STATIC ANALYSIS OF CROSS - PLY LAMINATED METHOD

STATIC ANALYSIS OF CROSS - PLY LAMINATED METHOD
STATIC ANALYSIS OF CROSS - PLY LAMINATED
COMPOSITE PLATE USING FINITE ELEMENT
METHOD
A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE
REQUIREMENTS FOR THE DEGREE OF
Master of Technology in Mechanical Engineering
(Machine Design and Analysis Specialization)
By
Venkata Sai Gopal . K
Department of Mechanical Engineering
National Institute of Technology, Rourkela
Rourkela-769008(Orissa)
May 2007
STATIC ANALYSIS OF CROSS-PLY LAMINATED
COMPOSITE PLATE USING FINITE ELEMENT
METHOD
A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE
REQUIREMENTS FOR THE DEGREE OF
Master of Technology in Mechanical Engineering
(Machine Design and Analysis Specialization)
BY
Venkata Sai Gopal . K
Under the esteemed guidance of
Dr. N.Kavi
Supervisor
Professor
Dept.of Mechanical Engg.
N.I.T. ,Rourkela
Shri. B.Rambabu
Co-Supervisor
Scientist
SDSC-SHAR
Sriharikota
Department of Mechanical Engineering
National Institute of Technology, Rourkela
Rourkela-769008(Orissa)
May 2007
National Institute of Technology
Rourkela
CERTIFICATE
This is to certify that the thesis entitled “static analysis of Cross-Ply laminated
composite plate using Finite Element Method” submitted by Mr. Venkata sai Gopal .K ,
in partial fulfillment of the requirements for the degree of Master of Technology in
Mechanical Engineering with specialization in Machine Design and Analysis during
session 2005-2007 in the department of Mechanical Engineering, National Institute of
Technology, Rourkela (Deemed University) is an authentic work carried out by him under
my supervision and guidance.
To the best of my knowledge, the matter embodied in the thesis has not been
submitted to any other University/Institute for the award of any Degree or Diploma.
(Dr.Niranjan Kavi)
Date:
Professor
Dept. of Mechanical Engg.
National Institute of Technology
Rourkela
CERTIFICATE
This is to certify that the work in this thesis entitled “static analysis of
Cross-Ply laminated composite plate using Finite Element Method” by Mr.
Venkata Sai Gopal .K, has been carried out under my supervision in partial
fulfillment of the requirements for the degree of Master of Technology in
Mechanical Engineering with specialization in Machine Design and Analysis
during session 2006-2007 in the department of Solid Propellant Plant, Satish
Dhawan Space Centre, SHAR, Sriharikota.
(B.Rambabu)
Date:
Scientist
Solid Propellant Space Booster Plant
Satish Dhawan Space Centre
Sriharikota
ACKNOWLEDGEMENT
The author expresses his sincere gratitude and indebtedness to the thesis guide Dr.
Niranjan Kavi, Professor, Department of Mechanical Engineering, N.I.T., Rourkela for
proposing this area for research, for his valuable guidance, encouragement and moral
support for the successful completion of this work. His kind attitude always encouraged
the author to carry out the present work firmly.
The author is thankful to his co-supervisor and colleague at SDSC, SHAR, Shri
B.Rambabu for his encouragement and invaluable suggestions in the enhancement of the
present work.
The author remains grateful to Dr.B.K.Nanda, Head of the Department, Department of
Mechanical Engineering, for his kind approval to continue the 4th semester thesis work at
Satish Dhawan Space Centre, ISRO.
Thanks are due to all the friends of the author, who are involved directly or indirectly in
successful completion of the present work.
Rourkela
Date:
(Venkata Sai Gopal. K)
ABSTRACT
Finite element Analysis is carried out to perform static analysis on a cross-ply laminated
composite square plate based on the First order Shear Deformation Theory (FSDT). The
theory accounts for constant variation of transverse shear stresses across the thickness of
the laminate; and it uses a shear correction factor .The element formulated is an 8-noded
iso-parametric quadratic (Serendipity) element. In this analysis, the square plate is
analyzed for transverse loading viz., sinusoidal varying load and uniformly distributed
load under simply supported boundary conditions. A program is written in MATLAB to
obtain the finite element solutions for transverse displacements, normal stresses and
transverse shear stresses. Solutions are obtained for 3, 4 and 5 layers of the laminate with
cross-ply orientation for different values of side to thickness ratios. Reduced integration
scheme is adopted to alleviate shear locking effects. Stresses are found at Gauss points
from constitutive relations. The solutions are compared with closed form solutions of
FSDT, 3D elasticity solutions and Classical Laminate Plate Theory (CLPT) solutions. It is
observed that the results are in close agreement with the available solutions. The element
being a C0 continuous element, it ensures the continuity of generalized displacements only;
not strains and thus stresses. The model is validated by its good convergence with the
analytical results. Analysis can be done on thin as well as moderately thick plates
satisfactorily by using this model.
CONTENTS
Page No
Abstract
i
List of Figures
iv
List of Tables
v
Chapter-1
INTRODUCTION
1.1 Introduction
2
1.2 Objective of thesis
3
Chapter -2 LITERATURE SURVEY
2.1 Literature Survey
5
Chapter - 3 FIRST ORDER SHEAR DEFORMATION THEORY
3.1 Introduction
8
3.2 Kinematic relations
10
3.3 Constitutive relations
12
3.4 Virtual work statement
15
Chapter - 4 FINITE ELEMENT METHOD
4.1 Introduction
20
4.2 Principle of virtual displacement
21
4.3 Stiffness matrix derivation
22
4.4 Post computation of stresses and strains
29
ii
Chapter – 5 RESULTS AND DISCUSSION
5.1 Results
32
5.2 Discussion
41
5.3 Observations
42
Chapter -6 CONCLUSIONS
45
Chapter -7 SCOPE FOR FUTURE WORK
46
REFERENCES
48
APPENDIX
51
iii
List of Figures
1)Non dimensionalized central transverse deflection versus side to thickness ratio for
simply supported cross ply (0/90/90/0) square laminate subjected to uniformly distributed
loading
2) Non dimensionalized normal stress sigma xx versus side to thickness ratio for simply
supported cross ply (0/90/90/0) square laminate subjected to uniformly distributed loading
3) Non dimensionalized normal stress sigma yy versus side to thickness ratio for simply
supported cross ply (0/90/90/0) square laminate subjected to uniformly distributed loading
4) Non dimensionalized transverse shear stress sigma xz versus side to thickness ratio for
simply supported cross ply (0/90/90/0) square laminate subjected to uniformly distributed
loading
5) Non dimensionalized transverse shear stress sigma yz versus side to thickness ratio for
simply supported cross ply (0/90/90/0) square laminate subjected to uniformly distributed
Loading
6) Non dimensionalized central transverse deflection versus side to thickness ratio for
simply supported cross ply (0/90/90/0) square laminate subjected to sinusoidal varying
load
7) Non dimensionalized normal stress sigma xx versus side to thickness ratio for simply
supported cross ply (0/90/90/0) square laminate subjected to sinusoidal varying load
8) Non dimensionalized normal stress sigma yy versus side to thickness ratio for simply
supported cross ply (0/90/90/0) square laminate subjected to sinusoidal varying load
9) Non dimensionalized transverse shear stress sigma xz versus side to thickness ratio for
simply supported cross ply (0/90/90/0) square laminate subjected to sinusoidal varying
load
10) Non dimensionalized transverse shear stress sigma yz versus side to thickness ratio for
simply supported cross ply (0/90/90/0) square laminate subjected to sinusoidal varying
load
iv
List of Tables
1) Non dimensionalized maximum deflection and stresses of simply supported cross-ply
(0/90/0) square plate subjected to uniformly distributed loading
2) Non dimensionalized maximum deflection and stresses of simply supported cross-ply
(0/90/90/0) square plate subjected to uniformly distributed loading
3) Non dimensionalized maximum deflection and stresses of simply supported cross-ply
(0/90/0/90/0) square plate subjected to uniformly distributed loading
4) Non dimensionalized maximum deflection and stresses of simply supported cross-ply
(0/90/0) square plate subjected to sinusoidal loading
5) Non dimensionalized maximum deflection and stresses of simply supported cross-ply
(0/90/90/0) square plate subjected to sinusoidal loading
6) Non dimensionalized maximum deflection and stresses of simply supported cross-ply
(0/90/0/90/0) square plate subjected to sinusoidal loading
v
Chapter-1
INTRODUCTION
1.1 INTRODUCTION:
Composite materials are increasingly used in aerospace, under water, and automotive
structures and space structures. The application of composite materials to engineering
components has spurred a major effort to analyze structural components made from them
Composite materials provide unique advantages over their metallic counterparts, but they also
present complex and challenging problems to analysts and designers. To take advantage of
full potential of composite materials, accurate models and design methods are required the
most common structural elements are plates and shells. An accurate modeling of stress fields
is of paramount importance in the design of such components. Laminated composites are one
of the classifications of the composites which are used in structural elements like leaf springs,
automobile drive shafts, and gears, and axles.
The Navier, Levy, and Rayleigh-Ritz developed solutions to composite beams and plate
problems. However, exact analytical or variational solutions to these problems cannot be
developed when complex geometries, arbitrary boundary conditions or nonlinearities ar
involved. Therefore one must resort to approximate methods of analysis that are capable of
solving such problems. There are several theories available to describe the kinematics of the
laminates. Classical Laminate Plate Theory and First order Shear Deformation Theory are
some among them.
The Finite element Method is such an approximate method and powerful numerical technique
for the solution of differential and integral equations that arise in various fields of engineering
and applied science. FEM is an effective method of obtaining numerical solutions to boundary
value, initial value and eigen value problems.
2
1.2Objective of the thesis:
The objective of the present work is to determine transverse displacements, normal stresses,
and transverse shear stresses of 3, 4 and 5 layered cross ply laminated composite square plate
subjected to transverse loading viz., sinusoidal varying load and uniformly distributed load
when it is simply supported at the edges.
In this present work, a displacement based finite element model is formulated based on First
order Shear Deformation Theory. It is an iso parametric element with 8 nodes and 3 degrees
of freedom at each node. The 3 degrees of freedom are; transverse displacement, rotation
about x and rotation about y axes.
3
Chapter- 2
LITERATURE SURVEY
4
2.1Literature Survey:
Closed form solutions to 3D elasticity problem of laminated structures are scarce and limited
in scope. Pagano developed solutions for simply supported rectangular plates with symmetric
lamination undergoing cylindrical and bidirectional bending; the fiber orientation is 0° and
90°. Ren (1987) has extended the cylindrical bending solution to infinitely long cylindrical
shells. Noor and Burton (1990a), (1992) have provided solutions for the bending, buckling
and vibration of anti-symmetrically laminated rectangular plates, periodic in the in-plane
directions. Savithri and Varadan (1992) studied plates under uniformly distributed and
concentrated loads. All these approaches use Fourier expansions in the in-plane directions
resulting in sets of ordinary differential equations with constant coefficients, which can be
solved exactly. The unknown coefficients of the solutions are determined by boundary and
interface conditions in thickness direction. J.N.Reddy and W.C.Chao studied and derived
Closed form solutions and Finite Element Solutions for Laminated Anisotropic Rectangular
Plates.[1]. AA.Khedier and J.N.Reddy derived Exact solutions for bending of thin and thick
cross-ply laminated beams[2]. T.J.R.Hughes and T.E.Tezduyar developed Four-Node Bilinear
Isoparametric Element based upon Mindlin Plate Theory [3]. The state-space concept in
conjunction with the Jordan canonical form is presented to solve the governing equations for
the bending of cross-ply laminated composite beams by J.N.Reddy1997[4].An elastic-plastic
stress analysis was carried out on simply supported and clamped aluminum metal-matrix
laminated plates. The thin plate model is also used for the study of rectangular plates with
practically important mixed edge constraints by Liew, Hua & Lim and Laura & Gutierrez.A
composite material model is presented to analyze progressive failure in composite structures
by Raimondo Luciano,Raffaele Zinno in 2000. Song Cen, Zhen-Han Yao developed a new
4-node quadrilateral finite element for the analysis of composite plates in 2002. A new
analytical method was developed by M.R.Khalili and R.K.Mittal to analyze the response of
laminated composite plates subjected to static and dynamic loading.(2005).B.N.Pandya and
T.Kant worked on Finite Element analysis of Laminated Composite Plates Using a Higher –
Order Displacement Model [5].B.R.Somashekar,G.Pratap and C.Ramesh Babu developed a
simple and efficient Four nodded, Laminated Anisotropic Plate Element[6]. Xiao-Ping
Shu,Kostas P.Soldatos determined Stress distributions in angle-ply laminated plates subjected
to cylindrical bending[7].
5
Exact solutions for rectangular bidirectional composites and sandwich plates were developed
by Pagano,N.J.,[8].Reddy, J.N., Khdeir, A.A. developed Levy type solutions for
symmetrically laminated rectangular plates using first order shear deformation theory[9].An
exact approach to the elastic stat of stress of shear deformable antisymmetic angle ply
laminated plates was developed by Khedeir A. A., [10]. He also compared shear deformable
and kirchoff theories for bending buckling and vibration of antisymmetric angle ply laminated
plates.[11].Srinivas and Jogarao got some results from exact analysis of thick laminates in
vibration and buckling[12]. A review is made on plate bending finite elements by Hrabok,
M.M. and Hrudey [13].Fraeijis de veubeke developed a conforming finite element for plate
bending [14]. A triangular refined plate bending element was suggested by Bell K. [15]. Irons
B.M. developed a conforming quartic triangular element for plate bending [16].Strcklin, J.a.
haisler, w developed a rapidly converging triangular plate element [17]. A study was made on
3 node triangular plate bending element by Batoz, J.L and Bathe, K.J. [18].
6
Chapter-3
FIRST ORDER SHEAR DEFORMATION THEORY
7
3.1Introduction:
The use of composite materials in structural components is increasing due to their
attractive properties such as high strength-to-weight ratio, ability to tailor the structural
properties, etc. Plate structures find numerous applications in the aerospace, military and
automotive industries. The effects of transverse shear deformation are considerable for
composite structures, because of their high ratio of extensional modulus to transverse shear
modulus. Most of the structural theories used to characterize the behavior of composite
laminates fall into the category of equivalent single layer (ESL) theories. In these theories, the
material properties of the constituent layers are combined to form a hypothetical single layer
whose properties are equivalent to through the thickness integrated sum of its constituents.
This category of theories has been found to be adequate in predicting global response
characteristics of laminates, like maximum deflections, maximum stresses, and fundamental
frequencies, or critical buckling loads
In the context of ESL theories, the simplest one is the CLT which neglects the shear
contribution in the laminate thickness. However, flat structures made of fiber-reinforced
composite materials are characterized by non negligible shear deformations in the thickness
direction, since the longitudinal elastic modulus of the lamina can much higher than the shear
and the transversal moduli; hence the use of a shear deformation laminate theory is
recommended. The extension of the Reissner and Mindlin model to the case of laminated
anisotropic plates, i.e. FSDT ,accounts for shear deformation along the thickness in the
simplest way. It gives satisfactory results for a wide class of structural problems, even for
moderately thick laminates, requiring only C0-continuity for the displacement field. The
transverse shearing strains (stresses) are assumed to be constant along the plate thickness so
that stress boundary conditions on the top and the bottom of the plate are violated; shear
correction factors must be introduced. The determination of shear correction factors is not a
trivial task, since they depend both on the lamination sequence and on the state of
deformation .
8
Assumptions:
1) The layers are perfectly bonded
2) The material of each layer is linearly elastic and has two planes of material symmetry
3) The strains and displacements are small
4) Deflection is wholly due to bending strains only
5) Plane sections originally perpendicular to the longitudinal plane of the plate remain
plane, but not necessarily perpendicular to longitudinal plane
6) The transverse shearing strains (stresses) are assumed to be constant along the plate
thickness
9
3.2Kinematic relations:
The displacement field of the first-order theory is of the form
u ( x , y , z , t ) = u 0 ( x , y , t ) + zφ x ( x , y , t )
v ( x , y , z , t ) = v 0 ( x , y , t ) + z φ y ( x , y , t ) ----------------------- (1)
w (x, y, z,t) = w0 (x, y,t)
Where (u0 , v0 , w0 , φ x , φ y ) are unknown functions, called the generalized displacements.
(u0 , v0 , w0 ) , denote the displacements of a point on the plane z = 0.
∂u
∂v
= φx,
= φy
∂z
∂z
Indicate that φx and φ y are the rotations of the transverse normal about the y- and x- axes,
respectively, owing to bending only.
The strains associated with the displacement field (1) are given by:
ε xx =
⎛ ∂u ∂v ⎞
∂v
∂w
∂u
∂w
∂w
, ε yy =
, ε zz =
, γ xy = ⎜ + ⎟ , γ xz =
+ φ y -------(2)
+ φx , γ yz =
∂y
∂y
∂x
∂z
∂x
⎝ ∂y ∂x ⎠
Substituting the expressions for u,v,w from eq. (1) in equation (2) gives:
ε xx =
∂φ y
∂v
∂u0
∂φ
, ε zz = 0
+ z x , ε yy = 0 + z
∂y
∂y
∂x
∂x
⎛ ∂u ∂v ⎞ ⎛ ∂φ ∂φ ⎞
∂w
∂w
γ xy = ⎜ 0 + 0 ⎟ + z ⎜ x + y ⎟ , γ xz = 0 + φx , γ yz = 0 + φ y -------------------- (3)
∂y
∂x
∂x ⎠
⎝ ∂y ∂x ⎠ ⎝ ∂y
In Matrix form the above equations are given by
10
⎧ ∂u0 ⎫
⎪ ∂x
⎪ ⎧
∂φx
⎪
⎪ ⎪
∂x
∂v0 ⎪ ⎪
⎧ε xx ⎫ ⎪
∂φ y
⎪ ⎪
⎪ε ⎪ ⎪ ∂y
⎪
⎪
⎪⎪ yy ⎪⎪ ⎪ ∂u ∂v ⎪ ⎪⎪
∂y
0
+ 0 ⎬+ z⎨
⎨γ xy ⎬ = ⎨
∂φ
⎪γ ⎪ ⎪ ∂y ∂x ⎪ ⎪⎜⎛ ∂φx + y
xz
⎪ ⎪ ⎪ ∂w0
⎪ ⎪ ∂y
∂x
+ φx ⎪ ⎪⎝
⎪⎩γ yz ⎪⎭ ⎪
0
⎪ ∂x
⎪ ⎪
∂
w
⎪ 0
⎪ ⎪
0
⎪ ∂y + φ y ⎪ ⎩
⎩
⎭
⎫
⎪
⎪
⎪
⎪
⎪
⎬
⎞⎪
⎟⎪
⎠
⎪
⎪
⎪⎭
The above matrix is in the form:
⎧ε xx ⎫ ⎧ε xx0 ⎫
⎧ kx ⎫
⎪ε ⎪ ⎪ 0 ⎪
⎪k ⎪
y
⎪⎪ yy ⎪⎪ ⎪⎪ε xx ⎪⎪
⎪⎪ ⎪⎪
0
γ
⎨ xy ⎬ = ⎨γ xy ⎬ + z ⎨k xy ⎬
⎪γ ⎪ ⎪γ 0 ⎪
⎪0⎪
⎪ xz ⎪ ⎪ xz0 ⎪
⎪ ⎪
⎪⎩γ yz ⎪⎭ ⎪⎩γ yz ⎪⎭
⎪⎩ 0 ⎪⎭
For plate bending problem, the in- plane displacements (u,v) are uncoupled from ( w0 , φ x , φ y ) .
Hence, the equations reduce as follows:
∂φ
⎧
z x
⎪
∂x
⎧ zk x ⎫ ⎪
∂
φ
⎪
z y
⎧ε xx ⎫ ⎪ zk y ⎪ ⎪
⎪
∂y
⎪ε ⎪ ⎪
⎪
⎪
⎪
yy
zk
⎪⎪ ⎪⎪ ⎪
xy
⎪ ⎪ ⎛ ∂φx ∂φ y
+
⎨γ xy ⎬ = ⎨
∂w0 ⎬ = ⎨ z ⎜
∂x
⎪γ ⎪ ⎪φx + ∂x ⎪ ⎪ ⎝ ∂y
xz
⎪ ⎪ ⎪
⎪ ⎪
∂w0 ⎪ ⎪ φx + ∂w0
⎪⎩γ yz ⎪⎭ ⎪
∂x
⎪φ y + ∂y ⎪ ⎪
⎩
⎭ ⎪
∂w0
⎪ φy +
∂y
⎩
11
⎫
⎪
⎪
⎪
⎪
⎪
⎞⎪
⎟ ⎬ ------------------(4)
⎠⎪
⎪
⎪
⎪
⎪
⎪
⎭
3.3Constitutive Relations:
The Stress-Strain relations for a typical lamina k with reference to the lamina co-ordinate axes
(1-2-3) are given by
k
k
k
⎧σ 1 ⎫ ⎡ Q11 Q12 0 ⎤ ⎧ ε1 ⎫
⎪ ⎪ ⎢
⎥ ⎪ ⎪
⎨σ 2 ⎬ = ⎢Q21 Q22 0 ⎥ ⎨ ε 2 ⎬
⎪τ ⎪ ⎢ 0
0 Q33 ⎥⎦ ⎩⎪γ 12 ⎭⎪
⎩ 12 ⎭ ⎣
k
⎧τ 23 ⎫
⎡Q44
⎨ ⎬ =K⎢
⎩τ 13 ⎭
⎣ 0
In which
(σ 1 , σ 2 ,τ12 ,τ 23 ,τ13 )
k
k
0 ⎤ ⎧γ 23 ⎫
⎨ ⎬ ---------------------(5)
Q55 ⎥⎦ ⎩γ 13 ⎭
are the stress and
(ε1 , ε 2 , γ 12 , γ 23 , γ 13 ) are
the linear strain
components referred to the lamina co-ordinate axes (1-2-3). The Qij’s are the plane stress
reduced elastic constants of the kth lamina and the following relations hold between these and
the engineering constants.
Q11 =
E1
1 −ν 12ν 21
Q22 =
,
E2
1 − ν 12ν 21
Q33 = G12 ,
Q12 =
,
ν 12 E2
1 − ν 12ν 21
=
ν 21 E1
1 − ν 12ν 21
,
Q44 = G23 , Q55 = G13 .
The stress-strain relations are the basis for the stiffness and stress analysis of an individual
lamina subjected to forces in its own plane. The relations are therefore indispensable in the
analysis of laminates. There are 4 independent material properties, E1 , E2 , G12 , and ν 12 the
reciprocal relation is given by
ν 12
E1
=
ν 21
E2
Stress – Strain relations for a lamina of arbitrary orientation
From elementary mechanics of materials the transformation equations for expressing stresses
in 1-2 coordinate system(principal coordinate system) in terms of stresses in x-y coordinate
system.
2
⎧σ 1 ⎫ ⎡ cos θ
⎪ ⎪ ⎢
2
⎨σ 2 ⎬ = ⎢ sin θ
⎪τ ⎪ ⎢ − sin θ cos θ
⎩ 12 ⎭ ⎣
sin 2 θ
cos 2 θ
sin θ cos θ
2sin θ cos θ ⎤ ⎧σ X ⎫
⎥⎪ ⎪
−2sin θ cos θ ⎥ ⎨σ y ⎬ -------------------(6)
cos 2 θ − sin 2 θ ⎥⎦ ⎪τ xy ⎪
⎩ ⎭
12
Where θ is the angle from the x-axis to the axis 1. Following the usual transformation of
Stress-Strain between the lamina and laminate coordinate systems, the Stress-Strain relations
for the kth lamina in the laminate coordinates (x,y,z) are written as:
k
⎧σ x ⎫ ⎡ Q11 Q12
⎪ ⎪ ⎢
⎨σ y ⎬ = ⎢Q21 Q22
⎪ ⎪ ⎢
⎩τ xy ⎭ ⎢⎣Q31 Q32
k
⎡Q44
⎧τ yz ⎫
⎨ ⎬ =K⎢
⎩τ xz ⎭
⎢⎣Q54
Q13 ⎤
⎥
Q23 ⎥
⎥
Q33 ⎦⎥
k
⎧εx ⎫
⎪ ⎪
⎨ε y ⎬
⎪γ ⎪
⎩ xy ⎭
k
k
k
Q45 ⎤ ⎧γ yz ⎫
⎥ ⎨ ⎬ -----------------------(7)
Q55 ⎥⎦ ⎩γ xz ⎭
in which , σ = (σ x , σ y , τ xy , τ yz , τ xz ) and ε = ( ε x , ε y , γ xy , γ yz , γ xz ) are the stress and linear
t
t
strain vectors with respect to the laminate axes and
Qij s are the plane stress reduced elastic
constants in the plate (laminate) axes of the kth lamina given in Appendix. The superscript t
denotes the transpose of a matrix. K refers to the Shear Correction Factor used in FSDT.
Normally its value is 5/6.
Stress Resultants:
The resultant forces and moments acting on a laminate are obtained by integration of the
stresses in each layer or lamina through the laminate thickness, for example,
t /2
N
x
=
∫
t/2
σ xdz
M
x
=
∫
σ x zd z
−t / 2
−t / 2
Actually, N x is a force per unit length (width) of the cross section of the laminate. Similarly,
M x is a moment per unit length. The entire collection of force and moment resultants for an
N-layered laminate is shown and is defined as
k
k
⎧ Nx ⎫
⎧σ x ⎫
⎧σ x ⎫
zk
t/2
N
⎪ ⎪
⎪ ⎪
⎪ ⎪
⎨Ny ⎬ = ∫ ⎨σ y ⎬ dz = ∑ ∫ ⎨σ y ⎬ dz
k =1 zk−1 ⎪
⎪ ⎪ −t / 2 ⎪ ⎪
⎪
N
τ
τ
⎩ xy ⎭
⎩ xy ⎭
⎩ xy ⎭
13
k
k
⎧M x ⎫
⎧σ x ⎫
⎧σ x ⎫
zk
t/2
N
⎪
⎪
⎪ ⎪
⎪ ⎪
M
zdz
σ
=
=
⎨ y ⎬ ∫ ⎨ y⎬
⎨σ y ⎬ zdz -----------------(8)
∑
∫
k =1 zk −1 ⎪
⎪
⎪ −t / 2 ⎪ ⎪
⎪
M
τ
τ
⎩ xy ⎭
⎩ xy ⎭
⎩ xy ⎭
Where, Z k and Z k-1 are defined in figure shown in appendix. Z 0 = - t/2.These force and
moment resultants do not depend on Z after integration, but are functions of x and y, the
coordinates in the plane of the laminate middle surface.
k
⎡Q11 Q12 Q13 ⎤ ⎧ ⎧ε 0 x ⎫
⎧Nx ⎫
⎧kx ⎫ ⎫
z
z
k
k
N
⎢
⎥ ⎪⎪ ⎪⎪ 0 ⎪⎪
⎪ ⎪
⎪ ⎪ ⎪⎪
N
Q
Q
Q
dz
=
+
ε
⎨ y ⎬ ∑ ⎢ 21
⎨k y ⎬zdz ⎬
22
23 ⎥ ⎨ ∫ ⎨
y ⎬
∫
⎥ ⎪ zk−1 ⎪ 0 ⎪
zk −1 ⎪
⎪ ⎪ k =1 ⎢
⎪ ⎪
N
k
Q
Q
Q
γ
xy
xy
32
33 ⎥
⎢⎣ 31
⎩ ⎭
⎩ ⎭ ⎪⎭
⎦ ⎪⎩ ⎪⎩ xy ⎪⎭
k
⎫
⎡Q11 Q12 Q13 ⎤ ⎧ ⎧ε 0 x ⎫
⎧M x ⎫
⎧kx ⎫
z
z
k
k
N
⎢
⎥ ⎪⎪ ⎪⎪ 0 ⎪⎪
⎪ ⎪
⎪ ⎪ 2 ⎪⎪
=
+
M
Q
Q
Q
zdz
ε
⎨ y ⎬ ∑ ⎢ 21 22
⎨ky ⎬z dz ⎬
23 ⎥ ⎨ ∫ ⎨ y ⎬
∫
⎥ ⎪zk−1 ⎪ 0 ⎪
zk −1 ⎪
⎪ ⎪ k =1 ⎢
⎪
⎪
M
k
Q
Q
Q
γ
xy
xy
31
32
33
⎢⎣
⎥⎦ ⎪⎩ ⎩⎪ xy ⎭⎪
⎩ ⎭
⎩ ⎭
⎪⎭
Thus the above equations can be written as
⎧Nx ⎫
⎡ A11
⎪ ⎪ N ⎢
⎨ N y ⎬ = ∑ ⎢ A12
⎪ ⎪ k =1 ⎢ A
⎣ 31
⎩ N xy ⎭
A12
A22
A32
⎧M x ⎫
⎡ B11 B12
⎪
⎪ N ⎢
⎨M y ⎬ = ∑ ⎢ B12 B22
⎪
⎪ k =1 ⎢ B B
M
32
⎣ 31
⎩ xy ⎭
0
A13 ⎤ ⎧ε x ⎫ ⎡ B11 B12
⎪⎪
⎪⎪
A23 ⎥⎥ ⎨ε 0 y ⎬ + ⎢⎢ B12 B22
A33 ⎥⎦ ⎪γ 0 ⎪ ⎢⎣ B31 B32
⎩⎪ xy ⎭⎪
B13 ⎤ ⎧kx ⎫
⎪ ⎪
B23 ⎥⎥ ⎨k y ⎬
B33 ⎥⎦ ⎪kxy ⎪
⎩ ⎭
0
B13 ⎤ ⎧ε x ⎫ ⎡ D11 D12
⎪⎪
⎪⎪
B23 ⎥⎥ ⎨ε 0 y ⎬ + ⎢⎢ D12 D22
B33 ⎥⎦ ⎪γ 0 ⎪ ⎢⎣ D31 D32
⎪⎩ xy ⎪⎭
D13 ⎤ ⎧kx ⎫
⎪ ⎪
D23 ⎥⎥ ⎨k y ⎬
D33 ⎥⎦ ⎪kxy ⎪
⎩ ⎭
Where
N
( )
Aij = ∑ Qij
k =1
( Z k − Z k −1 )
k
Bij =
( ) (Z
1 N
∑ Qij
2 k =1
2
k
− Z 2 k −1 ) Dij =
k
Where
Aij Coefficients are called extensional stiff nesses,
Bij Coefficients are called coupling stiff nesses,
14
( ) (Z
1 N
∑ Qij
3 k =1
k
3
k
− Z 3k −1 ) ---(9)
Dij Coefficients are called bending stiff nesses.
In this present work , as only cross ply laminates only analyzed, the Bij terms get vanished.
D16=D26=0.
As there present no inplane forces, they are uncoupled from the equations. So only D matrix
is used in this plate bending analysis.
3.4Virtual Work Statement
The variational formuations form a powerful basis for obtaining approximate solutions
to real world/practical problems. The variational method uses the variational principles, such
as the Principle of Virtual displacements, to determine approximate displacements as
continuous functions of position in the domain. In the Classical sense, variational principle
has to do with the minimization of a functional, which includes all the intrinsic features of the
problem, such as the governing equations, boundary and /or initial conditions, and constraint
conditions.
One of the concepts of Variational formulation is Principle of virtual work. It is the
work done on a particle or a deformable body by actual forces in displacing the particle or the
body through a hypothetical displacement that is consistent with the geometric constraints.
The applied forces are kept constant during the virtual displacement. The Principle of virtual
displacement states that the virtual work done by actual forces in moving through virtual
displacements is zero if the body is in equilibrium.
The principle of virtual work states that “a continuous body is in equilibrium if the
virtual work of all forces acting on the body is zero in a virtual displacement”. The principle
of virtual work is independent of any constitutive law and applies to elastic (linear and non
linear) and in elastic continuum problems. The plate to be analyzed may have curved or
straight boundaries as well as different boundary conditions. The principle of virtual work
statement for the plate can be stated as
δ WI + δ WE = 0
Where
δWI = Virtual work resulting from internal forces
15
δWE = Virtual work resulting from external forces
The governing equations of the first order shear deformation theory are derived using the
dynamic version of the principle of virtual displacements for the displacements (w, φx , φ y ):
T
0 =
∫ (δ K
− ( δ U + δ V ) )d t --------------(10)
0
δWI = δU and
Here
δWE = δK+δV
Where δU is virtual strain energy, δV is virtual work done by applied forces and δK is virtual
kinetic energy. On substituting the expressions for δU, δV and δK in the equation noA. we
get
⎛ ⎪⎧
∂ 2φ y
∂ 2φ x
∂ 2 w ⎪⎫
2
2
+ ρ z δφ y
+ ρδ w
⎜ ⎨ ρ z δφ x
⎬
∂t
∂t
∂ t ⎭⎪
⎜ ⎩⎪
⎜
0 = ∫ ⎜ + {δ ε x x σ x x + δ ε y y σ y y + 2 δ ε x y σ x y + 2 δ ε x z σ x z + 2 δ ε y z σ
Ve
⎜
⎜ − ⎧ δ w qdxdy ⎫
⎬
⎜ ⎨∫
e
Ω
⎩
⎭
⎝
Carrying out integration with respect to z, we get
0=
∫[
h/2
∫
Ωe − h / 2
ρ z 2δφx
∂ 2φ y h / 2
∂ 2φx h / 2 2
∂2w
z
w
+
ρ
δφ
+
ρδ
y
∂t 2 − h∫/ 2
∂t 2 − h∫/ 2
∂t 2
h/2
h/2
∂δφ y
⎛ ∂δφx ∂δφ y
∂δφx
z
z
σ
σ
+
+
+
⎜
xx
yy
∫
∫
∫
∂x
∂y
∂y
∂x
−h / 2
−h / 2
−h/ 2 ⎝
h/2
+
⎞
⎟σ xy z
⎠
h/2
h/2
⎛
∂δ w ⎞
∂δ w ⎞
⎛
+ ∫ ⎜ δφx +
⎟σ yz z − ∫ qδ w]dxdydz
⎟σ xz z + ∫ ⎜ δφ y +
∂x ⎠
∂y ⎠
−h / 2 ⎝
−h / 2 ⎝
−h / 2
h/2
On simplification the above equation yields to
16
⎞
⎟
⎟
⎟
yz } ⎟
⎟
⎟
⎟
⎠
⎡
⎤
⎛
∂ 2φ y ⎞
∂δφ y
∂ 2φx
∂δφx
∂2w
+ M yy
⎢ I 0δ w 2 + I 2 ⎜⎜ δφx 2 + δφ y 2 ⎟⎟ + M xx
⎥
∂t
∂t
∂t ⎠
∂x
∂y
⎢
⎥
⎝
0= ∫ ⎢
⎥dxdy
δφ
∂
⎛ ∂δφx
⎛
Ωe ⎢
∂δ w ⎞
∂δ w ⎞
⎛
y ⎞
⎥
+ M xy ⎜
+
⎟ + Qx ⎜ δφx +
⎟ − qδ w⎥
⎟ + Qy ⎜ δφ y +
⎢
y
x
x
y
∂
∂
∂
∂
⎝
⎠
⎝
⎠
⎝
⎠
⎣
⎦
Where
h/2
I0 =
∫
h/2
ρ dz , I 2 =
−h / 2
ρ z dz , Qx = K
−h / 2
h/2
M xx =
∫
∫
∫
σ xz dz , Qy = K
−h / 2
h/2
σ xx zdz ,
∫
M yy =
−h / 2
Where,
h/2
2
σ yy zdz ,
h/2
∫
σ yz dz ,
−h / 2
h/2
M xy =
−h / 2
∫
σ xy zdz
−h / 2
I0 is the mass moment of inertia term,
I2 is the rotary inertia term and
K is the shear correction factor.
For Static case, the above virtual displacement equation becomes,
0=
⎡
∫ ⎣⎢ M
Ωe
xx
∂δφ y
⎛ ∂δφx ∂δφ y
∂δφx
+ M yy
+ M xy ⎜
+
∂x
∂y
∂x
⎝ ∂y
⎤
⎞
⎛
∂δ w ⎞
∂δ w ⎞
⎛
⎟ + Qx ⎜ δφx +
⎟ − qδ w⎥dxdy
⎟ + Qy ⎜ δφ y +
∂x ⎠
∂y ⎠
⎝
⎝
⎠
⎦
The Virtual Work Statement contains 3 weak forms for the 3 displacements
(w, φx , φ y ).They are identified by collecting the terms involving δw , δ φx and δ φ y separately
and equating them to zero:
0=
⎛
∫ ⎜⎝ Q
x
Ωe
0=
⎛
∫ ⎜⎝ M
xx
Ωe
0=
Ωe
⎞
∂δφx
∂δφ x
+ M xy
+ Qxδφx ⎟dxdy
∂x
∂y
⎠
∂δφ y
⎛
∫ ⎜⎝ M
⎞
∂δ w
∂δ w
+ Qy
− qδ w ⎟dxdy
∂x
∂y
⎠
xy
∂x
+ M yy
∂δφ y
⎞
+ Qyδφ y ⎟dxdy
∂y
⎠
The governing equations of FSDT are obtained from the weak forms given above
17
∂Qx ∂Qy
+
+q =0
∂x
∂y
∂M xx ∂M xy
+
− Qx = 0 -----------------(11)
∂x
∂y
∂M xy ∂M yy
+
− Qy = 0
∂x
∂y
18
Chapter-4
FINITE ELEMENT METHOD
19
4.1Introduction:
The finite Element Method is a powerful computational technique for the solution of
differential and integral equations that arise in various fields of engineering and applied
science. The method is a generalization of the Classical Variational (i.e., the Rayleith-Ritz)
and weighted – residual (Galerkin, Least-squares etc.)methods. Since most real-world
problems are defined on domains that are geometrically complex and may have different
typers of boundary conditions on different portions of the boundary of the domain, it is
difficult to generate approximation functions required in the traditional variational methods.
The basic idea of the finite element method is to view a given domain as an assemblage of
simple geometric shapes called finite elements, for which it is possible to systematically
generate the approximation functions needed in the solution methods. The ability to represent
domains with irregular geometries by a collection of finite elements makes the method a
valuable practical tool for the solution of boundary, initial, and eigen value problems arising
in various fields of engineering. The approximation functions are often constructed using
ideas from interpolation theory, and hence they are also called interpolation functions. Thus,
the finite element method is a piecewise application of the variational and weighted –residual
methods.
Finite Element Modelling
In the FEM, the total solution domain is discretized in to N elements (sub-domains).Then, the
finite element model of the problem is developed using variational method .The variational
formuations form a powerful basis for obtaining approximate solutions to practical problems.
The variational method uses the variational principles, such as the Principle of Virtual
displacements, to determine approximate displacements as continuous functions of position in
the domain. In the Classical sense, variational principle has to do with the minimization of a
functional, which includes all the intrinsic features of the problem, such as the governing
equations, boundary and /or initial conditions, and constraint conditions.
20
4.2Principle of Virtual Displacements: It states that “a deformable body is in equilibrium if
the total external virtual work is equal to the total internal virtual work for every virtual
displacement satisfying the kinematic boundary conditions”.
δ WI = δ WE
Where
δWI = Virtual work resulting from internal forces
δWE = Virtual work resulting from external forces
The Principle of Virtual work for the plate can be stated as
∫ δε
V
T
σ dv + ∫ δγ Tτ dv = ∫ qδ ddA ------------(12)
V
V
The integration over the thickness reduces eq 12 as follows:
∫ {δε
σ xx + δε Tyyσ yy + δγ xyT σ xy + δγ xzT σ xz + δγ Tyzσ yz }dA -------------------(13)
T
xx
A
Where A is the cross-sectional area and V is the volume of the plate. Using the lamina
constitutive relation eq 12 leads to the following form:
∫ ⎡⎣ ε {M } + Φ {Q }⎤⎦ d A
t
t
A
Replacing the stress resultants by the product of rigidity matrix and strains in the strain energy
expression in equation-------------, we get
∫ {δε
T
xx
DBε xx + δε Tyy DBε yy + δγ xyT DBγ xy + δγ xzT Dsγ xz + δγ Tyz Dsγ yz }dA -----------------(14)
A
Which is the final form of the virtual work principle as it is required for finite element
calculations.
21
4.3Stiffness matrix derivation :
In this work, an eight noded isoparametric element (Serendipity Element) is chosen to
discretize the plate domain. The variation of displacement u is expressed by the polynomial in
natural coordinates as:
u = α1 + α 2 r + α 3 s + α 4 r 2 + α 5 rs + α 6 s 2 + α 7 r 2 s + α 8 rs 2 ----------(15)
In the above polynomial, cubic terms have been omitted.
The nodal displacement vector {d } is obtained by substituting the coordinates for the nodes
as:
⎧ u1 ⎫ ⎡1 −1 −1
⎪u ⎪ ⎢1 1 −1
⎪ 2⎪ ⎢
⎪u3 ⎪ ⎢1 1 1
⎪ ⎪ ⎢
u
1 −1 1
{d } = ⎨⎪ 4 ⎬⎪ = ⎢⎢
⎪u5 ⎪ ⎢1 0 −1
⎪u6 ⎪ ⎢1 1 0
⎪ ⎪ ⎢
⎪u7 ⎪ ⎢1 0 1
⎪u ⎪ ⎢1 −1 0
⎩ 8⎭ ⎣
1 −1 −1⎤ ⎧α1 ⎫
1 −1 1 −1 1 ⎥⎥ ⎪⎪α 2 ⎪⎪
1 1 1 1 1 ⎥ ⎪α 3 ⎪
⎥⎪ ⎪
1 −1 1 1 −1⎥ ⎪α 4 ⎪
⎨ ⎬
0 0 1 0 0 ⎥ ⎪α 5 ⎪
⎥
1 0 0 0 0 ⎥ ⎪α 6 ⎪
⎪ ⎪
0 0 1 0 0 ⎥ ⎪α 7 ⎪
⎥
1 0 0 0 0 ⎥⎦ ⎪⎩α 8 ⎪⎭
1
1
{α } = [ A]−1 {d }
where [ A]−1 is as given below:
⎡ −1 −1 −1 −1 2 2 2 2 ⎤
⎢ 0 0 0 0 0 2 0 −2 ⎥
⎢
⎥
⎢ 0 0 0 0 −2 0 2 0 ⎥
⎢
⎥
1 ⎢ 1 1 1 1 −2 0 −2 0 ⎥
−1
And [ A] =
4 ⎢ 1 −1 1 −1 0 0 0 0 ⎥
⎢
⎥
⎢ 1 1 1 1 0 − 2 0 −2 ⎥
⎢ −1 −1 1 1 2 0 −2 0 ⎥
⎢
⎥
⎢⎣ −1 1 1 −1 0 −2 0 2 ⎥⎦
{δ }
T
= [1, r , s, r 2 , rs, s 2 , r 2 s, rs 2 ]
22
⎡ −1 −1 −1 −1 2 2 2 2 ⎤
⎢
⎥
⎢ 0 0 0 0 0 2 0 −2 ⎥
⎢ 0 0 0 0 −2 0 2 0 ⎥
⎢
⎥
1 ⎢ 1 1 1 1 −2 0 −2 0 ⎥
−1
[ A] =
4 ⎢ 1 −1 1 −1 0 0 0 0 ⎥
⎢
⎥
⎢ 1 1 1 1 0 −2 0 −2 ⎥
⎢ −1 −1 1 1 2 0 −2 0 ⎥
⎢
⎥
⎢⎣ −1 1 1 −1 0 −2 0 2 ⎥⎦
Let δ be expressed as {δ } = [1, r , s, r 2 , rs, s 2 , r 2 s, rs 2 ] then, the shape functions for this
T
element are given by
[N ]
T
= {δ } [ A]−1 which is given as
T
⎧(1 − r )(1 − s )(−r − s − 1) ⎫
⎪(1 + r )(1 − s )(r − s − 1) ⎪
⎪
⎪
⎪(1 + r )(1 + s )(r + s − 1) ⎪
⎪
⎪
1 ⎪(1 − r )(1 + s )(− r + s − 1) ⎪
[N ] = ⎨
⎬ ------------------(16)
4 ⎪2(1 + r )(1 − r )(1 − s) ⎪
⎪2(1 + r )(1 + s )(1 − s) ⎪
⎪
⎪
⎪2(1 + r )(1 − r )(1 + s) ⎪
⎪2(1 − r )(1 + s )(1 − s) ⎪
⎩
⎭
The shape functions can be expressed in concise form as follows
[N ]
T
1
= [ N1 N 2 N 3 N 4 N 5 N 6 N 7 N8 ]
4
In the present work, of laminated plate bending, the transverse displacement w, the rotation
about x-axis φx and the rotation about y-axis φ y are considered as the only 3 degrees of
freedom at each node of the element. The in- plane displacements (u, v) are neglected in this
study.
The displacement vector d is given as
d = ( w, φ x , φ y )
The generalized displacements at any point (x, y) in the element are expressed in terms of the
nodal values of displacements and shape functions as given below:
23
8
8
w = ∑ N i wi
8
φx = ∑ N iφxi
i =1
φ y = ∑ N iφ yi
i =1
i =1
Adopting the same shape function ‘N’ to define all the components of the generalized
displacement vector, d, we can write
N
d = ∑ N i d i -----------(17)
i =1
The nodal displacements are given by
{d }
T
= ⎡⎣ w1θ x1θ y1w2θ x 2θ y 2 w3θ x 3θ y 3 w4θ x 4θ y 4 w5θ x 5θ y 5 w6θ x 6θ y 6 w7θ x 7θ y 7 w8θ x8θ y 8 ⎤⎦
In which, N is the number of nodes in the element. Now, referring to the expressions in
equation (4), the bending curvatures and the shear strains can be written in terms of nodal
displacements d using the matrix notation as follows:
{Φ} = LS d
{ε } = LB d
In which the subscripts B and S refer to bending and shear respectively and the matrices Lb
And Ls attain the following form
⎡
∂
⎢0
∂x
⎢
⎢
LB = ⎢0 0
⎢
⎢
∂
⎢0
∂y
⎣
⎡∂
⎢ ∂x
LS = ⎢
⎢∂
⎢⎣ ∂y
⎤
0⎥
⎥
∂⎥
∂y ⎥⎥
∂⎥
⎥
∂x ⎦
⎤
1 0⎥
⎥ -------(18)
0 1⎥
⎥⎦
Knowing the generalized displacement vector, d, at all points within the element, the
generalized strain vectors at any point are determined with the aid of equations (17)
and (18) as follows:
24
N
N
i =1
i =1
N
N
i =1
i =1
{ε } = LB d = LB ∑ Ni di = ∑ BiB di = BB a
{Φ} = LS d = LS ∑ Ni di = ∑ BiS di = BS a
----------------------------(19)
In which
The B matrix for the i th node can be written as
⎡B ⎤
Bi = ⎢ iB ⎥
⎣ BiS ⎦
⎡
⎢0
⎢
⎢
[ BiB ] = [ LB ][ Ni ] = ⎢0
⎢
⎢
⎢0
⎣
⎤
0 ⎥
⎥
∂N i ⎥
⎥
∂y ⎥
∂N i ⎥
⎥
∂x ⎦
∂N i
∂x
0
∂N i
∂y
⎡ ∂N i
⎢ ∂x
[ BiS ] = [ LS ][ N i ] = ⎢
⎢ ∂N i
⎢⎣ ∂y
⎤
1 0⎥
⎥
0 1⎥
⎥⎦
N
, BB = ∑ BiB
i =1
N
, BS = ∑ BiS
i =1
and a = (d1T , d 2T , d 3T ,............, d NT )
Substituting the above strain-displacement matrix B, in the virtual work statement derived
above results in
∫ ⎡⎣a B D
t
A
t
B
B
BB a + a t Bst D s Bs a ⎤⎦dA
or
∫ (a K a)dA
t
e
A
e
In which K is the element stiffness matrix and is expressed as
K e = ∫ ⎡⎣ BBt D B BB + Bst D s Bs ⎤⎦dA
A
Because of the symmetry of the stiffness matrix, only the blocks Kij lying on one side of the
main diagonal are formed for simplification. The integral is evaluated numerically using the
Gauss quadrature rule, in the limits of -1 to +1
Kije = ∫
1
∫
1
−1 −1
Bit DB j J drds
25
g
g
K ije = ∑∑ WaWb Bit DB j J
a =1 b =1
We = a t Fc + a t ∫ ( N iT q + N iT P)dA
A
mπ x
nπ y ⎞
T ⎛
sin
Pi = ∑∑ WaWb J N iT {100} ⎜ q + Pmn sin
⎟
a
b ⎠
⎝
a =1 b =1
g
g
Where, Wa and Wb are the weights of the Gauss points determined using Legendre
Polynomials. g’s are the Gauss sampling points at which numerical integration is carried out.
J is the determinant of the Jacobian matrix [J].Subscripts i and j vary from one to the
number of odes per element. The matrices Bi and D are given above and Bj is obtained by
replacing i by j.
For this flexural analysis , the total external work done by the applied external loads for an
element e, is given by
We = a t Fc + at ∫ ( NiT q + NiT P)dA
A
In which suffix, i , varies from one to number of nodes per element. Fc is the vector of
concentrated nodal loads corresponding to nodal degrees-of-freedom. q and P are the uniform
and sinusoidal distributed load intensities acting over an element e in the z-direction.
The integral of the above equation is evaluated numerically using the Gauss quadrature rule as
follows
mπ x
nπ y ⎞
T ⎛
Pi = ∑∑ WaWb J N iT {100} ⎜ q + Pmn sin
sin
⎟ in which a and b are the plate
a
b ⎠
⎝
a =1 b =1
g
g
dimensions, x and y are the Gauss point coordinates and m and n ar the usual harmonic
numbers.
26
Present work:
In the present study of Cross-Ply laminated composite plate, using finite element
method, a square laminated composite plate is taken as focus of study. Let, the side of the
square plate is ‘a’ unit. And the Laminate consists of ‘N’ number of laminas. The laminas
have either 0 degree or 90 degree orientation with respect to the material coordinates, i.e., the
lamina’s are cross-plied.
Here, the laminated square plate is considered as the domain. The domain is
discretized in to sub-domains/finite elements using 8- noded isoparametric quadratic element
(Serendipity Element). As the Plate is symmetric about both the axes in its plane, a quarter of
the plate is considered for the study. This quarter plate model is again discretized with 2×2
mesh. So, there are 4 numbers of elements in the quarter plate. The 2×2 mesh in quarter plate
model is equivalent to full plate model with 4×4 mesh.
After discretizing the domain into sub-domains, the finite element model of the
problem is developed using classical variational method as explained above. Finally, the
element stiffness matrix is obtained as:
⎡ K 11
⎢
[K e ] = ⎢
⎢
⎣
K 12
K 22
K 13 ⎤
⎥
K 23 ⎥
K 33 ⎥⎦
Where,
⎛
∂N ∂N j
∂N ∂N j ⎞
Kij11 = ∫ ⎜ A55 i
+ A44 i
⎟dxdy
A
∂x ∂x
∂y ∂y ⎠
⎝
∂N
⎛
⎞
K ij12 = ∫ ⎜ A55 i N j ⎟dxdy
A
∂x
⎝
⎠
⎛
⎞
∂N
K ij13 = ∫ ⎜ A44 i N j ⎟dxdy
A
∂y
⎝
⎠
⎛
⎞
∂N ∂N j
∂N ∂N j
Kij22 = ∫ ⎜ D11 i
+ D33 i
+ A55 Ni N j ⎟dxdy
A
∂x ∂x
∂y ∂y
⎝
⎠
27
⎛
∂N ∂N j
∂N ∂N j
Kij23 = ∫ ⎜ D12 i
+ D33 i
A
∂x ∂y
∂y ∂x
⎝
⎞
⎟dxdy
⎠
⎛
⎞
∂N ∂N j
∂N ∂N j
K ij33 = ∫ ⎜ D33 i
+ D22 i
+ A44 Ni N j ⎟dxdy
A
∂x ∂x
∂y ∂y
⎝
⎠
Because of the symmetry of the stiffness matrix, only the blocks Kij lying on one side
of the main diagonal are formed for simplification.
Subscripts i and j vary from one to the number of nodes per element. The integral is evaluated
numerically using the Gauss quadrature rule, in the limits of -1 to +1.
Kije = ∫
1
∫
1
−1 −1
g
Bit DB j J drds
g
K ije = ∑∑ WaWb Bit DB j J
a =1 b =1
σ yz = 0
φx = 0
φy = 0
Here, the bending stiffness and shear stiffness values are evaluated separately, to avoid shear
locking of problem. The bending terms are evaluated using 3×3 order of integration i.e. at 9
sampling points and the shear terms are evaluated using 2×2 order of integration i.e., at 4
sampling points for each element. Reduced integration scheme is adopted for shear terms.
The values of sampling points and weights for each order are given as below:
For 3×3 order, First weight is 0.8888888889 and the corresponding sampling point is 0
The other weight is 0.555555555 and the corresponding sampling point is +/- 0.7745966692
For 2×2 order, the weights are 1 and sampling points are +/- 0.5773502692
Since there are 3 degrees of freedom per node of an element viz., transverse
displacement w, rotation about x- axis φx and rotation about y- axis φ y and the domain is
discretized in to 2×2 meshes i.e., there are 4 elements present. And there are a total of 21
nodes and correspondingly, 21×3=63 degrees of freedom will be present. After getting the
28
Element stiffness matrix for all the elements, they are assembled to obtain the Global stiffness
matrix. Therefore, there present 63 simultaneous equations in [K][d]=[Q] form.
When the assembly of the element stiffness matrices is over, the boundary conditions
are applied at the boundaries of the plate. Initially the plate is simply supported on all the
edges of the plate. Since, a quarter plate model is considered for the analysis, only two of the
edges are simply supported and the remaining are free. The applied boundary conditions are
described as below:
w = 0 at x=0 and y=0
φx = 0 at x=0 and y= 0
φ y = 0 at x=0 and y=0
After applying the boundary conditions, the plate is transversely loaded. The loading
can be with a uniformly distributed load of magnitude ‘q’ units and a sinusoidal varying load
on the plate surface acting individually. So, the stiffness matrix after applying boundary
conditions is reduced from 63×63 matrix to 44×44 matrix. So, there are a total of 44
simultaneous equations present that are to be solved in the reduced stiffness matrix. These
equations are solved to obtain the displacement vector matrix [d] like, [d] = [K]-1[Q].The
inverse of the reduced stiffness matrix and the solution to obtain generalized displacement
vector is carried out using the program written in MATLAB.
4.4Post computation of Stresses and Strains:
Once the generalized displacements at the nodes are determined by solving the assembled
equations of the problem , the transverse displacements and slopes at any (x,y) can be
calculated using eq-----------.Strains at any point (x , y, z) in a typical element ‘e’ can be
computed from the strain-displacement relations stated above. It is to be noted here that, only
displacements are continuous across the element boundaries. Strain continuity across the
boundaries is not ensured as we are using a C0 continuous element. That is, along a boundary
common to two elements, the strains and hence stresses take different values on the two sides
29
of the interface. However, strains and hence stresses are continuous within an element. The
stesses can be calculated using the constitutive relations stated above.
Since the displacements in the finite element models are referred to the global
coordinates (x,y,z), the stresses are computed in the global coordinates using the relations at
the sampling points ; not at the nodes. The stresses can be transformed to principal material
coordinates using the stress transformation relations. Similarly, strains can also be
transformed to principle material coordinates.
30
Chapter-5
RESULTS AND DISCUSSION
31
5.1Results :
Numerical results are obtained for a specific problem whose data is given below:
Material: Graphite-Epoxy composite with material properties
E1=175 GPa
E2=7 GPa
G12=G13=0.5 E2=3.5 GPa
G23=0.2 E2= 1.4 GPa
ν 12=0.25
Shear correction factor K=5/6
Boundary conditions: Simply supported on all edges
Loading :
a)Sinusiodal varying load and
b) Uniformly distributed load acting individually
The Non-dimensionalized displacement, and stresses results are tabulated and given below.
Side to
Thickness
ratio a/h
10
Type of
solution
w
σ xx
σ yy
σ xy
σ xz
σ yz
FEM
100.35
0.7568
0.2865
0.0487
0.7157
0.2869
Closed
102.19
0.7719
0.3072
0.0514
0.7548
0.3107
FEM
69.25
0.7138
0.218
0.0423
0.7869
0.2654
Closed
75.72
0.7983
0.227
0.0453
0.7697
0.2902
FEM
62.78
0.7895
0.1856
0.0409
0.7496
0.2687
Closed
66.97
0.8072
0.1925
0.0426
0.7744
0.2842
66.00
0.8075
0.1912
0.0425
0.7191
0.3791
form
20
form
100
form
CLPT
Table 1: Non dimensionalized maximum deflection and stresses of simply supported cross-ply
(0/90/0) square plate subjected to uniformly distributed loading
32
Side to
Thickness
ratio a/h
10
20
100
Type of
solution
w
σ xx
σ yy
σ xy
σ xz
σ yz
FEM
101.80
0.7493
0.4989
0.0438
0.7896
0.3406
Closed form
102.50
0.7577
0.5006
0.0470
0.7986
0.3499
FEM
76.77
0.7959
0.3905
0.0408
0.8298
0.3184
Closed form
76.94
0.8045
0.3968
0.0420
0.8305
0.3228
FEM
65.19
0.8394
0.3527
0.0315
0.8395
0.3092
Closed form
68.33
0.8420
0.3558
0.0396
0.8420
0.3140
CLPT
67.96
0.8236
0.3540
0.0395
0.6404
0.4548
Table 2: Non dimensionalized maximum deflection and stresses of simply supported cross-ply
(0/90/90/0) square plate subjected to uniformly distributed loading
Side to
Thickness
ratio a/h
10
20
100
Type of
solution
w
σ xx
σ yy
σ xy
σ xz
σ yz
FEM
93.5
0.7459
0.5268
0.0394
0.6751
0.3658
Closed form
97.27
0.7649
0.5525
0.0436
0.6901
0.4410
FEM
71.59
0.7952
0.4685
0.0401
0.6927
0.4089
Closed form
75.81
0.8080
0.4844
0.0403
0.7166
0.4188
FEM
65.35
0.7995
0.4227
0.0351
0.7186
0.3982
Closed form
68.74
0.8264
0.4559
0.0386
0.7267
0.4108
CLPT
68.44
0.8272
0.4546
0.0385
----------
----------
Table 3: Non dimensionalized maximum deflection and stresses of simply supported cross-ply
(0/90/0/90/0) square plate subjected to uniformly distributed loading
33
Side to
Thickness
ratio a/h
10
20
100
Type of
solution
w
σ yy
σ xx
σ xy
σ xz
σ yz
FEM
66.92
0.5098
0.2518
0.0250
0.4060
0.0908
Closed form
66.93
0.5134
0.2536
0.0252
0.4089
0.0915
3D-Elasticity
---------
0.590
0.288
0.029
0.357
0.123
FEM
49.21
0.5281
0.1983
0.0222
0.4176
0.0754
Closed form
49.21
0.5318
0.1997
0.0223
0.4205
0.0759
3D-Elasticity
----------
0.552
0.210
0.234
0.385
0.092
FEM
43.36
0.5346
0.1791
0.0212
0.4215
0.0699
Closed form
43.37
0.5384
0.1804
0.0213
0.4247
0.0703
3D-Elasticity
---------
0.539
0.181
0.0213
0.395
0.083
43.13
0.5387
0.1796
0.0213
0.3951
0.0823
CLPT
Table 4: Non dimensionalized maximum deflection and stresses of simply supported cross-ply
(0/90/0) square plate subjected to sinusoidal loading
Side to
Thickness
ratio a/h
10
20
100
Type of
w
σ xx
σ yy
σ xy
σ xz
σ yz
FEM
66.75
0.4906
0.3512
0.0258
0.398
0.112
Closed form
66.27
0.4989
0.3614
0.0241
0.416
0.129
3D-Elasticity
73.70
0.5590
0.4010
0.0276
0.301
0.196
FEM
49.25
0.5198
0.2906
0.0228
0.430
0.126
Closed form
49.12
0.5273
0.2956
0.0221
0.437
0.109
3D-Elasticity
51.28
0.5430
0.3080
0.0230
0.328
0.156
FEM
43.18
0.5215
0.2598
0.0208
0.448
0.118
Closed form
43.37
0.5382
0.2704
0.0213
0.445
0.101
3D-Elasticity
43.47
0.5390
0.2710
0.0214
0.339
0.139
CLPT
43.13
0.5387
0.2667
0.0213
0.339
0.138
solution
Table 5 : Non dimensionalized maximum deflection and stresses of simply supported crossply (0/90/90/0) square plate subjected to sinusoidal loading
34
Side to
Thickness
ratio a/h
10
20
100
Type of
solution
w
σ xx
σ yy
σ xy
σ xz
σ yz
FEM
62.12
0.4986
0.4078
0.0219
0.3435
0.1984
Closed form
62.13
0.5021
0.4107
0.0221
0.3459
0.1998
3D-Elasticity
67.71
0.545
0.430
0.0247
0.258
0.223
FEM
47.96
0.5239
0.3722
0.0214
0.3592
0.1827
Closed form
47.96
0.5276
0.3748
0.0215
0.3617
0.1840
3D-Elasticity
49.38
0.539
0.380
0.0222
0.268
0.212
FEM
43.31
0.5345
0.3573
0.0211
0.3655
0.1761
Closed form
43.32
0.532
0.3598
0.0213
0.3683
0.1774
3D-Elasticity
43.38
0.539
0.360
0.0213
0.272
0.205
CLPT
43.13
0.5387
0.3591
0.0213
0.2722
0.2052
Table 6 :Non dimensionalized maximum deflection and stresses of simply supported cross-ply
(0/90/0/90/0) square plate subjected to sinusoidal loading
35
Graphs:
120
Deflection w
MaximumTransverse
Maximum Deflection Vs a/h ratio
100
80
FEM
60
CFS
40
CLPT
20
0
0
50
100
150
Side to thickness ratio a/h
Graph 1: Non dimensionalized central transverse deflection versus side to thickness ratio for
simply supported cross ply (0/90/90/0) square laminate subjected to uniformly distributed
loading
Normal Stress Sigma xx
VsThickness
FEM
Sigma xx
Normal stress
1
FSDT
CLPT
0.5
0
-1
-0.5
0
-0.5
0.5
1
-1
Thickness z/h
Graph 2: Non dimensionalized normal stress sigma xx versus side to thickness ratio for
simply supported cross ply (0/90/90/0) square laminate subjected to uniformly distributed
loading
36
Normal Stress Sigma yy
Normal Stress Sigma yy Vs z/h
0.8
0.6
FEM
0.4
FSDT
CLPT
0.2
0
-1
-0.5 -0.2 0
0.5
1
-0.4
-0.6
-0.8
Thickness z/h
Graph 3: Non dimensionalized normal stress sigma yy versus side to thickness ratio for
simply supported cross ply (0/90/90/0) square laminate subjected to uniformly distributed
Sigma xz
loading
-1
Sigma xz Vs Thickness z/h
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-0.5
0
0.5
1
Thickness
FEM
FSDT
CLPT
Graph 4: Non dimensionalized transverse shear stress sigma xz versus side to thickness ratio
for simply supported cross ply (0/90/90/0) square laminate subjected to uniformly distributed
loading
37
Sigma yz Vs Thickness z/h
0.6
FEM
0.5
FSDT
Sigma yz
0.4
CLPT
0.3
0.2
0.1
0
-1
-0.5
0
0.5
Thickness z/h
1
Graph 5: Non dimensionalized transverse shear stress sigma yz versus side to thickness ratio
for simply supported cross ply (0/90/90/0) square laminate subjected to uniformly distributed
Loading
Maximum Deflection Vs a/h ratio
100
Deflection w
Maximum Transverse
120
80
60
FEM
CFS
40
CLPT
20
0
0
50
100
150
Side to thickness ratio a/h
Graph 6: Non dimensionalized central transverse deflection versus side to thickness ratio for
simply supported cross ply (0/90/90/0) square laminate subjected to sinusoidal varying load
38
Normal Stress Sigma xx
VsThickness
1
Normal stress
Sigma xx
FEM
FSDT
0.5
CLPT
0
-1
-0.5
0
-0.5
0.5
1
-1
Thickness z/h
Graph 7: Non dimensionalized normal stress sigma xx versus side to thickness ratio for
simply supported cross ply (0/90/90/0) square laminate subjected to sinusoidal varying load
Normal Stress Sigma yy
Normal Stress Sigma yy Vs z/h
0.8
0.6
FEM
0.4
FSDT
CLPT
0.2
0
-1
-0.5 -0.2 0
0.5
1
-0.4
-0.6
-0.8
Thickness z/h
Graph 8: Non dimensionalized normal stress sigma yy versus side to thickness ratio for
simply supported cross ply (0/90/90/0) square laminate subjected to sinusoidal varying load
39
Sigma xz Vs z/h
0.45
0.44
Sigma xz
0.43
FEM
0.42
FSDT
0.41
CLPT
0.4
0.39
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
Thickness z/h
Graph 9: Non dimensionalized transverse shear stress sigma xz versus side to thickness ratio
for simply supported cross ply (0/90/90/0) square laminate subjected to sinusoidal varying
load
Sigma yz Vs z/h
1.2
FEM
1
FSDT
CLPT
Sigma yz
0.8
0.6
0.4
0.2
0
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
Thickness z/h
Graph 10: Non dimensionalized transverse shear stress sigma yz versus side to thickness ratio
for simply supported cross ply (0/90/90/0) square laminate subjected to sinusoidal varying
load
40
5.2Discussion:
The following non-dimensional quantities are used to get the non dimensionalized stresses
and deflections from the actual ones.
w = w0 (0, 0)
E2 h 2
a 4 q0
σ xy = σ xy (a / 2, a / 2, −h / 2)
h2
σ xx = σ xx (0, 0, h / 2) 2
a q0
σ xz = σ xz (a / 2, 0, k = 1, 4)
σ yy
h
aq0
h2
a 2 q0
h2
= σ yy (0, 0, h / 4) 2
a q0
σ yz = σ yz (0, a / 2, k = 1, 4)
h
aq0
The origin of the coordinate system is taken at the centre f the plate, -a/2<(x,y)<a/2 and
–h/2<z<h/2. As mentioned earlier, the stresses are computed at the reduced Gauss points. The
Gauss coordinates are mentioned like (A, A). The finite element solutions of the present study
is compared with the closed form solutions obtained using FSDT and that of Classical
Laminated Plate Theory (CLPT) for uniformly distributed loading case when the edges of the
plate are simply supported. Similarly, the results obtained from present finite element model
are compared with the closed form solutions of FSDT and CLPT as well as 3-D elasticity
solutions for sinusoidal variation of loading case when all the edges of the plate are simply
supported.
Comparison is made, between non dimensional quantities of transverse displacements,
for different values of side to thickness ratios. Comparison is also made with respect to the
lamina orientation in the laminate for different side to thickness values between the transverse
displacements.
The results of non dimensional quantities of normal stresses, in-plane shear stresses
and transverse shear stresses are compared, at various thickness co -ordinates for different
values of side to thickness ratios. The results are obtained for different lamina orientation
schemes (0 or 90 degrees i.e., cross – ply orientation with symmetry) in the laminate as well
as for varying number of layers i.e., for 3 layer (0/90/0), 4 layer (0/90/90/0) and 5 layer
(0/90/0/90/0) orientations.
41
5.3Observations:
It is observed from the results presented in the tables that, the FEM results obtained using 8noded isoparametric Serendipity element are giving near approximations for a/h ratios <20,
i.e., for thick plates. And as the ratio increases, the values are not that much satisfactory. That
is, the present finite element model using First order shear deformation theory can well
predict the results for a thick plate. As the plate a/h ratio is increasing i.e. when the plate is
becoming thin, the results are not that much in good comparison as that for the thick plates.
Reduced integration alleviated this phenomenon called shear locking to some extent.
From the results it can be observed that the Finite Element Solutions are in well agreement
with the results of closed form solutions of FSDT. The displacements converge faster than
stresses. This is expected because; the rate of convergence of gradients of the solution is one
order less than the rate of convergence of the solution. However, the results based on the
finite element solutions should not be expected to agree well with the 3-D elasticity solutions.
The finite element solutions should only converge to the closed form solutions of the FSDT.
It is also observed that, the normal stresses are varying non-linearly across the thickness of the
laminate. However, they are varying linearly for an individual lamina. The stresses are
discontinuous across the thickness of the laminate. That means, there exist different values of
stresses at the interfaces of the laminate. The stress at the bottom surface of a lamina zk, is
different from that at the top surface of the adjacent lamina zk-1.This is obvious from the First
order shear deformation theory. The element is a C0 continuous element. The generalized
displacements only are continuous across the thickness of the laminate. But, the strains and
thus the stresses are not continuous at the boundaries.
The stress values are maximum at the top and bottom of the laminate with + ve and – ve signs
respectively. The transverse shear stresses are constant throughout the thickness. It is because
of the use of a constant in calculating the shear stresses, the shear correction factor. Its value
42
varies with lamina orientation and stacking sequence. In the present study it’s value is taken
as 5/6.
Since the stresses in the finite element analysis are computed at locations different from the
analytical solutions, they are expected to be different.
43
Chapter -6
CONCLUSIONS
44
Conclusions
Finite element analysis of cross ply laminated composite square plate is carried out,
using a 8 – noded isoparametric quadratic element to predict the transverse displacements ,
normal stresses and transverse shear stresses, when it is subjected to transverse loading under
simply supported boundary conditions. The present model is developed based on the First
order Shear Deformation Theory (FSDT). This theory uses a shear correction factor to
approximate the transverse shear stresses. A computer program is written in MATLAB to get
various results. The accuracy of results obtained using the present formulation is
demonstrated by comparing the results with three-dimensional elasticity solution ,closed form
solutions of FSDT and Classical Laminated Plate Theory. The present analysis gives accurate
values for displacements and stresses compared to Classical Laminated Plate Theory. It is
observed that the results are in close agreement with closed form solutions of FSDT and 3-D
elasticity solutions. It is found that, the transverse shear stresses vary constantly through the
thickness. This is attributed to the use of shear correction factor in the theory. But, the actual
variation of the transverse shear stresses is parabolic according to 3D elasticity using
equilibrium relations in predicting the same. More over, the results of stresses are calculated
at Gauss points and they are expected to differ from the analytical solutions. Adoption of
reduced integration scheme alleviated the shear locking effects. The present model accurately
predicts the transverse displacements and various stresses for thin as well as thick laminated
composite plates. As the present model is developed using a non-conforming element, the
results can be further improved using a conforming element with improved mesh size thereby
increased no of elements. Infact , the FEM results approach the true solutions, with the
increase in the number of elements.
.
45
Chapter-7
SCOPE FOR FUTURE WORK
46
Scope for future work:
Results can be expected with excellent agreement with the analytical/experimental solutions
by using a conforming element with increased mesh size.
Analysis can be done for different loading conditions with various boundary conditions.
There is a need to develop a mechanics based theory to find out the optimal stacking
sequence, which will significantly help the designers.
Study can be made on real life problems pertaining to stress concentration, non-linearity and
complicated geometries
47
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Journal of solids and structures, 4(1), 95-108 (1968)
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numerical methods for engineering vol 1,101-122, (1969)
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49
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50
APPENDIX
Composite: A combination of two or more materials on a macroscopic scale which are
physically distinct.
Classification:
1) Fiber reinforced composites which consists of fibers in a matrix
2) Laminated composites which consists of layers of various materials
3) Particulate composites which consists of particles in a matrix
Advantages of composites:
They usually exhibit the best qualities of their constituents and often some qualities that
neither constituent possesses. The properties that can be improved by forming a composite
material include:
Strength
Fatigue life
Temperature-dependent behavior
stiffness
Wear resistance
Corrosion resistance
Weight
Mechanical behavior of composite materials
Composite materials unlike isotropic materials are often both inhomogeneous and nonisotropic
Anisotropic body has material properties that are different in all directions at a point in the
body. There are no planes of material property symmetry. Again, the properties are a function
of orientation at a point in the body.
Orthotropic body has material properties that are different in three mutually perpendicular
planes of material symmetry. Thus, the properties are a function of orientation at a point in the
body.
Micromechanics are the study of composite material behavior where in the interaction of the
constituent materials is examined on a microscopic scale.
51
Macro-mechanics is the study of composite material behavior wherein the material is
presumed homogeneous and the effects of the constituent materials are detected only as
averaged apparent properties of the composite.
Basic terminology of laminated fiber-reinforced composite materials:
Lamina: It is a flat (sometimes curved as in a shell) arrangement of unidirectional l fibers or
woven fibers in a matrix.
Laminate: It is a stack of lamina with various orientations of principal material direction in the
lamina as shown .fig.
The major purpose of lamination is to tailor the directional dependence of strength and
stiffness of a material to match the loading environment of the structural element. Laminates
are uniquely suited to this objective since the principal material direction of each layer can be
oriented according to need.
Geometry of the N- layered laminate is as shown:
52
The transformed reduced stiffness matrix terms can be expressed as
Q11 = Q11 cos 4 θ + 2( Q12 +2 Q33 ) sin 2 θ cos 2 θ + Q22 sin 4 θ
Q12 = ( Q11 + Q22 -4 Q33 ) sin 2 θ cos 2 θ + Q12 ( sin 4 θ + cos 4 θ )
Q22 = Q11 sin 4 θ + 2( Q12 +2 Q33 ) sin 2 θ cos 2 θ + Q22 cos 4 θ
Q13 = ( Q11 - Q12 -2 Q33 ) sin θ cos3 θ + ( Q12 - Q22 +2 Q33 ) sin 3 θ cos θ
Q23 = ( Q11 - Q12 -2 Q33 ) sin 3 θ cos θ + ( Q12 - Q22 +2 Q33 ) sin θ cos3 θ
Q33 = ( Q11 + Q22 -2 Q12 -2 Q33 ) sin 2 θ cos 2 θ + Q33 ( sin 4 θ + cos 4 θ )
Jacobian Matrix:
The derivatives of the shape functions with regard to x and y can be obtained by
transformation from natural coordinate’s r and s, using Jacobian matrix. The Jacobian matrix
is given by
⎡ ∂x
⎢
[ J ] = ⎢ ∂r
⎢ ∂x
⎣⎢ ∂r
∂y ⎤
∂s ⎥
⎥
∂y ⎥
∂s ⎦⎥
The transformation from (r, s) coordinates to (x, y) coordinate system is done by
⎧∂⎫
⎧∂⎫
⎪⎪ ∂x ⎪⎪
⎪ ∂r ⎪⎪
−1 ⎪
⎨ ∂ ⎬ = [J ] ⎨ ⎬
⎪ ⎪
⎪∂ ⎪
⎩⎪ ∂s ⎭⎪
⎩⎪ ∂y ⎭⎪
For the present problem with 8-noded Serendipity element, the transformation is as follows:
⎡ ∂N1
⎢ ∂x
⎢
⎢ ∂N1
⎣⎢ ∂y
∂N 2
∂x
∂N 2
∂y
∂N 3
∂N ⎤
⎡ ∂N1
.......................... 8 ⎥
⎢
∂x
∂x
−1
⎥ = [ J ] ⎢ ∂r
∂N3
∂N
⎢ ∂N1
.......................... 8 ⎥
⎥
⎢⎣ ∂s
∂y
∂y ⎦
53
∂N 2
∂r
∂N 2
∂s
∂N3
∂N ⎤
.......................... 8 ⎥
∂r
∂r
⎥
∂N3
∂N8 ⎥
..........................
∂s
∂s ⎥⎦
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