THREE DIMENSIONAL ANALYSIS OF FIBRE REINFORCED …

THREE DIMENSIONAL ANALYSIS OF FIBRE

REINFORCED POLYMER LAMINATED

COMPOSITES

A thesis submitted to

THE UNIVERSITY OF MANCHESTER

for the degree of

DOCTOR OF PHILOSOPHY (PhD)

in the Faculty of Engineering and Physical Sciences

2012

Haji Elmi Haji Kamis

School of Mechanical, Aerospace and Civil Engineering

Contents

2

Contents

CONTENTS….....................................................................................................2

LIST OF FIGURES .............................................................................................4

LIST OF TABLES..................................... ..........................................................8

ABSTRACT….......................................... .........................................................10

DECLARATION........................................ ........................................................11

COPYRIGHT STATEMENT..............................................................................12

ACKNOWLEDGEMENTS ................................... .............................................13

CHAPTER 1 .INTRODUCTION ....................................................................15

1.1 Introduction.........................................................................................15 1.2 Research Aim and Objectives ............................................................17 1.3 Outline of Thesis.................................................................................18

CHAPTER 2 .LITERATURE REVIEW .................................. ........................21

2.1 Plate Theories ....................................................................................21 2.1.1 Rectangular Kirchhoff-Love plates...............................................24 2.1.2 Development of plate theories.....................................................27

2.2 FRP ....................................................................................................31 2.2.1 Advantages of FRP .....................................................................32 2.2.2 Disadvantages of FRP.................................................................32 2.2.3 Constitutive Material Properties of FRP.......................................33

2.3 FRP strengthening RC structures.......................................................36 2.4 Conclusions ........................................................................................56

CHAPTER 3 .STATE SPACE METHOD OF 3D ELASTICITY................ .....58

3.1 The Concept of State Space Method of 3D Elasticity .........................58 3.2 Governing Equations of Elasticity Problems .......................................61 3.3 State Equations for Simply Supported Orthotropic Plate ....................63 3.4 Conclusions ........................................................................................80

CHAPTER 4 .STATE SPACE SOLUTION OF CLAMPED EDGES LAMINATED PLATE .................................... ....................................................82

4.1 State Space Solution of a Single Layer Plate .....................................83 4.2 State Space Solution of Laminated Plate ...........................................85 4.3 Application of State Space Method to Laminated Plate....................116 4.4 Conclusions ......................................................................................126

Contents

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CHAPTER 5 .NUMERICAL ANALYSIS OF LAMINATED PLATES ..........128

5.1 Introduction.......................................................................................128 5.2 Modelling of Composite Plate Using FEM ........................................129

5.2.1 Element types............................................................................130 5.2.2 Assembling the model ...............................................................131 5.2.3 Boundary Conditions .................................................................133 5.2.4 Analysis Type ............................................................................134

5.3 Material Properties............................................................................134 5.4 Mesh.................................................................................................135 5.5 Partially Clamped Edges Composite Plate with Variable Thickness to

Width Ratio .......................................................................................136 5.6 Fully Clamped Edges Composite Plate ............................................141 5.7 Laminated Plate Subjected To Hydrostatic Loading .........................161 5.8 Laminated Plate with increasing number of sub-layers.....................163 5.9 Conclusions ......................................................................................167

CHAPTER 6 .FLEXURAL DEFORMATION OF RC SLAB WITH FRP ......170

6.1 Numerical Modelling .........................................................................170 6.2 Geometric Properties of the Model ...................................................175 6.3 Element Type ...................................................................................178

6.3.1 Concrete Slab............................................................................179 6.3.2 Steel reinforcement bars ...........................................................179 6.3.3 FRP sheets................................................................................180 6.3.4 Interaction of FRP and concrete slab ........................................180 6.3.5 Boundary condition....................................................................182

6.4 Tension Stiffening.............................................................................183 6.4.1 Non-linear tension stiffening stress-strain relation.....................185 6.4.2 Linear tension stiffening stress-strain relation ...........................186 6.4.3 Multi-linear tension stiffening stress-strain relation....................187

6.5 FEM Against Experimental Test Results ..........................................188 6.6 Concrete slab reinforced with FRP ...................................................191 6.7 Conclusions ......................................................................................200

CHAPTER 7 .CONCLUSIONS AND RECOMMENDATIONS .................... 201

7.1 Conclusions ......................................................................................201 7.2 Future Works Recommendations .....................................................203

REFERENCES.. .............................................................................................205

LIST OF PUBLICATIONS ............................... ...............................................209

APPENDIX A… ........................................ ......................................................210

Word count: 37,568 words

List of Figures

4

List of Figures Figure 1.1: Layout of a UD lamina ........................................................................16

Figure 1.2: View of unstacked laminate ................................................................16

Figure 2.1: Normal and shear stresses .................................................................22

Figure 2.2: A simply supported rectangular plate under sinusoidal loading ..........25

Figure 2.3: An Ashby property map for composites ..............................................36

Figure 2.4: A test specimen with end-anchorage..................................................38

Figure 2.5: Mechanically anchored RC slab tested to failure in the laboratory .....39

Figure 2.6: RC Slab Section Strengthened with NSM CFRP rods ........................40

Figure 2.7: Loading system and slab dimensions .................................................41

Figure 2.8: Two way CFRP strips being installed..................................................41

Figure 2.9: (a) Slab dimension ; (b) Test set-up....................................................43

Figure 2.10: Dimensions and test set-up scheme.................................................44

Figure 2.11: Stress distribution for the analysed slabs..........................................45

Figure 2.12: Flexural Test Setup of Concrete Slabs .............................................46

Figure 2.13: Typical crack pattern of the CFRP grid reinforced slabs ...................46

Figure 2.14: Typical theoretical and experimental load versus midspan

deflection curve.....................................................................................................47

Figure 2.15: Strengthened RC beam details.........................................................50

Figure 2.16: Test setup and instrumentation of the two-way slab specimens .......51

Figure 2.17: slab specimen details and loading ....................................................52

Figure 2.18: Details of beam specimens (a) Cross Section and reinforcement

details (b) Zone of FRP repair...............................................................................53

Figure 2.19: Details of FRP retrofitting of slab specimens ....................................53

Figure 2.20: slab specimens tested to failure........................................................54

Figure 3.1: Spring - damper - mass system .........................................................59

Figure 3.2: Coordinate system and plate dimension.............................................63

Figure 4.1: A single layer plate..............................................................................83

Figure 4.2: Geometry and coordinate systems of the laminate.............................86

Figure 4.3: View of a clamped edges laminated plate.........................................117

Figure 5.1: Table used to model composite solid and shell in Abaqus................132

Figure 5.2: Geometry of plate consists of three plies..........................................137

List of Figures

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Figure 5.3: Mesh sensitivity test results for stresses at x=y=z=0 for h/a = 0.2 ....138

Figure 5.4: Displacement (W C11/qh) distribution through the thickness of the

plate at x = a/2 and y = b/2 when h/a = 0.2 .........................................................143

Figure 5.5: Stress (σx/q) x = a/2 and y = b/2 when h/a = 0.2...............................143

Figure 5.6: Stress (σy/q) x = a/2 and y = b/2 when h/a = 0.2...............................144

Figure 5.7: Stress (σx/q) x = 0 and y = b/2 when h/a = 0.2..................................144

Figure 5.8: Stress (σy/q) x = 0 and y = b/2 when h/a = 0.2..................................145

Figure 5.9: Stress (σx/q) x = a/2 and y = 0 when h/a = 0.2..................................145

Figure 5.10: Stress (σy/q) x = a/2 and y = 0 when h/a = 0.2................................146

Figure 5.11: Stress (τxz/q) at x = 0 and y = b/2 when h/a = 0.2 ..........................146

Figure 5.12: Stress (τyz/q) at x = a/2 and y = 0 when h/a = 0.2 ..........................147



Figure 5.14: Stress (σx/q) x = a/2 and y = b/2 when h/a = 0.4.............................149

Figure 5.15: Stress (σy/q) x = a/2 and y = b/2 when h/a = 0.4.............................150

Figure 5.16: Stress (σx/q) at x = 0 and y = b/2 when h/a = 0.4............................150

Figure 5.17: Stress (σy/q) at x = 0 and y = b/2 when h/a = 0.4............................151


Figure 5.19: Stress (σx/q) at x = a/2 and y = 0 when h/a = 0.4............................152

Figure 5.20: Stress (σy/q) at x = a/2 and y = 0 when h/a = 0.4............................152




Figure 5.23: Stress (σx/q) at x = a/2 and y = b/2 when h/a = 0.6.........................155

Figure 5.24: Stress (σy/q) at x = a/2 and y = b/2 when h/a = 0.6.........................156

Figure 5.25: Stress (σx/q) at x = 0 and y = b/2 when h/a = 0.6............................156

Figure 5.26: Stress (σy/q) at x = 0 and y = b/2 when h/a = 0.6............................157

Figure 5.27: Stress (σx/q) at x = a/2 and y = 0 when h/a = 0.6............................157

Figure 5.28: Stress (σy/q) at x = a/2 and y = 0 when h/a = 0.6............................158



Figure 5.31: Partially clamped plate subjected to hydrostatic loading for h/a=0.4161

List of Figures

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Figure 5.32: Central deflection across the thickness of plate h/a=0.4.................166

Figure 5.33: Stress,sx of plate h/a = 0.4 at x = a/2 y = b/2 .................................166

Figure 5.34: Stress,sy of plate h/a = 0.4 at x = a/2 y = b/2 .................................167

Figure 6.1: Stress and strain relationship of (a) concrete (b) steel and (c) FRP .172

Figure 6.2: View of RC slab model (a) without FRP (b) with FRP.......................177

Figure 6.3: FEM of CFRP strengthened RC slab................................................178

Figure 6.4: Damage traction-separation response used in FEM.........................181

Figure 6.5: Concrete tensile response characterized by damaged plasticity ......183

Figure 6.6: Exponential tension stiffening (Hordijk).............................................185

Figure 6.7: Linear tension stiffening....................................................................186

Figure 6.8: Multi-linear tension stiffening ............................................................187

Figure 6.9: Relationship of load against central deflection of RC slab by variable

tension stiffening response .................................................................................189

Figure 6.10: Relationship of load against central deflection of RC slab by

variable element types ........................................................................................189

Figure 6.11: Load against central deflection of un-strengthened and

strengthened of RC slab with CFRP ...................................................................190

Figure 6.12: Geometry and coordinate systems of the layered slab ...................192

Figure 6.13: Deflection distribution (W C11/qh) at x=a/2, y = b/2 h/a = 0.1 ..........197





Figure A-1: Deflection (WC11 / qh) at x = a/2 , y = b/2 at z = 0............................221

Figure A-2: Stress (σx/q) at x = a/2 and y = b/2 Ply1 top.....................................221

Figure A-3: Stress (σx/q) at x = a/2 and y = b/2 Ply1 bottom...............................222





Figure A-8: Stress (σy/q) at x = a/2 and y = b/2 Ply1 top.....................................224

Figure A-9: Stress (σy/q) at x = a/2 and y = b/2 Ply1 bottom...............................225

Figure A-10: Stress (σy/q) at x = a/2 and y = b/2 Ply2 top...................................225

List of Figures

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Figure A-11: Stress (σy/q) at x = a/2 and y = b/2 Ply2 bottom.............................226

Figure A-12: Stress (σy/q) at x = a/2 and y = b/2 Ply3 top...................................226

Figure A-13: Stress (σy/q) at x = a/2 and y = b/2 Ply3 bottom.............................227

Figure A-14: Stress (σx/q) at x = 0 and y = b/2 Ply1 top......................................227

Figure A-15: Stress (σx/q) at x = 0 and y = b/2 Ply1 bottom................................228





Figure A-20: Stress (σy/q) at x = 0 and y = b/2 Ply1 top......................................230

Figure A-21: Stress (σy/q) at x = 0 and y = b/2 Ply1 bottom................................231





Figure A-26: Stress (τxz/q) at x = 0 and y = b/2 Ply1 bottom ..............................233

Figure A-27: Stress (τxz/q) at x = 0 and y = b/2 Ply2 top ....................................234

Figure A-28: Stress (τxz/q) at x = 0 and y = b/2 Ply2 bottom ..............................234

Figure A-29: Stress (τxz/q) at x = 0 and y = b/2 Ply3 top ....................................235

List of Tables

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List of Tables Table 2-1: Material properties ...............................................................................34

Table 5-1: Displacement and stresses distribution of fully clamped laminate

when h/a = 0.2 ....................................................................................................142


when h/a = 0.4 ....................................................................................................148


when h/a = 0.6 ....................................................................................................154

Table 5-4: Exact solution versus FEM for plate subjected to hydrostatic loading162

Table 5-5: The exact solutions and FEM of Case 1 ............................................163



Table 6-1: The evolution of the damage variable for compression and tension ..173

Table 6-2: Basic material properties used in FEM ..............................................175

Table 6-3: Deflection distribution (WC11/qh) for h/a = 0.1 ...................................195





Table A-1: Maximum deflection at x = a/2, y = b/2 and z = 0 (W C11/ qh) ...........210

Table A-2: Stress (σx/q) at x = a/2, y = b/2 at Ply1 top surface ...........................210

Table A-3: Stress (σx/q) at x = a/2, y = b/2 at Ply1 bottom surface .....................210





Table A-8: Stress (σy/q) at x = a/2, y = b/2 at Ply1 top surface ...........................212

Table A-9: Stress (σy/q) at x = a/2, y = b/2 at Ply1 bottom surface .....................212

Table A-10: Stress (σy/q) at x = a/2, y = b/2 at Ply2 top surface .........................213

Table A-11: Stress (σy/q) at x = a/2, y = b/2 at Ply2 bottom surface ..................213

Table A-12: Stress (σy/q) at x = a/2, y = b/2 at Ply3 top surface .........................213

List of Tables

9

Table A-13: Stress (σy/q) at x = a/2, y = b/2 at Ply3 bottom surface ...................214

Table A-14: Stress (σx/q) at x = 0, y = b/2 at Ply1 top surface ............................214

Table A-15: Stress (σx/q) at x = 0, y = b/2 at Ply1 bottom surface ......................214





Table A-20: Stress (σy/q) at x = 0, y = b/2 at Ply1 top surface ............................216

Table A-21: Stress (σy/q) at x = 0, y = b/2 at Ply1 bottom surface ......................216





Table A-26: Stress (τxz/q) at x = 0, y = b/2 at Ply1 top surface...........................218

Table A-27: Stress (τxz/q) at x = 0, y = b/2 at Ply1 bottom surface.....................218





Table A-32: Stress (τxy/q) at x = 0, y = b/2 across the thickness........................220

Abstract

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Abstract

Name of the University: The University of Manchester Candidate’s full name: Haji Elmi Haji Kamis Degree Title: Doctor of Philosophy (PhD) Thesis Title: Three Dimensional Analysis of Fibre Reinforced Polymer Laminated Composites Date: 08th July 2012 The thesis presents the structural behaviour of fibre reinforced polymer (FRP) laminated composites based on 3D elasticity formulation and finite element modeling using Abaqus. This investigation into the performance of the laminate included subjecting it to various parameters i.e. different boundary conditions, material properties and loading conditions to examine the structural responses of deformation and stress. Both analytical and numerical investigations were performed to determine the stress and displacement distributions at any point of the laminates. Other investigative work undertaken in this study includes the numerical analysis of the effect of flexural deformation of the FRP strengthened RC slab. The formulation of 3D elasticity and enforced boundary conditions were applied to establish the state equation of the laminated composites. Transfer matrix and recursive solutions were then used to produce analytical solutions which satisfied all the boundary conditions throughout all the layers of the composites. These analytical solutions were then compared with numerical analysis through one of the commercial finite element analysis programs, Abaqus. Out of wide variety of element types available in the Abaqus element library, shells and solids elements are chosen to model the composites. From these FEM results, comparison can be made to the solution obtained from the analytical. The novel work and results presented in this thesis are the analysis of fully clamped laminated composite plates. The breakthrough results of fully clamped laminated composite plate can be used as a benchmark for further investigation. These analytical solutions were verified with FEM solutions which showed that only the solid element (C3D20) exhibited close results to the exact solutions. However, FEM gave poor results on the transverse shear stresses particularly at the boundary edges. As an application of the work above, it is noticed that the FEM results for the FRP strengthened RC slab, agreed well with the experimental work conducted in the laboratory. The flexural capacity of the RC slab showed significant increase, both at service and ultimate limit states, after FRP sheets were applied at the bottom surface of the slab. Given the established and developed programming codes, exact solutions of deflection and stresses can be determined for any reduced material properties, boundary and loading conditions, using Mathematica.

Declaration

11

Declaration

No portion of the work referred to in this thesis has been submitted in support of

an application for another degree of qualification of this or any other university,

or other institution of learning.

Copyright Statement

12

Copyright Statement

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thesis) owns certain copyright or related right in it (the “Copyright”) and s/he has

given The University of Manchester certain rights to use such Copyright,

including for administrative purposes.

Copies of this thesis, either in full or in extracts and whether in hard or

electronic copy, may be made only in accordance with the Copyright, Designs

and Patents Act 1988 (as amended) and regulations issued under it or, where

appropriate, in accordance with licensing agreements which the University has

from time to time. This page must form part of any such copies made.

The ownership of certain Copyright, patents, designs, trade marks and other

intellectual property (the “Intellectual property”) and any reproductions of

copyright works in the thesis, for example graphs and tables (“Reproductions”),

which may be described in this thesis, may not be owned by the author and

may be owned by third parties. Such Intellectual property and Reproductions

cannot and must not be made available for use without the prior written

permission of the owner(s) of the relevant Intellectual Property Rights and/or

Reproductions.

Further information on the conditions under which disclosure, publication and

commercialisation of this thesis, the Copyright and any Intellectual Property

and/or Reproductions described in it may take place is available in the

University IP Policy (see

http://www.campus.manchester.ac.uk/medialibrary/policies/intellectualproperty.

pdf), in any relevant Thesis restriction declarations deposited in the University

Library, the University Library’s regulations (see

http:www.manchester.ac.uk/library/aboutus/regulations) and in The University’s

policy on presentation of Theses.

Acknowledgments

13

Acknowledgements

Firstly, I am grateful to Allah S.W.T. for His continuous blessings and devotions

such that I could not see, hear nor sense anything other than His sole presence

guiding me throughout my journey.

My first academic year had not been a smooth sailing start for me unlike for

most students. Due to unforeseen circumstances, I had to commute to Cardiff

three days a week after spending the day work at the University. Having to

travel in chilled evenings by train, I continued my research work on my laptop

during the three and a half hours journeys. Cardiff University was where my wife

pursued her Master degree and where my family lived. On the following

mornings while my wife attended her lectures I had to sent my two young

children to school and took care of my one year old son at home. Such three

days a week tasks became a routine throughout my wife’s one and a half year

Master Degree course.

From this difficult and challenging experience, I had learnt on how tough it was

to pursue PhD study especially for those who are with families. I had

understood that the ability to juggle with time between family and study surely

needed sheer patience. However, with strong will and determination, I managed

to complete my study after significant amount of time and efforts were sacrificed.

Knowing the limited time I had left to complete my study after my family

returned home during my third year of study, I had pushed myself to the limits,

going through ups and downs of mixed emotions, until at some points I had to

stop spending the night work at the University where it had jeopardised my

health condition.

Nevertheless, I would like to send my deepest gratitude to my family especially

my beloved wife, Suzan for her strengths and patience for taking care of all our

children while I was away in Manchester. To all my lovely sons, Md Naim Bazli,

Md Nabil Wafie, Md Muazzam and Khairul Harisin, letting them grown up

Acknowledgments

14

without my presence and missing the full opportunities to observe every single

of their actions was heart- breaking. I really miss and love you all. My

appreciation also goes to my parents whom continuously give support and pray

for my success.

Special thank also goes to The Brunei Government whom provided me the

great opportunities to pursue my study by awarding me the sponsorship for the

third time.

Last but certainly not the least, I wish to thank Dr Jack Wu, my supervisor for

his assistance throughout my study. I can not also forget to thank all my friends,

Dr Ahmad Abdullah, Dr Hisham, Dr Ashfaq Khan, Dr Rao Krishnamoorthy, Dr

Sabri Jamil, Dr Yakub, Dr Febian, Dr Maturose Suchatawat, Dr Ramadan

Eghlio, Amin, Ahmad Sabri, Abdul Rahman, Moustafa, Aftab, Israr, Kwan Sete,

Noorhafiza, Sutham, Salwan, Michael, Barbie, Chao Han, and Dominic for all

their help and support throughout my study.

One of the sweet memories that I gained in Manchester was when my team (My

B team comprises of Dr Sabri, Dr Nasrun, Aizul, Abdul Latip and myself) won

the Badminton competition organised by North West Tri-Badminton Series 2012

at Belle Vue, Longsight, Manchester last January.

After going through all the experiences and mixed emotions, I am pretty sure

that I will miss Manchester especially the life as a student once I return back

home and get back to work. I really hope one day I can have the chance to

come and visit Manchester again, the birth place of my youngest son, Khairul

Harisin.

Chapter 1 Introduction

15


1.1 Introduction

Fibre reinforced polymer (FRP) composite materials have been widely used in

construction for buildings and bridges since 1980’s [1]. They have been used in

structural engineering for new construction and for strengthening and repair

purposes of existing building and bridge structures. FRP materials come in a

variety of shapes and sizes, ranging from internal reinforcing bars for concrete

members and sheets or strips for external strengthening of concrete or other

structures. The use of FRP applications has also been beneficial in other

industries including aerospace, manufacturing, automotive, biomedical,

infrastructure, sports equipments and others [2].

Typically, the use of FRP materials in civil engineering are ready made and

supplied with relevant dimensions and specifications to meet the desired design

requirements. In other words, FRPs material can be tailored according to the

designer’s or engineer’s requirement.

One typical type of composite laminae which consists of a single layer of FRP is

known as uni-directional (UD) fibre lamina as shown in Figure 1.1. This type of

lamina is commonly flat shape with the fibres being arranged parallel along their

principal material axes. These fibres are the main load carrying medium and

they are hold together by matrix. By combining the fibres and matrix together,

they make the UD lamina become strong and stiff. This type of UD lamina is

used in this study as the main type of FRP.

Furthermore, if there is a number of lamina being stacked and bonded together

with various fibre orientations in each lamina, they become a laminate as shown

in Figure 1.2. By forming this laminate, the material properties of each lamina

can be varied and specified to the designer’s requirement.


16

Figure 1.1: Layout of a UD lamina [3]

Figure 1.2: View of unstacked laminate [3]


17

The major purpose of forming the lamination is to meet the required strength

and stiffness of the composite material. Due to such great significant

advantages of laminate’s inherent property over the conventional metal, it has

inspired the author to investigate the structural response of composite

laminated plate by subjecting it to various parameters including boundary,

loading conditions and material properties.

Reinforced concrete (RC) slab is also another type of composite material in

such a way that steel reinforcement bars are the fibres and concrete is the

matrix. These materials are also being investigated in this study as the main

material properties for application purposes. Using the available experimental

test results of FRP strengthened RC slab, numerical analysis of the same

specimens are also reviewed using one of the finite element method (FEM),

Abaqus. For the analytical part, composite laminated plates are investigated

using classical plate theory and compared with the three dimensional analysis

of elasticity with state space method.

1.2 Research Aim and Objectives

The aim of this study is to investigate the structural performance such as

stresses and displacements of composite laminated plates subjected to various

parameters including loading, boundary conditions and material properties. Both

analytical and numerical analyses were carried out to examine the responses of

the structures. The analytical study has focused on the application of three

dimensional elasticity and state space method whereas numerical method of

investigation was carried out based on FEM analysis.

In order to achieve this aim, the following objectives of this research studies

have been identified:

• To investigate the structural response of laminated composite plate by

analytical method based on the 3D elasticity and state space method.

• The use of several different FEM element types to find a suitable element

for comparison with the analytical method.


18

• To carry out further case studies of laminated plate by changing various

parameters including material properties, boundary and loading

conditions to reflect the true response of composite laminated plate.

• To review the effect of flexural deformation of RC slab with and without

FRP. This task is performed by modeling and simulating the specimen’s

deflection, numerically. The available experimental test results are used

and compared with the numerical solutions by Abaqus/CAE.

• To realise the analytical solutions for the structural response of any

reduced material properties, boundary and loading conditions by

developing program codes with Mathematica.

• To compare the results of slab’s deflection between classical plate theory

and the exact solution of three dimensional elasticity approach.

1.3 Outline of Thesis

Thesis content can be briefly summarized as follows,

Chapter 1 gives the introduction of the FRP composites material. The definition

and layout view of UD laminate has been outlined and shown, respectively. The

function of laminate has also been briefly described in this chapter. The other

main content on this chapter is to state the research’s aim and objectives in

which it describes how and what tasks will be involved during the investigation.

Chapter 2 briefs the literature review of plate theory and it’s development from

two to three dimensional analyses. Some theoretical formulations are also

shown in this chapter to calculate the deflection of a simply supported plate

subjected to various loading conditions. These fundamental formulation of

deflection are based on the assumption of the Kirchhoff-Love theory. The uses

of FRP towards strengthening of structures have also been reviewed. This

chapter also summaries on why the novelty of work presented in this thesis are

carried out in connection to the available literature.


19

Chapter 3 deals with the concept of state space method for plate structures and

the governing equations of elasticity. A classical (continuum) or analytical

method based on the application of the equations of equilibrium and kinematics,

together with the stress and strain relations for the material are applied to

produce governing equations which should be solved to obtain displacements

and stresses. From these governing equations in state space, further

development into various applications having different boundary and loading

conditions can be explored. It is importance to understand thoroughly on this

chapter since the concept is extended to the following chapter.

Chapter 4 shows the application of state space method for clamped plates with

various loading and boundary conditions. It is clearly showed that the concept of

this clamped edges plate is applied from the simply supported plate. The exact

solution approach consists of determining the displacements of rectangular

plate by setting a general expression of the displacement field according to the

available boundary and loading conditions. State space methods together with

state transfer matrix are presented with the aid of programming code, in this

chapter. To understand the process of transfer matrix involved, initially only a

single layer of plate considered, and then followed by a plate consisting of many

sub-layers. The derivations of expressions are also shown in this chapter for

clarification purpose.

Chapter 5 demonstrates the use of numerical solutions by means of finite

element method in order to compare with the analytical works based on the 3D

elasticity and state space method. The objective of this method is to simulate

the composite plate behavior in such a way that the model (a continuum) is

discretized into simple geometric shapes which can be used to perform some

other parametric studies. FEM is an alternative approach to solving the

governing equations of a complicated structural problem particularly beyond the

elastic phase. The other important results presented here are the numerical

solutions of FEM verified the exact solutions of laminated plate subjected to

different loading conditions and material properties. The novelty solutions and

comparison of fully clamped laminated plate are presented in this chapter.


20

Chapter 6 illustrates the work undertaken during the first and second year of

PhD study. The deflection of RC slab strengthened with FRP using FEM is

reviewed. Modelling of FRP strengthened RC slab by Abaqus is explained in

this chapter. Numerical flexural deformation performance can be compared

against the behaviour of composition materials of the structure. The accuracy of

the modelling of deflection of the structures is verified with the available

experimental test results obtained from the full scale FRP strengthened RC

slabs tested in the University of Manchester two years ago. The results of

plate’s deflection are also compared between classical plate theory and the

exact solution based on the material properties of concrete and FRP lamina.

Chapter 7 outlines the conclusion and future work recommendations. This

chapter also summaries the ideal method of analysis that is to be used in the

future investigation of laminated plate.

Chapter 2 Literature Review

21

Chapter 2

LITERATURE REVIEW

2.1 Plate Theories

For the analysis of plate, classical finite difference and finite element methods, a

starting point of historical development of the theory and analysis of laminated

plates, are primarily based on the theory of thin plate, known as Kirchhoff-Love

plate theory. It is a two-dimensional theory developed by Kirchhoff-Love [4] from

the extension of the hypothesis of Euler-Bernoulli beam. Such a theory is

established on a few assumptions which neglect some parameters for the

analysis, namely, transverse shear deformations and rotatory inertia. Because

of the assumptions, this theory is inaccurate and will be unable to give a

solution precisely. More errors for the analysis of thick plate are expected,

particularly for the analysis of anisotropic laminated plate structures.

Classical plate bending theory makes errors when the ratio of the elastic

modulus to shear modulus becomes large. For instance, graphite epoxy and

boron epoxy have these modulus ratios of about 25 and 45, respectively against

ratio of 2.6 for isotropic materials [5]. Composite plate can leads to a

complicated coupling effect and significant changes of stresses in magnitude at

the interface of the laminate. As a result, classical plate theory is not accurate

for the analysis of anisotropic composite plates.

The assumptions that are made in this theory [6]:

• In-plane deformations in the x and y directions at the mid surface are

zero.

• straight lines normal to the mid-surface remain normal to the mid-surface

after deformation


22

• the stress σz is negligible compared with other stresses in the transverse

cross section, i.e. σx, σy and τxy as shown in Figure 2.1.

where σx,σy and σz are the in-plane and transverse stresses while τxy is the

shear stress as shown in Figure 2.1

These assumptions give approximate solutions and the following conditions.

Figure 2.1: Normal and shear stresses [6]

The first assumption implies that at the mid surface, the in-plane displacements

U(x,y,0) = 0 and V(x,y,0) = 0.

Second and third assumptions have lead transverse shear strains

yzxzγandγ and the normal strain,

zzε to be disappeared, i.e. [6]

0

0

0

=∂

∂=

=∂∂+

∂∂=

=∂∂+

∂∂=

z

Wε

z

V

y

Wγ

z

U

x

Wγ

zz

yz

xz

(2-1)

The remaining strains to be considered are xyyyxx

γandε,ε such that [6]

z(W)

x(U)

y(V)

h

σz

σx

σy

τxy


23

yx

Wz

yx

Wz

xy

Wz

x

V

y

Uγ

y

Wz

y

Vε

x

Wz

x

Uε

xy

yy

xx

∂∂∂−=

∂∂∂−

∂∂∂−=

∂∂+

∂∂=

∂∂−=

∂∂=

∂∂−=

∂∂=

222

2

2

2

2

2

(2-2)

The relationship of stress and strain is presented as follows:

For isotropic [6]:

−−=

xy

yy

xx

xy

yy

xx E

γεε

µµ

µ

µτσσ

2

100

01

01

1 2 and

( )µ+=

12

EG (2-3)

where

E is the Elastic modulus

G is the shear modulus

µ is the Poisson ratio

The above formulation is said to be true assuming the plate thickness is

relatively thin compared to its other dimensions. The typical ratio of thickness to

the plate plane dimension is less than 0.1, i.e. 1.0<a

h. Further formulations can

be obtained from the above equations including:

Strains from eqn.(2.2),

∂∂∂−

∂∂−

∂∂−

=

yx

W

y

Wx

W

z

γ

ε

ε

xy

yy

xx

2

2

2

2

2

2

(2-4)


24

Moments [7],

∂∂∂−

∂∂−

∂∂−

−=

=

∫−

yx

W

y

Wx

W

µ

µ

µ

Ddzz

τ

σ

σ

M

M

M h

h

xy

yy

xx

xy

yy

xx

2

2

2

2

2

2

2

22

100

01

01

where ( )2

3

112 µ−= Eh

D

(2-5)

D is also known as the flexural rigidity of the plate [7].

Shears [6],

∂∂∂+

∂∂−=

2

3

3

3

yx

W

x

WDQ

x and

∂∂∂+

∂∂−=

2

3

3

3

xy

W

y

WDQ

y (2-6)

Based on the Love-Kirchhoff assumptions and the equations of equilibrium at a

point for a plate, the following governing equations can be deduced [6]:

WD

q

y

W

yx

W

x

W 44

4

22

42

4

4∇==

∂

∂+

∂∂

∂+

∂

∂ (2-7)

The equation (2-7) is well known as bi-harmonic equation. For a specified

loading, q, and specified boundary conditions, the solution of the plate problem

is reduced to finding a deflection, W. Subsequently, the values of bending

moments and shears can be calculated. The following case is illustrated to

determine deflection of plate.

2.1.1 Rectangular Kirchhoff-Love plates

Plate bending refers to the small deflection of a plate out of its original plane

under the action of external transverse loading. The plate has relatively smaller

thickness than its other dimensions (less than 0.1 of it’s width). Classical plate

theory can be applied to the plate where shear strains are neglected across the

thickness. Only in-plane direct and shear stresses are considered.


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For a simply supported rectangular plate subjected to an applied loading as

shown in Figure 2.2, Navier [6] established the displacement and stress

expressed in terms of Fourier expansion.

Figure 2.2: A simply supported rectangular plate un der sinusoidal loading

For a single distributed lateral load of the form

( )b

yπnSin

a

xπmSinqy,xq

o= (2-8)

where qo is the constant (with the dimension of pressure), substitute

b

yπnSin

a

xπmSinWW

o= into equation (2-7) and verifying the boundary

conditions w = 0 and x = 0 and a; w = 0 at y = 0 and b, are satisfied, the

deflection can be determined as

b

yπnSin

a

xπmSin

b

n

a

mDπ

qW o

2

2

2

2

24

+

= , where 2

2

2

2

24

+

=

b

n

a

mDπ

qW o

o

z

x

y

a

b

h

q


26

Therefore, the general expression of displacement can be deduced as follows

[6]:

For a plate subjected to uniform loading, the deflection is given by

( )∞=

+

= ∑∑∞

=

∞

=......,.........,,n,m

b

n

a

mmn

b

yπnSin

a

xπmSin

Dπ

qW

m n

o 53116

1 12

2

2

2

26 (2-9)

For the case of plate subjected to concentrated load at the centre, its deflection

is given by

b

yπnSin

a

xπmSin

b

n

a

m

b

πηnSin

a

πξmSin

abDπ

PW

m n∑∑∞

=

∞

=

+

=1 1

2

2

2

2

24

4. (2-10)

where ;2

;2

by

ax ==== ηξ if the load is applied at the middle.

Further expression of theoretical formulations for various loading and boundary

conditions can be referred to Timoshenko [6].

It is important to note that classical laminate theory provides a simple and direct

method to determine stresses and strains. However, it is not very accurate as it

does not satisfy the equations of elasticity at every point of laminated plate. It

also ignores shear deformations of layers because of the assumed bond

between two laminae which are non-shear deformable. During loading, shear

stresses are developed at the interfaces. The transverse stresses

yzxzzand ττσ , are negligible in the regions away from the plate edges.

Therefore, laminate theory is only sufficient in the regions away from the plate

boundary. For the case of regions near the boundary, however, a plane stress

is no longer true, a 3D stress state would become more appropriate.


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2.1.2 Development of plate theories

More rigorous studies have been performed to improve such theory without

neglecting the transverse shear deformations and rotatory inertia. The study

that relates governing equation of plates and incorporates the effect of shear

was first established by Reissner [8]. The assumptions made by Reissner gave

a linear bending stress and a parabolic shear stress distribution through the

thickness of the plate. Since Reissner’s work, there have been a number of

refinement and further generalization beyond the classical plate theory. This

includes further improvement by Mindlin [9] which allowed shear strains to occur.

His theory assumed that there was a linear variation of displacements (U and V)

across the plate thickness, however, the deflection through the thickness did not

change during the loading, i.e. W had no relation with the thickness direction (z-

coordinate). This leads to the normal stress through the thickness is being

ignored, which is similar situation to the plane stress condition. Both Mindlin and

Reissner theories however have some similarities in such ways that their works

are based on the extension of Kirchhoff-Love plate theory and taken into

account of shear deformations through the thickness of a plate where the

normal to the mid-surface remains straight but not necessarily perpendicular to

the mid surface and rotatory inertia. Both theories satisfy the three boundary

conditions on the edge but do not satisfy the differential governing equations of

three dimensional elasticity. The form of Mindlin-Reissner plate theory also

includes in-plane shear strains and is often known as a first-order shear

deformation plate theory with a linear displacement variation through the

thickness. More details about the relationship between Reissner’s and Mindlin’s

theories are presented by Wang et al. [10]

In reality, the transverse shear strains cannot be constant through the thickness

of the plate. They are frequently, but not always, zero at the top and bottom

surfaces of the plate and they are generally undergo in parabolic form across

the thickness. Therefore, Mindlin theory represents average transverse shears

through the thickness.


28

Ambartsumyam’s theory [11] improved the mid-thick plate theories by

introducing a quadratic function to represent the variation of shear stresses

yzxzand ττ (or the corresponding deformations

yzxzand γγ ) across the

thickness.

Some related studies of interlaminar stress distributions of laminate which led to

important development on this field were reported by Pagano and Halpin [12],

Whitney and Leissa [13] and Whitney [14]. However, the first direct approach to

the problem was made by Puppo and Evenson [15] who derived an

approximate formulation of the laminate made of surface layers and an isotropic

shear layer. The anisotropic layers were assumed to carry only in-plane loads

hence resulting plane stress whereas the isotropic shear layers were assumed

to carry interlaminar shear stress.

The experimental work of Pipes and Daniels [16] confirmed the theoretical

results of Pipes and Pagano [17] by using the Moire technique. The Moire effect

used the arrays of images viewed when lights were transmitted to the specimen.

The observed images known as the Moire patterns, revealed the axial

displacement and strain fields of the laminate subjected to axial loading. The

results obtained from the test were accurately shown as compared to the

theoretical solutions.

Pagano and Pipes [18] extended their work on laminated plate to study the

effect of arranging the orientation direction of the laminate across the thickness

to provide optimum laminate strength. However, the results obtained were

merely approximate particularly at interlaminar stresses and at the boundary

edges.

Apart from classical solutions, numerical techniques had been developed for

comparison. This includes the work of Rybicki [19] on 3D FEM for the analysis

of finite width laminate which confirmed the sign changes of σz, as per Pagano

work.


29

Many researchers have continued to develop the theories showing that the

deflection W has no relationship with z-coordinate. Some works however have

considered the relation of deflection w and z coordinate by higher order theories

such as Lo et al [20]. However, these refined theories are still insufficient and do

not meet all governing equations of three dimensional elasticity. These theories

produce some form of approximation but not exact solutions.

Further investigations have been performed to study the bending, vibration and

buckling analyses of laminated rectangular plates associated with the three

dimensional elasticity.

The work of Pagano [21] on a cylindrical bending of a simply supported

orthotropic strip under sinusoidal transverse load showed that classical

laminated plate theory based on Kirchhoff-Love theory underestimated the plate

deflection and gave a very poor result as the span to depth ratio of plate

reduced.

Pagano [22] extended his study of three-dimensional elasticity solutions for the

case of rectangular laminates composed of orthotropic layers with pinned edges

under static loading. He concluded that the accuracy of classical thin plate

theory for a particular problem depended on the material properties, lamination

orientation and span to depth ratios.

Srinivas and Rao [23] have also established an exact analysis of three

dimensional elasticity theory solutions for the analysis of simply supported thick

orthotropic rectangular laminates subjected to sinusoidal normal pressures.

From the result obtained, they showed that as the thickness of the plate

increased, the stress and displacement distributions across thickness became

more complex and great computation process was required.

Bahar [24] presented a fundamental formulation for associated classical 2-D

elasticity by use of the transfer matrix approach in state space.


30

Based on the same approach, Iyengar and Pandya [25] combined it with the

method of initial functions (MIF) which was initially proposed by Vlasov [26] to

formulate the general solutions of orthotropic rectangular thick plates in three

dimensional elasticity. The result proved to be a kind of higher order theory.

They formed equations using a Taylor series expansion with respect to z-

coordinate. The main criticism, if any, concentrates mainly on the error in this

method by an expansion which cannot be avoided.

For orthotropic, transversely isotropic and isotropic 3-D simply supported elastic

plates, Wu [27] has developed the methods mentioned above and gave the

various close-form solutions explicitly, for the first time in this subject.

Fan and Ye [28-29] extended the work of Wu [27] and Srinivas and Rao [23] to

a three ply orthotropic thick plate. The subsequent equations were expressed in

terms of the initial value when z = 0 (top surface). A compatibility equation was

used at each interface of the plies, resulting from the continuity of the interface

displacements and stresses. A set of algebraic equations were then established

following the successive formulation from top layer until the bottom layer of the

laminated plate.

Fan and Sheng [30] have further presented the state equation for thick laminate

with clamped edges by introducing delta function to establish appropriate

boundary conditions along the edges of a plate. The results presented in their

study were incomplete, i.e. the exact solutions for a partially clamped laminated

plate.

Rogers et al. [31] presented general expressions for the deformation and stress

distribution for some particular cases such as an elliptical plate with moderate

thickness, a semi-infinite strip clamped along its two edges etc.

Agbossou and Mougin [32] investigated the static and dynamic analysis of

rectangular reinforced concrete slabs based on the laminated theory. Non linear

behaviour of simply supported slabs were designed and analysed from the

basis of laminated theory by changing the neutral axis position and steel and


31

concrete effective modulus when concrete slab started to crack. The objective

of the work is to analyse and design the slab to resist impact loading from rock

falls. After reaching the ultimate tensile strength of concrete and when the steel

reinforcement begins to resist further tensile loading, bending stiffness matrix

has been modified to take into account of the effect of the damaged concrete

and plasticity of steel bars within the laminate. The results provide good

relationship between analytical, FEM and experimental tests.

Recent work related to the application of state space equation to clamped thick

laminated plates included that of Sheng et al. [33]. Analytical solution was

presented for the analysis of laminated piezoelectric plate with clamped and

electric open-circuited boundary conditions. These results were also verified

with numerical outcomes from FEM.

Furthermore, Li et al. [34] used finite integral transform method to establish an

exact bending solutions for fully clamped orthotropic rectangular thin plates

subjected to arbitrary loading. The mathematical method did not require any

displacement functions to satisfy the governing equations of 2D elasticity and

the boundary conditions. Further works need to be done are to validate the

formulations including the case of 3D orthotropic plate on various boundary

conditions.

2.2 FRP

FRP offers an effective, sustainable method of structural strengthening and

rehabilitation. FRP composites consist of fibres of high tensile strength and high

modulus within a polymer matrix. The two or more materials combine together

to produce desirable properties that cannot be achieved with any single

constituent acting alone. The bonding of these aligned fibres into the matrix

material results in a fibre reinforced composite material with superior properties

in the fibre direction. [35]

Due to the fact that fibres are highly directional, the resultant composite will

exhibit anisotropic behaviour. One common example of such behaviour is steel


32

reinforced concrete members. The steel bars are the fibres, and the concrete

material is the matrix that keeps the fibres together. In this application, when

structural members are subjected to a load, fibres act as the principal load

carrying members and the concrete provides a load transfer medium between

fibres and protect fibres from being exposed to the environment such as

humidity.

Typical FRP composite material properties include low specific gravity, high

strength to weight ratio and a high modulus to weight ratio. Most FRP materials

require minimum protection and are very resistant to corrosion. Generally, FRP

materials behave in a linear elastic stress strain curve until failure under tension.

Brittle failure is the common mode of failure of FRP under excessive stress.

2.2.1 Advantages of FRP

In the case of repair of concrete structures, the use of fibre reinforced polymer

composites has the following advantages over conventional materials:

• Very light weight

• Superior toughness

• Low thermal conductivity (ability to conduct heat)

• Durability

• Ease of installation and transportation

• Can be wrapped over curved surfaces such as columns

The obvious advantages of using composite materials such as greater strength

and stiffness combined with lightness and durability are the reasons why they

are commonly used in a wide variety of applications.

2.2.2 Disadvantages of FRP

The significant disadvantage of externally strengthening structures with FRP is

the risk of fire, vandalism or accidental damage especially if they are not

protected. Poor workmanship during the installation of the FRP can lead to

unfavourable results. The performance of such a strengthening system can be


33

reduced if the adhesive layer does not achieve the desired quality. Another

obvious disadvantage of using FRP for strengthening purposes is the relatively

expensive cost of the material. However, this is balanced out given the

significant advantages FRP has over other systems such as steel plate. For

instance, FRP can be installed speedily without delaying and disrupting to the

user of the structure.

2.2.3 Constitutive Material Properties of FRP

Composite materials are often both non homogenous and non isotropic

(orthotropic or anisotropic).

A non homogenous body has non uniform properties over the body, i.e. the

properties depend on position in the body.

An orthotropic body has material properties that are different in three mutually

perpendicular directions at a point in the body and has three mutually

perpendicular planes of material property symmetry. Thus, the properties

depend on orientation at a point in the body.

An anisotropic body has material properties that are different in all directions at

a point in the body. No plane’s symmetry of material property exists. The

properties depend on orientation at a point in the body.

Composite materials have many mechanical behaviour characteristics that are

different from other typical engineering materials such as steel, aluminum and

other metals [36]. The desired properties of FRPs are achieved by the favorable

characteristics of the two major constituents, namely the fibre and the matrix.

Properties of Fibre

A fibre is characterised generally by its very high length-to-diameter ratio having

high strength and stiffness properties with low density when compared to

common materials such as aluminium, titanium, steel and others. Recall that,


34

fibres must have a surrounding matrix to achieve desired structural performance.

Typically, fibres comprise of 60 – 70% (by volume) of the composite.

Table 2-1 shows some materials properties arranged in increasing values of

ρS

and ρE

, where S is the tensile strength, E is the tensile stiffness and ρ is

the density. Fibres tend to have low transverse strength however they primarily

act in tension. Various form of fibres finish are manufactured and readily

available in the market including in ‘bundles’ called tows or rovings which can

be then further processed into tow sheets, fabrics or mats. In some types of

composite, the fibres are oriented randomly within a plane, while in others the

material is made up of a stack of differently-oriented “plies” to form a laminate,

each ply containing an aligned set of parallel fibres.

Table 2-1: Material properties [3]

FIBRE or

WIRE

Density, ρ

(kN/m3)

Tensile

Strength, S

(GPa)

ρS

(km)

Tensile

Stiffness,

E

(GPa)

ρE

(Mm or

x 106 m)

Aluminium 26.3 0.62 24 73 2.8

Titanium 46.1 1.9 41 115 2.5

Steel 76.6 0.5 * 6.5 * 207 2.7

E-glass 25.0 2.0 * 80 * 72 2.9

S-glass 24.4 4.8 197 86 3.5

Carbon 13.8 1.7 123 190 14

Beryllium 18.2 0.6 * 33 * 300 16

Boron 25.2 3.4 135 400 16

Graphite 13.8 1.7 123 250 18

(Note *: the typical values of tensile strength and specific tensile strength ( )ρ

S

of the material rather than as stated in the reference).


35

For advanced FRP used in civil engineering, the fibres are generally made of

carbon or glass in a polymer matrix such as vinylester or epoxy. From the

above Table 2-1, it shows that Graphite and S-glass have the higher stiffness

and strength to density ratios, respectively. For these reasons, these materials

are prominent type of materials used in today’s composite structures.

Note that the meanings of some basic mechanical properties of any material

such as the following:

Stiffness - The ability of material to resist elastic deformation. It is characterized

by the Young's modulus.

Strength - A measure of a material's resistance to failure. It depends of details

on how it is measured, specimen geometry and, for brittle materials, on the

presence of flaws.

Toughness – The ability of a material to absorb energy and plastically deform

without fracturing.

Properties of Matrix

By nature, matrix materials are at least an order of magnitude weaker than the

reinforcing embedded fibres. All matrix materials exhibit significant magnitude of

creep and have large coefficient of thermal expansion compared to traditional

construction materials. However, composite laminates could not exist without

matrix materials. In fact, the roles of a matrix are to support and protect the

fibres and transfer stress between broken fibres through shear. The matrix of a

composite acts as a binder that bond fibres together.

The most common type of matrix used in structural strengthening purposes is

polymers which consist of rubbers, thermoplastics and thermosets.

Typical examples of thermoplastics are nylon, polyethylene and polysulfone.

Epoxies, phenolics and polymides are common examples of thermosets.


36

The choices of composition and of the materials used as matrix and fibre are

dependent on the required properties. This can be deduced by deriving a merit

index for the performance required followed by the use of Ashby property maps

as shown in Figure 2.3.

Figure 2.3: An Ashby property map for composites [ 37]

2.3 FRP strengthening RC structures

In civil engineering, a growing level of activity in the repair and rehabilitation of

structures has been certainly acknowledged. Over many years, scientific

investigation of decay and deterioration of historical buildings in general and

concrete structures in particular has contributed significant results to gain

understanding of structural defects and finding ways to improve its durability. As

a consequence, service life assessment and rehabilitation are now widely

recognised as important issues.

Building or construction is considered to be among the earliest human activities.

To build and own a house has been essential to human needs and survival for

thousand of years. For any house constructed with building materials such as

concrete, timber, bricks, stones, mortars, plaster and steel, deterioration has

required the process of repairing and rebuilding to be an ongoing activity.


37

The load bearing capacity and the durability of all structures, ranging from

bridges to buildings, are susceptible to deterioration can either locally or globally,

by various mechanisms. The strength of the structures that are severely

exposed to environmental factors such as weather, aggressive chemical attack,

changes of usage or imposed loading etc can be adversely impacted. In the

21st century, concern about the decay and deterioration of buildings has grown

in tandem with the impact of industrial activities on the environment. Increasing

air pollution is believed to be responsible for the acceleration of building defects

and deterioration, particularly reinforced concrete structures. Corrosion of

reinforcement is the most common structural defect, having significant impact

on the economic process of repair and maintenance. The structural integrity of

the reinforced concrete members such as beams, columns, slabs or masonry

walls, deteriorates when they experience prolonged contact with acid rain,

sulphate or chloride attack. Sometimes the function of a structural member is

altered by the owner of the building. The departure from the intended use of

floor slab of a building results in it being overloaded. Furthermore, the capacity

and performance of the obsolete heritage buildings, landmark structures, ageing

prominent bridges and others are often degraded over time.

Due to the above factors, the needs for strengthening and rehabilitation of the

building structures are inevitably higher as the number of structures in the world

continues to increase. Properly designed and constructed RC structures which

operate under normal conditions of exposure and use, normally require

minimum maintenance. However, it is wrong to suggest that they are

maintenance free. The maintenance work for restoration or for increasing the

capacity of the structures has been in high demand. It is statistically published

that about 60% of investments are concerned with the maintenance and repair

of existing structures while only about 40% relate to the building of new

structures [38]. In addition, the performance of the strengthened structural

members has to be evaluated and monitored constantly, as safety is not an

issue which can be compromised. It is also equally important that the lessons

which can be learnt from the remedial work and correct treatment of the

strengthened (existing) structures should be put to practical use in the design

and construction of new structures. Lines of communication and exchange of


38

information are essential between repair parties involved and the design team

so that durability of structures can be enhanced.

In order to increase the loading capacity of the concrete structures, they have

been strengthened using bonding steel plates to the tension surface with

adhesives or bolts since 1960s. Towards the late of 1980s, the use of FRP has

been developing rapidly in the application of strengthening of concrete

structures [35].

Limited number of investigations on the behaviour of the strengthened RC slabs

with FRP has been carried out. The research activities have been increasing

with the application of modern fibre reinforced composite materials such as

CFRP and GRP [39].

The development of strengthening methods has evolved as there is strong

interest to the whole concrete repair community for the benefits of sustainable

life.

El Maaddawy and Soudki [40] investigated the use of mechanically anchored

un-bonded FRP (MA-UFRP) system to strengthen RC slabs as shown in Figure

2.4

Figure 2.4: A test specimen with end-anchorage [40]

A CFRP strengthened RC slab, with dimension 500mm width, 100mm thickness

and 1800mm length and reinforced with three 10mm diameter deformed steel


39

bars at the tension side, was tested to failure in the laboratory as shown in

Figure 2.5.

One CFRP strip having a width of 50mm and a thickness of 1.2mm was bonded

to the tension face of the slab. This CFRP composite strip had an elasticity

modulus of 155 GPa, tensile strength of 3.1 GPa at failure, ultimate elongation

of 1.9%. These correspond to steel reinforcement ratio (ρs) and FRP

reinforcement ratio (ρf) of 0.8% and 0.12% respectively. The average

compressive strength of concrete was 28 MPa and the steel reinforcement of

Grade 400 were noted during the investigation. Test results showed that the

ultimate load capacity and the mid span deflection of FRP strengthened RC

slabs was increased by 46% and reduced by 45%, respectively relative to a

control slab.

Figure 2.5: Mechanically anchored RC slab tested to failure in the laboratory [40]

The work of Foret and Limam [41] has also contributed to the significant use of

composite materials for strengthening purpose. They carried out an

experimental investigation to examine RC two-way slabs strengthened with

near surface mounted (NSM) CFRP rods as shown in Figure 2.6.


40

Figure 2.6: RC Slab Section Strengthened with NSM C FRP rods [41]

Two CFRP strengthened RC slabs, each measuring 1650mm length, 1150mm

width and 70mm thickness, reinforced with 6mm diameter of steel bars

positioned at two orthogonal directions were tested in the laboratory. The

concrete had compressive strength of about 40 MPa and the average elastic

modulus of steel was about 200GPa. CFRP strips had elastic modulus of

163GPa, 50mm width and a thickness of 1.4mm. These correspond to ρs and ρf

of 0.35% and 1.0% respectively. Results from the experimental tests had shown

the increase of flexural load capacity of CFRP strengthened slab up to 81% as

compared to the un-strengthened slab and the decrease of mid span deflection

of about 76% that of the control slab.

Michel et al [42] investigated the effective use of composite materials for

ultimate punching load of concrete slabs strengthened by CFRP. CFRP

strengthening RC slabs with 1200 x 1200 mm square geometry having ρs of

0.636% and ρf of 0.35% for one cross layer and 1.05% with three cross layers,

were loaded until failure as shown in Figure 2.7. The compressive strength of

concrete slab was 36 MPa and the steel yield strength was 500 MPa. Two-way

CFRP strips with elastic modulus of 240 GPa, ultimate tensile strength of 4 GPa

and ultimate elongation at break of 1.6%, were bonded at the bottom surface of

slab as shown in Figure 2.8. Experimental results show CFRP strips have

increased the slab ultimate punching load. The punching load for the slab

bonded with one cross layer CFRP has increased to about 15% and by 30% for


41

three cross layers. Reduction of the mid span deflection up to 35% with one

cross layers and 45% on three cross layers were also recorded.

Figure 2.7: Loading system and slab dimensions [42]

Figure 2.8: Two way CFRP strips being installed [42 ]


42

El-Sayed, El-Salakawy and Benmokrane [43] assessed the shear strength of

one-way concrete slabs reinforced with different types of FRP bars. All the eight

slabs of each size 3100mm long, 1000mm wide and 200mm deep were tested

under four point bending over a simply supported clear span of 2500mm and

shear span of 1000mm as shown in Figure 2.9. Five slabs were reinforced with

GRP bars and with FRP reinforcement ratio of 0.86%, 1.70%, 1.71%, 2.44%

and 2.63%. Three other slabs were reinforced with CFRP with reinforcement

ratio of 0.39%, 0.78% and 1.18%. The average concrete compressive strength

was 40 MPa, a modulus of elasticity of 30 GPa and average concrete tensile

strength of 3.5 MPa. The properties of reinforcing bars used include modulus of

elasticity of CFRP and GRP were 114 GPa and 40 GPa, respectively and

tensile strength of 1536 MPa and 597 MPa, respectively. From the tests, it

showed that the flexural stiffness of the slabs reinforced either with glass or

carbon FRP bars increased with an increase in the reinforcement ratio. The test

results indicate that the shear strength increases as the FRPs reinforcements

increases. With CFRP bars, the increase of ρf from 0.39 to 0.78% and 0.39 to

1.18% resulted shear capacity increased by 19% to 36%, respectively. For GRP,

an increase of 44% to 49% was obtained by increasing ρf from 0.86 to 1.71%

and 0.86 to 2.63% respectively. All tested slabs were failed in shear.


43

Figure 2.9: (a) Slab dimension ; (b) Test set-up [4 3]

Rochdi et al [44] investigated an analytical and experimental evaluation of the

strength of the two-way concrete slabs externally bonded with CFRP composite

sheets.

Eight slabs, each of in-plane size 600mm by 600mm and 50mm in thickness

(Figure 2.10) were tested to failure. Six slabs were strengthened with externally

bonded CFRP with variable thickness and the remaining two slabs were just

ordinary RC slabs with no FRP installed and acted as control specimens.

Material properties of Uni-directional (UD) CFRP used were elastic modulus of

82.6 GPa and tensile strength of 1140 MPa. RC slabs consist of an average

concrete compressive strength of about 25 MPa and were reinforced with steel

bars having reinforcement ratio of 0.12%. The steel bars had an average tensile

strength of 770MPa. The test showed that central deflections of all strengthened


44

slabs were considerably lower than the control slabs. The ultimate punching

shear strength of slab increases with increasing composite section in the range

of 67% to 177% over the control slabs. The average deflection at the ultimate

load was also reduced by 28% that of the corresponding reference specimens.

Shear failure was observed by the forming of the flexural cracks that appeared

at the slab top directly below the concentrated load. These cracks propagated

towards the sides of the slabs as the load increased. Finally, punching shear

failure occurred and the FRP reinforcement was also delaminated.

Figure 2.10: Dimensions and test set-up scheme [44]

These experimental results were also compared with the finite element

simulation. Using a 3D element, one quarter of the slab was modelled due to

the geometrical and loading symmetry as shown in Figure 2.11. The predicted

failure load presented was successfully showed a good comparison with the

experimental results.

The numerical study concluded that CFRP strengthening produced significant

improvements in punching shear strength.


45

Figure 2.11: Stress distribution for the analysed s labs [44]

Zhang et al [45] studied the flexural behaviour of one way concrete slabs

reinforced with CFRP grid reinforcements. The results of this behaviour were

then compared with conventional steel reinforcing rebars. All three slabs were

reinforced with CFRP grid reinforcement and one slab with steel rebars had the

same measurement of 3300 x 1000 x 250mm but with different reinforcement

ratios. Each of them was simply supported and tested under both static and

cyclic loading conditions in the laboratory to evaluate their flexural and shear

limit states (Figure 2.12).

Two types of CFRP were used in the experiment, namely, New Fibre

Composite Material for Reinforcing Concrete, NEFMAC C16 and NEFMAC

C19-R2. These materials comprised of modulus of elasticity of 98.1 and 90 GPa

with tensile strength of 1180 MPa and 1400 MPa and 0.49% and 0.99% of

reinforcement ratio, respectively. The slab’s concrete average compressive

strength of 45 MPa and modulus of elasticity of 37 GPa. One of the slabs that

acts as the control specimen was only reinforced with steel bars of grade 400

with reinforcement ratio of 0.69%.


46

Figure 2.12: Flexural Test Setup of Concrete Slabs [45]

A fairly regular crack pattern was observed at the location of the transverse bars

of the FRP grid as shown in Figure 2.13. It was also observed that the crack

width was larger, generally, in FRP reinforced slab than the steel reinforced slab

under the same applied loads. The deflection behaviour of steel reinforced slab

was generally characterised by plastic deformation as the steel yielded and non

plastic deformation behaviour was observed in FRP reinforced slabs followed

by brittle fracture due to crushing of the concrete.

Figure 2.13: Typical crack pattern of the CFRP grid reinforced slabs [45]


47

From the research work investigated, the experimental ultimate moment

recorded for steel bars reinforced slab was 184 kNm, and after post-

strengthening with CFRP 0.49% reinforcement ratio, the ultimate moment was

197 kNm. FRP reinforced slab also characterised by bi-linearly elastic until

failure as shown in Figure 2.14. This showed that FRP reinforcement structures

had greater ultimate flexural capacity than steel reinforced.

Figure 2.14: Typical theoretical and experimental l oad versus midspan deflection curve [45]

Ebead and Marzouk [46] investigated the effect of FRP strengthening on the

tensile behaviour of concrete slabs using finite element analysis (FEA). The

available experimental results of the strengthened reinforced concrete slabs

were used to calibrate the finite element model based on the ultimate load

carrying capacity of the two-way slabs. An overall increase in the post-peak

stiffness based on the tensile stress-strain relationship was observed. The

comparison of study between the tension-stiffening model of FRP strengthened

and un-strengthened concrete was the main focus of the research.

Slab specimens were measured by 1900 x 1900 x 150mm thick. Column stubs

were cast at the centre of the slab and of size 250 x 250mm. Two un-

strengthened specimens with variable reinforcement ratio ρs of 0.35% and 0.5%


48

were tested to failure and the ultimate load carrying capacity were 250kN and

330kN, respectively. Both GRP and CFRP strips were used at individual slab

and located at the tension side of the slab with reinforcement ratio ρs of 0.35%

and 0.5% for each type of FRPs. The modulus of elasticity of concrete, Ec, was

26.6 GPa and the compressive strength of concrete was 35 MPa. The tensile

strength of concrete was 2.8 MPa. The steel reinforcement’s yield stress and

the modulus of elasticity of 440 MPa and 210 GPa, respectively, were defined in

the simulation process.

The test results had been used to calibrate the FEA inputs data. The test

showed that GRP reinforced slabs exhibited an average gain in the load

carrying capacity of about 31% over that of the reference un-strengthened slabs.

Meanwhile, about an average of 40% increase with CFRP strips over the

control specimen was found.

From the finite analysis results, both CFRP and GRP strengthened slabs

showed the post behaviour of slabs was stiffened. The slope of the tensile

stress-tensile strain was decreased in the post-peak zone indicating the

contribution of the FRP strengthening materials in increasing the post-peak

stiffness of concrete in tension. The results of the load carrying capacity

between the FEA and the experimental test were compared and a good

agreement was found. FEA also exhibited a stiffer deformational behaviour

compared to the experimental results.

The study concluded that FRP strengthened concrete showed a stiffer post-

peak response than conventional reinforced concrete. Experimental test results

show that the use of FRP strengthened strips or laminates could lead to an

average gain in the load carrying capacity of about 36% over the un-

strengthened specimens. However, a decrease in ductility and energy

absorption was recorded due to the brittle nature of the FRP composites. It was

also observed that de-bonding between FRP composites and concrete was the

main cause of failure. None of the FRP composites experienced rupture. Slabs

were failed after exceeding flexural capacity.


49

Pesic and Pilakoutas [47] developed a numerical method for the computation of

the bending moment capacity and deflection of FRP plate reinforced concrete

beams and prediction of the flexural failure modes. The numerical procedure

was validated with available experimental results.

Material properties of concrete beam were defined such that concrete

compressive strength was 35 MPa, steel reinforcement ratio and yield strength

of 1.8% and 456 MPa, respectively, GRP elastic modulus, tensile strength and

reinforcement ratio of 32.7 GPa, 400 MPa and 1.11% respectively. The

experimental tests result showed that an increase of 33.3% of loading capacity

was achieved after post strengthening of RC beam with GRP plate.

Ashour et al [48] tested 16 RC continuous beams with different arrangements of

internal steel bars and external CFRP laminates (Figure 2.15). All test

specimens had the same geometrical dimensions and were classified into three

groups according to the amount of internal steel reinforcement. Each group

included one un-strengthened control beam designed to fail in flexure. Different

parameters including the length, thickness, position and form of the CFRP

laminates were investigated. Material properties used in the tests consist of

concrete compressive strength ranging from 24.0 MPa to 47.8 MPa and

average Young’s modulus, yield strength and ultimate strength of steel bars

were 200 GPa, 512 MPa and 616 MPa, respectively. The uni-directional CFRP

plates and sheets were used in the experiment. According to the manufacturer’s

recommendations, tensile strengths of CFRP plates of 1.2 mm thickness and

100 mm width and CFRP sheets of 0.1117 mm thickness and 110 mm width

were 3900 MPa and 2500 MPa. They have Young modulus of 240 GPa and

150 GPa, respectively. Results from the tests showed that an increase of

ultimate load was recorded after post strengthening with CFRP sheets and

plates. With CFRP sheets, failure load was increased up to 57% and 37% after

post strengthening. Three failure modes of beams with external CFRP

laminates were observed, namely, laminate rupture, laminate separation and

peeling failure of the concrete cover attached to the laminate. The ductility of all

strengthened beams was reduced compared with that of the respective un-

strengthened control beam.


50

Figure 2.15: Strengthened RC beam details [48]

The results from the experimental and the simplified methods were also

compared which showed that most beams were close to achieving their full

flexural capacity.

Mosallam and Mosalam [49] investigated the performance of two-way RC slabs

retrofitted with FRP composite laminates. CFRP strengthened RC slabs, each

measuring 2640mm width, 2640mm length and 76mm thick, were tested to the

ultimate load and compared with the control slab as shown in Figure 2.16.

Concrete slab’s average compressive strength was 32.87 MPa, steel bars of

grade 400 and CFRP having a thickness 0.58mm and 457mm width and elastic

modulus of 1.1GPa. These correspond to ρs and ρf of 0.64% and 0.79%

respectively. Results from the tests showed that CFRP strengthened RC slab

provided about 198% enhancement in flexural strength relative to the control

slab.


51

Figure 2.16: Test setup and instrumentation of the two-way slab specimens [49]


52

Sheikh [50] investigated the structural performance with the application of FRP

to strengthen and repair damaged slabs (Figure 2.17) and beams (Figure 2.18).

Figure 2.17: slab specimen details and loading [50]


53

Figure 2.18: Details of beam specimens (a) Cross Se ction and reinforcement details (b) Zone of FRP repair [50]

For each reinforced slab specimen of 1200mm wide, 2100mm long and 250mm

thick to be repaired with FRP (Figure 2.19 and Figure 2.20), three strips of

CFRP of about 600mm wide were used for slab2 and GRP for slab3. Slab2 was

loaded to failure in shear at a load of 478kN while slab3 shear failure occurred

at an applied load of 442kN.

Compressive strength of concrete was 30 MPa and grade 60 and grade 400

steel bars were used in the test.

Figure 2.19: Details of FRP retrofitting of slab sp ecimens [50]


54

Figure 2.20: slab specimens tested to failure [50]

Beam1 which considered as a control specimen formed crack widths exceeded

2.0mm at 1600kN loading. Beam1 failed at a load of 1700kN with a

corresponding deflection of 14mm. Beam2 was wrapped around the damaged

zone with three strips of 610mm wide carbon fabric. The maximum load applied

to this beam was 2528kN with corresponding deflection of 143mm.


55

Sheikh found out that FRP resulted a substantial increase in the ultimate

capacity of the slabs. The tests showed that 119% increase for GRP and 148%

for CFRP. The load corresponding to the shear capacity was much lower than

that for the enhanced flexural capacity. The failure in both repaired slabs was

caused by shear.

Sheikh concluded his investigation from experimental works that retrofitting with

FRP provided a feasible rehabilitation technique for repair as well as

strengthening. FRP wrapping was very effective in enhancing flexural strength

of the damage slabs, shear resistance of the damaged beams. Both CFRP and

GRP provided approximately 150% enhancement in flexural strength. Installing

one layer of CFRP had increased the beam failure capacity from 1700kN to

2528kN (about 49% enhancement).

Based on one of the experimental work conducted recently in the laboratory

within the school of civil engineering, University of Manchester [51], CFRP

strengthened RC slabs of dimension 1800mm length, 1800mm width and

150mm thick were tested to failure. Two CFRP strips, each measuring 100mm

width and 1.2mm thickness, having an elastic modulus of 150GPa and average

tensile strength of about 3 GPa, were bonded to the tensile face of the slabs in

each direction. The average concrete compressive strength was about 41 MPa

and steel reinforcement bars grade 600 were observed for the tests. The

corresponding ρs and ρf were 0.75% and 0.2% respectively. Results from the

experimental tests showed the increase of ultimate loading capacity of FRP

strengthened RC slab to about 43% and the reduction of central deflection up to

58%. These deflection results obtained from experimental tests were verified

with FEM which will be explained in the next section.

From the literature review stated above, obviously, different results of flexural

loading capacity enhancement and reduction of central deflection of post-

strengthening RC slab with FRP from the experimental works had been noted.

These variable results were affected due to several parameters including

material properties, geometric characteristics, loading and boundary conditions.


56

2.4 Conclusions

Plate theory remains one of the interesting topics to the present day particularly

for the applied mechanicians and structural engineers. It is interesting to note

that the subject of bending of plates was extended to shell where such element

was applied for finite element method. The thickness of the plate is very small

when compared to the other dimensions and full three dimensional numerical

analysis becomes not only costly but it often leads to serious numerical ill

conditioning problems.

In order to ease the solution, several classical assumptions with regard to the

behaviour of the structures were introduced which resulting a series of

approximations. Both Reissner and Mindlin have improved the classical thin

plate theory by incorporating the effect of transverse shear deformation,

however, the normal stress had been neglected in the Mindlin but it was

accounted for Reissner plate theory. From the literature, it clearly shows the

weakness of plate theory due to the assumptions made. The main problems on

the classical plate bending are due to the fact that the deflection is not a

constant across the thickness of the plate and it is only suitable for isotropic

plate having thickness to side length ratio of less than 1/10.

From the shortcomings of the plate theory, it brings interest and attention to

make comparison of classical plate theory over the exact three dimensional

elastic analysis. For this reason, the application of concrete slab with FRP as

laminated plate is investigated in this study to illustrate their differences on the

flexural deformation of the structure. The other application of the exact elasticity

analysis was conducted by Fan et. al. on partially clamped edges laminated

plate. Such problem with the given exact solutions is not complete, i.e. the plate

is not fully fixed at all edges. For this main reason, it attracts a new challenge

for the author to further investigate analytical exact analysis of a fully clamped

laminated plate. The novel solutions of such findings are presented in this thesis.


57

From the numerous researches conducted which are available in the literature

with regard to FRP strengthened RC slab, numerical analysis of FEM is also

carried out in this study for review purpose. This study aims to complement and

review on the practical significant enhancement of structural loading capacity of

RC slab when FRP is used.

The numerical results of this modeling and simulation using Abaqus are then

compared to the existing available experimental test results of similar

specimens. The results showed that punching shear strength and the stiffness

of the RC slabs have increased after FRPs are bonded to the slab.

FEM results also agreed well with the experimental test that the stiffness and

deflection of RC slabs have significantly increased and reduced, respectively

with the application of FRP.

Chapter 3 State Space Method of 3D Elasticity

58

Chapter 3

STATE SPACE METHOD OF 3D

ELASTICITY

The objective of this chapter is to review the concept of state space method and

the governing equations of elasticity. A classical (continuum) or analytical

method based on the application of the equations of equilibrium and

compatibility, together with the stress and strain relations for the material are

applied to produce governing equations which must be solved to obtain

displacements and stresses. It also shows the fundamental equations for

orthotropic plates based on the theory of plates and elasticity with respect to a

rectangular Cartesian coordinate system, x,y and z only. It is important to note

that it can also be represented in other coordinate system such as polar

coordinates and cylindrical coordinate systems.

From these governing equations for state space method, further development

into various applications having different boundary and loading conditions can

be explored. In the next chapter, state space equations are solved for the case

of partially fixed edges laminated plates subjected to various loading conditions.

Based on the formulation applied to a simply supported rectangular laminated

plate, state space methods are extended to the case of clamped edges

laminated plates.

3.1 The Concept of State Space Method of 3D Elastic ity

The state space method of 3D elasticity was initiated from Vlasov’s method of

initial function (MIF) [26] and Bahar’s transfer matrix approach [24]. The term

‘state space’ deals with a linear control system between the relationships of

actions and the responses of the related system. This system can be


59

represented an electrical, hydraulic, mechanical, thermal systems or others, for

instance, a spring-damper-mass system when subjected to excitation as shown

in Figure 3.1. The concept involved solving linear time-invariant system of the

elastic mechanical system subjected to an external source of loading. By using

the boundary conditions as the initial state of the system, a solution can be

determined in terms of a set of arbitrary constants which satisfy these boundary

conditions. From this, all the remaining outputs can be found at any time

between initial stage and the subsequent stages.

As an example to illustrate state space variables, a governing second order

linear differential equation of motion of such system can be represented by [52]:

( ) ( ) ( ) ( )tFtkxtxctxm =++ &&& (3-1)

where

m is the mass of the object

x is the displacement

t is the time

F is the force applied

c is the damping coefficient

k is spring constant

Figure 3.1: Spring - damper - mass system [52]

The equation (3-1) can be converted into a first order linear differential equation

system in a matrix form of [52]:

( )( )

( )( ) ( )

+

−−=

mtFtx

tx

m

c

m

ktx

tx

dt

d

/

010

&& (3-2)

k

c

m F ( t )

x(t)


60

or

( ){ } [ ] ( ){ } [ ]{ }tBtxAtx +=& (3-3)

where A is an n x n matrix and B(t) is an n x 1 vector. Notice that constant

matrix A is independent of time, hence equation (3-3) is called linear time-

invariant system. If matrix A is a function of time, i.e. A(t), the solution of this

linear time-variant system becomes more complicated. If equation (3-3) were

treated as a scalar differential equation, i.e. n = 1, then it would be expressed

as the following:

( ) ( ) ( )tbtaxtx +=& (3-4)

where a and b are constant. The solution of equation (3-4) can be found as:

( ) ( ) ( ) τττ dbeexetxt

aatat∫

−+=0

0 (3-5)

where to = 0 is assumed and x(0) = xo = 0 at to = 0.

Similar to the solution for the scalar differential equation, if to ≠ 0, the general

solution in terms of n x n matrix would become

( ) [ ]( ) ( ) [ ]( ) [ ] ( ) τττ dBeetxetxt

t

AttAttA

∫−−−

+=0

00

0 (3-6)

Hence, the final solution of equation (3-3) is given by

( ) [ ] ( ) [ ]( ) ( ) τdτBexetxt

τtAtA∫

−+=0

0 (3-7)

The concept can be extended to the three dimensional analysis of plate. The

application of state equations is adopted if knowing the initial boundary and

loading conditions of the plate as the initial state of the system. The

displacements and stresses at an arbitrary position across thickness direction of

the plate can be solved subsequently by using transfer matrix approach.

The application of the above state space equations for plate will be discussed in

more details in the following chapter.


61

3.2 Governing Equations of Elasticity Problems

The general governing equations for elasticity problem are as follows:

Equilibrium equations [7]:

2

2

2

2

2

2

t

Wρf

z

σ

y

τ

x

τ

t

Vρf

z

τ

y

σ

x

τ

t

Uρf

z

τ

y

τ

x

σ

zzyzxz

y

yzyxy

x

xzxyx

∂∂=+

∂

∂+

∂

∂+

∂

∂∂∂=+

∂

∂+

∂

∂+

∂

∂∂∂=+

∂

∂+

∂

∂+

∂

∂

(3-8)

where fx, fy and fz are the body force per unit volume in the x, y and z direction,

respectively. ρ is the density coefficients of the material and U, V and W are the

in-plane and transverse displacement respectively.

Kinematic (strain-displacement relations) equations [7]:

y

U

x

Vγ

x

W

z

Uγ

z

V

y

Wγ

z

Wε

y

Vε

x

Uε

xy

zx

yz

z

y

x

∂∂+

∂∂=

∂∂+

∂∂=

∂∂+

∂∂=

∂∂=

∂∂=

∂∂=

(3-9)

where U, V and W represent the displacements along the coordinate axes x, y

and z, respectively.


62

Stress – strain relationship for a general anisotropic linearly elastic material [7]:

=

xy

zx

yz

z

y

x

xy

zx

yz

z

y

x

γ

γ

γ

ε

ε

ε

CCCCCC

CCCCCC

CCCCCC

CCCCCC

CCCCCC

CCCCCC

τ

τ

τ

σ

σ

σ

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

(3-10)

where Cij (i,j = 1,2,3…..,6) are the material stiffness coefficients of anisotropic

body.

Compatibility equations:

In the strain-displacement relationships, there are six strain measures

xzxyzyxε,ε,ε,ε,ε and

yzε but only three independent displacements.

This means that there are six unknowns for only three independent variables.

As a result, there are constraints or compatibility equations exist, such as

xz

γ

z

ε

x

ε

zy

γ

y

ε

z

ε

yx

γ

x

ε

y

ε

zxxz

yzzy

xyyx

∂∂

∂=

∂

∂+

∂

∂

∂∂

∂=

∂

∂+

∂

∂

∂∂

∂=

∂

∂+

∂

∂

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

(3-11)

By using all the above equations, there are three methods to solve the elasticity

problem. Firstly, applying the displacements approach, secondly, using the

stresses approach and lastly using both approaches simultaneously which is

also known as hybrid method [7].


63

Displacement approach expresses stresses in terms of displacements following

by substitution of these stresses into the equilibrium equations. The solutions

expressed in terms of displacements are initially determined and followed by

strains and the stresses.

Stress approach determines the solution of six stresses firstly and satisfies the

compatibility conditions, equilibrium equations and stress boundary conditions.

Strains can be then determined afterwards. It is essential to note that in this

approach, the determination of displacements becomes more complicated as it

involves integration functions of the kinematic equation.

Hybrid approach seeks the solution for both displacement and some stress

components simultaneously.

3.3 State Equations for Simply Supported Orthotropi c Plate

A simply supported rectangular orthotropic thick plate is shown Figure 3.2

Figure 3.2: Coordinate system and plate dimension

a

b

1,U,x

3,W,z

2,V,y

h


64

Letting zyzxz

σZ,τY,τX === (X and Y are the transverse shear stresses. Z

is the normal stress).

And eliminating xyyx

τσσ ,, and the strains, by using the strain-displacement

and the equilibrium equations, the following expression can be obtained:

( )( )

−−+−

+−−−−−

−−

=

∂∂

W

Y

X

Z

V

U

CβCαC

βCβCαCξαβCC

αCαβCCβCαCξ

ξβα

βC

αC

W

Y

X

Z

V

U

z

000

000

000

000

0000

0000

751

5

2

4

2

6

2

63

163

2

6

2

2

2

29

8 (3-12)

where

2

22

449

558

337666

33

235

33

2

23224

33

2313123

33

2

13112

33

131

111

tρξ,

yβ

,x

α,C

C,C

C,C

C,CC,C

CC

,C

CCC,

C

CCCC,

C

CCC,

C

CC

∂∂=

∂∂=

∂∂=====−=

−=−=−=−=

where


65

( ) ( ) ( )

( ) ( ) ( )

yzzzyyxzzzxxxyyyxx

zxxzxyxzzxzyyzyxxy

xyzxyz

yxxyzzxxyzyyzxxzy

zyyxzxxyzzxyxxzyyzx

µEµE,µEµE,µEµE

,µµµµµµµµµQ

,GC,GC,GC

Q

µµEC,

Q

µµµEC,

Q

µµEC

Q

µµµEC,

Q

µµµEC,

Q

µµEC

===

−−−−=

===

−=

+=

−=

+=

+=

−=

21

11

1

665544

332322

131211

The derivation of the above equation (3-12) is shown as follows:

To eliminate three in-plane stresses, xyyx

τandσ,σ , and obtain the state

equation for elasticity, the following tasks are performed.

For orthotropic linear elasticity, from equation (3-9), stress-strain relationship

reduces to:

∂∂+

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂∂∂∂∂

=

=

y

U

x

Vx

W

z

Uz

V

y

Wz

Wy

Vx

U

C

C

C

CCC

CCC

CCC

γ

γ

γ

ε

ε

ε

C

C

C

CCC

CCC

CCC

τ

τ

τ

σ

σ

σ

xy

zx

yz

z

y

x

xy

zx

yz

z

y

x

66

55

44

332313

232212

131211

66

55

44

332313

232212

131211

00000

00000

00000

000

000

000

00000

00000

00000

000

000

000

(3-13)


66

From the third equation of (3-8) and neglecting the body force, fz,

y

τ

x

τ

t

Wρ

z

σ yzxzz

∂

∂−

∂

∂−

∂∂=

∂

∂2

2

(3-14)

From the third, fourth and fifth equations of (3-13), we can get,

x

Wτ

Cz

U

y

Wτ

Cz

V

σCy

V

C

C

x

U

C

C

z

W

xz

yz

z

∂∂−=

∂∂

∂∂−=

∂∂

+∂∂−

∂∂−=

∂∂

55

44

3333

23

33

13

1

1

1

(3-15)

Substitute xyx

τandσ from equation (3-13) and considering the first equation of

(3-15) into (3-8), we obtain,

x

σ

C

CV

yxC

yxC

CCC

Uy

CxC

CC

tρ

z

τ

z

xz

∂

∂−

∂∂∂+

∂∂∂

−

−

∂∂−

∂∂

−−

∂∂=

∂

∂

33

132

66

2

33

231312

2

2

662

2

33

2

13112

2

(3-16)

Substitute xyy

τandσ from equation (3-13) and considering the first equation

of (3-15) into (3-8), we obtain,

y

σ

C

CV

xC

yC

CC

tρ

Uyx

CyxC

CCC

z

τ

z

yz

∂

∂−

∂∂−

∂∂

−−

∂∂

+

∂∂∂+

∂∂∂

−−=

∂

∂

33

232

2

662

2

33

2

23222

2

2

66

2

33

231312

(3-17)


67

Let 2

22

tξ,

yβ,

xα,σZ,τY,τX

zyzxz ∂∂=

∂∂=

∂∂==== and use

449

558

337666

33

235

33

2

23224

33

2313123

33

2

13112

33

131

111

CC,

CC,

CC,CC,

C

CC

,C

CCC,

C

CCCC,

C

CCC,

C

CC

====−=

−=−=−=−=

Eqns. (3-14) to (3-17) can be simplified in the state equation as

( )( )

−−+−

+−−−−−

−−

=

∂∂

W

Y

X

Z

V

U

CβCαC

βCβCαCξαβCC

αCαβCCβCαCξ

ξβα

βC

αC

W

Y

X

Z

V

U

z

000

000

000

000

0000

0000

751

5

2

4

2

6

2

63

163

2

6

2

2

2

29

8 (3-18)

The eliminated stress components can be determined from (3-13) as

−−

=

Z

V

U

αCβC

CβCαC

CβCαC

τ

σ

σ

xy

y

x

066

543

132

(3-19)

For the derivation of eqns. (3-19), it can be shown below,

Substitute the first eqn.(3-15) into (3-13),


68

ZCVβCUαCσ

σC

C

y

V

C

CCC

x

U

C

CCσ

σCy

V

C

C

x

U

C

CC

y

VC

x

UCσ

z

WC

y

VC

x

UCσ

x

zx

zx

x

132

33

13

33

231312

33

2

1311

3333

23

33

13131211

131211

1

−+=∴

+∂∂

−+

∂∂

−=

+

∂∂−

∂∂−+

∂∂+

∂∂=

∂∂+

∂∂+

∂∂=

Similarly for σy, substitute the first eqn.(3-15) into (3-13),

ZCVβCUαCσ

σC

C

y

V

C

CC

x

U

C

CCCσ

σCy

V

C

C

x

U

C

CC

y

VC

x

UCσ

z

WC

y

VC

x

UCσ

y

zy

zy

y

543

33

23

33

2

2322

33

231312

3333

23

33

13232212

232212

1

−+=∴

+∂∂

−+

∂∂

−=

+

∂∂−

∂∂−+

∂∂+

∂∂=

∂∂+

∂∂+

∂∂=

and taking the last expression of eqn. (3-13),

VαCUβCτ

y

U

x

VCτ

xy

xy

66

66

+=∴

∂∂+

∂∂=

The above elastic constants have a relation with engineering elastic stiffness

coefficients in the form of [27]:


69

( ) ( ) ( )

( ) ( ) ( )

yzzzyyxzzzxxxyyyxx

zxxzxyxzzxzyyzyxxy

xyzxyz

yxxyzzxxyzyyzxxzy

zyyxzxxyzzxyxxzyyzx

µEµE,µEµE,µEµE

,µµµµµµµµµQ

,GC,GC,GC

Q

µµEC,

Q

µµµEC,

Q

µµEC

Q

µµµEC,

Q

µµµEC,

Q

µµEC

===

−−−−=

===

−=

+=

−=

+=

+=

−=

21

11

1

665544

332322

131211

(3-20)

For Figure 3.2, the case of a simply supported plate, the boundary conditions

may be expressed as [27]:

x = 0 and a ; σx = 0, W = 0 and V = 0,

y = 0 and b; σy = 0, W = 0 and U = 0. (3-21)

Letting

( )

( )

( )

∑∑∞

=

∞

=

=

1 1m n

mn

mn

mn

b

yπnSin

a

xπmSinzW

b

yπnCos

a

xπmSinzV

b

yπnSin

a

xπmCoszU

W

V

U

(3-22)

and

( )

( )

( )

∑∑∞

=

∞

=

=

1 1m n

mn

mn

mn

b

yπnSin

a

xπmSinzZ

b

yπnCos

a

xπmSinzY

b

yπnSin

a

xπmCoszX

Z

Y

X

(3-23)

It can be noted that from eqn.(3-19), (3-21) and (3-22), the boundary conditions

of eqn. (3-21) and the state eqn. (3-18) are satisfied.


70

And finally, substitute equation (3-22) and (3-23) into (3-18), the following first-

order homogeneous ordinary differential equation can be found for each

combination of m and n [27]:

( )( )

−−++

++

−−

=

mn

mn

mn

mn

mn

mn

mn

mn

mn

mn

mn

mn

W

Y

X

Z

V

U

CηCξC

ηCηCξCξηCC

ξCξηCCηCξC

ηξ

ηC

ξC

W

Y

X

Z

V

U

dz

d

000

000

000

0000

0000

0000

751

5

2

4

2

663

163

2

6

2

2

9

8

(3-24)

or

( ) ( )zDRzRdz

dmnmn

= (3-25)

where b

n

a

m πηπξ == ,


71

( )

=

mn

mn

mn

mn

mn

mn

mn

W

Y

X

Z

V

U

zR and

( )( )

−−++

++

−−

=

000

000

000

0000

0000

0000

751

5

2

4

2

663

163

2

6

2

2

9

8

CηCξC

ηCηCξCξηCC

ξCξηCCηCξC

ηξ

ηC

ξC

D

From eqn. (3-24), a sixth order differential equation of any of the six

components can be obtained such as the equation of transverse displacement

Wmn [27],

02

2

4

4

6

6

=+++mno

mn

o

mn

o

mn WCdz

WdB

dz

WdA

dz

Wd (3-26)

where Ao, Bo and Co are the coefficient matrix.

The derivation of eqn. (3-26) is clearly shown below:

Initially, eliminate Xmn and Ymn from eqn. (3-24) by letting,

=

=

⇒

=

Z

V

U

B

W

Y

X

dz

dand

W

Y

X

A

Z

V

U

dz

d

W

Y

X

Z

V

U

B

A

W

Y

X

Z

V

U

dz

d

0

0


72

where

( )( )

−−++

++

=

−−

=

751

5

2

4

2

663

163

2

6

2

2

9

8

0

0

0

CηCξC

ηCηCξCξηCC

ξCξηCCηCξC

Band

ηξ

ηC

ξC

A

Differentiate the second part and substitute it into the first part, i.e.

=

=

=

=

=

W

Y

X

KKK

KKK

KKK

W

Y

X

dz

d

A.BKwhere

W

Y

X

.K

W

Y

X

.A.B

Z

V

U

dz

dB

W

Y

X

dz

d

333231

232221

131211

2

2

2

2

Therefore

WKYKXKdz

Wd

WKYKXKdz

Yd

WKYKXKdz

Xd

3332312

2

2322212

2

1312112

2

++=

++=

++=

(3-27)

Substitute Y from the third into the first eqn. (3-27), gives

WK

KKK

dz

Wd

K

KX

K

KKK

dz

Xd

WKWKXKdz

Wd

KKXK

dz

Xd

−++

−=∴

+

−−+=

32

3312132

2

32

12

32

3112112

2

1333312

2

3212112

2 1

(3-28)

Again, substitute Y from the third into the second eqn. (3-27), yields


73

( )( )WKKKK

dz

WdKXKKKK

dz

WdK

dz

XdK

dz

Wd

WKWKXKdz

Wd

KKXK

WKXKdz

Wd

Kdz

d

33223223

2

2

22312221322

2

332

2

314

4

2333312

2

322221

33312

2

322

2

1

1

−

++−=−−∴

+

−−+

=

−−

(3-29)

Multiply the term K31 to the both sides of eqn.(3-28) and then adding the

expression to eqn.(3-29). By doing this, we can eliminate the term 2

2

dz

Xd from

eqn.(3-27),

WK

KKKK

dz

Wd

K

KKX

K

KKKK

dz

XdK

−++

−=

32

331213312

2

32

1231

32

311211312

2

31

Add this expression to eqn.(3-29),


74

( ) ( )

−

+−

+

++

−

−+−=∴

−+−+

++

−=

−+−⇒

−++

−

+−++−=−

WKK

K

KKK

KKKK

dz

Wd

K

KKKK

dz

Wd

K

KK

KKKKKKX

WKKK

KKKKKKK

dz

Wd

K

KKKK

dz

Wd

K

KKKKKKKKX

WK

KKKK

dz

Wd

K

KKX

K

KKKK

WKKKKdz

WdKXKKKK

dz

WdK

dz

Wd

133132

313312

32233322

2

2

32

31122233

4

4

32

2

3112

113131222132

133132

313312322333222

2

32

31122233

4

4

32

2

3112113131222132

32

331213312

2

32

1231

32

31121131

332232232

2

22312221322

2

334

4

1

(3-30)

Express Y from the third eqn.(3-27), then substitute X from eqn.(3-30),

−−=

++=

WKXKdz

Wd

KY

WKYKXKdz

Wd

33312

2

32

3332312

2

1


75

WK

K

WKK

K

KKK

KKKK

dz

Wd

K

KKKK

dz

Wd

K

KK

KKKKKKK

K

dz

Wd

KY

32

33

133132

313312

32233322

2

2

32

3112

2233

4

4

32

2

3112

11313122213232

31

2

2

32

1

1

−

−

+−

+

++

−

−+−

−=

(3-31)

And finally substitute eqns.(3-30) and (3-31) into the first expression of eqn.(3-

27),

WKYKXKdz

Xd1312112

2

++=


76

WK

WK

K

WKK

K

KKK

KKKK

dz

Wd

K

KKKK

dz

Wd

K

KK

KKKKKKK

K

dz

Wd

K

K

WKK

K

KKK

KKKK

dz

Wd

K

KKKK

dz

Wd

K

KK

KKKKKKK

WKK

K

KKK

KKKK

dz

Wd

K

KKKK

dz

Wd

K

KK

KKKKKKdz

d

13

32

33

133132

313312

32233322

2

2

32

31122233

4

4

32

2

3112

11313122213232

31

2

2

32

12

133132

313312

32233322

2

2

32

31122233

4

4

32

2

3112

11313122213211

133132

313312

32233322

2

2

32

31122233

4

4

32

2

3112

1131312221322

2

1

1

1

1

+

−

−

+−

+

++

−

−+−

−

+

−

+−

+

++

−

−+−

=

−

+−

+

++

−

−+−


77

Rearrange the above expression, gives

WKWK

KK

WKK

K

KKK

KKKK

dz

Wd

K

KKKK

dz

Wd

K

KKKK

KKKK

K

KK

dz

Wd

K

K

WKK

K

KKK

KKKK

dz

Wd

K

KKKK

dz

Wd

K

KK

KKKKKK

K

dz

Wd

KKK

KKK

KKKK

dz

Wd

K

KKKK

dz

Wd

K

KK

KKKKKK

1332

1233

133132

313312

32233322

2

2

32

31122233

4

4

32

2

31121131

31222132

32

31122

2

32

12

133132

313312

32233322

2

2

32

311222334

4

32

2

3112

113131222132

11

2

2

133132

313312

32233322

4

4

32

311222336

6

32

2

3112

113131222132

1

+−

−

+−

+

++

−

−+

−−

+

−

+−

+

++−

−

+−

=

−

+−

+

++−

−

+−


78

W

K

KKKK

KKKK

WK

KK

K

KKKKKKKK

WKK

K

KKK

KKKK

dz

Wd

K

KKKK

dz

Wd

K

KK

dz

Wd

K

K

K

KKKKKKKK

WKK

K

KKK

KKKK

dz

Wd

K

KKKK

dz

WdK

dz

Wd

KKK

KKK

KKKK

dz

Wd

K

KKKK

dz

Wd

−

+−

+

−+−

−

−

+−

+

++−

−

−+−

+

−

+−

+

++−

=

−

+−

+

++−⇒

32

2

31121131

31222132

32

1233

32

2

3112113131222132

133132

313312

32233322

2

2

32

311222334

4

32

3112

2

2

32

12

32

2

3112113131222132

133132

313312

32233322

2

2

32

311222334

4

11

2

2

133132

313312

32233322

4

4

32

311222336

6

( )

WK

KKK

K

KKKKKKKK

WKK

K

KKK

KKKK

K

KK

dz

Wd

K

KKKK

K

KK

dz

Wd

K

KK

dz

Wd

K

K

K

KKKKKKKK

WKK

K

KKK

KKKK

Kdz

Wd

K

KKKKK

dz

WdK

dz

Wd

KKK

KKK

KKKK

dz

WdKKK

dz

Wd

−

−+−

+

−

+−

−

++

+−

−+−

+

−

+−

+

++−

=

−

+−

+−−−+⇒

32

331213

32

2

3112113131222132

133132

313312

32233322

32

31122

2

32

31122233

32

3112

4

4

32

31122

2

32

12

32

2

3112113131222132

133132

313312

32233322

112

2

32

31122233114

4

11

2

2

133132

313312

32233322

4

4

2233116

6


79

( )

0

1332

1233

32

2

3112113131222132

1132

31121331

32

31331232233322

2

2

32

12

32

2

3112113131222132

32

311211

32

31122233

133132

31331232233322

4

4

2233116

6

=

−

−+−

−

−

−+−

+

−+−

−

−

++

+

−+−

+−−−+

W

KK

KK

K

KKKKKKKK

KK

KKKK

K

KKKKKKK

dz

Wd

K

K

K

KKKKKKKK

K

KKK

K

KKKK

KKK

KKKKKKK

dz

WdKKK

dz

Wd

Therefore, for simplification,

02

2

4

4

6

6

=+++mno

mn

o

mn

o

mn WCdz

WdB

dz

WdA

dz

Wd

where

−

−+−

−

−

−+−=

−+−

−

−

++

+

−+−=

−−−=

1332

1233

32

2

3112113131222132

1132

31121331

32

31331232233322

32

12

32

2

3112113131222132

32

311211

32

31122233

133132

313312

32233322

223311

KK

KK

K

KKKKKKKK

KK

KKKK

K

KKKKKKKC

K

K

K

KKKKKKKK

K

KKK

K

KKKK

KKK

KKKKKKKB

KKKA

o

o

o


80

Since

( )( )

−−++

++

=

−−

=

751

5

2

4

2

663

163

2

6

2

2

9

8

0

0

0

CηCξC

ηCηCξCξηCC

ξCξηCCηCξC

Band

ηξ

ηC

ξC

A

( ) ( )[ ] ( )

( )[ ] ( )

( ) ( )

+−−

++−++++

++−++++

=

==

2

5

2

1957817

2

6

2

4

2

32

6

2

49

2

56385

63

2

2

2

6391

2

6

2

28

2

1

333231

232221

131211

2

2

ηCξCCCCηCCCξ

ξCηC

ξCηξCηCCηCCCCCξη

CCη

ξCξCCCCξηηCξCCξC

KKK

KKK

KKK

A.BK

The numerical results for the above equation (3-26) can be referred to Wu [27]

where the solutions of a sixth order differential equation governing the

transverse displacement Wmn and stresses can be obtained. It is also

interesting to note that various expressions of the solutions are also provided.

3.4 Conclusions

The first section of this chapter introduces the basic introduction of state space

method. It represents a physical system in the form of mathematical model

where a set of input, output and state variables are related by a first order

differential equation.

The idea of expression of a spring-mass damper system as one of the

application of ‘state space’ is that when the initial conditions of the system are

known, i.e. external loading and displacements when time = 0, the subsequent

displacements can be determined at any given time. The importance expression


81

of such spring mass damper system which express in-terms of a linear time-

invariant function, makes it suitable for the analysis of the three dimensional

analysis of laminated plate. Applying this concept to the laminated composite

plate, the plate itself is divided firstly, into several layers. For this particular case,

when the external loading is applied to the surface of the plate, and knowing the

initial boundary conditions of the overall plate, the subsequent displacements

and stresses can be determined throughout the thickness of the plate. The

larger the number of sublayers within the plate, the more accurate solutions will

be produced. The great advantage of this concept for the application of

laminated plate is due to fact that as each layer of the laminated plate consists

of different material properties, the displacements and stresses can be found

simultaneously. There will be continuity at the interfaces of each layer which

means that the displacements will be exactly the same between the bottom

surfaces of the layer to the top surface of the adjacent layer. However, the in-

plane stresses will not be the same at the interfaces because of the different

elastic modulus of that layer.

The second part of this chapter reviews and rederives the governing equations

of elasticity which are essential equations for solving the stresses and

displacements of any laminated structures.

A simply supported orthotropic plate is shown in the last section of this chapter.

It is importance to understand this section thoroughly as it links to the following

chapter where a clamped edges laminated plate is presented. The relationship

is such that the boundary conditions of the simply supported plate are treated or

modified by using a traction in the form of mathematical function so that all

edges of the simply supported are satisfied in x and y directions.

Chapter 4 State Space Solution of Clamped Edges Laminated Plate

82

Chapter 4

STATE SPACE SOLUTION OF

CLAMPED EDGES LAMINATE D

PLATE

The investigation of 3D elasticity using state space method has been increased

in the past two decades and the application of such approach has been widely

applied to various fields. One of the significant works was performed by Fan

and Sheng on exact elasticity solution of partially clamped thick laminate in the

early 1990’s. The exact solutions obtained from their works have inspired the

author to explore further studies on various boundary and loading conditions of

laminate plate. In this and the next chapters, the state space method for

clamped plates with various loading and boundary conditions are presented.

The exact solution approach consists of determining the displacements of

rectangular plate by setting a general expression of displacement field

according to the boundary and loading conditions. The displacement field is

introduced to the equations of equilibrium which are then solved. The solving

technique can be applied to various loading and boundary conditions. In this

chapter, clamped edges plate with uniformly distributed loading is considered.

State space methods together with state transfer matrix, and with the aid of

programming code, are presented in this chapter to investigate the plate

behavior with various boundary, loading conditions and material properties. The

objective is to determine the exact elasticity solution of orthotropic plate with

various parameters by analytical analysis. The idea is to analyse the fixed

edges plate simply by applying traction to the edges of a simply supported plate


83

by superposition principle of elasticity. This principle is further explained in the

following section.

Knowing and understanding programming code is highly essential and utmost

importance in this work as massive iteration process is involved in order to get

the solutions. In this case, one of the available programming program, namely,

Mathematica (version 8), is used to create the codes and run the iteration

process.

4.1 State Space Solution of a Single Layer Plate

To understand briefly the process involved in the state transfer matrix, a single

layer of plate is considered as shown in Figure 4.1. Suppose a rectangular plate

of length b, width a and uniform thickness h with all edges being clamped.

Figure 4.1: A single layer plate [52]

The boundary conditions of the clamped plate are:

x = 0 and a ; U = V = W = 0,

y = 0 and b; U = V = W = 0.

a

b

x , U , 1

z , W , 3

y , V , 2

h


84

In order to satisfy the boundary conditions for clamped plates, the expressions

(3-22) which show the displacements expression for a simply supported plate

have to be treated in such a way it will satisfy the behaviour of clamped edges

plate. The treatment considers the combination of the bending of a simply

supported single lamina subjected to external transverse loading and in-plane

normal tractions along the simply supported edges. These tractions cause the

in-plane displacements of the simply supported to be nullified, resulting no

deformation to occur at all edges along the x and y directions. In another way to

explain this phenomenon, the in-plane normal tractions superpose the in-plane

displacements caused by the external transverse loading of the simply

supported plate. Consequently, U and V are zero at all edges of the plate.

Therefore, when all edges of the simply supported plate become fixed, the plate

is then behaves like a fully clamped edges structure. This means that only in-

plane displacements, U and V along the edges of a simply supported plate are

treated to behave as a fully clamped structure.

Such treatment can be expressed mathematically as,

V

U

fVV

fUU

+=

+=

where

VandU are the assumed in-plane displacements for a simply supported, i.e.

( )

( )

( )∑∑

∑∑

∑∑

∞ ∞

∞ ∞

∞ ∞

=

=

=

m nmn

m nmn

m nmn

b

yπnSin

a

xπmSinzWW

b

yπnCos

a

xπmSinzVV

b

yπnSin

a

xπmCoszUU

(4-1)

fU and fV are the specified functions that suppressed the in-plane

displacements of the plate along the edges. U and V are the resulted in-plane

displacements of a clamped edges plate.


85

The above plate is subjected to a uniformly distributed loading on the top

surface and all edges are simply supported and tractions are applied along the

edges so that all the in-plane displacements are zero along the edges. The

motivation for such a treatment for the boundaries is based on an application of

the superposition principle of elasticity: taking the corresponding simply support

plate as a template released structure, then applying the relevant tractions

(which are unknown now) along the edges of the released structure, and finally

ensuring the necessary deformation compatibility conditions to be satisfied.

For a single lamina plate, the specified functions fu and fv, can be assumed to

be linear to achieve excellent approximations of the unknowns. Obviously, for

greater accuracy, the plate must be divided into many thin layers as described

in more details in the following section. Providing all the interfaces preserve

continuity conditions of the state variables, external transverse loading are

applied on the surfaces and satisfies the clamped edges boundary conditions,

the solution for the laminated plate can be determined by solving a system of

linear algebra equation.

The following case of laminated plate comprises of many thin layers after

division process. The treatment is similar to a single thick homogenous plate as

described above.

4.2 State Space Solution of Laminated Plate

For the case of laminated plate having different material properties at each layer,

the following illustration is shown.

Considering a laminated plate composed of a number of different material

laminae. For an arbitrary ply in the laminated plate with clamped edges: jth ply

(Figure 4.2), we can assume that jth ply with clamped edges is equivalent to the

simply supported jth ply by adding assumed displacements function (traction) to

the simply supported plate. By applying this traction, all in-plane displacements

at the simply supported plate edges would become nullified, i.e. clamped edges.


86

A single ply is considered first and the lamination theory is taken into account

afterwards by using transfer method and recursive formulation.

The laminated composite structure comprises of layers which are bonded

together at the interface and still maintains the continuity. In this study, the

method used to solve the state space equations with respect to variable z-

coordinate is state transfer matrix approach which produces an analytical

solution satisfying all boundary conditions throughout all plies of the structure.

The idea is to create the relationship between the top and bottom surfaces of

the plate based on 3D elasticity and state space method, applying the existing

boundary conditions so that the initial displacements U, V and W can be

determined. Only knowing the values of these displacements, all the unknowns

constants can be solved which in turn gives the final exact solutions in term of

stresses and displacement at any z-location across the thickness of the plate.

Figure 4.2: Geometry and coordinate systems of the laminate [52]

(1)

(2)

( j )

( N )

b

x ,U,1

z , W,3

y ,V,2

h

(j+1)

a


87

Consider the jth ply of the laminate as shown in Figure 4.2, adding the in-plane

displacement (traction) functions ( ) ( ) ( ) ( ) ( ) ( )zxVzyUzyUj

ajj

,,,,, 00 and

( ) ( ),, zxV bj

along all edges, the displacement functions of the plate are

assumed as

( )

( )z,y,xfVV

z,y,xfUU

j

Vjj

j

Ujj

+=

+= (4-2)

where jj

VandU are the displacements assumed for simply supported plate

and W remains the same as the stated in eqn.(3-22), and [30]

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )zxVygzxVygzyxf

zyUxfzyUxfzyxf

bjj

jV

ajj

jU

,,,,

,,,,

20

1

20

1

+=

+= (4-3)

So the displacements of the jth ply, after applying traction, are assumed as [30]

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )z,xVygz,xVygVz,y,xfVV

z,yUxfz,yUxfUz,y,xfUU

b

jjj

j

Vjj

a

jjj

j

Ujj

2

0

1

2

0

1

++=+=

++=+= (4-4)

where f1(x), f2(x), g1(y) and g2(y) can be of any functions.

In this study, it is assumed that

( ) ( ) ( ) ( )b

yyg

b

yyg

a

xxf

a

xxf =−==−=

212111

These functions are substituted into eqns. (4-4) and yield the in-plane

displacement as [30]

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )z,xVb

yz,xV

b

yVV

z,yUa

xz,yU

a

xUU

b

jjjj

a

jjjj

+

−+=

+

−+=

0

0

1

1

(4-5)


88

Let

( )

( )

( )j

m n

mn

mn

mn

b

yπnSin

a

xπmSinzZ

b

yπnCos

a

xπmSinzY

b

yπnSin

a

xπmCoszX

Z

Y

X

∑∑∞

=

∞

=

=

1 1

and

( )

( ) ( ) ( )

( )

( ) ( ) ( )

( )

( ) ( ) ( )

( )

( ) ( ) ( )jm

b

m

mm

n

a

n

nn

a

xπmSinzVz,xV

a

xπmSinzVz,xV

b

yπnSinzUz,yU

b

yπnSinzUz,yU

b

a

=

=

=

=

∑

∑

∑

∑

0

0

0

0

(4-6)

To verify the boundary conditions of the clamped edges plate are satisfied,

substitute x = 0 and a and y = 0 and b into eqns. (4-1), (4-5) and (4-6), they all

show that U ,V and W are equal to 0 along the edges of the plate, hence they

satisfy the boundary conditions of a clamped edges plate.

By adding equation (4-4) to (3-12) of simply supported plate as discussed in the

previous chapter, the state equation can be deduced to [30]

( ){ } ( ){ } ( ){ } jmnjmnjjmn

zBzRDzRz

+=∂∂

(4-7)

The derivation expression of eqn.(4-7) is shown below,

By differentiating


89

( ) ( )

( ) ( )

( ) ( )

( ) ( )ygz

Vyg

z

V

z

V

z

V

VygVygVV

and

xfz

Uxf

z

U

z

U

z

U

UxfUxfUU

bo

b

ao

a

21

2

0

1

21

2

0

1

∂∂+

∂∂+

∂∂=

∂∂

++=

∂∂+

∂∂+

∂∂=

∂∂

++=

(4-8)

Substitute the above eqn.(4-8) into eqn.(3-12), give

( ) ( )

( ) ( )xfz

Uxf

z

UWαXC

z

U

WαXCxfz

Uxf

z

U

z

U

ao

ao

218

821

∂∂−

∂∂−−=

∂∂∴

−=∂

∂+∂

∂+∂∂

(4-9)

Similarly,

( ) ( )

( ) ( )ygz

Vyg

z

VWβYC

z

V

WβYCygz

Vyg

z

V

z

V

bo

bo

219

921

∂∂−

∂∂−−=

∂∂∴

−=∂

∂+∂

∂+∂∂

(4-10)

0+−−=∂∂

YβXαz

Z (4-11)

Substitute eqn.(4-8) into the fourth expression of eqn.(3-12), yields


90

( ) ( )

( ) ( ) ( )( )( ) ( ) ( )( )

( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( )( )( ) ( ) ( )( ) ( )( ) ( )( ) ( ) ( )( )

( ) ( ) ( ) ( )( )( ) ( ) ( )( ) ( )( ) ( )( ) ( ) ( )( )

( ) ( )( ) ( ) ( )( ) ( ) ( ) ( )( )( ) ( ) ( )[ ]b''

aa""

b''

a

a""

b

a

a

b

a

b

a

VαygVαygCC

UβxfUβxfCUxfUxfC

ZαCVαβCCUβCαCz

X

ZαCVygαCCVygαCC

VαβCCUxfUxfβC

UxfUxfCUβCαCz

X

ZαCVygαβCCVygαβCC

VαβCCUxfUxfβC

UxfUxfαCUβCαCz

X

ZαCVygVygαβCCVαβCC

UxfUxfβCαCUβCαCz

X

ZαCVygVygVαβCC

UxfUxfUβCαCz

X

ZαCVαβCCUβCαCz

X

20

163

2

2

02

162

0

12

163

2

6

2

2

12

63

0

163

632

0

1

2

6

2

0

12

2

6

2

2

1263

0

163

632

0

1

2

6

2

0

1

2

2

2

6

2

2

12

0

16363

2

0

1

2

6

2

2

2

6

2

2

12

0

163

2

0

1

2

6

2

2

163

2

6

2

2

++

−+−+−

+++−−−=∂∂∴

++−+

−+−+−

++−+−−=∂∂

++−+

−+−+−

++−+−−=∂∂

+++−+

−+−−+−−=∂∂

++++

−++−−=∂∂

++−−−=∂∂

(4-12)


91

Substitute eqn.(4-8) into the fifth expression of eqn.(3-12), yields

( ) ( )

( ) ( ) ( )( )( ) ( ) ( )( )

( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )( ) ( ) ( )( )

( ) ( )( ) ( ) ( )( )

( ) ( ) ( )( ) ( ) ( ) ( )( )bb""

a''

b""

b

a''

b

a

b

a

VαygVαygCVygVygC

UβxfUβxfCC

ZβCVβCαCUαβCCz

Y

ZβCVygVygC

VαygVαygCVβCαC

UβxfUβxfCCUαβCCz

Y

ZβCVygVygβCαCVβCαC

UxfUxfαβCCUαβCCz

Y

ZβCVygVygVβCαC

UxfUxfUαβCCz

Y

ZβCVβCαCUαβCCz

Y

2

2

02

162

0

14

2

0

163

5

2

4

2

663

52

0

14

2

2

02

16

2

4

2

6

2

0

16363

52

0

1

2

4

2

6

2

4

2

6

2

0

16363

52

0

1

2

4

2

6

2

0

163

5

2

4

2

663

+−+

−++−

+−−++−=∂∂∴

++

−+−−−

+++−+−=∂∂

++−−+−−

+++−+−=∂∂

+++−−

++++−=∂∂

+−−++−=∂∂

(4-13)


92

Substitute eqn.(4-8) into the sixth expression of eqn.(3-12), yields

( ) ( )( ) ( ) ( )( )

( ) ( )( ) ( ) ( )( )

( ) ( )( ) ( ) ( )( )

( ) ( )( )( ) ( )( )b''

a''

b''a''

ba

ba

VygVygC

UxfUxfCZCVβCUαCz

W

ZC

VygVygCVβCUxfUxfCUαCz

W

ZC

VygVygβCVβCUxfUxfαCUαCz

W

ZCVygVygVβCUxfUxfUαCz

W

ZCVβCUαCz

W

2

0

15

2

0

11751

7

2

0

1552

0

111

7

2

0

1552

0

111

72

0

152

0

11

751

+

+++++=∂

∂∴

+

+++++=∂

∂

+

+++++=∂

∂

++++++=∂

∂

++=∂

∂

(4-14)

Rearrange eqns.(4-9) to (4-14) in matrix form gives,


93

( )( )

( ) ( )

( ) ( )

( ) ( )[ ] ( ) ( )[ ]( ) ( ) ( )[ ]

( ) ( ) ( )[ ] ( ) ( )[ ]( ) ( )[ ]

( ) ( )[ ] ( ) ( )[ ]

+++

+−

+−++−

++

−+−+−

∂∂−

∂∂−

∂∂−

∂∂−

+

−−+−

+−−−−−

−−

=

∂∂

b''a''

b

b""a''

b''

aa""

bo

ao

VygVygCUxfUxfC

VαygVαygC

VygVygCUβxfUβxfCC

VαygVαygCC

UβxfUβxfCUxfUxfC

ygz

Vyg

z

V

xfz

Uxf

z

U

W

Y

X

Z

V

U

CβCαC

βCβCαCξαβCC

αCαβCCβCαCξ

ξβα

βC

αC

W

Y

X

Z

V

U

z

2

0

152

0

11

2

2

02

16

2

0

142

0

163

20

163

2

2

02

162

0

12

21

21

751

5

2

4

2

6

2

63

163

2

6

2

2

2

29

8

0

000

000

000

000

0000

0000

(4-15)

Or the above expression can be simplified as

B

W

Y

X

Z

V

U

.D

W

Y

X

Z

V

U

z+

=

∂∂


94

Substituting equations (4-1), (3-23) and (4-6) into (4-7), the following first order

non-homogeneous ordinary differential equation of jth ply can be determined for

each combination of m and n [30]:

( ){ } ( ){ } ( ){ }jmnjmnjjmn

zBzRDzRdz

d += (4-16)

where

( ){ } ( ) ( ) ( ) ( ) ( ) ( )[ ]Tjmnmnmnmnmnmn

jmnzWzYzXzZzVzUzR =

( )( )

−−++

++−−

−−

=

000

000

000

0000

0000

0000

751

52

42

663

1632

62

2

9

8

CCC

CCCCC

CCCCC

C

C

Dj

ηξηηξξη

ξξηηξηξ

ηξ

( ){ }

( ) ( )

( ) ( )[ ]

j

a

nn

a

nn

jmn

UUηC

dz

dU

dz

dU

zB

+

+−

=

0

02

10

0

2

1

02

6

0

when m = 0 and n ≠ 0 (4-17)


95

( ){ }

( ) ( )

( ) ( )[ ]j

b

mm

b

mm

jmn

VVξC

dz

dV

dz

dV

zB

+

+−

=

02

10

0

2

1

0

02

6

0

when m ≠ 0 and n = 0 (4-18)

( ){ }

( )( ) ( )

( )( ) ( )

( ) ( ) ( )[ ] ( )( ) ( ) ( )[ ]

( )( ) ( ) ( )[ ] ( ) ( ) ( )[ ]

( ) ( ) ( )[ ] ( ) ( ) ( )[ ]

−−−−−−

−−+−−+

−−+

+−−

−−−

−−−

=

b

mm

a

nn

b

mm

a

nn

b

mm

a

nn

b

mm

a

nn

jmn

VVπncosbπn

CUUπmcos

aπm

C

VVπncosπn

ξCUUπmcos

aπm

ηCC

VVπncosbπn

ξCCUUπmcos

πm

ηC

dz

dV

dz

dVπncos

πn

dz

dU

dz

dUπmcos

πm

zB

0501

0

22

2

6063

0630

22

2

6

0

22

0

22

12

12

12

12

12

12

0

12

12

when m ≠ 0 and n ≠ 0 (4-19)

( ){ } { }0=jmn

zB when m = 0 and n = 0 (4-20)

The derivation of eqns.(4-17) to (4-20) are explained in a very detail manner at

the end of this section.

Each lamina is divided into a number of layers to ensure that each layer is thin.

For each thin layer, the approximations solutions can be obtained by assuming

a linear relationship with respect to z. After applying the external loads on the

top surface of the plate, considering the boundary conditions and continuity

conditions of the state variables at the interfaces of these thin layers, the


96

solution for thick plate fully clamped along four edges is deduced by solving a

linear algebra equation system.

As a result, the laminated can be composed of many laminae which have

different material properties and each of the lamina is divided into a number of

layers to ensure that each layer is thin. Then the approximate solution obtained

can arbitrarily approach the exact three dimensional solution if the number of

layers and the sinusoidal displacement mode numbers m and n are sufficiently

large.

Taking a thin layer of thickness hj, the solution of non-homogenous eqn. (4-16)

is given by

( ){ } ( )[ ] ( ){ } ( ){ }jmnjmnjjmn

zCRzGzR += 0 (4-21)

where z = 0 (top surface) and z = h (bottom surface)

( ){ } ( ) ( ) ( ) ( ) ( ) ( )[ ]Tjmnmnmnmnmnmn

jmnzWzYzXzZzVzUzR =

( ){ } ( ) ( ) ( ) ( ) ( ) ( )[ ]Tjmnmnmnmnmnmn

jmnWYXZVUR 0000000 =

( )[ ][ ]( )zD

jjezG = , [G(z)]j is the transfer matrix of the homogeneous plate

( ){ } [ ]( )( ){ } ττ

τdBezC

jmn

zzD

jmnj

∫−

=0

Dj is the same as eqn.(4-16)

Notice that the non-homogeneous state eqn. (4-21) is analogous to the eqn.(3-7)

in terms of the through thickness z-coordinate, for this reason, the application of

state space method of spring-mass damper system is applied to the laminated

composite plate.

It is clearly notice that the non-homogenous vector ( ){ }jmn

zB contains the

unknown constants which need to be solved initially. To determine these


97

unknown constants, boundary conditions of the plate are applied to the state

equation of the clamped edges plate which consists of many thin layers for each

ply. Only when the thickness of these layers are thin enough, it can be assumed

that all the unknown functions are linearly distributed across the thickness of the

thin layers and subsequently through the thickness of the plate.

By considering the continuity between the bottom surface of jth layer and the

subsequent top surface of j+1th layer , the relationship of displacement functions

as shown in equation (4-5) can be expressed as [30]:

( ) ( )[ ] ( ) ( )

( ) ( )[ ] ( ) ( )

( ) ( )[ ] ( ) ( )

( ) ( )[ ] ( ) ( )

+

−=

+

−=

+

−=

+

−=

+

+

+

+

j

jb

jj

jb

jjj

b

m

j

j

jj

j

jjjm

j

ja

jj

ja

jjj

a

n

j

j

jj

j

jjjn

d

zB

d

zBzV

d

zB

d

zBzV

d

zA

d

zAzU

d

zA

d

zAzU

1

0

1

00

1

0

1

00

1

1

1

1

(4-22)

where ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )bjj

ajj

bjj

ajj

BandBAAandBBAA1

011

01

00 ,,,,,++++

are the end

unknown values of linear functions at the edges of the plate at jth and j+1th

layers, respectively and zj and dj are the local z-coordinate and thickness of the

jth layer, respectively.

Substitute eqn.(4-22) into (4-17), (4-18) and (4-19) [30]:


98

( ){ }

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )( )

j

a

j

a

jjjj

ja

jj

j

a

j

a

jjj

jmn

AAAAd

zAAηC

d

AAAA

zB

−+−++

−+−

=

++

++

0

0

2

1

0

0

2

1

1

00

1

02

6

1

0

1

0

(4-23)

when m = 0 and n ≠ 0

( ){ }

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )( )j

b

j

b

jjjj

jb

jj

j

b

j

b

jjj

jmn

BBBBd

zBBζC

d

BBBB

zB

−+−++

−+−

=

++

++

0

2

1

0

0

2

1

0

1

00

1

02

6

1

0

1

0

(4-24)

when m ≠ 0 and n = 0


99

( ){ }

( )( ) ( ) ( ) ( )

( )( ) ( ) ( ) ( )

( )

( ) ( )

( ) ( ) ( ) ( )( )

( )( )

( ) ( )

( ) ( ) ( ) ( )( )

( )( )

( ) ( )

( ) ( ) ( ) ( )( )

( )

( ) ( )

( ) ( ) ( ) ( )( )

( )

( ) ( )

( ) ( ) ( ) ( )( )

( )

( ) ( )

( ) ( ) ( ) ( )( )j

j

b

jj

b

jj

j

b

jj

j

a

jj

a

jj

j

a

jj

j

b

jj

b

jj

j

b

jj

j

a

jj

a

jj

j

a

jj

j

b

jj

b

jj

j

b

jj

j

a

jj

a

jj

j

a

jj

j

b

j

b

jjj

j

a

j

a

jjj

jmn

BBBBd

z

BB

πCosnbπn

C

AAAAd

z

AA

πCosmaπm

C

BBBBd

z

BB

πCosnπn

ξC

AAAAd

z

AA

πCosmaπm

ηCC

BBBBd

z

BB

πCosnbπn

ξCC

AAAAd

z

AA

πCosmπm

ηC

d

BBBBπCosn

πn

d

AAAAπCosm

πm

zB

−−+

+−

−

−

−−+

+−

−−

−−+

+−

−

+

−−+

+−

−+

−−+

+−

−+

+

−−+

+−

−

−+−−−

−+−−−

=

++

++

++

++

++

++

++

++

0

1

0

1

0

5

0

1

0

1

0

1

0

1

0

1

0

22

2

6

0

1

0

1

0

63

0

1

0

1

0

63

0

1

0

1

0

22

2

6

1

00

1

22

1

00

1

22

12

12

12

12

12

12

0

12

12

(4-25) when m ≠ 0 and n ≠ 0


100

Following the recursive formulation and applying the boundary conditions, we

assume that all layers are interconnected to each other by bonding the

interfaces to preserve continuity i.e. the state vectors at the bottom of the jth

layer is the same as those to the top of subsequent layer, j+1th layer.

For instance, let zj = dj (the thickness of the layer) at layer jth,

( ){ } ( )[ ] ( ){ } ( ){ }jmnjmnjjjjmn

dCRdGdR += 0 (4-26)

For the subsequent j+1th layer, the state vector is expressed as

( ){ } ( )[ ] ( ){ } ( ){ }111111

0++++++

+=jmnjmnjjjjmn

dCRdGdR (4-27)

By virtue of continuity conditions at the interface between the layers, this gives

( ){ } ( ){ }1

0+

=jmnjjmn

RdR (4-28)

Therefore,

( ){ } ( )[ ] ( )[ ] ( ){ } ( ){ }( ){ }

1

11110

+

+++++

+=

jmn

jmnjmnjjjjjjmn

dC

dCRdGdGdR (4-29)

Eqn.(4-29) can summarized as

( ){ } [ ] ( ){ } [ ]Π+Π=+

01 mnjmn

RdR (4-30)

where

( )[ ] ( )[ ]( )[ ] ( ){ } ( ){ }

111

11

+++

++

+=Π

=Π

jmnjmnjj

jjjj

dCdC.dG

dG.dG


101

Repeating the above eqns.(4-26), (4-27) and (4-28) from the top surface to the

bottom surface of the laminate, the relationship of the top surface through the

bottom surface of the laminate can be obtained by using recursive formulation.

The non homogeneous vector Π contains the boundary unknown coefficients,

.BandB,A,A b

j

o

j

a

j

o

j

After doing the process of recursive formulation from the top surface, i.e. z = 0

through the bottom surface of the plate at z = h, we can get the expression

collectively as,

( )( )( )( )( )( )

( )( )( )( )( )( )

ΠΠΠΠΠΠ

+

ΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠΠ

=

6

5

4

3

2

1

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

0

0

0

0

0

0

mn

mn

mn

mn

mn

mn

mn

mn

mn

mn

mn

mn

W

Y

X

Z

V

U

hW

hY

hX

hZ

hV

hU

(4-31)

Knowing the boundary and loading conditions of the laminated plate, the

relationship of top and bottom surfaces of the plate can now be established for

a uniformly distributed loading, as follows,

( ) ( ) ( ) ( )

( ) ( ) ( ) 0000

53116

000002

===

∞=−===

mnmnmn

mnmnmn

Z;hY;hX

............,,n,m,πmn

qZ;Y;X

(4-32)

Substitute eqns.(4-32) into the third, fourth and fifth eqns.(4-31), the top surface

displacements can be solved initially, i.e.

( )( )( )

ΠΠΠ

−

ΠΠΠ

ΠΠΠΠΠΠΠΠΠ

=

−

5

4

3

53

43

33

2

1

565251

464241

363231 16

0

0

0

πmn

q.

W

V

U

mn

mn

mn

(4-33)


102

Therefore, the initial vector of ( ){ }0mn

R must be determined as this vector give

the initial displacement vector of ( ) ( ) ( )000mn

mnmn WandV,U . Considering

the first and second rows of eqn.(4-31), we can get the relationship of the state

vectors of Umn(z) and Vmn(z) in terms of ( ) ( ) ( )000mn

mnmn WandV,U at any

layers of the plate, i.e.

( )( )

( ) ( ) ( )( ) ( ) ( )

( )( )( )

( ) ( ) ( )( ) ( ) ( )

( )( )( )

( )( )

ΠΠ+

ΠΠΠΠΠΠ

+

ΠΠΠΠΠΠ

=

z

z

Y

X

Z

zzz

zzz

W

V

U

.zzz

zzz

zV

zU

mn

mn

mn

mn

mn

mn

mn

mn

1

1

252423

151413

262221

161211

0

0

0

0

0

0

(4-34)

Substitute eqn.(4-33) into (4-34) and considering eqn. (4-32), gives

( )( )

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( )

( )( )( )

( )( )

ΠΠ+

ΠΠΠΠΠΠ

+

ΠΠΠ

−

ΠΠΠ

ΠΠΠΠΠΠΠΠΠ

ΠΠΠΠΠΠ

=

−

z

z

Y

X

Z

zzz

zzz

πmn

q

.

.zzz

zzz

zV

zU

mn

mn

mn

mn

mn

1

1

252423

151413

5

4

3

53

43

33

2

1

565251

464241

363231

262221

161211

0

0

0

16


103

( )( )

( ) ( ) ( )( ) ( ) ( )

( )( )

( )( )

ΠΠ+

ΠΠ

−

ΠΠΠ

−

ΠΠΠ

ΠΠΠΠΠΠΠΠΠ

ΠΠΠΠΠΠ

=

⇒

−

z

zz

z

πmn

q

πmn

q

.

.zzz

zzz

zV

zU

mn

mn

1

1

23

132

5

4

3

53

43

33

2

1

565251

464241

363231

262221

161211

16

16

(4-35)

To solve these ( ) ( )zVandzU mnmn which contains the unknowns, they must

be solved simultaneously within all the layers of the plate as further explained

below.

The following boundary conditions for clamped edges should be satisfied,

bandywhenWVU

aandxwhenWVU

00

00

========

(4-36)

Therefore, considering equations (4-1), (4-5), (4-6) and (4-36), we can easily

established the following expressions, [30]

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )∑∑ ∑

∑∑ ∑

∑∑ ∑

∑∑ ∑

=

+

−=

=

+

=

=

+

−=

=

+

=

m n m

b

mmn

nb

z

m n mm

mnz

m n n

a

nmn

ma

z

m n nn

mnz

a

xπmSinzV

a

xπmSinzVV

a

xπmSinzV

a

xπmSinzVV

b

yπnSinzU

b

yπnSinzUU

b

yπnSinzU

b

yπnSinzUU

01

0

01

0

00

00

(4-37)

By simplifying the above eqns.(4-37), for each m and n, the above equations

yields [30]


104

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )∑

∑ ∑

∑

∑ ∑

∑

∑ ∑

∑

∑ ∑

=−+⇒

=

−+

=+⇒

=

+

=−+⇒

=

−+

=+⇒

=

+

nmn

nb

m

m nmn

nb

m

nmn

m

m nmn

m

mmn

ma

n

n mmn

ma

n

mmn

n

n mmn

n

zVzV

a

xπmSinzVzV

zVzV

a

xπmSinzVzV

zUzU

b

yπnSinzUzU

zUzU

b

yπnSinzUzU

01

01

0

0

01

01

0

0

0

0

0

0

(4-38)

Finally, we can solved Umn(z) and Vmn(z) with the relationship of eqns.(4-22) and

(4-38) at all layers across the thickness of the laminated plate. Consequently, all

the unknowns ( ) ( ) ( ) ( )bjj

ajj

BBAA ,,, 00 can be determined for each sub-layer by

substituting the displacements ( ) ( ) ( ) ( )bmm

ann

VandVUU 00 ,, from eqn.(4-22) into

(4-38). This can be achieved by forming the algebraic equations that can be

solved with the same number of unknown coefficients.

It is also interesting to note that when the applied loading is symmetric, we can

have further relationship such that ( ) ( ) ( ) ( )z,yUz,yU a

nn−=0

and ( ) ( ) ( ) ( )z,xVz,xV b

mm−=0 which in turn decreases the number of unknowns

and reduces the number of algebraic equations. This means that only the first

and third expressions on eqns.(4-38) are required to determine these unknowns.

After considering the applied external loading, boundary conditions and the

continuity conditions at the interfaces of the laminated plate, the analytical

solution of the plate with clamped edges is now formulated in this section by a


105

set of linear algebra equations which can be solved by any of mathematical

tools such as Mathematica, Matlab or Fortran etc. The final result would then

depend on the number of thin layers and the sinusoidal displacement mode

numbers of m and n. The larger the number of thin layers and of m and n, the

greater the accuracy of the exact three dimensional elasticity solution. In this

study the values of m = 1,3,5,7,9,…..99 and n = 1,3,5,7,9,…..29 are considered.

The results of this study will be verified with the numerical analysis which are

presented in the following chapter.

The derivation of non-homogeneous vector B mn(z)

In this section, the derivation of eqns.(4-17) to (4-20) are shown clearly. The

method used consists of an expansion of a function f(x) in terms of an infinite

sum of sines and cosines, which is known as Fourier series.

Generally, the Fourier series of a function f(x) is defined as [53],

( )

+

+= ∑∑∞

=

∞

= a

xπmSinb

a

xπmCosaaxf

mm

mmo

112

1 (4-39)

If f(x) is odd, then the series is reduced to a Fourier sine series, i.e.

( )

= ∑∞

= a

xπmSinbxf

mm

1

(4-40)

If f(x) is even, then the series is reduced to a Fourier cosine series, i.e.

( )

+= ∑∞

= a

xπmCosaaxf

mmo

12

1 (4-41)

Since

( )

−=a

xxf 1

1 , ( )

a

xxf =

2 (4-42)

and


106

( ) ( )

( ) ( )

( ) ( )[ ] ( ) ( )[ ]( ) ( ) ( )[ ]

( ) ( ) ( )[ ] ( ) ( )[ ]( ) ( )[ ]

( ) ( )[ ] ( ) ( )[ ]

+++

+−

+−++−

++

−+−+−

−−

−−

=

b''a''

b

b""a''

b''

aa""

bo

ao

mn

VygVygCUxfUxfC

VαygVαygC

VygVygCUβxfUβxfCC

VαygVαygCC

UβxfUβxfCUxfUxfC

ygdz

dVyg

dz

dV

xfdz

dUxf

dz

dU

B

2

0

152

0

11

2

2

02

16

2

0

142

0

163

20

163

2

2

02

162

0

12

21

21

0

(4-43)

Integrate and expand f1(x) using Fourier series, i.e.,

( )

( )( )[ ] ( ) ( )

( )[ ] ( ) ( )

++=

−

++=

−

+

=

−

+=

−

+=

−=

∫ ∑∫∫

∫ ∑∫

∑

∞

=

∞

=

∞

=

πm

πmSinπma

πm

a.πmSin

πm

πmCos

πm

πmSinπma.a

πm

a.πmSin.a

πm

πmCosa

dxa

xπmCos

a

xπmCosadx

a

xπmCosadx

a

xπmCos

a

x

dxa

xπmCos

a

xπmCosaadx

a

xπmCos

a

x

a

xπmCosaa

a

xxf

m

m

a

mm

aa

a

mm

a

mm

4

221

4

221

1

1

1

022

022

0 100

0

0 10

0

101

( )( )22

0

120

2

10

πm

πmCosa:mwhen

a:mwhen

m

−=≠

=→ (4-44)


107

Now repeat the above procedures for f2(x), i.e.

( )

( )( ) ( )[ ] ( ) ( )[ ]

πm

b.πmSinπma

πm

b.πmSin.a

πm

πmSinπmπmCosa

dxa

xπmCos

a

xπmCosbdx

a

xπmCosbdx

a

xπmCos

a

x

dxa

xπmCos

a

xπmCosbbdx

a

xπmCos

a

x

a

xπmCosbb

a

xxf

m

a

mm

aa

a

mm

a

mm

4

221 022

0 100

0

0 10

0

102

++=

++−

+

=

+=

+=

=

∫ ∑∫∫

∫ ∑∫

∑

∞

=

∞

=

∞

=

( )[ ]πmCosπm

b:mwhen

b:mwhen

m−−=≠

=→

12

0

2

10

22

0 (4-45)

Therefore, taking the first row of eqn.(4-43),

( )[ ] ( ) ( )xfdz

dUxf

dz

dUzB

a

n

o

n

mn 211−−=

when m = 0, n ≠ 0 :

( )[ ]

+−=

−

−=dz

dU

dz

dU

dz

dU

dz

dUzB

a

n

o

n

a

n

o

n

mn 2

1

2

1

2

11

(4-46)

when m ≠ 0, n = 0 : ( )[ ] 01

=zBmn

(4-47)

when m ≠ 0, n ≠ 0:

( )[ ] ( )[ ] ( )[ ]

( )[ ] ( )[ ]

−−−=

−−−

−−=

dz

dU

dz

dUπmCos

πmzB

πmCosπmdz

dUπmCos

πmdz

dUzB

a

n

o

n

mn

a

n

o

n

mn

12

12

12

221

22221

(4-48)


108

Similarly, repeat the above steps for the case of g1(y) and g2(y), as follows:

Integrate and expand g1(y) using Fourier series, i.e.,

( )

( )( )[ ] ( ) ( )

( )[ ] ( ) ( )

++=

−

++=

−

+

=

−

+=

−

+=

−=

∫ ∑∫∫

∫ ∑∫

∑

∞

=

∞

=

∞

=

πn

πnSinπna

πn

a.πnSin

πn

πnCos

πn

πnSinπnb.a

πn

a.πnSin.b

πn

πnCosb

dyb

yπnCos

b

yπnCosady

b

yπnCosady

b

yπnCos

b

y

dyb

yπnCos

b

yπnCosaady

b

yπnCos

b

y

b

yπnCosaa

b

yyg

n

n

b

nn

bb

b

nn

b

nn

4

221

4

221

1

1

1

022

022

0 100

0

0 10

0

101

( )( )22

0

120

2

10

πn

πnCosa:nwhen

a:nwhen

n

−=≠

=→ (4-49)

Now repeat the above procedures for g2(y), i.e.

( )

( )( ) ( )[ ] ( ) ( )[ ]

πn

b.πnSinπnb

πn

b.πnSin.b

πn

πnSinπnπnCosb

dyb

yπnCos

b

yπnCosbdy

b

yπnCosbdy

b

yπnCos

b

y

dyb

yπnCos

b

yπnCosbbdy

b

yπnCos

b

y

b

yπnCosbb

b

yyg

n

b

nn

bb

b

nn

b

nn

4

221 022

0 100

0

0 10

0

102

++=

++−

+

=

+=

+=

=

∫ ∑∫∫

∫ ∑∫

∑

∞

=

∞

=

∞

=

( )[ ]πnCosπn

b:nwhen

b:nwhen

n−−=≠

=→

12

0

2

10

22

0 (4-50)


109

Therefore, taking the second row of eqn.(4-43),

( )[ ] ( ) ( )ygdz

dVyg

dz

dVzB

b

m

o

m

mn 212−−=

when m = 0, n ≠ 0 : ( )[ ] 02

=zBmn

(4-51)

when m ≠ 0, n = 0 :

( )[ ]

+−=

−

−=dz

dV

dz

dV

dz

dV

dz

dVzB

b

m

o

m

b

m

o

m

mn 2

1

2

1

2

12

(4-52)

when m ≠ 0, n ≠ 0:

( )[ ] ( )[ ] ( )[ ]

( )[ ] ( )[ ]

−−−=

−−−

−−=

dz

dV

dz

dVπnCos

πnzB

πnCosπndz

dVπnCos

πndz

dVzB

b

m

o

m

mn

b

m

o

m

mn

12

12

12

222

22222

(4-53)

From the third row of eqn.(4-43),

( )[ ] 03

=zBmn

(4-54)

Now consider the fourth row of eqn.(4-43), i.e.

( )[ ] ( ) ( )[ ] ( ) ( )[ ]( ) ( ) ( )[ ]b''

aa""

mn

VαygVαygCC

UβxfUβxfCUxfUxfCzB

2

0

163

2

2

02

162

0

124

++

−+−+−=

(4-55)

As

( )

( )

( ) 0

1

1

1

1

1

=

−=

−=

xf

axf

a

xxf

"

' and

( )

( )

( ) 0

1

2

2

2

=

=

=

xf

axf

a

xxf

"

' (4-56)


110

( )

( )

( ) 0

1

1

1

1

1

=

−=

−=

yg

byg

b

yyg

"

' and

( )

( )

( ) 0

1

2

2

2

=

=

=

yg

byg

b

yyg

"

' (4-57)

Substitute eqns.(4-63) and (4-64) into (4-62), gives

( )[ ]( ) ( )

( ) ( )

( )

( )

( ) ( ) ( )

( )[ ] ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( )[ ] ( )

( )( ) ( ) ( )

+

−+

−

+

−=

+

−+

−

−

+

−

−−=

+

−+

−

+

−

−

+

−=

∑∑

∑∑

∑∑

∑∑

∑∑

∑

∑

∑

∑

m

b

mm

o

m

n

a

nn

o

nmn

m

b

mm

o

m

n

a

nn

o

nmn

m

b

mm

o

m

n

a

n

n

o

n

n

a

n

n

o

n

mn

xξCosVb

xξCosVb

ξCC

b

yπnSinU

a

x

b

yπnSinU

a

xηCzB

xξCosVξb

xξCosVξb

CC

yηSinUηa

xyηSinUη

a

xCzB

xξSinVdx

d

bxξSinV

dx

d

bCC

yηSinUdy

d

a

x

yηSinUdy

d

a

x

C

yηSinU

yηSinU

CzB

11

1

11

1

11

1

0

0

63

2

64

63

22

64

63

2

2

2

2

624

(4-58)

Consider the first part of eqn.(4-58), i.e.

( )[ ] ( )

+

−= ∑∑n

a

nn

o

nmn b

yπnSinU

a

x

b

yπnSinU

a

xηCzB 12

64

( )[ ] ( ) ( ) ( )

+

−= ao

mnU

a

xU

a

xηCzB 12

64 (4-59)

Now, take the second part of eqn.(4-58), i.e.

( )[ ] ( )( ) ( ) ( )

+

−+−= ∑∑m

b

mm

o

mmnxξCosV

bxξCosV

bξCCzB

11634


111

Applying the Fourier series of

( ) ( )( ) ( )[ ] ( )

n

b

nn

b

b

nn

b

nn

aπn

πnSinb

πn

aπnCosb

πn

πnCos

dyb

yπnSin

b

yπnSinady

b

yπnSina

πn

πnCos

dyb

yπnSin

b

yπnSinaady

b

yπnSin

b

b

yπnSinaa

b

−+−

=−

+

=−

+=

+=

∫ ∑∫

∫ ∑∫

∑

∞

=

∞

=

∞

=

2

21

2

11

1

1

1

0

0 100

0 10

0

10

(4-60)

( )[ ]( )[ ]

( )[ ]bπn

πnCos

πnSinbπn

πnCosa:nwhen

ba:nwhen

n

−=

−−

=≠

=→

12

21

120

10

0 (4-61)

By using eqn. (4-44), we can get

when m = 0, n ≠ 0 :

( )[ ] ( ) ( ) ( )( )[ ] ( ) ( ) ( )[ ]a

n

o

nmn

a

n

o

nmn

UUηCzB

UUηCzB

+

=

+

=

2

64

2

64

2

12

1

2

1

(4-62)

when m ≠ 0, n = 0 :

( )[ ] 04

=zBmn

(4-63)

By using eqns. (4-44), (4-45) and (4-61), we can get

when m ≠ 0, n ≠ 0 :


112

( )[ ] ( ) ( ) ( ) ( )( )( )( )

( )[ ] ( ) ( )[ ]( ) ( )[ ]( )

( )( ) ( )[ ] ( ) ( )[ ] ( )

( )[ ] ( )[ ] ( ) ( )[ ]( )

( )[ ] ( ) ( )[ ]b

m

o

m

a

n

o

nmn

b

m

o

m

a

n

o

nmn

b

m

o

ma

n

o

nmn

VVπnCosbπn

ξCC

UUπmCosπm

ηCzB

Vbπn

πnCosV

bπn

πnCosξCC

UπmCosπm

UπmCosπm

ηCzB

Vb

Vb

ξCCUa

xU

a

xηCzB

−−+

+−−=∴

−+

−−+

−

−−−=

+

−+−

+

−=

12

12

1212

12

12

1

1

1

63

22

2

6

4

63

2222

2

64

63

2

64

(4-64)

Now consider the fifth row of eqn.(4-43), i.e.

( )[ ] ( ) ( ) ( )[ ] ( ) ( )[ ]( ) ( )[ ]b

b""a''

mn

VαygVαygC

VygVygCUβxfUβxfCCzB

2

2

02

16

2

0

142

0

1635

+−

+−++−=

(4-65)

Substitute eqns.(4-56) and (4-57) into (4-65), gives

( )[ ] ( ) ( ) ( )[ ] ( ) ( )[ ]( ) ( )[ ]

( )[ ] ( ) ( ) ( )

( ) ( )[ ]( )

( )

+

−−+−

+

−+−=

+−

+−++−=

∑

∑

∑∑

m

b

m

m

o

mb

n

a

nn

o

nmn

b

b""a''

mn

xξSinVdx

d

b

y

xξSinVdx

d

b

y

CVVC

yηSinUdy

d

ayηSinU

dy

d

aCCzB

VαygVαygC

VygVygCUβxfUβxfCCzB

2

2

2

2

6

0

4

635

2

2

02

16

2

0

142

0

1635

1

00

11


113

( )[ ] ( ) ( ) ( )

( ) ( )

( )[ ] ( ) ( ) ( )

( ) ( )

+

−+

+

−+−=

+

−−

+

−+−=

∑∑

∑∑

∑∑

∑∑

m

b

mm

o

m

n

a

nn

o

nmn

m

b

mm

o

m

n

a

nn

o

nmn

xξSinVξb

yxξSinVξ

b

yC

yηCosUηa

yηCosUηa

CCzB

xξSinVdx

d

b

yxξSinV

dx

d

b

yC

yηSinUdy

d

ayηSinU

dy

d

aCCzB

22

6

635

2

2

2

2

6

635

1

11

1

11

( )[ ] ( ) ( ) ( )

( ) ( )

+

−+

+

−+−=

∑∑

∑∑

m

b

mm

o

m

n

a

nn

o

nmn

xξSinVb

yxξSinV

b

yξC

yηCosUa

yηCosUa

ηCCzB

1

11

2

6

635

Therefore

when m = 0, n ≠ 0: ( )[ ] 05

=zBmn

(4-66)

when m ≠ 0, n = 0: from eqn.(4-49) and (4-50),

( )[ ] ( ) ( )( )[ ] [ ]b

m

o

mmn

b

m

o

mmn

VVξCzB

VVξCzB

+=

+

+=

2

65

2

65

2

12

1

2

10

(4-67)

when m ≠ 0, n ≠ 0: from eqns.(4-44), (4-45), (4-49) and (4-50),

( )[ ] ( ) ( )[ ] ( )( )

( )[ ] ( )[ ]

( )[ ] ( ) ( )[ ][ ] ( )[ ][ ]b

m

o

m

a

n

o

nmn

b

m

o

m

a

n

o

nmn

VVπnCosπn

ξCUU

aπm

πmCosηCCzB

VπnCosπn

VπnCosπn

ξC

Uaπm

πmCosU

aπm

πmCosηCCzB

−−+−−+

=

−−−+

−+

−−+−=

1212

12

12

1212

22

2

663

5

2222

2

6

635

(4-68)


114

Now, take the final sixth row of eqn.(4-43):

( )[ ] ( ) ( )[ ] ( ) ( )[ ]( )[ ]

+

−+

+

−=

+++=

ba

mn

b''a''

mn

Vb

Vb

CUa

Ua

CzB

VygVygCUxfUxfCzB

1111 0

5

0

16

2

0

152

0

116 (4-69)

Therefore

when m = 0, n ≠ 0: ( )[ ] 06

=zBmn

(4-70)

when m ≠ 0, n = 0: ( )[ ] 06

=zBmn

(4-71)

when m ≠ 0, n ≠ 0: using eqn.(4-61),

( )[ ] ( )[ ] ( )[ ]

( )[ ] ( )[ ]

−+

−−

+

−+

−−=

b

m

o

m

a

n

o

nmn

Vbπn

πnCosV

bπn

πnCosC

Uaπm

πmCosU

aπm

πmCosCzB

1212

1212

5

16

( )[ ] ( )[ ][ ] ( )[ ][ ]b

m

o

m

a

n

o

nmnVVπnCos

bπn

CUUπmCos

aπm

CzB −−−−−−= 1

21

251

6

(4-72)

Re-arrange the above expressions of eqns.(4-46) to (4-48), (4-51) to (4-53), (4-

54), (4-62) to (4-64) and (4-70) to (4-72), the non-homogeneous vectors Bmn(z)

are given by,


115

( ){ }

( ) ( )

( ) ( )[ ]

j

a

nn

a

nn

jmn

UUηC

dz

dU

dz

dU

zB

+

+−

=

0

02

10

0

2

1

02

6

0

m = 0, n ≠ 0

( ){ }

( ) ( )

( ) ( )[ ]j

b

mm

b

mm

jmn

VVξC

dz

dV

dz

dV

zB

+

+−

=

02

10

0

2

1

0

02

6

0

m ≠ 0, n = 0

( ){ }

( )( ) ( )

( )( ) ( )

( ) ( ) ( )[ ] ( )( ) ( ) ( )[ ]

( )( ) ( ) ( )[ ] ( ) ( ) ( )[ ]

( ) ( ) ( )[ ] ( ) ( ) ( )[ ]j

b

mm

a

nn

b

mm

a

nn

b

mm

a

nn

b

mm

a

nn

jmn

VVπCosnbπn

CUUπCosm

aπm

C

VVπCosnπn

ξCUUπCosm

aπm

ηCC

VVπCosnbπn

ξCCUUπCosm

πm

ηC

dz

dV

dz

dVπCosn

πn

dz

dU

dz

dUπCosm

πm

zB

−−−−−−

−−+−−+

−−+

+−−

−−−

−−−

=

0501

0

22

2

6063

0630

22

2

6

0

22

0

22

12

12

12

12

12

12

0

12

12

m ≠ 0, n ≠ 0


116

4.3 Application of State Space Method to Laminated Plate

In order to clarify the formulations as stated in the previous section, the

following example of the application of state space method to laminated plate is

shown for clarity of the method being used. In this section, the formation of

algebraic equations are illustrated in the detailed manner for clarification

purpose, however, in order to solve the equations, they require an advance

computer programming software since they involve massive matrix problem

calculation and iteration processes. They are many different types of

programming softwares available such as Matlab, Fortran and so on. In this

study the latest version of Mathematica version 8 is used.

Consider a laminated plate subjected to uniformly distributed loading at the top

surface of the plate as shown in Figure 4.3. The plate consists of 3 plies in

which ply 1 is the same as ply 3, i.e. same material properties and thickness of

d1 with comprised of 3 no. of sublayers. For ply 2, there are 10 no. of sublayers

having a thickness of d2 each. The relationship of material properties between

ply1 and ply 2 is in the ratio of 2

11

1

11ply

ply

C

C. The overall thickness of the plate is h and

ply1 is equal to ply3 = 0.1h while ply 2 thickness is 0.8h.


117

Figure 4.3: View of a clamped edges laminated plate

For ply 1: the thickness of each sublayer is d1,

Sublayer 1: when z = 0, the unknown coefficient is A1













118







Sublayer 16: when z = d1, the unknown coefficient is A17

Because of continuity and they are perfectly bonded at the interfaces, the

relationship of displacement functions as stated in eqns.(4-22), the following

unknown coefficients can be established at x = 0,

Since ( ) ( )[ ] ( ) ( )

+

−=

+j

j

jj

j

jjjn d

zA

d

zAzU 0

1

00 1

( )[ ]

( )[ ]

( )[ ] 0

31

0

41

0

33

0

0

21

0

31

0

22

0

0

11

0

21

0

11

0

00103

00102

00101

01

Ad

Ad

AU:sublayer

Ad

Ad

AU:sublayer

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AU:sublayer

zwhen:PlyFor

n

n

n

=

+

−=

=

+

−=

=

+

−=

=


119

( )[ ]

( )[ ]

( )[ ]

( )[ ]

( )[ ]

( )[ ]

( )[ ]

( )[ ]

( )[ ]

( )[ ] 0

132

0

142

0

1313

0

0

122

0

132

0

1212

0

0

112

0

122

0

1111

0

0

102

0

112

0

1010

0

0

92

0

102

0

99

0

0

82

0

92

0

88

0

0

72

0

82

0

77

0

0

62

0

72

0

66

0

0

52

0

62

0

55

0

0

42

0

52

0

44

0

001013

001012

001011

001010

00109

00108

00107

00106

00105

00104

02

Ad

Ad

AU:sublayer

Ad

Ad

AU:sublayer

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Ad

AU:sublayer

Ad

Ad

AU:sublayer

Ad

Ad

AU:sublayer

Ad

Ad

AU:sublayer

Ad

Ad

AU:sublayer

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Ad

AU:sublayer

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Ad

AU:sublayer

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Ad

AU:sublayer

zwhen:PlyFor

n

n

n

n

n

n

n

n

n

n

=

+

−=

=

+

−=

=

+

−=

=

+

−=

=

+

−=

=

+

−=

=

+

−=

=

+

−=

=

+

−=

=

+

−=

=


120

( )[ ]

( )[ ]

( )[ ]

( )[ ] 0

171

10

171

10

16171

0

0

161

0

171

0

1616

0

0

151

0

161

0

1515

0

0

141

0

151

0

1414

0

1

116

001016

001015

001014

03

Ad

dA

d

dAdU:sublayer

Ad

Ad

AU:sublayer

Ad

Ad

AU:sublayer

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AU:sublayer

dorzwhen:Ply

n

n

n

n

=

+

−=

=

+

−=

=

+

−=

=

+

−=

=

Now, applying the first algebraic expression as stated in eqn.(4-38), i.e.

( ) ( ) ( )

( )

( )

( ) 003

002

001

1

0

3

0

3

2

0

2

1

0

1

0

=

+

=

+

=

+

=

+

∑

∑

∑

∑

mmn

mmn

mmn

jmmn

n

UA:sublayer

UA:sublayer

UA:sublayer

:PlyFor

zUzU


121

( )

( )

( )

( )

( )

( )

( )

( )

( )

( ) 0013

0012

0011

0010

009

008

007

006

005

004

2

13

0

13

12

0

12

11

0

11

10

0

10

9

0

9

8

0

8

7

0

7

6

0

6

5

0

5

4

0

4

=

+

=

+

=

+

=

+

=

+

=

+

=

+

=

+

=

+

=

+

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

mmn

mmn

mmn

mmn

mmn

mmn

mmn

mmn

mmn

mmn

UA:sublayer

UA:sublayer

UA:sublayer

UA:sublayer

UA:sublayer

UA:sublayer

UA:sublayer

UA:sublayer

UA:sublayer

UA:sublayer

:PlyFor

( )

( )

( )

( ) 016

0016

0015

0014

3

171

0

17

16

0

16

15

0

15

14

0

14

=

+

=

+

=

+

=

+

∑

∑

∑

∑

mmn

mmn

mmn

mmn

dUA:sublayer

UA:sublayer

UA:sublayer

UA:sublayer

:Ply


122

Now, applying the first expression as stated in eqn.(4-35) into the above

expression, i.e.

[ ]

[ ]

[ ]0

1616

3

0

1616

2

0

1616

1

1

3

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

3

2

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

2

1

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

1

=

Π+Π−

ΠΠ

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

=

Π+Π−

Π

Π

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

=

Π+Π−

ΠΠ

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

−

−

−

πmn

q

πmn

q

A:sublayer

πmn

q

πmn

q

A:sublayer

πmn

q

πmn

q

A:sublayer

:PlyFor


123

[ ]

[ ]

[ ]

[ ]0

1616

7

0

1616

6

0

1616

5

0

1616

4

2

7

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

7

6

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

6

5

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

5

4

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

4

=

Π+Π−

Π

Π

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

=

Π+Π−

Π

Π

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

=

Π+Π−

Π

Π

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

=

Π+Π−

Π

Π

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

−

−

−

−

πmn

q

πmn

q

A:sublayer

πmn

q

πmn

q

A:sublayer

πmn

q

πmn

q

A:sublayer

πmn

q

πmn

q

A:sublayer

:PlyFor


124

[ ]

[ ]

[ ]

[ ]0

1616

11

0

1616

10

0

1616

9

0

1616

8

11

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

11

10

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

10

9

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

9

8

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

8

=

Π+Π−

Π

Π

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

=

Π+Π−

ΠΠ

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

=

Π+Π−

ΠΠ

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

=

Π+Π−

ΠΠ

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

−

−

−

−

πmn

q

πmn

q

A:sublayer

πmn

q

πmn

q

A:sublayer

πmn

q

πmn

q

A:sublayer

πmn

q

πmn

q

A:sublayer


125

[ ]

[ ]0

1616

13

0

1616

12

13

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

13

12

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

12

=

Π+Π−

ΠΠ

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

=

Π+Π−

Π

Π

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

−

−

πmn

q

πmn

q

A:sublayer

πmn

q

πmn

q

A:sublayer

[ ]

[ ]0

1616

15

0

1616

14

3

15

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

15

14

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

14

=

Π+Π−

ΠΠ

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

=

Π+Π−

ΠΠ

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

−

−

πmn

q

πmn

q

A:sublayer

πmn

q

πmn

q

A:sublayer

:PlyFor


126

[ ]

[ ]0

1616

61

0

1616

61

17

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

17

16

1132

5

4

3

53

43

33

2

1

565251

464241

363231

161211

0

16

=

Π+Π−

ΠΠ

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

=

Π+Π−

ΠΠ

Π

−

ΠΠΠ

×

ΠΠΠΠΠΠΠΠΠ

ΠΠΠ

+

−

−

πmn

q

πmn

q

A:sublayer

πmn

q

πmn

q

A:sublayer

From the above expressions, it clearly shows that there are 17 unknown

coefficients with 17 available equations, therefore they can be solved

simultaneously to determine their solutions. Since the applied loading is

symmetric, i.e. ( ) ( ) ( ) ( )z,yUz,yU a

nn−=0 , this means that

0

22

0

11AA,AA aa −=−= and so on. Otherwise, the relationship is not applicable.

Notice also that solving the above equations is depending on the term m and n.

Once the unknowns are determined, the final solutions can be obtained in-terms

of displacements and stresses at one time. Depending on the computer

specifications, a standard desktop PC can takes up to about 5 hours to solve

the above problems.

4.4 Conclusions

This chapter deals with the application of state space method to the clamped

edges laminated plate. The mathematical concept of clamped edges laminated

plate is clearly derived and developed from the state equation of a simply

supported plate. The details of the process or procedures involved in the


127

analysis of clamped edges laminated plate have been explained in a very detail

and systematic manner. For clarification and understanding purposes, thorough

steps have been presented with regard to the method of analyzing clamped

edges plate in determination of solving the unknown coefficients. Some

derivations of formulae or expressions have also been shown in this chapter.

It is clearly showed in this chapter that the nature of work of solving the problem

involves massive matrix calculations and iteration process. Therefore, the

knowledge and skills of using a programming software is highly prerequisite.

Since this task is based on the previous research performed by Fan [30], i.e. the

case of a partially clamped edges plate with 2 sides simply supported and 2

sides clamped, it brings the attention and interest to the author to solve a further

challenging task, i.e. to investigate a fully clamped edges laminated plate. This

is the novel solution presented in this study.

Further challenging works have been performed to analyse a fully clamped

edges laminated plate. This is because of the fact that there are new additional

terms need to be considered and that is solving the unknowns coefficient of B’s.

Applying the same principle as before but now the term

( ) ( ) ( )∑ =+n

mnm

zVzV 00 is added. As a result, there are 34 unknowns to be

solved for the problem, 17 unknowns each for A and B. The results of these

new analytical exact solutions are verified with numerical analysis and

presented in the next chapter.

Further cases related to the investigation on the exact solution of clamped

edges laminated plate are also presented in the following chapter including

different loading conditions and material properties.

Some of the derivations of expressions are also clearly presented in very detail

manner this chapter.

Chapter 5 Numerical Analysis of Laminated Plates

128

Chapter 5

NUMERICAL ANALYSIS OF

LAMINATED PLATES

In this chapter, numerical solutions by means of finite element method are

presented in order to validate the analytical works based on the 3D elasticity

and state space method. The objectives of this approach are to gain further

insight of composite plate behavior in such a way that the model (a continuum)

is discretized into simple geometric shapes called finite elements and to perform

some other parametric studies. Material properties and the governing

relationships are considered over these elements. FEM is an alternative

approach to solving the governing equations of a structural problem particularly

at the elastic phase.

The application together with their results of FEM to composites is discussed in

detail in the following chapter. Various particular structural problems are

illustrated to show the effectiveness of FEM, verifying the analytical solutions

etc.

5.1 Introduction

The finite element method has become a powerful tool to generate the

numerical solutions for a wide range of engineering problems since the

existence of the finite element method in the 1950s [54]. Solutions ranging from

deformation and stress analysis for various engineering structures such as

building and bridges, aircraft and others to temperature and flows problems, can

be determined. With the significant advancement of computer technology, one

can model and simulate the behavior using finite element method with relative

ease, efficient and economical manner. By assembling the model appropriately


129

with the precise material properties and considering the loading and constraints,

the response of the model can be accurately simulated and numerical solutions

would be obtained.

In this study, the shape of the model is only limited to rectangular plate,

however, if a more complicated geometry such as a rectangular plate with a cut-

out is analysed, the governing equations become increasingly more

complicated and thus it is difficult to solve with ordinary mathematical program.

In such cases, finite element method is very much useful.

One of the available commercial finite element packages is known as Abaqus. It

is a very versatile program with broad application. This program is used in this

study due to its great flexibility of choices in predicting the linear or non linear

behavior of structures and varieties options in material selection. Numerous

investigations have been performed towards the successful application of finite

element simulation to study the behaviour of composite structures for many

years. The scope of this study is to model and simulate the performance of

various composite rectangular plates. The details of work that have been

carried out towards producing the acquired results in this study are presented in

the subsequent sections.

5.2 Modelling of Composite Plate Using FEM

Creating the part of plate is the initial task in modeling composite plate using

FEM. Part module is used to build different parts of the model. In this study, the

models are created and divided into a number of different layers of composite

plate to represent the laminate having different material properties for each

layer. Later all the layers can be assembled to form the entire model.

Laminated composite plate has two significant distinctions over a metal. Firstly,

it comprises of layered material built up by stacking the plies of different

material properties and secondly, for each ply, it is not isotropic. It depends on

the direction of the fibre to give the desired stiffness which varies from each ply.

Careful considerations had been paid prior to the modeling of finite elements in


130

order to avoid mistakes and wrong interpretation such as sign conventions,

coordinate systems, selecting the right face of the member at the right

placement, applying the correct boundary conditions, etc. Otherwise, these

would lead to misleading results. Therefore, accurate modeling was a

prerequisite. So the tasks which involve modeling and analysing, require

considerable efforts and time consuming to arrive the correct element types,

analysis procedures, material properties and other factors. This procedure

requires meticulous thought in order to determine the precise results. The

following sections give the details explanation to determine the numerical

simulation behaviour.

Generally, finite element method is an approximate technique as such the

output results should be carefully evaluated before relied upon them for a

design application. Considering the number of elements, or the fineness of the

meshing, the type of elements to be used, choosing the appropriate analysis

type are some of the factors that can significantly affect the accuracy of the

model. As the models are increasingly created with more refined mesh or as

the number of elements is increased, the results should converge to the true

numerical solution.

5.2.1 Element types

There were many different type of elements used during the initial stages of the

analysis which in turn affect the quality of results. However, in Abaqus, there

are many different types of elements available, depending on the nature of the

problem to be simulate. These include solid, shell, beam, rigid, membrane

infinite, connector or truss elements. For each of these elements, there are

further types of elements depending on the number of nodes, linear or quadratic

elements, full or reduced integrations. The degrees of freedom are the

fundamental variables calculated during the analysis. The use of solid elements

is limited to three-dimensional brick elements that have only displacement

degrees of freedom. In contrast, conventional shell elements have displacement

and rotational degrees of freedom. Continuum shell elements look like three-

dimensional continuum solids, but their kinematic and constitutive behaviour is


131

similar to conventional shell elements. Continuum shell elements have only

displacement degrees of freedom. Various types of finite elements, implies

different assumptions made and various results are derived. For instance,

Kirchhoff plate elements are derived for the application of thin plates which only

account for flexural deformation. Mindlin plate elements can be used for the

application of moderately thick plates accounting for both flexural and shear

deformations.

There are numerous other applications or examples which can be simulated by

the numerical analysis using FEM including heat transfer or fluid mechanics

simulations. A very good result will also be achieved if using elements with

higher order shape function rather than those with lower order. Elements having

many nodes give accurate results than less number of nodes. Using Gaussian

quadrature numerical techniques, full integration type of element can give more

precise results than reduced integration type of element.

Reduced integration uses a lower-order integration to form the element stiffness,

hence reduces running time in comparison with full integration elements.

5.2.2 Assembling the model

All the parts of the composite plate created earlier with either composite solids

or shells can be put together (assembly) to get the required model. After doing

this, we can apply the necessary constraints and loads on the assembly. The

thickness, number of section points required for numerical integration through

the thickness of each layer, material properties, orientation associated with

each layer, correct boundary conditions are the basic specifications needed for

assembling and defining the composite model. This is very easily done in

Abaqus using a special tool from composite layup where the common inputs are

put in one table together as shown in Figure 5.1 . This tool is very useful and

convenient to use particularly for the case of composite consisting of many

layers and having various fibre orientations. The number of integration points

are also easily be specified from the same tool.


132

The unidirectional laminate plate is assembled by stacking up each lamina in

predetermined directions and thicknesses to obtain the desired stiffness. The

behaviour is still considered as transversely isotropic in a cross section

perpendicular to the fibres.

Figure 5.1: Table used to model composite solid and shell in Abaqus

The other beneficial use of assembling the model using composite lay-up tool is

that the model can be checked whether the layers are in the right order using

the ply stack plot option. If they are not positioned correctly, these can be

corrected from the composite lay-up tool table.

For all layers of elements, full bond is assumed at the interface which means

that delamination cannot be resulted during the loading. To do these constraints,

designated surface areas that need to be bonded must be assigned initially.

This can be done easily in Abaqus by creating a set of surface area in the

Assembly module. By assigning tie connection at every interface of the


133

elements after merging them together, the relationship between the bottom

surface of jth layer and the top surface of the j+1th layer can maintain continuity

right from the top surface through the bottom surface of the laminate.

5.2.3 Boundary Conditions

For the in-plane displacements, u and v, and transverse displacement, w

assigning the boundary conditions are straightforward in FEM. Depending on

the cases, we can set the following boundary conditions for the laminate plate:

• Clamped, restraining U, V and W and rotations

• Simply supported, restraining U or V, W and σy or σx

Accurate boundary conditions are one of the important parts of modeling

composite structures in finite element method. Establishing the right boundary

conditions is truly vital to determine the required results. If the boundary

conditions are not properly assigned, the results would be inaccurate even

though using the precise element type and fine meshing. Depending on the

loading and boundary conditions also, if the laminated plate is subjected to both

in-plane and transverse loading, the plate will experiencing both stretching and

bending. From Figure 4.2, partially clamped edges laminate means that both

edges or faces when x = 0 and a , are fixed in all directions. The remaining two

edges or faces are simply supported at y = 0 and b. For the case of fully

clamped, obviously, all four sides or faces are fixed in all directions. This implies

that both translation and rotations are constraint.

As far as the analytical calculation is concerned, if all the edges of the laminate

plate are fully clamped, and because of the loading distribution is symmetrical, it

can be simplified that ( ) ( ) ( ) ( ) 00 =−= z,xVz,xV b

and ( ) ( ) ( ) ( )z,yUz,yU a 0−= . For the case of partially clamped, however,

( ) ( ) ( ) ( ) 0,,0 == zxVzxV b (the edges are simply supported).


134

5.2.4 Analysis Type

In this study, one of the important assumptions considered in the analysis of

composite plate using FEM is the type of analysis. It is vital to be acknowledged

if accurate results are to be obtained. The type of analysis chosen in this study

is assumed to be a linear static elastic analysis. In practice, the behaviour of

composite structures may exhibit non-linear after they reach the elastic limit

whereby the matrix begins to cracking at a tensile stress or the fibre starts to

fracture. It is only considered that elastic macro-mechanics investigation is the

prime study in which failure criteria is not included. Six degrees of freedom are

chosen at each node for shell elements, three translations and three rotations.

This permits the transverse shear deformation to be occurred. However, only 3

rotations degree of freedom are available by default for continuum and solid

elements. Three Simpson-type integration points are used along the thickness

of each shell element.

5.3 Material Properties

Selecting the appropriate material properties of the composite structures into

Abaqus is an important task which is required a careful consideration.

Laminated plate, in particular, has different moduli and strength different

directions. It is essential to note that in this study, composite plate has been

assigned as orthogonal type of material as an input in FEM. This behaviour is

different from other materials such as isotropic metal which has the same

elastic properties in all directions.

The type of material considered in the finite element modeling is orthotropic.

This implies that the properties along the fibre direction (x) are significantly

different from the other two normal directions (y and z).

The basic type of material used in Abaqus is isotropic material which can be

applied to the majority of elements available in the library. For the case of

composite plate, anisotropic or orthotropic behaviour, different strength and

stiffness at different directions will be assigned to elements selected.


135

Here, the values of material properties that are inserted into Abaqus are in

terms of ratio between one layer to another layer with respect to their elastic

properties. This can be clearly shown in the following:

The reduced material properties used in the numerical and the analytical

analyses are [30],

262931.0,159914.0,266810.0,530172.0

,115017.0,543103.0,0831715.0,246269.0

11

66

11

55

11

44

11

33

11

23

11

22

11

13

11

12

====

====

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

There are 3 laminae considered in the laminate where material properties of top

lamina (ply 1) are the same as the bottom lamina (ply 3). However, the material

property of lamina 2 (middle ply) is such that stiffness property ratio of ply1/ply2

= 5 or

( )

( ) .52

11

111 =

C

C

5.4 Mesh

Meshing is one of the most important modules since accuracy of the results

may depend on the meshing of the assemblies. This module can generate

meshes and verify them. Before meshing can be done, sometimes, partitioning

of the part is required for accuracy and convenience purposes particularly for

the case of complex shape.

After partition tasks have been done, normally the entire model shall be turned

green to indicate that meshing can be proceed.

Another great advantage of using Abaqus is the flexibility of selecting the

procedure prior to meshing. One can choose meshing by the number, size,

shape and type of element.


136

In this study, since the shape of the entire model is dominantly rectangular, the

shape of the mesh chosen is hex-dominated type. Before the model is

submitted for analysis in the job module, mesh verification and sensitivity

checks can also be performed. More details explanation about FEM and the

results of FEM for various boundary, loading conditions and material properties

are further shown in the following chapter.

5.5 Partially Clamped Edges Composite Plate with Va riable

Thickness to Width Ratio

Here, a three plies laminate plate is modeled with a full 3D model, stacked on

each other, subjected to a uniformly distributed transverse loading and is

clamped on two sides (at x = 0 and a) and the other two edges are simply

supported (at y = 0 and b). Ply 1 is the top layer, ply 2 is the middle layer and

the bottom layer is ply3 as shown in Figure 5.2. Material properties are as

mentioned in the previous chapter, Section 5.3. As the loading distribution and

boundary conditions are symmetrical, it can be simplified that

( ) ( ) ( ) ( )z,yUz,yU a 0−= , i.e. clamped edges and ( ) ( ) ( ) ( ) 00 == z,xVz,xV b ,

i.e. simply supported edges. The thickness of ply 1 is the same as ply 3 and

equal to 0.1h while ply 2 thickness is equal to 0.8h.


137

Figure 5.2: Geometry of plate consists of three pli es

Given the wide variety of element types that are available in Abaqus, solid and

shell elements are selected and run for the analysis of composite plate. Based

on the results obtained for each type of element, they are then compared with

those results from the analytical calculations. From this comparison, only the

element type that behaves as close as the analytical is considered and selected

for the subsequent case studies. Element type for FE analysis is based on

specific characteristics including first (linear) or second (quadratic), full or

reduced integration, and hexahedra/quadrilaterals.

The three-dimensional brick elements (solid), C3D8 and C3D20, were used to

model the laminated composite solids. Three layers of different materials having

same orientations were specified in each solid element. All layers were stacked

in the z-coordinates.

The overall in-plane dimension of the plate, i.e. a = 1 and b = 1, while the

thickness, h was varied. The in-plane mesh size for each element was varied

from coarse (0.5 x 0.5) to fine (0.03 x 0.03) meshes in order to improve

convergence and accuracy of the results. However, the mesh size in the z-

a

b

x , u , 1

z , w , 3

y , v , 2

h

Ply 1

Ply 2

Ply 3


138

direction was kept constant, i.e. for ply 1 and ply 3, the mesh thickness was

assigned to 0.1h/3 (3 sublayers) and the thickness of ply 2 meshes was taken

as 0.8h/10 (10 sublayers). Mesh sensitivity tests were taken to confirm which

mesh size would be appropriate to use for particular cases. The results from a

particular case of laminate with h/a ratio of 0.2 are shown in Figure 5.3. The

results show that FEM results converged to the exact solutions when the in-

plane element’s mesh size reduced to 0.03 x 0.03 from 0.5 x 0.5.

-18

-16

-14

-12

-10

-8

-600.10.20.30.40.50.6mesh

size

stre

ss, s/

q

sx FEMsy FEMsx Exactsy Exact

Figure 5.3: Mesh sensitivity test results for stres ses at x = y = z = 0 for h/a = 0.2

The shell elements used in the FE analysis include conventional and continuum

shells. Shell elements that were used to model structures had thickness

significantly smaller than the other dimensions. Conventional shell elements use

this condition to discretize a body by defining the geometry at a reference

surface. In this case the thickness was defined through the section property

definition. Conventional shell elements generally have displacement and

rotational degrees of freedom.

In contrast, continuum shell elements discretize an entire three-dimensional

body. The thickness is determined from the element nodal geometry.

Continuum shell elements have only displacement degrees of freedom. From a


139

modeling point of view continuum shell elements look like three-dimensional

continuum solids, but their kinematic and constitutive behavior is similar to

conventional shell elements.

The interaction between one layer of continuum shell to the next layer had to be

defined like modelling the plate using solid element. Tie connection was chosen

to simulate such interaction to merge the nodes at the interfaces. In this study,

three section points was defined for Simpson’s integration points through the

shell thickness.

The non-dimensional results are presented in the Appendix A with varying the

thickness to width ratio (h/a) of 0.2, 0.3, 0.4, 0.5, 0.6, 0.8 and 1. Table A-1 to A-

31 and Figure A-1 to A-29, generally show that solid elements exhibit better

results than shell elements as a matter of fact that for stress/displacement

(bending) simulation, the degrees of freedom in solid consider only translations

whereas conventional shell elements calculate both translation and rotations at

each node. Solid elements agree well to the exact solutions in terms of the

displacement and stress distributions at the centre of the plate particularly solid

element of C3D20. Solid elements have two through-thickness translational

freedoms which allow through-thickness direct strains (and stresses) to occur

whilst shell element does not have. Thin shell elements also neglect transverse

shear stresses in Abaqus and these elements use Kirchhoff thin shell theory.

These hypotheses occur when the thickness to width ratio is less than 1/10 [55].

This ratio is used to separate the full three-dimensional plate bending problem

with a two-dimensional problem. Table A-1 also shows that shell elements do

not give precise deflection results as the thickness to width ratio increases. This

is due to the fact that shell elements are used to simulate structural behaviour in

which the thickness is significantly smaller than the in-plane dimensions.

Continuum shells also do not show good results for stresses as compared to

solid and the exact solution, as shown in Table A-3. As the thickness to width

ratio increases, continuum shells give poor stress results. This is because even

though continuum shells look like three-dimensional continuum solids from a


140

modeling point of view, their kinematic and constitutive behaviour is similar to

conventional shell elements. It is important to note that Abaqus assumes

transverse constant shear strains through the shell’s thickness. The transverse

shear stresses are zero at the shell surface and they are continuous between

layers.

However, FEM cannot give precise stress results by shell elements compared

to exact solutions at the boundary edges. FEM also gives poor result for

transverse shear stresses at the edges when solid elements are used. Instead,

shell elements exhibit good response compared to the exact solutions as shown

in Figure A-31. Conventional shell elements allow transverse shear deformation.

They use thick shell theory as the shell thickness increases and become

discrete Kirchhoff thin shell elements as the thickness decreases; the

transverse shear deformation becomes very small as the shell thickness

decreases. In exact solution, shear stresses are zero at the top and bottom

surfaces of the composite laminate at the edges and increases in magnitude

dramatically afterwards. However, FEM only takes average stresses near the

top and bottom surfaces of the plate.

In Abaqus, second order solid element (C3D20) provides higher accuracy than

first order solid (C3D8) since they capture stress concentrations more

effectively particularly in this case of bending-dominated problems. Fully

integrated linear isoparametric elements (C3D8) suffer from ‘shear locking’, they

cannot provide pure bending solution because they must shear at the numerical

integration points to respond with an appropriate kinematic behaviour

corresponding to the bending. This shear locks the element which resulting

them far too stiff response. Element type C3D20 can overcome this shear

locking deficiency.

From this analysis, it can also show that the maximum deflection and stresses

at the centre of the plate are reduced gradually as the thickness to width ratio

(h/a) reaches to about 0.6. Beyond this ratio, increasing the thickness of the

laminate is no longer effective as the plate deflection tends to become

horizontal as shown in Figure A-1.


141

Equally importance to be noted from this analysis is how to choose an

appropriate element type in the subsequent cases. It is clearly shown that only

solid element C3D20 exhibits a closer result to the exact solution and therefore,

this element would be selected hereafter.

5.6 Fully Clamped Edges Composite Plate

In this case, all geometries and loading conditions are the same as the previous

model (Figure 5.2) except that all the edges are fully clamped. This means that

along all edges, i.e. at x = 0, a and y = 0, b, then U = V = W = 0. Fully clamped

edges plate also means that the relationship of the unknowns are

( ) ( ) ( ) ( )z,yUz,yU a 0−= and ( ) ( ) ( ) ( )z,xVz,xV b−=0 .

There are no exact analytical solutions are available in the literature. As

indicated in the previous chapter, modelling the laminate plate using FEM with

the first order element such as C3D8 with fully integrated elements, suffer from

“locking” behaviour (both shear and volumetric locking). Shear locking occurs in

first-order, fully integrated elements which are subjected to bending. The

numerical formulation of such element gives rise to shear strains that do not

really exist. Therefore, this element is too stiff in bending. Volumetric locking

occurs in fully integrated elements for incompressible material. Although, there

is no incompressible materials to be used in this studies. Previous section has

shown that the element C3D8 is not as good as C3D20 in laminated analysis.

Due to these reasons, the results of FEM are based on solid element type

C3D20 which will be used from this point onwards.

The new breakthrough non-dimensional analytical and numerical results are

shown as follows:

(Note that the deflection curves show continuity across the thickness of the

plate. However, no continuity for in-plane stresses across the thickness

because of the different material properties. Therefore, three stress cases are

shown for FEM – x for ply 1; □ for ply 2 and ○ for ply 3)


142

Table 5-1: Displacement and stresses distribution o f fully clamped laminate when h/a = 0.2

Exact FEM z = 0 14.41962 14.57520

2211 b

yanda

xatqh

WC== z = h 13.62818 13.78035

Ply1 Top -5.70218 -7.27574 Ply1 Bottom -5.75164 -4.44513

Ply2 Top -1.26698 -1.00880 Ply2 Bottom 1.11187 0.85842

Ply3 Top 5.60194 4.32076 22

byand

axat

qx ==

σ

Ply3 Bottom 5.54484 7.15232 Ply1 Top -5.29157 -6.67196

Ply1 Bottom -4.95703 -4.68203 Ply2 Top -1.15272 -1.10204

Ply2 Bottom 0.93845 0.89455 Ply3 Top 4.75117 4.51241

22

byand

axat

q

y ==σ

Ply3 Bottom 5.10697 6.50327

Ply1 Top 44.80379 32.29940 Ply1 Bottom -12.27516 -8.49153


Ply3 Top 6.81822 5.19473 2

0b

yandxatqx ==

σ

Ply3 Bottom -34.59811 -26.48580 Ply1 Top 10.36055 7.95430



20

byandxat

q

y ==σ

Ply3 Bottom -8.00056 -6.52261 Ply1 Top 0 6.07206



20

byandxat

qxz ==

τ

Ply3 Bottom 0 4.02236

Ply1 Top 13.58259 10.79870 Ply1 Bottom -1.02771 0.75091

Ply2 Top -0.20554 0.15018 Ply2 Bottom -0.10236 -0.32949

Ply3 Top -0.51180 -1.64745

02

== yanda

xatqx

σ


Ply1 Bottom -2.33325 1.65600 Ply2 Top -0.46665 0.33120


02

== yanda

xatq

yσ




02

== yanda

xatq

yzτ



143

0.0

0.2

0.4

0.6

0.8

1.0

13.6 13.8 14.0 14.2 14.4 14.6

W C11/qhz/

h

Exact FEM

Figure 5.4: Displacement (W C 11/qh) distribution through the thickness of

the plate at x = a/2 and y = b/2 when h/a = 0.2

0

0.2

0.4

0.6

0.8

1

-8 -6 -4 -2 0 2 4 6 8

sx/q

z/h

ExactFEMFEMFEM

Figure 5.5: Stress ( σx/q) x = a/2 and y = b/2 when h/a = 0.2


144

0

0.2

0.4

0.6

0.8

1

-8 -6 -4 -2 0 2 4 6 8sy/q

z/h

ExactFEMFEMFEM

Figure 5.6: Stress ( σy/q) x = a/2 and y = b/2 when h/a = 0.2

0

0.2

0.4

0.6

0.8

1

-40 -20 0 20 40sx/q

z/h

ExactFEMFEMFEM

Figure 5.7: Stress ( σx/q) x = 0 and y = b/2 when h/a = 0.2


145

0

0.2

0.4

0.6

0.8

1

-10 -5 0 5 10 15sy/q

z/h

ExactFEMFEMFEM

Figure 5.8: Stress ( σy/q) x = 0 and y = b/2 when h/a = 0.2

0.0

0.2

0.4

0.6

0.8

1.0

-15 -10 -5 0 5 10 15

sx/q

z/h

ExactFEMFEMFEM

Figure 5.9: Stress ( σx/q) x = a/2 and y = 0 when h/a = 0.2


146

0

0.2

0.4

0.6

0.8

1

-30 -20 -10 0 10 20 30 40sy/q

z/h

ExactFEMFEMFEM

Figure 5.10: Stress ( σy/q) x = a/2 and y = 0 when h/a = 0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7txz/q

z/h

ExactFEMFEMFEM

Figure 5.11: Stress ( τxz/q) at x = 0 and y = b/2 when h/a = 0.2


147

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10tyz/q

z/h

ExactFEMFEMFEM

Figure 5.12: Stress ( τyz/q) at x = a/2 and y = 0 when h/a = 0.2


148


Exact FEM z = 0 2.94212 2.96443

2211 b

yanda

xatqh

WC== z = h 2.11877 2.13772



Ply3 Top 0.85980 0.53519 22

byand

axat

qx ==

σ




22

byand

axat

q

y ==σ

Ply3 Bottom 1.78259 2.14072



Ply3 Top 4.23392 3.88103 2

0b

yandxatqx ==

σ




20

byandxat

q

y ==σ




20

byandxat

qxz ==

τ




Ply3 Top 0.36194 0.364467

02

== yanda

xatqx

σ




02

== yanda

xatq

yσ




02

== yanda

xatq

yzτ



149

0

0.2

0.4

0.6

0.8

1

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0

W C11/qhz/

h

Analytical FEM



0

0.2

0.4

0.6

0.8

1

-3 -2 -1 0 1 2 3sx/q

z/h

ExactFEMFEMFEM

Figure 5.14: Stress ( σx/q) x = a/2 and y = b/2 when h/a = 0.4


150

0

0.2

0.4

0.6

0.8

1

-3 -2 -1 0 1 2 3sy/q

z/h

ExactFEMFEMFEM

Figure 5.15: Stress ( σy/q) x = a/2 and y = b/2 when h/a = 0.4

0

0.2

0.4

0.6

0.8

1

-20 -10 0 10 20 30sx/q

z/h

ExactFEMFEMFEM

Figure 5.16: Stress ( σx/q) at x = 0 and y = b/2 when h/a = 0.4


151

0

0.2

0.4

0.6

0.8

1

-4 -2 0 2 4 6sy/q

z/h

ExactFEMFEMFEM

Figure 5.17: Stress ( σy/q) at x = 0 and y = b/2 when h/a = 0.4

0

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3 3.5txz/q

z/h

ExactFEMFEMFEM



152

0

0.2

0.4

0.6

0.8

1

-6 -4 -2 0 2 4 6 8sx/q

z/h

ExactFEMFEMFEM

Figure 5.19: Stress ( σx/q) at x = a/2 and y = 0 when h/a = 0.4

0

0.2

0.4

0.6

0.8

1

-10 -5 0 5 10 15 20sy/q

z/h

ExactFEMFEMFEM

Figure 5.20: Stress ( σy/q) at x = a/2 and y = 0 when h/a = 0.4


153

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5tyz/q

z/h

ExactFEMFEMFEM



154


Exact FEM z = 0 1.48584 1.48930

2211 b

yanda

xatqh

WC== z = h 0.66226 0.66219



Ply3 Top 0.00582 -0.07443 22

byand

axat

qx ==

σ




22

byand

axat

q

y ==σ

Ply3 Bottom 0.90977 1.03978



Ply3 Top 2.31299 2.10680 2

0b

yandxatqx ==

σ




20

byandxat

q

y ==σ




20

byandxat

qxz ==

τ




Ply3 Top 0.32580 0.38627

02

== yanda

xatqx

σ




02

== yanda

xatq

yσ




02

== yanda

xatq

yzτ



155

0.0

0.2

0.4

0.6

0.8

1.0

0.6 0.8 1.0 1.2 1.4 1.6

W C11/qh

z/h

Exact sol FEM



0

0.2

0.4

0.6

0.8

1

-2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5sx/q

z/h

ExactFEMFEMFEM

Figure 5.23: Stress ( σx/q) at x = a/2 and y = b/2 when h/a = 0.6


156

0

0.2

0.4

0.6

0.8

1

-2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5sy/q

z/h

ExactFEMFEMFEM

Figure 5.24: Stress ( σy/q) at x = a/2 and y = b/2 when h/a = 0.6

0

0.2

0.4

0.6

0.8

1

-10 -5 0 5 10 15 20sx/q

z/h

ExactFEMFEMFEM

Figure 5.25: Stress ( σx/q) at x = 0 and y = b/2 when h/a = 0.6


157

0

0.2

0.4

0.6

0.8

1

-3 -2 -1 0 1 2 3 4 5

sy/q

z/h

ExactFEMFEMFEM

Figure 5.26: Stress ( σy/q) at x = 0 and y = b/2 when h/a = 0.6

0

0.2

0.4

0.6

0.8

1

-4 -2 0 2 4 6sx/q

z/h

ExactFEMFEMFEM

Figure 5.27: Stress ( σx/q) at x = a/2 and y = 0 when h/a = 0.6


158

0

0.2

0.4

0.6

0.8

1

-10 -5 0 5 10 15sy/q

z/h

ExactFEMFEMFEM

Figure 5.28: Stress ( σy/q) at x = a/2 and y = 0 when h/a = 0.6

0

0.2

0.4

0.6

0.8

1

0 0.5 1 1.5 2 2.5 3

txz/q

z/h

ExactFEMFEMFEM



159

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4tyz

z/h

ExactFEMFEMFEM


The tasks of seeking analytical solutions of fully clamped laminate plate were

very challenging indeed, due to the limitation of computer capacity used. Unless

a supercomputer having a massive memory and fast processor is used, this

task cannot be completed with cost-effective manner. However, using a

standard desktop computer, the author had divided the tasks into smaller

components in order to get the intermediate results (to solve the unknowns).

Once all these smaller components were solved, they were then combined

altogether to get the final results. Careful considerations were taken during this

laborious, lengthy and time consuming process of breaking the smaller tasks as

human errors are very high. A standard PC can only generate up to about

twelve loops (cycle per run) which takes to about one hour. The analysis

precision was set up to the accuracy of 1000 decimal places. Once the

unknown results are determined, the prime results of displacements and

stresses can be found and it took about five to six hours to get them by using

the machine with Linux operation system and ten to eleven hours if Windows

operating machine is used.


160

In general, the results obtained from the exact solutions and FEM agree well

except the transverse shear stresses τxz and τyz at x = 0, y = b/2 and x = a/2,

y = b/2, respectively as shown in Figure 5.29 and Figure 5.30. FEM exhibits

poor results of transverse shear stresses at the boundary. FEM does not

preserve the continuity of the interfaces from the top through the bottom of the

laminated plate.

Stress results at ply1 and ply 3 when x = a/2 and y = b/2 at h/a = 0.2, do not

show smooth curve compared to FEM results as shown in Figure 5.5. More

results need to be obtained to verify this behaviour such as stress results for

thickness to width ratio of 0.1 and also by increasing the number of sub-division

(sub-layers) within the ply1 and ply3.

As the thickness to width ratio increases from 0.2 to 0.6, the deflection and

stresses curves between exact and FEM solutions agree well to each other

apart from the ratio of 0.2 as shown in Figure 5.4, Figure 5.5, Figure 5.13 and

Figure 5.22. Further investigation needs to be carried out to check the exact

solutions including for the case of thickness to width ratio of 0.1 or less and

most importantly increasing the number of sub-layers to capture the deflection

and stress results. For each of these curve, since loading is applied at the top

surface of the laminated plate, generally, the maximum and minimum deflection

are located at the top and bottom surfaces, respectively, except h/a = 0.2.

Transverse shear stresses txz and tyz are maximum at top and bottom

surfaces in FEM while they are zero values obtained by the exact solutions.

This is because of FEM treatment over the nodes of the element near the top

and bottom surfaces of the plate. FEM may calculates the average stresses of

the top and bottom nodes within the same individual element to give the

resultant stress at the outer surface’s nodes. Conversely, the exact solutions

give zero stress at the top fibre and increases dramatically immediately below

the top fibre. This also applies the same response for the bottom fibre of the

plate. These stresses distribution can be seen from Figure 5.11, Figure 5.18

and Figure 5.30


161

5.7 Laminated Plate Subjected To Hydrostatic Loadin g

In this model, laminated plate is subjected to hydrostatic loading having

boundary conditions of 2 sides fixed (at x = 0 and a) and 2 sides simply

supported (at y = 0 and b) with h/a = 0.4.

Figure 5.31: Partially clamped plate subjected to h ydrostatic loading for

h/a = 0.4

The loading is applied linearly increase from zero at x = 0 and qo at x = a as

shown in Figure 5.31, or this can be expressed as a

xqq o= . From this

unsymmetrical loading, it is important to be noted and applied in the analytical

method of exact solution that ( ) ( ) ( ) ( )z,yUz,yU a 0−≠ . This means that all the

unknown coefficients of oa AA11

−≠ and so on. However the relationship of

( ) ( ) ( ) ( ) 00 == z,xVz,xV b are still used in this model since the edges are

simply supported, hence they are no unknown coefficients exists.

The results obtained from the analysis are as follows:

x

z

y qo


162

Table 5-4: Exact solution versus FEM for plate subj ected to hydrostatic

loading

4.0=a

h

Exact Solution FEM z = 0 1.77429 1.77587

2211 b

yanda

xathq

WC

o

== z = h 1.35222 1.35399

Ply1 Top -1.67846 -1.52918



22

byand

axat

q

σ

o

x ==

Ply3 Bottom 1.43574 1.43890



Ply3 Top 1.03720 1.04311 22

byand

axat

q

σ

o

y ==

Ply3 Bottom 1.71577 1.72043



Ply3 Top 3.01603 3.52264 2

0b

yandxatq

σ

o

x ==

Ply3 Bottom -5.52563 -6.18245

Ply1 Top 0 0.60958 Ply1 Bottom 0.62205 0.62353

Ply2 Top 0.62205 0.12471 Ply2 Bottom 0.31860 0.10641

Ply3 Top 0.31860 0.53204 2

0b

yandxatq

τ

o

xz ==


Table 5-4 shows that FEM produces inaccurate results for stresses at the edges

of the plate especially transverse shear stresses. However, in-plane stresses

away from the edges agree well.


163

5.8 Laminated Plate with increasing number of sub-l ayers

In this section, a laminated plate which is similar to the model as shown in

Figure 5.2, i.e. the plate is partially clamped, subjected to uniformly distributed

loading on the top surface of the plate and having material properties as stated

in section 5.3, is analysed. The only variable considered in this investigation is

the number of sub-layers within the plies.

There are three cases taken into investigation:

Ply 1 Ply 2 Ply 3

Case No. of

sub-

layers

Thickness

(mm)

No. of

sub-

layers

Thickness

(mm)

No. of

sub-

layers

Thickness

(mm)

Total

no. of

sub-

layers

1 3 0.013 10 0.032 3 0.013 16

2 4 0.01 11 0.0291 4 0.01 19

3 5 0.008 12 0.0267 5 0.008 22

All the three cases, the plate’s thickness to width ratio are kept the same value

of 0.4. This means that the thickness of ply 1 (= ply 3) and ply 2 are 0.1h and

0.8h, respectively.

From the investigation, the results are obtained at the centre of the plate, i.e. x =

a/2 and y = b/2, as follows:

Table 5-5: The exact solutions and FEM of Case 1

Deflection, qh

WC11 Stress

q

σx Stress

q

σy

Ply h

z

Exact FEM Exact FEM Exact FEM 0 3.54859 3.55175 -3.05251 -3.05836 -3.28738 -3.27664

0.0333 3.54438 3.54768 -2.32003 -2.31467 -2.82134 -2.82669 0.0667 3.53883 3.54228 -1.59978 -1.60413 -2.37990 -2.38977

1

0.1 3.53216 3.53568 -0.92631 -0.92682 -1.95543 -1.96488


164

0.1 3.53216 3.53568 -0.30544 -0.30566 -0.55727 -0.55933 0.18 3.40284 3.40640 -0.20759 -0.20854 -0.43974 -0.44132 0.26 3.28177 3.28535 -0.14124 -0.14177 -0.33269 -0.33398 0.34 3.17055 3.17413 -0.09571 -0.09619 -0.23384 -0.23496 0.42 3.07040 3.07398 -0.06342 -0.06389 -0.14113 -0.14204 0.5 2.98221 2.98580 -0.03859 -0.03911 -0.05259 -0.05324

0.58 2.90658 2.91018 -0.01692 -0.01746 0.03367 0.03332 0.66 2.84378 2.84738 0.00569 0.00523 0.11973 0.11977 0.74 2.79372 2.79730 0.03447 0.03422 0.20818 0.20868 0.82 2.75584 2.75943 0.07690 0.07720 0.30229 0.30339

2

0.9 2.72896 2.73253 0.14479 0.14510 0.40652 0.40806 0.9 2.72896 2.73253 0.75795 0.75870 2.07964 2.08622

0.9333 2.72202 2.72558 1.42482 1.42813 2.51492 2.52268 0.9667 2.71392 2.71745 2.13358 2.13352 2.96670 2.97367

3

1 2.70444 2.70798 2.87148 2.87780 3.43154 3.44086


Deflection, qh

WC11 Stress

q

σx Stress

q

σy

Ply h

z


0.025 3.54733 3.54908 -2.50739 -2.49741 -2.93711 -2.93806 0.05 3.54350 3.54538 -1.95572 -1.95522 -2.59958 -2.60681

0.075 3.53903 3.54098 -1.42821 -1.43161 -2.27391 -2.28255 1

0.1 3.53393 3.53593 -0.92889 -0.92678 -1.95703 -1.96495 0.1 3.53393 3.53593 -0.30597 -0.30564 -0.55761 -0.55933

0.1727 3.41605 3.41810 -0.21532 -0.21598 -0.45018 -0.45157 0.2455 3.30488 3.30693 -0.15181 -0.15208 -0.35165 -0.35280 0.3182 3.20161 3.20368 -0.10676 -0.10703 -0.26021 -0.26127 0.3909 3.10725 3.10933 -0.07423 -0.07453 -0.17432 -0.17524 0.4636 3.02255 3.02463 -0.04939 -0.04977 -0.09243 -0.09320 0.5364 2.94804 2.95010 -0.02869 -0.02914 -0.01316 -0.01373 0.6091 2.88404 2.88610 -0.00910 -0.00957 0.06497 0.06466 0.6818 2.83065 2.83273 0.01264 0.01224 0.14365 0.14366 0.7545 2.78773 2.78980 0.04094 0.04071 0.22497 0.22537 0.8273 2.75473 2.75680 0.08188 0.08214 0.31153 0.31244

2

0.9 2.73072 2.73280 0.14502 0.14509 0.40688 0.40808 0.9 2.73072 2.73280 0.75906 0.75866 2.08140 2.08632

0.925 2.72562 2.72768 1.25452 1.25747 2.40628 2.41239 0.95 2.71988 2.72193 1.77472 1.77609 2.74070 2.74641

0.975 2.71344 2.71548 2.31708 2.31573 3.08409 3.08903 3

1 2.70619 2.70825 2.87128 2.87768 3.43359 3.44102


165


Deflection, qh

WC11 Stress

q

σx Stress

q

σy

Ply h

z


0.02 3.54897 3.54988 -2.62132 -2.60804 -3.00783 -3.00528 0.04 3.54607 3.54710 -2.17144 -2.16988 -2.73310 -2.73856 0.06 3.54275 3.54385 -1.74416 -1.74349 -2.46962 -2.47635 0.08 3.53902 3.54015 -1.32509 -1.32914 -2.21035 -2.21858

1

0.1 3.53489 3.53605 -0.93035 -0.92675 -1.95804 -1.96500 0.1 3.53489 3.53605 -0.30627 -0.30562 -0.55783 -0.55932

0.1667 3.42660 3.42783 -0.22178 -0.22237 -0.45888 -0.46017 0.2333 3.32385 3.32508 -0.16105 -0.16124 -0.36765 -0.36870

0.3 3.22760 3.22883 -0.11672 -0.11691 -0.28256 -0.28357 0.3667 3.13864 3.13988 -0.08409 -0.08431 -0.20247 -0.20340 0.4333 3.05760 3.05885 -0.05912 -0.05940 -0.12617 -0.12699

0.5 2.98497 2.98620 -0.03878 -0.03914 -0.05255 -0.05323 0.5667 2.92105 2.92230 -0.02062 -0.02104 0.01947 0.01898 0.6333 2.86603 2.86728 -0.00232 -0.00274 0.09110 0.09084

0.7 2.81992 2.82118 0.01887 0.01852 0.16372 0.16374 0.7667 2.78252 2.78378 0.04669 0.04650 0.23908 0.23943 0.8333 2.75337 2.75463 0.08613 0.08642 0.31923 0.32003

2

0.9 2.73169 2.73293 0.14515 0.14507 0.40709 0.40807 0.9 2.73169 2.73293 0.75969 0.75865 2.08240 2.08639

0.92 2.72765 2.72888 1.15291 1.15615 2.34130 2.34665 0.94 2.72321 2.72445 1.56490 1.56616 2.60709 2.61191 0.96 2.71835 2.71958 1.98812 1.98933 2.87778 2.88248 0.98 2.71302 2.71425 2.42837 2.42623 3.15499 3.15875

3

1 2.70714 2.70838 2.87048 2.87763 3.43459 3.44111


166

0

0.2

0.4

0.6

0.8

1

2.7 2.9 3.1 3.3 3.5 3.7

Deflection WC 11/qh

z/h

Case 1Case 2Case 3FEM

Figure 5.32: Central deflection across the thicknes s of plate h/a=0.4

0

0.2

0.4

0.6

0.8

1

-4 -3 -2 -1 0 1 2 3 4sx/q

z/h

Case 1

Case 2

Case 3

FEM

Figure 5.33: Stress,sx of plate h/a = 0.4 at x = a/2 y = b/2


167

0

0.2

0.4

0.6

0.8

1

-4 -3 -2 -1 0 1 2 3 4sy/q

z/h

Case 1Case 2Case 3FEM

Figure 5.34: Stress,sy of plate h/a = 0.4 at x = a/2 y = b/2

The above Tables 5-5 to 5-7 and Figures 5.32 to 5.34 show the results of exact

solutions and FEM of laminated plate of h/a = 0.4 with the variable of number of

sub-layers within the plies. From this investigation, it clearly shows that by

increasing the number of sub-layers, the values of deflection and stresses are

slightly enhanced. If more sub-layers are provided in the analysis, further

precise results can be obtained.

5.9 Conclusions

The comparisons of results between the exact solution and FEM analysis for

the laminated composite plate subjected to various parameters have been

presented in this chapter. The investigation of laminated plate having different

boundary, loading conditions and dimension of the laminated plate, have been

carried out to emphasize the effectiveness of the exact solution method. To

verify the analytical solutions, finite element program, Abaqus/CAE is used

interms of displacement and stresses results.


168

In the beginning of the modelling and analysis of the laminated plate with FEM,

various type of elements which are available in Abaqus library are used

including solids, conventional and continuum shells. The purpose of this initial

task is to obtain which element type best describe of the laminated plate

deformation and stresses as compared to the exact solution. From the findings,

solid elements C3D20 have proven close relationship behaviour of the

laminated plate corresponding to the exact solutions. From the outcome of this

investigation also verified that the remaining analysis would be based on

modeling of the laminated plate with solid element C3D20.

The novelty of solutions presented in this study is based from the work of

analytical solution of a fully clamped edges laminated plate. It is interesting to

note all the exact solutions agree well with the numerical solutions except when

the plate thickness to width ratio of 0.2 and the transverse shear stresses at the

outer surfaces of the plate edges. Both normal stresses and deflection at the

centre of the plate shows significant difference between the exact and the

numerical solutions. From this observation, further investigations are required to

best describe the response particularly for the case of thickness to width ratio is

less than 0.2 and also by increasing the number of sublayers of the laminate.

The other case presented in this chapter is the analysis of laminated plate

subjected to hydrostatic loading. The importance point to note from this case is

regarding the unsymmetrical loading type which shows that the relationship of

the unknown coefficients are no longer related.

It has been shown clearly that better and more precise results of displacements

and stresses would be obtained when the numbers of sub-layers have

increased within the plies. The results of exact solutions closely agree with the

FEM.


169

The other most important point that needs to be highlighted from these works is

that the analytical method involves massive matrix calculations whereby it is

common to encounter a problem at one point. There is a point of an exceptional

stage where the equations fail to be solved. This problem is known as

singularities.

For instance, the function ( ) ,x

xf1= has a singularity at x = 0, where the

solution is not defined or it falls to ∞± .

For the case of laminated plate, singularities are extremely important since

matrix calculations involve inverse matrix problem solving, where complex

analysis or solutions can be encountered. The issue of singularity can be

reduced or overcome by introducing high precision analysis. All the analytical

methods carried out in this investigation have been set to a very high precision,

between 400 (for a partially clamped edges plate) to 1000 decimal places (for a

fully clamped edges plate).

Chapter 6 Flexural Deformation of RC Slab with FRP

170

Chapter 6

FLEXURAL DEFORMATION OF RC

SLAB WITH FRP

In this chapter, the deflection of RC slab strengthened with FRP using FEM is

reviewed. This task was investigated during my first and second year of PhD

study. The objective of this study is to model and analyse FRP strengthened RC

slab using FEM program, Abaqus, so that its numerical flexural deformation

behaviour would correspond to the performance of composition materials of the

structure tested experimentally in the laboratory. To verify the accuracy of the

modelling deflection of the structures, the results from FEM simulation were

then compared with those of full scale FRP strengthened RC slabs testing

which was carried out in the University of Manchester [51] by others.

6.1 Numerical Modelling

Modelling of FRP strengthened RC slab using FEM is not as easy and straight

forward as compared with modeling composite laminates as a matter of fact of

their inherent material properties particularly for concrete and steel. This is

because concrete is a brittle material and steel is ductile when subjected to

tensile loading. By combining both materials allow them to behave beyond their

elastic region before they reach failure stage. Unlike FRP which dominantly

behaves in linear elastic only, the input data of plastic behaviour for concrete

and steel are mandatory to capture the structural response beyond elastic limit.

The full input data that was inserted into Abaqus and actual behaviour of these

materials which are tested and recorded in the laboratory are shown in Figure

6.1. The concrete in compression data is shown in Figure 6.1(a), steel bars in

tension in Figure 6.1(b) and FRP in tension in Figure 6.1(c).


171

Damaged plasticity model was selected in the modeling of FRP strengthened

RC slab because it provides different characteristics of concrete in tension and

compression.

The basic specimens parameters recorded in the laboratory are required for the

input data of modeling the RC slab with FEM. The concrete 28 days cube

compressive strength of 35 MPa and the average Poisson’s ratio of 0.2 were

noted [51]. The concrete was designed having a slump of 50mm with 10 mm

aggregate size, free water-cement ratio of 0.48 and 410kg/m3 of cement content

[51].

The ultimate tensile strength, modulus of elasticity and Poisson’s ratio of FRP

were 2970 N/mm2 ,172000 N/mm2 and 0.29 (average), respectively [51].

Tensile test of steel bar reinforcement noted in the laboratory showed that the

yield strength and ultimate strength reached up to 570 N/mm2 and 655 N/mm2,

respectively [51].

0

5

10

15

20

25

30

35

40

45

0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004

Strain

Str

ess

(MP

a)

(a)


172

0

100

200

300

400

500

600

700

0 0.05 0.1 0.15 0.2

Strain

Str

ess

(MP

a)

(b)

0

500

1000

1500

2000

2500

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016

Strain

Str

ess

(MP

a)

(c)

Figure 6.1: Stress and strain relationship of (a) c oncrete (b) steel and (c) FRP [51]


173

Damage states of concrete in tension and compression are characterized

independently by two hardening variables, the plastic strains in tension and

compression, respectively. Cracking (tension) and crushing (compression) in

concrete are represented by increasing the values of the hardening (softening)

variables. These variables control the evolution of the yield surface and the

degradation of the elastic stiffness. Damage is calculated by the reduction of

strength against cracking and crushing strain as proposed by Jankowiak [56] as

shown in Table 6-1.

Table 6-1: The evolution of the damage variable for compression and tension

Compression Tension

Stress

(MPa) Plastic strain

Damage in

Compression

Stress

(MPa)

Cracking

Strain

Damage

in

Tension

16.39928 0 0 2.12947 0 0 16.61235 0.0000290544 0 1.50804 0.0008027 0.2918 18.15366 0.0000388282 0 1.08781 0.0016054 0.4892 21.66902 0.0000677221 0 0.81292 0.0024081 0.6183 24.92055 0.00010442 0 0.63589 0.0032108 0.7014 27.90273 0.000149086 0 0.52087 0.0040136 0.7554 30.60989 0.000201887 0 0.44294 0.0048163 0.792 33.03621 0.000262997 0 0.38591 0.005619 0.8188 35.17569 0.000332593 0 0.33996 0.0064217 0.8404 37.02219 0.000410858 0 0.29963 0.0072244 0.8593 38.56936 0.000497979 0 0.26220 0.0080271 0.8769 39.81070 0.000594149 0 0.22657 0.0088298 0.8936 40.73951 0.000699567 0 0.19253 0.0096325 0.9096 41.34890 0.000814438 0 0.16027 0.0104353 0.9247 41.63178 0.000938971 0 0.13006 0.011238 0.9389 41.65226 0.000982667 0 0.10219 0.0120407 0.952 41.50121 0.001114853 0.0036 0.07680 0.0128434 0.9639 41.03288 0.00125643 0.0149 0.05397 0.0136461 0.9747 40.24036 0.001407601 0.0339 0.03366 0.0144488 0.9842 39.11654 0.001568579 0.0609 0.01572 0.0152515 0.9926 37.65408 0.00173958 0.096 0 0.0160542 1 35.84544 0.001920828 0.1394 33.68287 0.002112553 0.1913 31.15837 0.002314991 0.2519 28.26372 0.002528387 0.3214


174

The above Table 6-1 values are obtained from the proposed expression by

Jankowiak [56] as shown below,

Damage in compression = stressecompressivimummax

stressecompressivstressecompressivimummax −

Damage in compression = stresstensileimummax

stresstensilestresstensileimummax −

The process of modeling of FRP strengthened RC slab is equivalently similar to

modeling composite laminate as described in the previous chapter. The

significant differences between them are the contribution of materials plastic

deformation and interaction of FRP and concrete interfaces. This will be

discussed in more detailed manner in the following section.

Having said that material properties provide major influence on the deformation

behaviour of the structures, these materials are modeled individually in Abaqus

to represent numerical simulation of the overall non linear structural behaviour.

Concrete material is conveniently modeled using Concrete Damaged Plasticity

in which Abaqus allows the flexibility of choosing the post-cracking behaviour

particularly the tensile property. It is also assumed that the main two failure

criteria are tensile cracking or compressive crushing of the concrete material.

Note that there is also another type of concrete modeling which is available in

Abaqus, namely, concrete smeared cracking. This model is intended for

concrete behaviour for relatively monotonic loadings under fairly low pressures.

Furthermore, damaged plasticity option is most suitable to predict flexural

effects but doest not consider for shear modeling in the post-cracking response.

The other important material properties inserted into Abaqus include:


175

Table 6-2: Basic material properties used in FEM

Concrete Steel CFRP

Density (kg/m3) 2400 7850 1700

Elastic Modulus

(MPa) 27385 200,000

150,000 (E1)

10,000 (E2)

Poisson ratio 0.2 0.3 0.3 ( υ12)

6.2 Geometric Properties of the Model

Schematic diagrams (not to scale) of the simply supported FRP strengthened

RC slab are shown in Figure 6.2

As we can see that the RC slab is symmetrical in geometric and loading

conditions, only a quarter of the original dimension of the simply supported RC

slab is modelled and analysed in FEM as shown in Figure 6.3. This means that

the model is simplified into a 800 x 800 x 150 mm dimension. A concentrated

loading of area 250 x 250 mm is applied at the centre of the top surface of the

slab. RC slab is also reinforced with eight numbers of 12mm diameter steel bars

uniformly distributed in each direction. Such RC slab is then strengthened with

four CFRP sheets, each of them has 100 mm width and 1.2 mm thick which are

bonded to the bottom surface of the slab. Note that the position of FRP shown

in Figure 6.3 is being lowered down for illustration purpose only.


176

(a) RC slab without CFRP

1600

1600

150

Steel bars

250

250


177

(b) RC slab with CFRP

Figure 6.2: View of RC slab model (a) without FRP ( b) with FRP

1600

1600

150

FRP


178

Figure 6.3: FEM of CFRP strengthened RC slab

6.3 Element Type

The tasks of meshing and choosing element types for FRP strengthened RC

slab are exactly similar to modeling composite laminates as described in the

previous chapter. As mentioned earlier, Abaqus has a vast element library to

choose from, depending on the model type, one can select any element to suit

the true model behaviour. Choosing an element can be based on the specific

element characteristics such as first or second order; full or reduced integration;

hexahedra/quadrilaterals etc.

In this case, a discrete modeling is adopted for RC slab in which all materials

are modelled separately and individually. There are five elements chosen to

simulate the behaviour of each material or component. These are described as

below,

CFRP Rigid support


179

6.3.1 Concrete Slab

Three dimensional solid elements are used to simulate non linear analysis of

concrete slab material. Abaqus Standard solid elements have two categories,

first category is first order (linear) interpolation and the other is second order

(quadratic) elements which use quadratic interpolation to calculate

displacements degree of freedom. Typically, an eight node first order brick

element is known as C3D8 while a quadratic element has 20 node, i.e. C3D20.

Solid (continuum) elements are more accurate and suitable for concrete slab, if

not distorted, particularly for quadrilaterals and hexahedra shape. In this study,

only solid element of C3D8 is used since this type of element is suitable to

simulate a structure with shear loading problem which resulting no difficulty with

‘shear locking’ phenomenon.

6.3.2 Steel reinforcement bars

Truss elements are assigned to model steel reinforcement bars as they are

suitable for line like structural components that carry axial loading along the axis

or at the centre line of the element. A two node three dimensional straight truss

element (T3D2) is selected which uses linear interpolation. An alternative

element to model steel bar is by using rebar element. However, Abaqus/CAE

does not provide results or outputs in the visualization module when using rebar

element.

All steel bar reinforcements which are assigned as truss elements, embedded

into the host element (solid) which is concrete slab. This means that the

translational degrees of freedom at each embedded node of truss elements are

eliminated and constrained to and shared with the three dimensional solid

element nodes. It is important to note that Abaqus only considers host elements

(concrete slab) which can have only translational degrees of freedom and the

number of translational degrees of freedom at a node of embedded element

(steel bars) must be identical to the number of translational degrees of freedom

at a node of the host elements (solid).


180

6.3.3 FRP sheets

As the thickness of the FRP sheets is quite smaller than the other dimensions, it

is more economic in computation cost to model them using a general

conventional shell element (S4). Due to its thin dimension, Abaqus conveniently

defined its thickness through the section property definition, rather than using

element nodal geometry with continuum shell elements. Conventional shell

elements S4 have three displacements and two rotational degrees of freedom

whereas continuum shell elements have only displacement degrees of freedom.

S4 is chosen in this study as it gives accurate solutions to out of plane bending

of FRP strengthened RC slab, it is suitable for large strain analysis and it

neglects transverse shear deformation as the shell thickness decreases. This

type of element is based on thin shell theory, i.e. the Kirchhoff theory.

6.3.4 Interaction of FRP and concrete slab

An interaction was allowed within this region to simulate any response during

loading particularly delamination of FRP. Like adhesive material, cohesive

element was used to model the delamination phenomenon at the interface of

concrete slab and FRP. This interaction is called the traction – separation laws

which describe the constitutive behaviour of cohesive elements. Cohesive

element COH3D8 was assigned to model such interaction in such a way that

the top nodes of cohesive elements were connected to the bottom nodes of

solid elements (concrete slab) whilst the bottom nodes of the cohesive elements

were joined together with the shell nodes (FRP). A cohesive element is an

independent element that is not embedded in another element. To assign this

element, a new thin layer was created between concrete and FRP surfaces.

The minimum thickness Abaqus allows in CAE is a layer of 0.0001 mm. After

partition of either concrete or FRP surface to allow for cohesive designated area,

then cohesive property can be assigned to this new section.

Damage of the interface was assumed to initiate when the maximum nominal

stress ratio reached a value of one in any direction. Damage initiation was

defined using the maximum nominal stress values and the progression of

damage at the interfaces was modeled using the mixed mode energy


181

independent behaviour. After damage initiation, the damage evolution variable

until failure was modeled using linear softening law. The strength values of

damaged property using Maximum damage option for nominal stresses can be

determined from either the material manufacturer or through experimental test

(peel test for normal and lap shear test for tangential). If only one value of

strength was provided, this value can be used for all three directions.

Figure 6.4: Damage traction-separation response use d in FEM [55]

Damage initiation is assumed when maximum nominal stress ratio reaches to

one as expressed below,

1=

o

t

to

s

so

n

n

t

t,

t

t,

t

tmax

where o

t

o

s

o

ntandt,t represent the peak values of the nominal stress when the

deformation is purely normal to the interface or purely in the first and second

shear direction, respectively. In this study, due to the lack of information, the

values of MPa.tandMPa.t,MPa.t o

t

o

s

o

n7537538641 === were assumed

[51].

Traction (stress)

Separation or slip (displacement)

damage initiation

damage evolution

Gf


182

Damage evolution was defined based on the energy that is required to

propagate a tensile crack of unit area, known as the fracture energy. The

fracture energy is equal to the area under the traction-separation curve.

Damage evolution with mixed mode ratio of energy is opted in the modeling and

fracture energy, Gf, of concrete can be estimated using the equation [57].

70.

fcmo

cm

Fof f

fGG

= (FEM input of Gf = 0.757 Nmm/mm2)

where fcmo = 10 MPa

GFo is the base value of fracture energy, depending on the maximum aggregate

size. GFo equal to 0.0275 Nmm/mm2 was used in the modeling.

fcm is the average compressive strength of concrete.

6.3.5 Boundary condition

In order to simulate the behaviour of RC slab using FEM close enough to the

test result, the slab’s symmetric boundary conditions are modeled using a rigid

support. The contact area between the surface of concrete and the support has

to be defined into tangential and normal surfaces. Frictionless is opted in the in

plane tangential interaction and the normal component only applies when the

slab comes into contact with the rigid support. This normal interaction will cease

when slab surface is no longer in contact with the rigid support and it can only

happen at the four corners of the slab being uplifted during the high magnitude

loading stage.

RC slab is positioned horizontally on a rigid support which is modeled using a

three dimensional four node bilinear quadrilateral surface element (R3D4). This

element is defined as master surfaces which in contact with the concrete slab.


183

6.4 Tension Stiffening

Throughout the overall process of modeling and analyzing FRP strengthened

RC slab using Abaqus, the task of getting the ideal plastic behaviour of the

structure to match closely with the experimental test results was in fact the most

difficult, challenging and time consuming job. There were many trials and error

jobs undertaken during this stage.

The model requires input data of the uniaxial tensile and compressive response

of plain concrete for damaged plasticity as shown in Figure 6.5 and Figure

6.1(a), respectively.

Figure 6.5: Concrete tensile response characterized by damaged plasticity

From Figure 6.5, it shows that at initial stage, the stress-strain response gives a

linear elastic behaviour until it reaches the failure stress, ft (about 2 MPa). This

failure stress corresponds to the beginning of micro-cracking in the concrete

slab. Since concrete is weak in tension, the failure strain of concrete is so low

and from this point, the formation of micro cracks is represented

macroscopically with a softening stress strain response which induces strain

localization in the concrete material.

cr

nnσ

ft

εt cr

nnε


184

In Abaqus, steel bars reinforcement are embedded into the concrete, however,

concrete behaviour is considered independently of the rebar. The effects related

to steel bar reinforcement and concrete interface such as bond slip and dowel

action, are modelled approximately by “tension stiffening” into the concrete

modeling. This stiffening simulates load transfer across the cracks through the

steel bars.

Such concrete strain softening response is the most crucial part to determine

the non elastic behaviour of RC slab by FEM in this study. This post failure

behaviour also allows for the effects of the steel bars reinforcement interaction

with the concrete to be simulated in a simpler manner.

As modeling approach, tension stiffening is required in the concrete damaged

plasticity and there are three different methods of tension stiffening available in

Abaqus. Firstly tension stiffening can define the post failure by stress-strain

relation, or stress-displacement or lastly by applying a fracture energy cracking

criterion.

There are three important parameters need to be defined for the descending

branch of the tensile stress-strain relation or tension stiffening:

1) The tensile strength of concrete at which a fracture zone started

2) The area under the stress-strain curve (fracture energy, Gf)

3) The shape of the descending branch

Out of these parameters, the first two are considered as material constants,

however, the shape of the descending branch varies depending on which

formulation to be used. Choosing the shape of this descending branch is the

utmost critical decision in modeling the non linear behaviour of RC slab, as it

influences the true behaviour of the structures as per experimental test results.


185

6.4.1 Non-linear tension stiffening stress-strain r elation

This relation gives the stiffening behaviour of concrete in terms of cracking

strain proposed by Hordijk [58] as shown in Figure 6.6, which shows an

exponential stress – strain descending path curve.

Figure 6.6: Exponential tension stiffening (Hordijk ) [58]

The above descending exponential curve can be plotted using the Hordijk

equation:

where the parameters c1 = 3 and c2 = 6.93

t

fcr

ult,nn f.h

G.ε 1365=

h is the crack bandwidth. For solid elements, the default is Vh = where V is

the volume of the element.

cr

nnσ

ft

εt cr

nnε

Gf

cr

ult,nnε


186

6.4.2 Linear tension stiffening stress-strain relat ion

In this case, the descending branch is linear as shown in Figure 6.7,

Figure 6.7: Linear tension stiffening [58]

The relation of the above linear descending stress-strain is given by

( )

−=

cr

ult,nn

cr

nn

t

cr

nn

cr

nnε

ε

fεσ 1

cr

nnσ

ft

εt cr

nnε

Gf

cr

ult,nnε


187

6.4.3 Multi-linear tension stiffening stress-strain relation

Multi-linear descending tension stiffening can also be used to define the plastic

behaviour of the model. This type of tension stiffening is used in this study as

shown in Figure 6.8,

Figure 6.8: Multi-linear tension stiffening [58]

The selection of the above shape of tension stiffening curve can be painstaking

tasks to do for modelling the inelastic behaviour of RC slab. Therefore choosing

the appropriate curve is vital and time consuming process in the modeling of

non linear analysis of RC slab.

In this study, multi-linear descending tension stiffening is used which exhibits

good relationship to the available test results.

The results of the numerical solutions are explained in the following section.

cr

nnσ

ft

εt cr

nnε

Gf

cr

ult,nnε


188

6.5 FEM Against Experimental Test Results

Getting the results of numerical non-linear analysis of FRP strengthened RC

slab using Abaqus took less time than obtaining analytical results using a

standard PC desktop. On average, Abaqus required about one quarter the time

of running the programming code using Mathematica. Depending on the mesh

size, element and model type, method of integration etc, Abaqus normally

needs about 3 to 4 hours for a complete run.

After spending tremendous time and effort in modeling the structures using

Abaqus, the most vital experience towards modeling FRP strengthened RC slab

using concrete damaged plasticity is gained using an ideal tension stiffening

behaviour of concrete. As there are many options available with regard to

tension stiffening including, linear, bi-linear, multi-linear and exponential,

choosing a suitable type of tension stiffening behaviour is crucial to ensure the

true response of the structure as close as to experimental test results.

Tension stiffening behaviour defines the post failure for cracked concrete which

allows for the effects of interaction between steel bars reinforcement and

concrete. Estimation of the tension stiffening effect depends on some factors

such as the ratio of reinforcement, the quality of bond between steel bars and

the concrete, the relative size of the concrete aggregate compared to the

diameter of steel bars and the mesh size. One can start to assume that

stiffening behaviour to be linear stress – strain relationship from maximum to

zero stress at a total strain of about 10 times the strain at failure. The typical

tensile strain value of concrete at failure is 0.00016. This means that tension

stiffening reduces the stress linearly from maximum to zero at a total strain of

about 0.0016 (or may be less than) can be sufficiently applied for initial

estimation. Numerical solutions are shown based on various tension stiffening

behaviour and element types as shown in Figure 6.9 and Figure 6.10,

respectively.


189

From Figure 6.9 shows the relationship of the applied loading to central

deflection of RC slab by varying the tension stiffening effect as described in

Section 6.4 against the results obtained from the experimental test [51].

0

50

100

150

200

250

300

0 10 20 30 40

Central Deflection (mm)

Load

(kN

) Exp TestBi-LinearLinearExponential

Figure 6.9: Relationship of load against central de flection of RC slab by variable tension stiffening response

0

50

100

150

200

250

300

0 10 20 30 40


Load

(kN

)

Exp TestFEM C3D8FEM C3D8RFEM C3D20

Figure 6.10: Relationship of load against central d eflection of RC slab by variable element types


190

0

50

100

150

200

250

300

350

400

450

0 10 20 30


Load

(kN

)RC slab FEM

RC slab experimental

RC slab with CFRPFEMRC slab with CFRPexperimental

Figure 6.11: Load against central deflection of un- strengthened and strengthened of RC slab with CFRP

From Figure 6.9 and Figure 6.10, the numerical solutions explicitly show that

multi-linear tension stiffening behaviour and three dimensional solid element

type C3D8 produce better results in modeling of RC slab with Abaqus as

compared to experimental test results. Rigid elements are used to allow the

contact interaction between the slab surface and the supports or the base. This

would simulate the corner of the slab being uplifted during the loading.

Same conclusion can be made with the analysis of RC slab strengthened with

CFRP as shown in Figure 6.11. The response generally agrees well for both

unstrengthened and strengthened RC slab with FRP.

This investigation shows the effectiveness of CFRP in strengthening of concrete

structures, both in serviceability and ultimate limit states. The use of CFRP in

this study has increased the ultimate RC slab flexural strength to about 43%

and also has reduced the central deflection to approximately 58%. In FEM, after

strengthening with CFRP, the serviceability load has increased to about 37%. It

was also noted that FRP strengthened RC slab was failed due to concrete

crushing whilst FRPs materials are still intact with concrete at failure.


191

6.6 Concrete slab reinforced with FRP

In this section, only linear analysis of layered plate or slab is considered and

analysed using the exact state space method. The purpose of this study is to

evaluate the effectiveness of state space method to concrete slab with CFRP

over the classical laminated plate theory. Their differences between these

approaches are being compared in-terms of determination of the central

deflection of the plate.

The effectiveness of a single sheet of typical CFRP reinforced concrete slab

subjected to uniformly distributed loading is investigated. For the given

mechanical properties of a single CFRP sheet, flexural deformation of concrete

slabs with various thickness is analysed.

The slab to be considered, as shown in Figure 6.12 (a), is originally designed

for the serviceability within the elastic limits. Due to slab degradation or some

other unspecified reasons, the structure is not able to bear more loading in

elastic range. In order to restore its elastic design and service, CFRP sheet is

bonded to the bottom of the slab. As a result of this retrofitting work, the whole

composite slab structure consists of two layers with different material properties.

These two material layers can be further divided many thin sub-layers.

The slab considered now consists of subdivided layers as shown in Figure 6.12

(b).


192

( a )

( b )

Figure 6.12: Geometry and coordinate systems of the layered slab

CFRP

a

b

x

z

y

h

concrete

hc

hCFRP

(1) d1

(2) d2

( j ) dj

( N ) dN

a

b

x ,u

z , w

y , v

h


193

Material properties used for both analytical and numerical analysis are shown

below:

Concrete: Compressive strength 30MPa

Elastic’s modulus 33GPa Poisson ratio 0.2 and Shear modulus 13.75GPa.

CFRP sheets: Longitudinal elastic modulus 135 GPa (E1)

Transverse elastic modulus 10 GPa (E2) Poisson’s ratio 0.3 (ν12) Shear moduli 5 GPa (G12) and 3.85 GPa (G23).

The calculated or reduced material parameters which will be inserted in the

analytical and numerical analyses, based on the above properties, are

For concrete:

GPaCC

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

c

c

c

c

c

c

c

c

c

c

c

c

c

c

c

c

c

67363750

3750375012502501250

11

11

44

11

55

11

66

11

33

11

23

11

13

11

22

11

12

.,.

,.,.,,.,.,,.

==

=======

For CFRP ply:

1048950136364013636403034830

0936930119153030348301191530753313

11

44

11

55

11

66

11

33

11

23

11

13

11

22

11

12

11

11

.,.,.,.

,.,.,.,.,.

====

=====

c

f

c

f

c

f

c

f

c

f

c

f

c

f

c

f

c

f

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

where cij

fij

CandC are the stiffness coefficients of CFRP and concrete,

respectively.

Boundary Conditions

In order to simplify the calculation, only a partly clamped slab is considered, two

opposite edges x = 0 and a are clamped and other two edges y = 0 and b, are

simply supported.


194

Geometric Conditions

In this case study, the in-plane dimensions are taken from the previous model,

however, the overall thickness are varies. The thickness of concrete slab are

varies from hc of 0.0988, 0.1988, 0.2988, 0.3988 to 0.5988 however FRP

thickness remains unchanged (hFRP = 0.0012). Therefore, the overall thickness

(h) of the composite slabs considered are 0.1, 0.2, 0.3. 0.4 and 0.6.

Considering the boundary conditions and the loading case, these imply that in

the current numerical analysis

( ) ( ) ( ) ( ) ( ) ( ) ( )zyUzyUandVzxV ab ,,0, 00 −=== .

Exact solution against FEM

In order to verify the analytical method of the slab strengthened with CFRP,

comparisons are provided with the Finite Element Method (FEM).

Since it is an elastic analysis, a proportional amount of the loading, qo, assigned

for the finite element analysis, is 1Pa. Quadratic elements with 20 nodes

(C3D20) were used to model the laminate structures and perfectly bonding (tie)

was also chosen at the interface between the concrete and CFRP material. All

other material, geometric parameters and boundary conditions were the same

as the analytical analysis. However, various cases with different ratio of

thickness to width (h/a) are considered.

The results are presented as follows:


195

Table 6-3: Deflection distribution (WC 11/qh) for h/a = 0.1 W C11/qh

No CFRP With FRP z/h

Exact FEM Exact FEM

0 27.52862 27.49960 26.34378 26.32090 0.076 27.57216 27.54290 26.38548 26.36240 0.152 27.60689 27.57790 26.41876 26.39580 0.228 27.63361 27.60470 26.44441 26.42160 0.304 27.65259 27.62370 26.46268 26.43990 0.38 27.66400 27.63520 26.47374 26.45100

0.456 27.66797 27.63920 26.47773 26.45500 0.532 27.66457 27.63580 26.47471 26.45200 0.608 27.65379 27.62500 26.46468 26.44200 0.684 27.63557 27.60680 26.44760 26.42490 0.76 27.60979 27.58100 26.42335 26.40070

0.836 27.57627 27.54750 26.39176 26.36910 0.912 27.53476 27.50590 26.35259 26.32990 0.988 27.48461 27.45600 26.30529 26.28280 0.988 27.48461 27.45600 26.30529 26.28280 0.992 27.48172 27.45310 26.30136 26.27890 0.996 27.47880 27.45020 26.29740 26.27500

1 27.47586 27.44730 26.29340 26.27110



Exact FEM Exact FEM

0 4.53008 4.53352 4.45659 4.45992 0.0765 4.54146 4.54503 4.46752 4.47098 0.1529 4.54829 4.55206 4.47400 4.47765 0.2294 4.55146 4.55530 4.47690 4.48062 0.3058 4.55142 4.55529 4.47665 4.48041 0.3823 4.54853 4.55242 4.47364 4.47741 0.4588 4.54307 4.54696 4.46811 4.47189 0.5352 4.53516 4.53905 4.46022 4.46400 0.6117 4.52480 4.52870 4.44997 4.45375 0.6882 4.51188 4.51577 4.43723 4.44101 0.7646 4.49613 4.50001 4.42178 4.42554 0.8411 4.47717 4.48103 4.40322 4.40697 0.9175 4.45448 4.45831 4.38107 4.38479 0.994 4.42730 4.43117 4.35458 4.35834 0.994 4.42730 4.43117 4.35458 4.35834 0.996 4.42651 4.43040 4.35350 4.35727

0.99800 4.42573 4.42962 4.35241 4.35619 1 4.42494 4.42883 4.35131 4.35511


196



Exact FEM Exact FEM

0 1.88921 1.89294 1.87608 1.87958 0.07662 1.88468 1.88870 1.87139 1.87517 0.15323 1.87713 1.88132 1.86370 1.86764 0.22985 1.86756 1.87180 1.85401 1.85800 0.30646 1.85667 1.86094 1.84301 1.84704 0.38308 1.84504 1.84933 1.83130 1.83535 0.45969 1.83308 1.83739 1.81927 1.82334 0.53631 1.82100 1.82531 1.80714 1.81121 0.61292 1.80881 1.81312 1.79493 1.79899 0.68954 1.79631 1.80061 1.78244 1.78650 0.76615 1.78312 1.78741 1.76931 1.77335 0.84277 1.76865 1.77291 1.75494 1.75897 0.91938 1.75211 1.75633 1.73859 1.74257 0.99600 1.73245 1.73666 1.71920 1.72318 0.99600 1.73245 1.73666 1.71920 1.72318 0.99733 1.73207 1.73628 1.71868 1.72266 0.9987 1.73169 1.73590 1.71815 1.72214

1 1.73130 1.73552 1.71763 1.72161



Exact FEM Exact FEM

0 1.15025 1.15331 1.14675 1.14964 0.0767 1.13383 1.13727 1.13025 1.13351 0.1534 1.11513 1.11870 1.11148 1.11488 0.2301 1.09526 1.09889 1.09155 1.09500 0.3068 1.07521 1.07888 1.07143 1.07492 0.3835 1.05578 1.05948 1.05193 1.05545 0.4602 1.03755 1.04127 1.03364 1.03718 0.5368 1.02087 1.02459 1.01688 1.02043 0.6135 1.00579 1.00951 1.00175 1.00529 0.6902 0.99212 0.99583 0.98803 0.99157 0.7669 0.97938 0.98307 0.97527 0.97878 0.8436 0.96678 0.97043 0.96268 0.96617 0.9203 0.95324 0.95684 0.94921 0.95265 0.997 0.93734 0.94091 0.93345 0.93685 0.997 0.93734 0.94091 0.93345 0.93685 0.998 0.93711 0.94068 0.93314 0.93654 0.999 0.93688 0.94045 0.93282 0.93622

1 0.93665 0.94022 0.93250 0.93590


197

Table 6-7: Deflection distribution (WC 11/qh) for h/a = 0.6

W C11/qh


Exact FEM Exact FEM

0 0.72483 0.72724 0.72445 0.72679 0.0768 0.68927 0.69217 0.68886 0.69169 0.1535 0.65159 0.65461 0.65117 0.65411 0.2303 0.61325 0.61633 0.61281 0.61581 0.3071 0.57587 0.57897 0.57540 0.57842 0.3838 0.54088 0.54396 0.54037 0.54337 0.4606 0.50939 0.51242 0.50883 0.51178 0.5374 0.48212 0.48510 0.48151 0.48441 0.6142 0.45941 0.46233 0.45874 0.46157 0.6909 0.44115 0.44401 0.44042 0.44319 0.7677 0.42677 0.42957 0.42599 0.42870 0.8445 0.41522 0.41795 0.41440 0.41705 0.9212 0.40487 0.40753 0.40405 0.40664 0.998 0.39350 0.39608 0.39273 0.39524 0.998 0.39350 0.39608 0.39273 0.39524

0.9987 0.39338 0.39597 0.39258 0.39509 0.9993 0.39327 0.39586 0.39243 0.39493

1 0.39316 0.39575 0.39227 0.39478

0

0.2

0.4

0.6

0.8

1

26.0 26.5 27.0 27.5 28.0

WC11/qh

z/h

Concrete (Exact)

Concrete with FRP(Exact)Concrete (FEM)

Concrete with FRP(FEM)

Figure 6.13: Deflection distribution (W C 11/qh) at x=a/2, y = b/2 h/a = 0.1


198

0

0.2

0.4

0.6

0.8

1

4.3 4.4 4.4 4.5 4.5 4.6 4.6

WC11/qh

z/h

Concrete (Exact)




0.0

0.2

0.4

0.6

0.8

1.0

1.70 1.75 1.80 1.85 1.90

W C11/qh

z/h

Concrete (Exact)





199

0

0.2

0.4

0.6

0.8

1

0.90 0.95 1.00 1.05 1.10 1.15 1.20

WC11/qh

z/h

Concrete (Exact)




0

0.2

0.4

0.6

0.8

1

0.35 0.45 0.55 0.65 0.75

WC11/qh

z/h

Concrete (Exact)





200

From Table 6-3 to Table 6-7 or Figure 6.13 to Figure 6.17, it can be seen that

CFRP can effectively be used to reduce the deflection of the concrete slab

under service load within the elastic stage. For the typical material property of

CFRP as mentioned before, a single sheet of CFRP can reduce the deflection

of the slab up to the thickness to width ratio reaches to 0.4. When the thickness

ratio of the slab increases beyond 0.4, a single sheet of CFRP is insufficient and

may require more sheets of CFRP in order to reduce the deflection. This can

be shown in Figure 6.16 and Figure 6.17 where the contribution from a single

CFRP can be neglected.

From Table 6-3 and Figure 6.13, the exact solutions show that the magnitudes

are slightly higher than the FEM results. However, the FEM outputs are slightly

overestimated than the exact solutions when h/a equals to 0.2 and higher.

It can also be shown that the maximum deflection for plate with thickness to

width ratio (h/a) of 0.1 is located at about half the thickness of the plate and

gradually, the position of maximum deflection moves towards the top surface of

the plate as the ratio increases.

6.7 Conclusions

Modelling of RC slab strengthening with FRP using finite element analysis

program, Abaqus has been reviewed in this chapter. The main objective of this

task is to review and simulate the flexural deflection of the structure

corresponding to the available experimental test result.

From the outcome of the simulation work, the numerical solution of FEM

matches well to the test result which shows the significant enhancement of

loading capacity of RC slab and reducing it’s central deflection after FRP is

applied.

The central deflection of concrete slab with and without FRP agrees well

between the exact solutions and FEM results.

Chapter 7 Conclusions and Recommendations

201

Chapter 7

CONCLUSIONS AND

RECOMMENDATIONS

7.1 Conclusions

The analysis of laminated composite plate based on the 3D elasticity and state

space method under general boundary conditions have been presented.

All analytical solutions are compared and verified with numerical solutions

obtained from the finite element program, Abaqus. Modeling of plate using solid

element is more sufficient than shell element as FEM solutions agree well in

general. However, FEM results obviously fail to satisfy the traction free

conditions on the top and bottom surfaces of the plate. This means that FEM

predicts a magnitude of transverse shear stresses on the top and bottom

surfaces of the plate’s boundary edges, whereas these stresses are still zero

values obtained from the exact solution. The accuracy of FEM solutions are

determined using solid elements with higher order shape function (quadratic)

rather than low order shape function (linear).

Solid elements provide a more accurate solution than shell elements as

compared to the exact solutions because of the following:

• The transverse shear effects of solid elements are predominant than

shell elements.

• Solid elements do not ignore the normal stress.

• Solid elements provide accurate interlaminate stresses including near

localized regions of complex loading or geometry.


202

By using transfer matrix method and preserving the continuity conditions at the

interfaces and imposing the in-plane traction along the edges which satisfies the

boundary conditions, exact elasticity solution is very accurate since it satisfies

the elasticity equations at every point of the laminate plate. Modeling the

composite plates using solids elements of C3D20 are more precise than using

shell elements. These are due to the fact that solids elements have more effect

on transverse stresses and interlaminar stresses which are obtained and

calculated by Abaqus based on the theory of linear elasticity.

From this study, it concludes that exact 3D elasticity solution can produces

exact solution of composite plate of any reduced material properties, boundary

conditions and loading conditions. Solid element C3D20 provides very good

result compared with the 3D exact elasticity solution with respect to deflection

and in-plane stresses. Meanwhile, shell element shows a better result than solid

elements for transverse shear stresses. FEM does not get an exact solution for

the transverse shear stresses.

The breakthrough and novel exact solutions of fully clamped composite plate

have been achieved in this study. Such new results can be used as a

benchmark solution for further analysis.

The use of computer machine with Linux operating system is an advantage in-

terms of computing efficiency rather than using machine with Windows

operating system.

FEM model was also enabled to predict the performance of FRP strengthened

RC slab that was tested in the laboratory. Both the numerical solutions and

experimental test results verified that ultimate flexural capacity of RC slab

strengthened with CFRP has increased by 43%. Solid element C3D8 and multi-

linear response are the ideal element type and best possible tension stiffening

behaviour to describe the performance of flexural deformation of FRP

strengthened RC slab.


203

The accuracy of the numerical results depends on the thickness to width ratio,

the point of interest and the material properties of the laminated plate.

7.2 Future Works Recommendations

The following tasks can be performed for future investigation with regard to 3D

elasticity of laminated composite plates:

• Increasing number of m and n as these would improve the accuracy of

the results of analytical solutions.

• Increasing the number of sublayers within each ply of laminate. The

thinner the sublayer, the better the results.

• In the FEM analysis, the solution can give only an approximate solution

such that equilibrium is satisfied on average over an element. FEM does

not preserve the continuity between the interfaces. More accuracy of the

results can be obtained when using fine meshes but keeping in mind that

fine meshes generate more elements, hence costing computing process

to run.

• The novel solutions of fully fixed clamped edges laminated plate can be

used as benchmark for further analysis with different loading conditions

and material properties.

• Numerical simulations of FRP strengthened RC slab using Abaqus were

only modeled and analysed using concrete damaged plasticity which

mainly accounts for load – deflection relationship. From this adopted

method, the choices of various tension stiffening greatly influence the

global flexural response of such model. Further analysis using smeared

crack modeling can be used to incorporate the shear modeling and crack

propagation.

• Having developed code of analytical method, the exact solutions of

structural performance of any reduced material properties and boundary

conditions can be further explored such as the graphene structures.

• Further investigation of the simulation can be performed to get better

accuracy such as laminated plate having different material properties.


204

• It is interesting to analyse further on the laminated plate having thickness

to width ratio of less than 0.2, particularly to compare the central

deflections and in-plane stresses at the centre of the plate.

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205

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List of Publications

209

LIST OF PUBLICATIONS

1. Wu, Z.J. and Kamis, E., Influence of Element Type on the Simulation of

Laminated Composite Plates with Clamped Edges, Conference of the

Association for Computational Mechanics in Engineering (ACME), 27-

28th March 2012, the University of Manchester, UK.

2. Kamis, E. and Wu, Z.J., Exact Elasticity Solution for Slabs Reinforced

with Fibre Reinforced Plastic Sheet. (to be submitted for journal

publication)

3. Kamis, E. and Wu, Z.J., Exact Elasticity Solution for Thick Laminate with

Fully Clamped Edges. (to be submitted for journal publication)

Appendix A

210

APPENDIX A

Table A-1: Maximum deflection at x = a/2, y = b/2 a nd z = 0 (W C11/ qh)

Shell Solid

Conventional Continuum

Linear Quadratic Linear Quadratic Linear h/a Exact

C3D8 C3D20 S4 S8R SC8R

0.2 19.85492 19.70885 19.85000 20.70275 20.69185 20.17570 0.3 6.90785 6.90893 6.90500 6.96923 6.96373 6.42797 0.4 3.54859 3.54093 3.54863 3.33113 3.32780 3.58780 0.5 2.26852 2.28758 2.26940 1.91869 1.91804 2.50204 0.6 1.65476 1.67465 1.65685 1.24354 1.24166 1.94725 0.8 1.08207 1.08860 1.08541 0.64374 0.64250 1.35578 1 0.79664 0.80336 0.80043 0.39407 0.39321 1.00998

Table A-2: Stress ( σx/q) at x = a/2, y = b/2 at Ply1 top surface

Shell Solid




0.2 -9.95082 -8.59663 -10.02990 -8.82711 -8.89627 -8.52921 0.3 -4.78541 -3.82362 -4.80254 -3.58893 -3.61438 -3.19393 0.4 -3.05251 -2.78390 -3.05836 -1.86807 -1.87838 -1.68732 0.5 -2.36148 -2.12438 -2.36150 -1.12867 -1.13209 -0.93771 0.6 -2.11374 -1.86817 -2.10372 -0.750821 -0.752578 -0.62842 0.8 -2.07925 -1.80791 -2.07229 -0.40063 -0.400703 -0.46830 1 -2.20737 -1.89060 -2.20740 -0.24901 -0.248722 -0.61819

Table A-3: Stress ( σx/q) at x = a/2, y = b/2 at Ply1 bottom surface Shell

Solid Conventional Continuum



0.2 -6.05797 -7.37290 -6.10997 -7.06169 -7.11702 -6.41338 0.3 -2.15883 -3.04784 -2.16349 -2.87114 -2.8915 -2.31356 0.4 -0.92630 -1.14816 -0.92682 -1.49445 -1.50270 -0.53669 0.5 -0.39640 -0.60238 -0.39780 -0.902934 -0.905671 -0.37342 0.6 -0.09078 -0.28668 -0.08217 -0.600657 -0.602062 -0.07150 0.8 0.39008 0.14918 0.385447 -0.320504 -0.320562 0.21217 1 0.75763 0.47419 0.751947 -0.199208 -0.198978 0.48004

Appendix A

211


Shell Solid




0.2 -1.33206 -0.89997 -1.34265 -1.41234 -1.42341 -1.15223 0.3 -0.55165 -0.39958 -0.55260 -0.57423 -0.57830 -0.51587 0.4 -0.30544 -0.24894 -0.30566 -0.29889 -0.30054 -0.29689 0.5 -0.20013 -0.15254 -0.20063 -0.18059 -0.18114 -0.18462 0.6 -0.13951 -0.09136 -0.13811 -0.12013 -0.12041 -0.12297 0.8 -0.04267 -0.00883 -0.04391 -0.06410 -0.06411 -0.06317 1 0.03440 0.06658 0.03281 -0.03984 -0.03980 0.01011

Table A-5: Stress ( σx/q) at x = a/2, y = b/2 at Ply2 bottom surface

Shell Solid




0.2 1.20446 0.75934 1.20057 1.41234 1.42341 1.26250 0.3 0.40391 0.25438 0.40489 0.57423 0.57830 0.63055 0.4 0.14479 0.09623 0.14510 0.29889 0.30054 0.35164 0.5 0.03422 0.00865 0.03436 0.18059 0.18114 0.24374 0.6 -0.01846 -0.02499 -0.01627 0.12013 0.12041 0.16136 0.8 -0.04070 -0.03512 -0.04044 0.06410 0.06411 0.07619 1 -0.03030 -0.02275 -0.03070 0.03984 0.03980 0.03883


Shell Solid




0.2 6.05059 7.29380 6.02718 7.06169 7.11702 7.50375 0.3 2.05010 2.95178 2.05395 2.87114 2.89150 3.76482 0.4 0.75795 0.97869 0.75870 1.49445 1.50270 1.77609 0.5 0.20891 0.39527 0.20889 0.90293 0.90567 1.46746 0.6 -0.05258 0.11370 -0.04073 0.60066 0.60206 0.97488 0.8 -0.16710 -0.07120 -0.16577 0.32050 0.32056 0.44430 1 -0.12414 -0.07534 -0.12603 0.19921 0.19898 0.19152

Appendix A

212

Table A-7: Stress ( σx/q) at x = a/2, y = b/2 at Ply3 bottom surface

Shell Solid




0.2 9.93898 8.51696 9.94639 8.82711 8.89627 9.86828 0.3 4.68826 3.71646 4.70426 3.58893 3.61438 4.67811 0.4 2.87148 2.59650 2.87780 1.86807 1.87838 2.88275 0.5 1.98177 1.75051 1.98580 1.12867 1.13209 2.35680 0.6 1.41810 1.23765 1.43358 0.75082 0.75258 1.45682 0.8 0.74686 0.62972 0.75264 0.40063 0.40070 0.74341 1 0.37058 0.30528 0.37364 0.24901 0.24872 0.35918

Table A-8: Stress ( σy/q) at x = a/2, y = b/2 at Ply1 top surface

Shell Solid




0.2 -8.39846 -7.42995 -8.36945 -8.43043 -8.44158 -7.61809 0.3 -4.88857 -4.24907 -4.87596 -4.90654 -4.91795 -2.92017 0.4 -3.28739 -3.08263 -3.27664 -3.21709 -3.22801 -2.06006 0.5 -2.46848 -2.29218 -2.45607 -2.25525 -2.26214 -1.15395 0.6 -2.04166 -1.87835 -2.03460 -1.65395 -1.66190 -0.97315 0.8 -1.74207 -1.55455 -1.73062 -0.98911 -0.99458 -0.87882 1 -1.69364 -1.47021 -1.68334 -0.65272 -0.65658 -0.74259

Table A-9: Stress ( σy/q) at x = a/2, y = b/2 at Ply1 bottom surface

Shell Solid




0.2 -6.03856 -6.87907 -6.04680 -6.74434 -6.75327 -6.03275 0.3 -3.23138 -3.83999 -3.24440 -3.92523 -3.93436 -1.93642 0.4 -1.95543 -2.10412 -1.96488 -2.57367 -2.58241 -1.21315 0.5 -1.27051 -1.41120 -1.27773 -1.80420 -1.80971 -0.28650 0.6 -0.85195 -1.00163 -0.86480 -1.32316 -1.32952 -0.20818 0.8 -0.36474 -0.53530 -0.37395 -0.79129 -0.79566 -0.19897 1 -0.04711 -0.24367 -0.05303 -0.52217 -0.52526 -0.10487

Appendix A

213


Shell Solid




0.2 -1.37431 -1.07036 -1.37621 -1.34887 -1.35066 -1.05234 0.3 -0.81207 -0.70828 -0.81470 -0.78505 -0.78687 -0.27633 0.4 -0.55728 -0.51682 -0.55933 -0.51474 -0.51648 -0.20335 0.5 -0.42122 -0.38748 -0.42298 -0.36084 -0.36194 -0.17546 0.6 -0.33821 -0.30733 -0.34091 -0.26463 -0.26590 -0.13176 0.8 -0.23984 -0.21748 -0.24216 -0.15826 -0.15913 -0.10985 1 -0.17140 -0.15009 -0.17323 -0.10444 -0.10505 -0.09614


Shell Solid




0.2 1.24154 0.93874 1.24407 1.34887 1.35066 1.39135 0.3 0.67748 0.57370 0.67873 0.78505 0.78687 0.89135 0.4 0.40652 0.36932 0.40806 0.51474 0.51648 0.47131 0.5 0.24954 0.22763 0.25095 0.36084 0.36194 0.23290 0.6 0.15422 0.14212 0.15411 0.26463 0.26590 0.11719 0.8 0.05594 0.05622 0.05607 0.15826 0.15913 0.05490 1 0.01900 0.02190 0.01925 0.10444 0.10505 0.03846


Shell Solid




0.2 6.24685 7.08642 6.25398 6.74434 6.75327 5.64233 0.3 3.42967 4.03891 3.43441 3.92523 3.93436 3.64233 0.4 2.07965 2.22285 2.08622 2.57367 2.58241 2.35653 0.5 1.30001 1.42405 1.30602 1.80420 1.80971 1.16448 0.6 0.82600 0.92383 0.82505 1.32316 1.32952 0.85562 0.8 0.33002 0.39179 0.33072 0.79129 0.79566 0.47691 1 0.13285 0.16716 0.13422 0.52217 0.52526 0.31045

Appendix A

214


Shell Solid




0.2 8.56515 7.63727 8.57717 8.43043 8.44158 7.20677 0.3 5.08548 4.46198 5.09223 4.90654 4.91795 5.20677 0.4 3.43155 3.24181 3.44086 3.21709 3.22801 3.38912 0.5 2.43217 2.27495 2.44150 2.25525 2.26214 1.92122 0.6 1.75310 1.61652 1.75459 1.65395 1.66190 1.44386 0.8 0.88926 0.80637 0.89382 0.98911 0.99458 0.92394 1 0.43075 0.38684 0.43557 0.65272 0.65658 0.14320

Table A-14: Stress ( σx/q) at x = 0, y = b/2 at Ply1 top surface

Shell Solid




0.2 53.55596 25.92480 51.06000 10.84380 12.84330 11.67580 0.3 30.38884 16.16450 28.61140 3.21713 3.85948 3.37407 0.4 21.73011 14.26170 20.16670 1.18875 1.45989 1.31707 0.5 17.59548 11.92760 16.13850 0.56742 0.63724 0.94346 0.6 15.26024 10.46910 14.44660 0.22809 0.30617 0.72718 0.8 12.67571 5.78926 11.60800 0.05030 0.08334 0.57393 1 11.11321 5.31766 10.48180 0.00617 0.02370 0.41177

Table A-15: Stress ( σx/q) at x = 0, y = b/2 at Ply1 bottom surface

Shell Solid




0.2 -16.62360 -8.72296 -15.55646 -8.67506 -10.27460 -5.46955 0.3 -13.80722 -5.30263 -13.34220 -2.57370 -3.08758 -1.40093 0.4 -11.33251 -4.24720 -10.71385 -0.95100 -1.16791 -1.18536 0.5 -9.71725 -4.22684 -9.72375 -0.45393 -0.50979 -1.17945 0.6 -8.65996 -4.18473 -8.52141 -0.18247 -0.24493 -1.15555 0.8 -7.37854 -4.06174 -7.30472 -0.04024 -0.06667 -0.91967 1 -6.54162 -3.87882 -6.68188 -0.00494 -0.01896 -0.15760

Appendix A

215


Shell Solid




0.2 -3.32472 -1.96878 -3.11129 -1.73501 -2.05492 -2.24324 0.3 -2.76144 -1.02041 -2.46845 -0.51474 -0.61752 -0.53707 0.4 -2.26650 -0.92410 -2.19277 -0.19020 -0.23358 -0.23589 0.5 -1.94345 -0.91762 -1.94475 -0.09079 -0.10196 -0.14060 0.6 -1.73199 -0.89218 -1.70428 -0.03649 -0.04899 -0.13660 0.8 -1.47571 -0.87093 -1.46094 -0.00805 -0.01333 -0.11322 1 -1.30832 -0.82207 -1.33638 -0.00099 -0.00379 -0.09172


Shell Solid




0.2 2.29242 1.21956 2.13553 1.73501 2.05492 0.80398 0.3 1.72281 0.47893 1.59960 0.51474 0.61752 0.14515 0.4 1.20642 0.40724 1.18510 0.19020 0.23358 0.13769 0.5 0.84555 0.34860 0.78371 0.09079 0.10196 0.11031 0.6 0.59353 0.27241 0.56496 0.03649 0.04899 0.09602 0.8 0.28733 0.15399 0.28961 0.00805 0.01333 0.00283 1 0.13305 0.07573 0.13778 0.00099 0.00379 0.00103


Shell Solid




0.2 11.46209 3.96288 10.67764 8.67506 10.27460 5.57556 0.3 8.61407 2.44794 7.62841 2.57370 3.08758 0.72574 0.4 6.03208 2.05442 5.92550 0.95100 1.16791 0.18843 0.5 4.22776 1.59429 3.91855 0.45393 0.50979 0.08405 0.6 2.96764 0.90020 2.82479 0.18247 0.24493 0.04370 0.8 1.43665 0.87276 1.44806 0.04024 0.06667 0.03115 1 0.66523 0.43092 0.68892 0.00494 0.01896 0.02500

Appendix A

216


Shell Solid




0.2 -43.27342 -24.67300 -41.38450 -10.84380 -12.84330 -10.33740 0.3 -19.89811 -14.10180 -18.93650 -3.21713 -3.85948 -2.83846 0.4 -11.05129 -5.32642 -10.87167 -1.18875 -1.45989 -1.54643 0.5 -6.74559 -3.39907 -6.20385 -0.56742 -0.63724 -1.38473 0.6 -4.31346 -2.25281 -4.05856 -0.22809 -0.30617 -0.90930 0.8 -1.84821 -1.01757 -1.80302 -0.05030 -0.08334 -0.38314 1 -0.78724 -0.44697 -0.79056 -0.00617 -0.02370 -0.05320

Table A-20: Stress ( σy/q) at x = 0, y = b/2 at Ply1 top surface

Shell Solid




0.2 12.38443 4.43123 10.58658 2.50754 2.95316 9.91810 0.3 7.02720 2.62853 5.59638 0.74393 0.88647 4.98424 0.4 5.02493 2.27565 4.47389 0.27489 0.33507 3.91270 0.5 4.06883 1.78167 3.72813 0.13121 0.14615 3.21063 0.6 3.52882 1.47681 3.31147 0.05274 0.07016 2.39296 0.8 2.93117 0.56568 2.85869 0.01163 0.01905 1.35899 1 2.56985 0.46908 2.58132 0.00143 0.00538 1.00101

Table A-21: Stress ( σy/q) at x = 0, y = b/2 at Ply1 bottom surface

Shell Solid




0.2 -3.84409 -2.86587 -3.50718 -2.00603 -2.36253 -3.42708 0.3 -3.19282 -1.26099 -3.04696 -0.59515 -0.70918 -2.93172 0.4 -2.62056 -0.70182 -2.39221 -0.21991 -0.26805 -2.25293 0.5 -2.24704 -0.63168 -2.14838 -0.10497 -0.11692 -1.91373 0.6 -2.00255 -0.62473 -1.97542 -0.04219 -0.05613 -1.63903 0.8 -1.70623 -0.60119 -1.79892 -0.00931 -0.01524 -1.05918 1 -1.51270 -0.56507 -1.51612 -0.00114 -0.00431 -0.55429

Appendix A

217


Shell Solid




0.2 -0.76882 -0.27429 -0.72144 -0.40121 -0.47251 -0.68447 0.3 -0.63857 -0.18152 -0.60940 -0.11903 -0.14184 -0.55059 0.4 -0.52412 -0.17116 -0.47844 -0.04398 -0.05361 -0.45600 0.5 -0.44941 -0.17005 -0.42968 -0.02099 -0.02338 -0.36924 0.6 -0.40051 -0.16253 -0.39509 -0.00844 -0.01123 -0.30263 0.8 -0.34125 -0.16009 -0.35979 -0.00186 -0.00305 -0.15917 1 -0.30254 -0.14452 -0.30323 -0.00023 -0.00086 -0.11266


Shell Solid




0.2 0.53011 0.24512 0.51965 0.40121 0.47251 0.49402 0.3 0.39839 0.19052 0.34468 0.11903 0.14184 0.33559 0.4 0.27898 0.10037 0.26389 0.04398 0.05361 0.23407 0.5 0.19553 0.05112 0.19300 0.02099 0.02338 0.17203 0.6 0.13725 0.01223 0.13913 0.00844 0.01123 0.10993 0.8 0.06644 0.01108 0.07132 0.00186 0.00305 0.04471 1 0.03077 0.00549 0.03393 0.00023 0.00086 0.04070


Shell Solid




0.2 2.65052 1.88650 2.69822 2.00603 2.36253 1.48415 0.3 1.99194 1.15356 1.93237 0.59515 0.70918 0.86714 0.4 1.39487 0.47822 1.31945 0.21991 0.26805 0.46858 0.5 0.97764 0.26768 0.96501 0.10497 0.11692 0.28540 0.6 0.68625 0.21913 0.69566 0.04219 0.05613 0.28068 0.8 0.33221 0.12534 0.33624 0.00931 0.01524 0.17606 1 0.15383 0.06299 0.16049 0.00114 0.00431 0.08370

Appendix A

218


Shell Solid




0.2 -10.00666 -3.23849 -9.21567 -2.50754 -2.95316 -9.43712 0.3 -4.60129 -1.30403 -3.95940 -0.74393 -0.88647 -4.02563 0.4 -2.55553 -0.80716 -2.43107 -0.27489 -0.33507 -2.09981 0.5 -1.55987 -0.48183 -1.52781 -0.13121 -0.14615 -1.43784 0.6 -0.99746 -0.30183 -0.99949 -0.05274 -0.07016 -0.96739 0.8 -0.42739 -0.12360 -0.41562 -0.01163 -0.01905 -0.94773 1 -0.18204 -0.04931 -0.18063 -0.00143 -0.00538 -0.14923

Table A-26: Stress ( τxz/q) at x = 0, y = b/2 at Ply1 top surface

Shell Solid




0.2 0 9.12096 7.15914 0 0 0 0.3 0 5.45091 4.41834 0 0 0 0.4 0 3.95007 3.51389 0 0 0 0.5 0 3.26464 3.02200 0 0 0 0.6 0 2.88679 2.75652 0 0 0 0.8 0 2.49997 2.49024 0 0 0 1 0 2.29488 2.34987 0 0 0

Table A-27: Stress ( τxz/q) at x = 0, y = b/2 at Ply1 bottom surface

Shell Solid




0.2 2.44791 8.95765 6.79592 1.89790 2.02112 0 0.3 1.61330 5.29573 4.12603 1.10793 1.18713 0 0.4 1.24783 3.77662 3.17580 0.75340 0.81141 0 0.5 1.05043 3.05966 2.63791 0.58592 0.60881 0 0.6 0.93774 2.65064 2.32812 0.44773 0.48542 0 0.8 0.81018 2.20789 1.98315 0.31728 0.34520 0 1 0.73559 1.95501 1.78877 0.24616 0.26835 0

Appendix A

219


Shell Solid




0.2 2.44791 1.79153 1.35918 1.89790 2.02112 0 0.3 1.61330 1.05915 0.82521 1.10793 1.18713 0 0.4 1.24783 0.75532 0.63516 0.75340 0.81141 0 0.5 1.05043 0.61193 0.52758 0.58592 0.60881 0 0.6 0.93774 0.53013 0.46562 0.44773 0.48542 0 0.8 0.81018 0.44158 0.39663 0.31728 0.34520 0 1 0.73559 0.39100 0.35775 0.24616 0.26835 0


Shell Solid




0.2 1.85606 1.49682 1.01324 1.89790 2.02112 0 0.3 1.01449 0.73564 0.49866 1.10793 1.18713 0 0.4 0.63348 0.41783 0.30542 0.75340 0.81141 0 0.5 0.42755 0.26235 0.19934 0.58592 0.60881 0 0.6 0.29410 0.17191 0.13566 0.44773 0.48542 0 0.8 0.14387 0.07752 0.06488 0.31728 0.34520 0 1 0.06902 0.03504 0.03078 0.24616 0.26835 0


Shell Solid




0.2 1.85606 7.48408 5.06619 1.89790 2.02112 0 0.3 1.01449 3.67822 2.49331 1.10793 1.18713 0 0.4 0.63348 2.08914 1.52711 0.75340 0.81141 0 0.5 0.42755 1.31175 0.99669 0.58592 0.60881 0 0.6 0.29410 0.85955 0.67830 0.44773 0.48542 0 0.8 0.14387 0.38762 0.32441 0.31728 0.34520 0 1 0.06902 0.17521 0.15388 0.24616 0.26835 0

Appendix A

220


Shell Solid




0.2 0 7.51593 5.14023 0 0 0 0.3 0 3.65195 2.41301 0 0 0 0.4 0 2.04101 1.41982 0 0 0 0.5 0 1.26171 0.88786 0 0 0 0.6 0 0.81381 0.58169 0 0 0 0.8 0 0.35591 0.26254 0 0 0 1 0 0.15608 0.11844 0 0 0

Table A-32: Stress ( τxy/q) at x = 0, y = b/2 across the thickness

Shell Solid




0.2 0 0 0 0 0 0 0.3 0 0 0 0 0 0 0.4 0 0 0 0 0 0 0.5 0 0 0 0 0 0 0.6 0 0 0 0 0 0 0.8 0 0 0 0 0 0 1 0 0 0 0 0 0

Appendix A

221

0

5

10

15

20

25

0.2 0.4 0.6 0.8 1

h/a

W C

11/q

h

Exactsolid C3D8solid C3D20shell S4shell S8Rshell SC8R

Figure A-1: Deflection (WC 11 / qh) at x = a/2 , y = b/2 at z = 0

-12

-10

-8

-6

-4

-2

00.2 0.4 0.6 0.8 1h/a

sx/

q Exactsolid C3D8solid C3D20shell S4shell S8Rshell SC8R

Figure A-2: Stress ( σx/q) at x = a/2 and y = b/2 Ply1 top

Appendix A

222

-8

-6

-4

-2

0

2

0.2 0.4 0.6 0.8 1h/a

sx/

q


Figure A-3: Stress ( σx/q) at x = a/2 and y = b/2 Ply1 bottom

-1.6

-1.4

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.2 0.4 0.6 0.8 1h/a

sx/

q



Appendix A

223

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

0.2 0.4 0.6 0.8 1

h/a

sx/

q



-1

0

1

2

3

4

5

6

7

8

0.2 0.4 0.6 0.8 1

h/a

sx/

q



Appendix A

224

0

2

4

6

8

10

12

0.2 0.4 0.6 0.8 1

h/a

sx/

q



-9

-8

-7

-6

-5

-4

-3

-2

-1

00.2 0.4 0.6 0.8 1

h/a

sy/


Figure A-8: Stress ( σy/q) at x = a/2 and y = b/2 Ply1 top

Appendix A

225

-8

-7

-6

-5

-4

-3

-2

-1

00.2 0.4 0.6 0.8 1

h/a

sy/


Figure A-9: Stress ( σy/q) at x = a/2 and y = b/2 Ply1 bottom

-1.6

-1.4

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.00.2 0.4 0.6 0.8 1

h/a

sy/



Appendix A

226

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

0.2 0.4 0.6 0.8 1

h/a

sy/

q



0

1

2

3

4

5

6

7

8

0.2 0.4 0.6 0.8 1

h/a

sy/

q



Appendix A

227

0

2

4

6

8

10

0.2 0.4 0.6 0.8 1.0

h/a

sy/

q



0

10

20

30

40

50

60

0.2 0.4 0.6 0.8 1

h/a

sx/

q


Figure A-14: Stress ( σx/q) at x = 0 and y = b/2 Ply1 top

Appendix A

228

-18

-16

-14

-12

-10

-8

-6

-4

-2

00.2 0.4 0.6 0.8 1

h/a

sx/

q


Figure A-15: Stress ( σx/q) at x = 0 and y = b/2 Ply1 bottom

-3.5

-3.0

-2.5

-2.0

-1.5

-1.0

-0.5

0.00.2 0.4 0.6 0.8 1

h/a

sx/

q


`


Appendix A

229

0.0

0.5

1.0

1.5

2.0

2.5

0.2 0.4 0.6 0.8 1h/a

sx/

q



0

2

4

6

8

10

12

14

0.2 0.4 0.6 0.8 1h/a

sx/

q



Appendix A

230

-50

-40

-30

-20

-10

00.2 0.4 0.6 0.8 1

h/a

sx/



0

2

4

6

8

10

12

14

0.2 0.4 0.6 0.8 1h/a

sy/

q


Figure A-20: Stress ( σy/q) at x = 0 and y = b/2 Ply1 top

Appendix A

231

-5.0

-4.0

-3.0

-2.0

-1.0

0.00.2 0.4 0.6 0.8 1

h/a

sy/

q


Figure A-21: Stress ( σy/q) at x = 0 and y = b/2 Ply1 bottom

-0.8

-0.6

-0.4

-0.2

0.00.2 0.4 0.6 0.8 1

h/a

sy/

q



Appendix A

232

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.2 0.4 0.6 0.8 1

h/a

sy/

q



0.0

0.5

1.0

1.5

2.0

2.5

3.0

0.2 0.4 0.6 0.8 1

h/a

sy/

q



Appendix A

233

-12

-10

-8

-6

-4

-2

00.2 0.4 0.6 0.8 1

h/a

sy/



0

2

4

6

8

10

0.2 0.4 0.6 0.8 1

h/a

txz

/q


Figure A-26: Stress ( τxz/q) at x = 0 and y = b/2 Ply1 bottom

Appendix A

234

0.0

0.5

1.0

1.5

2.0

2.5

0.2 0.4 0.6 0.8 1.0

h/a

txz

/q


Figure A-27: Stress ( τxz/q) at x = 0 and y = b/2 Ply2 top

0.0

0.5

1.0

1.5

2.0

2.5

0.2 0.4 0.6 0.8 1

h/a

txz

/q


Figure A-28: Stress ( τxz/q) at x = 0 and y = b/2 Ply2 bottom

Appendix A

235

0

1

2

3

4

5

6

7

8

0.2 0.4 0.6 0.8 1

h/a

txz

/q


Figure A-29: Stress ( τxz/q) at x = 0 and y = b/2 Ply3 top

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THREE DIMENSIONAL ANALYSIS OF FIBRE REINFORCED …