Introduction to modelization of thick and heterogeneous plates

(1)

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Introduction to modelization of thick and heterogeneous

plates

Arthur Lebée, Karam Sab

To cite this version:

Arthur Lebée, Karam Sab. Introduction to modelization of thick and heterogeneous plates. Rencontres

franciliennes de mécanique, Jun 2014, Paris, France. �hal-01134568�

(2)

’

Introduction to modelization of thick and

heterogeneous plates

Arthur Leb´

ee, Karam Sab

Laboratoire Navier (UMR CNRS 8205)

(3)

The 3D Problem











σ

t ij,j

= 0

on Ω.

σ

t ij

= C

ijkl

(x

3

)

ε

tkl

on Ω

.

σ

t i3

=

±T

i±

on

ω

±

_.

ε

t ij

= u

t (i,j)

on Ω

.

u

t i

= 0

on

∂ω

×] − t/2, t/2[

I

monoclinic

C

_∼ ∼

:

C

αβγ3

= C

333α

= 0

I

symmetrically laminated plate

I

symmetric transverse load

T

−

±

₌

p

3

2 e

−

₃

C

∼∼

(x

3

)

: even

Ω

T

− +

T

− −

ω

+

ω

−

∂ω

e

−₃

e

−₂

e

−₁

⇒ pure bending:

I

u

3 and

σ

α3

even / x

3 I

u

α

,

σ

αβ

and

σ

33 odd / x

3

(4)

Introduction

Building a plate model?

For typical width L let t

_{→ 0}

I

Solve a 2D problem

I

“fair” 3D displacement localization

(5)

Some energy principles...

I

Statically admissible fields:

SA

3D

=

n

σ

ij

/

σ

ij ,j

= 0

and

σ

∼

±

t

2 · ±e

−

₃

=

P

3

2 e

−

₃

o

SA

3D,0

_{⇔ P}

3 = 0

I

Kinematically admissible fields:

KA

3D

=

u

i

/

u

i

= 0 on

∂ω

×

−

₂

t

,

t

₂

KA

3D,0

= KA

3D

I

Orthogonality between KA

3D,0

and SA

3D,0

:

∀u

∈ KA

3D,0

, σ

_{∈ SA}

3D,0

,

Z

(6)

Introduction

Some energy principles...

I

Potential energy:

u

−

3D

_{= argmin}

u

−

∈KA

3D

W

3D

ε

_∼

(

u

−

)

−

Z

ω

P

3 u

+

₃

+ u

₃

−

2 d

ω

where W

3D

=

1 ₂

R

Ω

ε

∼

(

u

−

) :

C

∼∼

: ε

∼

(

u

−

)d Ω

I

Complementary energy:

σ

3D

= argmin

n

W

∗3D

σ

o

where

W

∗3D

=

1 Z

σ

:

S

: σ

d Ω

(7)

The 2-energy principle

∀ˆu

−

∈ KA

3D

_,

_∀ˆ

_σ

∼

∈ SA

3D

_:

W

3D

ε

∼

u

−

ˆ

− S

∼∼

: ˆ

σ

∼

= W

∗3D

ˆ

σ

_∼

− C

∼∼

: ε

∼

u

−

ˆ

= W

∗3D

ˆ

σ

_∼

− σ

∼

3D

_{+ W}

3D

_ε

∼

ˆ

u

−

− u

−

3D

⇒

W

∗3D

σ

ˆ

_∼

_{− C}

_∼ ∼

: ε

∼

u

−

ˆ

provides an error estimate in terms of constitutive

equation.

(8)

Introduction

The case of homogeneous and isotropic plates

The case of laminated plates

Applications

Cylindrical bending of laminates

Extension to periodic plates

The case of cellular sandwich panels

(9)

The case of homogeneous and isotropic plates

The case of laminated plates

(10)

The case of homogeneous and isotropic plates

Natural scaling of the stress

SA

3D











σ

αβ,β

+

σ

α3,3

= 0

σ

α3,α

+

σ

33,3

= 0

σ

33 (

±t/2) = ±P

3 /2

σ

α3

(

±t/2) = 0

⇒











σ

α3

=

−

Z

x

3

−t/2

σ

αβ,β

dz

σ

33 =

−

Z

x

3

t/2

σ

α3,α

dz

− P

3 /2

σ

αβ

∼ t

0 ⇒

σ

α3

∼ t

1 ,

σ

33 ∼ t

2 and

P

3 ∼ t

2

(11)

From 3D equilibrium to 2D

Plate generalized stresses:

(

M

αβ

(x

1 , x

2 ) =

hx

3 σ

αβ

i ∼ t

2 Q

α

(x

1 , x

2 ) =

hσ

α3

i

∼ t

2 h•i =

Z

t 2

−

t 2

• dx

3 2D equilibrium equations:

hσ

α3,α

+

σ

33,3

i = 0

hx

3 (σ

αβ,β

+

σ

α3,3

)

i = 0

⇒

_Q

α,α

+ P

3 = 0

M

αβ,β

− Q

α

= 0

Boussinesq (1871); Mindlin (1951)...

(12)

“Kirchhoff’s assumption”

At leading order in t:

_ε

αβ

= x

3 K

αβ

where

K

αβ

=

−U

3,αβ

σ

i3

' 0 + O(t

1 )

plane stress

ε

33

6= 0!!

In-plane stress:

σ

αβ

= C

αβγδ

σ

ε

δγ

= x

3 C

αβγδ

σ

K

δγ

where C

σ

(13)

The Kirchhoff-Love plate problem

Bending constitutive equation:

M

αβ

=

hx

3 σ

αβ

i =

x

3

2 C

αβγδ

σ

K

δγ

=

−D

αβγδ

U

3,δγ

D

∼∼

=

t

3

12 C

∼∼

σ

Statically admissible fields:

SA

KL

:

{M

αβ

/M

αβ,αβ

+ P

3 = 0

}

Kinematically compatible fields:

KA

KL

:

U

3 /U

3 = 0

and

U

3,n

= 0 on

∂ω,

n

−

outer normal to

ω

(14)

Building

SA

3D

fields

ˆ

σ

αβ

= x

3 C

αβγδ

σ

d

δγζ

M

ζ

=

12x

3 t

3 M

αβ

∼ t

0 ˆ

σ

α3

=

−

Z

x

₃

−t/2

σ

αβ,β

dz

=

3 2t

1 −

4x

2

3 t

2 M

αβ,β

∼ t

1 ˆ

σ

33 =

−

Z

x

₃

−t/2

σ

α3,α

dz

− P

3 /2 =

3x

3 2t

1 ₋

4x

2

3 3t

2 P

3 ∼ t

2 ⇔

σ

ˆ

_∼

=

s

_∼ ∼

M

_(x

3 ) :

M

∼

+

s

∼−

Q

_(x

3 )

· M

∼

· ∇

−

+

s

∼

P

3

_(x

3 )P

3

(15)

Building

KA

3D

fields

Strains

ˆ

ε

∼

=











ˆ

ε

αβ

= x

3 d

αβγδ

M

γδ

= S

αβγδ

σ

ˆ

δγ

+

S

αβ33

σ

ˆ

33 ˆ

ε

α3

=

3 4Gt

1 ₋

4x

2

3 t

2 M

αβ,β

= 2S

α3β3

σ

ˆ

β3

ˆ

ε

33 =

−

12νx

3 Et

3 M

αα

= S

33αβ

σ

ˆ

αβ

+

S

3333

σ

ˆ

33

(16)

Building

KA

3D

fields

Integration

ˆ

u

3 =

Z

x

₃

ˆ

ε

33 (z)dz + U

3 =

u

M

3αβ

(x

3 )M

βα

|

{z

}

∼t

1

+ U

3 |{z}

∼t

−1

where:

u

M

_3αβ

(x

3 ) =

ν

2Et

12x

₃

2 t

2 − 1

δ

αβ

and

Du

M

3αβ

(x

3 )

E

= 0

(17)

Building

KA

3D

fields

Integration

ˆ

u

3 =

Z

x

₃

ˆ

ε

33 (z)dz + U

3 =

u

M

3αβ

(x

3 )M

βα

|

{z

}

∼t

1

+ U

3 |{z}

∼t

−1

ˆ

u

α

=

Z

x

3

2ˆ

ε

α3

(z)

− ˆ

u

3,α

dz =

u

M

αβγδ

⊗

∇

(x

3 )M

δγ,β

|

{z

}

∼t

2

− x

3 U

3,α

|

{z

}

∼t

0

where:

u

M

⊗

∇

αβγδ

(x

3 )M

δγβ

=

x

3 2Et

6(1 +

ν)

1 ₋

4x

2

3 3t

2 M

αβ,β

+

ν

1 ₋

4x

2

3 t

2 M

ββ,α

(18)

Building

KA

3D

fields

Integration

ˆ

u

3 =

Z

x

₃

ˆ

ε

33 (z)dz + U

3 =

u

M

3αβ

(x

3 )M

βα

|

{z

}

∼t

1

+ U

3 |{z}

∼t

−1

ˆ

u

α

=

Z

x

3

2ˆ

ε

α3

(z)

− ˆ

u

3,α

dz =

u

M

αβγδ

⊗

∇

(x

3 )M

δγ,β

|

{z

}

∼t

2

− x

3 U

3,α

|

{z

}

∼t

0

⇒

u

ˆ

−

= U

3 e

−

₃

− x

3 U

3,α

e

−

_α

+

u

M

3αβ

(x

3 )M

βα

e

−

₃

+

u

M

⊗

∇

αβγδ

(x

3 )M

δγ,β

e

−

_α

(19)

Application of the Two-Energy principle

Consider U

₃

KL

and

M

_∼

KL

the solution of the Kirchhoff-Love plate problem

and define:

σ

_∼

KL

=

s

_∼ ∼

M

_(x

3 ) :

M

∼

KL

₊

_s

∼ −

Q

_(x

3 )

· M

∼

KL

_{· ∇}

−

+

s

_∼

P

3

_(x

3 )P

3 u

−

KL

_{= U}

KL

3 e

−

₃

− x

3 U

KL

3,α

e

−

_α

+

u

M

3αβ

(x

3 )M

βα

KL

e

−

₃

+

u

M

⊗

∇

αβγδ

(x

3 )M

KL

δγ,β

e

−

_α

We have:

ε

∼

(

u

−

KL

₎

_{− S}

∼ ∼

: σ

∼

KL

₌

_s

P

3

33 P

3 



S

αβ33

0 ₀

0

0 S

3333





∼ t

2

(20)

Application of the Two-Energy principle

ε

∼

(

u

−

KL

)

− S

∼∼

: σ

∼

KL

_{∼ t}

2 _⇒

_W

3D

_ε

∼

(

u

−

KL

)

− S

∼∼

: σ

∼

KL

_{∼ t}

5 Would lead to a relative error in t

2 ...

σ

_∼

KL

_{∈ SA}

3D

but

u

−

KL

∈ KA

/

3D

(21)

Reissner’s original plate model (1945)

min of

W

∗3D

Let us consider:

ˆ

σ

_∼

=

s

_∼ ∼

M

_(x

3 ) :

M

∼

+

s

∼₋

Q

_(x

3 )

· Q

−

+

s

∼

P

3

_(x

3 )P

3 with

SA

RM

=

n

(

M

_∼

,

Q

₋

)

/ Q

α,α

+ P

3 = 0

and

M

αβ,β

− Q

α

= 0

on

ω

o

W

∗3D

σ

_∼

3D

_≤

min

(M

∼

,

Q

−

)∈SA

RM

W

∗3D

σ

ˆ

_∼

≤ W

∗3D

σ

_∼

KL

is a better approximation of W

∗3D

σ

3D

(22)

Reissner’s original plate model (1945)

Dualization

Q

α,α

+ P

3 = 0

×U

3 M

αβ,β

− Q

α

= 0

×ϕ

α

⇒

Q

α

↔

γ

α

=

ϕ

α

+ U

3,α

M

αβ

↔

χ

αβ

=

ϕ

(α,β)

KA

RM

=

_{(U

3 , ϕ

α

)

/ U

3 = 0

and

ϕ

α

= 0 on

∂ω

}

(23)

Reissner’s original plate model (1945)

Constitutive equation

W

∗RM

M

_∼

,

Q

₋

= W

∗3D

σ

ˆ

_∼

=

1

2 Z

ω

T





M

∼

Q

−

P

3 







d

∼∼

0 6ν

5Et

δ

∼

0 _5Gt

6 δ

_∼

0 6ν

5Et

δ

∼

0 17t

140E









M

∼

Q

−

P

3 



d

ω

⇒











χ

αβ

|{z}

∼t

−1

= d

αβγδ

M

δγ

|

{z

}

∼t

−1

+

6ν

5Et

δ

αβ

P

3 |

{z

}

∼t

1

γ

α

=

6 5Gt

Q

α

∼ t

1

(24)

The case of laminated plates

The case of homogeneous and isotropic plates

The case of laminated plates

(25)

Field Localization

Following the same procedure leads to:

ˆ

σ

_∼

=

s

_∼ ∼

M

_(x

3 ) :

M

∼

+

_

s

_∼

R

_(x

3 ) ·

·· M

∼

⊗

∇

−

|

{z

}

∼t

1

+

_{_}

s

_

T

_(x

3 ) ·

··· M

∼

⊗

∇

−

2 ₊

_s

∼

P

3

_(x

3 )P

3 |

{z

}

∼t

2

ˆ

u

−

= U

3 e

−

₃

− x

3 U

3,α

e

−

_α

+

u

M

3αβ

(x

3 )M

βα

e

−

₃

+

u

R

αβγδ

(x

3 )M

δγ,β

e

−

_α

where

R

_{_}

=

M

_∼ ⊗

∇

−

and

T

∼∼

=

R

_⊗

∇

−

(26)

Building

SA

3D

fields

ˆ

σ

αβ

= x

3 C

αβγδ

σ

d

δγζ

M

ζ

=

s

M

αβγδ

(x

3 )M

δγ

ˆ

σ

α3

=

−

Z

x

3

−t/2

σ

αβ,β

dz

=

s

R

α3βγδ

(x

3 )M

δγ,β

ˆ

σ

33 =

−

Z

x

₃

−t/2

σ

α3,α

dz

− P

3 /2 =

s

T

33αβγδ

(x

3 )M

δγ,βα

+

s

P

33

3

(x

3 )P

3

(27)

The Bending-Gradient constitutive equation

Extending Reissner’s approach?:

ˆ

σ

_∼

=

s

_∼ ∼

M

_(x

3 ) :

M

∼

+

_∼

s

R

_(x

3 ) ·

·· R

_

+

s

_ _

T

_(x

3 ) ·

··· T

∼∼

+

s

∼

P

3

_(x

3 )P

3 Let us define:

ˆ

σ

_∼

∗

=

s

_∼ ∼

M

_(x

3 ) :

M

∼

+

_

s

_∼

R

_(x

3 ) ·

·· R

_

with: ˆ

σ

_∼

∗

_{· ∇}

−

= 0 + O(t

1 ) only.

W

∗BG

M

_∼

,

R

_{_}

= W

∗3D

_σ

_ˆ

∼

∗

₌

1

2 Z

ω

M

∼

:

d

∼∼

:

M

∼

+

R

_

··· f

__

··· R

_

d

ω

(28)

The Bending-Gradient theory for thick plates

I

Equilibrium equations:

BG:

R

αβγ

= M

αβ,γ

× Φ

αβγ

R

αββ,α

+ P

3 = 0

× U

3 RM:

Q

α

= M

αβ,β

× ϕ

α

Q

α,α

+ P

3 = 0

× U

3 I

Mechanical meaning of

R

_{_}

Q

α

= R

αββ

⇔

Q

1

= R

111

+ R

122

= M

11,1

+ M

12,2

Q

2

= R

121

+ R

222

= M

21,1

+ M

22,2

(29)

The Bending-Gradient theory for thick plates

I

Equilibrium equations:

BG:

R

αβγ

= M

αβ,γ

× Φ

αβγ

R

αββ,α

+ P

3 = 0

× U

3 RM:

Q

α

= M

αβ,β

× ϕ

α

Q

α,α

+ P

3 = 0

× U

3

(30)

The Bending-Gradient theory for thick plates

I

Equilibrium equations:

BG:

R

αβγ

= M

αβ,γ

× Φ

αβγ

R

αββ,α

+ P

3 = 0

× U

3 RM:

Q

α

= M

αβ,β

× ϕ

α

Q

α,α

+ P

3 = 0

× U

3 Equilibrium

Work of internal forces

Work on Boundary

BG:

(

R

αβγ

= M

αβ,γ

R

αββ,α

+ P

3

= 0

M

αβ

Φ

αβ,

+

R

αβγ

(Φ

αβγ

+ I

αβγ

U

3,

)

M

αβ

Φ

αβγ

n

γ

+

R

αββ

n

α

U

3

Q

α

= M

αβ,β

M

αβ

ϕ

(α,β)

+

M

αβ

n

β

ϕ

α

+

(31)

The Bending-Gradient theory for thick plates

I

Equilibrium equations:

BG:

R

αβγ

= M

αβ,γ

× Φ

αβγ

R

αββ,α

+ P

3 = 0

× U

3 RM:

Q

α

= M

αβ,β

× ϕ

α

Q

α,α

+ P

3 = 0

× U

3 Equilibrium

Work of internal forces

Work on Boundary

BG:

(

R

αβγ

= M

αβ,γ

R

αββ,α

+ P

3

= 0

M

αβ

Φ

αβ,

+

R

αβγ

(Φ

αβγ

+ I

αβγ

U

3,

)

M

αβ

Φ

αβγ

n

γ

+

R

αββ

n

α

U

3

RM:

Q

α

= M

αβ,β

Q

α,α

+ P

3

= 0

M

αβ

ϕ

(α,β)

+

Q

α

(ϕ

α

+ U

3,α

)

M

αβ

n

β

ϕ

α

+

Q

α

n

α

U

3

(32)

The Bending-Gradient theory for thick plates

I

Equilibrium equations:

BG:

R

αβγ

= M

αβ,γ

× Φ

αβγ

R

αββ,α

+ P

3 = 0

× U

3 RM:

Q

α

= M

αβ,β

× ϕ

α

Q

α,α

+ P

3 = 0

× U

3 Equilibrium

Work of internal forces

Work on Boundary

BG:

(

R

αβγ

= M

αβ,γ

R

αββ,α

+ P

3

= 0

M

αβ

Φ

αβ,

+

R

αβγ

(Φ

αβγ

+ I

αβγ

U

3,

)

M

αβ

Φ

αβγ

n

γ

+

R

αββ

n

α

U

3

Q

α

= M

αβ,β

M

αβ

ϕ

(α,β)

+

M

αβ

n

β

ϕ

α

+

(33)

Local Fields reconstruction

Once the plate problem is solved (U

₃

BG

, Φ

_{_}

BG

,

M

_∼

BG

,

R

_{_}

BG

known), we

suggest the following field reconstruction:

I

σ

_∼

BG

=

s

_∼ ∼

M

_(x

3 ) :

M

∼

+

_

s

∼

R

_(x

3 ) ·

·· R

_

+

__

s

T

_(x

3 ) ·

··· R

_

⊗

∇

−

+

s

∼

P

3

_(x

3 )P

3 I

u

₋

BG

= U

3 e

−

₃

− x

3 U

3,α

e

−

_α

+

u

_3αβ

M

(x

3 )M

βα

e

−

₃

+

u

R

_αβγδ

(x

3 )R

δγβ

e

−

_α

(34)

Applications

The case of homogeneous and isotropic plates

The case of laminated plates

Applications

Cylindrical bending of laminates

Extension to periodic plates

The case of cellular sandwich panels

(35)

The case of homogeneous and isotropic plates

The case of laminated plates

Applications

Cylindrical bending of laminates

Extension to periodic plates

The case of cellular sandwich panels

(36)

Applications Cylindrical bending of laminates

Pagano’s boundary value problem

(Pagano, 1969)

CFRP layers with different orientiations:

x

3 p

3 /2

p

3 /2







σ

11

(x

3

) = 0

σ

12

(x

3

) = 0

u

₃

(x

₃

) = 0

x

1 x

2 L

(37)

Practical Localization...

Kirchhoff-Love

I

σ

_∼

KL

=

s

_∼ ∼

M

_(x

3 ) :

M

∼

KL

₊

((

(

s

∼ −

Q

_(x

3 )

· M

∼

KL

· ∇

−

+

s

∼

P

3

_(x

3 )P

3 I

u

₋

KL

= U

₃

KL

e

−

₃

− x

3 U

KL

3,α

e

−

_α

+

u

M

3αβ

(x

3 )M

βα

KL

e

−

₃

+

((

(

u

M

⊗

∇

αβγδ

(x

3 )M

KL

δγ,β

e

−

_α

Bending-Gradient

I

σ

_∼

BG

=

s

_∼ ∼

M

_(x

3 ) :

M

∼

+

_

s

∼

R

_(x

3 ) ·

·· R

_

+

s

_ _

T

_(x

3 ) ·

··· R

_

⊗

∇

−

+

s

∼

P

3

_(x

3 )P

3 I

u

₋

BG

= U

3 e

−

₃

− x

3 U

3,α

e

−

_α

+

u

M

3αβ

(x

3 )M

βα

e

−

₃

+

u

R

αβγδ

(x

3 )R

δγβ

e

−

_α

(38)

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 0.0 0.2 0.4 KL BG Pagano 0.0 0.2 0.4

L/t = 1.00

KL BG Pagano 0.0 0.2 0.4 /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

(39)

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 KL BG Pagano −0.4 −0.2 0.0 0.2 0.4

L/t = 1.39

KL BG Pagano −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

(40)

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

L/t = 1.95

(41)

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

L/t = 2.71

(42)

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

L/t = 3.79

(43)

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

L/t = 5.28

(44)

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

L/t = 7.37

(45)

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

L/t = 10.28

(46)

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

L/t = 14.34

(47)

Stress distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

L/t = 20.00

(48)

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 u(0, b/2, x)/(pλ3₎ −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 u(a/2, 0, x)/(pλ) −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 u(a/2, b/2, x)/huP ag_i −0.4 −0.2 0.0 0.2 0.4 KL_BG Pagano

(49)

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

(50)

Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(51)

Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(52)

Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(53)

Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(54)

Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(55)

Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(56)

Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(57)

Displacement distributions for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

(58)

Stress distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

L/t = 1.00

(59)

Stress distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

L/t = 1.39

(60)

Stress distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

L/t = 1.95

(61)

Stress distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

L/t = 2.71

(62)

Stress distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

L/t = 3.79

(63)

Stress distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

L/t = 5.28

(64)

Stress distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

L/t = 7.37

(65)

Stress distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

L/t = 10.28

(66)

Stress distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

L/t = 14.34

(67)

Stress distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

L/t = 20.00

(68)

Displacement distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

−2.0−1.5−1.0−0.50.0 0.5 1.0 1.5 2.0 u(0, b/2, x)/(pλ3₎ −0.4 −0.2 0.0 0.2 0.4 x3 /t KL BG Pagano −0.4−0.3−0.2−0.10.0 0.1 0.2 0.3 0.4 u(a/2, 0, x)/(pλ) −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 u(a/2, b/2, x)/huP ag_i −0.4 −0.2 0.0 0.2 0.4 KL_BG Pagano

(69)

Displacement distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

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 −0.4−0.3−0.2−0.10.0 0.1 0.2 0.3 0.4 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

(70)

Displacement distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

(71)

Displacement distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

(72)

Displacement distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

(73)

Displacement distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

(74)

Displacement distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

(75)

Displacement distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

(76)

Displacement distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

(77)

Displacement distributions for a

[45

◦

,

₋₄₅

◦

]

4 , 45

◦

stack

(78)

Convergence for a

[30

◦

,

₋₃₀

◦

, 30

◦

]

stack

100 ₁₀1 ₁₀2 ₁₀3 Slenderness: L/t 10−6 10−5 10−4 10−3 10−2 10−1 100 101 Stress Error KL BG 100 ₁₀1 ₁₀2 ₁₀3 Slenderness: L/t 10−7 10−6 10−5 10−4 10−3 10−2 10−1 100 Deflection Error KL BG

(79)

The case of homogeneous and isotropic plates

The case of laminated plates

Applications

Cylindrical bending of laminates

Extension to periodic plates

The case of cellular sandwich panels

(80)

Applications Extension to periodic plates

Extension to periodic plates

I

Unit-cell and average estimates

I

Bending auxiliary problem (Caillerie, 1984)

(81)

Extension to periodic plates

I

Unit-cell and average estimates

I

Bending auxiliary problem (Caillerie, 1984)

P

K











σ

_∼K

_{· ∇}

−

= 0

σ

_∼K

=

C

_∼ ∼

y

−

: ε

_∼K

ε

∼ K

_{= y}

3

K

∼

+ ∇

−

⊗

s

_u

− per

σ

_∼K

_{· e}

−₃

= 0 on free faces

∂Y

± 3

σ

_∼K

_{· n}

−

skew-periodic on lateral edge

∂Y

l

u

−

per

₍

_y

−

) (y

1

, y

2

)-periodic on lateral edge

∂Y

l

→ gives:

Localization

u

₋K

_σ

∼

K

_{related to the curvature}

_K

(82)

Extension to periodic plates

I

Unit-cell and average estimates

I

Bending auxiliary problem (Caillerie, 1984)

I

Shear auxiliary problem

P

R











σ

_∼R

_{· ∇}

₋

+

σ

_∼M

(

y

−

) = 0

σ

_∼R

=

C

_∼ ∼

y

−

: δ

_∼⊗s

u

− M

_{+ ∇}

−⊗ s

_u

− R

σ

_∼R

_{· e}

₋

3

= 0 on free faces

∂Y

± 3

σ

_∼R

_{· n}

−

skew-periodic on lateral edge

∂Y

l

u

−

R

₍

_y

(83)

The case of homogeneous and isotropic plates

The case of laminated plates

Applications

Cylindrical bending of laminates

Extension to periodic plates

The case of cellular sandwich panels

(84)

Applications The case of cellular sandwich panels

Justification of the Sandwich Theory

I

Divide in 3 layers

(homogeneous skins and heterogeneous core)

I

Bending auxiliary problem

(85)

Justification of the Sandwich Theory

I

Divide in 3 layers

(homogeneous skins and heterogeneous core)

I

Bending auxiliary problem

I

Contrast assumption

_{⇔ t}

f

t

s

:

→ t

s

/t

f

Contrast ratio

⇒ Skins under traction/compression

⇒ Core not involved in Bending stiffness

(86)

Justification of the Sandwich Theory

I

Divide in 3 layers

(homogeneous skins and heterogeneous core)

I

Bending auxiliary problem

I

Shear auxiliary problem

I

f

−

R

becomes

f

₋(Q)

+ Direct homogenization scheme

I

The BG is degenerated into RM model

I

f

₋(Q)

confirms the classical intuition

(87)

(88)

Application to the chevron pattern

(89)

Application to the chevron pattern

Shear forces

localization σ

_∼

(Q)

I

Overall shearing

of the core

skins distorsion

I

Critically

influence shear

force stiffness

(90)

Application to the chevron pattern

Shear forces

localization σ

_∼

(Q)

I

Overall shearing

of the core

I

Out-of-plane

skins distorsion

I

Critically

influence shear

force stiffness

&

(91)

Application to the chevron pattern

Shear forces

localization σ

_∼

(Q)

I

Overall shearing

of the core

I

Out-of-plane

skins distorsion

I

Critically

influence shear

force stiffness

0.2 0.5 1 2 5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 k2 sF11 ρGmh Kelsey− Kelsey+ BG, tf= 0.1 and ts= 0.1: ts/tf= 1 BG, tf= 0.1 and ts= 0.2: ts/tf= 2 BG, tf= 0.1 and ts= 0.5: ts/tf= 5 BG, tf= 0.1 and ts= 1: ts/tf= 10 BG, tf= 0.1 and ts= 2: ts/tf= 20 BG, tf= 0.1 and ts= 5: ts/tf= 50

(92)

Applications Why all periodic plates are not “Reissner” like...

The case of homogeneous and isotropic plates

The case of laminated plates

Applications

Cylindrical bending of laminates

Extension to periodic plates

The case of cellular sandwich panels

(93)

Homogenizing an orthogonal beam lattice?

=

+

Thick-plate model

(macro)

2 St-Venant Beams

(micro)

Localization

e

−2

e

−1

1

2 _b

b

p

3₋₃

e

−1

e

−2

e

−3

ω

∂ω

(94)

Field localization

−bM12 bM11 bM₁₂ bM22

e

−2

e

−1

1

2 Bending moment

r

−

(M)

,

m

−

(M)

:

Apply

M

_∼

”on average” on the unit-cell

(Caillerie, 1984) 1

_r

−(M)

=

2−

r

(M)

= 0

− 1

_m

−(M)

=





−bM

12

bM

11

0 



1

and

2

m

−(M)

=





bM

12

bM

22

0 



2

(95)

Field localization

e

−2

e

−1

1

2

−bR122 (s − b 2) bR121(s − b 2)

e

−2

e

−1 bR₁₂₂(s −b 2) bR₁₂₁(s −b 2) bQ₁ bQ2

Bending moment

r

−

(M)

,

m

−

(M)

:

Apply

M

_∼

”on average” on the unit-cell

(Caillerie, 1984)

Bending gradient

r

−

(R)

,

m

−

(R)

:

Assume M

αβ

= R

αβγ

x

γ (Leb´ee and Sab, 2013a)

1

_r

−(R)

=







0

0 b

(R

111

+ R

122

)

|

{z

}

Q1







1 1

_m

−(R)

=







bR

121

s −

b₂

bR

122

s −

b₂

0 





1 2

_r

−(R)

=







0

0 b

(R

121

+ R

222

)

|

{z

}

Q2







2 2

_m

−(R)

=







−bR

122

s −

b₂

bR

121

s −

b₂

0 





2

(96)

Field localization

e

−2

e

−1

1

2

−bR122 (s − b 2) bR121(s − b 2)

e

−2

e

−1 bR₁₂₂(s −b 2) bQ2

Bending moment

r

−

(M)

,

m

−

(M)

:

Apply

M

_∼

”on average” on the unit-cell

(Caillerie, 1984)

Bending gradient

r

−

(R)

,

m

−

(R)

:

Assume M

αβ

= R

αβγ

x

γ (Leb´ee and Sab, 2013a)

Reissner-Mindlin

r

−

(Q)

_,

_m

−

(Q)

_:

Assume cylindrical bending

(Whitney, 1969; Cecchi and Sab, 2007)

Q

1

= R

111

, Q

2

= R

222

,

R

121

= R

122

= R

221

= R

112

= 0

1

_r

(Q)

₌



0

0 

and

1

m

(Q)

=



0

0 

(97)

Application: lattice rotated

45 ◦

and cylindrical bending

I

Exact solution

I

Plate solution + Localization

(98)

Application: lattice rotated

45 ◦

and cylindrical bending

I

Exact solution

I

Plate solution + Localization

(RM and BG)

-0.3 -0.2 -0.1 0 0.1 Bending Moment m2∗m Exact BG RM=KL -0.2 -0.2 -0.15 -0.1 -0.05 0 T orsion m1 ∗m Exact BG RM=KL