27919257 m rades finite element analysis

Preface

This textbook represents the Finite Element Analysis lecture course given to students in the third year at the Department of Engineering Sciences (now F.I.L.S.), English Stream, University Politehnica of Bucharest, since 1992.

It grew in time along with a course taught in Romanian to students in the Faculty of Transports, helped by the emergence of microcomputer networks and integration of the object into mechanical engineering curricula. The syllabus of the 28-hour course, supplemented by 28-hour tutorial and lab. classes, was structured along the NAFEMS recommendations published in the October 1988 issue of BENCHmark. The course represents only an introduction to the finite element analysis, for which we wrote simple stand-alone single-element programs to assist students in solving problems as homework. It is followed by an advanced course in the fourth year at F.I.L.S., called Computational Structural Mechanics, where students are supposed to use commercial programs.

In designing the course, our aim was to produce students capable of: (a) understanding the theoretical background, (b) appreciating the structure of finite element programs for potential amendment and development, (c) running packages and assessing their limitations, (d) taking a detached view in checking output, and (e) understanding failure messages and finding ways of rectifying the errors.

The course syllabus was restricted to 2D linear elastic structural problems. It has been found advantageous to divide the finite element analysis into two parts. Firstly, the assembly process without any approximations (illustrated by frameworks) followed by the true finite element process which involves approximations. This is achieved starting with trusses, then with beams and plane frames, and progressively dealing with membrane and plate-bending elements. Solid elements and shells are not treated. Our objective was to ensure that students have achieved: (a) a familiarity in working with matrix methods and developing stiffness matrices, (b) an understanding of global versus local coordinate systems, (c) the abilty to use the minimum potential energy theorem and virtual work equations, (d) the mapping from isoparametric space to real geometrics and the need for numerical integration, (e) an insight in numerical techniques for linear equation solving (Gauss elimination, frontal solvers etc), and (f) the use of equilibrium, compatibility, stress/strain relations and boundary conditions.

As a course taught for non-native speakers, it has been considered useful to reproduce as language patterns some sentences from English texts.

November 2006 Mircea Radeş

Prefaţă

Lucrarea reprezintă cursul Analiza cu elemente finite predat studenţilor anului III al Facultăţii de Inginerie în Limbi Străine, Filiera Engleză, la Universitatea Politehnica Bucureşti, începând cu anul 1992.

Conţinutul cursului s-a lărgit în timp, fiind predat din 1992 şi studenţilor de la facultatea de Transporturi, favorizat de apariţia reţelelor de calculatoare şi de includerea sa în planul de învăţământ al facultăţilor cu profil mecanic. Programa cursului, care prevede 28 ore de curs şi 28 ore de seminar/laborator, a fost structurată în conformiatate cu recomandările NAFEMS publicate în numărul din Octombrie 1988 al revistei BENCHmark. Cursul reprezintă doar o introducere în analiza cu elemente finite, pentru care am scris programe simple, cu un singur tip de element finit, care să fie utilizate de studenţi la rezolvarea unor teme de casă. Nu se tratează învelişuri şi elemente tridimensionale. În anul IV, planul de învăţământ de la F.I.L.S. conţine cursul Computational Structural Mechanics, la care studenţii aprofundează modelarea cu elemente finite şi utilizează un program de firmă.

La structurarea cursului am avut în vedere necesitatea formării unor studenţi capabili: (a) să înţeleagă baza teoretică, (b) să desluşească structura programelor cu elemente finite pentru eventuale corecţii şi dezvoltări, (c) să ruleze programe şi să recunoască limitele acestora, (d) să poată verifica rezultatele şi (e) să înţeleagă mesajele de eroare şi să găsească modalităţi de corectare a erorilor.

Programa cursului a fost limitată la structuri elastice liniare bidimensionale. S-a considerat potrivit să se prezinte analiza cu elemente finite în două etape: întâi procesul de asamblare fără nici o aproximare (aplicat la grinzi cu zăbrele), apoi modelarea cu elemente finite, care presupune aproximarea câmpului de deplasări, de la triunghiul cu deformaţii specifice constante la elemente patrulatere izoparametrice, incluzând integrarea numerică. S-a urmărit ca studenţii să dobândească: (a) familiaritate cu metodele matriciale şi calculul matricelor de rigiditate; (b) înţelegerea utilităţii coordonatelor locale şi globale; (c) abilitatea folosirii principiului energiei potenţiale minime şi a principiului lucrului mecanic virtual; (d) trecerea de la coordonate naturale la coordonate fizice şi necesitatea integrării numerice; (e) o vedere de ansamblu asupra rezolvării sistemelor algebrice liniare (eliminarea Gauss, metoda frontală etc.) şi (f) utilizarea celor patru tipuri de ecuaţii – echilibru, compatibilitate, constitutive şi condiţii la limită.

Fiind un curs predat unor studenţi a căror limbă maternă nu este limba engleză, au fost reproduse expresii şi fraze din cărţi scrise de vorbitori nativi ai acestei limbi.

Noiembrie 2006 Mircea Radeş

Contents

Preface i

Contents iii

1. Introduction 1 1.1 Object of FEA 1

1.2 Finite element displacement method 3

1.3 Historical view 4

1.4 Stages of FEA 5

2. Displacement Method 9 2.1 Equilibrium equations 9

2.2 Conditions for geometric compatibility 10

2.3 Force/elongation relations 11

2.4 Boundary conditions 12

2.5 Solving for displacements 12

2.6 Comparison of the force method and displacement method 13

3. Direct Stiffness Method 17 3.1 Stiffness matrix for a bar element 17

3.2 Transformation from local to global coordinates 19 3.2.1 Coordinate transformation 19

3.2.2 Force transformation 20

3.2.3 Element stiffness matrix in global coordinates 21

3.2.4 Properties of the element stiffness matrix 22

3.3 Link’s truss 25

3.4 Direct method 26

3.5 Compatibility of nodal displacements 28

3.6 Expanded element stiffness matrix 29

3.7 Unreduced global stiffness matrix 30

FINITE ELEMENT ANALYSIS iv

3.8 Joint force equilibrium equations 31

3.9 Reduced global stiffness matrix 33

3.10 Reactions and internal forces 35

3.11 Thermal loads and stresses 36

3.12 Node numbering 37

Exercises 41

4. Bars and shafts 47 4.1 Plane bar elements 47

4.1.1 Differential equation of equilibrium 47

4.1.2 Coordinates and shape functions 48

4.1.3 Bar not loaded between ends 49

4.1.4 Element stiffness matrix in local coordinates 51

4.1.5 Bar loaded between ends 52

4.1.6 Vector of element nodal forces 55

4.1.7 Assembly of the global stiffness matrix and load vector 56

4.1.8 Initial strain effects 59

4.2 Plane shaft elements 60

Exercises 63

5. Beams, frames and grids 79 5.1 Finite element discretization 79

5.2 Static analysis of a uniform beam 81

5.3 Uniform beam not loaded between ends 83 5.3.1 Shape functions 84

5.3.2 Stiffness matrix 86

5.3.3 Physical significance of the stiffness matrix 88

5.4 Uniform beam loaded between ends 89 5.4.1 Consistent vector of nodal forces 89

5.4.2 Higher degree interpolation functions 92

5.4.3 Bending moment and shear force 95

5.5 Basic convergence requirements 96

5.6 Frame element 97 5.6.1 Axial effects 97

5.6.2 Stiffness matrix and load vector in local coordinates 98

CONTENTS v

5.6.3 Coordinate transformation 98

5.6.4 Stiffness matrix and load vector in global coordinates 100

5.7 Assembly of the global stiffness matrix 100

5.8 Grids 111

5.9 Deep beam bending element 116 5.9.1 Static analysis of a uniform beam 117

5.9.2 Shape functions 118

5.9.3 Stiffness matrix 121

6. Linear elasticity 123 6.1 Matrix notation for loads, stresses and strain 123

6.2 Equations of equilibrium inside V 125

6.3 Equations of equilibrium on the surface σS 126

6.4 Strain-displacement relations 127

6.5 Stress-strain relations 128

6.6 Temperature effects 130

6.7 Strain energy 130

7. Energy methods 131 7.1 Principle of virtual work 131 7.1.1 Virtual displacements 131

7.1.2 Virtual work of external forces 133

7.1.3 Virtual work of internal forces 133

7.1.4 Principle of virtual displacements 134

7.1.5 Proof that PDV is equivalent to equilibrium equations 137

7.2 Principle of minimum total potential energy 139

7.2.1 Strain energy 139

7.2.2 External potential energy 140

7.2.3 Total potential energy 140

7.3 The Rayleigh-Ritz method 143

7.4 FEM – a localized version of the Rayleigh-Ritz method 148 7.4.1 FEM in Structural Mechanics 148

7.4.2 Discretization 149

7.4.3 Principle of virtual displacements 149

FINITE ELEMENT ANALYSIS vi

7.4.4 Approximating functions for the element 149

7.4.5 Compatibility between strains and nodal displacements 150

7.4.6 Element stiffness matrix and load vector 151

7.4.7 Assembly of the global stiffness matrix and load vector 151

7.4.8 Solution and back-substitution 152

8 Two-dimensional elements 153 8.1 The plane constant-strain triangle (CST) 153

8.1.1. Discretization of structure 153

8.1.2 Polynomial approximation of the displacement field 154

8.1.3 Nodal approximation of the displacement field 155

8.1.4 The matrix [ ] B 158

8.1.5 Element stiffness matrix and load vector 159

8.1.6 Remarks 160

8.2 Rectangular elements 176

8.2.1 The four-node rectangle (linear) 176

8.2.2 The eight-node rectangle (quadratic) 178

8.3 Triangular elements 180

8.3.1 Area coordinates 180

8.3.2 Linear strain triangle (LST) 182

8.3.3 Quadratic strain triangle 185

8.4 Equilibrium, convergence and compatibility 187

8.4.1 Equilibrium vs. compatibility 187

8.4.2 Convergence and compatibility 188

9 Isoparametric elements 191 9.1 Linear quadrilateral element 191 9.1.1 Natural coordinates 192

9.1.2 Shape functions 193

9.1.3 The displacement field 194

9.1.4 Mapping from natural to Cartesian coordinates 195

9.1.5 Element stiffness matrix 198

9.1.6 Element load vectors 199

9.2 Numerical integration 200

9.2.1 One dimensional Gauss quadrature 200

CONTENTS vii

9.2.2 Two dimensional Gauss quadrature 203

9.2.3 Stiffness integration 204

9.2.4 Stress calculations 207

9.3 Eight-node quadrilateral 208 9.3.1 Shape functions 209

9.3.2 Shape function derivatives 210

9.3.3 Determinant of the Jacobian matrix 211

9.3.4 Element stiffness matrix 211

9.3.5 Stress calculation 213

9.3.6 Consistent nodal forces 214

9.4 Nine-node quadrilateral 219

9.5 Six-node triangle 221

9.6 Jacobian positiveness 223

10 Plate bending 225 10.1 Thin plate theory (Kirchhoff) 225

10.2 Thick plate theory (Reissner-Mindlin) 229

10.3 Rectangular plate-bending elements 232 10.3.1 ACM element (non-conforming) 232

10.3.2 BFS element (conforming) 238

10.3.3 HTK thick rectangular element 239

10.4 Triangular plate-bending elements 244

10.4.1 Thin triangular element (non-conforming) 245

10.4.2 Thick triangular element (conforming) 248

10.4.3 Discrete Kirchhoff triangles (DKT) 250

References 257

Index 265

1. INTRODUCTION

Finite Element Analysis (FEA) as applied to structures is a multidisciplinary technique, based on knowledge from three fields: (1) Structural Mechanics, encompassing elasticity, strength of materials, dynamics, plasticity, etc, (2) Numerical Analysis, involving approximation methods, solving linear sets of equations, eigenproblems, etc, and (3) Applied Computer Science, dealing with the development and maintenance of large computer codes.

FEA is used to solve large-scale analytical problems. Its task is to model and describe the mechanical behaviour of geometrically complex structures. The procedure is a discretized approach: the geometric shape or the internal stress-strain-displacement field are described by a series of discrete quantities (like coordinates) distributed through the structure. This requires a matrix notation. The tools are the computers, able to store long lists of numbers and manipulate them.

1.1 Object of FEA

The object of FEA is to replace the infinite degree of freedom system in continuum applications by a finite system exhibiting the same basis as discrete analysis.

The aim is finding an approximate solution to a boundary- and initial-value problem by dividing the domain of the system into a set of interconnected finite-sized subdomains of different size and shape, and defining the unknown state variable approximately, within each element, by means of a linear combination of trial functions. The subdomains are called finite elements, the set of finite elements is known as the mesh and the trial functions are referred to as interpolation functions. With the individually defined functions matching each other at certain points called nodes, the unknown function is approximated over the entire domain.

The primary difference between the FEA and other approximate methods for the solution of boundary-value problems (finite-difference, weighted-residual,

FINITE ELEMENT ANALYSIS 2

Rayleigh-Ritz, Galerkin) is that in the FEA the approximation is confined to relatively small subdomains.

FEA is a localized version of the Rayleigh-Ritz method. Instead of finding an admissible function satisfying the boundary conditions for the entire domain, which is often difficult, in the FEA the admissible functions (called shape functions) are defined over element domains with simple geometry and pay no attention to complications at the boundaries.

Since the entire domain is divided into numerous elements and the function is approximated in terms of its values at the element nodes, the evaluation of such a function will require the solution of simultaneous equations. This was possible only at the time the computers became available. The outstanding success of the finite element method can be attributed to a large extent to timing. While the finite element method was being developed, so were increasingly powerful digital computers, which led to automation. The computer is not only able to solve the discretized equations of equilibrium, but also to carry out such diverse tasks as the formulation of equations, by making decisions concerning the finite element mesh and the assembly of stiffness matrices.

Perhaps more important is the fact that the finite element method can accommodate systems with complicated geometries and parameter distributions. The wide use of the classical Rayleigh-Ritz method has been limited by the inability to generate suitable admissible functions for a large number of practical problems. Indeed, systems with complex boundary conditions or complex geometry cannot be described easily by global admissible functions, which tend to have complicated expressions, difficult to handle on a routine basis. In turn, in the FEA an approximate solution is constructed using local admissible functions, defined over small subdomains of the structure. In order to match a given irregular boundary, or to handle parameter non-uniformities, the FEA can change not only the size of the finite elements but also their shape. This extreme versatility, coupled with the development of powerful computer codes based on the method, some of them made available as open source free software, has made the FEA the method of choice for the analysis of structures.

In FEA, the equations of equilibrium are obtained from variational principles implying the stationarity of the functional defined by the total potential energy. While solving differential equations with complicated boundary conditions may be difficult, integrating known polynomial functions, even approximately, should be easier. Mathematically, solving [ ]{ } { }bxA = is equivalent to

minimizing ( ) { } [ ]{ } { } { }bxxAxxP TT −=21 . This is the heart of the FEA when

applied to structures.

1. INTRODUCTION 3

1.2 Finite element displacement method

In the finite element modeling, a structure is discretized (hypothetically) into finite elements and points named nodes are selected on the inter-element boundaries or in the interior of the elements. Displacements at the nodes are taken as the primary discrete variables. Displacements within the elements are expressed in terms of these nodal displacements using interpolation functions referred to as shape functions. Finite elements are so small that the shape of the displacement field can be approximated without too much error, leaving only the magnitude to be found. The “shapes” are polynomials, but may be trigonometric functions as well.

All individual elements are assembled together in such a way that the displacements are continuous in some fashion across element interfaces, the internal stresses are in equilibrium with the applied loads, and the prescribed boundary conditions are satisfied. Finally, the governing discrete equations are generated by a variational approach.

The first part of the finite element modeling process involves choosing the correct and appropriate types of elements, understanding the pedigree of elements and spotting wrong answers due to the use of inadequate elements. The second part of the process is the assembly of the elements and the solution of the complete structural equations. This involves recognizing error messages when this process breaks down or when it simply becomes inefficient because the structure has been modeled inconveniently.

The six basic steps of FEA are the following: (1) discretize the continuum, (2) select interpolation functions, (3) find the element properties, (4) assemble the element properties, (5) solve the system of equations, and (6) make additional computations if desired.

The three main sources of approximation are: (1) the definition of the domain (physically or geometrically), (2) the discretization of the domain (cutting the corners, making curved lines straight, and curved elements flat), and (3) the solution algorithms. Modeling the joints and the contact between structural parts as well as the damping in dynamic problems are the most difficult tasks. Mesh refinements (and automatic mesh generation) do not bring necessarily increased accuracy. Finer mesh yields also larger stiffness matrices, a larger number of equations to be solved, hence larger computer storage space and running time.

Among the reasons why the FEA has gained such universal acceptance are: (1) the routine choice of shape functions, (2) the easiness of producing stiffness matrices (and load vectors), by just assembling predetermined element matrices, and (3) the versatility. Developed originally as a method for analyzing stresses in complex aircraft structures, FEA has evolved into a technique that can be applied


to a large variety of linear and nonlinear, static, stability and dynamic engineering problems.

1.3 Historical view

The idea of representing a given domain as a collection of discrete elements is not novel with the finite element method. Ancient mathematicians estimated the value of π approximating the circumference of a circle by the perimeter of a polygon inscribed in the circle. In modern times, the idea found application in aircraft structural analysis, where wings and fuselages are treated as assemblages of stringers, panels, ribs, stiffeners and spars.

The use of piecewise continuous functions defined over a subdomain to approximate the unknown function dates back to Courant (1943), who used an assemblage of small triangular elements and the principle of minimum potential energy to study Saint Venant’s torsion problem. The reason why Courant’s paper did not attract more attention can be attributed to poor timing. In the early 1940s, computers capable of solving large sets of equations of equilibrium did not exist, so that the method was not practical then.

The theoretical background of FEA laids on the energy approach of Structural Mechanics and on the approximation techniques. The first energy theorems have been established by Maxwell (1864) and Castigliano (1875). The approximation methods have been developed by Ritz (1908) and Galerkin (1915). Ostenfeld (1926) is credited with the first book on the deformation method.

After the Second World War, the Force Method (Flexibility Method) was sustained by Levy (1947) and Garvey (1951) and the matrix Displacement Method (Stiffness Method) was used by Levy (1953) in the sweptback wing analysis. Turner formulated and perfected the Direct Stiffness Method at Boeing (1959). The development of the Force Method ended in 1969.

The development of delta wing structures revived the interest in stiffness methods. Modeling delta wings required two-dimensional panel elements of arbitrary geometry. After a first attempt by Levy (1953) with triangular elements, the article series by Argyris in four issues of Aircraft Engineering (1954, 1955), collected later in a book by Argyris and Kelsey (1960), contains the derivation of the stiffness matrix of a flat rectangular panel using bilinear displacement interpolation. But that geometry was inadequate to model delta wings.

The formal presentation of the finite element method is attributed to Turner, Clough, Martin, and Topp (1956), who during 1952-1953 succeeded to directly derive the stiffness of a triangular panel at Boeing. The term “finite element” was first used by Clough (1960).

1. INTRODUCTION 5

The first book devoted to FEA was written by Zienkiewicz and Cheung (1967), followed by books by Przemieniecki (1968) and Gallagher (1964). Influential papers have been written by Argyris (1965), Fraeijs de Veubeke (1964) and Irons and coworkers (1964, 1966, 1970). Research developed in the Civil Engineering Department at Berkeley, directed by Clough, at Washington University, under Martin, and at Swansea University, lead by Zienkiewicz. Since 1963, finite element computer programs were freely disseminated into the non-aerospace community.

Major contributions are due to B. M. Irons, the inventor of isoparametric models, shape functions, frontal solvers and the patch test (1964-1980), R. J. Melosh, who systematized the variational derivation of stiffness matrices and recognized that FEA is a Rayleigh-Ritz method applied on small size elements (1963), J. S. Archer, who introduced the consistent mass matrix concept (1963), and E. L. Wilson, who studied the sparse matrix assembly and solution techniques (1963), developed the static condensation algorithm (1974) and three SAP computer programs (the first open source FEA software). He was joined later by K.-J. Bathe to develop the finite element codes SAP4 (1973), SAP5 and NONSAP.

Starting with 1965 the NASTRAN finite element system was developed by COSMIC, MacNeal Schwendler, Martin Baltimore and Bell Aero Systems under contract to NASA, completed in 1968 and first revised in 1972. Other known finite element codes are ANSYS, developed by Swanson Analysis Systems (1970), STRUDL - by the Civil Engineering Department at Massachusetts Institute of Technology and McDonnell Douglas Automation Company (1967), STARDYNE - by Mechanics Research Inc, ADINA – developed by K.-J. Bathe at M.I.T. (1975), SESAM – by Det Norske Veritas, NISA – by Engineering Mechanics Research Corporation, MARC – by Marc Analysis Research Corporation, ABAQUS - by Hibbitt, Karlsson @ Sorensen, Inc. (1978), COSMOS-M – by Structural Research & Analysis Corp. (1985), SAMCEF - by SAMTECH (1965), IDEAS-MS, PATRAN, ALGOR etc.

General purpose programs have capabilities of linear dynamic response, including computation of natural frequencies, nonlinear static and dynamic response, crashworthiness, static and dynamic stability, and thermal loading. After 1967 the FEA has been applied to non-structural field problems (thermal, fluids, electromagnetics etc).

1.4 Stages of FEA

FEA involves three stages of activity: (1) preprocessing, (2) processing, and (3) postprocessing.

Preprocessing involves the input and preparation of data, such as nodal coordinates, element connectivity, boundary conditions, material properties and


loading. Data input can be carried out either in an interactive way, through a user-friendly interface, or reading from a data file. Alternatively, some input data can be imported from other F.E.A. or C.A.D. programs.

Automatic mesh generation can be used to produce nodal coordinate data and optimal node numbering, as well as element connectivity data. Mesh plotting is a convenient and useful way of checking the input data. Badly placed nodes or improper blocking of boundary nodes can be easily traced.

Fig. 1.1

Three-dimensional finite element meshes, represented with hidden line removal, are presented for a connecting rod (Fig. 1.1) and a car engine piston (Fig. 1.2), as obtained using the program SIMPAT developed by I.T.I. Italia.

Fig. 1.2

1. INTRODUCTION 7

The finite element model of a vehicle cabin frame obtained with MSC/NASTRAN is shown in Fig. 1.3.

Fig. 1.3

In the processing stage, the finite element program processes the input data and calculates the nodal variable quantities such as displacements and temperatures (equation solving), and element quantities such as stresses and gradients (back-calculation).

Fig. 1.4

In dynamic analyses, processing involves solving an eigenproblem or determining the transient response by incremental techniques. The cost in terms of computer resource increases with the cube of the problem size. In static analyses, the cost of the solution of the linear set of equations increases linearly with the problem size. It would be obviously convenient to use in dynamic analyses the same finite element model that was built for the static analysis. Often this contains much more detail than the dynamic analysis requires, so that condensation and dynamic substructuring are used to reduce the size of the dynamic problem before the processing stage.


Postprocessing deals with the presentation of results. Early programs used tabular presentations. Most programs produce displays of the deformed configuration, vibration mode shapes and stress distributions. Fig. 1.4 shows the initial mesh and the deformed shape of a cooling tower under the wind action, as obtained using ALGOR SUPERSAP.

Fig. 1.5

Scalar nodal variables such as temperatures or pressures are presented in the form of contour plots of isotherms or isobars.

More recent finite element programs show animated displays of the deformed configuration, as in Fig. 1.5 for a crankshaft.

a b

Fig. 1.6

Fig. 1.6, a shows the two-dimensional initial mesh for the analysis of a gear tooth, with 1174 triangular 6-node elements. Fig. 1.6, b shows the optimized mesh obtained with the postprocessor ESTEREF, containing only 814 elements and a four times reduced global discretization error.

2. DISPLACEMENT METHOD

In solving any structural problem, there are four types of equations that should be used: equilibrium equations, geometric compatibility conditions, constitutive relationships and boundary conditions. In order to illustrate the usual longhand analytical procedure, a relatively simple pin-jointed framework will be used. Variables include reaction forces at the supports and internal forces, displacements of the bar ends and bar extensions (elongations). If forces and elongations are eliminated and the displacements are the variables which are solved first, the procedure is referred to as the displacement method. It works whether the structure is statically determinate or not. Once the displacements are determined, they are back-substituted into the compatibility equations to obtain bar extensions, and hence strains, then stresses from the constitutive relationship.

2.1 Equilibrium equations

Consider the truss shown in Fig. 2.1 [74], assume that all members are in tension and write the equilibrium of each node in turn. , , are the tensions in members and , , , are the reaction forces at the supports.

1T 2T 3T

1F 2F 4F 5F

Fig. 2.1


At Node 1 (Fig. 2.2, a), resolving forces horizontally and vertically leads to

.F/T

,FTT

022

022

22

121

=+

=++ (2.1)

At Node 2 (Fig. 2.2, b), equilibrium gives

.F/T

TTF

022

0,/226

43

31

=+

=−− (2.2)

At Node 3 (Fig. 2.2, c)

.T/TF

TFT

0/22229

0,/22/22

32

253

=−−

=−+ (2.3)

The six equations (2.1) to (2.3) have seven unknowns. The system is statically indeterminate. The solution is not possible by using only equilibrium, and consideration must be given to the geometry of deformation.

a b c Fig. 2.2

2.2 Conditions for geometric compatibility

The compatibility equations relate a bar extension lΔ to the displacements of the ends of the bar.

Consider a typical pin-jointed element 1-2 of a frame, Fig. 2.3, inclined an angle θ with respect to the X axis of the global coordinate system.

The displacements in the local coordinate frame can be expressed in terms of the displacements in the global coordinate frame as

xOy

.VUu

,VUu

θθ

θθ

sincos

sincos

222

111

+=

+= (2.4)

The change in length of the member is

2. DISPLACEMENT METHOD 11

( ) ( ) θθΔ sincos 121212 VVUUuu −+−=−=l . (2.5)

Fig. 2.3

Applying equation (2.5) to each member in turn leads to

- 1212 UU=lΔ , (2.6)

( ) ( )22

22 131313 VVUU −+−=lΔ , (2.7)

( ) ( )22

22 232323 VVUU −+⎟⎟⎠

⎞⎜⎜⎝

⎛−−=lΔ . (2.8)

The three equations (2.6) to (2.8) have nine unknowns, six displacements and three elongations.

2.3 Force/elongation relations

Truss members are in either simple tension or compression. Starting from the Hooke’s law for uniaxial stress-strain conditions, the force/elongation relations can be written

112 AE

T ll =Δ ,

AE

T

22

2 213

ll =Δ ,

AE

T

22

2 323

ll =Δ

or

112 AE

T ll =Δ ,

AET2

213

ll =Δ ,

AET2

323

ll =Δ . (2.9)

Three more equations have been added and so there are now 12 equations for 16 unknowns.


2.4 Boundary conditions

The discrepancy of 4 equations is made up by adding the boundary conditions

03211 ==== UVVU (2.10)

which complete the set of 16 linear algebraic equations.

2.5 Solving for displacements

Equations (2.9) can be used to convert the compatibility equations (2.6)-(2.8) to force/displacement equations as follows

( ) ,UUAEAET 12121 −==l

ll

Δ

( ) ,VVUUAEAET 1313132 22−+−==

ll

lΔ

( ).VVUUAEAET 2323233 22−++−==

ll

lΔ

Substitution of the expressions for to into the equilibrium equations (2.1) - (2.3) yields 6 equations

1T 3T

,AE

FVUUVU l 2 133211 =−−−+

,AE

FVUVU l 23311 =−−+

,AE

FVUVUU l 62 33221 =+−−+−

,AE

FVUVU l 43322 =−++−

,AE

FUVUVU l 2 532211 =++−−−

.AE

FVVUVU l92 32211 =+−+−−

Taking into account the boundary conditions (2.10), this set of six equations can be written in matrix form as


⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⋅

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−−−

−−−−−

−−−−−

FFFF

FF

V

UEA

9

6

00

00

201111021111111100

111201110011110112

5

4

2

1

3

2

l. (2.11)

The third and the sixth equation can be decoupled

⎭⎬⎫

⎩⎨⎧

=⎥⎦

⎤⎢⎣

⎡⋅⎥

⎦

⎤⎢⎣

⎡FF

VUEA

96

2112

3

2

l

and solved to give

AE

FU l 2 = , AE

FV l 43 = . (2.12)

Substituting and into the remaining equations yields the reaction forces

2U 3V

FF 51 −= , FF 42 −= , FF 54 −= , FF 5 −= . (2.13)

Substituting these forces into the equilibrium equations (2.1)-(2.3) yields the tensions in the members, which are equal and opposite to the forces acting on the joints. Divided by the corresponding area they give the stresses.

2.6 Comparison of the force method and displacement method

Navier’s Problem. Consider the 7-bar pin-jointed framework shown in Figure 2.4. The joint 8 is subjected to a force of components and . Determine the internal bar forces and the displacement of joint 8.

xF yF

Force Method

Denoting ( the forces applied by each bar to the end nodes, the equilibrium equations of joint 8 can be written

iT )71,...,i =

(2.14)

∑

∑

=

=

=−

=−

7

1

7

1

0sin

0cos

iiiy

iiix

.TF

,TF

θ

θ


This is a set of two equations with seven unknowns, so the framework is statically indeterminate. We choose 31 TX = , 42 TX = , 53 TX = , , and

as redundant forces. 64 TX =

75 TX =

Fig. 2.4

The strain energy is

⎟⎟⎠

⎞⎜⎜⎝

⎛++= ∑

=

5

1

22

221

212

1

iiiXTT

EAU lll (2.15)

where and are functions of to , according to (2.14). 1T 2T 1X 5X

Using Menabrea’s theorem, the five deformation conditions can be written

0=∂∂

iXU , ( )51,...,i = (2.16)

since we are assuming no support movement. They are of the form

0222

111 =+

∂∂

+∂∂

iiii

XXTT

XTT lll . ( )51,...,i = (2.17)

This is a set of five linear equations wherefrom the five redundant forces to are determined. 1X 5X

The components of the displacement of joint 8 are given by Castigliano’s second theorem

xF

Uu∂∂

=8 , yF

U∂∂

=8v . (2.18)


In the force method, the larger the number of bars, the larger the number of statically indeterminate forces, hence the number of equations (2.16).

Displacement Method

The elongation-displacement (compatibility) equations (2.5) are

iii u θθΔ sincos 88 v+=l . ( )71,...,i = (2.19)

The bar length is

i

iaθsin

=l . ( )71,...,i = (2.20)

Substituting (2.19) into the force-elongation equations

i

ii EAT

l

lΔ= , ( )71,...,i = (2.21)

gives the force-displacement relations

( ) iiii ua

EAT θθθ sinsincos 88 v+= . ( )71,...,i = (2.22)

Inserting (2.22) into the equilibrium equations (2.14) of joint 8 yields

∑∑

∑∑

==

==

=−−

=−−

7

1

38

7

1

28

7

1

28

7

1

28

0sinsincos

0cossinsincos

ii

iiiy

iii

iiix

,uFEAa

,uFEAa

θθθ

θθθθ

v

v (2.23)

which is solved for and . 8u 8v

Regardless the number of concurrent bars, only two equations are obtained for the two joint displacements. In matrix form they can be written

(2.24) ⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

8

8

2221

1211vu

KKKK

FF

y

x

where the stiffness coefficients are


.a

EA.a

EAK

,a

EAKK

,a

EA.a

EAK

ii

iii

iii

5674sin

0sincos

4091sincos

7

1

322

7

1

22112

7

1

211

==

===

==

∑

∑

∑

=

=

=

θ

θθ

θθ

(2.25)

Solving (2.24) gives

AEaF. x709708 =u ,

AEaF

. y21908 =v . (2.26)

The approach used in the displacement method is the same whether the structure is statically determinate or not.

3. DIRECT STIFFNESS METHOD

The Finite Element Method (FEM) started as an extension of the stiffness method or displacement method. In the stiffness method for skeletal structures, the elements of the actual structure are connected together at discrete joints. The relationship between the end forces and end displacements of each member is represented by an element stiffness matrix.

We may imagine that the structure is built by adding elements one by one, with each element being placed in a preassigned location. As elements are added to the structure, contributions are made to the structure load carrying capacity, hence to the structure stiffness matrix, which relates all joint displacements to all joint forces. If the members are pin-ended bars they are real distinct elements requiring no approximation. They are natural finite elements. Assembly and solution for displacements are of main concern.

Starting with simple planar frameworks it is possible to explain the assembly process and to make an introduction into the matrix stiffness method. In the following, the basic steps of the Direct Stiffness Method (DSM) are shown using a pin-jointed plane truss.

3.1 Stiffness matrix for a bar element

In the FEM, the names “joint” and “member” are replaced by node and element, respectively. Consider a two-noded pin-jointed element in the own local coordinate system (Fig. 3.1). It has length , cross section area and Young’s modulus . Nodes are conveniently numbered 1 and 2. It is acted upon by the nodal forces , . The nodal displacements are , .

el eA

eE

1f 2f 1q 2q

Generally, bar elements are assumed to be uniform ( ).constEAe = , pin-connected at the ends, linearly elastic, axially loaded (no bending) and with no forces between ends.


Fig. 3.1

Both the nodal displacements , and the nodal forces , are positive in the positive x direction.

1q 2q 1f 2f

The equilibrium equation for the bar element is

.ff 021 =+ (3.1)

Next, the force/elongation relations are used, which incorporate compatibility and stress/strain relations. If end 1 is fixed and end 2 is allowed to

move, then for 0 221

ee

eAE

fq,q l== and

221 qAEffe

eel

−=−= . (3.2)

Similarly, if end 2 is now fixed but end 1 allowed to move,

0 112

ee

eAE

fq,q l== and

121 qAEffe

eel

=−= . (3.3)

Combining equations (3.2) and (3.3), the complete stiffness relationship is obtained as

1 111

ntsdisplacemenodal

e2

1

matrix stiffnesselement

forcesnodal

e2

1

32144 344 21l

321⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

qqAE

ff

e

ee (3.4)

or

{ } [ ] { }eee qkf = (3.5)

where the element stiffness matrix in local coordinates is

[ ] ⎥⎦

⎤⎢⎣

⎡−

−=

1 111

e

eee AEkl

. (3.6)

3. DIRECT STIFFNESS METHOD 19

3.2 Transformation from local to global coordinates

Bar elements in a truss have different orientations in space and it is necessary to define their stiffness properties with respect to a single global coordinate system attached to the whole structure. End forces and displacements have two components at each node, so that nodal forces and nodal displacements can be arranged into 4-element column vectors related by a 44× stiffness matrix.

3.2.1 Coordinate transformation

A typical bar element 1-2 is shown in Figure 3.2, where both the local coordinate system xOy and the global coordinate system XOY are drawn. Nodal displacements are denoted by lower case letters in the local coordinate system and by upper case letters in the global coordinate system.

Fig. 3.2

Let the bar be inclined an angle eθ with respect to the X-axis of the global coordinate system. In fact, the angle eθ is the angle between the positive X-axis and the positive direction of the beam (defined as 1 to 2).

Displacements in the local coordinate frame can be expressed in terms of the displacements in the global coordinate frame as (2.4)

xOy

.QQq

,QQq

eyex

eyex

θθ

θθ

sincos

sincos

222

111

+=

+= (3.7)

In matrix form


(3.8)

434214444444 34444444 2143421

ntsdisplacemeglobal

2

2

1

1

matrix tiontransforma

ntsdisplacemelocal

2

1

sincos0000sincos

ey

x

y

x

ee

ee

QQQQ

qq

e ⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎦

⎤⎢⎣

⎡=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

θθθθ

or

{ } [ ] { }eee QTq = , (3.9)

where

[ ] (3.10) ⎥⎦

⎤⎢⎣

⎡=

ee

eeeTθθ

θθsincos00

00sincos

is a coordinate transformation matrix.

From nodal coordinate data, denoting ( )11 Y,X and ( )22 Y,X the coordinates of nodes 1 and 2, respectively, we obtain

e

eXX

l

12cos−

=θ , e

eYY

l

12sin−

=θ , ( ) ( )212

212 YYXXe −+−=l .

The entries in the matrix (3.10) are calculated based on the above equations.

3.2.2 Force transformation

Consider the pin-jointed member (Fig. 3.3) subjected to forces f1 and f2 applied at the ends 1 and 2.

Fig. 3.3

The force components in the global coordinate system are


.sin sin

cos cos

21

21

21

21

eyey

exex

fF,fF

,fF,fF

θθ

θθ

==

== (3.11)

In matrix form

, (3.12)

321444 3444 2143421

forceslocal

2

1

matrix tiontransforma

componentsforce

global

sin0cos0

0sin0cos

2

2

1

1

e

e

e

e

e

ey

x

y

x

ff

FFFF

⎭⎬⎫

⎩⎨⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

θθ

θθ

or

{ } [ ] { }eTee fTF = . (3.13)

Equation (3.13) can be directly obtained from consideration of mechanical work. Work is a scalar quantity, having the same value regardless the coordinate system

{ } { } { } { }eTeeTe QFqf = . (3.14)

Substituting (3.9) for the local displacements, equation (3.14) becomes

{ } { } { } [ ] { } { } { }eTeeeTeeTe QFQTfqf == ,

hence

{ } [ ] { }TeeTe FTf =

which by transposition becomes (3.13).

3.2.3 Element stiffness matrix in global coordinates

Inserting equation (3.5) into (3.13), then equation (3.9) into the resulting matrix product, and comparing with (3.13)

{ } [ ] { } [ ] [ ] { } [ ] [ ] [ ] { }eeeTeeeTeeTee QTkTqkTfTF444 3444 21

===

we find

{ } [ ] { }eee QKF = (3.15) where

[ ] [ ] [ ] [ ].TkTK eeTee = . (3.16)


Carrying out the matrix multiplications and using (3.10) we find the element stiffness matrix in global coordinates

[ ]⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−

=

22

22

22

22

sscsscsccsccsscsscsccscc

AEKe

eee

l, (3.17)

where ec θcos= and es θsin= .

Remember that the element stiffness matrix is a proportionality factor between the components of forces applied by the nodes to the element and the components of nodal displacements in global coordinates

ey

x

x

e

ee

ey

x

y

x

QQQyQ

ssccsscssccscc

AE

FFFF

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⋅

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−−−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

2

2

1

1

2

2

22

22

2

2

1

1

SYMl

. (3.17, a)

3.2.4 Properties of the element stiffness matrix

The element stiffness matrix is symmetric, with positive diagonal elements, singular (order 4, rank 1) and each column (row) sums to zero.

3.2.4.1 Symmetric matrix

Stiffness matrices are symmetrical

[ ] [ ]Tee KK = .

Expressing displacements in terms of forces gives

{ } [ ] { } [ ] { }eeeee FFKQ δ== −1

so that the inverse of the stiffness matrix is the flexibility matrix

[ ] [ ]eeK δ=−1

. (3.18)

According to Maxwell’s reciprocity theorem, the flexibility matrix must be symmetrical about the leading diagonal. And so must be its inverse, the stiffness matrix.

[ ]eδ


The same result can be obtained following strain energy arguments. For linear structures, displacements are proportional to the applied loads. As forces are increased from zero to their final values, the total work done by these forces is

{ } { }eTee FQW21

= . (3.19)

In the absence of dynamic effects, this work is absorbed by the structure as strain energy. Substituting (3.15) into (3.19) we obtain the strain energy which is a scalar

{ } [ ] { }eeTee QKQU

21

= . (3.20)

It is equal to its transpose

{ } [ ] { }eTeTee QKQU

21

= (3.20, a)

so that

[ ] [ ]Tee KK = , which defines the symmetry.

3.2.4.2 Singular matrix

The element stiffness matrix is of order 4 and rank 1. The rank deficiency is 3 and this corresponds to the three possible and independent forms of rigid body motion in plane for the unsupported bar: two translations and one rotation.

A single ungrounded bar can be moved in space as a rigid body without straining it and hence with zero strain energy. This means that there exist a set of

rigid body displacements { }eQ for which { } [ ] { } 02 == eeTee QKQU

wherefrom . This can only be true if the determinant of [ ] { } { }0=ee QK [ ]eK

vanishes. The matrix [ ]eK is said to be singular. The rigid body modes are defined by the eigenvectors corresponding to its zero eigenvalues.

The zero determinant implies that there are linear relationships between its columns (rows). The rank of a matrix is the size of the largest sub-matrix with a non-zero determinant. One can verify that the determinants of the 33× and reduced sets are still zero, so that the stiffness matrix has rank 1 (or its rank deficiency is 3).

22×

3.2.4.3 Positive diagonal elements

Each diagonal entry of the matrix [ ]eK is positive. If this were not so, a force and its corresponding displacement would be oppositely directed, which is


physically unsound. Moreover, the matrix [ ]eK is positive semidefinite. That is, the quadratic form that represents strain energy (3.20) is either positive or zero.

3.2.4.4 Each column (row) sums to zero

Consider a bar element with end 2 fixed ( )022 == yx QQ and end 1 having

a unit displacement along the global X-axis ( )0 1 11 == yx Q,Q as in Fig. 3.4.

Fig. 3.4

Equation (3.17, a) can be written

(3.21)

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

====

⋅

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

0001

2

2

1

1

41

31

21

11

2

2

1

1

y

x

y

x

y

x

y

x

QQQQ

...k

...k

...k

...k

FFFF

which yields

, 111 kFx = 211 kFy = , 312 kFx = , 412 kFy = . (3.22)

This shows that the first column of the stiffness matrix represents the forces that must be applied to the element to preserve static equilibrium when

and all other displacements are zero. 11 =xQ

The displacement 11 =xQ produces an axial shortening eexQ θθ coscos1 = ,

which corresponds to a compressive force ee

ee AE θcosl

, whose components must

be equilibrated by the external forces , , and . Equilibrium of horizontal and vertical forces yields

11k 21k 31k 41k

03111 =+ kk , 04121 =+ kk ,


so that the first column sums to zero. The same applies for the other columns.

In the element stiffness matrix, each column represents an equilibrium set of nodal forces produced by a unit displacement of one nodal degree of freedom.

3.3 Link’s truss

In order to illustrate the assembly of the global stiffness matrix from the elemental stiffness matrices, consider Link’s truss [74] shown in Fig. 3.5, already analyzed in Chapter 2, Fig. 2.1.

Fig. 3.5

The truss comprises 3 elements and 3 nodes. It is simply supported in 2 and 3, firmly located in 1, and acted upon by forces 6F and 9F. The global displacements and nodal forces are shown in Fig. 3.6. A node whose global index is i has associated with it the global displacements and forces ( )12 −i and . i2

a b


Fig. 3.6

Element data are given in Table 3.1 together with information useful for the computation. The first three columns define the element connectivities, i.e. their localization within the structure. Element numbering can be arbitrary.

Table 3.1

Nodes Member i j

eθ eθcos

eθsin

2c 2s

cs el eEA e

eEAl

1 1 2 0 1 0 1 0 0 l EA l

EA

2 1 3 45 22

22

21

21

21

l22 EA2

l

EA2

3 2 3 135 22

−

22

21

21

21

−

l

22

EA2 l

EA2

3.4 Direct method

Using the data from Table 3.1, the element stiffness matrices (3.17) are calculates as

[ ]4321

AEK

4321

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

=

0000010100000101

1l

T ,

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

••••••••••••••••

KKKKKK

KKKKKK

KK

KK

KK

KK

[ ]65

1111111111111111

65

2 21

AEK

21

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−

=l

T ,

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

••••••••

••••••••

KK

KK

KKKKKK

KKKKKK

KK

KK


and

[ ]65

1111111111111111

65

3 43

AEK

43

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−−−

−−

=l

T .

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

••••••••••••••••

KK

KK

KK

KK

KKKKKK

KKKKKK

At the top and on the right of the element stiffness matrices, the numbering of coordinates in the global stiffness matrix is shown, according to Table 3.2.

Table 3.2

Node 1 2

Direction X Y X Y Local nodal coordinate

1 2 3 4 Element Global nodal coordinate

1 1 2 3 4 2 1 2 5 6 3 3 4 5 6

The assembly of the unreduced global stiffness matrix (3.28) is done systematically, locating each coefficient of the element stiffness matrices into the appropriate place in the global 66× matrix (for this example), as indicated by dots above, eventually adding it to the coefficients already accumulated at that location. This is referred to as the direct matrix method.

For instance, element 2 is located in the truss between the left end node and the right end node 1=i 3=j (nodal labels i and j may be assigned arbitrarily).

In the global matrix, the two displacement and force components (along X and Y) for node are numbered 1=i 112 =−i and 22 =i , and those for the node are numbered

3=j512 =−j and . 62 =j

The simple addition of different stiffness coefficients in a location is based on the fact that finite element equations are in fact nodal equilibrium equations, so that if a node is common to several elements, each member will contribute with a force component to maintain equilibrium under an arbitrary set of nodal displacements. An alternative algebraic explanation of the assembly of system stiffness matrices is presented in the following.


3.5 Compatibility of nodal displacements

The compatibility of nodal displacements at element level, with the nodal displacements at the whole truss structure level, can be expressed by equations of the form

{ } [ ] { }QT~Q ee = , (3.23)

where { }eQ is the element displacement vector in global coordinates, { }Q is the

full displacement vector of the truss structure and [ ]eT~ is referred to as a connectivity or localization matrix, containing ones at the nodal displacements of element nodes and zeros elsewhere.

For the truss from Fig. 3.5, equation (3.23) applied to the three elements yields

{ } [ ] [ ]QT~

QQQQQQ

QQQQ

Q 1

6

5

4

3

2

1

4

3

2

1

1

001000000100000010000001

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⋅

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

= ,

{ } [ ] [ ]QT~

QQQQQQ

QQQQ

Q 2

6

5

4

3

2

1

6

5

2

1

2

100000010000000010000001

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⋅

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

= ,

{ } [ ] [ ]QT~

QQQQQQ

QQQQ

Q 3

6

5

4

3

2

1

6

5

4

3

3

100000010000001000000100

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⋅

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

= .


3.6 Expanded element stiffness matrix

The element strain energy in global coordinates can be written in terms of the global displacement vector, substituting (3.23) into (3.20, a)

{ } [ ] [ ] [ ]{ }QT~KT~QU eeTeTe 2

1=

or

{ } [ ]{ }QK~QU eTe 2

1= ,

where the expanded element stiffness matrix

[ ] [ ] [ ] [ ]eeTee T~KT~K~ = (3.24)

has the size of the system matrix.

For the truss from Fig. 3.5, equation (3.24) yields

[ ] [ ] [ ][ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⋅

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

==

001000000100000010000001

0000010100000101

000000001000010000100001

1111l

EAT~KT~K~T

,

or

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

=

000000000000000000000101000000000101

1l

EAK~ .

[ ] [ ] [ ][ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⋅

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

==

100000010000000010000001

1111111111111111

100001000000000000100001

2222l

EAT~KT~K~T

,


or

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−

=

110011110011000000000000110011110011

2l

EAK~

and

[ ] [ ] [ ][ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⋅

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−−−

−−

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

==

100000010000001000000100

1111111111111111

100001000010000100000000

3333l

EAT~KT~K~T

,

or

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−−−

−−=

111100111100111100111100000000000000

3l

EAK~ .

3.7 Unreduced global stiffness matrix

The strain energy for the complete truss structure

{ } [ ]{QKQU T

21

= }, (3.25)

can be calculated by simply adding the element strain energies

{ } [ ]{ } { } [ ] { QK~QQK~QUUe

eT

e

eT

ee ∑∑∑ ===

21

21 }. (3.26)

Comparing (3.25) and (3.26) we get

[ ] [ ]∑=e

eK~K . (3.27)

The global stiffness matrix is equal to the sum of the expanded element stiffness matrices.


The unreduced stiffness matrix [ ]K is symmetric, singular (for a plane truss, the deficiency is 3), has positive elements along the main diagonal and each column (row) sums to zero. It corresponds to the free-free system. For a grounded system, this matrix is condensed using the boundary conditions.

The effect of elastic supports modelled as lumped springs can be accounted for by adding their stiffnesses along the main diagonal at the appropriate locations in the stiffness matrix.

For the truss from Fig. 3.5, equation (3.27) yields

[ ] [ ] [ ] [ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−−−

−−−−−

−−−−−

=++=

201111021111111100111201110011110112

321l

EAK~K~K~K . (3.28)

The expanded element stiffness matrices have been used above only to show algebraically how to assemble a global stiffness matrix; they are never used in practice. The expensive product (3.24) is never formed. The global stiffness assembly is a simple book-keeping exercise and is done by directly placing the nonzero coefficients of element stiffness matrices in the right locations of the global stiffness matrix based on element connectivity.

3.8 Joint force equilibrium equations

The assembly of the global stiffness matrix has been based on strain energy considerations. An alternative presentation is given below, based on joint equilibrium equations, using for convenience the truss from Fig. 3.5.

Note that element equilibrium equations (3.1) used so far involved only forces applied by nodes to the elements. The joint equilibrium equations, involving forces applied by elements to nodes, are used in the following.

An exploded layout of the truss is shown in Fig. 3.7. Apart from external forces and support reactions, nodes are acted upon by forces equal and in opposite direction to those applied to elements. Equal forces are labelled only once for clarity.

Resolving nodal forces horizontally and vertically, we obtain 6 equilibrium equations


.FFF,FFF,FFF

,FFF,FFF,FFF

yyyyyy

xxxxxx32

226

31

124

21

112

32

225

31

123

21

111

+=+=+=

+=+=+=

Fig. 3.7

In matrix form

[ ] [ ] [ ]

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪

⎨

⎧

⋅

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−−−

−−−

32

32

31

31

22

22

21

21

12

12

11

11

6

5

4

3

2

1

321

100001000010000100000000

100001000000000000100001

000000001000010000100001

y

x

y

x

y

x

y

x

y

x

y

x

T~T~T~

FF

FF

FF

FF

FF

FF

FFFFFF

TTT443442144 344 214434421

whose partitioned form can be written in scalar product form as

{ } [ ] { } [ ] { } [ ] { }332211 FT~FT~FT~FTTT

++=

or generally

{ } [ ] { }∑=e

eTe FT~F . (3.29)


Substituting (3.15) and (3.23), equation (3.29) becomes

{ } [ ] [ ]{ } [ ] [ ] [ ] { }∑∑ ==e

eeTe

e

eeTe QT~KT~QKT~F444 3444 21

, (3.30)

then, using equations (3.24) and (3.27), it can be written

{ } [ ] { } [ ] { }QKQK~Fe

e ==∑ . (3.31)

The unreduced global stiffness matrix [ ]K relates the unreduced vector of nodal forces { F } to the unreduced vector of nodal displacements { }Q .

3.9 Reduced global stiffness matrix

The global truss should be supported adequately and there should be no internal mechanism. Boundary conditions eliminate the possibility of the structure to move as a rigid body.

A general type of boundary conditions include specified displacements of the support nodes, of the type ii aQ = ( 1=i to ), where is the number of supports. Most often the support nodal displacements are zero, . Another type are the multipoint constraints, encountered in inclined roller supports and rigid connections, of the type

sn sn0=iQ

0bQbQb jjii =+ , where , and are known constants.

ib jb 0b

For an N-degree-of-freedom structure, the finite element equations (3.31) are of the form

.

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⋅

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

NNNNNN

N

N

F

FF

Q

QQ

KKK

KKKKKK

LL

L

LLLL

L

L

2

1

2

1

21

22221

11211

Consider a single boundary condition, 11 aQ = . As is known, the finite element equations in the remaining

1Q1−N unknowns are

. (3.32)

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⋅

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

11

1313

1212

3

2

32

33332

22322

aKF

aKFaKF

Q

QQ

KKK

KKKKKK

NNNNNNN

N

N

LL

L

LLLL

L

L


The ( ) stiffness matrix above is obtained simply by deleting or eliminating the first row and column from the original

( 11 −×− NN )( )NN× stiffness matrix.

Equation (3.32) may be written in condensed form

[ ]{ } { }FQK = , (3.33)

where is a reduced stiffness matrix, obtained by eliminating the row and column corresponding to the specified or “support” degrees of freedom, { and

are the reduced vectors of global displacements and forces, respectively.

[ ]K}Q

{ }F

For a truss supported on many supports, the reduced global stiffness matrix is obtained deleting as many rows and columns as the number of specified displacements. The final equations (3.33) can be solved for the displacement vector

using Gauss elimination. The matrix { Q } [ ]K is not singular.

It is instructive now to consider the truss from Fig. 3.5. Having assembled the unreduced global stiffness matrix, after substitution of the boundary conditions

05421 ==== QQQQ and the values of the applied forces FF 63 = and , the following set of linear equations is obtained

FF 96 =

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⋅

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−−−

−−−−−

−−−−−

FFFF

FF

Q

QEA

9

6

00

00

201111021111111100111201110011110112

5

4

2

1

6

3l

. (3.34)

It can be seen that where the displacements are specified, the nodal forces are unknown, and where the forces are specified, the nodal displacements are unknown. Equations (3.34) can be rearranged as follows

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⋅

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−

−−−−−−

−−−−−−

5

4

2

1

6

396

0000

2111110010111012

01111011

01111101

2112

FFFFFF

QQ

EAl

. (3.35)

The first two equations can be written

⎭⎬⎫

⎩⎨⎧

=⎥⎦

⎤⎢⎣

⎡⋅⎥

⎦

⎤⎢⎣

⎡FF

QQEA

96

2112

6

3

l


and solved to give

AE

FQ l 3 = , AE

FQ l 46 = .

In general, for non-zero boundary conditions, equations (3.33) can be written in partitioned form

[ ] [ ][ ] [ ]

{ }{ }

{ }{ }⎭

⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧⋅⎥

⎦

⎤⎢⎣

⎡

b

a

b

a

bbba

abaaFF

QQ

KKKK

, (3.36)

where is the vector of known forces, { aF } { }bQ is the vector of known

displacements and [ ] . [ ]Tabba KK =

The first set of equations can be written

{ } [ ] { } [ ] { }babaaaa QKQKF += , (3.37)

then multiplied to the left by to yield the unknown displacements [ ] 1−aaK

{ } [ ] { } [ ] { }( )babaaaa QKFKQ −= −1 , (3.38)

3.10 Reactions and internal forces

The unknown forces, which in this case are the external reactions, are given by { } [ ] { } [ ] { }bbba

Tabb QKQKF += . (3.39)

For the truss from Fig. 3.5, the support displacements are zero, so that

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−−−−

=⎭⎬⎫

⎩⎨⎧⋅

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

FFFF

QQEA

FFFF

545

01111011

6

3

5

4

2

1

l.

The elongation of an element (2.5) is

( ) ( ) eyyexxe QQQQqq θθΔ sincos 121212 −+−=−=l (3.40)

which can be computed when all displacements have been determined.

The axial force in a truss element is


ey

x

y

x

e

eee

e

eee

QQQQ

scscAEAET

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−−==

2

2

1

1

cbl

ll

Δ . (3.41)

For the truss from Fig. 3.5, the member forces are

FQEA

QQQQ

EAT ==

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−= 3

4

3

2

1

1 0101ll

cb ,

FQEA

QQQQ

EAT 2421111222

6

6

5

2

1

2 ==

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−−=ll

cb ,

( ) FQQEA

QQQQ

EAT 2521111222

63

6

5

4

3

3 =+=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−=ll

cb .

Stresses can be determined dividing these forces by the element cross-section areas.

3.11 Thermal loads and stresses

Thermal stresses are calculated using the “restraining method” suggested by J. M. C. Duhamel (1838).

Suppose the node displacements are completely restrained (blocked). This produces thermal strains TT αε −= , where α is the coefficient of thermal expansion and T is the amount of uniform heating (temperature difference). The restrained state is equivalent to a pre-stressing with compressive stresses

TET ασ −= , where E is Young’s modulus.

Restraining produces a compressive axial force TEAα in the element, where A is the cross-section area. Accordingly, the element acts upon its nodes with equal and opposite forces

{ } TT scscTEAF cb −−=α , (3.42)


or, in local coordinates,

{ } TT TEAf cb 11−=α . (3.42, a)

These forces should be included in the vector of nodal forces (added to the external forces, if the case). After determining the displacements produced by these forces, the element elongations are determined from (3.37) and the stresses are calculated as

⎟⎟⎠

⎞⎜⎜⎝

⎛−= TE

e

eee αΔσ

l

l , (3.43)

i.e. adding to the stresses produced by the thermal (and external) loads, the initial stresses produced by restraining. It is as if unrestraining forces are applied at the ends of the element to free it from the initial restraining.

3.12 Node numbering

Stiffness matrices are symmetric and sparse. They are also banded, i.e. with the nonzero elements clustered in a band along the main diagonal. This is obvious for one-directional structures and can be achieved by rational node numbering in other structures.

Consider the unreduced global stiffness matrix of the truss from Fig. 3.8, a, with nonzero elements in the upper triangle identified by the symbol X

1212×

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

×××××××××××××××××××××

×××××××××××

×××××××××××

×××××××××××

=

→←

cirtemmys

0

K

B

. (3.44)

Because of symmetry, only the diagonal elements and those from one side of the main diagonal are retained. Even so, due to the sparseness, there are many zero elements in the upper triangle. The half-bandwidth is denoted B.


Since only nonzero elements need to be stored, the information in the above matrix can be compactly stored in the 612× matrix below

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

×××

×××××××

×××××××××××

×××××××××××

×××××××××××

×××××××××××

=

↓↓↓

0

21

bK

B

.

The first column contains the diagonal elements of the full matrix (3.44). The second column contains the elements from the second diagonal. In general, the mth diagonal of the original matrix is stored as the mth column. The number of columns in the banded-form storage is equal to B - the half-bandwidth of the original matrix. The efficiency of band-storage increases with the order of the matrix.

a b

Fig. 3.8

For plane trusses, B is equal to 2 plus twice the maximum node number difference in an element. For the numbering scheme of Fig. 3.8, a , . However, for the numbering from Fig. 3.8, b,

6=B12=B , i.e. equal to the size of the

original matrix.

As a general rule, a small bandwidth can be obtained by numbering nodes along the shorter dimension of a structure, then progressing along the longer dimension.


The half-bandwidth is automatically determined within the finite element program from the node numbering. Apart from storage savings, Gaussian elimination algorithms are used for symmetric banded matrices which enable also the reduction of computer time.

For large sparse stiffness matrices, efficient reduction of both storage and computing time can be achieved using the skyline storage and a skyline equation solver. In this case, the columns of the matrix upper triangle are stored serially and concatenated in a column vector { }sK . If there are zeros at the top of a column, only the elements between the diagonal term and the first nonzero term need be stored. The line separating the top zeroes from the first nonzero element is called the skyline.

Consider the following matrix

Skyline

kkk

kkk

kkkkkk

kkkkkk

heightColumn

0000

00000000000000

5 4 1 3 4 2 2 1

88

7877

6766

55

484744

353433

242322

141211

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

→

The active columns are stored in the column vector { }sK and a diagonal pointer vector { }ID is built up with the indices of diagonal elements in { } sK

{ }

9

5

3

1

88

44

34

24

14

33

23

22

12

11

←

←

←

←

⎪⎪⎪⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪⎪⎪⎪

⎨

⎧

=

k

kkkkkkkkk

Ks

L

L

. { }

⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪

⎨

⎧

=

221713129531

ID


The height of the jth column is given by ( ) ( )1−− jIDjID . For the solution of the finite element equations, Gaussian elimination can be applied using a skyline solver program.

A generic flow chart of the Matrix Displacement Method is given in the following.

Matrix Displacement Method

Input Data Geometric data of truss (Nodal coordinates) Material properties + Cross-section area of members Connectivity table of elements Boundary conditions

1

↓

Element stiffness matrix in local coordinates Transformation to global coordinates

2

↓

Assembly of the global stiffness matrix 3

↓

Introduction of boundary conditions 4

↓

Solving the linear set of equations 5

↓

Back-calculation External reactions Member forces (and stresses)

6


Exercises E3.1. For the truss in Fig. E3.1, a, determine: a) the maximum nodal

displacement and its location b) the maximum stress and its location, and c) the support reactions. Plot the deformed shape.

Fig. E3.1, a

Answer. a) The finite element model consists of 8 nodes and 13 elements.

Taking , 1=l 1=A , 1=E and 1=F , we obtain the following:

Nodal data Node

nr Restr

X Restr

Y Coord

X Coord

Y Displ

X Displ

Y 1 1 1 0 0 0 0 2 0 0 4 0 64 -361.1 3 0 0 4 3 208.3 -361.1 4 0 0 8 0 128 -391.8 5 0 0 8 6 127.6 -343.7 6 0 0 12 0 176 -339.7 7 0 0 12 3 62.9 -339.7 8 0 1 16 0 224 0

Element data Element

nr Nodes Axial stress

Element nr Nodes

Axial stress

1 1, 2 16 8 4, 6 12 2 1, 3 -10 9 4, 7 -4.16 3 2, 3 0 10 5, 7 -10.83 4 2, 4 16 11 6, 7 0 5 3, 4 -9.16 12 6, 8 12 6 3, 6 -10.83 13 7, 8 -15 7 4, 5 8


The support reactions are 81 −=R , 62 =R , and 93 =R . The deformed shape is shown in Fig. E3.1, b.

Fig. E3.1, b

E3.2. Consider the pin-jointed framework shown in Fig. E3.2, a. Determine: a) the maximum nodal displacement, b) the maximum stress, and c) the support reactions. Plot the deformed shape.

Fig. E3.2, a

Answer. The finite element model consists of 7 nodes and 11 elements. a)

The vertical displacement of point 7 is EAF. l7872197 −=v . b) The axial stress in

element 7 is AFN 67 −= . c) The support reactions are F.RR 4531 =−= , and . FR 62 =

The deformed shape is presented in Fig. E3.2, b.


Fig. E3.2, b

E3.3. Consider the truss shown in Fig. E3.3, a. Determine the location and the value of: a) the maximum nodal displacement, b) the maximum stress. Calculate c) the support reactions. Plot the deformed shape.

Fig. E3.3, a

Answer. The finite element model consists of 6 nodes and 9 elements. a)

The vertical displacement of point 3 is EAF. l831373 −=v . b) The axial stress in

element 1 is AF.N 2571 −= . c) The support reactions are 01 =R , and .

F.R 432 =F.R 623 =



Fig. E3.3, b

E3.4. Consider the truss shown in Fig. E3.4, a. Determine: a) the maximum nodal displacement, b) the maximum stress, and c) the support reactions. Plot the deformed shape.

Fig. E3.4, a

Answer. The finite element model consists of 4 nodes and 5 elements. Taking , 1=l 1=A , 1=E and 1=F , we obtain the following:

Nodal data Node

nr Restr

X Restr

Y Coord

X Coord

Y Displ

X Displ

Y 1 1 1 0 0 0 0 2 1 1 0 1 0 0 3 0 0 1 0 -1.3294 -3.2095 4 0 0 2 1 2.6705 -9.0896


Element data Element

nr Nodes Axial stress

1 3, 4 - 0.940 2 1, 3 - 1.329 3 2, 3 0.940 4 2, 4 1.335 5 1, 4 - 0.749

a) The vertical displacement of point 4 is AEF. l0994 −=v . b) The axial stress in element 4 is AF.N 33514 = . c) The support reactions are

, and FRR 2=−= F.R 3350= F.R 665031 2 4 = . The deformed shape is presented in Fig. E3.4, b.

Fig. E3.4, b

E3.5. Consider the pin-jointed framework shown in Fig. E3.5, a where point 6 is displaced . Determine the location and the value of: a) the maximum nodal displacement; b) the maximum stress. Plot the deformed shape.

56 −=v

Fig. E3.5, a


Answer. The finite element model consists of 14 nodes and 25 elements. Taking , 1=l 1=A and 1=E , the vertical displacements of points 7 and 8 are

, and the axial stress in element 15 is 538487 .−== vv 366015 .N = . The deformed shape is presented in Fig. E3.5, b.

Fig. E3.5, b

4. BARS AND SHAFTS

This chapter deals with simple one-dimensional structural elements, having one degree of freedom per node. The displacement within the element is expressed in terms of the nodal displacements using shape functions. The unknown displacement field within an element is usually interpolated by a linear distribution. This approximation becomes increasingly accurate as more elements are considered in the model. For a bar without loads between ends the linear interpolation is exact. The compatibility of adjacent elements requires only continuity. Displacements must be continuous across the element boundary.

0C

For bars with distributed loads, true displacements are described by higher order polynomials. It is shown that their use is tantamount to adding internal nodes. However, it is common practice to use linear shape functions and two-node elements without loads between ends. This implies replacement of the distributed loads by equivalent forces applied to nodes. These kinematically equivalent forces are determined using the appropriate shape functions from the condition to perform the same mechanical work as the actual loading. In this section, the corresponding element stiffness matrix and load vectors will be derived.

4.1 Plane bar elements

Bars are structural elements used to model truss elements, cables, chains and ropes. Their longitudinal dimension is much larger than the transverse dimensions. Bars are loaded only by axial forces. They are modeled by elements having one-degree-of-freedom per node.

4.1.1 Differential equation of equilibrium

In a thin uniform rod of cross-section area A and Young’s modulus E, there are axial displacements ( )xuu = due to axial loads ( )xp . The dimensions of p are force/length. The displacement at xx d+ will be uu d+ .


The axial strain is given by the strain-displacement relation

xux dd=ε . (4.1)

The normal stresses result from Hooke’s law

xuEx dd=σ . (4.2)

Fig. 4.1

The internal axial force N is

xuAEAN x d

d== σ . (4.3)

The equation of equilibrium of an infinitesimal element of length dx of the rod (Fig. 4.1) is , 0dd =+ xpN

( ) 0dd

=+ xpxN ,

or, using (4.3),

( )xpxuEA −=2

2

dd . (4.4)

4.1.2 Coordinates and shape functions

Consider a two-node pin-jointed element in the own or local coordinate system. Nodes are conveniently numbered 1 and 2, their coordinates in the physical (Cartesian) reference system being and respectively (Fig. 4.2, a). 1x 2x

We define a natural or intrinsic reference system which permits the specification of a point within the element by a dimensionless number

⎟⎟⎠

⎞⎜⎜⎝

⎛ +−

−=

22 21

12

xxx

xxr (4.5)

so that 1−=r at node 1 and 1+=r at node 2 (Fig. 4.2, b).

4. BARS AND SHAFTS 49

Fig. 4.2

Expressing the physical coordinate in terms of the natural coordinate yields

( ) ( ) 2211 xrNxrNx += , (4.6) where

) and ( ) ( )rrN += 121

2 (4.7) ( ) ( rrN −= 121

1

can be considered as geometric interpolation functions. The graphs of these functions are shown in Figs. 4.3, a,b. They have a unit value at the node of the same index and zero at the other node.

Fig. 4.3

4.1.3 Bar not loaded between ends

For a prismatic bar not loaded between ends, 0=p , 0dd 22 =xu , .constxu =dd , so that the displacement field within the element may be expressed

as a linear polynomial


( ) xbaxu += . (4.8)

The two integration constants above may be determined from the nodal displacements and the geometry of the element. With 11 xbaq += at node 1 and

at node 2, equation (4.8) becomes 22 xbaq +=

12

1221

12

12xx

xqxqxxxqqu

−−

+−−

= . (4.9)

Equation (4.9) can also be written

212

11

12

2 qxx

xxqxxxxu

−−

+−+−

= ,

or, using (4.5),

21 21

21 qrqru +

+−

= .

The displacement of an arbitrary point within the element can be expressed in terms of the nodal displacements and as 1q 2q

( ) ( ) 2211 qrNqrNu += , (4.10)

where ( ) ( rrN −= 121

1 ) and ( ) ( rrN += 121

2 )

}

are the shape functions of the

element.

The polynomial form (4.6) is simpler, but the integration constants, a and b, have no simple physical meaning. The nodal expansion (4.10) is more complicated, but the integration constants, and , are the nodal displacements.

1q 2q

In matrix form

, (4.11) ⎣ ⎦ { e

iii qNqNu ==∑

=

2

1where

and ⎣ ⎦ ⎣ ⎦21 NNN = { } { }Te qqq 21= . (4.12)

Thus, the displacement at any point within an element can be found by multiplying the matrix of shape functions by the vector of nodal displacements.

In (4.11), { }eq is the column vector of element nodal displacements and is the row vector of displacement interpolation functions also named shape ⎣ ⎦N


functions. It is easy to check that 1qu = at node 1 and 2qu = at node 2, and that u varies linearly (Fig. 4.3, c).

Equations (4.6) and (4.10) show that both the element geometry and the displacement field are interpolated using the same shape functions, which is referred to as the isoparametric formulation.

4.1.4 Element stiffness matrix in local coordinates

Strains can be expressed in terms of the shape functions as

⎣ ⎦ { } ⎣ ⎦ { }eex qBqN

xxu

===dd

ddε (4.13)

where

⎣ ⎦ ⎣ ⎦Nx

Bdd

= (4.14)

is the row vector of the derivatives of shape functions, generally called the element strain-displacement matrix. It gives the strain at any point due to unit nodal displacement.

The transformation from x to r in equation (4.5) yields

rrxx

x e d2

d2

d 12 l=

−= , (4.15)

where 11 +≤≤− r and the length of the element is 12 xxe −=l .

The element strain energy is eU

∫∫ ==e

xe

e

xxe VEVU d21d

21 2εεσ . (4.16)

Substituting (4.13) in (4.16) we obtain

{ } ⎣ ⎦ ⎣ ⎦ { e

Ve

TTee qVBEBqU

e 444 3444 21

∫= d21 }. (4.17)

The above equation is of the form

{ } [ ] { }eeTee qkqU

21

= , (4.18)

where the element stiffness matrix [ ]ek is given by


[ ] . (4.19) ⎣ ⎦ ⎣ ⎦∫=eV

eTe VBEBk d

This form guarantees that [ ]ek will be a symmetric matrix.

In terms of the shape functions

[ ] ∫ ⎥⎦

⎥⎢⎣

⎢⎥⎦

⎥⎢⎣

⎢=

e

xxN

xNAEk

T

eee

l

ddd

dd . (4.20)

Because rN

xr

rN

xN

e dd2

dd

dd

dd

l== , we can write

[ ] ∫+

−

⎥⎦

⎥⎢⎣

⎢⎥⎦

⎥⎢⎣

⎢=

1

1

ddd

dd2 r

rN

rNAEk

T

e

eee

l. (4.21)

Substituting 21

dd 1 −=

rN and

21

dd 2 +=

rN , equation (4.21) yields

[ ] ∫+

−⎥⎦⎥

⎢⎣⎢−

⎭⎬⎫

⎩⎨⎧−

=1

1

d21

21

21212 rAEk

e

eee

l

or

[ ] ⎥⎦

⎤⎢⎣

⎡−

−=

1 111

e

eee AEkl

. (4.22)

which is the same as equation (3.6).

4.1.5 Bar loaded between ends

For a prismatic bar acted upon by a uniformly distributed axial load, .constp = , .constxu =22 dd , so that the displacement field within the element

may be expressed as a quadratic polynomial

( ) 2xcxbaxu ++= . (4.23)

The three integration constants, a, b and c, have to be determined from three boundary conditions. This can be done if we use a three-node one-dimensional element. An internal node is added at the midpoint to comply with the requirement of a quadratic fit (Fig. 4.4, a).


For a linearly distributed axial load (as in a bar rotating at constant angular speed around an end, acted upon by a distributed centrifugal load proportional to the distance to the rotation centre), the true displacement field is given by a cubic polynomial, involving four integration constants (see Example 4.8).

For an exact solution, this implies using a four-node element (adding two internal nodes). The usual practice is to assume an approximate lower order linear displacement field, i.e. a two-node element and to replace the actual linearly distributed load by equivalent nodal forces, having thus an element not loaded between ends describable by linear shape functions.

Fig. 4.4

Consider the three-node quadratic element from Fig. 4.4, a, with node 3 at the midpoint. Nodal coordinates are , , , and the vector of element nodal

displacements is { } . The x-coordinate system is mapped onto an intrinsic r-coordinate system, given by the transformation

1x 2x 3x

{ TTe qqqq 321= }

( )12

32xxxxr

−−

= . (4.24)

It comes out that 1−=r , 0, and 1+ at nodes 1, 3, and 2 (Fig. 4.4, b).

The displacements within the element can be written in terms of the three nodal displacements , , and as 1q 2q 3q

( ) ( ) ( ) 332211 qrNqrNqrNu ++= , (4.25)

where the quadratic shape functions iN ( )3 2 1 ,,i = are

( ) ( ) ( ) ( )

( ) ( )( ).rrrN

,rrrN,rrrN

−+=

+=−−=

11

121 1

21

3

21 (4.26)

The graphs of the shape functions (4.26) are shown in Fig. 4.5. They have a unit value at the node with the same index and zero at the other nodes. This is a general property of the shape functions.


The expressions for these shape functions can be written down by inspection. For example, since 01 =N at 0=r and 1=r , we know that has to contain the product , i. e. the left hand part of the equations of the vertical lines passing through nodes 3 and 2. That is, is of the form . The constant C is obtained from the condition

1N( rr −1 )

1N ( )rrCN −= 11

11 =N at 1−=r , which yields 21−=C , resulting in the expression given in (4.26).

Fig. 4.5

The displacement field within the element is written in matrix form as

, (4.27) ⎣ ⎦ { e

iii qNqNu ==∑

=

3

1

}where

⎣ ⎦ ⎣ ⎦321 NNNN = and { } { }Te qqqq 321= . (4.28)

At any point within an element the axial displacement can be found by multiplying the matrix of shape functions by the vector of nodal displacements, as in (4.27). It is easy to check that 1qu = at node 1, because and

. Similarly, at node 2, and

11 =N032 == NN 2qu = 3qu = at node 3. Thus, u is a

quadratic polynomial passing through , , and (Fig. 4.6). It is obtained by interpolation, using quadratic shape functions.

1q 2q 3q


The element strain-displacement row vector ⎣ ⎦B in (4.14) is given by

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎥⎦⎥

⎢⎣⎢ −

+−−=== rrr

xrN

rN

xB

e2

221

2212

dd

dd

dd

l. (4.29)

Fig. 4.6

The element stiffness matrix (4.19) is

[ ] ⎣ ⎦ ⎣ ⎦∫+

−

=

1

1

d2

rBEBAk eTeee l

or, substituting (4.29),

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

=1688

871817

3 e

eee AEkl

. (4.30)

4.1.6 Vector of element nodal forces

Consider an axial load ( )xp , having the units of force per unit length, distributed along the bar element. The mechanical work of such a force is

. (4.31) ∫∫ ==

ee

xpuxpuW T

ll

dd

Substituting (4.11), equation (4.31) becomes

. (4.32) { } ⎣ ⎦∫=e

xpNqW TTe

l

d

It has the form


{ } { }eTe fqW = (4.33)

where the element equivalent load vector is

{ } ⎣ ⎦ ⎣ ⎦∫∫+

−

==1

1

d2

d rpNxpNf TeTe

e

l

l

. (4.34)

For the linear two-node element, if the axial force is uniformly distributed, .constp = , then

{ } ⎣ ⎦∫+

−

=

1

1

d2

rNpf Tee l (4.35)


{ }⎭⎬⎫

⎩⎨⎧

=11

2ee pf l . (4.36)

A force 2

ep l , equal to half the total force on the element, is applied at each node.

For the quadratic three-node element, if .constp = , substituting the shape functions (4.26) into (4.35), gives

{ } ⎣ Te

e pf 326161l= ⎦ . (4.37)

4.1.7 Assembly of the global stiffness matrix and load vector

Assembly of the system stiffness matrix for one-dimensional structures modelled as bars is carried out as shown in sections 3.4 to 3.7 for trusses.

Consider the five-node finite element model in Fig. 4.7, a. Each node has only one degree of freedom in the x-direction. The nodal displacements are ,

,…, (Fig. 4.7, b). The global vector of nodal displacements is denoted by 1Q

2Q 5Q

{ } ⎣ ⎦TQQQQQQ 54321= .

The nodal forces are , ,…, . The global load vector is denoted by 1F 2F 5F

{ } ⎣ ⎦TFFFFFF 54321= .

The unreduced global stiffness matrix [ ]K plays the role of a proportionality factor between the two global vectors


{ } [ ] { }QKF = .

Fig. 4.7

The assembly of [ ]K from the element stiffness matrices can be explained by an energy approach. Consider the strain energy in, say, element 3. We have

{ } [ ] { }3333 2

1 qkqUT

=

or

⎣ ⎦⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−=

4

3

3

3433 1 1

11 21

QQEAQQU

l.

We can write as 3U

⎣ ⎦

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−=

5

4

3

2

1

3

3

3

33

3

3

3543213

00000

000

000

000 0000000

21

QQQQQ

EAEA

EAEAQQQQQU

ll

ll

or

{ } [ ] { }QK~QU T 33 2

1= ,


where [ ]3K~ is the expanded stiffness matrix of element 3. It has the size of the

global stiffness matrix, but the elements of the matrix [ ]3k occupy the third and

fourth rows and columns of the [ ]K matrix.

The strain energy of the entire structure is equal to the sum of the element strain energies

{ } [ ]{ } { } [ ] { }QK~QQK~QUUe

eT

e

eT

ee ∑∑∑ ===

21

21

so that, as in (3.27), the global stiffness matrix is equal to the sum of the expanded element stiffness matrices

[ ] [ ]∑=e

eK~K .

This is a convenient algebraic explanation of the assembly process, which is never done in practice. The entries of the element matrices [ ]ek are placed in

the appropriate locations of the global [ ]K matrix by the so-called direct method, based on element connectivity. Overlapping elements are simply added as already shown in section 3.4 for truss elements. This is based on the simple addition of element strain energies.

The element matrices can be written

[ ] ,21AEk

21

1 111

1

11⎥⎦

⎤⎢⎣

⎡−

−=

l

[ ] , 1 111

2

22

32AEk

3 2

⎥⎦

⎤⎢⎣

⎡−

−=

l

[ ] , 1 111

3

33

43AEk

4 3

⎥⎦

⎤⎢⎣

⎡−

−=

l

[ ] . 1 111

4

44

54AEk

5 4

⎥⎦

⎤⎢⎣

⎡−

−=

l

At the top and on the right of the element stiffness matrices, the numbering of coordinates in the global stiffness matrix is shown, according to the connectivity Table 4.1. Table 4.1

Node Element i j 1 1 2 2 2 3 3 3 4 4 4 5


The result is the following unreduced global stiffness matrix

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−+−

−+−

−+−

−

=

4

4

4

44

4

4

4

3

3

3

33

3

3

3

2

2

2

22

2

2

2

1

1

1

11

1

1

1

000

00

00

00

000

ll

llll

llll

llll

ll

AA

AAAA

AAAA

AAAA

AA

EK .

The global load vector is assembled as

{ } ⎣ ⎦TFFRF 521 00−= .

The final finite element equations are obtained, as shown in section 3.5, using the boundary condition, 01 =Q . The reduced (non-singular) stiffness matrix is obtained deleting the first row and column of the unreduced matrix, and the reduced load vector - by deleting the first element. In fact, this information is stored for the subsequent calculation of the reaction . Stresses are then calculated from the axial forces obtained using equation (3.41).

1R

4.1.8 Initial strain effects

Let an initial strain 0ε be induced in a bar element. It may arise from thermal action or by forcing members into place that are either too short or too long, due to fabrication errors.

The stress-strain law in the presence of 0ε is of the form

( )0εεσ −= E . (4.38)

The mechanical work of external nodal forces applied to suppress the initial prestressing due to 0ε is

∫∫∫ ===

eee

xAEVVW Tee

V

T

V l

ddd 000 εεεσεσ . (4.39)

Substituting (4.13), the expression (4.39) becomes


. (4.40) { } ⎣ ⎦∫=e

xBAEqW Tee

Te

l

d0ε

It has the form (4.33), where the element load vector due to initial straining is

{ } ⎣ ⎦ ⎣ ⎦∫∫+

−

==1

100 d

2d rBAExBAEf Te

eeT

eee

e

l

l

εε (4.41)

or

{ } ∫∫+

−

+

−⎭⎬⎫

⎩⎨⎧−

=⎥⎦

⎥⎢⎣

⎢=

1

1

0

1

1

0 d11

2d

dd rAEr

rNAEf e

ee

T

eee lεε ,

hence

{ } . (4.42) ⎭⎬⎫

⎩⎨⎧ −

=11

0 eee AEf ε

After solving for nodal displacements, stresses are computed as

( 012 εσ e

ee

e EqqE −+−

=l

). (4.43)

In the case of thermal loading,

Tαε =0 , (4.44)

where α is the coefficient of thermal expansion and T is the average change in temperature within the element.

4.2 Plane shaft elements

From Mechanics of Materials it is known that a uniform shaft of diameter d and length , from a material with shear modulus of elasticity G, acted upon by a

torque will twist an angle

l

tMp

tIG

M l=θ , where

32π 4d

I p = is the polar second

moment of area of the shaft cross section. The shaft torsional stiffness is then

l

pt IGMK ==θ

.

A two-node shaft finite element, of length and torsional rigidity , is

shown in Fig. 4.8. The nodal torques and can be related to the nodal rotation angles

l pIG

1M 2M

1θ and 2θ using the equilibrium and the torque/rotation equations


.KMM

,KMM

0 when

0 when

1221

2121

=−=−=

==−=

θθ

θθ (4.45)

Equations (4.45) may be written in matrix form as

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−=

⎭⎬⎫

⎩⎨⎧

2

1

2

1θθ

KKKK

MM

(4.46)

or in shorthand form

{ } [ ] { }eee kM θ= ,

where

[ ] ⎥⎦

⎤⎢⎣

⎡−

−=

1111

l

pe IGk (4.47)

is the stiffness matrix of the shaft element.

Fig. 4.8

The derivation of the shaft stiffness matrix is essentially identical to the derivation of the stiffness matrix for an axially loaded bar element. Similarity between these two derivations occurs because the differential equations for both problems have the same mathematical form. The differential equation for torsional displacement is

p

tIG

Mx=

ddθ , (4.48)

while for the axial displacement is (4.3)

AE

Nxu=

dd . (4.49)

The rotation angle of an arbitrary section within the shaft element can be expressed in terms of the nodal rotations 1θ and 2θ as

( ) ( ) 2211 θθθ rNrN += . (4.50)


where

( ) ( rrN −= 121

1 ) and ( ) ( )rrN += 121

2 (4.51)

are the shape functions of the shaft element, the same as for the axially loaded bar element.

Analogous to equation (4.20), the stiffness matrix for the shaft element can be calculated from

[ ] ∫ ⎥⎦

⎥⎢⎣

⎢⎥⎦

⎥⎢⎣

⎢=

e

xxN

xNIGk

T

epee

l

ddd

dd (4.52)

or

[ ] ∫+

−

⎥⎦

⎥⎢⎣

⎢⎥⎦

⎥⎢⎣

⎢=

1

1

ddd

dd2

rrN

rNIG

kT

e

epee

l (4.53)

which yields

[ ] ⎥⎦

⎤⎢⎣

⎡−

−=

1111

e

pee eIG

kl

(4.54)

Equation (4.54) is also used to account for torsional effects in grid finite elements, as equation (4.22) is used to account for axial effects in inclined beam finite elements. For non-axially-symmetric cross sections, the polar second moment of area is replaced by the torsional constant . pI tI

Example 4.1 Consider the bar in Fig. E4.1 with mm5=d , m20.=l , ,

loaded by an axial load . Determine: a) the displacement of point 2, b) the stresses in bar, and c) the support reactions.

GPa200=EkN2=F

Fig. E4.1

Solution. a) The bar is divided into three bar finite elements.


The areas of the cross sections are

22

1 mm 62194 .dA ==

π , ( ) 22

32 mm 47384

41 .d,AA ===π .

The element stiffness matrices are

[ ]mmN

1 111

10819 1 111

4006219102 3

51

⎥⎦

⎤⎢⎣

⎡−

−⋅=⎥

⎦

⎤⎢⎣

⎡−

−⋅⋅= ..k ,

[ ] [ ]mmN

1 111

1047381 111

2004738102 3

532

⎥⎦

⎤⎢⎣

⎡−

−⋅=⎥

⎦

⎤⎢⎣

⎡−

−⋅⋅== ..kk .

The global unreduced stiffness matrix is

[ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−+−

−+−−

=

473847380047384738473847380

04738473881981900819819

103

......

......

K .

Including the boundary conditions (degrees of freedom 1 and 4 are fixed), the finite element equations can be written

.

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−⋅

−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

−−−

4

3

1

3

23

1020

0

0

47384738004738947647380

04738284881900819819

10

R

R

QQ

.....

.....

Retaining only the second and third equation yields

⎭⎬⎫

⎩⎨⎧

⋅=

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−3

3

23

1020

9476473847382848

10QQ

....

with solutions

, mm42342 .Q = mm21433 .Q = .

b) The reactions are obtained from the first and fourth equation

, N8337423410819819 321 ...Q.R =⋅⋅==

. N2166221431047384738 334 ...Q.R =⋅⋅==

c) Stresses are


252

1

121212 mm

N217400

42341022

..QEQQEE =⋅==−

==ll

εσ ,

2523

2

232323 mm

N798200

798102 ..QQEQQEE =⋅=−

=−

==ll

εσ ,

253

3

343434 mm

N243200

2143102 ..QEQQEE −=⋅−=−=−

==ll

εσ .

Example 4.2 A support, consisting of a steel bar 1 fitted inside a cast iron tube 2, is loaded by a axial force, as in Fig. E4.2. Consider ,

, ,

kN60 21 mm 200=A

22 mm 800=A GPa2101 =E GPa1202 =E , m20.=l . Determine: a) the

elongation of the assembly, b) the stress in each material, and c) the internal forces in bar and tube.

Fig. E4.2

Solution. a) Using a two-node two-element finite element model, the element stiffness matrices are

[ ]mmN

1 111

1012 1 111

2002001012 5

51

⎥⎦

⎤⎢⎣

⎡−

−⋅=⎥

⎦

⎤⎢⎣

⎡−

−⋅⋅= ..k ,

[ ]mmN

1 111

10841 111

2008001021 5

52

⎥⎦

⎤⎢⎣

⎡−

−⋅=⎥

⎦

⎤⎢⎣

⎡−

−⋅⋅= ..k .

The global unreduced stiffness matrix is

[ ]mmN

1 111

10968412. 8412.8412.8412.

10 55⎥⎦

⎤⎢⎣

⎡−

−⋅=⎥

⎦

⎤⎢⎣

⎡+−−−−+

= .....

K .


If point 1 is fixed, the finite element equations are

, ⎭⎬⎫

⎩⎨⎧

⋅−=

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−⋅ 4

1

2

5

1060

1111

1096R

Q.

mm08702 .Q −= ,

which is equal to the beam shortening.

b) Stresses are

252

112

11

11

mmN391

20008701012 ...QEQQEE −=

−⋅==

−==

llεσ ,

252

212

22

22

mmN1452

20008701021 ...QEQQEE −=

−⋅==

−==

llεσ .

c) Internal forces are

, kN261820039111

1 ..AN −=⋅−==σ

. kN7441800145222

2 ..AN −=⋅−== σ

Example 4.3

The temperature of the bar in Fig. E4.3 is raised . If

and , determine the thermal stresses.

C080 GPa130=E61017 −⋅=α

Fig. E4.3

Solution. We should first determine whether contact occurs between the bar and the wall. If the wall does not exist, the displacement of point 2 is

mm2mm44828010171800 6 >=⋅⋅⋅== − .TαΔ ll

so that the contact does occur.

Denoting by A the cross section area, the temperature forces (4.42) are


{ } . N11

817611

103180101711 56

⎭⎬⎫

⎩⎨⎧ −

⋅=⎭⎬⎫

⎩⎨⎧ −

⋅⋅⋅⋅⋅=⎭⎬⎫

⎩⎨⎧ −

= − A.A.AETfT α

The stiffness matrix is

[ ]mmN

1 111

18001031 5

⎥⎦

⎤⎢⎣

⎡−

−⋅⋅=

A.k .

As point 1 is fixed, the finite element equations are

⎭⎬⎫

⎩⎨⎧

+−

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−⋅⋅A.RA.RA.

81768176

20

1111

18001031

2

15

wherefrom we get the reactions

A.A.A.R 432414481761 =−= , A.R 4322 −= ,

so that the normal stress is

22

mmN432.

AR

−==σ .

Example 4.4 A prismatic bar is made of two different materials, as shown in Fig. E4.4. The temperature of the central part is raised . Determine the axial stress in the bar.

C0T

Fig. E4.4

Solution. The bar is divided into three linear finite elements. The element stiffness matrices are

[ ] [ ] 1 111 231⎥⎦

⎤⎢⎣

⎡−

−==

l

AEkk , [ ] 1 111

212

⎥⎦

⎤⎢⎣

⎡−

−=

l

AEk .

For element 2, the vector of nodal thermal forces is

{ } ⎣ ⎦TAETf 1112 −= α .


Using the boundary conditions 041 ==QQ , the finite element equations can be written

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−+−

−+−

−

4

1

1

1

3

2

22

2211

1122

22

0

0

0022

0

022

00

RAETAET

R

QQ

EE

EEEE

EEEE

EE

Aαα

l.

The second and third equation yield

⎭⎬⎫

⎩⎨⎧ −

=⎭⎬⎫

⎩⎨⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−

−+

11

22

221

3

2

12

1

112

AETQQ

EEE

EEEA αl

with solutions

21

12 EE

ETQ+

−= lα , 21

13 EE

ETQ+

= lα .

The strains in elements are

21

14231

EEETQQ+

−=−=== αεεll

, 21

1232 2 EE

ETQQ+

=−

= αεl

.

Stresses are

21

12

3111

1

EE

TE+

−=== αεσσ , 1

21

12

12

111 σααεσ =+

−=−=

EE

TTEE .

Example 4.5 A steel bolt of active length mm100=l and diameter mm10=δ is single threaded with a pitch. It is mounted inside a copper tube with diameters

and (Fig. E4.5, a). After the nut has been fitted smugly, it is tightened one-quarter of a full turn. Determine stresses in bolt and tube, if for steel and for copper

mm61.mm12=d mm18=D

GPa2081 =E GPa1002 =E .

Solution. The assembly is modeled by two bar finite elements as in Fig. E4.5, b. Both elements have fixed ends at points 1 and 4 so that 041 ==QQ .

The problem has a multipoint constraint


mm4023 .QQ =− .

Such conditions are programmed using a so-called penalty approach. Herein a simpler straightforward solution is given.

Fig. E4.5

The cross section areas are

22

1 mm 54784 .A ==δπ , ( ) 2

22

2 mm 371414

.dDA =−

=π .


[ ]mmN

1 111

10631 1 111

100547810082 5

51

⎥⎦

⎤⎢⎣

⎡−

−⋅=⎥

⎦

⎤⎢⎣

⎡−

−⋅⋅= ...k ,

[ ]mmN

1 111

104111 111

1003714110 5

52

⎥⎦

⎤⎢⎣

⎡−

−⋅=⎥

⎦

⎤⎢⎣

⎡−

−⋅= ..k .

The finite element equations are

.

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

4

1

3

25

0

0

4114110041141100

0063163100631631

10

RFF

R

QQ

....

....

Retaining only the second and third equation yields

⎭⎬⎫

⎩⎨⎧−−

=⎭⎬⎫

⎩⎨⎧−⎥

⎦

⎤⎢⎣

⎡FF

QQ

..

3

25

41100631

10

with solutions


52 10631 ⋅−=

.FQ , 53 10411 ⋅

=.

FQ .

Substituting in the multipoint constraint condition gives

4010411

110631

155 .

..F =⎟

⎠⎞

⎜⎝⎛

⋅+

⋅,

or N30314=F .

Stresses are given by

21

1 mmN386

547830316

===.A

Fσ ,

22

2 mmN5214

3714130316 .

.AF

−=−=−=σ .

Example 4.6 For the bar of Fig. E4.6, with linearly variable cross-section

( ) ⎟⎠⎞

⎜⎝⎛ +=

l210

xAxA , find the displacement at the free end under the action of force

F, using two tapered bar finite elements.

Fig. E4.6

Solution. The stiffness matrix for a bar element with variable cross section is

[ ] ∫⎥⎦⎤

⎢⎣

⎡−

−=

e

xAEke

e

ll

d1 111

2 .

For the bar of Fig. E4.6, divided into two equal length tapered elements, the element stiffness matrices are


[ ] ⎥⎦

⎤⎢⎣

⎡−

−=⎟

⎠⎞

⎜⎝⎛ +⎥

⎦

⎤⎢⎣

⎡−

−= ∫ 1 1

11 4

5d2

11 111 0

0

021

lll

l

AExxAEk ,

[ ] ⎥⎦

⎤⎢⎣

⎡−

−=⎟

⎠⎞

⎜⎝⎛ +⎥

⎦

⎤⎢⎣

⎡−

−= ∫ 1 1

11 4

7d2

11 111 0

2

022

lll

l

l

AExxAEk .

With 0 , the finite element equations can be written 1 =Q

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−

F

R

QQEA 00

7707125

055

4

1

3

20l

.

From the second and third equation

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−FQ

QEA 077712

4 3

20l

,

wherefrom

0

2 54

AEFQ l

= , 0

3 3548

AEFQ l

= .

The approximate assumed linear variation of the displacement field yields constant strain elements, hence a stepwise variation of stresses along the bar.

Example 4.7

Write the stiffness matrix of the bar shown in Fig. E4.7, condensing from condition . Model the bar by: a) two two-node linear elements; b) a three-node quadratic element. Comment the results.

3Q03 =F

Fig. E4.7

Solution. a) Consider the bar divided into two linear elements. The element stiffness matrices are


[ ] [ ] 1 111 221⎥⎦

⎤⎢⎣

⎡−

−==

l

EAkk .

The unreduced global stiffness matrix is

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−=

110121

0112l

EAK .

The finite element equations can be written

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

3

2

1

3

2

1

211110101

2

FFF

QQQ

EAl

.

For 03 =F , the last equation yields

⎣ ⎦⎭⎬⎫

⎩⎨⎧

=+

=2

1213 2121

2 QQQQQ

which assumes a linear displacement field within the bar.

The first two equations give

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧−−

+⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡

2

13

2

1

112

10012

FF

QEAQQEA

ll

and upon substitution of 3Q

⎣ ⎦⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧−−

+⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡

2

1

2

1

2

1 2121112

10012

FF

QQEA

QQEA

ll

which can be written

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−

2

1

2

1

1111

FF

QQEA

l

where the left hand side contains the ( )22× stiffness matrix of the two-node linear element.

b) Consider the bar modeled by a 3-node quadratic bar element. The finite element equations can be written


⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

3

2

1

3

2

1

1688871817

3FFF

QQQ

EAl

.

Condensing from condition 3Q 03 =F yields

⎣ ⎦⎭⎬⎫

⎩⎨⎧

=+

=2

1213 2121

2 QQQQQ .

The first two equations give

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧−−

+⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡

2

13

2

1

88

37117

3 FF

QEAQQEA

ll

and upon substitution of 3Q

⎣ ⎦⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧−−

+⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡

2

1

2

1

2

1 212188

37117

3 FF

QQEA

QQEA

ll

which can be written again

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−

2

1

2

1

1111

FF

QQEA

l,

where the matrix in the left hand can be recognized as the conventional bar matrix developed from linear polynomials.

c) Comments. Substituting

⎭⎬⎫

⎩⎨⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

2

1

3

2

1

21211001

QQ

QQQ

into the expression of the strain energy (4.18) gives

⎣ ⎦⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

=

3

2

1

321

1688871817

321

QQQ

EAQQQUel

,

⎣ ⎦⎭⎬⎫

⎩⎨⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

⎥⎦

⎤⎢⎣

⎡=

2

121

21211001

1688871817

321102101

21

QQEAQQUe

l,


⎣ ⎦⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−=

2

121 11

1121

QQEAQQUe

l.

For .constp = , the work of nodal forces is

⎣ ⎦ ⎣ ⎦ ee pQQQ

FFF

QQQW l⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

326161

321

3

2

1

321 ,

⎣ ⎦ ⎣ ⎦ 211

326161

21102101

2121e

ee pQQpQQW l

l⎭⎬⎫

⎩⎨⎧

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎦

⎤⎢⎣

⎡= .

We may conclude that the introduction of the quadratic term in equation (4.23) does not bring about a change in the conventional stiffness matrix and load vector. Whenever the assumed functions, used to describe the displacement field, form the complete homogeneous solution of the differential equation of equilibrium (4.4), the developed stiffness matrix and the equivalent load vector will be exact. This is because, as it is shown in a next chapter, only the homogeneous part of the solution contains the free parameters with respect to which the total potential energy is minimized. The parameters in the particular part of the solution are prescribed and do not take part in the process of minimization.

The exactness of the stiffness matrix and load vector also implies that the computed nodal displacements will also be exact. However, displacements within elements depend upon the general (homogeneous plus the particular) solution. The conventional formulation based on a linear polynomial will yield exact displacements within the elements only when 0=p . For the case .constp = , exact displacements within the elements may be obtained from equation (4.25). However, before using equation (4.25), the variable must be computed from the exact nodal displacements (computed for the conventional linear element), via a constraint equation between and the remaining nodal variables, obtained by a minimization of the total potential energy with respect to at element level.

3Q

3Q

3Q

Example 4.8 Consider a prismatic robot arm, rotating at constant angular velocity

secrad30=ω (Fig. E4.8, a). Determine the axial stress distribution due to the centrifugal force in the rod using: a) two quadratic elements, and b) three linear elements.

Solution. The beam is acted upon by a centrifugal linearly distributed load


( )l

xpxp 0= ,

where . 20 ωρ lAp =

Fig. E4.8


a) A finite element model with two quadratic elements is shown in Fig. E4.8, b. The model has five degrees of freedom.


[ ] [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−==

7818168

187

3221

l

EAkk .


[ ]

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−+−−−

−

=

78100816800

18778100816800187

32l

EAK .

The axial load acting on the two elements can be decomposed as in Fig. E4.8, c. The nodal forces, equivalent to a linearly distributed load, are given by equation (4.34)

{ } ⎣ ⎦( )( )

( )( ) ⎪

⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−−−−−

== ∫∫3161

0d

1421

121d 0

1

0

0 eeTe ppxpNf

e

ll

l

ξξξξξξξξ

,

where ex l=ξ and the shape functions (4.26) are defined for convenience over an interval . For a uniformly distributed load they are given by equation (4.37). [ ]10,

The distributed loads from Fig. E4.8, c are replaced by the kinematically equivalent nodal forces shown in Fig. E4.8, d. The element nodal load vectors are

{ } ⎣ ⎦Tpf 211012

01 l= , { } ⎣ ⎦Tpf 1321

1202 l

= .

Using the boundary condition 01 =Q , the finite element equations can be written

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

+=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−−−

−

12123

242412

0

78100816800

18148100816800187

32

0

0

00

0

1

5

4

3

2

l

l

ll

l

l

pp

ppp

R

QQQQ

EA .


Deleting the first row and column, we obtain the reduced global stiffness matrix and reduced global load vector which yield

AE

pQ2

02 384

47 l= ,

AEpQ

20

3 38488 l

= , AE

pQ2

04 384

117 l= ,

AEpQ

20

5 384128 l

= .

The nodal displacements can be compared with the exact solution, given by

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−= 3

320

32 ll

l xxAE

pxu . (a)

We obtain

AE

p2384

474

20ll

=⎟⎠⎞

⎜⎝⎛u ,

AEp248

112

20ll

=⎟⎠⎞

⎜⎝⎛u ,

AEp2384

11743 2

0ll=⎟

⎠⎞

⎜⎝⎛u , ( )

AEp23

1 20ll =u .

The finite element values of the nodal displacements are exact. This is due to the nodal equivalence of forces. While the nodal displacements are exact, the displacements within the elements are approximate because the exact cubic distribution (a) has been replaced by a quadratic law.


⎣ ⎦ ⎣ ⎦ ⎣ ⎦ξξξ 8441431dd

−+−+−==l

Nx

B .

The strains are calculated as

( ) ( )ξε 62548

2 02332

1 −=′+′=EA

pQNQN l

l,

( ) ( )ξε 181948

2 0435231

2 −=′+′+′=EA

pQNQNQN l

l,

which yields the following nodal values

EAp l0

1 4825

=ε , EAp l0

2 4822

=ε , EAp l0

3 4819

=ε , EAp l0

4 4810

=ε , EAp l0

5 481

=ε .

The corresponding axial stresses are

A

p l01 48

25=σ ,

Ap l0

2 4822

=σ , A

p l03 48

19=σ ,

Ap l0

4 4810

=σ , A

p l05 48

1=σ .

The stresses can be compared with the exact solution


( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−= 2

20 1

2 l

l xA

pxσ , (b)

which yields the following nodal values

( )A

p l048240 =σ ,

Ap. ll 0

48522

4=⎟

⎠⎞

⎜⎝⎛σ ,

A

p ll 04818

2=⎟

⎠⎞

⎜⎝⎛σ ,

Ap. ll 0

48510

43

=⎟⎠⎞

⎜⎝⎛σ , ( ) 0=lσ .

A graph of the stress distribution is shown in Fig. E4.8, e.

b) A finite element model comprised of three linear elements is shown in Fig. E4.8, f. The model has four degrees of freedom.


[ ] [ ] [ ] 1 111 3321⎥⎦

⎤⎢⎣

⎡−

−===

l

EAkkk .

The axial load acting on the three elements can be decomposed as in Fig. E4.8, g. The nodal forces, equivalent to a load per unit length, are given by equation (4.34). If and are the intensities of a linearly distributed load at nodes 1 and 2, respectively,

1p 2p

( ) ( )ξξ 121 pppp −+= ,

{ } ⎣ ⎦ ( )[ ]⎭⎬⎫

⎩⎨⎧

++

=−+⎭⎬⎫

⎩⎨⎧ −

== ∫∫ 21

21121

1

02

26

d1

dpppp

pppxpNf ee

Te

e

ll

l

ξξξξ

,

where ex l=ξ and the shape functions (4.26) are defined for convenience over an interval . For , the nodal forces are given by equation (4.36). [ 10, ] ppp == 21

The distributed loads from Fig. E4.8, g are replaced by the kinematically equivalent nodal forces shown in Fig. E4.8, h. The element nodal load vectors are

{ } ⎣ ⎦Tpf 2154

01 l= , { } ⎣ ⎦Tpf 54

5402 l

= , { } ⎣ ⎦Tpf 8754

03 l= .

Using the boundary condition 01 =Q , the finite element equations can be written


⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧ +

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

−−−

8126

541

54

0

11001210

01210011

3 0

1

0

4

3

2l

l

l

pR

p

QQQEA .

The last three equations yield

AE

pQ2

02 81

13 l= ,

AEpQ

20

3 8123 l

= , AE

pAE

pQ2

02

04 3

18127 ll

== .


⎣ ⎦ ⎣ ⎦ ⎣ ⎦111dd

−==e

Nx

Bl

.

The strains are calculated as

EApQ l

l0

21

27133

==ε , ( )EApQQ l

l0

232

27103

=−=ε , ( )EApQQ l

l0

343

2743

=−=ε ,

and are constant within each element.

The corresponding axial stresses are

A

p l01

2713

=σ , A

p l02

2710

=σ , A

p l03

274

=σ .

In Fig. E4.8, i they are compared to those given by equation (b).

Example 4.9 Show that the shape functions of the four-node cubic bar element, in local natural coordinates, are the following

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) .rrrrN

,rrrrN

,rrrrN

,rrrrN

+⎟⎠⎞

⎜⎝⎛ −⎟

⎠⎞

⎜⎝⎛ +−=

⎟⎠⎞

⎜⎝⎛ +−+=

⎟⎠⎞

⎜⎝⎛ −−+=

⎟⎠⎞

⎜⎝⎛ −⎟

⎠⎞

⎜⎝⎛ +−−=

131

31

169

3111

1627

3111

1627

31

311

169

4

3

2

1

5. BEAMS, FRAMES AND GRIDS

Frames are structures with rigidly connected members called beams. Beams are slender members used to support transverse loading. They are connected by rigid joints that have determinate rotations and, apart from forces, transmit bending moments from member to member.

One-dimensional mathematical models of structural beams are constructed on two beam theories: the Bernoulli-Euler beam theory, that neglects transverse shear deformations, and the Timoshenko beam theory, that incorporates a first order correction for transverse shear effects. Beam behaviour is described by fourth order differential equations and require 1C continuity. This requires that both transverse displacements and slopes must be continuous over the entire member and, in particular, between adjacent beam elements. The Timoshenko beam model pertains to the class of 0C elements. It is based on the assumption that plane sections remain plane but not necessarily normal to the deformed neutral surface. This leads to the introduction of a mean shear distortion, which is constant over the element.

In this section, we first present the finite element formulation for plane Bernoulli-Euler beams, then extend it to plane frames, and for grids. An inclined beam element will be referred to as a frame element. Grids are planar frames subjected to loads applied normally to their plane.

5.1 Finite element discretization

A plane frame is divided into elements, as shown in Fig. 5.1. Each node has three degrees of freedom, two linear displacements and a rotation. Typically, the degrees of freedom of node i are 23 −iQ , 13 −iQ and iQ3 , defined as the displacement along the X axis, the displacement along the Y axis and the rotation about the Z axis, respectively.


Nodes are located by their coordinates in the global reference frame XOY and element connectivity is defined by the indices of the end nodes. Elements are modelled as uniform beams without shear deformations and not loaded between ends. Their properties are the bending rigidity IE and the length l .

Fig. 5.1

In the following, the shape functions are established for the plane Bernoulli-Euler beam element, then the element stiffness matrix is calculated first in the local coordinate system, then in the global coordinate system. The latter are expanded to the structure size, then simply added to get the global uncondensed stiffness matrix. Imposing the boundary conditions, the reduced stiffness matrix and load vector are calculated and used in the static analysis.

Fig. 5.2

5. BEAMS, FRAMES AND GRIDS 81

Consider an inclined beam element, as illustrated in Fig. 5.2, a, where the nodal displacements are also shown. In a local physical coordinate system, the x axis, oriented along the beam, is inclined an angle θ with respect to the global X axis. Alternatively, an intrinsic (natural) coordinate system can be used.

The vector of nodal displacements in the local coordinate system is

{ } ⎣ ⎦Te qqqqqqq 654321= (5.1)

and the corresponding vector of element nodal forces can be written

{ } ⎣ ⎦Te fffffff 654321= . (5.2)

Forces 6532 f,f,f,f and the corresponding displacements

6532 q,q,q,q describe the element bending (Fig. 5.2, b) while axial forces ,f,f 41 and axial displacements ,q,q 41 describe the element stretching (Fig.

5.2, c). Their action is decoupled so that the respective stiffness matrices can be calculated separately.

5.2 Static analysis of a uniform beam

Beams with cross sections that are symmetric with respect to the plane of loading are considered herein (Fig. 5.3). Transverse shear deformations are neglected, as in the Bernoulli-Euler classical beam theory. Only transverse loads act upon the beam, axial forces are ignored.

Fig. 5.3

The axial displacement of any point on the section, at a distance y from the neutral axis, is approximated by

yyxvu

dd

−=−= ϕ , (5.3)


where v is the deflection of the centroidal axis at x and v′=ϕ is the cross section rotation (or slope) at x . Axial strains are

yχy −=−== 2x xv

xu

dd

dd 2

ε , (5.4)

where v ′′≈χ denotes the deformed beam axis curvature.

Normal stresses on the cross section are given by Hooke’s law

yEE 2xx xv

dd2

−== εσ , (5.5)

where E is Young’s modulus of the material.

The bending moment is the resultant of the stress distribution on the cross section

( ) χσ zzA

IEIEAyM ==−= ∫ 2x xvx

ddd

2 (5.6)

where zI is the second moment of area of the section with respect to the neutral axis z. The negative sign above is introduced because M is considered positive if it compresses the upper portion of the beam cross section. The product zIE is called the bending rigidity of the beam.

The shear force is given by

( ) III3 vxv

xx zz IEIEMT −=−=−=

dd

dd 3

. (5.7)

The transverse load per unit length is

( ) IV4 vxv

xx zz IEIETp ==−=

dd

dd 4

. (5.8)

The differential equation of equilibrium is

( )xpIE z =4xv

dd4

. (5.9)

This is a fourth order differential equation and consequently four boundary conditions are required, two at each end. They can be geometric or kinematic boundary conditions, involving the transverse displacement and slope, and physical boundary conditions, involving the shear and bending moment.


5.3 Uniform beam not loaded between ends

For a uniform beam not loaded between ends, 0=p and equation (5.9)

yields 0dd 44 =xv . Integrating four times, we obtain the deflection v described by a third order polynomial

( ) 432

23

1 aaaa +++= xxxxv . (5.10) For a free-free beam element, the four integration constants 4321 ,,, aaaa

in (5.10) can be determined from the geometric boundary conditions at each end (Fig. 5.4):

at 1xx = , 21 q== vv , and 31dd q==ϕxv ; (5.11, a)

at 2xx = , 52 q== vv , and 62dd q==ϕxv . (5.11, b)

Fig. 5.4

This gives

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⋅

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

4

3

2

1

222

222

32

121

121

31

2

2

1

1

0123101231

aaaa

xxxxx

xxxxx

ϕ

ϕv

v

.

On inversion, the integration constants 4321 ,,, aaaa can be expressed in

terms of the nodal displacements 2211 ϕϕ ,,, vv , so that the beam deflection can be expressed in terms of the nodal displacements.

5.3.1 Shape functions

The transverse displacement v can be expressed in terms of the nodal displacements as


( ) ⎣ ⎦ { }eqN=xv , (5.12)

where ⎣ ⎦N is a row vector containing the shape functions, called Hermitian cubic polynomials, and

{ } ⎣ ⎦ Teq 2211 ϕϕ vv= . (5.13)

Using natural coordinates, with 1−=r at node 1 and 1+=r at node 2,

⎟⎟⎠

⎞⎜⎜⎝

⎛ +−

−=

22 21

12

xxx

xxr (5.14)

the beam transverse displacement can be written

( ) ( ) ( ) ( ) ( )2

4231

211 dd

dd

⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

rNN

rNN vrvrvrvrrv . (5.15)

Because the coordinates transform by the relationship (5.14)

rxxxx

x22

1221 −+

+=

and since 12 xxe −=l is the length of the element,

rdxd e2l

= . (5.16)

Using the differentiation chain rule

xvv

dd

2dd e

rl

= , (5.17)

equation (5.15) becomes

( ) ( ) ( ) ( ) ( )2

4231

211 dd

2dd

2 ⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

xNN

xNN ee vrvrvrvrrv ll (5.18)

or

( ) 64533221 22qNqNqNqN ee ⋅+⋅+⋅+⋅=

llrv . (5.19)

In (5.12) the row vector of shape functions is

⎣ ⎦ ⎥⎦

⎥⎢⎣

⎢= 4321 22NNNNN ee ll

. (5.20)


The Hermitian shape functions are cubic polynomials which should satisfy the boundary conditions given in Table 5.1, where primes indicate differentiation with respect to r.

Table 5.1

1N 1N ′ 2N 2N ′ 3N 3N ′ 4N 4N ′

1−=r 1 0 0 1 0 0 0 0 1+=r 0 0 0 0 1 0 0 1

Imposing the above conditions to cubic polynomials with four arbitrary constants, we obtain the expressions of the beam element shape functions in terms of r, graphically shown in Fig. 5.5

( ) ( ) ( ) ( )321 32

4121

41 rrrrrN +−=+−= ,

( ) ( ) ( ) ( )3222 1

4111

41 rrrrrrN +−−=+−= , (5.21)

( ) ( ) ( ) ( )323 32

4121

41 rrrrrN −+=−+= ,

( ) ( ) ( ) ( )3224 1

4111

41 rrrrrrN −−+−=−+−= .

Fig. 5.5


It is easy to check that at node 1, 2q=v and 32dd q

rel=

v , while at node 2,

5q=v and 62dd q

rel=

v .

5.3.2 Stiffness matrix

The strain energy eU of a beam element is

xIEU

e

ee d

dd

2

2

2

2

∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛=

xv . (5.22)

Equation (5.17) yields

re d

d2dd vxv

l= and 2

2

2

2

dd4

dd

re

vxv2 l= .

Substituting (5.12) we obtain

{ }e

eq

rN⎥⎥⎦

⎥

⎢⎢⎣

⎢= 2

2

2

2

dd4

dd

l2xv . (5.23)

The square of the above quantity is calculated as

{ } { }eT

e

TeT

qrN

rNq

⎥⎥⎦

⎥

⎢⎢⎣

⎢

⎥⎥⎦

⎥

⎢⎢⎣

⎢=⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛2

2

2

2

4

2222

dd

dd16

dd

dd

dd

l222 xv

xv

xv ,

which can also be written

{ } ⎣ ⎦ ⎣ ⎦ { }er

Tr

e

Te qNNq ′′′′=⎟⎟⎠

⎞⎜⎜⎝

⎛4

22 16dd

l2xv . (5.24)

On substituting (5.16) and (5.24) into (5.22) we get the element strain energy

{ } ⎣ ⎦ ⎣ ⎦ { }er

Tr

e

eTee qrdNNIEqU ∫

+

−

′′′′=1

13

821

l (5.25)

which has the form


{ } [ ] { }eeB

Tee qkqU

21

= . (5.26)

Comparing (5.25) with (5.26) we obtain the element stiffness matrix due to bending

[ ] ⎣ ⎦ ⎣ ⎦∫+

−

′′′′=1

13

8 rdNNEIk rT

re

eeB

l (5.27)

or

[ ]( )

( )( )

( )

r

NNNNNNNNNNNNNNNNNNNNNNNNNNNN

EIke

eeB d8

1

1 24342414

432

32313

42322

212

4131212

1

3 ∫+

−⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′′

=l

. (5.28)

Substituting the shape functions (5.21) and performing the integration yields the stiffness matrix due to bending in local coordinates

[ ]e

e

eeB

EIk⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−

=

22

22

3

4626612612

2646612612

llll

ll

llll

ll

l. (5.29)

An alternative way of deriving the matrix (5.29) is based on the general formula (4.19)

[ ] ⎣ ⎦ ⎣ ⎦∫=eV

eTe VBEBk d . (5.30)

The curvature χ (5.23) can be expressed in terms of the nodal displacements as

{ } ⎣ ⎦ { }ee

eqBq

rN

=⎥⎥⎦

⎥

⎢⎢⎣

⎢== 2

2

2

2

dd4

dd

l2xvχ , (5.31)

where the curvature-displacement row vector ⎣ ⎦B is

⎣ ⎦ ⎥⎦

⎥⎢⎣

⎢+−−= 1361361 rrrrB

eee lll. (5.32)

Substituting (5.32) and xAV e dd = into (5.30) gives the matrix (5.29).


5.3.3 Physical significance of the stiffness matrix

The element stiffness matrix (5.29) relates the vector of nodal forces to the vector of nodal displacements

ee

e

e

e qqqq

EI

ffff

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

6

5

3

2

22

22

3

6

5

3

2

4626612612

2646612612

llll

ll

llll

ll

l. (5.33)

Consider a beam element with end 2 fixed ( )065 == qq and end 1 having a unit displacement along the global Y-axis ( )0 1 32 == q,q and zero rotation, as in Fig. 5.6.

Fig. 5.6

Equation (5.33) can be written

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

====

⋅

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

0001

6

5

3

2

41

31

21

11

6

5

3

2

qqqq

...k

...k

...k

...k

ffff

(5.34)

which gives

112 kf = , 213 kf = , 315 kf = , 416 kf = . (5.35)

This shows that the first column of the stiffness matrix represents the forces and moments that must be applied to the beam element to preserve static equilibrium when 12 =q and all other displacements are zero.

For equilibrium

03111 =+ kk , 0314121 =++ lkkk . (5.36)


5.4 Uniform beam loaded between ends

For a uniform beam loaded between ends, 0≠p , 0dd 44 ≠xv in equation (5.9) and the beam deflected shape is no more a cubic polynomial. However, it is the homogeneous solution of the differential equation. Using cubic polynomials as admissible functions, the computed nodal displacements are exact. Within the elements, the displacements, moments and shear forces are in error.

When the transverse load is uniformly distributed, .constp = , the general solution of equation (5.9) is a quartic polynomial. The corresponding five constants have to be determined from five boundary conditions. An internal node added at the centre of element will solve the problem, introducing its nodal displacement as the fifth nodal coordinate.

For a linearly distributed transverse load, ( )xv will be a quintic with six arbitrary constants. They can be determined, adding the transverse displacement and slope at the element midpoint to the element nodal coordinates. As already shown in Chapter 4, rising the power of the function describing the displacement within a beam element is tantamount to introducing additional internal nodes.

However, the current practice is to use lower order approximate assumed shape functions that ensure the minimal convergence requirements, as shown in the following. The cubic shape functions do the job. But they can be used only if the element has uniform rigidity zIE and is not loaded between nodes. For beams with transverse forces, the solution is to replace the actual distributed load by equivalent nodal forces.

5.4.1 Consistent vector of nodal forces

Consider a transverse load ( )xp , having the units of force per unit length, distributed along the beam element. The mechanical work of such a force is

∫∫ ==

ee

xpxpW T

ll

dd vv . (5.37)

Substituting (5.12), equation (5.37) becomes

{ } ⎣ ⎦∫=

e

xpNqW TTe

l

d . (5.38)

It has the form

{ } { }eTe fqW = (5.39)

where the element load vector is


{ } ⎣ ⎦ ( ) ⎣ ⎦ ( )∫∫+

−

==1

1

d2

d rrpNxxpNf TeTe

e

l

l

. (5.40)

For the Hermitian two-node element, if the transverse force is uniformly distributed, .constp = , the vector of element consistent nodal forces is

{ } ⎣ ⎦∫+

−

=1

1

d2

rNpf Tee l (5.41)


{ }T

eeeee ppppf⎥⎥⎦

⎥

⎢⎢⎣

⎢−=

122122

22 llll . (5.42)

In Fig. 5.7, a it is seen that ef2 is a shear force and ef3 is a moment. They are called “kinematically equivalent” nodal forces since they replace a distributed load ( )rp weighted with the shape functions ( )rNi so that the correct work is simulated.

a b

c d

Fig. 5.7

The kinematically equivalent loads are those which, if applied in the opposite direction as constraints, would keep all nodal displacements zero in the presence of the true loading. To replace .constp= by statically equivalent forces (Fig. 5.6, b) would be incorrect since the beam element has the ends rigidly jointed,


i.e. it is rigidly built in the adjacent beam elements. In order to ensure the 1C continuity across elements, nodal forces must include moments, not only shear forces. Equivalent nodal forces for linearly distributed loads are given in Figs. 5.6, c and d.

Kinematically equivalent loads yield displacements which do not coincide with those produced by actual loading, as shown in Example 5.1. Assuming approximate deflected shapes instead of the true ones may be imagined as the result of application of a fake loading, forcing the beam to maintain the approximate deflection. This is equivalent to applying additional constraints to the beam, i.e. stiffening it. The deflections of this over-stiff finite element model are smaller “in the mean” than the true deflections of the actual structure.

The source of error comes from the arbitrary selection of the shape functions. Even if these functions are built up to satisfy the geometric boundary conditions at the ends, the equilibrium within the elements is broken, due to the difference between the applied load ( )xp and the resistance IVvzIE which gives rise to a sort of unbalanced residual force.

The smaller the element, the smaller the error, so we would expect to increase the accuracy by increasing the number of elements modeling the same structure, or refining the mesh. A correct solution will approach the true value with monotonically increasing values of displacements. The finite element solution is therefore referred to as a lower bound. This applies only to the strain energy and not to the displacement or stress at a point. Local stresses may be higher than the true ones.

Assuming cubic displacement functions implies linearly varying bending moments (and hence stresses) in uniform beams, even if it is known that, for uniform loading, for instance, they have a quadratic distribution.

Example 5.1 Calculate the transverse displacement at the centre of the simply supported

beam shown in Fig. E5.1.

Fig. E5.1


Solution. Using a single beam element, the distributed load is replaced by two concentrated moments at the ends.

Using the boundary conditions and the equivalent nodal loads, the equations of equilibrium can be written

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−

12

2

12

2

0

0

4626612612

2646612612

2

2

6

5

3

2

22

22

3

l

l

l

l

llll

ll

llll

ll

l

p

p

p

p

qq

qq

IE z .

The set of equations in the unknown displacements is

( )

( ) ,pqqEI

,pqqEI

1242

1224

2

63

2

63

l

l

l

l

−=+

=+

with solutions

IE

pqq24

3

36l

−=−= .

The displacement at the middle is

IE

pqNqN96

4

6432llll

=⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

222v .

The true solution is

IE

ptrue 384

5 4l=v

so that the approximate finite element solution is, as expected, smaller

truev2

v54

=⎟⎠⎞

⎜⎝⎛ l .

5.4.2 Higher-degree interpolation functions

Let explore the possibility of approximating the displacements of a fourth order system by a quartic function of the form


( ) 5433

24

1 axaaaa ++++= 2xxxxv . (5.43)

The five integration constants 54321 ,,, a,aaaa in (5.43) can be determined from the end displacements and slopes, and the displacement of the internal node at the centre of the element (Fig. 5.8):

at 1xx = , 1vv = , and 1dd ϕ=xv ,

at 2xx = , 2vv = , and 2dd ϕ=xv , (5.44)

at 3xx = , 3vv = .

This gives the nodal coordinates in terms of the polynomial coefficients

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⋅

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

1

2

3

4

5

432

32

432

3

2

2

1

1

168421

432101

0001000001

aaaaa

llll

lll

llll

v

v

v

ϕ

ϕ.

On inversion

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⋅

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−

−−−=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

3

2

2

1

1

43434

32323

222

1

2

3

4

5

16 2 828

32314 5 18

16 1 54110 0 0 1 0 0 0 0 0 1

v

v

v

ϕ

ϕ

lllll

lllll

lllll

aaaaa

.

The beam deflection is given by an expression of the form (5.12)

( ) ⎣ ⎦ { } ⎣ ⎦

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

==

3

2

2

1

1

54321

v

v

v

xvϕ

ϕNNNNNqN e , (5.45)

where the quartic shape functions are


( ) ( ) ( )rrrrN 23141 2

1 +−−= ,

( ) ( ) ( )lrrrrN +−−= 1181 2

2 ,

( ) ( ) ( )23 123

41 rrrrN +−= , (5.46)

( ) ( ) ( ) l24 11

81 rrrrN +−−= ,

( ) ( ) ( )225 11 rrrN +−= .

In equations (5.46), 5N is called a “bubble function”, having zero displacements and slopes at the ends.

Fig. 5.8

Substituting the shape functions (5.46) into equation (5.27) and performing the integration, we obtain the element stiffness matrix

[ ]⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−−−−−−

=

1024SYM12836

5129431612863436

5123419694316

5 2

22

3

ee

eeeee

ee

e

eeB

EIkll

lllll

ll

l. (5.47)

For .constp = , the element consistent load vector (5.41) is


{ }T

eeeeee pppppf⎥⎥⎦

⎥

⎢⎢⎣

⎢−=

158

60307

60307 22 lllll . (5.48)

5.4.3 Bending moment and shear force

Using the bending moment expression (5.6) and equation (5.12) we get

⎣ ⎦ { }e

ez

ezz qNIEIEIEM ′′=== 2

2

2

2 4dd4

dd

ll 22 rv

xv ,

( ) ( )⎣ ⎦ { }eee

e

z qrrrrIEM lll

1361362 +−−= . (5.49)

The shear force is given by equation (5.7)

⎣ ⎦ { }e

ez

ezz qNIEIEIET ′′′−=−=−= 3

3

3

3 8dd8

dd

ll 33 rv

xv ,

⎣ ⎦ { }eee

e

z qIET lll

−−−= 2263 . (5.50)

For elements with uniformly distributed load, the end equilibrium loads are given by

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−

−

−

+

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

12

2

12

2

4626612612

2646612612

2

2

6

5

3

2

22

22

3

6

5

3

2

e

e

e

e

ee

e

e

e p

p

p

p

qqqq

EI

RRRR

l

l

l

l

llll

ll

llll

ll

l. (5.51)

The first term on the right is [ ]{ }eeB qk . The second term consists of

elements that are called fixed-end reactions.

The shear forces at the two ends of the beam element are 21 RT −= and

52 RT = . The end bending moments are 31 RM −= and 62 RM = . For 0=p , they are obtained substituting 1−=r and 1+=r in (5.49) and (5.50).

When .constp = , the exact moment and shear force within the element are

( ) ( )222

142

12212

++++−= rprppMM eeeep

llll ,


rpTT ep 2

l−= ,

where M and T are the expressions (5.49) and (5.50), respectively, valid for 0=p .

5.5 Basic convergence requirements

As element sizes are reduced, the sequence of solutions to a problem is expected to converge to the correct result if the assumed element displacement fields satisfy the following criteria:

1. An element should describe rigid body modes exactly.

Although equilibrium is not satisfied exactly at every interior point or across interfaces, an element as a whole should be in equilibrium, because the structure as a whole should be in equilibrium. When nodal displacements are given values corresponding to a state of rigid body motion, the element must exhibit zero strain and therefore zero nodal forces, and the interior points should correspond to the assumed rigid body displacement.

a. Vertical translation. If the nodes are given unit vertical displacements (Fig. 5.9, a), equation (5.12) gives

( ) ⎣ ⎦ ( ) ( ) .constrrrrNNN ==−+++−=+=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

= 1324132

41

0101

3331xv

which shows that the element has indeed a vertical displacement as a rigid body.

b. Rotation. If the nodes are given unit rotations, with node 1 fixed and node 2 having a vertical displacement el (Fig. 5.9, b), equation (5.12) gives

( ) ⎣ ⎦ ( ) linearrNNNN eee

e =+=++=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

= 1222

1

10

432ll

ll

lxv ,

which shows that the element has indeed an anticlockwise rotation as a rigid body.


a b c

Fig. 5.9

2. An element should simulate constant strain states.

In the case of beams, when element sizes shrink to zero, they must have at least constant curvature. Assuming zero nodal vertical displacements and unit rotations in opposite directions (Fig. 5.9, c)

( ) ⎣ ⎦ ( )242 1

4221

010

rNNN eee −=−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−

=lllxv ,

so that the second derivative (curvature) is constant

.const==′′ 2-v l

5.6 Frame element

As shown in Fig. 5.2, an inclined beam element should include longitudinal displacements since, apart from moments and shear forces, it is acted upon by axial forces. Because there is no coupling between the bending and stretching displacements, the two stiffness matrices can be added taking into account the proper location of their elements.

5.6.1 Axial effects

The axial nodal forces are related to the nodal displacements by equation

[ ]⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

4

1

4

1

qq

kff e

S (5.52)

where the stiffness matrix for stretching (4.22) is

[ ] ⎥⎦

⎤⎢⎣

⎡−

−=

1 111

e

eeS

AEkl

. (5.53)


5.6.2 Stiffness matrix and load vector in local coordinates

For the frame element, combining equations (5.53) and (5.29), and arranging the elements in proper locations we get the element stiffness matrix given by

[ ]

e

e

IEIEIEIE

IEIEIEIE

AEAE

IEIEIEIE

IEIEIEIE

AEAE

k

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−−−

−

−

−

−

=

llll

llll

ll

llll

llll

ll

460260

61206120

0000

260460

61206120

0000

22

2323

22

2323

. (5.54)

The ratio of the bending terms to the stretching terms in (5.54) is of order ( ) 2li , where ‘i’ is the relevant radius of gyration. For slender beams, this ratio may be as small as 201 or 501 , so the stiffness matrix may possibly be numerically ill-conditioned.

If there is a uniformly distributed load on a member, the vector of consistent nodal forces is

{ }T

eeeee ppppf⎥⎥⎦

⎥

⎢⎢⎣

⎢−=

1220

1220

22 llll . (5.55)


A frame element is shown in Fig. 5.10 both in the initial and deformed state. For node 1, the local linear displacements 1q and 2q are related to the global linear displacements 1Q and 2Q by the equations

.QQq

,QQq

θθ

θθ

cossin

sincos

212

211

+−=

+=. (5.56)

Equations (5.56) can be written in matrix form as

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

=⎭⎬⎫

⎩⎨⎧

2

1

2

1

QQ

cssc

qq

(5.57)


where θcos=c and θsin=s .

The angular displacements (rotations) are the same in both coordinate systems

33 Qq = . (5.58)

Fig. 5.10

Adding the similar relationships for node 2

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

=⎭⎬⎫

⎩⎨⎧

5

4

5

4

QQ

cssc

qq

, 66 Qq = , (5.59)

we obtain

{ } [ ] { }eee QTq = . (5.60)

In (5.60), { }eq is the beam element displacement vector in the local

coordinate system, { }eQ is the beam element displacement vector in the global coordinate system and

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

=

1000000000000000010000000000

cssc

cssc

T e (5.61)

is the local-to-global coordinate transformation matrix for plane frames.


5.6.4 Stiffness matrix and load vector in global coordinates

Using the same procedure as in section 3.7, the stiffness matrix of the frame element in global coordinates is obtained as

[ ] [ ] [ ] [ ]eeTee TkTK = . (5.62)

The nodal loads due to a uniformly distributed load p are given by

{ } [ ] { }eTee fTF = (5.63)

The values of { }eF are added to the global load vector.

5.7 Assembly of the global stiffness matrix

The global stiffness matrix, [ ]K , is assembled from element matrices

[ ]eK using element connectivity matrices [ ]eT~ , that relate the nodal displacements at element level with the nodal displacements at the entire structure level, by equations of the form

{ } [ ] { }QT~Q ee = . (5.64)

The global uncondensed stiffness matrix is equal to the sum of the expanded element stiffness matrices

[ ] [ ]∑=e

eK~K , (5.65)

where [ ] [ ] [ ] [ ]eeTee T~KT~K~ = . (5.66)

The equations of equilibrium can be regarded as having been derived as soon as the reduced stiffness matrix and load vector have been calculated using the boundary conditions.

Having solved [ ]{ } { }FQK = it is routine to backtrack and recover the eth element strains, using equations (5.4), (5.12) and (5.60)

⎣ ⎦ { } ⎣ ⎦ [ ] { }eeee QTNyqNyy ′′−=′′−=′′−= vε . (5.67)


Strains, and hence stresses, are not accurate. Strains are derivatives of an approximate displacement (in beams - second derivatives) and differentiation inevitably decreases accuracy.

Example 5.2 Calculate the transverse displacement at the free end of the cantilever stepped beam shown in Fig. E5.2.

Fig. E5.2

Solution. Consider the beam divided into two cubic Hermitian elements. The element stiffness matrices are

[ ]

4626612612

2646612612

2

22

22

31

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−

=

llll

ll

llll

ll

l

IEk , [ ]

4626612612

2646612612

22

22

32

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−

=

llll

ll

llll

ll

l

IEk .


[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−−−

−−

=

22

222

22

3

46260061261200

261264126126361224004128120012241224

llll

ll

llllll

lll

llll

ll

l

IEK .

Using the boundary conditions at the fixed end 021 == QQ , and omitting the first two rows and columns, we obtain the finite element equations


⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−

0

00

4626612612

26126612636

6

5

4

3

22

22

3 FQQQQ

EI

llll

ll

llll

ll

l.

The three equations with zero right hand side can be written

5

6

4

3

22

22

6612

42621266636

QQQQ

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−

l

l

lll

lll

ll

.

or

55

2

2

6

4

3

18010860

2161

6612

9927212727921911

2161 QQ

QQQ

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧ l

ll

l

l

l

lll

l.

By substitution into the equation with non-zero right hand side

( ) FQQQQEI−=−+−− 65433 612612 ll

l

the transverse displacement in 3 is obtained as

IE

F.Q3

53 51 l−==v .

Example 5.3 Calculate the transverse displacement at 2 and the support reactions for the beam shown in Fig. E5.3.

Solution. Consider the beam modeled by two Bernoulli-Euler beam elements. Assembling the element stiffness matrices (5.29), the unreduced global stiffness matrix is obtained as

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−

−−

=

22

222

22

3

46260061261200

26802661202461200264600612612

llll

ll

lllll

ll

llll

ll

l

IEK .


At the fixed end 021 ==QQ ; at the simply supported end 05 =Q . Omitting the corresponding rows and columns, we obtain the finite element equations

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧ −=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

00

4262806024

6

4

3

22

223

F

QQQ

IE

lll

ll

l

l.

The last two equations give

36

422

22

60

4228 Q

QQ

⎭⎬⎫

⎩⎨⎧

−=⎭⎬⎫

⎩⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡

lll

ll

or

34 73 QQl

= , 36 712 QQl

−= .

Fig. E5.3

Substitution of 4Q and 6Q into the first equation gives

⎣ ⎦ IEFQQ

3

33 71273

6024 l

l

ll −=

⎭⎬⎫

⎩⎨⎧−

+

or

IE

FQ3

32 967 l

−==v .

The rotations are

IE

FQ2

42 963 l

−==ϕ , IE

FQ2

63 9612 l

==ϕ .

The support reactions are given by


⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−

3

1

1

6

4

32

366120260612

VMV

QQQ

IE

ll

ll

l

l

which yield

FV1611

1 = , lFM83

1 −= , FV165

3 = .

Example 5.4 Find the transverse displacement at 2 and the support reactions for the beam with fixed ends shown in Fig. E5.4.

Fig. E5.4

Solution. Consider the beam modeled by two Bernoulli-Euler beam elements. Using the boundary conditions 06521 ==== QQQQ , the finite element equations can be written

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−−−

−−

3

3

1

1

4

3

22

222

22

3 0

00

00

25151005151515100

516542651515451361200264600612612

MV

FMV

QQ

......

......IE

llll

ll

llllll

lll

llll

ll

l.

From the third and fourth rows we obtain

⎭⎬⎫

⎩⎨⎧ −

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−0654

54513

4

323

FQQ

...IEll

l

l

with solutions

IE

FQ3

32 818 l

−==v , IE

FQ2

42 272 l

−==ϕ .


Substituting the displacements into the other four equations yields

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−

−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

3

3

1

12

2

2

3

272

818

515151

26612

MVMV

IEF

...

IE ll

ll

l

ll

l

l.

The support reactions are

FV2720

1 = , lFM94

1 = , FV277

3 = , lFM92

3 −= .

Example 5.5 Calculate the rotations at supports and the reaction forces for the two-span beam shown in Fig. E5.5 where the right span carries a uniformly distributed load.

Fig. E5.5

Solution. The continuous beam is modeled by two Bernoulli-Euler beam elements. Using the boundary conditions 0531 === QQQ , the finite element equations can be written

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−−−

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−

−−

122

122

0

0

0

0

46260061261200

26802661202461200264600612612

23

22

1

6

4

2

22

222

22

3

l

l

l

l

llll

ll

lllll

ll

llll

ll

l

ppV

ppV

V

Q

Q

QIE .

Omitting the first, third and fifth rows and columns, we obtain

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

1212

0

420282024

2

2

6

4

2

22

222

22

3l

l

ll

lll

ll

l pp

QQQ

IE .


The first equation above yields

⎣ ⎦⎭⎬⎫

⎩⎨⎧

−=6

42 021

QQ

Q (a)

which, substituted into the second and third equation

⎭⎬⎫

⎩⎨⎧−

=⎟⎟⎠

⎞⎜⎜⎝

⎛

⎭⎬⎫

⎩⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

11

124228

02 2

6

422

22

2

2

3l

ll

lll

l

pQQ

QIE ,

gives

IE

pQ48

3

42l

−==ϕ , IE

pQ32

3

63l

==ϕ . (b)

Substituting the rotations (b) in (a) we obtain

IE

pQ96

3

21l

==ϕ .

The reaction forces are calculated from

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

22

660606066

3

2

1

6

4

2

3l

l

ll

ll

ll

l pVpV

V

QQQ

IE

obtaining

161lpV −= , lpV

85

2 = , lpV167

3 = .

Example 5.6 Find the transverse displacement and the angle of rotation at 2 for the beam shown in Fig. E5.6.

Fig. E5.6


Solution. Two Bernoulli-Euler beam elements are used to model the system. Using the boundary conditions 0521 === QQQ , the finite element equations can be written

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−−−

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−

−−

0

13330302070

0

00

46260061261200

266545161254513515100512510051515151

3

20

0

201

01

6

4

3

22

222

22

3

Vp.p.

p.Mp.V

Q

QQ

......

......

IEl

l

l

l

llll

ll

llllll

lll

llll

ll

l.

Deleting the first, second and fifth rows and columns we obtain

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧ −=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

013330

30

4262654654513

20

0

6

4

3

22

223 l

l

lll

lll

ll

lp.p.

QQQ

...

IE .

The last equation yields

436 21

23 QQQ −−=l

which, substituted into the first two equations

⎭⎬⎫

⎩⎨⎧ −

=⎟⎟⎠

⎞⎜⎜⎝

⎛

⎭⎬⎫

⎩⎨⎧

+⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡2

0

062

4

323 13330

3026

65454513

l

l

l

l

ll

l

l p.p.

QQQ

...IE ,

gives

IE

p.Q4

032 0840 l

−==v , IE

p.Q3

042 05180 l==ϕ .

Example 5.7 Find the shape functions for the 3-node beam element shown in Fig. E5.7.

Answer.

( ) ( )3221 3254

41 rrrrrN +−−= , ( ) ( )322

2 141 rrrrrN +−−= ,


( ) ( )223 1 rrN −= , ( ) ( )22

4 1 rrrN −= ,

( ) ( )3225 3254

41 rrrrrN −−+= , ( ) ( )322

6 141 rrrrrN ++−−= .

Fig. E5.7, a

Fig. E5.7, b


Example 5.8 The planar frame shown in Fig. E5.8, a has pinned ends and is loaded in 2 by a force N1000=F . It has 211 mN102 ⋅=E , 2mm1600=A and

45 mm102 ⋅=I . Determine the nodal displacements, plot the deformed shape and the diagrams of axial force, shear and bending moment.

Answer. The frame is modeled with 6 elements and 7 nodes.

Nodal data Node

nr Restr

X Restr

Y Restr

Z Coord

X Coord

Y Displ

X Displ

Y Rotation

Z 1 1 1 0 0 0 0 0 -1.232e-2 2 0 0 0 0 1 0.010 8.500e-7 -5.531e-3 3 0 0 0 0 2 1.0699e-2 1.700e-6 2.353e-3 4 0 0 0 0.5 2 1.0698e-2 1.309e-3 2.592e-3 5 0 0 0 1.4 2 1.0697e-2 2.321e-3 -1.260e-3 6 0 0 0 2 2 1.0696e-2 -8.500e-7 -6.889e-3 7 1 1 0 2 1 0 0 -1.260e-2

The deformed shape is shown in Fig. E5.8, b.

The diagrams of the axial force N , shear T and bending moment M are shown in Figs. E5.8, c, d, e.

a b

c d e

Fig. E5.8


Example 5.9

The planar frame shown in Fig. E5.9, a has fixed ends in 1, 5, 10, 13. It is loaded by a couple mmN102 6

3 ⋅=M and two point loads N102 46 ⋅−=F and

N10412 −=F . For 25 mmN102 ⋅=E , 2mm400=A , 44 mm102 ⋅=I ,

mm10021 =−l and mm200119109 == −− ll , find the displacements in 6 and plot the deformed shape.

Answer. The frame is modeled with 12 elements and 13 nodes.

Fig. E5.9, a

The displacements in 6 are

mm247606 .h = , mm967306 .−=v , rad0062306 .−=ϕ .


Fig. E5.9, b


5.8 Grids

Grids or grillages are planar structural systems subjected to loads applied normally to their plane. They are special cases of tree-dimensional frames in which each joint has only three nodal displacements, a translation and two rotations, describing bending and torsional effects.

Fig. 5.11

5.8.1 Finite element discretization

The grid is divided into finite elements, as shown in Fig. 5.11. Each node has three degrees of freedom, two rotations and a linear displacement. Typically, the degrees of freedom of node i are 23 −iQ , 13 −iQ and iQ3 , defined as the rotation about the X axis, the rotation about the Y axis and the displacement along the Z axis, respectively.

Nodes are located by their coordinates in the global reference frame XOY and element connectivity is defined by the indices of the end nodes. Elements are modelled as uniform rods with bending and torsional flexibility, without shear deformations and not loaded between ends. Their properties are the flexural rigidity

IE , the torsional rigidity tIG and the length l . Only cross sections whose shear centre coincides with the centroid are considered.

5.8.2 Element stiffness matrix in local coordinates

Consider an inclined grid element, as illustrated in Fig. 5.12, a, where the nodal displacements are also shown.


In a local physical coordinate system, the x axis, oriented along the beam, is inclined an angle α with respect to the global X axis. The z axis for the local coordinate system is collinear with the Z axis for the global system. Alternatively, an intrinsic (natural) coordinate system can be used.

Fig. 5.12

The vector of element nodal displacements is

{ } ⎣ ⎦ Te qqqqqqq 654321= (5.68)

and the corresponding vector of element nodal forces can be written

{ } ⎣ ⎦ Te fffffff 654321= . (5.69)

In (5.69) 3 f and 6 f are transverse forces, while 2f and 5 f are couples producing bending (Fig. 5.12, b). The corresponding displacements

63 q,q are translations, while 52 q,q are rotations. Their column vectors are related by the flexural stiffness matrix.

Rearranging the matrix (5.29) we obtain

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−−−

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

6

5

3

2

22

22

3

6

5

3

2

1261266462126126

6264

qqqq

EI

ffff

e

e

e

ll

llll

ll

llll

l. (5.70)

The axial nodal forces 41 f,f are torques and the nodal displacements

41 q,q are twist angles. They describe torsional effects so that their action is decoupled from bending. The respective stiffness matrix can be calculated


separately. The derivation of this matrix is essentially identical to the derivation of the stiffness matrix for axial effects in a frame element or in a truss element.

The twist angle can be expressed in terms of the shape functions (4.51) as

( ) ( ) ( ) 4211 qrNqrNr +=θ , (5.71)

which substituted into the strain energy

xx

IGU

e

ete d

2

2

∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=θ (5.72)

yields, after the change of coordinates, the element stiffness matrix

[ ] ⎣ ⎦ ⎣ ⎦∫+

−

′′=1

1

d2

rNNIG

k rT

re

etet

l. (5.73)

As a consequence of this analogy, the nodal forces are related to the nodal displacements by equation

[ ]⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

4

1

4

1

qq

kff e

t (5.74)

where the stiffness matrix for torsional effects is

[ ] ⎥⎦

⎤⎢⎣

⎡−

−=

1 111

e

etet

IGk

l. (5.75)

In (5.75), G is the shear modulus of elasticity and etI is the torsional constant of the cross section. For axially symmetrical cross sections the latter is the polar second moment of area.

For the grid element, combining the stiffness matrices from equations (5.70) and (5.75), we get the stiffness matrix in local coordinates relating the nodal forces (5.69) and the nodal displacements (5.68)

[ ]

e

e

eeaa

aa

EIk

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−−−

−−

=

12601260640620000012601260

6206400000

22

22

3

ll

llll

ll

llll

l (5.76)

where eeet IEIGa 2l= .



It is necessary to transform the matrix (5.76) from the local to the global system of coordinates before its assemblage in the stiffness matrix for the complete grid. As has been indicated, the z direction for local axes coincides with the Z direction for the global axes, so that only the rotational components of displacements should be converted. The transformation of coordinates is defined by equation

{ } [ ] { }eee QTq = , (5.77)

where { }eq is the element displacement vector (5.68) in the local coordinate system,

{ } ⎣ ⎦ Te QQQQQQQ 654321=

is the element displacement vector in the global coordinate system (Fig. 5.12) and

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

=

1000000000000000010000000000

cssc

cssc

T e , (5.78)

where αcos=c and αsin=s , is the local-to-global coordinate transformation matrix. The same transformation matrix (5.78) serves to transform the nodal forces from local to global coordinates.

5.8.4 Element stiffness matrix in global coordinates

Using the same procedure as for frame elements, we obtain the stiffness matrix of the grid element in global coordinates as

[ ] [ ] [ ] [ ]eeTee TkTK = . (5.79)

It is used to assemble the unreduced global stiffness matrix [ ]K using

element connectivity matrices [ ]eT~ that relate the nodal displacements at element level with the nodal displacements at the complete structure level, by equations of the form (5.64).


For grounded systems the unreduced matrix [ ]K is then condensed using the boundary conditions. The effect of lumped springs can be accounted for by adding their values along the main diagonal at the appropriate locations in the global stiffness matrix.

Example 5.10 The grid shown in Fig. E5.10 is fixed at points 1 and 2, has GPa210=E ,

GPa81=G , m1=l and diameter mm20=d . Find the vertical displacement of point 7 when the grid is loaded by forces N50087 −== FF and draw the spatial deflected shape.

a b

Fig. E5.10

Answer. The grid is modeled with 14 elements and 8 nodes, having 18 dof’s. The deflected shape is presented in Fig. E5.10, b. The deflection is

m43607 .−=w .

Example 5.11

The grid shown in Fig. E5.11, a is fixed at points 1 and 2, and has m1=l , 48 m107850 −⋅= .I , 48 m10571 −⋅= .It , GPa210=E and GPa81=G . Find the

vertical displacement of point 5 when the grid is loaded by a force N1035 −=F .

Draw the deflected shape and the diagrams of the bending moment and torque.


a b

c d

Fig. E5.11

Answer. The grid is modeled with 5 elements and 5 nodes, having 9 dof’s. The largest displacement is m405 .−=w . The deflected shape is presented in Fig. E5.11, b. The bending moment and torque diagrams are shown in Figs. E5.11, c, d.

5.9 Deep beam bending element

Shear deformation becomes important when analyzing deep beams, for which Bernoulli’s hypothesis is no more valid. The nonlinear distribution of shear stresses produces the warping of the cross section.

A simplifying hypothesis (Bresse, 1859) considers an average shear strain, constant over the cross section. This way, planar cross sections remain undistorted and plane (warping neglected) but no more perpendicular to the centroidal axis. The assumption is adopted in the formulation of the Timoshenko beam element used in vibration studies.


5.9.1 Static analysis of a uniform beam

Beams with cross sections that are symmetric with respect to the plane of loading are considered herein (Fig. 5.13, a). Only transverse loads act upon the beam, axial forces are ignored.

Fig. 5.13

The axial displacement of any point on the section, at a distance y from the neutral axis, is approximated by

( ) ( )xyy ϕ−=x,u , (5.80)

where ϕ is the cross section rotation at position x .

The strain components xε and xyγ are given by

ϕϕε ′−=−== yyxx

ux d

ddd , (5.81)

vxvvu

xy ′+−=+−=∂∂

+∂∂

= ϕϕγdd

xy, (5.82)

where v′ is the slope of the deformed beam axis.

Note that the slope v′ is no more equal to the rotation ϕ , as in the Bernoulli-Euler theory.

Normal stresses on the cross section are given by Hooke’s law

yEExxx d

dϕεσ −== , (5.83)

where E is Young’s modulus of the material.

The bending moment is the resultant of the normal stress distribution on the cross section


( ) ϕϕσ ′==−= ∫ zz

A

IEIEAyMx

x x ddd (5.84)

where zI is the second moment of area of the cross section.

The sign convention used here (Fig. 5.13, b) is that positive internal forces and moments act in positive (negative) coordinate directions on beam cross sections with a positive (negative) outward normal.

The shear force is given by

( )x

xddMT −= . (5.85)

The transverse load per unit length is

( )x

xddTp −= . (5.86)

The average shear strain is

GA

TAG

T

syx κ

γ == (5.87)

where G is the shear modulus of elasticity, κ is a shear factor and sA is the ‘effective shear area’ calculated as

[ ]∫∫=

A

AAs d

d2

2

τ

τ. (5.88)

Equations (5.82) and (5.87) give

( )ϕ−′= vsAGT . (5.89)

For a uniform beam not loaded between ends ( )0=p , elimination of M and T gives the differential equations of equilibrium

0=′−′′ ϕss AGAG v , (5.90)

0=−′′+′ ϕϕ szs AGIEAG v . (5.91)


Consider a prismatic beam element (Fig. 5.14) of length l (the index e is omitted) not loaded between ends ( )0=p . Using natural coordinates, lxr 2= .


Fig. 5.14

Eliminating ϕ and v in turn in (5.90) and (5.91) gives

0dd

4

4=

xv , and 0

dd

3

3=

xϕ . (5.92)

Changing to the r coordinate yields

0dd

4

4=

rv , and 0

dd

3

3=

rϕ . (5.93)

The general solutions of these equations are

( ) 122

33

4 aaaa +++= rrrrv , (5.94)

( ) 122

3 bbb ++= rrrϕ . (5.95)

The seven constants of integration are not independent since the above solutions must also satisfy equation (5.91) which, in the new variable r, becomes

0dd

dd2

2

2=−+ ϕϕβ

rrv

l, (5.96)

where

24

ls

z

AGIE

=β . (5.97)

Substituting (5.94) and (5.95) into (5.96) gives

436 abl

= , 324 abl

= , 421122 aabll

β+= . (5.98)

This leaves only four independent constants which can be determined by evaluating (5.94) and (5.95) at 1±=r . The resulting displacement functions are


{ } ⎣ ⎦ { }ee qNqNNNN =⎥⎦⎥

⎢⎣⎢= 4321 22

llv , (5.99)

{ } ⎣ ⎦ { }ee qN~qN~N~N~N~ =⎥⎦⎥

⎢⎣⎢= 4321

22ll

ϕ , (5.100)

where

{ } ⎣ ⎦Teq 2211 ϕϕ vv= . (5.101)

The shape functions are

( ) ( ) ( )[ ]rrrrN −++−+

= 1632314

1 31 β

β,

( ) ( ) ( )[ ]2322 131

3141 rrrrrN −++−−+

= ββ

,

( ) ( ) ( )[ ]rrrrN ++−++

= 1632314

1 33 β

β,

( ) ( ) ( )[ ]2324 131

3141 rrrrrN +−+++−−+

= ββ

,

( ) ( ) ( )21 33

3141 rrN~ +−+

=β

, (5.102)

( ) ( ) ( )[ ]rrrrN~ −++−−+

= 16321314

1 22 β

β,

( ) ( ) ( )23 33

3141 rrN~ −+

=β

,

( ) ( ) ( )[ ]rrrrN~ ++++−+

= 16321314

1 24 β

β.

For 0=β , the first four functions (5.102) become the third degree Hermitic polynomials (5.21) and the last four functions satisfy the relationship

( )r

NrN~ ii ∂

∂=l

2 ( )41,...,i = . (5.103)

It is useful to introduce a third set of shape functions which are used in the derivation of the element stiffness matrix, and defined by


⎣ ⎦ ⎣ ⎦ ⎥⎦

⎥⎢⎣

⎢−=

rNN~N̂

dd2

l. (5.104)

They are

( ) ( ) ( ) ( )l

131

34321 β

β+

==−== rN̂rN̂rN̂rN̂ . (5.105)

5.9.3 Stiffness matrix

The strain energy for a beam element with shear effects included is

∫∫−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛=

2

2

22

2

2

ddd

2d

dd

2

l

l

l

l

xAGxIEU sze x

vx

ϕϕ . (5.106)

As

rd

d2dd ϕϕ

l=

x and rx d

2d l

= , (5.107)

the contribution due to bending is

∫+

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

1

2

ddd2

2rIEU zB

e rϕ

l, (5.108)

and the contribution due to shear is

∫+

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

1

1

2

ddd2

22rAGU sS

e rv

l

l ϕ . (5.109)

On substituting (5.100) into (5.108) we get

{ } ⎣ ⎦ ⎣ ⎦ { }eTz

TeBe qrN~N~IEqU ∫

+

−

′′=

1

1

d221

l (5.110)

wherefrom we obtain the element stiffness matrix due to bending

[ ] ⎣ ⎦ ⎣ ⎦∫+

−

′′=1

1

d2 rN~N~IEk Tz

eB

l. (5.111)

On substituting (5.100) and (5.99) into (5.109) we get


{ } ⎣ ⎦ ⎣ ⎦ { }eTs

TeSe qrN̂N̂AGqU ∫

+

−

=

1

1

d22

1 l (5.112)

wherefrom we obtain the element stiffness matrix due to shear

[ ] ⎣ ⎦ ⎣ ⎦∫+

−

=1

1

d2

rN̂N̂AGkT

seS

l . (5.113)

Substituting the shape functions (5.102) and (5.105) and performing the integration yields the stiffness matrix

[ ] ( ) ( )

( ) ( ) ⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+−−−−−−−+

−

+=

22

22

3

346326612612

326346612612

311

llll

ll

llll

ll

l

ββ

βββ

ze EIk . (5.114)

The vector of consistent nodal forces is identical to the corresponding vector (5.42) derived for a slender beam.

6. LINEAR ELASTICITY

In this chapter the fundamental concepts from the linear theory of elasticity are recalled, with emphasis on two-dimensional problems. The four main groups of equations are written in the matrix notation used in FEA: (a) equations of equilibrium, (b) equations of compatibility or strain/displacement relations, (c) stress/strain relations or Hooke’s law, and (d) boundary conditions.

6.1 Matrix notation for loads, stresses and strains

An arbitrarily shaped tree-dimensional body of volume V, in equilibrium under the action of external loads and the reactions in supports, is shown in Fig. 6.1. The total surface S of the body has two distinct parts: uS , the portion of the boundary on which displacements are prescribed, and σS , the portion on which surface forces are prescribed.

Points in the body are located by x, y, z coordinates. Any point on the surface has a local outward-pointing normal n whose orientation is usually described by its three direction cosines xn ∂∂ , yn ∂∂ , zn ∂∂ . In general there may be three sets of applied forces: (a) internal body forces, (b) surface forces, and (c) concentrated forces.

Internal body forces

Internal body forces inside the volume V can be inertial forces, like centrifugal or gravity forces. Their the magnitude per unit volume is denoted by components xpv , ypv , zpv . It is convenient to write these components as a single body force vector

{ } ⎣ ⎦ Tzyx pppp vvvv = . (6.1)


Surface tractions

Likewise there could be surface forces (not necessarily normal pressures) on the surface σS , defined by the magnitude per unit surface area, also having three components

{ } ⎣ ⎦ Tzsysss pppp x= , (6.2)

Concentrated loads

Concentrated loads are defined by their three components

{ } ⎣ ⎦ Tziyixii FFFF = . (6.3)

Any system of loads has to fall into categories (6.1) to (6.3).

Fig. 6.1

Displacements

It is natural to form the single displacement vector

{ } ⎣ ⎦ Twvuu = , (6.4)

where wvu ,, are the displacement components inside the body or on the surface σS with unprescribed displacements.

Stresses and strains

The stresses inside V will have two types of component, the direct stress components zyx ,, σσσ and the shear stresses zxyzxy ,, τττ . It is convenient to represent both stress and strain components as single column matrices. Thus

{ } ⎣ ⎦ Tzxyzxyzyx τττσσσσ = . (6.5)

6. LINEAR ELASTICITY 125

In two-dimensional problems

{ } ⎣ ⎦ Txyyx τσσσ = . (6.5, a)

Strains are represented in vector form as

{ } ⎣ ⎦ Tzxyzxyzyx γγγεεεε = . (6.6)

In two-dimensional problems

{ } ⎣ ⎦ Txyyx γεεε = . (6.6, a)

6.2 Equations of equilibrium inside V

Figure 6.2 shows stresses acting on an infinitesimal element in the plane xOy. The small linear increments in stresses are equilibrated by the applied body forces yielding the following equilibrium equations, where yxxy ττ =

.pyx

,pyx

yyxy

xyxx

0

0

=+∂

∂+

∂

∂

=+∂

∂+

∂∂

v

v

στ

τσ

(6.7)

Fig. 6.2

In matrix form, equations (6.7) can be written


⎭⎬⎫

⎩⎨⎧

−=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

y

x

xy

y

x

pp

xy

yxv

v

τσσ

0

0 (6.8)

or, denoting the matrix of differential operators

[ ]

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=∂

xy

y

x0

0

, (6.9)

and using (6.1) [ ] { } { } { }0=+∂ vpT σ . (6.10)

6.2 Equations of equilibrium on the surface σS

Let now consider an element near the loaded boundary σS . The equilibrium along the two axes directions yields

.pm

,pm

ysyxy

xsyxx

=+

=+

στ

τσ

l

l (6.11)

Fig. 6.3

In (6.11) the direction cosines for the outward normal n are


( )

( ) .nyny,nm

,nxnx,n

y

x

=∂∂

==

=∂∂

==

cos

cosl

(6.12)

In matrix form, equations (6.11) can be written

⎭⎬⎫

⎩⎨⎧

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎦

⎤⎢⎣

⎡ys

xs

xy

yx

xy

yxpp

nnnn

τσσ

00

, (6.13)

or in condensed form

[ ] { } { }sT pn =σ , (6.14)

sometimes written

[ ]{ } { }sT pn =∂ σ , (6.15)

where nT∂ implies that the operators in [ ]T∂ are acting upon n.

6.3 Strain-displacement relations

Figure 6.4 gives the deformation of the dydx − face for small deformations.

The equations of compatibility have the familiar form

xx ∂

∂=uε ;

yy ∂∂

=vε ,

yxxy ∂∂

+∂∂

=uvγ . (6.16)

In matrix form

⎭⎬⎫

⎩⎨⎧

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

vu

xy

y

x

xy

y

x

0

0

γεε

, (6.17)

and can be summarized as

{ } [ ]{ }u∂=ε . (6.18)


Fig. 6.4

6.4 Stress-strain relations

For linear isotropic elastic materials, the stress-strain relations come from the generalized Hooke’s law

{ } [ ]{ }σε C= , (6.19)

where the material elastic compliance matrix is

[ ] ( )( )

( )⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

++

+

−−−

=

νν

ν

ννν

12012SYM0012000100010001

1E

C . (6.20)

The inverse relation is

{ } [ ] { } [ ]{ }εεσ DC == −1 . (6.21)

The inverse of [ ]C is the material stiffness matrix


[ ] ( )( )

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

−

−−

−

−+=

ν

ν

ν

νννννν

νν

21

021SYM

0021

000100010001

211ED . (6.22)

Plane stress

A thin planar body subjected to in-plane loading on its edge surface is said to be in plane stress. If stresses zσ , xzτ , and yzτ are set as zero, discarding rows and columns 3, 5 and 6 in (6.20), the Hooke’s law can be written

( ) ⎪

⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−

−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

xy

y

x

xy

y

x

Eτσσ

νν

ν

γεε

12000101

1 . (6.23)

The inverse relations are given by

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

xy

y

x

xy

y

x E

γεε

νν

ν

ντσσ

2100

0101

1 2 (6.24)

which is used as { } [ ]{ }εσ D= .

Plane strain

If a long body of uniform cross section is subjected to transverse loading along its length, a small thickness in the loaded area can be treated as subjected to plane strain. If strains zε , xzγ , and yzγ are set as zero, from (6.21) and discarding rows and columns 3, 5 and 6 in (6.22), we obtain

( )( ) ⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−

−+=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

xy

y

x

xy

y

x E

γεε

ννν

νν

νντσσ

22100

0101

211 (6.25)


which is also used as { } [ ]{ }εσ D= but here [ ]D has a different expression as in (6.24).

6.5 Temperature effects

The temperature strains are represented as initial strains

{ } ⎣ ⎦ TTTT 0000 αααε = , (6.26)

where T is the temperature rise and α is the coefficient of linear expansion of the material.

The stress-strain relations become

{ } [ ] { } { }( )0εεσ −= D . (6.27)

In plane stress

{ } ⎣ ⎦ TTT 00 ααε = . (6.28)

In plane strain

{ } ( ) ⎣ ⎦ TTT 010 αανε += . (6.29)

6.6 Strain energy

For linear elastic materials, the strain energy per unit volume in the body is

{ } { } { } { }σεεσ TTU21

21

0 == . (6.30)

For the elastic body shown in Fig. 6.1, the total strain energy is given by

{ } { }∫=

V

T VU d21 σε . (6.31)

Substituting (6.21) we obtain the strain energy in terms of strains

{ } [ ]{ }∫=

V

T VDU d21 εε . (6.32)

7. ENERGY METHODS

The finite element method can be considered a Rayleigh-Ritz method.

The classical Rayleigh-Ritz technique represents a variational approach whereby a distributed system is approximated by a discrete one by assuming a solution of the differential boundary-value problem as a finite series of admissible functions. Unfortunately, systems with complex geometry or complex boundary conditions cannot be accomodated easily by global admissible functions.

In the finite element method, the approximate solution is constructed using local admissible functions, defined over small subdomains of the structure. Good approximations can be realized with low-degree polynomials. Displacements are calculated by methods based on the principle of virtual work and/or the principle of minimum total potential energy.

Instead of solving differential equations with complicated boundary conditions, the finite element method evaluates integrals of relatively simple polynomial functions. Variational methods put less strict conditions on the functions approximating the displacement field than the analytical methods based on differential equations.

7.1 Principle of virtual work (PVW)

PVW is basically a statement of the static equilibrium of a mechanical system. In the following, the form known as the principle of virtual displacements (PVD) will be used, as applied to elastic bodies.

7.1.1 Virtual displacements

By definition, virtual displacements are:

a) arbitrary (fictitious, virtual);

b) infinitesimal (follow the rules of differential calculus);


c) not related to either the actual displacements or to the forces producing them;

d) continuous in the interior and on the surface of the body;

A continuity 0C is generally required for bars and elasticity problems, while a continuity 1C is imposed for beams, plates and shells. Exceptions do exist. Remember that a function of several variables is said to be of class mC in a domain V if all its partial derivatives, up to the mth order inclusive, exist and are continuous in the domain V.

e) kinematically admissible, i.e. consistent with the system kinematic boundary conditions (geometric constraints).

If the differential equation of the problem is of order nm 2= , the

admissible functions must have continuity 1−nC , i.e. the geometrical boundary conditions must be satisfied to the ( )thn 1− derivative. For bars, 2=m and the

assumed functions must have continuity 0C .

For beams 4=m and the approximating functions must have continuity 1C . Because the continuity required is reduced from 2C in the governing

differential equation to 1C in the variational equation, the functional is said to have a “weak form”.

A virtual displacement will be denoted by ‘ δ ’ in front of a letter, e.g. uδ . The symbol ‘ δ ’ was introduced by Lagrange to emphasize the virtual character of the variations, as opposed to the symbol d which designates actual differentials of position coordinates.

Denoting by

{ } ⎣ ⎦ Twvuu = ,

the displacement vector (6.4) inside the body or on the surface σS with unprescribed displacements (Fig. 6.1), the vector of virtual displacements is

{ } ⎣ ⎦ Twvuu δδδδ = . (7.1)

The vector of the corresponding virtual strains will be

{ } [ ]{ }uδδ B=ε . (7.2)

where [ ]B is the strain-displacement matrix.

7. ENERGY METHODS 133

7.1.2 Virtual work of external loads

For a bar in tension (Fig. 7.1, a), the virtual work of the external force F is

uFWE δδ ⋅= . (7.3)

It has the same value whether the bar material is linear elastic (Fig. 7.1, b) or nonlinear elastic (Fig. 7.1, c). Note that it is simply (force × displacement), because the force is constant along the virtual displacement, the latter being arbitrary, hence independent of the force.

In the general case of loading by conservative body forces (6.1), surface tractions (6.2) and point forces (6.3), the virtual work of external loads is

{ } { } { } { } { } { }∑∫∫ ++=i

iT

iS

T

V

TE FuApuVpuW δdδdδδ

σ

sv . (7.4)

Note that the scalar product under the first integral is

{ } { } zyxT ppppu vvvv wvu ⋅+⋅+⋅= δδδδ .

a b c

Fig. 7.1

Note also the absence of the factor 21 which occurs in the expression of the work of elastic forces, because the external loads remain constant during the action along the virtual displacements.

7.1.3 Virtual work of internal forces

For a three-dimensional continuum, the virtual work of internal stresses is

{ } { }∫=V

TI VW dδδ σε (7.5)

as shown in Fig. 7.2 for the uniaxial case.


Fig. 7.2

Again, stresses remain constant during the action on virtual strains.

7.1.4 Principle of virtual displacements

For elastic bodies, the principle of virtual displacements states that:

If a system is in equilibrium, then during an arbitrary small displacement from the equilibrium position, the virtual work of applied loads equals the virtual work of internal forces

IE WW δδ = . (7.6)

Also: A body is in equilibrium if the internal virtual work equals the external virtual work for every kinematically admissible displacement field.

{ } { } { } { } { } { } { } { } 0δdδdδdδ =−−− ∑∫∫∫i

iT

iS

T

V

T

V

T FuApuVpuVσ

σε sv .

(7.6, a)

In (7.6) EWδ is the work of external loads on the virtual displacements { }uδ which are independent of loading and kinematically admissible.

If stresses are expressed in terms of a set of parameters defining completely the displacement pattern – the nodal displacements, then equilibrium relations can be obtained and the displacement parameters determined. The nodal displacements do not permit the fully equilibrating position to be reached, so that the PVD will ensure approximate equilibrium.

Note that the virtual work of reaction forces at supports is zero. Since the principle of virtual displacements is an equilibrium requirement, it is independent of material behaviour, i.e. whether the material is elastic or inelastic. It applies only


for loading by conservative forces, which do not change direction during the action on the virtual displacements. The external work is independent of the path taken.

Example 7.1 For the three-bar pin-jointed framework shown in Fig. 7.3, loaded by a

force F, find the internal bar forces and the displacement of point 4.

Fig. 7.3

Solution. Consider three states of the analyzed system:

1. The initial state, in which bars are not loaded by external forces and are not prestressed (Fig. 7.4, a).

2. The final state of static equilibrium, in which the external force F, of components αsin 1 FF = and αcos 2 FF = , produces a displacement of the joint 4, of components 1u and 2u (Fig. 7.4, b).

The joint 4 is acted upon by the external forces 1F , 2F and by internal forces 1T , 2T , 3T (Fig. 7.4, c). The joint reacts with forces equal in magnitude but of opposite sign, producing the elongations 1Δ , 2Δ , 3Δ (Fig. 7.4, d).

3. An imaginary state, in which the joint 4 is given a virtual displacement of components 1δu and 2δu (Fig. 7.4, e), which produce virtual elongations in bars

1δΔ , 2δΔ , 3δΔ (Fig. 7.4, f), the applied forces remaining constant.

The virtual displacements 1δu and 2δu and the virtual bar extensions 1δΔ ,

2δΔ , 3δΔ satisfy the compatibility equations (2.19)


.uu,u

,uu

cos δsin δδ δδ

cos δsin δδ

213

22

211

θθΔΔ

θθΔ

+−==

+= (7.7)

For the three bars, the force-elongation equations (2.21) can be written

11 AE

T l=Δ ,

AET θΔ cos 2

2l

= , AE

T l 33=Δ . (7.8)

Fig. 7.4

Equating internal work to external work (7.6) using the products of real forces and virtual displacements we obtain

2211332211 δδδδδ uFuFTTT +=++ ΔΔΔ . (7.9)

Substituting (7.7) into (7.9) and collecting coefficients of 1δu and 2δu gives


( ) ( ) 0cos cos δsin sin δ 232121311 =−+++−− FθTTθTuFθTθTu . (7.10)

Since 1δu and 2δu are unrelated to each other, we could put either to zero. Equation (7.10) must be zero whatever the values of 1δu and 2δu . This can only be true if their coefficients vanish

.FTTT,FTT

cos cos sin sin

2321

131

=++=−

θθθθ

(7.11)

These are indeed the equations of equilibrium. It is confirmed that the principle of virtual work is an equivalent statement of statical equilibrium.

Substituting (7.8) in the finite form of (7.7), then in (7.11), the components of the displacement of point 4 can be determined from the following equations

.FuAE

,FuAE

cos

1cos 2

sin 2

222

112

=⎟⎠⎞

⎜⎝⎛ +

=⋅

θθ

θ

l

l (7.12)

7.1.5 Proof that PVD is equivalent to equilibrium equations

Consider a form of equation (7.6, a) without point forces

{ } { } { } { } { } { }∫∫∫ +=σ

σεS

T

V

T

V

T ApuVpuV dδdδdδ sv . (7.6, b)

Convert { }εδ to { }uδ using the integration by parts

( )

( ) ,zyxx

uAuAu

Axux

zyxxuV

V V

x

Sx

Sx

Sx

Vx

Vxx

∫∫∫

∫∫∫

∂∂

−==

=∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=

ddddδdδdδd

ddδdddδdδ

σσσ

σσεσ

σσ

σ

ll

l

∫∫∫ ∂

∂−=

V

y

Sy

Vyy V

yAmV dδdδdδ

σσεσ

σ

vv ,

( ) ∫∫∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂+

∂

∂−⋅+⋅=

V

xyxy

Sxy

Vxyxy V

yxAmV dδδdδδdδ uvuv

τττγτ

σ

l ,


where l and m are direction cosines of the outward normal at the surface.

Adding together

( ) ( ) ( )[ ]

.Vyxyx

AmmV

V

yxyxyx

Syxyxy

Vxyxyyyxx

∫

∫∫

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂+

∂

∂+⎟⎟

⎠

⎞⎜⎜⎝

⎛∂

∂+

∂∂

−

−+++=++

dδδ

dδδdδδδ

σττσ

σττστγσεσεσ

vu

vu x ll

In matrix form (Gauss’ theorem)

{ } { } { } [ ] { } { } [ ] { }∫∫∫ ∂−=V

TT

S

TT

V

T VuAnuV dδdδdδ σσσεσ

,

which is true provided [ ] { }σT∂ and { }ε are finite in V, that is the stress and displacement fields are continuous. This applies only within a single element and up to its surface.

The componets of stresses at element interfaces may not balance at a point, but only in the mean. Most finite elements in use today do not achieve continuous stresses across interfaces.

On substituting in (7.6, b)

{ } [ ] { } { }( ) { } [ ] { } { }( ) 0dδdδ =−−+∂ ∫∫σ

σσS

TT

V

TT ApnuVpu sv .

Because { }uδ are arbitrary, its coefficients must vanish. The equations of equilibrium emerge from the brackets

[ ] { } { } { }0=+∂ vpT σ in V, [ ] { } { } { }0=− spn T σ on σS .

The PVD supplies equilibrium conditions both within and on the surface of the body. So, when using approximate functions for { }u , it is not necessary to worry about the equilibrium boundary conditions. Part of the surface, i.e. σS , is supported in some way and there the tractions { }sp will be unknown reactions and not specified loads. It is conventional to remove the unknown reactions by choosing the virtual displacements { }uδ to be zero over σS .


7.2 Principle of minimum total potential energy

The total potential energy Π of an elastic body is defined as the sum of the strain energy U and the work potential of external loads PW

PWU +=Π . (7.13)

7.2.1 Strain energy

Consider the strain energy (6.32)

{ } [ ]{ }∫=V

T VDU d21 εε .

For a virtual strain { }εδ , the virtual increase of strain energy is

{ } [ ]{ } { } [ ]{ }∫∫ +=V

T

V

T VDVDU dδ21dδ

21δ εεεε .

Because

{ } [ ]{ }( ) { } [ ]{ }εεεε δδ DD TTT = ,

{ } [ ]{ } { } { } IV

T

V

T WVVDU δdδdδδ === ∫∫ σεεε .

IWU δδ = . (7.14)

For an elastic body, the virtual variation of the strain energy is equal to the virtual work of the internal stresses on virtual strains.

For a bar in tension, substituting εσ E= and xu

dd

=ε , yields

∫∫∫ ===ll

xxuAE

xuxAEVU

V

ddd

dδddδdδδ εεσε . (7.15)

For a beam in bending, substituting 2

2

dd

xy v

−=ε (5.4), gives

∫∫ ∫ =⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

∂∂

∂∂

=l

xx

IEx

xAyx

Ex

Ul A

ddd

dδdddδδ 2

2

2

22

2

2

2

2 vvvv . (7.16)

The above expressions can be obtained directly from the strain energies


for a bar ∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛=

l

xxuAEU d

dd

2

2

, (7.17)

for a beam ∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛=

l

xx

IEU ddd

2

2

2

2v . (7.18)

7.2.2 External potential energy

The work potential of external loads PW is equal to the negative product of external forces by the corresponding displacement

EP WW −= . (7.19)

The negative sign appears because the external loads lose some of their capacity for doing work when displaced in the direction they act.

For example, a gravitational force gmF = acts in the opposite direction to a vertical displacement h and the potential becomes hgm .

An external point load jF has potential energy ( )jj uF− instead of

⎟⎠⎞

⎜⎝⎛− jj uF

21 , because this potential arises from the magnitude of force and its

capacity to do work when it moves, being independent of the linear properties of the body on which it acts.

For a three-dimensional continuum

{ } { } { } { } { } { }∑∫∫ −−−=i

iT

iS

T

V

TP FuApuVpuW

σ

dd sv . (7.20)

For virtual displacements { }uδ

EP WW δδ −= . (7.21)

7.2.3 Total potential energy

The total potential energy (7.13) can be written

{ } { } { } { } { } { } { } { }∑∫∫∫ −−−=i

iT

iS

T

V

T

V

T FuApuVpuVσ

σεΠ dddδ sv .

(7.13, a) Its variation is


EP WWWU δδδδδ I −=+=Π . (7.22)

Based on equation (7.6) it follows that

0δ =Π , (7.22, a)

hence, at equilibrium, the total potential energy has a stationary value. If 0δ2 >Π , the stationary value is a minimum, the equilibrium is stable.

The principle of minimum total potential energy states that:

If a deformable body is in equilibrium under the action of external loads and reaction forces, then the total potential energy has a minimum value.

Reciprocally, if under the action of external loads and reaction forces the total potential energy of a deformable body is a minimum, then it is in a stable equilibrium state.

Thus, it can be considered that (7.22, a) is a condition that establishes or defines the equilibrium, rather than a result of the equilibrium.

An equivalent statement is: For conservative systems, of all possible kinematically admissible displacement fields, the one satisfying equilibrium corresponds to a minimum value of the total potential energy.

Reciprocally, any kinematically admissible displacement field which minimizes the total potential energy represents a stable equilibrium configuration.

Example 7.2

For the truss shown in Fig. 7.3, the strain energy for a bar is

21

21 2

ii

iiii l

AETU ΔΔ == ,

and the external potential energy is

∑−=i

iiP uFW .

Expressing the elongations in terms of displacements, according to the compatibility relations, the total potential energy can be written


( )

( ) .uFuFuuEA

uos

EAuuEA

22112

21

22

221

cossin2

c2cossin

2

−−+−+

+++=

θθ

θθθΠ

l

ll

Cancelling the derivatives of Π with respect to each independent variable

01=

∂∂

uΠ , 0

2=

∂∂

uΠ ,

we obtain the equilibrium equations (7.11).

Example 7.3

Apply the principle of minimum total potential energy to a beam in bending, subjected to a distributed load and to the end bending moments and shear forces as shown in Fig. 7.5. Show that PMTPE is equivalent to the equilibrium conditions inside and at the ends of the beam.

Fig. 7.5

Solution. For a beam segment loaded as shown, the total potential energy is

( ) llll

ll

vvvvvv 00 TTMMxpxIE +−′−′+−′′= ∫∫ 002 dd

21Π .

At equilibrium, Π is stationary, 0δ =Π or

( ) 0δδδδdδdδ 00 =+−′−′+−′′′′ ∫∫ llll

ll

vvvvvvv 00 TTMMxpxIE . (a)

Integrating by parts the first term gives


( ) ( )

( ) ( ) .xIEIExIEx

IE

IExxx

IExIE

dδδδdddδ

δddddδ

dddδ

00 vvvvvvvv

vvvvvv

′′′′−′′′=′′′−′′′=

=′′′=⎟⎟⎠

⎞⎜⎜⎝

⎛′′=′′′′

∫∫

∫∫∫

l

l

l

l

lll

Integrating by parts the last term gives

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) .xIEIExIEx

IE

IExx

IExIE

dδδδdddδ

δddδdddδ

00 vvvvvvvv

vvvvvv

∫∫

∫∫∫″′′−′′′=′′′−′′′=

=′′′=′′′=′′′′

l

l

l

l

lll

Equation ( )a becomes

( ) ( ) 0δδdδdδδδ 0000 =+′−−″′′+′′′−′′′ ∫∫ ll

ll

ll vvvvvvvvv TMxpxIEIEIE

or

( )( ) ( ) ( )( ) 0δδdδ00 =−′′′−′′′+−″′′∫l

l

l

vvv-vvv TIEMIExpIE .

The coefficient of vδ in the integrand gives the equation of equilibrium

( ) ( )xpIE =″′′v .

As vδ is arbitrary, the other terms deliver the equilibrium conditions at the beam ends

( ) 00 MIE =′′v , or 0δ =′0v , ( ) 00 TIE =′′′v , or 0δ =0v ,

( ) ll MIE =′′v , or 0δ =′lv , and ( ) ll TIE =′′′v , or 0δ =lv ,

which are the boundary conditions.

7.3 The Rayleigh-Ritz method

The Rayleigh-Ritz method involves the construction of an assumed displacement field. For a beam, the transverse displacement ( )xv is approximated by a finite series


( ) ( )∑=

≅n

jjj xax

1

ϕv (7.23)

where ja are undetermined constants called generalized coordinates, and ( )xjϕ are prescribed functions of x , called admissible functions, that satisfy the kinematic (geometric) boundary conditions and are continuous within the definition interval.

Substituting the displacements (7.23) into the expression of the total potential energy Π , the latter becomes a function of the parameters ja , whose values are determined from the stationarity conditions

0δδ =∂∂

=∑j

jj

aaΠΠ .

Because jaδ are arbitrary,

0=∂∂

jaΠ , ( )n,...,j 1= , (7.24)

which is a linear algebraic set of equations in the constants ja . The solutions are back-substituted into (7.23) which represents an

approximate deflected shape, which is more accurate the more terms are selected in the respective series.

The necessary requirements for the convergence of the Rayleigh-Ritz method are the following:

a) The approximating functions must be continuous to one order less the highest derivative in the integrand.

b) The functions must individually satisfy the geometric boundary conditions, i.e. to be admissible functions.

c) The sequence of functions must be complete.

If the functions are not selected from the domain space of the operator of the equation being solved (completeness property) the resulting solution could be either zero or wrong.

Example 7.4 For the beam shown in Fig. 7.6 find the vertical displacement of point 2.

Consider: MPa210=E , 4mm 1600=I , m3=l , N100=F and mN200=q .

Solution. The total potential energy is


( ) ⎟⎠⎞

⎜⎝⎛−−⎟

⎟⎠

⎞⎜⎜⎝

⎛= ∫∫ 2

d ddd

21

320

2

2

2 ll

l

l

vvv Fxxqxx

IEΠ . (7.25)

The geometric boundary conditions are

( ) 00 =v , ( ) 00 =′v , 03

2=⎟

⎠⎞

⎜⎝⎛ lv , ( ) 0=lv . (7.26)

In the following, for simplicity, the transverse displacements are approximated by a series consisting of only two terms

( ) ( ) ( )xaxax 2211 ϕϕ +=v , (7.27) where the functions

( ) ( )( )4

2

123l

ll −−=

xxxxϕ , ( ) ( )( )5

3

223l

ll −−=

xxxxϕ , (7.28)

satisfy all geometrical boundary conditions (7.26).

Fig. 7.6

Substituting (7.27) into (7.25) we obtain the functional

( ) ( )

( ) ( )[ ].aaF

xaaqxaaIE

22

dd21

2211

322211

0

22211

ll

l

l

l

ϕϕ

ϕϕϕϕΠ

+−

−+−′′+′′= ∫∫ .

Requiring Π to be stationary with respect to 1a and 2a , leads to two equations relating the generalized coordinates

( ) ( ) 02dd 132

10

122111

=−−′′′′+′′=∂∂

∫∫ l

l

l

l

ϕϕϕϕϕΠ FxqxaaIEa

,


( ) ( ) 02dd 232

20

222112

=−−′′′′+′′=∂∂

∫∫ l

l

l

l

ϕϕϕϕϕΠ FxqxaaIEa

.

Because Π is of second order in ϕ and "ϕ , the equations are linear in ja .

Substituting the functions (7.28), for 6lqF = , we obtain

lll

qaIEaIE43201110

556

2313 −=+ ,

lll

qaIEaIE46656

26977210 2313 −=+ .

For example, the coefficient of 1a in the first equation is

( ) ( ) 30

2228

0

21 5

56d430361dl

lll

llIExxxIExIE =+−=′′ ∫∫ ϕ .

The solutions are

IEq.a

4

1 002070 l= ,

IEq.a

4

2 002570 l−= . (7.29)

In 2, the displacement is

( ) ( ) ( )

mm. 4821094321

161

2224

521

2211

.IE

q.aa

aa

y=⋅=+=

=+=

− l

lll ϕϕv

Example 7.5 Consider a linear spring of stiffness k (Fig. 7.7, a) subjected to a load F.

Comment on the approximations of the Rayleigh-Ritz method.

Answer. The total potential energy (Fig. 7.7, b) is

uFuk −= 2

21Π . (7.30)

For a small virtual displacement uδ , the variation of the total potential energy is

( ) uFukuFuuk δδδδ −=−=Π .


The equilibrium equation is obtained by requiring Πδ to be zero (Π be stationary) for arbitrary uδ .

Fig. 7.7

For 0δ =Π we obtain

0=− Fuk eq . (7.31)

As seen in Fig. 7.7, b, the exact solution corresponds to an absolute minimum value of Π .

The total potential energy of the equilibrium configuration is

k

FuFuFuF eqeqeqeq

2

21

21

21

−=−=−=Π . (7.32)

The stiffness is

eqeq

FFkΠΠ

22

21

21

=−= . (7.33)

The strain energy is

eqeqeq kFukU Π−===

22

21

21 . (7.34)

If F is prescribed and the resulting u has been approximated by a Rayleigh-Ritz solution eqapp uu ≠ , equations (7.32) – (7.34) and Fig. 7.7, b indicate that:

a) Because eqΠ is a minimum, the potential energy for an approximate displacement which satisfies the kinematic boundary conditions is greater than the true value

eqapp ΠΠ > . In magnitude eqapp ΠΠ < ,


the approximate total potential energy is underestimated.

b) The approximate stiffness is overestimated

↓

↑=eq

FkΠ

2

21 . (7.35)

c) The approximate displacement is underestimated

↑

↓=kFu . (7.36)

An approximate compatible displacement field corresponds to a structure which is stiffer than the actual structure and therefore will give a lower bound on displacement.

7.4 F.E.M. - a localized version of the Rayleigh-Ritz method

Instead of finding an admissible function satisfying the boundary conditions for the entire domain, which is often difficult, in the FEM the admissible functions are defined over small size subdomains.

7.4.1 F.E.M. in Structural Mechanics

a) Problem. Given a geometrically complex structure (including the boundary conditions) and the external loads { }vp , { }sp , { }iF , find the displacement field { }u within V and on the surface σS (Fig. 6.1). Then determine stresses, internal forces, reaction forces, etc.

b) Solution approach. Use PVD or PMTPE as an approximate method for solving the boundary-value problem. Admissible functions are defined over small size finite elements, with simple geometry and well identified structural behaviour. With these individually defined functions matching each other at certain points (nodes) at the element interfaces, the unknown function is approximated piecewise over the entire domain (continuity at global level).

c) Procedure. The geometric shape and the internal displacement field are described by a series of discrete quantities (like nodal coordinates and nodal displacements) distributed through the structure. For this a matrix notation is used.

d) Tools. Computers are used to store long lists of separate numbers and to manipulate them, to present output data in an engineering format, taking advantage of graphical and animation facilities.


7.4.2 Discretization

The structure is divided into finite elements (Fig. 7.8) that define the mesh. Elements are defined by their nodal coordinates and some physical parameters.

Fig. 7.8

7.4.3 Principle of virtual displacements

For the entire structure, equation (7.6, a) can be written (considering only surface tractions)

{ } { } { } { } 0dδdδδ =−= ∫∫σ

σεΠS

T

V

T ApuV s . (7.37)

As a summation of virtual works on all elements, PVD yields

{ } { } { } { } 0dδdδδδ =⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−== ∑ ∫∫∑

e S

T

V

T

e

e

ee

ApuVσ

σεΠΠ s . (7.38)

In the following, only eΠδ will be considered. The aim is the calculation of the element stiffness matrix and load vector.

7.4.4 Approximating functions for the element

In the Rayleigh-Ritz method, the trial function is expressed as a finite expansion


( ) ( ) ⎣ ⎦ ⎣ ⎦{ }a

a

aa

xax

n

n

n

jjj ϕϕϕϕϕ =

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=≅∑=

LL 2

1

211

u (7.39)

where the undetermined constants ja have no direct evident signification.

The basic idea of FEM is to choose the constants - the displacement unknowns at the nodes { } { }eQa = and to prescribe admissible functions denoted ⎣ ⎦ ⎣ ⎦N=ϕ so that

⎣ ⎦

⎣ ⎦⎣ ⎦

{ }{ }{ }⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

e

e

euu

QQQ

NN

N

w

v

w

v

wvu

000000

or { } [ ]{ }eQNu = , (7.40)

where [ ]N is the matrix of shape functions (interpolation functions).

The reason is that elements are small enough so that the shape of the displacement field can be approximated without too much error and only the magnitude, defined by { }eQ remains to be found.

The proper selection of shape functions ensure the continuity of the displacement field at global level. A finite element described by admissible shape functions (integrable in the interior and with equal values of generalized coordinates at element interfaces) is referred to as co-deformable or conforming.

7.4.5 Compatibility between strains and nodal displacements

From the compatibility relationship (6.18)

{ } [ ]{ } [ ][ ]{ } [ ]{ }ee QBQNu =∂=∂=ε , (7.41)

where [ ]B is the matrix of differentiated shape functions.

The strain virtual variation is

{ } [ ]{ }eQB δδ =ε .


7.4.6 Element stiffness matrix and load vector

Using the constitutive equation { } [ ]{ }εσ D= , the virtual work for an element is

{ } [ ] [ ][ ]{ } { } [ ] { } 0dδdδδ =−= ∫∫ee A

TTe

V

eTTee ApNQVQBDBQ sΠ

or

{ } [ ] [ ][ ] { } [ ] { } 0ddδδ =⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−= ∫∫

ee A

Te

V

TTee ApNQVBDBQ sΠ .

As { }eQδ are arbitrary and non-zero, cancelation of the bracket yields the element equilibrium equation

[ ] { } { }eee FQK = , (7.42)

where the element stiffness matrix is

[ ] [ ] [ ][ ]∫=eV

Te VBDBK d (7.43)

and the vector of consistent nodal forces is

{ } [ ] { }∫=eA

Te ApNF ds . (7.44)

7.4.7 Assembly of the global stiffness matrix and load vector

In the next step, all individual elements are assembled together so that the displacements are continuous across element interfaces and the boundary conditions are satisfied.

The kinematic connectivity is expressed by the relationship between element and global displacements

{ } [ ] { }QT~Q ee = , (7.45)

where { }eQ is the vector of nodal element displacements, { }Q is the vector of the

global displacements of the structure and [ ]eT~ is a connectivity matrix, containing ones at the nodal displacements of element nodes and zeros elsewhere.

The variation of element displacements is


{ } [ ] { }QT~Q ee δδ = . (7.46)

The PVD equation for the entire structure is

{ } [ ] { } { } { }∑∑ =e

eTe

e

eeTe FQQKQ δδ ,

or using (7.45) and (7.46)

{ } [ ] [ ] [ ] { } { } [ ] { }∑∑ =e

eTeT

e

eeTeT QT~QQT~KT~Q δδ .

As { }Qδ are arbitrary and non-zero, the unreduced global equilibrium equations are

[ ]{ } { }FQK = , (7.47)

where the global stiffness matrix is

[ ] [ ] [ ] [ ]∑=e

eeTe T~KT~K (7.48)

and the global load vector is

{ } [ ] { }∑=e

eTe FTF ~ . (7.49)

Applying the boundary conditions, the condensed equilibrium equations are

[ ]{ } { }FQK = . (7.50)

The above procedure is never used in practice. It has been used only to show algebraically how to assemble a global stiffness matrix. The assembly is done by directly placing the nonzero entries of element stiffness matrices in the right locations of the global stiffness matrix based on element connectivity.

7.4.8 Solution and back-substitution

Nodal displacements are determined by solving the linear equations (7.50).

In the back-substitution phase, element stresses are evaluated as

{ } [ ]{ } [ ][ ]{ } [ ][ ][ ] { }QT~BDQBDD ee === εσ

where [ ]D is given by (6.24) or (6.25).

8. TWO-DIMENSIONAL ELEMENTS

Many engineering structures can be modeled as two-dimensional flat plates, designed to be primarily loaded in their plane and to resist loads by membrane action rather than bending. In-plane displacement, strain and stress components are uniform through the plate thickness, which is considered constant. Only transversely homogeneous plates will be analyzed herein, composite and sandwich plates being studied in other courses. This chapter presents the element stiffness matrices and consistent force vectors for triangular and rectangular elements, that allow closed form derivations, without the need for numerical integration. The very first approximate finite element developed in 1956 to model delta wing skin panels, the three-noded triangle with constant strain field, is treated separately.

8.1 The plane constant-strain triangle (CST)

Before the advent of arbitrarily shaped isoparametric elements, discussed in the next chapter, the CST was one of the most widely used elements and is still available in systems today. It is a much more adaptable shape than the rectangle and it allows the user to tailor the element mesh to suit any structural geometry. A large number of small elements can be densely packed into a region of expected high stress gradients, and uniformly stressed regions can be left with a small number of larger triangles.

8.1.1 Discretization of structure

The plate is divided into a number of straight-sided triangles (Fig. 8.1, a), joined together at their corners (nodes), so that the corners of adjacent elements have common displacements. The elements fill the entire region except of a small region at the boundary. This unfilled region exists for curved boundaries and it can be reduced by choosing smaller elements.


The three nodes of the isolated element from Fig. 8.1, b are numbered locally as 1, 2 and 3. The corresponding nodal coordinates are designated as ( )11 y,x , ( )22 y,x and ( )33 y,x . The numbering is in anticlockwise direction to avoid calculating a negative area.

a b Fig. 8.1

Each node is permitted to displace in the two directions x and y. Thus, each node has two degrees of freedom. The displacement components of a local node j are denoted as ju in the x direction and jv in the y direction. The vector of element nodal displacements is defined as

{ } ⎣ ⎦Teq 332211 vuvuvu= . (8.1)

8.1.2 Polynomial approximation of the displacement field

The displacements u and v of a point within the triangle are expressed in terms of the nodal displacements. Because for two displacements there are six boundary conditions, the assumed displacement field is linear

( )( ) yaxaay,x

,yaxaay,x

654

321

++=++=

vu

(8.2)

with six arbitrary parameters.

8. TWO-DIMENSIONAL MEMBRANES 155

The strains (6.16) are

.constaxx ==

∂∂

= 2uε , .consta

yy ==∂∂

= 6vε ,

.constaaxyxy =+=

∂∂

+∂∂

= 53vuγ ,

hence the name “constant strain triangle”.

8.1.3 Nodal approximation of the displacement field

In the polynomial approximation (8.2) the constants ia have no physical meaning. A nodal approximation is preferred, in which the constants are the nodal displacements and the displacement field is obtained by interpolation based on values of corner displacements.

Since the functions for u and v are of the same form, only one need be considered in detail. We can write

⎣ ⎦ ⎣ ⎦ { }ayxaaa

yxu 11

3

2

1

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧= . (8.3)

Evaluating the expression for u at the three nodes gives the nodal displacements in terms of the polynomial coefficients

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

3

2

1

33

22

11

3

2

1

111

aaa

yxyxyx

uuu

, (8.4)

or

{ } [ ]{ }aAue = , (8.4, a)

where

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

33

22

11

111

yxyxyx

A . (8.5)

By inversion

{ } [ ] { }euAa 1−= . (8.6) where


[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=−

321

321

3211

21

γγγβββααα

AA . (8.7)

in which

jkkji yxyx −=α , kji yy −=β , jki xx −=γ (8.8)

and the subscripts k,j,i permute in a natural order.

The area of the triangle A is equal to one half the magnitude of the determinant of [ ]A

( ) ( ) ( )122131132332

33

22

11

111

2 yxyxyxyxyxyxyxyxyx

A −+−+−== . (8.9)

The determinant is positive if nodes 1, 2, 3 are labeled anticlockwise around the element.

Substituting (8.6) into (8.3) gives

⎣ ⎦ { }euNu = (8.10)

where the row vector of shape functions

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ [ ] 1321 1 −== AyNNNN x . (8.11)

By transposition

⎣ ⎦ [ ] ⎣ ⎦TTT yxAN 1−= ,

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

yx

ANNN 1

21

333

222

111

3

2

1

γβαγβαγβα

(8.12)

or

( )yxA

N iiii γβα ++=21 . ( )321 ,,i = (8.12, a)

Similarly

⎣ ⎦ { }eN vv = . (8.13)


The shape functions (8.12, a) can also be written

( ) ( ) ( )[ ]yxxxyyyxyxA

y,xN 233223321 21

−+−+−= ,


y,xN 311331132 21

−+−+−= , (8.12, b)


y,xN 122112213 21

−+−+−= .

Fig. 8.2

The shape functions iN vary linearly, have a unit value at node i and zero values at the other two nodes, as illustrated in Fig. 8.2:

( ) jiiii y,xN δ= , ( )321 ,,j,i = (8.14)

and

13

1=∑

=iiN . (8.15)

Combining (8.10) and (8.14) yields


⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

3

3

2

2

1

1

321

321

000000

vuvuvu

vu

NNNNNN

(8.16)

or

[ ]{ }eqN=⎭⎬⎫

⎩⎨⎧vu

. (8.16, a)

Fig. 8.3

The displacement ( ) ( ) ( ) ( )yx,Nyx,Nyx,Nyx, 332211 uuuu ++= determines a plane surface passing through 1u , 2u and 3u , as shown in Fig. 8.3.

8.1.4 The matrix [ ]B

Strains are expressed in terms of displacements as

{ } [ ]{ } [ ]{ } [ ]{ }eee qBqN

xy

y

x

xy

y

x

=

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=⎭⎬⎫

⎩⎨⎧

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=∂= 0

0

0

0

vu

uε .


The matrix of the derivatives of shape functions is

[ ]

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=

xN

yN

xN

yN

xN

yN

yN

yN

yN

xN

xN

xN

B

332211

321

321

000

000

,

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

332211

321

321

000000

21

βγβγβγγγγ

βββ

AB ,

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−−−−−

−−−=

211213313223

123123

211332

000000

21

yyxxyyxxyyxxxxxxxx

yyyyyy

AB . (8.17)

The matrix [ ]B is constant for a given CST element. Sometimes it is written simply

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

122131132332

211332

123123

000000

21

yxyxyxxxx

yyy

AB (8.17, a)

where the notation is obvious.

8.1.5 Element stiffness matrix and load vector


[ ] [ ] [ ][ ] [ ] [ ][ ] AtBDBVBDBk eT

V

Te

e

== ∫ d , (8.18)

where et is the element thickness, A is the element area and [ ]D is the material stiffness matrix given by (6.24) for plane stress and by (6.25) for plane strain conditions.

The consistent nodal forces due to traction loads acting on a portion of the boundary are calculated as for a linear two-node element. Along an edge 1-2, the shape function 3N is zero while 1N and 2N are similar to the shape functions in one dimension, satisfying 121 =+ NN . When the surface load distribution (per unit


area) is linear, varying from 1p at node 1 to 2p at node 2 (Fig. 8.4), the nodal forces are

( )211 26

pptfe

e +=l , ( )212 2

6pptf

ee +=

l (8.19)

where et is the thickness of the element. Because of linearity they coincide with the static resultants.

Fig. 8.4

The nodal forces associated with the weight of an element are equally distributed at the nodes.

8.1.6 Remarks

A mesh like in Fig. 8.5, a is clearly a directionally sensitive assembly, and this could be corrected by using the “union jack” pattern of Fig. 8.5, b which however produces a larger bandwidth. Benchmark tests using triangular elements have shown that CST elements, even in a fine mesh, are much inferior to higher order elements in a coarse mesh.

Fig. 8.5


A drawback of the displacement form of the finite element method is that equilibrium is only satisfied in the mean or over the element. This means that along an edge which is common to two elements the stresses are different across the edge, where they should be continuous. Most programs contain facilities for averaging the stresses. The simplest form of averaging consists of simply connecting the centroids of two adjacent triangles and to assign the mean stress value to the crossing point of this line with the common edge.

A simple example of stress averaging is shown in Fig. 8.6 for a square plate with a circular hole. Taking advantage of symmetry, only one quarter of the plate is considered. The diagram compares the theoretical stress distribution along the marked line with the averaged values calculated using CSTs.

Fig. 8.6


The procedure will be used in Example E8.3

Example 8.1 A square plate of thickness t and Young’s modulus E is pin-jointed at

three corners and subjected to a force F at the free corner (Fig. E8.1, a). Let 0=ν . Divide the plate into four CST elements and find: a) the nodal

displacements; b) stresses in elements; and c) the support reactions.

a b

Fig. E8.1

Solution. The material stiffness matrix (6.24) is

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

2100010001

ED . (a)

With the origin of coordinate axes at the plate centre, the nodal coordinates of element 1 are

011 == yx , l=2x , 02 =y , 03 =x , l=3y .

The area is 22l=A and the matrix (8.17) is

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−=

011011100010000101

11

lB . (b)



[ ] [ ] [ ] [ ] [ ] AtBDBkKT 1111 == , (c)

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡−−−

−−−

=

1021SYM02121000112121023

0212112123

21 tEK . (d)

Element 2 can be obtained by rotating 090 anticlockwise element 1. In general, the transformation matrix for a rotation with an angle θ is

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

−

=

cssc

cssc

cssc

T e

00000000

0000000000000000

(e)

where θcos=c and θsin=s .

The stiffness matrix of the rotated element is

[ ] [ ] [ ] [ ].TkTK eeTee = . (f)

For 090=θ we obtain

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

−

=

010000100000000100001000000001000010

2T (g)

The stiffness matrix of element 2 is

[ ] [ ] [ ] [ ].TkTKT 2122 = ,


[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−−

=

2101SYM001

21002121012123

2110212123

22 tEK . (h)

The same matrix is obtained by substituting the nodal coordinates

011 == yx , 02 =x , 12 =y , 13 −=x , 03 =y

in (8.17) and performing the product (8.18).

Due to symmetry, only the upper half of the plate can be considered (Fig. E8.1, b). Using the appropriate boundary conditions, the finite element equations can be written

⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪

⎨

⎧

=

⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−−

−−−

−−−−−−−

∗

4

4

3

3

2

1

2

1

2

0

00000

0

2100210021210100000100200020

210012100100021210212100000101

21020210302110121103

2

VHVHVFV

u

u

tE .

Solving

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−2

01113

2 2

1

FuutE

we obtain the nodal displacements

tE

Fu21 = ,

tEFu

23

2 = .

The reaction forces for half the plate are obtained as

43 FH −= , 44 FH −=∗ ,

so that for the entire plate

24 FH −= , 45 FH −= .


Strains and stresses in elements are

{ } [ ]{ }

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−==

000

0

011011100010000101

1 2

1

111 u

u

qBl

ε ,

{ }⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−=

⎭⎬⎫

⎩⎨⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−=

102

2010011

12

11

tEF

uu

llε .

[ ]{ }⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

2102

21

02

22100

010001

1

1

tF

tEFED

xy

y

x

llε

τσσ

,

{ } [ ]{ }⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−==

101

2

00000

110011001010010001

1

1

222

tEF

u

qBll

ε ,

[ ]{ }⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

2101

21

01

22100

010001

2

2

tF

tEFED

xy

y

x

llε

τσσ

.

Example 8.2

A thin triangular plate is fixed along the edge 5-4 and loaded by forces 11 −=F and 22 −=F along the upper edge in its plane (Fig. E8.2). Assume the

cantilever to have unit thickness 1=t and be in plane stress, with Young’s modulus 1=E and Poisson’s ratio 30.=ν . Divide the plate into three CST elements and find: a) the nodal displacements; b) stresses in elements; and c) the support reactions. The units are coherent.


Fig. E8.2

Solution. The input data are given below

For each element, the stiffness matrix is calculated longhand from equation (8.18), where the matrix [ ]B is obtained from (8.17), based on the nodal coordinates, and the matrix [ ]D is given by (6.24).


The element stiffness matrices are the following:

The condensed global stiffness matrix is

The condensed finite element equations can be written

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

−

−

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−−−

−−

002

01

0

125209031001650030110666019200903101252011101650066600301119203170019201110192011100165001650317003170

3

3

2

2

1

1

vuvuvu

......

............

....

.

The solution to these equations gives

71271 .=u , 54262 .=u , 68623 .−=u , 823401 .−=v , 835152 .−=v , 582133 .−=v .


The output data are presented below:

The reaction forces at nodes 4 and 5 are obtained from the ‘unused’ unreduced equations.

Note that the adopted discretization is very crude, to allow longhand calculation, and this leads to misleading results. Along the edge 4-5, a distribution of bending stresses from compressive at 4 to tensile at 5 is expected. However, element 3, being a constant strain element, gives only one value of xσ , which is wrong. Normally, the plate should be modeled by many more elements. Ascribing stress values to the element centroids and averaging them as shown in section 8.1.6, a more realistic stress distribution along the edge 4-5 is obtained.

Example 8.3

A thin rectangular plate, containing a circular hole of radius mm10=a , is subjected to loads that produce uniform tensile stresses MPa50 =σ at its ends (Fig. E8.3, a). The plate has length mm60=l , width mm40=b , thickness

mm5=t , GPa200=E and 30.=ν . Determine: a) the deformed shape of the hole; b) the location and magnitude of the maximum von Mises stress in the plate; c) the distribution of xσ stresses in the midsection. Compare the stress values at the periphery of the hole obtained by FEM and from the theory of elasticity.

Solution. Taking advantage of the symmetry of geometry and symmetry of loading, we can analyze only one-quarter of the plate (upper right).


Fig. E8.3, a

A 55-node, 81-element mesh is created as shown in Fig. E8.3, b. Let x and y represent the axes of symmetry. The points along the x axis are constrained in the y direction, and points along the y axis are constrained along the x direction.

Fig. E8.3, b

The applied nodal forces are shown, but the element numbering is omitted for clarity. The centroids of the elements near the midsection are marked, and the crossing points of the lines connecting the centroids with the common sides (where stresses are averaged), are denoted a to f.

The deformed shape is shown in Fig. E8.3, c. The hole is elongated in the direction of the loading axis.


Fig. E8.3, c

The calculation of stresses is summarized in Fig. E8.3, d. Elements are hatched according to the value of von Mises stresses, using five intervals with limits shown in the legend.

Fig. E8.3, d

The maximum von Mises stress is MPa619. and occurs in element 1. Stress values in elements near the midsection are given in Table E8.3.


Table E8.3

Element xσ yσ xyτ eqσ

1 20.0178 0.8845 -0.5706 19.6155 2 11.4577 1.9228 -0.6221 10.6821 3 11.3275 0.815 0.0595 10.9433 4 8.068 1.5768 0.927 7.5786 5 7.6445 0.1651 0.174 7.5693 6 5.7162 0.2632 0.6667 5.7073 7 4.0752 0.763 0.3595 3.8037

Stresses xσ , averaged at points a to f, have the following values:

Point a b c d e f

MPa,xσ 15.78 11.39 9.69 7.85 6.68 4.89

The theoretical distribution of stresses xσ in the midsection of a plate containing a small circular hole and subjected to uniaxial tension is approximated by

⎟⎟⎠

⎞⎜⎜⎝

⎛++= 4

4

2

2

0 23

21

ra

ra

x σσ ,

where r is the distance from the x axis (Fig. E8.3, a).

The maximum stress occurs at the periphery of the hole ( )ar = and is

MPa153 0 == σσmaxx .

The averaged value of elements 1 and 2, in point a, is MPa7815.ax =σ ,

which is surprisingly accurate for the rough mesh used in the exercise.

Example 8.4

An axially loaded thin plate with a circular fillet (Fig. E8.4, a) has length mm150 , width of the reduced portion mm40 , width of the extended portion

mm80 , thickness mm5 , fillet radius mm20 , MPa102 5⋅=E and 250.=ν . The normal stress at a point far to the right of the fillet has a uniformly distributed value


of MPa440 . Find the location and magnitude of the maximum von Mises stress in the plate. Determine the stress concentration factor for the circular fillet.

Fig. E8.4, a

Answer. Taking advantage of the symmetry, only half of the plate is considered. The points along the symmetry axis are constrained in the vertical direction. Points along the left edge are constrained in the horizontal direction. A model comprising 95 nodes and 144 CST elements is constructed as shown in Fig. E8.4, b.

Fig. E8.4, b

The nodal loads are calculated from equation (8.19). The right edge has four elements, each with a surface of 2mm25 , subjected to a normal stress of

2mmN440 . The nodal loads for each element are N5500225440 =× / . The resulting five nodal forces, for the quarter plate, have magnitudes of N5500 at the upper and lower node, and N11000 at the three middle nodes.

The distribution of von Mises stresses is shown in Fig. E8.4, c for five stress intervals given in the legend. The largest equivalent stress is .. MPa7604


Fig. E8.4, c

Values of the principal stress 1σ around the fillet are presented in the upper part. The largest principal stress 1σ is .. MPa7622 The stress concentration factor relative to the value MPa440 is ..4151 This value is reasonably close to the theoretical true value of 421. given by Singer (1962).

Example 8.5

The nodal coordinates and the nodal displacements of element 96 in Fig. E8.4, b are given below:

359260 .x = , 522160 .y = , 9564 =x , 1864 =y , 59565 .x = , 52065 .y = ;

11224060 .u = , 397360 −−= e.v , 12367064 .u = , 363364 −−= e.v , 12247065 .u = , 392465 −−= e.v .

MPa102 5⋅=E and 250.=ν . Calculate the element stresses.

Solution. The triangle area is 296 mm19254.A = .

The matrix [ ]B is (8.17)

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

−−=

419803160012160375702982005960316000375700059600041980012160029820

......

......

B .


The matrix [ ]D is (6.24)

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

80000133325333005333013332

105

.....

D .

The vector of nodal displacements (8.1) is

{ } ⎣ ⎦T......q 3e9241224703e6331236703e973112240 −−−−−−= .

Element stresses are given by

{ } [ ][ ]{ }qBD=σ ,

{ } ⎣ ⎦ ⎣ ⎦TTxyyx ... 08120586372596 −== τσσσ .

Example 8.6

Find the deformed shape and the stress distribution in the loaded gear tooth shown in Fig. E8.6, a. The load is not applied at the tooth tip, but is distributed over a larger area than for the mating contact of two teeth, to concentrate only on the influence of fillet geometry.

Fig. E8.6, a


Answer. The adopted finite element mesh has 126 nodes and 212 CST elements. Note the restraints which model the tooth built-in end. The deformed shape is presented in Fig. E8.6, b.

Fig. E8.6, b

The stress distribution is shown in Fig. E8.6, c. Note the stress concentration at both fillet areas, where the equivalent von Mises stresses are indicated.

Fig. E8.6, c


8.2 Rectangular elements

Rectangular elements are the easiest to discuss and are used to model stiffened thinwalled panel structures, built up beams and boxes.

8.2.1 The four-node rectangle (linear)

Consider the 4-node element in Fig. 8.7.

Fig. 8.7

We have to find two-dimensional shape functions which have unit values at one selected node, but zero at all other nodes on the element

( ) ijjji y,xN δ= . (8.20)

They can be generated using the equations of the sides. For example, 1N is identically zero on lines ax = and by = , and has a unit value at the node with the same index ( ) 1111 =y,xN , hence it must be of the form

( ) ( ) ( )ybxacy,xN −−= 11 .

From condition ( ) ( ) 1001111 == ,Ny,xN we get ab

c 11 = , hence

( ) ( ) ( ) ⎟⎠⎞

⎜⎝⎛ −⎟

⎠⎞

⎜⎝⎛ −=−−=

by

axybxa

bay,xN 111

1 .

Likewise, the remaining three shape functions are

( ) ⎟⎠⎞

⎜⎝⎛ −=

by

axy,xN 12 , ( )

by

axy,xN =3 , ( ) ⎟

⎠⎞

⎜⎝⎛ −=

ax

byy,xN 14 .

It is convenient to transform to dimensionless central coordinates


12−=

axr , 12

−=bys . (8.21)

The equations of element sides become

01 =± r ,…. 01 =± s . (8.22)

The required shape functions are

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) .srN,srN

,srN,srN

1141 11

41

1141 11

41

43

21

+−=++=

−+=−−= (8.23)

The displacements inside the element can be expressed in terms of the nodal displacements as

{ }eqNNNN

NNNN⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

4321

4321

00000000

vu

, (8.24)

where

{ } ⎣ ⎦44332211 vuvuvuvu=Teq . (8.25)

The use of the shape functions (8.23) has one drawback – they are not complete. That is, all like powers in x and y are not present. In this case yx is

present but 2x and 2y are not. Thus the variation of strain does not have the same order in all directions.

The displacements u and v of a point within the rectangle can be expressed in terms of the nodal displacements by a polynomial approximation. Because, for two displacements, there are eight boundary conditions, the assumed displacement field is

( )( ) .yxbybxbby,x

,yxayaxaay,x

4321

4321

+++=

+++=

vu

(8.26)

The direct strain yaax 42 +=ε is constant in the x direction whereas the shear strain ybbxaaxy 4243 +++=γ varies linearly with x and y .

Some improvement can be gained by diminishing the shear variation through the use of reduced integration on the shear contribution in the element stiffness matrix. Many users prefer to use higher order elements rather than use a larger number of small 4-noded rectangles. The 4-noded rectangular element is exploited in non-linear problems like elasto-plastic behaviour, where the stiffness has to be reevaluated as the load increments are applied and plasticity spreads.



[ ] [ ] [ ][ ]∫=eV

Te VBDBk d ,

where [ ] [ ][ ]NB ∂= .

The poor performance of the 4-noded element is shown below, in an example where it is expected to behave like a slender beam in which bending stresses are dominant.

Fig. 8.8

Figure 8.8, a shows the deformed shape of a single element in pure bending. In comparison, Fig. 8.8, b shows the expected circular deformed shape free of shear deformations. The 4-node element has flat sides. At the element ends the shear strain is bdxy =γ and only the centre has no shear deformation. The direct strains are adx =ε on the upper and lower surface. The ratio

ratioaspect strains bendingshears spurious

==ba

.

This explains the poor results obtained with high aspect ratio elements ( )3≥ which have also badly conditioned stiffness matrices. The stress discontinuities at element interfaces can be comparable with the mean values and the free edge stresses are not zero. Generally, because the strain energy is underestimated, stresses are lower than the true values.

8.2.2 The eight-node rectangle (quadratic)

The higher order 8-node rectangular element is shown in Fig. 8.9. Adding four nodes means adding four supplementary boundary conditions (or nodal displacements) so that in the polynomial approximation we can add four more (higher order) terms.


The assumed displacement field is cubic

( )( ) ,yxbyxbybyxbxbybxbby,x

,yxayxayayxaxayaxaay,x

2

82

72

652

4321

28

27

265

24321

+++++++=

+++++++=

v

u (8.27)

so that the strains are quadratic. Note that choosing the last terms in 3x and 3y would make the element more anisotropic. The 16 constants can be expressed (interpolated) in terms of the 16 nodal displacements and the nodal coordinates.

Using the nodal approximation, the displacement field is expressed as

,N...NNN,N...NNN

88332211

88332211

vvvvvuuuuu

++++=++++=

(8.28)

where the shape functions iN can be easily built up based on the equations of the lines passing through the nodes (serendipity approach).

Fig. 8.9

In Fig. 8.9 only the lines through the new nodes are shown for clarity, the old ones are presented in Fig. 8.7. In the natural system of coordinates r and s, the shape functions have the following expressions

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ).11121 , 111

21

, 11121 , 111

21

, 11141 ,111

41

, 11141 , 111

41

87

65

43

21

ssrNsrrN

ssrNsrrN

srsrNsrsrN

srsrNsrsrN

+−−=++−=

+−+=−+−=

−++−−=−−++−=

+−−+−=++−−−=

(8.29)


The interpolation function 1N should be zero at nodes 82, 3,..., and have a value of unity at node 1. Equivalently, 1N should vanish on the sides defined by the equations of lines passing through nodes 2 to 8: 01 =− s , 01 =− r ,

01 =++ sr . Therefore 1N is of the form

( ) ( ) ( ) 11111 ,srsrcN ++−−=

where 1c is a constant which is determined so as to yield ( ) 1111 =−− ,N , i.e. 411 −=c , which gives the expression of 1N in (8.29).


{ }eqNNN

NNN⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

821

821

000000

LL

LL

vu

, (8.30)

where

{ } ⎣ ⎦Teq 882211 vuvuvu LL= . (8.31)

Having generated the shape functions [ ]N , the functions [ ] [ ][ ]NB ∂=

follow and the element stiffness matrix [ ]ek can be evaluated using [ ]D .

8.3 Triangular elements

In order to obtain accurate results for stresses with a constant strain discretization, one has to use a large number of elements. Considerable effort has been devoted to develop ‘refined’ elements, i.e. elements having linear, quadratic or higher-order strain expansions.

8.3.1 Area coordinates

For a triangle it is possible to define a completely symmetrical coordinate system known as area coordinates.

The position of any point M in the triangle is identified by the perpendicular distances 1h , 2h , 3h from the three sides (Fig. 8.10, a), and non-dimensionalized, by division to the triangle heights, as

1

11 H

hζ = ,

2

22 H

hζ = ,

3

33 H

hζ = ,


or

AA

ζ 11 = ,

AA

ζ 22 = ,

AA

ζ 33 = . (8.32)

As AAAA =++ 321 (Fig. 8.10, b), one obtains

1321 =++ ζζζ , (8.33)

so the three coordinates are not independent and they behave like the shape functions.

Fig. 8.10

The physical coordinates of point M can be expressed in terms of the nodal coordinates as

,yζyζyζy,ζζζ

332211

332211

++=++= xxxx

(8.34)

or in matrix form

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

3

2

1

321

321

1111

ζζζ

yyyxxx

yx .

By inversion, we obtain equation (8.12)

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

yx

A

1

21

ζζζ

333

222

111

3

2

1

γβαγβαγβα

or


( )yxA iiii γβα ++=

21ζ . ( )321 ,,i = (8.35)

where A is given by (8.9) and the expressions (8.8) are

jkkji yxyx −=α , kji yy −=β , jki xx −=γ .

8.3.2 Linear strain triangle (LST)

The linear strain triangle is a six-noded element, obtained adding three mid-side nodes to the CST. In Fig. 8.11, the equations of the three sides and the lines through the mid-side nodes are shown using area coordinates.

Fig. 8.11

The shape functions are readily seen to be

. ζ ζ4 , ζ ζ4 , ζ ζ4

, ζ21ζ2 , ζ

21ζ2 , ζ

21ζ2

136325214

333222111

===

⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛ −=

NNN

NNN (8.36)

The displacements in terms of the nodal displacements are

{ }eqNNN

NNN⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

621

621

000000

LL

LL

vu

, (8.37)

where { } ⎣ ⎦Teq 662211 vuvuvu LL= . (8.38)


To form [ ] [ ][ ]NB ∂= we need to find the derivatives of shape functions with respect to the physical coordinates, so we must transform from y,x to

321 ζ ζ ζ ,, . Using (8.33) we can eliminate 1ζ in (8.34) obtaining

( ) ( )( ) ( ) .yζyyζyyy

,ζxζx

1313212

1313212

+−+−=+−+−= xxxx

(8.39)

Using the chain rule for partial derivatives

[ ]⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

y

xJ

y

xyx

yx

33

22

3

2

ζζ

ζζ

ζ

ζ , (8.40)

where [ ]J is the Jacobian of the transformation.

The inverse relationship is

[ ]⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

⎥⎦

⎤⎢⎣

⎡=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

−

3

2

32

32

3

21

ζ

ζ21

ζ

ζγγββ

AJ

y

x , (8.41)

where A is the triangle area (8.9) and iβ , iγ are defined by (8.8).

The element stiffness matrix is

[ ] [ ] [ ][ ]∫=A

eTe AtBDBk d , (8.42)

where

323223 dd2sinddd ζζθζζ AA == ll .

The integration is frequently performed numerically. However, it can be solved explicitly in terms of a, b and [ ]D using the integration formula for monomials

( )!2!!2dζζba

baAAbj

A

ai ++

=∫ . (8.43)

The results for first- and second-degree terms are

3dζ AA

Ai =∫ ,

12dζζ AAj

Ai =∫ ,

6dζ2 AA

Ai =∫ . (8.44)


The LST element is based on expanding the displacements in a complete second-degree polynomial in 2ζ and 3ζ

.bbbbbb

,aaaaaa236325

22433221

236325

22433221

ζζζζζζ

ζζζζζζ

+++++=

+++++=

v

u (8.45)

The displacement field is complete, containing all possible products, so the element is truly isotropic, without recourse to computationally inefficient internal nodes. There are twelve nodal displacements, six for each component. In order to satisfy inter-element displacement compatibility, the displacement expansion on a boundary must involve only the nodal quantities for that boundary. There are three constants for this case (the function is quadratic) and an additional interior node is required for each boundary. It is convenient to locate these interior nodes at the mid-points of the sides.

Solving for the constants in terms of the nodal displacements, the nodal approximation (8.37) is obtained.

Denoting

,rζ =2 ,sζ =3 srζζζ −−=−−= 11 321 (8.46)

we can introduce oblique triangular coordinates (Fig. 8.12).

The physical coordinates of a point within the triangle are

( )( ) .ysyrysry

,srsr

321

321

11

++−−=++−−= xxxx

(8.47)

Fig. 8.12

The coefficients of the nodal coordinates are genuine geometric interpolation functions

srN −−=11 , ,rN =2 ,sN =3 (8.48)


The nodal displacements for the 3-node triangle can also be written

( )( ) ,srsr

,srsr

321

321

11

vvvvuuuu

++−−=++−−=

(8.49)

so that the same functions can be used as interpolation functions. This is the basic idea behind the isoparametric formulation, treated in the next chapter.

8.3.3 Quadratic strain triangle

A triangular element with a quadratic strain field can be built up in two ways. One possibility is to work with corner point nodes and two interior nodes per side, taking as nodal quantities the displacement components. This element is shown in Fig. 8.13, a.

Fig. 8.13

The displacements are expressed as complete cubic polynomials. In order to satisfy inter-element displacement compatibility, the displacement function for a side must depend only on the nodal displacement quantities for the side. Since the function is cubic, four nodal quantities are required to define the distribution on a side. The side nodes are located at the third points. An additional interior node is needed to maintain completeness of the polynomial since, if the polynomial is not complete, the stiffness will have preferred direction which is not desirable. It is convenient to take the interior node at the centroid.

The displacement nodal expansion has the form

cci

ii uNuNu +=∑=

9

1, cc

iii NN vvv +=∑

=

9

1, (8.50)

where


( ) ( ) ( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

.N

NN

NN

NN

N

NN

c 321

11393138

33272326

22151214

3333

22221111

ζ ζ ζ27

,1ζ3 ζ ζ29 ,1ζ3 ζ ζ

29

,1ζ3 ζ ζ29 ,1ζ3 ζ ζ

29

,1ζ3 ζ ζ29 ,1ζ3 ζ ζ

29

, 2ζ31ζ3ζ21

, 2ζ31ζ3ζ21 , 2ζ31ζ3ζ

21

=

−=−=

−=−=

−=−=

−−=

−−=−−=

(8.51)

Another possibility is to work only with corner nodes, in order to reduce the bandwidth of the stiffness matrix. The solution is to include displacement

derivatives xu∂∂ ,

yu

∂∂ ,

x∂∂v ,

y∂∂v as nodal quantities. At each corner, there are six

nodal variables, two displacements and four first derivatives, a total of 18 parameters (Fig. 8.13, b). Two additional displacement quantities not associated with the boundaries are required for completeness. It is convenient to take the displacement components of the centroid ( )cc , vu as the remaining parameters.

The nodal expansion for u has the form

ccyxyx uNuN...uNuNuNuNuNu +′+++′++′+′+= 392524131211 . (8.52)

In (8.52) the interpolation polynomials, expressed in terms of triangular coordinates, are

( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )

,NN

N

N

N

N

N

N

N

N

c 321

321121221239

321212112238

321213237

321313113226

321131331225

321132224

321232332213

321323223212

321321211

ζ ζ ζ27 , ζ ζ ζζζζ

, ζ ζ ζζζζ

, ζ ζ ζ7ζ3ζ3ζζ

, ζ ζ ζζζζ

, ζ ζ ζζζζ


, ζ ζ ζζζζ

, ζ ζ ζζζζ


=−+−=

−+−=

−++=

−+−=

−+−=

−++=

−+−=

−+−=

−++=

ββββ

γγγγ

ββββ

γγγγ

ββββ

γγγγ

(8.53)


where iγ , iβ are defined by (8.8) and the subscript c refers to the centroid.

The same interpolation functions are valid for v .

8.4 Equilibrium, convergence and compatibility

It is useful to make some general comments on the fulfilment of the equilibrium and compatibility conditions in a finite element solution based on assumed displacement fields [33].

8.4.1 Equilibrium vs. compatibility

In a FEM solution based on assumed displacement fields, one can say that:

a) Within elements, compatibility is satisfied if the assumed element displacement field is continuous, but equilibrium is usually not satisfied.

Let write the equilibrium equations in terms of displacements for the case 0== yx pp vv . Substituting the stress-strain relations (6.24) into the equilibrium

equations (6.7) we obtain

( )

( ) .yx

ν

,y

νx

xyxy

xyyx

02

1

02

1

=+∂∂

+∂∂−

=∂∂−

++∂∂

ενεγ

γενε (8.54)

Substituting the strain-displacement relations (6.16) in (8.54) gives

.yxx

νyx

,yxy

uνyu

xu

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂∂

−∂∂+

=∂∂

+∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂∂

−∂∂+

=∂∂

+∂∂

uvvv

v

2

2

2

2

2

2

2

2

2

2

2

2

2

2

21

21

(8.55)

Equations (8.55) are not satisfied by the assumed displacement field (8.26)

( )( ) .yxbybxbby,x

,yxayaxaay,x

4321

4321

+++=+++=

vu

(8.56)

of the rectangular element.

For example,


yaaxu

42 +=∂∂ , xaa

yu

43 +=∂∂ , 02

2

2

2=

∂∂

=∂∂

yu

xu , and 4

2b

yx=

∂∂∂ v

which is not zero, so the first equation (8.55) is not satisfied.

The rectangle would satisfy equilibrium if 044 == ba , but this is the case only in a field of constant strain. But as already shown, the direct strain xε is linear in the y direction whereas the shear strain xyγ varies linearly with x and y .

However, equilibrium is satisfied within the CST because of its extreme simplicity. One cannot conclude from this that, in general, the rectangle is inferior to the triangle. In some cases it can give better results.

b) Between elements, compatibility may or may not be satisfied, and equilibrium is usually not satisfied.

For example, for both the 3-node triangle and the 4-node rectangle, u and v are linear in x ( )yor along element edges. So, for any nodal displacement, the edges remain straight, and adjacent elements do not overlap or separate, the elements fit together. Inter-element equilibrium is obviously violated in the CST, where stresses are constant within the element but differ from one element to another. However, inter-element stress continuity may exist, as in the case of uniform beams loaded only at nodes.

When inter-element compatibility is satisfied, the finite element solution gives an upper bound on the total potential energy. As the discretization is refined, the solution will converge monotonically to the true solution, provided that the new mesh contains all the nodes of the previous meshes.

c) At nodes, compatibility is enforced by joining elements at these locations, and equilibrium of nodal forces and moments is satisfied.

The finite element equations are a set of equilibrium equations and the solution is such that resultant forces and moments acting on each node are zero.

Happily, in a ‘proper’ finite element solution, any violations of equilibrium and compatibility tend to vanish as more and more elements are used in the mesh. Moreover, the convergence of displacements with the mesh refinement must be monotonic from below.

8.4.2 Convergence and compatibility

As the mesh of elements is refined, the sequence of solutions to a problem is expected to converge to the correct result if the assumed element displacement fields satisfy the following criteria:


a) The displacement field within an element must be continuous. This is normally ensured by the selection of shape functions.

b) When the nodal degrees of freedom are given values corresponding to a state of constant strain, the displacement field must produce the constant strain state throughout the element.

The check is done using the so called “patch test”. The model consists of an assembly of several elements arranged so that at least one node is completely surrounded by elements. The boundary nodes are then given either displacements or forces consistent with a constant strain state. Internal nodes are to be neither loaded nor restrained. The computed displacements, strains, and stresses within elements should be consistent with the constant strain state. If not, the element type is invalid or at least suspect (it may happen that an element is valid in certain configurations only).

c) Rigid body modes must be represented. When nodal degrees of freedom are given values corresponding to a state of rigid body motion, the element must exhibit zero strain and therefore zero nodal forces.

If this requirement is violated, extraneous nodal forces appear, and the equations of nodal equilibrium are altered. To satisfy the requirements on both rigid body modes and constant strain rates, the expansion must be at least a complete polynomial of order equal to the highest derivative occurring in the strain-displacement relations.

d) Compatibility must exist between elements. Elements must not overlap or separate. In the case of beam, plate and shell elements, the slope must be continuous across interelement boundaries.

This requirement is violated by many successful non-conforming elements. Such elements do satisfy inter-element compatibility in the limit of mesh refinement, as each element approaches a state of constant strain. However, non-compatible elements do not provide a bound on the potential energy, i.e. we do not know whether the potential energy corresponding to a particular non-conforming element is higher or lower than the true value. Also, it is not possible to construct a minimizing sequence with non-compatible elements. This means that by suitably specializing the nodal displacements for the nth discretization, we will not be able to reproduce the displacement patterns corresponding to the 1−n previous discretizations.

e) The element should have no preferred directions. Elements should be invariant with respect to the orientation of the load system. Invariance exists if complete polynomials are used for element displacement fields. It is achieved even when based on incomplete polynomials, if there is a ‘balanced’ representation of terms in the polynomial expansion.


Example 8.7

For the assembly of four triangular elements shown in Fig. E8.6, a simple patch test is carried out as follows.

Fig. E8.7

Assume a displacement field yx +== vu

for which the strains are

{ } ⎣ ⎦T211=ε .

The resulting nodal displacements are

01 =u , 12 =u , 43 =u , 14 =u , 01 =v , 12 =v , 43 =v , 14 =v .

If these are the boundary conditions imposed to the model, the internal node must behave accordingly.

The condensed set of finite element equations has the form

[ ] { }FK =⎭⎬⎫

⎩⎨⎧

5

5

vu

,

where [ ]K depends on the material properties and { }F depends on both the material properties and the prescribed nodal displacements.

Its solution must be

25155 .== vu

and the strains at any point must be { } ⎣ ⎦T211=ε .

9. ISOPARAMETRIC ELEMENTS

Simple triangular and rectangular elements allow closed form derivations of stiffness matrices and load vectors. The construction of shape functions and evaluation of stiffness matrices for quadrilateral and higher-order elements with curved sides faces difficulties which are overcome by the use of isoparametric elements and numerical integration. Elements with curved sides provide a better fit to curved edges of an actual structure.

For isoparametric elements, the same interpolation functions are used to define the element shape as are used to define the displacement field within the element. The constant strain triangle is an isoparametric element though it was not treated like that. In fact, the shapes of the interpolation functions and not the parameters are the same. It is possible to construct subparametric elements whose geometry is determined by a lower order model than the displacements. When the eight node rectangle is transformed into a quadrilateral with straight sides, we obtain a subparametric element. When it is mapped into a quadrilateral with parabolically curved sides, the result is an isoparametric element. If we develop higher order triangular elements while keeping straight sides, they are subparametric elements, because only the displacement expansion is refined whereas the geometry definition remains the same.

9.1 Linear quadrilateral element

Consider the general quadrilateral element shown in Fig. 9.1, a. The local nodes 1, 2, 3 and 4 are labelled counterclockwise. The coordinates of node i are ( )ii y,x . Each node has two degrees of freedom. The displacement components of a local node i are denoted as iu in the x direction and iv in the y direction. The vector of element nodal displacements is defined as

{ } ⎣ ⎦Teq 44332211 vuvuvuvu= . (9.1)


9.1.1 Natural coordinates

A natural coordinate system can be attached to a quadrilateral element as illustrated in Fig. 9.1, b. The coordinates r and s vary from 1− on one side to 1+ at the other, taking the value zero over the quadrilateral medians. They are called quadrilateral coordinates.

a b

Fig. 9.1

In the development of isoparametric elements it is useful to visualize the quadrilateral coordinates plotted as cartesian coordinates in the { }s,r plane. This is called the reference plane. In the reference plane, quadrilateral elements become a square of side 2 (Fig. 9.2, a), called the reference element (or master element), which extends over [ ]11,r −∈ , [ ]11,s −∈ . The transformation between the natural coordinates { }s,r in the reference plane and the cartesian coordinates { }y,x is called the isoparametric mapping.

a b

Fig. 9.2

9. ISOPARAMETRIC ELEMENTS 193

Each quadrilateral ‘child’ in the { }y,x is generated by the ‘parent’ or ‘reference’ element from the { }s,r plane (Fig. 9.2, b). The advantage is that the interpolation functions defined for the reference element are the same for all actual elements and have simple expressions. The drawback is that the mapping is a change of coordinates which implies complications in the evaluation of the integral in the element stiffness matrix.


The interpolation functions (8.23) developed for the rectangle are directly applicable

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) .srN,srN

,srN,srN

1141 11

41

1141 11

41

43

21

+−=++=

−+=−−= (9.2)

The following compact representation is useful for the implementation in a computer program

( ) ( ) ( ) 1141 ,ssrrs,rN iii ++= (9.3)

where ( )ii s,r are the coordinates of node i .

The coordinates of a point within the element are expressed in terms of the nodal coordinates as

.yNyNyNyNyNy

,NNNNN

iii

iii

∑

∑

=

=

=+++=

=+++=

4

144332211

4

144332211 xxxxxx

(9.4)

In matrix form

⎣ ⎦ { } ⎣ ⎦

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

==

4

3

2

1

4321

xxxx

NNNNN exx , (9.4, a)

⎣ ⎦ { } ⎣ ⎦

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

==

4

3

2

1

4321

yyyy

NNNNyNy e . (9.4, b)


9.1.3 The displacement field

In the isoparametric formulation, the same shape functions are used to express the displacements within the element in terms of the nodal values as are used to define the element shape.

If u and v are the displacement components of a point located at ( )s,r , then

,NNNNN

,NNNNN

iii

iii

∑

∑

=

=

=+++=

=+++=

4

144332211

4

144332211

vvvvvv

uuuuuu (9.5)

which can be written in matrix form

{ } [ ]{ }eqNu = , (9.6) where

[ ] ⎥⎦

⎤⎢⎣

⎡=

4321

4321

00000000

NNNNNNNN

N . (9.7)

As the shape functions in natural coordinates (9.2) fulfil continuity of geometry and displacements both within the element and between adjacent elements, the compatibility requirement is satisfied in cartesian coordinates too. As mentioned in Chapter 8, polynomial models are inherently continuous within the element. Then it is easy to show that the displacements along a side of the element depend only upon the displacements of the nodes occurring on that side.

If the u displacement is approximated as

( ) srasaraas,ru 4321 +++= , (9.8)

along each side the approximation is linear, because r (or s) is constant.

For example, along the side 2-3 (Fig. 9.2, a), 1=r and

( ) saaaau r 43211 +++== ,

2

1 2

1321 ususu r

++

−== .

The four-node quadrilateral is termed a “linear” (sometimes bi-linear) element because its sides remain straight during deformation. This ensures inter-element compatibility, i.e. there are no openings, overlaps or discontinuities


between the elements. The 4-node isoparametric quadrilateral is a conforming element.

Also, because the interpolation displacement model provides rigid body displacements in the natural coordinate system, the conditions of both rigid body displacements and constant strain states are satisfied in cartesian coordinates too.

On the contrary, the 4-node quadrilateral is a non-conforming element.

If the u displacement is approximated as

( ) yxbybxbby,xu 4321 +++= , (9.9)

along each side the approximation is quadratic.

For example, the equation of the side 2-3 (Fig. 9.1, a) has the form cxmy += , where m is the slope of 2-3, so that

( ) ( ) 244323132 xbmxbcbmbbcbxu +++++=− .

Along 2-3, ( )xu varies quadratically and cannot be uniquely defined as a function of 2u and 3u . The approximation (9.9) is non-conforming.

9.1.4 Mapping from natural to cartesian coordinates

Subsequently, we need to express the derivatives of a function in { }y,x coordinates in terms of its derivatives in { }s,r coordinates. This is done considering the function ( )y,xff = to be an implicit function of r and s as

( ) ( )[ ]s,ry,s,rxff = .

Transformation of differential operators

Using the chain rule of differentiation, we have

[ ]⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

y

xJ

y

x

sy

sx

ry

rx

s

r , (9.10)

where

[ ] ⎥⎦

⎤⎢⎣

⎡=

2221

1211

JJJJ

J (9.11)

is the Jacobian matrix. Using (9.10) and (9.4), it can also be expressed as


[ ] ⎣ ⎦ ⎣ ⎦ { } ⎣ ⎦ { }⎣ ⎦

⎣ ⎦

⎣ ⎦{ } { }⎣ ⎦ .y

Ns

Nr

yNN

s

ry

s

rJ

ee

ee

x

xx

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

=

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

=

(9.12)

For the four-node quadrilateral element

[ ] ( ) ( ) ( ) ( )( ) ( ) ( ) ( )

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎦

⎤⎢⎣

⎡−++−−−+−+−−−

=

4

3

2

11

11111111

41

yyyy

rrrrssss

J

4

3

2

xxxx

, (9.13)

[ ]( ) ( )

( ) ( ) ⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−+−++++−−

−+−++++−−

−+−++−++−

−+−++−++−

=

4321

4321

4321

4321

4321

43214321

4321

41

yyyyryyyy

r

yyyysyyyy

sJ

xxxxxxxx

xxxxxxxx

. (9.13, a)

Analogously, the inverse relationship is

[ ]⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

s

rj

s

r

ys

yr

xs

xr

y

x , (9.14)

where [ ] [ ] 1−= Jj is the inverse of the Jacobian matrix

[ ] [ ] ⎥⎦

⎤⎢⎣

⎡−

−=⎥

⎦

⎤⎢⎣

⎡=

1121

1222

2221

1211

det1

JJJJ

Jjjjj

j . (9.15)

For the four-node quadrilateral element

[ ] sArAAJ 210det ++= , (9.16)

where

( )( ) ( )( )[ ]241313240 81 xxxx −−−−−= yyyyA ,


( )( ) ( )( )[ ]431212431 81 xxxx −−−−−= yyyyA ,

( )( ) ( )( )[ ]142323142 81 xxxx −−−−−= yyyyA .

Usually [ ]j is used because we want to express xf

∂∂ ,

yf

∂∂ in terms of

rf

∂∂ ,

sf

∂∂ , determined on the reference element. Because ( )y,xrr = , ( )y,xss =

are not explicitly known, we need [ ]j .

The above expressions will be used in the derivation of the element stiffness matrix.

Transformation of an infinitesimal area

An additional result that will be needed is the relation

[ ] srJyx dddetdd = . (9.17)

It is needed because, for the evaluation of the element stiffness matrix, the integration on the real element is replaced by the simpler integration over the reference element.

In cartesian coordinates { }y,x , the elementary area Ad is given by the modulus of the cross product yx dd × , where ixx ⋅= dd , jyy ⋅= dd , and i , j are base vectors.

In a curvilinear system { }s,r , the elementary area Ad is given by the modulus of the cross product sr dd × . It is equal to the area of the elemental parallelogram enclosed by the two vectors rd and sd directed tangentially to the

.constr = and .consts = contours respectively. The components of these vectors in a cartesian coordinate system are

( ) rjJiJjrryir

rxr dddd 1211 +=⋅

∂∂

+⋅∂∂

= , ( ) sjJiJs dd 2221 += .

By equating the moduli of the cross products

kyxyx ⋅=× dddd ,


[ ] ksrJsrJJJJ

kjisr ⋅==× dddetdd

00dd

2221

1211

we obtain equation (9.17).

9.1.5 Element stiffness matrix


[ ] [ ] [ ] [ ]∫=A

Tee ABDBtk d , (9.18)

where [ ]D is the material stiffness matrix and et is the element thickness.

The matrix of differentiated shape functions [ ]B is (7.41)

[ ] [ ][ ]NB ∂= , (9.19)

where [ ]∂ is the matrix of differential operators (6.9) and [ ]N is the matrix of shape functions (9.7).

Substituting (6.9) into (7.41) gives the strain-displacement matrix

[ ] [ ] [ ] [ ]N

xy

y

xNB

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=∂= 0

0

. (9.20)

Using the transformation (9.14) we obtain

[ ] [ ][ ]21 BBB = , (9.21)

where

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

12112221

2221

1211

1 0000

jjjjjj

jjB , (9.22)


and [ ] [ ]N

s

r

s

r

B

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂∂∂

∂∂∂∂

=

0

0

0

0

2 , (9.23)

or

[ ]

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ⎥

⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−++−−−+−+−−−

−++−−−+−+−−−

=

rrrrssss

rrrrssss

B

10101010101010100101010101010101

41

2 .

Using equation (9.17), the element stiffness matrix can be written

[ ] [ ] [ ][ ] [ ]∫ ∫+

−

+

−

=1

1

1

1

dddet srJBDBtk Tee . (9.24)

Because [ ]B and [ ]Jdet are involved functions of r and s, the above integration has to be performed numerically, as shown in the following.

9.1.6 Element load vectors

Because the displacements along a side of the quadrilateral isoparametric element are linear, the consistent nodal forces applied along that side are calculated as for a linear two-node element.

If there is a distributed load having components ( )yx p,p per unit length along side 2-3 in Fig. 9.1, then the equivalent nodal force vector is

{ } ⎣ ⎦Tyxyx

e ppppf 000021

32−= l , (9.25)

if xp and yp are constants and 32−l is the length of the side 2-3.


9.2 Numerical integration

The numerical evaluation of stiffness integrals is usually done by Gaussian quadrature. It is more efficient than other methods, as for example the Newton-Cotes integration, because it involves only a half of the sample values of the integrand required by the latter. For higher order elements, the stiffness integrals become progressively more complicated. As the order of the displacement field increases, the differentiated shape functions in the matrix [ ]B grow algebraically more cumbersome. The isoparametric mapping introduces [ ]Jdet in the integrand so that closed form evaluation of stiffness integrals becomes impossible.

9.2.1 One dimensional Gauss quadrature

The one-dimensional version of the Gauss method relies on the concept that any function ( )rf can be represented approximately over the interval

[ ]11,r −∈ by a polynomial which can be integrated exactly. A polynomial of order ( )12 −n can be ‘fitted’ to ( )rf by imposing n weights iw and n sampling points in such a way that the summation

( ) ( )∑∫=

+

−

=n

iii rfrrf

1

1

1

d w (9.26)

is exact up to the chosen order.

The particular positions of these sampling points are known as Gauss Points. They turn out to be the roots of Legendre polynomials and so the method is referred to as a Gauss-Legendre integration.

The actual area represented by the integral is replaced by a series of rectangles of unequal widths, whose heights are equal to the function values at the sampling points. In other words, the integral of a polynomial function is replaced by a linear combination of its values at the integration points ir :

( ) ( ) ( ) ( ) ( )nnii rf...rf...rfrfrrf wwww +++++=∫+

−2211

1

1

d .

The n2 coefficients are determined from the condition that the above equation is satisfied for a polynomial of order 12 −n of the form

( ) 122

2321

−++++= nn ra...raraarf . (9.27)


On substitution above we get

( )

( ) ( ) .r...rra...r...rra

...arra...rrara

nnn

nnnnn

nn

n

ddd

121222

1211222112

211

1

1

122

1

12

1

11

−−−

+

−

−+

−

+

−

+++++++++

++++=+++ ∫∫∫

wwwwww

www(9.28)

Identifying the terms

∑∫=

+

−

=n

iir

1

1

1

d w , ∑∫=

+

−

=n

iii rrr

1

1

1

d w , ∑∫=

+

−

=n

iii rrr

1

1

1

d αα w .

Generally

⎪⎩

⎪⎨⎧

+=∫

+

−

even is if

12

odd is if 0 d

1

1i

irr

α

α (9.29)

so that

n... www +++= 212 , nn r...rr www +++= 22110 ,

2222

2113

2nn r...rr www +++= , (9.30)

LLLLLL 1212

2212

110 −−− +++= nnn

nn r...rr www .

This is a set of n2 equations, linear in iw and nonlinear in ir . The n2 parameters are determined from conditions

n,...,,iri

i 2111

0=

⎭⎬⎫

+<<−>w

.

One-point formula

For 1=n we have the midpoint rule (Fig. 9.3, a)

( ) ( )11

1

1

d rfrrf w≈∫+

−


which is exact only when ( )rf is a polynomial of order 1. However, it is employed in some selective integration schemes, e.g. to separate shear from extension effects, avoiding the so-called shear locking by using reduced integration for shearing.

Two-point formula

For a two-point integration

( ) ( ) ( )2211

1

1

d rfrfrrf ww +=∫+

−

. (9.31)

For a polynomial of order 312212 =−⋅=−n , equations (9.30) give

212 ww += , 22110 rr ww += ,

222

2113

2 rr ww += ,

322

3110 rr ww += .

The solution to this set of equations is

121 == ww , 577350269103

121 .r ==−=r

so that (Fig. 9.3, b)

( ) ( ) ( )57735015773501d1

1

.f.frrf −⋅+⋅=∫+

−

.

Fig. 9.3


Three-point formula

For a three-point integration (Fig. 9.3, c)

( ) ( ) ( ) ( )7740555008880774055503

1.f.f..f.rf

iii ⋅+⋅+−⋅=∑

=w .

Gauss points and weights are given in Table 9.1 for the first four orders.

Table 9.1

Number of Gauss points

Order of polynomial integrated exactly

Gauss Points, ir

Weights, iw

1 1 0 2.0 2 3 577350.± 1.0

3 5 0 7745960.±

0.8888 0.5555

4 7 3399810.± 8611360.±

0.652145 0.347854

9.2.2 Two dimensional Gauss quadrature

The one dimensional version of Gaussian quadrature is readily extended to two dimensions if we use the rectangles or isoparametric quadrilaterals of section 9.1:

( ) ( ) ( )∑∑∫ ∑∫∫==

+

− =

+

−

+

−

=⎥⎥⎦

⎤

⎢⎢⎣

⎡≅=

n

jjij

n

ii

n

jjj s,rfdss,rfsrs,rfI

11

1

1 1

1

1

1

1

dd www ,

( ) ( )∑∑∫∫= =

+

−

+

−

==n

i

n

jjiji s,rfsrs,rfI

1 1

1

1

1

1

dd ww . (9.32)

For 2=n

121 == ww , 57735011 .s −==r , 57735022 .s ==r ,

so that

( ) ( ) ( ) ( )22122111 s,rfs,rfs,rfs,rfI +++= .

Gauss quadrature formulae for quadrilateral elements are shown in Table. 9.2.


Table 9.2

n

Number of Gauss points

nn× Gauss Points,

ir , js Weights,

iw , jw

1

1 ( )11×

( )224 ×= at centre

2

4 ( )22×

( )111 ×= at points 4321 ,,,

3

9 ( )33×

⎟⎠⎞

⎜⎝⎛ ×=

95

95

8125

at points 4321 ,,,

⎟⎠⎞

⎜⎝⎛ ×=

98

95

8140

at points 4321 ,,,

⎟⎠⎞

⎜⎝⎛ ×=

98

98

8164

at points 4321 ,,,

9.2.3 Stiffness integration The element stiffness matrix (9.24) is

[ ] [ ] [ ][ ] [ ]∫ ∫− −

=1

1

1

1

dddet srJBDBtk Tee

which means that the integral above actually consists of the integral of each element (below or above the main diagonal) in an ( )88× matrix.

In general, [ ]Jdet is a linear function of r and s . For a rectangle or parallelogram it is a constant. The elements of [ ]B are obtained by dividing a bi-linear function of r and s by a linear function. Therefore, the elements of


[ ] [ ][ ] [ ]JBDB T det are bi-quadratic functions divided by a linear function. This

means that [ ]ek cannot be evaluated exactly using numerical integration.

From practical considerations, it is best to use as few integration points as is possible without causing numerical difficulties. An alternative is to use reduced integration at fewer points than necessary, with the aim of decreasing the stiffness and so compensating for the overstiff finite element model. This is cheaper as well.

A lower limit on the number of integration points can be obtained by observing that as the mesh is refined, the state of constant strain is reached within an element. In this case, the stiffness matrix equation (9.24) becomes

[ ] [ ] [ ][ ] [ ]∫ ∫+

−

+

−

≈1

1

1

1

dddet srJtBDBk eTe . (9.33)

The integral in (9.33) represents the volume of the element. Therefore, the minimum number of integration points, is the number required to evaluate exactly the volume of the element. Taking the thickness, et , to be constant and noting that

[ ]Jdet is linear, indicates that the volume can be evaluated exactly using one integration point.

Fig. 9.4

However, in the present case, one integration point is unacceptable since it gives rise to zero-energy deformation modes (Fig. 9.4). These are modes of deformation which give rise to zero strain energy. This will be the case if one of these modes gives zero strain at the integration point. The existence of these modes is indicated by the stiffness matrix having more zero eigenvalues than rigid body modes.

Experience has shown that the best order of integration for the 4-node quadrilateral element is a ( )22× array of points.

Using the 22× rule (9.33), we can write


[ ] ( )[ ] [ ] ( )[ ][ ] ( )[ ]( )[ ] [ ] ( )[ ][ ] ( )[ ]( )[ ] [ ] ( )[ ][ ] ( )[ ]( )[ ] [ ] ( )[ ][ ] ( )[ ] .s,rBs,rJDs,rBt

s,rBs,rJDs,rBt

s,rBs,rJDs,rBt

s,rBs,rJDs,rBtk

Te

Te

Te

Tee

222222

121212

212121

111111

det

det

det

det

+

++

++

+≅

(9.34)

The calculation consists of four basic steps:

a) Calculation of [ ]J at the Gauss points (9.11)

( )[ ] ( ) ( ) ( ) ( )( ) ( ) ( ) ( )

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−++−−−+−+−−−=

4

3

2

11

11111111

41

yyyy

rrrrsssss,rJiiii

jjjjji

4

3

2

xxxx

and its determinant.

b) Calculation of the inverse [ ] [ ] 1−= Jj at the Gauss points.

c) Calculation of the [ ]B matrix (9.21)

[ ]

( ) ( )( ) ( )

( ) ( )( ) ( ) ⎥

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−++−−−+−+−−−

−++−−−+−+−−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

iriririrjsjsjsjs

iriririrjsjsjsjs

jjjjjj

jjB

1010101010101010

0101010101010101

12112221222100001211

41 .

d) Calculation of [ ]ek using the 22× rule (9.34).

Example 9.1 Consider the quadrilateral element from Fig. 9.1 with the following nodal coordinates

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

258302515201010

4

3

2

11

yyyy

4

3

2

xxxx

.

For convenience, consider approximately the Gauss Points at 50.s,r ji ±= instead of 577350.s,r ji ±= . For the first term in (9.34) we need

[ ]Jdet and [ ]B calculated at GP1.


At GP1, the Jacobian matrix (9.11) is

[ ] ⎥⎦

⎤⎢⎣

⎡=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎦

⎤⎢⎣

⎡−−

−−=

60132061

81

258302515201010

5051515051515050

41

........

J .

Its determinant and inverse are

[ ] 12553det .J = , [ ] [ ] ⎥⎦

⎤⎢⎣

⎡−

−== −

61132060

42511Jj .

At GP1, the matrix [ ]B is given by (9.21)

[ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−=

500510510500510510500500

050051051050051051050050

20606113611300002060

17001

........

........

B ,

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−

−−=

5025303630491012250360490120050030030010

8501B .

9.2.4 Stress calculations

In the quadrilateral element, stresses

{ } [ ] [ ] { }eqBD=σ (9.35)

vary within the element.

In practice, stresses are evaluated at the Gauss points, where they are found to be accurate (Barlow, 1976). Some programs which use Gauss point stresses extrapolate to the element nodes and then output the mean value if several elements meet at a node.

Maximum stresses usually occur at edges of plates or at other discontinuities at the element boundaries. In order not to miss these peak stresses, a refined mesh should be used in these regions. However, Gauss point values are ideal for constructing internal stress contours.


9.3 Eight-node quadrilateral

This element is the isoparametric version of the eight-node rectangle presented in section 8.2.2. The difference is that in cartesian coordinates it has curved sides (Fig. 9.5, a) and the master element is a square (Fig. 9.5, b).

Fig. 9.5

The displacement functions in simple polynomial form (8.27) are

.

2

82

72

652

4321

28

27

265

24321

srbsrbsbsrbrbsbrbb

,srasrasasrarasaraa

+++++++=

+++++++=

v

u (9.36)

The element has three nodes along one edge, so that the displacement variation should be parabolic (three constants) to satisfy compatibility.

The eight-node quadrilateral employs displacement fields quadratic in r and s , as are the stresses. The product [ ] [ ][ ]BDB T is therefore fourth order and

[ ]Jdet is cubic. It would seem necessary to use a 33× Gauss quadrature for exact stiffness integration.

For a single 8-node rectangular element, even if it is supported conventionally, using a 22× integration scheme would result in a singular stiffness matrix. In practice, large groups of elements will not suffer from such mechanisms, and 22× integration is normally used for this popular element. The optimal sampling points for this element are the Gauss points for the 22× scheme


and not those for the higher order 33× scheme, so the advantages of reduced integration are compounded.

For an 8-node quadrilateral element, better displacements are obtained using a 33× Gauss point mesh, nine points in all.

It was shown that, in order to evaluate the minimum order of the numerical integration, it is sufficient to examine the Jacobian determinant. For the 8-node quadratic element, [ ]Jdet is of third order hence the minimum number of integration points is 4 ( )22× .

The eight-node curved ‘quad’ is ideal for nonlinear problems and can be easily adapted to model cracks by moving the midside nodes. Care is needed because the accuracy declines with excessive shape distortion.


The shape functions (8.29) have the following expressions

( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( ) ( )

.1121 , 111

41

, 1121 ,111

41

, 1121 , 111

41

, 1121 , 111

41

284

273

262

251

srNsrsrN

srNsrsrN

srNsrsrN

srNsrsrN

−−=−++−−=

+−=−−++−=

−+=+−−+−=

−−=++−−−=

(9.37)

As they are constructed based on the equations of the lines passing through the nodal points, they belong to the serendipity *) family of elements.

For ease in programming, one can also use the following equivalent formulae in which ( )ii s,r are the natural coordinates of node i:

for corner nodes

( ) ( ) ( )

( ) ( ) ( ) ( ); 214

214

11141

iiiii

iiiii

iiiii

rrssrrss

N,ssrrssrr

N

,ssrrssrrN

++=∂∂

++=∂∂

−+++= (9.38)

___________

*) After the extraordinary discoveries of the princes of Serendip from the horror novels by Horace Walpole (1717-1797).


for mid-side nodes, when 0=ir

( )( )

( ) ( ) ; 121 1

1121

2

2

rss

N,ssrr

N

,ssrN

ii

ii

ii

−=∂∂

+−=∂∂

+−= (9.39)

for mid-side nodes, when 0=is

( )( )

( ) ( ) . 1 121

1121

2

2

ii

ii

ii

rrss

N,srr

N

,rrsN

+−=∂∂

−=∂∂

+−= (9.40)


[ ]{ }eqN=⎭⎬⎫

⎩⎨⎧vu

, (9.41)

where

[ ] ⎥⎦

⎤⎢⎣

⎡=

821

821

000000

NNNNNN

NLL

LL, (9.42)

{ } ⎣ ⎦Teq 882211 vuvuvu LL= . (9.43)

Having generated the shape functions [ ]N , the matrix [ ] [ ][ ]NB ∂= is

[ ]

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=

xN

yN

yN

xN

xN

yN

yN

xN

xN

yN

yN

xN

B

88

8

8

22

2

2

11

1

1

0

0

0

0

0

0

L

L

L

. (9.44)

9.3.2 Shape function derivatives

Use of the isoparametric formulation

( ) ( )∑∑==

==8

1

8

1

iii

iii ys,rNy,s,rN xx (9.45)

leads to


[ ] [ ]⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

−

∂∂

−∂∂

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

sNr

N

rx

sx

ry

sy

Js

Nr

N

j

yNx

N

i

i

i

i

i

i

det1 , (9.46)

where

( ) ( )

( ) ( ) .ys

s,rNsy,y

rs,rN

ry

,s

s,rNsx,

rs,rN

rx

ii

ii

i

i

ii

ii

i

i

8

1

8

1

8

1

8

1

∑∑

∑∑

==

==

∂∂

=∂∂

∂∂

=∂∂

∂∂

=∂∂

∂∂

=∂∂ xx

(9.47)

9.3.3 Determinant of the Jacobian matrix

The Jacobian matrix is

[ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=

sy

sx

ry

rx

J , (9.48)

and its determinant is

[ ]ry

sx

sy

rxJ

∂∂

∂∂

−∂∂

∂∂

=det . (9.49)

9.3.4 Element stiffness matrix

The element stiffness matrix is

[ ] [ ] [ ][ ] [ ]∫ ∫+

−

+

−

=1

1

1

1

dddet srJBDBtk Tee (9.50)

where et is the thickness of the plate and the material stiffness matrix [ ]D has the form

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

5

42

21

0000

DDDDD

D . (9.51)


Substituting (9.44), the stiffness matrix (9.50) can be written

[ ] [ ]∫ ∫+

−

+

−⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=⎥⎦⎤

⎢⎣⎡

1

1

1

1

dddet

88

80

08

22

20

02

11

10

01

880

808

220

202

110

101

srJ

xN

yN

yN

xN

xN

yN

yN

xN

xN

yN

yN

xN

D

T

xN

yN

yN

xN

xN

yN

yN

xN

xN

yN

yN

xN

etek

L

L

L

LLL

.

(9.52)

The ( )1616× element stiffness matrix contains submatrices [ ] jik which,

evaluated by numerical integration, have the form

[ ] [ ] [ ]J

xN

yN

yN

xN

D

xN

yN

yN

xN

tk

jj

j

j

n

p

n

qii

ii

qpe

ji det0

0

0

0

1 1

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂

∂

∂

∂∂

∂∂

∂

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

= ∑∑= =

ww , (9.53)

or [ ]ji

ji kkkk

k ⎥⎦

⎤⎢⎣

⎡=

43

21 . (9.53, a)

Denoting

[ ]Jt qpe detwww = , (9.54)

the elements of the matrix (9.53, a) can be determined as

∑∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂

∂∂

+∂

∂

∂∂

=y

Ny

NDx

Nx

NDk jijiji 511 w ,

∑∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂

∂∂

+∂

∂

∂∂

=x

Ny

NDy

Nx

NDk jijiji 522 w , (9.55)

∑∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂

∂∂

+∂

∂

∂∂

=y

Nx

NDx

Ny

NDk jijiji 523 w ,


∑∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂

∂∂

+∂

∂

∂∂

=x

Nx

NDy

Ny

NDk jijiji 544 w .

When ji = , 32 kk = .

A close inspection of equations (9.55) shows that all the terms are of the form

( ) ( ) ( )sderivativefunctionshapeofproductpropertymaterialttancons ×× .

Therefore, providing that the material properties are constant throughout the element, it is useful to store separately in an array [ ] jiS the sum of the

products of the shape function derivatives multiplied by w , such that

[ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∂

∂

∂∂

∂

∂

∂∂

∂

∂

∂∂

∂

∂

∂∂

=

∑∑∑∑

∑∑∑∑

yN

yN

xN

yN

yN

xN

xN

xN

Sjiji

jiji

jiww

ww (9.56)

and then to work out equations (9.55) by incorporating the relevant material constants.

This method (K. A. Gupta & B. Mohraz, 1972) reduces the computing time by a factor of nine over the full matrix multiplication method.

9.3.5 Stress calculation

The vector of element stresses is

{ } [ ][ ]{ }eqBD=σ . (9.57)

The explicit form of equation (9.57) for the quadratic 8-node element is

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

8

2

1

1

88

8

8

22

2

2

11

1

1

5

42

21

0

0

0

0

0

0

0000

v

v

LL

L

L

u

u

xN

yN

yN

xN

xN

yN

yN

xN

xN

yN

yN

xN

DDDDD

xy

y

x

τσσ

wherefrom

∑∑== ∂

∂+

∂∂

=8

12

8

11

ii

i

ii

ix y

NDux

ND vσ ,


∑∑== ∂

∂+

∂∂

=8

14

8

12

ii

i

ii

iy y

NDux

ND vσ , (9.58)

∑=

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

=8

15

ii

ii

ixy x

Nuy

ND vτ ,

In some programs the stresses are determined at nodes, since the nodal positions are readily located and it is convenient to output the displacements and stresses at the same points. It has been found that nodal stresses from an 8-node quadratic element are usually incorrect. However, if the stresses of all elements meeting at a node are averaged, closer results to the true values are obtained.

A better alternative is to calculate stresses at the Gauss points, in which superior accuracy is obtained and averaging is not necessary. This is due to the fact that the element stiffness is calculated by sampling at the Gauss points, and it is therefore reasonable to expect the most accurate stresses and strains occurring at the same points.

9.3.6 Consistent nodal forces

Unlike the triangular element, in which all loads can be reduced to nodes intuitively or by statics, for a quadratic isoparametric element the nodal forces due to distributed loads must be computed in accordance with equation (7.44)

{ } [ ] { }∫=eV

Te VpNf d . (9.59)

The equivalent nodal forces are added, element by element, into the global load vector which represents the right hand side of the linear set of equations to be solved for displacements.

Edge pressure

When coding the load vectors in a computer program, the actual pressure distribution along an element edge is replaced by a parabolic distribution defined by the pressure values at each of the three nodes along that edge. All intermediate values can be calculated using the shape functions. It is usual to use all the nodes of the element in the computation, so that there is no need to sort out the appropriate shape functions for the three nodes with given pressure values.

Consider a distributed load p , specified in force per unit length, acting along the 1+=s edge of an element. The components of the force rp d acting upon an elemental length rd are


r

rx

ry

ppp

y

x d

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂

−

∂∂

=⎭⎬⎫

⎩⎨⎧

. (9.60)

The consistent load for p is given by a modified form of equation (9.59) in which the volume integral has been reduced to a line integral

{ } [ ] r

rx

ry

Npf Te d1

1 ⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂

−

∂∂

= ∫+

−

(9.61)

where [ ]N is given by (9.42).

The above vector is usually integrated numerically and is written as

{ } [ ]

i

n

ii

Ti

e

rx

ry

pNf

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂

−

∂∂

=∑=1

w (9.62)

where ip is the pressure at the Gauss point i along the 1+=s edge and is computed by equation

( )∑=

=8

1kkk ps,rNp . (9.63)

If the same pressure acts on the edge 1−=s , the sign of the force vector { }ef is reversed, since a positive pressure is assumed to act towards the centre of the element. A more general equation applicable to the two edges 1±=s is

{ } [ ]

i

n

i kkk

Tsi

e

rx

ry

pNNf

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂

−

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛=∑ ∑

= =1

8

1ww (9.64)

where sw takes up the value of the s coordinate for the loaded edge.

A similar expression for pressure loads on the 1±=r edges is

{ } [ ]

i

n

i kkk

Tri

e

sxsy

pNNf

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂∂∂

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=∑ ∑

= =1

8

1ww . (9.65)


For a rectangular isoparametric element with edges parallel to the y,x axes, it is possible to integrate the equations for the equivalent nodal forces explicitly.

Example 9.2 Consider the element of Fig. E9.2, a with a uniform pressure load along the top edge. Find the equivalent nodal forces.

Fig. E9.2

Solution.

For the edge 1+=s we have 086251 ===== NNNNN .

For the rectangular element 0=∂∂

ry and a

rx=

∂∂ . Equation (9.61) gives

{ } ( ) raNNN

pfff

f

y

y

ye d

4

7

31

14

7

3

−⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

= ∫+

−

. (9.66)

Denoting the total pressure load apP 2= , we can write

( ) ( ) ( ) rssrrssrrPf iiiiy d11141

2

1

13 ∫

+

−

−+++−= ,

( ) ( ) rrssPf iy d1141

2

1

1

27 ∫

+

−

−+−= ,


( ) ( ) ( ) rssrrssrrPf iiiiy d11141

2

1

14 ∫

+

−

−+++−= .

In the above expressions, substituting 1=ir and 1=is for node 3, 0=ir and 1=is for node 7, and 1−=ir and 1=is for node 4, yields

( ) ( ) ( )6

d11 2141

2

1

13

PrrrPf y −=−++−= ∫+

−

,

( ) ( )3

2d1241

2

1

1

27

PrrPf y −=−−= ∫+

−

,

( ) ( ) ( )6

d112141

2

1

14

PrrrPf y −=−+−−−= ∫+

−

.

Hence a uniform load acting along an edge is not distributed in the intuitive ratio of 4

1 : 21 : 4

1 , but in the ratio 61 : 3

2 : 61 (Fig. E9.2, b).

The same result could have been obtained noticing that the integrands in the above equations are the shape functions of the three-node isoparametric one-dimensional element (4.26).

Gravity loading

For gravity loading (downwards negative) the vector of equivalent nodal forces (9.59) is

{ } [ ] [ ] [ ]∫ ∫∫+

−

+

− ⎭⎬⎫

⎩⎨⎧−

=⎭⎬⎫

⎩⎨⎧−

=1

1

1

1

dddet0

dd0

srJg

Nmyxg

Nmf TTe (9.67)

where m is the mass per unit area and g the acceleration in the y direction.

The integration is carried out numerically and equation (9.67) takes the form

{ } [ ] [ ]Jg

Nmfn

i

n

j

Tji

e det0

1 1 ⎭⎬⎫

⎩⎨⎧−

=∑∑= =

ww . (9.68)

Example 9.3 Consider the rectangular element of Fig. E9.3, a with gravity loading. Find the equivalent nodal forces.

Solution. Because all the equivalent nodal forces act in the y direction,


[ ]∫ ∫+

−

+

−⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧1

1

1

18

2

1

8

2

1

dddet srJ

N

NN

gm

f

ff

y

y

y

LL. (9.69)

For a rectangular element [ ] baJ =det .

For a corner node i the integral becomes

( ) ( ) ( ) [ ]∫ ∫+

−

+

−

−+++−=1

1

1

1

dddet11141 srJssrrssrrgmf iiiiyi . (9.70)

Denoting the total weight of the element bagmW 22 ⋅⋅= , the force (9.70) can be written

( )32116IIIWf yi ++−= , (9.71)

where

( ) ( )34dd11

1

1

1

11 =++= ∫ ∫

+

−

+

−

srrrssrrI iii (9.72)

since 1±=ir for all corner nodes,

( ) ( )34dd11

1

1

1

12 =++= ∫ ∫

+

−

+

−

srssssrrI iii (9.73)

since 1±=is for all corner nodes, and

( ) ( ) 4dd111

1

1

13 −=++−= ∫ ∫

+

−

+

−

srssrrI ii . (9.74)

Therefore, the equivalent nodal forces for gravity loading at a corner node i are given by

( )12

434

34

1616 321WWIIIWf yi =⎟

⎠⎞

⎜⎝⎛ −+−=++−= . (9.75)

For a midside node i with 0=ir , the integral is

( )( ) [ ]3

dddet1121

1

1

1

1

2 WsrJssrgmf iyi −=+−−= ∫ ∫+

−

+

−

. (9.76)

For a midside node i with 0=is , the integral is

( )( ) [ ]3

dddet1121

1

1

1

1

2 WsrJrrsgmf iyi −=+−−= ∫ ∫− −

. (9.77)


Fig. E9.3

Hence for gravity loading the equivalent nodal forces are as shown in Fig. E9.3, b. Again, they are different from the values of 12W− and 6W− , which would have been assigned intuitively to the corner and midside nodes respectively.

9.4 Nine-node quadrilateral

This element belongs to the Lagrange family of elements. Apart from the eight nodes located on the boundary, it contains an internal node. The local node numbers for this element are shown in Fig. 9.6, a. The master element is presented in Fig. 9.6, b.

Fig. 9.6


The associated displacement functions in polynomial form are

.

22

92

82

72

652

4321

229

28

27

265

24321

srbsrbsrbsbsrbrbsbrbb

,srasrasrasasrarasaraa

++++++++=

++++++++=

v

u(9.78)

Along the sides of the element, the polynomial is quadratic (with three terms - as can be seen setting 0=s in u ), and is determined by its values at the three nodes on that side.

Higher-order rectangular elements can be systematically developed with the help of the so-called Pascal’s triangle, which contains the terms of polynomials of various degrees in the two variables r and s , as shown in Fig. 9.7.

Fig. 9.7

Since a linear quadrilateral element has four nodes, the polynomial should have the first four terms 1, r, s, and rs . In general, a pth-order Lagrange rectangular element has ( ) 21+p nodes ( )...,,p 10= . The quadratic quadrilateral element has 9 nodes. The polynomial is incomplete. It contains the complete polynomial of the second degree (6 terms) plus other three terms which have to be located symmetrically: the third degree terms sr2 and 2sr and also an 22sr term.

The shape functions are defined as follows. Considering the r axis alone, as shown at the bottom of Fig. 9.6, b, we can define generic shape functions 1L ,

2L and 3L having unit value at the node with the same index and zero at the other two nodes

( ) ( )2

11

rrrL −−= , ( ) ( ) ( )rrrL −+= 112 , ( ) ( )

21

3rrrL +

= . (9.79)

They turn out to be Lagrange polynomials, which, for three points, have the expressions


( ) ( ) ( )( ) ( )3121

321

rrrr

rrrrrL

−−

−−= , ( ) ( ) ( )

( ) ( )3212

312

rrrr

rrrrrL

−−

−−= , ( ) ( ) ( )

( ) ( )2313

213

rrrr

rrrrrL

−−

−−= .

Similarly, generic shape functions can be defined along the s axis, as shown at the right of Fig. 9.6, b

( ) ( )2

11

sssL −−= , ( ) ( ) ( )sssL −+= 112 , ( ) ( )

21

3sssL +

= . (9.80)

The shape functions iN can be constructed as products of the above one-dimensional functions

( )( )( )

( ) ( ) ( )⎣ ⎦sLsLsLrLrLrL

NNNNNNNNN

321

3

2

1

362

795

481

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡. (9.81)

It should be cautioned that the subscripts of iN refer to the node numbering used in Fig. 9.6, a.

Since the internal nodes of the higher-order elements of the Lagrange family do not contribute to the inter-element connectivity, they can be condensed out at the element level to reduce the size of the element matrices. Alternatively, one can use the serendipity elements, but their shape functions cannot be obtained using products of one-dimensional interpolation functions.

9.5 Six-node triangle

Higher-order triangular elements can be developed using Pascal’s triangle (Fig. 9.8). One can view the position of the terms as the nodes of the triangle, with the constant term and the first and last terms of a given row being the vertices of the triangle. The six node triangle (Fig. 9.9, a) is an element of order 2 (i.e., the degree of the polynomial is 2), as can be seen from the top three rows of Pascal’s triangle. The polynomial involves six constants, which can be expressed in terms of the nodal values of the variable being interpolated.

By referring to the master element in Fig. 9.9, b, the shape functions can be expressed in terms of area coordinates as for the linear strain triangle (8.36)

( ) ( ) ( )

, ζ ζ4 , ζ ζ4 , ζ ζ4 , 1ζ2ζ , 1ζ2ζ , 1ζ2ζ

136325214

333222111

===−=−=−=

NNNNNN

(9.82)


where 321 1 ζζζ −−= . Because of terms 22ζ , 2

3ζ in the shape functions, this element is also called a quadratic triangle.

Fig. 9.8

The isoparametric representation is (8.37)

{ }eqNNN

NNN⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧

621

621

000000

LL

LL

vu

, (9.83)

where { } ⎣ ⎦Teq 662211 vuvuvu LL= , and

.yNy,Ni

iii

ii ∑∑==

==6

1

6

1 xx (9.84)

Fig. 9.9


The element stiffness matrix, which has to be integrated numerically, can be evaluated using (9.24) if the ( )s,r coordinates are defined by (8.46). This gives

12 NNr ∂

∂−

∂∂

=∂∂ ,

13 NNs ∂∂

−∂∂

=∂∂ . (9.85)

Details on numerical integration schemes for triangles are given in [18, 39]. The Gauss quadrature formulas for a triangle differ from those considered earlier for the rectangle.

9.6 Jacobian positiveness

In the higher-order isoparametric elements discussed above, we note the presence of ‘midside’ nodes. The midside node should be as near as possible to the centre of the side. It must be placed inside the middle third of a side. This condition ensures that [ ]Jdet does not attain a value of zero in the element.

The sign of [ ]Jdet should be checked. If the Jacobian becomes negative at any location, a warning message will be signaled indicating the nonuniqueness of mapping. Note that while the Jacobian is always computed at Gauss points, it is more likely to be negative at corners, where stresses may be computed.

A necessary requirement for applying equation (9.14) to shape functions is that [ ]J can be inverted. This inverse exists if there is no excessive distortion of the element such that lines of constant r or s intersect inside or on the element boundaries or there are re-entrant angles. When the element is degenerated into a triangle by increasing an internal angle to 0180 then [ ]J is singular at that corner. A similar situation occurs when two adjacent corner nodes are made coincident to produce a triangular element. Therefore to ensure that [ ]J can be inverted, any

internal angle of each corner node of the element should be less than 0180 , and, as an internal angle approaches 0180 there is a loss of accuracy in the element stress, particularly at that corner.

If the determinant [ ] 0det →J , then [ ]j and the operators in [ ]∂ will increase without limit and consequently produce infinite strains.

The above statements are illustrated for the one-dimensional quadratic isoparametric element shown in Fig. 4.4. The row vector of shape functions is

⎣ ⎦ ⎣ ⎦ ( ) ( )( ) ( ) ⎥⎦⎥

⎢⎣⎢ +−+−== 1

21111

21

321 rrrrrrNNNN .


The Jacobian matrix is

[ ]⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎦⎥

⎢⎣⎢ +

−−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎦

⎥⎢⎣

⎢∂∂

∂∂

∂∂

==l2

3

2

1321

11

0

2122

212 xrrr

xxx

rN

rN

rNJJ .

The determinant of the Jacobian matrix is

[ ] ( ) 211 250det xr.rJJ −+== l .

For any 2x this determinant vanishes at the point

l

l

−=

20 2

50x.r .

Fig. 9.10

This point lies within the element (Fig. 9.10) if 11 0 ≤≤− r , i.e. if

12

5012

≤−

≤−l

l

x. ,

where l≤≤ 20 x .

If 43

4 2ll

<< x , then [ ]Jdet does not vanish on the element. This explains

why it is recommended to place precautiously the ‘midside’ node inside the middle third of a side.

10. PLATE BENDING

Flat plate structures, such as the floors of buildings and aircraft, enclosures surrounding machinery and bridge decks are subject to loads normal to their plane. Such structures can be analyzed by dividing the plate into an assemblage of two-dimensional finite elements called plate bending elements. These elements may be either triangular, rectangular or quadrilateral in shape. In this chapter, finite element displacement models for the flat plate bending problem are discussed.

Thin plates with transverse shear neglected are analyzed based on Kirchhoff’s classical theory. Plates with a constant shear deformation through the plate thickness are treated in the Reissner-Mindlin plate theory. For thick or composite plates some higher-order shear deformation plate theories are available, in which on the faces of plate the shear strains are equal to zero.

10.1 Thin plate theory (Kirchhoff)

A plate is described as a structure in which the thickness is very small compared with the other dimensions, that is the thickness-to-span is 10.h ≤l . For this case, it can be assumed that the plate deformation may be expressed by the deformation state at the middle surface, which is the plane midway between the faces of the plate.

For thin plates, Kirchhoff’s hypotheses are adopted: a) there is no deformation (stretching) in the middle plane of the plate; b) normals to the middle plane of the undeformed plate remain straight and normal to the middle surface of the plate during deformation; and c) the direct stress in the transverse direction can be disregarded.

The x-y plane is taken to coincide with the middle surface of the plate (Fig. 10.1) and the positive z-direction is upwards. The plate has constant thickness h and is subject to distributed surface loads.


The displacements parallel to the undeformed middle surface are given by

( )x

zz,y,xu∂∂

−=w , ( )

yzz,y,x∂∂

−=wv , (10.1)

where denotes the displacement of the middle surface in the z-direction (Fig. 10.2, a).

( yx,w )

Fig. 10.1

The components of the strain are given as follows:

2

2

xz

xu

x∂∂

−=∂∂

=wε , 2

2

yz

yy∂∂

−=∂∂

=wvε ,

yxz

xyyx ∂∂∂

−=∂∂

+∂∂

=wvu 2

2γ ,

0=∂∂

+∂∂

=yzzywvγ , 0=

∂∂

+∂∂

=zxxzuwγ . (10.2)

a b

Fig. 10.2

The strain vector can be written in the form

{ } ⎣ ⎦ { }χγεεε zTyxyx −== , (10.3)

where the curvature vector

10. PLATE BENDING 227

{ }T

yxyx ⎥⎥⎦

⎥

⎢⎢⎣

⎢

∂∂∂

∂∂

∂∂

=www 2

2

2

2

22χ . (10.4)

Since 0=zσ , the stress-strain relations take the form

{ } [ ]{ } [ ]{ }χεσ DzD −== , (10.5) where

{ } ⎣ ⎦Tyxyx τσσσ = , (10.6)

and

[ ] . (10.7) ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

662616

262212

161211

ddddddddd

D

The strain energy in the plate can be written as

{ } { } { } [ ]{ } ADhVU T

AV

T d122

1d21 3

χχεσ ∫∫ == . (10.8)

Fig. 10.3

The positive sense of internal bending and twisting moments (per unit length) is shown in Fig. 10.3. Their definig equations are

⎣ ⎦ ⎣ ⎦∫−

=

2

2

dh

h

yxyxyxyx zzMMM τσσ . (10.9)

Denoting

∫−

=

2

2

2 dh

h

jiji zzdD (10.10)


the moment - curvature equations can be expressed as follows

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

∂∂∂∂∂∂∂

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

yx

y

x

DDDDDDDDD

MMM

xy

y

x

w

w

w

2

2

2

2

2

662616

262212

161211

(10.11)

or { } [ ]{ }χDM −= . (10.12)

For isotropic materials

[ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−=

2100

0101

νν

νDD . (10.13)

where

( )2

3

112 ν−=

hED , (10.14)

E is Young’s modulus and ν is the Poisson’s ratio.

The equilibrium equations are

,py

Qx

Q

,Qy

Mx

M

,Qy

Mx

M

zyx

yyyx

xxyx

0

0

0

=+∂

∂+

∂∂

=−∂

∂+

∂

∂

=−∂

∂+

∂∂

(10.15)

where is the lateral distributed load (per unit surface) and , are the shear forces (per unit length).

zp xQ yQ

From (10.11) and (10.15) it can be shown that

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+∂∂

∂∂

−= 2

2

2

2

yxxDQx

ww , ⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+∂∂

∂∂

−= 2

2

2

2

yxyDQy

ww . (10.16)


By substituting (10.16) into the last equation (10.15), the differential equilibrium equation for isotropic materials can be obtained as

Dp

yyxxz=

∂∂

+∂∂

∂+

∂∂

4

4

22

4

4

42 www . (10.17)

The bending and torsion stresses are given by (10.9)

⎣ ⎦ ⎣ yxyxyxyx MMMh

z3

12=τσσ ⎦ (10.18)

and the shear stresses by

⎣ ⎦ ⎣ yxzyxz QQhz

h ⎟⎟⎠

⎞⎜⎜⎝

⎛−= 2

24123ττ ⎦, (10.19)

where a parabolic distribution is assumed.

The strain energy (10.8) can be expressed as

( ) yxyxyxyx

DUA

dd122 2

2

2

2222

2

2

2

2

∫⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

∂∂

∂∂

−⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂∂

−+⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+∂∂

=wwwww ν .

(10.20)

The potential of the external load is

yxpWA

zP dd∫−= w . (10.21)

10.2 Thick plate theory (Reissner-Mindlin)

For moderately thick plates, the thickness-to-span ratio is not small enough to neglect transverse shear deformations and Kirchhoff’s assumption is no longer valid. To overcome this problem, the classical hypothesis of zero transverse shear strain is relaxed. First, Reissner proposed to introduce the rotations of the normal to the plate midsurface in the xOz and yOz plane as independent variables. Then, Mindlin simplified Reissner’s assumption considering that normals to the plate midsurface before deformation remain straight but not necessarily normal to the plate after deformation. The displacements of the middle surface are independent of the rotations of the normal (Fig. 10.2, b). The transverse normal stress is disregarded as in the Kirchhoff’s theory.

According to Reissner-Mindlin assumptions, the displacements parallel to the middle surface can be expressed as


( ) ( )y,xzz,y,xu yθ= , ( ) ( )y,xzz,y,x xθ−=v , (10.22)

where xθ , yθ are the rotations about the Ox and Oy axes of lines originally normal to the middle plane before deformation.

The in-plane strains are now given by

{ } { }χε z−= , (10.23)

where the curvature vector is

{ }T

yxxy

yxyx ⎥⎦

⎥⎢⎣

⎢∂

∂−

∂∂

∂∂

∂

∂−=

θθθθχ . (10.24)

The transverse shear strains are

yzzy ∂

∂+

∂∂

=wvγ ,

zxxz ∂∂

+∂∂

=uwγ . (10.25)

Substituting for the displacements from (10.22) gives

{ }⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂

+

∂∂

+−=

⎭⎬⎫

⎩⎨⎧

=

x

y

y

x

xz

zyw

w

θ

θ

γγ

γ . (10.26)

Note that when 0=zyγ , 0=xzγ , equation (10.26) gives yx ∂

∂=wθ ,

xy ∂∂

−=wθ , and equation (10.24) reduces to (10.4).

The average shear stresses are given by

{ } [ ]{ }γκττ

τ S

xz

zy D=⎭⎬⎫

⎩⎨⎧

= . (10.27)

where κ is a shear correction factor and

[ ] ( ) ⎥⎦

⎤⎢⎣

⎡+

=⎥⎦

⎤⎢⎣

⎡=

1001

1200

νE

GG

DS . (10.28)

The strain energy in the plate can be written as the sum of the energies due to bending and shear deformation

{ } { } { } { } VVU T

VV

T d21d

21 γτεσ ∫∫ += . (10.29)


or, integrating through the thickness

{ } [ ]{ } { } [ ]{ } ADhADhU ST

A

T

A

d21d

1221 3

γγκχχ ∫∫ += . (10.30)

For isotropic materials, equation (10.11) becomes

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

∂

∂−

∂∂

∂∂∂

∂−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

yx

y

x

DMMM

yx

x

y

xy

y

x

θθ

θ

θ

νν

ν

2100

0101

(10.31)

or (10.12), where { }χ is given by (10.24).

The transverse shear constitutive equation is

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∂∂

+

∂∂

+−=

⎭⎬⎫

⎩⎨⎧

x

yGQQ

y

x

x

yw

w

θ

θκ . (10.32)

The strain energy (10.30) can be expressed as

.yxy

Qx

Q

yxy

Myx

Mx

MU

A

xyyx

A

xy

yxyx

yx

dd21

dd21

∫

∫⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+−+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

++

+⎥⎥⎦

⎤

⎢⎢⎣

⎡

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂−

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂−=

ww θθ

θθθθ

If yx ∂

∂=wθ and

xy ∂∂

−=wθ , the above expression reduces to (10.20).

The main drawback of this theory is the arbitrary averaging of the shear strains. If a constant shear strain is considered through the plate thickness, on the faces of plate the shear strains are not zero, which is not true. This assumption is equivalent to the introduction of ‘parasitic’ shear stresses that force the normal to remain a straight line. As the thickness of the plate becomes extremely thin, the shear strain energy predicted by the finite element analysis can be magnified unreasonably and a ‘shear locking’ occurs for large span-to-thickness ratios.


10.3 Rectangular plate bending elements

For a thin plate bending element, the strain energy is given by (10.20) and the potential of the external load by (10.21). The highest derivative appearing in these expressions is the second. Hence, for convergence, it will be necessary to

ensure that and its first derivatives wx∂

∂w and y∂

∂w are continuous between

elements. These three quantities are, therefore, taken as degrees of freedom at each node. Also, complete polynomials of at least degree two should be used, according to the convergence criteria of the Rayleigh-Ritz method. The assumed form of the displacement function, irrespective of the element shape, is

higher degree terms. (10.33) ++++++= 265

24321 yayxaxayaxaaw

Figure 10.4 shows a thin rectangular element of thickness h, having four node points, one at each corner.

Fig. 10.4

The dimensionless coordinates ax=ξ and by=η will be used in the following with the origin at the plate centre.

10.3.1 ACM element (non-conforming)

Figure 10.5 shows the ACM element (Adini, Clough, Melosh - 1961, 1963). There are three degrees of freedom at each node, namely, the transverse

displacement , and the two rotations wyx ∂

∂=

wθ and xy ∂

∂−=wθ . In terms of the

dimensionless coordinates ξ and η , these become


η

θ∂∂

=w

bx1 ,

ξθ

∂∂

−=w

ay1 . (10.34)

Since the element has 12 degrees of freedom, the displacement function can be represented by a polynomial having twelve terms, that is

.312

311

310

29

28

37

265

24321

ξηαηξαηαξηαηξαξα

ηαηξαξαηαξαα

++++++

++++++=w (10.35)

Note that this function is a complete cubic to which two quartic terms

and have been added, which are symetrically located in Pascal’s triangle. This ensures that the element is geometrically invariant.

ηξ 3

3ξη

Fig. 10.5

The expression (10.35) can be written in the following matrix form

⎣ ⎦ { }αξηηξηηξηξξηηξξηξ 333223221=w , (10.36)

( )⎣ ⎦ { }αηξ ,P=w , (10.36, a)

where

{ } ⎣ 12321 ααααα L=T ⎦ . (10.37)

Differentiating (10.36) gives

⎣ ⎦ { }αηηξηξηξηξξ

3222 302302010=∂∂w , (10.38)

⎣ ⎦ { }αξηξηξηξηξη

2322 332020100=∂∂w . (10.39)


Evaluating (10.36), (10.38) and (10.39) at 1±=ξ , 1±=η gives

{ } [ ]{ }αee Aq = , (10.40) where

{ } ⎣ ⎦ { }αθθθθ 444111 yxyxTe ababq ww L= , (10.41)

and the matrix [ ]eA is given in [87] but not reproduced here, for conciseness.

Solving (10.40) for { }α gives

{ } [ ] { }ee qA1−

=α . (10.42)

Substituting (10.42) into (10.36, a) yields

( )⎣ ⎦ { } ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ { }ee qNNNNq,N 4321== ηξw , (10.43)

where the vector of nodal displacements is

, (10.44) { } ⎣ ⎦444111 yxyxTeq θθθθ ww L=

and the shape functions are defined as

( )⎣ ⎦

( )( )( )( )( )( )( )( )( )

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

+−+

−++

−−++++

=

ηηξξξ

ηηηξξ

ηξηηξξηηξξ

ηξ

ii

ii

iiji

Ti

a

b,N

118

118

21181

2

2

22

, (10.45)

where ( ii , )ηξ are the coordinates of node i.

The element is non-conforming. Indeed, evaluating (10.45) on the side 2-3 (for 1+=ξ ) gives

, ⎣ ⎦⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

000

1TN ⎣ ⎦

( )( )( )( )( )( )

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−+−−−−

=0

1142141

2

2

2 ηηηηη

bN T ,

⎣ ⎦( )( )( )( )( )( )

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−+−++

=0

1142141

2

2

3 ηηηηη

bN T , . ⎣ ⎦⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

000

4TN


This indicates that the displacement , and hence the rotation w xθ , are uniquely determined by their values at nodes 2 and 3. If the element is attached to another rectangular element at nodes 2 and 3, then and w xθ will be continuous along the common side. Unfortunately, this is not the case with the rotation yθ .

The rotation yθ is given by (10.34)

⎣ ⎦ ⎣ ⎦ { }ey qNN

aa ⎥⎦

⎥⎢⎣

⎢∂

∂∂

∂−=

∂∂

−=ξξξ

θ 4111L

w . (10.46)

Substituting (10.45) into (10.46) gives

( )⎣ ⎦( )

( )( )( ) ( ⎪

⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

++−

−+

−

=∂

∂

ηηξ

ηηηξ

ηηηξ

ξηξ

ii

ii

iiTi

a

b,N

1228

18

2

)

. (10.47)

Evaluating (10.47) along 1+=ξ gives

⎣ ⎦( )

( ) ( )( )⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−+−−

−−−

=∂

∂

0118

121 ηη

ηη

ξbN T

, ⎣ ⎦( )

( ) ( )( )( ) ( ) ⎪

⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−−−+−

−−

=∂

∂

ηηη

ηη

ξ12

118

122

abN T

,

⎣ ⎦( )

( ) ( )( )( ) ( ) ⎪

⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

+−−+

−−

=∂

∂

ηηη

ηη

ξ12

118

123

abN T

, ⎣ ⎦( )

( ) ( )( )⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−+−

−−

=∂

∂

0118

124 ηη

ηη

ξbN T

.

The above expressions indicate that yθ is determined by the values of

and

w

xθ at nodes 1, 2, 3, and 4 as well as by yθ at nodes 2 and 3. The rotation yθ is not continuous across the side 2-3 and the element is non-conforming.

Substituting (10.43) into (10.4) and (10.8) gives

{ } [ ]{ }eeTee qkqU

21

= , (10.48)

where the element stiffness matrix is

[ ] [ ] [ ] [ ] ABDBhk T

A

e d12

3

∫= (10.49)

and the strain-displacement matrix is


[ ] ⎣ ⎦ ( )⎣ ⎦ηξ

ηξ

η

ξ

,N

ba

b

a

N

yx

y

x

B

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

∂∂∂∂∂∂∂

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

∂∂∂

∂∂∂∂

=

2

2

2

2

2

2

2

2

2

2

2

2

2

1

1

2

. (10.50)

Substituting the shape functions from (10.45) and integrating gives the element stiffness matrix

[ ] ( )

[ ] [ ] [ ] [ ][ ] [ ] [ ]

[ ] [ ][ ]⎥

⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−=

44

3433

242322

14131211

2

3

SYM148

kkkkkkkkkk

ba

hEke

ν (10.51)

where the submatrices have the following expressions

[ ]

( ) ( ) ( ) ( )

( )

( )⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎭⎬⎫

⎩⎨⎧ −+

−⎭⎬⎫

⎩⎨⎧ −+

⎭⎬⎫

⎩⎨⎧ +−−

⎭⎬⎫

⎩⎨⎧ ++−++

=

22

22

2222

11

1154

34SYM

1154

34

4151241

51227

524

a

bab

ab

k

νβ

ννα

νβναναβ

,

[ ]

( ) ( ) ( ) ( )

( ) ( )

( ) ( )⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎭⎬⎫

⎩⎨⎧ −−

⎭⎬⎫

⎩⎨⎧ −+

⎭⎬⎫

⎩⎨⎧ −−

⎭⎬⎫

⎩⎨⎧ +−

⎭⎬⎫

⎩⎨⎧ −+−

⎭⎬⎫

⎩⎨⎧ +−

⎭⎬⎫

⎩⎨⎧ −+−−

=

222

222

2222

12

1151

3201

512

01154

3241

51

151241

5127

5222

aa

bb

ab

k

νβνβ

νανα

νβναναβ

,

[ ]

( ) ( ) ( ) ( )

( ) ( )

( ) ( )⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎭⎬⎫

⎩⎨⎧ −+

⎭⎬⎫

⎩⎨⎧ −−

⎭⎬⎫

⎩⎨⎧ −+

⎭⎬⎫

⎩⎨⎧ −+−

⎭⎬⎫

⎩⎨⎧ −+−

⎭⎬⎫

⎩⎨⎧ −−

⎭⎬⎫

⎩⎨⎧ −++−

=

222

222

2222

13

1151

3101

51

01151

311

51

1511

5127

522

aa

bb

ab

k

νβνβ

νανα

νβναναβ

,


[ ]

( ) ( ) ( ) ( )

( ) ( )

( ) ( )⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎭⎬⎫

⎩⎨⎧ −−

⎭⎬⎫

⎩⎨⎧ ++−

⎭⎬⎫

⎩⎨⎧ −−

⎭⎬⎫

⎩⎨⎧ −−−

⎭⎬⎫

⎩⎨⎧ ++−

⎭⎬⎫

⎩⎨⎧ −+−−−

=

222

222

2222

41

1154

32041

51

01151

321

512

41511

51227

5222

aa

bb

ab

k

νβνβ

νανα

νβναναβ

and

ba

=α , ab

=β .

The remaining submatrices of (10.51) are

[ ] [ ] [ ][ ]311322 IkIk T= , [ ] [ ] [ ][ ]314323 IkIk T= , [ ] [ ] [ ][ ]323324 IkIk T= ,

[ ] [ ] [ ][ ]111133 IkIk T= , [ ] [ ] [ ][ ]112134 IkIk T= ,

[ ] [ ] [ ][ ]211244 IkIk T= , (10.52)

where

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

100010001

1I , , . (10.53) [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

100010001

2I [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

100010001

3I

The above relationships are presented in reference [87].

In deriving this result, it is simpler to use the expression (10.36) for and substitute for

w{ }α after performing the integration. A typical integration is then of

the form

( )( )⎪⎩

⎪⎨⎧

++=∫ ∫

+

−

+

−

even and 11

4

odd or 0 dd

1

1

1

1nm

nm

nmnm ηξηξ

For =zp constant , substituting the shape functions into

{ } ⎣ ⎦ ApNfA

zTe d∫= (10.54)

and integrating, the vector of equivalent nodal forces is obtained as

{ } ⎣ ⎦Tz

e ababababbapf −−−−= 33333

.

Stresses at any point in the plate are given by (10.5). In terms of the nodal displacements they can be expressed as


{ } [ ][ ]{ }eqBDz−=σ , (10.55)

where is defined in (10.50). The most accurate values are at the Gauss points of a numerical integration scheme.

[ ]B( 22× )

10.3.2 BFS element (conforming)

A conforming rectangular element, of the form shown in Fig. 10.4, commonly referred to as the BFS element (Bogner, Fox and Schmit - 1966), can be obtained using products of separate one-dimensional Hermitian shape functions (5.21) as for uniform slender beams. It is a four-node thin plate bending element.

The nodal expansion of the displacement function has the form (10.43) with

( )⎣ ⎦( ) ( )( ) ( )( ) ( ) ⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−=

ηξηξηξ

ηξ

ii

ii

iiT

i

fgagfbff

,N , (10.56)

where

( ) ( ) ( ) ( )( ) ( ) ( ) ( ),g,f

,g,f

iiiiii

iiiiii

41 32

41

41 32

41

323

323

ηηηηηηηηηηη

ξξξξξξξξξξξ

++−−=−+=

++−−=−+= (10.57)

in which ( )ii ,ηξ are the coordinates of node i.

Unfortunately, when we examine the derivatives of these products we find that the twist ηξ ∂∂∂ w2 is zero at all four corners and that there is no constant component to this second derivative. As this controls the shear strain in equation (10.2), this violates the fundamental requirement that the element can represent all constant strain states. In the limit, as an increasing number of elements is used, the plate will tend towards a zero twist condition.

A compromise is to use the cubic Hermitian polynomials for the deflection shape functions but reduce the order to linear for the rotation shape functions. This element does have constant strain but is non-conforming, and discontinuities in slope occur at interfaces. However, as the mesh is refined and the element size is decreased, the results are convergent.

The solution used for the BFS element is the introduction of ηξ ∂∂∂ w2 as an additional degree of freedom at each node. In this case, the displacement function is of the form (10.43) but with 16 terms


, (10.58) { } ⎣ ⎦44441111 yxyxyxyxTeq wwww ′′′′= θθθθ L

where yxyx ∂∂∂≡′′ ww 2 and

( )⎣ ⎦

( ) ( )( ) ( )( ) ( )( ) ( ) ⎪

⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−=

ηξηξηξηξ

ηξ

ii

ii

ii

ii

Ti

ggbafga

gfbff

,N . (10.59)

It can be shown that this element can perform rigid body movements without deformation and can describe pure bending behaviour in the x- and y-directions. This is ensured by the presence in the functions (10.59) of the first six terms in (10.33).

The element stiffness matrix and consistent vector of nodal forces are given by (10.49) and (10.53) where the matrix ⎣ ⎦N is defined by (10.43) and (10.59).

Although the BFS element is more accurate than the ACM element, it is difficult to use in conjunction with other types of elements in built-up structures due to the presence of the degree of freedom yx∂∂∂ w2 . This is overcome in the WB element (Wilson, Brebbia – 1971) by introducing the approximations

( )411 21

yyyx bθθ −=′′w , ( )122 2

1xxyx a

θθ −=′′w ,

( )323 21

yyyx bθθ −=′′w , ( )434 2

1xxyx a

θθ −=′′w .

Applying the above constraints to the BFS element makes it a non-conforming one. The transverse displacement and tangential slope are continuous between elements but the normal slope is not.

10.3.3 HTK thick rectangular element

When the thickness is greater than about a tenth of the plate width, the shear deformations become important and a Reissner-Mindlin plate model is adopted. For a thick plate-bending element, the strain energy expression is (10.30) and the potential of the external load is (10.21), with { }χ and { }γ given by (10.24) and (10.26). The highest derivative of , w xθ and yθ appearing in these

expressions is the first. Hence, for convergence, , w xθ and yθ are the only degrees of freedom required at the nodal points (Fig. 10.5).


The four-node HTK element (Hughes, Taylor and Kanoknukulcha - 1977) expands separately , w xθ and yθ in terms of their nodal values. It represents the shear deformation directly, without having to infer it as the derivative of the bending moment.

The displacement functions are of the form

, , , (10.60) ∑=

=4

1iiiN ww ∑

==

4

1iiN ixx θθ ∑

==

4

1iyiy N iθθ

where the functions are defined by (9.3), that is iN

( )( )ηηξξ iiiN ++= 1141 . (10.61)

These functions ensure that , w xθ and yθ are continuous between elements.

In matrix form, expressions (10.60) can be written

, (10.62) [ ]{ e

y

x qN=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

θθw

}

where

, (10.44) { } ⎣ ⎦444111 yxyxTeq θθθθ ww L=

and

. (10.63) [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

41

41

41

000000000000

NNNN

NNN

L

L

L

Substituting (10.62) into (10.24), (10.26) and (10.30) gives

{ } [ ] { }eeTee qkqU

21

= , (10.64)

where the element stiffness matrix [ ]ek can be written as the sum of the matrices due to bending and shear

[ ] [ ] [ ]SBe kkk += . (10.65)

In (10.65)


[ ] [ ] [ ] [ ] ABDBhk BTB

A

B d12

3

∫= (10.66)

and

[ ] [ ] [ ] [ ] ABDBhk SSTS

A

S d∫= κ . (10.67)

The strain-displacement matrix [ ]BB is of the form

[ ] [ ] [ ] [ ] [ ][ ]BBBBB BBBBB 4321= (10.68)

where

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∂∂−∂∂∂∂

∂∂−=

yNxNyN

xNB

ii

i

iBi

000

00. (10.69)

The strain-displacement matrix [ ]SB is of the form

[ ] [ ] [ ] [ ] [ ][ ]SSSSS BBBBB 4321= (10.70)

where

[ ] ⎥⎦

⎤⎢⎣

⎡−∂∂

∂∂=

00

ii

iiSi NyN

NxNB . (10.71)


[ ]( )

( )( ) ( ) ⎥

⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−++

+−=

bab

aB

iiii

ii

iiBi

414100410

4100

ηξξηηξηξξ

ηηξ (10.72)

and

[ ] ( ) ( )( )( ) ( )( ) ⎥

⎦

⎤⎢⎣

⎡++−+

+++=

041141411041

ηηξξηξξηηξξηηξ

iiii

iiiiSi b

aB .(10.73)

Substituting (10.72) into (10.68) and the resulting matrices into (10.66) gives the element stiffness matrix due to bending


[ ] ( )

[ ] [ ] [ ] [ ][ ] [ ] [ ]

[ ] [ ][ ] ⎥⎥

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−=

B

BB

BBB

BBBB

B

k

kk

kkk

kkkk

ba

hEk

44

33

242322

14131211

SYM

148 342

3

ν (10.74)

where the submatrices are

[ ] ( ) ( )

( ) ( )⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎭⎬⎫

⎩⎨⎧ −++−

+−⎭⎬⎫

⎩⎨⎧ −+=

22

22

121

341

210

1211

21

340

000

11

aba

babk B

νβν

ννα ,

[ ] ( ){ } ( )

( ) ( ){ } ⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−+−−−

−−−=

22

22

143113

210

13211

320

000

12

aba

babk B

νβν

ννα ,

[ ] ( ) ( )

( ) ( )⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎭⎬⎫

⎩⎨⎧ −−−+

+⎭⎬⎫

⎩⎨⎧ −−−=

22

22

121

321

210

1211

21

320

000

13

aba

babk B

νβν

ννα ,

[ ] ( ){ } ( )

( ) ( ){ } ⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−−+−=

22

22

13213

210

132114

310

000

14

aba

babk B

νβν

ννα ,

and

ba

=α , ab

=β .


The remaining submatrices of (10.74) are given by relationships corresponding to (10.52).

Substituting (10.73) into (10.71) and the resulting matrices into (10.67) gives the element stiffness matrix due to shear

[ ][ ] [ ] [ ] [ ]

[ ] [ ] [ ][ ] [ ]

[ ] ⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

S

SS

SSS

SSSS

S

S

k

kk

kkk

kkkk

ba

hEk

44

33

242322

14131211

SYM

γ48 34

3, (10.75)

where 23 12γ bGhES κ= is a shear parameter similar to (5.97) and

, , [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−+=

2

222

22

00

1

11

aabb

abk S αα

αα[ ]

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −+−=

2

222

22

00

1

12

aabb

abk S αα

αα

, . [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

=2

222

22

00

1

13

aabb

abk S αα

αα[ ]

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−−=

2

222

22

00

1

14

aabb

abk S αα

αα

The remaining submatrices of (10.75) are given by relationships corresponding to (10.52).

The above expressions are presented in reference [87].

The vector of equivalent nodal forces is

{ } (10.76) ⎣ ⎦ Ap

NfA

zTe d

00∫ ⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

For = constant, substituting the shape functions from (10.63) and (10.61) into (10.76) and integrating gives

zp

{ } ⎣ ⎦Tz

e bapf 001001001001= (10.77)

which means that one quarter of the total force is concentrated at each node.

The bending and twisting moments per unit length are given by


[ ] [ ] [ ]{ eB

xy

y

x

qBDIh

MMM

3

3

12−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

}

])

, (10.78)

where is defined by (10.53). The most accurate values are at the Gauss points of a numerical integration scheme.

[ 3I( 22×

The shear forces per unit length are

[ ] [ ]{ eSS

y

x qBDhQQ

κ=⎭⎬⎫

⎩⎨⎧ }

)

. (10.79)

The most accurate values are at the centre of the element.

Benchmark problems have shown that the HTK element yields accurate solutions for simply supported or clamped plates. However, large errors can occur in the case of cantilever plates.

The above analysis can be easily extended to 8-node or higher-order plate elements. The bending strains in the 8-node version are recovered accurately if sampled at the reduced Gauss points. In very thin plates, the shear strains become very small and near-zero values in (10.26) thus imply a dependent relationship between ,

( 22×

w xθ and yθ which is not true for thick plates. This is avoided by selective integration, lowering the integration order for the stiffness matrix due to shear. However, the standard ( )22× integration is reported to give good results for width-to-thickness ratios up to 50.

There is no perfect rectangular plate-bending element. There are different formulations using mixtures of corner and mid-side freedoms in order to achieve near complete polynomials. Alternative methods, considering rather difficult to generate a displacement function valid over the entire rectangular element, suggest to subdivide the element into regions (e.g., into four triangles) and work with different displacement functions over each region. Obviously, the individual functions must be continuous (up to the first derivatives) across the interior boundaries as well as the exterior boundaries.

10.4 Triangular plate bending elements

In this section we outline the development of some plate-bending triangular elements.


10.4.1 Thin triangular element (non-conforming)

Figure 10.6 shows the T element (Tocher - 1962). The local x-axis lies along the side 1-2 and the local y-axis is perpendicular to it. Nodes 1, 2 and 3 have coordinates , and ( )00, ( 02 ,x ) ( )33 y,x . There are three degrees of freedom at each

node, namely, the transverse displacement , and the two rotations wyx ∂

∂=

wθ and

xy ∂∂

−=wθ with respect to the local axes.

Since the element has 9 degrees of freedom, the displacement function can be represented by a polynomial having nine terms, that is

(10.80) ( ) .yyxyxx

yyxxyx3

922

83

7

265

24321

ααα

αααααα

++++

++++++=w

Note that a complete cubic has ten terms so that the polynomial (10.80) is incomplete. In order to maintain symmetry, the coefficients of and are taken to be equal.

yx2 2yx

Fig. 10.6

The expression (10.80) can be written in the following matrix form

( )⎣ ⎦ { }α3223221 yyxyxxyyxxyx +=w , (10.81)

( )⎣ ⎦ { }αy,xP=w , (10.81, a)

where

{ } . (10.82) ⎣ 9321 ααααα L=T ⎦

Differentiating (10.81) gives


{ }αθθ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−−++

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

0230201032020100

1

22

22

322322

yyxxyxyyxxyxyyxyxxyyxxyx

y

x

w. (10.83)

Evaluating (10.83) at the nodal points gives

{ } [ ]{ }αee Aq = , (10.84) where

, (10.85) { } ⎣ ⎦333111 yxyxTeq θθθθ ww L=

and the matrix

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−−−++

−−

−

=

0230201032020100

10030020100000100000001000000010000000100000000001

2333

2333

2333

2333

33

2333

23

33

2333

2333

222

222

32

222

yyxxyxyyxxyxyyxyxxyyxxyx

xxxx

xxxAe . (10.86)

Solving (10.84) for { }α gives

{ } [ ] { }ee qA1−

=α . (10.87)

Substituting (10.87) into (10.81, a) yields

( )⎣ ⎦ [ ] { }ee qAy,xP1−

=w . (10.88)

Unfortunately, the matrix [ ]eA is singular whenever

02 332 =−− yxx (10.89)

and, therefore, cannot be inverted. If this occurs, the locations of the nodes should be altered to avoid this condition.

This element is non-conforming. Evaluating (10.83) along the side 1-2, of equation , gives 0=y


. (10.90) { }αθθ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−0030020100000100000001

2

2

32

21xx

xxxxx

y

x

w

The rotation xθ is a quadratic function having coefficients 3α , 5α and

8α . These cannot be determined using the values of xθ at nodes 1 and 2 only. Therefore the normal slope is not continuous between elements. Moreover, the assumed function (10.80) is not invariant with respect to the choice of coordinate axes, due to the and terms. yx2 2yx

The edge 1-2 was taken along the x-axis. In this case xθ is a tangential component and yθ is a normal component of rotation. The transverse displacement

and the normal component w yθ are continuous between elements, while the

tangential component xθ is not. To alleviate this, in some plate-bending elements a linear variation of the tangential component of rotation is adopted.


{ } [ ]{ }eeTee qkqU

21

= (10.91)

where the element stiffness matrix in the local coordinate system is

[ ] [ ] [ ] [ ] [ ] [ ] 13d

12−−

∫= eT

A

Tee AABDBhAk (10.92)

and the strain-displacement matrix is

[ ]( ) ⎥

⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+=

040020000620200000026002000

yxyx

yxB . (10.93)

A typical element of the integrand in (10.92) is of the form ,

which can be evaluated analytically.

∫A

nm Ayx d

The vector of consistent nodal forces is given by

{ } [ ] ⎣ ⎦ ApPAf zT

A

Tee d∫−

= . (10.94)

Note that nodal coordinates are usually given in global axes. Calculation of the element stiffness matrix referred to local axes requires the local coordinates of


nodes 2 and 3. These can be obtained from their global coordinates using the corresponding transformation matrix.

In order to assemble the global stiffness matrix, the element matrices have to be first transformed from local to global axes through a matrix triple product which is one of the more time-consuming procedures in finite element analysis.

10.4.2 Thick triangular element (conforming)

Consider a thick triangular element (THT element – Henschel, Tocher, 1969) referred to a global coordinate system. Nodes 1, 2 and 3 have area coordinates 1ζ , 2ζ and 3ζ (8.35). There are three independent degrees of freedom at each node, namely, the transverse displacement , and the two rotations w Xθ and Yθ with respect to the global axes.

The displacement functions are assumed to be

∑=

=3

1ii iww ζ , , , (10.95) ∑

==

3

1iXiX iθζθ ∑

==

3

1iYiY iθζθ

where , iw iXθ and iYθ are the degrees of freedom at node i. They ensure that , w

Xθ and Yθ are continuous between elements.

In matrix form, expressions (10.95) are written

, (10.96) [ ]{ e

Y

X qN=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

θθw

}

where the vector of nodal displacements is

, (10.97) { } ⎣ ⎦333111 YXYXTeq θθθθ ww L=

and

. (10.98) [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

321

321

321

000000000000000000

ζζζζζζ

ζζζN

Functions expressed in area coordinates can be differentiated with respect to Cartesian coordinates using


∑= ∂

∂=

∂∂ 3

121

i iiAx ζ

β , ∑= ∂

∂=

∂∂ 3

121

i iiAy ζγ , (10.99)

where iβ and iγ are defined in (8.8).

Substituting (10.96) into (10.24) and using (10.99) gives the strain-displacement matrix due to bending

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−−−=

332211

321

321

000000000

000000

21

γβγβγβγγγ

βββ

ABB . (10.100)

As this matrix is constant, the stiffness matrix due to bending is

[ ] [ ] [ ] [ ]BTBB BDBAhk12

3= . (10.101)

Substituting (10.96) into (10.24) and using (10.99) shows that the strain-displacement matrix due to shear is

[ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−=

02

02

02

02

02

02

33

22

11

33

22

11

ζγ

ζγ

ζγ

ζβ

ζβ

ζβ

AAA

AAABS . (10.102)

Substituting (10.102) into (10.67) and integrating using the formula

( ) ApnmpnmA

A

pnm 2!2

!!!d321∫ +++=ζζζ (10.103)

gives the stiffness matrix due to shear

[ ][ ] [ ] [ ]

[ ] [ ][ ] ⎥

⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

SYM33

2322

131211

S

SS

SSS

S

k

kk

kkk

hGk κ , (10.104)

where


[ ]

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−+

=

60

6

066

664

1

1

1121

21

11

A

AA

k S

β

γ

βγγβ

, [ ]

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−+

=

120

6

0126

664

2

2

111212

12

A

AA

k S

β

γ

βγγγββ

,

[ ]

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−+

=

120

6

0126

664

3

3

111313

13

A

AA

k S

β

γ

βγγγββ

, [ ]

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−+

=

60

6

066

664

2

2

2222

22

22

A

AA

k S

β

γ

βγγβ

,

[ ]

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−+

=

120

6

0126

664

3

3

222323

23

A

AA

k S

β

γ

βγγγββ

, [ ]⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−+

=

60

6

066

664

3

3

3323

23

33

A

AA

k S

β

γ

βγγβ

.

The complete element stiffness matrix is the sum (10.65).

The above relationships are presented in reference [87].

For = constant, substituting the shape functions from (10.98) into (10.76) and integrating gives the vector of equivalent nodal forces

zp

{ } ⎣ ⎦Tz

e Apf 0010010013

= (10.105)

which shows that one third of the total force is concentrated at each node.

10.4.3 Discrete Kirchhoff triangles (DKT)

The difficulties to formulate simple (minimum degrees of freedom) and high-performing elements based on the Kirchhoff theory at the continuum level (which requires a continuity for ) led to the development of the so-called Discrete Kirchhoff (DK) technique (Wempner-1969, Stricklin et al.-1969, Dhatt-1967).

1C w


In the formulation of various DK plate elements, only the bending strain energy is considered, in which the curvatures are expressed, as in the Reissner-Mindlin plate theory, in terms of the first derivatives of the rotations (10.24). In this case only approximations of these rotations can be considered. The shear strain energy is neglected.

0C

The Kirchhoff constraints are imposed in a discrete manner on the element or/and on the sides. For instance, the transverse shear strains are either taken equal to zero at the mid-side points (collocation on sides) or their integral along each edge is taken equal to zero. The aim is to preserve the continuity of the tangential components of rotations (normal slopes). Constant curvature patch-tests are needed to check the validity of the elements.

0C

The T element presented in section 10.4.1 can be called a ‘continuous’ Kirchhoff triangle. It was shown that, along a side, the rotations vary quadratically and their tangential component (normal slope) is not continuous between elements.

To remedy this, in the discrete Kirchhoff triangles the two components of rotations are assumed independent of one another. The tangential component sθ is assumed to vary linearly along each edge, while the normal component nθ varies quadratically (Fig. 10.7, a). The latter condition means that the transverse displacement can vary cubically. w

a b

Fig. 10.7

For a side ji − , of length , the tangential and normal components of rotations

kl

ksθ and knθ at the mid-side node k, are defined in terms of the components along the coordinate axes by

(10.106) ⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

=⎭⎬⎫

⎩⎨⎧

ky

kx

kk

kk

kn

ks

cssc

θθ

θθ

where


( ) kijk xxc l−= , ( ) kijk yys l−= . (10.107)

Consider a discrete Kirchhoff triangle (DKT) with three nodes and three degrees of freedom per node (sometimes called DKT9).

Initially, it is assumed that it has six nodes (Fig. 10.7, b). Then, the degrees of freedom at mid-side nodes are eliminated.

The rotation nθ varies parabolically (Fig. 10.8, a). Its expression in terms of the dimensionless coordinate

kss l=′ can be written as

( ) ( ) kjninn ssss αθθθ ′−′+′+′−= 141 .

At the mid-side node k

( ) kjninkn αθθθ ++=21 ,

where the first term in the right hand side is expressed in terms of the rotations at the corner nodes. The parameter kα has to be eliminated.

The rotation sθ is assumed to vary linearly (Fig. 10.8, b)

( ) jsiss ss θθθ ′+′−= 1 .

Fig. 10.8

In order to eliminate the degrees of freedom at mid-side nodes, the three parameters 4α , 5α , 6α must be expressed in terms of the degrees of freedom at the corner nodes. This is done requiring the shear strains zsγ to vanish along each side. One kind of ‘discrete’ Kirchhoff constraint is formulated in integral form as

0dd

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

∂∂

= ∫∫j

i

n

j

i

zs ss

s θγ w . (10.108)


Equation (10.108) can be written

, 0dd

0

=+ ∫∫k

sn

j

i

l

θw

( ) ( ) ( )[ ] 0d1411

0

=′′−′+′+′−+− ∫ sssss kjnink αθθlij ww ,

wherefrom we obtain

( ) ( )jninjik

k θθα +−−=43

23 wwl

,

or, in terms of the degrees of freedom of the corner points

( ) ( )jykjxkiykixkjik

k cscs θθθθα +−+−−−=43

23 wwl

. (10.109)

The equation (10.109) can also be obtained by adopting a Hermitian cubic polynomial for and using the discrete Kirchhoff collocation constraint (zero

shear strain)

( )sw

0=+∂∂

= nzs sθγ w at points i, j, k. Then, the condition 0=zsγ will be

satisfied at all points along the contour, because is cubical and w nθ is quadratic.

The rotations xθ and yθ can be expressed as

⎣ ⎦ { } ⎣ ⎦ { }⎣ ⎦ { } ⎣ ⎦ { },PN

,PN

yyy

xxx

αθθ

αθθ

+=

+= (10.110)

where

⎣ ⎦ ⎣ ⎦ ⎣ ⎦ηξηξ −−== 1321 NNNN ,

, { } ⎣ ⎦Txxxx 321 θθθθ = { } ⎣ ⎦T

yyyy 321θθθθ = ,

{ } , (10.111) ⎣ T654 αααα = ⎦

⎣ ⎦ ⎣ ⎦665544 sPsPsPPx −−−= , ⎣ ⎦ ⎣ ⎦665544 cPcPcPPy = ,

⎣ ⎦ ( ) ( )⎣ ⎦ηξηηξηξξ −−−−= 14414 654 PPP .

Substituting (10.109) into (10.110) we obtain the explicit expressions of xθ and yθ in terms of the corner nodal variables


, (10.112) [ ]{ e

y

x qN=⎭⎬⎫

⎩⎨⎧θθ }

where

, (10.113) { } ⎣ ⎦333111 yxyxTeq θθθθ ww L=

[ ] , (10.114) ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎥⎥

⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

yyy

xxx

NNN

NNNN

321

321

⎣ ⎦ ⎣ ⎦xi

xi

xi

x NNNN 3211 = , ⎣ ⎦ ⎣ ⎦yi

yi

yi

y NNNN 3211 = . (10.115)

In (10.115), the shape functions have the following expressions

mmm

kkk

xi sPsPN

ll 23

23

1 +−= , mmm

kkk

yi cPcPN

ll 23

23

1 −= ,

222 4

343

mmkkixi sPsPNN −−= , , (10.116) x

iyi NN 32 =

mmmkkkxi csPcsPN

43

43

3 += , 223 4

343

mmkkiyi cPcPNN −−= .

In the above expressions the indices k and m relate to the two edges having the common corner point i, as shown in Table 10.1.

Taking into account the hypotheses adopted in formulating the DKT, the nodal rotations ixθ and iyθ have the expressions used in Kirchhoff’s theory

i

ix y ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=wθ ,

iiy x ⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

−=wθ , ( )3 2 1 ,,i =

allowing the introduction of Kirchhoff-type boundary conditions.

Table 10.1

Corner point, i Side k ( )ji − Side m ( )ji − 1 4 ( )21− 6 ( )13− 2 5 ( )32 − 4 ( )21− 3 6 ( )13− 5 ( )32 −


Because the parameters kα are eliminated using an expression which is a

function of the nodal variables of the side k only, the continuity C of 0nθ is

maintained.

The stiffness matrix of the DKT element can be expressed in explicit form, in a local coordinate system [18], using a Hammer integration rule. In the following, it will be derived in global coordinates as in [68].

Equation (10.112) can be split as

⎣ ⎦{ }ex qG=θ , ⎣ ⎦{ }e

y qH=θ . (10.117)

The shape function row vectors ⎣ ⎦G and ⎣ ⎦H , presented explicitly in terms of the local oblique coordinates ξ and η , can be rewritten as

⎣ ⎦ ⎣ ⎦ [ ]⎣ ⎦ ⎣ ⎦ [ ] .HH

,GG22

22

1

1

ηξηξηξ

ηξηξηξ

=

= (10.118)

The curvature vector (10.24) is written for convenience as

{ } ⎣ TT

yxxy

yxyx 1221 χχχθθθθ

χ =⎥⎦

⎥⎢⎣

⎢∂

∂−

∂∂

∂∂

∂

∂−= ⎦

) ]

. (10.119)

The elements of the curvature vector can be expressed in terms of the rows of the matrices [ and ( 96× G [ ]H as follows

⎣ ⎦ { }{ }eqXA

ηξχ 121

1 = ,

⎣ ⎦ { } { }eqYA

ηξχ 121

2 = , (10.120)

⎣ ⎦ { } { }eqZA

ηξχ 121

12 = ,

where A is the area of the triangle with vertices ( )11 y,x , ( )22 y,x and ( )33 y,x .

In (10.120)

{ }⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎪

⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

+++

=

6352

5342

3322

22

GGGG

GGX

ββββββ

, { } , ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎪

⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

+++

=

6352

5342

3322

22

HHHH

HHY

γγγγγγ


{ }⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎪

⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

+++++++++

=

63526352

53425342

33223322

2222

HHGGHHGG

HHGGZ

ββγγββγγββγγ

, (10.121)

where and represent the ith row of ⎣ ⎦ iG ⎣ ⎦ iH [ ]G and [ ]H , respectively, and

kji yy −=β and jki xx −=γ as in (8.8), with i, j, k taking values 1, 2, 3 cyclically.

Substituting for the curvature terms into the bending strain energy expression

{ } [ ]{ }∫=

A

Te yxDU dd

21 χχ , (10.122)

where the material matrix [ ]D has the form (10.13), can be written as eU

{ } [ ]{ }eeTee qKqU

21

= ,

where [ ]eK is the element stiffness matrix in global coordinates, given by

[ ][ ][ ][ ]

[ ] [ ] [ ][ ] [ ]

[ ]

[ ][ ][ ] ⎥

⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

ZYX

RDRDRDRDRDRD

ZYX

AK

T

e

66

2622

161211

SYM21 , (10.123)

with

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

2141244412

241R . (10.124)

The matrix [ ]eK is of rank 6, so there are 3 deformation modes with zero strain energy.

The integrations in (10.122) have the form

( )!2!!dd

0

1

0++

==∫ ∫−

nmnmnm

ξ ξ

ηξηξ .

The presented procedure avoids the matrix triple products needed when the element stiffness matrix is given in the local coordinate system.

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Index ACM plate-bending element 232 Admissible functions 144 Area coordinates 180 Aspect ratio 178 Assembly of global matrices 100, 151 Axial effects 97

Band storage 38 Bars 47 Bar element 47 Beams 79 Beam element 83

− stiffness matrix 87 − vector of nodal forces 89

Bernoulli-Euler beam theory 79 BFS plate-bending element 238

Compatibility 28, 150, 187 Conforming elements 150, 238, 248 Consistent nodal forces 214 Consistent vector of nodal forces 89 Constant strain triangle 153 Continuity 132 Convergence 97, 187 Coordinate transformation 19, 98

Deep beam 117 Direct method 26 Direct stiffness method 17 Discrete Kirchhoff constraints 252 Discrete Kirchhoff triangle 250 Displacement method 9

Effective shear area 118 Eight-node quad 208 Eight-node rectangle 178 Energy methods 131 Equation of equilibrium 199 Equivalent nodal forces, vector of

− beam 90, 95 − frame 98 − constant strain triangle 160 − linear quad 199

− 8-node quad 214 − ACM plate-bending element 237 − T plate-bending element 247 − THT plate-bending element 249 Expanded stiffness matrix 29 External potential energy 140

Force transformation 20 Four-node rectangle 176 Frame element 97

Gauss points 200 Gauss quadrature 200 − one-dimensional 200 − two-dimensional 203 Gravity loading 217 Grids 111

Hermitian polynomials 85 HTK plate-bending element 239

Initial strain effects 59 Internal forces 35 Inverse of Jacobian matrix 196 Isoparametric elements 191

Jacobian matrix 195 Jacobian positiveness 223 Joint force equilibrium equations 31

Kinematic equivalent forces 47 Kirchhoff’s hypotheses 225

Lagrange polynomials 220 Linear elasticity 123 Linear strain triangle 182

Master element 192

Natural coordinates 192 Nine-node quad 219 Nodal approximation 155 Nodal forces 55 Node numbering 37 Non-conforming elements 232, 245 Numerical integration 200


Object of FEA 1

Pascal’s triangle 220 Patch test 190 Plate bending elements 225

− rectangular 232 − triangular 244

Polynomial approximation 154 Principle of

− minimum total potential energy 139 − virtual displacements 134 − virtual work 131

Quadratic strain triangle 185 − triangle 222

Quadrilateral membrane element 191 − eight-node 208 − linear 191 − nine-node 219

Rayleigh-Ritz method 143 Reactions 35 Rectangular membrane element 176

− thin plate element 232 − thick plate elements 239

Reduced global stiffness matrix 33 − integration 205

Reference element 192 Reissner-Mindlin plate theory 229

Serendipity elements 209 Shaft elements 60 Shape functions

− bars 50, 53 − shaft 62 − beam 85, 107, 120 − constant strain triangle 157 − four-node rectangle 177 − eight-node rectangle 179 − linear strain triangle 182 − quadratic strain triangle 186 − linear quadrilateral 193 − eight-node quad 209 − nine-node quad 221 − six-node triangle 221 − ACM plate-bending element 234 − BFS plate-bending element 238 − HTK plate-bending element 240 − THT plate-bending element 248

Shear correction factor 118, 230 Six-node triangle 221

Skyline storage 39 Stiffness matrix 17 − bars 17, 22, 52 − shaft 61 − beam 87, 94, 122 − constant strain triangle 159 − frame 98 − grid 113 − linear strain triangle 183 − linear quadrilateral 199 − eight-node quad 211 − T plate-bending element 247 − ACM plate-bending element 236 − HTK plate-bending element 242 − THT plate-bending element 249 Strain-displacement matrix 158, 236 − relations 121 Strain energy 124 ..139 − bending 231 − shear 232 Stress averaging 161 Stress-strain relations 122 Subparametric elements 191

T plate-bending element 245 Temperature effects 124 Thick plate theory 229 Thermal loads 36 Thin plate theory 225 Total potential energy 140 Transformation of − differential operators 195 − infinitesimal area 197 Triangle area 156 − quadratic strain triangle 185 Triangular membrane element 180 − linear strain triangle 182 − quadratic strain triangle 185 − six-node 221 Triangular thin plate element 245 − thick plate element 248 Two-dimensional elements 153 Two-point formula 202

Unreduced global stiffness matrix 30

Virtual displacements 131 − work of external loads 133 − work of internal forces 193

WB plate-bending element 239

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