analysis of fiber-reinforced composite materials by using M. W. Hyers codes

ME518

01/26/2004

Chia-Wei Wang

macro-scale mechanics analysis

micro-scale mechanics analysis

ABBD matrix

maximum stress failure criterion

outlines

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

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

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:

macro-scale mechanics analysis

microscale mechanics analysis

ABBD matrix

maximum stress failure criterion

outlines

micromechanics approaches

finite element approach theory of elasticityapproachStrength-of materialapproach

Information:stress-strain Information:

prediction of effective material properties

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

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

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( )

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

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

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

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

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

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.