influence of anisotropy on flexural optimal design of ...p. vannucci [email protected]...
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Università di PisaDipartimento di Ingegneria Strutturale26 maggio 2008
Influence of anisotropy on flexural optimal design of plates and laminates
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
Analysis of the objective function: cross-ply solutions
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
Analysis of the objective function: cross-ply solutions
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
Analysis of the objective function: angle-ply solutions
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
Analysis of the objective function: angle-ply solutions
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
Analysis of the objective function: angle-ply solutions
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
Bounds on optimal and anti-optimal solutions
38
Bounds on optimal and anti-optimal solutions
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
( )
( )
( ) .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
39
Bounds on optimal and anti-optimal solutions
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
χχρρ
ρτρρ
ρτ
χχρρ
ρτρρ
ρτ
δδ
δδ
+
−−=
+
−+=
+
++=
+
+−=
;and,,0on
;and0on;and0on
21
22
11
ststq
χχχχχχ
χχχχ
θθ ==∞→=
∞→=
∞→=
40
Bounds on optimal and anti-optimal solutions
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
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
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
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
Bounds on optimal and anti-optimal solutions
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
The case of a non-sinusoidal load
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
The case of a non-sinusoidal load
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 optimal critical load of buckling
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 optimal critical load of buckling
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
Some examples of exact optimal solutions
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
Some examples of exact optimal solutions
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.