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The Finite Element Method

Contents

1. Introduction

2. A Simple Example

3. Trusses

4. Linear Systems of Equations

5. Basic Equations of Elasticity Theory

6. Finite Elements for Plane Stress Problems

7. Finite Elements for Three-Dimensional Problems

8. Dynamical Problems

9. Beam Elements

GerhardMercatorUniversitätDuisburg

The Finite Element Method Manfred BraunFEM 0.0-1

Literature

[1] Johannes Altenbach, Udo Fischer: Finite-Elemente-Praxis. Fachbuchverlag, Leipzig 1991.ISBN 3-343-00686-6

[2] Klaus-Jürgen Bathe: Finite-Elemente-Methoden. Matrizen und lineare Algebra, die Methodeder finiten Elemente, Lösung von Gleichgewichtsbedingungen und Bewegungsgleichungen.Springer-Verlag, Berlin ·Heidelberg ·New York ·Tokyo 1986. ISBN 3-540-15602-X

[3] Josef Betten: Finite Elemente für Ingenieure 1. Grundlagen, Matrixmethoden, ElastischesKontinuum. Springer-Verlag, Berlin ·Heidelberg ·New York 1997. ISBN 3-540-63239-5

[4] Josef Betten: Finite Elemente für Ingenieure 2. Variationsrechnung, Energiemethoden,Näherungsverfahren, Nichtlinearitäten. Springer-Verlag, Berlin ·Heidelberg ·New York 1998.

ISBN 3-540-63240-9

[5] Richard H. Gallagher: Finite Element Analysis: Fundamentals. Prentice-Hall, EnglewoodCliffs, N. J., 1975. ISBN 0-13-317248-1


Literature Manfred BraunFEM 0.1-1

Literature (cont’d)

[6] Dietmar Gross, Werner Hauger, Walter Schnell, Peter Wriggers: Technische Mechanik.Band 4: Hydromechanik, Elemente der Höheren Mechanik, Numerische Methoden. Springer-Verlag, Berlin ·Heidelberg ·New York 1993. ISBN 3-540-56629-5

[7] Bernd Klein: FEM. Grundlagen und Anwendungen der Finite-Elemente-Methode. Vieweg,Braunschweig ·Wiesbaden, dritte, überarbeitete Auflage, 1999. ISBN 3-528-25125-5

[8] Günther Müller, Clemens Groth: FEM für Praktiker. Die Methode der Finiten Elemente mitdem FE-Programm ANSYS. expert-verlag, Renningen-Malmsheim, dritte, völlig neubear-beitete Auflage, 1997. ISBN 3-8169-1525-6

[9] Douglas H. Norrie, Gerard de Vries: The Finite Element Method. Academic Press, New York1973. ISBN 0-12-521650-5

[10] J. Tinsley Oden: Finite Elements of Nonlinear Continua. McGraw-Hill, New York 1972.


Literature (cont’d) Manfred BraunFEM 0.1-2

Literature (cont’d)

[11] Hans Rudolf Schwarz: Methode der finiten Elemente. Eine Einführung unter besondererBerücksichtigung der Rechenpraxis. B. G. Teubner, Stuttgart 1980. ISBN 3-519-02349-0

[12] Hans Rudolf Schwarz: FORTRAN-Programme zur Methode der finiten Elemente. B. G. Teub-ner, Stuttgart 1981. ISBN 3-519-02064-5

[13] Gilbert Strang and George J. Fix: An Analysis of the Finite Element Method. Prentice-Hall,Englewood-Cliffs, N. J., 1973. ISBN 0-13-032946-0

[14] Olgierd C. Zienkiewicz: Methode der finiten Elemente. Hanser Verlag, München, zweite,erweiterte und völlig neubearbeitete Auflage, 1984. ISBN 3-446-12525-6

[15] Olgierd C. Zienkiewicz and Robert L. Taylor: The Finite Element Method. McGraw-Hill,London, fourth edition, 1989.


Literature (cont’d) Manfred BraunFEM 0.1-3

Contents

1. Introduction

1.1 What is the Finite Element Method

1.2 Brief History

2. A Simple Example

3. Trusses






9. Beam Elements


Contents Manfred BraunFEM 1.0-1

Introduction

What is the Finite Element Method?

• The finite element method (FEM) is a numerical method for solving problems of engineeringand mathematical physics. Its primary application is in Strength of Materials.

• The FEM is useful for problems with complicated geometries, loadings, and material propertieswhere analytical solutions cannot be obtained.

• The model body is divided into a system of small but finite bodies, the finite elements, inter-connected at nodal points or nodes.

• In each of the finite element the unknown fields are approximated by simple functions, whichare determined by their nodal values.

• The discretization by finite elements yields a large system of equations for the unknown nodalvalues.


Introduction Manfred BraunFEM 1.1-1

Brief History

• A. Hrennikoff (1941), Solutions of problems in elasticity by the framework method

• D. McHenry (1943), A lattice analogy for the solution of plane stress problems

• R. Courant (1943), Variational methods for the solutions of problems of equilibrium and vibra-tion

• J. H. Argyris (1954–55), Energy theorems and structural analysis

• M. J. Turner, R. W. Clough, H. C. Martin, and L. P. Topp (1956), Stiffness and deflectionanalysis of complex structures

• R. W. Clough (1960), The finite element method in plane stress analysis

Some Names

John H. Argyris, Ivo Babuška, Klaus-Jürgen Bathe, Philipe G. Ciarlet, Richard H. Gallagher,Erwin Stein, Robert L. Taylor, Peter Wriggers, Olek C. Zienkiewicz


Brief History Manfred BraunFEM 1.2-1


Manfred BraunFEM 1.2-2

Contents

1. Introduction

2. A Simple Example

2.1 Statement of Problem and Exact Solution

2.2 Approximate Solution Using Finite Elements

2.3 New Approach: Strain Energy

3. Trusses






9. Beam Elements



Elastic Rod Loaded by Self-Weight and End Load

x

ρ , E , A

F`

0

x

`

u(x)

g

Elongation or strain

ε =du

dx

Stressσ = Eε

Tensile forceF = Aσ = AE

du

dx

Equilibrium conditiondF

dx= −ρgA


Elastic Rod Loaded by Self-Weight and End Load Manfred BraunFEM 2.1-1

Boundary Value Problem and Solution

x

ρ , E , A

F`

0

x

`

u(x)

g

Differential equation

d

dx

(

AEdu

dx

)

+ ρgA = 0

Boundary conditions

u(0) = 0,du

dx

∣

∣

∣

∣

x=`

=F`

AE

Assumption: Constant tensile stiffness, AE = const

Closed-form solution of the boundary-value problem

u =

[

ρg

E

(

`− x

2

)

+F`

EA

]

x


Boundary Value Problem and Solution Manfred BraunFEM 2.1-2

Exact Solution

0 G`2EA

G`EA

0

`

u

x

F` = 0 G/2 G

0 G 2G0

`

F

x

G/2 G

u =Gx

EA

(

1 +F`

G− x

2`

)

F = F` + G(

1− x

`

)


Exact Solution Manfred BraunFEM 2.1-3

Discretization by Finite Elements

Total system

nodeelement

x

0

h

`

0

1

2

3

4

1

2

3

4

Single element

node

ξ

0

1

0

1

coordinatetransformation

x = x0 + hξ

Displacement ansatz

u = (1− ξ)u0 + ξu1

Nodal displacements

u0 , u1

Interpolation functions

N0 = 1− ξ , N1 = ξ

Strain in element

ε =du

dx=

1

h· du

dξ=

u1 − u0

h

Stress resultant

F =EA

h(u1 − u0)


Discretization by Finite Elements Manfred BraunFEM 2.2-1

Collecting the Elements

Overall system

nodeelement

x

0

h

`

0

1

2

3

4

1

2

3

4

Single element

node

ξ

0

1

0

1

coordinatetransformation

x = x0 + hξ

Return to global numbering within the overall system

Fi =EA

h(ui − ui−1)

i = 1, 2, 3, 4

Global vector of stress resultants

F1

F2

F3

F4

=EA

h

−1 1

−1 1

−1 1

−1 1

u0

u1

u2

u3

u4


Collecting the Elements Manfred BraunFEM 2.2-2

Equilibrium ConditionsF0

12ρgAh

F1

0

Fk

12ρgAh

12ρgAh

Fk+1

k

F4

12ρgAh

F`

4

Node with adjacent half elements

F0 = F1 +1

2ρgAh

Fk = Fk+1 + ρgAh 0 < k < 4

F4 = F` +1

2ρgAh

Equilibrium conditions in matrix form

−1

1 −1

1 −1

1 −1

1

F1

F2

F3

F4

= ρgAh

1/21

1

11/2

+

−F0

0

0

0

F`


Equilibrium Conditions Manfred BraunFEM 2.2-3

Resulting System of Equations

Equilibrium conditions

−11 −1

1 −11 −1

1

F1

F2

F3

F4

= ρgAh

1/21

111/2

+

−F0

0

00

F`

Stress resultants

F1

F2

F3

F4

=EA

h

−1 1−1 1

−1 1−1 1

u0

u1

u2

u3

u4

System of equations

EA

h

1 −1

−1 2 −1

−1 2 −1

−1 2 −1

−1 1

u0

u1

u2

u3

u4

= ρgAh

1/21

1

11/2

+

− F0

0

0

0

F`

· unknown


Resulting System of Equations Manfred BraunFEM 2.2-4

Comparison Between Exact and Approximate Solutions

Displacement

0 G`2EA

G`EA

0

h

2h

3h

4h

u

x

�

�

�

�

�

Stress resultant

G 2G0

h

2h

3h

4h

F

x


Comparison Between Exact and Approximate Solutions Manfred BraunFEM 2.2-5

Strain Energy

Displacement within single element

u = (1− ξ)u0 + ξu1

Strain

ε =1

h(u1 − u0)

Strain energy of single element

Πelem =1

2

∫ 1

ξ=0

EAε2 h dξ =EA

2h(u 2

0 − 2u0u1 + u 21 )

Strain energy of total system

Π =EA

2h

[

u 20 − 2u0u1 + u 2

1 +

+ u 21 − 2u1u2 + u 2

2 +

+ u 22 − 2u2u3 + u 2

3 +

+ u 23 − 2u3u4 + u 2

4

]


Strain Energy Manfred BraunFEM 2.3-1

Strain Energy (contd.)

Matrix representation of strain energy

Π =1

2uTKu

Global nodal displacement vector

u =

u0

u1

u2

u3

u4

Global stiffness matrix

K =EA

h

1 −1

−1 2 −1

−1 2 −1

−1 2 −1

−1 1


Strain Energy (contd.) Manfred BraunFEM 2.3-2

Principle of Virtual Work

Virtual work of the external forces

(δW ) = pTδu

Global load vector, vector of virtual displacements

p = ρgAh

1/21

1

11/2

+

−F0

0

0

0

F`

, δu =

δu0

δu1

δu2

δu3

δu4

Principle of virtual work

δΠ = (δW ) for arbitrary virtual displacements δu


Principle of Virtual Work Manfred BraunFEM 2.3-3

Principle of Virtual Work (contd.)

Strain energy

Π =1

2uTKu

Variation of strain energy

δΠ =1

2uT(

K + KT)

δu = uTK δu

due to symmetry of the global stiffness matrix K

Consequence of the Principle of Virtual Work

(Ku− p)T δu = 0 for arbitrary δu

Linear system of equations

Ku = p


Principle of Virtual Work (contd.) Manfred BraunFEM 2.3-4

Contents

1. Introduction

2. A Simple Example

3. Trusses

3.1 Data of a Truss

3.2 Element Stiffness Matrix

3.3 Global Stiffness Matrix

3.4 Supports and Reactive Forces

3.5 How to Develop a Truss Program




7. Finite Elements for Three-Dimensional Prob-lems


9. Beam Elements



Trusses

��

��

��

��

��

��

�

� �

�

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

Characteristics of a truss

• Assembly of pin-jointed members

• External forces applied to nodes

• Members loaded in axial direction

Prescribed:

– Geometry– Material data– Support– Load

Demanded:

– Nodal displacements– Member forces– Reactive forces


Trusses Manfred BraunFEM 3.1-1

Elongation of a Rod

Rod in undeformed and deformed configuration

e

`

` + ∆`u 1

u 2

Elongation∆` = e

. (u 2 − u 1)

Matrix representation

∆` =[

−eT +e

T]

u1

u2

=[

uT1 u

T2

]

−e

+e


Elongation of a Rod Manfred BraunFEM 3.2-1

Element Stiffness Matrix of a Single Member

Strain energy stored in a single member

Π =1

2

EA

`(∆`)2

Strain energy represented in terms of nodal displacements

Π =1

2

EA

`

[

u1 u2

]

eeT −ee

T

−eeT

eeT

u1

u2

=

1

2uTKu

Nodal displacement vector and element stiffness matrix

u =

u1

u2

K =

EA

`

eeT −ee

T

−eeT

eeT


Element Stiffness Matrix of a Single Member Manfred BraunFEM 3.2-2

Contribution of a Single Rod to the Global Stiffness Matrix

1 2 (1)

3(2) 4Element stiffness matrix of rod (2→ 4 )

K(2→4) =

K11 K12

K21 K22

Global nodal displacement vector and contribution of rod (2 → 4) to the global stiffness matrix

u =

u1

u2

u3

u4

K =

K11 K12

K21 K22


Contribution of a Single Rod to the Global Stiffness Matrix Manfred BraunFEM 3.3-1

Support and Reaction Forces

Types of support

Fx

ux = 0

Fy

uy = 0

Fx

Fy

ux = uy = 0

Decomposition of the nodal displacement and force vectors

u =

ue

ux

ue free displacementsux fixed displacements

p =

pe

px

pe given forcespx reactive forces


Support and Reaction Forces Manfred BraunFEM 3.4-1

Decomposition of the System of Equations

System of equations

Kee Kex

Kxe Kxx

ue

ux

=

pe

px

· unknown quantity

Decomposition

1. Kee ue = pe −Kexux =⇒ ue free displacements

2. px = Kxeue + Kxxux =⇒ px reactive forces


Decomposition of the System of Equations Manfred BraunFEM 3.4-2

How to Develop a Truss Program

1. Input nodal coordinates. Reserve memory for

� Nodal displacement vector u

� Nodal force vector p

� Global stiffness matrix K (clear matrix to 0 )

2. Input and process member data:

� Compute element stiffness matrices KElem

=

K11 K12

K21 K22

, split them up, and

� accumulate them in the global stiffness matrix K

3. Allow for supports:

� Split up the nodal vectors u =

ue

ux

and p =

pe

px

� Enter the fixed nodal displacements ux


How to Develop a Truss Program Manfred BraunFEM 3.5-1

How to Develop a Truss Program (contd.)

4. Input loads

� Enter the vector pe

of given loads

5. Solve the reduced system of equations for the unknown free displacements

Kee

ue

= pe−K

exu

x=⇒ u

e

6. Compute the reaction forces

px = Kxeue + Kxxux =⇒ px

7. Compute member forces

F =EA

`e

. (u 2 − u 1)


How to Develop a Truss Program (contd.) Manfred BraunFEM 3.5-2

Contents

1. Introduction

2. A Simple Example

3. Trusses


4.1 Some Mathematical Foundations

4.2 Cholesky Decomposition

4.3 How to Store Sparse Matrices

4.4 Other Methods





9. Beam Elements



Linear Systems of Equations

Ax = b

detA 6= 0

Solution exists, is unique.

��

��

detA = 0

yTb = 0

for all y satisfying

yTA = 0

Solution exists, is not unique.

��

��

yTb 6= 0

for some y satisfying

yTA = 0

Solution does not exist.

��

��


Linear Systems of Equations Manfred BraunFEM 4.1-1

Cholesky’s Method

Solution of the linear system Ax = b with a symmetric, positiv definite matrix A .

1. Factorization

Decompose the symmetric matrix A into the product A = UTU

where U is an upper triangular matrix.

2. Forward Substitution

Solve the lower triangular system UTy = b for the auxiliary vector y .

3. Backward Substitution

Solve the upper triangular system Ux = y for the requested vector x .

André-Louis Cholesky (1875–1918): French military officer involved in geodesy and surveyingin Crete and North Africa


Cholesky’s Method Manfred BraunFEM 4.2-1

Cholesky’s Method: Factorization

1. Factorization

A = UT U

s = Aij −i−1∑

k=1

Uki Ukj

Uij =

s

Uii

if i < j

√s if i = j

i = 1, . . . , j

j = 1, . . . , n


Cholesky’s Method: Factorization Manfred BraunFEM 4.2-2

Cholesky’s Method: Forward and Backward Substitution

2. Forward Substitution

UT y = b yi =1

Uii

(

bi −i−1∑

k=1

Uki yk

)

i = 1, . . . , n

3. Backward Substitution

U x = y xi =1

Uii

(

yi −n∑

k=i+1

Uik xk

)

i = n, . . . , 1


Cholesky’s Method: Forward and Backward Substitution Manfred BraunFEM 4.2-3

Contents

1. Introduction

2. A Simple Example

3. Trusses



5.1 Displacements

5.2 Strain

5.3 Stress

5.4 Equilibrium

5.5 Strain Energy




9. Beam Elements



Strain

x

y

x

ydx

dy

x

y

(

1 +∂ux

∂x

)

dx

∂uy

∂xdx

(

1 +∂uy

∂y

)

dy

∂ux

∂ydy

ux

uy

Elongations

εx =∂ux

∂xεy =

∂uy

∂y

Shear

γxy =∂ux

∂y+

∂uy

∂x


Strain Manfred BraunFEM 5.2-1

Hooke’s Law

Uniaxial stress

σ

ε =1

Eσ ε

t= − ν

Eσ E Young’s modulus

ν Poisson’s ratio

Shear stress

τ

γ =1

Gτ G Shear modulus


Hooke’s Law Manfred BraunFEM 5.3-1

Simple Shear: Principal Stresses

τxy

τxy

τxy

π/4

σ1 = τxy

σ2 = −τxy

π/2σ

τ

τxy

−τxy

σ1 = τxy

σ2 = −τxy


Simple Shear: Principal Stresses Manfred BraunFEM 5.3-2

Simple Shear: Deformation

π/4

a

a

σ1 = τ

σ2 = −τ

π / 4−

γ / 2

a(1 + ε)

a(1

−ε)

τττ


Simple Shear: Deformation Manfred BraunFEM 5.3-3

Relation Among Material Constants

Geometry1− ε

1 + ε= tan

(π

4− γ

2

)

=

√

1− sin γ

1 + sin γ

Approximation for small strain:γ ≈ 2ε

Hooke’s Law

ε =1

E(σ1 − νσ2) =

1 + ν

Eτ γ =

1

Gτ

Shear modulus expressed in terms of Young’s modulus and Poisson’s ratio

G =E

2(1 + ν)


Relation Among Material Constants Manfred BraunFEM 5.3-4

Governing Equations of Plane Stress Problems

Definition of strain

εx

εy

γxy

=

∂∂x

0

0 ∂∂y

∂∂y

∂∂x

[

ux

uy

]

Stress-strain relation (Hooke’s law)

σx

σy

τxy

=E

1− ν2

1 ν 0

ν 1 0

0 0 1−ν2

εx

εy

γxy

Specific strain energy

W =1

2

[

εx εy γxy

]

σx

σy

τxy


Governing Equations of Plane Stress Problems Manfred BraunFEM 5.3-5

Contents

1. Introduction

2. A Simple Example

3. Trusses




6.1 Linear Triangular Elements

6.2 Bilinear Quadrilateral Elements

6.3 Higher Elements



9. Beam Elements



Linear Interpolation

ξ

η

0 10

1

1 2

3Problem: Determine the linear function u = u(ξ, η) satisfying

u(0, 0) = u1 u(1, 0) = u2 u(0, 1) = u3

Solution:

u(ξ, η) = (1− ξ − η)u1 + ξu2 + ηu3

Interpolation or shape functions

N1(ξ, η) = 1− ξ − η

N2(ξ, η) = ξ

N3(ξ, η) = η

Characteristic property:

Ni(ξk, ηk) =

1 i = k

if0 i 6= k


Linear Interpolation Manfred BraunFEM 6.1-1

Affine Transformation

x

y

x1 x2x3

y1

y2

y3

1

2

3

⇐=

ξ

η

0 10

1

1 2

3

Coordinate transformation[

x

y

]

=

[

x1 x2 x3

y1 y2 y3

]

1− ξ − η

ξ

η


Affine Transformation Manfred BraunFEM 6.1-2

Derivatives

Jacobi matrix

∂x∂ξ

∂x∂η

∂y∂ξ

∂y∂η

=

[

x1 x2 x3

y1 y2 y3

]

−1 −1

1 0

0 1

=

[

x2 − x1 x3 − x1

y2 − y1 y3 − y1

]

Determinant of Jacobi matrix, “Jacobian”

J = (x2 − x1)(y3 − y1)− (x3 − x1)(y2 − y1)

Inverse Jacobi matrix

∂ξ∂x

∂ξ∂y

∂η∂x

∂η∂y

=1

J

[

y3 − y1 x1 − x3

y1 − y2 x2 − x1

]

Transformation of derivatives

[

∂ϕ∂x

∂ϕ∂y

]

=[

∂ϕ∂ξ

∂ϕ∂η

]

∂ξ∂x

∂ξ∂y

∂η∂x

∂η∂y


Derivatives Manfred BraunFEM 6.1-3

Linear Triangular Element

x

y

x1 x2x3

y1

y2

y3

1

2

3Shape functions

N1(ξ, η)

N2(ξ, η)

N3(ξ, η)

=

1− ξ − η

ξ

η

Derivatives of the shape functions with respect to the global coordinates (x, y)

∂N1

∂x

∂N1

∂y

∂N2

∂x

∂N2

∂y

∂N3

∂x

∂N3

∂y

=

∂N1

∂ξ

∂N1

∂η

∂N2

∂ξ

∂N2

∂η

∂N3

∂ξ

∂N3

∂η

∂ξ∂x

∂ξ∂y

∂η∂x

∂η∂y

=

−1 −1

1 0

0 1

· 1

J

[

y3 − y1 x1 − x3

y1 − y2 x2 − x1

]

=1

J

y2 − y3 x3 − x2

y3 − y1 x1 − x3

y1 − y2 x2 − x1

Jacobian J = (x2 − x1)(y3 − y1)− (x3 − x1)(y2 − y1)


Linear Triangular Element Manfred BraunFEM 6.1-4

Linear Triangular Element: Stiffness Matrix

Matrix of derivatives of shape functions

DN =1

J

y2 − y3 0 y3 − y1 0 y1 − y2 0

0 x3 − x2 0 x1 − x3 0 x2 − x1

x3 − x2 y2 − y3 x1 − x3 y3 − y1 x2 − x1 y1 − y2

Matrix of material constants

E =E

1− ν2

1 ν 0

ν 1 0

0 0 1−ν2

Element stiffness matrix of the linear triangular element

K =1

2Jh (DN)T E (DN)


Linear Triangular Element: Stiffness Matrix Manfred BraunFEM 6.1-5

Bilinear Interpolation

ξ

η

0 10

1

1 2

3 4Problem: Determine the bilinear function u = u(ξ, η) satisfying

u(0, 0) = u1 u(1, 0) = u2 u(0, 1) = u3 u(1, 1) = u4

Solution:

u(ξ, η) = (1− ξ)(1− η)u1 + ξ(1− η)u2 + (1− ξ)ηu3 + ξηu4

Interpolation or shape functions

N1(ξ, η) = (1− ξ)(1− η)

N2(ξ, η) = ξ(1− η)

N3(ξ, η) = (1− ξ)η

N4(ξ, η) = ξη

Characteristic property:

Ni(ξk, ηk) =

1 i = k

if0 i 6= k


Bilinear Interpolation Manfred BraunFEM 6.2-1

Bilinear Parallelogram Element

x

y

x1 x2x3

y1

y2

y3

12

34

Shape functions

N1(ξ, η)

N2(ξ, η)

N3(ξ, η)

N4(ξ, η)

=

(1− ξ)(1− η)

ξ(1− η)

(1− ξ)η

ξη

Derivatives of the shape functions with respect to the global coordinates (x, y)

∂N1

∂x

∂N1

∂y∂N2

∂x

∂N2

∂y∂N3

∂x

∂N3

∂y∂N4

∂x

∂N4

∂y

=

∂N1

∂ξ

∂N1

∂η∂N2

∂ξ

∂N2

∂η∂N3

∂ξ

∂N3

∂η∂N4

∂ξ

∂N4

∂η

[

∂ξ∂x

∂ξ∂y

∂η∂x

∂η∂y

]

=

−(1− η) −(1− ξ)

1− η −ξ

−η 1− ξ

η ξ

· 1

J

[

y3 − y1 x1 − x3

y1 − y2 x2 − x1

]

Jacobian J = (x2 − x1)(y3 − y1)− (x3 − x1)(y2 − y1)


Bilinear Parallelogram Element Manfred BraunFEM 6.2-2

Bilinear Transformation

x

y

x1 x2x3 x4

y1

y2

y3

y4

1

2

34

⇐=

ξ

η

0 10

1

1 2

3 4

Coordinate transformation

[

x

y

]

=

[

x1 x2 x3 x4

y1 y2 y3 y4

]

(1− ξ)(1− η)

ξ (1− η)

(1− ξ) η

ξ η


Bilinear Transformation Manfred BraunFEM 6.2-3

Derivatives

Jacobi matrix

∂x∂ξ

∂x∂η

∂y∂ξ

∂y∂η

=

[

x1 x2 x3 x4

y1 y2 y3 y4

]

−(1− η) −(1− ξ)

1− η −ξ

−η 1− ξ

η ξ

Determinant of Jacobi matrix, “Jacobian”

J = ∂x∂ξ

∂y∂η− ∂y

∂ξ∂x∂η

Inverse Jacobi matrix

∂ξ∂x

∂ξ∂y

∂η∂x

∂η∂y

=

∂x∂ξ

∂x∂η

∂y∂ξ

∂y∂η

−1

=1

J

∂x∂η−∂x

∂η

−∂y∂ξ

∂x∂ξ

matrix elements arerational functions of ξ, η

Transformation of derivatives

[

∂ϕ∂x

∂ϕ∂y

]

=[

∂ϕ∂ξ

∂ϕ∂η

]

∂ξ∂x

∂ξ∂y

∂η∂x

∂η∂y


Derivatives Manfred BraunFEM 6.2-4

Bilinear Quadrilateral Element: Stiffness Matrix

Derivatives of shape functions

[

∂Nk

∂x

∂Nk

∂y

]

=[

∂Nk

∂ξ

∂Nk

∂η

]

∂ξ∂x

∂ξ∂y

∂η∂x

∂η∂y

. . . rational functions of ξ , η

Matrix of derivatives of shape functions

DN =

∂N1

∂x0 ∂N2

∂x0 ∂N3

∂x0 ∂N4

∂x0

0 ∂N1

∂y0 ∂N2

∂y0 ∂N3

∂y0 ∂N4

∂y

∂N1

∂y

∂N1

∂x

∂N2

∂y

∂N2

∂x

∂N3

∂y

∂N3

∂x

∂N4

∂y

∂N4

∂x

Element stiffness matrix of the bilinear quadrilateral element

K =

1∫

η=0

1∫

ξ=0

(DN)T E (DN) Jh dξ dη numericalintegration

(Gauss)


Bilinear Quadrilateral Element: Stiffness Matrix Manfred BraunFEM 6.2-5

Quadratic Triangular Elements

ξ

η

0 10

1

1 2

3

45

6

N1 = (1− ξ − η)(1− 2ξ − 2η)

N2 = ξ(2ξ − 1)

N3 = η(2η − 1)

N4 = 4ξη

N5 = 4η(1− ξ − η)

N6 = 4ξ(1− ξ − η)


x

y

1

2

3

Isoparametric Transformation

x

y

1

2

3

4

5

6


Quadratic Triangular Elements Manfred BraunFEM 6.3-1

Quadrilateral Elements of Lagrange Type

ξ

η

0 1

1

1 2 3

4

567

8 9

N1 =1

4ξ(1− ξ)η(1− η)

N2 = −1

2(1− ξ2)η(1− η)

N3 = −1

4ξ(1 + ξ)η(1− η)

N4 =1

2ξ(1 + ξ)(1− η2)

N5 =1

4ξ(1 + ξ)η(1 + η)

N6 =1

2(1− ξ2)η(1 + η)

N7 = −1

4ξ(1− ξ)η(1 + η)

N8 = −1

2ξ(1− ξ)(1− η2)

N9 = (1− ξ2)(1− η2)


x

y

1

3

7


x

y

1 2

3

4

56

7

89


Quadrilateral Elements of Lagrange Type Manfred BraunFEM 6.3-2

Quadrilateral Elements of Serendipity Type

ξ

η

0 1

1

1 2 3

4

567

8

N1 = −1

4(1− ξ)(1− η)(1 + ξ + η)

N3 = −1

4(1 + ξ)(1− η)(1− ξ + η)

N5 = −1

4(1 + ξ)(1 + η)(1− ξ − η)

N7 = −1

4(1− ξ)(1 + η)(1 + ξ − η)

N2 =1

2(1− ξ2)(1− η)

N4 =1

2(1 + ξ)(1− η2)

N6 =1

2(1− ξ2)(1 + η)

N8 =1

2(1− ξ)(1− η2)


x

y

1

3

7


x

y

1 2

3

4

56

7

8


Quadrilateral Elements of Serendipity Type Manfred BraunFEM 6.3-3

Deriving the Stiffness Matrix of a Finite Element

Displacement vector

u (ξ, η) =

n∑

i=1

Ni(ξ, η) ui

Strain

ε = D u =n∑

i=1

(DNi) ui

Stress

σ = E ε =n∑

k=1

E (DNk) uk

Strain energy

Π =1

2

∫

V

εTσ dV =

1

2

n∑

i=1

n∑

k=1

uT

i

∫

V

(DNi)T E (DNk) dV uk


Deriving the Stiffness Matrix of a Finite Element Manfred BraunFEM 6.3-4

Element Stiffness Matrices of Finite Elements for Plane Stress Problems

General structure

K =

K11 K12 . . . K1n

K21 K22 . . . K2n

... ... . . . ...

Kn1 Kn2 . . . Knn

Submatrices

Kik =

∫

A

(DNi)T E (DNk) h dA

Kik =E

1− ν2

∫

A

∂Ni

∂x

∂Nk

∂x+

1− ν

2

∂Ni

∂y

∂Nk

∂yν∂Ni

∂x

∂Nk

∂y+

1− ν

2

∂Ni

∂y

∂Nk

∂x

ν∂Ni

∂y

∂Nk

∂x+

1− ν

2

∂Ni

∂x

∂Nk

∂y

∂Ni

∂y

∂Nk

∂y+

1− ν

2

∂Ni

∂x

∂Nk

∂x

h dA


Element Stiffness Matrices of Finite Elements for Plane Stress Problems Manfred BraunFEM 6.3-5

Contents

1. Introduction

2. A Simple Example

3. Trusses





7.1 Linear Tetrahedral Elements

7.2 Trilinear Hexahedral Elements

7.3 Higher Elements


9. Beam Elements



Contents

1. Introduction

2. A Simple Example

3. Trusses






8.1 Natural and Forced Vibrations

8.2 Mass Matrices

8.3 Natural frequencies and modes

9. Beam Elements



Contents

1. Introduction

2. A Simple Example

3. Trusses






9. Beam Elements

9.1 Hermite Interpolation

9.2 Mass and Stiffness Matrices of Beam Elements

9.3 Cubic Splines



Beam Elements: Hermite Interpolation

��

��

w0

w′0 w1

w′1

x

w

hx0

x = x0 + hξ w = N00(ξ)w0 + hN01(ξ)w′0 + N10(ξ)w1 + hN11(ξ)w′

1


Beam Elements: Hermite Interpolation Manfred BraunFEM 9.1-1

Shape Functions

0

1

0 1

N00 = (1− ξ)2(1 + 2ξ)

N01 = ξ(1− ξ)20

1

0

1

N10 = ξ2(3− 2ξ)

N11 = −ξ2(1− ξ)


Shape Functions Manfred BraunFEM 9.1-2

Element Stiffness Matrix

Strain energy of beam element

Π =1

2

x0+h∫

x=x0

EI

(

d2w

dx2

)2

dx =1

2h3

1∫

ξ=0

EI

(

d2w

dξ2

)2

dξ

EI bending stiffness

Displacementw = N00(ξ)w0 + hN01(ξ)w′

0 + N10(ξ)w1 + hN11(ξ)w′1

Nodal displacement and load vectors, element stiffness matrix

u =

w0

w′0

w1

w′1

p =

F0

M0

F1

M1

K =2EI

h3

6 3h −6 3h

3h 2h2 −3h h2

−6 −3h 6 −3h

3h h2 −3h 2h2


Element Stiffness Matrix Manfred BraunFEM 9.2-1

Element Mass Matrix

Kinetic energy of beam element

Π =1

2

x0+h∫

x=x0

ρAw2dx =

h

2

1∫

ξ=0

ρAw2dξ

ρA mass per unit length

Displacement velocity

w = N00(ξ)w0 + hN01(ξ)w′0 + N10(ξ)w1 + hN11(ξ)w′

1

Nodal velocity vector and element mass matrix

u =

w0

w′0

w1

w′1

M =m

420

156 22h 54 −13h

22h 4h2 13h −3h2

54 13h 156 −22h

−13h −3h2 −22h 4h2


Element Mass Matrix Manfred BraunFEM 9.2-2

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