The Bending-Gradient theory for laminates and in-plane periodic plates

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HAL Id: cel-01266716

https://hal-enpc.archives-ouvertes.fr/cel-01266716

Submitted on 3 Feb 2016

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The Bending-Gradient theory for laminates and in-plane

periodic plates

Arthur Lebée

To cite this version:

Arthur Lebée. The Bending-Gradient theory for laminates and in-plane periodic plates. Doctoral. Arpino, Italy. 2015. �cel-01266716�

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The Bending-Gradient theory for laminates and

in-plane periodic plates

Arthur Leb´ee

Laboratoire Navier (UMR CNRS 8205)

Universit´e Paris-Est - ´Ecole des Ponts ParisTech - IFSTTAR

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The open question of shear forces in heterogeneous plates

I Thick or Thin plates?

I Thin Plate – Kirchhoff-Love –ϕα= U3,α:

asymptotic derivation, transverse shear effects neglected (Kirchhoff (1850); Love (1888); Ciarlet and Destuynder (1979))

I Thick Plate – Reissner-Mindlin –ϕα6= U3,α:

axiomatic and controversial. Natural boundary conditions! (Reissner (1944); Hencky (1947); Mindlin (1951))

. . . plates are generalized continua!

I Deriving formally Reissner-Mindlin plate model from asymptotic expansions?

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The open question of shear forces in heterogeneous plates

I Thick or Thin plates?

I Thin Plate – Kirchhoff-Love –ϕα= U3,α:

asymptotic derivation, transverse shear effects neglected (Kirchhoff (1850); Love (1888); Ciarlet and Destuynder (1979))

I Thick Plate – Reissner-Mindlin –ϕα6= U3,α:

axiomatic and controversial. Natural boundary conditions! (Reissner (1944); Hencky (1947); Mindlin (1951))

I Deriving formally Reissner-Mindlin plate model from asymptotic expansions?

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The 3D Problem

The 3D problem configuration

C

_∼_∼t (

x

3): even Ωt

f

-f

-ωL+ ωL ωL− ∂ωL

e

-₃

e

-₂

e

-1

t

L

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The 3D Problem

The 3D problem equations

               σt ij,j = 0 on Ωt. σt ij= C t ijkl(x3)εtkl on Ω t . σt i3=±fi onωL±. εt ij = u t (i,j) on Ω t_. ut i = 0 on∂ω L ×] − t/2, t/2[

I monoclinic and even C_∼

∼

t_:

Ct

αβγ3= C333αt = 0,

α, β, γ, ... = 1, 2.

I symmetrically laminated plate

I symmetric transverse load f -= f3-e₃                    σ ∼t· ∇- = 0 on Ω t_. σ ∼t -x =C∼ ∼ t (x₃) : ε_∼t x - on Ωt_. σ ∼t· ±e-₃ =-f onω L±_. ε ∼ t₌u -t ⊗s∇- on Ω t_. u -t _{= 0} _on_∂ωL ×] − t/2, t/2[ ⇒ pure bending: I ut 3 andσα3t even / x3 I ut α,σαβt and σ t 33 odd / x3

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The 3D Problem

Proof of skew symmetry

Let-u

t0 _{be the image of} u

-t _{by the symmetry with respect to the}

(x1, x2)-plane u -t0_(x 1, x2, x3) =   +ut 1(x1, x2,−x3) +ut 2(x1, x2,−x3) −ut 3(x1, x2,−x3)   Obviously,u -t0 _{= 0} _on_∂ωL

×] − t/2, t/2[. Its corresponding strain ε ∼ t0 ₌u -t0 ⊗s∇- is: ε ∼ t0_(x 1, x2, x3) =   εt 11 εt12 −εt13 εt 12 εt22 −εt23 −εt 13 −εt23 εt33  (x1, x2,−x3)

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The 3D Problem

Proof of skew symmetry

Its corresponding stress σ_∼t0 _{- =}x C

∼ ∼ t (x3) : ε_∼t0 - is:x         σt0 11 σt0 22 σt0 12 σt0 13 σt0 23 σt0 33         =         Ct 1111 C1122t C1112t 0 0 C1133t Ct 2222 C2212t 0 0 C2233t Ct 1212 0 0 C1233t Ct 1313 C1323t 0 SYM Ct 2323 0 Ct 3333                 εt0 11 εt0 22 2εt0 12 2εt0 13 2εt0 23 εt0 33         Because C_∼ ∼ t

(x3) is even and monoclinic, then:

σ_∼t0_(x 1, x2, x3) =   σt 11 σ12t −σ13t σt 12 σ22t −σ23t −σt 13 −σ23t σ33t  (x₁, x₂,−x₃)

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The 3D Problem

Proof of skew symmetry

Finally, the balance equation σ_∼t0

· ∇- = 0 is easy to check and we have:

σ_∼t ·±e-₃ =−f-=−f3-e₃ onω L± . Therefore, u -t0 x - = −u -t _{- , ε}x ∼ t0 x - = −ε∼ t _{- , σ}x ∼ t0 x - = −σ∼ t x - Hence,   +ut 1(x1, x2,−x3) +ut 2(x1, x2,−x3) −ut 3(x1, x2,−x3)  =   −ut 1(x1, x2, x3) −ut 2(x1, x2, x3) −ut 3(x1, x2, x3)  ...

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The 3D Problem

Variational formulation

The set of statically compatible stress fields is:

SC3D,t :    σ_∼t · ∇- = 0 on Ω t σ_∼t ·±e-₃ =-f onω L± ,

The set of kinematically compatible displacement fields is:

KC3D,t : ( ε∼ t ₌u -t ⊗s∇- on Ω t u -t = 0 on∂ωL ×] − t/2, t/2[

The strain and stress energy density w3D and w∗3D are respectively given by: w3D ε ∼ = 1 2ε∼:C∼∼ t : ε_∼, w∗3D σ_∼ = 1 2σ∼:S∼∼ t : σ_∼ with: S ∼ ∼ t ₌C ∼ ∼ t−1

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The 3D Problem

Potential energy

P3D_∼εt = min ε ∼∈KC3D,t n P3D_∼εo

The potential energy P3D is given by:

P3Dε_∼= Z Ωt w3D_∼εd Ωt − Z ωL f3 u3++ u3− dωL u -±₌_u_-(x

1, x2,±t/2) are the 3D displacement fields on the upper and

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The 3D Problem

Complementary energy

P∗3Dσ_∼t₌ _min σ_∼∈SC3D,t n P∗3Dσ_∼o

The complementary potential energy P∗3D given by:

P∗3Dσ_∼= Z

Ωt

w∗3Dσ_∼d Ωt

At the solution (Clapeyron’s formula):

P3D_∼εt_{+ P}∗3D_σ

∼

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The 3D Problem

Building a plate model?

For typical width L and thickness t, let _Lt → 0

I Solve a 2D problem, called the “plate problem”

I “fair” 3D displacement localization

I “fair” 3D stress localization

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The asymptotic expansions for a laminated plate

Differential system depending on a small parameter

We want to solve the following differential equation on [0, 1]:

u00(x )₋ηu (x ) = 0, u (0) = 0, u (1) = a

where η> 0 is a small parameter.

The solution is trivial:

uη(x ) = asinh √ηx sinh √η

The limit of uη_{(x ) as} _η _{goes to 0}+ _is:

lim

η→0+u

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Taylor’s series

Using Taylor’s series:

sinh √ηx = √_η_x1 1! + √_η_x3 3! +· · · sinh √η = √_η1 1! + √_η3 3! +· · · We obtain: uη(x ) = ax + aη1x 3_{− x} 3! + aη 2 x5− x 5! − x3_{− x} 3!3! +_{· · ·}

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The method

I Write uη_{(x ) as a series:}

uη(x ) = u0(x ) +η1u1(x ) +...· · · +ηiui(x ) +· · ·

where ui are unknown functions.

I Inject this series in the differential system

u00(x )₋ηu (x ) = 0, u (0) = 0, u (1) = a

and make null all the terms inηi.

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Resolution

The cascade system is:

Term in η0 : u000(x ) = 0, u0(0) = 0, u0(1) = a

Term in η1 : u100(x ) = u0, u1(0) = 0, u1(1) = 0

....

Term in ηi : ui 00(x ) = ui −1, ui(0) = 0, ui(1) = 0

The solution is obtained by mathematical induction:

u0(x ) = ax, u1(x ) = ax 3_{− x} 3! , u 2_{(x ) = a}x5− x 5! − a x3_{− x} 3!3! ,· · ·

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The asymptotic expansions for a laminated plate Scaling of the 3D problem

Change of variables

Ωt f -+ f -− ωL+ ωL _ωL− ∂ωL e -₃,x3 e -₂,x2 e -₁,x1 t L Ω η2F₂3 η2F3 2 ω+ ω ω− ∂ω e -₃,z e -₂,Y2 e -₁,Y1 1 1 I Yα= xα

L for the in-plane variables, Yα ∈ ω

I z = x3

t for the out-of-plane variable, z ∈] −

1 2,12[

I η= t

L is the small parameter

The fourth-order elasticity tensor can be rewritten as: C_∼t

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Non-dimensional fields

We define the non-dimensional fields (u-, ε_∼, σ_∼) as follows:    u -t _(x 1, x2, x3) = Lu- (x1/L, x2/L, x3/t) = Lu- (Y1, Y2, z) ε ∼t (x1, x2, x3) = ∼ε (x1/L, x2/L, x3/t) = ∼ε (Y1, Y2, z) σ_∼t _(x 1, x2, x3) = σ_∼ (x1/L, x2/L, x3/t) = σ_∼ (Y1, Y2, z)

The derivation rule for these fields is:

∇- = d dx1 , d dx2 , d dx3 = L−1 ∂ ∂Y1, ∂ ∂Y2, 0 + t−1 0, 0, ∂ ∂z = L−1∇ -Y + t −1_∇ -z = L−1∇ -η (Y ,z) where ∇ -η (Y ,z)..= ∇ -Y + 1 η ∇-z

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Natural scaling of the stress

         σt αβ,β+σ t α3,3 = 0 σt α3,α+σ33,3t = 0 σt 33(±t/2) = ±f3 σt α3(±t/2) = 0 ⇒          σt α3 =− Z x3 −t/2 σt αβ,βdu σt 33=− Z x3 −t/2 σt α3,αdu− f3 σt αβ ∼η0 ⇒ σtα3∼η1, σ33t ∼η2 and f3 ∼η2

The out-of-plane loading is scaled as:

f

-(x1, x2) =η2F3(Y1, Y2)

2 -e

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The non-dimensional 3D problem

The set of statically compatible fields can be rewritten as:

SC3D :      σ_∼_{· ∇} -η (Y ,z)= 0 on Ω =ω×] − 1 2, + 1 2[, σ_∼_·_±e -3 = η2 2F3e -3 onω ±

The kinematically compatible fields becomes:

KC3D :    ε ∼=-u⊗ s_∇ -η (Y ,z) on Ω, u -= 0 on∂ω×] − 1 2, + 1 2[ The constitutive law becomes:

σ_∼(Y1, Y2, z) =C_∼_∼(z) : ε_∼(Y1, Y2, z) ∇ -η (Y ,z)= ∇ -Y + 1 η ∇ -z

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Properties of the non-dimensional solution

For given ω,C_∼

∼, F3,η

whereC_∼

∼ 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 x3 → z:

I u3 andσα3 are even in z

I uα,σαβ andσ33 are odd in z

We have the following new properties:

I u3 andσα3 are odd in η

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Proof

New change of variable z0 =₋x3

t for the out-of-plane variable. The new

non-dimensional fields (-u

0_{, ε}

∼0, σ∼0) are defined by:

   u -t _(x 1, x2, x3) = L-u 0 _(x 1/L, x2/L,−x3/t) = Lu -0 _(Y 1, Y2, z0) ε ∼ t _(x 1, x2, x3) = ε_∼0 (x1/L, x2/L,−x3/t) = _∼ε0 (Y1, Y2, z0) σ_∼t _(x 1, x2, x3) = σ_∼0 (x1/L, x2/L,−x3/t) = σ_∼0 (Y1, Y2, z0)

The new derivation rule for these fields is: ∇- = d dx1 , d dx2 , d dx3 = L−1 ∂ ∂Y1 , ∂ ∂Y2 , 0 − t−1 0, 0, ∂ ∂z0 = L−1∇ -Y − t −1_∇ -z0 = L−1∇ -−η (Y ,z0₎ where ∇ -−η (Y ,z0₎= ∇ -Y − 1 η ∇ -z0

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The new equations

SC3D0 :      σ_∼0_{· ∇} -−η (Y ,z0₎= 0 on Ω =ω×] − 1 2, + 1 2[, σ_∼0_·_±e -3 =₋η2 2F3-e 3 onω ± KC3D0 :    ε ∼ 0 ₌_u -0 ⊗s∇ -−η (Y ,z0₎ on Ω, u -0 = 0 on∂ω×] −1 2, + 1 2[ σ_∼0 Y1, Y2, z0 =C_∼ ∼ z 0_{: ε} ∼ 0 _Y 1, Y2, z0

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Effect of the transformation

η

→ −

η

on the

non-dimensional solution

The new non-dimensional fields (u

-0_{, ε}

∼0, σ∼0) are solutions of the same

equations as for (u-, ε_∼, σ_∼) where F3 → −F3 andη → −η:

(u

-0_{, ε}

∼0, σ∼0) Y1, Y2, z0 = (-, εu _∼, σ_∼)(−F3,−η) Y1, Y2, z0

Moreover, by definition, the new non-dimensional fields coincide with the initial ones with z =_−z0:

(u -0 , ε_∼0, σ_∼0) Y1, Y2, z0 = (-, εu _∼, σ_∼)(F3,η) Y1, Y2,−z0 Hence, we have: (-, εu _∼, σ_∼)(−η)(Y1, Y2, z) =−(u-, ε_∼, σ_∼)(η)(Y1, Y2,−z)

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The asymptotic expansions for a laminated plate The expansion

Expansion

We assume    u - = η −1_u -−1 ₊ _η0u -0 ₊ _η1u -1 ₊_{· · ·} ε ∼ = η0∼ε0 + η1∼ε1 +· · · σ ∼ = η0σ∼0 + η1σ∼1 +· · · u -p_{, ε}

∼p and σ∼p, p =−1, 0, 1, 2..., are functions of (Y1, Y2, z)

Because:

I u3 andσα3 are odd in η

I uα,σαβ andσ33 are even in η

we have:

I up

3 andσ

p

α3 are null for even p and even in z for odd p.

I up

α,σ

p

αβ andσ

p

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Statics

The normalized 3D equilibrium equation becomes:

σ_∼_{· ∇} -η (Y ,z)=η −1_σ ∼ 0 · ∇ -z +η0 σ_∼0 · ∇ -Y + σ∼ 1 · ∇ -z +_{· · · = 0.} Hence, σ_∼0 · ∇-_z = 0 and σ∼ p · ∇ -Y + σ∼ p+1 · ∇-_z = 0, p≥ 0. Or in components: σ0 i3,3= 0 and σ p i α,α+σ p+1 i3,3= 0, p≥ 0.

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Statics

The static boundary conditions on ω± _writes:

σ_∼p ·±e -3 = 0 when p_{6= 2 and σ}_∼2 ·±e -3 = F3 2-e 3. Or in components: σp i3 Y1, Y2,± 1 2 = 0 when p _{6= 2} σ2 α3 Y1, Y2,± 1 2 = 0 and σ2 33 Y1, Y2,± 1 2 =_±1 2F3(Y1, Y2)

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Kinematics

The non-dimensional displacement field is: u -=η −1_u -−1₊ η0-u 0₊ η1-u 1₊_{· · ·}

The non-dimensional strain field is: ε ∼=u-⊗ s_∇ -η (Y ,z)=η −2_ε ∼ −2₊ η−1ε ∼ −1₊ η0ε ∼ 0₊ · · · with: ε ∼−2=u -−1_⊗s_∇ -_z and ε∼ p =-u p+1_⊗s_∇ -_z+u -p ⊗s∇-_Y, p≥ −1 In components: ε−2 αβ = 0, ε −2 α3= 1 2u −1 α,3 and ε −2 33= u −1 3,3

and for all p_{≥ −1:} εp αβ = 1 2 up α,β+ u p β,α , εp α3 = 1 2 up+1_α,3+ up 3,α and εp 33= u p+1 3,3

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Kinematics

The kinematic condition on the lateral boundary leads to:

∀p ≥ −1 and ∀ (Y1, Y2)∈ ∂ω, u

-p

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The Kirchhoff-Love plate model

Lower order displacements

We set ε_∼−2_{= 0 which leads to:}

ε−2 αβ = 0, ε −2 α3= 1 2u −1 α,3= 0 and ε −2 33= u −1 3,3 = 0 Hence, u -−1 _{is a function of (Y} 1, Y2). Moreover, u−1

α is null sinceη=−1 is odd:

u -−1_{= U}−1 3 (Y1, Y2)-e 3=   0 0 U−1 3  

with the boundary conditions:

U−1

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Lower order displacements

We set ε_∼−1_{= 0 which leads to:}

ε−1 αβ = 1 2 u−1 α,β+ u −1 β,α = 0, ε−1 α3= 1 2 u0 α,3+ U3,α−1 = 0, ε−1 33 = u03,3= 0 Hence, u

-0_{has the following form:}

u -0₌ −zU−1 3 ⊗∇ -Y =   −zU−1 3,1 −zU−1 3,2 0  

with the boundary conditions:

U_3,α−1nα = 0 ∀ (Y1, Y2)∈ ∂ω

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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 _{∈ [−}1₂,1₂]):

           σ_∼0 · ∇-_z = 0. σ_∼0₌_C ∼ ∼(z) : ε∼ 0_. ε ∼0=u -1_⊗s_∇ -_z+u -0_⊗s_∇ -_Y. σ_∼0 _{z =} ±12 · ±e -3 = 0                σ0 i3,3= 0 σ0 ij =Cijklε0kl ε0 αβ = zK −1 αβ ε0 α3= 12uα,31 and ε033= u13,3 σ0 i3 z =±12 = 0

The lowest-order curvature is: K_∼−1_.. =−U−1 3 ∇ -Y⊗∇ -Y or K −1 αβ ..=−U −1 3,αβ

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Resolution

From σ0 i3,3 = 0 and σi03 z =_±1 2 = 0 we obtain plane-stress: σ0 i3 = 0

The constitutive equation writes:

        σ0 11 σ0 22 σ0 12 0 0 0         =         C1111 C1122 C1112 0 0 C1133 C1122 C2222 C2212 0 0 C2233 C1112 C2212 C1212 0 0 C1233 0 0 0 C1313 C1323 0 0 0 0 C1323 C2323 0 C1133 C2233 C12331 0 0 C3333                 zK−1 11 zK−1 22 2zK₁₂−1 2ε0 13 2ε0 23 ε0 33        

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Resolution

The strain is given by:

ε0 αβ = zKαβ−1(Y1, Y2), ε 0 α3 = 0 and ε033=− zC33αβ(z) C3333(z) K−1 αβ(Y1, Y2)

The stress is given by:

σ ∼ 0 =_∼s ∼ K_{(z) :}_K ∼ −1 (Y1, Y2) or in components σij0 =sKij γδK −1 δγ

where the fourth-order stress localization tensor is:

sK αβγδ(z)..= zC σ αβγδ(z) and sKi3γδ ..= 0 and Cσ αβγδ=Cαβγδ− Cαβ33C33γδ/C3333

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Resolution

By integrating of ε0 α3 = 0 = 1 2u 1 α,3 and ε033=− zC33αβ C3333 K−1 αβ = u 1 3,3 We find: u -1₌_u ∼ -K _:_K ∼ −1_{+ U}1 3-e₃=   0 0 uK 3αβKβα−1+ U31  .

where the displacement localization tensor u_∼

-K_{(z) related to the curvature}

is given by: uK 3αβ(z)..=− " Z z −1₂ rC33αβ C3333 dr #∗ and uK_αβγ ..= 0 ∗ _{∗ ..}

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Isotropic materials

The leading order strain:

ε0 αβ = zK −1 αβ, ε 0 α3 = 0 and ε033=− ν 1− νzK −1 αα

The leading order stress is derived through C_∼

∼ σ_:   σ0 11 σ0 22 σ0 12  =    E 1−ν2 _1−νE ν2 0 E 1−ν2 0 SYM _2(1+ν)E      zK₁₁−1 zK−1 22 2zK−1 12  

The displacement corrector is:

u -1₌_u ∼ -K _:_K ∼−1+ U31-e 3=   0 0 ν 2(1−ν) 121 − z2 Kαα−1+ U31  .

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The Kirchhoff-Love plate model Macroscopic problem

Determination of

U

−1

3

Resultants

The zeroth-order bending moment is defined as

M0

αβ(Y1, Y2)..=zσ

0

αβ ,

The first-order shear force is:

Q1

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Determination of

U

−1 3 Equilibrium Recall that: σ0 i3,3= 0 and σ p i α,α+σ p+1 i3,3= 0, p≥ 0.

The bending equilibrium equations are:

z σ0 αβ,β+σ 1 α3,3 = 0 = Mαβ,β0 − Q 1 α

The out-of-plane equilibrium equation is:

σ1

3α,α+σ233,3 = 0 = Qα,α1 + F3

Finally, we have:

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Determination of

U

−1 3 Constitutive equation From D zσ_∼0E₌D_z_s ∼ ∼ K_{(z) :}_K ∼ −1E₌D_z2_C ∼ ∼ σ_{(z) :}K ∼ −1E

we obtain the Kirchhoff’s constitutive equation:

M_∼0₌_D ∼ ∼ :K∼ −1 _where: _D ∼ ∼ = D z2C_∼ ∼ σE

The Kirchhoff-Love plate equations are:                M_∼0_:_∇ -Y⊗∇ -Y + F3 = 0, on ω M_∼0₌_D ∼ ∼ :K∼ −1 , on ω K_∼−1₌ −U−1 3 ∇ -Y ⊗∇ -Y, on ω U−1 3 = 0 and U−1 3 ⊗∇ -Y · n-= 0 on ∂ω

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Summary of the Kirchhoff-Love model

The displacement is approximated by:

u -≈η −1_u -−1₊_u -0₌   −zU−1 3,1 −zU−1 3,2 η−1U₃−1  =u -LK

The strain ε_∼is approximated by ε_∼0_{6= u}

-LK_⊗s_∇ -η (Y ,z)with: ε0 αβ = zK −1 αβ(Y1, Y2), ε 0 α3 = 0 and ε033=− zC33αβ(z) C3333(z) K−1 αβ(Y1, Y2) where K_αβ−1 =−U−1 3,αβ

The stress σ_∼ is approximated by σ_∼0_{such that σ}

∼0· ∇ -η (Y ,z)6= 0 and σ0 αβ = zC σ αβγδ(z) Kδγ−1 and σ 0 i3 = 0

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Lateral boundary conditions

It should be emphasized that the assumed expansion is not compatible with clamped lateral boundary conditions. Indeed,

u -1₌_u ∼ -K _:_K ∼−1+ U31e-₃ =   0 0 uK 3αβK −1 βα+ U 1 3  6= 0.

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First-order auxiliary problem

Equilibrium equation for order 0, compatibility equation, boundary conditions and constitutive equations of order 1 lead to (for z _{∈ [−}1₂,1₂]):

           σ_∼0 · ∇-_Y + σ∼ 1 · ∇-_z = 0 σ_∼1₌_C ∼ ∼ (z) : ε∼ 1 ε ∼ 1 =u -2 ⊗s∇-_z+u -1 ⊗s∇-_Y σ_∼1 z =±1₂ · ±e-₃= 0                        σ0 i α,α+σi3,31 = 0 σ1 ij =Cijklε1kl ε1 αβ = u 1 (α,β) = 0 ε1 α3= 12 u2 α,3+ u13,α ε1 33= u3,32 = 0 σ1 i3 z =±12 = 0

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Resolution

Transverse stress From        σ0 βα,α+σ 1 β3,3= 0 σ0 βα,α=sKβαγδ(z) K −1 δγ ,α= zC σ βαγδ(z) K −1 δγ,α σ1 β3 z =±12 = 0

we obtain the first-order transverse shear stress:

σ1 α3=− Z z −1 2 rCσ αβγδ dr Kδγ,β−1

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Resolution

In-plane stress Fromε1 αβ = 0 and ε 1

33= 0 and the constitutive equation:         σ1 11 σ1 22 σ1 12 σ1 13 σ1 23 0         =         C1111 C1122 C1112 0 0 C1133 C2222 C2212 0 0 C2233 C1212 0 0 C1233 C1313 C1323 0 SYM C₂₃₂₃ 0 C3333                 0 0 0 2ε1 13 2ε1 23 0         we haveσ1

αβ = 0 and the first-order stress localization writes as:

σ ∼ 1₌_s _ ∼ K ∇_(z)_...K ∼ −1 ⊗∇ -Y

where we defined the fifth-order localization tensor as: sK ∇ αβγδη..= 0, sK ∇α3γδη(z)..=− Z z −1 rCσ αγδη dr and sK ∇33γδη ..= 0

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Resolution

Displacement

We find that the second-order displacement field writes as:

u -2₌_u _ -K ∇_(z)_...K ∼−1⊗∇ -Y − zU1 3 ⊗∇ -Y =   −zU1 3,1+uK ∇1βγδKδγ,β−1 −zU1 3,2+uK ∇2βγδK −1 δγ,β 0  

where the displacement localization tensor related to the curvature gradient writes as:

uK ∇ αβγδ(z)..=− Z z 0 4Sα3η3 Z y −1 2 vCσ ηβγδ dv +δαβuK3γδ ! dy and uK ∇ 3βγδ..= 0

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The Bending-Gradient plate model Higher orders?

The general form of the expansion

It can be formally shown that we have:

u -= U3 η -e3− zU3 ⊗ ∇ -Y +ηu∼ -K _:_K ∼ +η2u_ -K ∇_..._K ∼⊗∇ -Y +. . . where U3..= ∞ X p=−1 ηp+1U₃p=ηhu₃i and K ∼ ..=−U3∇ -Y⊗∇ -Y

We have also for the stress:

σ_∼=_∼s

∼ K _:_K

∼ +ηs__∼K ∇...K∼⊗∇

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A higher order plate model from asymptotic expansions?

Including shear effects...:

I ... from asymptotic expansion?: U3 ∈ C6(ω)

I ... from the approach from Smyshlyaev and Cherednichenko (2000)?: U3 ∈ C4(ω)

I ... with the Bending-Gradient theory: U3 ∈ C1(ω)

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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:

σ_∼BG=_∼s ∼ K _{: χ} ∼+η_∼s K ∇_{... χ} ∼⊗∇ -Y χ

∼ = (χαβ) (Y1, Y2) is an unknown symmetric second-order tensor field.

Choice of χ

∼?: The minimum of complementary energy!

The corresponding bending moment is:

M_∼BG₌_D ∼ ∼ : χ∼ where D∼∼ = D z2C_∼ ∼ σE and d_∼ ∼=D∼∼ −1

Its gradient is:

R_{_} =M_∼BG⊗∇

-Y or Rαβγ = M

BG

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Stress localization as function of static variables

It is possible to rewrite σ_∼BG in terms ofM_∼BG andR_{_}: σ_∼BG=_∼s ∼ K _:d ∼ ∼:M∼BG +ηs_{_} ∼ K ∇_...d ∼ ∼:M∼BG ⊗∇ -Y and σ ∼BG =∼_∼sM :M∼BG+η__∼s R _..._R _

where the localizations tensors are given by:

s ∼ ∼ M ₌_s ∼ ∼ K _:_d ∼ ∼, __∼s R ₌_s _ ∼ K ∇_:_d ∼ ∼

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The Bending-Gradient stress energy

Plugging σ_∼BG into the complementary energy of the full 3D problem leads to the following functional:

P∗BGM_∼BG,R_{_}= Z ω w∗KLM_∼BG +η2w∗BGR_{_} dω

where the stress elastic energies are defined as:

w∗KLM_∼BG = 1 2M∼ BG_:_d ∼ ∼ :M∼ BG _and _w∗BGR _ = 1 2 T_R _...__h...R_ with: h _ _= D_T s _ ∼ R _:_S ∼ ∼ :_∼s RE

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.

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Extended plate equilibrium equations

Exact plate equilibrium equations

The total bending moment and the total shear force are defined as

Mαβ(Y1, Y2) =hzσαβi , and Qα(Y1, Y2) =η−1hσ_3αi .

Moment equilibrium equations:

z σαβ,β+η−1σα3,3 = Mαβ,β− Qα = 0 or M_∼ · ∇

-Y − Q- = 0 The out-of-plane equilibrium equation:

η−1σ_3α,α+η−1σ_33,3 = Q_α,α+ F₃ = 0 or Q

-· ∇

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Extended plate equilibrium equations

Link between shear forces and generalized shear forces

Q- =M∼ · ∇

-Y is now replaced byR_ =M∼

BG_⊗_∇

-Y

We have the following relation:

i ∼ ∼...R_ =M∼BG· ∇ -Y =Q -BG _{or R} αββ = Mαβ,βBG = QαBG where iαβγδ = 1 2(δαγδβδ+δαδδβγ) Mechanical meaning of R_{_} Qα= Rαββ ⇔ Q1= R111+ R122 = M11,1+ M12,2 Q2= R121+ R222 = M21,1+ M22,2

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The Bending-Gradient statically compatible fields

The Bending-Gradient stress energy must be minimized over the set:

SCBG: ( R_{_} =M_∼BG⊗∇ -Y or Rαβγ = M BG αβ,γ i_∼_∼...R_{_}_{· ∇} -Y + F3 = 0 or Rαββ,α+ F3 = 0

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The Bending-Gradient constitutive equations

Now we define the generalized strains as:

χ ∼ = ∂w∗KL ∂M_∼BG and _Γ= ∂w∗BG ∂R_{_} Note that the third-order tensor has the symmetry:

Γαβγ = Γβαγ

This leads to the following constitutive equations: ( χ ∼ =d∼∼:M∼ BG _or _χ αβ = dαβγδM_δγBG Γ _=h__...R_ or Γαβγ = hαβγδµνRνµδ

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Dualization of equilibrium equations

Multiplying Rαβγ = Mαβ,γBG with Φαβγ and integrating by parts on the plate

domain ω yield: Z ω M_αβBGΦαβγ,γ+ RαβγΦαβγdω = Z ∂ω M_αβBGΦαβγnγdl

Multiplying Rαββ,α+ F3= 0 with U3BG and integrating by parts on the

plate domain ω yield: Z ω RαββU3,αBGdω = Z ∂ω RαββnαU3BGdl + Z ω F3U3BGdω

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Weak formulation

Adding these equations leads to the following expression: Z ω M_αβBGΦαβγ,γ + Rαβγ Φαβγ + 1 2 δβγU3,αBG+δαγU3,βBG dω = Z ω F3U3BGdω + Z ∂ω M_αβBGΦαβγnγ+ RαββnαU3BGdl

Therefore, we have obtained the weak formulation of this plate theory: V_intBG= V_extBG where V_intBG= Z ω M_∼BG_:_Φ _· ∇ -Y +TR _... Φ_{_}+_∼i ∼· ∇ -YU BG 3 dω V_extBG= Z ω F3U3BGdω + Z ∂ω M_∼BG _:_Φ _· n - +i_∼ ∼...R_· n - U₃BGdl n

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Kinematic compatibility conditions

We identify the internal power obtained by dualization

V_intBG= Z ω M_∼BG_:_Φ _· ∇ -Y +T_R _... Φ_{_}+_∼i ∼· ∇ -YU BG 3 dω

with the one obtained with the constitutive equations

V_intBG= Z ω M_∼BG_{: χ} ∼+η 2TR _ ... Γ_dω

Finally, we define the set of kinematically compatible fields as

KCBG: (_χ ∼ = Φ_· ∇ -Y η2Γ_{_}= Φ_{_}+i ∼ ∼· ∇ -YU BG 3

to which the following boundary conditions must be added for a clamped plate:

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Summary

The Bending-Gradient plate theory equations are the following:

               R_{_} =M_∼BG⊗∇ -Y and i∼_∼...R_ · ∇ -Y + F3 = 0 on ω χ ∼ =d∼∼ :M∼BG and Γ_=h__...R_ on ω χ ∼ = Φ_ · ∇ -Y and η 2_Γ _= Φ_ +∼∼i · ∇ -YU BG 3 on ω U₃BG= 0 and Φ_{_}_{· n}-= 0 on∂ω Note that: χ ∼ =K∼ BG₊ η2Γ_{_}· ∇ -Y where K∼ BG₌ −U3BG∇ -Y⊗∇ -Y

Setting η2= 0 in the Bending-Gradient model leads exactly to

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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 UBG, Φ_{_} are the fields solution of the plate problem: u -BG₌ U3BG η -e3− zU BG 3 ⊗∇ -Y +η u∼ -K _{: χ} ∼+η 2 _u _ -K ∇_..._χ ∼⊗∇ -Y

Defining the strain as

ε

∼BG=S∼∼ : σ∼BG

it is possible to check that:

εu -BG (Y ,z)− ε∼ BG₌ η2 δ ∼⊗su_ -K ∇_::_χ ∼⊗∇ -2 Y + zΓ_{_}· ∇ -Y

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The Bending-Gradient plate model The Reissner-Mindlin theory for homogeneous plates

Homogeneous plates

In this case, we have:

σ ∼BG =          σ_αβBG = 12z iαβγδM_δγBG = 12zM_αβBG σ_α3BG =η3₂ 1− 4z2 iαβγδRδγβ =η3₂ 1− 4z2 Q_αBG σBG 33 = 0

which is a function of M_∼BG and Q

-BG₌_i ∼

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The constitutive equations

The Bending-Gradient part of the stress energy becomes:

w∗BGR_{_}= 1 2 T_R _...h__...R_ = 1 2Q -BG_{· h} ∼ RM · Q -BG with: h _ _=∼∼i · h∼ RM · i_∼_∼

where the Reissner’s shear forces stiffness is given by:

h_αβRM = 6 5Sα3β3

(it is equal to _5G6 δαβ with G the shear modulus for isotropic plates). The

Bending-Gradient constitutive equation becomes: Γ _=h__...R_ =∼_∼i· γ -with γ -=∼h RM · Q -BG

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The kinematics

Using the kinematic compatibility

η2Γ_{_}= Φ_{_}+i ∼ ∼· ∇ -YU BG 3 ,

we find that Φ_{_} is also of the form:

Φ_{_} =_∼i

∼· ϕ -where ϕ

- is the classical rotation vector of the Reissner theory. Therefore, the kinematic unknowns are U₃BG and ϕ

-, and we have: ( χ ∼ = ϕ -⊗s∇ -Y = d∼_∼ :M∼ BG η2γ - = ϕ-+ U BG 3 ⊗∇ -Y = η 2h ∼ RM · Q -BG

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Static

The following boundary conditions must be added for a clamped plate:

U₃BG= 0 and ϕ

- = 0 on∂ω Finally, the balance equations are:

( M_∼BG_{· ∇} -Y − Q -BG _{= 0 on}_ω Q -BG_{· ∇} -Y + F3 = 0 onω

In conclusion: the Bending-Gradient theory completely coincides for homogeneous plates with the Reissner-Mindlin model.

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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:

∆RM/BG= kh__ W_k kh_{_}_{_}k where kh_{_}_{_}k =qT_h _ _...h__

is the norm for Bending-Gradient compliance tensors and h_{_}_{_}W is the pure warping part of h_{_}_{_}: h _ _ W₌_h _ _− 4 9∼∼i · i∼∼...__h...∼_∼i · i∼_∼

∆RM/BG gives an estimate of the pure warping fraction of the shear stress energy. When the plate constitutive equation is restricted to a

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Distance between the BG and the RM models

x1 x2 x3 θ Stack [0◦] [30◦,−30◦_] s [0◦,−45◦, 90◦, 45◦]s ∆RM/BG 0 16.0% 12.4%

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Applications of the Bending-Gradient theory to laminates

Voigt Notations

We introduce the linear operator h

•

∼

i

reallocating tensor components. For instance, the bending moment and the curvature are reallocated in a vector form: hM_∼i =   M11 M22 √ 2M12   and h χ ∼ i =   χ11 χ22 √ 2χ12  

and the fourth-order compliance tensor d_∼

∼ is reallocated in a matrix form

hd_∼_∼i =   d1111 d2211 √ 2d1211 d2211 d2222 √ 2d1222 √ 2d1211 √ 2d1222 2d1212  

so that the constitutive equation χ ∼ =d∼∼ :M∼ becomes h χ ∼ i =hd_∼ ∼ i ·hM_∼i D Cσ

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Voigt Notations

The constitutive sixth-order tensorh_{_}_{_} is turned into the 6_{× 6 matrix}hh_{_}_{_} i :          h111111 h111122 √ 2h111121 h111211 h111222 √ 2h111221 h221111 h221122 √ 2h221121 h221211 h221222 √ 2h221221 √ 2h121111 √ 2h121122 2h121121 √ 2h121211 √ 2h121222 2h121221 h112111 h112122 √ 2h112121 h112211 h112222 √ 2h112221 h222111 h222122 √ 2h222121 h222211 h222222 √ 2h222221 √ 2h122111 √ 2h122122 2h122121 √ 2h122211 √ 2h122222 2h122221         

The third-order tensors Γ_{_}and R_{_} are reallocated in a vector form:

h Γ _ i =         Γ111 Γ221 √ 2Γ121 Γ112 Γ222 √ 2Γ122         , hR_{_}i=         R111 R221 √ 2R121 R112 R222 √ 2R122         and hΓ_{_}i=hh_{_} _ i ·hR_{_}i

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Applications of the Bending-Gradient theory to laminates Cylindrical bending of laminates

Pagano’s boundary value problem

(Pagano, 1969)

CFRP layers with different orientiations: F3(Y1) =−F0sinκY1 where

λ = 1/κ = _nπ1 , n _{∈ N}+∗ _{is the non-dimensional wavelength of the loading.}

z η2F₃/2 η2F₃/2    σ11(z) = 0 σ12(z) = 0 u₃(z) = 0 Y1 Y2 1

Invariant in x2-Direction, “periodic” in x1-Direction

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Resolution of the Bending-Gradient problem

All the non-dimensional fields are invariant in Y2-Direction

Fromχαβ = Φαβγ,γ, we obtain: h χ ∼ i =   χ11 χ22 √ 2χ12  =   Φ111,1 Φ221,1 √ 2Φ121,1  =   Φ1,1 Φ2,1 Φ3,1  

The equilibrium equations write as:

hR_{_}i =         R111 R221 √ 2R121 R112 R222 √ 2R122         =         M11,1 M22,1 √ 2M12,1 0 0 0         and M11,11=−F3(Y1)

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Shear constitutive equation

Taking into account R112 = R222= R122= 0, U3,2= 0, shear constitutive

equation is rewritten in two parts.

A first part with unknowns involving active boundary conditions:

  Φ1 Φ2 Φ3  =η2   h11 h12 h13 h12 h22 h23 h13 h23 h33  ·   M11,1 M22,1 √ 2M12,1  −   U3,1 0 0  

and a second part which enables the derivation of Φ4 = Φ112, Φ5 = Φ222,

Φ6 =

√

2Φ122 on which no boundary condition applies:

  Φ4 Φ5 Φ6  =η 2   h41 h42 h43 h51 h52 h53 h61 h62 h63  ·   M11,1 M22,1 √ 2M12,1  −   0 0 U3,1/ √ 2  

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Final System

Finally, combining the above equations leads to the following set of equations which fully determines the problem:

                   M11,11= F0sinκY1 hd_∼_∼i ·hM_∼i−η2h_∼·hM_∼ i ,11=   U3,11 0 0   hM_∼i = 0 for Y1 = 0 and Y1 = 1 U3 = 0 for Y1 = 0 and Y1 = 1

where for convenience,_∼h is the 3_{× 3 submatrix of}hh_{_}_{_} i : h ∼=   h11 h12 h13 h12 h22 h23 h13 h23 h33  

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Solution

This differential system is well-posed and the solution is unique. Its is of the form:

hM_∼i

=hM_∼∗ i

sinκY1 and U3 = U3∗sinκY1

where hM_∼∗iand U₃∗ are constants explicitly known in terms of the problem inputs.

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Applications of the Bending-Gradient theory to laminates Numerical illustrations

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

−10 −5 0 5 10 t2_σ 11(a/2, b/2, x3)/(p3λ2) −0.4 −0.2 0.0 0.2 0.4 x3 /t KL BG Pagano −10 −5 0 5 10 t2_σ 22(a/2, b/2, x3)/(p3λ2) −0.4 −0.2 0.0 0.2 0.4 KL BG Pagano −0.4 −0.2 0.0 0.2 0.4 σ33(a/2, b/2, x3)/p3 −0.4 −0.2 0.0 0.2 0.4 KL BG Pagano 0.0 0.5 1.0 1.5 2.0 tσ23(a/2, 0, x3)/(p3λ) −0.4 −0.2 0.0 0.2 0.4 L/t = 1.00 KL BG Pagano 0.0 0.5 1.0 1.5 2.0 tσ13(0, b/2, x3)/(p3λ) −0.4 −0.2 0.0 0.2 0.4 x3 /t KL BG Pagano −10 −5 0 5 10 t2_σ 12(a/2, b/2, x3)/(p3λ2) −0.4 −0.2 0.0 0.2 0.4 KL BG Pagano

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Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

−10 −5 0 5 10 t2_σ 11(a/2, b/2, x3)/(p3λ2) −0.4 −0.2 0.0 0.2 0.4 x3 /t KL BG Pagano −10 −5 0 5 10 t2_σ 22(a/2, b/2, x3)/(p3λ2) −0.4 −0.2 0.0 0.2 0.4 KL BG Pagano −0.4 −0.2 0.0 0.2 0.4 −0.4 −0.2 0.0 0.2 0.4 KL BG Pagano 0.0 0.5 1.0 1.5 2.0 −0.4 −0.2 0.0 0.2 0.4 L/t = 1.39 KL BG Pagano 0.0 0.5 1.0 1.5 2.0 −0.4 −0.2 0.0 0.2 0.4 x3 /t KL BG Pagano −10 −5 0 5 10 t2_σ 12(a/2, b/2, x3)/(p3λ2) −0.4 −0.2 0.0 0.2 0.4 KL BG Pagano

(95)

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(96)

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(97)

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(98)

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(99)

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(100)

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(101)

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(102)

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(103)

Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

−2.0−1.5−1.0−0.50.0 0.5 1.0 1.5 2.0 u1(0, b/2, x3)/(p3λ3) −0.4 −0.2 0.0 0.2 0.4 x3 /t KL BG Pagano −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 u2(a/2, 0, x3)/(p3λ) −0.4 −0.2 0.0 0.2 0.4 KL_BG Pagano 0.0 0.2 0.4 0.6 0.8 1.0 1.2 u3(a/2, b/2, x3)/huP ag3 i −0.4 −0.2 0.0 0.2 0.4 KL_BG Pagano

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Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(105)

Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(106)

Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

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Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(108)

Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

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Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(110)

Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(111)

Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(112)

Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(113)

Stress distributions for a

[45

◦

,

₋₄₅

◦

]

4

, 45

◦

stack

(114)

Stress distributions for a

[45

◦

,

₋₄₅

◦

]

4

, 45

◦

stack

(115)

Stress distributions for a

[45

◦

,

₋₄₅

◦

]

4

, 45

◦

The Bending-Gradient theory for laminates and in-plane periodic plates

HAL Id: cel-01266716

https://hal-enpc.archives-ouvertes.fr/cel-01266716

The Bending-Gradient theory for laminates and in-plane

periodic plates

Arthur Lebée

To cite this version:

The Bending-Gradient theory for laminates and

in-plane periodic plates

The open question of shear forces in heterogeneous plates

The open question of shear forces in heterogeneous plates

Contents

The 3D problem configuration

C

x

f

-f

e

e

e

t

L

The 3D problem equations

Proof of skew symmetry

Proof of skew symmetry

Proof of skew symmetry

Variational formulation

Potential energy

Complementary energy

Building a plate model?

Contents

Contents

Differential system depending on a small parameter

Taylor’s series

The method

Resolution

Contents

Change of variables

Non-dimensional fields

Natural scaling of the stress

The non-dimensional 3D problem

Properties of the non-dimensional solution

Proof

The new equations

Effect of the transformation

→ −

on the

non-dimensional solution

Contents

Expansion

Statics

Statics

Kinematics

Kinematics

Contents

Contents

Lower order displacements

Lower order displacements

Contents

Zeroth-order auxiliary problem

Resolution

Resolution

Resolution

Isotropic materials

Contents

Determination of

U

Determination of

U

Determination of

U

Summary of the Kirchhoff-Love model

Lateral boundary conditions

Contents

Contents

First-order auxiliary problem

Resolution

Resolution

Resolution

Contents

₋₃₀

₋₃₀

₋₃₀

₋₃₀

₋₃₀