influence of anisotropy on flexural optimal design of ...p. vannucci [email protected]...

Università di PisaDipartimento di Ingegneria Strutturale26 maggio 2008

Influence of anisotropy on flexural optimal design of plates and laminates

P. [email protected]

UVSQ – IJLRDA Institut Jean Le Rond d'AlembertUMR CNRS 7190 - Paris

2

Foreword

The use of composite materials forces designers to use optimal procedures for obtaining non intuitive suitable solutions.

Designing of laminates with respect to flexural properties is the most cumbersome task in the design of laminates; few researches have been carried on in this field, and the most part of them lead to only approximate solutions.

The first task of this research was to find exact optimal solutions to some classical flexural problems of plates, when such plates are laminates composed of anisotropic identical plies.

3

A second task was that of assessing the influence of the anisotropy of the material on the optimal solutions so found: this is the topic of this talk.

Dimensionless invariant material properties have been chosen to represent the layer elastic properties, along with other dimensionless parameters describing geometry, deformation and/or loading.

Some unattended features and pathological cases have been so found.

4

Content

Governing equationsDimensionless design parametersA common problem for bending, buckling and vibrationsAnalysis of the objective functionEffectiveness of the optimal solutionBounds on optimal and anti-optimal solutionsThe case of a non-sinusoidal loadThe optimal critical load of bucklingThe optimal fundamental frequencySome examples of exact optimal solutions

5

Governing equations: the mechanical model

Simply supported rectangular laminate made of identical layers.Classical lamination theory (Kirchhoff model etc.).

Bending stiffness tensor:

=

κε

DBBA

MN o

∑ == pnj jj

pδd

nh

13

3),(

121 QD

.and)2(34)1(12 13∑ = =+++−−= pn

j pjpppj ndnnnjjd

b

ax

y h

znp

1

...

.DDDDDDDDD

ssysxs

ysyyxy

xsxyxx

=D

6

Governing equations: supplementary assumptions

In order to dispose of an analytical solution, the laminate is assumed to be specially orthotropic in bending:

B=O; Dxs=Dys=0.

In this way, the equilibrium equation for deflection w is not coupled to the equations of in-plane displacements and the separation of variables is possible: the Navier's method can be applied.

For buckling, a further assumption is that

N=(Nx, Ny, 0)

i.e. no shearing in-plane forces (if not, the Navier's method does not apply).

7

Governing equations: transverse equilibrium equation

[ ] equation; mequilibriu)(33 →= zpwL

.)2(2 4

4

22

4

4

433

yD

yxDD

xDL yyssxyxx

∂

∂+

∂∂

∂++

∂

∂=

,,2 2

2

2

22

2

2

tL

yN

yxN

xNL ysx

∂

∂=

∂

∂+

∂∂∂

+∂

∂= µωλ

[ ] equation; vibration0)(33 →=− wLL ω

[ ] equation; buckling0)(33 →=− wLL λ

8

Governing equations: Navier's solution method

[ ].sinsinsin),(

,sinsin),(

1,

1,

∑

∑∞

=

∞=

=

=

nm mnmn

nm mnz

tbyn

axmayxw

byn

axmpyxp

ωππ

ππ

center. the in load edconcentrat a for2

sin2

sin and4

, load uniform a for1 and16

with,

*

*2

*

Pnmp

abPp

mnp

pabPp

mn

mn

mnmn

ππψ

πψ

ψ

==

===

=

n

m

ab

x

y

9

Governing equations: polar parameters of the material

Φ0−Φ1=k π/4, k=0, 1: common orthotropyR0=0: R0-orthotropyR1=0: square symmetric orthotropy

.2cos44cos2,2sin24sin

,4cos,4cos2

,2sin2 4sin,2cos44cos2

11 0010

11 00

000

0010

1100

11 0010

ΦΦΦΦ

ΦΦ

ΦΦΦΦ

RRTTTRRT

RTTRTTT

RRTRRTTT

yy

ys

ss

xy

xs

xx

−++=+−=

−=−+−=

+=+++=

10

Governing equations: laminate bending stiffness

Separation of geometry (lamination parameters) and material (polar constants):

Specially orthotropic laminates: ξ0 and ξ1 are sufficient to completely describe bending stiffness.

.)1(

200200

001021

421421

12

1

0

1

0

32

32

0

0

10

10

3

−

−

−−−

−

=

RR

TT

h

DDDDDD

k

ys

xs

ss

xy

yy

xx

ξξξξ

ξξ

ξξξξ

∑∑

∑∑

==

==

==

==

pp

pp

nj jj

p

nj jj

p

nj jj

p

nj jj

p

dn

dn

dn

dn

133132

131130

.2sin1,4sin1

,2cos1,4cos1

δξδξ

δξδξ

.112,11 0211 ≤≤−≤≤− ξξξ

ξ0

ξ11 -1 0

1

-1

A C

B Arc of the angle-ply laminates

Segment of the cross-ply laminates

Feasible domain

11

Governing equations: polar form of the equations

Separation of the mean isotropic part from the pure anisotropic part:

Mean isotropic part: meaning of the polar isotropy constants

[ ]( ) ,)(~)2(12 3310

3zpwLLLwTTh

=−−++ ωλ∆∆

( )[ ] ( ) ( )[ ]

∂

∂−−+

∂∂

∂−−

∂

∂+−= 4

4110022

4004

41100

333 411641

12~

yRR

yxR

xRRhL kkk ξξξξξ

.1

2 210ν−

=+ETT

12

Dimensionless parameters: material

ρ : anisotropy ratio: ρ =0, R0-orthotropic materials;ρ =∞ : square-symmetric materials.

τ : isotropy-to-anisotropy ratio; τ >1k: orthotropy index; k=0 : low shear modulus orthotropy

k=1 : high shear modulus orthotropy

.4,2, 1021

20

10

1

0π

ΦΦτρ −=

+

+== k

RR

TTRR

ρ<1ρ>1

Qxx Qxx k=1

k=1 k=0k=0

13

Dimensionless parameters: Cartesian expression

For a generic orthotropic elasticity tensor T it is:

.

,

)T(T)TTT)(TT(T

)T(T)T(T

TT)T(TTT

yyxxyyxxssxyssxy

ssxyyyxx

yyxx

ssxyyyxx

222224

223

22

++−−++

+++=

−

+−+=

τ

ρ

).2T(TTTk

),2T(TTTk

ssxyyyxx

ssxyyyxx

+<+=

+>+=

2if1

2if0

14

Dimensionless parameters: geometry and mode

Aspect ratio:

Mode ratio:

Wave-length ratio: .

Force ratio:

If m=n, χ=0 → χ=1 → χ=∞ →

.nm

=γ

.ba

=η

γηχ ==

mbna

.x

yNN

=ν

15

A common problem for bending, buckling and vibrations: bending stiffness

Maximization of the bending stiffness= minimization of the compliance JD

Navier's solution of equilibrium equation

Replacing the Dij by their polar expressions we get

.0 0∫ ∫=a b

z dydxwpJD

,)2(2

1224 βαβαπ yyssxyxx

mnmn

DDDDpa

+++= ., 2

2

2

2

bn

am

== βα

.)(4)6()1())(2(

31, 22

1122

002

10

2

34 ∑∞= −+−+−+++

= nm kmn

RRTTp

habJ

βαξαββαξβαπD

16

A common problem for…: bending stiffness

Using the dimensionless parameters above, we get

with

Further simplification: for a material, geometry and a sinusoidal load given (i.e. for fixed m and n), the optimization problem is reduced to the maximization of the function ϕ(ξ0, ξ1).

( ),

,)1(3

1,10

224

2*

21

20

3

22

4

2∑∞

= ++= nm

mnm

p

RRh

aPJξξϕχπ

ηψD

.114

)1(16)1(

1

1),( 2

2122

240210

+

−+

+

+−−

++=

χχξ

χχχρξ

ρτξξϕ k

17

A common problem for…: buckling loads

Be N=λ (Nx, Ny, 0), λ = load multiplier. We want to maximise λmn, the buckling load multiplier for the mode (m, n). The Navier's non-trivial solution of the buckling equation for the mode (m, n) is

Once again, replacing the Dij by their polar expressions and using the dimensionless parameters above, we get

.)2(2 22

2βNαN

βDαβDDαD

yx

yyssxyxxmn +

+++= πλ

).,(1

1)1(12

102

222

22

21

20

2

322ξξϕ

χννχπλ

+

++

+

+=

yxmn

NNRR

ahm

18

A common problem for…: natural frequencies

Finally, we consider the problem of maximising the natural frequency ωmn of a given mode (m, n). The Navier's non-trivial solution of the vibration equation for the mode (m, n) is

µ: mass of the laminate per unit area of the plate's surface.As usual, replacing the Dij by their polar expressions and using the dimensionless parameters above, we get

[ ].)2(2 224

2 βDαβDDαD yyssxyxxmn +++=µ

πω

.),()1(12

10222

1204

3442 ξξϕχ

µπω ++= RR

ahm

mn

19

A common problem for…: the objective function

Finally, the three problems above, concerning the flexural behaviour of the laminate for a precise mode, are reduced to the same non linear optimization problem:

To remark that also the opposite problem (that we will call the anti-optimization one) is physically meaningful, as the objective function can be proved to be always positive:

.112,11 to subjected

),,( maximise

0211

10

≤≤−≤≤− ξξξ

ξξϕ

.112,11 to subjected

),,( minimise

0211

10

≤≤−≤≤− ξξξ

ξξϕ

20

Analysis of the objective function: separation of material and mode

We rewrite the objective function ϕ as

The functions c0(χ) and c1(χ) give the influence of the mode and geometry. Their roots are of some importance.

ρ, τ and k give the influence of the materialτ : gives the influence of the ply's isotropyρ : gives the influence of the ply's anisotropyk : gives the influence of the orthotropy's type

[ ]

.11)(,

)1(16)( with

,)(4)()1(1

1),(

2

2122

240

1100210

χχχ

χχχχ

ξχξχρρ

τξξϕ

+

−=

+

+−=

+−+

+=

cc

cck

c0(χ)

c1(χ)

χ

12 + 12 − 1

21

Analysis of the objective function: pathological solutions

ϕ(ξ0,ξ1) is linear with respect to ξ0, ξ1 → maxima and minima are located on the boundary of the feasible domain.Nonetheless, it is useful to analyse the gradient of ϕ(ξ0,ξ1):

∇ϕ(ξ0,ξ1)=0 ↔

ρ=0 and χ=1: this is the case of laminates made of R0 -orthotropic materials (R0=0) and with equal wave-length of the mode along x and y, (e.g. square plates and modes with m=n);

or

,)(1

4),(1

)1(,),( 12021010

++

−=

∂∂

∂∂

=∇ χρ

χρ

ρξϕ

ξϕξξϕ cc

k

22

Analysis of the objective function: pathological solutions

ρ=∞ and : this is the case of laminates made of square-symmetric materials (R1=0), i.e. reinforced by balanced fabrics, and, if for instance m=n, having anaspect ratio .

In these two circumstances, it is not possible to optimize the laminate, because the objective function is constant and reduces to only its isotropic part, τ. Actually, in such cases, the contribution of the anisotropic part disappears, due to special combinations of geometry, mode and anisotropy properties of the layer: the laminate behaves like it was made of isotropic layers, and any possible stacking sequence give the same result.

12 ±=χ

12 ±=η

23

Analysis of the objective function: cross-ply solutions

Cross-ply laminates are represented by lamination points of the type ξ0=1, −1≤ ξ1≤1; this is possible ↔

For the anti-optimal problem, it is sufficient in the first condition above to change k=0 into k=1 and vice-versa, i.e., k changes maxima into minima and vice-versa: this is typical.

{ }.0,)0,(),( 210 −∈=∇ Raaξξϕ

.or101

)(;1212and1

,12or)12,0[ and00)()1(

21

0

∞==⇔=+

+<<−=+>−∈=

⇔>−

ρχρ

χχ

χχχ

ck

kck

24


A remark: cross-ply solutions exist only in the presence of a generalised square-symmetry: of the material, condition ρ =∞, or of the geometry and mode, condition χ=1 (e.g., m=n and a square plate). The values of the solutions are

To notice that in the first case ρ influences the extreme values, while χ in the second:

).()1(,for

,1

)1(,1for

0

2

χτϕρ

ρ

ρτϕχ

ck

k

−+=∞=

+

−−==

25


Optimal and anti-optimal cross-ply solutions are not unique, as ξ1 disappears from the different expressions above: any laminate combination of layers at 0° and at 90° is an optimal (or anti-optimal) solution if conditions above are satisfied.

[ ]{ }( )

( )

[ ]{ }( )

( ).1)()1(min

1

)1(min

,1)()1(max1

)1(max

1,0 or,0,10

,02

1

1,1 or,0,00

,12

1

−=−+=

+

−−=

+=−+=

+

−−=

==∞==

∞==

==∞==

∞==

τχτρ

ρτϕ

τχτρ

ρτϕ

χχ

ρ

χχ

ρ

kk

k

k

kmin

kk

k

k

kmax

c

c

26

Analysis of the objective function: angle-ply solutions

Angle-ply laminates are located on the boundary of the feasible domain, where

whose maxima and minima can be only

12 210 −= ξξ

[ ],)(4)12)(()1(1

1)( 1121021 ξχξχρ

ρτξϕ cck +−−

++=

[ ]

[ ]

,)(

)(2)(

1

)1()(

,)(4)()1(1

1)1(

,)(4)()1(1

1)1(

0

21

20

2

211

102122

102111

χρχχρ

ρτξξϕϕ

χχρρ

τξϕϕ

χχρρ

τξϕϕ

δδ ccc

cc

cc

kst

k

k

+

+

−−===

−−+

+=−==

+−+

+===

27


The corresponding orientation angles δ are

with

Remark: it is easy to verify that for two plates having reciprocal wave-length ratios, the respective solution angles δ are complementary.

plyangle truearccos21 to correponds

onalunidirecti,2

to correponds 1

onalunidirecti,0 toscorrespond 1

111

1

1

−→==

→=−=

→==

stst ξδξξ

πδξ

δξ

,)()()1(0

0

11

1 1χχ

ρξ

ξϕ

ξ cck

stst

−−=⇒=

∂∂

28


Angle-ply solutions exist ↔This conditions give link the influence of the material part to that of the mode on the existence of angle-ply optimal laminates:

.1)()()1(1

0

1 ≤−

≤−χχ

ρ cck

);(and1),()(and0

),(0and1

4

32

1

ρχχρρχχρχρ

ρχχρ

≥>

≤<≥

≤<>

.1

183)(,1

183)(

,1

183)(,1

183)(

24

23

22

21

−++

=+

++=

−+−

=+

+−=

ρρρρχ

ρρρρχ

ρρρρχ

ρρρρχ

3

31

0

1

2

3

4

5

6

0 1 2 3 4 5 6

12 −

12 +

χ

ρ χ1(ρ) χ2(ρ)

χ3(ρ)

χ4(ρ)

29


Once more k changes minima into maxima:

In particular, using the expression of c0(χ) we find that

Remark: the isotropy parameter τ does not affect the solution.

).(1

)1(4 0221

2χ

ρ

ρξϕ ck

+−=

∂

∂

• if ( )(0,1 1 ρχχρ ≤<> ) or ( )(,1 4 ρχχρ ≥> ), then ,1for == kmaxϕϕδδ0for == kminϕϕδδ ;

• if ( )()(,0 32 ρχχρχρ ≤<≥ ), then 1for ,0for ==== kk minmax ϕϕϕϕ δδδδ .

30

Analysis of the objective function: map of the solutions

Considering the hierarchy of ϕ11, ϕ22 and ϕδδ we see that

Crossing all these results, we can trace a map of the optimal and anti-optimal solutions in the plane (ρ, χ):

[ ]

=+>−<<=+<<−

⇔

>>±−−⇔>

>⇔>⇔>

−

.1 if12or120,0 if1212

0)(0)()()1()1(

;11)(

0

210

1

22

11

12211

kk

ccc

c

kk

χχχ

χχχρ

ϕϕ

ϕ

χχϕϕ

δδ

31

Analysis of the objective function: map of the solutions

0

1

2

3

4

5

6

0 1 2 3 4 5 6

δ=0°0°<δ ≤45°

45°<δ ≤90°

δ=90°

η

ρ

0

1

2

3

4

5

6

0 1 2 3 4 5 6

δ=90°

η

ρ

cross-plyδ=0°

0°<δ ≤45°

45°<δ ≤90° χ χ

12 +12 +

12 −12 −

k=0: optimal sol.

0°<δ <45°

k=0: anti-optimal sol.

δ =90°

δ =0°

45°<δ ≤90°

0

1

2

3

4

5

6

0 1 2 3 4 5 6

δ=0°0°<δ ≤45°

45°<δ ≤90°

δ=90°

η

ρ

0

1

2

3

4

5

6

0 1 2 3 4 5 6

δ=90°

η

ρ

cross-plyδ=0°

0°<δ ≤45°

45°<δ ≤90° χ χ

12 +12 +

12 −12 −

k=1: anti-optimal sol.

45°<δ <90°

k=1: optimal sol.

δ =0°

δ =90°

45°<δ<90°

δ=0°

0°<δ<45°

δ=90°

ϕ22

ϕ22

ϕ22

ϕ22

ϕ11

ϕ11

ϕ11

ϕ11

ϕδδ

ϕδδ

ϕδδ

ϕδδ

ϕδδ

ϕδδ

32

Effectiveness of the optimal solution

It is interesting to evaluate the gain of the true (angle-ply) optimal or anti-optimal solution with respect to the intuitive(unidirectional) one, i.e. the ratios

;1if2,1if1,),,( >=<== χχϕϕχρτζ δδ ii

iimax

ζmin

χ

ρρ

χ

ζmax

;1if1,1if2,),,( >=<== χχϕϕχρτζ δδ ii

iimin

33

Effectiveness of the optimal solution: extreme value

These ratios get their extreme value for ρ →∞ and χ=0, χ=1 or χ→∞, i.e., if m=n, for square plates or infinite strips composed by square symmetric layers (R1=0):

The effectiveness of the solution is determined by the value of τ.

( ) ,11max

min

maxmax Q

Q=

−+

=ττζ

( ) .11min

max

minmin Q

Q=

+−

=ττζ

Qmin

Qmax

x

y Qxx

34

Effectiveness of the optimal solution: real materials

It can be shown that materials with k=1 are less diffused than those with k=0, but they do exist.

Apart from square symmetric layers, ρ =∞, the most part of composite layers have ρ<1 and k=0 → only the left part of the map of solutions is usually of concern. If ρ<1, the range of χ where optimisation is meaningful, i.e. where the solution is not 0° or 90°, increases with ρ and, for ρ=1, it is comprised between and .

For current materials (ρ≈1, k=0), it is ζmax=2.

Some examples of materials are in the following table:

31=χ 3=χ

35

Effectiveness of the optimal solution: real materials

(modules in GPa)

Material Fir wood Ice Titanium-Boride TiB2

Boron-epoxy B(4)-55054

Carbon-epoxy T300-5208

Kevlar-epoxy 149

S-Glass-epoxy S2-449/SP

381

Glass-epoxy balanced

fabric

Braided carbon-epoxy

BR45a

Braided carbon-epoxy

BR60

Reference Lekhnitskii, 1950

Cazzani & Rovati, 2003

Cazzani & Rovati, 2003

Tsai & Hahn, 1980

Tsai & Hahn, 1980

Daniel & Ishai, 1994

MIL-HDBK-17-2F, 2002

Daniel & Ishai, 1994

Falzon & Herszberg,

1998

Falzon & Herszberg,

1998 E1 10 11.75 387.60 205.00 181.00 86.90 47.66 29.70 40.40 30.90 E2 0.42 9.61 253.81 18.50 10.30 5.52 13.31 29.70 19.60 42.60 G12 0.75 3.00 250.00 5.59 7.17 2.14 4.75 5.30 25.00 14.00 ν12 0.01 0.27 0.44 0.23 0.28 0.34 0.27 0.17 0.75 0.34 Q11 10 12.51 445.62 206.00 181.81 87.54 48.65 30.58 55.56 36.76 Q22 0.42 10.22 291.80 18.59 10.35 5.56 13.59 30.58 26.96 50.68 Q66 0.75 3.00 250.00 5.59 7.17 2.14 4.75 5.30 25 14.00 Q12 0.004 2.78 130.12 4.27 2.89 1.89 3.67 5.20 20.22 17.23 T0 1.68 3.65 184.65 29.80 26.88 12.23 92.38 8.99 17.76 13.62 T1 1.30 3.54 124.71 29.14 24.74 12.11 86.97 8.94 15.37 15.24 R0 0.93 0.65 65.35 24.21 19.71 10.09 44.86 3.70 7.24 0.38 R1 1.19 0.28 19.23 23.42 21.43 10.25 43.82 0 3.57 1.74 Φ0 0 0 π/4 0 0 0 0 0 π/4 π/4 Φ1 0 0 0 0 0 0 0 0 0 π/2 k 0 0 1 0 0 0 0 0 1 (−)1 ρ 0.78 2.32 3.40 1.03 0.92 0.98 1.02 ∞ 2.03 0.22 τ 2.83 15.16 6.37 2.61 2.62 2.53 4.25 7.26 6.01 24.76 Vf - - - 0.50 0.70 0.60 0.50 0.45 0.60 0.60

36

Bounds on optimal and anti-optimal solutions

We consider the influence of ρ and χ upon ϕmax and ϕmin, i.e. we look for the curves χ=χ(ρ) in the plane (ρ, χ) where the surfaces ϕ11, ϕ22 and ϕδδ have a local or absolute maximum (minimum) with respect to χ:

These curves are:

;,2,1,0 δχ

ϕ==

∂∂ iii

.1

)1(813,1

)1(813

,1

)1(813,1

)1(813

,11,

11,,1,0

65

43

21

−+++

=+

−+−=

+−−−

=−

+−+=

−+

=+−

=∞→==

ρρρρχ

ρρρρχ

ρρρρχ

ρρρρχ

ρρχ

ρρχχχχ

stst

stst

stst

37


38


Comparison with qxx(θ):

So, actually ϕ is similar to qxx(θ), the isotropic part is the same (τ), and only functions c0(χ) and c1(χ) introduce the influence of geometry and mode in ϕ. In particular, the maximum and minimum can be only

also similar to ϕ11, ϕ22 and ϕδδ.

[ ] .2cos,4cos,4)1(1

1)()( 1010221

20

θξθξξρξρ

τθθ ==+−+

+=+

= kxxxx

RR

Qq

( )

( )

( ) .1foronly,1

2)1(/)1(

,1

4)1(1

,1

4)1(1

2

21

2122

2111

>+

+−−=−−==

+

−−+=−==

+

+−+===

ρρρ

ρτρξ

ρ

ρτξ

ρ

ρτξ

θθkk

xx

kxx

kxx

qq

qq

qq

39


In particular, we have that the extreme values of ϕ and qxx(θ) are equal on some of the preceding curves:

To complete the comparison with ϕ, let us introduce the following intermediate values of qxx(θ):

;1on,1

,1 2221 =

++=

+−= χ

ρ

ρτρ

ρτ ss qq

.andon,1

12,1

12

;andon,1

12,1

12

542423

632221

stst

stst

qq

qq

χχρρ

ρτρρ

ρτ

χχρρ

ρτρρ

ρτ

δδ

δδ

+

−−=

+

−+=

+

++=

+

+−=

;and,,0on

;and0on;and0on

21

22

11

ststq

qq

χχχχχχ

χχχχ

θθ ==∞→=

∞→=

∞→=

40


The diagrams of ϕa, the anisotropic part of the optimal solution:

χ ρ

ϕa

χ ρ ϕa

a) k=0, optimal solution b) k=0, anti-optimal solution

c) k=1, optimal solution d) k=1, anti-optimal solutionχ

ρ

ϕa ϕ a

ρ

χ

q11

q11

q22

q22

q22

q22

q11

qθθ

qθθ

qδ3

qδ2

qδ1qδ4

qs1

qs1

qs2

qs2

q11

qθθ

qθθ

qθθ

qδ1

qδ2

ϕ11ϕ11

ϕ22ϕ22

ϕδδϕδδ ϕδδϕδδ

ϕδδϕδδ

ϕδδϕδδϕδδϕδδ

ϕδδϕδδ

ϕ11ϕ11

ϕ11ϕ11

ϕ22ϕ22

ϕ22ϕ22

ϕ11ϕ11ϕ22ϕ22

χ ρ

ϕa

χ ρ ϕa



ρ

ϕa ϕ a

ρ

χ

χ ρ

ϕa

χ ρ ϕa



ρ

ϕa ϕ a

ρ

χϕ a

ρ

χ

q11

q11

q22

q22

q22

q22

q11

qθθ

qθθ

qδ3

qδ2

qδ1qδ4

qs1

qs1

qs2

qs2

q11

qθθ

qθθ

qθθ

qδ1

qδ2

ϕ11ϕ11

ϕ22ϕ22

ϕδδϕδδ ϕδδϕδδ

ϕδδϕδδ

ϕδδϕδδϕδδϕδδ

ϕδδϕδδ

ϕ11ϕ11

ϕ11ϕ11

ϕ22ϕ22

ϕ22ϕ22

ϕ11ϕ11ϕ22ϕ22

41


Remark: q11, q22 and qθθ, bound the optimal and anti-optimal values of ϕ(ξ0, ξ1), ϕ11, ϕ22 and ϕδδ.

So, global maxima and minima can be taken only for χ =0, χ→∞, and , while local maxima and minima only for χ=1 and on to .

In addition, χ=1 and to can be absolute maxima of the anti-optimal solution and absolute minima of the optimal solution.

Finally, the optimal and anti-optimal values of the objective function, can be interpreted as a sort of normal stiffness, that takes its highest or lowest possible value in some special cases.

st1χχ = st

2χχ =st3χ st

6χ

st3χ st

6χ

42

The case of a non-sinusoidal load

In this case the problem of stiffness maximization is

Now, the objective function is no more linear and the isotropic part, τ, influences the location of the solution.Nevertheless, being

the solution is again on the boundary (except some pathological situations, see below)

( ).112,11 to subjected

,,)1(

),(minimize

0211

1,10

224

2*

10

≤≤−≤≤−

+= ∑∞

=

ξξξ

ξξϕχξξϕ nm

mnT

mp

( )),0,0(

1

)(4;1

)()1(,)1(

),(2

12

01,

102224

2*

10 ≠

++

−

+=∇ ∑∞

= ρ

χ

ρ

χρξξϕχ

ξξϕ ccm

p k

nmmnT

ξ0 ξ0ξ0

ξ1 ξ1 ξ1

ϕ ϕΤ ϕΤ

43


A fundamental question is : if we consider the case m=n=1, for which we can find the solution as seen, how much does the optimal solution change when the whole series is considered? To evaluate this, we introduce the ratio

It can be seen that for m→∞, σ→0 ∀γ; in addition, for γ=1, i.e. if m=n, then

For instance, for a uniform load σ =1/m8, viz. the term m=n=3 is 1/6561≈1.5 ×10-4 the first term, while for a concentrated load it is 1/324≈3.1 ×10-3.

,

1

)1(** 11

2

2

24

22

211

2

mn

mn

mpp

ϕϕ

γη

ησ

+

+=

,*

*211

4

2

pmp mn=σ

44


Generally, though not quite identical, optimal solutions for a generic load do not differ substantially form those found for a sinusoidal load; this means also that optimal solutions for buckling and natural frequencies for χ=η are similar to those of bending stiffness for a generic load.

(mmax=nmax=50)

Optimal orientation

angle δ Fir wood

Boron-epoxy B(4)-55054

Carbon-epoxy

T300-5208

Glass-epoxy

balanced fabric

Braided carbon-epoxy BR45a

η=1.2 η=6.0 δsin 52.16° 50.40° 51.05° 45° 70.64° δunif 51.86° 50.22° 50.82° 45° 75.68° δconc 49.52° 48.65° 48.92° 45° 90.00°

δuni f − δsin −0.30° −0.18° −0.23° 0° 5.04° δconc − δsin −2.66° −1.75° −2.12° 0° 19.36°

45

The case of a non-sinusoidal load: pathological solutions

The previous pathological solutions do not exist, as χchanges with m and n. Nevertheless, if ρ =0, functions ϕ and ϕT do not depend upon ξ0. In such a case, ∇ϕT can be null along a line parallel to the ξ0 axis.

ξ0 ξ1

ϕΤ

For η=1,possible solutions are a balanced cross-ply laminate as well as an angle-ply with δ=45°, but also any other laminate of the type (ξ1=0, −1≤ ξ0≤1), for instance isotropic laminates in bending(ξ1=0, ξ0=0).

46

The case of a non-sinusoidal load: pathological solutions

The other case is ρ →∞; ϕ and ϕT do not depend upon ξ1.∇ϕT=0 for ξ0=±1, that is for an angle ply with δ=45°, if k=0, or for a cross-ply laminate, if k=1. As for a sinusoidal load, all the cross-ply laminates are possible solutions, if k=1.

ξ0 ξ1

ϕΤ

ξ0 ξ1

ϕΤ

47

The optimal critical load of buckling

The problem is to determine the mode (m, n) that leads to the lowest buckling load.To this purpose, we consider the ratio

where λa is the buckling load for m=ma, n=na, and λb for m=mb, n=nb; ϕa is ϕ calculated for η/γa and ϕb for η/γb.

If γa=γb, then → we can consider mb=nb=1 and

analyse what happens for a given ma and for γa≠1.

,22

222

22

222

b

a

a

b

b

a

b

a

b

amm

ϕϕ

νηγνηγ

ηγηγ

λλε

+

+

+

+

==

2

=

b

ammε

48


The two buckling loads mb=nb=1 and (ma, na) can be found and ε computed; so, for a given plate and mode (ma), we can trace the surface representing ε(γa, ν) and look when it is lower or greater than 1. The curves separating the domains can be put in explicit form; they are:

b1 → solution of ε →∞:

b2 → solution of ε=1:

b3 → trivial solution γa=1 if ma=1.

;2

2

ηγν a−=

;)(

)1(,1222

22

2

22

2 ηγη

ϕϕγ

ην

+

+=

−

−−=

aa

aa AAm

Am

49


The areas where ε>1 are those where the buckling load of the case a is greater than the one of the case b.In this way all the significant cases can be easily verified.

Examples: carbon-epoxy, τ=2.62, ρ=0.92, k=0 (ε>1 in blue).

η=1.2, ma=2

γa

ν η=1.7, ma=2

γa

νη=0.6, ma=1

γa

ν

50

The optimal fundamental frequency

Like for the critical load, the problem is to determine the mode (m, n) that leads to the lowest natural frequency.

To this purpose, we consider the ratio

where ωa is the natural frequency for m=ma, n=na, and ωbfor m=mb, n=nb; ϕ(χ) is ϕ calculated for χ=η/γb and ϕ(χ∗) for χ∗=η/γa=χ/γ∗, where γ∗=γa/γb and m∗=ma/mb. We can trace the curve and see where .

,)(*)(

1*1*

2

2

24

2

χϕχϕ

χχ

ωωϖ

+

+=

= m

b

a

1*)*,( =γϖ m 1>ϖ

51

The optimal fundamental frequency

The equation of the curve is simply

Examples: carbon-epoxy, τ=2.62, ρ=0.92, k=0 (ϖ>1 in blue).

.)(*)(

1*1*

42

2

2−

+

+=

χϕχϕ

χχm

χ=1

γ *

m*χ=2

γ *

m*

52

Some examples of exact optimal solutions

A simple strategy for obtaining exact specially orthotropic laminates in bending: to choose a quasi-homogeneous solution (A/h=12D/h3, B=O) of the quasi-trivial set.This ensures that for the angle-ply laminate will be not only B=O, but also Dxs=Dys=0 as they are equal to Axs and Ayswhich are automatically null for angle-ply laminates. So, for this class of laminates, the Navier's solutions are exact.

Example 1: carbon-epoxy laminate, τ=2.62, ρ=0.92, k=0, η=1.2, γ=1, ν=1.As 1<χ=1.2<χ3=1.70 and k=0, the optimal value of the objective function is ϕmax=ϕδδ=3.31.

53


The solution angle is δ=51°.

The gain with respect to the intuitive solution ϕ22=2.57 is ζmax=1.29.

The optimal solution for bending stiffness in case of uniform load is δ=50.8° and for concentrated load 48.9°.

Example 2: braided carbon-epoxy BR45a laminate, τ=6.01, ρ=2.03, k=1, η=10, γ=1, ν=1.

As χ=10>χ4=3.40, ρ>1 and k=1, the optimal value of the objective function is ϕmax=ϕδδ=7.29.

54


The solution angle is δ=60.8°.

The gain with respect to the intuitive solution ϕ22=6.06 is ζmax=1.2.

The optimal solution for bending stiffness in case of uniform load is δ=61.5° and for concentrated load 65.9°.

Possible exact (unsymmetrical) solutions: 8 plies: [δ, −δ, −δ, δ, −δ, δ, δ, −δ]12 plies: [δ, −δ, δ, −δ3, δ3, −δ, δ, −δ]16 plies: [δ, −δ, δ, −δ2, δ, −δ2, δ4, −δ3, δ] etc.

55

Conclusion

The design analysis made with dimensionless invariant parameters helps in some way the understanding of the flexural problems in presence of anisotropy: it puts in evidence some pathological situations and characterizes the localization of the different types of optimal solutions as well as their effectiveness.

This study is merely qualitative, but can help in similar studies with other geometries and conditions.

Thank you very much for your attention.

influence of anisotropy on flexural optimal design of ...p. vannucci [email protected]...

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