A Finite Element Study of the Deflection of Simply Supported
Composite Plates Subject to Uniform Load
by
Mer Arnel Manahan
An Engineering Project Submitted to the Graduate
Faculty of Rensselaer Polytechnic Institute
In Partial Fulfillment of the
Requirements for the degree of
MASTER OF ENGINEERING IN MECHANICAL ENGINEERING
Approved:
_________________________________________________
Ernesto Gutierrez-Miravete, Engineering Project Adviser
Rensselaer Polytechnic Institute
Hartford, Connecticut
December 2011
© Copyright 2011
By
Mer Arnel Manahan
All Rights Reserved
ii
CONTENTS
LIST OF TABLES ............................................................................................................. v
LIST OF FIGURES .......................................................................................................... vi
LIST OF SYMBOLS ....................................................................................................... vii
ACKNOWLEDGMENT ................................................................................................ viii
ABSTRACT ..................................................................................................................... ix
1. Introduction.................................................................................................................. 1
2. Methodology ................................................................................................................ 3
2.1
Classical Lamination Theory ............................................................................. 3
2.1.1
Kirchhoff’s Hypothesis .......................................................................... 4
2.2
Governing Equations for the Response of a Laminated Plate............................ 6
2.3
Plate Modeling ................................................................................................. 11
3. Results........................................................................................................................ 14
3.1
Introduction ...................................................................................................... 14
3.2
MATLAB Validation ....................................................................................... 15
3.2.1
3.3
3.4
Deflection Function .............................................................................. 16
4-Ply Deflection Results .................................................................................. 18
3.3.1
Optimization of 4-Ply Composite Plate and Effect of D16 and D26 ...... 19
3.3.2
MATLAB Inaccuracy When Calculating Symmetric Angle-Ply
Laminate............................................................................................... 20
20-Ply Deflection Results ................................................................................ 21
4. Conclusion ................................................................................................................. 25
5. References.................................................................................................................. 27
Appendix A: ANSYS Analysis code for 4-Ply Composite Plate .................................... 28
Appendix B: MATLAB Validation code for 4-Ply Composite Plate .............................. 30
Appendix C: ANSYS Analysis code for 20-Ply Composite Plate .................................. 32
Appendix D: MATLAB Validation code for 10-Ply Composite Plate............................ 34
iii
Appendix E: Derivation of Deflection Solution for a Uniform Load .............................. 37
iv
LIST OF TABLES
Table 1: Material properties for VM 824 ......................................................................... 12
Table 2: Summary of Nineteen Cases Investigated ......................................................... 14
Table 3: ANSYS and MATLAB Results for 4-Ply Composite Plates ............................ 18
Table 4: ANSYS and MATLAB results for 20-Ply Composite Plates ............................ 21
v
LIST OF FIGURES
Figure 1: Cross-section of cross-ply laminate2 .................................................................. 1
Figure 2: Plate Dimension1 ................................................................................................ 4
Figure 3: Kirchhoff’s straight line2 .................................................................................... 4
Figure 4: Showing Geometry, Nomenclature and Loading2 ............................................. 6
Figure 5: Resultants at the x = a/2 boundary2 .................................................................... 6
Figure 6: Forces in the Z-direction2 ................................................................................... 7
Figure 7: Moments on all sides about the x-axis2 .............................................................. 8
Figure 8: SHELL181 Geometry4 ..................................................................................... 11
Figure 9: Visual Representation of Boundary Conditions ............................................... 13
Figure 10: Fourier Expansion for a Uniform Load8 ........................................................ 17
Figure 11: Deflection of 4-Ply [0 90]S Plate .................................................................... 19
Figure 12: ANSYS Deflection Results for 4-Ply Laminate............................................. 20
Figure 13: ANSYS Deflection Results for 20-Ply Plates ................................................ 23
Figure 14: % Error between MATLAB and ANSYS as 60º Ply Changes Location ....... 24
vi
LIST OF SYMBOLS
N = Normal force resultant (N/m)
M = Bending moment resultant (N-m/m)
Q = Transverse shear force resultant (N/m)
E = Elastic Modulus (N/m2)
G = Shear Modulus (N/m2)
υ = Poisson’s Ratio
q = Load applied (N/m2)
S = Compliance matrix (N/m2)
Q = Stiffness matrix (N/m2)
A = Extensional stiffness matrix (N/m)
B = Coupling stiffness matrix (N)
D = Bending stiffness matrix (N-m)
Ɵ = Fiber orientation angle (o)
e0,κ0, γ0 = reference surface extensional strains
u0 = displacement in the x-direction (m)
v0 = displacement in the y-direction (m)
w0 = displacement in the z-direction (m)
t = Thickness (m)
vii
ACKNOWLEDGMENT
To my parents Arnel and Susan Manahan: Thank you for your support not only in my
pursuit of higher academic achievements, but in life in general. Thank you to all my
professors at SUNY Binghamton and RPI for their time and effort spent to help me grow
academically. A special thanks to Professor Ernesto Gutierrez-Miravete for his patience,
understanding and guidance throughout the masters project.
“Give thanks to the LORD, for he is good; his love endures forever”
- 1 Chronicles 16:14
viii
ABSTRACT
This paper discusses the response of simply supported, symmetrically laminated
composite plates subjected to a uniformly distributed load. ANSYS is used to both
model and perform FEA analysis on the plate of a 4-ply composite at first. The response
of the plate to the uniformly distributed load was analyzed and validated through
calculations using MATLAB in coherence with the classical lamination theory. Once the
FEA had been validated, an investigation on the affects of the fiber orientation to the
displacement value was conducted for different cases of symmetric 4-ply layup. The
same analysis was completed and validated for a 20-ply plate and in addition will reveal
the effect of stacking sequence on bending stiffness.
ix
1. Introduction
Fiber-reinforced composites (FRC) have an extensive array of applications. They
range from structural to recreational use. The aerospace and automotive industries look
to composites to improve fuel economy due to its high strength to weight ratio. The
sports industry looks to composites to improve sports equipment technologies. The fact
that composites offer increased strength without sacrificing additional weight is what
gives composites the advantage from most structural and recreational materials1.
FRC’s are often manufactured in laminates. A laminate consists of individual
lamina or plies as seen in Figure 1. These plies consist of different combinations of fiber
materials embedded in a matrix material, usually of a polymer resin.
Figure 1: Cross-section of cross-ply laminate2
When designing for the use of composites, there are many aspects to investigate as there
are many variables that affect a laminate’s response to load. Fiber and matrix material,
fiber orientation, layer stacking sequence are some of the variables affecting the
response of a laminate. The virtually limitless combinations of ply materials, ply
1
orientations and ply-stacking sequences increase the design flexibility of composite
structures. This paper will investigate the response of composite plates to uniformly
distributed load.
2
2. Methodology
This project develops FEA models of a simply supported, symmetric, composite
laminate plate under a uniformly distributed load. The first step is to develop an FEA
model of a composite plate that consists of 4 layers. The model will be subjected to a
uniformly distributed load. Deflection results will be analyzed and validated with hand
calculations using MATLAB. The hand calculations will apply classical lamination
theory to arrive at a deflection value of the plate. Once the hand calculations validate the
4-ply FEA model, an investigation of the affects of fiber orientations on deflection
results will be conducted. Deflection results will then be discussed, and a case of a 20ply laminate, symmetric plate will be modeled and analyzed. This model will also be
validated through hand calculations with investigations on how additional plies and
layup sequence affects the deflection results.
2.1 Classical Lamination Theory
The analysis of laminate plates involves the Classical Lamination Theory which is
almost identical to the classical plate theory except for anisotropy. Both require that the
laminate be thin where the span of a and b is greater than 10 times the thickness t and is
visualized through Figure 2 below. They also require a small displacement in the
transverse direction w, where w is significantly smaller than t. They both share the
assumptions made by Kirchhoff’s Hypothesis.2
3
Figure 2: Plate Dimension1
2.1.1
Kirchhoff’s Hypothesis
Kirchhoff’s Hypothesis assumes that normals remain straight, normals remain
unstretched, and normals remain normal. Figure 3(a) below shows the cross-section of a
laminate plate. The midplane is located on the x-axis, and line AA’ is perpendicular to
the midplane.
Figure 3: Kirchhoff’s straight line2
4
Kirchhoff proposed that the straight line AA’ in Figure 3 above, will remain straight,
will not stretch and will remain perpendicular to the midplane under various loading
conditions. Figure 3(a) shows an undeformed laminate plate. The laminate plate,
depending on loading conditions may bend or twist, but the line AA’ will remain straight
and perpendicular to the midplane. Line AA’ will translate and rotate accordingly as
seen in Figure 3(b). The rotations and translations can be described by the following
equations where u, v and w are the displacements in the x, y and z directions
respectively. uo, vo and wo are the translations at any point on the AA’ line in the x, y,
and z directions respectively and
w o
w o
and
are the rotations about the x and y axes
y
x
respectively:
u ( x, y , z )  u o ( x, y )  z
v ( x, y , z )  v o ( x, y )  z
w o ( x, y )
x
w o ( x, y )
y
w( x, y, z )  w o ( x, y)
(1)
(2)
(3)
Another important assumption is that the line AA’ does not change in length. Line
segment tt’ is the same length in both Figure 3(a) and Figure 3(b). This indicates that
there is zero strain in the z-direction.
In addition to Kirchhoff’s Hypothesis, perfectly bonded layers are assumed. This
proposes that there is no flaw or gap between the layers, and no slipping between the
layers. No slipping between the layers means that the layers will remain parallel with
one another and hence remain perpendicular to the normal AA’.
5
2.2 Governing Equations for the Response of a Laminated Plate
The governing equations consist of the behavior of the plate internally as well as
the behavior of the boundary conditions. The governing equations will be derived below
using the Newtonian approach where summing the forces and moments on the plate is
used to develop the differential equations. Figure 4 below shows a composite plate in the
x-y-z orientations, a loading of q(x,y) and a width ‘a’ and a length ‘b’ with the reference
point at the center of the plate.
Figure 4: Showing Geometry, Nomenclature and Loading 2
Figure 5: Resultants at the x = a/2 boundary2
6
Figure 5 above shows the resultant forces N and moments M acting on the x = a/2 side of
the plate. The results are then required from the x = - a/2 and y = +- b/2 sides. Once
these resultant forces are drawn, the sum of the forces in the x and y directions can be
found. This results in two of the three governing equations.
Nx Nxy

0
x
y
(4)
Nxy Ny

0
x
y
(5)
The third governing equation is derived by the summation of the shear force resultants Q
acting on the interior of the plate. In Figure 6 below, q(x,y) is the applied load and the
resultant forces are shown.
Figure 6: Forces in the Z-direction2
7
Summation of these forces results in:
Qx Qy

q0
x
y
(6)
The moment equilibrium is now taken about the x-axis as seen in Figure 7 below. The
sum of the moments results in
Qy 
My Mxy

y
x
(7)
Summing up the moments about the y-axis results in:
Qx 
Mx Mxy

x
y
(8)
Substituting equations (7) and (8) in equation (6), results in the third governing equation.
 2 Mx
 2 Mxy  2 My

2

q 0
xy
x 2
y 2
Figure 7: Moments on all sides about the x-axis2
8
(9)
Since the purpose of this paper is to investigate a composite laminate plate’s deflection
in response to a uniformly distributed load, it is beneficial to express the governing
equations in terms of displacements. Fortunately, the stress resultants can be expressed
in terms of strains and curvatures by
 Nx   A11
 Ny   A

  12
 Nxy   A16


 Mx   B11
 My   B12

 
Mxy  B16
A12
A16
B11
B12
A21
A26
B11
B11
A26
A66
B11
B11
B12
B16
D11
D12
B22
B26
D12
D22
B11
B11
D16
D26
B16   e x0 
 
B26   e 0y 
B66   xy0 
 
D16    x0 
D26    y0 
 
D66   xy0 
(10)
The strains and curvatures can be expressed in terms of displacements.
 2 w 0 ( x, y )
x 2
 2 w 0 ( x, y )
 y0 ( x, y ) 
y 2
u 0 ( x, y )
e ( x, y ) 
x
0
v ( x, y )
e 0y ( x, y ) 
y
 x0 ( x, y )  
0
x
 xy0 ( x, y ) 
v 0 ( x, y ) u 0 ( x, y )

x
y
 xy0 ( x, y )  2
(11)
 2 w 0 ( x, y )
xy
Likewise, the stress resultants can be expressed in terms of displacements3. Substituting
equation (11) into equation (10), we get equation (12)-(17):

 2 wo
 2 wo
 2 wo
  B11

B

2
B
12
16
xy
x 2
y 2

(12)
 u o v o 
u o
v o
 2 wo
 2 wo
 2 wo


Ny  A12
 A22
 A26 

 B12
 B22
 2 B26
x
y
x 
xy
x 2
y 2
 y
(13)
Nx  A11
 u o v o
u o
v o
 A12
 A16 

x
y
x
 y
Nxy  A16
 u o v o 
u o
v o
 2 wo
 2 wo
 2 wo
  B16
 A26
 A66 


B

2
B
(14)
26
66
x
y
x 
xy
x 2
y 2
 y
9
Mx  B11
 u o v o 
u o
v o
 2 wo
 2 wo
 2 wo
  D11
 B12
 B16 


D

DB
12
16
x
y
x 
xy
x 2
y 2
 y
My  B12
 u o v o
u o
v o
 B22
 B26 

x
y
x
 y
(15)

 2 wo
 2 wo
 2 wo
  D12
(16)

D

2
D
22
26
xy
x 2
y 2

 u o v o 
u o
v o
 2 wo
 2 wo
 2 wo
  D16
(17)
Mxy  B16
 B26
 B66 

 D26
 2 D66
x
y
x 
xy
x 2
y 2
 y
Substituting expressions in equations (12)-(17) into the governing equations (4), (5) and
(9), we get the equilibrium equations that govern the response of a laminated plate in
terms of displacements.3
 2u o
 2u o
 2u o
 2vo
 2vo
 2v o
A11
 2 A16
 A66
 A16
 ( A12  A66 )
 A26
xy
xy
x 2
y 2
x 2
y 2
 3 wo
 3 wo
 3 wo
 3 wo
 B11

3
B

(
B

2
B
)

B
0
16
12
66
26
x 3
x 2 y
xy 2
y 3
(18)
 2u o
 2u o
 2u o
 2v o
 2v o
 2v o
A16
 ( A12  A66 )
 A26
 A66 2  2 A26
 A22 2
x 2
xy
y 2
x
xy
y
 3 wo
 3 wo
 3wo
 B16
 ( B12  2 B66 ) 2  3B26
0
x3
x y
y 3
D11
(19)
 4 wo
 4 wo
 4 wo
 4 wo
 4 wo

4
D

2
(
D

2
D
)

4
D

D
16
12
66
26
22
x 4
x3y
x 2y 2
xy 3
y 4
 B11
 3u o
 3u o
 3u o
 3u o
 3v o

3
B

(
B

2
B
)

B

B
16
12
66
26
16
x3
x 2y
xy 2
y 3
x3
 3v o
 3v o
 3v o
 ( B12  2 B66 ) 2  3B26
 B22 3  0
x y
xy 3
y
10
(20)
Equations (18)-(20) are the equations in terms of displacements that govern the response
of a composite plate. In this project, we will limit the investigation to a simply
supported, symmetrically laminated composite plate.
2.3 Plate Modeling
Verification Manual 82 (VM82) ANSYS code was used for the purposes of this paper.
VM82 is a solved solution to a simply supported, composite plate subjected to a
uniformly distributed load. VM82 was utilized to validate the MATLAB code and
modified to run several different cases of laminate plates discussed in a later section.
VM82 uses several elements to perform the analysis on the plate. The SHELL181 was
chosen as the element to use for this paper because the input parameters of a composite
plate can be easily modified due to it being a shell element. SHELL181 was used as the
element type because it is suitable for thin to moderately thick shell structures and
commonly used for laminated composite shells. SHELL181 is a 4 node element with six
degrees of freedom at each node.4 Figure 8 below shows its geometry.
Figure 8: SHELL181 Geometry4
11
VM82 analyzes a [0 90]S , 5m by 5m composite laminate plate subjected to a uniformly
distributed load in the transverse direction with the material properties defined in Table
1. The ANSYS code can be found in APPENDIX A.
Table 1: Material properties for VM 824
The code in APPENDIX A shows that nine material property inputs were entered to run
the analysis, however Table 1 only consists of five.
VM82 assumes that Ey = Ez,  xy   xz , Gxy = Gxz, and  yz =  yx   xy (
Ey
Ex
) . The plate
is modeled using four key points at (0,0), (5,0), (5,5) and (0,5) xy coordinates. An area is
then made from these key points where the mesh is applied. Considering that the origin
is at (0,0), the boundary conditions are such that the side at x = 5, the plate is constrained
in the X and Z directions and at y = 5 it is constrained in the Y and Z directions. This
means that there will be no deflection and moments along these sides. A pressure of 1.0
N/m2 is then applied along the surface area of the plate. Figure 9 below provides a visual
representation of the boundary constraints applied in ANSYS. Figure 9 is one of the four
quadrants that the plate is divided into. Since the original plate is simply supported on all
sides, only two sides of the quadrant need to be constrained. This reduces the complexity
of the model.
12
Figure 9: Visual Representation of Boundary Conditions
13
3. Results
3.1 Introduction
This section discusses the results found during the FEA analysis and validation of
the composite plate subjected to a uniformly distributed load. Several cases were
analyzed and grouped by the amount of layers existent for each case. One group consists
of symmetric plates with only four layers. The other group consists of symmetric plates
with twenty layers. The group with four layer plate analyses ranged from a symmetric
plate to a symmetric, angle ply plates in which the fiber orientations of the outer layers
were changed to varying angles ranging from 0º to 90º. The group of twenty layer plates
ranged from various stack up sequences of symmetric plates to symmetric, angle ply
plates also. Overall, there were nineteen cases performed. All of these cases were
modeled and analyzed in ANSYS and were also attempted to be validated in MATLAB.
The layup sequences of the different cases analyzed are tabulated in Table 2 below.
Table 2: Summary of Nineteen Cases Investigated
Layup (4-Ply Group)
[0 90]s
[10 90]s
[20 90]s
[30 90]s
[40 90]s
[45 90]s
[50 90]s
[60 90]s
[70 90]s
[80 90]s
[90 90]s
Layup (20-Ply Group)
[60 0 90 90 0 90 90 0 90 0]s
[0 60 90 90 0 90 90 0 90 0]s
[0 0 60 90 0 90 90 0 90 0]s
[0 0 90 60 0 90 90 0 90 0]s
[0 0 90 90 60 90 90 0 90 0]s
[0 0 90 90 0 60 90 0 90 0]s
[0 0 90 90 0 90 60 0 90 0]s
[0 0 90 90 0 90 90 60 90 0]s
[0 0 90 90 0 90 90 0 60 0]s
[0 0 90 90 0 90 90 0 90 60]s
14
The first step of this project in the investigation of a composite plate’s response to
a uniformly distributed load begins with the validation of the FEA analysis through hand
calculations involving the Classical Lamination Theory as described in Section 2.1 of
this paper. Section 2.1.1 describes the assumptions applied by Kirchhoff’s Hypothesis
and Section 2.2 shows the derivations of the equations that govern the response of a
composite plate in terms of displacements. The MATLAB code that applies the classical
lamination theory to a 4-ply symmetric laminate plate is found in APPENDIX B.
3.2 MATLAB Validation
The MATLAB code starts off by stating the material properties for each layer. In
this case, each layer has the same material properties listed in Table 1. In this section, the
code only enters in five material constants, E1, E2, G12, v12 and v21 (calculated), but
ANSYS requires nine material property inputs. This is because of the plane stress
assumption. In applications such as the plate analyzed in this project, three of the six
components of stress are generally much smaller than the other three. In the example of
a plate, the stresses acting in the plane of the plate are much larger than the stresses
perpendicular to that plane. In all calculations then, the stress components perpendicular
to the plane of the structure can be set to zero.2
The code then computes the values of the stiffness matrix [ Qij ] for each ply. It does
this by first computing the compliance matrix [ S ij ] for each ply. This matrix is then
transformed and inverted to acquire the values of [ Qij ]. Next, the code calculates the z
distances of each plate from the midplane which will provide the distances to calculate
the bending stiffness matrix [D]. This bending stiffness matrix is used to compute the
15
deflection of the plate. Since the composite that is analyzed by VM82 is a symmetric,
simply supported plate with the boundary conditions set to limit deflection in the x and y
directions, the governing equations (18)-(20) reduce to:
 4 wo
 4 wo
 4 wo
D11
 2( D12  2 D66 ) 2 2  D22
 q ( x, y )
x 4
x y
y 4
(21)
This reduction is due to the fact that for symmetric, orthotropic plates, D16 and D26 are
zero. The boundary conditions described cause the displacements uo and yo to be zero,
therefore those components with uo and yo go to zero.
3.2.1
Deflection Function
The deflection function for a symmetric composite rectangular plate, simply
supported on all sides under a uniformly distributed load is expressed by equation (22)5

w ( x, y )  
m 1
ny
mx
sin
a
b

m
m
n
n
n 1
 6 mn[ D11 ( ) 4  2( D12  2 D66 )( ) 2 ( ) 2  D22 ( ) 4 ]
a
a
b
b
16q 0 sin

(22)5
For m,n = 1,3,5…
The method in APPENDIX E to derive equation (22) utilized the Navier solution for a
simply supported rectangular plate6. For any kind of loading given by q  q(xy) , load
q(x,y) is represented by:

q ( x, y )  
m 1


n 1
q mn sin(
my
mx
) sin(
)
a
b
Then the differential equation (21) and boundary conditions can be satisfied by:
16
(23)7


w( x, y )  

m 1
n 1
wmn sin(
my
mx
) sin(
)
a
b
(24)7
The derivation in APPENDIX E shows that when substituting equation (23) and (24)
into equation (21), we get:
w 
a 4 q mn
 [ D11m  2( D12  2 D66 (mnR)  D22 (nR) ]
4
4
2
4
(25)7
Where R = a/b, the plate aspect ratio
APPENDIX E continues to solve for qmn, which for a uniform load, q(x,y) = qo, a
constant, the Fourier coefficients are7:
q mn 
16q o
 2 mn
for m,n = 1,3,5…
(26)6,7,8
Figure 10 below shows that the Fourier expansion equation (26) represents a uniform
load distributed along the entire surface of the plate.
Figure 10: Fourier Expansion for a Uniform Load8
17
Equation (25) and (26) are now plugged into equation (24) to get the deflection equation
for a simply supported orthotropic laminated plate under a uniformly distributed load
expressed in equation (22).
3.3 4-Ply Deflection Results
The following results pertain to the analyses run through MATLAB and the FEA
through ANSYS. Table 3 below shows the results of not only analyses run for a [0 90] S
but for a symmetric angle-ply plate in which the inner layers are kept at 90º while the
fiber orientation of the outer layers are increased ranging from 0º to 90º. The bending
stiffness values for D16 and D26 shown in Table 3 below were computed in the
MATLAB code in APPENDIX B. The code outputs the [D] matrix where the D16 and
D26 values were acquired.
Table 3: ANSYS and MATLAB Results for 4-Ply Composite Plates
Layup
[0 90]s
[10 90]s
[20 90]s
[30 90]s
[40 90]s
[45 90]s
[50 90]s
[60 90]s
[70 90]s
[80 90]s
[90 90]s
ANSYS
Matlab
Deflection (m) Deflection (m) % Error D16 (N-m)
D26 (N-m)
0.0681
0.0680 0.1468
0.0000
0.0000
0.0883
0.0626 29.1053
288.1000
12.0000
0.1078
0.0523 51.4842
493.4000
70.4000
0.1164
0.0440 62.1993
565.8000
193.8000
0.1160
0.0399 65.6034
505.3907
358.4758
0.1154
0.0393 65.9445
438.5965
438.5965
0.1155
0.0398 65.5411
358.4758
505.3907
0.1165
0.0436 62.5751
193.8000
565.8000
0.1100
0.0514 53.2727
70.4000
493.4000
0.0889
0.0605 31.9460
12.0000
288.1000
0.0651
0.0650 0.1536
0.0000
0.0000
Figure 11 below is the graphical output of the ANSYS FEA result for a [0 90]S plate.
The graphic as well as the result in Table 3 show that the MATLAB code is validated
18
through the FEA result. The MATLAB code resulted in a 0.0680m deflection while the
FEA resulted in 0.0681m deflection, a 0.14% error.
Figure 11: Deflection of 4-Ply [0 90]S Plate
3.3.1
Optimization of 4-Ply Composite Plate and Effect of D16 and D26
After the validation of the MATLAB code for a [0 90]S, an investigation on the
effects of the fiber orientation of the outer layers of the plate was conducted. This
investigation will use the same material properties in Table 1 and will keep the inner
layers of the plate arranged at 90º while the outer layers are changed ranging from 10º to
90º. For this investigation the MATLAB code, as well as the ANSYS VM82 code was
modified for each layup. Table 2 provides the results for the deflections acquired from
MATLAB and ANSYS, as well as the value of the D16 and D26 of the bending stiffness
matrix. Figure 12 below is a graph showing how changing the orientation angle of outer
19
layers a 4-layer composite plate affects its deflection response. As the angles of the outer
layers are increased, the deflection of the plate increases as well until the angle of 60º.
This angle resulted in the maximum deflection of the plate. After this angle, the behavior
of the plate starts to be that of an isotropic material until it the layup was [90 90]S.
Figure 12: ANSYS Deflection Results for 4-Ply Laminate
3.3.2
MATLAB Inaccuracy When Calculating Symmetric Angle-Ply Laminate
As can be seen in the results of Table 2, the MATLAB code only agreed with the
ANSYS results when the outer layers were either 0º or 90º. This is because of the elastic
coupling that exists when D16 and D26 of the bending stiffness matrix [D] are non-zeros.
Equation (21) becomes:
D11
 4 wo
 4 wo
 4 wo

2
(
D

2
D
)

D

12
66
22
x 4
x 2 y 2
y 4
4 D16
 4 wo
 4 wo

4
D
 q ( x, y )
26
x 3 y
xy 3
(27)
Because of the separation of variables involving D16 and D26, the proposed solution to
equation (22) above and derived in APPENDIX E does not satisfy the governing
20
differential equation (27). Thus, the variables are not actually separable. Moreover, the
deflection expansion equation (24) also does not satisfy the boundary conditions5. This
problem has been solved using the Rayleigh-Ritz method5. However, this paper does not
cover this method. Instead, an investigation on how additional plies and stacking
sequence affects deflection results and the values of D16 and D26 was conducted.
3.4 20-Ply Deflection Results
Campbell states in Reference 1 that the values of D16 and D26 become small when a large
number of plies are stacked and become insignificant for thicknesses of more than
sixteen plies. This section investigates how the values of D16 and D26 become
insignificant when the number of plies is increased to more than sixteen plies. FEA
analysis and MATLAB calculations were conducted for 20-ply plates. These plates
consist of the same material properties as those in Table 1 and same loading is applied.
Thickness of each layer however was reduced from 0.025m to 0.0025m to keep the
requirement of the classical lamination theory where thickness is ten times smaller than
the length or width of the plate. The results of the analyses are tabulated in Table 4
Table 4: ANSYS and MATLAB results for 20-Ply Composite Plates
Layup
[60 0 90 90 0 90 90 0 90 0]s
[0 60 90 90 0 90 90 0 90 0]s
[0 0 60 90 0 90 90 0 90 0]s
[0 0 90 60 0 90 90 0 90 0]s
[0 0 90 90 60 90 90 0 90 0]s
[0 0 90 90 0 60 90 0 90 0]s
[0 0 90 90 0 90 60 0 90 0]s
[0 0 90 90 0 90 90 60 90 0]s
[0 0 90 90 0 90 90 0 60 0]s
[0 0 90 90 0 90 90 0 90 60]s
ANSYS
MATLAB
Deflection (m) Deflection (m)
0.5829
0.4698
0.5733
0.4842
0.5657
0.4970
0.5606
0.5093
0.5576
0.5208
0.5547
0.5300
0.5533
0.5379
0.5525
0.5440
0.5519
0.5481
0.5517
0.5501
21
% Error D16 (N-m)
D26 (N-m)
19.4030
7.5042
21.9060
15.5416
6.0089
15.1954
12.1442
4.6797
13.6609
9.1509
3.5167
10.2659
6.5997
2.5199
7.3559
4.4529
1.6891
4.9309
2.7833
1.0246
2.9909
1.5385
0.5261
1.5358
0.6885
0.1938
0.5658
0.2900
0.0277
0.0808
The analyses for the 20-ply plates were conducted to see how additional plies along with
layup sequencing would affect the deflection results. There are many possible stacking
sequences when a laminate plate involves twenty plies. The stacking sequence chosen
and displayed in the layup column of Table 4 was to keep the layup symmetric and
orthotropic. This section also investigated the effects on the values of D16 and D26 when
the plate is made of twenty plies. Comparing Table 3 with Table 4, there are noticeable
differences in the value of D16 and D26, as well as the % error between the ANSYS and
the MATLAB deflection results. The increase to 20-plies certainly decreased the value
of D16 and D26 when compared to the values of D16 and D26 of the 4-ply plates in Table
3. Sequentially moving the 60º ply closer to the midplane caused a decrease in the
deflection results as can be seen in Figure 13 below. When the layup had the 60º ply in
layers straddling the midplane, the laminate plate was at its stiffest point. Figure 14
below shows the decrease in % error between the ANSYS and MATLAB deflection
results. The % error decreased as the 60º ply moved closer to the midplane due to the
decreasing value of D16 and D26 which eventually got closer to zero.
22
Figure 13: ANSYS Deflection Results for 20-Ply Plates
23
Figure 14: % Error between MATLAB and ANSYS as 60º Ply Changes Location
24
4. Conclusion
An investigation on the response of a symmetric composite laminate plate was
conducted. ANSYS was utilized to model and analyze the response of a simply
supported symmetric plate subjected to a uniform load. VM82 was used for the purposes
of this project because it is a solved simply supported, symmetric plate problem. It was
used to validate the hand calculations performed using MATLAB that invoked the
classical lamination theory and modified accordingly to analyze the different cases in
Table 2. The MATLAB calculations along with the ANSYS analyses were performed to
investigate how fiber orientation, lay-up sequencing and total number of plies affect the
deflection results.
The response of a symmetric 4-ply plate with a layup of [0 90]S was performed using
VM82 to validate the MATLAB code. The effect of fiber orientation of the outer plies of
the 4-ply plate was then investigated. The results are tabulated in Table 3 and the
following conclusions are observed:
1. The minimum deflection of the 4-ply plate is when the lay-up is [90 90]S
2. The maximum deflection occurs when the outer layers of are oriented at 60º.
3. The large % error between the MATLAB and ANSYS deflection results is
due to values of D16 and D26 being non-zeros for plies whose angles are those
other than 0º or 90º, thus changing the governing differential equation.
25
The response of a symmetric 20-ply plate with an original layup up of
[0 0 90 90 0 90 90 0 90 0]S was conducted to investigate how an increase in the number
of plies affects the deflection results. Table 4, Figure 13 and Figure 14 tabulate and show
the results and thus the following conclusions are observed:
1. The % error between the MATLAB and ANSYS deflection results has
decreased when sixteen plies were added to the original four. This is due to
the decrease of D16 and D26 values in Table 4. These values are significantly
smaller than those in Table 3 when there were only four plies to the plate.
2. As the 60º ply was re-located from the outer plates to the inner plates, the
stiffness of the plate increased. The min deflection in Table 4 is when the 60º
ply was in layers ten and eleven, straddling the midplane as shown in Figure
13.
When designing for composite materials with a specific purpose, there are several
variables an engineer should investigate. Fiber and matrix material, fiber orientation,
layer stacking sequence and so on affect the response of a laminate. The virtually
limitless combinations of ply materials, ply orientations and ply-stacking sequences
increase the design flexibility of composite structures.
26
5. References
1.
Campbell, F.C. “Structural Composite Materials” ASM International. Copyright
2010
2.
Hyer, Michael W. “Stress Analysis of Fiber-Reinforced Composite Materials”.
Updated Edition. 2009.
3.
MANE-6180 Mechanics of Composite Materials –Prof. R. Naik, Lecture Notes
2010.
4.
ANSYS Element Reference. pp 1171-1186. Release 12.0. April 2009
5.
Jones, Robert M. “Mechanics of Composite Materials”. First Edition. McGraw-Hill
Book Company. Copyright 1975
6.
Timoshenko P., Stephen, S. Woinowsky-Kreiger. “Theory of Plates and Shells”.
Second Edition. McGraw-Hill, Inc. Copyright 1959
7.
Gibson, F. Ronald. “Principles of Composite Material Mechanics. Second Edition.
Taylor and Francis Group, LLC. Copyright 2007
8.
Szilard, R. “Theory and Analysis of Plates: Classical and Numerical Methods”.
Prentice Hall 1974, pp 58-59.
27
Appendix A: ANSYS Analysis code for 4-Ply Composite Plate
/TITLE, 4-Ply SIMPLY SUPPORTED LAMINATED PLATE UNDER PRESSURE
C*** USING SHELL181
/PREP7
SMRT,OFF
ANTYPE,STATIC
ET,1,SHELL181
! 4 NODE LAYERED SHELL
KEYOPT,1,3,2
! FULL INTEGRATION
KEYOPT,1,8,1
! WRITE LAYER RESULTS
SECTYPE,1,SHELL
SECDATA,0.025,1,0
! LAYER 1: 0.025 THK, THETA 0
SECDATA,0.025,1,90
! LAYER 2: 0.025 THK, THETA 90
SECDATA,0.025,1,90
! LAYER 3: 0.025 THK, THETA 90
SECDATA,0.025,1,0
! LAYER 4: 0.025 THK, THETA 0
MP,EX,1,25E6
! ORTHOTROPIC MATERIAL PROPERTIES
MP,EY,1,1E6
MP,EZ,1,1E6
! EZ=EY ASSUMED
MP,GXY,1,5E5
MP,GYZ,1,2E5
MP,GXZ,1,5E5
MP,PRXY,1,0.25
! MAJOR POISSONS RATIO
MP,PRYZ,1,0.01
! MAJOR POISSONS RATIO
MP,PRXZ,1,0.25
! MAJOR POISSONS RATIO
K,1
! CORNER KEYPOINTS OF QUADRANT (AREA)
K,2,5
K,3,5,5
K,4,,5
A,1,2,3,4
ESIZE,,20
! 20x20 MESH USING QUARTER SYMMETRY
AMESH,1
NSEL,S,LOC,X,0
DSYM,SYMM,X
NSEL,S,LOC,Y,0
DSYM,SYMM,Y
NSEL,S,LOC,X,5
! APPLY FREELY SUPPORTED B.C.
D,ALL,UZ,,,,,UY
NSEL,S,LOC,Y,5
D,ALL,UZ,,,,,UX
NSEL,ALL
/USER, 1
/VIEW, 1, 0.429637382252
/ANG, 1, -0.384698175032
/REPLO
EPLOT
FLST,2,1,5,ORDE,1
FITEM,2,1
/GO
!*
SFA,P51X,2,PRES,1
FINISH
/SOL
/STATUS,SOLU
SOLVE
, 0.886870752090E-01, 0.898635811919
28
FINISH
/POST1
!*
/EFACET,1
PLNSOL, U,Z, 0,1.0
29
Appendix B: MATLAB Validation code for 4-Ply Composite Plate
% This Matlab script calculates the laminate bending stiffness and then
% calculates the deflection of a simply supported plate.
%
clear; close all;
%
% Ply lay-up for a four ply laminate [(0/90)]s
%
theta(1)=60;
E1(1)=25e6; E2(1)=1e6; G12(1)=500000; v12(1)=0.25;
t(1)=0.025;
theta(2)=90;
E1(2)=25e6; E2(2)=1e6; G12(2)=500000; v12(2)=0.25;
t(2)=0.025;
theta(3)=90;
E1(3)=25e6; E2(3)=1e6; G12(3)=500000; v12(3)=0.25;
t(3)=0.025;
theta(4)=60;
E1(4)=25e6; E2(4)=1e6; G12(4)=500000; v12(4)=0.25;
t(4)=0.025;
%
%
% Define cell arrays to store the transformed stiffness matrices of %
%each
% individual ply
%
Qbar=cell(length(theta),1);
%
% Loop through all plies, calculate Qbar for each ply
%
for i=1:length(theta)
v21(i)=v12(i)*(E2(i)/E1(i));
S=[1/E1(i) -v21(i)/E2(i) 0; -v12(i)/E1(i) 1/E2(i) 0; 0 0 1/G12(i)];
c=cosd(theta(i));
s=sind(theta(i));
T=[c^2 s^2 2*c*s; s^2 c^2 -2*c*s; -c*s c*s (c^2 - s^2)];
Sbar=T'*S*T;
Qbar{i,1}=inv(Sbar);
end
%
% Calulate the z distances for each ply interface
%
total_thick=sum(t);
z(1)=-total_thick/2;
for k=2:length(theta)+1
z(k)=z(k-1)+t(k-1);
end
%
% Calculate the D matrix
%
D=[0 0 0; 0 0 0; 0 0 0];
for i=1:length(theta)
k=i+1;
Dply=(1/3)*Qbar{i,1}*(z(k)^3-z(k-1)^3);
D=D+Dply;
end
%
% Applied Transverse Loading
q0=1;
30
%
% Plate dimensions
a=10; b=10;
%
%**********************************************************************
***
m=[1 3 5 7];
n=[1 3 5 7];
%
% The deflection function expressed in Fourier Expansion
for i=1:length(m)
for j=1:length(n)
den=D(1,1)*(m(i)/a)^4 + 2*(D(1,2) +
2*D(3,3))*((m(i)*n(j))/(a*b))^2 + D(2,2)*(n(j)/b)^4;
wm(i,j)=16*q0/((pi^6)*m(i)*n(j)*den);
dw(i,j)=wm(i,j)*sin(m(i)*pi/2)*sin(n(j)*pi/2);
end
end
%
for k=1:4
wc(k)=sum(sum(dw(1:k,1:k)));
end
% Outputs the Deflection value
wc(k)
% Outputs the Bending Stiffness Matrix to get D16 and D26 values
D
31
Appendix C: ANSYS Analysis code for 20-Ply Composite Plate
/TITLE, 20-Ply SIMPLY SUPPORTED LAMINATED PLATE UNDER PRESSURE with Angle Ply at
Layer 1 & 20
C*** USING SHELL181
/PREP7
SMRT,OFF
ANTYPE,STATIC
ET,1,SHELL181
! 4 NODE LAYERED SHELL
SECTYPE,1,SHELL
SECDATA,0.0025,1,0
SECDATA,0.0025,1,0
SECDATA,0.0025,1,90
SECDATA,0.0025,1,90
SECDATA,0.0025,1,0
SECDATA,0.0025,1,90
SECDATA,0.0025,1,90
SECDATA,0.0025,1,0
SECDATA,0.0025,1,90
SECDATA,0.0025,1,60
! LAYER 1: 0.025 THK, THETA 0
! LAYER 2: 0.025 THK, THETA 90
! LAYER 3: 0.025 THK, THETA 90
! LAYER 4: 0.025 THK, THETA 45
! LAYER 5: 0.025 THK, THETA 0
! LAYER 6: 0.025 THK, THETA 0
! LAYER 7: 0.025 THK, THETA 45
! LAYER 8: 0.025 THK, THETA 90
! LAYER 9: 0.025 THK, THETA 90
! LAYER 10: 0.025 THK, THETA 0
SECDATA,0.0025,1,60
SECDATA,0.0025,1,90
SECDATA,0.0025,1,0
SECDATA,0.0025,1,90
SECDATA,0.0025,1,90
SECDATA,0.0025,1,0
SECDATA,0.0025,1,90
SECDATA,0.0025,1,90
SECDATA,0.0025,1,0
SECDATA,0.0025,1,0
! LAYER 6: 0.025 THK, THETA 0
! LAYER 7: 0.025 THK, THETA 45
! LAYER 8: 0.025 THK, THETA 90
! LAYER 9: 0.025 THK, THETA 90
! LAYER 10: 0.025 THK, THETA 0
! LAYER 6: 0.025 THK, THETA 0
! LAYER 7: 0.025 THK, THETA 45
! LAYER 8: 0.025 THK, THETA 90
! LAYER 9: 0.025 THK, THETA 90
! LAYER 10: 0.025 THK, THETA 0
MP,EX,1,25E6
MP,EY,1,1E6
MP,EZ,1,1E6
MP,GXY,1,5E5
MP,GYZ,1,2E5
MP,GXZ,1,5E5
MP,PRXY,1,0.25
MP,PRYZ,1,0.01
MP,PRXZ,1,0.25
! ORTHOTROPIC MATERIAL PROPERTIES
! EZ=EY ASSUMED
! MAJOR POISSONS RATIO
! MAJOR POISSONS RATIO
! MAJOR POISSONS RATIO
K,1
! CORNER KEYPOINTS OF QUADRANT (AREA)
K,2,5
K,3,5,5
K,4,,5
A,1,2,3,4
ESIZE,,6
! 20X20 MESH USING QUARTER SYMMETRY
AMESH,1
NSEL,S,LOC,X,0
DSYM,SYMM,X
NSEL,S,LOC,Y,0
DSYM,SYMM,Y
NSEL,S,LOC,X,5
! APPLY FREELY SUPPORTED B.C.
32
D,ALL,UZ,,,,,UY
NSEL,S,LOC,Y,5
D,ALL,UZ,,,,,UX
NSEL,ALL
FLST,2,1,5,ORDE,1
FITEM,2,1
/GO
!*
SFA,P51X,2,PRES,1
FINISH
/SOL
/STATUS,SOLU
SOLVE
FINISH
/POST1
!*
/EFACET,1
PLNSOL, U,Z, 0,1.0
33
Appendix D: MATLAB Validation code for 10-Ply Composite Plate
% This Matlab script calculates the laminate bending stiffness and then
% calculates the deflection of a simply supported plate.
%
clear; close all;
%
% Ply lay-up for a four ply laminate [(0/90)]s
%
theta(1)=0;
E1(1)=25e6; E2(1)=1e6; G12(1)=500000; v12(1)=0.25;
t(1)=0.0025;
theta(2)=0;
E1(2)=25e6; E2(2)=1e6; G12(2)=500000; v12(2)=0.25;
t(2)=0.0025;
theta(3)=90;
E1(3)=25e6; E2(3)=1e6; G12(3)=500000; v12(3)=0.25;
t(3)=0.0025;
theta(4)=90;
E1(4)=25e6; E2(4)=1e6; G12(4)=500000; v12(4)=0.25;
t(4)=0.0025;
theta(5)=0;
E1(5)=25e6; E2(5)=1e6; G12(5)=500000; v12(5)=0.25;
t(5)=0.0025;
theta(6)=90;
E1(6)=25e6; E2(6)=1e6; G12(6)=500000; v12(6)=0.25;
t(6)=0.0025;
theta(7)=90;
E1(7)=25e6; E2(7)=1e6; G12(7)=500000; v12(7)=0.25;
t(7)=0.0025;
theta(8)=0;
E1(8)=25e6; E2(8)=1e6; G12(8)=500000; v12(8)=0.25;
t(8)=0.0025;
theta(9)=90;
E1(9)=25e6; E2(9)=1e6; G12(9)=500000; v12(9)=0.25;
t(9)=0.0025;
theta(10)=0;
E1(10)=25e6; E2(10)=1e6; G12(10)=500000; v12(10)=0.25;
t(10)=0.0025;
% MIDPLANE
theta(11)=0;
t(11)=0.0025;
theta(12)=90;
t(12)=0.0025;
theta(13)=0;
t(13)=0.0025;
theta(14)=90;
t(14)=0.0025;
theta(15)=90;
t(15)=0.0025;
theta(16)=0;
t(16)=0.0025;
theta(17)=90;
t(17)=0.0025;
theta(18)=90;
t(18)=0.0025;
theta(19)=0;
t(19)=0.0025;
theta(20)=0;
t(20)=0.0025;
%
%
E1(11)=25e6; E2(11)=1e6; G12(11)=500000; v12(11)=0.25;
E1(12)=25e6; E2(12)=1e6; G12(12)=500000; v12(12)=0.25;
E1(13)=25e6; E2(13)=1e6; G12(13)=500000; v12(13)=0.25;
E1(14)=25e6; E2(14)=1e6; G12(14)=500000; v12(14)=0.25;
E1(15)=25e6; E2(15)=1e6; G12(15)=500000; v12(15)=0.25;
E1(16)=25e6; E2(16)=1e6; G12(16)=500000; v12(16)=0.25;
E1(17)=25e6; E2(17)=1e6; G12(17)=500000; v12(17)=0.25;
E1(18)=25e6; E2(18)=1e6; G12(18)=500000; v12(18)=0.25;
E1(19)=25e6; E2(19)=1e6; G12(19)=500000; v12(19)=0.25;
E1(20)=25e6; E2(20)=1e6; G12(20)=500000; v12(20)=0.25;
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% Define cell arrays to store the transformed stiffness matrices of
each
% individual ply
%
Qbar=cell(length(theta),1);
%
% Loop through all plies, calculate Qbar for each ply
%
for i=1:length(theta)
v21(i)=v12(i)*(E2(i)/E1(i));
S=[1/E1(i) -v21(i)/E2(i) 0; -v12(i)/E1(i) 1/E2(i) 0; 0 0 1/G12(i)];
c=cosd(theta(i));
s=sind(theta(i));
T=[c^2 s^2 2*c*s; s^2 c^2 -2*c*s; -c*s c*s (c^2 - s^2)];
Sbar=T'*S*T;
Qbar{i,1}=inv(Sbar);
end
%
% Calulate the z distances for each ply interface
%
total_thick=sum(t);
z(1)=-total_thick/2;
for k=2:length(theta)+1
z(k)=z(k-1)+t(k-1);
end
%
% Calculate the D matrix
%
D=[0 0 0; 0 0 0; 0 0 0];
for i=1:length(theta)
k=i+1;
Dply=(1/3)*Qbar{i,1}*(z(k)^3-z(k-1)^3);
D=D+Dply;
end
%
% Applied Transverse Loading
q0=1;
%
% Plate dimensions
a=10; b=10;
%
%**********************************************************************
***
m=[1 3 5 7 11];
n=[1 3 5 7 11];
% The deflection function expressed in Fourier Expansion
for i=1:length(m)
for j=1:length(n)
den=D(1,1)*(m(i)/a)^4 + 2*(D(1,2) +
2*D(3,3))*((m(i)*n(j))/(a*b))^2 + D(2,2)*(n(j)/b)^4;
wm(i,j)=16*q0/((pi^6)*m(i)*n(j)*den);
dw(i,j)=wm(i,j)*sin(m(i)*pi/2)*sin(n(j)*pi/2);
end
end
%
for k=1:4
wc(k)=sum(sum(dw(1:k,1:k)));
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end
%Outputs the deflection
wc(k)
% Outputs the Bending Stiffness Matrix to get D16 and D26 values
D
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Appendix E: Derivation of Deflection Solution for a Uniform Load
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