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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
%
% J.A. Mayugo, N. Blanco
clear all,close all, clc
% 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)
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex102.m
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.
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex102.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex102.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex102.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex102.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex102.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex102.m -
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Chapter 1. Constitutive equations for transversely isotropic materials 3
%
% DACFE. Solution to Example 1.3
%
% J.A. Mayugo, N. Blanco
clear all,close all, clc
% 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)
This file can be found at:
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
%
% J.A. Mayugo, N. Blanco
clear all,close all, clc
% 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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4 Disseny i Anlisi de Compsits amb Elements Finits
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
fprintf(fidout,%10.4e\t %10.4e\t %10.4e\t %10.4e\t %10.4e\t %10.4e\n,...
S(row,1),S(row,2),S(row,3),S(row,4),S(row,5),S(row,6));
end
fclose(fidinp);
fclose(fidout);
This file can be found at:
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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Chapter 1. Constitutive equations for transversely isotropic materials 5
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.
%
% DACFE. Solution to Example 1.4
%
% J.A. Mayugo, N. Blanco
clear all,close all, clc
% 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)
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex104.m
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex104.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex104.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex104.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex104.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex104.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex104.m -
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6 Disseny i Anlisi de Compsits amb Elements Finits
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.
%
% DACFE. Solution to Example 1.5
%
% J.A. Mayugo, N. Blanco
clear all,close all, clc
% 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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Chapter 1. Constitutive equations for transversely isotropic materials 7
% 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
fprintf(fidout,%10.4e\t %10.4e\t %10.4e\t %10.4e\t %10.4e\t %10.4e\n,...
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
fprintf(fidout,%10.4e\t %10.4e\t %10.4e\t %10.4e\t %10.4e\t %10.4e\n,...
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);
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex105.m
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
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex105.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex105.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex105.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex105.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex105.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex105.m -
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8 Disseny i Anlisi de Compsits amb Elements Finits
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.
%
% DACFE. Solution to Example 1.6
%
% J.A. Mayugo, N. Blanco
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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Chapter 1. Constitutive equations for transversely isotropic materials 9
% 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)
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex106.m
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
Next, we present the solution implemented using the calculation capabilities of MATLABTM.
%
% DACFE. Solution to Example 1.7
%
% J.A. Mayugo, N. Blanco
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;
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex106.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex106.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex106.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex106.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex106.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex106.m -
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10 Disseny i Anlisi de Compsits amb Elements Finits
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
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex107.m
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.
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex107.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex107.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex107.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex107.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex107.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex107.m -
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Chapter 1. Constitutive equations for transversely isotropic materials 11
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
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12 Disseny i Anlisi de Compsits amb Elements Finits
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
This file can be found at:
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 -
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Chapter 1. Constitutive equations for transversely isotropic materials 13
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
!Elements and options
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
VMESH,ALL !mesh geometry
CSYS,0 !original coord. syst.
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
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex110.dat
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.
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex110.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex110.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex110.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex110.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex110.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex110.dat -
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14 Disseny i Anlisi de Compsits amb Elements Finits
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
!Elements and options
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
VMESH,ALL !mesh geometry
CSYS,0 !original coord. syst.
FINISH
/SOLU
!Boundary conditions
NSEL,S,LOC,X,0
D,ALL,UX
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Chapter 1. Constitutive equations for transversely isotropic materials 15
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
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
RSYS,11 !plot results local coord. systPLNSL,U,Y,2,1 !vertical displacement
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex111.dat
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
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex111.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex111.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex111.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex111.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex111.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T1/DACFE_Ex111.dat -
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16 Disseny i Anlisi de Compsits amb Elements Finits
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.
%
% DACFE. Solution to Example 2.1
%
% J.A. Mayugo, N. Blanco
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
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18 Disseny i Anlisi de Compsits amb Elements Finits
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)
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex201.m
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
Next, we present the solution implemented using the calculation capabilities of MATLABTM.
%
% DACFE. Solution to Example 2.2
%
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex201.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex201.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex201.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex201.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex201.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex201.m -
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Chapter 2. Laminate theory 19
% J.A. Mayugo, N. Blanco
clear all,close all, clc;
% Transversely isotropic material need 5 constants:
% 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(1,3,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(2,3,i)=-Mat(i_mat,4)/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;
% Compute transformation matrices T and Tgamma
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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20 Disseny i Anlisi de Compsits amb Elements Finits
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)
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex202.m
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.
%
% DACFE. Solution to Example 2.3
%
% J.A. Mayugo, N. Blanco
clear all,close all, clc;
% Transversely isotropic material need 4 constants:
% E1, E2, nu12, G12 (MPa)
Mat(1,:)=[181000, 10300, 0.28, 7170];
[n_mat,n_prop]=size(Mat);
% The lamina need to define material, angle and thickness
% mat, theta, t
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex202.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex202.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex202.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex202.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex202.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex202.m -
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Chapter 2. Laminate theory 21
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(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(3,3,i)=1/Mat(i_mat,4);
Q(:,:,i)=inv(S(:,:,i));
end;
% Compute transformation matrices T and Tgamma
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(1,3,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)));
% Compute stiffness and compliance transformed matrices
Sb(:,:,i)=inv(Tg(:,:,i))*S(:,:,i)*T(:,:,i);
Qb(:,:,i)=inv(Sb(:,:,i));
end;
Qb,Sb
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex203.m
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
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex203.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex203.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex203.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex203.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex203.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex203.m -
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22 Disseny i Anlisi de Compsits amb Elements Finits
Next, we present the solution implemented using the symbolic calculation capabilities of
MATLABTM.
%
% DACFE. Solution of Example 2.4
%
% J.A. Mayugo, N. Blanco
clear all,close all, clc;
clear all,close all, clc;
% Transversely isotropic material need 4 constants:
% E1, E2, nu12, G12 (MPa)
Mat(1,:)=[181000, 10300, 0.28, 7170];
[n_mat,n_prop]=size(Mat);
% The lamina need to define material, angle and thickness
% 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(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(3,3,i)=1/Mat(i_mat,4);
Q(:,:,i)=inv(S(:,:,i));
end;
% Compute transformation matrices T and Tgamma
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(1,3,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)));
% Compute stiffness and compliance transformed matrices
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;
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Chapter 2. Laminate theory 23
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
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex204.m
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
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex204.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex204.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex204.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex204.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex204.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex204.m -
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24 Disseny i Anlisi de Compsits amb Elements Finits
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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Chapter 2. Laminate theory 25
[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
!Elements and options
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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26 Disseny i Anlisi de Compsits amb Elements Finits
AMESH,ALL !mesh geometry
FINISH
/SOLU
!Boundary conditions
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
/PBC,ALL !to show BCs when solve
SOLVE
FINISH
/POST1
/VIEW,1,1,1,1 !iso-view
PLNSOL,U,Z,2,1 !vertical displacement
This file can be found at:
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
/TITLE, Composite plate
!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
!Elements and options
ET,1,SHELL181 !element type: 8-node laminated shell
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex206.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex206.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex206.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex206.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex206.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex206.dat -
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Chapter 2. Laminate theory 27
!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
SECTYPE,2,SHELL !section type: shell
SECDATA,lt*4,1,0 !layer thickness, material and orientation
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
SECOFFSET,MID !nodes on mid-thickness of elements
!Laminate #3: [0_2/90_2]_10
SECTYPE,3,SHELL !section type: shell
SECDATA,lt*2,1,0 !layer thickness, material and orientation
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
SECOFFSET,MID !nodes on mid-thickness of elements
!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
AMESH,ALL !mesh geometry
FINISH
/SOLU
!Boundary conditions
DL,2,1,UZ,0 !simply supported
DL,3,1,UZ,0
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28 Disseny i Anlisi de Compsits amb Elements Finits
DL,1,1,SYMM !symmetry
DL,4,1,SYMM
!Apply load
SFL,2,PRES,P !apply pressure on line 2 in N/mm
/PBC,ALL !to show BCs when solve
SOLVEFINISH
/POST1
/VIEW,1,1,1,1 !iso-view
PLNSOL,U,Z,2,1 !vertical displacement
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex207.dat
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
/TITLE, Composite plate
!3D composite laminate LSS
/PREP7
!Parameters
P=10 !applied stress
l=1000 !length
w=200 !width
lt=0.125 !layer thickness
nl=8 !number of layers
h=lt*nl !plate thickness
!Elements and options
ET,1,SHELL181 !element type: 8-node laminated shell
KEYOPT,1,8,2 !store bottom, mid and top data for all layers
!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,1,0.28
MP,PRYZ,1,0.42
MP,PRXZ,1,0.28
!Section properties for the laminate (up to 250 layers)
!Laminate #1: [0/90/60/75]2
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex207.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex207.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex207.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex207.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex207.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex207.dat -
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Chapter 2. Laminate theory 29
SECTYPE,1,SHELL !section type: shell
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
SECOFFSET,MID !nodes on mid-thickness of elements
!Geometry
RECTNG,0,l,0,w !rectangle x1,x2,y1,y2
!Mesh
SECNUM,1 !section type #1
LESIZE,ALL,w/10 !element size
AMESH,ALL !mesh geometry
FINISH
/SOLU
!Boundary conditions
DL,4,1,ALL !encastred line 4
!Apply load
SFL,2,PRES,-P*h !apply pressure on line 2 in N/mm
/PBC,ALL !to show BCs when solve
SOLVE
FINISH
/POST1
/VIEW,1,1,1,1 !iso-view
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
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex208.dat
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
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex208.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex208.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex208.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex208.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex208.datftp://amade.udg.edu/amade/mme/DACFE/input_files/T2/DACFE_Ex208.dat -
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Chapter 2. Laminate theory 31
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
Show all work in a report.
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
Show all work in a report.
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.
%
% DACFE. Solution to Example 3.1
%% J.A. Mayugo, N. Blanco
clear all,close all, clc;
% 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];
% Compute transformation matrices T and Tgamma
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
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex301.m
33
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex301.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex301.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex301.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex301.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex301.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex301.m -
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34 Disseny i Anlisi de Compsits amb Elements Finits
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.
%
% DACFE. Solution to Example 3.2
%
% J.A. Mayugo, N. Blanco
clear all,close all, clc;
% 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];
% Compute transformation matrices T and Tgamma
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));
% Compute transformed expansion coefficients
alfa_xyz=inv(Tg)*alfa
beta_xyz=inv(Tg)*beta
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex302.m
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex302.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex302.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex302.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex302.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex302.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex302.m -
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36 Disseny i Anlisi de Compsits amb Elements Finits
% Compute transformed expansion coefficients
alfa_xyz=inv(Tg)*alfa;
beta_xyz=inv(Tg)*beta;
% Compute the resulting strains
eps_T=alfa_xyz*DeltaT
eps_H=beta_xyz*hr
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex303.m
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
Next, we present the solution implemented using the calculation capabilities of MATLABTM.
%
% DACFE. Solution to Example 3.4
%
% J.A. Mayugo, N. Blanco
clear all,close all, clc;
% A transversally isotropic material needs 4 elastic and 2
% thermal and hygroscopic expansion coefficients
% 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];
% Compute transformation matrices T and Tgamma
m=cos(theta);
n=sin(theta);
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex303.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex303.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex303.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex303.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex303.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex303.m -
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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));
% Compute transformed expansion coefficients
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
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex304.m
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
Next, we present the solution implemented using the calculation capabilities of MATLABTM.
%
% DACFE. Solution to Example 3.5
%
% J.A. Mayugo, N. Blanco
clear all,close all, clc;
% A transversally isotropic material needs 4 elastic and 2
% thermal and hygroscopic expansion coefficients
% 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];
[n_mat,n_prop]=size(Mat);
% The lamina need to define material, angle and thickness
% 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;
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex304.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex304.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex304.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex304.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex304.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex304.m -
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38 Disseny i Anlisi de Compsits amb Elements Finits
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(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(3,3,i)=1/Mat(i_mat,4);
Q(:,:,i)=inv(S(:,:,i));
end;
% Compute transformation matrices T and Tgamma
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(1,3,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)));
% Compute stiffness and compliance transformed matrices
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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Chapter 3. Hygro-thermal effects 39
% 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];
% Compute transformed expansion coefficients
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
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex305.m
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.
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex305.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex305.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex305.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex305.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex305.mftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex305.m -
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40 Disseny i Anlisi de Compsits amb Elements Finits
FINISH
/CLEAR
/TITLE, Residual stress composite plate
!Thermal residual stress composite laminate
/PREP7
!Parameters
l=1000 !length
w=100 !width
lt=0.125 !layer thickness
nl=24 !number of layers
h=lt*nl !plate thickness
Tc=122 !curing temperature
Tr=22 !room temperature
!Elements and options
ET,1,SHELL181 !element type: 8-node laminated shell
KEYOPT,1,8,2 !store bottom, mid and top data for all layers
!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,1,0.28
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
!Section properties for the laminate (up to 250 layers)
!Laminate #1: [0_16/45_8]
SECTYPE,1,SHELL !section type: shell
SECDATA,lt*16,1,0 !layer thickness, material and orientation
SECDATA,lt*8,1,45
SECOFFSET,MID !nodes on mid-thickness of elements
!Geometry
RECTNG,0,l,0,w !rectangle x1,x2,y1,y2
!Mesh
SECNUM,1 !section type #1
LESIZE,ALL,w/10 !element size
AMESH,ALL !mesh geometry
FINISH
/SOLU
!Boundary conditions
DL,4,1,ALL !encastred line 4
!Apply load
TUNIF,Tr-Tc !apply constant temperature to nodes
/PBC,ALL !to show BCs when solve
SOLVE
FINISH
/POST1
/VIEW,1,1,1,1 !iso-view
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
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Chapter 3. Hygro-thermal effects 41
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
This file can be found at:
ftp://amade.udg.edu/amade/mme/DACFE/input_files/T3/DACFE_Ex306.dat
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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