definition: Any geometrical figure which can be brought to coincidence with itself, by an operation which changes the position of any of its points, is said to possess symmetry. Rotation and Reflection

Form of W: Quadratic in strain components: Note that the strains used are engineering strains.

Material Symmetry: One Plane of Material Symmetry: Monoclinic Materials

Material Symmetry: One Plane of Material Symmetry Transformation of axes: Transformation matrix: Transformation of strains:

Material Symmetry: One Plane of Material Symmetry Transformation of stresses:

Material Symmetry: One Plane of Material Symmetry Transformation of stiffness: C ij' = C ij Comparison of stress components: C14 = C15 = 0

Material Symmetry: One Plane of Material Symmetry Similarly, 13 independent constants

Material Symmetry: One Plane of Material Symmetry Second Approach: Invariance of W W for Hyperelastic material

Material Symmetry: One Plane of Material Symmetry Second Approach: Invariance of W For W to be invariant the product terms ε 1ε 4, ε 1ε 5, ε ε 4, ε ε 5, ε 3ε 4, ε 3ε 5, ε 4 ε 6, ε 5ε 6 must vanish, that is, C14, C15, C 4, C 5, C 34, C 35, C 46, C 56 are zero

Material Symmetry: Two Orthogonal Planes of Material Symmetry Two Orthogonal Planes of Material Symmetry: Orthotropic Materials Transformation of axes: Transformation matrix: Transformation of strains:

Material Symmetry: Two Orthogonal Planes of Material Symmetry Transformation of stresses: Transformation of stiffness: C ij' = C ij

Material Symmetry: Two Orthogonal Planes of Material Symmetry Comparison of stresses: C16 = 0 Similarly,

Material Symmetry: Two Orthogonal Planes of Material Symmetry Stiffness Tensor: 9 independent constants

Material Symmetry: Two Orthogonal Planes of Material Symmetry Second approach: Invariance of w W for monoclinic material: C11ε 1 + C ε + C33ε 3 + C 44 ε 4 + 1 C55ε 5 + C 66ε 6 + C1ε 1ε + C13ε 1ε 3 + W = C16ε 1ε 6 + C 3ε ε 3 + C 6 ε ε 6 + C36ε 3ε 6 C 45ε 4 ε 5 or form of W: ε 1, ε, ε 3, ε 4, ε 5, ε 6, W = W ε 1ε, ε 1ε 3, ε 1ε 6, ε ε, ε ε, ε ε, ε ε 3 6 3 6 4 5 +

Material Symmetry: Two Orthogonal Planes of Material Symmetry For W to be invariant under the strain transformations the product terms ε 1ε 6, ε ε 6, ε 3ε 6, ε 4 ε 5 must vanish, which is possible when C16, C 6, C36, C 45 are zero W for orthotropic material: 1 C11ε 1 + C ε + C33ε 3 + C 44 ε 4 + C55ε 5 + C 66ε 6 W = C1 ε 1ε + C13ε 1ε 3 + C 3ε ε 3 +

Material Symmetry: Two Orthogonal Planes of Material Symmetry When material has two orthogonal planes of symmetry then it also symmetric about a plane which is mutually orthogonal earlier two planes! Such materials are called Orthotropic Materials. Select now the remaining plane x1-x3 as the third orthogonal plane of material symmetry in addition to earlier two planes Follow the same procedure, either comparing the stresses or invariance of W There will be no change in the final stiffness tensor. Number of independent constants will be still 9!

Isotropy in a Plane: Isotropic behaviour of UD lamina in the cross-sectional plane (perpendicular to fibre s length) Transformation matrix:

Isotropy in a Plane: Transformation of strains:

Material Symmetry: Two Orthogonal Planes of Material Symmetry W for monoclinic material: ε 1, ε, ε 3, ε 4, ε 5, ε 6, W = W ε 1ε, ε 1ε 3, ε 1ε 6, ε ε, ε ε, ε ε, ε ε 3 6 3 6 4 5 For invariance ε 1ε 6, ε ε 6, ε 3ε 6, ε 4 ε 5 = 0 C16, C 6, C 36, C 45 are zero

Isotropy in a Plane: Trigonometric identities for the strains: ' ' ε + ε 33 = ε + ε 33, ( ) = (ε ) + (ε ) ε ε 33 ( ε 3 ) = (ε1 ) + ( ε 13 ) ' ε ' 1 ' + ε 33 ' ε 3 ' 13, Form of W: W = W ε11, ε + ε 33, ε ε 33 ( ε 3 ), ( ε 1 ) + ( ε 13 ) ) ' ' ' ' ' ' ε 33 W = W ε 11, ε + ε 33, ε ε 3 ( ( ),( ) + ( ) ' ε1 ' ε13

Isotropy in a Plane: Strain energy density function for orthotropic material: W = C11ε 11 + C1 ε 11ε + C13ε 11ε 33 + C ε + C 33ε 33 + C 3ε ε 33 + 4C 44 ε 3 + 4C 55ε 13 + 4C 66 ε 1 rearranging W = C11ε 11 + ε 11 ( C1 ε + C13ε 33 ) + 4C55ε 13 + 4C 66 ε 1 + + C 33ε 33 + C 3ε ε 33 + 4C 44 ε 3 + C ε

Isotropy in a Plane: W = C11ε 11 + ε 11 ( C1 ε + C13ε 33 ) + 4C55ε 13 + 4C 66 ε 1 + + C ε + C 33ε 33 + C 3ε ε 33 + 4C 44 ε 3 In the second bracket, we take C1 = C13 In the third bracket, we take C55 = C 66 C = C33 and C 3 + C 33ε 33 + C 3ε ε 33 + 4C 44ε 3 Rearrange the last bracket with C ε unchanged = C ( ε + ε 33 ) C ε ε 33 + C 3ε ε 33 + 4C 44 ε 3

Isotropy in a Plane: Rearrange the last bracket further as C ε + C 33ε 33 + C 3ε ε 33 + 4C 44 ε 3 = C ( ε + ε 33 ) we need C 44 ( C C 3 ) ε ε 33 C 44 ε 3 C C 3 =

Isotropy in a Plane: Stiffness tensor C11 C ij = C1 C1 0 0 C C 3 0 0 C 0 0 C C 3 0 Sym C 66 5 independent constants Such materials are called Transversely isotropic materials. Define: where 0 0 0 0 0 C 66

Transverse Isotropy with an Additional Orthogonal Plane: Consider isotropy in x1-x plane as well strain transformation

Transverse Isotropy with an Additional Orthogonal Plane: Trigonometric identities: ε 11 + ε = ' ε 11 ' + ε, ( ) = (ε ) + (ε ) ε 11ε ( ε 1 ) = (ε13 ) + ( ε 3 ) ' ε 11 ' + ε ' 13 ' ε 1 ' 3, Form of W: W = W ε11 + ε, ε 33, ε 11ε ( ε 1 ), ( ε13 ) + ( ε 3 ) ) ' ' ' ' ' ' ε ε1 W = W ε 11 + ε, ε 33, ε 11 ( ( ),( ) + ( ) ' ε13 ' ε 3

Transverse Isotropy with an Additional Orthogonal Plane: In the second bracket, we take C1 = C 3 In the third bracket, we take Rearrange the last bracket with C11 = C and C1 unchanged

Isotropy: Two independent constants! C11 Cij = C1 C1 0 0 C11 C1 0 0 C11 0 0 C11 C1 0 Define: where C11 C1 Sym 0 0 0 0 C11 C1 0

3D Constitutive Relations: Quick Review Generalized Hooke s Law: 81 independent constants Stress tensor symmetry: 54 independent constants Strain tensor symmetry: 36 independent constants Existence of W (Hyperelastic/Aelotropic): 1 independent constants Existence of one plane of material symmetry: 13 independent constants Existence of two/three mutually perpendicular planes of symmetry: (Orthotropic Material) 9 independent constants One plane of isotropy: 5 independent constants Two/three/infinite planes of isotropy: independent constants

3D Constitutive Relations for Orthotropic Materials

Constitutive Relations for Orthotropic Materials: Strain-stress Relations Normal stresses and strains

Constitutive Relations for Orthotropic Materials: Shear stresses and strains Poisson s ratio: (no sum over i, j) In general, ν ij ν ji

Constitutive Relations for Orthotropic Materials: Determination of Engineering Constants:

Constitutive Relations for Orthotropic Materials: Matrix-vector form: where, Always work with compliance tensor. It is easy to remember.

Constitutive Relations for Orthotropic Materials: Important Relations: Reciprocal relation where,

Constitutive Relations for Orthotropic Materials: Stiffness Relations: where,

Constraints on Engineering Constants: For strain energy to be positive definite both Compliance and Stiffness tensors must be positive definite. Strain energy to be positive definite the diagonal entries of the Compliance tensor must be positive. Similarly, the diagonal entries of the Stiffness tensor must be positive and the determinant must also be positive

Constraints on Engineering Constants: Constraint on Poisson s ratio: From constraint on determinant:

Constraints on Engineering Constants: For transverse isotropic material: with we get, Finally, leads to the condition For isotropic materials: and

Constitutive Relations: Transformations 13 Principal material directions xyz global reference directions Transformation matrix for rotation about z-axis:

Constitutive Relations: Stress Transformations Stress transformation: ' σ ij For example, that is, Transformation matrix: = a ki aljσ kl

Constitutive Relations: Stress and Strain Transformations Transformation matrix: where, m=cosθ and n=sinθ

Constitutive Relations: Stress and Strain Transformations Stress transformation: ' ε ij For example, that is, Now using = a ki alj ε kl

Constitutive Relations: Stress and Strain Transformations Strain transformation: Transformation matrix:

Constitutive Relations: Stress and Strain Transformations Stress Transformation: Strain Transformation:

Constitutive Relations: Stiffness Transformations From the first principles: Writing in global coordinates leads to and

Constitutive Relations: Stiffness Transformations Results in monoclinic behaviour! Constitutive relations:

Constitutive Relations: Stiffness Transformations

Constitutive Relations: Compliance Transformations From the first principles: Writing in global coordinates Or and

Constitutive Relations: Compliance Transformations The transformed Stiffness and Compliance tensors are symmetric! From invariance of W one can show 1 [T1 ] T = [T ] and 1 [T ] T = [T1 ]

Constitutive Relations: Compliance Transformations

Constitutive Relations: Thermal Effects Coefficient of thermal expansion is different in 3 directions!

Constitutive Relations: Thermal Effects Thermal strain in principal material directions where, These strains will not produce stresses unless restricted! Transforming strains into global coordinates We get,

Constitutive Relations: Thermal Effects That is, where, and

Constitutive Relations: Thermo-Elastic Equations Total strains: Mechanical strains: Thus, gives the thermo-elastic constitutive equations as

Constitutive Relations: Thermo-Elastic Equations Stresses in global direction: where,

Constitutive Relations: Hygral Effects Hygral strains: where, are coefficient s of hygral expansion

Constitutive Relations: Hygro-Thermal Effects Total strains: with, The hygro-thermo-elastic constitutive equation:

D Constitutive Relations for Orthotropic Materials

Constitutive Relations: Planar Equations Constitutive relation in 3D: Transverse stresses are zero: σ zz = τ xz = τ yz = 0 γ yz = S 44τ yz + S 45τ xz = 0 leads to γ xz = S 45τ xz + S 55τ yz = 0

Constitutive Relations: Transverse Strain Transverse normal strain: ε zz = S 13σ xx + S 3σ yy + S 36τ xy 0

Constitutive Relations: In-Plane Stresses

Constitutive Relations: Planar Equations Constitutive relation in 3D (Principal Directions): Transverse stresses are zero: leads to

Planar Relations: Principal Directions Transverse normal strain: ε 33 = S13σ 11 + S 3σ + S 36 τ 1 0 that is, ε 33 = S13σ 11 + S 3σ 0 Therefore, for planar case

Planar Relations: Principal Directions Transverse normal strain from stiffness relations: Transverse normal stress: gives

Planar Relations: Principal Directions Stresses in principal directions: Putting ε 33 in terms of ε 11, ε

Planar Relations: Principal Directions Stresses in principal directions: And can be written in a form as Qij and Cij are not same Inverse f orm: Sij are same as in 3D relations

Planar Relations: Principal Directions Reduced Stiffness Matrix: Stiffness Compliance in terms of engineering constants

Planar Relations: Principal Directions Stresses in principal directions:

Planar Relations: Transformation of Stresses and Strains Stresses in principal directions: Strains in principal directions:

Planar Relations: Transformation of Stresses and Strains Stresses in principal directions:

Planar Relations: Transformation of Stresses and Strains

Planar Relations: Transformation of Stresses and Strains Stresses in principal directions:

Planar Relations: Hygro-thermo-elastic Relations Stresses in principal directions: Stresses in principal directions Stresses in global directions:

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