FE of Composites

8/13/2019 FE of Composites

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Universitat de Girona

DESIGN AND ANALYSIS OF COMPOSITES

WITH FINITE ELEMENTS

N. Blanco, D. Trias

February 2012


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Contents

1 Constitutive equations for transversely isotropic materials 1

1.1 Examples. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2 Laminate theory 17

2.1 Examples. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3 Hygro-thermal effects 33

3.1 Examples. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4 Failure criteria 43

4.1 Examples. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

5 Micromechanics 55

5.1 Example: Modeling a periodic RVE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

5.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

iii


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Chapter 1

Constitutive equations for

transversely isotropic materials

1.1 Examples

Example 1.1. Write a computer program to evaluate the compliance and stiffness matrices in

terms of the engineering properties for an orthotropic material. Consider a 3D analysis.

Solution to Example 1.1. Both the stiffness and compliance matrix for an orthotropic material

have 12 non-zero terms and need to be defined by means of 9 independent properties: E11,E22,

E33,12,13,23,G23,G13and G12. Next, we present the solution implemented using the symbolic

calculation capabilities of MATLABTM. Execute the program and observe that the stiffness matrix

is much more complicated than the flexibility matrix.

%

% DACFE. Solution to Example 1.1%

% J.A. Mayugo, N. Blanco

clear all,close all, clc

% Orthotropic material need 9 constants:

% E11, E22, E33, nu12, nu13, nu23, G12, G13, G23

syms E11 E22 E33 nu12 nu13 nu23 G12 G13 G23

% Compute orthotropic properties

nu21 = nu12*E22/E11;

nu31 = nu13*E33/E11;

nu32 = nu23*E33/E22;

% Compute S and CS = [1/E11 -nu21/E22 -nu31/E33 0 0 0;

-nu12/E11 1/E22 -nu32/E33 0 0 0;

-nu13/E11 -nu23/E22 1/E33 0 0 0;

0 0 0 1/G23 0 0;

0 0 0 0 1/G13 0;

0 0 0 0 0 1/G12];

C = inv(S);

pretty(S)

pretty(C)

This file can be found at:

ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex101.m

Example 1.2. Write a computer program to evaluate the compliance and stiffness matrices in

terms of the engineering properties for an orthotropic material. Consider plane stress analysis.

1
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex101.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex101.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex101.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex101.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex101.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex101.m


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2 Disseny i Anlisi de Compsits amb Elements Finits

Solution to Example 1.2. Both the stiffness and compliance matrix for an orthotropic material

under plane stress conditions have 5 non-zero terms and need to be defined by 4 independent

properties: E11,E22,12 andG12. Next, we present the solution implemented using the symbolic

calculation capabilities of MATLABTM. Execute the program and observe that the stiffness matrix

is much more complicated than the flexibility matrix.

%

% DACFE. Solution to Example 1.2

%



% Plane stress orthotropic material

% need 4 constants:

% E11, E22, nu12, G12

syms E11 E22 nu12 G12

% Compute orthotropic properties

nu21 = nu12*E22/E11;

% Compute S and Q

S = [1/E11 -nu21/E22 0;

-nu12/E11 1/E22 0;

0 0 1/G12];

Q = inv(S);

pretty(S)

pretty(Q)



Example 1.3. Write a computer program to evaluate the compliance and stiffness matrices of a

CFRP with the following properties and assume that the material is transversely isotropic.

E11= 190 GPa E22= 50 GPa G12= 74 GPa

12= 0.24 23= 0.42

Solution to Example 1.3. The stiffness and compliance matrices for a transversely isotropic

material have 12 non-zero terms and need to be defined by 5 independent properties: E11, E22,

12,23andG12. The resultingSandCmatrices are

[S] =

0.0053 0.0013 0.0013 0 0 0

0.0013 0.02 0.0084 0 0 00.0013 0.0084 0.02 0 0 0

0 0 0 0.0568 0 0

0 0 0 0 0.0135 0

0 0 0 0 0 0.0135

GPa1

[C] =

200.48 21.83 21.83 0 0 0

21.83 63.09 27.81 0 0 0

21.83 27.87 63.09 0 0 0

0 0 0 17.61 0 0

0 0 0 0 74 0

0 0 0 0 0 74

GPa

Next, we present the solution implemented in MATLABTM.


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Chapter 1. Constitutive equations for transversely isotropic materials 3

%


%



% The parameters are: E11,E22,nu12,nu23,G12prop = [190,50,0.24,0.42,74]

% Transversaly isotropic need 5 constants:

% E11, E22=E33, nu12=nu13, nu23, G12=G13, G23=E22/(2*(1+nu23))

% Extract and compute transversally isotropic properties

E11 = prop(1);

E22 = prop(2);

E33 = E22;

nu12 = prop(3);

nu21 = nu12*E22/E11;

nu13 = nu12;

nu31 = nu21;

nu23 = prop(4);

nu32 = nu23*E33/E22;

G12 = prop(5);G13 = G12;

G23 = E22/(2*(1+nu23));

% Compute S and C

S = [1/E11 -nu21/E22 -nu31/E33 0 0 0;

-nu12/E11 1/E22 -nu32/E33 0 0 0;

-nu13/E11 -nu23/E22 1/E33 0 0 0;

0 0 0 1/G23 0 0;

0 0 0 0 1/G13 0;

0 0 0 0 0 1/G12]

C = inv(S)


ftp://amade.udg.edu/amade/mme/DACFE/input_

files/T1/DACFE_

Ex103a.m

The previous code can be modified so the material properties are read from an ASCII file

and the computed matrices are written in another ASCII file. This MATLABTM code can be

implemented as follows:

%

% DACFE. Alternative solution to Example 1.3

%



% Material-file identification

n_file = DACFE_Ex103_CFRP % file name

% Open I/O files

fidinp = fopen([n_file,.dat],r);

fidout = fopen([n_file,_CS_123.dat],w);

% Read properties input file

prop = fscanf(fidinp,%g)

% Transversaly isotropic need 5 constants:

% E11, E22=E33, nu12=nu13, nu23, G12=G13, G23=E22/(2*(1+nu23))

% Extract and compute transversally isotropic properties

E11 = prop(1);

E22 = prop(2);

E33 = E22;

nu12 = prop(3);

nu21 = nu12*E22/E11;

nu13 = nu12;
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex103a.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex103a.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex103a.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex103a.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex103a.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex103a.m


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nu31 = nu21;

nu23 = prop(4);

nu32 = nu23*E33/E22;

G12 = prop(5);

G13 = G12;

G23 = E22/(2*(1+nu23));

% Compute S and CS = [1/E11 -nu21/E22 -nu31/E33 0 0 0;

-nu12/E11 1/E22 -nu32/E33 0 0 0;

-nu13/E11 -nu23/E22 1/E33 0 0 0;

0 0 0 1/G23 0 0;

0 0 0 0 1/G13 0;

0 0 0 0 0 1/G12]

C = inv(S)

% Write solution in file

%fprintf(fidout,%10s\n,SIFFNESS MATRIX);

for row=1:6

fprintf(fidout,%10.4e\t %10.4e\t %10.4e\t %10.4e\t %10.4e\t %10.4e\n,...

C(row,1),C(row,2),C(row,3),C(row,4),C(row,5),C(row,6));

end

%fprintf(fidout,%10s\n,COMPLIANCE MATRIX);

for row=1:6


S(row,1),S(row,2),S(row,3),S(row,4),S(row,5),S(row,6));

end

fclose(fidinp);

fclose(fidout);


ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex103b.m

Example 1.4. Construct the[a]rotation matrix for a rotation = 30 around thez-axis. Computethe[T] and[T]in 3D space.

Solution to Example 1.4. The[a=30]rotation matrix around z-axis is

[a=30] =

cos sin 0 sin cos 0

0 0 1

=

3

2

1

2 0

12

3

2 0

0 0 1

Considering the general expression for the coordinates transformation of the second order

stress tensorij =aipajqpq (1.1)

The following algorithm is used to obtain a 6 6 transformation matrix [T] in contracted

notation as

= T (1.2)

If 3 and 3 theni = j andp = q, so

T =aipaip= a2ip (1.3)

If 3 and > 3 then i = j butp = q, and taking into account that switching p by qyields

the same value of= 9p q, we have

T=aipaiq+ aiqaip = 2aipaiq (1.4)
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex103b.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex103b.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex103b.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex103b.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex103b.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex103b.m


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If >3,theni =j , but we want only one stress, say ij, notji because they are numerically

equal. In fact= ij =ji with = 9 ij. If in addition 3thenp = qand we get

T =aipajp (1.5)

When >3 and >3,i =j andp =qso we get

T =aipajq + aiqajp (1.6)

which completes the derivation ofT. Expanding (1.3-1.6) we get

[T] =

l21 m21 n

21 2m1n1 2l1n1 2l1m1

l22 m22 n

22 2m2n2 2l2n2 2l2m2

l23

m23

n23

2m3n3 2l3n3 2l3m3l2l3 m2m3 n2n3 m2n3+ n2m3 l2n3+ n2l3 l2m3+ m2l3l1l3 m1m3 n1n3 m1n3+ n1m3 l1n3+ n1l3 l1m3+ m1l3l1l2 m1m2 n1n2 m1n2+ n1m2 l1n2+ n1l2 l1m2+ m1l2

(1.7)

Next, there is the solution implemented using MATLABTM.

%


%



% Rotation angle around z-axis

theta = 30/180*pi

% Compute transformation matrix (Clokwise rotation of 12 respect to xy)

a = [cos(theta) sin(theta) 0;

-sin(theta) cos(theta) 0;0 0 1];

for i1=1:3

for j1=1:3

if i1==j1

alpha=i1;

else

alpha=9-i1-j1;

end

for p1=1:3

for q1=1:3

if p1==q1

beta=p1;

else

beta=9-p1-q1;end

if (alpha3)

T(alpha,beta)=a(i1,p1)*a(j1,q1)+a(i1,q1)*a(j1,p1);

end

end

end

end

end

TTg = inv(T)




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Example 1.5. Write a computer program to transform the stiffness and compliance matrices

from material coordinates, [C123] and [S123], to another coordinate system, [Cxyz ] and [Sxyz ], by

a rotation of 90, 30 and -30 around the z-axis. The data, [C123] and[S123], should be read froma file. The output, [Cxyz ] and [Sxyz ], should be written to another file. Use the same material

properties as in Example1.3.

Solution to Example 1.5. Both the stiffness and compliance matrices for a transversally isotropic

material 12 non-zero terms and need to be defined by 5 independent properties. The resulting

stiffness and compliance matrices are 66 as the transformation matrices must be. However, as

a consequence of the transformation, the resulting stiffness and compliance matrices have more

than 12 non-zero terms and there coupling between normal stress and shear strains and vice

versa.

[Cxyz ] =

180.4 7.56 23.34 0 0 21.51

8.56 111.7 26.36 0 0 37.98

23.34 26.36 63.09 0 0 2.6170 0 0 31.70 24.42 0

0 0 0 24.42 59.9 0

21.51 37.98 2.62 0 0 59.73

GPa

[Sxyz ] =

0.0063 0.0014 0.003 0 0 0.0033

0.0014 0.0136 0.0066 0 0 0.0095

0.003 0.0066 0.02 0 0 0.0062

0 0 0 0.046 0.0187 0

0 0 0 0.0187 0.0243 0

0.0033 0.0095 0.0062 0 0 0.0242

GPa1

Next, we present the solution implemented using the calculation capabilities of MATLABTM.

%


%



% Run first CManaEx103.m with the same material-file identification

% The theta value is the last value in the material-file

% Material-file identification

n_file = DACFE_Ex103_CFRP % file name

% Rotation angle in degrees around Z-axis

theta_deg = 30;

theta = theta_deg*pi/180;

% Open I/O files

fidinp = fopen([n_file,_CS_123.dat],r);

fidout = fopen([n_file,_CS_xyz.dat],w);

% Read properties input file

prop = (fscanf(fidinp,%g %g %g %g %g %g,[6 inf]))

% First 6x6 data are C matrix

% Second 6x6 data are S matrix

% next escalar is theta in radians

% Extract C and S in material coordinate system

C_123=prop(1:6,:)

S_123=prop(7:12,:)


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% Compute transformation matrix (Clokwise rotation of 12 respect to xy)

a = [cos(theta) sin(theta) 0;

-sin(theta) cos(theta) 0;

0 0 1];

for i1=1:3

for j1=1:3

if i1==j1alpha=i1;

else

alpha=9-i1-j1;

end

for p1=1:3

for q1=1:3

if p1==q1

beta=p1;

else

beta=9-p1-q1;

end

% te(alpha,beta)=0

if (alpha3)

T(alpha,beta)=a(i1,p1)*a(j1,q1)+a(i1,q1)*a(j1,p1);

end

end

end

end

end

% Compute C and S in new coordiante system

C_xyz=inv(T)*C_123*inv(T)

% First method (using T_gamma)

S_xyz=(T)*S_123*(T)

% Second method (invert C matrix)

S_xyz=inv(C_xyz)

% Write solution file

%fprintf(fidout,%10s\n,SIFFNESS MATRIX);

for row=1:6


C_xyz(row,1),C_xyz(row,2),C_xyz(row,3),C_xyz(row,4),C_xyz(row,5),C_xyz(row,6));

end

%fprintf(fidout,%10s\n,COMPLIANCE MATRIX);

for row=1:6


S_xyz(row,1),S_xyz(row,2),S_xyz(row,3),S_xyz(row,4),S_xyz(row,5),S_xyz(row,6));

end

fclose(fidinp);

fclose(fidout);



Example 1.6. Write a computer program to evaluate the stiffness matrices of a unidirectional

CFRP T300/5208 lamina when the reinforcement is oriented according to 0, 30 and 90. Con-sider plane stress. The elastic properties of the material are: E11 = 181 GPa, E22 = 10.3 GPa,

G12 = 7.17 GPa and12 = 0.28.

Solution to Example 1.6. Both the transformed stiffness and compliance matrix for a transver-

sally isotropic material under plane stress conditions have 5 non-zero terms and need to be


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defined by 5 independent properties. So resulting stiffness and compliance matrices are3 3as

the transformation matrices must be.

The obtained stiffness matrices for the considered orientations are

[Q]0

= Q =

181.81 2.90 0

2.90 10.35 00 0 7.17

GPa

[Q]90=

10.35 2.90 02.90 181.81 0

0 0 7.17

GPa

[Q]30=

109.38 32.46 54.1932.46 56.66 20.0554.19 20.05 36.74

GPa

Observe that a rotation of 90 is equivalent to interchange the matrix terms 11 and 22 and

there is no coupling between normal and shear stress and strain. However, when the rotation isof30, all the terms in the matrix are different to zero and there is coupling between normaland shear stress and strain. Next, we present the solution implemented using the calculation

capabilities of MATLABTM.

%


%


clear all;close all; clc;

% Transversaly isotropic under plane stress

% only 4 material constants are needed

% E11 (MPa), E22 (MPa), nu12, G12 (MPa), angle ()

prop=[181000, 10300, 0.28, 7170, -30];

E11 = prop(1);

E22 = prop(2);

nu12 = prop(3);

nu21 = nu12*E22/E11;

G12 = prop(4);

theta = prop(5)*pi/180;

% Calculation of the compliance and stiffness matrices

S=zeros(3,3);

S(1,1)=1/E11;

S(1,2)=-nu12/E11;

S(2,1)=S(1,2);

S(2,2)=1/E22;

S(3,3)=1/G12;

Q=inv(S);

% Calculation of the stress transformation matrix

m=cos(theta);

n=sin(theta);

T=zeros(3,3);

T(1,1)=m^2;

T(1,2)=n^2;

T(1,3)=2*m*n;

T(2,1)=n^2;

T(2,2)=m^2;

T(2,3)=-2*m*n;

T(3,1)=-m*n;

T(3,2)=m*n;

T(3,3)=m^2-n^2;


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% Calculation of the strain transformation matrix (T_gamma)

Tg=(inv(T));

% Calculation of the transformed stiffness and compliance matrices

Qb=inv(T)*Q*Tg

Sb=inv(Qb)



Example 1.7. Based on the program used in Ex.1.6, write a computer program to calculate the

resulting stresses in the lamina coordinates and the resulting strains in the global coordinates.

Consider the unidirectional lamina used in Ex.1.6when the reinforcement is oriented at45 andthe applied stress is xx = 5 MPa,yy = -6.5 MPa andxy = -2.5 MPa. Consider plane stress.

Solution to Example 1.7. The resulting stress and strain tensors when the reinforcement is

oriented at45 and the applied stress is xx = 5 MPa,yy = -6.5 MPa and xy = -2.5 MPa are

112212

=

3.25

1.75

5.75

MPa,

xxyyxy

=

478.1

323.8

195.6

106


%


%


clear all;close all; clc;

% Transversaly isotropic under plane stress

% only 4 material constants are needed

% E11 (MPa), E22 (MPa), nu12, G12 (MPa), angle ()

prop=[181000, 10300, 0.28, 7170, 45];

% Applied stress in the global coordinates system

% sigma_xx (MPa), sigma_yy (MPa), sigma_xy (MPa)

sigma_xyz=[5, -6.5, -2.5];

E11 = prop(1);

E22 = prop(2);

nu12 = prop(3);

nu21 = nu12*E22/E11;

G12 = prop(4);

theta = prop(5)*pi/180;

% Calculation of the compliance and stiffness matrices

S=zeros(3,3);

S(1,1)=1/E11;

S(1,2)=-nu12/E11;

S(2,1)=S(1,2);

S(2,2)=1/E22;

S(3,3)=1/G12;

Q=inv(S);

% Calculation of the stress transformation matrix

m=cos(theta);

n=sin(theta);

T=zeros(3,3);

T(1,1)=m^2;

T(1,2)=n^2;


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T(1,3)=2*m*n;

T(2,1)=n^2;

T(2,2)=m^2;

T(2,3)=-2*m*n;

T(3,1)=-m*n;

T(3,2)=m*n;

T(3,3)=m^2-n^2;

% Calculation of the strain transformation matrix (T_gamma)

Tg=(inv(T));

% Calculation of the transformed stiffness and compliance matrices

Qb=inv(T)*Q*Tg

Sb=inv(Qb)

% Calculation of stress and strain in the lamina coordinate system

sigma_123=T*sigma_xyz

epsilon_123=S*sigma_123

% Calculation of the strain in the global coordinate system

% epsilon_xyz=Sb*sigma_xyz

epsilon_xyz=inv(Tg)*epsilon_123



Example 1.8. Using computer program created in Ex.1.6, calculate the strains in the global

coordinates of the unidirectional lamina when the applied stress is xx = 1 MPa. What can be

concluded after the resulting strains?

Solution to Example 1.8. The resulting strains when the applied stress is xx = 1 MPa are

xx

yyxy

= 59.75

9.9945.78

106

From the resulting strains, it can be observed that for an off-axis lamina the normal and shear

stresses and strains are no longer uncoupled. In fact, a normal stress produces a shear strain

and vice versa.

Example 1.9. Generate the ANSYSTM input file to simulate and analyse a rectangular bar 100

mm long, 10 mm high and 10 mm wide. The bar is encastred in one end and subjected to a

longitudinal tension of 1000 MPa at the other end. The material is an anisotropic material with

the elastic properties summarised in the following stiffness matrix:

[C] =

118890 5210 5180 540 8370 465505210 88770 5350 8100 30 45910

5180 5350 20840 8740 9480 30

540 8100 8740 25780 24710 4540

8370 30 9480 24710 33660 3820

46550 45910 30 4540 3820 50750

MPa

Solution to Example 1.9. To conduct the analysis, first it is necessary to use the elastic anisotropic

linear behaviour and define the elastic properties of the material. To do so, it must be taken into

account that some finite element programs do follow a different ordering of the constitutive

equations and the different terms of the constitutive matrices must be reordered. For instance,

ANSYSTM and ABAQUSTM interchange the order of the shear terms with respect each other and

with respect the ordering considered in this course. Table 1.1summarises the conventions in the

notation used in this course (Standard), ANSYSTM and ABAQUSTM.


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Table 1.1: Convention in the contracted notation for different FE programs and Standard.

Contracted Standard ANSYSTM ABAQUSTM

1 11 11 11

2 22 22 223 33 33 33

4 23 12 12

5 13 23 13

6 12 13 23

According to the ANSYSTM convention, the previous stiffness matrix of the considered mate-

rial is transformed to:

[C] =

118890 5210 5180 46550 540 8370

5210 88770 5350 45910 8100 30

5180 5350 20840 30 8740 9480

46550 45910 30 50750 4540 3820

540 8100 8740 4540 25780 24710

8370 30 9480 3820 24710 33660

MPa

The ANSYSTM command sequence for this example is listed below. You can either type these

commands on the command window, or you can type them on a file, then, on the command

window enter /input, file, ext.

FINISH

/CLEAR

/TITLE, Anisotropic material

!3D anisotropic bar

/PREP7

!Parameters

P=1000 !applied pressure

l=100 !length

h=10 !height

w=10 !width

!Elements and options

ET,1,SOLID185 !element type: 8-node anisotropic brick

TB,ANEL,1,1,21,0 !table material properties

!anisotropic-elastic,material 1, 1 temperature,

!21 properties, stiffness matrix

!TBTEMP,0 !temperature material table

!material properties table: D11,D12,D13,D14,D15,D16

!D22,D23,D24,D25,D26,D33,

!D34,D35,D36,D44,D45,D46,

!D55,D56,D66

!Terms must be entered according to ANSYS convention

TBDATA,,118890,5210,5180,46550,540,-8370

TBDATA,,88770,5350,45910,-8100,30,20840

TBDATA,,-30,-8740,-9480,50750,-4540,-3820

TBDATA,,25780,24710,33660

!Geometry

BLOCK,0,l,0,h,0,w !3D block x1,x2,y1,y2,z1,z2

!Mesh

LESIZE,all,h/4 !element size

VMESH,ALL !mesh geometry


16/62


FINISH

/SOLU

!Boundary conditions

DA,5,ALL,0 !encastred area 5

!Apply loads

SFA,6,1,PRES,-P !apply pressure area 6

/PBC,ALL !to show BCs when solve

SOLVE

FINISH

/POST1

/VIEW,1,1,1,1 !iso-view

PLNSL,U,Z,2,1 !vertical displacement


ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex109.dat

Once the problem has been simulated, it can be observed that although the bar is loaded uni-

axially, it also deflects in the two transversal directions. Indeed, its the deflection is higher in oneof the transversal directions than in the longitudinal direction, even if the applied load is longitu-

dinal. Moreover, shear deformations appear on the bar. Obviously, none of these effects would be

present if the bar was made of an equivalent isotropic material. The resulting displacements and

strains obtained for the anisotropic bar are compared to those of a bar made of isotropic material

withE= 41000 MPa and= 0.3. The comparison is established in Table1.2.

Table 1.2: Comparison of the resulting displacements and strains of an anisotropic and isotropic

encastred bars subjected to uniaxial tension.

Anisotropic Isotropic

Component Displacement Strain Displacement Strain

(mm) (, ) (mm) (, )

xx 2.45 0.026 2.44 0.024

yy -3.53 0.02 0 0

zz 0.725 -0.014 0 0

xy - -0.046 - 0

yz - -0.014 - 0

xz - 0.01 - 0

Example 1.10. Generate the ANSYSTM input file to simulate and analyse the rectangular bar

considered in Ex. 1.9but made of an orthotropic material with the following mechanical proper-

ties:E11= 118560 MPa E22= 54100 MPa E33= 12050 MPa

G12= 15320 MPa G23= 4720 MPa G13= 6080 MPa

12= 0.2 23= 0.381 13= 0.324

Compare the resulting displacements and strains with those obtained when the material is

rotated 30 around the z-axis.

Solution to Example 1.10. The ANSYSTM command sequence for this example when the ma-

terial is rotated 30 around the z-axis is listed below. You can either type these commands on thecommand window, or you can type them on a file, then, on the command window enter /input,

file, ext.
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex109.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex109.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex109.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex109.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex109.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex109.dat


17/62


FINISH

/CLEAR

/TITLE, Orthotropic material

!3D orthotropic bar

/PREP7

!Parameters

P=1000 !applied pressure

l=100 !length

h=10 !height

w=10 !width

ang=30 !orientation angle


ET,1,SOLID185 !element type: 8-node brick

!Material properties

MP,EX,1,118560

MP,EY,1,54100

MP,EZ,1,12050

MP,GXY,1,15320

MP,GYZ,1,4720

MP,GXZ,1,6080

MP,PRXY,1,0.2

MP,PRYZ,1,0.381

MP,PRXZ,1,0.324

!Geometry

BLOCK,0,l,0,h,0,w !3D block x1,x2,y1,y2,z1,z2

!Define material orientation

LOCAL,11,0,0,0,0,ang,0,0 !local coord. syst., origin, 30deg Z

ESYS,11 !use coord. syst. for elements

!Mesh

LESIZE,all,h/4 !element size


CSYS,0 !original coord. syst.

FINISH

/SOLU


DA,5,ALL,0 !encastred area 5

!Apply loads



SOLVE

FINISH

/POST1


PLNSL,U,Z,2,1 !vertical displacement



Once the problem has been simulated, it can be observed that although the bar is loaded

uniaxially, it also deflects in the y-direction when the material is rotated. Moreover, there are

shear deformations on the bar in the xy-plane. None of these effects appear when the material is

oriented according to the directions of the bar. A comparison is established in Table1.3.

Example 1.11.Generate the ANSYSTM input file to simulate and analyse the transverally isotropic

material considered in Ex. 1.6. Assume a rectangular plate 100 mm long, 100 mm wide and 1

mm thick subjected to uniaxial loading. The reinforcement is in the xy-plane and oriented with

respect the x-axis. Assume that23= 0.42.


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Table 1.3: Comparison of the resulting displacements and strains of an orthotropic encastred bar

subjected to uniaxial tension with different material orientations.

Orthotropic 30 Orthotropic 0

Component Displacement Strain Displacement Strain(mm) (, ) (mm) (, )

xx 1.75 0.017 0.843 0.0085

yy -1.355 -0.009 0 -0.0017

zz 0 -0.004 0 -0.0027

xy - -0.012 - 0

yz - 0 - 0

xz - 0 - 0

Solution to Example 1.11. The ANSYSTM

command sequence for this example when the ma-terial is rotated 45 around the z-axis is listed below. You can either type these commands on thecommand window, or you can type them on a file, then, on the command window enter /input,

file, ext.

FINISH

/CLEAR

/TITLE, Transversally isotropic material

!3D transversally isotropic plate

/PREP7

!Parameters

P=1 !applied stress

l=100 !length

w=100 !width

h=1 !thickness


ET,1,SOLID185 !element type: 8-node brick

!Material properties

MP,EX,1,181000

MP,EY,1,10300

MP,EZ,1,10300

MP,GXY,1,7170

MP,GYZ,1,3627

MP,GXZ,1,7170

MP,PRXY,1,0.28

MP,PRYZ,1,0.42

MP,PRXZ,1,0.28

!Geometry

BLOCK,0,l,0,w,0,h !3D block x1,x2,y1,y2,z1,z2

!Define material orientation

LOCAL,11,0,0,0,0,45,0,0 !local coord. syst., origin, 45deg Z

ESYS,11 !use coord. syst. for elements

!Mesh

LESIZE,all,h*5 !element size


CSYS,0 !original coord. syst.

FINISH

/SOLU


NSEL,S,LOC,X,0

D,ALL,UX


19/62


NSEL,S,LOC,X,0

NSEL,R,LOC,Y,0

NSEL,R,LOC,Z,0

D,ALL,ALL

NSEL,S,LOC,X,0

NSEL,R,LOC,Y,wNSEL,R,LOC,Z,0

D,ALL,UZ

NSEL,ALL

!Apply loads



SOLVE

FINISH

/POST1


RSYS,11 !plot results local coord. systPLNSL,U,Y,2,1 !vertical displacement



The resulting strains in the x-y-z coordinate system when the applied stress is xx = 1 MPa

are summarised next. The resulting strains in the 1-2-3 coordinate system are also included for

comparison. From the resulting strains, it can be observed that for an off-axis lamina the normal

and shear stresses and strains are no longer uncoupled. In fact, a normal stress produces a shear

strain and vice versa.

xxyyxy

= 59.7

9.99

45.8

106, 112212

= 1.99

47.8

69.7

106

1.2 Problems

Problem 1.1. Using the same procedure as in Ex. 1.1and1.2, obtain the mathematical expres-

sions for the stiffness and compliance matrices of a transversally isotropic material in 3D and

plane stress analysis. Show all work in a report.

Problem 1.2. Calculate the [S] matrix and the [C] matrices in the material coordinate system

(1-2-3) of a 3D braided carbon-carbon composite material that can be considered as transversely

isotropic. The engineering elastic properties of the material are:

E11= 200 GPa E22= E33= 100

G12= G13= 50 GPa 12= 0.2 23= 0.3

Show all work in a report.

Problem 1.3. At a particular point of a component made with the material considered in Problem

1.2, the fibre directionx1 is oriented with = 30 and = 60 (see Figure1.1). Calculate the[S]

matrix in global coordinate system (x-y-z). Show all work in a report.

Problem 1.4. Compute the [C] matrix for next material for different rotations around 1-axis.

Verify numerically if the material is or not transversally isotropic. Show all work in a report.E11= 145.9 GPa E22= E33= 13.3 GPa G12= G13= 4.39 GPa

G23= 4.53 GPa 23= 0.470 12= 13= 0.263


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Figure 1.1: Fibre orientation with respect the global coordinate system

Problem 1.5. Consider a similar plate to that simulated in Ex. 1.11 but 50 mm long, 50 mm

wide and 5 mm thick. The plate is made of glass-reinforced Polyester, which can be consideredtransversally isotropic, and is subjected to a uniaxial tensile stressxx = 50 MPa. Use a FE script

to determine the orientation of the reinforcement, in the xy-plane, if the resulting strains in the

material coordinate system are:

112212

=

116

3484

5054

106

The elastic properties of the material are: E11 = 19981 MPa, E22 = 11389 MPa, G12 = 3789

MPa,12 = 0.274 and23= 0.3. Show all work in a report.


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Chapter 2

Laminate theory

2.1 Examples

Example 2.1. Write a computer program to evaluate the transformed compliance and stiffnessmatrices in terms of the engineering properties for a composite laminate made of two trans-

versely isotropic laminae. Consider plane stress analysis.

Solution to Example 2.1. The plane stress 33 stiffness and compliance matrices for a com-

posite laminate can have 9 non-zero terms and need to be defined by 4 independent properties:

E11, E22, 12 and G12 per lamina material. Next, we present the solution implemented using

the symbolic calculation capabilities of MATLABTM. Execute the program and observe that the

resulting stiffness matrix is much more complicated than the flexibility matrix.

%


%


clear all,close all, clc;

% Transversely isotropic material need 4 constants:

% E1, E2, nu12, G12

syms E1_1 E2_1 nu12_1 G12_1

syms E1_2 E2_2 nu12_2 G12_2

% The angles for both laminae are also required

syms theta_1 theta_2

% Compute S and Q

S_1 = [1/E1_1 -nu12_1/E1_1 0;

-nu12_1/E1_1 1/E2_1 0;

0 0 1/G12_1];

Q_1 = inv(S_1);

S_2 = [1/E1_2 -nu12_2/E1_2 0;

-nu12_2/E1_2 1/E2_2 0;

0 0 1/G12_2];

Q_2 = inv(S_2);

% Compute transformation matrices T and Tgamma

T_1 = [(cos(theta_1))^2 (sin(theta_1))^2 2*cos(theta_1)*sin(theta_1);

(sin(theta_1))^2 (cos(theta_1))^2 -2*cos(theta_1)*sin(theta_1);

2*

cos(theta_1)*

sin(theta_1) -2*

cos(theta_1)*

sin(theta_1) (cos(theta_1))^2-(sin(theta_1))^2];

T_2 = [(cos(theta_2))^2 (sin(theta_2))^2 2*cos(theta_2)*sin(theta_2);

(sin(theta_2))^2 (cos(theta_2))^2 -2*cos(theta_2)*sin(theta_2);

2*cos(theta_2)*sin(theta_2) -2*cos(theta_2)*sin(theta_2) (cos(theta_2))^2-(sin(theta_2))^2];

17


22/62


T_1gamma = simplify((inv(T_1)).);

T_2gamma = simplify((inv(T_2)).);

% Compute transformed stiffness and compliance matrices

Qb_1 = inv(T_1)*Q_1*T_1gamma

Qb_2 = inv(T_2)*Q_2*T_2gamma

Sb_1 = inv(Qb_1)

Sb_2 = inv(Qb_2)



Example 2.2. Write a computer program to obtain the 3D compliance and stiffness matrices of

a [0/20/-20/90]s laminate in terms of the apparent engineering properties. Also obtain the value

of the nine apparent properties for the considered laminate. All the laminae are 1 mm thick and

are made of unidirectional T300/5208 carbon-epoxy: E11= 181 GPa,E22= 10.3 GPa,12 = 0.28,

23= 0.42 andG12 = 7.17 GPa.Solution to Example 2.2. Both the stiffness and compliance matrix for an orthotropic material

have 12 non-zero terms and need to be defined by 9 independent properties: E11, E22, E33, 12,

13, 23, G23, G13 and G12. The global stiffness matrix of the laminate is obtained by adding the

global stiffness matrices of the layers multiplied by the thickness ratio. The global compliance

matrix is obtained inverting the global stiffness matrix.

[C] =

122.9 13.19 5.16 0 0 0

13.19 57.31 5.29 0 0 0

5.16 5.29 12.64 0 0 0

0 0 0 4.72 0 0

0 0 0 0 6.08 0

0 0 0 0 0 15.31

GPa

[S] =

8.4 1.7 2.7 0 0 0

1.7 18.5 7 0 0 0

2.7 7 83.1 0 0 0

0 0 0 211.9 0 0

0 0 0 0 164.6 0

0 0 0 0 0 65.3

106 GPa1

Taking into account the following expressions that relate the stiffness matrix terms with the

apparent propertiesExx= 1/S11 Gyz = 1/S44 xy = S12/S11Eyy = 1/S22 Gzx = 1/S55 yz = S23/S22Ezz = 1/S33 Gxy = 1/S66 zx = S31/S33

the apparent properties of the considered laminate are

Exx= 118560 MPa Gyz = 4720 MPa xy = 0.2

Eyy = 54090 MPa Gzx = 6077MPa yz = 0.381

Ezz = 12028 MPa Gxy = 15314 MPa zx = 0.033


%


%


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Chapter 2. Laminate theory 19




% E1, E2, nu12, nu23, G12 (MPa)

Mat(1,:)=[181000, 10300, 0.28, 0.42, 7170];

[n_mat,n_prop]=size(Mat);

% The lamina need to define material, angle and thickness

% mat, theta, t

L(1,:)=[1, 0, 1.5];

L(2,:)=[1, 20, 1.5];

L(3,:)=[1, -20, 1.5];

L(4,:)=[1, 90, 1.5];

L(5,:)=[1, 90, 1.5];

L(6,:)=[1, -20, 1.5];

L(7,:)=[1, 20, 1.5];

L(8,:)=[1, 0, 1.5];

n_lam=size(L,1);

% Compute S and Q

for i=1:n_lam;

i_mat=L(i,1);

S(1,1,i)=1/Mat(i_mat,1);

S(1,2,i)=-Mat(i_mat,3)/Mat(i_mat,1);


S(2,1,i)=S(1,2,i);

S(2,2,i)=1/Mat(i_mat,2);


S(3,1,i)=S(1,3,i);

S(3,2,i)=S(2,3,i);

S(3,3,i)=S(2,2,i);

S(4,4,i)=2*(1+Mat(i_mat,4))/Mat(i_mat,2);

S(5,5,i)=1/Mat(i_mat,5);

S(6,6,i)=S(5,5,i);

C(:,:,i)=inv(S(:,:,i));

end;


for i=1:n_lam;

theta=L(i,2)*pi/180;

m=cos(theta);

n=sin(theta);

T(1,1,i)=m^2;

T(1,2,i)=n^2;

T(1,6,i)=2*m*n;

T(2,1,i)=n^2;

T(2,2,i)=m^2;

T(2,6,i)=-2*m*n;

T(3,3,i)=1;

T(4,4,i)=m;

T(4,5,i)=-n;

T(5,4,i)=n;

T(5,5,i)=m;

T(6,1,i)=-m*n;

T(6,2,i)=m*n;

T(6,6,i)=m^2-n^2;

Tg(:,:,i)=(inv(T(:,:,i)));

% Compute stiffness and compliance transformed matrices

Sb(:,:,i)=inv(Tg(:,:,i))*S(:,:,i)*T(:,:,i);

Cb(:,:,i)=inv(Sb(:,:,i));

end;

% Compute apparent stiffness and compliance matrices


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t_lam=0;

for i=1:n_lam;

t_lam=t_lam+L(i,3);

end;

C_lam=zeros(6,6);

for i=1:n_lam;

C_lam=C_lam+L(i,3)/t_lam*Cb(:,:,i);end;

S_lam=inv(C_lam)

C_lam

% Obtain apparent engineering properties

Exx=1/S_lam(1,1)

Eyy=1/S_lam(2,2)

Ezz=1/S_lam(3,3)

Gyz=1/S_lam(4,4)

Gzx=1/S_lam(5,5)

Gxy=1/S_lam(6,6)

nuxy=-S_lam(1,2)/S_lam(1,1)

nuyz=-S_lam(2,3)/S_lam(2,2)

nuzx=-S_lam(3,1)/S_lam(3,3)



Example 2.3. Based on the procedure employed in Ex. 2.2,write a computer program to evalu-

ate the transformed compliance and stiffness matrices of the two transversely isotropic laminae

of a laminate. Both laminae are made of unidirectional T300/5208 CFRP (the elastic properties

of the material are listed in Ex.2.2). The top layer is 1 mm thick and oriented at 45. The bottomlayer is 2 mm thick and oriented at 0. Make the program general enough so it can be used withdifferent laminates. Considerer plane stress analysis.

Solution to Example 2.3. The plane stress 33 stiffness and compliance matrices for a com-posite laminate can have 9 non-zero terms and need to be defined by 4 independent properties:

E11,E22,12 andG12 per lamina material.

[Q]1 =

181810 2900 02900 10350 0

0 0 7170

MPa, [S]1 =

5.5 1.5 01.5 97.1 0

0 0 139.5

106 GPa1

[Q]2 =

56658 42318 42866

42318 56658 42866

42866 42866 46591

MPa, [S]2 =

59.7 10 45.8

10 59.7 45.8

45.8 45.8 105.7

106 GPa1

Next, we present the solution implemented using the symbolic calculation capabilities of

MATLABTM.

%


%




% E1, E2, nu12, G12 (MPa)

Mat(1,:)=[181000, 10300, 0.28, 7170];



% mat, theta, t


25/62


L(1,:)=[1, 0, 2];

L(2,:)=[1, 45, 1];

n_lam=size(L,1);

% Compute S and Q

for i=1:n_lam;

i_mat=L(i,1);S(1,1,i)=1/Mat(i_mat,1);


S(2,1,i)=S(1,2,i);

S(2,2,i)=1/Mat(i_mat,2);

S(3,3,i)=1/Mat(i_mat,4);

Q(:,:,i)=inv(S(:,:,i));

end;


for i=1:n_lam;


m=cos(theta);

n=sin(theta);

T(1,1,i)=m^2;

T(1,2,i)=n^2;

T(1,3,i)=2*m*n;

T(2,1,i)=n^2;

T(2,2,i)=m^2;

T(2,3,i)=-2*m*n;

T(3,1,i)=-m*n;

T(3,2,i)=m*n;

T(3,3,i)=m^2-n^2;

Tg(:,:,i)=(inv(T(:,:,i)));


Sb(:,:,i)=inv(Tg(:,:,i))*S(:,:,i)*T(:,:,i);

Qb(:,:,i)=inv(Sb(:,:,i));

end;

Qb,Sb



Example 2.4. Extend the computer program generated in Ex. 2.3 to calculate the constitutive

matrix of the laminate considered in the same example. Make the program general enough so it

can be used with different laminates.

Solution to Example 2.4. The constitutive matrix of a laminate is determined as a function of

the in-plane stiffness matrix[A], the coupling matrix[B]and the bending stiffness matrix[D]. For

the calculation of the[A],[B]and [D]matrices, the transformed stiffness matrix of each lamina of

the laminate have to be first evaluated. The[A],[B]and [D]matrices for the considered laminate

are

[A] =

420.3 48.11 42.8748.11 77.35 42.87

42.87 42.87 60.93

103 N/mm

[B] =

125.2 39.42 42.87

39.42 46.31 42.87

42.87 42.87 39.42

103 N

[D] =

273.5 49.22 46.4449.22 73.45 46.44

46.44 46.44 58.84

103 Nmm


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Next, we present the solution implemented using the symbolic calculation capabilities of

MATLABTM.

%

% DACFE. Solution of Example 2.4

%





% E1, E2, nu12, G12 (MPa)

Mat(1,:)=[181000, 10300, 0.28, 7170];



% mat, theta, t

L(1,:)=[1, 0, 2];

L(2,:)=[1, 45, 1];

n_lam=size(L,1);

% Compute S and Q

for i=1:n_lam;

i_mat=L(i,1);

S(1,1,i)=1/Mat(i_mat,1);


S(2,1,i)=S(1,2,i);

S(2,2,i)=1/Mat(i_mat,2);

S(3,3,i)=1/Mat(i_mat,4);

Q(:,:,i)=inv(S(:,:,i));

end;


for i=1:n_lam;


m=cos(theta);

n=sin(theta);

T(1,1,i)=m^2;

T(1,2,i)=n^2;

T(1,3,i)=2*m*n;

T(2,1,i)=n^2;

T(2,2,i)=m^2;

T(2,3,i)=-2*m*n;

T(3,1,i)=-m*n;

T(3,2,i)=m*n;

T(3,3,i)=m^2-n^2;

Tg(:,:,i)=(inv(T(:,:,i)));


Sb(:,:,i)=inv(Tg(:,:,i))*S(:,:,i)*T(:,:,i);

Qb(:,:,i)=inv(Sb(:,:,i));

end;

% Laminate constitutive matrix (ABD)

% Location of the bounds of the laminae and laminates midplane

TH=0;

Z=zeros(n_lam+1,1);

for i=1:n_lam;

TH=TH+L(i,3);

end;

th=TH/2;

TH=0;

Z(1)=-th;

for i=1:n_lam;


27/62


TH=TH+L(i,3);

Z(i+1)=TH-th;

end;

% Compute A matrix

A=zeros(3,3);

for i=1:n_lam;

for j=1:3;for k=1:3;

A(j,k)=A(j,k)+Qb(j,k,i)*(Z(i+1)-Z(i));

end;

end;

end;

% Compute B matrix

B=zeros(3,3);

for i=1:n_lam;

for j=1:3;

for k=1:3;

B(j,k)=B(j,k)+Qb(j,k,i)*(Z(i+1)^2-Z(i)^2)*(1/2);

end;

end;

end;

% Compute D matrix

D=zeros(3,3);

for i=1:n_lam;

for j=1:3;

for k=1:3;

D(j,k)=D(j,k)+Qb(j,k,i)*(Z(i+1)^3-Z(i)^3)*(1/3);

end;

end;

end;

A,B,D



Example 2.5. Use the computer program generated in Ex. 2.4to compare the[A], [B] and[D]

matrices of the following cross-ply laminates: [90/02/90]s, [0/902/0]s, [0/90/0/90]s, [902/02]s and

[02/902]s. Consider the same material as in Ex. 2.2and a ply thickness of 0.125 mm.

Solution to Example 2.5. Taking into account that the five laminates considered are cross-ply

laminates, only the transformed stiffness matrices for laminae at 0 and 90 are required:

[Q]0= 181.81 2.90 02.90 10.35 0

0 0 7.17

GPa

[Q]90=

10.35 2.90 02.90 181.81 0

0 0 7.17

GPa

The five cross-ply laminates have the same number of 0 and 90 plies with the same proper-ties. Therefore, the in-plane behaviour will be the same for the five laminates and the in-plane

stiffness matrix[A] will be exactly the same for the five laminates

[A] =

96079 2897 02897 96079 0

0 0 7170

N/mm


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The five stacking sequences considered are symmetric. Therefore, the coupling matrix [B]

will be zero for the five laminates.

[B] = [0]N

As the five laminates are cross-ply laminates, the terms D16 = D61 and D26 = D62 of the

[D]matrix are zero. Thus, there is no coupling between bending moments and plate twisting or

torque and plate curvatures. The resulting[D] matrices for the five laminates are

[90/02/90]s= [D] =

6667 241.4 0241.4 9346 0

0 0 597.5

Nmm

[0/902/0]s= [D] =

9346 241.4 0241.4 6667 0

0 0 597.5

Nmm

[0/90/0/90]s= [D] = 10686 241.4 0241.4 5327 0

0 0 597.5

Nmm

[902/02]s= [D] =

2648 241.4 0241.4 13365 0

0 0 597.5

Nmm

[02/902]s= [D] =

13365 241.4 0241.4 2648 0

0 0 597.5

Nmm

Example 2.6. Consider a simply supported composite plate 1000 mm long, 1000 mm wide and 5

mm thick. The plate is made of unidirectional T300/5208 plies (0.125 mm thick) with a cross-ply

stacking sequence and is loaded in compression with an edge loadNxx = -1 N/mm (Nyy =Nxy =

Mxx =Myy = Mxy =Vyz =Vxz = 0). Generate the[A], [B],[D] and[H] matrices to simulate and

calculate the centre deflection of the plate with ANSYSTM for [020/9020], [04/904]5 and [02/902]10

stacking sequences.

Solution to Example 2.6. For the three stacking sequences considered, the total thickness of

material with the reinforcement oriented at 0 and 90 is always the same. Thus, the [A] matrixdoes not vary.

[A] =

480.4 14.48 014.48 480.4 0

0 0 35.85

103 N/mm

The three cross-ply laminates considered are not symmetric. Therefore, the [B] matrices

are not zero and different for every stacking sequence. The coupling between moments and

in-plane deformation, and vice versa, is less important when differently oriented plies are more

distributed.

[020/9020] = [B] =

535.8 0 0

0 535.8 0

0 0 0

103 N

[04/904]5= [B] =

107.2 0 00 107.2 0

0 0 0

103 N


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[02/902]10= [B] =

53.58 0 00 53.58 0

0 0 0

103 N

For the three stacking sequences, the thickness of material with the reinforcement oriented

at 0 and 90 is the same. Thus, the [D]matrix does not vary.

[D] =

1001 30.2 030.2 1001 0

0 0 74.7

103 Nmm

Similarly, the[H]matrix does not vary and is the same for the three laminates considered.

[H] =

22.5 0

0 22.5

103 N/mm

As the geometry, the load and the material configuration are symmetric with respect the x

and y axis, only a quarter of the plate is simulated. The ANSYSTM command sequence for thecross-ply plate is listed below. You can either type these commands on the command window, or

you can type them on a file, then, on the command window enter /input, file, ext.

FINISH

/CLEAR

/TITLE, Composite plate

!3D composite laminate ABDH

/PREP7

!Parameters

P=1 !applied load in N/mm

l=1000 !length

w=1000 !width

h=5 !thickness


ET,1,SHELL99,,2 !element type: 8-node laminated shell

!keyopt(2)=2 to enter ABDH matrices 6x6

!Material properties through ABDH matrices [0_20/90_20]

R,1,480400,14480,0,0,0,0 !mat1,A11,A12,0,A16,0,0

RMODIF,1,7,480400,0,0,0,0 !mat1,loc7,A22,0,A26,0,0

RMODIF,1,16,35850,0,0 !mat1,loc16,A66,0,0

RMODIF,1,19,22500,0 !mat1,loc19,H44,0

RMODIF,1,21,22500 !mat1,loc21,H55

RMODIF,1,22,-535.8e+3,0,0,0,0,0 !mat1,loc22,B11,B12,0,B16,0,0

RMODIF,1,28,535.8e+3,0,0,0,0 !mat1,loc28,B22,0,B26,0,0

RMODIF,1,37,0,0,0 !mat1,loc37,B66,0,0

RMODIF,1,43,1001e+3,30.2e+3,0,0,0,0 !mat1,loc43,D11,D12,0,D16,0,0

RMODIF,1,49,1001e+3,0,0,0,0 !mat1,loc49,D22,0,D26,0,0

RMODIF,1,58,74.7e+3,0,0 !mat1,loc58,D66,0,0

RMODIF,1,77,h !mat1,loc77,average thickness

!Uncomment for laminate [0_4/90_4]_5

!RMODIF,1,22,-107.2e+3,0,0,0,0,0 !mat1,loc22,B11,B12,0,B16,0,0

!RMODIF,1,28,107.2e+3,0,0,0,0 !mat1,loc28,B22,0,B26,0,0

!Uncomment for laminate [0_2/90_2]_10

!RMODIF,1,22,-53.58e+3,0,0,0,0,0 !mat1,loc22,B11,B12,0,B16,0,0

!RMODIF,1,28,53.58e+3,0,0,0,0 !mat1,loc28,B22,0,B26,0,0

!Geometry

RECTNG,0,l/2,0,w/2 !rectangle x1,x2,y1,y2

!Mesh

LESIZE,ALL,h*5 !element size


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AMESH,ALL !mesh geometry

FINISH

/SOLU


DL,2,1,UZ,0 !simply supported

DL,3,1,UZ,0

DL,1,1,SYMM !symmetryDL,4,1,SYMM

!Apply load

SFL,2,PRES,P !apply pressure on line 2 in N/mm


SOLVE

FINISH

/POST1


PLNSOL,U,Z,2,1 !vertical displacement


ftp://amade.udg.edu/amade/mme/DACFE/input_

files/T2/DACFE_

Ex206.dat

After the simulation, the resulting centre deflection for the [020/9020] plate is 154.7m, 15.2

m for the [04/904]5 case and 7.5m for the [02/902]10 plate. As the bending-extension coupling

diminishes when the plies are more distributed, as the terms of matrix [B] do, the resulting

deflection is smaller for the later case, even if the thickness of the plate is the same. Actually,

changing from [020/9020] to [04/904]5 reduces the deflection to one tenth.

Observe that if the plate is simulated in this way (defining the [A],[B],[D]and[H]matrices),

the stress-state in the material cannot be obtained.

Example 2.7. Consider the same simply supported composite plate as in Ex. 2.6with the same

load. Generate the ANSYSTMprogram to simulate and calculate the centre deflection of the plate

for the same three stacking sequences but defining the ply sequence into the program.

Solution to Example 2.7. In order to simplify the definition of the laminates in ANSYSTM, all

groups of plies with the same orientation are modelled as one ply with the total thickness of

the group. Thus, the [020/9020] stacking sequence with 0.125 mm plies is modelled as a [0/90]

stacking sequence with 0.125 20 mm plies and similarly for the other two stacking sequences.

As in the previous example, the geometry, the load and the material configuration are sym-

metric with respect the x andy axis and only a quarter of the plate is simulated. The ANSYSTM

command sequence for the cross-ply plate is listed below. You can either type these commands

on the command window, or you can type them on a file, then, on the command window enter/input, file, ext.

FINISH

/CLEAR


!3D composite plate ABDH-LSS

/PREP7

!Parameters

P=1 !applied load in N/mm

l=1000 !length

w=1000 !width

h=5 !thickness

lt=0.125 !layer thickness


ET,1,SHELL181 !element type: 8-node laminated shell


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!Material properties for the T300/5208 UD lamina

MP,EX,1,181000

MP,EY,1,10300

MP,EZ,1,10300

MP,GXY,1,7170

MP,GYZ,1,3627

MP,GXZ,1,7170

MP,PRXY,1,0.28MP,PRYZ,1,0.42

MP,PRXZ,1,0.28

!Section properties for the laminate (up to 250 layers)

!Laminate #1: [0_20/90_20]

SECTYPE,1,SHELL !section type: shell

SECDATA,lt*20,1,0 !layer thickness, material and orientation

SECDATA,lt*20,1,90

SECOFFSET,MID !nodes on mid-thickness of elements

!Laminate #2: [0_4/90_4]_5



SECDATA,lt*4,1,90

SECDATA,lt*4,1,0SECDATA,lt*4,1,90

SECDATA,lt*4,1,0

SECDATA,lt*4,1,90

SECDATA,lt*4,1,0

SECDATA,lt*4,1,90

SECDATA,lt*4,1,0

SECDATA,lt*4,1,90


!Laminate #3: [0_2/90_2]_10



SECDATA,lt*2,1,90

SECDATA,lt*2,1,0

SECDATA,lt*2,1,90

SECDATA,lt*2,1,0

SECDATA,lt*2,1,90

SECDATA,lt*2,1,0

SECDATA,lt*2,1,90

SECDATA,lt*2,1,0

SECDATA,lt*2,1,90

SECDATA,lt*2,1,0

SECDATA,lt*2,1,90

SECDATA,lt*2,1,0

SECDATA,lt*2,1,90

SECDATA,lt*2,1,0

SECDATA,lt*2,1,90

SECDATA,lt*2,1,0

SECDATA,lt*2,1,90

SECDATA,lt*2,1,0

SECDATA,lt*2,1,90


!Geometry

RECTNG,0,l/2,0,w/2 !rectangle x1,x2,y1,y2

!Mesh

SECNUM,1 !section type #1

!SECNUM,2 !uncomment for laminate #2

!SECNUM,3 !uncomment for laminate #3

LESIZE,ALL,l/50 !element size


FINISH

/SOLU


DL,2,1,UZ,0 !simply supported

DL,3,1,UZ,0


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DL,1,1,SYMM !symmetry

DL,4,1,SYMM

!Apply load

SFL,2,PRES,P !apply pressure on line 2 in N/mm


SOLVEFINISH

/POST1


PLNSOL,U,Z,2,1 !vertical displacement



After the simulation, the resulting centre deflection for the [020/9020] plate is 154.4m, 15.2

m for the [04/904]5 case and 7.5 m for the [02/902]10 plate. As it happened in the previous

example, the deflection of the plate decreases if the plies are more distributed. In fact, the

calculated deflections are the same in both cases. However, when the stacking sequence of thelaminate is specified, the stress-state in the material can be obtained.

Example 2.8. Consider an encastred composite plate subjected toxx = 10 MPa. The plate is

1000 mm long, 200 mm wide and is made of 0.125 mm T300/5208 unidirectional laminae with

the following stacking sequence [0/90/60/75]2. Generate the ANSYSTM program to simulate and

analyse the plate. Obtain the maximum displacements and the in-plane stress components of the

third and fourth plies (60 and 75) at top and bottom locations according to the laminate (xyz)and ply (1-2-3) coordinate systems.

Solution to Example 2.8. The ANSYSTM command sequence to simulate the [0/90/60/75]2

laminated plate is listed below. You can either type these commands on the command window, oryou can type them on a file, then, on the command window enter /input, file, ext.

FINISH

/CLEAR


!3D composite laminate LSS

/PREP7

!Parameters

P=10 !applied stress

l=1000 !length

w=200 !width


nl=8 !number of layers

h=lt*nl !plate thickness



KEYOPT,1,8,2 !store bottom, mid and top data for all layers


MP,EX,1,181000

MP,EY,1,10300

MP,EZ,1,10300

MP,GXY,1,7170

MP,GYZ,1,3627

MP,GXZ,1,7170

MP,PRXY,1,0.28

MP,PRYZ,1,0.42

MP,PRXZ,1,0.28


!Laminate #1: [0/90/60/75]2


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SECDATA,lt,1,0 !layer thickness, material and orientation

SECDATA,lt,1,90

SECDATA,lt,1,60

SECDATA,lt,1,75

SECDATA,lt,1,0

SECDATA,lt,1,90

SECDATA,lt,1,60SECDATA,lt,1,75


!Geometry

RECTNG,0,l,0,w !rectangle x1,x2,y1,y2

!Mesh


LESIZE,ALL,w/10 !element size


FINISH

/SOLU


DL,4,1,ALL !encastred line 4

!Apply load

SFL,2,PRES,-P*h !apply pressure on line 2 in N/mm


SOLVE

FINISH

/POST1


RSYS,SOLU !lamina coordinate system

LAYER,1 !specify layer 1 results

MID !specify midthickness results

PLNSOL,S,X,2,1 !vertical displacement

!Obtain results selecting layer and coordinate system

ESEL,,,,250 !select element #250

LAYER,3

RSYS,SOLU

TOP

ETABLE,TSXXloc,S,X !write X-stress in table TSXX

ETABLE,TSYYloc,S,Y

ETABLE,TSXYloc,S,XY

BOT

ETABLE,BSXXloc,S,X

ETABLE,BSYYloc,S,Y

ETABLE,BSXYloc,S,XY

PRETAB,BSXXloc,BSYYloc,BSXYloc,TSXXloc,TSYYloc,TSXYloc !print results table

RSYS,0

TOP

ETABLE,TSXXglb,S,X !write X-stress in table TSXX

ETABLE,TSYYglb,S,Y

ETABLE,TSXYglb,S,XY

BOT

ETABLE,BSXXglb,S,X

ETABLE,BSYYglb,S,Y

ETABLE,BSXYglb,S,XY

PRETAB,BSXXglb,BSYYglb,BSXYglb,TSXXglb,TSYYglb,TSXYglb !print results table



To specify a particular layer in which the results should be plotted, type LAYER,layer_number

in the command window, where layer_number should be replaced by desired the number of

layer. Iflayer_numberis replaced by 0 (default), the bottom surface of the bottom layer and the

top surface of the top layer are shown. To specify a particular position in the layer where the


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Problem 2.3. Based on the procedure of Ex. 2.4, write a computer program to calculate the

laminae strains of the laminate in the same example according to thexy z coordinate system

when the following forces and moments are applied:

Nxx= 1 N/mm Nyy = 10 N/mm Nxy = 20 N/mm

Mxx= 60 N Myy = 20 N Mxy = 5 N


Problem 2.4. Combine the computer program generated in Pb. 2.2 and 2.3 to calculate the

laminae strains according to the x y z coordinate system of the same laminate when the

following forces and moments are applied:

Nxx= 1 N/mm Nyy = 0N/mm Nxy = 0N/mm

Mxx = 0N Myy = 0 N Mxy = 0 N

Vxz = 1 N/mm Vyz = 1 N/mm


Problem 2.5. Extend the computer program generated in Pb. 2.3 to calculate the laminae

strains according to the lamina coordinate system in the laminate of Ex. 2.4when the loads and

moments of Pb. 2.3are applied. Calculate the strains in the bottom, middle and top locations of

each layer. Show all work in a report.

Problem 2.6. Extend the computer program generated in the previous problem to obtain the

stresses in the lamina coordinate system of the previously calculated strains. Show all work in a

report.

Problem 2.7.Modify the computer program generated in Ex.2.8to simulate the same plate with

a stacking sequence [0/90/60/75]s. Calculate the maximum displacements and the in-plane stress

components of the third and fourth plies (60 and 75) at top and bottom locations according tothe laminate (x y z) and ply (1-2-3) coordinate systems. Show all work in a report.

Problem 2.8. Modify the computer program generated in Ex. 2.7 to simulate the same plate

with stacking sequences [0/0/90/90/45/45/-45/-45] and [0/90/45/-45/0/90/45/-45]. Comment on

the results and justify the deformed shape analysing the matrices you can obtain with MATLAB

using the code created for Pb. 2.2.


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Chapter 3

Hygro-thermal effects

3.1 Examples

Example 3.1. Write a computer program to evaluate the transformed thermal and hygroscopicexpansion coefficients of an orthotropic lamina rotated an angle around thez-axis.

Solution to Example 3.1. The thermal and hygroscopic coefficients can be written in the form

of 1 6 vectors. When expressed in the coordinate system of the lamina, the last three therms

of each vector are zero and only the thermal and hygroscopic coefficients of the material in the

1-2-3 directions are required. After rotating the coordinate system using matrices [T] and [T],

the last three terms of each vector might not be zero. Next, we present the solution implemented

using the symbolic calculation capabilities of MATLABTM.

%


%% J.A. Mayugo, N. Blanco


% An orthotropic material needs 3 thermal and hygroscopic expansion coefficients

syms alfa_1 alfa_2 alfa_3 beta_1 beta_2 beta_3

% The angle for the lamina is

syms theta

alfa=[alfa_1; alfa_2; alfa_3; 0; 0; 0];

beta=[beta_1; beta_2; beta_3; 0; 0; 0];


m=cos(theta);n=sin(theta);

T=[m^2, n^2, 0, 0, 0, 2*m*n;

n^2, m^2, 0, 0, 0, -2*m*n;

0, 0, 1, 0, 0, 0;

0, 0, 0, m, -n, 0;

0, 0, 0, -n, m, 0;

-m*n, m*n, 0, 0, 0, m^2-n 2];

Tg = simplify((inv(T)).);

% Compute transformed expansion coefficients

alfa_xyz=inv(Tg)*alfa

beta_xyz=inv(Tg)*beta



33


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Example 3.2. Based on the procedure employed in Ex. 3.1, write a computer program to

evaluate the transformed thermal and hygroscopic expansion coefficients of an unidirectional

T300/5208 CFRP lamina when the reinforcement is oriented at = 30 around thez-axis. Theexpansion coefficients of the material in the lamina directions are given by: 11 = 0.02 10

6

K1, 22= 33= 22.5 106 K1, 11= 0.0and22= 33= 0.5.

Solution to Example 3.2. As in the previous example, the thermal and hygroscopic coefficients

can be written in the form of 1 6 vectors. After rotating the coordinate system using matrices

[T] and [T], the last three terms of each vector might not be zero. In fact, the resulting thermal

and hygroscopic coefficients for a T300/5208 lamina rotated at 30 are:

30=

5.64

16.89

22.5

0

0

19.47

106 K1, 30=

125

375

500

0

0

433

103

Next is the solution implemented using the calculation capabilities of MATLABTM.

%


%



% A transversally isotropic material needs 2

% thermal and hygroscopic expansion coefficients

% alfa_1, alfa_2, beta_1, beta_2

prop=[0.02e-6, 22.5e-6, 0.0, 0.5];

% The angle of the reinforcement is also defined

theta=30;

theta=theta*pi/180;

alfa_1=prop(1,1);

alfa_2=prop(1,2);

alfa_3=alfa_2;

beta_1=prop(1,3);

beta_2=prop(1,4);

beta_3=beta_2;

alfa=[alfa_1; alfa_2; alfa_3; 0; 0; 0];

beta=[beta_1; beta_2; beta_3; 0; 0; 0];


m=cos(theta);

n=sin(theta);

T=[m^2, n^2, 0, 0, 0, 2*m*n;

n^2, m^2, 0, 0, 0, -2*m*n;

0, 0, 1, 0, 0, 0;

0, 0, 0, m, -n, 0;

0, 0, 0, -n, m, 0;

-m*n, m*n, 0, 0, 0, m^2-n 2];

Tg=(inv(T));


alfa_xyz=inv(Tg)*alfa

beta_xyz=inv(Tg)*beta




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alfa_xyz=inv(Tg)*alfa;

beta_xyz=inv(Tg)*beta;

% Compute the resulting strains

eps_T=alfa_xyz*DeltaT

eps_H=beta_xyz*hr



Example 3.4. Extend the computer program generated in Ex. 3.3to calculate the total strains

of the lamina in thex y z coordinate system at room temperature (22C), a moisture contentof 0.5 % and a cure temperature of 122C when the following stresses are applied: xx = 25MPa,yy = 50 MPa andxy = 20 MPa. Assume plane-stress analysis.

Solution to Example 3.4. The resulting in-plain strains of the lamina in thexy zcoordinate

system are:

xyz =

4033370

726


%


%



% A transversally isotropic material needs 4 elastic and 2


% E11, E22, nu12, G12, alfa_1, alfa_2, beta_1, beta_2

prop=[181000, 10300, 0.28, 7170, 0.02e-6, 22.5e-6, 0.0, 0.5];

% The angle of the reinforcement is also defined

theta=30;

theta=theta*pi/180;

% Variation of temperature and moisture

Tc=122;

Tr=22;

hr=0.005;

DeltaT=Tr-Tc;

% Applied stresses

sigma=[25, 50, 20];

E11=prop(1,1);

E22=prop(1,2);

nu12=prop(1,3);

G12=prop(1,4);

alfa_1=prop(1,5);

alfa_2=prop(1,6);

beta_1=prop(1,7);

beta_2=prop(1,8);

alfa=[alfa_1; alfa_2; 0];

beta=[beta_1; beta_2; 0];


m=cos(theta);

n=sin(theta);


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Chapter 3. Hygro-thermal effects 37

T=[m^2, n^2, 2*m*n;

n^2, m^2, -2*m*n;

-m*n, m*n, m^2-n 2];

Tg=(inv(T));


alfa_xyz=inv(Tg)*alfa;

beta_xyz=inv(Tg)*beta;

% Compute the hygro-thermal strains

eps_T=alfa_xyz*DeltaT

eps_H=beta_xyz*hr

% Compute compliance matrices and mechanical strains

S(1,1)=1/E11;

S(1,2)=-nu12/E11;

S(2,1)=S(1,2);

S(2,2)=1/E22;

S(3,3)=1/G12;

Sb=inv(Tg)*S*T;

eps=Sb*sigma

eps_tot=eps+eps_T+eps_H



Example 3.5. Consider the laminate in Ex. 2.3with the same material properties as in Ex. 3.3.

The laminate has been cured at 122C and it is at room temperature (22C) with a moisturecontent of 0.5 %. Write a computer program to determine the applied in-plane unit forces and

unit moments that result in the following total strains at the mid-plane of the laminate: 0xx = 0.5

%,0yy = 0.25 % and0xy = 0.1 %.

Solution to Example 3.5. To obtain the previous mid-plane strains (curvatures are assumed tobe zero) in the laminate under the considered conditions, the applied in-plane unit forces and

unit moments must be:

N =

2262

470

384

N/mm, M=

485

357

362

N


%


%



% A transversally isotropic material needs 4 elastic and 2


% E11, E22, nu12, G12, alfa_1, alfa_2, beta_1, beta_2

Mat(1,:)=[181000, 10300, 0.28, 7170, 0.02e-6, 22.5e-6, 0.0, 0.5];



% mat, theta, t

L(1,:)=[1, 0, 2];

L(2,:)=[1, 45, 1];

n_lam=size(L,1);

% Variation of temperature and moisture

Tc=122;


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Tr=22;

hr=0.005;

DeltaT=Tr-Tc;

% Resulting strains

% Only in-plane strains for the whole laminate

% the curvatures are assumed to be zeroeps=[0.005; 0.0025; 0.001];

kappa=[0; 0; 0];

% Compute S and Q

for i=1:n_lam;

i_mat=L(i,1);

S(1,1,i)=1/Mat(i_mat,1);


S(2,1,i)=S(1,2,i);

S(2,2,i)=1/Mat(i_mat,2);

S(3,3,i)=1/Mat(i_mat,4);

Q(:,:,i)=inv(S(:,:,i));

end;


for i=1:n_lam;


m=cos(theta);

n=sin(theta);

T(1,1,i)=m^2;

T(1,2,i)=n^2;

T(1,3,i)=2*m*n;

T(2,1,i)=n^2;

T(2,2,i)=m^2;

T(2,3,i)=-2*m*n;

T(3,1,i)=-m*n;

T(3,2,i)=m*n;

T(3,3,i)=m^2-n^2;

Tg(:,:,i)=(inv(T(:,:,i)));


Sb(:,:,i)=inv(Tg(:,:,i))*S(:,:,i)*T(:,:,i);

Qb(:,:,i)=inv(Sb(:,:,i));

end;

% Laminate constitutive matrix (ABD)

% Location of the bounds of the laminae and laminates midplane

TH=0;

Z=zeros(n_lam+1,1);

for i=1:n_lam;

TH=TH+L(i,3);

end;

th=TH/2;

TH=0;

Z(1)=-th;

for i=1:n_lam;

TH=TH+L(i,3);

Z(i+1)=TH-th;

end;

% Compute A matrix

A=zeros(3,3);

for i=1:n_lam;

for j=1:3;

for k=1:3;

A(j,k)=A(j,k)+Qb(j,k,i)*(Z(i+1)-Z(i));

end;

end;

end;


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% Compute B matrix

B=zeros(3,3);

for i=1:n_lam;

for j=1:3;

for k=1:3;

B(j,k)=B(j,k)+Qb(j,k,i)*(Z(i+1)^2-Z(i)^2)*(1/2);

end;

end;end;

% Compute D matrix

D=zeros(3,3);

for i=1:n_lam;

for j=1:3;

for k=1:3;

D(j,k)=D(j,k)+Qb(j,k,i)*(Z(i+1)^3-Z(i)^3)*(1/3);

end;

end;

end;

ABD=[A,B;B,D];

NM_tot=ABD*[eps;kappa];


for i=1:n_lam;

i_mat=L(i,1);

alfa_123(:,:,i)=[Mat(i_mat,5); Mat(i_mat,6); 0];

beta_123(:,:,i)=[Mat(i_mat,7); Mat(i_mat,8); 0];

alfa_xyz(:,:,i)=inv(Tg(:,:,i))*alfa_123(:,:,i);

beta_xyz(:,:,i)=inv(Tg(:,:,i))*beta_123(:,:,i);

end;

% Compute unit hygro-thermal forces and moments

N_T=zeros(3,1);

N_H=zeros(3,1);

M_T=zeros(3,1);

M_H=zeros(3,1);

for i=1:n_lam;

N_T=N_T+Qb(:,:,i)*alfa_xyz(:,:,i)*DeltaT*(Z(i+1)-Z(i));

N_H=N_H+Qb(:,:,i)*beta_xyz(:,:,i)*hr*(Z(i+1)-Z(i));

M_T=M_T+Qb(:,:,i)*alfa_xyz(:,:,i)*DeltaT*(Z(i+1)^2-Z(i)^2)*(1/2);

M_H=M_H+Qb(:,:,i)*beta_xyz(:,:,i)*hr*(Z(i+1)^2-Z(i)^2)*(1/2);

end;

% Compute unit mechanical forces and moments

N=NM_tot(1:3)-N_T-N_H

M=NM_tot(4:6)-M_T-M_H



Example 3.6. Consider an encastred composite plate 1000 mm long and 100 mm. The plate

is made of 0.125 mm thick T300/5208 unidirectional laminae according to a [016/458] stacking

sequence. The laminate has been cured at 122C and it is at room temperature (22C). Consid-ering the mechanical and thermal properties of the material used in previous examples, generate

the ANSYSTM program to simulate the plate and determine the residual stresses present in the

plate.

Solution to Example 3.6. As no mechanical loads are applied to the laminate, the resulting

stresses after the simulation of the plate correspond to the residual stresses generated during

the manufacturing process when cooling the part from the curing temperature to the room tem-

perature.

The ANSYSTM command sequence to simulate the [016/458] laminated plate is listed below.

You can either type these commands on the command window, or you can type them on a file,

then, on the command window enter /input, file, ext.


44/62


FINISH

/CLEAR

/TITLE, Residual stress composite plate

!Thermal residual stress composite laminate

/PREP7

!Parameters

l=1000 !length

w=100 !width


nl=24 !number of layers

h=lt*nl !plate thickness

Tc=122 !curing temperature

Tr=22 !room temperature



KEYOPT,1,8,2 !store bottom, mid and top data for all layers


MP,EX,1,181000

MP,EY,1,10300

MP,EZ,1,10300

MP,GXY,1,7170

MP,GYZ,1,3627

MP,GXZ,1,7170

MP,PRXY,1,0.28

MP,PRYZ,1,0.42

MP,PRXZ,1,0.28

MP,ALPX,1,0.02E-6 !thermal expansion coefficients

MP,ALPY,1,22.5E-6

MP,ALPZ,1,22.5E-6


!Laminate #1: [0_16/45_8]



SECDATA,lt*8,1,45


!Geometry

RECTNG,0,l,0,w !rectangle x1,x2,y1,y2

!Mesh


LESIZE,ALL,w/10 !element size


FINISH

/SOLU


DL,4,1,ALL !encastred line 4

!Apply load

TUNIF,Tr-Tc !apply constant temperature to nodes


SOLVE

FINISH

/POST1


RSYS,SOLU !lamina coordinate system

LAYER,1 !specify layer 1 results

MID !specify midthickness results

PLNSOL,S,X,2,1 !vertical displacement

!Obtain results selecting layer and coordinate system

ESEL,,,,500 !select element #250

LAYER,1

RSYS,SOLU


45/62


TOP

ETABLE,TSXXloc,S,X !write X-stress in table TSXX

ETABLE,TSYYloc,S,Y

ETABLE,TSXYloc,S,XY

BOT

ETABLE,BSXXloc,S,X

ETABLE,BSYYloc,S,Y

ETABLE,BSXYloc,S,XYPRETAB,BSXXloc,BSYYloc,BSXYloc,TSXXloc,TSYYloc,TSXYloc !print results table

RSYS,0

TOP

ETABLE,TSXXglb,S,X !write X-stress in table TSXX

ETABLE,TSYYglb,S,Y

ETABLE,TSXYglb,S,XY

BOT

ETABLE,BSXXglb,S,X

ETABLE,BSYYglb,S,Y

ETABLE,BSXYglb,S,XY

PRETAB,BSXXglb,BSYYglb,BSXYglb,TSXXglb,TSYYglb,TSXYglb !print results table



After the simulation, the resulting maximum displacements are x = -0.119 mm, y = 0.658

mm and z = 94.4 mm. It is observed that even if thermal expansion is an in-plane effect, the

higher displacement is out-of-plane. This due to the fact that th

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