Laminate Analysis
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Transcript of Laminate Analysis
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analysis of fiber-reinforced composite materials by using M. W. Hyers codes
ME518
01/26/2004
Chia-Wei Wang
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macro-scale mechanics analysis
micro-scale mechanics analysis
ABBD matrix
maximum stress failure criterion
outlines
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1fiber direction
2
3
principal material coordinate system
1-direction is aligned with the fiber direction the strain of an individual fiber or element of matrix is of no consequence at this level of analysis assume that the two-material fiber-matrix system is replaced with single homogeneous material
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112
13
2
2312
2
31
32
1
2
3
stress-strain relationship for orthotropic material
123 231312
=
1E1
21E2
31E3
0 0 0
12E1
1E2
32E3
0 0 0
13E1
23E2
1E3
0 0 0
0 0 0 1G23
0 0
0 0 0 0 1G13
0
0 0 0 0 0 1G12
1 2 3 231312
1 2 3 231312
=
C11 C12 C13 0 0 0C12 C22 C23 0 0 0C13 C23 C33 0 0 00 0 0 C44 0 00 0 0 0 C55 00 0 0 0 0 C66
123 231312
inverse relationship
in the principal direction
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free thermal strains
1
23
1
3
2
free thermal strains do not involve shear deformation coefficients of thermal expansion (CET) in three principal
directions are constant in range of interest
1T T,Tref( ) = 11
=1T
2T T,Tref( ) =
2 2
=2T
3T T,Tref( ) = 3 3
=3T
stress-strain relationship include free thermal strains interms of compliance matrix can be expressed as
1 2 3 231312
=
C11 C12 C13 0 0 0C12 C22 C23 0 0 0C13 C23 C33 0 0 00 0 0 C44 0 00 0 0 0 C55 00 0 0 0 0 C66
1 1T2 2T3 3T
231312
assumptions:
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macro-scale mechanics analysis
microscale mechanics analysis
ABBD matrix
maximum stress failure criterion
outlines
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micromechanics approaches
finite element approach theory of elasticityapproachStrength-of materialapproach
Information:stress-strain Information:
prediction of effective material properties
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finite element approach
2
3
2
3fibermatrix
unit cellunit cell is the simplest repeating unit
finite element models of unit cells1
Stresses of interest within unit cell1
nf = n
m
sf s
m
1. Hyer, M.W., Stress Analysis of Fiber-Reinforced Composite Materials, WCB/McGraw-Hill, New York, 1997
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theory of elasticity concentric cylindrical model
V F = b2
c 2=bc
2
volume fraction of fiber
2
3
r
r = c
r = b
concentric cylinder model
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strength-of material approach rule of mixtures models
WfWm
unit cell
WfWm
Am Af
1
2
1
1f = E1
f1f = E1
f LL
=F1
f
A f
1m = E1
m1m = E1
m LL
=F1
m
Am
1 =F1
f + F1m
A= E1
f A f
A+ E1
m Am
A
LL
= E1LL
E1 = E1f A f
A+ E1
m Am
A= E1
fV f + E1mV m = E1
fV f + E1m 1V f( )
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Hyers codes
fiber-reinforced composite materials is usually made in plate form. therefore, the plane stressassumption is usually made for analysis of effective material properties of fiber-reinforcedcomposite material.
E1, E2, 12, 1, 2 and G12 are the most important material properties determined for laminatedcomposite materials.
E1.F and E2.F E1 and E2 Hyers code material property predicted
G12.F G12
NU12.F 12
ALPHA12.F 1 and 2
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plane stress assumption
=
12
13
23
3
2
1
66
55
44
332313
232212
131211
12
13
23
3
2
1
00000
00000
00000
000
000
000
S
S
S
SSS
SSS
SSS
1212
=
1E1
12E1
0
12E1
1E2
0
0 0 1G12
1 212
reduced
xy12 xy
=
cos2 sin2 2cos sinsin2 cos2 2cos sin
cos sin cos sin cos2 sin2
1E1
12E1
0
12E1
1E2
0
0 0 12G12
cos2 sin2 2cos sinsin2 cos2 2cos sin
cos sin cos sin cos2 sin2
x y xy
=
S 11 S 12 S 13S 21 S 22 S 23S 31 S 32 S 33
x y xy
rotated
x y xy
=
Q 11 Q 12 Q 13Q 21 Q 22 Q 23Q 31 Q 32 Q 33
1212
or
3 is not zero 1, 2, 23 and 13 are zero
SBAR2D.F
QBAR.F
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E1 = E1f 1+( )V f + Em 1+( ) 1 V f( )
=221
f Em 1 V f( ) 12f m( )E2f 1+ vm( ) 1+V f 1 2vm( )( ) + Em 1 23f 212f 21f( ) 1 V f( )
=2E2
f mV f m 12f( )
E2f 1+ vm( ) 1+V f 1 2vm( )( ) + Em 1 23f 212f 21f( ) 1V f( )
E1 = E1fV f + Em 1 V f( )
v12 = 12f V f + 1V f( ) m
E1, E2, 12 and 21
rule of mixture
rule of mixture
1E2
=V f
E2f +
1 V f( )Em
rule of mixture
modified rule of mixture
1E2
=
V f
E2f +
1 V f( )Em
V f + 1 V f( )
alternative rule of mixture
1E2
=
fV f
E2f +
m 1 V f( )Em
V f + 1 V f( )
f =E1fV f + 1 v12
f v21f( )Em + vmv21f E1f[ ] 1 V f( )
E1fV f + Em 1 V f( )
m =1 vm
2( )E1f 1 vmv12f( )Em[ ]V f + Em 1 V f( )E1fV f + Em 1 V f( )
concentric cylinders model
(3.32)
(3.29)
(3.30a)
(3.30b)
(3.77)
(3.87)
(3.93)
(3.106)
(3.107)
Because fiber composite materials aretransversely isotropic; hence, 13=12. But,n12n21
use E1.F to obtain E1 use E2.F to obtain E2 use NU12.F to obtain 12E1:
E2:
12:
Please referred to Hyer [1] for derivation details.1. Hyer, M.W., Stress Analysis of Fiber-Reinforced Composite Materials, WCB/McGraw-Hill, New York, 1997
12E1
= 21E2
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G12, G13, G23
: partitioning factor used to correct the volumefraction of matrix subjected to load
rule of mixtures
concentric cylinders model
( )m
f
f
f
G
V
G
V
G
+=11
1212
(3.117)
( ) ( )( ) ( )
++
+=
fmffm
fmffmm
GGVGG
GGVGGGG12 (3.60)
modified rule of mixtures
1G12
=
V f
G12f +
' 1V f( )Gm
V f +' 1V f( )(3.118)
Here we assumed composite materialsare transversely isotropic; hence,G12=G13.G23 can be calculated from
( )232
23 12 +=
EG
use G12.F to obtain G12
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alternative rule of mixtures
mmff
mmmfff
VEVE
VEVE
+
+=
1
111
(3.130)
( ) ( ) fmfmfmmf
fmfm VVE
EEV 1
1
1122
++= (3.140)
rule of mixturesmmff VV += 22 (3.139)
Simple rule of mixtures does not make sense for 1, since it is clear that the matrix andfiber must expand the same amount in the 1 direction when the temperature ischanged. The simple rule of mixtures in the fiber direction ignores this fact!
use ALFA12.F
1 and 2
-
=
12
13
23
3
2
1
66
55
44
332313
232212
131211
12
13
23
3
2
1
00000
00000
00000
000
000
000
S
S
S
SSS
SSS
SSS
1266
1355
2344
333
2
2323
222
1
1313
1
1212
111
111
11
1
GS
GS
GS
ES
E
vS
ES
E
vS
E
vS
ES
===
===
===
where
=
12
13
23
3
2
1
66
55
44
332313
232212
131211
12
13
23
3
2
1
00000
00000
00000
000
000
000
C
C
C
CCC
CCC
CCC
132312121233131322232311332211
6666
5555
4444
1123131223
1212221133
2213231213
1313331122
3312231312
2323332211
2
1
11
SSSSSSSSSSSSSSSS
SC
SC
SC
S
SSSSC
S
SSSSC
S
SSSSC
S
SSSSC
S
SSSSC
S
SSSSC
+=
=
==
=
=
=
=
=
=
where
use THREESC.F
the compliance matrix, Sij, and the stiffness matrix, Cij
-
( )
3
21
22
21
22
21
sincos2
cossin
sincos
=
=
+=
+=
z
xy
y
x
use ALPHA.F
x, y, xy and z
-
macro-scale mechanics analysis
micro-scale mechanics analysis
ABBD matrix
maximum stress failure criterion
outlines
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Nx = xdzH 2H 2
Ny = ydzH 2H 2
Nxy = xydzH 2H 2
Mx = xzdzH 2H 2
My = yzdzH 2H 2
Mxy = xyzdzH 2H 2
x y xy
=
Q 11 Q 12 Q 16Q 12 Q 22 Q 26Q 16 Q 26 Q 66
xyxy
x y xy
=
Q 11 Q 12 Q 16Q 12 Q 22 Q 26Q 16 Q 26 Q 66
xo + z x
o
yo + z x
o
xyo + z x
o
NxNyNxy
=
A11 A12 A16A12 A22 A26A61 A26 A66
xo
yo
xyo
+
B11 B12 B16B12 B22 B26B16 B26 B66
xo
yo
xyo
MxMyMxy
=
B11 B12 B16B12 B22 B26B16 B26 B66
xo
yo
xyo
+
D11 D12 D16D12 D22 D26D16 D26 D66
xo
yo
xyo
NxNxNxyMxMyMxy
=
A11 A12 A16 B11 B12 B16A12 A22 A26 B12 B22 B26A16 A26 A66 B16 B26 B66B11 B12 B16 D11 D12 D16B12 B22 B26 D12 D22 D26B16 B26 B66 D16 D26 D66
xo
yo
xyo
xo
xo
xyo
stress resultants (Nij) and moment resultants (Mij)
plane stress assumption
strain (x, y, xy) can be expressed in terms of and at reference surface
xo ,yo ,xyo( )
xo ,yo ,xyo( )
combine ABD matrix together
stress resultants (Nij) and moment resultants (Mij)expressed in terms of and
xo ,yo ,xyo( )
xo ,yo ,xyo( )
laminate theory: ABBD matrix relations
-
NxNyNxy
=
A11 A12 A13A12 A22 A23A13 A23 A33
xo
yo
xyo
+
B11 B12 B13B12 B22 B23B13 B23 B33
xo
yo
xyo
MxMyMxy
=
B11 B12 B13B12 B22 B23B13 B23 B33
xo
yo
xyo
+
D11 D12 D13D12 D22 D23D13 D23 D33
xo
yo
xyo
laminate theory: ABBD matrix relations
symmetric laminated materials - ABBD matrix reduced to:
NxNyNxy
=
A11 A12 A13A12 A22 A23A13 A23 A33
xo
yo
xyo
MxMyMxy
=
D11 D12 D13D12 D22 D23D13 D23 D33
xo
yo
xyo
-
material properties of IM6/3501-6 (graphite fiber/epoxy)
Base properties of (IM6) graphite fibersProperty Units Nominal values
E11 GPa 258.59E22, E33 GPa 13.90G12, G13 GPa 50.85
G23 GPa 8.2612, 13 - 0.2623 - 0.331 1/oC -0.855x10-6
2, 3 1/oC 3.24x10-6Vf - 63%
Base properties of (3501-6) EpoxyE GPa 4.34G GPa 1.59 - 0.36 1/oC 40.5x10-6Vm - 37%
Base properties of IM6/3501-6 graphite fiber/epoxy1 GPa 164.532,3 GPa 7.6583
G12,G13 Gpa 4.0937G23 GPa 2.853312,13 - 0.296923 - 0.3421, 1/oC -0.4513x10-62,3 1/oC 17.03x10-6
-
If the graphite-epoxy material is cooled (T= 75oC), and no initial strainis applied, what are the resulting residual stresses in each direction?
1 2 3
=
C11 C12 C13C12 C22 C23C13 C23 C33
0 1T0 2T0 3T
=
166.79 3.499 3.4993.499 8.7458 3.03933.499 3.0393 8.7458
0.00003384750.001276950.000127695
=
3.314.914.9
(MPa)
solution:1. use THREESC.F to obtain Cij2. use ALFA12.F to obtain 1 and 23. use the matrix operation:
example I
-
example II. stretch-twist coupling.
For a symmetric laminated composite material with lay up [0/45/90]4s - if a tensile loadis applied in the x-direction, what would happen if one of the layers is (inadvertantly)dropped?z
x
0o45o
-45o90o
90o-45o45o0o
90o-45o45o0o
0o45o
-45o90o
x
0o45o
-45o90o
90o-45o45o0o
90o-45o45o0o
0o
-45o90o
-
Solution: use STRESS.F equal thicknesses for each layer assumed input E1, E2, G12, 12, and layer thickness the difference of ABBD matrix between the two cases
A =108.0 34.08 034.08 108.0 00 0 36.97
B =0 0 00 0 00 0 0
D =30.92 6.966 1.2626.966 15.78 1.2621.262 1.262 7.583
[0/45/90/ 0/45/90]S [0/45/90/0/-45/90/90/45/0/90/45/0]
A =103.2 30.05 3.94230.05 103.2 3.9423.942 3.942 32.76
B =0.4081 0.3804 0.39420.3804 1.169 0.39420.3942 0.3942 0.3804
D =25.88 5.647 0.27925.647 12.79 0.2792
0.2792 0.2792 6.155
by applying 100MN in x-direction, we find for the curvatures: x=0, y=0, xy=0 x=0.1518e-4, y=-0.44293e-3, xy=0 .3149e-3vs.
example II. stretch-twist coupling.
-
-Fx Fx
symmetric case
-
-Fx Fx
asymmetric case
-
macro-scale mechanics analysis
micro-scale mechanics analysis
ABBD matrix
maximum stress failure criterion
outlines
-
maximum stress failure criterion:
the maximum stress in the fiber direction equals the maximumstress in a uniaxial specimen of the same material loaded inthe fiber direction when it fails;
the maximum stress perpendicular to the fiber direction equalsthe maximum stress in a uniaxial specimen of the samematerial loaded perpendicular to the fiber direction when itfails;
the maximum shear stress in the 1-2 plane equals themaximum shear stress in a specimen of the same materialloaded in shear in the 1-2 plane when it fails.
1C
-
example III.
A symmetric graphite-reinforced composite plate as [45/02]S is subjected toonly thermal loading (below). At what temperature will the composite platefail? In what mode? Base your answer on the maximum stress criterion.
10012F502T
-2002c15001T-12501c
failure stresses (MPa)graphite-reinforced composites
24.3x10-6/oC2, 3-0.01800x10-6/oC1
4.40GPaG13,G12
3.20GPaG23
0.24812, 130.45823
12.10GPaE2, E3
155.0GPaE1
typical engineeringproperties of graphite-
polymer composite
-
solution:
0-0.1959n-0.11879n0o-0.038577n-0.23602n0.55075n-45o0.038577n-0.23602n0.55075n45o
1221layer
Failure is due to tensile stress, perpendicular to fibers in the 45o layers bydecreasing the temperature by 211.84oC!
use STRESS.F
example III
-255.231021-12627105230o-25922592-211.84847.4-27232270-45o
2592-2592-211.84847.4-2723227045o
12F- 12F2T2c1T1clayer
failure mode
-
download and compile instructions
1. Login into any CAEN UNIX machines2. Make a directory, which is a directory for storage the file, by using %mkdir
hyer_directory. You can use other than hyer_directory for your favor.3. Under %, type in cd ~chiawei/Public4. You will need to copy a file Hyer.tar.gz into the directory you created at step 2 by using
the following command %cp Hyer.tar.gz ~/hyer_directory/5. Then go to your hyer_directory by %cd ~/hyer_directory6. Then you will need to uncompressed the the file with the following two commands
% gzip -d Hyer.tar.gz % tar -xvf Hyer.tar7. You will see a directory named Hyer created at your hyer_directory.8. Goto Hyer directory by using % cd Hyer9. There are 15 files and one directory SUN. Under SUN direcotry are all executable files
of the 14 FOTRAN codes. Sometimes, depending your account setup, you might need touse % chmod +x *.exe to make all executable files work.
10. If you want to compile the FORTRAN codes by yourself. Please use the followingcommand % f77 -o e1.exe E1.F for example to compile E1.F.