M-Rades-FiniteElementAnalysis

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

iv

FINITE ELEMENT ANALYSIS 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

v

CONTENTS 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


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 Sσ

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

vi

FINITE ELEMENT ANALYSIS 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 8.1 The plane constant-strain triangle (CST)

153 153

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


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 9.2.1 One dimensional Gauss quadrature

200 200

vii

CONTENTS 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


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 stressstrain-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 initialvalue 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,

2

FINITE ELEMENT ANALYSIS

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 [ A ] { x } = { b } is equivalent to 1 minimizing P (x ) = { x }T [ A ]{ x } − { x }T { b }. This is the heart of the FEA when 2 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

4


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 nonaerospace 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

6


loading. Data input can be carried out either in an interactive way, through a userfriendly 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

7

1. INTRODUCTION

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 (backcalculation).

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.

8


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. T1 , T2 , T3 are the tensions in members and F1 , F2 , F4 , F5 are the reaction forces at the supports.

Fig. 2.1


10

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

T1 + T2 2 2 + F1 = 0,

(2.1)

T2 2 / 2 + F2 = 0. At Node 2 (Fig. 2.2, b), equilibrium gives 6 F − T1 − T3 2 /2 = 0,

(2.2)

T3 2 / 2 + F4 = 0. At Node 3 (Fig. 2.2, c) T3 2 /2 + F5 − T2 2 /2 = 0,

(2.3)

9 F − T2 2 / 2 − T3 2 /2 = 0.

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 Fig. 2.2

c

2.2 Conditions for geometric compatibility The compatibility equations relate a bar extension Δl 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 xOy can be expressed in terms of the displacements in the global coordinate frame as

u 1 = U 1 cos θ + V 1 sin θ , u 2 = U 2 cos θ + V 2 sin θ . The change in length of the member is

(2.4)


11

(

)

(

)

Δl = u 2 − u1 = U 2 − U1 cosθ + V 2 − V1 sin θ .

(2.5)

Fig. 2.3 Applying equation (2.5) to each member in turn leads to

Δl12 = U 2 - U1 ,

(2.6)

Δl13 = (U 3 − U1 )

2 2 + (V3 − V1 ) , 2 2

⎛ 2 ⎞⎟ Δl 23 = (U 3 − U 2 ) ⎜⎜ − ⎟ + (V3 − V2 ) ⎝

2 ⎠

(2.7) 2 . 2

(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

Δl12

T l = 1 , EA

Δl 13 =

T2

2 l 2 , 2E A

Δl 23 =

T3

2 l 2 2E A

or T l T1 l T l , Δl13 = 2 , Δl 23 = 3 . (2.9) EA 2E A 2E A Three more equations have been added and so there are now 12 equations for 16 unknowns.

Δl12 =


12

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

U1 = V1 = V2 = U 3 = 0

(2.10)

which complete the set of 22 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 EA EA ( U 2 − U1 ) , Δl 12 = l l 2E A EA T2 = Δl 13 = 2 ( U 3 − U1 + V3 − V1 ) , l l 2E A EA T3 = Δl 23 = 2 ( − U 3 + U 2 + V3 − V2 ) . l l

T1 =

Substitution of the expressions for T1 to T3 into the equilibrium equations (2.1) - (2.3) yields 6 equations 2 U1 + V1 − U 2 − U 3 − V3 = U1 + V1 − U 3 − V3 =

F1 l , EA

F2 l , EA

− U1 + 2 U 2 − V2 − U 3 + V3 = − U 2 + V2 + U 3 − V3 =

6F l , EA

F4 l , EA

− U1 − V1 − U 2 + V2 + 2 U 3 =

F5 l , EA

− U1 − V1 + U 2 − V2 + 2 V3 =

9 Fl . EA

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


⎡2 ⎢1 ⎢ EA ⎢ − 1 ⎢ l ⎢0 ⎢− 1 ⎢ ⎣⎢ − 1

13

1 1 0 0

−1 0 2 −1

0 0 −1 1

−1 −1

−1 1

1 −1

− 1 − 1⎤ ⎡ 0 ⎤ ⎡ F1 ⎤ − 1 − 1⎥⎥ ⎢⎢ 0 ⎥⎥ ⎢⎢ F2 ⎥⎥ − 1 1 ⎥ ⎢U 2 ⎥ ⎢ 6 F ⎥ ⎥ ⋅ ⎢ ⎥ = ⎢ ⎥ . (2.11) 1 − 1⎥ ⎢ 0 ⎥ ⎢ F4 ⎥ 2 0 ⎥ ⎢ 0 ⎥ ⎢ F5 ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ 0 2 ⎦⎥ ⎣⎢ V3 ⎦⎥ ⎢⎣ 9 F ⎥⎦

The third and the sixth equation can be decoupled EA l

⎡2 1 ⎤ ⎡U 2 ⎤ ⎧6 F ⎫ ⎢1 2⎥ ⋅ ⎢ V ⎥ = ⎨9 F ⎬ ⎣ ⎦ ⎣ 3⎦ ⎩ ⎭

and solved to give U2 =

Fl , EA

V3 = 4

Fl . EA

(2.12)

Substituting U 2 and V3 into the remaining equations yields the reaction forces

F1 = −5F , F2 = −4 F , F4 = −5 F , F5 = − F .

(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 Fx and Fy . Determine the internal bar forces and the displacement of joint 8. Force Method

Denoting Ti ( i = 1,...,7 ) the forces applied by each bar to the end nodes, the equilibrium equations of joint 8 can be written

Fx − Fy −

7

∑ Ti cosθi = 0, i =1 7

∑ Ti sinθi = 0. i =1

(2.14)


14

This is a set of two equations with seven unknowns, so the framework is statically indeterminate. We choose X1 = T3 , X 2 = T4 , X 3 = T5 , X 4 = T6 , and X 5 = T7 as redundant forces.

Fig. 2.4 The strain energy is

U=

5 ⎞ 1 ⎛⎜ 2 T1 l 1 + T22l 2 + X i2l i ⎟ ⎟ 2 EA ⎜⎝ i =1 ⎠

∑

(2.15)

where T1 and T2 are functions of X 1 to X 5 , according to (2.14). Using Menabrea’s theorem, the five deformation conditions can be written ∂U = 0, ∂X i

( i = 1,...,5 )

(2.16)

since we are assuming no support movement. They are of the form l1 T1

∂T1 ∂T + l 2 T2 2 + l i X i = 0 . ∂X i ∂X i

( i = 1,...,5 )

(2.17)

This is a set of five linear equations wherefrom the five redundant forces X 1 to X 5 are determined. The components of the displacement of joint 8 are given by Castigliano’s second theorem u8 =

∂U , ∂Fx

v8 =

∂U . ∂Fy

(2.18)


15

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

Δl i = u 8 cosθi + v 8 sin θi .

( i = 1,...,7 )

(2.19)

( i = 1,...,7 )

(2.20)

The bar length is li =

a . sinθ i

Substituting (2.19) into the force-elongation equations Ti = EA

Δl i li

,

( i = 1,...,7 )

(2.21)

( i = 1,...,7 )

(2.22)

gives the force-displacement relations Ti =

(

)

EA u 8 cos θ i + v 8 sin θ i sin θ i . a

Inserting (2.22) into the equilibrium equations (2.14) of joint 8 yields a Fx − u8 EA

7

∑

a Fy − u8 EA

i =1 7

∑

cos 2θ i sinθ i − v 8 2

7

∑ i =1 7

cosθ i sin θ i − v 8

i =1

∑

sin 2θ i cosθ i = 0 , (2.23) 3

sin θ i = 0 ,

i =1

which is solved for u8 and v 8 . Regardless the number of concurrent bars, only two equations are obtained for the two joint displacements. In matrix form they can be written ⎧ Fx ⎫ ⎡ K11 ⎨ ⎬=⎢ ⎩ Fy ⎭ ⎣ K 21

where the stiffness coefficients are

K12 ⎤ ⎧ u8 ⎫ ⎨ ⎬ K 22 ⎥⎦ ⎩ v 8 ⎭

(2.24)


16

K11 =

EA a

7

∑ cos2θi sinθi = 1.409 i =1

K12 = K 21 = K 22 =

EA a

EA a

7

∑

EA , a

7

∑ cosθi sin 2θi = 0,

(2.25)

i =1

sin 3θ i = 4.567

i =1

EA . a

Solving (2.24) gives u 8 = 0.7097

Fx a , EA

v 8 = 0.219

Fy a EA

.

(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 l e , cross section area Ae and Young’s modulus Ee . Nodes are conveniently numbered 1 and 2. It is acted upon by the nodal forces f1 , f 2 . The nodal displacements are q 1 , q 2 . Generally, bar elements are assumed to be uniform (EAe = const.) , pinconnected at the ends, linearly elastic, axially loaded (no bending) and with no forces between ends.

18


Fig. 3.1 Both the nodal displacements q 1 , q 2 and the nodal forces f1 , f 2 are positive in the positive x direction. The equilibrium equation for the bar element is

f1 + f 2 = 0.

(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 f l move, then for q 1 = 0 , q 2 = 2 e and Ee Ae E A f1 = − f 2 = − e e q 2 . (3.2) le

Similarly, if end 2 is now fixed but end 1 allowed to move, f l q 2 = 0 , q 1 = 1 e and Ee Ae Ee Ae q1 . le

f1 = − f 2 =

(3.3)

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

⎧⎪ f 1 ⎫⎪ Ee Ae ⎡ 1 − 1⎤ ⎨ ⎬ = l e ⎢⎣− 1 1 ⎥⎦ ⎪⎩ f 2 ⎪⎭e 123 1442443 element stiffness matrix

nodal forces

⎧q1 ⎫ ⎨q ⎬ 2 ⎭e ⎩12 3

(3.4)

nodal displacements

or

{ f }= [ k ] { q } e

e

e

(3.5)

where the element stiffness matrix in local coordinates is

[ k ] = El A e

e e e

⎡ 1 − 1⎤ ⎢− 1 1 ⎥ . ⎣ ⎦

(3.6)

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 4 × 4 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 θ e with respect to the X-axis of the global coordinate system. In fact, the angle θ e 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 xOy can be expressed in terms of the displacements in the global coordinate frame as (2.4)

q 1 = Q x1 cos θ e + Q y1 sin θ e , q 2 = Q x 2 cos θ e + Q y 2 sin θ e . In matrix form

(3.7)

20

FINITE ELEMENT ANALYSIS ⎧ Qx1 ⎫ ⎪Q ⎪ ⎪ y1 ⎪ ⎨ ⎬ ⎪ Qx 2 ⎪ ⎪ Qy2 ⎪ ⎩1 ⎭e 424 3

⎧⎪ q 1 ⎫⎪ 0 0 ⎤ ⎡ cosθ e sinθ e = ⎢ ⎨q ⎬ 0 0 cos θ sin θ e ⎥⎦ ⎪⎩ 2 ⎪⎭ e ⎣ e 1 424 3 1444444 424444444 3

local displacements

transformation matrix

or

global displacements

{ q }= [T ] { Q }, e

where

[T ] = ⎡⎢ cos0θ e

⎣

e

e

e

sinθ e 0

(3.8)

0 cosθ e

(3.9) 0 ⎤ sinθ e ⎥⎦

is a coordinate transformation matrix. From nodal coordinate data, denoting

( X 1 ,Y1 )

(3.10)

and

( X 2 ,Y 2 )

the

coordinates of nodes 1 and 2, respectively, we obtain cos θ e =

X 2 − X1 le

,

sin θ e =

Y 2 − Y1 le

,

le =

(X 2 − X 1 ) 2 + (Y 2 − Y1 ) 2 .

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

21


Fx1 = f1 cosθ e ,

Fx 2 = f 2 cosθ e ,

Fy1 = f1 sinθ e ,

Fy 2 = f 2 sinθ e .

(3.11)

In matrix form ⎧ Fx1 ⎫ 0 ⎤ ⎡cosθ e ⎪F ⎪ ⎢ sinθ 0 ⎥⎥ ⎧ f1 ⎫ ⎪ y1 ⎪ e ⎢ = ⎨F ⎬ ⎨ ⎬ , ⎢ 0 cosθ e ⎥ ⎩ f 2 ⎭ e ⎪ x2 ⎪ ⎢ ⎥ ⎪Fy ⎪ sinθ e ⎦ ⎣ 0 2 ⎭e ⎩1 424 3 144 42444 3 123

global force components

or

(3.12)

local forces

transformation matrix

{ F } = [ T ] { f }. e T

e

e

(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

{ f } { q }= {F } { Q } . e T

e T

e

e

(3.14)

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

{ f } { q } = { f } [T ] { Q } = {F } { Q } , e T

hence

e T

e

e

e T

e

e

{ f } [T ] = { F } e T

e T

e

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)

{F }= [T ] { f }= [T ] [ k ] { q }= [1T 4]44[2k 4][4T43] { Q } e

we find

e T

e

e

e T

e

e

e

{F } = [ K ] { Q }

(3.15)

[ K ] = [T ] [ k ] [T ]. .

(3.16)

e

where

e T

e

e

e

e T

e

e

e

22


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

[K ] e

⎡ c2 cs − c2 − cs⎤ ⎢ ⎥ E A cs s2 − cs − s2 ⎥ = e e⎢ 2 , l e ⎢− c − cs c2 cs ⎥ ⎢ ⎥ 2 cs s 2 ⎦⎥ ⎣⎢ − c s − s

(3.17)

where c = cosθ e and s = sin θ e . 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 ⎧ Fx1 ⎫ ⎪F ⎪ Ee Ae ⎪ y1 ⎪ ⎨ ⎬ = F le ⎪ x2 ⎪ ⎪ Fy 2 ⎪ ⎩ ⎭e

⎡ c2 c s − c 2 − c s ⎤ ⎧ Qx1 ⎫ ⎢ ⎥ ⎪ ⎪ s 2 − c s − s 2 ⎥ ⎪ Qy1 ⎪ ⎢ ⋅ ⎨ ⎬ . ⎢ c2 c s ⎥ ⎪Q x 2 ⎪ ⎢ ⎥ s 2 ⎦⎥ ⎪⎩Q y 2 ⎪⎭e ⎣⎢SYM

(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

[ K ]= [K ]

e T

e

.

Expressing displacements in terms of forces gives

{ Q }= [ K ] { F }= [ δ ] { F } e

e

−1

e

e

e

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

[K ] e

−1

[ ]

= δe .

(3.18)

[ ]

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

23


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

{ }{ }

T 1 Qe Fe . (3.19) 2 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 T 1 Ue = Qe K e Qe . (3.20) 2

We =

{ } [ ]{ }

It is equal to its transpose Ue =

so that

1 2

{ Q } [K ] { Q } e T

e

T

[ K ]= [K ] e

e T

e

(3.20, a)

,

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

{ } [ K ] { Q }= 0 wherefrom [ K ] { Q } = { 0 } . This can only be true if the determinant of [ K ] vanishes. The matrix [ K ] is said to be singular. The rigid body modes are defined

rigid body displacements e

{Q } e

for which

2U e = Q e

e

T

e

e

e

e

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 3× 3 and 2 × 2 reduced sets are still zero, so that the stiffness matrix has rank 1 (or its rank deficiency is 3). 3.2.4.3 Positive diagonal elements

[ ]

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

24


[ ]

physically unsound. Moreover, the matrix K e 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 Qx 2 = Q y 2 = 0 and end 1 having

(

)

a unit displacement along the global X-axis Qx1 = 1, Q y1 = 0 as in Fig. 3.4.

Fig. 3.4 Equation (3.17, a) can be written

. . .⎤ ⎧ Qx1 = 1 ⎫ ⎪ ⎪ . . .⎥⎥ ⎪ Q y1 = 0 ⎪ ⋅⎨ ⎬ . . .⎥ ⎪Qx 2 = 0 ⎪ ⎥ . . .⎦ ⎪⎩Q y 2 = 0⎪⎭

(3.21)

Fx1 = k11 , Fy1 = k21 , Fx 2 = k31 , Fy 2 = k 41 .

(3.22)

⎧ Fx1 ⎫ ⎡ k11 ⎪F ⎪ ⎢ ⎪ y1 ⎪ ⎢k 21 ⎨ ⎬= ⎪ Fx 2 ⎪ ⎢ k31 ⎪ Fy 2 ⎪ ⎢⎣k 41 ⎩ ⎭ which yields

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 Qx1 = 1 and all other displacements are zero. The displacement Qx1 = 1 produces an axial shortening Qx1 cosθ e = cosθ e , E A which corresponds to a compressive force e e cosθ e , whose components must le be equilibrated by the external forces k11 , k 21 , k31 and k 41 . Equilibrium of horizontal and vertical forces yields

k11 + k31 = 0 ,

k21 + k41 = 0 ,

25


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 (2 i − 1) and 2 i .

a

b

26


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

θe

cosθ e

sinθ e

c2

s2

cs

le

EAe

EAe le

EA

EA l

i

j

1

1

2

0

1

0

1

0

0

l

2

1

3

45

2 2

2 2

1 2

1 2

1 2

2 l 2

2 EA 2

EA l

3

2

3

135

2 2

1 2

1 2

−

2 l 2

2 EA 2

EA l

−

2 2

1 2

3.4 Direct method Using the data from Table 3.1, the element stiffness matrices (3.17) are calculates as 1

[K ] 1

⎡ 1 ⎢ EA ⎢ 0 = l ⎢− 1 ⎢ ⎣ 0

1

2

3 4

0 − 1 0⎤ 1 0 0 0⎥⎥ 2 0 1 0⎥ 3 ⎥ 0 0 0⎦ 4

2

5

T

6

⎡ 1 1 − 1 − 1⎤ 1 ⎢ ⎥ E A ⎢ 1 1 − 1 − 1⎥ 2 T K2 = 1⎥ 5 l ⎢− 1 − 1 1 ⎢ ⎥ 1⎦ 6 ⎣− 1 − 1 1

[ ]

⎡• • • • K ⎢• • • • K ⎢ ⎢• • • • K ⎢ ⎢• • • • K ⎢K K K K K ⎢ ⎣⎢K K K K K

K⎤ K⎥⎥ K⎥ ⎥, K⎥ K⎥ ⎥ K⎦⎥

⎡• • K K • •⎤ ⎢• • K K • •⎥ ⎢ ⎥ ⎢K K K K K K⎥ ⎢ ⎥, ⎢K K K K K K⎥ ⎢• • K K • •⎥ ⎢ ⎥ ⎢⎣ • • K K • • ⎥⎦

27


and 3

4

5

6

⎡ 1 − 1 − 1 1⎤ 3 ⎢ 1 − 1⎥⎥ 4 T E A ⎢− 1 1 K3 = 1 − 1⎥ 5 l ⎢− 1 1 ⎢ ⎥ ⎣ 1 − 1 − 1 1⎦ 6

[ ]

⎡K ⎢K ⎢ ⎢K ⎢ ⎢K ⎢K ⎢ ⎣⎢K

K K K K K⎤ K K K K K⎥⎥ K • • • •⎥ ⎥. K • • • •⎥ K • • • •⎥ ⎥ K • • • • ⎦⎥

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

Element 1 2 3

2 Direction X Y X Y Local nodal coordinate 1 2 3 4 Global nodal coordinate 1 2 3 4 1 2 5 6 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 6 × 6 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 i = 1 and the right end node j = 3 (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 i = 1 are numbered 2 i − 1 = 1 and 2 i = 2 , and those for the node j = 3 are numbered 2 j − 1 = 5 and 2 j = 6 . 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.

28


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 ~ (3.23) Q e = T e { Q },

{ } [ ]

{ Q } is the element displacement vector in global coordinates, {Q } is the ~ displacement vector of the truss structure and [T ] is referred to as a

where

e

e full 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

{Q } 1

⎧ Q1 ⎫ ⎡1 ⎪ ⎪ ⎢ ⎪Q ⎪ 0 = ⎨ 2⎬ = ⎢ ⎪Q3 ⎪ ⎢⎢0 ⎪⎩Q4 ⎪⎭ ⎣0

⎧ Q1 ⎫ ⎡1 ⎪Q ⎪ ⎢0 ⎪ ⎪ 2 Q = ⎨ 2⎬ = ⎢ ⎪Q5 ⎪ ⎢0 ⎪⎩Q6 ⎪⎭ ⎢⎣0

{ }

⎧Q3 ⎫ ⎡0 ⎪Q ⎪ ⎢ ⎪ ⎪ 0 Q3 = ⎨ 4 ⎬ = ⎢ ⎪Q5 ⎪ ⎢⎢0 ⎪⎩Q6 ⎪⎭ ⎣0

{ }

0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0

0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0

0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0

⎧ Q1 ⎫ ⎪ ⎪ 0⎤ ⎪Q2 ⎪ 0⎥⎥ ⎪⎪Q3 ⎪⎪ ~ ⋅⎨ ⎬ = T1 0⎥ ⎪Q4 ⎪ ⎥ 0⎦ ⎪Q5 ⎪ ⎪ ⎪ ⎩⎪Q6 ⎭⎪

[ ] [Q ] ,

⎧ Q1 ⎫ ⎪ ⎪ 0⎤ ⎪Q2 ⎪ ⎥ 0⎥ ⎪⎪Q3 ⎪⎪ ~ ⋅⎨ ⎬ = T 2 0⎥ ⎪Q4 ⎪ ⎥ 1⎦ ⎪Q5 ⎪ ⎪ ⎪ ⎪⎩Q6 ⎪⎭

[ ] [ Q ],

⎧ Q1 ⎫ ⎪ ⎪ 0⎤ ⎪Q2 ⎪ 0⎥⎥ ⎪⎪Q3 ⎪⎪ ~ ⋅⎨ ⎬ = T 3 0⎥ ⎪Q4 ⎪ ⎥ 1⎦ ⎪Q5 ⎪ ⎪ ⎪ ⎩⎪Q6 ⎭⎪

[ ] [ Q ].

29


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) Ue =

1 2

{ Q }T [T~ e ] T [ K e ][T~ e ]{ Q }

or Ue =

1 2

{ Q }T [ K~ e ]{ Q },

where the expanded element stiffness matrix

[ K~ ] = [T~ ] [ K ] [T~ ] e

e

T

e

e

(3.24)

has the size of the system matrix. For the truss from Fig. 3.5, equation (3.24) yields

[K~ ] = [T~ ] [ K ][T~ ] 1

1 T

1

1

⎡1 ⎢0 ⎢ ⎢0 =⎢ ⎢0 ⎢0 ⎢ ⎢⎣0

0 0 0⎤ 1 0 0⎥⎥ ⎡ 1 ⎥ 0 1 0 EA ⎢⎢ 0 ⎥ 0 0 1⎥ l ⎢− 1 ⎢ 0 0 0⎥ ⎣ 0 ⎥ 0 0 0⎥⎦

0 − 1 0 ⎤ ⎡1 0 0 0⎥⎥ ⎢⎢0 ⋅ 0 1 0⎥ ⎢0 ⎥ ⎢ 0 0 0⎦ ⎣0

0 0 0 0 0⎤ ⎥ 1 0 0 0 0⎥ , 0 1 0 0 0⎥ ⎥ 0 0 1 0 0⎦

or ⎡ 1 ⎢ 0 ⎢ ~ 1 EA ⎢− 1 K = ⎢ l ⎢ 0 ⎢ 0 ⎢ ⎣⎢ 0

[ ]

[K~ ] = [T~ ] [K ][T~ ] 2

2 T

2

2

⎡1 ⎢0 ⎢ ⎢0 =⎢ ⎢0 ⎢0 ⎢ ⎣⎢0

0 − 1 0 0 0⎤ 0 0 0 0 0⎥⎥ 0 1 0 0 0⎥ ⎥. 0 0 0 0 0⎥ 0 0 0 0 0⎥ ⎥ 0 0 0 0 0⎦⎥ 0 0 0⎤ 1 0 0⎥⎥ 1 − 1 − 1⎤ ⎡1 0 0 ⎡ 1 ⎢ 0 0 0⎥ EA ⎢ 1 1 − 1 − 1⎥⎥ ⎢⎢0 1 0 ⋅ ⎥ 0 0 0⎥ l ⎢ − 1 − 1 1 1 ⎥ ⎢0 0 0 ⎥ ⎢ ⎢ 0 1 0⎥ 1 ⎦ ⎣0 0 0 ⎣− 1 − 1 1 ⎥ 0 0 1⎦⎥

0 0 0⎤ 0 0 0⎥⎥ , 0 1 0⎥ ⎥ 0 0 1⎦

30


or

[K~ ] 2

0 0 − 1 − 1⎤ 0 0 − 1 − 1⎥⎥ 0 0 0 0⎥ ⎥ 0 0 0 0⎥ 0 0 1 1⎥ ⎥ 0 0 1 1 ⎦⎥

1 ⎡ 1 ⎢ 1 1 ⎢ 0 EA ⎢ 0 = ⎢ 0 l ⎢ 0 ⎢− 1 − 1 ⎢ ⎣⎢− 1 − 1

and

[K~ ] = [T~ ] [ K ][T~ ] 3 T

3

3

3

⎡0 ⎢0 ⎢ ⎢1 =⎢ ⎢0 ⎢0 ⎢ ⎢⎣0

0 0 0⎤ 0 0 0⎥⎥ ⎡ 1 − 1 − 1 1 ⎤ ⎡0 0 0 0⎥ EA ⎢⎢− 1 1 1 − 1⎥⎥ ⎢⎢0 ⋅ ⎥ 1 0 0⎥ l ⎢− 1 1 1 − 1⎥ ⎢0 ⎥ ⎢ ⎢ 0 1 0⎥ ⎣ 1 − 1 − 1 1 ⎦ ⎣0 ⎥ 0 0 1⎥⎦

0 1 0 0 0⎤ ⎥ 0 0 1 0 0⎥ , 0 0 0 1 0⎥ ⎥ 0 0 0 0 1⎦

or

[K~ ] 3

⎡0 ⎢0 ⎢ EA ⎢0 = ⎢ l ⎢0 ⎢0 ⎢ ⎢⎣0

0⎤ 0 ⎥⎥ 0 1 −1 − 1 1⎥ ⎥. 0 −1 1 1 − 1⎥ 0 −1 1 1 − 1⎥ ⎥ 0 1 − 1 − 1 1 ⎥⎦

0 0

0 0

0 0

0 0

3.7 Unreduced global stiffness matrix The strain energy for the complete truss structure 1 T { Q } [ K ] { Q }, 2 can be calculated by simply adding the element strain energies

U=

U=

1 1 ~ ~ ∑U = ∑ 2 { Q } [ K ]{ Q } = 2 { Q } ∑ [ K ] { Q }. T

T

e

e

e

e

e

(3.25)

(3.26)

e

Comparing (3.25) and (3.26) we get

[ K ] = ∑ [ K~ e ]

.

(3.27)

e

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

31


[ ]

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

[ K ] = [ K~ 1 ]+ [ K~ 2 ]+ [ K~ 3 ]

⎡ 2 1 − 1 0 − 1 − 1⎤ ⎢ 1 1 0 0 − 1 − 1⎥⎥ ⎢ 2 − 1 − 1 1⎥ EA ⎢− 1 0 = ⎥. ⎢ 0 − 1 1 1 − 1⎥ l ⎢ 0 ⎢− 1 − 1 − 1 1 2 0⎥ ⎥ ⎢ 2 ⎦⎥ ⎣⎢− 1 − 1 1 − 1 0

(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

32


F1 = Fx11 + Fx21 ,

F3 = Fx12 + Fx31 ,

F2 = Fy11 + Fy21 ,

F4 = Fy12 + Fy31 ,

F5 = Fx22 + Fx32 , F6 = Fy22 + Fy32 .

Fig. 3.7 In matrix form

⎧ F1 ⎫ ⎡ 1 0 0 0 ⎪F ⎪ ⎢ ⎪ 2⎪ ⎢ 0 1 0 0 ⎪⎪ F3 ⎪⎪ ⎢ 0 0 1 0 ⎨ ⎬=⎢ ⎪ F4 ⎪ ⎢ 0 0 0 1 ⎪ F5 ⎪ ⎢ 0 0 0 0 ⎪ ⎪ ⎢ ⎪⎩ F6 ⎪⎭ ⎢⎣ 0 0 0 0 14 4244 3

[ T~ ]

1 T

⎧ Fx11 ⎫ ⎪ 1 ⎪ ⎪ Fy1 ⎪ ⎪ 1 ⎪ ⎪ Fx 2 ⎪ ⎪ Fy12 ⎪ 1 0 0 0 0 0 0 0 ⎤ ⎪−−− ⎪ ⎪ ⎪ 0 1 0 0 0 0 0 0 ⎥⎥ ⎪ Fx21 ⎪ 0 0 0 0 1 0 0 0 ⎥ ⎪⎪ Fy21 ⎪⎪ ⎥⋅⎨ ⎬ 0 0 0 0 0 1 0 0 ⎥ ⎪ Fx22 ⎪ 0 0 1 0 0 0 1 0 ⎥ ⎪ Fy22 ⎪ ⎥ ⎪ ⎪ 0 0 0 1 0 0 0 1 ⎥⎦ ⎪ − − − ⎪ 144244 3 14 4244 3 F3 ⎪ x1 ⎪ ~2 T ~3 T T T ⎪F3 ⎪ ⎪ y1 ⎪ ⎪ Fx32 ⎪ ⎪ 3 ⎪ ⎪⎩ Fy 2 ⎪⎭

[ ]

[ ]

whose partitioned form can be written in scalar product form as

{ F } = [ T~1 ] T {F 1 }+ [ T~ 2 ] T {F 2 }+ [ T~ 3 ] T {F 3 } or generally

{ F } = ∑ [ T~ e ] T {F e }. e

(3.29)

33


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

{ F } = ∑ [ T~ e ] T [ K e ]{Q e }= ∑ e

e

[1T~4]44[ K24][4T~43] {Q }, e T

e

e

(3.30)

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

{ F } = ∑ [ K~ e ] {Q } = [ K ] {Q }. 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 Qi = ai ( i = 1 to ns ), where ns is the number of supports. Most often the support nodal displacements are zero, Qi = 0 . Another type are the multipoint constraints, encountered in inclined roller supports and rigid connections, of the type bi Qi + b j Q j = b0 , where bi , b j and b0 are known constants. For an N-degree-of-freedom structure, the finite element equations (3.31) are of the form ⎡ K11 K12 L K1N ⎤ ⎧ Q1 ⎫ ⎧ F1 ⎫ ⎢K ⎥ ⎪ ⎪ ⎪ ⎪ ⎢ 21 K 22 L K 2 N ⎥ ⋅ ⎪⎨ Q2 ⎪⎬ = ⎪⎨ F2 ⎪⎬ . ⎢L L L L ⎥ ⎪ L ⎪ ⎪L⎪ ⎢ ⎥ ⎣ K N 1 K N 2 L K NN ⎦ ⎪⎩QN ⎪⎭ ⎪⎩ FN ⎪⎭ Consider a single boundary condition, Q1 = a1 . As Q1 is known, the finite element equations in the remaining N − 1 unknowns are ⎡ K 22 ⎢K ⎢ 32 ⎢ L ⎢ ⎣K N 2

K 23 K 33 L KN3

L K 2 N ⎤ ⎧ Q2 ⎫ ⎧ F2 − K 21 a1 ⎫ L K 3 N ⎥⎥ ⎪⎪ Q3 ⎪⎪ ⎪⎪ F3 − K 31 a1 ⎪⎪ ⋅⎨ ⎬ = ⎨ ⎬. L L ⎥ ⎪L⎪ ⎪ L ⎪ ⎥ L K NN ⎦ ⎪⎩QN ⎪⎭ ⎪⎩ FN − K N 1 a1 ⎪⎭

(3.32)

34


The ( N − 1 ) × (N − 1) stiffness matrix above is obtained simply by deleting or eliminating the first row and column from the original ( N × N ) stiffness matrix. Equation (3.32) may be written in condensed form

[ K ] { Q } = { F },

(3.33)

where [K ] is a reduced stiffness matrix, obtained by eliminating the row and column corresponding to the specified or “support” degrees of freedom, { Q } and { F } are the reduced vectors of global displacements and forces, respectively. 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 { Q } using Gauss elimination. The matrix [ 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 Q1 = Q2 = Q4 = Q5 = 0 and the values of the applied forces F3 = 6 F and F6 = 9 F , the following set of linear equations is obtained 1 −1 0 − 1 − 1⎤ ⎡ 0 ⎤ ⎡ F1 ⎤ ⎡ 2 ⎢ ⎥ ⎢ 1 1 0 0 − 1 − 1⎥⎥ ⎢ 0 ⎥ ⎢⎢ F2 ⎥⎥ ⎢ 0 2 −1 −1 1 ⎥ ⎢Q3 ⎥ ⎢6 F ⎥ EA ⎢ − 1 ⎢ ⎥ ⋅⎢ ⎥ = ⎢ ⎥ . 0 −1 1 1 − 1⎥ ⎢ 0 ⎥ ⎢ F4 ⎥ l ⎢ 0 ⎢− 1 − 1 − 1 1 2 0⎥ ⎢ 0 ⎥ ⎢ F5 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 2⎥⎦ ⎢⎣Q6 ⎥⎦ ⎢⎣9 F ⎥⎦ ⎢⎣ − 1 − 1 1 − 1

(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 ⎡ 2 1 ⎢ 2 ⎢ 1 ⎢ EA ⎢− 1 − 1 l ⎢ 0 −1 ⎢ ⎢− 1 − 1 ⎢ ⎣− 1 0

0 − 1 − 1 ⎤ ⎡Q3 ⎤ ⎡6 F ⎤ ⎥ − 1 − 1 − 1 0 ⎥ ⎢Q6 ⎥ ⎢9 F ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ 2 1 0 − 1⎥ ⋅ ⎢ 0 ⎥ = ⎢ F1 ⎥ . ⎢ ⎥ ⎢ ⎥ 1 1 0 − 1⎥ ⎢ 0 ⎥ ⎢ F2 ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ 0 0 1 1⎥ ⎢ 0 ⎥ ⎢ F4 ⎥ ⎥ −1 −1 1 2⎦ ⎣⎢ 0 ⎦⎥ ⎢⎣ F5 ⎥⎦ −1

The first two equations can be written EA ⎡2 1 ⎤ ⎡Q3 ⎤ ⎧6 F ⎫ ⋅⎢ ⎥ = ⎨ ⎬ l ⎢⎣1 2⎥⎦ ⎣Q6 ⎦ ⎩9 F ⎭

(3.35)

35


and solved to give Fl Fl Q6 = 4 , . EA EA In general, for non-zero boundary conditions, equations (3.33) can be written in partitioned form Q3 =

⎡ [ K aa ] ⎢[K ] ⎣ ba

where

{ Fa }

[ K ab ]⎤ ⎧ { Qa }⎫ ⎧ { Fa }⎫ ⋅ = , [ K bb ]⎥⎦ ⎨⎩ { Qb }⎬⎭ ⎨⎩ { Fb }⎬⎭

is the vector of known forces,

displacements and [ K ba ] = [ K ab ] .

{ Qb }

(3.36) is the vector of known

T

The first set of equations can be written

{ Fa } = [ K aa ] { Qa }+ [ K ab ] { Qb } ,

(3.37)

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

{ Qa } = [ K aa ] −1 ( { Fa } − [ K ab ] { Qb } ) ,

(3.38)

3.10 Reactions and internal forces The unknown forces, which in this case are the external reactions, are given by { Fb } = [ K ab ] T { Qa }+ [ K bb ] { Qb } . (3.39) For the truss from Fig. 3.5, the support displacements are zero, so that ⎧ F1 ⎫ ⎧− 5F ⎫ ⎡ − 1 − 1⎤ ⎪F ⎪ ⎢ 0 − 1⎥ Q ⎪ ⎪ ⎧ 3 ⎫ ⎪− 4 F ⎪ ⎪ 2 ⎪ EA ⎢ ⎥ ⋅ = = ⎨ ⎬ ⎨ ⎬ ⎨ ⎬. l ⎢ − 1 − 1⎥ ⎩Q6 ⎭ ⎪ − 5 F ⎪ ⎪ F4 ⎪ ⎥ ⎢ ⎪⎩ F5 ⎪⎭ ⎪⎩ − F ⎪⎭ ⎣− 1 0 ⎦ The elongation of an element (2.5) is

Δl e = q2 − q1 = ( Qx 2 − Qx1 ) cosθ e + ( Q y 2 − Q y1 ) sinθ e

which can be computed when all displacements have been determined. The axial force in a truss element is

(3.40)

36


⎧ Qx1 ⎫ ⎪Q ⎪ E A E A ⎪ y1 ⎪ Te = e e Δl e = e e b − c − s c s c ⎨ ⎬ . le le ⎪Q x 2 ⎪ ⎪Q y 2 ⎪ ⎭e ⎩ For the truss from Fig. 3.5, the member forces are

(3.41)

⎧ Q1 ⎫ ⎪Q ⎪ EA ⎪ ⎪ EA T1 = b−1 0 1 0c ⎨ 2 ⎬ = Q3 = F , Q l l ⎪ 3⎪ ⎪⎩ Q4 ⎪⎭ ⎧ Q1 ⎫ ⎪Q ⎪ 2 EA 2 ⎪ 2⎪ T2 = b − 1 − 1 1 1c ⎨ ⎬= l 2 ⎪ Q5 ⎪ ⎪⎩ Q6 ⎪⎭ ⎧ Q3 ⎫ ⎪Q ⎪ 2 EA 2 ⎪ 4⎪ T3 = b1 1 − 1 1 c ⎨ ⎬= l 2 ⎪ Q5 ⎪ ⎪⎩ Q6 ⎪⎭

2 EA Q6 = 4 2 F , l

2 EA ( Q3 + Q6 ) = 5 2 F . l

Stresses can be determined dividing these forces by the element crosssection 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 ε T = −α T , 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 σ T = −α ET , where E is Young’s modulus. Restraining produces a compressive axial force α EAT in the element, where A is the cross-section area. Accordingly, the element acts upon its nodes with equal and opposite forces

{ FT } = α EAT b − c

− s c s cT ,

(3.42)

37


or, in local coordinates,

{ fT } = α EAT b − 1

1cT .

(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 ⎛ Δl ⎞ (3.43) σ e = Ee ⎜⎜ e − α T ⎟⎟ , ⎝ le ⎠ 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 12 × 12 unreduced global stiffness matrix of the truss from Fig. 3.8, a, with nonzero elements in the upper triangle identified by the symbol X ←

B

⎡× × × × ⎢ × × × ⎢ ⎢ × × ⎢ × ⎢ ⎢ ⎢ ⎢ K =⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢s y m m ⎣

[ ]

→ × × × ×

× × × × × × × ×

0

× × × × × × × × × × × × × × × × × × × × × × × × × × e

t

r

i

c

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ×⎥ ⎥ ×⎥ ⎥ ×⎥ ×⎥ ⎥ ×⎥ × ⎥⎦

(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.

38


Since only nonzero elements need to be stored, the information in the above matrix can be compactly stored in the 12× 6 matrix below B

1 2 ↓ ↓

[ Kb ]

⎡× ⎢ ⎢× ⎢× ⎢ ⎢× ⎢× ⎢ ⎢× =⎢ × ⎢ ⎢× ⎢ ⎢× ⎢× ⎢ ⎢× ⎢× ⎣

× × × × × × × × × × ×

↓ × × × × × × × × × ×

× × × × × × × × ×

× × × × × × × ×

×⎤ ⎥ ⎥ ×⎥ ⎥ ⎥ ×⎥ ⎥. ⎥ ×⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 0 ⎥ ⎥ ⎦

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 , B = 6 . However, for the numbering from Fig. 3.8, b, B = 12 , 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.

39


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 { K s } . 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 Column height

→ 1

2

⎡k11 k12 ⎢ k 22 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣

2

4

3

1

4

5

0 k 23

k14 k 24

0 0

0 0

0 0

k33

k34 k 44

k35 0 k55

0 0 0

0

0 ⎤ ⎥ 0 ⎥ 0 ⎥ ⎥ k 48 ⎥ 0 ⎥ ⎥ 0 ⎥ k 78 ⎥ ⎥ k88 ⎥⎦

k 47 0

k 66

k67 k77

Skyline

The active columns are stored in the column vector { K s } and a diagonal pointer vector { ID } is built up with the indices of diagonal elements in { K s } ⎧ k11 ⎫ ⎪ k ⎪←1 ⎪ 12 ⎪ ⎪ k22 ⎪ ⎪ ⎪←3 ⎪ k23 ⎪ ⎪ k33 ⎪ ⎪ ⎪←5 ⎪ k14 ⎪ {Ks } = ⎨ ⎬ ⎪ k24 ⎪ ⎪ k34 ⎪ ⎪ ⎪ ⎪ k44 ⎪ ⎪ L ⎪←9 ⎪ ⎪ ⎪L⎪ ⎪k ⎪ ⎩ 88 ⎭

⎧1⎫ ⎪3⎪ ⎪ ⎪ ⎪5⎪ ⎪ ⎪ 9 { ID } = ⎪⎨ ⎪⎬ . ⎪12 ⎪ ⎪13 ⎪ ⎪ ⎪ ⎪17 ⎪ ⎪22⎪ ⎩ ⎭

40


The height of the jth column is given by ID ( j ) − ID ( j − 1) . 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

41


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 l = 1 , A = 1 , E = 1 and F = 1 , we obtain the following: Nodal data Node nr 1 2 3 4 5 6 7 8

Restr X 1 0 0 0 0 0 0 0

Restr Y 1 0 0 0 0 0 0 1

Coord X 0 4 4 8 8 12 12 16

Coord Y 0 0 3 0 6 0 3 0

Axial stress 16 -10 0 16 -9.16 -10.83 8

Element nr 8 9 10 11 12 13

Displ X 0 64 208.3 128 127.6 176 62.9 224

Displ Y 0 -361.1 -361.1 -391.8 -343.7 -339.7 -339.7 0

Element data Element nr 1 2 3 4 5 6 7

Nodes 1, 2 1, 3 2, 3 2, 4 3, 4 3, 6 4, 5

Nodes 4, 6 4, 7 5, 7 6, 7 6, 8 7, 8

Axial stress 12 -4.16 -10.83 0 12 -15

42


The support reactions are R1 = −8 , R2 = 6 , and R3 = 9 . 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) Fl . b) The axial stress in The vertical displacement of point 7 is v7 = −219.787 EA element 7 is N 7 = −6 F A . c) The support reactions are R1 = − R3 = 5.4 F , and R2 = 6 F . The deformed shape is presented in Fig. E3.2, b.

43


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) Fl . b) The axial stress in The vertical displacement of point 3 is v3 = −137.83 EA element 1 is N1 = −7.25 F A . c) The support reactions are R1 = 0 , R2 = 3.4 F and R3 = 2.6 F . The deformed shape is presented in Fig. E3.3, b.

44


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 l = 1 , A = 1 , E = 1 and F = 1 , we obtain the following: Nodal data Node nr 1 2 3 4

Restr X 1 1 0 0

Restr Y 1 1 0 0

Coord X 0 0 1 2

Coord Y 0 1 0 1

Displ X 0 0 -1.3294 2.6705

Displ Y 0 0 -3.2095 -9.0896

45


Element data Element nr 1 2 3 4 5

Nodes 3, 4 1, 3 2, 3 2, 4 1, 4

Axial stress - 0.940 - 1.329 0.940 1.335 - 0.749

a) The vertical displacement of point 4 is v4 = −9.09 F l E A . b) The axial stress in element 4 is N 4 = 1.335 F A . c) The support reactions are R1 = − R3 = 2 F , R2 = 0.335F and R4 = 0.665F . 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 v6 = −5 . Determine the location and the value of: a) the maximum nodal displacement; b) the maximum stress. Plot the deformed shape.

Fig. E3.5, a

46


Answer. The finite element model consists of 14 nodes and 25 elements. Taking l = 1 , A = 1 and E = 1 , the vertical displacements of points 7 and 8 are v7 = v8 = −4.538 , and the axial stress in element 15 is N15 = 0.366 . 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 C 0 continuity. Displacements must be continuous across the element boundary. 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 u = u (x ) due to axial loads p(x ) . The dimensions of p are force/length. The displacement at x + d x will be u + du .

48


The axial strain is given by the strain-displacement relation

ε x = du d x .

(4.1)

The normal stresses result from Hooke’s law

σ x = E du d x .

(4.2)

Fig. 4.1 The internal axial force N is du . (4.3) dx The equation of equilibrium of an infinitesimal element of length dx of the rod (Fig. 4.1) is d N + p d x = 0 , N = Aσ x = E A

dN + p (x ) = 0 , dx or, using (4.3),

EA

d2 u = − p (x ) . d x2

(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 x 1 and x 2 respectively (Fig. 4.2, a). We define a natural or intrinsic reference system which permits the specification of a point within the element by a dimensionless number

r=

x + x2 ⎞ 2 ⎛ ⎜x− 1 ⎟ x 2 − x 1 ⎜⎝ 2 ⎟⎠

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

(4.5)

49

4. BARS AND SHAFTS

Fig. 4.2 Expressing the physical coordinate in terms of the natural coordinate yields

x = N1 (r ) x 1 + N 2 (r ) x 2 ,

(4.6)

where 1 ( 1 − r ) and N 2 (r ) = 1 ( 1 + r ) (4.7) 2 2 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. N1 (r ) =

Fig. 4.3

4.1.3 Bar not loaded between ends For a prismatic bar not loaded between ends, p = 0 , d 2u d x 2 = 0 , d u d x = const . , so that the displacement field within the element may be expressed as a linear polynomial

50


u (x ) = a + b x .

(4.8)

The two integration constants above may be determined from the nodal displacements and the geometry of the element. With q1 = a + b x1 at node 1 and q2 = a + b x2 at node 2, equation (4.8) becomes u=

q2 − q1 q x − q2 x1 x+ 1 2 . x2 − x1 x2 − x1

(4.9)

Equation (4.9) can also be written u=

x − x1 − x + x2 q1 + q2 , x2 − x1 x2 − x1

u=

1− r 1+ r q1 + q2 . 2 2

or, using (4.5),

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

u = N1 (r ) q1 + N 2 (r ) q2 , where N1 (r ) =

(4.10)

1 ( 1 − r ) and N 2 (r ) = 1 ( 1 + r ) are the shape functions of the 2 2

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, q 1 and q 2 , are the nodal displacements. In matrix form 2

u=

∑ N q = ⎣N ⎦ { q }, e

i

(4.11)

i

i =1

where

⎣N ⎦ = ⎣N1 N 2 ⎦ and

{q }= { q e

1

q2

}T .

(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.

{ }

⎣N ⎦

In (4.11), q e is the column vector of element nodal displacements and is the row vector of displacement interpolation functions also named shape

51

4. BARS AND SHAFTS

functions. It is easy to check that u = q 1 at node 1 and u = q 2 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

εx =

{ }

{ }

du d e e = ⎣N ⎦ q = ⎣B ⎦ q dx dx

(4.13)

where d

⎣B ⎦ = d x ⎣N ⎦

(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 dx =

x 2 − x1 2

dr =

le dr , 2

(4.15)

where − 1 ≤ r ≤ +1 and the length of the element is l e = x 2 − x 1 . The element strain energy U e is

Ue =

1 2

∫σ

x εx

dV =

e

1 2

∫E ε e

2 x

dV .

(4.16)

e

Substituting (4.13) in (4.16) we obtain Ue =

1 2

{ q } ∫ ⎣B ⎦ e T

T

Ee ⎣B ⎦ dV

{ q }. e

(4.17)

Ve

144 42444 3

The above equation is of the form Ue =

1 2

{ q } [ k ] { q }, e T

[ ]

e

e

where the element stiffness matrix k e is given by

(4.18)

52


[ k ] = ∫ ⎣B⎦ e

[ ]

T

Ee ⎣B ⎦ dV .

(4.19)

Ve

This form guarantees that k e will be a symmetric matrix. In terms of the shape functions

[k ]= E A ∫ e

e e

le

Because

(4.20)

dN dN d r 2 dN = = , we can write dx dr d x l e dr

[ ke ] = 2 Ee Ae le Substituting

+1

∫

−1

T

⎢dN ⎥ ⎢dN ⎥ ⎢ ⎥ ⎢ ⎥ dr . ⎣ dr ⎦ ⎣ dr ⎦

(4.21)

d N2 1 d N1 1 = − and = + , equation (4.21) yields dr 2 dr 2

[k ] e

or

T

⎢dN ⎥ ⎢dN ⎥ ⎢ ⎥ ⎢ ⎥ dx . ⎣ dx ⎦ ⎣ dx ⎦

2E A = e e le

+1

−1

[ k ] = El A ⎡⎢−11 e

e e e

⎣

⎧− 1 2 ⎫ ⎢ 1

∫ ⎨⎩ 1 2 ⎬⎭ ⎢⎣− 2 − 1⎤ . 1 ⎥⎦

1⎥ dr 2 ⎥⎦

(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, p = const. , d 2u d x 2 = const . , so that the displacement field within the element may be expressed as a quadratic polynomial u (x ) = a + b x + c x 2 .

(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 onedimensional element. An internal node is added at the midpoint to comply with the requirement of a quadratic fit (Fig. 4.4, a).

53

4. BARS AND SHAFTS

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 x1 , x2 , x3 , and the vector of element nodal

{ }

displacements is q e = { q1 q2 q3 }T . The x-coordinate system is mapped onto an intrinsic r-coordinate system, given by the transformation T

2 ( x − x3 ) . x2 − x1 It comes out that r = −1 , 0, and + 1 at nodes 1, 3, and 2 (Fig. 4.4, b). r=

(4.24)

The displacements within the element can be written in terms of the three nodal displacements q 1 , q2 , and q 3 as

u = N1 (r ) q 1 + N 2 (r ) q 2 + N 3 (r ) q 3 , where the quadratic shape functions N i N1 (r ) = −

(4.25)

( i = 1, 2, 3 ) are

1 1 r ( 1 − r ) , N 2 (r ) = r ( 1 + r ) , 2 2 N 3 (r ) = ( 1 + r )( 1 − r ) .

(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.

54


The expressions for these shape functions can be written down by inspection. For example, since N1 = 0 at r = 0 and r = 1 , we know that N1 has to contain the product r ( 1 − r ) , i. e. the left hand part of the equations of the vertical lines passing through nodes 3 and 2. That is, N1 is of the form N1 = C r ( 1 − r ) . The constant C is obtained from the condition N1 = 1 at r = −1 , which yields C = −1 2 , resulting in the expression given in (4.26).

Fig. 4.5 The displacement field within the element is written in matrix form as 3

u=

∑ N q = ⎣N ⎦ { q } , e

i

(4.27)

i

i =1

where

⎣N ⎦ = ⎣N1 N 2 N3 ⎦ and

{ q }= { q e

1

q2

q3

}T .

(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 u = q 1 at node 1, because N1 = 1 and

N 2 = N3 = 0 . Similarly, u = q 2 at node 2, and u = q 3 at node 3. Thus, u is a quadratic polynomial passing through q1 , q2 , and q3 (Fig. 4.6). It is obtained by interpolation, using quadratic shape functions.

55

4. BARS AND SHAFTS

The element strain-displacement row vector ⎣B ⎦ in (4.14) is given by 1− 2r

⎢ ⎣B ⎦ = d x ⎣N ⎦ = d r ⎣N ⎦ d x = l ⎢− 2 e ⎣ d

d

dr

2

1+ 2r 2

⎥ − 2 r⎥ . ⎦

(4.29)

Fig. 4.6 The element stiffness matrix (4.19) is

[k ] e

+1

Al = e e 2

∫⎣⎦ B

T

Ee ⎣B ⎦ d r

−1

or, substituting (4.29),

[k ] e

1 − 8⎤ ⎡7 Ee Ae ⎢ = 1 7 − 8⎥⎥ . ⎢ 3l e ⎢⎣− 8 − 8 16 ⎥⎦

(4.30)

4.1.6 Vector of element nodal forces Consider an axial load p (x ) , having the units of force per unit length, distributed along the bar element. The mechanical work of such a force is W=

∫ u p dx = ∫ u

le

T

p dx .

(4.31)

le

Substituting (4.11), equation (4.31) becomes

{ } ∫ ⎣N ⎦

W = qe

T

le

It has the form

T

p dx .

(4.32)

56


{ } {f }

W = qe

T

e

(4.33)

where the element equivalent load vector is

{ f }= ∫ ⎣N ⎦ e

T

p dx =

le

le 2

+1

∫−1 ⎣N ⎦

T

p dr .

(4.34)

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

{f } e

+1

l = e p 2

∫

T ⎣N ⎦ d r

(4.35)

−1


{ f }= p2l e

A force

e

⎧ 1⎫ ⎨ ⎬. ⎩ 1⎭

(4.36)

p le , equal to half the total force on the element, is applied at each node. 2

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

{ f }= p l e

e

T ⎣1 6 1 6 2 3 ⎦ .

(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 Q1 , Q2 ,…, Q5 (Fig. 4.7, b). The global vector of nodal displacements is denoted by

{Q } = ⎣ Q1

Q2

Q3

Q4

Q5 ⎦ T .

The nodal forces are F1 , F2 ,…, F5 . The global load vector is denoted by

{ F } = ⎣ F1

F2

F3

F4

F5 ⎦ T .

The unreduced global stiffness matrix [ K ] proportionality factor between the two global vectors

plays the role of a

57

4. BARS AND SHAFTS

{ F } = [ K ] {Q }.

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 U3 =

1 2

{ q } [k ] { q } 3 T

3

3

or U3 =

EA3 ⎡ 1 − 1 ⎤ ⎧ Q3 ⎫ 1 Q3 Q4 ⎦ ⎨ ⎬. ⎣ 1⎥⎦ ⎩ Q4 ⎭ l 3 ⎢⎣ − 1 2

We can write U 3 as

U3 =

1 ⎣ Q1 Q2 2

Q3 Q4

⎡0 ⎢0 ⎢ ⎢ 0 Q5 ⎦ ⎢ ⎢ ⎢0 ⎢ ⎢0 ⎣

0 0

0 0 EA3 0 l3 EA 0 − 3 l3 0 0

or U3 =

1 2

{ Q }T [ K~ 3 ] { Q },

0 0 EA − 3 l3 EA3 l3 0

0⎤ 0 ⎥⎥ ⎥ 0⎥ ⎥ 0⎥ ⎥ 0 ⎥⎦

⎧ ⎪ ⎪⎪ ⎨ ⎪ ⎪ ⎪⎩

Q1 ⎫ Q2 ⎪⎪ ⎪ Q3 ⎬ Q4 ⎪ ⎪ Q5 ⎪⎭

58


[ ]

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

[ ]

global stiffness matrix, but the elements of the matrix k 3 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 U=

1 1 ~ T ~ T ∑e U e = ∑e 2 { Q } [ K e ]{ Q }= 2 { Q } ∑e [ K e ] { Q }

so that, as in (3.27), the global stiffness matrix is equal to the sum of the expanded element stiffness matrices [ K ] = K~ e .

∑[

]

e

This is a convenient algebraic explanation of the assembly process, which is never done in practice. The entries of the element matrices k e 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

1

[ k ] = ElA ⎡⎢− 11 1

1

⎣

1

3

[ k ] = ElA 3

3

3

2

2

[ k ] = ElA

− 1⎤ 1 , 1⎥⎦ 2

2

2

2

4

⎡ 1 − 1⎤ 2 ⎢ − 1 1⎥ 3 , ⎣ ⎦ 4

[ k ] = ElA

⎡ 1 − 1⎤ 3 , ⎢− 1 1⎥⎦ 4 ⎣

4

4

4

3

5

⎡ 1 − 1⎤ 4 . ⎢− 1 1⎥⎦ 5 ⎣

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 Element 1 2 3 4

Node i 1 2 3 4

j 2 3 4 5

59

4. BARS AND SHAFTS

The result is the following unreduced global stiffness matrix ⎡ A1 ⎢ l ⎢ 1 ⎢− A1 ⎢ l1 ⎢ [ K ]= E ⎢ 0 ⎢ ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢⎣

A1 l1 A1 A2 + l1 l 2 A − 2 l2 −

0

0

A2 l2 A2 A3 + l2 l3 A − 3 l3

0

−

0 0

⎤ 0 ⎥ ⎥ 0 ⎥ ⎥ ⎥ 0 ⎥. ⎥ A4 ⎥ − ⎥ l4 ⎥ A4 ⎥ l 4 ⎥⎦

A3 l3 A3 A4 + l3 l4 A − 4 l4 −

0

The global load vector is assembled as

{F }= ⎣ − R1

F2

0 0 F5 ⎦ T .

The final finite element equations are obtained, as shown in section 3.5, using the boundary condition, Q 1 = 0 . 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 R 1 . Stresses are then calculated from the axial forces obtained using equation (3.41).

4.1.8 Initial strain effects Let an initial strain ε 0 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 ε 0 is of the form

σ = E ( ε − ε0 ) .

(4.38)

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

∫σε

Ve

0 dV

=

∫σ

Ve

T

ε 0 dV = ε 0 Ee Ae

∫ε

le

Substituting (4.13), the expression (4.39) becomes

T

dx .

(4.39)

60


{ }ε T

W = qe

0 Ee Ae

∫ ⎣B⎦

T

dx .

(4.40)

le

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

{ f }= ε e

0 Ee

Ae

∫

⎣B ⎦

le

T

l dx = ε 0 Ee Ae e 2

+1

∫ ⎣B⎦

T

dr

(4.41)

−1

or

{ f }= ε e

+1 0 Ee Ae

∫

−1

hence

{ f }= ε e

T

⎢dN ⎥ le ⎢ ⎥ d r = ε 0 Ee Ae 2 ⎣ dr ⎦

0 Ee

+1

∫

−1

⎧− 1⎫ ⎨ ⎬ dr , ⎩1⎭

⎧ − 1⎫ Ae ⎨ ⎬ . ⎩ 1 ⎭

(4.42)

After solving for nodal displacements, stresses are computed as

σ e = Ee

q2 − q1 + ( − Ee ε 0 ) . le

(4.43)

In the case of thermal loading,

ε0 = αT ,

(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 l , from a material with shear modulus of elasticity G, acted upon by a torque M t will twist an angle θ =

πd4 Mt l , where I p = is the polar second GIp 32

moment of area of the shaft cross section. The shaft torsional stiffness is then GIp M K= t = . θ l A two-node shaft finite element, of length l and torsional rigidity G I p , is shown in Fig. 4.8. The nodal torques M 1 and M 2 can be related to the nodal rotation angles θ 1 and θ 2 using the equilibrium and the torque/rotation equations

61

4. BARS AND SHAFTS

M1 = − M 2 = K θ 1

when

θ 2 = 0,

M 1 = − M 2 = − K θ 2 when

θ 1 = 0.

(4.45)

Equations (4.45) may be written in matrix form as ⎧ M1 ⎫ ⎡ K ⎨ ⎬=⎢ ⎩ M 2 ⎭ ⎣− K or in shorthand form

− K⎤ ⎧ θ1 ⎫ ⎨ ⎬ K ⎥⎦ ⎩ θ 2 ⎭

(4.46)

{ M }= [ k ] { θ }, e

e

e

where

[ k ] = GlI e

p

⎡ 1 − 1⎤ ⎢− 1 1⎥ ⎣ ⎦

(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 Mt dθ , (4.48) = dx GI p while for the axial displacement is (4.3) du N = . (4.49) dx E A The rotation angle of an arbitrary section within the shaft element can be expressed in terms of the nodal rotations θ 1 and θ 2 as

θ = N1 (r ) θ1 + N 2 (r ) θ 2 .

(4.50)

62


where N1 (r ) =

1 ( 1 − r ) and N 2 (r ) = 1 ( 1 + r ) 2 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

[ k ] = G I ∫ ⎢⎢ ddNx ⎥⎥ ⎣ ⎦ e

e pe

le

T

⎢dN ⎥ ⎢ dx ⎥ dx ⎣ ⎦

(4.52)

or

[ ke ] = which yields

2 Ge I p e le

[ k ] = G lI e

e

e

+1

∫

−1

pe

T

⎢dN ⎥ ⎢dN ⎥ ⎢ dr ⎥ ⎢ dr ⎥ dr ⎣ ⎦ ⎣ ⎦

⎡ 1 − 1⎤ ⎢− 1 1⎥ ⎣ ⎦

(4.53)

(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 I p is replaced by the torsional constant I t .

Example 4.1 Consider the bar in Fig. E4.1 with d = 5 mm , l = 0.2 m , E = 200 GPa , loaded by an axial load F = 2 kN . Determine: a) the displacement of point 2, b) the stresses in bar, and c) the support reactions.

Fig. E4.1 Solution. a) The bar is divided into three bar finite elements.

63

4. BARS AND SHAFTS

The areas of the cross sections are A1 =

π d2 4

= 19.62 mm 2 , A2 = A3 =

π ( 1,4 d

)2

4

= 38.47 mm 2 .

The element stiffness matrices are

[ k ] = 2 ⋅10400⋅19.62 ⎡⎢− 11 1

5

⎣

[ k ] = [ k ] = 2 ⋅10200⋅ 38.47 ⎡⎢− 11 2

3

5

⎣

− 1⎤ = 9.81 ⋅ 103 1⎥⎦

⎡ 1 − 1⎤ N , ⎢− 1 1⎥⎦ mm ⎣

− 1⎤ = 38.47 ⋅ 103 1⎥⎦

⎡ 1 − 1⎤ N . ⎢− 1 1⎥⎦ mm ⎣

The global unreduced stiffness matrix is

− 9.81 0 0 ⎤ ⎡ 9.81 ⎢− 9.81 9.81 + 38.47 − 38.47 0 ⎥⎥ 3 ⎢ [ K ] = 10 ⎢ 0 . − 38.47 38.47 + 38.47 − 38.47 ⎥ ⎢ ⎥ 0 − 38.47 38.47 ⎦ ⎣ 0 Including the boundary conditions (degrees of freedom 1 and 4 are fixed), the finite element equations can be written

− 9.81 0 0 ⎤⎧ 0 ⎡ 9.81 ⎢− 9.81 48.28 − 38.47 0 ⎥⎥ ⎪⎪ Q2 103 ⎢ ⎨ ⎢ 0 − 38.47 76.94 − 38.47 ⎥ ⎪ Q3 ⎥ ⎢ − 38.47 38.47 ⎦ ⎪⎩ 0 0 ⎣ 0

⎫ ⎧ − R1 ⎫ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎬=⎨ 3 ⎬. ⋅ 2 10 ⎪ ⎪ ⎪ ⎪⎭ ⎪⎩ − R4 ⎪⎭

Retaining only the second and third equation yields ⎡ 48.28 − 38.47 ⎤ ⎧ Q2 ⎫ ⎧ 0 ⎫ 103 ⎢ ⎬=⎨ ⎥⎨ 3 ⎬ ⎣− 38.47 76.94 ⎦ ⎩ Q3 ⎭ ⎩ 2 ⋅ 10 ⎭ with solutions

Q2 = 34.42 mm ,

Q3 = 43.21 mm .

b) The reactions are obtained from the first and fourth equation R1 = 9.81 Q2 = 9.81 ⋅ 103 ⋅ 34.42 = 337.8 N , R4 = 38.47 Q3 = 38.47 ⋅103 ⋅ 43.21 = 1662 .2 N .

c) Stresses are

64


σ 12 = E ε12 = E

Q2 − Q1 Q 34.42 N = E 2 = 2 ⋅ 105 = 17.2 , l1 2l 400 mm 2

σ 23 = E ε 23 = E

Q3 − Q2 Q − Q2 8.79 N =E 3 = 2 ⋅105 = 8.79 , 200 l2 l mm 2

σ 34 = E ε 34 = E

Q4 − Q3 Q 43.21 N = − E 3 = −2 ⋅ 105 = − 43.2 . l3 l 200 mm 2

Example 4.2 A support, consisting of a steel bar 1 fitted inside a cast iron tube 2, is loaded by a 60 kN axial force, as in Fig. E4.2. Consider A1 = 200 mm 2 , A2 = 800 mm 2 , E1 = 210 GPa , E2 = 120 GPa , l = 0.2 m . 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

[ k ] = 2.1⋅10200⋅ 200 ⎡⎢− 11 5

1

⎣

[ k ] = 1.2 ⋅10200⋅ 800 ⎡⎢− 11 2

5

⎣

− 1⎤ = 2.1 ⋅ 105 1⎥⎦

⎡ 1 − 1⎤ N , ⎢− 1 1⎥⎦ mm ⎣

− 1⎤ = 4.8 ⋅ 105 1⎥⎦

⎡ 1 − 1⎤ N . ⎢− 1 1⎥⎦ mm ⎣

The global unreduced stiffness matrix is

[ K ] = 105 ⎡⎢− 2.2.11 +− 44..88 ⎣

− 2.1 − 4.8⎤ = 6.9 ⋅ 105 ⎥ 2.1 + 4.8⎦

⎡ 1 − 1⎤ N . ⎢− 1 1⎥⎦ mm ⎣

65

4. BARS AND SHAFTS

If point 1 is fixed, the finite element equations are ⎡ 1 − 1⎤ ⎧ 0 ⎫ ⎧ R1 ⎫ 6.9 ⋅105 ⎢ ⎬=⎨ ⎥⎨ 4 ⎬, ⎣− 1 1 ⎦ ⎩ Q2 ⎭ ⎩ − 6 ⋅10 ⎭

Q2 = − 0.087 mm , which is equal to the beam shortening. b) Stresses are

σ 1 = E1 ε 1 = E1

Q2 − Q1 Q − 0.087 N = E1 2 = 2.1 ⋅105 = −91.3 , 200 l l mm 2

σ 2 = E2 ε 2 = E2

Q2 − Q1 Q − 0.087 N = E2 2 = 1.2 ⋅ 105 = −52.14 . l l 200 mm 2

c) Internal forces are N1 = σ 1 A1 = − 91.3 ⋅ 200 = − 18.26 kN , N 2 = σ 2 A2 = − 52.14 ⋅ 800 = − 41.74 kN .

Example 4.3 The temperature of the bar in Fig. E4.3 is raised 800 C . If E = 130 GPa and α = 17 ⋅10−6 , determine the thermal stresses.

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

Δl = lα T = 1800 ⋅ 17 ⋅ 10 −6 ⋅ 80 = 2.448 mm > 2 mm so that the contact does occur. Denoting by A the cross section area, the temperature forces (4.42) are

66


− 1⎫ ⎧ − 1⎫ ⎧ − 1⎫ −6 5 ⎬ = 17 ⋅ 10 ⋅ 80 ⋅ 1.3 ⋅ 10 ⋅ A ⎨ ⎬ = 176.8 ⋅ A⎨ ⎬ N . ⎩ 1⎭ ⎩ 1 ⎭ ⎩ 1⎭

{ fT } = α T E A ⎧⎨

The stiffness matrix is ⋅ A ⎡ 1 − 1⎤ N . 1⎥⎦ mm 1800 ⎢⎣ − 1

[ k ] = 1.3 ⋅10

5

As point 1 is fixed, the finite element equations are 1.3 ⋅ 105 ⋅ A ⎡ 1 − 1⎤ ⎧ 0 ⎫ ⎧ R 1 − 176.8 A ⎫ ⎢− 1 1 ⎥ ⎨ 2 ⎬ = ⎨ R + 176.8 A ⎬ 1800 ⎣ ⎦⎩ ⎭ ⎩ 2 ⎭

wherefrom we get the reactions

R 1 = 176.8 A − 144.4 A = 32.4 A ,

R 2 = − 32.4 A ,

so that the normal stress is

σ=

R2 N . = − 32.4 A mm 2

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 T 0C . Determine the axial stress in the bar.

Fig. E4.4 Solution. The bar is divided into three linear finite elements. The element stiffness matrices are

[ k ] = [ k ] = El A ⎡⎢− 11 1

3

2

⎣

− 1⎤ , 1⎥⎦

[ k ] = E2lA ⎡⎢− 11 2

⎣

For element 2, the vector of nodal thermal forces is

{ f }= α T E A ⎣ − 1 2

1

1

1⎦T .

− 1⎤ . 1⎥⎦

67

4. BARS AND SHAFTS

Using the boundary conditions Q1 = Q4 = 0 , the finite element equations can be written

⎡ E2 ⎢ A ⎢ − E2 ⎢ l ⎢ 0 ⎢ ⎣⎢ 0

− E2 E E2 + 1 2 E1 − 2 0

0 E − 1 2 E1 + E2 2 − E2

0 ⎤ ⎥⎧ 0 0 ⎥ ⎪Q ⎪ 2 ⎥⎨ − E2 ⎥ ⎪ Q3 ⎥⎪ 0 E2 ⎦⎥ ⎩

R1 ⎫ ⎫ ⎧ ⎪ ⎪ −α T E A ⎪ ⎪ ⎪ 1 ⎪ ⎬. ⎬=⎨ ⎪ ⎪ α T E1 A ⎪ ⎪⎭ ⎪⎩ ⎪⎭ R4

The second and third equation yield

A l

E1 ⎡ ⎢ E2 + 2 ⎢ E ⎢ − 1 2 ⎣

E1 ⎤ 2 ⎥ ⎧ Q2 ⎫ = α T E A ⎧ − 1 ⎫ ⎬ ⎬ 1 ⎨ E ⎥⎨ ⎩ 1 ⎭ E2 + 1 ⎥ ⎩ Q3 ⎭ 2⎦ −

with solutions Q2 = − α T l

E1 , E1 + E2

Q3 = α T l

E1 . E1 + E2

The strains in elements are

ε1 = ε 3 =

Q2 Q E1 , = − 4 = −α T l l E1 + E2

ε2 =

Q3 − Q2 E1 . =αT 2l E1 + E2

Stresses are

σ 1 = σ 3 = E2 ε 1 = −α T

1 , 1 1 + E1 E2

σ 2 = E1 ε 2 − E1α T = −α T

1 = σ1. 1 1 + E1 E2

Example 4.5 A steel bolt of active length l = 100 mm and diameter δ = 10 mm is single threaded with a 1.6 mm pitch. It is mounted inside a copper tube with diameters d = 12 mm and D = 18 mm (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 E1 = 208 GPa and for copper E2 = 100 GPa . 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 Q1 = Q4 = 0 . The problem has a multipoint constraint

68


Q3 − Q2 = 0.4 mm . 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 A1 =

π δ2 4

= 78.54 mm 2 , A2 =

(

π D2 − d 2 4

) = 141.37 mm

2

.

The element stiffness matrices are

[ k ] = 2.08 ⋅10100⋅ 78.54 ⎡⎢− 11 5

1

⎣

[ k ] = 10 2

5

− 1⎤ ⎡ 1 − 1⎤ N , = 1.63 ⋅ 105 ⎢ ⎥ 1⎥⎦ mm 1⎦ ⎣− 1

⎡ 1 − 1⎤ N ⋅ 141.37 ⎡ 1 − 1⎤ . = 1.41 ⋅ 105 ⎢ ⎥ ⎢ 1⎥⎦ mm 100 ⎣− 1 ⎣ − 1 1⎦

The finite element equations are 0 0 ⎤ ⎧ 0 ⎫ ⎧ R1 ⎡ 1.63 − 1.63 ⎢− 1.63 1.63 0 0 ⎥⎥ ⎪⎪ Q2 ⎪⎪ ⎪⎪ − F 105 ⎢ ⎨ ⎬=⎨ ⎢ 0 0 1.41 − 1.41⎥ ⎪ − Q3 ⎪ ⎪ − F ⎢ ⎥ − 1.41 1.41 ⎦ ⎪⎩ 0 ⎪⎭ ⎪⎩ R4 0 ⎣ 0 Retaining only the second and third equation yields ⎡1.63 0 ⎤ ⎧ Q2 ⎫ ⎧ − F ⎫ 105 ⎢ ⎬=⎨ ⎬ ⎥⎨ ⎣ 0 1.41⎦ ⎩ − Q3 ⎭ ⎩ − F ⎭

with solutions

⎫ ⎪ ⎪ ⎬. ⎪ ⎪⎭

69

4. BARS AND SHAFTS

Q2 = −

F 1.63 ⋅ 10

5

,

Q3 =

F 1.41 ⋅ 105

.

Substituting in the multipoint constraint condition gives 1 1 ⎛ ⎞ F⎜ + = 0.4 , 5 5⎟ ⎝ 1.63 ⋅ 10 1.41 ⋅ 10 ⎠ or

F = 30314 N . Stresses are given by

σ1 =

F 30316 N = = 386 , A1 78.54 mm 2

σ2 = −

F 30316 N . =− = −214.5 A2 141.37 mm 2

Example 4.6 For the bar of Fig. E4.6, with linearly variable cross-section x ⎞ ⎛ A ( x ) = A0 ⎜1 + ⎟ , find the displacement at the free end under the action of force ⎝ 2l ⎠ F, using two tapered bar finite elements.

Fig. E4.6 Solution. The stiffness matrix for a bar element with variable cross section is

[ k ] = lE e

2 e

⎡ 1 − 1⎤ ⎢− 1 1⎥⎦ ⎣

∫

A dx .

le

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

70


[k ] 1

l

∫

E ⎡ 1 − 1⎤ = 2⎢ ⎥ A0 l ⎣− 1 1⎦

[k ] 2

0

E ⎡ 1 − 1⎤ A0 = 2⎢ 1⎥⎦ l ⎣− 1

2l

∫ l

5 E A0 ⎡ 1 − 1⎤ x ⎞ ⎛ , ⎜ 1 + ⎟ dx = 4l ⎢⎣− 1 1⎥⎦ ⎝ 2l ⎠ 7 E A0 x ⎞ ⎛ ⎜ 1 + ⎟ dx = 4l ⎝ 2l ⎠

⎡ 1 − 1⎤ . ⎢− 1 1⎥⎦ ⎣

With Q1 = 0 , the finite element equations can be written ⎡ 5 − 5 0 ⎤ ⎧ 0 ⎫ ⎧ R1 ⎫ EA0 ⎢ ⎪ ⎪ ⎪ ⎪ − 5 12 − 7 ⎥⎥ ⎨ Q2 ⎬ = ⎨ 0 ⎬ . ⎢ 4l ⎢⎣ 0 − 7 7 ⎥⎦ ⎪⎩ Q3 ⎪⎭ ⎪⎩ F ⎪⎭ From the second and third equation EA0 4l

⎡ 12 − 7 ⎤ ⎧ Q2 ⎫ ⎧ 0 ⎫ ⎢− 7 7 ⎥ ⎨ Q ⎬ = ⎨ F ⎬ , ⎣ ⎦⎩ 3⎭ ⎩ ⎭

Q2 =

4 Fl , 5 E A0

wherefrom Q3 =

48 F l . 35 E A0

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 Q3 from condition F3 = 0 . Model the bar by: a) two two-node linear elements; b) a three-node quadratic element. Comment the results.

Fig. E4.7 Solution. a) Consider the bar divided into two linear elements. The element stiffness matrices are

71

4. BARS AND SHAFTS

⎡ 1 [ k ] = [ k ] = 2EA l ⎢− 1 1

2

⎣

− 1⎤ . 1⎥⎦

The unreduced global stiffness matrix is ⎡ 1 −1 0 ⎤ 2 EA ⎢ [ K ] = l ⎢ − 1 2 − 1 ⎥⎥ . ⎢⎣ 0 − 1 1 ⎥⎦ The finite element equations can be written ⎡ 1 0 − 1 ⎤ ⎧ Q1 ⎫ ⎧ F1 2 EA ⎢ ⎪ ⎪ ⎪ 0 1 − 1 ⎥⎥ ⎨ Q2 ⎬ = ⎨ F2 ⎢ l ⎢⎣ − 1 − 1 2 ⎥⎦ ⎪⎩ Q3 ⎪⎭ ⎪⎩ F3

⎫ ⎪ ⎬. ⎪ ⎭

For F3 = 0 , the last equation yields Q3 =

⎧Q ⎫ Q1 + Q2 =⎣ 1 2 1 2 ⎦⎨ 1 ⎬ 2 ⎩ Q2 ⎭

which assumes a linear displacement field within the bar. The first two equations give 2 EA l

⎧ F1 ⎫ ⎡ 1 0 ⎤ ⎧ Q1 ⎫ 2 EA ⎧ − 1 ⎫ ⎢ 0 1 ⎥ ⎨ Q ⎬ + l ⎨ − 1 ⎬ Q3 = ⎨ F ⎬ ⎣ ⎦⎩ 2⎭ ⎩ ⎭ ⎩ 2⎭

and upon substitution of Q3 2 EA l

⎧ Q1 ⎫ ⎧ F1 ⎫ ⎡ 1 0 ⎤ ⎧ Q1 ⎫ 2 EA ⎧ − 1 ⎫ ⎢ 0 1 ⎥ ⎨ Q ⎬ + l ⎨ − 1 ⎬ ⎣ 1 2 1 2 ⎦ ⎨ Q ⎬ = ⎨F ⎬ ⎣ ⎦⎩ 2⎭ ⎩ ⎭ ⎩ 2 ⎭ ⎩ 2⎭

which can be written EA l

⎡ 1 − 1 ⎤ ⎧ Q1 ⎫ ⎧ F1 ⎫ ⎢ − 1 1 ⎥ ⎨ Q ⎬ = ⎨F ⎬ ⎣ ⎦ ⎩ 2 ⎭ ⎩ 2⎭

where the left hand side contains the (2 × 2) 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

72


1 − 8 ⎤ ⎧ Q1 ⎫ ⎧ F1 ⎡ 7 EA ⎢ ⎪ ⎪ ⎪ 1 7 − 8 ⎥⎥ ⎨ Q2 ⎬ = ⎨ F2 ⎢ 3l ⎢⎣ − 8 − 8 16 ⎥⎦ ⎪⎩ Q3 ⎪⎭ ⎪⎩ F3

⎫ ⎪ ⎬. ⎪ ⎭

Condensing Q3 from condition F3 = 0 yields Q3 =

⎧Q ⎫ Q1 + Q2 =⎣ 1 2 1 2 ⎦⎨ 1 ⎬. 2 ⎩ Q2 ⎭

The first two equations give ⎧ F1 ⎫ EA ⎡ 7 1 ⎤ ⎧ Q1 ⎫ EA ⎧ − 8 ⎫ ⎨ ⎨ ⎬+ ⎬ Q3 = ⎨ ⎬ ⎢ ⎥ 3l ⎣ 1 7 ⎦ ⎩ Q2 ⎭ 3l ⎩ − 8 ⎭ ⎩ F2 ⎭ and upon substitution of Q3 ⎧ Q1 ⎫ ⎧ F1 ⎫ EA ⎡ 7 1 ⎤ ⎧ Q1 ⎫ EA ⎧ − 8 ⎫ ⎨ ⎨ ⎬+ ⎬⎣1 2 1 2 ⎦⎨ ⎬=⎨ ⎬ ⎢ ⎥ 3l ⎣ 1 7 ⎦ ⎩ Q2 ⎭ 3l ⎩ − 8 ⎭ ⎩ Q2 ⎭ ⎩ F2 ⎭ which can be written again EA ⎡ 1 − 1 ⎤ ⎧ Q1 ⎫ ⎧ F1 ⎫ ⎨ ⎬ = ⎨ ⎬, l ⎢⎣ − 1 1 ⎥⎦ ⎩ Q2 ⎭ ⎩ F2 ⎭ where the matrix in the left hand can be recognized as the conventional bar matrix developed from linear polynomials. c) Comments.

Substituting ⎧ Q1 ⎪ ⎨ Q2 ⎪Q ⎩ 3

0 ⎤ ⎫ ⎡ 1 ⎧Q ⎫ ⎪ ⎢ 1 ⎥⎥ ⎨ 1 ⎬ ⎬=⎢ 0 ⎪ ⎢ 1 2 1 2 ⎥ ⎩ Q2 ⎭ ⎭ ⎣ ⎦

into the expression of the strain energy (4.18) gives 1 U e = ⎣Q1 Q2 2

1 U e = ⎣ Q1 Q2 2

⎦

1 − 8 ⎤ ⎧ Q1 ⎫ ⎡ 7 EA ⎢ ⎪ ⎪ 1 7 − 8 ⎥⎥ ⎨ Q2 ⎬ , Q3 ⎦ 3l ⎢ ⎢⎣ − 8 − 8 16 ⎥⎦ ⎪⎩ Q3 ⎪⎭

⎡1 0 1 ⎢0 1 1 ⎣

0 ⎤ 1 −8 ⎤ ⎡ 1 ⎡ 7 2⎤ EA ⎢ ⎥ ⎢ ⎥ ⎧ Q1 − 0 1 1 7 8 ⎥ ⎥ ⎢ ⎥ ⎨Q 2⎦ 3l ⎢ ⎢⎣ − 8 − 8 16 ⎥⎦ ⎢⎣ 1 2 1 2 ⎥⎦ ⎩ 2

⎫ ⎬, ⎭

73

4. BARS AND SHAFTS

Ue =

1 ⎣ Q1 Q2 2

EA ⎡ 1

− 1⎤ ⎧ Q1 ⎫

⎦ l ⎢− 1 1 ⎥ ⎨ Q ⎬ . ⎣ ⎦ ⎩ 2⎭

For p = const . , the work of nodal forces is W = ⎣Q1 Q2 e

W = ⎣Q1 Q2 e

Q3

⎦

⎦

⎧ F1 ⎪ ⎨ F2 ⎪F ⎩ 3

⎫ ⎪ ⎬ = ⎣Q1 Q2 ⎪ ⎭

Q3

⎦

⎧1 6 ⎫ ⎪ ⎪ ⎨ 1 6 ⎬ pl e , ⎪2 3⎪ ⎭ ⎩

⎧1 6 ⎫ ⎡1 0 1 2 ⎤ ⎪ ⎪ ⎢0 1 1 2⎥ ⎨ 1 6 ⎬ pl e = ⎣Q1 Q2 ⎣ ⎦ ⎪2 3⎪ ⎭ ⎩

⎧1⎫ pl

⎦ ⎨1⎬ 2 e . ⎩⎭

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 p = 0 . For the case p = const. , exact displacements within the elements may be obtained from equation (4.25). However, before using equation (4.25), the variable Q3 must be computed from the exact nodal displacements (computed for the conventional linear element), via a constraint equation between Q3 and the remaining nodal variables, obtained by a minimization of the total potential energy with respect to Q3 at element level.

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

ω = 30 rad sec (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

74


p ( x ) = p0

where p0 = ρ Al ω 2 .

x , l

Fig. E4.8

75

4. BARS AND SHAFTS

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

[k ]= [k ] 1

2

⎡ 7 −8 1 ⎤ 2 EA ⎢ = − 8 16 − 8 ⎥⎥ . ⎢ 3l ⎢⎣ 1 − 8 7 ⎥⎦

The unreduced global stiffness matrix is

1 0 0 ⎤ ⎡ 7 −8 ⎢ − 8 16 − 8 0 0 ⎥⎥ ⎢ ⎢ 1 −8 7 + 7 −8 1 ⎥ . [ K ] = 23EA l ⎢ ⎥ − 8 16 − 8 ⎥ 0 ⎢ 0 ⎢⎣ 0 − 8 7 ⎥⎦ 0 1 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)

{ f }= ∫ ⎣N ⎦ e

1

T

p dx = p0 l e

∫ 0

le

⎧(1 − 2ξ )(1 − ξ )⎫ ⎧0⎫ ⎪ ⎪ ⎪ ⎪ ⎨ − ξ (1 − 2ξ ) ⎬ ξ dξ = p0 l e ⎨1 6⎬ , ⎪ 4 ξ (1 −ξ ) ⎪ ⎪1 3⎪ ⎩ ⎭ ⎩ ⎭

where ξ = x l e and the shape functions (4.26) are defined for convenience over an

interval [0, 1] . For a uniformly distributed load they are given by equation (4.37).

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

{ f }= p12l ⎣ 0 1

0

1 1 2 ⎦T ,

{ f }= p12l ⎣1 2 2

0

3 1⎦T .

Using the boundary condition Q1 = 0 , the finite element equations can be written R1 0 0 ⎤⎧ 0 ⎫ ⎧ ⎫ ⎡ 7 −8 1 ⎪ ⎪Q ⎪ ⎪ ⎢ − 8 16 − 8 0 ⎥ p0l 12 0 ⎥⎪ 2 ⎪ ⎪ ⎪⎪ ⎢ 2 EA ⎪ ⎪ ⎪ ⎢ 1 − 8 14 − 8 1 ⎥ ⎨ Q3 ⎬ = ⎨ p0l 24 + p0l 24 ⎬ . 3l ⎢ ⎥ ⎪ 3 p0l 12 0 − 8 16 − 8 ⎥ ⎪ Q4 ⎪ ⎪ ⎢ 0 ⎪ ⎪ ⎪ ⎪ ⎢⎣ 0 ⎪⎭ p0l 12 0 1 − 8 7 ⎥⎦ ⎪⎩ Q5 ⎪⎭ ⎪⎩

76


Deleting the first row and column, we obtain the reduced global stiffness matrix and reduced global load vector which yield Q2 =

47 p0 l 2 88 p0 l 2 117 p0 l 2 128 p0 l 2 , Q3 = , Q4 = , Q5 = . 384 E A 384 E A 384 E A 384 E A

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

u( x ) =

p0 l 2 2E A

⎛ x x3 ⎞ ⎜ − ⎟. ⎜ l 3l 3 ⎟ ⎝ ⎠

(a)

We obtain 2 2 2 1 p0 l 2 ⎛ l ⎞ 47 p0 l ⎛ l ⎞ 11 p0 l ⎛ 3l ⎞ 117 p0 l u⎜ ⎟ = , u⎜ ⎟ = , u⎜ ⎟ = , u (l ) = . 3 2E A ⎝ 4 ⎠ 384 2 E A ⎝ 2 ⎠ 48 2 E A ⎝ 4 ⎠ 384 2 E A

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. The element strain-displacement row vector ⎣B ⎦ in (4.29) is given by d

1

⎣B ⎦ = d x ⎣N ⎦ = l ⎣− 3 + 4ξ − 1 + 4ξ 4 − 8 ξ ⎦ . The strains are calculated as

ε1 =

2 (N 2′ Q3 + N 3′ Q2 ) = p0l ( 25 − 6ξ ) , 48 EA l

ε2 =

2 (N1′ Q3 + N 2′ Q5 + N 3′ Q4 ) = p0l ( 19 − 18ξ ) , l 48 EA

which yields the following nodal values

ε1 =

25 p0l 22 p0 l 19 p0 l 10 p0 l 1 p0 l , ε2 = , ε3 = , ε4 = , ε5 = . 48 EA 48 EA 48 EA 48 EA 48 EA

The corresponding axial stresses are

σ1 =

25 p0l 22 p0 l 19 p0 l 10 p0 l 1 p0 l , σ2 = , σ3 = , σ4 = , σ5 = . 48 A 48 A 48 A 48 A 48 A

The stresses can be compared with the exact solution

77

4. BARS AND SHAFTS

σ ( x )=

p0 l 2A

2⎞ ⎛ ⎜ 1− x ⎟ , ⎜ l 2 ⎟⎠ ⎝

(b)

which yields the following nodal values

σ (0) = ⎛l⎞ ⎝2⎠

σ⎜ ⎟=

24 p0 l ⎛ l ⎞ 22.5 p0 l , σ⎜ ⎟= , 48 A ⎝ 4 ⎠ 48 A

18 p0 l ⎛ 3l ⎞ 10.5 p0 l , σ⎜ ⎟= , σ (l ) = 0 . 48 A ⎝ 4 ⎠ 48 A

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. The element stiffness matrices are ⎡ 1 [ k ] = [ k ] = [ k ] = 3EA l ⎢− 1 1

2

3

⎣

− 1⎤ . 1⎥⎦

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 p1 and p2 are the intensities of a linearly distributed load at nodes 1 and 2, respectively,

p ( ξ ) = p1 + ( p2 − p1 )ξ ,

{ f }= ∫ ⎣ N ⎦ e

1

T

p dx = l e

∫ 0

le

⎧1 − ξ ⎫ l e ⎧ 2 p1 + p2 ⎨ ⎬ [ p1 + ( p2 − p1 ) ξ ] dξ = ⎨ 6 ⎩ p1 + 2 p2 ⎩ ξ ⎭

⎫ ⎬, ⎭

where ξ = x l e and the shape functions (4.26) are defined for convenience over an interval [ 0, 1 ] . For p1 = p2 = p , the nodal forces are given by equation (4.36).

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

{ f }= p54l ⎣1 1

0

2⎦ T ,

{ f }= p54l ⎣4 2

0

5⎦ T ,

{ f }= p54l ⎣7 3

0

8⎦ T .

Using the boundary condition Q 1 = 0 , the finite element equations can be written

78


⎡ 1 −1 0 0 ⎤ ⎧ 0 ⎢ ⎥⎪ 3EA ⎢ − 1 2 − 1 0 ⎥ ⎪ Q2 ⎨ l ⎢ 0 − 1 2 − 1 ⎥ ⎪ Q3 ⎢ ⎥ ⎣ 0 0 − 1 1 ⎦ ⎪⎩ Q4

54 R1 ⎧ ⎫ ⎪1+ p l ⎪ 0 ⎪ p0 l ⎪⎪ 6 = ⎬ ⎨ ⎪ 54 ⎪ 12 ⎪⎭ ⎪ ⎪⎩ 8

⎫ ⎪ ⎪⎪ ⎬. ⎪ ⎪ ⎪⎭

The last three equations yield Q2 =

13 p0 l 2 23 p0 l 2 27 p0 l 2 1 p0 l 2 = , Q3 = , Q4 = . 81 E A 81 E A 81 E A 3 E A

The element strain-displacement row vector ⎣B ⎦ in (4.29) is given by d

1

⎣B ⎦ = d x ⎣N ⎦ = l ⎣− 1 1⎦ . e

The strains are calculated as 3 l

ε 1 = Q2 =

13 p0 l 3 4 p0 l 3 10 p0 l , ε 2 = ( Q3 − Q2 ) = , ε 3 = ( Q4 − Q3 ) = , l 27 EA 27 EA l 27 EA

and are constant within each element. The corresponding axial stresses are

σ1 =

13 p0l , 27 A

σ2 =

10 p0 l , 27 A

σ3 =

4 p0 l . 27 A

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 9 ( 1 − r ) ⎛⎜ 1 + r ⎞⎟ ⎛⎜ 1 − r ⎞⎟ , 16 ⎝3 ⎠⎝3 ⎠ 27 ( 1 + r ) ( 1 − r ) ⎛⎜ 1 − r ⎞⎟ , N 2 (r ) = 16 ⎝3 ⎠ N1 (r ) = −

27 ( 1 + r ) ( 1 − r ) ⎛⎜ 1 + r ⎞⎟ , 16 ⎝3 ⎠ 9 ⎛1 ⎞⎛1 ⎞ N 4 (r ) = − ⎜ + r ⎟ ⎜ − r ⎟ (1 + r ) . 16 ⎝ 3 ⎠⎝3 ⎠ N 3 (r ) =

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 C1 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 C 0 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 Q3 i − 2 , Q3 i −1 and Q3 i , defined as the displacement along the X axis, the displacement along the Y axis and the rotation about the Z axis, respectively.


80

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 E I 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


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

{ q }= ⎣ q e

1

q2

q3

q4

q5

q6 ⎦ T

(5.1)

and the corresponding vector of element nodal forces can be written

{ f }= ⎣ f e

Forces

f 2, f 3, f 5, f 6

1

f2

and

f3

f4

the

f6 ⎦ T .

(5.2)

corresponding

displacements

f5

q 2 , q 3 , q 5 , q 6 describe the element bending (Fig. 5.2, b) while axial forces f 1 , f 4 , and axial displacements q 1 , q 4 , 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 dv u = −ϕ y = − y, (5.3) dx


82

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

εx =

du d2 v = − 2 y = −χ y , dx dx

(5.4)

where χ ≈ v′′ denotes the deformed beam axis curvature. Normal stresses on the cross section are given by Hooke’s law d2 v y, dx2 where E is Young’s modulus of the material.

σ x = E ε x = −E

(5.5)

The bending moment is the resultant of the stress distribution on the cross section M (x ) = − σ x y dA = E I z

∫

A

d2 v dx2

= EIz χ

(5.6)

where I z 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 E I z is called the bending rigidity of the beam. The shear force is given by

T (x ) = −

dM d3 v = −E I z = − E I z v III . dx dx 3

(5.7)

The transverse load per unit length is

p (x ) = −

dT d4 v = EIz = E I z v IV . dx dx4

(5.8)

The differential equation of equilibrium is

EIz

d4 v = p (x ) . dx4

(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.


83

5.3 Uniform beam not loaded between ends For a uniform beam not loaded between ends, p = 0 and equation (5.9) yields d 4 v d x 4 = 0 . Integrating four times, we obtain the deflection v described by a third order polynomial

v (x ) = a 1 x 3 + a 2 x 2 + a 3 x + a 4 .

(5.10)

For a free-free beam element, the four integration constants a 1 , a 2 , a 3 , a 4 in (5.10) can be determined from the geometric boundary conditions at each end (Fig. 5.4): (5.11, a) at x = x1 , v = v1 = q2 , and d v d x = ϕ1 = q3 ; at x = x 2 , v = v2 = q5 , and d v d x = ϕ 2 = q6 .

(5.11, b)

Fig. 5.4 This gives

⎧ v1 ⎪ϕ ⎪ 1 ⎨ ⎪ v2 ⎪⎩ ϕ 2

⎫ ⎡ x13 ⎪ ⎢ 2 ⎪ ⎢ 3 x1 ⎬=⎢ 3 ⎪ ⎢ x2 ⎪⎭ ⎢ 3 x 2 ⎣ 2

x12 2 x1

x1 1

x 22 2 x2

x2 1

1 ⎤ ⎧ a1 ⎥ ⎪ 0 ⎥ ⎪ a2 ⋅⎨ 1 ⎥ ⎪ a3 ⎥ 0 ⎥⎦ ⎪⎩ a 4

⎫ ⎪ ⎪ ⎬. ⎪ ⎪⎭

On inversion, the integration constants a 1 , a 2 , a 3 , a 4 can be expressed in terms of the nodal displacements v1 , ϕ1 , v2 , ϕ 2 , 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


84

{ }

v (x ) = ⎣N ⎦ q e ,

(5.12)

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

{ q }= ⎣ v e

ϕ1 v2 ϕ 2 ⎦ T .

1

(5.13)

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

x + x2 ⎞ 2 ⎛ ⎟ ⎜⎜ x − 1 2 ⎟⎠ x 2 − x1 ⎝

(5.14)

the beam transverse displacement can be written ⎛dv⎞ ⎛dv⎞ ⎟⎟ + N 3 (r ) v 2 + N 4 (r ) ⎜⎜ ⎟⎟ . v (r ) = N1 (r ) v 1 + N 2 (r ) ⎜⎜ ⎝ d r ⎠1 ⎝ d r ⎠2

(5.15)

Because the coordinates transform by the relationship (5.14) x=

x1 + x 2 2

+

x 2 − x1 2

r

and since l e = x 2 − x 1 is the length of the element, dx =

le dr . 2

(5.16)

Using the differentiation chain rule

d v le d v = , dr 2 dx

(5.17)

equation (5.15) becomes v (r ) = N1 (r ) v 1 + N 2 (r )

le 2

⎛dv⎞ l ⎜⎜ ⎟⎟ + N 3 (r ) v 2 + N 4 (r ) e 2 ⎝ d x ⎠1

⎛dv⎞ ⎜⎜ ⎟⎟ ⎝ d x ⎠2

(5.18)

or v (r ) = N1 ⋅ q2 +

le l N 2 ⋅ q3 + N 3 ⋅ q5 + e N 4 ⋅ q6 . 2 2

(5.19)

In (5.12) the row vector of shape functions is ⎢

⎣N ⎦ = ⎢ N1 ⎣

le N2 2

N3

le ⎥ N4 ⎥ 2 ⎦

.

(5.20)


85

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

r = −1 r = +1

N1

N ′1

N2

N ′2

N3

N ′3

N4

N ′4

1 0

0 0

0 0

1 0

0 1

0 0

0 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

(

)

N 1(r ) =

1 ( 1 − r ) 2 ( 2 + r ) = 1 2 − 3r + r 3 , 4 4

N 2 (r ) =

1 ( 1 − r ) 2 ( 1 + r ) = 1 1 −r − r 2 + r 3 , 4 4

N 3(r ) =

1 ( 1 + r ) 2 ( 2 − r ) = 1 2 + 3r − r 3 , 4 4

(

N 4 (r ) = −

)

(

(5.21)

)

(

)

1 ( 1 + r ) 2 ( 1 − r ) = − 1 1 + r −r 2 − r 3 . 4 4

Fig. 5.5


86

It is easy to check that at node 1, v = q 2 and v = q 5 and

d v le = q 3 , while at node 2, dr 2

d v le = q6 . dr 2

5.3.2 Stiffness matrix The strain energy U e of a beam element is

E Ie Ue = 2

∫ e

2

⎛ d 2v ⎞ ⎜ ⎟ dx . ⎜ d x2 ⎟ ⎝ ⎠

(5.22)

Equation (5.17) yields dv 2 d v = d x le d r

d 2v 4 d 2 v = . d x 2 l 2e d r 2

and

Substituting (5.12) we obtain

d 2v dx

2

=

4 ⎢d 2 N ⎥ ⎥ ⎢ l 2e ⎣⎢ d r 2 ⎦⎥

{ q }. e

(5.23)

The square of the above quantity is calculated as T

2

⎛ 2 ⎞ ⎛ d 2v ⎞ ⎟ =⎜d v⎟ ⎜ ⎜ d x2 ⎟ ⎜ d x2 ⎟ ⎠ ⎝ ⎠ ⎝

{ }

⎛ d 2v ⎞ ⎟ = qe ⎜ ⎜ d x2 ⎟ ⎠ ⎝

T

16 l 4e

⎢d 2 N ⎥ ⎢ 2⎥ ⎢⎣ d r ⎥⎦

T

{ }

⎢d 2 N ⎥ e ⎢ 2 ⎥ q , ⎢⎣ d r ⎥⎦

which can also be written 2

{ }

⎛ d 2v ⎞ ⎟ = qe ⎜ ⎜ d x2 ⎟ ⎠ ⎝

T

{ }

16 N ′′ T N ′′ q e . 4 ⎣ r⎦ ⎣ r⎦ le

(5.24)

On substituting (5.16) and (5.24) into (5.22) we get the element strain energy 1 Ue = 2 which has the form

{q }

e T

8E I e l 3e

+1

∫

−1

T e ⎣N r′′ ⎦ ⎣N r′′ ⎦ d r { q }

(5.25)


Ue =

87

{ q } [ k ] { q }. e T

1 2

e B

e

(5.26)

Comparing (5.25) with (5.26) we obtain the element stiffness matrix due to bending +1

[ k ] = 8lEI ∫ e B

e

3 e

T ⎣N r′′ ⎦ ⎣N r′′ ⎦ d r

(5.27)

−1

or +1

[ k ] = 8lEI ∫ e B

e

3 e

−1

⎡ (N1′′)2 N1′′N 2′′ ⎢ 2 ⎢ N 2′′ N1′′ (N 2′′ ) ⎢ N ′′N ′′ N ′′N ′′ 3 2 ⎢ 3 1 ′ ′ ′ ′ N N N 4′′ N 2′′ ⎣⎢ 4 1

N1′′N 3′′ N 2′′ N 3′′ (N 3′′ )2 N 4′′ N 3′′

N1′′N 4′′ ⎤ ⎥ N 2′′ N 4′′ ⎥ dr . N 3′′N 4′′ ⎥ ⎥ (N 4′′ )2 ⎦⎥

(5.28)

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

[k ] e B

6l − 12 6l ⎤ ⎡ 12 ⎢ 6l 4l 2 − 6l 2l 2 ⎥ EI ⎥ . = 3e ⎢ l e ⎢− 12 − 6l 12 − 6l ⎥ ⎥ ⎢ 2 − 6l 4l 2 ⎦ e ⎣ 6l 2l

(5.29)

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

[ k ] = ∫ ⎣B⎦ e

T

Ee ⎣B ⎦ dV .

(5.30)

Ve

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

χ=

d 2v dx

2

=

4 ⎢d 2 N ⎥ ⎢ ⎥ l 2e ⎣⎢ d r 2 ⎦⎥

{ q }= ⎣B⎦ { q }, e

e

(5.31)

where the curvature-displacement row vector ⎣B ⎦ is 1 ⎢ r

r

⎥

⎣B ⎦ = l ⎢6 l 3r − 1 − 6 l 3r + 1⎥ . e ⎣ e e ⎦

(5.32)

Substituting (5.32) and d V = Ae d x into (5.30) gives the matrix (5.29).


88

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 ⎧ ⎪ ⎪ ⎨ ⎪ ⎪⎩

f2 ⎫ f 3 ⎪⎪ EI e ⎬ = 3 f5 ⎪ le f 6 ⎪⎭e

6l − 12 6l ⎤ ⎧ q2 ⎡ 12 ⎢ 6l 4l 2 − 6l 2l 2 ⎥ ⎪ q ⎢ ⎥ ⎪⎨ 3 ⎢− 12 − 6l 12 − 6l ⎥ ⎪ q5 ⎢ ⎥ 2 − 6l 4l 2 ⎦ e ⎪⎩ q6 ⎣ 6l 2 l

⎫ ⎪ ⎪ ⎬ . ⎪ ⎪⎭ e

(5.33)

Consider a beam element with end 2 fixed ( q5 = q6 = 0 ) and end 1 having a unit displacement along the global X-axis ( q2 = 1, q3 = 0) and zero rotation, as in Fig. 5.6.

Fig. 5.6 Equation (5.33) can be written

⎧ f 2 ⎫ ⎡ k11 ⎪ f ⎪ ⎢k ⎪ 3 ⎪ ⎢ 21 ⎨ ⎬= ⎪ f 5 ⎪ ⎢ k31 ⎪⎩ f 6 ⎪⎭ ⎢⎣k 41

. . . .

. . . .

.⎤ ⎧ q2 = 1⎫ .⎥⎥ ⎪⎪q3 = 0 ⎪⎪ ⋅⎨ ⎬ .⎥ ⎪q5 = 0 ⎪ ⎥ .⎦ ⎪⎩q6 = 0⎪⎭

(5.34)

which gives f 2 = k11 ,

f 3 = k 21 ,

f 5 = k31 ,

f 6 = k 41 .

(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 q2 = 1 and all other displacements are zero. For equilibrium

k11 + k31 = 0 ,

k 21 + k 41 + k31 l = 0 .

(5.36)


89

5.4 Uniform beam loaded between ends For a uniform beam loaded between ends, p ≠ 0 , d 4 v d x 4 ≠ 0 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, p = const . , 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, v (x ) 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 E I z 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 p (x ) , having the units of force per unit length, distributed along the beam element. The mechanical work of such a force is W=

∫ v p dx = ∫ v

le

T

p dx .

(5.37)

le

Substituting (5.12), equation (5.37) becomes

{ } ∫ ⎣N ⎦

W = qe

T

T

p dx .

(5.38)

le

It has the form

{ } {f }

W = qe where the element load vector is

T

e

(5.39)


90

{ f } = ∫ ⎣N ⎦ e

le

T

l p ( x ) dx = e 2

+1

∫ ⎣N ⎦

T

p (r ) d r .

(5.40)

−1

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

{f } e

l = e p 2

+1

∫ ⎣N ⎦

T

dr

(5.41)

−1


{f } e

⎢ pl =⎢ e ⎢⎣ 2

p l 2e 12

p le 2

T

p l 2e ⎥ − ⎥ . 12 ⎥⎦

(5.42)

In Fig. 5.7, a it is seen that f 2e is a shear force and f 3e is a moment. They are called “kinematically equivalent” nodal forces since they replace a distributed load p (r ) weighted with the shape functions N i (r ) 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 p = const . by statically equivalent forces (Fig. 5.6, b) would be incorrect since the beam element has the ends rigidly jointed,


91

i.e. it is rigidly built in the adjacent beam elements. In order to ensure the C1 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 p (x ) and the resistance E I z v IV 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


92

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 ⎧ pl ⎫ ⎪ 2 ⎪ ⎪ 6l − 12 6l ⎤ ⎧ q2 = 0 ⎫ ⎪ ⎡ 12 pl 2 ⎪ ⎢ 6l 4l 2 − 6l 2l 2 ⎥ ⎪ q ⎪ ⎪ EIz ⎢ ⎥ ⎪⎨ 3 ⎪⎬ = ⎪⎨ 12 ⎪⎬ . 3 ⎢ − 12 − 6l 12 − 6l ⎥ ⎪ q5 = 0 ⎪ ⎪ pl ⎪ l ⎥ ⎢ 2 − 6l 4l 2 ⎦ ⎪⎩ q6 ⎪⎭ ⎪ 2 2 ⎪ ⎣ 6l 2l ⎪ pl ⎪ ⎪− ⎪ ⎩ 12 ⎭ The set of equations in the unknown displacements is 2 EI ( 4q3 + 2q6 ) = pl , 12 l 2 EI ( 2q3 + 4q6 ) = − pl , 12 l

with solutions

q6 = − q3 = −

pl3 . 24 E I

The displacement at the middle is pl 4 ⎛l⎞ ⎛l⎞ ⎛l⎞ v ⎜ ⎟ = N 2 ⎜ ⎟ q3 + N 4 ⎜ ⎟ q6 = . 96 E I ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ The true solution is

v true =

5 pl 4 384 E I

so that the approximate finite element solution is, as expected, smaller ⎛l⎞ 4 v ⎜ ⎟ = vtrue . ⎝ 2⎠ 5

5.4.2 Higher-degree interpolation functions Let explore the possibility of approximating the displacements of a fourth order system by a quadratic function of the form


93

v (x ) = a 1 x 4 + a 2 x 3 + a 3 x 2 + a 4 x + a5 .

(5.43)

The five integration constants a 1 , a 2 , a 3 , a 4 , a5 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 x = x1 , v = v1 ,

and

d v d x = ϕ1 ,

at x = x 2 , v = v2 ,

and

d v d x = ϕ2 ,

(5.44)

at x = x3 , v = v3 . This gives the nodal coordinates in terms of the polynomial coefficients ⎧ v1 ⎪ϕ ⎪⎪ 1 ⎨ v2 ⎪ϕ ⎪ 2 ⎩⎪ v3

⎡1 ⎫ ⎢ ⎪ ⎢0 ⎪⎪ ⎢ 1 ⎬=⎢ ⎪ ⎢0 ⎪ ⎢ ⎭⎪ ⎢ 1 ⎣

0 1 l 1 l 2

0 0 l2 2l l2 4

0 0 l3 3l 2 l3 8

0 ⎤ ⎧ a5 ⎫ 0 ⎥⎥ ⎪ ⎪ ⎪ a4 ⎪ l4 ⎥ ⎪ ⎪ ⎥ ⋅ ⎨ a3 ⎬ . 4l 3 ⎥ ⎪ ⎪ a2 l4 ⎥ ⎪ ⎪ ⎥ ⎩⎪ a1 ⎭⎪ 16 ⎦

On inversion ⎡ 1 ⎢ ⎧ a5 ⎫ ⎢ 0 ⎪ a ⎪ ⎢ 11 ⎪⎪ 4 ⎪⎪ ⎢ − 2 ⎨ a3 ⎬ = ⎢ l ⎪ a ⎪ ⎢ 18 ⎪ 2 ⎪ ⎢ l3 ⎪⎩ a1 ⎪⎭ ⎢ 8 ⎢− ⎣⎢ l 4

−

0 1

0 0

0 0

4 l 5

5 − 2 l 14

1 l 3

l2 2

l3 8

− 3 l

− 4 l

− 2 l 2 l3

0 ⎤ 0 ⎥⎥ ⎧ v1 ⎫ 16 ⎥ ⎪ ϕ1 ⎪ ⎥ ⎪ ⎪ l 2 ⎥ ⋅ ⎪⎨ v 2 ⎪⎬ . 32 − 3 ⎥ ⎪ ϕ2 ⎪ l ⎥ ⎪ ⎪ 16 ⎥ ⎪⎩ v3 ⎪⎭ ⎥ l 4 ⎦⎥

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

{ }

v (x ) = ⎣N ⎦ q e = ⎣N1

N2

where the quartic shape functions are

N3

N4

⎧ v1 ⎪ϕ ⎪⎪ 1 N 5 ⎦ ⎨ v2 ⎪ϕ ⎪ 2 ⎪⎩ v3

⎫ ⎪ ⎪⎪ ⎬, ⎪ ⎪ ⎪⎭

(5.45)


94 N 1( r ) = −

1 ( 1 − r ) 2 r ( 3 + 2r ) , 4 1 N 2 ( r ) = − ( 1 −r ) 2 r ( 1 + r )l , 8 1 N 3 ( r ) = ( 3 − 2r ) r ( 1 + r ) 2 , 4 1 N 4 (r ) = − (1 − r ) r (1 + r ) 2 l , 8

(5.46)

N 5 (r ) = (1 − r ) 2 (1 + r ) 2 . In equations (5.46), N 5 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

[k ] e B

⎡ 316 94l e 196 − 34l e ⎢ 36l 2e 34l e − 6l 2e EI e ⎢ = 3⎢ 316 − 94l e 5l e ⎢ 36l 2e ⎢SYM ⎣

− 512 ⎤ − 128l e ⎥ ⎥ − 512 ⎥ . 128l e ⎥ 1024 ⎥⎦

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

(5.47)


{f } e

⎢7 ple =⎢ ⎣⎢ 30

95

p l 2e 60

7 p le 30

p l 2e − 60

T

8 p le ⎥ ⎥ . 15 ⎦⎥

(5.48)

5.4.3 Bending moment and shear force Using the bending moment expression (5.6) and equation (5.12) we get M = EIz M=

{ }

d2 v 4 d2 v 4 = E I = E I z 2 ⎣N ′′⎦ q e , z 2 2 2 dx l e dr le

EIz ⎣6 r l 2e

(3r − 1) l e

(3r + 1) l e ⎦ { q e }.

− 6r

(5.49)

The shear force is given by equation (5.7) T = −E I z T=

{ }

d3 v 8 d3 v 8 = − E I = − E I z 3 ⎣N ′′′⎦ q e , z 3 3 3 dx l e dr le

6E I z ⎣− 2 − l e l 3e

{ }

2 − l e ⎦ qe .

(5.50)

For elements with uniformly distributed load, the end equilibrium loads are given by ⎧ pl e ⎫ ⎪− 2 ⎪ ⎪ ⎧ R2 ⎫ 6l − 12 6l ⎤ ⎧ q2 ⎫ ⎪ ⎡ 12 pl2 ⎪R ⎪ ⎢ 6l 4l 2 − 6l 2l 2 ⎥ ⎪ q ⎪ ⎪− e ⎪ EI e ⎪ 3⎪ ⎥ ⎪⎨ 3 ⎪⎬ + ⎪⎨ 12 ⎪⎬ . (5.51) ⎨ ⎬ = 3 ⎢ R ⎢ ⎥ ⎪ q5 ⎪ ⎪ p l e ⎪ 12 6 12 6 − − − l l le ⎪ 5⎪ − ⎢ ⎥ 2 2 ⎪ R6 ⎪ − 6l 4l ⎦ e ⎪⎩ q6 ⎪⎭e ⎪ 22 ⎪ ⎣ 6l 2l ⎩ ⎭e ⎪ pl e ⎪ ⎪ ⎪ ⎩ 12 ⎭

[ ]{ q }. The second term consists of

The first term on the right is k Be elements that are called fixed-end reactions.

e

The shear forces at the two ends of the beam element are T1 = − R2 and T2 = R5 . The end bending moments are M 1 = − R3 and M 2 = R 6 . For p = 0 , they are obtained substituting r = −1 and r = +1 in (5.49) and (5.50). When p = const . , the exact moment and shear force within the element are

Mp =M −

2 p l 2e p l e l e ( r + 1) + p l e ( r + 1) 2 , + 12 2 2 2 4


96 Tp = T −

pl e r, 2

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

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 ⎧ ⎪ ⎪ v (x ) = ⎣N ⎦ ⎨ ⎪ ⎪⎩

1⎫ 0 ⎪⎪ 1 1 3 3 ⎬ = N1 + N 3 = 2 − 3 r + r + 2 + 3 r − r = 1 = const . 1⎪ 4 4 ⎪ 0⎭

(

) (

)

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 l e (Fig. 5.9, b), equation (5.12) gives ⎧ ⎪ ⎪ v (x ) = ⎣N ⎦ ⎨ ⎪ ⎪⎩

0⎫ 1 ⎪⎪ l e l l N 2 + l e N 3 + e N 4 = e ( 1 + r ) = linear , ⎬= l⎪ 2 2 2 1 ⎪⎭

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


97

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)

⎧ 0 ⎫ ⎪ 1 ⎪ l l ⎪ ⎪ le v (x ) = ⎣N ⎦ ⎨ N2 − e N4 = e 1 − r 2 , ⎬= 2 4 ⎪ 0 ⎪ 2 ⎪⎩ − 1 ⎪⎭ so that the second derivative (curvature) is constant

(

)

v′′ = - l 2 = const .

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

[ ]

⎧ f1 ⎫ e ⎧ q1 ⎫ ⎨ ⎬ = kS ⎨ ⎬ ⎩ f4 ⎭ ⎩ q4 ⎭

(5.52)

where the stiffness matrix for stretching (4.22) is

[ k ]= ElA ⎡⎢−11 e S

e

e

⎣

− 1⎤ . 1 ⎥⎦

(5.53)


98

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

[k ] e

(i l )

⎡ EA ⎢ l ⎢ ⎢ 0 ⎢ ⎢ ⎢ 0 =⎢ ⎢− E A ⎢ l ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢⎣

0

0

12 E I l3 6E I l2

6E I l2 4E I l

0

0

−

12 E I l3 6E I l2

−

EA l 0

0 12 E I l3 6E I − 2 l

−

0 EA l

6E I l2 2E I l

−

0 12 E I l3 6E I − 2 l

0 0

⎤ ⎥ 6E I ⎥ ⎥ l2 ⎥ 2E I ⎥ l ⎥ . (5.54) ⎥ 0 ⎥ ⎥ 6E I ⎥ − 2 ⎥ l 4 E I ⎥⎥ l ⎥⎦ e 0

The ratio of the bending terms to the stretching terms in (5.54) is of order 2

, where ‘i’ is the relevant radius of gyration. For slender beams, this ratio may be as small as 1 20 or 1 50 , 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

{ f }= ⎢⎢0 e

⎢⎣

p le 2

p l 2e 12

0

T

p le 2

−

p l 2e ⎥ ⎥ . 12 ⎥⎦

(5.55)

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

q 1 = Q 1 cos θ + Q 2 sin θ , q 2 = −Q 1 sin θ + Q 2 cos θ .

.

(5.56)

Equations (5.56) can be written in matrix form as ⎧ q1 ⎫ ⎡ c ⎨q ⎬=⎢ ⎩ 2 ⎭ ⎣− s

s ⎤ ⎧ Q1 ⎫ ⎨ ⎬ c ⎥⎦ ⎩ Q 2 ⎭

(5.57)


99

where c = cos θ and s = sin θ . The angular displacements (rotations) are the same in both coordinate systems q 3 = Q3 .

(5.58)

Fig. 5.10 Adding the similar relationships for node 2 ⎧ q4 ⎫ ⎡ c ⎨q ⎬=⎢ ⎩ 5 ⎭ ⎣− s

s ⎤ ⎧ Q4 ⎫ ⎨ ⎬, c ⎥⎦ ⎩ Q 5 ⎭

q6 = Q6 ,

(5.59)

we obtain

{ q } = [ T ] { Q }. (5.60) In (5.60), { q } is the beam element displacement vector in the local coordinate system, { Q } is the beam element displacement vector in the global e

e

e

e

e

coordinate system and

[T ] e

⎡c ⎢− s ⎢ ⎢0 =⎢ ⎢0 ⎢0 ⎢ ⎣⎢ 0

s 0 c 0 0 0 0 0

1 0 0 0

0 0⎤ 0 0⎥⎥ 0 0 0⎥ ⎥ c s 0⎥ − s c 0⎥ ⎥ 0 0 1⎦⎥ 0 0

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

(5.61)


100

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

[ K ]= [T ] [ k ][T ] . e

e

T

e

e

(5.62)

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

{F }= [T ] { f } e

e

T

e

(5.63)

{ } are added to the global load vector.

The values of F e

5.7 Assembly of the global stiffness matrix

[K ],

is assembled from element matrices ~ K e using element connectivity matrices T e , that relate the nodal displacements at element level with the nodal displacements at the entire structure level, by equations of the form ~ Q e = T e { Q }. (5.64)

[ ]

The global stiffness matrix,

[ ]

{ } [ ]

The global uncondensed stiffness matrix is equal to the sum of the expanded element stiffness matrices [ K ] = K~ e , (5.65)

∑[ ] e

where

[ K~ ] = [T~ ] [ K ][T~ ] . e

e

T

e

e

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

{ }

[ ] { Q }.

ε e = − y v′′ = − y ⎣N ′′⎦ q e = − y ⎣N ′′⎦ T e

e

(5.67)


101

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

[k ] 1

6l − 12 6l ⎤ ⎡ 12 ⎢ 6l 4l 2 − 6l 2l 2 ⎥ 2E I ⎥ , = 3 ⎢ ⎢ ⎥ l l − − − 12 6 12 6 l ⎢ 2 2⎥ − 6l 4l ⎦ ⎣ 6l 2l

[k ] 2

6l − 12 6l ⎤ ⎡ 12 ⎢ 6l 4l 2 − 6l 2l 2 ⎥ EI ⎥ . = 3 ⎢ ⎢ ⎥ l l − − − 12 6 12 6 l ⎢ 2 2⎥ − 6l 4l ⎦ ⎣ 6l 2l

The unreduced global stiffness matrix is

[

12l − 24 12l ⎡ 24 ⎢ 12l 8l 2 − 12l 4l 2 ⎢ − 6l E I ⎢− 24 − 12l 36 K ]= 3 ⎢ 2 − 6l 12l 2 l ⎢ 12l 4l ⎢ 0 0 − 12 − 6l ⎢ 0 6l 2l 2 ⎣⎢ 0

0 ⎤ 0 ⎥⎥ − 12 6l ⎥ ⎥. − 6l 2l 2 ⎥ 12 − 6l ⎥ ⎥ − 6l 4l 2 ⎦⎥ 0 0

Using the boundary conditions at the fixed end Q1 = Q2 = 0 , and omitting the first two rows and columns, we obtain the finite element equations


102

⎡ 36 − 6l ⎢ 2 EI ⎢ − 6l 12l l 3 ⎢ − 12 − 6l ⎢ 2l 2 ⎣ 6l

− 12 6l ⎤ ⎧ Q3 − 6l 2l 2 ⎥⎥ ⎪⎪ Q4 ⎨ 12 − 6l ⎥ ⎪ Q5 ⎥ − 6l 4l 2 ⎦ ⎪⎩ Q6

⎫ ⎧ 0 ⎪ ⎪ 0 ⎪ ⎪ ⎬=⎨ ⎪ ⎪−F ⎪⎭ ⎪⎩ 0

⎫ ⎪ ⎪ ⎬. ⎪ ⎪⎭

The three equations with zero right hand side can be written ⎡ 36 − 6l ⎢ − 6l 12l 2 ⎢ ⎢⎣ 6l 2l 2

6l 2l 2 4l 2

⎤ ⎧ Q3 ⎥ ⎪Q ⎥⎨ 4 ⎥⎦ ⎪⎩ Q6

⎫ ⎧ 12 ⎫ ⎪ ⎪ ⎪ ⎬ = ⎨ 6l ⎬ Q5 . ⎪ ⎪ 6l ⎪ ⎭ ⎩ ⎭

or ⎧ Q3 ⎪ ⎨ Q4 ⎪Q ⎩ 6

⎫ 1 ⎪ ⎬= 2 ⎪ 216l ⎭

⎡ 11l 2 9l − 21l ⎤ ⎧ 12 ⎫ ⎧ 60l ⎫ ⎥⎪ ⎪ ⎢ 1 ⎪ ⎪ 27 − 27 ⎥ ⎨ 6l ⎬ Q5 = ⎨ 108 ⎬ Q5 . ⎢ 9l 216 l ⎪ ⎪ ⎢ − 21l − 27 99 ⎥ ⎪ 6l ⎪ ⎩ 180 ⎭ ⎦⎩ ⎭ ⎣

By substitution into the equation with non-zero right hand side EI (− 12 Q3 − 6l Q4 + 12 Q5 − 6l Q6 ) = − F l3 the transverse displacement in 3 is obtained as

v3 = Q5 = −1.5

F l3 . EI

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 6l − 12 6l 0 0 ⎤ ⎡ 12 ⎢ 6l 4l 2 − 6l 2l 2 0 0 ⎥⎥ ⎢ ⎢ 0 − 6l 24 − 12 6l ⎥ [ K ] = E3I ⎢−612 ⎥. 2 2 0 8l − 6l 2l 2 ⎥ l ⎢ l 2l ⎢ 0 0 − 12 − 6l 12 − 6l ⎥ ⎥ ⎢ 0 6l 2l 2 − 6l 4l 2 ⎦⎥ ⎣⎢ 0


103

At the fixed end Q1 = Q2 = 0 ; at the simply supported end Q5 = 0 . Omitting the corresponding rows and columns, we obtain the finite element equations 6l ⎤ ⎧ Q3 ⎫ ⎧ − F ⎫ ⎡ 24 0 EI ⎢ ⎪ ⎪ ⎪ ⎪ 2 0 8l 2l 2 ⎥⎥ ⎨ Q4 ⎬ = ⎨ 0 ⎬ . 3 ⎢ l ⎢⎣ 6l 2l 2 4l 2 ⎥⎦ ⎪⎩ Q6 ⎪⎭ ⎪⎩ 0 ⎪⎭ The last two equations give ⎡ 8l 2 ⎢ 2 ⎢⎣ 2l

⎧0 ⎫ 2l 2 ⎤ ⎧ Q4 ⎫ = − ⎥ ⎨ ⎬ Q3 ⎨ ⎬ 4l 2 ⎥⎦ ⎩ Q6 ⎭ ⎩ 6l ⎭

or Q4 =

3 Q3 , 7l

Q6 = −

12 Q3 . 7l

Fig. E5.3 Substitution of Q4 and Q6 into the first equation gives ⎧ 3 7l ⎫ F l3 24 Q3 + ⎣ 0 6l ⎦ ⎨ ⎬ Q3 = − EI ⎩ − 12 7l ⎭ or

v2 = Q3 = −

7 F l3 . 96 E I

The rotations are

ϕ 2 = Q4 = −

3 F l2 , 96 E I

ϕ3 = Q6 =

The support reactions are given by

12 F l 2 . 96 E I


104

0 ⎤ ⎧ Q3 ⎡ − 12 6l EI ⎢ ⎪ 2 0 ⎥⎥ ⎨ Q4 − 6l 2l 3 ⎢ l ⎢⎣ − 12 − 6l − 6l ⎥⎦ ⎪⎩ Q6

⎫ ⎧ V1 ⎫ ⎪ ⎪ ⎪ ⎬ = ⎨ M1 ⎬ ⎪ ⎪V ⎪ ⎭ ⎩ 3 ⎭

which yield V1 =

11 F, 16

3 M1 = − F l , 8

V3 =

5 F. 16

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 Q1 = Q2 = Q5 = Q6 = 0 , the finite element equations can be written − 12 6l 6l 0 0 ⎤ ⎧ 0 ⎫ ⎧ V1 ⎫ ⎡ 12 ⎪ ⎪ ⎢ 6l 4l 2 2 − 6l 2l 0 0 ⎥⎥ ⎪⎪ 0 ⎪⎪ ⎪ M 1 ⎪ ⎢ E I ⎢− 12 − 6l 13.5 − 4.5l − 1.5 1.5l ⎥ ⎪⎪Q3 ⎪⎪ ⎪⎪− F ⎪⎪ ⎢ ⎥⎨ ⎬=⎨ ⎬. 2 − 4.5l 6l 2 − 1.5l l 2 ⎥ ⎪Q4 ⎪ ⎪ 0 ⎪ l 3 ⎢ 6l 2l ⎢ 0 − 1.5 − 1.5l 1.5 − 1.5l ⎥ ⎪ 0 ⎪ ⎪ V3 ⎪ 0 ⎥⎪ ⎪ ⎪ ⎢ ⎪ 2 − 1.5l 2l 2 ⎥⎦ ⎪⎩ 0 ⎪⎭ ⎪⎩ M 3 ⎪⎭ 0 1.5l l ⎢⎣ 0 From the third and fourth rows we obtain E I ⎡ 13.5 − 4.5l ⎤ ⎧ Q3 ⎫ ⎧ − F ⎫ ⎬=⎨ ⎬ ⎢ 2 ⎥⎨ l 3 ⎣ − 4.5l 6l ⎦ ⎩ Q4 ⎭ ⎩ 0 ⎭ with solutions

v2 = Q3 = −

8 F l3 , 81 E I

ϕ 2 = Q4 = −

2 F l2 . 27 E I


105

Substituting the displacements into the other four equations yields 6l ⎤ ⎧ V1 ⎫ ⎡ − 12 8 ⎪ ⎪ ⎢ − 6l 2l 2 ⎥ ⎧⎪− l ⎫⎪ 2 EI ⎢ ⎥ ⎨ 81 ⎬ F l = ⎪⎨ M 1 ⎪⎬ . l 3 ⎢− 1.5 − 1.5l ⎥ ⎪ − 2 ⎪ E I ⎪ V3 ⎪ ⎢ ⎥ ⎪⎩M 3 ⎪⎭ l 2 ⎦ ⎩ 27 ⎭ ⎣ 1.5l The support reactions are

V1 =

20 4 7 2 F , M 1 = F l , V3 = F , M3 = − Fl . 27 9 27 9

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 Q1 = Q3 = Q5 = 0 , the finite element equations can be written 6l − 12 6l 0 0 ⎤ ⎧ 0 ⎫ ⎧ V1 ⎫ ⎡ 12 ⎪Q ⎪ ⎪ ⎢ 6l 4l 2 − 6l 2l 2 ⎥ ⎪ 0 0 0 ⎥ ⎪ 2⎪ ⎪ ⎢ ⎪ − 12 6l ⎥ ⎪⎪ 0 ⎪⎪ ⎪⎪V2 − p l 2⎪⎪ 0 E I ⎢− 12 − 6l 24 ⎢ ⎥⎨ ⎬=⎨ ⎬. 2 0 8l 2 − 6l 2l 2 ⎥ ⎪Q4 ⎪ ⎪ − p l 2 12 ⎪ l 3 ⎢ 6l 2l ⎢ 0 0 − 12 − 6l 12 − 6l ⎥ ⎪ 0 ⎪ ⎪V3 − p l 2 ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪ 0 6l 2l 2 − 6l 4l 2 ⎥⎦ ⎪⎩Q6 ⎪⎭ ⎪⎩ p l 2 12 ⎪⎭ ⎣⎢ 0 Omitting the first, third and fifth rows and columns, we obtain

EI l3

⎡ 4l 2 ⎢ 2 ⎢ 2l ⎢ 0 ⎣

2l 2 8l 2 2l 2

0 ⎤ ⎧ Q2 ⎥⎪ 2l 2 ⎥ ⎨ Q4 4l 2 ⎥ ⎪⎩ Q6 ⎦

0 ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ 2 ⎬ = ⎨ − p l 12 ⎬ . ⎪ ⎪ p l 2 12 ⎪ ⎭ ⎩ ⎭


106

The first equation above yields ⎧Q ⎫ Q2 = ⎣ − 1 2 0 ⎦ ⎨ 4 ⎬ ⎩ Q6 ⎭

(a)

which, substituted into the second and third equation

EI l3

⎛ ⎜ ⎜ ⎝

⎧⎪ 2l 2 ⎨ ⎪⎩ 0

⎫⎪ ⎡ 8l 2 ⎬ Q2 + ⎢ 2 ⎪⎭ ⎢⎣ 2l

2l 2 ⎤ ⎧ Q4 ⎫ ⎞⎟ p l 2 ⎧− 1⎫ ⎥⎨ ⎬ = ⎨ ⎬, 4l 2 ⎥⎦ ⎩ Q6 ⎭ ⎟⎠ 12 ⎩ 1 ⎭

gives

ϕ 2 = Q4 = −

pl3 , 48 E I

ϕ3 = Q6 =

pl3 . 32 E I

(b)

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

pl3 . 96 E I The reaction forces are calculated from

ϕ1 = Q2 =

6l 0 ⎤ ⎧ Q2 ⎡ 6l EI ⎢ ⎪ − 6l 0 6l ⎥⎥ ⎨ Q4 3 ⎢ l ⎢⎣ 0 − 6l − 6l ⎥⎦ ⎪⎩ Q6

V1 ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎬ = ⎨ V2 − p l 2 ⎬ ⎪ ⎪ V − pl 2 ⎪ ⎭ ⎩ 3 ⎭

obtaining

V1 = −

pl , 16

V2 =

5 pl , 8

V3 =

7 pl . 16

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


107

Solution. Two Bernoulli-Euler beam elements are used to model the system. Using the boundary conditions Q1 = Q2 = Q5 = 0 , the finite element equations can be written 1.5l 0 0 ⎤ ⎧ 0 ⎫ ⎧ V1 − 0.7 p0 l ⎫ − 1.5 1.5l ⎡ 1.5 ⎪ ⎪ ⎢ 1.5l 2 2 2l 0 0 ⎥⎥ ⎪ 0 ⎪ ⎪⎪M 1 − 0.2 p0 l 2 ⎪⎪ − 1.5l l ⎢ E I ⎢− 1.5 − 1.5l 13.5 4.5l − 12 6l ⎥ ⎪⎪Q3 ⎪⎪ ⎪⎪ − 0.3 p0 l ⎪⎪ ⎢ ⎥⎨ ⎬=⎨ ⎬. l2 4.5l 6l 2 − 6l 2l 2 ⎥ ⎪Q4 ⎪ ⎪ 0.1333 p0 l 2 ⎪ l 3 ⎢ 1.5l ⎢ 0 ⎪ 0 V3 − 12 − 6l 12 − 6l ⎥ ⎪ 0 ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎪⎭ 0 6l 2l 2 − 6l 4l 2 ⎥⎦ ⎪⎩Q6 ⎪⎭ ⎪⎩ 0 ⎢⎣ 0 Deleting the first, second and fifth rows and columns we obtain ⎡ 13.5 4.5l 6l ⎤ ⎧ Q3 EI ⎢ ⎪ 4.5l 6l 2 2l 2 ⎥⎥ ⎨ Q4 3 ⎢ l ⎢⎣ 6l 2l 2 4l 2 ⎥⎦ ⎪⎩ Q6

⎫ ⎧ − 0 .3 p 0 l ⎪ ⎪ 2 ⎬ = ⎨ 0.1333 p0l ⎪ ⎪ 0 ⎭ ⎩

⎫ ⎪ ⎬. ⎪ ⎭

The last equation yields

Q6 = −

3 1 Q3 − Q4 2l 2

which, substituted into the first two equations

EI l3

⎛ ⎜ ⎜ ⎝

⎞ ⎧ − 0.3 p0l ⎫ ⎡ 13.5 4.5l ⎤ ⎧ Q3 ⎫ ⎧ 6l ⎫ ⎢ 4.5l 6l 2 ⎥ ⎨ Q ⎬ + ⎨ 2l 2 ⎬ Q6 ⎟⎟ = ⎨0.1333 p l 2 ⎬ , 0 ⎭ ⎦⎩ 4⎭ ⎩ ⎭ ⎣ ⎠ ⎩

gives

v2 = Q3 = −0.084

p0 l 4 , EI

ϕ 2 = Q4 = 0.0518

p0 l 3 . EI

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

Answer. N 1( r ) =

(

)

1 2 r 4 − 5r − 2r 2 + 3r 3 , 4

N 2 (r ) =

(

)

1 2 r 1 −r − r 2 + r 3 , 4


108

(

N 3(r ) = 1 − r 2 N 5 (r ) =

(

)

2

,

(

N 4 (r ) = r 1− r 2

)

1 2 r 4 + 5r − 2r 2 − 3r 3 , 4

Fig. E5.7, a

Fig. E5.7, b

), 2

N 6 (r ) =

(

)

1 2 r −1− r +r 2 + r3 . 4


109

Example 5.8 The planar frame shown in Fig. E5.8, a has pinned ends and is loaded in 2

F = 1000 N . It has

by a force

E = 2 ⋅ 1011 N m 2 ,

A = 1600 mm 2

and

I = 2 ⋅ 10 mm . Determine the nodal displacements, plot the deformed shape and the diagrams of axial force, shear and bending moment. 5

4

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

Nodal data Node nr 1 2 3 4 5 6 7

Restr X 1 0 0 0 0 0 1

Restr Y 1 0 0 0 0 0 1

Restr Z 0 0 0 0 0 0 0

Coord X 0 0 0 0.5 1.4 2 2

Coord Y 0 1 2 2 2 2 1

Displ X 0 0.010 1.0699e-2 1.0698e-2 1.0697e-2 1.0696e-2 0

Displ Y 0 8.500e-7 1.700e-6 1.309e-3 2.321e-3 -8.500e-7 0

Rotation Z -1.232e-2 -5.531e-3 2.353e-3 2.592e-3 -1.260e-3 -6.889e-3 -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

Fig. E5.8

e


110

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 M 3 = 2 ⋅ 106 N mm and two point loads F6 = −2 ⋅ 10 4 N and F12 = −10 4 N . For E = 2 ⋅105 N mm 2 , A = 400 mm 2 , I = 2 ⋅104 mm 4 , l1− 2 = 100 mm and l 9 −10 = l 9 −11 = 200 mm , 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 h 6 = 0.2476 mm , v 6 = −0.9673 mm , ϕ 6 = −0.00623 rad .

The deformed shape is presented in Fig. E5.9, b.

Fig. E5.9, b


111

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 Q3 i − 2 , Q3 i −1 and Q3 i , 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 E I , the torsional rigidity G I t 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.


112

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

{ q }= ⎣ q e

1

q2

q3

q4

q5

q6

⎦T

(5.68)

and the corresponding vector of element nodal forces can be written

{ f }= ⎣ f e

In (5.69)

f 3 and

1

f2

f3

f4

f5

f6

⎦T .

f 6 are transverse forces, while f 2 and

(5.69) f 5 are

couples producing bending (Fig. 5.12, b). The corresponding displacements q 3 , q 6 are translations, while q 2 , q 5 are rotations. Their column vectors are related by the flexural stiffness matrix. Rearranging the matrix (5.29) we obtain

⎧ ⎪ ⎪ ⎨ ⎪ ⎪⎩

⎡ 4l 2 6l 2l 2 − 6l ⎤ ⎧q 2 ⎫ ⎫ ⎢ ⎥ ⎪ ⎪ ⎪ 12 6l − 12 ⎥ ⎪q 3 ⎪ ⎪ EI e ⎢ 6l ⎨ ⎬. ⎬= 3 ⎢ 2 6l 4l 2 − 6l ⎥ ⎪q 5 ⎪ ⎪ l e ⎢ 2l ⎥ f 6 ⎪⎭ ⎢⎣− 6l − 12 − 6l 12 ⎥⎦ e ⎪⎩q 6 ⎪⎭

f2 f3 f5

(5.70)

The axial nodal forces f 1 , f 4 are torques and the nodal displacements q 1 , q 4 are twist angles. They describe torsional effects so that their action is decoupled from bending. The respective stiffness matrix can be calculated


113

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

θ (r ) = N1 (r ) q 1 + N 2 (r ) q 4 ,

(5.71)

which substituted into the strain energy Ue =

G It e

2

⎛ ∂θ ⎞ ⎜⎜ ⎟⎟ d x ⎝ ∂x ⎠

∫

2

e

(5.72)

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

[ ]= kte

2 G It e

le

+1

∫ ⎣N ′ ⎦ r

T

⎣N r′ ⎦ d r .

(5.73)

−1

As a consequence of this analogy, the nodal forces are related to the nodal displacements by equation ⎧ f1 ⎫ e ⎧ q1 ⎫ (5.74) ⎨ ⎬ = kt ⎨ ⎬ ⎩ f4 ⎭ ⎩ q4 ⎭

[ ]

where the stiffness matrix for torsional effects is

[ k ] = GlI e t

te

e

⎡ 1 − 1⎤ ⎢− 1 1 ⎥ . ⎣ ⎦

(5.75)

In (5.75), G is the shear modulus of elasticity and I t e 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)

[k ] e

−a 0 0 0 0 ⎤ ⎡ a ⎢ 0 4l 2 6l 2 − 6l ⎥⎥ 0 2l ⎢ 6l 12 0 6l − 12 ⎥ EI ⎢ 0 = 3e ⎢ ⎥ a 0 0 0 0 ⎥ l e ⎢− a ⎢ 0 2l 2 6l 0 4l 2 − 6l ⎥ ⎥ ⎢ ⎢⎣ 0 − 6l − 12 0 − 6l 12 ⎥⎦ e

where a = G I t e l 2e E I e .

(5.76)


114

5.8.3 Coordinate transformation 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

{ q } = [ T ] { Q }, e

e

e

(5.77)

{ } is the element displacement vector (5.68) in the local coordinate

where q e system,

{ Q } = ⎣Q e

1

Q2

Q3

Q4

Q5

Q6 ⎦ T

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

[T ] e

⎡c ⎢− s ⎢ ⎢0 =⎢ ⎢0 ⎢0 ⎢ ⎢⎣ 0

0 0⎤ 0 0 0⎥⎥ 0 0 0⎥ ⎥, c s 0⎥ − s c 0⎥ ⎥ 0 0 1 ⎥⎦

s 0 c 0

0

0 1 0 0 0 0 0 0

(5.78)

where c = cos α and s = sin α , 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

[ K ]= [T ] [ k ][T ] . e

e

T

e

e

(5.79)

It is used to assemble the unreduced global stiffness matrix [ K ] using ~ element connectivity matrices T e that relate the nodal displacements at element level with the nodal displacements at the complete structure level, by equations of the form (5.64).

[ ]


115

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 E = 210 GPa , G = 81 GPa , l = 1 m and diameter d = 20 mm . Find the vertical displacement of point 7 when the grid is loaded by forces F7 = F8 = −500 N 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 w7 = − 0.436 m .

Example 5.11 The grid shown in Fig. E5.11, a is fixed at points 1 and 2, and has l = 1 m , I = 0.785 ⋅ 10 −8 m 4 , I t = 1.57 ⋅ 10 −8 m 4 , E = 210 GPa and G = 81 GPa . Find the vertical displacement of point 5 when the grid is loaded by a force F5 = −103 N . Draw the deflected shape and the diagrams of the bending moment and torque.


116

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 w5 = − 0.4 m . 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 (Poncelet, 1825) 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.


117

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

u (x, y ) = − y ϕ (x ) ,

(5.80)

where ϕ is the cross section rotation at position x . The strain components ε x and γ xy are given by

εx =

γ xy =

du dϕ = −y = − yϕ ′ , dx dx

dv ∂u ∂ v + = −ϕ + = −ϕ + v′ , dx ∂ y ∂x

(5.81) (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 dϕ y, dx where E is Young’s modulus of the material.

σ x = E ε x = −E

(5.83)

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


118

∫

M (x ) = − σ x y dA = E I z A

dϕ = E I zϕ′ dx

(5.84)

where I z 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 T (x ) = −

dM . dx

(5.85)

The transverse load per unit length is p (x ) = −

dT . dx

(5.86)

The average shear strain is

γ xy =

T T = G As κ AG

(5.87)

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

As

[ ∫τ dA ] = ∫τ

2

2

dA

.

(5.88)

Equations (5.82) and (5.87) give

T = G As ( v′ − ϕ ) .

(5.89)

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

G As v′′ − G Asϕ ′ = 0 ,

(5.90)

G As v′ + E I z ϕ ′′ − G Asϕ = 0 .

(5.91)

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


119

Fig. 5.14 Eliminating ϕ and v in turn in (5.90) and (5.91) gives d4v =0, d x4

and

d 3ϕ = 0. d x3

(5.92)

d 3ϕ = 0. d r3

(5.93)

Changing to the r coordinate yields d4v =0, d r4

and

The general solutions of these equations are v (r ) = a 4 r 3 + a 3 r 2 + a 2 r + a 1 ,

(5.94)

ϕ (r ) = b 3 r 2 + b 2 r + b 1 .

(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 2 dv d 2ϕ +β −ϕ = 0, l dr d r2

(5.96)

where

β=

4E I z . G As l 2

(5.97)

Substituting (5.94) and (5.95) into (5.96) gives b3 =

6 a4 , l

b2 =

4 a3 , l

b1 =

2 12 a2 + β a 4 . l l

(5.98)

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


120

⎢ v = ⎢ N1 ⎣

l N2 2

⎢2 ~ ⎣l

N3

~

2 ~ N3 l

ϕ = ⎢ N1 N 2 where

{ q } = ⎣N ⎦ { q },

(5.99)

{ q } = ⎣N~ ⎦ { q },

(5.100)

ϕ 1 v2 ϕ 2 ⎦ T .

(5.101)

l ⎥ N4 ⎥ 2 ⎦

e

~ ⎥ N4 ⎥ ⎦

{ q }= ⎣v e

1

e

e

e

The shape functions are

[ 2 − 3r + r

N 1( r ) =

1 4 ( 1 + 3β

)

N 2 (r ) =

1 4 ( 1 + 3β

[1 −r − r )

N 3(r ) =

1 4 ( 1 + 3β

)

N 4 (r ) =

1

4 ( 1 + 3β )

2

3

]

+ 6β ( 1 − r ) ,

)]

(

+ r 3 + 3β 1 − r 2 ,

[ 2 + 3r − r

3

[ − 1 −r + r

2

]

+ 6β ( 1 + r ) ,

)]

(

+ r 3 + 3β − 1 + r 2 ,

~ N1 ( r ) =

1 4 ( 1 + 3β

)

( − 3 + 3r ) ,

~ N 2 (r ) =

1 4 ( 1 + 3β

)

[ − 1 −2r + 3r

~ N3 ( r ) =

1 4 ( 1 + 3β

)

( 3 − 3r ),

~ N 4 (r ) =

1 4 ( 1 + 3β

)

[ − 1 +2r + 3r

2

(5.102)

]

2

+ 6β ( 1 − r ) ,

2

+ 6β ( 1 + r ) .

2

]

For β = 0 , the first four functions (5.102) become the third degree Hermitic polynomials (5.21) and the last four functions satisfy the relationship 2 ∂ Ni ~ N i (r ) = l ∂r

( i = 1,..., 4 ) .

(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


121 2 ⎢ dN ⎥ ⎢ ⎥ . ⎣ dr ⎦

⎣ Nˆ ⎦= ⎣ N~ ⎦ − l

(5.104)

They are 3β 1 Nˆ1 (r ) = Nˆ 2 (r ) = − Nˆ 3 (r ) = Nˆ 4 (r ) = . 1 + 3β l

(5.105)

5.9.3 Stiffness matrix The strain energy for a beam element with shear effects included is E Iz Ue = 2 As

l2

∫

2

⎛dϕ ⎞ G As ⎜⎜ ⎟⎟ d x + 2 ⎝dx⎠

−l 2

dϕ 2 d ϕ = dx l dr

l 2

∫

2

⎛ dv ⎞ ⎜⎜ ϕ − ⎟ dx . d x ⎟⎠ ⎝

−l 2

dx =

and

(5.106)

l dr , 2

(5.107)

the contribution due to bending is

U eB

E Iz 2 = 2 l

+1

∫

−1

2

⎛dϕ ⎞ ⎜⎜ ⎟⎟ d r , ⎝dr ⎠

(5.108)

and the contribution due to shear is G As l U eS = 2 2

+1

∫

−1

2

⎛ 2 dv ⎞ ⎟ dr . ⎜⎜ ϕ − l d r ⎟⎠ ⎝

(5.109)

On substituting (5.100) into (5.108) we get U eB

1 = 2

{q }

e T

+1

2 EIz l

∫ ⎣N ′⎦ ⎣N ′⎦ dr { q } ~

T

~

e

(5.110)

−1

wherefrom we obtain the element stiffness matrix due to bending +1

[ k ] = E I 2l ∫ ⎣N~ ′⎦ ⎣N~ ′⎦ dr . e B

T

z

−1

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

(5.111)


122

U eS

1 = 2

{q }

e T

+1

l G As 2

∫ ⎣Nˆ ⎦

T

⎣Nˆ ⎦ d r { q e }

(5.112)

−1

wherefrom we obtain the element stiffness matrix due to shear +1

[ k ] = G A 2l ∫ ⎣Nˆ ⎦ ⎣Nˆ ⎦ dr . e S

T

(5.113)

s

−1

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

[k ] e

⎡ 12 ⎢ 1 EI z ⎢ 6l = 1 + 3β l 3 ⎢− 12 ⎢ ⎣ 6l

− 12

6l

(4 + 3β ) l − 6l

2

(2 − 3β ) l 2

− 6l 12 − 6l

6l

(2 − 3β ) l

⎤

2⎥

⎥. ⎥ 2⎥ (4 + 3β ) l ⎦ − 6l

(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: S u , the portion of the boundary on which displacements are prescribed, and S σ , 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 ∂n ∂x , ∂n ∂y , ∂n ∂z . 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 pv x , pv y , pv z . It is convenient to write these components as a single body force vector

{ pv } = ⎣ pv x

pv y

pv z ⎦ T .

(6.1)


124 Surface tractions

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

{ ps } = ⎣ ps x

ps y

ps z

⎦T ,

(6.2)

Concentrated loads Concentrated loads are defined by their three components

{ Fi } = ⎣ Fi x

Fi y

Fi z

⎦T .

(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

{ u} = ⎣u

v w⎦ T ,

(6.4)

where u , v , w are the displacement components inside the body or on the surface Sσ with unprescribed displacements. Stresses and strains

The stresses inside V will have two types of component, the direct stress components σ x , σ y ,σ z and the shear stresses τ xy ,τ yz ,τ zx . It is convenient to represent both stress and strain components as single column matrices. Thus

{σ } = ⎣σ x

σ y σ z τ xy τ yz τ zx

⎦T .

(6.5)


125

In two-dimensional problems

{σ } = ⎣ σ x

σ y τ xy

⎦T .

(6.5, a)

Strains are represented in vector form as

{ε } = ⎣ε x

ε y ε z γ xy γ yz γ zx

⎦T .

(6.6)

In two-dimensional problems

{ε } = ⎣ε x

ε y γ xy

⎦T .

(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 τ xy = τ yx ∂σ x ∂τ yx + pv x = 0, + ∂y ∂x ∂τ xy ∂σ y + + p v y = 0. ∂x ∂y

Fig. 6.2 In matrix form, equations (6.7) can be written

(6.7)


126

⎡ ∂ ⎢ ∂x ⎢ ⎢ 0 ⎢⎣

0 ∂ ∂y

∂ ∂y ∂ ∂x

⎤ ⎧σ ⎥⎪ x ⎥ ⎨σy ⎥⎪ ⎥⎦ ⎩ τ xy

⎫ ⎧ p vx ⎫ ⎪ ⎬ ⎬ = −⎨ ⎩ pvy ⎭ ⎪ ⎭

(6.8)

or, denoting the matrix of differential operators ⎡ ∂ ⎢ ⎢ ∂x [ ∂ ] = ⎢⎢ 0 ⎢ ∂ ⎢ ⎣⎢ ∂y and using (6.1)

⎤ 0 ⎥ ⎥ ∂ ⎥ , ∂y ⎥ ∂ ⎥ ⎥ ∂x ⎥⎦

(6.9)

[ ∂ ]T {σ } + { pv } = { 0 } .

6.2 Equations of equilibrium on the surface

(6.10)

Sσ

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

σ x l + τ yx m = ps x , τ xy l + σ y m = ps y .

Fig. 6.3 In (6.11) the direction cosines for the outward normal n are

(6.11)


127

∂n = nx , ∂x ∂n m = cos (n , y ) = = ny . ∂y

l = cos (n , x ) =

(6.12)

In matrix form, equations (6.11) can be written ⎧σ ⎡ nx 0 n y ⎤ ⎪ x ⎢ 0 n n ⎥ ⎨σy y x ⎦⎪ ⎣ ⎩ τ xy

⎫ ⎪ ⎧ ps x ⎫ ⎬ =⎨ p ⎬ , ⎪ ⎩ sy ⎭ ⎭

(6.13)

or in condensed form

sometimes written

[ n ] T {σ } = { p s } ,

(6.14)

[ ∂ n ] {σ } = { p

(6.15)

T

s

},

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

6.3 Strain-displacement relations Figure 6.4 gives the deformation of the dx − dy face for small deformations. The equations of compatibility have the familiar form

εx =

∂u ; ∂x

εy =

∂v , ∂y

γ xy =

∂ v ∂u + . ∂x ∂ y

(6.16)

In matrix form ⎡ ∂ ⎢ ⎧ εx ⎫ ⎢ ∂x ⎪ ⎪ ⎢ ⎨ εy ⎬ = ⎢ 0 ⎪γ ⎪ ⎢ ⎩ xy ⎭ ∂ ⎢ ⎢⎣ ∂ y and can be summarized as

⎤ 0 ⎥ ⎥ ∂ ⎥⎧u⎫ ⎨ ⎬, ∂y ⎥⎩ v ⎭ ∂ ⎥ ⎥ ∂ x ⎦⎥

{ε } = [ ∂ ] { u } .

(6.17)

(6.18)


128

Fig. 6.4

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

{ε } = [ C ] {σ } ,

(6.19)

where the material elastic compliance matrix is

−ν ⎡1 − ν ⎢ 1 −ν ⎢ ⎢ 1 [C ]= 1 ⎢ E ⎢ ⎢ SYM ⎢ ⎢⎣

0 0 0 2 (1 +ν

0 0

)

0 0 2 (1 +ν

⎤ ⎥ ⎥ 0 ⎥ ⎥. 0 ⎥ 0 ⎥ ⎥ 2 ( 1 + ν )⎥⎦ 0 0

)

(6.20)

The inverse relation is

{σ } = [ C ] −1{ε } = [ D ] {ε } . The inverse of [ C ] is the material stiffness matrix

(6.21)


129

⎡1 − ν ⎢ ⎢ ⎢ ⎢ E ⎢ [D] = (1 + ν )(1 − 2ν ) ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

ν 1 −ν

ν ν

0

0

0

0

1 −ν

0 1 −ν 2

0 0 1 −ν 2

SYM

0 ⎤ 0 ⎥⎥ 0 ⎥ ⎥ 0 ⎥ .(6.22) ⎥ 0 ⎥ ⎥ ⎥ 1 −ν ⎥ 2 ⎦

Plane stress

A thin planar body subjected to in-plane loading on its edge surface is said to be in plane stress. If stresses σ z , τ xz , and τ yz are set as zero, discarding rows and columns 3, 5 and 6 in (6.20), the Hooke’s law can be written ⎧ εx ⎪ ⎨ εy ⎪γ ⎩ xy

⎫ ⎡ 1 ⎪ 1 ⎢ ⎬ = ⎢ −ν ⎪ E⎢ 0 ⎣ ⎭

−ν 1 0

⎤ ⎧σx ⎪ 0 ⎥⎥ ⎨ σ y 2 (1 + ν ) ⎥⎦ ⎪⎩ τ xy 0

⎫ ⎪ ⎬. ⎪ ⎭

(6.23)

The inverse relations are given by ⎧σx ⎪ ⎨σy ⎪τ ⎩ xy

⎫ ⎪ E ⎬= 2 ⎪ 1 −ν ⎭

⎡ 0 ⎢1 ν ⎢ν 1 0 ⎢ 1 −ν ⎢0 0 2 ⎣

⎤⎧ ⎥ ⎪ εx ⎥ ⎨ εy ⎥⎪ ⎥ ⎩ γ xy ⎦

⎫ ⎪ ⎬ ⎪ ⎭

(6.24)

which is used as {σ } = [ D ] {ε }. 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 ε z , γ xz , and γ yz are set as zero, from (6.21) and discarding rows and columns 3, 5 and 6 in (6.22), we obtain ⎧σx ⎪ ⎨σy ⎪τ ⎩ xy

⎡ ⎫ ν ⎢ 1 −ν ⎪ E ⎢ ν 1 −ν ⎬= ( )( ) + − ν ν 1 1 2 ⎢ ⎪ 0 ⎢ 0 ⎭ ⎣

0 0 1 −ν 2

⎤⎧ ⎥ ⎪ εx ⎥ ⎨ εy ⎥⎪ ⎥ ⎩ γ xy ⎦

⎫ ⎪ ⎬ ⎪ ⎭

(6.25)


130

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

6.5 Temperature effects The temperature strains are represented as initial strains

{ε 0 } = ⎣ α T

αT αT 0 0 0 ⎦ T ,

(6.26)

where T is the temperature rise and α is the coefficient of linear expansion of the material. The stress-strain relations become

{σ } = [ D ] ( {ε } − {ε 0 } ) .

(6.27)

{ε 0 } = ⎣α T

(6.28)

In plane stress αT 0 ⎦ T .

In plane strain

{ε 0 } = (1 + ν ) ⎣α T

αT

0⎦ T .

(6.29)

6.6 Strain energy For linear elastic materials, the strain energy per unit volume in the body is U0 =

1 {σ 2

}T {ε } = 1 {ε }T {σ }. 2

(6.30)

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

1 2

∫ {ε } {σ } dV . T

(6.31)

V

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

1 2

∫ {ε } [ D ] {ε } dV . T

V

(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 eigenvalue 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);

132


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 C 0 is generally required for bars and elasticity problems, while a continuity C1 is imposed for beams, plates and shells. Exceptions do exist. Remember that a function of several variables is said to be of class C m 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 m = 2n , the admissible functions must have continuity C n −1 , i.e. the geometrical boundary conditions must be satisfied to the (n − 1)th derivative. For bars, m = 2 and the assumed functions must have continuity C 0 . For beams m = 4 and the approximating functions must have continuity 1

C . Because the continuity required is reduced from C 2 in the governing differential equation to C1 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. δu . 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

{ u} = ⎣u

v w⎦ T ,

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

{δu } = ⎣ δu

δv δw ⎦ T .

(7.1)

The vector of the corresponding virtual strains will be

{ δε } = [ B ] { δu } . where [ B ] is the strain-displacement matrix.

(7.2)

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7. ENERGY METHODS

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

δWE = F ⋅ δu .

(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 δ WE =

∫ { δu } { pv }dV + ∫ { δu } { ps }d A + ∑i { δui } { Fi } . T

T

V

T

(7.4)

Sσ

Note that the scalar product under the first integral is

{ δu }T { pv } = δu ⋅ pv x + δ v ⋅ pv y + δ w ⋅ pv z .

a

b

c

Fig. 7.1 Note also the absence of the factor 1 2 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 δ WI =

∫ { δε } {σ }dV . T

V

as shown in Fig. 7.2 for the uniaxial case.

(7.5)

134


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 δWE = δWI . (7.6) Also: A body is in equilibrium if the internal virtual work equals the external virtual work for every kinematically admissible displacement field.

∫ { δε } {σ }dV − ∫ { δu } { pv }dV − ∫ { δu } { ps }d A − ∑i { δui } { Fi } = 0 . T

V

T

V

T

T

Sσ

(7.6, a) In (7.6) δWE is the work of external loads on the virtual displacements { δu } 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

135

7. ENERGY METHODS

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 F1 = F sinα and F2 = F cosα , produces a displacement of the joint 4, of components u1 and u2 (Fig. 7.4, b). The joint 4 is acted upon by the external forces F1 , F2 and by internal forces T1 , T2 , T3 (Fig. 7.4, c). The joint reacts with forces equal in magnitude but of opposite sign, producing the elongations Δ1 , Δ2 , Δ3 (Fig. 7.4, d). 3. An imaginary state, in which the joint 4 is given a virtual displacement of components δu1 and δu2 (Fig. 7.4, e), which produce virtual elongations in bars δΔ1 , δΔ2 , δΔ3 (Fig. 7.4, f), the applied forces remaining constant. The virtual displacements δu1 and δu2 and the virtual bar extensions δΔ 1 , δΔ 2 , δΔ3 satisfy the compatibility equations (2.19)

136


δΔ1 = δu1 sinθ + δu 2 cosθ , δΔ2 = δu 2 ,

(7.7)

δΔ3 = −δu1 sinθ + δu 2 cosθ . For the three bars, the force-elongation equations (2.21) can be written

Δ 1=

T1 l , EA

Δ 2=

T2 l cosθ , EA

Δ 3=

T3 l . EA

(7.8)

Fig. 7.4 Equating internal work to external work (7.6) using the products of real forces and virtual displacements we obtain

T1 δ Δ 1+ T2 δ Δ 2 + T3 δ Δ 3 = F1 δu1 + F2 δu2 .

(7.9)

Substituting (7.7) into (7.9) and collecting coefficients of δu1 and δu2 gives

137

7. ENERGY METHODS

δu1 (T1 sin θ − T3 sin θ − F1 ) + δu2 (T1 cos θ + T2 + T3 cos θ − F2 ) = 0 .

(7.10)

Since δu1 and δu2 are unrelated to each other, we could put either to zero. Equation (7.10) must be zero whatever the values of δu1 and δu2 . This can only be true if their coefficients vanish T1 sinθ − T3 sinθ = F1 ,

(7.11)

T1 cosθ + T2 + T3 cosθ = F2 .

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 2 EA 2 sin θ ⋅ u 1 = F1 , l EA ⎛ 1 ⎞ 2 ⎜ 2 cos θ + ⎟ u 2 = F2 . cosθ ⎠ l ⎝

(7.12)

7.1.5 Proof that PDV is equivalent to equilibrium equations Consider a form of equation (7.6, a) without point forces

∫ { δε } {σ }dV = ∫ { δu } { pv }dV + ∫ { δu } { ps }d A . T

T

V

T

V

(7.6, b)

Sσ

Convert { δε } to { δu } using the integration by parts

⎛ ∂u ⎞

∫ σ x δε x dV = ∫ σ x δ ⎜⎜⎝ ∂ x ⎟⎟⎠ d x dy dz = ∫

V

V

=

Sσ

V

Sσ

∫ σ y δv m d A − ∫ δv

Sσ

V

V

∂σ y ∂y

Sσ

δu

∂σ x dx dy dz , ∂x dV

dV ,

∫ τ xy δγ xy dV = ∫ τ xy (δ v ⋅ l + δu ⋅ m)d A − ∫

V

∂ (δu ) d x l d A = ∂x

∫ σ x d(δu ) l d A = ∫ σ x δu l d A − ∫

Sσ

∫ σ y δε y dV =

σx

V

∂τ xy ⎞ ⎛ ∂τ xy ⎜⎜ δv + δu ⎟⎟ dV , ∂y ⎝ ∂x ⎠

138


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

∫ ( δε x σ x + δε y σ y + δγ xy τ xy )dV = ∫ [ δu (σ x l + τ xy m)+ δ v (τ xy l + σ y m) ] d A −

V

Sσ

−

∫

V

⎡ ⎛ ∂σ x ∂τ xy + ⎢δu ⎜⎜ ∂y ⎣⎢ ⎝ ∂ x

∂σ y ⎞ ⎛ ∂τ ⎟⎟ + δ v ⎜⎜ xy + ∂y ⎠ ⎝ ∂x

⎞⎤ ⎟⎟⎥ dV . ⎠⎦⎥

In matrix form (Gauss’ theorem) T T T T T ∫ { δε } {σ }dV = ∫ { δu } [ n ] {σ }d A − ∫ { δu } [ ∂ ] {σ }dV ,

V

Sσ

V

which is true provided [ ∂ ] T {σ } 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) T T T T ∫ { δu } ( [ ∂ ] {σ } + { pv }) dV − ∫ { δu } ( [ n ] {σ } − { ps }) d A = 0 .

V

Sσ

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

[ ∂ ]T {σ } + { pv } = { 0 } in V, [ n ]T {σ } − { ps } = { 0 } on

Sσ .

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. Sσ , is supported in some way and there the tractions { ps } will be unknown reactions and not specified loads. It is conventional to remove the unknown reactions by choosing the virtual displacements { δu } to be zero over Sσ .

139

7. ENERGY METHODS

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 WP

Π = U + WP .

(7.13)

7.2.1 Strain energy Consider the strain energy (6.32) U=

1 2

∫ {ε }

T

[ D ] { ε } dV .

V

For a virtual strain { δε } , the virtual increase of strain energy is δU =

1 2

∫ { δε }

T

[ D ] {ε } dV + 1 ∫ { ε }T [ D ] { δε } dV . 2

V

V

Because

({ δε } δU =

∫V { δε }

T

T

[ D ] {ε } )

T

= { ε } T [ D ] { δε } ,

[ D ] {ε } dV = ∫ { δε }T {σ } dV = δWI . V

δ U = δWI .

(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 σ = E ε and ε =

δU =

d δu du EA dx . dx dx

∫ δε σ dV = ∫ δε Eε A d x = ∫

V

l

l

For a beam in bending, substituting ε = − y

δU =

∫ l

⎛ ⎞ ∂ 2δv ∂ 2 v ⎜ 2 ⎟ E y d A ⎟⎟ dx = ∂ x2 ∂ x 2 ⎜⎜ ⎝A ⎠

∫

du , yields dx

∫ l

(7.15)

d2v (5.4), gives d x2

d 2δv d2v E I dx . d x2 d x2

(7.16)

The above expressions can be obtained directly from the strain energies

140


U=

for a bar

∫

∫

EI 2

l

U=

for a beam

2

E A ⎛ du ⎞ ⎜ ⎟ dx , 2 ⎜⎝ d x ⎟⎠

l

(7.17)

2

⎛ d2v ⎞ ⎜ ⎟ x. ⎜ d x2 ⎟ d ⎝ ⎠

(7.18)

7.2.2 External potential energy The work potential of external loads WP is equal to the negative product of external forces by the corresponding displacement

WP = − WE .

(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 F = m g acts in the opposite direction to a vertical displacement h and the potential becomes m g h .

(

An external point load F j has potential energy − F j u j

)

instead of

⎛ 1 ⎞ ⎜ − F j u j ⎟ , because this potential arises from the magnitude of force and its ⎝ 2 ⎠ 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 WP = −

∫ {u } { pv }dV − ∫ {u } { ps }d A − ∑i {ui } { Fi } . T

T

V

T

(7.20)

Sσ

For virtual displacements {δu }

δWP = − δWE .

(7.21)

7.2.3 Total potential energy The total potential energy (7.13) can be written

Π = ∫ { δε }T {σ } dV − ∫ { u }T { pv }dV − V

V

∫ {u } { ps }d A − ∑i { ui } { Fi } . T

T

Sσ

(7.13, a) Its variation is

141

7. ENERGY METHODS

δΠ = δU + δWP = δWI − δWE .

(7.22)

Based on equation (7.6) it follows that

δΠ = 0 ,

(7.22, a)

hence, at equilibrium, the total potential energy has a stationary value. If

δ 2 Π > 0 , 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 1 1 E Ai 2 Δi , U i = Ti Δ i = 2 2 li and the external potential energy is WP = −

∑F u . i i

i

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

142


Π =

EA ( u1 sinθ + u2 cosθ 2l

)2 +

EA u 22 + 2 l cosθ

+

EA (− u1 sinθ + u2 cosθ 2l

)2 − F1 u1 − F2 u2 .

Cancelling the derivatives of Π with respect to each independent variable ∂Π = 0, ∂ u1

∂Π =0, ∂ u2

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

Π =

1 2

∫ E I (v′′) d x − ∫ p v dx + M 2

l

0 v′0

− M l v′l − T0 v0 + Tl vl .

l

At equilibrium, Π is stationary, δΠ = 0 or

∫ E I v′′ δ(v′′)d x − ∫ p δv dx + M l

0

δ v0′ − M l δ v′l − T0 δ v0 + Tl δ vl = 0 .

l

Integrating by parts the first term gives

(a)

143

7. ENERGY METHODS

∫

∫

E I v′′ δ(v′′) d x =

l

E I v′′

l

l

= E I v′′ δ v′ 0 −

d ⎛ dv ⎞ ⎜ δ ⎟ dx = d x ⎜⎝ d x ⎟⎠

d

∫ E I v′′ d (δv′) = l

′

∫ d x (E I v′′) dx δv′ = E I v′′ δv′ − ∫ (E I v′′) δv′ dx. l 0

l

l

Integrating by parts the last term gives ′

′ d

l 0

d (E I v′′)′ dx δ v = (E I v′′)′ δ v dx

′

∫ (E I v′′) δv′ dx = ∫ (E I v′′) d x (δv)dx = ∫ (E I v′′) d (δv)= l

= (E I v′′)′ δ v

l

−

∫ l

l

l 0

−

″

∫ (E I v′′) δv dx. l

Equation (a ) becomes l E I v′′ δ v′ 0 − (E I v′′)′ δ v

l 0

+

″

∫ (E I v′′) δv dx − ∫ p δv dx − M δv′ l

or

l 0

+ T δv

l 0

=0

l

″ ′ ∫ ((E I v′′) − p ) δv dx + ( E I v′′ - M ) δv′ − (( E I v′′ ) − T )δv l 0

l

l 0

=0.

The coefficient of δ v in the integrand gives the equation of equilibrium

(E I v′′)″ = p (x ) . As δ v is arbitrary, the other terms deliver the equilibrium conditions at the beam ends

( E I v′′)0′ = T0 ,

( E I v′′) 0 = M 0 ,

or δ v′0 = 0 ,

( E I v′′) l = M l ,

or δ v′l = 0 , and

or δ v0 = 0 ,

( E I v′′)l′ = Tl ,

or δ vl = 0 ,

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 v (x ) is approximated by a finite series

144


v (x ) ≅

n

∑ a ϕ (x ) j

j

(7.23)

j =1

where a j are undetermined constants called generalized coordinates, and ϕ j ( x ) 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 a j , whose values are determined from the stationarity conditions δΠ =

∂Π

∑ ∂a j

j

δa j = 0 .

Because δa j are arbitrary, ∂Π =0, ∂aj

( j = 1,..., n ) ,

(7.24)

which is a linear algebraic set of equations in the constants a j . 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: E = 210 MPa , I = 1600 mm4 , l = 3 m , F = 100 N and q = 200 N m . Solution. The total potential energy is

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7. ENERGY METHODS

1 Π = EI 2

l

∫ 0

2

l

⎛ d2v ⎞ ⎜ ⎟ d x − q v (x ) d x − F v ⎛⎜ l ⎞⎟ . ⎜ d x2 ⎟ ⎝2⎠ ⎝ ⎠ 2l 3

∫

(7.25)

The geometric boundary conditions are

v (0) = 0 ,

⎛ 2l ⎞ v ⎜ ⎟=0, ⎝ 3 ⎠

v′ (0) = 0 ,

v (l ) = 0 .

(7.26)

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

v ( x ) = a1 ϕ1 (x ) + a2 ϕ 2 (x ) ,

(7.27)

where the functions

ϕ1 (x ) =

x 2 (3 x − 2 l )( x − l ) l4

,

ϕ 2 (x ) =

x 3 (3 x − 2 l )( x − l ) l5

,

(7.28)

satisfy all geometrical boundary conditions (7.26).

Fig. 7.6 Substituting (7.27) into (7.25) we obtain the functional 1 Π = EI 2

l

l

∫ ( a1 ϕ1′′ + a2 ϕ2′′ ) dx − ∫ q ( a1 ϕ1 + a2 ϕ2 ) dx − . 2

0

2l 3

−F

[ a1 ϕ1(l 2) + a2 ϕ2 (l 2) ].

Requiring Π to be stationary with respect to a1 and a 2 , leads to two equations relating the generalized coordinates ∂Π = EI ∂ a1

l

l

0

2l 3

∫ ( a1 ϕ1′′ + a2 ϕ2′′ ) ϕ1′′ d x − ∫ q ϕ1 d x − F ϕ1(l 2) = 0 ,

146


∂Π = EI ∂ a2

l

l

∫ ( a ϕ ′′ + a ϕ ′′ ) ϕ ′′ d x − ∫ q ϕ 1 1

2

2

2

0

2

d x − F ϕ 2 (l 2 ) = 0 .

2l 3

Because Π is of second order in ϕ and ϕ ′ , the equations are linear in

aj . Substituting the functions (7.28), for F = q l 6 , we obtain 56 E I 11 EI a 1 + 10 3 a2 = − ql, 3 5 l 4320 l 10

72 E I 269 EI a + a2 = − ql. 3 1 3 7 l 46656 l

For example, the coefficient of a1 in the first equation is l

EI

∫ ( ϕ ′′ ) 1

2

l

dx = EI

0

1 ∫ l ( 36 x 8

2

− 30 l x + 4 l 2

)

2

dx =

0

56 E I . 5 l3

The solutions are a1 = 0.00207

q l4 , EI

a2 = −0.00257

q l4 . E I

(7.29)

In 2, the displacement is

v (l 2 ) = a1 ϕ1 (l 2) + a2 ϕ 2 (l 2) = =

1 1 q l4 = 2.48 mm. a1 + a2 = 4.9 ⋅10− 5 16 32 E Iy

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

Π =

1 2 ku − Fu . 2

(7.30)

For a small virtual displacement δu , the variation of the total potential energy is δ Π = k u δu − F δ u = ( k u − F ) δ u .

147

7. ENERGY METHODS

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

Fig. 7.7 For δΠ =0 we obtain

k ueq − F = 0 .

(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

Π eq =

1 1 1 F2 . F ueq − F ueq = − F ueq = − 2 2 2 k

(7.32)

The stiffness is 1 F2 1 F2 . = 2 Π eq 2 Π eq

(7.33)

1 2 1 F2 = −Π eq . k ueq = 2 2 k

(7.34)

k =− The strain energy is

U eq =

If F is prescribed and the resulting u has been approximated by a RayleighRitz solution uapp ≠ ueq , equations (7.32) – (7.34) and Fig. 7.7, b indicate that a) Because Π eq is a minimum, the potential energy for an approximate displacement which satisfies the kinematic boundary conditions is greater than the true value Π app > Π eq . In magnitude

Π app < Π eq ,

148


the approximate total potential energy is underestimated. b) The approximate stiffness is overestimated k ↑=

1 F2 . 2 Π eq ↓

(7.35)

c) The approximate displacement is underestimated u ↓=

F . k↑

(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 displcement.

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 { pv } , { ps } , { Fi } , find the displacement field { u } within V and on the surface Sσ (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.

149

7. ENERGY METHODS

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) δΠ =

∫ { δε } {σ }dV − ∫ { δu } { ps }d A = 0 . T

V

T

(7.37)

Sσ

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

δΠ =

∑ δΠ e

e

=

⎛ ⎞ ⎜ ⎟ T T ⎜ { δε } {σ }dV − { δu } { ps }d A ⎟ = 0 . ⎜ Ve ⎟ Sσ e ⎝ ⎠

∑ ∫ e

∫

(7.38)

In the following, only δ Π e 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

150


⎧ a1 ⎫ ⎪a ⎪ ⎪ ⎪ (7.39) u (x ) ≅ a j ϕ j (x ) = ⎣ϕ1 ϕ 2 L ϕ n ⎦ ⎨ 2 ⎬ = ⎣ϕ ⎦ { a } L ⎪ ⎪ j =1 ⎪⎩ an ⎪⎭ where the undetermined constants a j have no direct evident signification. n

∑

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

{ }

⎧ u ⎫ ⎡ ⎣N u ⎦ ⎪ ⎪ ⎢ ⎨ v ⎬=⎢ 0 ⎪w⎪ ⎢ 0 ⎩ ⎭ ⎣

or

0

⎣N v ⎦ 0

{ } ⎫⎪ { }⎬ { }⎪⎭

e ⎤ ⎧ Qu ⎪ 0 ⎥⎥ ⎨ Qve e ⎣N w ⎦ ⎥⎦ ⎪⎩ Qw

0

{ u } = [ N ] {Q e },

(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 Q e 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)

{ε } = [ ∂ ] { u } = [ ∂ ][ N ] {Q e }= [ B ] {Q e }, where [ B ] is the matrix of differentiated shape functions. The strain virtual variation is

{ δε } = [ B ] {δQ e }.

(7.41)

151

7. ENERGY METHODS

7.4.6 Element stiffness matrix and load vector Using the constitutive equation element is δΠ e =

{σ } = [ D ] {ε } ,

the virtual work for an

T T e T e e T ∫ {δ Q } [ B ] [ D ][ B ]{Q }dV − ∫ { δQ } [ N ] { ps }d A = 0

Ve

Ae

or δΠ

e

⎛

⎜ = { δQ } ⎜ ∫ [ B ] ⎜ e T

T

[ D ][ B ] dV

⎝ Ve

{

{Q }− ∫ [ N ] e

Ae

T

⎞

{ ps }d A ⎟⎟ = 0 . ⎟ ⎠

}

As δQ e are arbitrary and non-zero, cancelation of the bracket yields the element equilibrium equation

[ K ] {Q }= {F }, e

e

e

(7.42)

where the element stiffness matrix is

[ K ]= ∫ [ B ] e

T

[ D ][ B ] dV

(7.43)

Ve

and the vector of consistent nodal forces is

{F }= ∫ [ N ] e

T

{ ps }d A .

(7.44)

Ae

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 ~ Q e = T e { Q }, (7.45)

{ }

{ } [ ]

where Q e is the vector of nodal element displacements, {Q } is the vector of the ~ global displacements of the structure and T e is a connectivity matrix, containing ones at the nodal displacements of element nodes and zeros elsewhere.

[ ]

The variation of element displacements is

152


{ δQ }= [ T~ ] { δQ }. e

e

(7.46)

The PVD equation for the entire structure is

∑ {δQ } [ K ] {Q }= ∑ {δQ } {F }, e

T

e

e

T

e

e

e

e

or using (7.45) and (7.46)

{δQ }T ∑ e

[ T~ ] [ K ][ T~ ] {Q } = {δQ } ∑ [ T~ ] {Q }. e T

e

e T

T

e

e

e

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

[ K ] {Q } = { F } ,

(7.47)

where the global stiffness matrix is

[ K ]= ∑

[ T~ ] [ K ][ T~ ] e T

e

e

(7.48)

e

and the global load vector is

{ F } = ∑ [ T~ e ] T {Q e }.

(7.49)

e

Applying the boundary conditions, the condensed equilibrium equations are

[ K ] {Q } = { F } .

(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 {σ } = [ D ] {ε } = [ D ][ B ] Q e = [ D ][ B ] T~ e { Q }

{ }

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.

154


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 ( x1 , y1 ) , ( x2 , y2 ) and ( x3 , y3 ) . 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 u j in the x direction and v j in the y direction. The vector of element nodal displacements is defined as

{ q }= ⎣ u e

1

v1 u2

v2

u3

v3

T ⎦ .

(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 u (x , y ) = a1 + a2 x + a3 y , v ( x , y ) = a4 + a5 x + a6 y

with six arbitrary parameters.

(8.2)

155

8. TWO-DIMENSIONAL MEMBRANES

The strains (6.16) are

εx =

∂u = a2 = const . , ∂x

γ xy =

εy =

∂v = a6 = const . , ∂y

∂u ∂ v + = a3 + a5 = const . , ∂ y ∂x

hence the name “constant strain triangle”.

8.1.3 Nodal approximation of the displacement field In the polynomial approximation (8.2) the constants ai 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

u = ⎣1 x

⎧ a1 ⎪ y ⎦ ⎨ a2 ⎪a ⎩ 3

⎫ ⎪ ⎬ = ⎣1 x ⎪ ⎭

y⎦ { a } .

(8.3)

Evaluating the expression for u at the three nodes gives the nodal displacements in terms of the polynomial coefficients ⎧ u1 ⎪ ⎨ u2 ⎪u ⎩ 3 or

⎫ ⎡ 1 x1 ⎪ ⎢ ⎬ = ⎢ 1 x2 ⎪ ⎢1 x 3 ⎭ ⎣

y1 y2 y3

{ u }= [ A ] { a } , e

⎤ ⎧ a1 ⎥ ⎪a ⎥⎨ 2 ⎥⎦ ⎪⎩ a3

⎫ ⎪ ⎬ , ⎪ ⎭

(8.4)

(8.4, a)

where ⎡ 1 x1 [A ] = ⎢⎢ 1 x2 ⎢⎣ 1 x3 By inversion where

y1 ⎤ y2 ⎥⎥ . y3 ⎥⎦

{ a } = [ A ] −1 { u e }.

(8.5)

(8.6)

156


[A ]

−1

⎡α1 α 2 α 3 ⎤ 1 ⎢ β1 β 2 β 3 ⎥⎥ . = 2A ⎢ ⎢⎣ γ 1 γ 2 γ 3 ⎥⎦

(8.7)

in which

α i = x j yk − xk y j ,

β i = y j − yk ,

γ i = xk − x j

(8.8)

and the subscripts i , j , k permute in a natural order. The area of the triangle A is equal to one half the magnitude of the determinant of [ A ] 1 x1 2 A = 1 x2 1 x3

y1

y2 = (x2 y3 − x3 y2 ) + ( x3 y1 − x1 y3 ) + (x1 y2 − x2 y1 ) . y3

(8.9)

The determinant is positive if nodes 1, 2, 3 are labeled anticlockwise around the element. Substituting (8.6) into (8.3) gives u = ⎣N ⎦

{u } e

(8.10)

where the row vector of shape functions −1 ⎣N ⎦ = ⎣N1 N 2 N 3 ⎦ = ⎣ 1 x y ⎦ [ A ] .

(8.11)

By transposition T −T T ⎣N ⎦ = [A ] ⎣1 x y ⎦ ,

⎧ N1 ⎪ ⎨ N2 ⎪N ⎩ 3

⎫ ⎡ α1 ⎪ 1 ⎢ ⎬= ⎢α 2 ⎪ 2 A ⎢α ⎭ ⎣ 3

β1 γ 1 ⎤ ⎧ 1 ⎫ ⎪ ⎪ β 2 γ 2 ⎥⎥ ⎨ x ⎬ β 3 γ 3 ⎥⎦ ⎪⎩ y ⎪⎭

(8.12)

or Ni =

Similarly

1 ( α i + βi x + γ i y ) . 2A

{ }

v = ⎣N ⎦ v e .

( i = 1, 2, 3 )

(8.12, a)

(8.13)

157


The shape functions (8.12, a) can also be written N1 (x , y ) =

1 [ x2 y3 − x3 y2 + ( y2 − y3 ) x + (x3 − x2 ) y ] , 2A

N 2 (x , y ) =

1 [ x3 y1 − x1 y3 + ( y3 − y1 ) x + (x1 − x3 ) y ] , 2A

N 3 (x , y ) =

1 [ x1 y2 − x2 y1 + ( y1 − y2 ) x + (x2 − x1 ) y ] . 2A

(8.12, b)

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

N i ( xi , yi ) = δ i j ,

( i , j = 1, 2, 3 )

(8.14)

and 3

∑ Ni = 1 .

i =1

Combining (8.10) and (8.14) yields

(8.15)

158


⎧ u ⎫ ⎡ N1 ⎨ ⎬=⎢ ⎩v⎭ ⎣ 0

0

N2

0

N3

N1

0

N2

0

⎧ u1 ⎪v ⎪ 1 0 ⎤ ⎪⎪ u2 ⎨ N 3 ⎥⎦ ⎪ v2 ⎪ u3 ⎪ ⎪⎩ v3

⎫ ⎪ ⎪ ⎪⎪ ⎬ ⎪ ⎪ ⎪ ⎪⎭

(8.16)

or ⎧u⎫ ⎨ ⎬=[N ⎩v⎭

] { q e }.

(8.16, a)

Fig. 8.3 The displacement u (x,y ) =u1 N1 (x,y ) + u2 N 2 (x,y ) + u3 N 3 (x,y ) determines a plane surface passing through u1 , u2 and u3 , as shown in Fig. 8.3.

8.1.4 The matrix [B ] Strains are expressed in terms of displacements as

{ε } e

⎡ ∂ ⎢ ⎢ ∂x = [ ∂ ] { u }= ⎢ 0 ⎢ ⎢ ∂ ⎢ ⎣⎢ ∂ y

⎤ ⎡ ∂ 0 ⎥ ⎢ ⎥ ⎢ ∂x u ⎧ ⎫ ∂ ⎥ ⎢ ⎨ ⎬= 0 ∂y ⎥ ⎩ v ⎭ ⎢ ⎢ ∂ ∂ ⎥ ⎥ ⎢ ∂ x ⎥⎦ ⎣⎢ ∂ y

⎤ 0 ⎥ ⎥ ∂ ⎥ [N ∂y ⎥ ∂ ⎥ ⎥ ∂ x ⎥⎦

] { q e }= [ B ] { q e }.

159


The matrix of the derivatives of shape functions is ⎡ ∂ N1 ⎢ ⎢ ∂x [ B ] = ⎢⎢ 0 ⎢ ∂N 1 ⎢ ⎣⎢ ∂ y

0 ∂ N1 ∂y ∂ N1 ∂x

∂ N2 ∂x

∂ N2 ∂y

⎡ β1 1 ⎢ [ B ]= ⎢ 0 2A ⎢⎣ γ 1 ⎡ y2 − y3 1 ⎢ [ B ]= ⎢ 0 2A ⎢⎣ x3 − x2

The matrix written simply

∂ N2 ∂y ∂ N2 ∂x

0

0 ∂ N3 ∂y

β2

0

β3

0

γ1 0 β1 γ 2

⎤ 0 ⎥ ⎥ ∂ N3 ⎥ , ∂y ⎥ ∂ N3 ⎥ ⎥ ∂ x ⎥⎦

∂ N3 ∂x

0

γ2 0 β2 γ 3

0 ⎤ γ 3 ⎥⎥ , β 3 ⎥⎦

0

y3 − y1

0

y1 − y2

x3 − x2

0

x1 − x3

0

y 2 − y3

x1 − x3

y3 − y1

x2 − x1

[B]

⎤ x2 − x1 ⎥⎥ . y1 − y2 ⎥⎦

0

(8.17)

is constant for a given CST element. Sometimes it is

⎡ y23 1 ⎢ [ B ]= ⎢ 0 2A ⎢⎣ x32

0

y31

0

y12

x32

0

x13

0

y23

x13

y31

x21

0 ⎤ x21 ⎥⎥ y12 ⎥⎦

(8.17, a)

where the notation is obvious.

8.1.5 Element stiffness matrix and load vector The element stiffness matrix (7.43) is

[k ]= ∫ [ B ] e

T

[ D ][ B ] dV = [ B ]T [ D ][ B ] t e

A,

(8.18)

Ve

where t e 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 N 3 is zero while N1 and N 2 are similar to the shape functions in one dimension, satisfying N1 + N 2 = 1 . When the surface load distribution (per unit

160


area) is linear, varying from p1 at node 1 to p2 at node 2 (Fig. 8.4), the nodal forces are f1e =

lte ( 2 p1 + p2 ) , 6

f 2e =

lte ( p1 + 2 p2 6

)

(8.19)

where t e 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

161


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

162


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 ν = 0 . 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 ⎡1 0 0 ⎤ [ D ] = E ⎢⎢ 0 1 0 ⎥⎥ . ⎢⎣ 0 0 1 2 ⎥⎦

(a)

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

x1 = y1 = 0 ,

x2 = l , y2 = 0 ,

x3 = 0 , y3 = l .

The area is A = l 2 2 and the matrix (8.17) is

[B ] 1

⎡ −1 0 1 0 0 0 ⎤ 1⎢ = ⎢ 0 − 1 0 0 0 1 ⎥⎥ . l ⎢⎣ − 1 − 1 0 1 1 0 ⎥⎦

The element stiffness matrix (8.18) is

(b)

163


[ K ]= [k ]= [ B ] 1

[K ] 1

1 T

1

[ D ] [ B1 ] t A ,

⎡ 3 2 1 2 −1 −1 2 −1 2 0 ⎤ ⎢ 32 0 − 1 2 − 1 2 − 1 ⎥⎥ ⎢ 1 0 0 0 ⎥ Et ⎢ = ⎢ ⎥. 12 12 0 ⎥ 2 ⎢ ⎢ SYM 12 0 ⎥ ⎢ ⎥ 1 ⎦⎥ ⎣⎢

(c)

(d)

Element 2 can be obtained by rotating 900 anticlockwise element 1. In general, the transformation matrix for a rotation with an angle θ is

[T ] e

⎡ c ⎢−s ⎢ ⎢ 0 =⎢ ⎢ 0 ⎢ 0 ⎢ ⎣⎢ 0

s

0

0

c 0

0 c

0 s

0 −s c 0 0

0 0

0 0

0⎤ 0 0 ⎥⎥ 0 0⎥ ⎥ 0 0⎥ c s⎥ ⎥ − s c ⎦⎥ 0

(e)

where c = cos θ and s = sin θ . The stiffness matrix of the rotated element is

[ K ] = [T ] [ k ] [T ]. . e T

e

e

e

(f)

For θ = 900 we obtain

[T ] 2

⎡ 0 ⎢ −1 ⎢ ⎢ 0 =⎢ ⎢ 0 ⎢ 0 ⎢ ⎣⎢ 0

1

0

0

0 0

0 0

0 1

0 −1 0 0 0

0 0

0 0

The stiffness matrix of element 2 is

[ K ] = [T ] [ k ] [T ]. , 2

2 T

1

2

0⎤ 0 0 ⎥⎥ 0 0⎥ ⎥ 0 0⎥ 0 1⎥ ⎥ − 1 0 ⎦⎥ 0

(g)

164


[K ] 2

⎡ 3 2 −1 2 −1 2 0 −1 1 2 ⎤ ⎢ 32 1 2 − 1 0 − 1 2 ⎥⎥ ⎢ 12 0 0 −1 2 ⎥ Et ⎢ = ⎢ ⎥. 1 0 0 ⎥ 2 ⎢ ⎢ SYM 1 0 ⎥ ⎢ ⎥ 1 2 ⎦⎥ ⎣⎢

(h)

The same matrix is obtained by substituting the nodal coordinates

x1 = y1 = 0 ,

x2 = 0 , y2 = 1 ,

x3 = −1 , y3 = 0

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 0 0 − 1 1 2 ⎤ ⎧ u1 ⎫ ⎧ 0 ⎫ −1 −1 2 −1 ⎡ 3 ⎢ 0 3 0 −1 2 0 − 2 0 − 1 2 ⎥⎥ ⎪⎪ 0 ⎪⎪ ⎪⎪ V1 ⎪⎪ ⎢ ⎢ −1 0 1 0 0 0 0 0 ⎥ ⎪ u2 ⎪ ⎪ F 2 ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ 12 0 0 0 ⎥ ⎪ 0 ⎪ ⎪ V2 ⎪ Et ⎢ − 1 2 −1 2 0 1 2 ⎨ ⎬=⎨ ⎬. 0 0 12 1 0 0 −1 2 ⎥ ⎪ 0 ⎪ ⎪ H3 ⎪ 2 ⎢ −1 ⎢ ⎥ 0 0 0 2 0 0 ⎥ ⎪ 0 ⎪ ⎪ V3 ⎪ −2 ⎢ 0 ⎪ ⎪ ⎪ ⎪ ⎢ −1 0 0 0 0 0 1 0 ⎥ ⎪ 0 ⎪ ⎪ H 4∗ ⎪ ⎢ ⎥ 0 0 1 2 ⎥⎦ ⎪⎩ 0 ⎪⎭ ⎪⎩ V4 ⎪⎭ −1 2 0 ⎢⎣ 1 2 − 1 2 0 Solving

E t ⎡ 3 − 1⎤ ⎧ u1 ⎫ ⎧ 0 ⎫ ⎨ ⎬=⎨ ⎬ 2 ⎢⎣ − 1 1 ⎥⎦ ⎩ u2 ⎭ ⎩ F 2 ⎭ we obtain the nodal displacements u1 =

F , 2E t

u2 =

3F . 2E t

The reaction forces for half the plate are obtained as

H3 = − F 4 ,

H 4∗ = − F 4 ,

so that for the entire plate H4 = − F 2 ,

H5 = − F 4 .

165


Strains and stresses in elements are

{ ε }= [ B ]{ q } 1

1

1

⎧ u1 ⎪ 0 ⎡ −1 0 1 0 0 0 ⎤ ⎪ ⎪⎪ u 1 = ⎢⎢ 0 − 1 0 0 0 1 ⎥⎥ ⎨ 2 0 l ⎢⎣ − 1 − 1 0 1 1 0 ⎥⎦ ⎪ ⎪ 0 ⎪ ⎩⎪ 0

{ε }

⎧2⎫ ⎡− 1 1⎤ ⎧ u1 ⎫ F ⎪ ⎪ 1⎢ ⎥ = ⎢ 0 0⎥ ⎨ ⎬ = ⎨ 0 ⎬. u 2 ⎭ 2l E t ⎪ ⎪ l ⎩ ⎢⎣− 1 0⎥⎦ ⎩− 1⎭

⎧ σx ⎪ ⎨ σy ⎪τ ⎩ xy

⎫ ⎪ 1 ⎬ =[D ] ε = E ⎪ ⎭1

1

{ }

{ ε }= [ B ] { q } 2

2

2

⎫ ⎪ ⎪ ⎪⎪ ⎬, ⎪ ⎪ ⎪ ⎭⎪

⎡1 0 0 ⎤ ⎧ 2 ⎫ ⎧2⎫ ⎢0 1 0 ⎥ F ⎪ 0 ⎪= F ⎪ 0 ⎪, ⎬ ⎢ ⎥ 2l E t ⎨ ⎬ 2 l t ⎨ ⎪ ⎪ ⎪ ⎪ ⎢⎣ 0 0 1 2 ⎥⎦ ⎩− 1 2 ⎭ ⎩− 1⎭

⎧u1 ⎫ ⎪0⎪ 0 0 0 −1 0 ⎤ ⎪ ⎪ ⎡ 1 ⎧1⎫ F ⎪ ⎪ 1⎢ ⎪⎪ 0 ⎪⎪ ⎥ = ⎢ 0 −1 0 1 0 0 ⎥ ⎨ ⎬ = ⎨ 0 ⎬, 0 ⎪ 2l E t ⎪ ⎪ l ⎪ ⎢⎣ − 1 1 0 0 1 − 1 ⎥⎦ ⎩− 1⎭ ⎪0⎪ ⎪ ⎪ ⎪⎩ 0 ⎪⎭

⎧ σx ⎫ ⎡1 0 0 ⎤ ⎧ 1 ⎫ ⎧1⎫ ⎪ ⎪ F ⎪ ⎪ F ⎪ ⎪ ⎢ ⎥ 2 ⎨ σy ⎬ =[D ] ε = E ⎢0 1 0 ⎥ ⎨ 0 ⎬. ⎨ 0 ⎬= 2l E t ⎪ ⎪ 2 l t ⎪ ⎪ ⎪ τ ⎪ ⎢⎣ 0 0 1 2 ⎥⎦ ⎩− 1 2 ⎭ ⎩− 1⎭ ⎩ xy ⎭2

{ }

Example 8.2 A thin triangular plate is fixed along the edge 5-4 and loaded by forces F1 = −1 and F2 = −2 along the upper edge in its plane (Fig. E8.2). Assume the cantilever to have unit thickness t = 1 and be in plane stress, with Young’s modulus E = 1 and Poisson’s ratio ν = 0.3 . 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.

166


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).

167


The element stiffness matrices are the following:

The condensed global stiffness matrix is

The condensed finite element equations can be written 0 0 − 0.317 0.165 − 0.165 ⎤ ⎧ u1 ⎡ 0.317 ⎢ 0 0.111 0.192 − 0.111 − 0.192 0 ⎥⎥ ⎪⎪ v1 ⎢ ⎢ − 0.317 0.192 1.301 0 0 ⎥ ⎪⎪ u2 − 0.666 ⎢ ⎥⎨ 0 2.125 0 − 1.903 ⎥ ⎪ v2 ⎢ 0.165 − 0.111 ⎢ 0 − 0.192 − 0.666 0 1.301 0 ⎥ ⎪ u3 ⎢ ⎥⎪ 0 0 − 1.903 0 2.125 ⎦⎥ ⎪⎩ v3 ⎣⎢ − 0.165

⎫ ⎧ 0 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ −1 ⎪ ⎪⎪ ⎪⎪ 0 ⎪⎪ ⎬=⎨ ⎬. ⎪ ⎪−2⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪⎭ ⎪⎩ 0 ⎪⎭

The solution to these equations gives

u1 = 7.712 , u2 = 6.542 , u3 = −2.686 , v1 = −40.823 , v2 = −15.835 , v3 = −13.582 .

168


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 σ x , 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 a = 10 mm , is subjected to loads that produce uniform tensile stresses σ 0 = 5 MPa at its ends (Fig. E8.3, a). The plate has length l = 60 mm , width b = 40 mm , thickness t = 5 mm , E = 210 GPa and ν = 0.3 . 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 σ x 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).


169

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.

170


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 19.6 MPa and occurs in element 1. Stress values in elements near the midsection are given in Table E8.3.

171


Table E8.3 Element

σx

σy

τ xy

σ eq

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 σ x , averaged at points a to f, have the following values: Point

a

b

c

d

e

f

σ x , MPa

15.78

11.39

9.69

7.85

6.68

4.89

The theoretical distribution of stresses σ x in the midsection of a plate containing a small circular hole and subjected to uniaxial tension is approximated by ⎛ a2 3 a4 ⎞ σ x = σ 0 ⎜⎜ 1 + 2 + 4 ⎟⎟ , 2r 2r ⎠ ⎝

where r is the distance from the x axis (Fig. E8.3, a). The maximum stress occurs at the periphery of the hole (r = a ) and is

σ xmax = 3σ 0 = 15 MPa . The averaged value of elements 1 and 2, in point a, is σ xa = 15.78 MPa , 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 150 mm , width of the reduced portion 40 mm , width of the extended portion

80 mm , thickness 5 mm , fillet radius 20 mm , E = 2 ⋅ 105 MPa and ν = 0.25 . The normal stress at a point far to the right of the fillet has a uniformly distributed value

172


of 440 MPa . 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 25 mm 2 , subjected to a normal stress of 440 N mm 2 . The nodal loads for each element are 440 × 25 / 2 = 5500 N . The resulting five nodal forces, for the quarter plate, have magnitudes of 5500 N at the upper and lower node, and 11000 N 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 604.7 MPa .

173


Fig. E8.4, c Values of the principal stress σ 1 around the fillet are presented in the upper part. The largest principal stress σ 1 is 622.7 MPa . The stress concentration factor relative to the value 440 MPa is 1.415. This value is reasonably close to the theoretical true value of 1.42 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:

x60 = 92.35 , y60 = 21.52 , x64 = 95 , y64 = 18 , x65 = 95.5 , y65 = 20.5 ; u65

u60 = 0.11224 , v60 = −3.97e − 3 , = 0.12247 , v65 = −4.92e − 3 .

u64 = 0.12367 ,

v64 = −3.63e − 3 ,

E = 2 ⋅ 105 MPa and ν = 0.25 . Calculate the element stresses.

Solution. The triangle area is A96 = 4.1925 mm 2 . The matrix [ B ] is (8.17) 0 − 0.1216 0 0.4198 0 ⎤ ⎡ − 0.2982 ⎢ [ B ]= ⎢ 0 0.0596 0 − 0.3757 0 0.3160 ⎥⎥ . ⎢⎣ 0.0596 − 0.2982 − 0.3757 − 0.1216 0.3160 0.4198 ⎥⎦

174


The matrix [ D ] is (6.24) ⎡ 2.1333 0.5333 0 ⎤ [ D ] = 10 ⎢⎢ 0.5333 2.1333 0 ⎥⎥ . ⎢⎣ 0 0 0.8 ⎥⎦ 5

The vector of nodal displacements (8.1) is

{ q } = ⎣ 0.11224

− 3.97e − 3 0.12367 − 3.63e − 3 0.12247 − 4.92e − 3 ⎦ T .

Element stresses are given by

{σ } = [ D ][ B ] { q }, {σ } = ⎣σ x

σ y τ xy ⎦ T = ⎣596.72 63.58 − 120.08⎦ T .

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


175

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

176


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

(

)

N i x j , y j = δ ij .

(8.20)

They can be generated using the equations of the sides. For example, N1 is identically zero on lines x = a and y = b , and has a unit value at the node with the same index N1 ( x1 , y1 ) = 1 , hence it must be of the form

N1 ( x , y ) = c1 ( a − x ) ( b − y ) . 1 , hence ab 1 ( a − x ) ( b − y ) = ⎛⎜ 1 − x ⎞⎟ ⎛⎜ 1 − y ⎞⎟ . N1 ( x , y ) = ab a ⎠⎝ b ⎠ ⎝

From condition N1 ( x1 , y1 ) = N1 ( 0,0 ) = 1 we get c1 =

Likewise, the remaining three shape functions are N2 ( x, y ) =

x a

y ⎞ ⎛ ⎜ 1− ⎟ , b ⎝ ⎠

N3 ( x, y ) =

x y , a b

N4 ( x, y ) =

y b

x⎞ ⎛ ⎜ 1− ⎟ . a ⎝ ⎠

It is convenient to transform to dimensionless central coordinates

177


r=

2x −1, a

s=

2y −1. b

(8.21)

The equations of element sides become

1 ± r = 0 ,…. 1 ± s = 0 .

(8.22)

The required shape functions are 1 (1 − r ) (1 − s ) , N2 = 1 (1 + r ) (1 − s ) , 4 4 1 1 N3 = ( 1 + r ) ( 1 + s ) , N 4 = ( 1 − r ) ( 1 + s ) . 4 4 N1 =

(8.23)

The displacements inside the element can be expressed in terms of the nodal displacements as ⎧ u ⎫ ⎡ N1 ⎨ ⎬=⎢ ⎩v⎭ ⎣ 0

where

{q }

0 N1

e T

N2 0

= ⎣u1

0 N2

v1 u2

0 N3

N3 0

v2

u3

N4 0

v3

0 ⎤ e q , N 4 ⎥⎦

{ }

(8.24)

v4 ⎦ .

(8.25)

u4

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 x y is present but x 2 and y 2 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

u (x , y ) = a 1 + a2 x + a3 y + a4 x y , v ( x , y ) = b1 + b 2 x + b 3 y + b4 x y .

(8.26)

The direct strain ε x = a2 + a4 y is constant in the x direction whereas the shear strain γ xy = a3 + a4 x + b2 + b4 y 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.

178


The element stiffness matrix (7.39) is

[k ]= ∫ [ B ] e

T

[ D ][ B ] dV ,

Ve

where

[ B ] = [ ∂ ][ N ] .

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 γ xy = d b and only the centre has no shear deformation. The direct strains are ε x = d a on the upper and lower surface. The ratio spurious shears a = = aspect ratio . bending strains b This explains the poor results obtained with high aspect ratio elements 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.

(≥ 3)

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.

179


The assumed displacement field is cubic

u (x , y ) = a 1 + a2 x + a3 y + a4 x 2 + a5 x y + a6 y 2 + a7 x 2 y + a8 x y 2 , v ( x , y ) = b1 + b2 x + b3 y + b4 x 2 + b5 x y + b6 y 2 + b7 x 2 y + b8 x y 2 ,

(8.27)

so that the strains are quadratic. Note that choosing the last terms in x 3 and y 3 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 u = N1 u1 + N 2 u2 + N 3 u3 + ... + N8 u8 , v = N1 v1 + N 2 v2 + N 3 v3 + ... + N8 v8 ,

(8.28)

where the shape functions N i 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 1 ( 1 − r ) ( 1 − s ) (1 + r + s ) , N 2 = − 1 ( 1 + r ) ( 1 − s ) (1 − r + s ) , 4 4 1 1 N 3 = − ( 1 + r ) ( 1 + s ) ( 1 − r − s ), N 4 = − ( 1 − r ) ( 1 + s ) (1 + r − s ) , 4 4 (8.29) 1 1 N5 = ( 1 − r ) ( 1 + r ) ( 1 − s ) , N6 = ( 1 + r ) ( 1 − s ) ( 1 + s ) , 2 2 1 1 N 7 = ( 1 − r ) ( 1 + r ) (1 + s ) , N 8 = ( 1 − r ) ( 1 − s ) ( 1 + s ). 2 4 N1 = −

180


The interpolation function N1 should be zero at nodes 2, 3,..., 8 and have a value of unity at node 1. Equivalently, N1 should vanish on the sides defined by the equations of lines passing through nodes 2 to 8: 1 − s = 0 , 1 − r = 0 , 1 + r + s = 0 . Therefore N1 is of the form

N1 = c1 ( 1 − r ) ( 1 − s ) (1 + r + s ) , where c1 is a constant which is determined so as to yield N1 (− 1,−1) = 1 , i.e. c1 = −1 4 , which gives the expression of N1 in (8.29). The displacements inside the element can be expressed in terms of the nodal displacements as ⎧ u ⎫ ⎡ N1 ⎨ ⎬=⎢ ⎩v⎭ ⎣ 0 where

0 N1

N2 0

{ q }= ⎣u e

1

v1

0 L L N8 N2 L L 0

u2

0 ⎤ e q , N 8 ⎥⎦

{ }

(8.30)

v8 ⎦ T .

(8.31)

v 2 L L u8

Having generated the shape functions [ N ] , the functions [ B ] = [ ∂ ][ N

[ ]

follow and the element stiffness matrix k e 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 h1 , h2 , h3 from the three sides (Fig. 8.10, a), and nondimensionalized, by division to the triangle heights, as h1 h2 h3 , ζ2 = , ζ3 = , ζ1 = H1 H2 H3

181


or ζ1 =

A1 A

, ζ2 =

A2 A

, ζ3 =

A3 A

.

(8.32)

As A1 + A2 + A3 = A (Fig. 8.10, b), one obtains

ζ1 + ζ 2 + ζ 3 = 1 ,

(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 x = ζ 1 x1 + ζ 2 x 2 + ζ 3 x3 ,

(8.34)

y = ζ 1 y1 + ζ 2 y2 + ζ 3 y3 , or in matrix form ⎧1 ⎫ ⎡1 ⎪ ⎪ ⎢ ⎨ x ⎬ = ⎢ x1 ⎪ y ⎪ ⎢y ⎩ ⎭ ⎣ 1

1 x2 y2

1 ⎤ ⎧ ζ1 ⎪ x3 ⎥⎥ ⎨ ζ 2 y3 ⎥⎦ ⎪⎩ ζ 3

⎫ ⎪ ⎬. ⎪ ⎭

By inversion, we obtain equation (8.12) ⎧ ζ1 ⎪ ⎨ ζ2 ⎪ζ ⎩ 3 or

⎫ ⎡α1 ⎪ 1 ⎢ ⎬= ⎢α 2 ⎪ 2 A ⎢α ⎭ ⎣ 3

β1 γ 1 ⎤ ⎧ 1 ⎫ ⎪ ⎪ β 2 γ 2 ⎥⎥ ⎨ x ⎬ β 3 γ 3 ⎥⎦ ⎪⎩ y ⎪⎭

182


ζi =

1 ( α i + βi x + γ i y ) . 2A

( i = 1, 2, 3 )

(8.35)

where A is given by (8.9) and the expressions (8.8) are

α i = x j yk − xk y j , β i = y j − yk , γ i = xk − x j .

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

1 ⎛ N1 = 2 ⎜ ζ1 − 2 ⎝ N 4 = 4 ζ1 ζ 2 ,

1⎞ 1⎞ ⎞ ⎛ ⎛ ⎟ ζ1 , N 2 = 2 ⎜ ζ 2 − ⎟ ζ 2 , N 3 = 2 ⎜ ζ 3 − ⎟ ζ 3 , 2⎠ 2⎠ ⎠ ⎝ ⎝ N 5 = 4 ζ 2 ζ 3 , N 6 = 4 ζ 3 ζ1 .

(8.36)

The displacements in terms of the nodal displacements are ⎧ u ⎫ ⎡ N1 ⎨ ⎬=⎢ ⎩v⎭ ⎣ 0

where

N2 0

0 N1

{ q }= ⎣u e

1

0 L L N6 N2 L L 0

v1 u2

0 ⎤ e q , N 6 ⎥⎦

{ }

(8.37)

v6 ⎦ T .

(8.38)

v 2 L L u6

183


To form [ B ] = [ ∂ ][ N ] we need to find the derivatives of shape functions with respect to the physical coordinates, so we must transform from x , y to ζ1 , ζ 2 , ζ 3 . Using (8.33) we can eliminate ζ1 in (8.34) obtaining x = (x 2 − x1 ) ζ 2 + (x3 − x1 ) ζ 3 + x1 ,

(8.39)

y = ( y2 − y1 ) ζ 2 + ( y3 − y1 ) ζ 3 + y1 .

Using the chain rule for partial derivatives ⎧ ∂ ⎪⎪ ∂ ζ 2 ⎨ ∂ ⎪ ⎪⎩ ∂ ζ 3

⎫ ⎡ ∂x ⎪⎪ ⎢ ∂ ζ 2 ⎬ = ⎢ ∂x ⎢ ⎪ ⎪⎭ ⎢⎣ ∂ ζ 3

∂y ∂ ζ2 ∂y ∂ ζ3

⎤⎧ ∂ ⎥ ⎪⎪ ∂ x ⎥⎨ ⎥⎪ ∂ ⎥⎦ ⎪⎩ ∂ y

⎫ ⎪⎪ ⎬=[J ⎪ ⎪⎭

⎧ ∂ ⎪ ] ⎪⎨ ∂∂x ⎪ ⎪⎩ ∂ y

⎫ ⎪⎪ ⎬, ⎪ ⎪⎭

(8.40)

where [ J ] is the Jacobian of the transformation. The inverse relationship is ⎧ ∂ ⎪⎪ ∂ x ⎨ ∂ ⎪ ⎪⎩ ∂ y

⎫ ⎪⎪ ⎬=[J ⎪ ⎪⎭

⎧ ∂ ⎪ ] −1 ⎪⎨ ∂∂ζ 2 ⎪ ⎪⎩ ∂ ζ 3

⎫ ⎪⎪ 1 ⎡ β 2 ⎬= ⎢ ⎪ 2A ⎣ γ 2 ⎪⎭

⎧ ∂ β 3 ⎤ ⎪⎪ ∂ ζ 2 ⎨ γ 3 ⎥⎦ ⎪ ∂ ⎪⎩ ∂ ζ 3

⎫ ⎪⎪ ⎬, ⎪ ⎪⎭

(8.41)

where A is the triangle area (8.9) and β i , γ i are defined by (8.8). The element stiffness matrix is

[k ]= ∫ [ B ] e

T

[ D ][ B ] t e d A ,

(8.42)

A

where

d A = l 3 dζ 2 l 2 dζ 3 sinθ = 2 A dζ 2 dζ 3 . 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

∫ζ

a b i ζj

dA = 2 A

A

a! b! . (2 + a + b ) !

(8.43)

The results for first- and second-degree terms are

∫ζ A

i

dA =

A , 3

∫ζ ζ i

A

j

dA =

A , 12

∫ζ A

2 i

dA =

A . 6

(8.44)

184


The LST element is based on expanding the displacements in a complete second-degree polynomial in ζ 2 and ζ 3

u = a1 + a2 ζ 2 + a3 ζ 3 + a4 ζ 22 + a5 ζ 2 ζ 3 + a6 ζ 32 , v = b1 + b2 ζ 2 + b3 ζ 3 + b4 ζ 22 + b5 ζ 2 ζ 3 + b6 ζ 32 .

(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

ζ 2 = r,

ζ 3 = s,

ζ1 = 1 − ζ 2 − ζ 3 = 1 − r − s

(8.46)

we can introduce oblique triangular coordinates (Fig. 8.12). The physical coordinates of a point within the triangle are x = ( 1 − r − s ) x1 + r x 2 + s x3 ,

y = ( 1 − r − s ) y1 + r y2 + s y3 .

(8.47)

Fig. 8.12 The coefficients of the nodal coordinates are genuine geometric interpolation functions

N1 = 1 − r − s ,

N2 = r ,

N3 = s ,

(8.48)

185


The nodal displacements for the 3-node triangle can also be written u = ( 1 − r − s ) u1 + r u2 + s u3 , v = ( 1 − r − s ) v1 + r v2 + s v3 ,

(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 displacement 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 9

u = ∑ N i ui + N c u c , i =1

where

9

v = ∑ N i vi + N c v c , i =1

(8.50)

186


1 1 ζ1 ( 3 ζ1 − 1 ) ( 3 ζ1 − 2 ) , N 2 = ζ 2 ( 3 ζ 2 − 1 ) ( 3 ζ 2 − 2 ) , 2 2 1 = ζ3 ( 3 ζ3 − 1 ) ( 3 ζ3 − 2 ) , 2 9 9 = ζ1 ζ 2 ( 3 ζ1 − 1 ) , N 5 = ζ1 ζ 2 ( 3 ζ 2 − 1 ), 2 2 9 9 = ζ 2 ζ 3 ( 3 ζ 2 − 1 ), N 7 = ζ 2 ζ 3 ( 3 ζ 3 − 1 ), 2 2 9 9 = ζ 3 ζ1 ( 3 ζ 3 − 1 ) , N 9 = ζ 3 ζ 1 ( 3 ζ 1 − 1 ) , 2 2 = 27 ζ1 ζ 2 ζ 3 .

N1 = N3 N4 N6 N8 Nc

(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 ∂ u ∂ u ∂v ∂v derivatives , , as nodal quantities. At each corner, there are six , ∂x ∂ y ∂x ∂ y 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 (uc , vc ) as the remaining parameters. The nodal expansion for u has the form

u = N1 u1 + N 2 u′x1 + N 3 u′y1 + N 4 u2 + N 5 u′x 2 + ... + + N 9 u′y 3 + N cuc .

(8.52)

In (8.52) the interpolation polynomials, expressed in terms of triangular coordinates, are

N1 = ζ12 ( ζ1 + 3 ζ 2 + 3 ζ 3 ) − 7 ζ1 ζ 2 ζ 3 , N 2 = ζ12 ( γ 3 ζ 2 − γ 2 ζ 3 ) + ( γ 2 − γ 3 ) ζ1 ζ 2 ζ 3 , N 3 = ζ12 ( β 2 ζ 3 − β 3 ζ 2 ) + ( β 3 − β 2 ) ζ1 ζ 2 ζ 3 , N 4 = ζ 22 ( ζ 2 + 3 ζ 3 + 3 ζ1 ) − 7 ζ1 ζ 2 ζ 3 , N 5 = ζ 22 ( γ 1 ζ 3 − γ 3 ζ1 ) + ( γ 3 − γ 1 ) ζ1 ζ 2 ζ 3 , N 6 = ζ 22 ( β 3 ζ1 − β1 ζ 3 ) + ( β1 − β 3 ) ζ1 ζ 2 ζ 3 , N 7 = ζ 32 ( ζ 3 + 3 ζ1 + 3 ζ 2 ) − 7 ζ1 ζ 2 ζ 3 , N8 = ζ 32 ( γ 2 ζ1 − γ 1 ζ 2 ) + ( γ 1 − γ 2 ) ζ1 ζ 2 ζ 3 , N 9 = ζ 32 ( β1 ζ 2 − β 2 ζ1 ) + ( β 2 − β1 ) ζ1 ζ 2 ζ 3 , N c = 27 ζ1 ζ 2 ζ 3 ,

(8.53)

187


where γ i , β i 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.

pv x

Let write the equilibrium equations in terms of displacements for the case = pv y = 0 . Substituting the stress-strain relations (6.24) into the equilibrium

equations (6.7) we obtain

(

)

∂ 1− ν ∂ ε x +ν ε y + γ xy = 0, ∂x 2 ∂y

(

)

1− ν ∂ ∂ γ xy + ε y + ν ε x = 0. 2 ∂x ∂y

(8.54)

Substituting the strain-displacement relations (6.16) in (8.54) gives ∂ 2u ∂ 2u 1 + ν ⎛⎜ ∂ 2u ∂ 2 v ⎞⎟ + = − , 2 ⎜⎝ ∂ y 2 ∂ x ∂y ⎟⎠ ∂ x2 ∂ y 2 ∂ 2 v ∂ 2 v 1 + ν ⎛⎜ ∂ 2 v ∂ 2 u ⎞⎟ + = − . 2 ⎜⎝ ∂ x 2 ∂ x ∂y ⎟⎠ ∂ x2 ∂ y 2

(8.55)

Equations (8.55) are not satisfied by the assumed displacement field (8.26) u ( x , y ) = a1 + a2 x + a3 y + a4 x y , v ( x , y ) = b1 + b2 x + b3 y + b4 x y .

of the rectangular element. For example,

(8.56)

188


∂u ∂u ∂ 2v ∂ 2u ∂ 2u = a2 + a4 y , = a3 + a4 x , , and = b4 = = 0 ∂x ∂y ∂ x ∂y ∂ x2 ∂ y2 which is not zero, so the first equation (8.55) is not satisfied. The rectangle would satisfy equilibrium if a4 = b4 = 0 , but this is the case only in a field of constant strain. But as already shown, the direct strain ε x is linear in the y direction whereas the shear strain γ xy 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 (or y ) 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:


189

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, noncompatible 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 n − 1 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.

190


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 u=v= x+ y for which the strains are

{ε } = ⎣1

1 2⎦ T .

The resulting nodal displacements are

u1 = 0 , u2 = 1 , u3 = 4 , u4 = 1 , v1 = 0 , v2 = 1 , v3 = 4 , v4 = 1 . 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

[K ] ⎧⎨

u5 ⎫ ⎬ = {F } , ⎩v5 ⎭

where [K ] depends on the material properties and {F } depends on both the material properties and the prescribed nodal displacements. Its solution must be

u5 = v5 = 1.25 and the strains at any point must be {ε } = ⎣1 1 2⎦ T .

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 ( xi , yi ) . Each node has two degrees of freedom. The displacement components of a local node i are denoted as ui in the x direction and vi in the y direction. The vector of element nodal displacements is defined as

{ q }= ⎣ u e

1

v1 u2

v2

u3

v3

u4

v4 ⎦ T .

(9.1)

192


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 { r , s } 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 r ∈ [ − 1, 1 ] , s ∈ [ − 1, 1 ] . The transformation between the natural coordinates { r , s } in the reference plane and the cartesian coordinates { x , y } is called the isoparametric mapping.

a

b Fig. 9.2

193


Each quadrilateral ‘child’ in the { x , y } is generated by the ‘parent’ or ‘reference’ element from the { r , s } 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.

9.1.2 Shape functions The interpolation functions (8.23) developed for the rectangle are directly applicable 1 1 N1 = ( 1 − r ) ( 1 − s ) , N 2 = ( 1 + r ) ( 1 − s ) , 4 4 (9.2) 1 1 N3 = ( 1 + r ) ( 1 + s ) , N 4 = ( 1 − r ) ( 1 + s ) . 4 4 The following compact representation is useful for the implementation in a computer program Ni ( r , s ) =

1 ( 1 + r ri )( 1 + s si ) , 4

(9.3)

where ( ri , si ) are the coordinates of node i . The coordinates of a point within the element are expressed in terms of the nodal coordinates as 4

x = N1 x1 + N 2 x 2 + N 3 x3 + N 4 x 4 = ∑ N i x i , i =1 4

(9.4)

y = N1 y1 + N 2 y2 + N 3 y3 + N 4 y4 = ∑ N i y i . i =1

In matrix form

{ }

x = ⎣N ⎦ x e = ⎣N1

{ }

y = ⎣N ⎦ y e = ⎣N1

N2

N2

N3

⎧ ⎪ ⎪ N4 ⎦ ⎨ ⎪ ⎪⎩

x1 ⎫ x2 ⎪⎪ ⎬, x3 ⎪ x4 ⎪⎭

(9.4, a)

N3

⎧ ⎪ ⎪ N4 ⎦ ⎨ ⎪ ⎪⎩

y1 y2 y3 y4

⎫ ⎪ ⎪ ⎬. ⎪ ⎪⎭

(9.4, b)

194


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 ( r , s ) , then 4

u = N1 u 1 + N 2 u 2 + N 3 u 3 + N 4 u 4 = ∑ N i u i , i =1 4

(9.5)

v = N1 v1 + N 2 v2 + N 3 v3 + N 4 v4 = ∑ N i v i , i =1

which can be written in matrix form

{ u } = [ N ] { q e },

(9.6)

where

[ N ] = ⎡⎢

N1

0

N2

0

N3

0

N4

⎣ 0

N1

0

N2

0

N3

0

0 ⎤ . N 4 ⎥⎦

(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

u ( r , s ) = a1 + a2 r + a3 s + a4 r s ,

(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), r = 1 and u u

r =1

r =1

= a1 + a2 + ( a3 + a4 ) s , =

1− s 1+ s u2 + u3 . 2 2

The four-node quadrilateral is termed a “linear” (sometimes bi-linear) element because its sides remain straight during deformation. This ensures interelement compatibility, i.e. there are no openings, overlaps or discontinuities

195


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

u ( x , y ) = b1 + b2 x + b3 y + b4 x y ,

(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 y = m x + c , where m is the slope of 2-3, so that u ( x)

2 −3

= b1 + cb3 + ( b2 + m b3 + c b4 ) x + m b4 x 2 .

Along 2-3, u ( x ) varies quadratically and cannot be uniquely defined as a function of u 2 and u 3 . 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 { x , y } coordinates in terms of its derivatives in { r , s } coordinates. This is done considering the function f = f ( x , y ) to be an implicit function of r and s as f = f [ x (r , s ) , y (r , s ) ] . Transformation of differential operators

Using the chain rule of differentiation, we have ⎧ ∂ ⎪⎪ ∂ r ⎨ ∂ ⎪ ⎪⎩ ∂ s

⎫ ⎡ ∂x ⎪⎪ ⎢ ∂ r ⎬ = ⎢ ∂x ⎪ ⎢ ⎪⎭ ⎢⎣ ∂ s

∂y ∂r ∂y ∂s

⎤⎧ ∂ ⎥ ⎪⎪ ∂ x ⎥⎨ ⎥⎪ ∂ ⎥⎦ ⎪⎩ ∂ y

J11

J12 ⎤ J 22 ⎥⎦

⎫ ⎪⎪ ⎬=[J ⎪ ⎪⎭

⎧ ∂ ⎪ ] ⎪⎨ ∂∂x ⎪ ⎪⎩ ∂ y

⎫ ⎪⎪ ⎬, ⎪ ⎪⎭

(9.10)

where

[ J ] = ⎡⎢

⎣ J 21

is the Jacobian matrix. Using (9.10) and (9.4), it can also be expressed as

(9.11)

196


⎧ ∂ ⎪ [ J ] = ⎪⎨ ∂∂r ⎪ ⎩⎪ ∂ s

⎫ ⎪⎪ ⎬ ⎣x ⎪ ⎭⎪

⎧ ∂ ⎪⎪ ∂ r y⎦ = ⎨ ∂ ⎪ ⎩⎪ ∂ s

⎫ ⎪⎪ ⎬ ⎪ ⎭⎪

⎣ ⎣N ⎦ { x e }

⎫ ⎧ ∂ ⎪⎪ ∂ r ⎣N ⎦ ⎪⎪ =⎨ ⎬ ∂ ⎪ N ⎣ ⎦ ⎪⎪ ⎪⎩ ∂ s ⎭

⎣N ⎦ { y

e

}⎦ = (9.12)

⎣{ x e } { y e }⎦ .

For the four-node quadrilateral element − (1 − s ) (1 − s ) 4 ⎣ − (1 − r ) − (1 + r )

[ J ] = 1 ⎡⎢

(1 + s ) (1 + r )

⎡ − x1 + x 2 + x3 − x 4 + ⎢ 1 ⎢ + s (x1 − x 2 + x3 − x 4 ) [ J ]= ⎢ 4 − x1 − x 2 + x3 + x 4 + ⎢ ⎢⎣ + r (x1 − x 2 + x3 − x 4 )

⎡ x1 − (1 + s ) ⎤ ⎢⎢ x 2 (1 − r ) ⎥⎦ ⎢ x3 ⎢ ⎣ x4

y1 y2 y3 y4

⎤ ⎥ ⎥, ⎥ ⎥ ⎦

− y1 + y2 + y3 − y4 + ⎤ ⎥ + s ( y1 − y2 + y3 − y4 ) ⎥ ⎥. − y1 − y2 + y3 + y4 + ⎥ + r ( y1 − y2 + y3 − y4 ) ⎥⎦

(9.13)

(9.13, a)

Analogously, the inverse relationship is ⎧ ∂ ⎪⎪ ∂ x ⎨ ∂ ⎪ ⎪⎩ ∂ y

where

⎫ ⎡ ⎪⎪ ⎢ ⎬=⎢ ⎪ ⎢ ⎪⎭ ⎢⎣

∂r ∂x ∂r ∂y

∂s ∂x ∂s ∂y

⎤⎧ ∂ ⎥ ⎪⎪ ∂ r ⎥⎨ ⎥⎪ ∂ ⎥⎦ ⎪⎩ ∂ s

⎫ ⎧ ∂ ⎫ ⎪⎪ ⎪⎪ ∂ r ⎪⎪ ⎬ = [ j ]⎨ ∂ ⎬ , ⎪ ⎪ ⎪ ⎪⎭ ⎪⎩ ∂ s ⎪⎭

(9.14)

[ j ] = [ J ] −1 is the inverse of the Jacobian matrix

[ j ] = ⎡⎢

j11 ⎣ j21

j12 ⎤ 1 ⎡ J 22 − J12 ⎤ = . ⎥ j22 ⎦ det [J ] ⎢⎣ − J 21 J11 ⎥⎦

(9.15)

For the four-node quadrilateral element

det [ J ] = A0 + A1 r + A2 s , where A0 =

1 8

[ ( y4 − y2 )( x3 − x1 ) − ( y3 − y1 )( x4 − x2 ) ] ,

(9.16)

197


Usually

A1 =

1 8

[ ( y3 − y4 )( x2 − x1 ) − ( y2 − y1 )( x3 − x4 ) ],

A2 =

1 8

[ ( y4 − y1 )( x3 − x2 ) − ( y3 − y2 )( x4 − x1 ) ] .

[ j]

is used because we want to express

∂f ∂f , in terms of ∂x ∂y

∂f ∂f , , determined on the reference element. Because r = r ( x , y ) , s = s ( x , y ) ∂r ∂s

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

dx dy = det [ J ] dr ds .

(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 { x , y }, the elementary area dA is given by the modulus of the cross product d x ×d y , where d x = d x ⋅ i , d y = d y ⋅ j , and i , j are base vectors. In a curvilinear system { r , s } , the elementary area dA is given by the modulus of the cross product d r ×d s . It is equal to the area of the elemental parallelogram enclosed by the two vectors d r and d s directed tangentially to the r = const . and s = const . contours respectively. The components of these vectors in a cartesian coordinate system are dr =

∂x ∂y d r ⋅ i + d r ⋅ j = ( J11 i + J12 j ) d r , ∂r ∂r

d s = ( J 21 i + J 22 j ) d s .

By equating the moduli of the cross products

d x ×d y = d x d y ⋅ k ,

198


i d r ×d s = J11 J 21

j

k

0 d r d s = det [J ] d r d s ⋅ k 0

J12 J 22

we obtain equation (9.17).

9.1.5 Element stiffness matrix The element stiffness matrix (7.43) is

[k ]= t ∫ [ B ] e

e

T

[ D ][ B ] d A ,

(9.18)

A

where [ D ] is the material stiffness matrix and t e is the element thickness. The matrix of differentiated shape functions [ B ] is (7.41)

[ B ] = [ ∂ ][ N ] , where [ ∂ ] is the matrix of differential operators (6.9) and shape functions (9.7).

(9.19)

[ N ] is the matrix of

Substituting (6.9) into (7.41) gives the strain-displacement matrix

⎡ ∂ ⎢ ⎢ ∂x [ B ] = [ ∂ ] [ N ] = ⎢⎢ 0 ⎢ ∂ ⎢ ⎢⎣ ∂y

⎤ 0 ⎥ ⎥ ∂ ⎥ [ N ]. ∂y ⎥ ∂ ⎥ ⎥ ∂x ⎥⎦

(9.20)

Using the transformation (9.14) we obtain

[ B ] = [ B1 ][ B2 ] ,

(9.21)

where

⎡ j11 [ B1 ] = ⎢⎢ 0 ⎢⎣ j21

j12 0 j22

0 j21 j11

0 j22 j12

⎤ ⎥, ⎥ ⎥⎦

(9.22)

199


⎡ ∂ ⎢ ∂r ⎢ ⎢ ∂ [ B2 ] = ⎢⎢ ∂ s ⎢ 0 ⎢ ⎢ ⎢ 0 ⎣

and

⎤ 0 ⎥ ⎥ 0 ⎥ ⎥ [ N ], ∂ ⎥ ⎥ ∂r ⎥ ∂ ⎥ ∂ s ⎥⎦

(9.23)

or 0 0 (1 − s ) ⎡ − (1 − s ) ⎢ − (1 − r ) 0 − (1 + r ) 0 [ B2 ] = 1 ⎢⎢ (1 − s ) 0 − (1 − s ) 0 4 ⎢ 0 − (1 − r ) 0 − (1 + r ) ⎣

(1 + s ) (1 + r ) 0 0

0

− (1 + s )

0 (1 + s ) (1 + r )

0 0

⎤ 0 ⎥⎥ . − (1 + s )⎥ (1 − r ) ⎥⎦ 0

(1 − r )

Using equation (9.17), the element stiffness matrix can be written +1 +1

[k ]= t ∫ ∫ [ B ] e

e

T

[ D ][ B ] det [ J ] dr ds .

(9.24)

−1 −1

Because [ B ] and det [ J ] 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

( px , p y ) per unit length

along side 2-3 in Fig. 9.1, then the equivalent nodal force vector is

{ f }= 12 l ⎣ 0 e

2−3

0

px

py

px

py

0 0 ⎦T ,

if p x and p y are constants and l 2 −3 is the length of the side 2-3.

(9.25)

200


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 NewtonCotes 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 det [ J ] 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 f ( r ) can be represented approximately over the interval r ∈ [ − 1, 1 ] by a polynomial which can be integrated exactly. A polynomial of order (2n − 1) can be ‘fitted’ to f ( r ) by imposing n weights wi and n sampling points in such a way that the summation +1

∫

f (r ) d r =

−1

n

∑ wi f ( r i )

(9.26)

i =1

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 ri : +1

∫ f ( r )d r = w1 f (r1 ) + w2 f (r 2 ) + ... + wi f (ri ) + ... + wn f (r n ) .

−1

The 2n coefficients are determined from the condition that the above equation is satisfied for a polynomial of order 2n − 1 of the form f ( r ) = a1 + a2 r + a3 r 2 + ... + a2 n r 2 n −1 .

(9.27)

201


On substitution above we get +1

a1

(

∫

−1

+1

+1

−1

−1

d r + a2 ∫ r d r + ... + a2 n

)

∫r

2 n −1

d r = a1 ( w1 + w2 + ... + wn ) +

(

)

(9.28)

+ a2 w1 r1 + w2 r 2 + ... + wn r n + ... + a2 n w1 r12 n −1 + w2 r 2 2 n −1 + ... + wn r n 2 n −1 .

Identifying the terms +1

∫

−1

n

+1

i =1

−1

d r = ∑ wi ,

∫

n

+1

i =1

−1

r d r = ∑ wi r i ,

∫

n

r α d r = ∑ wi riα . i =1

Generally +1

0 ⎧ ⎪ rα dr = ⎨ 2 ⎪ +1 α ⎩ −1

∫

if i is odd if i is even

(9.29)

so that

2 = w1 + w2 + ... + wn , 0 = w1 r1 + w2 r 2 + ... + wn r n , 2 = w1 r12 + w2 r 22 + ... + wn r n2 , 3

(9.30)

L L L L L L 0 = w1 r12 n −1 + w2 r 22 n −1 + ... + wn r n2 n −1 . This is a set of 2 n equations, linear in wi and nonlinear in r i . The 2 n parameters are determined from conditions wi > 0 ⎫ i = 1, 2 ,..., n . − 1 < r i < +1 ⎬⎭

One-point formula For n = 1 we have the midpoint rule (Fig. 9.3, a) +1

∫ f ( r )d r ≈ w1 f (r1 )

−1

202


which is exact only when f (r ) 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 +1

∫ f ( r )d r = w1 f (r1 ) + w2 f (r 2 ) .

(9.31)

−1

For a polynomial of order 2n − 1 = 2 ⋅ 2 − 1 = 3 , equations (9.30) give

2 = w1 + w2 , 0 = w1 r1 + w2 r 2 , 2 = w1 r12 + w2 r 22 , 3 0 = w1 r13 + w2 r 23 .

The solution to this set of equations is

w1 = w2 = 1 ,

r1 = − r 2 =

1 = 0.5773502691 3

so that (Fig. 9.3, b) +1

∫ f ( r )d r = 1 ⋅ f ( 0.57735 ) + 1 ⋅ f ( − 0.57735 ) .

−1

Fig. 9.3

203


Three-point formula For a three-point integration (Fig. 9.3, c)

∑ wi f (ri ) = 0.555 ⋅ f ( − 0.774) + 0.888 ⋅ f ( 0 ) + 0.555 ⋅ f ( 0.774 ) . 3

i =1

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

1 2

1 3

3

5

4

7

Gauss Points,

Weights,

ri

wi

0

2.0 1.0 0.8888 0.5555 0.652145 0.347854

± 0.57735 0

± 0.774596 ± 0.339981 ± 0.861136

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:

I=

+1 +1

∫ ∫

f ( r , s )d r d s ≅

−1 −1

I=

+1 +1

∫ ∫

−1 −1

n n ⎡ n ⎤ w f r , s ds = w ⎢ ⎥ ∑ i ∑ w j f ri , s j , j ∫ ⎢∑ j ⎥⎦ j =1 i =1 j =1 −1 ⎣

+1

(

)

n

n

(

(

)

f ( r , s ) d r d s = ∑ ∑ wi w j f ri , s j . i =1 j =1

)

(9.32)

For n = 2

w1 = w2 = 1 , r1 = s 1 = − 0.57735 , r 2 = s 2 = 0.57735 , so that

I = f ( r1 , s1 ) + f ( r1 , s2 ) + f ( r2 , s1 ) + f ( r2 , s2 ) . Gauss quadrature formulae for quadrilateral elements are shown in Table. 9.2.

204


Table 9.2 n

Number of Gauss points

1

1

(1×1)

4 (= 2 × 2) at centre

2

4 (2 × 2)

1 (= 1 × 1) at points 1, 2, 3, 4

Gauss Points, ri , s j

n×n

Weights, wi , w j

25 ⎛ 5 5 ⎞ ⎜ = × ⎟ at points 1, 2, 3, 4 81 ⎝ 9 9 ⎠

3

9 (3 × 3)

40 ⎛ 5 8 ⎞ ⎜ = × ⎟ at points 1, 2 , 3, 4 81 ⎝ 9 9 ⎠

64 ⎛ 8 8 ⎞ ⎜ = × ⎟ at points 1, 2, 3, 4 81 ⎝ 9 9 ⎠

9.2.3 Stiffness integration The element stiffness matrix (9.24) is

[k ]= t ∫ ∫ [ B ] e

e

1 1

T

[ D ][ B ] det [ J ] dr ds

−1 −1

which means that the integral above actually consists of the integral of each element (below or above the main diagonal) in an ( 8 × 8 ) matrix. In general, det [ J ] 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 bilinear function of r and s by a linear function. Therefore, the elements of

205


[ B ]T [ D ][ B ] det [ J ] are bi-quadratic functions divided by a linear function. This

[ ]

means that k e 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

[k ]≈ [ B ] e

+1 +1

T

[ D ][ B ] ∫

∫ t det [ J ] dr ds . e

(9.33)

−1 −1

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, t e , to be constant and noting that

det [ J ] 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 ( 2 × 2 ) array of points. Using the 2 × 2 rule (9.33), we can write

206


[ k ] ≅ t [ B ( r , s ) ] [ [ D ] det [J ( r , s )] ] [ B ( r , s ) ]+ e

e

1

T

1

1

1

1

1

[ B ( r1 , s2 ) ] [ [ D ] det [J ( r1 , s2 )] ] [ B ( r1 , s2 ) ]+ + t e [ B ( r2 , s 1 ) ]T [ [ D ] det [ J ( r2 , s 1 )] ] [ B ( r2 , s 1 ) ]+ +t

T

e

+ t e [ B ( r2 , s2 ) ] T

(9.34)

[ [ D ] det [ J ( r2 , s2 )] ] [ B ( r2 , s2 ) ] .

The calculation consists of four basic steps: a) Calculation of [ J ] at the Gauss points (9.11)

[ J ( ri ,s j ) ] = 14 ⎡⎢ −( 1 − s j ) ( 1 − s j ) ( 1 + s j ) − ( 1 + s j ) ⎣⎢ − ( 1 − ri )

− ( 1 + ri )

( 1 + ri )

( 1 − ri )

⎡ x1 ⎤ ⎢ x2 ⎥ ⎢ ⎢ ⎦⎥ ⎢ x 3 ⎣ x4

y1 ⎤ y 2 ⎥⎥ y3 ⎥ ⎥ y4 ⎦

and its determinant. b) Calculation of the inverse

[ j ] = [ J ] −1 at the Gauss points.

c) Calculation of the [ B ] matrix (9.21) ⎡ j11 [ B ] = 1 ⎢⎢ 0 4 ⎢⎣ j 21

j12 0 j 22

0 j 21 j11

0 ⎤ ⎥ j 22 ⎥ j12 ⎥⎦

d) Calculation of

(

)

0 ⎡− 1 − s j ⎢ ( ) 1 0 r − − i ⎢ ⎢ 0 − 1− s j ⎢ 0 − (1 − ri ) ⎣⎢

(

[k ] e

)

1− s j 0 1+ s j 0 1 + ri − (1 + ri ) 0 1− s j 0 0 0 − (1 + ri )

0 0 1+ s j 1 + ri

(

− 1+ s j 1 − ri 0 0

)

⎤ ⎥ ⎥ − 1+ s j ⎥ ⎥ 1 − ri ⎦⎥

(

0 0

using the 2 × 2 rule (9.34).

Example 9.1 Consider the quadrilateral element from Fig. 9.1 with the following nodal coordinates ⎡ x1 ⎢x ⎢ 2 ⎢ x3 ⎢ ⎣ x4

y1 y2 y3 y4

⎤ ⎡ 10 ⎥ ⎢ 20 ⎥ =⎢ ⎥ ⎢ 25 ⎥ ⎢ ⎦ ⎣ 8

10 ⎤ 15 ⎥⎥ . 30 ⎥ ⎥ 25 ⎦

For convenience, consider approximately the Gauss Points at r i , s j = ± 0.5 instead of r i , s j = ± 0.57735 . For the first term in (9.34) we need

det [ J ] and [ B ] calculated at GP1.

)

.

207


At GP1, the Jacobian matrix (9.11) is ⎡ 10 −0.5 0.5 1.5 − 1.5⎤ ⎢⎢ 20 [ J ] = 1 ⎡⎢ 4 ⎣− 0.5 − 1.5 1.5 0.5 ⎥⎦ ⎢ 25 ⎢ ⎣ 8

10 ⎤ 15 ⎥⎥ 1 = 30 ⎥ 8 ⎥ 25 ⎦

⎡ 61 20 ⎤ ⎢ 13 60 ⎥ . ⎣ ⎦

Its determinant and inverse are

det [ J ]= 53.125 ,

[ j ] = [ J ] −1 =

1 ⎡ 60 − 20 ⎤ . 425 ⎢⎣ − 13 61 ⎥⎦

At GP1, the matrix [ B ] is given by (9.21) 0 ⎤ ⎡ 60 − 20 0 [ B ] = 1 ⎢⎢ 0 0 − 13 61 ⎥⎥ 1700 ⎢⎣− 13 61 60 − 20⎥⎦

0 ⎡− 0.5 ⎢− 0.5 0 ⎢ ⎢ 0 − 0.5 ⎢ 0 − 0. 5 ⎣

0.5

0

− 1.5 0

0 0.5

0

− 1.5

1. 5 1.5

0

0

0 1. 5

0

1.5

− 1. 5 0.5 0 0

0 ⎤ 0 ⎥⎥ , − 1. 5 ⎥ ⎥ 0.5 ⎦

0 30 0 30 0 − 50 0 ⎤ ⎡− 10 1 ⎢ [ B ] = ⎢ 0 − 12 0 − 49 0 36 0 25 ⎥⎥ . 850 ⎢⎣− 12 10 − 49 30 36 30 25 − 50⎥⎦

9.2.4 Stress calculations In the quadrilateral element, stresses

{σ } = [ D ] [ B ] {q e }

(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.

208


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

u = a1 + a2 r + a3 s + a4 r 2 + a5 r s + a6 s 2 + a7 r 2 s + a8 r s 2 , v = b1 + b2 r + b3 s + b4 r 2 + b5 r s + b6 s 2 + b7 r 2 s + b8 r s 2 .

(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 [ B ] T [ D ][ B ] is therefore fourth order and det [ J ] is cubic. It would seem necessary to use a 4 × 4 Gauss quadrature for exact stiffness integration. For a single 8-node rectangular element, even if it is supported conventionally, using a 2 × 2 integration scheme would result in a singular stiffness matrix. In practice, large groups of elements will not suffer from such mechanisms, and 2 × 2 integration is normally used for this popular element. The optimal sampling points for this element are the Gauss points for the 2 × 2 scheme

209


and not those for the higher order 3 × 3 scheme, so the advantages of reduced integration are compounded. For an 8-node quadrilateral element, better displacements are obtained using a 3 × 3 Gauss point mesh, due to reduced integration. 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, det [ J ] is of third order hence the minimum number of integration points is 4 (2 × 2) . 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.

9.3.1 Shape functions The shape functions (8.29) have the following expressions 1 ( 1 − r ) ( 1 − s ) (1 + r + s ) , 4 1 N 2 = − ( 1 + r ) ( 1 − s ) (1 − r + s ) , 4 1 N3 = − ( 1 + r ) ( 1 + s ) ( 1 − r − s ), 4 1 N 4 = − ( 1 − r ) ( 1 + s ) (1 + r − s ) , 4 N1 = −

(

) ( ) (

1 1 − r 2 (1 − s ) , 2 1 N6 = (1 + r ) 1 − s 2 , 2 1 N 7 = 1 − r 2 ( 1 + s ) , (9.37) 2 1 N8 = ( 1 − r ) 1 − s 2 . 4 N5 =

(

)

)

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 ( ri , si ) are the natural coordinates of node i: for corner nodes 1 ( 1 + r ri ) ( 1 + s si ) (r ri + s si − 1) , 4 ∂ N i ri ( 1 + s si ) (2r ri + s si ) , ∂ N i = si ( 1 + r ri ) (2s si + r ri ) ; = ∂r ∂s 4 4

Ni =

(9.38)

___________

*) After the extraordinary discoveries of the princes of Serendip from the horror novels by Horace Walpole (1717-1797).

210


for mid-side nodes, when ri = 0

(

)

1 1 − r 2 ( 1 + s si ) , 2 ∂ Ni ∂ Ni 1 = − r ( 1 + s si ) , = si 1 − r 2 ; ∂r ∂s 2 Ni =

(

(9.39)

)

for mid-side nodes, when si = 0

(

)

1 1 − s 2 ( 1 + r ri ) , 2 ∂ Ni 1 ∂ Ni = ri 1 − s 2 , = − s ( 1 + r ri ∂r ∂s 2 Ni =

(

)

(9.40)

).

The displacements inside the element can be expressed in terms of the nodal displacements as ⎧u⎫ e (9.41) ⎨ ⎬ =[N ] q , v ⎩ ⎭ where N 0 N 2 0 L L N8 0 ⎤ [ N ] = ⎡⎢ 1 (9.42) ⎥, ⎣ 0 N1 0 N 2 L L 0 N 8 ⎦

{ }

{ q }= ⎣u e

1

v1

u2

v8 ⎦ T .

v 2 L L u8

(9.43)

Having generated the shape functions [ N ] , the matrix [ B ] = [ ∂ ][ N ⎡ ∂ N1 ⎢ ⎢ ∂x [ B ] = ⎢⎢ 0 ⎢∂N ⎢ 1 ⎢⎣ ∂ y

0 ∂ N1 ∂y ∂ N1 ∂x

∂ N2 ∂x 0 ∂ N2 ∂y

0

L

∂ N8 ∂x

∂ N2 L 0 ∂y ∂ N2 ∂ N8 L ∂x ∂y

0 ∂ N8 ∂y ∂ N8 ∂x

⎤ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥⎦

]

(9.44)

9.3.2 Shape function derivatives Use of the isoparametric formulation 8

x = ∑ N i (r , s ) x i , i =1

leads to

8

y = ∑ N i (r , s ) y i i =1

(9.45)

is

211


⎧ ∂ Ni ⎪⎪ ∂ x ⎨ ∂N i ⎪ ⎪⎩ ∂ y

⎫ ⎧ ∂ Ni ⎫ ⎡ ∂y ⎪⎪ ⎪⎪ ∂ r ⎪⎪ 1 ⎢ ∂s ⎢ ⎬ = [ j ]⎨ ∂N ⎬ = i ⎪ det [ J ] ⎢ ∂ x ⎪ ⎪ − ⎢⎣ ∂ s ⎪⎭ ⎪⎩ ∂ s ⎪⎭

∂y ∂r ∂x ∂r

−

⎤ ⎧ ∂ Ni ⎥ ⎪⎪ ∂ r ⎥⎨ ⎥ ⎪ ∂ Ni ⎥⎦ ⎪⎩ ∂ s

⎫ ⎪⎪ ⎬, ⎪ ⎪⎭

(9.46)

where

∂x = ∂r ∂y = ∂r

8

∑

∂ N i (r , s ) xi , ∂r

∂x = ∂s

∑

∂ N i (r , s ) yi , ∂r

∂y = ∂s

i =1 8

i =1

8

∑ ∂ N∂(sr , s ) x i

i =1 8

∑ i =1

i

,

∂ N i (r , s ) yi . ∂s

(9.47)

9.3.3 Determinant of the Jacobian matrix The Jacobian matrix is ⎡ ∂x ⎢ [ J ] = ⎢ ∂∂ xr ⎢ ⎢⎣ ∂ s

∂y ∂r ∂y ∂s

⎤ ⎥ ⎥ , ⎥ ⎥⎦

(9.48)

∂x ∂ y ∂x ∂ y . − ∂r ∂s ∂s ∂r

(9.49)

and its determinant is det [ J ] =

9.3.4 Element stiffness matrix The element stiffness matrix is +1 +1

[k ]= t ∫ ∫ [ B ] e

e

−1 −1

T

[ D ][ B ] det [ J ] d r d s

(9.50)

where t is the thickness of the plate and the material stiffness matrix [ D ] has the form ⎡ D1 D 2 0 ⎤ (9.51) [ D ] = ⎢⎢ D 2 D 4 0 ⎥⎥ . ⎢ 0 0 D 5 ⎥⎦ ⎣ e

212


Substituting (9.44), the stiffness matrix (9.50) can be written ⎡ ∂ N1 ⎢ ∂x ⎢ ⎢ 0 ⎢ ⎢ ⎢ ∂ N2 +1 +1 ⎢ ∂x ⎢ ⎡ ke ⎤ = te ⎢ 0 ⎥⎦ ⎣⎢ ⎢ −1 −1 ⎢ ⎢ L ⎢∂N ⎢ 8 ⎢ ∂x ⎢ ⎢ 0 ⎣⎢

0 ∂ N1 ∂y

0

∫∫

∂ N2 ∂y L

0 ∂ N8 ∂y

∂ N1 ⎤ ∂ y ⎥⎥ ∂ N1 ⎥ ∂x ⎥ ⎥ ∂ N2 ⎥ ∂y ⎥ ⎥ ∂ N2 ⎥ ∂x ⎥ ⎥ L ⎥ ∂ N8 ⎥ ⎥ ∂y ⎥ ∂ N8 ⎥ ⎥ ∂ x ⎥⎦

T

⎡ ∂ N1 ⎢ ⎢ ∂x [D ] ⎢⎢ 0 ⎢ ⎢ ∂ N1 ⎢⎣ ∂ y

0 ∂ N1 ∂y ∂ N1 ∂x

∂ N2 ∂x

0 ∂ N2 ∂y

0 ∂ N2 ∂y ∂ N2 ∂x

L

∂ N8 ∂x

L

0

∂ N8 L ∂y

0 ∂ N8 ∂y ∂ N8 ∂x

⎤ ⎥ ⎥ ⎥ ⎥ det [ J ] d r d s ⎥ ⎥ ⎥⎦

.

(9.52) The (16 × 16) element stiffness matrix contains submatrices [ k ] i j which, evaluated by numerical integration, have the form

[k ] i j = t

n

e

⎡ ∂ Ni ⎢ ∂x w p wq ⎢ ⎢ 0 q =1 ⎢⎣

∑∑ p =1

0

n

∂ Ni ∂y

⎡k

[ k ] i j = ⎢k 1

or

⎣

3

∂ Ni ∂y ∂ Ni ∂x

k 2⎤ k 4 ⎥⎦

⎡ ∂N j ⎢ ⎤ ⎢ ∂x ⎥ ⎢ ⎥ [D]⎢ 0 ⎥ ⎢ ⎥⎦ ⎢ ∂N j ⎢ ∂y ⎣

⎤ 0 ⎥ ⎥ ∂N j ⎥ det [ J ] , (9.53) ∂ y ⎥⎥ ∂N j ⎥ ∂ x ⎥⎦

.

(9.53, a)

ij

Denoting w = t e w p wq det [ J ] ,

(9.54)

the elements of the matrix (9.53, a) can be determined as ⎛ ∂ Ni ∂ N j ∂ Ni ∂ N j ⎞ ⎟, k 1 i j = ∑∑ w ⎜⎜ D 1 + D5 ∂x ∂x ∂ y ∂ y ⎟⎠ ⎝

k2 k3

ij

ij

⎛ ∂ Ni ∂ N j ∂ Ni ∂ N j ⎞ ⎟, = ∑∑ w ⎜⎜ D 2 + D5 ∂x ∂ y ∂ y ∂ x ⎟⎠ ⎝ ⎛ ∂ Ni ∂ N j ∂ Ni ∂ N j ⎞ ⎟, = ∑∑ w ⎜⎜ D 2 + D5 ∂ y ∂x ∂ x ∂ y ⎟⎠ ⎝

(9.55)

213


⎛ ∂ Ni ∂ N j ∂ Ni ∂ N j ⎞ ⎟. k 4 i j = ∑∑ w ⎜⎜ D 4 + D5 ∂y ∂y ∂ x ∂ x ⎟⎠ ⎝

When i = j , k 2 = k 3 . A close inspection of equations (9.55) shows that all the terms are of the form

(cons tan t ) × (material

property) × ( product of shape function derivatives ) .

Therefore, providing that the material properties are constant throughout the element, it is useful to store separately in an array [ S ] i j the sum of the products of the shape function derivatives multiplied by w , such that

[ S ]i j

⎡ ⎢∑∑ w =⎢ ⎢ ⎢∑∑ w ⎣

∂ Ni ∂x ∂ Ni ∂y

∂N j ∂x ∂N j ∂x

∂ Ni ∂x ∂N ∑∑ w ∂ yi

∑∑ w

∂N j ⎤ ⎥ ∂y ⎥ ∂N j ⎥ ∂ y ⎥⎦

(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

{σ } = [ D ][ B ]{ q e }.

(9.57)

The explicit form of equation (9.57) for the quadratic 8-node element is ⎧σx ⎪ ⎨σy ⎪τ ⎩ xy

⎫ ⎡ D1 ⎪ ⎢ ⎬ = ⎢D 2 ⎪ ⎢ 0 ⎭ ⎣

D2 D4 0

⎡ ∂ N1 ⎢ 0 ⎤ ⎢ ∂x ⎥ 0 ⎥⎢ 0 ⎢ D 5 ⎥⎦ ⎢ ∂ N ⎢ 1 ⎢⎣ ∂ y

0 ∂ N1 ∂y ∂ N1 ∂x

∂ N2 ∂x 0 ∂ N2 ∂y

0

8

i =1

∂ N8 ∂x

∂ N2 L 0 ∂y ∂ N2 ∂ N8 L ∂x ∂y

wherefrom

σ x = D1 ∑

L

8 ∂ Ni ∂ Ni ui + D2 ∑ vi , ∂x i =1 ∂ y

0 ∂ N8 ∂y ∂ N8 ∂x

⎤ ⎧u 1 ⎫ ⎥⎪ ⎪ ⎥ ⎪v1 ⎪ ⎥ ⎪u ⎪ ⎥ ⎨ 2⎬ ⎥ ⎪L ⎪ ⎥ ⎪v ⎪ ⎥⎦ ⎪⎩ 8 ⎪⎭

214

FINITE ELEMENT ANALYSIS 8

σ y = D2 ∑

i =1

8 ∂ Ni ∂ Ni vi , ui + D4 ∑ ∂x i =1 ∂ y

(9.58)

8

⎛ ∂ Ni ∂ Ni ⎞ v i ⎟⎟ , ui + ∂x ⎠ i =1 ⎝ ∂ y

τ xy = D 5 ∑ ⎜⎜

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)

{ f }= ∫ [ N ] e

T

{ p }d V

.

(9.59)

Ve

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 s = +1 edge of an element. The components of the force p d r acting upon an elemental length d r are

215


⎧ ∂y ⎫ ⎪⎪ ∂ r ⎪⎪ ⎧ px ⎫ ⎨ ⎬ = p ⎨ ∂ x ⎬ dr . ⎩ py ⎭ ⎪− ⎪ ⎪⎩ ∂ r ⎪⎭

(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 ⎧ ∂y ⎪⎪ ∂ r p [ N ]T ⎨ ∂x ⎪− ⎪⎩ ∂ r

+1

{ f }= ∫ e

−1

where [ N ] is given by (9.42).

⎫ ⎪⎪ ⎬ dr ⎪ ⎪⎭

(9.61)

The above vector is usually integrated numerically and is written as ⎧ ∂y ⎪⎪ ∂ r pi ⎨ ∂x ⎪− ⎪⎩ ∂ r

⎫ ⎪⎪ (9.62) wi [ N ] T fe = ⎬ ⎪ i =1 ⎪⎭i where pi is the pressure at the Gauss point i along the s = +1 edge and is computed by equation n

{ } ∑

8

p=

∑ N (r ,s ) p k

k

.

(9.63)

k =1

If the same pressure acts on the edge s = −1 , the sign of the force vector

{ f } is reversed, since a positive pressure is assumed to act towards the centre of e

the element. A more general equation applicable to the two edges s = ±1 is ⎧ ∂y ⎫ 8 ⎪⎪ ∂ r ⎪⎪ ⎞ ⎛ f e = ∑ wi ws [N ]T ⎜⎜ ∑ N k p k ⎟⎟ ⎨ ∂x ⎬ i =1 ⎠⎪− ⎝ k =1 ⎪ ⎪⎩ ∂ r ⎪⎭i where w s takes up the value of the s coordinate for the loaded edge.

{ }

n

(9.64)

A similar expression for pressure loads on the r = ±1 edges is

{ f }= ∑ e

n

i =1

⎧ ∂y 8 ⎪⎪ − ∂ s ⎞ ⎛ T⎜ wi wr [N ] ⎜ ∑ N k p k ⎟⎟ ⎨ ∂x ⎠⎪ ⎝ k =1 ⎪⎩ ∂ s

⎫ ⎪⎪ ⎬ . ⎪ ⎪⎭i

(9.65)

216


For a rectangular isoparametric element with edges parallel to the x , y 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 s = +1 we have N1 = N 5 = N 2 = N 6 = N8 = 0 . For the rectangular element

{f } e

∂y ∂x = 0 and = a . Equation (9.61) gives ∂r ∂r

⎧ f y3 ⎪ = ⎨ f y7 ⎪f ⎩ y4

⎫ +1 ⎪ ⎬= p ⎪ −1 ⎭

∫

⎧ N3 ⎪ ⎨ N7 ⎪N ⎩ 4

⎫ ⎪ ⎬ (− a ) d r . ⎪ ⎭

Denoting the total pressure load P = 2 p a , we can write f y3 = − f y7 = −

P 2 P 2

+1

1 ∫ 4 ( 1 + r ri ) ( 1 + s si ) ( r ri + s si − 1) d r ,

−1 +1

1 2 ∫ 4 ( 1 + s si ) ( 1 − r ) d r ,

−1

(9.66)

217


f y4 = −

P 2

+1

∫

−1

1 (1 + r ri 4

) (1 + s si ) ( r ri + s si − 1) d r .

In the above expressions, substituting r i = 1 and s i = 1 for node 3, r i = 0 and s i = 1 for node 7, and r i = −1 and s i = 1 for node 4, yields P 2

f y3 = − f y7 = − f y4 = −

P 2 P 2

+1

∫

1 (1 + r )( 2)( r + 1 − 1) d r = − P , 4 6

∫

1 ( 2 ) 1 − r 2 dr = − 2P , 4 3

∫

1 (1 − r )(2)(− r + 1 − 1) d r = − P . 4 6

−1 +1 −1 +1 −1

(

)

Hence a uniform load acting along an edge is not distributed in the intuitive ratio of 1 4 : 1 2 : 1 4 , but in the ratio 1 6 : 2 3 : 1 6 (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 onedimensional element (4.26). Gravity loading

For gravity loading (downwards negative) the vector of equivalent nodal forces (9.59) is

{ f }= ∫ m [ N ] e

T

⎧ 0 ⎫ ⎨ ⎬d x d y = ⎩− g ⎭

+1 +1

∫ ∫ m [N ]

−1 −1

T

⎧ 0 ⎫ ⎨ ⎬ det [ J ] d r d s ⎩− g ⎭

(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

{ f }= ∑∑ w w e

n

n

i

i =1 j =1

j

⎧ 0 ⎫ m [ N ] T ⎨ ⎬ det [ J ] . ⎩− g ⎭

(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,

218

FINITE ELEMENT ANALYSIS ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

f y1 ⎫ f y 2 ⎪⎪ = L ⎬⎪ f y8 ⎪⎭

+1 +1

∫ ∫

−1 −1

⎧ N1 ⎫ ⎪N ⎪ ⎪ ⎪ m g ⎨ 2 ⎬ det [ J ] d r d s . L ⎪ ⎪ ⎪⎩ N 8 ⎪⎭

(9.69)

For a rectangular element det [ J ] = ab .

For a corner node i the integral becomes +1 +1

f yi = −

1

∫ ∫ m g 4 (1 + r ri ) (1 + s si ) ( r ri + s si − 1) det [ J ] d r d s .

(9.70)

−1 −1

Denoting the total weight of the element W = m g ⋅ 2a ⋅ 2b , the force (9.70) can be written f yi = −

W ( I1 + I 2 + I 3 ) , 16

(9.71)

where +1 +1

I1 =

4

∫ ∫ (1 + r ri ) (1 + s si ) r ri d r d s = 3

(9.72)

−1 −1

since ri = ±1 for all corner nodes, +1 +1

I2 =

4

∫ ∫ (1 + r ri ) (1 + s si ) s si d r d s = 3

(9.73)

−1 −1

since si = ±1 for all corner nodes, and +1 +1

I3 = −

∫ ∫ (1 + r ri ) (1 + s si ) d r d s = − 4 .

(9.74)

−1 −1

Therefore, the equivalent nodal forces for gravity loading at a corner node i are given by f yi = −

W ( I1 + I 2 + I 3 ) = − W ⎛⎜ 4 + 4 − 4 ⎞⎟ = W . 16 16 ⎝ 3 3 ⎠ 12

(9.75)

For a midside node i with ri = 0 , the integral is +1 +1

f yi = −

1 W 2 ∫ ∫ m g 2 (1 − r ) (1 + s si ) det [ J ] d r d s = − 3 .

(9.76)

−1 −1

For a midside node i with si = 0 , the integral is 1 1

f yi = −

1 W 2 ∫ ∫ m g 2 (1 − s ) (1 + r ri ) det [ J ] d r d s = − 3 .

−1 −1

(9.77)

219


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 − W 12 and − W 6 , 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

220


The associated displacement functions in polynomial form are u = a1 + a2 r + a3 s + a4 r 2 + a5 r s + a6 s 2 + a7 r 2 s + a8 r s 2 + a9 r 2 s 2 , v = b1 + b2 r + b3 s + b4 r 2 + b5 r s + b6 s 2 + b7 r 2 s + b8 r s 2 + b9 r 2 s 2 .

(9.78)

Along the sides of the element, the polynomial is quadratic (with three terms - as can be seen setting s = 0 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 ( p + 1) 2 nodes ( p = 0,1, ...) . 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 r 2 s and r s 2 and also an r 2 s 2 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 L1 , L2 and L3 having unit value at the node with the same index and zero at the other two nodes L1 (r ) = −

r (1 − r ) , 2

L 2 (r ) = ( 1 + r )( 1 − r ) ,

L 3 (r ) =

r (1 + r ) . 2

(9.79)

They turn out to be Lagrange polynomials, which, for three points, have the expressions

221

9. ISOPARAMETRIC ELEMENTS L 1 (r ) =

( r − r 2 ) ( r − r3 ) ( r1 − r 2 ) ( r1 − r3 )

, L 2 (r ) =

( r − r1 ) ( r − r 3 ) ( r 2 − r1 ) ( r 2 − r 3 )

, L 3 (r ) =

( r − r1 ) ( r − r 2 ) ( r 3 − r1 ) ( r 3 − r 2 )

.

Similarly, generic shape functions can be defined along the s axis, as shown at the right of Fig. 9.6, b L1 (s ) = −

s (1 − s ) , 2

L 2 (s ) = ( 1 + s ) ( 1 − s ) ,

L 3 (s ) =

s (1 + s ) . 2

(9.80)

The shape functions N i can be constructed as products of the above onedimensional functions ⎡ N1 ⎢N ⎢ 5 ⎢⎣ N 2

N8 N9 N6

N 4 ⎤ ⎧ L 1 (r ) ⎫ ⎪ ⎪ N 7 ⎥⎥ = ⎨ L 2 (r ) ⎬ ⎣ L 1 (s ) L 2 (s ) L 3 (s ) ⎦ . N 3 ⎥⎦ ⎪⎩ L 3 (r ) ⎪⎭

(9.81)

It should be cautioned that the subscripts of N i 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) N1 = ζ1 ( 2 ζ1 − 1 ) , N 2 = ζ 2 ( 2 ζ 2 − 1 ) , N 3 = ζ 3 ( 2 ζ 3 − 1 ) , N 4 = 4 ζ1 ζ 2 ,

N5 = 4 ζ 2 ζ 3 ,

N 6 = 4 ζ 3 ζ1 ,

(9.82)

222


where ζ 1 = 1 − ζ 2 − ζ 3 . Because of terms ζ 22 , ζ 32 in the shape functions, this element is also called a quadratic triangle.

Fig. 9.8 The isoparametric representation is (8.37) ⎧ u ⎫ ⎡ N1 ⎨ ⎬=⎢ ⎩v⎭ ⎣ 0

where

{ q }= ⎣u e

1

0

N2

0

N1

0

N2

v1 u2

L L N6 L L 0

v 2 L L u6 6

x = ∑ Ni x i , i =1

6

Fig. 9.9

(9.83)

v6 ⎦ T , and

y = ∑ Ni y i . i =1

{ }

0 ⎤ e q , N 6 ⎥⎦

(9.84)

223


The element stiffness matrix, which has to be integrated numerically, can be evaluated using (9.24) if the (r , s ) coordinates are defined by (8.46). This gives ∂ ∂ ∂ = − , ∂ r ∂ N 2 ∂ N1

∂ ∂ ∂ = − . ∂ s ∂ N 3 ∂ N1

(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 det [ J ] does not attain a value of zero in the element. The sign of det [ J ] 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 (9.13) 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 1800 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 1800 , and, as an internal angle approaches 1800 there is a loss of accuracy in the element stress, particularly at that corner. If the determinant det [ J ] → 0 , then [ j ] and the operators in increase without limit and consequently produce infinite strains.

[∂ ]

will

The above statements are illustrated for the one-dimensional quadratic isoparametric element shown in Fig. 4.4. The row vector of shape functions is ⎢ ⎥ ⎣N ⎦ = ⎣N1 N 2 N 3 ⎦ = ⎢ 2 r ( r − 1) ( 1 + r )( 1 − r ) 2 r ( r + 1) ⎥ . ⎦ ⎣

1

1

224


The Jacobian matrix is

[ J ] = J11 = ⎢⎢ ∂N1 ⎣ ∂r

∂N 2 ∂r

⎧ x1 ∂N 3 ⎥ ⎪ ⎥ ⎨ x2 ∂r ⎦ ⎪ ⎩ x3

⎫ ⎪ ⎢ 2r − 1 − 2r ⎬=⎢ ⎪ ⎣ 2 ⎭

⎧ 0 2r + 1 ⎥ ⎪ ⎨ x2 2 ⎥⎦ ⎪ ⎩ l

⎫ ⎪ ⎬. ⎪ ⎭

The determinant of the Jacobian matrix is

det [ J ] = J11 = ( r + 0.5 ) l − 2 r x2 . For any x2 this determinant vanishes at the point 0.5 l . 2x2 −l

r0 =

Fig. 9.10 This point lies within the element (Fig. 9.10) if − 1 ≤ r 0 ≤ 1 , i.e. if −1 ≤

0 .5 l ≤1, 2x2 −l

where 0 ≤ x 2 ≤ l . l 3l < x 2 < , then det [ J ] does not vanish on the element. This explains 4 4 why it is recommended to place precautiously the ‘midside’ node inside the middle third of a side.

If

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 twodimensional 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 h l ≤ 0.1 . 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.

226


The displacements parallel to the undeformed middle surface are given by u ( x, y, z ) = − z

∂w , ∂x

v ( x, y, z ) = − z

∂w , ∂y

(10.1)

where w ( x, y ) denotes the displacement of the middle surface in the z-direction (Fig. 10.2, a).

Fig. 10.1 The components of the strain are given as follows:

εx =

∂2w ∂u ∂v ∂u ∂2w ∂v ∂2w , = −z 2 , ε y = = −z 2 , γ x y = = −2 z + ∂x ∂y ∂x ∂ y ∂ y ∂x ∂x ∂y

γ yz =

∂v ∂w + =0, ∂z ∂ y

γ zx =

∂w ∂u + = 0. ∂x ∂z

a

(10.2)

b Fig. 10.2

The strain vector can be written in the form

{ε } = ⎣ ε x where the curvature vector

ε y γ x y ⎦ T = − z { χ },

(10.3)

227

10. PLATE BENDING

⎢ 2 { χ } = ⎢ ∂ w2 ⎣⎢ ∂ x

∂2w ⎥ 2 ⎥ ∂ x ∂ y ⎦⎥

∂2w ∂ y2

T

.

(10.4)

Since σ z = 0 , the stress-strain relations take the form

{σ } = [ D ] {ε } = − z [ D ] { χ } ,

(10.5)

{σ } = ⎣σ x

σ y τ x y ⎦T ,

(10.6)

⎤ ⎥. ⎥ ⎥⎦

(10.7)

where and

⎡ d11 [ D ] = ⎢⎢ d12 ⎢⎣ d16

d12 d 22 d 26

d16 d 26 d 66

The strain energy in the plate can be written as

U=

1 2

∫

{σ }T {ε }dV = 1

2

V

∫

h3 { χ }T [ D ] { χ }d A . 12

(10.8)

A

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 h2

⎣M x

My

Mxy ⎦=

∫ z ⎣σ

x

σ y τ xy ⎦ dz .

(10.9)

−h 2

Denoting h2

Di j =

∫d

−h 2

ij

z2 d z

(10.10)

228


the moment - curvature equations can be expressed as follows

⎧ Mx ⎪ ⎨ My ⎪M ⎩ xy

⎫ ⎡ D11 ⎪ ⎢ ⎬ = ⎢ D12 ⎪ ⎢D ⎭ ⎣ 16

or

D12 D22 D26

⎧ ∂2w ⎪ ∂ x2 D16 ⎤ ⎪ 2 ⎥ ⎪⎪ ∂ w D26 ⎥ ⎨ ∂ y2 D66 ⎥⎦ ⎪ 2 ⎪ ∂ w ⎪ ⎪⎩ ∂ x ∂ y

{ M } = − [ D ]{ χ }

⎫ ⎪ ⎪ ⎪⎪ ⎬ ⎪ ⎪ ⎪ ⎪⎭

(10.11)

.

(10.12)

For isotropic materials ⎡ 0 ⎢1 ν [ D ] = D ⎢ν 1 0 ⎢ 1 −ν ⎢0 0 2 ⎣

⎤ ⎥ ⎥. ⎥ ⎥ ⎦

(10.13)

,

(10.14)

where

D=

(

E h3

12 1 − ν

2

)

E is Young’s modulus and ν is the Poisson’s ratio. The equilibrium equations are ∂M x ∂M y x − Qx = 0 , + ∂y ∂x ∂M x y ∂M y + − Qy = 0 , ∂x ∂y ∂ Qx ∂ Q y + + pz = 0 , ∂x ∂y

(10.15)

where p z is the lateral distributed load (per unit surface) and Qx , Q y are the shear forces (per unit length). From (10.11) and (10.15) it can be shown that

Qx = − D

∂ ⎛⎜ ∂ 2 w ∂ 2 w + ∂ x ⎜⎝ ∂ x 2 ∂ y 2

⎞ ⎟, ⎟ ⎠

Qy = − D

∂ ⎛⎜ ∂ 2 w ∂ 2 w + ∂ y ⎜⎝ ∂ x 2 ∂ y 2

⎞ ⎟. ⎟ ⎠

(10.16)

229

10. PLATE BENDING

By substituting (10.16) into the last equation (10.15), the differential equilibrium equation for isotropic materials can be obtained as

∂ 4 w pz ∂4w ∂4w 2 = . + + D ∂ x2 ∂ y2 ∂ y4 ∂ x4

(10.17)

The bending and torsion stresses are given by (10.9)

⎣σ x

σ y τxy ⎦=

12 z ⎣M x h3

My

Mxy ⎦

(10.18)

Qy ⎦,

(10.19)

and the shear stresses by 2 ⎛ ⎜ 1 − 4z ⎜ h2 ⎝ where a parabolic distribution is assumed.

⎣τ z x

τ yz ⎦=

3 2h

⎞ ⎟ ⎣ Qx ⎟ ⎠

The strain energy (10.8) can be expressed as D U= 2

∫ A

2 ⎡⎛ 2 2 ⎢ ⎜ ∂ w + ∂ w ⎞⎟ + 2 ( 1 − ν ⎢ ⎜ ∂ x2 ∂ y 2 ⎟ ⎠ ⎣⎢ ⎝

⎞⎤ ⎛ ⎛ 2 ⎞2 ∂2w ∂ 2w ⎟ ⎥ ⎜⎜ ∂ w ⎟ )⎜ ⎜ dx dy . − 2 ∂ x ∂ y 2 ⎟⎟ ⎥ ⎜ ⎝ ∂ x ∂ y ⎟⎠ ⎠ ⎦⎥ ⎝ (10.20)

The potential of the external load is

∫

W P = − p z w dx dy .

(10.21)

A

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

230


v (x , y , z ) = − zθ x (x , y ) ,

u (x , y , z ) = z θ y (x , y ) ,

(10.22)

where θ x , θ y 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

{ ε } = − z { χ },

(10.23)

where the curvature vector is

⎢ ∂θ {χ }= ⎢ − y ⎣ ∂x

T

∂θ x ∂θ y ⎥ − ⎥ ∂x ∂y ⎦

∂θ x ∂y

.

(10.24)

The transverse shear strains are

γ yz =

∂v ∂w + , ∂z ∂ y

γ zx =

∂w ∂u . + ∂x ∂z

(10.25)

Substituting for the displacements from (10.22) gives ∂w ⎧ −θ + ⎧ γ y z ⎫ ⎪⎪ x ∂ y {γ } = ⎨ ⎬ = ⎨ ∂w ⎩ γ zx ⎭ ⎪ θy + ⎪⎩ ∂x

⎫ ⎪⎪ ⎬. ⎪ ⎪⎭

(10.26)

Note that when γ y z = 0 , γ z x = 0 , equation (10.26) gives θ x =

θy = −

∂w , ∂y

∂w , and equation (10.24) reduces to (10.4). ∂x The average shear stresses are given by

[ ]{ γ } .

⎧τ yz ⎫ S ⎬=κ D τ ⎩ zx ⎭

{τ } = ⎨

(10.27)

where κ is a shear correction factor and

[ D ] = ⎡⎢ G0 S

⎣

0 G

⎡1 0⎤ ⎤ E ⎥ = 2 (1 +ν ) ⎢ 0 1 ⎥ . ⎣ ⎦ ⎦

(10.28)

The strain energy in the plate can be written as the sum of the energies due to bending and shear deformation U=

1 2

∫ {σ } {ε }dV + 2 ∫ T

V

1

V

{τ }T {γ }dV .

(10.29)

231

10. PLATE BENDING

or, integrating through the thickness U=

1 2

∫

h3 { χ }T [ D ] { χ }d A + 1 12 2

A

∫ κ h {γ } [ D ]{γ }d A . T

S

(10.30)

⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭

(10.31)

A

For isotropic materials, equation (10.11) becomes

⎧ Mx ⎪ ⎨ My ⎪M ⎩ xy

⎡ ⎫ 0 ⎢1 ν ⎪ ⎢ 0 ⎬=D ν 1 ⎢ 1 −ν ⎪ ⎢0 0 ⎭ 2 ⎣

⎧ ∂θ y − ⎤⎪ ∂x ⎥ ⎪⎪ ∂θ x ⎥⎨ ∂y ⎥⎪ ⎥ ⎪ ∂θ ∂θ y x ⎦ ⎪ ∂x − ∂ y ⎩

or (10.12), where { χ } is given by (10.24). The transverse shear constitutive equation is ∂w ⎫ ⎧ −θx + ⎪ ⎧ Qy ⎫ ⎪ ∂ y ⎪⎪ . ⎬=κG ⎨ ⎨ ∂w ⎬ ⎩ Qx ⎭ ⎪ ⎪ θy + ⎪⎩ ∂ x ⎪⎭

(10.32)

The strain energy (10.30) can be expressed as

U=

1 2

∫

⎡ ⎛ ∂θ y ⎢ M x ⎜⎜ − ⎝ ∂x ⎣⎢

∫

⎡ ⎛ ∂w ⎢Qx ⎜⎜ θ y + ∂x ⎣ ⎝

A

+

1 2

A

∂θ ⎞ ⎛ ∂θ ⎟⎟ + M x y ⎜⎜ x − y ∂y ⎠ ⎝ ∂x ⎛ ⎞ ∂w ⎟⎟ + Q y ⎜⎜ − θ x + ∂y ⎝ ⎠

⎞ ∂θ ⎤ ⎟⎟ + M y x ⎥ dx dy + ∂ y ⎥⎦ ⎠ ⎞⎤ ⎟⎟⎥ dx dy . ⎠⎦

∂w ∂w and θ y = − , the above expression reduces to (10.20). ∂y ∂x 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. If θ x =

232


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 ∂w ∂w and are continuous between ensure that w and its first derivatives ∂x ∂y 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 w = a1 + a2 x + a3 y + a4 x 2 + a5 x y + a6 y 2 + higher degree terms.

(10.33)

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 ξ = x a and η = y b 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 ∂w ∂w displacement w , and the two rotations θ x = and θ y = − . In terms of the ∂y ∂x dimensionless coordinates ξ and η , these become

233

10. PLATE BENDING

θx =

1 ∂w , b ∂η

θy = −

1 ∂w . a ∂ξ

(10.34)

Since the element has 12 degrees of freedom, the displacement function can be represented by a polynomial having twelve terms, that is

w = α1 + α 2 ξ + α 3η + α 4 ξ 2 + α 5 ξ η + α 6η 2 +

+ α 7 ξ 3 + α 8 ξ 2η + α 9 ξη 2 + α10η 3 + α11ξ 3η + α12 ξη 3 .

(10.35)

Note that this function is a complete cubic to which two quartic terms ξ 3η and ξη 3 have been added, which are symetrically located in Pascal’s triangle. This ensures that the element is geometrically invariant.

Fig. 10.5 The expression (10.35) can be written in the following matrix form

⎣

w = 1 ξ η ξ 2 ξ η η 2 ξ 3 ξ 2η ξ η 2 η 3 ξ 3η ξη 3

⎦ {α },

w = ⎣P ( ξ ,η )⎦ {α } ,

(10.36) (10.36, a)

where

{α }T = ⎣ α1

α 2 α 3 L α12 ⎦ .

(10.37)

Differentiating (10.36) gives

∂w = 0 1 0 2ξ ∂ξ

⎣

∂w = 0 0 1 0 ξ ∂η

⎣

η 0 3ξ 2 2ξη η 2 0 3ξ 2η η 3 ⎦ {α } ,

(10.38)

⎦ {α }.

(10.39)

2η

0 ξ2

2ξη 3η 2 ξ 3 3ξη 2

234


Evaluating (10.36), (10.38) and (10.39) at ξ = ±1 , η = ±1 gives

{ q }= [ A ]{α }, e

e

(10.40)

where

{ q } = ⎣w bθ aθ L w bθ aθ ⎦ {α }, (10.41) and the matrix [ A ] is given in [87] but not reproduced here, for conciseness. e T

1

x1

y1

4

x4

y4

e

Solving (10.40) for {α } gives

{α } = [ Ae ]

−1

{q } . e

(10.42)

Substituting (10.42) into (10.36, a) yields

{ }

w = ⎣N (ξ ,η )⎦ q e = ⎣ ⎣N1 ⎦

e ⎣N 2 ⎦ ⎣N 3 ⎦ ⎣N 4 ⎦ ⎦ {q },

(10.43)

where the vector of nodal displacements is

{ q } = ⎣w e T

θ x1 θ y1 L w4 θ x 4 θ y 4 ⎦ ,

1

(10.44)

and the shape functions are defined as

)(

(

⎣N i (ξ ,η )⎦

T

⎧1 2 2 ⎪ 8 ( 1 + ξ i ξ ) 1 + η jη 2 + ξ i ξ + ηi η − ξ − η ⎪⎪ b ( 1 + ξi ξ )( ηi + η ) η 2 − 1 =⎨ 8 ⎪ a ⎪ ( ξi + ξ ) ξ 2 − 1 ( 1 + ηi η ) ⎪⎩ 8

(

(

where ( ξi ,ηi

)

)

) ⎫⎪

⎪⎪ ⎬, ⎪ ⎪ ⎪⎭

(10.45)

) are the coordinates of node i.

The element is non-conforming. Indeed, evaluating (10.45) on the side 2-3 (for ξ = +1 ) gives

⎣N 1 ⎦

T

(

⎧ 0 ⎫ ⎪ ⎪ = ⎨ 0 ⎬ , ⎪ 0 ⎪ ⎩ ⎭

⎣N 2 ⎦

(

⎣N 3 ⎦

T

⎧ (1 4 )( 1 + η ) 2 + η − η 2 ⎪ = ⎨ (b 4 )( 1 + η ) η 2 − 1 ⎪ 0 ⎩

(

)

)⎫⎪

⎬, ⎪ ⎭

T

)⎫⎪ ) ⎬,

⎧ (1 4 )( 1 − η ) 2 − η − η 2 ⎪ = ⎨ (b 4 )( − 1 + η ) η 2 − 1 ⎪ 0 ⎩

⎣N 4 ⎦

T

⎧0⎫ ⎪ ⎪ = ⎨ 0 ⎬. ⎪0⎪ ⎩ ⎭

(

⎪ ⎭

235

10. PLATE BENDING

This indicates that the displacement w , and hence the rotation θ x , 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 w and θ x will be continuous along the common side. Unfortunately, this is not the case with the rotation θ y . The rotation θ y is given by (10.34)

θy = −

{ }

∂ ⎣N 4 ⎦ ⎥ e 1 ∂w 1⎢∂ N = − ⎢ ⎣ 1⎦ L ⎥ q . ∂ξ ⎦ a ∂ξ a ⎣ ∂ξ

(10.46)

Substituting (10.45) into (10.46) gives ∂ ⎣N i (ξ ,η )⎦ T ∂ξ

⎫ ⎧ ⎪ ⎪ ξ i η ( ηi − η ) ⎪⎪ ⎪⎪ b ξ i ( ηi + η ) η 2 − 1 ⎬ . =⎨ ⎪ ⎪ 8 ⎪ − a ( 2 + 2ξ ) ( 1 + η η ) ⎪ i i ⎪⎭ ⎪⎩ 8

(

)

(10.47)

Evaluating (10.47) along ξ = +1 gives −η ( − 1 − η ) ⎧ ∂ ⎣N1 ⎦ T ⎪ = ⎨− (b 8) ( − 1 + η ) η 2 − 1 ∂ξ ⎪ 0 ⎩

(

η (− 1 − η ) ⎧ ∂ ⎣N 3 ⎦ T ⎪ = ⎨ (b 8) ( 1 + η ) η 2 − 1 ∂ξ ⎪ − (a 2 ) ( 1 + η ) ⎩

(

)

)

η (− 1 − η ) ⎫ ⎧ ⎪ ∂ ⎣N 2 ⎦ T ⎪ = ⎨ (b 8) ( − 1 + η ) η 2 − 1 ⎬, ∂ ξ ⎪ ⎪ − (a 2 ) ( 1 − η ) ⎭ ⎩

⎫ ⎪ ⎬, ⎪ ⎭

(

−η ( 1 − η ) ⎧ ∂ ⎣N 4 ⎦ T ⎪ = ⎨ − (b 8) ( 1 + η ) η 2 − 1 ∂ξ ⎪ 0 ⎩

(

)

)

⎫ ⎪ ⎬, ⎪ ⎭ ⎫ ⎪ ⎬. ⎪ ⎭

The above expressions indicate that θ y is determined by the values of w and θ x at nodes 1, 2, 3, and 4 as well as by θ y at nodes 2 and 3. The rotation θ y is not continuous across the side 2-3 and the element is non-conforming. Substituting (10.43) into (10.4) and (10.8) gives Ue =

1 2

{ q } [k ] { q }, e T

e

e

(10.48)

where the element stiffness matrix is

[ k ] = ∫ 12h [ B ] 3

e

A

and the strain-displacement matrix is

T

[ D ] [ B ]dA

(10.49)

236


⎧ ∂2 ⎪ 2 ⎪ ∂x 2 ⎪ [ B ] = ⎪⎨ ∂ 2 ⎪ ∂y ⎪ ∂2 ⎪2 ⎪⎩ ∂ x ∂ y

⎫ ⎧ 1 ∂2 ⎪ ⎪ 2 2 ⎪ ⎪ a ∂ξ ⎪⎪ ⎪⎪ 1 ∂ 2 N = ⎬⎣ ⎦ ⎨ 2 2 ⎪ ⎪ b ∂η ⎪ ⎪ 2 ∂2 ⎪ ⎪ ⎪⎭ ⎪⎩ ab ∂ξ ∂η

⎫ ⎪ ⎪ ⎪⎪ ⎬ ⎣N (ξ ,η )⎦ . ⎪ ⎪ ⎪ ⎪⎭

(10.50)

Substituting the shape functions from (10.45) and integrating gives the element stiffness matrix

[k ] e

⎡ [ k11 ] ⎢ Eh ⎢ = 2 48 1 −ν a b ⎢ ⎢ ⎣SYM 3

(

)

[ k12 ] [ k13 ] [ k14 ]⎤ [ k22 ] [ k23 ] [ k24 ]⎥⎥ [ k33 ] [ k34 ]⎥ [ k44 ]⎥⎦

(10.51)

where the submatrices have the following expressions

(

)

⎡ 2 ⎧ 2 1 ⎫ 2 2 ⎢4 β + α + 5 (7 − 2ν ) ⎨2α + 5 (1 + 4ν )⎬ b ⎩ ⎭ ⎢ ⎧4 2 4 ⎫ 2 ⎢ [ k11 ] = ⎢ ⎨ α + (1 − ν )⎬ b 15 ⎩3 ⎭ ⎢ ⎢ SYM ⎢⎣

(

⎧ ⎫ ⎤ 2 1 ⎨− 2β − (1 + 4ν )⎬ a ⎥ 5 ⎩ ⎭ ⎥ ⎥, −ν a b ⎥ ⎥ 4 4 ⎧ ⎫ 2 2 ⎨ β + ( 1 − ν )⎬ a ⎥ 15 ⎩3 ⎭ ⎥⎦

)

⎡ ⎧ 2 ⎧ 2 1 ⎫ ⎫ 2 2 ⎨α − (1 + 4ν )⎬ b ⎢− ⎨2 2 β − α + 5 (7 − 2ν )⎬ 5 ⎩ ⎭ ⎭ ⎢ ⎩ ⎧2 2 4 ⎫ 2 ⎧ 2 1 ⎫ ⎢ [ k12 ] = ⎢ ⎨ α − (1 − ν )⎬ b ⎨α − (1 + 4ν )⎬ b 5 15 ⎩3 ⎭ ⎩ ⎭ ⎢ 1 ⎧ ⎫ 2 ⎢ 0 ⎨ 2 β + (1 − ν )⎬ a 5 ⎩ ⎭ ⎣⎢

(

)

⎡ ⎧ 2 ⎧ 2 1 ⎫ ⎫ 2 2 ⎨α − ( 1 − ν )⎬ b ⎢− ⎨2 β + α + 5 (7 − 2ν )⎬ 5 ⎩ ⎭ ⎭ ⎢ ⎩ ⎧1 2 1 ⎫ 2 ⎧ 2 1 ⎫ ⎢ [ k13 ] = ⎢ ⎨− α + ( 1 − ν )⎬ b ⎨ α + ( 1 − ν )⎬ b 5 15 ⎩3 ⎭ ⎩ ⎭ ⎢ 1 ⎧ ⎫ 2 ⎢ 0 ⎨ β − ( 1 − ν )⎬ a ⎢⎣ 5 ⎩ ⎭

1 ⎧ ⎫ ⎤ − ⎨2β 2 + (1 − ν )⎬ a ⎥ 5 ⎩ ⎭ ⎥ ⎥, 0 ⎥ ⎥ ⎧2 2 1 ⎫ 2 ⎨ β − (1 − ν )⎬ a ⎥ 15 ⎩3 ⎭ ⎦⎥ ⎧ 2 1 ⎫ ⎤ ⎨− β + ( 1 − ν )⎬ a ⎥ 5 ⎩ ⎭ ⎥ ⎥, 0 ⎥ ⎥ ⎧1 2 1 ⎫ 2 ⎨ β + ( 1 − ν )⎬ a ⎥ 15 ⎩3 ⎭ ⎥⎦

237

10. PLATE BENDING

(

)

⎡ 2 ⎧ 2 1 ⎫ 2 2 ⎢2 β − 2α − 5 (7 − 2ν ) ⎨ 2α + 5 (1 − ν )⎬ b ⎩ ⎭ ⎢ ⎧2 2 1 ⎫ 2 ⎧ ⎫ ⎢ 2 1 [ k 41 ] = ⎢ ⎨− 2α − (1 − ν )⎬ b ⎨ α − (1 − ν )⎬ b 5 15 ⎩3 ⎭ ⎩ ⎭ ⎢ 1 ⎧ ⎫ ⎢ ⎨− β 2 + (1 + 4ν )⎬ a 0 5 ⎭ ⎣⎢ ⎩

⎧ ⎫ ⎤ 2 1 ⎨− β + (1 + 4ν )⎬ a ⎥ 5 ⎩ ⎭ ⎥ ⎥ 0 ⎥ ⎥ ⎧2 2 4 ⎫ 2 ⎨ β − (1 − ν )⎬ a ⎥ 15 ⎩3 ⎭ ⎦⎥

and

α=

a , b

β=

b . a

The remaining submatrices of (10.51) are

[ k22 ] = [ I 3 ]T [ k11 ][ I 3 ] , [ k23 ] = [ I 3 ]T [ k14 ][ I 3 ] , [ k24 ] = [ I 3 ]T [ k23 ][ I 3 ] , [ k33 ] = [ I1 ]T [ k11 ][ I1 ] , [ k34 ] = [ I1 ]T [ k12 ][ I1 ] ,

[ k44 ] = [ I 2 ]T [ k11 ][ I 2 ] , (10.52) where ⎡ − 1 0 0⎤ [ I1 ] = ⎢⎢ 0 1 0⎥⎥ , ⎢⎣ 0 0 1⎥⎦

⎡1 0 0⎤ [ I 2 ] = ⎢⎢0 − 1 0⎥⎥ , ⎢⎣0 0 1⎥⎦

⎡1 0 0 ⎤ [ I 3 ] = ⎢⎢0 1 0 ⎥⎥ . ⎢⎣0 0 − 1⎥⎦

(10.53)

The above relationships are presented in reference [87]. In deriving this result, it is simpler to use the expression (10.36) for w and substitute for {α } after performing the integration. A typical integration is then of the form ⎧ ⎪

+1 +1

∫∫

0 4 ⎪ (m + 1)(n + 1) ⎩

ξ mη n dξ dη = ⎨

−1 −1

m or n odd m and n even

For p z = constant , substituting the shape functions into

{ f } = ∫ ⎣N ⎦ e

T

pz d A

(10.54)

A

and integrating, the vector of equivalent nodal forces is obtained as

{ f }= p

ab T ⎣ 3 b − a 3 b a 3 − b a 3 − b − a⎦ . 3 Stresses at any point in the plate are given by (10.5). In terms of the nodal displacements they can be expressed as e

z

238


{σ } = − z [ D ][ B ]{q e },

(10.55)

where [ B ] is defined in (10.50). The most accurate values are at the Gauss points of a (2 × 2) numerical integration scheme.

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

⎣N i (ξ ,η )⎦

T

⎧ f i (ξ ) f i (η ) ⎫ ⎪ ⎪ = ⎨ b f i (ξ ) g i (η ) ⎬ , ⎪ − a g (ξ ) f (η ) ⎪ i i ⎩ ⎭

(10.56)

where

( (

) )

( (

) )

1 1 2 + 3ξ i ξ − ξ i ξ 3 , g i (ξ ) = − ξ i − ξ +ξ i ξ 2 + ξ 3 , 4 4 1 1 f i (η ) = 2 + 3ηi η − ηi η 3 , g i (η ) = − ηi − η +ηi η 2 + η 3 , 4 4 f i (ξ ) =

(10.57)

in which ( ξi ,ηi ) are the coordinates of node i. Unfortunately, when we examine the derivatives of these products we find that the twist ∂ 2 w ∂ξ ∂η 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 ∂ 2 w ∂ξ ∂η 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

239

10. PLATE BENDING

{ q } = ⎣w e T

1

w′x′ y 4 ⎦ , (10.58)

θ x1 θ y1 w′x′ y1 L w4 θ x 4 θ y 4

where w′x′ y ≡ ∂ 2 w ∂x ∂y and

⎣N i (ξ ,η )⎦

T

⎧ f i (ξ ) f i (η ) ⎫ ⎪ b f (ξ ) g (η ) ⎪ ⎪ ⎪ i i =⎨ ⎬. ⎪ − a g i (ξ ) f i (η ) ⎪ ⎪⎩ ab g i (ξ ) g i (η ) ⎪⎭

(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 ydirections. 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 ∂ 2 w ∂x ∂y . This is overcome in the WB element (Wilson, Brebbia – 1971) by introducing the approximations

(

)

w′x′ y 2 =

1 θ x 2 − θ x1 , 2a

(

)

w′x′ y 4 =

1 θ x3 − θ x 4 . 2a

w′x′ y1 =

1 θ y1 − θ y 4 , 2b

w′x′ y 3 =

1 θ y 2 − θ y3 , 2b

(

)

(

)

Applying the above constraints to the BFS element makes it a nonconforming 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 , θ x and θ y appearing in these expressions is the first. Hence, for convergence, w , θ x and θ y are the only degrees of freedom required at the nodal points (Fig. 10.5).

240


The four-node HTK element (Hughes, Taylor and Kanoknukulcha - 1977) expands separately w , θ x and θ y 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 4

4

4

i =1

i =1

i =1

w = ∑ Ni w i , θ x = ∑ Ni θ x i , θ y = ∑ Ni θ y i ,

(10.60)

where the functions N i are defined by (9.3), that is

(

)(

)

1 1 + ξ i ξ 1 + η iη . 4

Ni =

(10.61)

These functions ensure that w , θ x and θ y are continuous between elements. In matrix form, expressions (10.60) can be written

⎧w ⎪ ⎨ θx ⎪θ ⎩ y

⎫ ⎪ e ⎬=[N] q , ⎪ ⎭

{ }

(10.62)

where

{ q } = ⎣w e T

θ x1 θ y1 L w4 θ x 4 θ y 4 ⎦ ,

1

(10.44)

and

⎡ N1 [ N ] = ⎢⎢ 0 ⎢⎣ 0

0

0

N1 0

0 L N1 L

0

L N4 0 0

N4 0

0 ⎤ 0 ⎥⎥ . N 4 ⎥⎦

(10.63)

Substituting (10.62) into (10.24), (10.26) and (10.30) gives Ue =

1 2

{ q } [ k ] { q }, e T

e

e

(10.64)

[ ]

where the element stiffness matrix k e can be written as the sum of the matrices due to bending and shear

[ k ] = [ k ] + [ k ]. e

In (10.65)

B

S

(10.65)

241

10. PLATE BENDING

[ k ] = ∫ 12h [ B ] 3

B

T

B

[ D ] [BB ] d A

(10.66)

A

and

[ k ] = ∫ κ h [ B ] [ D ][ B ] d A . The strain-displacement matrix [ B ] is of the form [ B ]= [ [ B ] [ B ] [ B ] [ B ] ] S

S

T

S

S

(10.67)

A

B

B

B 1

B 2

B 3

B 4

(10.68)

0 − ∂ Ni ∂ x ⎤ ⎡0 ⎥. ⎢ 0 = ⎢ 0 ∂ Ni ∂ y ⎥ ⎢⎣ 0 ∂ N i ∂ x − ∂ N i ∂ y ⎥⎦

(10.69)

where

[ ] BiB

[ ] [ B ]= [ [ B ] [ B ] [ B ] [ B ] ]

The strain-displacement matrix B S is of the form S

S 1

S 2

S 3

S 4

(10.70)

where

[ B ] = ⎡⎢ ∂∂ NN S i

⎣

i i

∂x 0 ∂ y − Ni

Ni ⎤ . 0 ⎥⎦

(10.71)

Substituting (10.61) into (10.69) and (10.71) gives

[ ] BiB

(

)

⎡0 − ξ i 1 + η iη 4a ⎤ 0 ⎢ ⎥ = ⎢ 0 1 + ξ i ξ η i 4b 0 ⎥ ⎢ 0 ξ i 1 + η i η 4 a − 1 + ξ i ξ η i 4b ⎥ ⎣ ⎦

(

)

(

)

(

(10.72)

)

and

[ B ] = ⎡⎢ ξ(1 +(1ξ+ ξη )ηη ) 44ba − (1 + ξ ξ )(01 + η η ) 4 (1 + ξ ξ )(01 + η η ) 4 ⎤⎥ .(10.73) ⎣ ⎦ S i

i

i

i

i

i

i

i

i

Substituting (10.72) into (10.68) and the resulting matrices into (10.66) gives the element stiffness matrix due to bending

242


[k ]= B

E h3

(

48 1 − ν 2

)

[ ] [ k ] [ k ] [ k ] ⎤⎥ [ k ] [ k ] [ k ] ⎥⎥ [ k ] [ k ] ⎥⎥ [ k ] ⎥⎦

⎡ kB ⎢ 11 ⎢ ⎢ ab ⎢ ⎢ ⎢⎣SYM

B

B

B

12

13

14

B

B

B

22

23

24

B

B 34

33

B

44

where the submatrices are

[k ] B

11

⎡0 ⎢ ⎢ = ⎢0 ⎢ ⎢ ⎢0 ⎣⎢

[k ] B

12

[k ] B

13

[k ] B

14

⎡0 ⎢ ⎢ = ⎢0 ⎢ ⎢0 ⎢⎣ ⎡0 ⎢ ⎢ = ⎢0 ⎢ ⎢ ⎢0 ⎣⎢

⎤ ⎥ ⎥ 1 ⎥, − (1 +ν )a b ⎥ 2 ⎥ 4⎧ 2 1 ⎫ 2⎥ ( ) + − β 1 ν a ⎨ ⎬ 3⎩ 2 ⎭ ⎦⎥

0 4 3

0

⎧ 2 1 ⎫ 2 ⎨α + ( 1 −ν )⎬ b 2 ⎩ ⎭ 1 − (1 +ν )a b 2

⎤ ⎥ ⎥ 1 ( 3ν − 1) a b ⎥ , 2 ⎥ 1 2 2⎥ − 4 β + ( 1 −ν ) a ⎥⎦ 3

0

{

0

}

2 α 2 − ( 1 −ν ) b 2 3 1 − ( 3ν − 1 ) a b 2

{

0

}

⎤ ⎥ ⎥ 1 ⎥, ( 1 +ν )a b ⎥ 2 ⎥ 2⎧ ⎫ 2⎥ 2 1 ( ) − − − β 1 ν a ⎨ ⎬ 3⎩ 2 ⎭ ⎦⎥ 0

2⎧ ⎫ 2 2 1 ⎨ − α − ( 1 −ν ) ⎬ b 3⎩ 2 ⎭ 1 (1 +ν )a b 2

⎡0 0 ⎢ ⎢ 1 = ⎢0 − 4α 2 + ( 1 − ν ) b 2 3 ⎢ 1 ⎢0 ( 3ν − 1 ) a b ⎢⎣ 2

{

}

⎤ ⎥ ⎥ 1 − ( 3ν − 1) a b ⎥ , 2 ⎥ 2 2 2⎥ β − ( 1 −ν ) a ⎥⎦ 3 0

{

}

and

α=

a , b

β=

b . a

(10.74)

243

10. PLATE BENDING

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

[k ] S

[ ] [ k ] [ k ] [ k ] ⎤⎥ [ k ] [ k ] [ k ] ⎥⎥ , [ k ] [ k ] ⎥⎥ [ k ] ⎥⎦

⎡ kS ⎢ 11 ⎢ E h3 ⎢ = 48 γ S a b ⎢ ⎢ ⎢⎣SYM

S

S

S

12

13

14

S

S

S

22

23

24

S

S 34

33

(10.75)

S

44

where γ S = E h 3 12κ G b 2 is a shear parameter similar to (5.97) and

[k ] S

11

⎡1 + α 2 α 2b ⎢ = ⎢ α 2b α 2 b 2 ⎢ −a 0 ⎣

[k ] S

13

− a⎤ ⎥ 0 ⎥, a2 ⎥ ⎦

⎡ − 1 − α 2 α 2b ⎢ = ⎢ − α 2b α 2 b 2 ⎢ a 0 ⎣

[k ] S

12

− a⎤ ⎥ 0 ⎥, a2 ⎥ ⎦

[k ] S

14

⎡ − 1 + α 2 α 2b ⎢ = ⎢ α 2b α 2 b2 ⎢ a 0 ⎣ ⎡1 − α 2 α 2b ⎢ = ⎢ − α 2b α 2 b 2 ⎢ −a 0 ⎣

− a⎤ ⎥ 0 ⎥, a2 ⎥ ⎦ − a⎤ ⎥ 0 ⎥. a2 ⎥ ⎦

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 ⎧ pz ⎫ ⎪ f = ⎣N ⎦ ⎨ 0 ⎬ d A (10.76) ⎪ ⎪ A ⎩ 0 ⎭ For pz = constant, substituting the shape functions from (10.63) and (10.61) into (10.76) and integrating gives

{ } ∫ e

{ f }= p a b ⎣ 1 e

z

T⎪

0 0 1 0 0 1 0 0 1 0 0⎦ T

(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

244


⎧ Mx ⎪ ⎨ My ⎪M ⎩ xy

⎫ ⎪ h3 [ I3 ] [ D ] B B = − ⎬ 12 ⎪ ⎭

[ ]{q }, e

(10.78)

where [ I 3 ] is defined by (10.53). The most accurate values are at the Gauss points of a (2 × 2) numerical integration scheme. The shear forces per unit length are ⎧ Qx ⎫ ⎨ ⎬=κ h ⎩ Qy ⎭

[ D ][ B ]{q }. S

S

e

(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 (2 × 2) 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 w , θ x and θ y 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 (2 × 2) 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.

245

10. PLATE BENDING

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 (0,0) , (x2 ,0) and (x3 , y3 ) . There are three degrees of freedom at each ∂w node, namely, the transverse displacement w , and the two rotations θ x = and ∂y

θy = −

∂w with respect to the local axes. ∂x

Since the element has 9 degrees of freedom, the displacement function can be represented by a polynomial having nine terms, that is w = α1 + α 2 x + α 3 y + α 4 x 2 + α 5 x y + α 6 y 2 +

(

)

+ α 7 x3 + α8 x 2 y + x y 2 + α 9 y 3 .

(10.80)

Note that a complete cubic has ten terms so that the polynomial (10.80) is incomplete. In order to maintain symmetry, the coefficients of x 2 y and x y 2 are taken to be equal.

Fig. 10.6 The expression (10.80) can be written in the following matrix form

⎣

w= 1 x

y

x2

xy

y2

x3

(x

2

y + x y2

w = ⎣P ( x , y )⎦ {α } ,

)

y3

⎦ {α } ,

(10.81) (10.81, a)

where

{α }T = ⎣ α1

α 2 α3 L α9 ⎦ .

Differentiating (10.81) gives

(10.82)

246 ⎧w ⎪ ⎨ θx ⎪θ ⎩ y


⎫ ⎡ 1 x y x2 x y y2 x3 ⎪ ⎢ 0 0 x 2y ⎬=⎢0 0 1 ⎪ ⎢ 0 − 1 0 − 2 x − y 0 − 3x 2 ⎭ ⎣

x2 y + x y2 x2 + 2x y − 2x y − y2

y3 3y2 0

⎤ ⎥ ⎥ {α }. (10.83) ⎥ ⎦

Evaluating (10.83) at the nodal points gives

{ q } = [ A ] {α } , e

e

(10.84)

where

{ q } = ⎣w e T

1

θ x1 θ y1 L w3 θ x 3 θ y 3 ⎦ ,

(10.85)

and the matrix

[A ] e

⎡1 0 0 ⎢0 0 1 ⎢ ⎢ 0 −1 0 ⎢ ⎢ 1 x2 0 =⎢0 0 1 ⎢ ⎢ 0 −1 0 ⎢1 x y3 3 ⎢ ⎢0 0 1 ⎢ ⎣ 0 −1 0

0 0 0 x22 0 − 2 x2 x32

0 − 2 x3

0 0 0

0 0 0

0 x2

0 0 0 y32 2 y3

0 x3 y3 x3 − y3

0 0 0 x23 0 3x22

0 0 0

0 x22

0 0 0 y33 3 y32

0

x33

x32 y3 + x3 y32 x32 + 2 x3 y3 − 2 x3 y3 − y32

0 − 3x32

0

0 0 0

0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ . (10.86) ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Solving (10.84) for {α } gives

{α } = [ Ae ]

−1

{q } . e

(10.87)

Substituting (10.87) into (10.81, a) yields

[ ] { q }. Unfortunately, the matrix [ A ] is singular whenever w = ⎣P (x , y )⎦ Ae

−1

e

(10.88)

e

x2 − 2 x3 − y3 = 0

(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 y = 0 , gives

247

10. PLATE BENDING

⎧w ⎪ ⎨ θx ⎪θ ⎩ y

⎡ 1 x 0 x2 0 0 ⎫ x3 ⎢ ⎪ 0 x 0 0 ⎬ =⎢0 0 1 ⎪ ⎢ 0 − 1 0 − 2 x 0 0 − 3x 2 ⎭1− 2 ⎣

0 x2 0

0⎤ ⎥ 0 ⎥ {α } . 0⎥ ⎦

(10.90)

The rotation θ x is a quadratic function having coefficients α 3 , α 5 and α 8 . These cannot be determined using the values of θ x 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 x 2 y and x y 2 terms. The edge 1-2 was taken along the x-axis. In this case θ x is a tangential component and θ y is a normal component of rotation. The transverse displacement

w and the normal component θ y are continuous between elements, while the tangential component θ x is not. To alleviate this, in some plate-bending elements a linear variation of the tangential component of rotation is adopted. Substituting (10.88) into (10.4) and (10.8) gives Ue =

1 2

{ q } [ k ]{ q } e T

e

e

(10.91)

where the element stiffness matrix in the local coordinate system is

[ k ] = [ A ] ∫ 12h [ B ] e

3

e −T

T

[ D ] [ B ]d A

[A ]

e −1

(10.92)

A

and the strain-displacement matrix is 2y 0 ⎤ ⎡ 0 0 0 2 0 0 6x ⎢ [ B ] = ⎢ 0 0 0 0 0 2 0 2 x 6 y ⎥⎥ . ⎢⎣ 0 0 0 0 2 0 0 4 ( x + y ) 0 ⎥⎦ A typical element of the integrand in (10.92) is of the form

(10.93)

∫x

m n

y dA ,

A

which can be evaluated analytically. The vector of consistent nodal forces is given by

{ f }= [ A ] ∫ e

e −T

T ⎣P ⎦ p z d A .

(10.94)

A

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

248


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 ζ 1 , ζ 2 and ζ 3 (8.35). There are three independent degrees of freedom at each node, namely, the transverse displacement w , and the two rotations θ X and θY with respect to the global axes. The displacement functions are assumed to be 3

3

3

i =1

i =1

i =1

w = ∑ ζ i wi , θ X = ∑ ζ i θ X i , θY = ∑ ζ i θY i ,

(10.95)

where wi , θ X i and θY i are the degrees of freedom at node i. They ensure that w ,

θ X and θY are continuous between elements. In matrix form, expressions (10.95) are written ⎧ w ⎪ ⎨θX ⎪θ ⎩ Y

⎫ ⎪ ⎬=[N ⎪ ⎭

]{ q e },

(10.96)

where the vector of nodal displacements is

{ q } = ⎣w e T

1

θ X 1 θ Y 1 L w3 θ X 3 θ Y 3 ⎦ ,

(10.97)

and ⎡ ζ1 0 0 ζ 2 0 0 ζ 3 0 0 ⎤ [ N ] = ⎢⎢ 0 ζ 1 0 0 ζ 2 0 0 ζ 3 0 ⎥⎥ . ⎢⎣ 0 0 ζ 1 0 0 ζ 2 0 0 ζ 3 ⎥⎦

(10.98)

Functions expressed in area coordinates can be differentiated with respect to Cartesian coordinates using

249

10. PLATE BENDING

∂ 1 3 ∂ = βi , ∑ ∂ x 2 A i =1 ∂ζ i

∂ 1 3 ∂ = γi , ∑ ∂ y 2 A i =1 ∂ζ i

(10.99)

where β i and γ i are defined in (8.8). Substituting (10.96) into (10.24) and using (10.99) gives the straindisplacement matrix due to bending

[B ] B

⎡ 0 0 − β1 0 0 1 ⎢ = 0 γ1 0 0 γ2 2 A ⎢⎢ ⎣ 0 β 1 − γ1 0 β 2

−β2 0 −γ2

0 0 0 γ3 0 β3

−β3 ⎤ ⎥ 0 ⎥. − γ 3 ⎥⎦

(10.100)

As this matrix is constant, the stiffness matrix due to bending is

[ k ] = 12h A [ B ] 3

B

B

T

[ D ] [ B B ].

(10.101)

Substituting (10.96) into (10.24) and using (10.99) shows that the straindisplacement matrix due to shear is

[B ] S

⎡ ⎢ =⎢ ⎢ ⎢ ⎣

β1 2A

γ1

2A

0 − ζ1

ζ1 0

β2 2A

γ2

2A

0 −ζ2

ζ2 0

β3 2A

γ3

2A

0 −ζ3

⎤

ζ3 ⎥

⎥. ⎥ 0 ⎥ ⎦

(10.102)

Substituting (10.102) into (10.67) and integrating using the formula

∫ζ

m n p 1 ζ2 ζ3 dA =

m ! n!p ! 2A (m + n + p + 2)!

(10.103)

A

gives the stiffness matrix due to shear

[k ] S

where

[ ] [ k ] [ k ] ⎤⎥ [ k ] [ k ] ⎥⎥ , [ k ] ⎥⎦

⎡ kS ⎢ 11 = κG h ⎢ ⎢ ⎢ SYM ⎣

S

S

12

13

S

S

22

23

S

33

(10.104)

250

[k ] S

11

[k ] S

13

[k ] S

23


⎡ β 12 + γ 12 ⎢ ⎢ 4A γ1 ⎢ =⎢ − 6 ⎢ β 1 ⎢ ⎢ 6 ⎣

−

β1 ⎤

γ1

⎥ 6⎥ ⎥ 0 ⎥, ⎥ A⎥ 6⎥ ⎦

6 A 6 0

⎡ β3 β 1 + γ 3 γ 1 ⎢ 4A ⎢ γ3 ⎢ =⎢ − 6 ⎢ β 3 ⎢ ⎢ 6 ⎣

−

⎡ β3 β 2 + γ 3 γ 2 ⎢ 4A ⎢ γ3 ⎢ =⎢ − 6 ⎢ β 3 ⎢ ⎢ 6 ⎣

γ1

6 A 12 0

−

γ2

6 A 12 0

[k ] S

12

β1 ⎤

⎥ 6⎥ ⎥ 0 ⎥, ⎥ A⎥ 12 ⎥ ⎦

[k ] S

22

β2 ⎤

⎥ 6 ⎥ ⎥ 0 ⎥, ⎥ A⎥ 12 ⎥ ⎦

[k ] S

33

⎡ β2 β 1 + γ 2 γ1 ⎢ 4A ⎢ γ2 ⎢ =⎢ − 6 ⎢ β 2 ⎢ ⎢ 6 ⎣ ⎡ β 22 + γ 22 ⎢ ⎢ 4A γ2 ⎢ =⎢ − 6 ⎢ β 2 ⎢ ⎢ 6 ⎣

⎡ β 23 + γ 32 ⎢ ⎢ 4A γ3 ⎢ =⎢ − 6 ⎢ β 3 ⎢ ⎢ 6 ⎣

−

−

6 A 12 0

γ2 6 A 6 0

−

γ1

γ3 6 A 6 0

β1 ⎤

⎥ 6⎥ ⎥ 0 ⎥, ⎥ A⎥ 12 ⎥ ⎦

β2 ⎤

⎥ 6 ⎥ ⎥ 0 ⎥, ⎥ A⎥ 6 ⎥ ⎦

β3 ⎤

⎥ 6 ⎥ ⎥ 0 ⎥. ⎥ A⎥ 6 ⎥ ⎦

The complete element stiffness matrix is the sum (10.65). The above relationships are presented in reference [87]. For pz = constant, substituting the shape functions from (10.98) into (10.76) and integrating gives the vector of equivalent nodal forces

{ f }= p e

z

A 1 0 0 1 0 0 1 0 0⎦ T ⎣ 3

(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 C1 continuity for w ) led to the development of the so-called Discrete Kirchhoff (DK) technique (Wempner-1969, Stricklin et al.-1969, Dhatt1967).

251

10. PLATE BENDING

In the formulation of various DK plate elements, only the bending strain energy is considered, in which the curvatures are expressed, as in the ReissnerMindlin plate theory, in terms of the first derivatives of the rotations (10.24). In this case only C 0 approximations of these rotations can be considered. The shear strain energy is neglected. 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 C 0 continuity of the tangential components of rotations (normal slopes). Constant curvature patch-tests are needed to check the validity of the elements. 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 θ s is assumed to vary linearly along each edge, while the normal component θ n varies quadratically (Fig. 10.7, a). The latter condition means that the transverse displacement w can vary cubically.

a

b Fig. 10.7

For a side i − j , of length l k , the tangential and normal components of rotations θ s k and θ n k at the mid-side node k, are defined in terms of the components along the coordinate axes by ⎧ θ s k ⎫ ⎡ ck ⎨θ ⎬ = ⎢ ⎩ n k ⎭ ⎣ − sk where

sk ⎤ ⎧ θ x k ⎫ ⎨ ⎬ ck ⎥⎦ ⎩ θ y k ⎭

(10.106)

252


(

)

(

ck = x j − x i l k ,

)

sk = y j − y i l k .

(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 θ n varies parabolically (Fig. 10.8, a). Its expression in terms of the dimensionless coordinate

s′ = s l k

can be written as

θ n = ( 1 − s′ )θ ni + s′θ n j + 4s′ ( 1 − s′ )α k . At the mid-side node k

θnk =

(

)

1 θ ni + θ n j + α k , 2

where the first term in the right hand side is expressed in terms of the rotations at the corner nodes. The parameter α k has to be eliminated. The rotation θ s is assumed to vary linearly (Fig. 10.8, b)

θ s = ( 1 − s′ )θ s i + s′θ s j .

Fig. 10.8 In order to eliminate the degrees of freedom at mid-side nodes, the three parameters α 4 , α 5 , α 6 must be expressed in terms of the degrees of freedom at the corner nodes. This is done requiring the shear strains γ s z to vanish along each side. One kind of ‘discrete’ Kirchhoff constraint is formulated in integral form as j

∫ i

j

γ s z ds =

∫ i

⎛ ∂w ⎞ ⎜⎜ + θ n ⎟⎟ d s = 0 . ⎝ ∂s ⎠

(10.108)

253

10. PLATE BENDING

Equation (10.108) can be written lk

j

∫ dw + ∫ θ

n

ds = 0 ,

0

i

1

(w j − wi ) + l k ∫ [ ( 1 − s′ )θ ni + s′θ n j + 4s′ ( 1 − s′ )α k ] d s′ = 0 , 0

wherefrom we obtain

αk =

) (

(

)

3 3 wi − w j − θ n i + θ n j , 2l k 4

or, in terms of the degrees of freedom of the corner points

αk =

(

) (

)

3 3 wi − w j − − s k θ x i + c k θ y i − s k θ x j + c k θ y j . 2l k 4

(10.109)

The equation (10.109) can also be obtained by adopting a Hermitian cubic polynomial for w (s ) and using the discrete Kirchhoff collocation constraint (zero ∂w shear strain) γ s z = + θ n = 0 at points i, j, k. Then, the condition γ s z = 0 will be ∂s satisfied at all points along the contour, because w is cubical and θ n is quadratic. The rotations θ x and θ y can be expressed as

θ x = ⎣N ⎦ {θ x } + ⎣Px ⎦ {α },

(10.110)

θ y = ⎣N ⎦ {θ y }+ ⎣Py ⎦ {α },

where

⎣N ⎦ = ⎣ N1 N 2 N3 ⎦ = ⎣1 − ξ − η ξ η ⎦ ,

{θ x } = ⎣θ x1

θ x 2 θ x3 ⎦T ,

{α } = ⎣ α 4

{θ y }= ⎣θ y1

θ y 2 θ y 3 ⎦T ,

α5 α6 ⎦ T ,

⎣Px ⎦ = ⎣ − P4 s4 − P5 s5 − P6 s6 ⎦ ,

⎣Py ⎦ = ⎣ P4c4

(10.111)

P5c5

P6 c6 ⎦ ,

⎣ P4 P5 P6 ⎦ = ⎣ 4ξ ( 1 − ξ − η ) 4ξη 4η ( 1 − ξ − η ) ⎦ . Substituting (10.109) into (10.110) we obtain the explicit expressions of

θ x and θ y in terms of the corner nodal variables

254


⎧ θx ⎫ ⎨ ⎬=[N ⎩θy ⎭ where

{ q } = ⎣w e T

1

] {q e },

(10.112)

θ x1 θ y1 L w3 θ x 3 θ y 3 ⎦ ,

(10.113)

⎣ ⎦ ⎣N 2x ⎦ ⎣N3x ⎦ ⎤⎥ , y y y ⎥ ⎢⎣ ⎣N1 ⎦ ⎣N 2 ⎦ ⎣N 3 ⎦ ⎥⎦

(10.114)

⎡ Nx 1

[ N ] = ⎢⎢

⎣N1x ⎦ = ⎣N ix1

N ix2

⎦

N ix3 ,

⎣N1y ⎦ = ⎣Niy1

N iy2

⎦

N iy3 .

(10.115)

In (10.115), the shape functions have the following expressions N ix1 = −

3 3 Pk sk + Pm sm , 2l k 2l m

3 3 Pk ck − Pm cm , 2l k 2l m

3 3 Pk sk2 − Pm sm2 , 4 4

N iy2 = N ix3 ,

3 3 Pk sk ck + Pm sm cm , 4 4

N iy3 = N i −

N ix2 = N i − N ix3 =

N iy1 =

(10.116) 3 3 Pk ck2 − Pm cm2 . 4 4

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 θ x i and θ y i have the expressions used in Kirchhoff’s theory ⎛∂w⎞

⎟⎟ , θ x i = ⎜⎜ ⎝ ∂ y ⎠i

⎛∂w⎞

( i = 1, 2, 3 )

⎟⎟ , θ y i = − ⎜⎜ ⎝ ∂ x ⎠i

allowing the introduction of Kirchhoff-type boundary conditions. Table 10.1 Corner point, i 1 2 3

Side k ( i − j )

Side m ( i − j )

5 (2−3 )

4 (1− 2 )

4 (1− 2 ) 6 (3 −1 )

6 (3 −1 )

5 (2−3 )

255

10. PLATE BENDING

Because the parameters α k are eliminated using an expression which is a function of the nodal variables of the side k only, the continuity C 0 of θ n 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

{ }

{ }

θ x = ⎣G ⎦ q e ,

θ y = ⎣H ⎦ q e .

(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

⎣G ⎦ = ⎣1 ⎣H ⎦ = ⎣1

ξ η ξ 2 ξη η 2 ⎦ [ G ] ,

(10.118)

ξ η ξ 2 ξη η 2 ⎦ [ H ] .

The curvature vector (10.24) is written for convenience as ⎢ ∂θ {χ }= ⎢ − y ⎣ ∂x

∂θ x ∂θ y ⎥ − ⎥ ∂x ∂y ⎦

∂θ x ∂y

T

= ⎣ χ1

χ2

χ12 ⎦ T .

(10.119)

The elements of the curvature vector can be expressed in terms of the rows of the ( 6 × 9 ) matrices [ G ] and [ H ] as follows

}{ q e },

χ1 =

1 ⎣1 ξ η ⎦ { X 2A

χ2 =

1 e ⎣1 ξ η ⎦ {Y } q , 2A

χ12 =

1 e ⎣1 ξ η ⎦ { Z } q , 2A

{ }

(10.120)

{ }

where A is the area of the triangle with vertices (x1 , y1 ) , (x2 , y2 ) and (x3 , y3 ) . In (10.120) ⎧ β 2 ⎣G ⎦ + β 3 ⎣G ⎦ ⎫ 2 3 ⎪ ⎪ { X } = ⎨ 2β 2 ⎣G ⎦ 4 + β 3 ⎣G ⎦ 5 ⎬ , ⎪ β G + 2β G ⎪ 3 ⎣ ⎦6 ⎭ ⎩ 2 ⎣ ⎦5

⎧ γ 2 ⎣H ⎦ + γ 3 ⎣H ⎦ ⎫ 2 3 ⎪ ⎪ {Y } = ⎨ 2γ 2 ⎣H ⎦ 4 + γ 3 ⎣H ⎦ 5 ⎬ , ⎪ γ H + 2γ H ⎪ 3 ⎣ ⎦6 ⎭ ⎩ 2 ⎣ ⎦5

256


⎧ γ 2 ⎣G ⎦ + γ 3 ⎣G ⎦ + β 2 ⎣H ⎦ + β 3 ⎣H ⎦ ⎫ 2 3 2 3 ⎪ ⎪ { Z } = ⎨ 2γ 2 ⎣G ⎦ 4 + γ 3 ⎣G ⎦ 5 + 2β 2 ⎣H ⎦ 4 + β 3 ⎣H ⎦ 5 ⎬ , ⎪ γ G + 2γ G + β H + 2 β H ⎪ 3 ⎣ ⎦6 2 ⎣ ⎦5 3 ⎣ ⎦6 ⎭ ⎩ 2 ⎣ ⎦5

(10.121)

where ⎣G ⎦ i and ⎣H ⎦ i represent the ith row of [ G ] and [ H ] , respectively, and

β i = y j − yk and γ i = xk − x j 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 1 { χ }T [ D ] { χ }d x d y , Ue = (10.122) 2

∫ A

where the material matrix [ D

]

Ue =

1 2

[ ]

has the form (10.13), U e can be written as

{ q } [ K ]{ q }, e T

e

e

where K e is the element stiffness matrix in global coordinates, given by

[K ] e

⎡ [ X ]⎤ 1 ⎢ = [ Y ] ⎥⎥ ⎢ 2A ⎢⎣ [ Z ] ⎥⎦

T

⎡ D11 [ R ] D12 [ R ] D16 [ R ]⎤ ⎡ [ X ] ⎤ ⎢ ⎥ D22 [ R ] D26 [ R ]⎥ ⎢⎢ [ Y ] ⎥⎥ , ⎢ ⎢ SYM D66 [ R ]⎥⎦ ⎢⎣ [ Z ] ⎥⎦ ⎣

(10.123)

with ⎡ 12 4 4 ⎤ 1 ⎢ [ R ] = ⎢ 4 2 1 ⎥⎥ . 24 ⎢⎣ 4 1 2 ⎥⎦

(10.124)

[ ]

The matrix K e is of rank 6, so there are 3 deformation modes with zero strain energy. The integrations in (10.122) have the form ξ 1−ξ

∫∫ 0 0

ξ mη n dξ dη ==

m ! n! . (m + n + 2)!

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 Admissible functions 144 Area coordinates 180 Aspect ratio 178 Assembly of global matrices Axial effects 97

232

100, 151

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

266 Object of FEA

FINITE ELEMENT ANALYSIS 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 − virtual displacements 134 − virtual work 131 Quadratic strain triangle 185 − triangle 222 Quadrilateral membrane element − eight-node 208 − linear 191 − nine-node 219

139

191

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 Virtual displacements 131 − work of external loads 133 − work of internal forces 193 WB plate-bending element

239

30

M-Rades-FiniteElementAnalysis

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