periodic plates The Bending-Gradient theory for laminates and - PDF document

The asymptotic expansions for a laminated plate Example Differential system depending on a small parameter We want to solve the following differential equation on [0 , 1]: u ′′ ( x ) − η u ( x ) = 0 , u (0) = 0 , u (1) = a where η > 0 is a small parameter. The solution is trivial: � √ � u η ( x ) = a sinh η x � √ � sinh η The limit of u η ( x ) as η goes to 0 + is: η → 0 + u η ( x ) = ax . lim A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 11 / 85

The asymptotic expansions for a laminated plate Example Taylor’s series Using Taylor’s series: � √ � √ � 1 � 3 η x η x � √ � sinh η x = + + · · · 1! 3! � √ � √ � 1 � 3 � √ η η � sinh = + + · · · η 1! 3! We obtain: u η ( x ) = ax + a η 1 x 3 − x � x 5 − x − x 3 − x � + a η 2 + · · · 3! 5! 3!3! A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 12 / 85

The asymptotic expansions for a laminated plate Example The method ◮ Write u η ( x ) as a series: u η ( x ) = u 0 ( x ) + η 1 u 1 ( x ) + ... · · · + η i u i ( x ) + · · · where u i are unknown functions. ◮ Inject this series in the differential system u ′′ ( x ) − η u ( x ) = 0 , u (0) = 0 , u (1) = a and make null all the terms in η i . ◮ Solve the cascade system which determines the u i A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 13 / 85

The asymptotic expansions for a laminated plate Example Resolution The cascade system is: u 0 ′′ ( x ) = 0 , u 0 (0) = 0 , u 0 (1) = a Term in η 0 : u 1 ′′ ( x ) = u 0 , u 1 (0) = 0 , u 1 (1) = 0 Term in η 1 : .... u i ′′ ( x ) = u i − 1 , u i (0) = 0 , u i (1) = 0 Term in η i : The solution is obtained by mathematical induction: u 0 ( x ) = ax , u 1 ( x ) = ax 3 − x , u 2 ( x ) = ax 5 − x − ax 3 − x 3!3! , · · · 3! 5! A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 14 / 85

The asymptotic expansions for a laminated plate Scaling of the 3D problem Contents The 3D Problem The asymptotic expansions for a laminated plate Example Scaling of the 3D problem The expansion The Kirchhoff-Love plate model The Bending-Gradient plate model Applications of the Bending-Gradient theory to laminates Periodic plates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 14 / 85

The asymptotic expansions for a laminated plate Scaling of the 3D problem Change of variables ❡ 3 , x 3 3 , z L + ❡ ω + ω - - ω L − ω L ω − ω ❡ 2 , Y 2 ❡ 2 , x 2 - - ∂ω L 1 , x 1 ∂ω ❡ + η 2 F 3 ❢ 1 , Y 1 ❡ - t - 1 - 2 Ω t Ω − η 2 F 3 ❢ - 2 L 1 ◮ Y α = x α L for the in-plane variables, Y α ∈ ω ◮ z = x 3 t for the out-of-plane variable, z ∈ ] − 1 2 , 1 2 [ ◮ η = t L is the small parameter The fourth-order elasticity tensor can be rewritten as: t ( x 3 ) = ❈ t − 1 x 3 � � ❈ = ❈ ∼ ( z ) ∼ ∼ ∼ ∼ ∼ A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 15 / 85

The asymptotic expansions for a laminated plate Scaling of the 3D problem Non-dimensional fields We define the non-dimensional fields ( ✉ - , ε ∼ , σ ∼ ) as follows: t  ✉ ( x 1 , x 2 , x 3 ) = L ✉ ( x 1 / L , x 2 / L , x 3 / t ) = L ✉ ( Y 1 , Y 2 , z ) - - -  t ( x 1 , x 2 , x 3 ) = ( x 1 / L , x 2 / L , x 3 / t ) = ( Y 1 , Y 2 , z ) ε ε ε ∼ ∼ ∼ σ t ( x 1 , x 2 , x 3 ) = σ ( x 1 / L , x 2 / L , x 3 / t ) = σ ( Y 1 , Y 2 , z )  ∼ ∼ ∼ The derivation rule for these fields is: � d � , d , d - = ∇ dx 1 dx 2 dx 3 � ∂ � � � ∂ 0 , 0 , ∂ = L − 1 + t − 1 = L − 1 ∇ - Y + t − 1 ∇ , , 0 - z ∂ Y 1 ∂ Y 2 ∂ z = L − 1 ∇ η ( Y , z ) - where - Y + 1 η . . = ∇ ∇ η ∇ - z ( Y , z ) - A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 16 / 85

The asymptotic expansions for a laminated plate Scaling of the 3D problem Natural scaling of the stress � x 3 σ t αβ,β + σ t α 3 , 3 = 0   σ t σ t α 3 = − αβ,β du     σ t α 3 ,α + σ t  33 , 3 = 0    − t / 2 ⇒ � x 3 σ t 33 ( ± t / 2) = ± f 3 σ t σ t   33 = − α 3 ,α du − f 3     σ t  α 3 ( ± t / 2) = 0  − t / 2 αβ ∼ η 0 α 3 ∼ η 1 , 33 ∼ η 2 f 3 ∼ η 2 σ t σ t σ t ⇒ and The out-of-plane loading is scaled as: - ( x 1 , x 2 ) = η 2 F 3 ( Y 1 , Y 2 ) ❢ ❡ - 3 2 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 17 / 85

The asymptotic expansions for a laminated plate Scaling of the 3D problem The non-dimensional 3D problem The set of statically compatible fields can be rewritten as: ( Y , z ) = 0 on Ω = ω × ] − 1 2 , +1  η σ ∼ · ∇ 2[ ,   - D : SC 3 η 2 � � on ω ±  σ ∼ · ± ❡ = 2 F 3 ❡  - 3 - 3 The kinematically compatible fields becomes: - ⊗ s ∇ η ∼ = ✉ ( Y , z ) on Ω ,  ε - D :  KC 3 on ∂ω × ] − 1 2 , +1 ✉ - = 0 2[  The constitutive law becomes: σ ∼ ( Y 1 , Y 2 , z ) = ❈ ∼ ( z ) : ε ∼ ( Y 1 , Y 2 , z ) ∼ - Y + 1 η ∇ ( Y , z ) = ∇ η ∇ - z - A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 18 / 85

The asymptotic expansions for a laminated plate Scaling of the 3D problem Properties of the non-dimensional solution � � For given ω, ❈ ∼ , F 3 , η where ❈ ∼ is monoclinic and even in z , and under ∼ ∼ some regularity conditions, the solution of the non-dimensional problem is unique. Obviously, due the change of variables x 3 → z : ◮ u 3 and σ α 3 are even in z ◮ u α , σ αβ and σ 33 are odd in z We have the following new properties: ◮ u 3 and σ α 3 are odd in η ◮ u α , σ αβ and σ 33 are even in η A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 19 / 85

The asymptotic expansions for a laminated plate Scaling of the 3D problem Proof New change of variable z ′ = − x 3 t for the out-of-plane variable. The new - ′ , ε ′ , σ ′ ) are defined by: non-dimensional fields ( ✉ ∼ ∼ - ′ - ′ ( Y 1 , Y 2 , z ′ ) t ( x 1 , x 2 , x 3 ) = ( x 1 / L , x 2 / L , − x 3 / t ) =  ✉ L ✉ L ✉ -  ′ ′ ( Y 1 , Y 2 , z ′ ) t ( x 1 , x 2 , x 3 ) = ( x 1 / L , x 2 / L , − x 3 / t ) = ε ε ε ∼ ∼ ∼ ′ ′ ( Y 1 , Y 2 , z ′ ) σ t ( x 1 , x 2 , x 3 ) = σ ( x 1 / L , x 2 / L , − x 3 / t ) = σ  ∼ ∼ ∼ The new derivation rule for these fields is: � d � , d , d ∇ - = dx 1 dx 2 dx 3 � ∂ ∂ � � 0 , 0 , ∂ � = L − 1 − t − 1 = L − 1 ∇ - Y − t − 1 ∇ , , 0 - z ′ ∂ Y 1 ∂ Y 2 ∂ z ′ = L − 1 ∇ − η ( Y , z ′ ) - where - Y − 1 − η ( Y , z ′ ) = ∇ ∇ η ∇ - z ′ - A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 20 / 85

The asymptotic expansions for a laminated plate Scaling of the 3D problem The new equations ( Y , z ′ ) = 0 on Ω = ω × ] − 1 2 , +1  ′ · ∇ − η 2[ , σ  D ′ : ∼ -  SC 3 η 2 ′ · � � on ω ±  ± ❡ = − 2 F 3 ❡ σ  ∼ - 3 - 3 ′ = ✉ ′ ⊗ s ∇ − η  ε ( Y , z ′ ) on Ω , ∼ D ′ : - -  KC 3 on ∂ω × ] − 1 2 , +1 ′ = 0 ✉ 2[  - ′ � Y 1 , Y 2 , z ′ � � z ′ � ′ � Y 1 , Y 2 , z ′ � = ❈ : ε σ ∼ ∼ ∼ ∼ A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 21 / 85

The asymptotic expansions for a laminated plate Scaling of the 3D problem Effect of the transformation η → − η on the non-dimensional solution - ′ , ε ′ , σ ′ ) are solutions of the same The new non-dimensional fields ( ✉ ∼ ∼ equations as for ( ✉ - , ε ∼ , σ ∼ ) where F 3 → − F 3 and η → − η : ′ , ε ′ , σ ′ ) Y 1 , Y 2 , z ′ � ∼ ) ( − F 3 , − η ) � Y 1 , Y 2 , z ′ � � ( ✉ = ( ✉ - , ε ∼ , σ ∼ ∼ - Moreover, by definition, the new non-dimensional fields coincide with the initial ones with z = − z ′ : ∼ ) ( F 3 ,η ) � ′ , ε ′ , σ ′ ) Y 1 , Y 2 , z ′ � Y 1 , Y 2 , − z ′ � � ( ✉ = ( ✉ - , ε ∼ , σ ∼ ∼ - Hence, we have: ∼ ) ( − η ) ( Y 1 , Y 2 , z ) = − ( ✉ ∼ ) ( η ) ( Y 1 , Y 2 , − z ) ( ✉ - , ε ∼ , σ - , ε ∼ , σ Even components in z are odd in η and odd components in z are even in η . A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 22 / 85

The asymptotic expansions for a laminated plate The expansion Contents The 3D Problem The asymptotic expansions for a laminated plate Example Scaling of the 3D problem The expansion The Kirchhoff-Love plate model The Bending-Gradient plate model Applications of the Bending-Gradient theory to laminates Periodic plates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 22 / 85

The asymptotic expansions for a laminated plate The expansion Expansion We assume η − 1 ✉ η 0 ✉ η 1 ✉  ✉ = − 1 + 0 + 1 + · · · - - - -  η 0 ε η 1 ε 0 1 ε = + + · · · ∼ ∼ ∼ η 0 σ η 1 σ = 0 + 1 + · · ·  σ ∼ ∼ ∼ p and σ ✉ p , ε p , p = − 1 , 0 , 1 , 2 ... , are functions of ( Y 1 , Y 2 , z ) ∼ ∼ - Because: ◮ u 3 and σ α 3 are odd in η ◮ u α , σ αβ and σ 33 are even in η we have: p p ◮ u 3 and σ α 3 are null for even p and even in z for odd p . ◮ u p p p α , σ αβ and σ 33 are null for odd p and odd in z for even p . A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 23 / 85

The asymptotic expansions for a laminated plate The expansion Statics The normalized 3D equilibrium equation becomes: ( Y , z ) = η − 1 � 0 · ∇ � + η 0 � 0 · ∇ 1 · ∇ � η ∼ · ∇ - Y + σ + · · · = 0 . σ σ σ - z - z ∼ ∼ ∼ - Hence, 0 · ∇ p · ∇ p +1 · ∇ σ - z = 0 and σ - Y + σ - z = 0 , p ≥ 0 . ∼ ∼ ∼ Or in components: σ 0 p p +1 i 3 , 3 = 0 and σ i α,α + σ i 3 , 3 = 0 , p ≥ 0 . A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 24 / 85

The asymptotic expansions for a laminated plate The expansion Statics The static boundary conditions on ω ± writes: = F 3 p · � � 2 · � � ± ❡ = 0 when p � = 2 and ± ❡ 2 ❡ - 3 . σ σ ∼ ∼ - 3 - 3 Or in components: � � Y 1 , Y 2 , ± 1 p σ = 0 when p � = 2 i 3 2 � Y 1 , Y 2 , ± 1 � � Y 1 , Y 2 , ± 1 � = ± 1 σ 2 = 0 and σ 2 2 F 3 ( Y 1 , Y 2 ) α 3 33 2 2 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 25 / 85

The asymptotic expansions for a laminated plate The expansion Kinematics The non-dimensional displacement field is: − 1 + η 0 ✉ 0 + η 1 ✉ 1 + · · · - = η − 1 ✉ ✉ - - - The non-dimensional strain field is: − 2 + η − 1 ε − 1 + η 0 ε 0 + · · · η ( Y , z ) = η − 2 ε - ⊗ s ∇ ∼ = ✉ ε ∼ ∼ ∼ - with: − 2 = ✉ p = ✉ − 1 ⊗ s ∇ p +1 ⊗ s ∇ p ⊗ s ∇ ε - z and ε - z + ✉ - Y , p ≥ − 1 ∼ ∼ - - - In components: α 3 = 1 ε − 2 ε − 2 2 u − 1 ε − 2 33 = u − 1 αβ = 0 , and α, 3 3 , 3 and for all p ≥ − 1: αβ = 1 α 3 = 1 � � � � p +1 p +1 p p p p p p ε u α,β + u , ε u α, 3 + u and ε 33 = u β,α 3 ,α 3 , 3 2 2 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 26 / 85

The asymptotic expansions for a laminated plate The expansion Kinematics The kinematic condition on the lateral boundary leads to: p ( Y 1 , Y 2 ) = 0 . ∀ p ≥ − 1 and ∀ ( Y 1 , Y 2 ) ∈ ∂ω, ✉ - A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 27 / 85

The Kirchhoff-Love plate model Contents The 3D Problem The asymptotic expansions for a laminated plate The Kirchhoff-Love plate model Lower order displacements Auxiliary Problem Macroscopic problem The Bending-Gradient plate model Applications of the Bending-Gradient theory to laminates Periodic plates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 27 / 85

The Kirchhoff-Love plate model Lower order displacements Contents The 3D Problem The asymptotic expansions for a laminated plate The Kirchhoff-Love plate model Lower order displacements Auxiliary Problem Macroscopic problem The Bending-Gradient plate model Applications of the Bending-Gradient theory to laminates Periodic plates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 27 / 85

The Kirchhoff-Love plate model Lower order displacements Lower order displacements − 2 = 0 which leads to: We set ε ∼ α 3 = 1 ε − 2 ε − 2 2 u − 1 ε − 2 33 = u − 1 αβ = 0 , α, 3 = 0 and 3 , 3 = 0 − 1 is a function of ( Y 1 , Y 2 ). Hence, ✉ - Moreover, u − 1 α is null since η = − 1 is odd:   0 − 1 = U − 1 ✉ 3 ( Y 1 , Y 2 ) ❡ - 3 = 0   - U − 1 3 with the boundary conditions: U − 1 = 0 ∀ ( Y 1 , Y 2 ) ∈ ∂ω 3 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 28 / 85

The Kirchhoff-Love plate model Lower order displacements Lower order displacements − 1 = 0 which leads to: We set ε ∼ αβ = 1 α 3 = 1 � � � � ε − 1 u − 1 α,β + u − 1 ε − 1 u 0 α, 3 + U − 1 = 0 , ε − 1 33 = u 0 = 0 , 3 , 3 = 0 β,α 3 ,α 2 2 0 has the following form: Hence, ✉ - − zU − 1   3 , 1 0 = − zU − 1 − zU − 1 ✉ - Y = ⊗ ∇ 3  3 , 2  - 0 with the boundary conditions: U − 1 3 ,α n α = 0 ∀ ( Y 1 , Y 2 ) ∈ ∂ω where ♥ - is the outer normal to ∂ω . A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 29 / 85

The Kirchhoff-Love plate model Auxiliary Problem Contents The 3D Problem The asymptotic expansions for a laminated plate The Kirchhoff-Love plate model Lower order displacements Auxiliary Problem Macroscopic problem The Bending-Gradient plate model Applications of the Bending-Gradient theory to laminates Periodic plates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 29 / 85

The Kirchhoff-Love plate model Auxiliary Problem Zeroth-order auxiliary problem Equilibrium equation of order -1, compatibility equation, boundary conditions and constitutive equations of order 0 lead to (for z ∈ [ − 1 2 , 1 2 ]): σ 0 i 3 , 3 = 0  0 · ∇  σ - z = 0 .  ∼  σ 0 ij = ❈ ijkl ε 0   0 = ❈   kl  ∼ ( z ) : ε 0 .  σ    ∼ ∼ ∼  αβ = zK − 1 ε 0 0 = ✉ αβ 1 ⊗ s ∇ 0 ⊗ s ∇ ε - z + ✉ - Y . ∼   α 3 = 1 - - ε 0 2 u 1 ε 0 33 = u 1 and     α, 3 3 , 3  0 � z = ± 1 �  σ · ± ❡ - 3 = 0    ∼ 2 z = ± 1 σ 0 � �  = 0 i 3 2 The lowest-order curvature is: − 1 . . . = − U − 1 K − 1 . = − U − 1 ❑ 3 ∇ - Y ⊗ ∇ or ∼ αβ 3 ,αβ - Y A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 30 / 85

The Kirchhoff-Love plate model Auxiliary Problem Resolution From � z = ± 1 � σ 0 σ 0 i 3 , 3 = 0 and = 0 i 3 2 we obtain plane-stress: σ 0 i 3 = 0 The constitutive equation writes: zK − 1  σ 0   0 0    ❈ 1111 ❈ 1122 ❈ 1112 ❈ 1133 11 11 σ 0 zK − 1 ❈ 1122 ❈ 2222 ❈ 2212 0 0 ❈ 2233  22     22       2 zK − 1  σ 0 0 0 ❈ 1112 ❈ 2212 ❈ 1212 ❈ 1233  12      12 =       0 0 0 0 ❈ 1313 ❈ 1323 0 2 ε 0       13       0 0 0 0 0 2 ε 0 ❈ 1323 ❈ 2323       23 0 ❈ 1133 ❈ 2233 ❈ 12331 0 0 ❈ 3333 ε 0 33 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 31 / 85

The Kirchhoff-Love plate model Auxiliary Problem Resolution The strain is given by: 33 = − z ❈ 33 αβ ( z ) ε 0 αβ = zK − 1 ε 0 ε 0 ❈ 3333 ( z ) K − 1 αβ ( Y 1 , Y 2 ) , α 3 = 0 and αβ ( Y 1 , Y 2 ) The stress is given by: 0 = s K ( z ) : ❑ − 1 ( Y 1 , Y 2 ) ij = s K ij γδ K − 1 or in components σ 0 σ ∼ ∼ ∼ δγ ∼ where the fourth-order stress localization tensor is: s K s K αβγδ ( z ) . i 3 γδ . . = z ❈ σ αβγδ ( z ) and . = 0 and ❈ σ αβγδ = ❈ αβγδ − ❈ αβ 33 ❈ 33 γδ / ❈ 3333 denotes the plane-stress elasticity tensor. A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 32 / 85

The Kirchhoff-Love plate model Auxiliary Problem Resolution By integrating of α 3 = 0 = 1 33 = − z ❈ 33 αβ ε 0 2 u 1 ε 0 K − 1 αβ = u 1 and α, 3 3 , 3 ❈ 3333 We find:  0  1 = ✉ K : ❑ − 1 + U 1  . ✉ 3 ❡ - 3 = 0 ∼ ∼  - - ✉ K 3 αβ K − 1 βα + U 1 3 K ( z ) related to the curvature where the displacement localization tensor ✉ ∼ - is given by: �� z � ∗ r ❈ 33 αβ ✉ K 3 αβ ( z ) . ✉ K αβγ . . = − dr and . = 0 ❈ 3333 − 1 2 where [ • ] ∗ denotes the averaged-out distribution: [ • ] ∗ . . = • − �•� . A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 33 / 85

The Kirchhoff-Love plate model Auxiliary Problem Isotropic materials The leading order strain: ν ε 0 αβ = zK − 1 ε 0 ε 0 1 − ν zK − 1 αβ , α 3 = 0 and 33 = − αα σ : The leading order stress is derived through ❈ ∼ ∼  E E ν  0  σ 0   zK − 1  1 − ν 2 1 − ν 2 11 11  = E zK − 1 σ 0 0   22 1 − ν 2   22    E 2 zK − 1 σ 0 SYM 12 12 2(1+ ν ) The displacement corrector is:  0  1 = ✉ K : ❑ − 1 + U 1  . 0 ✉ 3 ❡ - 3 = ∼ ∼  � 1 - - ν 12 − z 2 � K − 1 αα + U 1 3 2(1 − ν ) A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 34 / 85

The Kirchhoff-Love plate model Macroscopic problem Contents The 3D Problem The asymptotic expansions for a laminated plate The Kirchhoff-Love plate model Lower order displacements Auxiliary Problem Macroscopic problem The Bending-Gradient plate model Applications of the Bending-Gradient theory to laminates Periodic plates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 34 / 85

The Kirchhoff-Love plate model Macroscopic problem Determination of U − 1 3 Resultants The zeroth-order bending moment is defined as αβ ( Y 1 , Y 2 ) . � � M 0 z σ 0 . = , αβ The first-order shear force is: α ( Y 1 , Y 2 ) . Q 1 . = � σ 1 3 α � . A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 35 / 85

The Kirchhoff-Love plate model Macroscopic problem Determination of U − 1 3 Equilibrium Recall that: σ 0 p p +1 i 3 , 3 = 0 and σ i α,α + σ i 3 , 3 = 0 , p ≥ 0 . The bending equilibrium equations are: � � σ 0 αβ,β + σ 1 �� = 0 = M 0 αβ,β − Q 1 z α 3 , 3 α The out-of-plane equilibrium equation is: � � σ 1 3 α,α + σ 2 = 0 = Q 1 α,α + F 3 33 , 3 Finally, we have: M 0 αβ,βα + F 3 = 0 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 36 / 85

The Kirchhoff-Love plate model Macroscopic problem Determination of U − 1 3 Constitutive equation From � 0 � � K ( z ) : ❑ − 1 � � σ ( z ) : ❑ − 1 � z 2 ❈ z σ = z s = ∼ ∼ ∼ ∼ ∼ ∼ ∼ we obtain the Kirchhoff’s constitutive equation: 0 = ❉ � σ � z 2 ❈ − 1 ▼ ∼ : ❑ where: ❉ ∼ = ∼ ∼ ∼ ∼ ∼ ∼ The Kirchhoff-Love plate equations are:  0 : � � + F 3 = 0 , on ω ▼ ∇ - Y ⊗ ∇  ∼ - Y   0 = ❉   − 1 , ▼ ∼ : ❑ on ω   ∼ ∼ ∼ − 1 = − U − 1 - Y , on ω ❑ 3 ∇ - Y ⊗ ∇ ∼     � �  U − 1 U − 1 = 0 and · ♥ - = 0 on ∂ω  ⊗ ∇  3 3 - Y A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 37 / 85

The Kirchhoff-Love plate model Macroscopic problem Summary of the Kirchhoff-Love model The displacement is approximated by: − zU − 1   3 , 1 − 1 + ✉ 0 = - ≈ η − 1 ✉  = ✉ − zU − 1 LK ✉  3 , 2 - - - η − 1 U − 1 3 0 � = ✉ η LK ⊗ s ∇ The strain ε ∼ is approximated by ε ( Y , z ) with: ∼ - - 33 = − z ❈ 33 αβ ( z ) ε 0 αβ = zK − 1 ε 0 ε 0 ❈ 3333 ( z ) K − 1 αβ ( Y 1 , Y 2 ) , α 3 = 0 and αβ ( Y 1 , Y 2 ) where K − 1 αβ = − U − 1 3 ,αβ 0 such that σ 0 · ∇ η The stress σ ∼ is approximated by σ ( Y , z ) � = 0 and ∼ ∼ - σ 0 αβ = z ❈ σ αβγδ ( z ) K − 1 and σ 0 i 3 = 0 δγ A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 38 / 85

The Kirchhoff-Love plate model Macroscopic problem Lateral boundary conditions It should be emphasized that the assumed expansion is not compatible with clamped lateral boundary conditions. Indeed,   0 1 = ✉ K : ❑ − 1 + U 1  � = 0 . 0 ✉ 3 ❡ - 3 =  ∼ ∼ - - ✉ K 3 αβ K − 1 βα + U 1 3 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 39 / 85

The Bending-Gradient plate model Contents The 3D Problem The asymptotic expansions for a laminated plate The Kirchhoff-Love plate model The Bending-Gradient plate model Shear effects Higher orders? Derivation of the Bending-Gradient theory The Reissner-Mindlin theory for homogeneous plates Distance between the BG and the RM models Applications of the Bending-Gradient theory to laminates Periodic plates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 39 / 85

The Bending-Gradient plate model Shear effects Contents The 3D Problem The asymptotic expansions for a laminated plate The Kirchhoff-Love plate model The Bending-Gradient plate model Shear effects Higher orders? Derivation of the Bending-Gradient theory The Reissner-Mindlin theory for homogeneous plates Distance between the BG and the RM models Applications of the Bending-Gradient theory to laminates Periodic plates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 39 / 85

The Bending-Gradient plate model Shear effects First-order auxiliary problem Equilibrium equation for order 0, compatibility equation, boundary conditions and constitutive equations of order 1 lead to (for z ∈ [ − 1 2 , 1 2 ]):  σ 0 i α,α + σ 1 i 3 , 3 = 0   0 · ∇ 1 · ∇  σ 1 ij = ❈ ijkl ε 1  σ - Y + σ - z = 0   kl ∼ ∼     1 = ❈  ε 1 αβ = u 1 ( α,β ) = 0 1  σ ∼ ( z ) : ε     ∼ ∼ ∼  1 = ✉ � � 2 ⊗ s ∇ 1 ⊗ s ∇ α 3 = 1 ε - z + ✉ ε 1 u 2 α, 3 + u 1 - Y ∼ 3 ,α  - -  2     z = ± 1  1 � �  · ± ❡ - 3 = 0 σ ε 1 33 = u 2   3 , 3 = 0  ∼ 2     z = ± 1 � � σ 1 = 0  i 3 2 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 40 / 85

The Bending-Gradient plate model Shear effects Resolution Transverse stress From σ 0 βα,α + σ 1 β 3 , 3 = 0    � �  s K σ 0 βαγδ ( z ) K − 1 ,α = z ❈ σ βαγδ ( z ) K − 1 βα,α = δγ δγ,α   z = ± 1 � � σ 1 = 0  β 3 2 we obtain the first-order transverse shear stress: � z αβγδ dr K − 1 σ 1 α 3 = − r ❈ σ δγ,β − 1 2 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 41 / 85

The Bending-Gradient plate model Shear effects Resolution In-plane stress From ε 1 ε 1 αβ = 0 and 33 = 0 and the constitutive equation:       σ 1 ❈ 1111 ❈ 1122 ❈ 1112 0 0 ❈ 1133 0 11 σ 1 0 0 0 ❈ 2222 ❈ 2212 ❈ 2233  22            σ 1 0 0 0 ❈ 1212 ❈ 1233  12      =       σ 1 0 2 ε 1 ❈ 1313 ❈ 1323  13     13        σ 1 SYM 0 2 ε 1 ❈ 2323  23     23  0 0 ❈ 3333 we have σ 1 αβ = 0 and the first-order stress localization writes as: 1 = s K ∇ ( z ) . . � � . − 1 ⊗ ∇ σ ❑ ∼ ⌢ ∼ ∼ - Y where we defined the fifth-order localization tensor as: � z s K ∇ αβγδη . s K ∇ α 3 γδη ( z ) . s K ∇ 33 γδη . r ❈ σ . = 0 , . = − αγδη dr and . = 0 − 1 2 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 42 / 85

The Bending-Gradient plate model Shear effects Resolution Displacement We find that the second-order displacement field writes as: 3 , 1 + ✉ K ∇ 1 βγδ K − 1  − zU 1  δγ,β 2 = ✉ K ∇ ( z ) . . � � . − 1 ⊗ ∇ 3 , 2 + ✉ K ∇ − zU 1 - Y = − zU 1 2 βγδ K − 1 ✉ ❑ 3 ⊗ ∇ ⌢ ∼  δγ,β  - - Y - 0 where the displacement localization tensor related to the curvature gradient writes as: � z � y � � ✉ K ∇ ηβγδ dv + δ αβ ✉ K αβγδ ( z ) . . = − 4 ❙ α 3 η 3 v ❈ σ dy 3 γδ − 1 0 2 and ✉ K ∇ 3 βγδ . . = 0 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 43 / 85

The Bending-Gradient plate model Higher orders? Contents The 3D Problem The asymptotic expansions for a laminated plate The Kirchhoff-Love plate model The Bending-Gradient plate model Shear effects Higher orders? Derivation of the Bending-Gradient theory The Reissner-Mindlin theory for homogeneous plates Distance between the BG and the RM models Applications of the Bending-Gradient theory to laminates Periodic plates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 43 / 85

The Bending-Gradient plate model Higher orders? The general form of the expansion It can be formally shown that we have: - = U 3 . ❑ K : ❑ ∼ + η 2 ✉ K ∇ . . ✉ η ❡ - 3 − zU 3 ⊗ ∇ - Y + η ✉ - Y + . . . ∼ ⊗ ∇ ∼ ⌢ - - where ∞ � η p +1 U U 3 . . p . = 3 = η � u 3 � and ❑ . = − U 3 ∇ - Y ⊗ ∇ ∼ - Y p = − 1 We have also for the stress: K : ❑ K ∇ . . ❑ . ∼ = s ∼ + η s - Y + . . . σ ∼ ⊗ ∇ ∼ ⌢ ∼ ∼ A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 44 / 85

The Bending-Gradient plate model Higher orders? A higher order plate model from asymptotic expansions? Including shear effects...: ◮ ... from asymptotic expansion?: U 3 ∈ C 6 ( ω ) ◮ ... from the approach from Smyshlyaev and Cherednichenko (2000)?: U 3 ∈ C 4 ( ω ) ◮ ... with the Bending-Gradient theory: U 3 ∈ C 1 ( ω ) A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 45 / 85

The Bending-Gradient plate model Derivation of the Bending-Gradient theory Contents The 3D Problem The asymptotic expansions for a laminated plate The Kirchhoff-Love plate model The Bending-Gradient plate model Shear effects Higher orders? Derivation of the Bending-Gradient theory The Reissner-Mindlin theory for homogeneous plates Distance between the BG and the RM models Applications of the Bending-Gradient theory to laminates Periodic plates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 45 / 85

The Bending-Gradient plate model Derivation of the Bending-Gradient theory Stress localization as function of static variables The stress field can be accurately approximated by: . χ BG = s K : χ K ∇ . . σ ∼ + η s ∼ ⊗ ∇ ∼ ∼ ⌢ ∼ ∼ - Y ∼ = ( χ αβ ) ( Y 1 , Y 2 ) is an unknown symmetric second-order tensor field. χ Choice of χ ∼ ?: The minimum of complementary energy! The corresponding bending moment is: BG = ❉ � σ � z 2 ❈ − 1 ▼ ∼ : χ where ❉ ∼ = and ❞ ∼ = ❉ ∼ ∼ ∼ ∼ ∼ ∼ ∼ ∼ ∼ Its gradient is: BG ⊗ ∇ R αβγ = M BG ❘ ⌢ = ▼ or with R αβγ = R βαγ αβ,γ ∼ - Y A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 46 / 85

The Bending-Gradient plate model Derivation of the Bending-Gradient theory Stress localization as function of static variables BG in terms of ▼ BG and ❘ It is possible to rewrite σ ⌢ : ∼ ∼ . � BG = s K : � BG � K ∇ . BG � . σ ❞ ∼ : ▼ + η s ∼ : ▼ ❞ ⊗ ∇ ∼ ∼ ∼ ∼ ⌢ ∼ ∼ ∼ ∼ - Y and BG = s M : ▼ BG + η s R . . ❘ . σ ∼ ∼ ∼ ⌢ ⌢ ∼ ∼ where the localizations tensors are given by: M = s K : ❞ R = s K ∇ : ❞ s ∼ , s ∼ ∼ ∼ ⌢ ⌢ ∼ ∼ ∼ ∼ ∼ ∼ A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 47 / 85

The Bending-Gradient plate model Derivation of the Bending-Gradient theory The Bending-Gradient stress energy BG into the complementary energy of the full 3D problem leads Plugging σ ∼ to the following functional: � P ∗ BG � � w ∗ KL � BG � + η 2 w ∗ BG � � BG , ❘ ▼ = ▼ ❘ d ω ∼ ⌢ ∼ ⌢ ω where the stress elastic energies are defined as: = 1 = 1 . ❤ . ❘ w ∗ KL � BG � BG : ❞ w ∗ BG � � BG . . ▼ 2 ▼ ∼ : ▼ and ❘ T ❘ ⌢ . ⌢ . ∼ ∼ ∼ ∼ ⌢ ⌢ ⌢ 2 with: � R : ❙ R � T s ❤ ⌢ = ∼ : s ⌢ ⌢ ∼ ⌢ ∼ ∼ This sixth-order tensor is the compliance related to the transverse shear of the plate. It is positive, symmetric, but not definite in the general case. A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 48 / 85

The Bending-Gradient plate model Derivation of the Bending-Gradient theory Extended plate equilibrium equations Exact plate equilibrium equations The total bending moment and the total shear force are defined as Q α ( Y 1 , Y 2 ) = η − 1 � σ 3 α � . M αβ ( Y 1 , Y 2 ) = � z σ αβ � , and Moment equilibrium equations: σ αβ,β + η − 1 σ α 3 , 3 � � �� z = M αβ,β − Q α = 0 or ▼ ∼ · ∇ - Y − ◗ - = 0 The out-of-plane equilibrium equation: η − 1 � σ 3 α,α + η − 1 σ 33 , 3 � = Q α,α + F 3 = 0 or ◗ - · ∇ - Y + F 3 = 0 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 49 / 85

The Bending-Gradient plate model Derivation of the Bending-Gradient theory Extended plate equilibrium equations Link between shear forces and generalized shear forces BG ⊗ ∇ ◗ - = ▼ ∼ · ∇ - Y is now replaced by ❘ ⌢ = ▼ ∼ - Y We have the following relation: . ❘ BG · ∇ BG or R αββ = M BG . αβ,β = Q BG ∼ . ✐ ⌢ = ▼ - Y = ◗ α ∼ ∼ - where i αβγδ = 1 2 ( δ αγ δ βδ + δ αδ δ βγ ) Mechanical meaning of ❘ ⌢ � Q 1 = R 111 + R 122 = M 11 , 1 + M 12 , 2 Q α = R αββ ⇔ Q 2 = R 121 + R 222 = M 21 , 1 + M 22 , 2 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 50 / 85

The Bending-Gradient plate model Derivation of the Bending-Gradient theory The Bending-Gradient statically compatible fields The Bending-Gradient stress energy must be minimized over the set: � ❘ BG ⊗ ∇ R αβγ = M BG ⌢ = ▼ or SC BG : αβ,γ ∼ - Y � . ❘ � . ∼ . ✐ · ∇ - Y + F 3 = 0 or R αββ,α + F 3 = 0 ∼ ⌢ A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 51 / 85

The Bending-Gradient plate model Derivation of the Bending-Gradient theory The Bending-Gradient constitutive equations Now we define the generalized strains as: ∼ = ∂ w ∗ KL ⌢ = ∂ w ∗ BG χ and Γ BG ∂ ❘ ∂ ▼ ⌢ ∼ Note that the third-order tensor has the symmetry: Γ αβγ = Γ βαγ This leads to the following constitutive equations: BG � χ αβ = d αβγδ M BG χ ∼ = ❞ ∼ : ▼ or ∼ ∼ δγ . ❘ . ⌢ = ❤ ⌢ . or Γ αβγ = h αβγδµν R νµδ Γ ⌢ ⌢ A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 52 / 85

The Bending-Gradient plate model Derivation of the Bending-Gradient theory Dualization of equilibrium equations Multiplying R αβγ = M BG αβ,γ with Φ αβγ and integrating by parts on the plate domain ω yield: � � M BG M BG αβ Φ αβγ,γ + R αβγ Φ αβγ d ω = αβ Φ αβγ n γ dl ω ∂ω Multiplying R αββ,α + F 3 = 0 with U BG and integrating by parts on the 3 plate domain ω yield: � � � R αββ U BG R αββ n α U BG F 3 U BG 3 ,α d ω = 3 dl + 3 d ω ω ∂ω ω A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 53 / 85

The Bending-Gradient plate model Derivation of the Bending-Gradient theory Weak formulation Adding these equations leads to the following expression: � � Φ αβγ + 1 �� � M BG δ βγ U BG 3 ,α + δ αγ U BG αβ Φ αβγ,γ + R αβγ d ω = 3 ,β 2 ω � � F 3 U BG M BG αβ Φ αβγ n γ + R αββ n α U BG 3 d ω + 3 dl ω ∂ω Therefore, we have obtained the weak formulation of this plate theory: V BG int = V BG ext where � BG : � � . � � V BG . - Y U BG int = ▼ ⌢ · ∇ + T ❘ ⌢ . ⌢ + ✐ ∼ · ∇ d ω Φ Φ 3 ∼ ∼ - Y ω � � BG : � � � . ❘ � V BG F 3 U BG . U BG ext = 3 d ω + ▼ ⌢ · ♥ + ∼ . ✐ ⌢ · ♥ Φ 3 dl ∼ ∼ - - ω ∂ω and ♥ - is the in-plane unit vector outwardly normal to ω . A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 54 / 85

The Bending-Gradient plate model Derivation of the Bending-Gradient theory Kinematic compatibility conditions We identify the internal power obtained by dualization � BG : � � . � � . V BG - Y U BG + T ❘ int = ▼ Φ ⌢ · ∇ ⌢ . Φ ⌢ + ✐ ∼ · ∇ d ω 3 ∼ ∼ - Y ω with the one obtained with the constitutive equations � BG : χ . Γ V BG ∼ + η 2 T ❘ . ⌢ . int = ▼ ⌢ d ω ∼ ω Finally, we define the set of kinematically compatible fields as � χ ∼ = Φ ⌢ · ∇ KC BG : - Y η 2 Γ - Y U BG ⌢ = Φ ⌢ + ✐ ∼ · ∇ 3 ∼ to which the following boundary conditions must be added for a clamped plate: U BG = 0 and Φ ⌢ · ♥ - = 0 on ∂ω 3 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 55 / 85

The Bending-Gradient plate model Derivation of the Bending-Gradient theory Summary The Bending-Gradient plate theory equations are the following: � . ❘ �  BG ⊗ ∇ . ❘ ⌢ = ▼ and ∼ . ✐ · ∇ - Y + F 3 = 0 on ω ∼ ∼ ⌢  - Y    . ❘ BG .  ∼ = ❞ ∼ : ▼ and ⌢ = ❤ ⌢ . on ω χ Γ   ∼ ∼ ⌢ ⌢ η 2 Γ - Y U BG χ ∼ = Φ ⌢ · ∇ and ⌢ = Φ ⌢ + ✐ ∼ · ∇ on ω  ∼ 3  - Y     U BG = 0 and Φ ⌢ · ♥ - = 0 on ∂ω  3 Note that: BG + η 2 Γ BG = − U BG ∼ = ❑ ⌢ · ∇ where ❑ χ 3 ∇ - Y ⊗ ∇ ∼ ∼ - Y - Y Setting η 2 = 0 in the Bending-Gradient model leads exactly to Kirchhoff-Love plate model. A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 56 / 85

The Bending-Gradient plate model Derivation of the Bending-Gradient theory 3D localization Once the exact solution of the macroscopic problem is derived, it is possible to reconstruct the local displacement field. We suggest the following 3D displacement field where U BG , Φ ⌢ are the fields solution of the plate problem: BG = U BG K : χ ∼ + η 2 ✉ K ∇ . . � � 3 - 3 − zU BG . ✉ ❡ - Y + η ✉ ⊗ ∇ χ ∼ ⊗ ∇ 3 ∼ ⌢ - - Y η - - Defining the strain as BG = ❙ BG ∼ : σ ε ∼ ∼ ∼ it is possible to check that: � BG � BG = η 2 �� K ∇ � � � � ∼ ⊗ s ✉ 2 ε ✉ ( Y , z ) − ε δ :: χ + z Γ ⌢ · ∇ ∼ ⊗ ∇ ∼ ⌢ Y - - - Y - A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 57 / 85

The Bending-Gradient plate model The Reissner-Mindlin theory for homogeneous plates Contents The 3D Problem The asymptotic expansions for a laminated plate The Kirchhoff-Love plate model The Bending-Gradient plate model Shear effects Higher orders? Derivation of the Bending-Gradient theory The Reissner-Mindlin theory for homogeneous plates Distance between the BG and the RM models Applications of the Bending-Gradient theory to laminates Periodic plates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 57 / 85

The Bending-Gradient plate model The Reissner-Mindlin theory for homogeneous plates Homogeneous plates In this case, we have:  σ BG = 12 z i αβγδ M BG = 12 zM BG αβ δγ αβ    BG =  σ BG = η 3 1 − 4 z 2 � = η 3 1 − 4 z 2 � Q BG � � σ i αβγδ R δγβ ∼ α 3 α 2 2    σ BG = 0  33 BG and ◗ BG = ✐ . ❘ . which is a function of ▼ ∼ . ⌢ instead of the whole ❘ ∼ ∼ ⌢ - A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 58 / 85

The Bending-Gradient plate model The Reissner-Mindlin theory for homogeneous plates The constitutive equations The Bending-Gradient part of the stress energy becomes: = 1 ⌢ = 1 w ∗ BG � � . ❤ . ❘ BG · ❤ RM · ◗ . . BG T ❘ ❘ ⌢ . ⌢ . 2 ◗ ⌢ ⌢ ∼ 2 - - with: RM · ✐ ⌢ = ✐ ❤ ∼ · ❤ ⌢ ∼ ∼ ∼ ∼ where the Reissner’s shear forces stiffness is given by: αβ = 6 h RM 5 ❙ α 3 β 3 6 (it is equal to 5 G δ αβ with G the shear modulus for isotropic plates). The Bending-Gradient constitutive equation becomes: . ❘ . Γ ⌢ = ❤ ⌢ . ⌢ = ✐ ∼ · γ ⌢ ∼ - with RM · ◗ BG - = ❤ γ ∼ - A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 59 / 85

The Bending-Gradient plate model The Reissner-Mindlin theory for homogeneous plates The kinematics Using the kinematic compatibility η 2 Γ - Y U BG ⌢ = Φ ⌢ + ✐ ∼ · ∇ 3 , ∼ we find that Φ ⌢ is also of the form: Φ ⌢ = ✐ ∼ · ϕ ∼ - where ϕ - is the classical rotation vector of the Reissner theory. Therefore, the kinematic unknowns are U BG and ϕ - , and we have: 3 BG � = - ⊗ s ∇ = ∼ : ▼ χ ϕ ❞ ∼ ∼ ∼ - Y RM · ◗ - + U BG BG η 2 γ = ϕ = η 2 ❤ ⊗ ∇ 3 ∼ - Y - - A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 60 / 85

The Bending-Gradient plate model The Reissner-Mindlin theory for homogeneous plates Static The following boundary conditions must be added for a clamped plate: U BG = 0 and - = 0 on ∂ω ϕ 3 Finally, the balance equations are: � ▼ BG · ∇ BG = 0 on ω - Y − ◗ ∼ - BG · ∇ ◗ - Y + F 3 = 0 on ω - In conclusion: the Bending-Gradient theory completely coincides for homogeneous plates with the Reissner-Mindlin model. A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 61 / 85

The Bending-Gradient plate model Distance between the BG and the RM models Contents The 3D Problem The asymptotic expansions for a laminated plate The Kirchhoff-Love plate model The Bending-Gradient plate model Shear effects Higher orders? Derivation of the Bending-Gradient theory The Reissner-Mindlin theory for homogeneous plates Distance between the BG and the RM models Applications of the Bending-Gradient theory to laminates Periodic plates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 61 / 85

The Bending-Gradient plate model Distance between the BG and the RM models Distance between the BG and the RM models We introduce the following relative distance: W � � ❤ ∆ RM/BG = ⌢ ⌢ � ❤ ⌢ � ⌢ where � .. . ❤ . . � ❤ ⌢ � = T ❤ ⌢ . ⌢ ⌢ ⌢ ⌢ W is the pure is the norm for Bending-Gradient compliance tensors and ❤ ⌢ ⌢ warping part of ❤ ⌢ : ⌢ ⌢ − 4 W = ❤ . ❤ . ✐ . . ∼ . ⌢ . ❤ 9 ✐ ∼ · ✐ ∼ · ✐ ⌢ ⌢ ∼ ∼ ⌢ ∼ ∼ ⌢ ∼ ∆ RM/BG gives an estimate of the pure warping fraction of the shear stress energy. When the plate constitutive equation is restricted to a Reissner-Mindlin one we have exactly ∆ RM/BG = 0. A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 62 / 85

The Bending-Gradient plate model Distance between the BG and the RM models Distance between the BG and the RM models x 3 x 2 x 1 θ [0 ◦ ] [30 ◦ , − 30 ◦ ] s [0 ◦ , − 45 ◦ , 90 ◦ , 45 ◦ ] s Stack ∆ RM/BG 0 16.0% 12.4% Table: The criterion ∆ RM/BG for several laminates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 63 / 85

Applications of the Bending-Gradient theory to laminates Contents The 3D Problem The asymptotic expansions for a laminated plate The Kirchhoff-Love plate model The Bending-Gradient plate model Applications of the Bending-Gradient theory to laminates Voigt Notations Cylindrical bending of laminates Numerical illustrations Periodic plates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 63 / 85

Applications of the Bending-Gradient theory to laminates Voigt Notations Contents The 3D Problem The asymptotic expansions for a laminated plate The Kirchhoff-Love plate model The Bending-Gradient plate model Applications of the Bending-Gradient theory to laminates Voigt Notations Cylindrical bending of laminates Numerical illustrations Periodic plates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 63 / 85

Applications of the Bending-Gradient theory to laminates Voigt Notations Voigt Notations � � We introduce the linear operator reallocating tensor components. For • ∼ instance, the bending moment and the curvature are reallocated in a vector form:     M 11 χ 11 � � � � χ 22 ▼ = M 22 and χ = √ √ ∼     ∼ 2 M 12 2 χ 12 and the fourth-order compliance tensor ❞ ∼ is reallocated in a matrix form ∼ √  2 d 1211  d 1111 d 2211 √ � � ❞ = 2 d 1222 d 2211 d 2222 √ √ ∼   ∼ 2 d 1211 2 d 1222 2 d 1212 so that the constitutive equation � � � � � � χ ∼ = ❞ ∼ : ▼ becomes χ = ❞ · ▼ ∼ ∼ ∼ ∼ ∼ ∼ σ . The same for ❉ ∼ and ❈ ∼ ∼ ∼ A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 64 / 85

Applications of the Bending-Gradient theory to laminates Voigt Notations Voigt Notations � � The constitutive sixth-order tensor ❤ ⌢ is turned into the 6 × 6 matrix ❤ : ⌢ ⌢ ⌢ √ √   h 111111 h 111122 2 h 111121 h 111211 h 111222 2 h 111221 √ √ h 221111 h 221122 2 h 221121 h 221211 h 221222 2 h 221221   √ √ √ √   2 h 121111 2 h 121122 2 h 121121 2 h 121211 2 h 121222 2 h 121221   √ √   h 112111 h 112122 2 h 112121 h 112211 h 112222 2 h 112221   √ √    h 222111 h 222122 2 h 222121 h 222211 h 222222 2 h 222221  √ √ √ √   2 h 122111 2 h 122122 2 h 122121 2 h 122211 2 h 122222 2 h 122221 The third-order tensors Γ ⌢ and ❘ ⌢ are reallocated in a vector form:  Γ 111   R 111  Γ 221 R 221 √ √         2Γ 121 2 R 121 � � � � � � � � � �     Γ = , ❘ = and Γ = ❤ · ❘     ⌢ ⌢ ⌢ ⌢ ⌢ Γ 112 R 112 ⌢         Γ 222 R 222  √   √  2Γ 122 2 R 122 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 65 / 85

Applications of the Bending-Gradient theory to laminates Cylindrical bending of laminates Contents The 3D Problem The asymptotic expansions for a laminated plate The Kirchhoff-Love plate model The Bending-Gradient plate model Applications of the Bending-Gradient theory to laminates Voigt Notations Cylindrical bending of laminates Numerical illustrations Periodic plates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 65 / 85

Applications of the Bending-Gradient theory to laminates Cylindrical bending of laminates Pagano’s boundary value problem (Pagano, 1969) CFRP layers with different orientiations: F 3 ( Y 1 ) = − F 0 sin κ Y 1 where n π , n ∈ N + ∗ is the non-dimensional wavelength of the loading. 1 λ = 1 /κ = η 2 F 3 / 2 z Y 2 Y 1 η 2 F 3 / 2  σ 11 ( z ) = 0  σ 12 ( z ) = 0 u 3 ( z ) = 0  1 Invariant in x 2 -Direction, “periodic” in x 1 -Direction ⇒ No boundary layer! A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 66 / 85

Applications of the Bending-Gradient theory to laminates Cylindrical bending of laminates Resolution of the Bending-Gradient problem All the non-dimensional fields are invariant in Y 2 -Direction From χ αβ = Φ αβγ,γ , we obtain:  χ 11   Φ 111 , 1   Φ 1 , 1  � �  =  = χ = χ 22 Φ 221 , 1 Φ 2 , 1  √  √   ∼ 2 χ 12 2Φ 121 , 1 Φ 3 , 1 The equilibrium equations write as:  R 111    M 11 , 1 R 221 M 22 , 1 √ √         2 R 121 2 M 12 , 1 � �     ❘ = = and M 11 , 11 = − F 3 ( Y 1 )     ⌢ R 112 0         R 222 0     √ 0 2 R 122 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 67 / 85

Applications of the Bending-Gradient theory to laminates Cylindrical bending of laminates Shear constitutive equation Taking into account R 112 = R 222 = R 122 = 0, U 3 , 2 = 0, shear constitutive equation is rewritten in two parts. A first part with unknowns involving active boundary conditions:         Φ 1 h 11 h 12 h 13 M 11 , 1 U 3 , 1  = η 2  ·  − Φ 2 0 h 12 h 22 h 23 M 22 , 1 √      Φ 3 h 13 h 23 h 33 2 M 12 , 1 0 and a second part which enables the derivation of Φ 4 = Φ 112 , Φ 5 = Φ 222 , √ Φ 6 = 2Φ 122 on which no boundary condition applies:         Φ 4 M 11 , 1 0 h 41 h 42 h 43  = η 2  ·  − Φ 5 h 51 h 52 h 53 M 22 , 1 0 √ √      Φ 6 h 61 h 62 h 63 2 M 12 , 1 U 3 , 1 / 2 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 68 / 85

Applications of the Bending-Gradient theory to laminates Cylindrical bending of laminates Final System Finally, combining the above equations leads to the following set of equations which fully determines the problem:  M 11 , 11 = F 0 sin κ Y 1      U 3 , 11   � � � � � �   − η 2 ❤ ❞ · ▼ ∼ · ▼ , 11 = 0     ∼ ∼ ∼ ∼ 0   � �  ▼ = 0 for Y 1 = 0 and Y 1 = 1   ∼     U 3 = 0 for Y 1 = 0 and Y 1 = 1 � � where for convenience, ❤ ∼ is the 3 × 3 submatrix of ❤ : ⌢ ⌢   h 11 h 12 h 13 ∼ = ❤ h 12 h 22 h 23   h 13 h 23 h 33 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 69 / 85

Applications of the Bending-Gradient theory to laminates Cylindrical bending of laminates Solution This differential system is well-posed and the solution is unique. Its is of the form: � � � � U 3 = U ∗ ▼ = ▼ ∗ sin κ Y 1 and 3 sin κ Y 1 ∼ ∼ � � and U ∗ where ▼ ∗ 3 are constants explicitly known in terms of the ∼ problem inputs. A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 70 / 85

Applications of the Bending-Gradient theory to laminates Numerical illustrations Contents The 3D Problem The asymptotic expansions for a laminated plate The Kirchhoff-Love plate model The Bending-Gradient plate model Applications of the Bending-Gradient theory to laminates Voigt Notations Cylindrical bending of laminates Numerical illustrations Periodic plates A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 70 / 85

Applications of the Bending-Gradient theory to laminates Numerical illustrations Stress distributions for a [30 ◦ , − 30 ◦ , 30 ◦ ] stack KL KL KL 0 . 4 0 . 4 0 . 4 BG BG BG Pagano Pagano Pagano 0 . 2 0 . 2 0 . 2 x 3 /t 0 . 0 0 . 0 0 . 0 − 0 . 2 − 0 . 2 − 0 . 2 − 0 . 4 − 0 . 4 − 0 . 4 − 10 − 5 0 5 10 − 10 − 5 0 5 10 − 10 − 5 0 5 10 t 2 σ 11 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) t 2 σ 22 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) t 2 σ 12 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) KL KL KL 0 . 4 0 . 4 0 . 4 BG BG BG Pagano Pagano Pagano 0 . 2 0 . 2 0 . 2 x 3 /t L/t = 1 . 00 0 . 0 0 . 0 0 . 0 − 0 . 2 − 0 . 2 − 0 . 2 − 0 . 4 − 0 . 4 − 0 . 4 0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 − 0 . 4 − 0 . 2 0 . 0 0 . 2 0 . 4 tσ 13 (0 , b/ 2 , x 3 ) / ( p 3 λ ) tσ 23 ( a/ 2 , 0 , x 3 ) / ( p 3 λ ) σ 33 ( a/ 2 , b/ 2 , x 3 ) /p 3 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 71 / 85

Applications of the Bending-Gradient theory to laminates Numerical illustrations Stress distributions for a [30 ◦ , − 30 ◦ , 30 ◦ ] stack KL KL KL 0 . 4 0 . 4 0 . 4 BG BG BG Pagano Pagano Pagano 0 . 2 0 . 2 0 . 2 x 3 /t 0 . 0 0 . 0 0 . 0 − 0 . 2 − 0 . 2 − 0 . 2 − 0 . 4 − 0 . 4 − 0 . 4 − 10 − 5 0 5 10 − 10 − 5 0 5 10 − 10 − 5 0 5 10 t 2 σ 11 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) t 2 σ 22 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) t 2 σ 12 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) KL KL KL 0 . 4 0 . 4 0 . 4 BG BG BG Pagano Pagano Pagano 0 . 2 0 . 2 0 . 2 x 3 /t L/t = 1 . 39 0 . 0 0 . 0 0 . 0 − 0 . 2 − 0 . 2 − 0 . 2 − 0 . 4 − 0 . 4 − 0 . 4 0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 − 0 . 4 − 0 . 2 0 . 0 0 . 2 0 . 4 tσ 13 (0 , b/ 2 , x 3 ) / ( p 3 λ ) tσ 23 ( a/ 2 , 0 , x 3 ) / ( p 3 λ ) σ 33 ( a/ 2 , b/ 2 , x 3 ) /p 3 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 71 / 85

Applications of the Bending-Gradient theory to laminates Numerical illustrations Stress distributions for a [30 ◦ , − 30 ◦ , 30 ◦ ] stack KL KL KL 0 . 4 0 . 4 0 . 4 BG BG BG Pagano Pagano Pagano 0 . 2 0 . 2 0 . 2 x 3 /t 0 . 0 0 . 0 0 . 0 − 0 . 2 − 0 . 2 − 0 . 2 − 0 . 4 − 0 . 4 − 0 . 4 − 10 − 5 0 5 10 − 10 − 5 0 5 10 − 10 − 5 0 5 10 t 2 σ 11 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) t 2 σ 22 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) t 2 σ 12 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) KL KL KL 0 . 4 0 . 4 0 . 4 BG BG BG Pagano Pagano Pagano 0 . 2 0 . 2 0 . 2 x 3 /t L/t = 1 . 95 0 . 0 0 . 0 0 . 0 − 0 . 2 − 0 . 2 − 0 . 2 − 0 . 4 − 0 . 4 − 0 . 4 0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 − 0 . 4 − 0 . 2 0 . 0 0 . 2 0 . 4 tσ 13 (0 , b/ 2 , x 3 ) / ( p 3 λ ) tσ 23 ( a/ 2 , 0 , x 3 ) / ( p 3 λ ) σ 33 ( a/ 2 , b/ 2 , x 3 ) /p 3 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 71 / 85

Applications of the Bending-Gradient theory to laminates Numerical illustrations Stress distributions for a [30 ◦ , − 30 ◦ , 30 ◦ ] stack KL KL KL 0 . 4 0 . 4 0 . 4 BG BG BG Pagano Pagano Pagano 0 . 2 0 . 2 0 . 2 x 3 /t 0 . 0 0 . 0 0 . 0 − 0 . 2 − 0 . 2 − 0 . 2 − 0 . 4 − 0 . 4 − 0 . 4 − 10 − 5 0 5 10 − 10 − 5 0 5 10 − 10 − 5 0 5 10 t 2 σ 11 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) t 2 σ 22 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) t 2 σ 12 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) KL KL KL 0 . 4 0 . 4 0 . 4 BG BG BG Pagano Pagano Pagano 0 . 2 0 . 2 0 . 2 x 3 /t L/t = 2 . 71 0 . 0 0 . 0 0 . 0 − 0 . 2 − 0 . 2 − 0 . 2 − 0 . 4 − 0 . 4 − 0 . 4 0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 − 0 . 4 − 0 . 2 0 . 0 0 . 2 0 . 4 tσ 13 (0 , b/ 2 , x 3 ) / ( p 3 λ ) tσ 23 ( a/ 2 , 0 , x 3 ) / ( p 3 λ ) σ 33 ( a/ 2 , b/ 2 , x 3 ) /p 3 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 71 / 85

Applications of the Bending-Gradient theory to laminates Numerical illustrations Stress distributions for a [30 ◦ , − 30 ◦ , 30 ◦ ] stack KL KL KL 0 . 4 0 . 4 0 . 4 BG BG BG Pagano Pagano Pagano 0 . 2 0 . 2 0 . 2 x 3 /t 0 . 0 0 . 0 0 . 0 − 0 . 2 − 0 . 2 − 0 . 2 − 0 . 4 − 0 . 4 − 0 . 4 − 10 − 5 0 5 10 − 10 − 5 0 5 10 − 10 − 5 0 5 10 t 2 σ 11 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) t 2 σ 22 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) t 2 σ 12 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) KL KL KL 0 . 4 0 . 4 0 . 4 BG BG BG Pagano Pagano Pagano 0 . 2 0 . 2 0 . 2 x 3 /t L/t = 3 . 79 0 . 0 0 . 0 0 . 0 − 0 . 2 − 0 . 2 − 0 . 2 − 0 . 4 − 0 . 4 − 0 . 4 0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 − 0 . 4 − 0 . 2 0 . 0 0 . 2 0 . 4 tσ 13 (0 , b/ 2 , x 3 ) / ( p 3 λ ) tσ 23 ( a/ 2 , 0 , x 3 ) / ( p 3 λ ) σ 33 ( a/ 2 , b/ 2 , x 3 ) /p 3 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 71 / 85

Applications of the Bending-Gradient theory to laminates Numerical illustrations Stress distributions for a [30 ◦ , − 30 ◦ , 30 ◦ ] stack KL KL KL 0 . 4 0 . 4 0 . 4 BG BG BG Pagano Pagano Pagano 0 . 2 0 . 2 0 . 2 x 3 /t 0 . 0 0 . 0 0 . 0 − 0 . 2 − 0 . 2 − 0 . 2 − 0 . 4 − 0 . 4 − 0 . 4 − 10 − 5 0 5 10 − 10 − 5 0 5 10 − 10 − 5 0 5 10 t 2 σ 11 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) t 2 σ 22 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) t 2 σ 12 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) KL KL KL 0 . 4 0 . 4 0 . 4 BG BG BG Pagano Pagano Pagano 0 . 2 0 . 2 0 . 2 x 3 /t L/t = 5 . 28 0 . 0 0 . 0 0 . 0 − 0 . 2 − 0 . 2 − 0 . 2 − 0 . 4 − 0 . 4 − 0 . 4 0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 − 0 . 4 − 0 . 2 0 . 0 0 . 2 0 . 4 tσ 13 (0 , b/ 2 , x 3 ) / ( p 3 λ ) tσ 23 ( a/ 2 , 0 , x 3 ) / ( p 3 λ ) σ 33 ( a/ 2 , b/ 2 , x 3 ) /p 3 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 71 / 85

Applications of the Bending-Gradient theory to laminates Numerical illustrations Stress distributions for a [30 ◦ , − 30 ◦ , 30 ◦ ] stack KL KL KL 0 . 4 0 . 4 0 . 4 BG BG BG Pagano Pagano Pagano 0 . 2 0 . 2 0 . 2 x 3 /t 0 . 0 0 . 0 0 . 0 − 0 . 2 − 0 . 2 − 0 . 2 − 0 . 4 − 0 . 4 − 0 . 4 − 10 − 5 0 5 10 − 10 − 5 0 5 10 − 10 − 5 0 5 10 t 2 σ 11 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) t 2 σ 22 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) t 2 σ 12 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) KL KL KL 0 . 4 0 . 4 0 . 4 BG BG BG Pagano Pagano Pagano 0 . 2 0 . 2 0 . 2 x 3 /t L/t = 7 . 37 0 . 0 0 . 0 0 . 0 − 0 . 2 − 0 . 2 − 0 . 2 − 0 . 4 − 0 . 4 − 0 . 4 0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 − 0 . 4 − 0 . 2 0 . 0 0 . 2 0 . 4 tσ 13 (0 , b/ 2 , x 3 ) / ( p 3 λ ) tσ 23 ( a/ 2 , 0 , x 3 ) / ( p 3 λ ) σ 33 ( a/ 2 , b/ 2 , x 3 ) /p 3 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 71 / 85

Applications of the Bending-Gradient theory to laminates Numerical illustrations Stress distributions for a [30 ◦ , − 30 ◦ , 30 ◦ ] stack KL KL KL 0 . 4 0 . 4 0 . 4 BG BG BG Pagano Pagano Pagano 0 . 2 0 . 2 0 . 2 x 3 /t 0 . 0 0 . 0 0 . 0 − 0 . 2 − 0 . 2 − 0 . 2 − 0 . 4 − 0 . 4 − 0 . 4 − 10 − 5 0 5 10 − 10 − 5 0 5 10 − 10 − 5 0 5 10 t 2 σ 11 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) t 2 σ 22 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) t 2 σ 12 ( a/ 2 , b/ 2 , x 3 ) / ( p 3 λ 2 ) KL KL KL 0 . 4 0 . 4 0 . 4 BG BG BG Pagano Pagano Pagano 0 . 2 0 . 2 0 . 2 x 3 /t L/t = 10 . 28 0 . 0 0 . 0 0 . 0 − 0 . 2 − 0 . 2 − 0 . 2 − 0 . 4 − 0 . 4 − 0 . 4 0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 − 0 . 4 − 0 . 2 0 . 0 0 . 2 0 . 4 tσ 13 (0 , b/ 2 , x 3 ) / ( p 3 λ ) tσ 23 ( a/ 2 , 0 , x 3 ) / ( p 3 λ ) σ 33 ( a/ 2 , b/ 2 , x 3 ) /p 3 A. Leb´ ee (Laboratoire Navier) The Bending-Gradient theory 20-26 July 2015 71 / 85