Labrosse_Finite Element Analysis

8/11/2019 Labrosse_Finite Element Analysis

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MCG 4102/5108 Finite ElementAnalysis

Michel R. Labrosse, Ph.D., P.Eng., Associate ProfessorDepartment of Mechanical Engineering

University of Ottawa

cM.R. Labrosse, Ottawa, Canada, 2012


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Contents

1 Introduction to FEA: Trusses 11.1 Discretization . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Element stiness relationship in local coordinates . . . . . . . 31.3 Transformation from local to global coordinates . . . . . . . . 51.4 Global element stiness relationship . . . . . . . . . . . . . . . 61.5 Assemblage . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.6 Application of loads . . . . . . . . . . . . . . . . . . . . . . . . 91.7 Application of restraints on nodal displacements and solution . 91.8 Processing of results . . . . . . . . . . . . . . . . . . . . . . . 111.9 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Linear elasticity - Principle of virtual work 132.1 Stress at a point . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2 Equations of static equilibrium . . . . . . . . . . . . . . . . . 142.3 Strain at a point . . . . . . . . . . . . . . . . . . . . . . . . . 152.4 Strain-displacement relations . . . . . . . . . . . . . . . . . . . 162.5 Compatibility equations . . . . . . . . . . . . . . . . . . . . . 172.6 A constitutive relationship - Hookes law . . . . . . . . . . . . 172.7 Principle of minimum potential energy . . . . . . . . . . . . . 182.8 Principle of virtual work . . . . . . . . . . . . . . . . . . . . . 19

3 Finite element for beams 213.1 Theory - Beam bending in one plane . . . . . . . . . . . . . . 213.2 Discretization . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.3 Element stiness and load vectors . . . . . . . . . . . . . . . . 253.4 Assemblage and solution . . . . . . . . . . . . . . . . . . . . . 263.5 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.6 Beam element with axial loading . . . . . . . . . . . . . . . . 29

i


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CONTENTS ii

3.7 General 3-D beam (or frame) element . . . . . . . . . . . . . . 31

4 Interpolation functions - Integration formulas 334.1 Compatibility and completeness requirements . . . . . . . . . 334.2 One dimensional (1-D) elements . . . . . . . . . . . . . . . . . 35

4.2.1 2-node C0-continuous element . . . . . . . . . . . . . . 354.2.2 2-node C1-continuous element . . . . . . . . . . . . . . 36

4.3 2-D elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.3.1 Triangular C0-continuous element . . . . . . . . . . . . 374.3.2 Rectangular (quad) C0-continuous element . . . . . . . 394.3.3 Curved elements . . . . . . . . . . . . . . . . . . . . . 40

4.4 3-D elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.4.1 Four-node, C0

-continuous tetrahedral element (pyramid) 424.4.2 Eight-node, C0-continuous brick element . . . . . . . . 424.4.3 Axisymmetric elements . . . . . . . . . . . . . . . . . . 43

4.5 Integration formulas . . . . . . . . . . . . . . . . . . . . . . . 434.5.1 Direct integration . . . . . . . . . . . . . . . . . . . . . 434.5.2 Numerical integration - Gaussian quadrature . . . . . . 44

4.6 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

5 Stress analysis in 2-D with linear triangular element 465.1 Plane stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

5.1.1 The shape function matrix[N]. . . . . . . . . . . . . . 47

5.1.2 The strain-nodal displacement matrix[B] . . . . . . . 475.1.3 Constitutive relationship . . . . . . . . . . . . . . . . . 485.1.4 Finite element formulation (triangular element) . . . . 485.1.5 The element stiness matrix[Ke] . . . . . . . . . . . . 495.1.6 The element nodal force vectorffeg . . . . . . . . . . 505.1.7 Assemblage . . . . . . . . . . . . . . . . . . . . . . . . 545.1.8 Prescribed displacements - Solution . . . . . . . . . . . 555.1.9 Element resultants . . . . . . . . . . . . . . . . . . . . 55

5.2 Plane strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565.3 Axisymmetric problems . . . . . . . . . . . . . . . . . . . . . . 575.4 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

6 Finite elements for plates and shells 596.1 Introduction - Bending of thin plates . . . . . . . . . . . . . . 59

6.1.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . 59


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CONTENTS iii

6.1.2 Kirchho assumptions . . . . . . . . . . . . . . . . . . 60

6.1.3 Constitutive relationships . . . . . . . . . . . . . . . . 616.1.4 Equilibrium equations . . . . . . . . . . . . . . . . . . 636.2 Finite element formulation (rectangular element) . . . . . . . . 64

6.2.1 Element type . . . . . . . . . . . . . . . . . . . . . . . 646.2.2 Displacement function . . . . . . . . . . . . . . . . . . 656.2.3 Principle of virtual work . . . . . . . . . . . . . . . . . 666.2.4 Element stiness matrix[Ke] . . . . . . . . . . . . . . 676.2.5 Element nodal force vectorffeg . . . . . . . . . . . . . 67

7 Method of weighted residuals 697.1 The method of weighted residuals . . . . . . . . . . . . . . . . 69

7.1.1 General concepts . . . . . . . . . . . . . . . . . . . . . 697.1.2 Point collocation . . . . . . . . . . . . . . . . . . . . . 707.1.3 Subdomain collocation . . . . . . . . . . . . . . . . . . 717.1.4 Least squares . . . . . . . . . . . . . . . . . . . . . . . 727.1.5 Galerkin . . . . . . . . . . . . . . . . . . . . . . . . . . 727.1.6 Comparison with exact solution . . . . . . . . . . . . . 73

7.2 The Galerkin nite element method . . . . . . . . . . . . . . . 747.2.1 Formulation . . . . . . . . . . . . . . . . . . . . . . . . 747.2.2 Application: 1-D heat transfer in a pin n . . . . . . . 76

8 Steady state thermal analysis 79

8.1 1-D heat conduction . . . . . . . . . . . . . . . . . . . . . . . 798.2 Governing equation . . . . . . . . . . . . . . . . . . . . . . . . 798.3 FE formulation . . . . . . . . . . . . . . . . . . . . . . . . . . 808.4 2-D heat conduction . . . . . . . . . . . . . . . . . . . . . . . 83

8.4.1 Governing equation . . . . . . . . . . . . . . . . . . . . 838.4.2 Green-Gauss theorem . . . . . . . . . . . . . . . . . . . 838.4.3 FE formulation . . . . . . . . . . . . . . . . . . . . . . 85

8.5 Axisymmetric heat conduction . . . . . . . . . . . . . . . . . . 87

9 Fluid ow analysis 899.1 2-D potential ow . . . . . . . . . . . . . . . . . . . . . . . . . 89

9.1.1 Velocity potential formulation . . . . . . . . . . . . . . 909.1.2 Stream function formulation . . . . . . . . . . . . . . . 92

9.2 2-D incompressible ow . . . . . . . . . . . . . . . . . . . . . . 93


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CONTENTS iv

10 Transient and dynamic analyses 96

10.1 Dynamic structural analysis . . . . . . . . . . . . . . . . . . . 9610.2 Transient thermal analysis . . . . . . . . . . . . . . . . . . . . 9710.3 Lumped versus consistent matrices . . . . . . . . . . . . . . . 9810.4 Solution methods . . . . . . . . . . . . . . . . . . . . . . . . . 99

10.4.1 Two-point recurrence scheme . . . . . . . . . . . . . . 9910.4.2 Three-point recurrence scheme . . . . . . . . . . . . . . 10210.4.3 Initial conditions . . . . . . . . . . . . . . . . . . . . . 103

10.5 Introduction to modal analysis . . . . . . . . . . . . . . . . . . 104


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Introduction

The present notes cover the theoretical material discussed in the MCG 4102/ 5108 Finite Element Analysis course, which is only introductory in nature.Although the nite element method is a mathematical approach to solve

many dierent types of partial dierential and integral equations, the pre-sentation is deeply rooted in mechanical engineering, as it is the backgroundof most students taking this course. The notes are not intended to replace therequired textbook (N-H Kim and BV Sankar, Introduction to Finite ElementAnalysis and Design, Wiley, 2009), but are meant to complement it. How-ever, for examination purposes, the notations herein prevail. The documentwas prepared with the help of several additional textbooks, in particular:

J Fish and T Belytschko, A rst course in nite elements, Wiley, 2007.FL Stasa, Applied nite element analysis for engineers, Saunders College

Publishing, 1985.

IH Shames and CL Dym, Energy and nite element methods in structuralmechanics, Taylor & Francis, 1985.DL Logan, A rst course in the nite element method, 4th edition, Thom-

son, 2007.JN Reddy, An introduction to the nite element method, 3rd edition,

McGraw Hill, 2004.Questions and comments from the students over the years have also

tremendously contributed to shape the document into its current form. Allcited and anonymous contributors are gratefully acknowledged, and futurecomments are welcome.

Finally, the reader should keep in mind that the present document is lim-

ited to linear elasticity. Nonlinear behaviours (related to non-innitesimallysmall displacements and/or deformations, material nonlinearities, such asplasticity, or changing boundary conditions/contact issues) are not covered,and require the use of yet other textbooks.

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Chapter 1

Introduction to FEA: Trusses

Trusses: structures composed of straight members connected at joints bypins (Fig. 1.1). Most or all members in a truss do not experience bendingor torsional moments.

Figure 1.1: Truss bridge example.

Given: external forces, geometry, material properties.Unknown: displacements at each joint; axial elongation, strain, stress,

force for each member.

1.1 Discretization

Let us consider each member of a truss as an element (Fig. 1.2):

Figure 1.2: Bar element.

1


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CHAPTER 1. INTRODUCTION TO FEA: TRUSSES 2

Elemente in Fig. 1.2 has two nodes, i andj; at either end. Forces are

transmitted from one element to the next at nodes. Bar elements

can onlytake axial forces (beam elementsalso allow bending moments, as will be seenlater). i andj (lowercase) are the local node numbers within the element.I and J (uppercase) denote global node numbers in the whole structure.Elements are also numbered (Fig. 1.3).

Figure 1.3: Example of discretization of a given truss.

Nodal coordinates:

Node number x coordinate (in) y coordinate (in)1 12 02 12 63 0 04 0 10

Element data (connectivity table):Element Nodei Node j Material ag

1 1 2 1 : 0:5-in (diameter) steel2 3 1 2 : 0:4-in aluminium3 3 2 14 4 2 2

5 3 4 1


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1.2 Element stiness relationship in local co-

ordinatesLet (x0; y0) be the local coordinate system for element e; with x0 along thelength of element fromi toj , andy 0 perpendicular tox0:

Figure 1.4: Bar element with degrees of freedom in local coordinate system.

The nodal displacements are noted asu0i andv0i inx

0 andy 0, respectively,at node i(Fig. 1.4). The corresponding forces are noted F0xiand F

0yiSimilarly,

the nodal displacements are noted as u0j andv0

j inx0 andy 0, respectively, at

nodej;with corresponding forcesF0xj andF0yj . From elementary strength ofmaterials,= P LAE; where is the axial elongation,L the member length, Pthe axial force, A the cross-sectional area and E the elastic modulus. It isassumed that the elastic range is not exceeded and that A is constant. Inother words, writing= u0j u

0i;

F0xi = AE

L (u0i u

0j)

F0xj = AE

L (u0j u

0i)

, whereF0xi= F0

xj for equilibrium.

Because bar elements do not withstand transverse forces, F0yi = F0

yj = 0:In matrix form,

AEL 2664

1 0 1 0

0 0 0 01 0 1 00 0 0 0

37758>>>:u0i

v0i

u0jv0j

9>>=>>; = 8>>>:F0xi

F0

yiF0xjF0yj

9>>=>>;With global nodal coordinates (xi; yi) and(xj ; yj ) for nodes i andj, re-

spectively, the element length can be computed as L=p

(xj xi)2 + (yj yi)2.


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The other two properties A andEcan be specied for each element. Con-

cisely,Ke0 Ue0 = Fe0 ; where Ke0 represents the local stiness matrix,Ue

0

the local nodal displacement vector, and

Fe0

the local nodal forcevector for the element. "e" denotes element, " 0 " denotes local coordinatesystem.


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Example 1 for Element 3:

K(3)

0

= A(3)E(3)

L(3)

26641 0 1 00 0 0 0

1 0 1 00 0 0 0

3775 :Material ag set to 1: 0.50-in steel: A(3) =

4(0:5)2 = 0:196in2

E(3) = 30 106 psi

L(3) =p

(0 12)2 + (0 6)2 = 13:42in

:

Then,

K(3)

0

= 103

2

664438 0 438 00 0 0 0438 0 438 0

0 0 0 0

3

775lbf/in

1.3 Transformation from local to global coor-

dinates

In the global(x; y)coordinate system shown in Fig. 1.5, position vector !r to

an arbitrary pointPcan be written as !r =rx!i + ry

!j :In the bar (rotated)

coordinate system(x0; y0);!r =rx0!i0 + ry0

!j0 :

Figure 1.5: Local and global coordinate systems.

Reminder: Scalar product ":" of any two vectors !u and!v:!u :!v = k!u k k!v k cos(!u ; !v): Therefore,

!r :!i =rx

!i :

!i + ry

!j :

!i =rx0

!i0 :

!i + ry0

!j0 :

!i

=rx+ 0 =rx0cos(x0; x) + ry0cos(y

0; x)


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!r :!j =rx

!i :

!j + ry

!j :

!j =rx0

!i0 :

!j + ry0

!j0 :

!j

= 0 + ry =rx0

cos(x0

; y) + ry0

cos(y0

; y)In matrix form, rxry

=

cos(x0; x) cos(y0; x)cos(x0; y) cos(y0; y)

rx0ry0

or frg = [T] fr0g ;with

[T] =

cos(x0; x) cos(y0; x)cos(x0; y) cos(y0; y)

=

cos cos(+

2) = sin

cos( 2 ) = sin cos

In the following, the notation [T] =

n11 n12n21 n22

is used, where coe-

cientsn11; n12; n21 andn22 are called direction cosines.It can be shown that [T] is orthogonal, i.e. [T]1 = [T]T ; therefore

fr0g = [T]T frg, with[T]T = n11 n21n12 n22 .1.4 Global element stiness relationship

The same transformation of coordinates as above holds for displacement vec-

tors

u0iv0i

at node i; and

u0jv0j

at node j: We get8>>>:u0iv0iu0j

v0

j

9>>=>>;

=

2

664n11 n21 0 0n12 n22 0 0

0 0 n11 n21

0 0 n12 n22

3

775

8>>>:

uiviuj

vj

9>>=>>;

or more concisely,

Ue

0

= [R] fUeg ;where[R] =

TT 00 TT

andfUeg =

8>>>:uiviujvj

9>>=>>; :

Similarly,

Fe0

= [R] fFeg ; withfFeg =

8>>>:

FxiFyiFxjFyj

9>>=>>;

:

Ke0 Ue0= Fe0 becomes Ke0 [R] fUeg = [R] fFeg ;or[R]T

Ke

0

[R] fUeg = fFeg ; since [R]T [R] = [R]1 [R] = [I] :

More simply,[Ke] fUeg = fFeg ;where[Ke] = [R]T

Ke0

[R]is the globalelement stiness matrix


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[Ke] = AE

L 2664n211 n21n11 n

211 n21n11

n11n21 n2

21

n11n21 n2

21n211 n21n11 n211 n21n11

n11n21 n221 n11n21 n

221

3775with n11 = cos x =

xjxiL

n21= cos y = yjyi

L

:

Note that computingn11 = xjxi

L andn21 =

yjyiL

is a very robust way toobtain the direction cosines (Fig. 1.6). It is highly recommended (instead ofusing rotation angles).

Figure 1.6: Direction cosines in 2-D.

Example 2 for Element 3:

n11 = x2x3

L(3) = 12013:42 = 0:8944

n21 = y2y3

L(3) = 6013:42 = 0:4472

K(3)

= 103

2664351 176 351 176176 88 176 88351 176 351 176176 88 176 88

3775 lbf/in

1.5 Assemblage

The original structure is put back together from individual elements. Thisis done based on the compatibility of nodal displacements: the x- and y-displacements at one node must be identical (in the intended design) tothose of the other nodes from other elements to be merged.


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Example 3 for the truss in Fig. 1.7, determine the assemblage stiness

matrix in terms of the 2x2 global stiness submatrices KeI;J :

Figure 1.7: Figure 1.3 repeated here for convenience.

Procedure: let us rst create a null assemblage stiness matrix[Ka] in-volving global node numbers 1 to 4.

[Ka] =

1 2 3 42664

02x2 02x2 02x2 02x202x2 02x2 02x2 02x202x2 02x2 02x2 02x202x2 02x2 02x2 02x2

3775

123

4From the connectivity table, element 1 has global node numbers 1 and 2,

so K(1)

=

" K

(1)1;1 K

(1)1;2

K(1)2;1 K

(1)2;2

#:

Element 2 has global node numbers 3 and 1, soK(2)

=

" K

(2)3;3 K

(2)3;1

K(2)1;3 K

(2)1;1

#:

Element 3 has global node numbers 3 and 2, so

K(3)= " K(3)3;3 K(3)3;2K(3)2;3 K(3)2;2 # :Element 4 has global node numbers 4 and 2, so

K(4)

=

" K

(4)4;4 K

(4)4;2

K(4)2;4 K

(4)2;2

#:


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Element 5 has global node numbers 3 and 4, so

K(5)= " K(5)3;3 K(5)3;4K

(5)4;3 K

(5)4;4

# :Finally,

[Ka] =

1 2 3 426664K

(1)1;1 + K

(2)1;1 K

(1)1;2 K

(2)1;3 02x2

K(1)2;1 K

(1)2;2 + K

(3)2;2 + K

(4)2;2 K

(3)2;3 K

(4)2;4

K(2)3;1 K

(3)3;2 K

(2)3;3 + K

(3)3;3 + K

(5)3;3 K

(5)3;4

02x2 K(4)4;2 K

(5)4;3 K

(4)4;4 + K

(5)4;4

377751234

Note that each individual 2x2 element stiness submatrix is symmetric,and[Ka]is symmetric. Also note that, depending on the connectivity table,

some submatrices may have to be transposed to t in the assemblage matrix.

1.6 Application of loads

By design, the (known) external loads in a truss can only occur at the joints(nodes).

Example 4 for the truss in Fig. 1.7, thex- andy- components of the ap-plied load areFx = 2; 000cos60

o = 1; 000 lbf andFy = 2; 000sin60o =

1; 732 lbf: The load is applied to Node 1. There is an unknown reactionforce in thex-direction at Node 3 (roller) and two unknown reaction forces

in thex- andy-directions at Node 4 (pin).

fFag =

8>>>>>>>>>>>>>>>>>>>:

1; 0001; 732

00

Rx30

Rx4Ry4

9>>>>>>>>>>=>>>>>>>>>>;

FxFyFxFyFxFyFxFy

Node 1

Node 2

Node 3

Node 4

1.7 Application of restraints on nodal displace-ments and solution

After assemblage, the system equation is of the form [Ka] fUag = fFag ;


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wherefUag is the vector of nodal displacements.

fU

a

g

T

= fu1 v1 u2 v2 ::: uN vNg

T

; where N is the maximumnumber of nodes and uI; vIare the x- and y- displacements at global nodenumber I : For now, no restraint on the nodal displacements have been con-sidered, and the whole structure can y into space!!! This is a rigid bodymotion, and fUag cannot be determined because [Ka] cannot be inverted(singular matrix, i.e. its determinant is zero). Restraints MUST be applied.

After the restraints corresponding to the boundary conditions are en-forced (u3= 0;andu4 = v4= 0for the truss in Fig. 1.7), the system equationbecomes [K] fUg =fFg ; where [K] is non-singular, therefore [K]1 exists,which implies a unique solution for fUg = [K]1 fFg :

There are dierent methods for enforcing displacement restraints:

Method 1(interesting for a technique called sub-structuring, but notused otherwise)

Consider the system k11u1+ k12u2+ k13u3 = f1k21u1+ k22u2+ k23u3 = f2k31u1+ k32u2+ k33u3 = f3

;

withkij =kji andi; j= 1:::3:Let us impose u2=uknown; therefore the set of equations above is equiv-

alent to e.g.

k11u1+ k12u2+ k13u3= f1u2 = uknown

k31u1+ k32u2+ k33u3= f3The symmetry has been destroyed but can be restored by transposing

terms involvingu2 to the right-hand side, and with uknown replacingu2 :24 k11 0 k130 1 0k31 0 k33

358


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24k11 k12 k13k21 (k22+ ) k23k31 k32 k33 358


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Internal forces

Internal forces can be determined as described above, i.e. from the elementstrains. Another, sometimes more direct, method consists in writing theequilibrium of one element of interest and using the known displacements atits nodes. Of course, all methods must yield the same results!

Example 5 determine the internal forces in Element 3 of the truss in Fig.1.7, from

K(3)8>>>:u3v3u2v2

9>>=>>; = 8>>>>>:F

(3)x3

F(3)

y3

F(3)x2F

(3)y2

9>>>=>>>; (in the global coordinate system)or from

K(3)0

8>>>:u03v03u02v02

9>>=>>; =8>>>>>:

F(3)0

x3

F(3)0

y3

F(3)0

x2

F(3)0

y2

9>>>=>>>; (in the local coordinate system).

1.9 Examples

Assignment 1.For additional problems to practice with, use the results in the statement

of Assignment 1, and nd yourself

K(2)

;

K(4)

and

K(5)

:


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Chapter 2

Linear elasticity - Principle ofvirtual work

This is a short review of linear elasticity in statics and equilibrium principles,including the principle of virtual work for use in nite element formulation.

2.1 Stress at a point

Consider a deformable body in static equilibrium, loaded as shown in Fig.2.1:

Figure 2.1: Static equilibrium of a body.

13


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CHAPTER 2. LINEAR ELASTICITY - PRINCIPLE OF VIRTUAL WORK14

The external forces!F1;

!F2; ... are transmitted through the deformable

body in a complex manner. When the body is cut along some plane, a force!R is required to maintain static equilibrium. !R has a normal component Rnand two tangential orthogonal components Rt1 andRt2:Consider small areaA instead of A: Then, Rn; Rt1; Rt2 act on A: The normal stressn is dened as n = lim

A!0

RnA

and the two shear (tangential) stresses as

t1 = limA!0

Rt1A and t2 = limA!0

Rt2A : For an innitesimal volume element

of a body positioned at a point in a global (x;y;z) coordinate system (Fig.2.2):

Figure 2.2: Some of the stresses acting on an innitesimal volume element.

(1st subscript: facet normal; 2nd subscript: direction)

2.2 Equations of static equilibrium

Consider an innitesimal 2-D volume element of thickness t (Fig. 2.3). Itis assumed that the normal and shear stresses vary from point to point inthe body in some continuous manner. Therefore, a rst order Taylor series

expansion is used.


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Figure 2.3: Stresses acting on a 2-D cross-section.

bx; by :components of body force per unit volume.Equilibrium inx-direction:(xx+

@xx@x

dx)dy xxdy+ (yx+ @yx

@y dy)dx yxdx + bxdxdy= 0

Equilibrium iny-direction:(yy+

@yy@y dy)dx yy dx + (xy+

@xy@x dx)dy xydy+ bydxdy= 0

Finally, in 2-D, @xx@x

+ @yx@y

+ bx= 0@xy

@x + @yy

@y + by = 0

andxy = yx for moment equi-

librium.

Similarly in 3-D, @xx@x + @yx

@y + @zx

@z + bx = 0@xy

@x + @yy

@y + @zy

@z + by = 0

@xz@x

+ @yz@y

+ @zz@z

+ bz = 0

and xy =yxxz =zxzy =yz

for mo-

ment equilibrium.

2.3 Strain at a point

Consider a square element whose sides are of unit length (Fig. 2.4). Underthe action of the external loading, the element will deform such that the sidesof the square are no longer perpendicular.


22/110


Figure 2.4: Strains in 2-D.

By denition,"xy=1+ 2 > 0 when angleB

0

OA

0

becomes smaller than2 : Strain is dimensionless.

2.4 Strain-displacement relations

For small deformations strains and displacements are related as follows:In 2-D, "xx=

@u@x

"yy = @v@y

"xy = @u

@y+ @v

@x

whereu; v are the displacements of a point in the x and y directions.

In 3-D, "xx = @u@x

"xy = @u

@y + @v@x

"yy = @v@y

"yz = @v@z +

@w@y

"zz = @w

@z

"zx = @w

@x + @u@z

whereu; v;w are the displacements of a point in the x; y;z directions.

In matrix form,f"g = [L] fUg ;where,

in 2-D, f"gT = ["xx; "yy; "xy] ;[L] =

24 @@x 00 @@y@@y

@@x

35and fUgT = [u; v]

in 3-D, f"gT

= ["xx; "yy ; "zz ; "xy; "yz ; "zx] ; [L] =

2

66666664

@@x

0 00 @@y 0

0 0 @@z@@y

@@x 0

0 @@z@@y

@@z 0

@@x

3

77777775 andfUgT = [u;v;w]


23/110


2.5 Compatibility equations

In 2-D, since"xy = @u@y + @v@x ; "xx=

@u@x and"yy =

@v@y ;

@2"xy@x@y

= @2@u

@x@y2+ @

2@v@x2@y

= @2"xx@y2

+ @2"yy@x2

; hence an extra relationship, calledcompatibility equation, between three strains and two displacements.

Similar compatibility equations are obtained in 3-D. Compatibility equa-tions are automatically satised in the stiness approach used in the displacement-based nite element method described in this course.

2.6 A constitutive relationship - Hookes law

The uniaxial Hookes law = E"; where E is the elastic modulus, can begeneralized, and is expressed in 3-D as fg = [D] (f"g f"0g) + f0g ;where[D] is the material property matrix, fg the stress vector, f"g the strainvector, f"0gthe initial strain vector, and f0gthe initial (or residual) stressvector.

fgT = [xx; yy; zz ; xy; yz ; zx]

f"gT = ["xx; "yy ; "zz ; "xy; "yz ; "zx]

[D] = E(1+)(12)

26666664

1 0 0 0 1 0 0 0 1 0 0 00 0 0 122 0 0

0 0 0 0 122 00 0 0 0 0 122

37777775

E: elastic modulus, : Poissons ratioOther specic[D]matrices will be found for plane stress, plane strain, or

axisymmetric analyses. [D] is symmetric for both isotropic and anisotropicmaterials. In the isotropic case, only E andare needed. f0g representsstresses that are known to exist in a material before it is loaded. Theymust be specied by the analyst. f"0g may be the result of crystal growth,shrinkage, or temperature changes.

Note that in dierent textbooks, shear stresses are often noted with

(e.g. xy is notedxy); and shear strains are often noted with (e.g. "xy isnotedxy):


24/110


2.7 Principle of minimum potential energy

How to nd the mechanical equilibrium of a structure other than by writingthe equilibrium equations for each component of the structure, and hopingto be able to solve for all of them? The principle of minimum potential en-ergy (presented in this section, and which youve encountered many timesin physics) and the principle of virtual displacements (presented in next sec-tion) are powerful tools to look for the equilibrium of a whole structure atonce.

The principle of minimum potential energy (PMPE) states that: "outof all the possible displacements elds that satisfy the geometric boundaryconditions (i.e. prescribed displacements), the one that also satises the

equations of static equilibrium results in the minimum of total potentialenergy for the structure (or body)."

The total potential energy is dened as the sum of the strain energy(internal potential energyUi)and the external potential energyUe from theexternal forces. =Ui+Ue: For conservative systems, the loss of externalpotential energy during the loading process must be equal to the work Wedone on the system by the external forces, or Ue = We; and therefore =Ui We: is a function of functions (strains and displacements) and iscalled a functional. Minimizing is called a variational problem. The rstvariation of the total potential energy = Ui We must be zero, i.e.Ui= We:

In a global Cartesian (x;y;z)coordinate system,Ui=

RV(xx"xx+ yy "yy + zz "zz + xy"xy+ yz "yz + zx"zx) dV:

Using matrix notation,Ui=R

Vf"gT fg dV:

For the work of external forces, considering a body force fbg (per unitvolume), a surface traction fsg (per unit area) andNpoint loads ffpg ;then

We=R

V(bxu + byv+ bzw) dV +

RA

(sxu + syv+ szw) dA

+Pp=N

p=1 (fpxu + fpyv+ fpzw):

Using matrix notation,We =R

VfUgT fbg dV +

RA

fUgT fsg dA

+

Pp=Np=1 fUg

T ffpg ; where

fbgT = [bx; by; bz] :body force vector

fsgT = [sx; sy; sz] : surface traction vector

ffpgT = [fpx; fpy; fpz] : point load vector

fUgT = [u;v;w] : rst variation of displacement vector


25/110


Eliminating fg using the linear elastic stress-strain relationship, the

PMPE becomesRVf"g

T [D] f"g dV =RVf"g

T [D] f"0g dV R

Vf"gT f0g dV

+R

VfUgT fbg dV +

RA fUg

T fsg dA +Pp=N

p=1 fUgT ffpg :

Going back to the total potential energy, it is given by = 12

RVf"g

T [D] f"g dV R

Vf"gT [D] f"0g dV +

RVf"g

T f0g dV

R

VfUgT fbg dV

RA

fUgT fsg dA Pp=N

p=1 fUgT ffpg

Example 6 consider the elongation and an axial force P in a uniaxialstress member (bar) of uniform cross-sectional areaA; lengthL and modulusof elasticityE(Fig. 2.5). One end of the bar is xed.

Figure 2.5: Extension of a bar.

"0 = 0; 0 = 0; b= 0and s = 0: = 1

2E"2AL P with" =

L;therefore = 1

2E(

L)2AL P:

We want for equilibrium, i.e. for minimum:

=EL2

AL P = 0() = P LAE

: this is well known!

Note that 2

2 = EA

E >0;therefore, is really at a minimum of.

2.8 Principle of virtual workThis principle, also known as the principle of virtual displacements, will bevery convenient and useful for nite element formulation of complex prob-lems. Work is the product of a displacement and the component of the force


26/110


in the direction of the displacement. Virtual work is imagined to occur when

the forces are real and the displacements are virtual (imagined), or vice-versa,but this not used herein.Statement of the principle of virtual work (PVW): if the work done by

the external forces on the structure is equal to the increase in strain energyfor any set of admissible virtual displacements (i.e. satisfying the prescribeddisplacements), then the system is in equilibrium.

Let us denote the virtual displacements in x; y; z directions as u; v;w (not variations!). The virtual displacements will cause virtual strains"xx; "yy; "zz ; "xy; "yz ; "zx:

In matrix form, the PVW becomes

RVf"gT fg dV = RVfUg

T fbg dV+RA fUgT fsg dA+P

p=Np=1 fUg

T ffpg ;

8 fUg ; f"g :with notations as in Section 2.7.This is known as a weak form of the equilibrium equations because this

equation only contains rst derivatives of displacements whereas the originalequilibrium equations (see Section 2.2 + Hookes law) contain second orderderivatives of the displacements.


27/110

Chapter 3

Finite element for beams

3.1 Theory - Beam bending in one plane

The deection of the neutral axis of a beam at any location xis represented byv(x)(bending in a plane). The deections are supposed to be small comparedto the length of the beam (typically less than 3% of length if this is not thecase, a more advanced theory must be used). The material is also supposedto be linearly elastic. Finally, it is assumed that the beam cross-section hasan axis of symmetry in the plane of bending, and that planar cross-sectionsremain planar during deformation. Lets take pointPon the beam neutral

axis, and point Q at distance y from the neutral axis (Fig. 3.1, left). Afterdeformation (Fig. 3.1, right, in which the deection v(x) and rotation (x)of the cross-section are grossly exaggerated), relationships between dierentvariables can be established as follows. Lets introduceu(x)the longitudinaldisplacement ofQ due to the deformation.

21


28/110

CHAPTER 3. FINITE ELEMENT FOR BEAMS 22

Figure 3.1: Beam deformation.

v: local deection of the beamdvdx

: local slope of the beam, physically interpreted as the rotation of thelocal cross-section of the beam, therefore dvdx =:

From trigonometry, tan = uy

; and because is very small, ' tan :Therefore, one can write

y dvdx =u(x) : longitudinal displacement due to v(x):Recalling that a longitudinal strain is obtained by taking the rst deriv-

ative of the longitudinal displacement, one obtainsy d

2vdx2

="x : longitudinal strain due to v(x):Because the material of the beam is assumed linearly elastic, the local

constitutive equation is such that the longitudinal (i.e. normal) stress in thebeam at point Q is x = E"x, where Eis the materials elastic modulus. Inother words, x= yE

d2vdx2

:

In beam cross-section A, the external normal force is N(x) =

ZA

xdA:

Because x varies linearly in the y-direction,N(x) = 0:

In beam cross-sectionA; the external bending moment isM(x) =

ZA

yxdA =Ed2vdx2

ZA

y2dA: RecallingZA

y2dA = I : second mo-

ment of area, one obtains, M(x) = EId2v

dx2: This is the global constitutive


29/110


equation for the beam, i.e. it expresses the relationship between the deec-

tion and the global external loads applied to the beam.Note that beam deection analysis also often aims at determining thebending stresses in the beam. To do so, one simply applies

x= yEd2vdx2 =

M(x)yI for each location of interest in the beam.

Considering the free-body diagram of an elemental beam segment oflengthdx (Fig.3.2), one can get additional relationships as follows.

Figure 3.2: Free-body diagram of an elemental segment of beam.

Lets introduce q(x) the transverse distributed load (force per unit length not necessarily constant).

XM = 0 : V dx + dM= 0;thereforeV(x) = dM(x)dx :XF = 0 :dV + qdx = 0;therefore dV(x)dx =q(x):

In the absence of distributed load, obviously,q(x) = 0:

Finally, the governing equations for the beam are:

EId2v

dx2 =M bending moment

EId3v

dx3 = dM

dx =V transverse shear force

EId4v

dx4 = dV

dx =q transverse distributed load

3.2 Discretization

Consider the beam element shown in Fig. 3.3:


30/110


Figure 3.3: Beam element and its degrees of freedom.

In the absence of distributed lads, the equilibrium equation for the ele-

ment is

d4v

dx4 = 0; for which the general solution is a third-order polynomialv(x) =c1+ c2x + c3x2 + c4x

3:The elements end conditions can be expressed in terms of the following

nodal values:Node 1: v(0) =v1 =c1

dv(0)dx =1 =c2

Node 2: v(L) =v2 =c1+ c2L + c3L2 + c4L

3

dv(L)dx

=2 =c2+ 2c3L + 3c4L2

In matrix form, the same relationships can be written as:8>>>:

v11

v22

9>>=>>; =

2664

1 0 0 00 1 0 0

1 L L2

L3

0 1 2L 3L2

37758>>>:

c1c2

c3c4

9>>=>>; orfUeg = [A] fQg ;where[A]is

called coecient matrix.By inverting[A] ; we can solve for fQg in terms offUeg :

[A]1 =

26641 0 0 00 1 0 0 3L2

2L

3L2

1L

2L3

1L2

2L3

1L2

3775Finally,v(x) can be written in matrix form as:v(x) =

1 x x2 x3

fQg =

1 x x2 x3

[A]1 fUeg ;

orv(x) = [N] fUeg with[N] = 1 x x2 x3 [A]1[N] = N1(x) N2(x) N3(x) N4(x) ; or more explicitly,[N] =

1 3x

2

L2 + 2x

3

L3 x 2x

2

L + x

3

L23x2

L2 2x

3

L3 x

2

L + x

3

L2

:


31/110


We note that [N] jx=0=

1 0 0 0

[N] jx=L= 0 0 1 0 dNdx jx=0= 0 1 0 0

dNdx

jx=L =

0 0 0 1

3.3 Element stiness and load vectors

Let us assume that at Node 1 of the beam element, nodal force F1 worksin displacementv1;nodal moment M1 works in rotation 1;and similarly atNode 2, F2 works in displacement v2; nodal moment M2 works in rotation2:

The principle of virtual work for the element is:ZV

f"gT fg dV = fUegT8>>>:

F1M1F2M2

9>>=>>; ; 8 fUeg ; 8 f"g :Here, f"gT = "x; and fg = x; as the other components are zero.

Then,ZV

"xxdV =

Z L0

ZA

(y(d2v(x)dx2

)(yEd2v(x)dx2

)dA

dx

= Z L

0

EI(d2v(x)dx2

)( d2v(x)dx2

)dx since ZA y2dA= I :

Fromv(x) = [N] fUeg, one can derive d2v(x)dx2

= d2

dx2[N] fUeg= [B] fUeg ;

andd2v(x)dx2 = [B] fU

eg ;where[B] =

6L2 +

12xL3

4L +

6xL2

6L2

12xL3

2L +

6xL2

:

Then,Z L0

EI(d2v(x)dx2

)(d2v(x)dx2

)dx=

Z L0

EIfUegT [B]T [B] fUeg dx

= fUegTZ L0

[B]T EI[B] dx fUeg :

Finally, the PVW yields:

fUegTZ L0

[B]T EI[B] dx fUeg = fUegT8>>>:

F1M1F2M2

9>>=>>; ; 8 fUeg :In other words,


32/110


Z L0

[B]T

EI[B] dx fUeg = 8>>>:F1M1F2M2

9>>=>>; ; or[Ke] fUeg = fFeg ;with[Ke] = EIL3

266412 6L 12 6L6L 4L2 6L 2L2

12 6L 12 6L6L 2L2 6L 4L2

3775 :So far, we only considered point forces and moments acting in the dis-

placements and rotations, respectively. For a distributed load w(x); thevirtual work of external forces is:

Z L

0

v(x)w(x)dx= fUegTZ L

0

[N]T w(x)dx= fUegT fFwg ;wherefFwg

is the nodal force vector representing a distributed load on the basis of workequivalence.

Forw(x) =q (constant),

fFwg =

8>>>>>>>>>>>>>>>>>>>:

Z L0

N1(x)w(x)dxZ L0

N2(x)w(x)dxZ L0

N3(x)w(x)dx

Z L0 N4(x)w(x)dx

9>>>>>>>>>>=>>>>>>>>>>;

=

8>>>>>:

qL2

qL2

12qL2

qL2

12

9>>>=>>>;

(Fig. 3.4).

Figure 3.4: Constant distributed load and its nodal equivalents.

3.4 Assemblage and solution

Consider the example shown in Fig. 3.5:


33/110


Figure 3.6: Example of beam problem.Elements 1 and 2 have stiness matrices

K(1)

= EIL3

266412 6L 12 6L6L 4L2 6L 2L2

12 6L 12 6L6L 2L2 6L 4L2

3775 and K(2) = K(1). This iswith respect to the degrees of freedom described in Fig. 3.7:

Figure 3.7: Discretization and degrees of freedom for problem in Fig. 3.6.

At this stage, the elements "ignore" each other. Let us nd the assemblagestiness matrix for the system.

The assemblage must enforce that v2 v3 and 2 3 (compatibilityof displacements and rotations between elements): This is done by simplyadding the corresponding matrix entries, and overlapping the element sti-ness matrices (see Chapter 1):

[Ka] = EIL3

2666666412 6L 12 6L 0 0

6L 4L2 6L 2L2 0 012 6L 24 0 12 6L6L 2L2 0 8L2 6L 2L2

0 0 12 6L 12 6L0 0 6L 2L2 6L 4L2

37777775vA

AvBBvCC

works in

FA

MAFBMBFCMC


34/110


The assemblage nodal displacement vector is

fUg

T

= vA A vB B vC C :Let us now determine the assemblage nodal force vector, or assemblageload vector. Can we identify some components in the generic load vectorfFgT =

FA MA FB MB FC MC

?

Consistent with what was done with the stiness matrix entries,F2+ F3= FBM2+ M3= MBFurthermore, a free-body diagram of the system (Fig. 3.8) gives FA RA;

FB = P; MB M; FC RCand MA 0and MC 0(obviously,HA 0).

Figure 3.8: Free-body diagram of beam in Fig. 3.6.

So far, we have 6 equations and 8 unknowns:RA; RC; and vA; A; vB; B; vC; C:However, the boundary conditions are such that vA =vC 0; therefore theactual number of unknowns is 6, and the problem can be solved.

BecausevA = vC 0; we eliminate the corresponding rows and columns

to solve for the unknowns. Then,

EIL3

26644L2 6L 2L2 06L 24 0 6L2L2 0 8L2 2L2

0 6L 2L2 4L2

37758>>>:

AvBBC

9>>=>>; =8>>>:

0PM0

9>>=>>;For the case where P = 1; 000 lbfandM= 0;A=

250L2

EI ; vB = 167L3

EI ; B = 0; C= 250L2

EI :

From these solutions, and extracting two rows of the assemblage system,one can determine the reaction forces RA and RC corresponding to vA =vC 0 as

EIL3

12 6L 12 6L 0 0

0 0 12 6L 12 6L

8>>>>>>>>>>>:

vA

AvBBvCC

9>>>>>>=>>>>>>;=

RARC

:


35/110


Finally,RA = 504lbf andRC = 504lbf: Note that a value of 504 is ob-

tained instead of the expected 500 due to round-o errors during calculation.Once the displacements and rotations are known along with all the shearforces and moments, the bending stresses can be calculated at critical lo-cations, using the same equations as in strength of materials or machinedesign.

3.5 Example

Assignment 2.

3.6 Beam element with axial loading

Outside of buckling and stress stiening (e.g. a taut guitar string), which arenonlinear cases, we can linearly superimpose the results from bar and beamelements, because displacements and strains are assumed to be small. Notethat in many textbooks and software packages, such combination describesa frame element (Fig. 3.9).

Figure 3.9: Combined bar and beam element, or frame element.

[ke] =

[keBar ] [0]

[0] [keBeam]

withfuegT =

u1 u2 v1 1 v2 2

:

With more convenient fuegT =

u1 v1 1 u2 v2 2

, [ke] can be

re-organized into:


36/110


[ke] =26666664

AEL 0 0

AEL 0 0

0 12EI

L3

6EI

L2 0

12EI

L3

6EI

L2

0 6EIL2

4EIL

0 6EIL2

2EIL

AEL 0 0 AE

L 0 00 12EIL3

6EIL2 0

12EIL3

6EIL2

0 6EIL2

2EIL

0 6EIL2

4EIL

37777775If the element is oriented at an arbitrary angle from the X axis of

the global reference frame, we have (Fig. 3.10)

Figure 3.10: Local and global degrees of freedom for a 2-D frame element.

u1 = U1cos + U2sin

v1 = U1sin + U2cos 1 = U3

u2 = U4cos + U5sin v2 = U4sin + U5cos

2 = U6

and in matrix form,

8>>>>>>>>>>>:

u1v11u2v22

9>>>>>>=

>>>>>>;=

26666664

cos sin 0 0 0 0 sin cos 0 0 0 0

0 0 1 0 0 00 0 0 cos sin 00 0 0 sin cos 00 0 0 0 0 1

37777775

8>>>>>>>>>>>:

U1U2U3U4U5U6

9>>>>>>=

>>>>>>;;

or fueg = [R] fUeg : Then, the element stiness matrix in the globalsystem is[Ke] = [R]T [ke] [R] :


37/110


3.7 General 3-D beam (or frame) element

We want to include

axial behaviour (bar)

bending behaviour inxyplane (beam)

bending behaviour inxzplane (beam)

torsional behaviour

Items 1 and 2 have been taken care of. Bending in xzplane needs

attention because it is similar to bending inxyplane, but instead ofz = dv

dx ;we have y =

dwdx

; and instead ofMz = EIzd2vdx2

; we have My = EIyd2wdx2

(Fig. 3.11):

Figure 3.11: Degrees of freedom for a general 3-D beam element.

Therefore,[ke]xz = EI

L3

266412 6L 12 6L

6L 4L2 6L 2L2

12 6L 12 6L6L 2L2 6L 4L2

3775 :Torsion is represented by stiness matrix [keTorsion] =

JGL

1 11 1

;

nodal displacement vector

x1x2

and load vector

Mx1Mx2

: Finally, for

the general 3-D beam element:


38/110


2664[keBar ] [0] [0] [0]

[0] [keBeam]xy [0] [0]

[0] [0] [keBeam]xz [0][0] [0] [0] [keTorsion]

3775

8>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>:

u1u2v1z1v2z2w1y1w2y2x1x2

9>>>>>>>>>>>>>>>>>>=>>>>>>>>>>>>>>>>>>;

=

8>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>:

Fx1Fx2Fy1Mz1Fy2Mz2Fz1My1Fz2My2Mx1Mx2

9>>>>>>>>>>>>>>>>>>=>>>>>>>>>>>>>>>>>>;

:


39/110

Chapter 4

Interpolation functions -Integration formulas

From the principle of virtual work, it is evident that in stress analysis prob-lems, the variables of interest are the displacements (vector). In thermalanalysis, it would be the temperature (scalar). In uid ow problems, itwould be the uid velocities (vector) and the pressure (scalar). Below areideas and results regarding interpolation functions for these variables andintegration over element length, surface or volume that can be used for yetother studies (electromagnetism, mass transfer, etc...).

4.1 Compatibility and completeness require-

ments

Compatibility

For C0-continuous problems, the interpolation function must be continuousalong the boundaries of the element. For C1-continuous problems, the func-tion and its rst derivative must be continuous (Fig. 4.1).

33


40/110

CHAPTER 4. INTERPOLATION FUNCTIONS - INTEGRATION FORMULAS34

Figure 4.1: C0-continuous function (left), and C1-continuous function(right).

Many problems are C0-continuous once formulated using the weak form(2-D stress, strain, axisymmetric stress, 3-D stress analyses), but some areC1-continuous or higher. Elements that obey the compatibility requirementare said to be conforming (vs. non-conforming). Some non-conforming ele-ments are useful, but they should be used with extra caution.

Completeness

For Cn-continuous problems, the interpolation function must be capable ofrepresenting a constant value of the variable as well as partial derivatives of

up to ordern+1 as the element size decreases to a point. Example in uniaxialstress analysis: let us assumeu= c1+c2x:Ifc2= 0; u = c1 = constant : rigidbody mode (displacement of the whole body without straining). Also, du

dx =

c2 = constant : this allows for a constant strain in the element. Therefore,the interpolation function u = c1 +c2x can be used for a C

0-continuousproblem.

Mesh sensitivity analysis

With both compatibility and completeness requirements satised, conver-gence of the solution during mesh renement can be achieved (Fig. 4.2).

Regardless of the expectations, convergence of the solution during mesh re-nement is a verication that MUST be done for every analysis (except forbar elements) when using a nite element code.


41/110


Figure 4.2: Convergence of solution upon mesh renement.

4.2 One dimensional (1-D) elements

Interpolation functions are based on polynomials or rational functions. The

functions can be developed with global, natural (or serendipity) or lengthcoordinates, depending on the base of integration. Many dierent types ofelements can be created, but not all will be useful.

4.2.1 2-node C0-continuous element

Global coordinates (Fig. 4.3)

Figure 4.3: Global coordinates.

For variable ; we pose(x) =c1+ c2x;

or(x) = [1 x]

c1c2

= [1 x] fQg :

At nodesi;j;we require (xi) =i = c1+ c2xi(xj) =j =c1+ c2xj

:

In matrix form, ij = 1 xi1 xj c1c2 or feg = [A] fQg :Solving for the vector of constants yields

c1c2

=

1 xi1 xj

1 ij

:


42/110


Then,(x) = [1 x] 1 xi

1 xj 1

i

j or(x) = [1 x] [A]1 feg = [N] feg with[N] = [ xjxxjxi xxixjxi ]:Natural coordinates (Fig. 4.4)

Figure 4.4: Natural coordinates.

Let us dene a local, normalized coordinate rdened relative to the globalcoordinatex:

r= 0 at x =x = 12(xi+ xj);and1 r +1withr = 2(xx)

xjxi:

With these coordinates,[N] = [12(1 r) 12(1 + r)]:

Length coordinates (Fig.4.5)

Figure 4.5: Length coordinates.

Let us consider an internal point p (not a node) in element e; and deneLi =

length pjlength ij andLj =

length iplength ij :

Obviously,0 Li 1; 0 Lj 1;andLi+ Lj = 1:Then,[N] = [Li Lj]:

4.2.2 2-node C1

-continuous elementSee development of beam element in global coordinates in Chapter 3 (Section3.2, Discretization). Note that this element is C2-continuous in w; but C1-continuous in:


43/110


4.3 2-D elements

4.3.1 Triangular C0-continuous element

Global coordinates (Fig. 4.6)

Figure 4.6: Global 2-D coordinates.

We pose(x; y) =c1+ c2x + c3y or

(x; y) = [1 x y]

8


44/110


Area coordinates (Fig. 4.7)

Figure 4.7: Area coordinates.

Let us consider an internal point p (not a node) in element e; and deneLi =

area pj karea ij k ; Lj =

area ipkarea ij k ; and Lk =

area ijparea ij k

Obviously,0 Ll 1forl = i; j ork;and Li+ Lj+ Lk = 1:Also, Li(xi; yi) = 1 Li(xj ; yj ) = 0 Li(xk; yk) = 0

Lj (xi; yi) = 0 Lj(xj; yj) = 1 Lj (xk; yk) = 0Lk(xi; yi) = 0 Lk(xj; yj) = 0 Lk(xk; yk) = 1

Figure 4.8: Area coordinates withLk = constant.

As can be seen in Fig. 4.8, trianglesp1ij andp2ij have the same area(they have the same base and height), therefore Lk = constant is a line


45/110


parallel to the opposite leg ij: Lk varies linearly between 0 and 1. It can be

shown that[N] = [Li Lj Lk]:

Example 7 show thatLi = Ni:

Li= area pjk

area ijk =

12det

26664

1 x y1 xj yj1 xk yk

37775

A

Li= (xjykxkyj)+(yjyk)x+(xkxj)y

2A

Li= m11+ m21x + m31y= Ni QED.

4.3.2 Rectangular (quad) C0

-continuous elementNatural coordinates (Fig. 4.9)

Figure 4.9: Natural 2-D coordinates.

We have1 r +1and 1 s +1withr = xxa ands = yy

b :Then,(x; y) = [Ni Nj Nk Nl] f

eg

with Ni = 14

(1 + r)(1 s)Nj =

14(1 + r)(1 +s)

Nk = 14(1 r)(1 +s)Nl =

14(1 r)(1 s)

andfeg =

8>>>:

ijk

l

9>>=>>;

:


46/110


Three rules

1. Nm = 1 at node m; and 0 at other nodes, form = i; j; k orl:

2. At nodem; onlyNm= 1;the others are 0.

3. Ni+ Nj+ Nk+ Nl = 1 8r; s

4.3.3 Curved elements

So far, all the elements considered have had straight edges. Curved ele-ments (with more nodes) may be needed to describe curved boundaries moreaccurately (or with fewer elements). A mapping can be used between the

straight-edged parent element and the curved element (Fig. 4.10). In anisoparametric mapping, the same interpolation functions are used both forthe variable of interest () and the description of the geometry.

Figure 4.10: Parent element (left), and curved element (right).

For simplicity of presentation, let us stick with linear elements (straightedges), as in Fig. 4.11:


47/110


Figure 4.11: Mapping between parent and real geometries.

To map the geometry, we want x = [Gi Gj Gk Gl]

8>>>:xixjxkxl

9>>=>>; and

y= [Gi Gj Gk Gl]

8>>>:yiyjykyl

9>>=>>; :We may want to use the interpolation functions dened previously, such

thatNm(r; s) =Gm; for m = i; j; k orl:With (x; y) = (r; s) = [N

i N

j N

k N

l] feg for the variable of in-

terest, this clearly leads to an isoparametric element.


48/110


4.4 3-D elements

4.4.1 Four-node, C0-continuous tetrahedral element (pyra-mid)

Figure 4.12: Tetrahedral element.

(x;y;z) =c1+ c2x + c3y+ c4z

4.4.2 Eight-node, C0-continuous brick element

Figure 4.13: Brick element.

(x;y;z) =c1+ c2x + c3y+ c4z+ c5xy+ c6yz + c7zx+ c8xyz


49/110


4.4.3 Axisymmetric elements

A problem is axisymmetric if the body of interest is a body of revolution ANDif the material properties, boundary conditions and loads do not change with in a global cylindrical coordinate system attached to the body (Fig. 4.14).

Figure 4.14: Cylindrical coordinate system for axisymmetric problem.

In an axisymmetric problem, the variable of interest is a function of randz only, therefore the problem is actually two-dimensional. For volumeintegrations,dV =rdrddz= 2rdrdz = 2rdA:

4.5 Integration formulasBuilding a stiness matrix or a force vector calls for many integrations of theinterpolation functions and their derivatives over the element length, surfaceor volume.

4.5.1 Direct integration

Length coordinates

Zl LiL

j dl=

!!(++1)!

l wheredl is an elemental length between nodes i and

j;andl is the length of line between nodes i andj: Exponents andmustbe positive integers. Recall that n! =n (n 1) 1:


50/110


Area coordinates

ZA

LiLj Lk dA = !!!(+++2)!2A where dA is an elemental area of the ele-

ment, andAis the area of the triangle formed by nodesi; jandk: Exponents; andmust be positive integers.

4.5.2 Numerical integration - Gaussian quadrature

1-D formulas

Consider I =

Z xjxi

f(x)dx: First we need to use natural coordinates, such

that, whenx =xi; r= 1 and whenx = xj; r= +1:

Take x = xi + 12(xj xi)(1 +r): Then, dx = dxdr dr = Jdr; where J isthe Jacobian of the transformation (Note: the notation is not useful in 1-D,but becomes very convenient in 2-D and 3-D). Here, J= 1

2(xj xi):Finally,

according to the Gauss-Legendre quadrature,

I =

Z xjxi

f(x)dx =

Z +11

f(x(r))Jdr =

Z +11

f(r)Jdr ' J

nXi=1

f(x(ri))wi

(indexi has nothing to do with the node index).wherewi are the weight factors, ri the base points, and n the number of

Gauss points (see numerical methods textbook for values ofwi andri). Apolynomial of degreepis integrated exactly by employing n= 1

2(p+1)Gauss

points or nearest larger integer. Note that the location of Gauss points hasnothing to do with that of the element nodes.

2-D and 3-D formulas:

Similarly in 2-D,

I=

Z +11

Z +11

f(r; s)drds 'mX

j=1

nXi=1

f(ri; sj)wri w

sj

and in 3-D,

I= Z +1

1 Z +1

1 Z +1

1

f(r;s;t)drdsdt 'l

Xk=1m

Xj=1n

Xi=1 f(ri; sj ; tk)wri wsj wtk:Therefore, in 2-D, with dxdy= det [J] drds;

[Ke] =

ZZA

[B]T [D] [B] dxdy=

Z +11

Z +11

[B]T [D] [B]det[J] drds;


51/110


and in 3-D, withdxdydz= det [J] drdsdt;

[Ke] = ZZZV

[B]T [D] [B] dxdydz = Z +11

Z +11

Z +11

[B]T [D] [B]det[J] drdsdt

where[J]is the Jacobian matrix of the transformation.[J] shows up in many other instances, anytime coordinate changes are

needed. For example, for each interpolationNi;

@Ni@r

= @Ni@x

@x@r

+ @Ni@y

@y@r

+ @Ni@z

@z@r

or, in matrix form,8


52/110

Chapter 5

Stress analysis in 2-D withlinear triangular element

5.1 Plane stress

The plane stress model is appropriate for a thin plate loaded uniformly acrossits thicknesst in a direction parallel to the mid-plate plane (Fig. 5.1). Thethickness need not be constant. Plane stress state implies: zz = 0 andxz =yz = 0:

Figure 5.1: Plane stress problem (top view, left; side view, right).

46


53/110

CHAPTER 5. STRESS ANALYSIS IN 2-D WITH LINEAR TRIANGULAR ELEMENT47

5.1.1 The shape function matrix [N]

In plane stress, element nodes have 2 degrees of freedom: the x andy com-ponents of the displacements. For a triangular element (see Chapter 4)

u(x; y)v(x; y)

=

Ni 0 Nj 0 Nk 0

0 Ni 0 Nj 0 Nk

8>>>>>>>>>>>:

uiviujvjukvk

9>>>>>>=>>>>>>;orfUg = [N] fUeg :

5.1.2 The strain-nodal displacement matrix [B]Recalling that for 2-D small deformations,

"xx = @u@x

; "yy = @v@y

and"xy = @u

@y+ @v

@x:In matrix form,8


54/110


5.1.3 Constitutive relationship

For plane stress, the constitutive relationship for a linear elastic material

fg = [D] (f"g f"0g) + f0g is written with fg =

8


55/110


Note that the previous derivation is general and not limited to 2-D cases.

It is done here for convenience.

5.1.5 The element stiness matrix [Ke]

From above,[Ke] =R

Ve[B]T [D] [B] dVwhere both[D]and[B](in our case)

are composed of constants, and dV =tdxdy:Therefore,[Ke] = [B]T [D] [B]

RAe

tdxdy:For reasonably small elements,t

may be taken as a constant average value. Then, [Ke] = [B]T [D] [B] tA:Notethat in our case,[Ke]is a 6x6 matrix because [B]T is 6x3,[D]is 3x3 and[B]is 3x6. This is consistent with the fact that there are 6 nodal displacementsper triangular element.

Example 8 determine[Ke] in the case shown in Fig. 5.2:

Figure 5.2: Triangular element.

Steel plate: E= 30 106 psi= 0:3t= 0:25 in (constant)

A= 12det

24 1 xi yi1 xj yj1 xk yk

35= 12det

24 1 4 21 0 21 0 0

35= 4 in2m21 = (yj yk)=2A=

2024 = 0:25in

1 m31 = (xk xj)=2A= 0024 = 0

m22 = (yk yi)=2A= 0224 = 0:25 in

1 m32 = (xi xk)=2A= 0:50in1

m23 = (yi yj )=2A= 0 m33 = (xj xi)=2A= 0:50 in

1

Then,[B] =

24 0:25 0 0:25 0 0 00 0 0 0:50 0 0:500 0:25 0:50 0:25 0:50 0

35in1:


56/110


[D] = E

12 24

1 0 1 00 0 1

235= 24

33:0 9:89 09:89 33:0 0

0 0 11:5 35 106 psi (lbf/in2)Finally,

[Ke] =

266666642:06 0 2:06 1:24 0 1:24

0:72 1:44 0:72 1:44 04:94 2:68 2:88 1:24

8:97 1:44 8:252:88 0

Sym 8:25

37777775 106 lbf/in.

5.1.6 The element nodal force vector ffeg

Self-strain

ffe"0g =R

Ve[B]T [D] f"0g dV =

RAe

[B]T [D] f"0g tdxdy:Considering thermal strains only, with t andT constant,thenffe"0g = [B]

T [D] f"0g tA:ffe"0g is 6x1 because[B]

T is 6x3,[D]is 3x3 and f"0gis 3x1.

Example 9 (continued from above): determine ffe"0g if t = 6:0 106

in/(in.F) and the temperature increases by 150 F.

f"0g = 8


57/110


Body forces

ffeb g =R

Ve[N]T fbg dV =

RAe

26666664Ni 00 Ni

Nj 00 Nj

Nk 00 Nk

37777775

bxby

tdxdy:

Recalling that for a triangular element, Nl =Ll for l = i; j ork; whereLl is an area coordinate,

ffeb g =

8>>>>>>>>>>>:

RAe

Libxtdxdy

RAe

Libytdxdy

RAeLj bxtdxdyRAeLj bytdxdyRAeLkbxtdxdyRAe

Lkbytdxdy

9>>>>>>=>>>>>>;

withR

AeLibxtdxdy=

1!0!0!(1+0+0+2)!

bxt2A= 13

bxtA:

Finally,ffeb g = tA3

8>>>>>>>>>>>:

bxbybxbybxby

9>>>>>>=>>>>>>;

(ifbx andby are constant).

Surface tractions

ffes g =R

Ae[N]T fsg dA=

RAe

26666664Ni 00 Ni

Nj 00 Nj

Nk 00 Nk

37777775

sxsy

dA:

Surface tractions are only present at the surface of the structure. Let usconsider an elemente with nodesiand j (but notk) on the global boundary

(Fig. 5.3).


58/110


Figure 5.3: Surface traction.

A surface traction is assumed to act on leg ij: Then, dA = tdl; andNk =Lk = 0on legij:

ffes g =

8>>>>>>>>>>>>>:

Rlij

LisxtdlRlij

LisytdlRlij

LjsxtdlRlij

Lj sytdl

00

9>>>>>>>=>>>>>>>;

:

Ifsx andsy are assumed constant, then ffes g =

tlij2

8>>>>>>>>>>>:

sxsysxsy00

9>>>>>>=>>>>>>;for legij on

the global boundary.

Example 10 determineffes g (dimensions from Fig. 5.2)


59/110


Figure 5.4: Surface traction on triangular element.

sy = 0

sx is not constant over leg j k: Eectivesx=

1;600+2;000

2 = 1; 800 psi.

ffes g = tljk2

8>>>>>>>>>>>:

00sxsysxsy

9>>>>>>=>>>>>>;= 0:252

2

8>>>>>>>>>>>:

00

1; 8000

1; 8000

9>>>>>>=>>>>>>;=

8>>>>>>>>>>>:

00

4500

4500

9>>>>>>=>>>>>>;lbf.

(This is with respect to the local node ordering i; j; k)

Point loads

fep

=Pp=N

p=1 [N]T ffpg =

Pp=Np=1

26666664Ni 00 NiNj 00 Nj

Nk 00 Nk

37777775

fpxfpy

fep

=Pp=N

p=1

8>>>>>>>>>>>:

Ni(xp; yp)fpxNi(xp; yp)fpyNj(xp; yp)fpxNj(xp; yp)fpyNk(xp; yp)fpx

Nk(xp; yp)fpy

9>>>>>>=>>>>>>;

:

Example 11 determine

fep

for a point load acting at coordinates(0:6; 1:6)for the element above, withfpx= 1; 500 lbf andfpy = 2; 300 lbf.


60/110


SinceNi(x; y) =m11+ m21x + m31yNj (x; y) =m12+ m22x + m32yNk(x; y) =m13+ m23x + m33y

; calculation of all mij s is needed.

The only ones that have not been calculated yet are:m11 = (xjyk xkyj )=2A= 0m12 = (xkyi xiyk)=2A= 0m13= (xiyj xj yi)=2A= 1

Therefore, at x = 0:6andy = 1:6;Ni= 0 + 0:25 0:6 + 0 1:6 = 0:15

Nj = 0 0:25 0:6 + 0:5 1:6 = 0:65Nk = 1 + 0 0:6 0:25 1:6 = 0:20

fep

=8>>>>>>>>>>>:

225

345975

1; 495300

460

9>>>>>>=>>>>>>;lbf.

5.1.7 Assemblage

Example 12 see Fig. 5.5

Figure 5.5: Assemblage of two triangular elements.

Connectivity table: Element # Nodei Node j Node k1 2 1 32 3 4 2


61/110


For Element 1,

K(1)

=264 K

(1)

22 K(1)

21 K(1)

23

K(1)12 K

(1)11 K

(1)13

K(1)32 K

(1)31 K

(1)33

375u2v2u1v1u3v3

For Element 2,

K(2)

=

264 K(2)33 K

(2)34 K

(2)32

K(2)43 K

(2)44 K

(2)42

K(2)23 K

(2)24 K

(2)22

375u3v3u4v4u2v2

By assemblage, with fUagT = [u1 v1 u2 v2 u3 v3 u4 v4] :[Ka] is rst zeroed out, and then entries are added.

[Ka] =

26664K

(1)11 K

(1)12 K

(1)13 0

K(1)21 K

(1)22 + K

(2)22 K

(1)23 + K

(2)23 K

(2)24

K(1)31 K

(1)32 + K

(2)32 K

(1)33 + K

(2)33 K

(2)34

0 K(2)42 K

(2)43 K

(2)44

37775 : [Ka]is symmetric.

ffag =

8>>>>>:

f(1)1

f(1)2 + f

(2)2

f(1)3 + f

(2)3

f

(2)

4

9>>>=>>>;

:

5.1.8 Prescribed displacements - Solution

[Ka] is singular, therefore prescribed displacements must be included forsolution. Finally, the system to be solved is [K] fUg = ffg ; from whichfUg = [K]1 ffg :

5.1.9 Element resultants

From the computed displacements fUeg ;one can determine the strains and

stresses in the element.f"g = [B] fUeg : For a triangular element, the average strains across the

element are:"xx = m21ui+ m22uj+ m23uk"yy = m31vi+ m32vj+ m33vk"xy =m31ui+ m21vi+ m32uj+ m22vj+ m33uk+ m23vk

:


62/110


The average stresses across the element are:

fg = [D] ([B] fU

e

g f"0g) + f0g :

5.2 Plane strain

The plane strain model is appropriate when a long prismatic member ofconstant cross section is held between two xed rigid planes (Fig. 5.6). Planestrain state implies: "zz = 0 and "xz ="yz = 0:

Figure 5.6: Plane strain problem (side view, left; end view, right).

All the results for plane stress apply, except that the constitutive rela-tionship is changed: fg = [D] (f"g f"0g) + f0g

with[D] = E(1+)(12)

24

1 0 1 0

0 0 12

2

35 :

For a temperature change T; f"0g =

8


63/110


5.3 Axisymmetric problems

Figure 5.7: Axisymmetric problem.

Strains considered (Fig. 5.7) f"gT = ["rr " "zz "rz ] ;stresses considered: fgT = [rr zz rz ] ;with "rr =

@u@r

" = u

r

"zz = @v

@z"rz = @u@z + @v@r

;

whereu and v are the radial and axial displacements, respectively.

In matrix form, f"g = [L] fUg with fUg =

uv

and[L] =

2664@@r 01r

00 @@z@@z

@@r

3775 :With the same shape function matrix [N] and the same nodal displace-

ments vectorfUegas in plane stress,f"g = [B] fUegwith[B] = [L] [N] :

[B] =

2664

m21 0 m22 0 m23 0Nir 0

Njr 0

Nkr 0

0 m31 0 m32 0 m33m31 m21 m32 m22 m33 m23

3775 :

Note that Nlr is not a constant, but a function ofr andz: The constitutiverelationship isfg = [D] (f"g f"0g) + f0g


64/110


65/110

Chapter 6

Finite elements for plates andshells

Plates are at; shells are curved plates.

6.1 Introduction - Bending of thin plates

6.1.1 Geometry

As shown in Fig. 6.1, the plate surfaces are at z = t=2 and its midsurface

atz = 0: The assumed basic geometry of the plate is as follows: t b andt c(ift >' 0:2bor0:2c;thick plate). The deectionw due toqis assumedto be much less than the thickness: w=t 1:

59


66/110

CHAPTER 6. FINITE ELEMENTS FOR PLATES AND SHELLS 60

Figure 6.1: Thin plate geometry.

6.1.2 Kirchho assumptions

They generalize ideas taken from beam bending. Consider a dierential sliceof plate (Fig. 6.2). Loading qcauses the plate to deform laterally in thez-direction, and the deection w of point P is assumed to be a function ofx andy only. That is, w = w(x; y); and the plate does not stretch in thez-direction.

Figure 6.2: Dierential slice of plate, before loading (left), and after loading(right). Similar displacements iny z plane.

The Kirchho assumptions are as follows:

1. Normals remain normals. This implies that shear strains "yz and"xzare zero. However, "xy 6= 0 :the plate may twist in its plane.


67/110


2. Thickness changes can be neglected, and normals undergo no extension,

i.e. "zz = 0:3. Normal stresszz has no eect on in-plane strains "xx and"yy; and is

considered negligible.

4. Membrane or in-plane forces are neglected here (they can be super-imposed at a later stage). Therefore, the in-plane deformations inthex- and y-directions at the midsurface are assumed to be zero, i.e.u(x; y) =v(x; y) = 0:

According to the above assumptions, any point Phas displacementsu= z = z( @w@x ) in thex-direction

v= z= z( @w@y) in they-directionwhere @w@x and

@w@y are the slopes at the midsurface.

"xx= @u@x = z

@2w@x2 ; "yy =

@v@y = z

@2w@y2 ; "zz =

@w@z 0

"xz = @w

@x+ @u

@z 0; "yz =

@w@y

+ @v@z

0; "xy = @u

@y+ @v

@x= 2z @

2w@x@y

Because in small deection theory, the square of a slope may be regarded

as negligible (recall that in 2-D, 1rx

=d2vdx2

[1+( dvdx

)2]3=2 ' d

2vdx2

, whererxis the radius

of curvature), the curvatures (1r ) of the plate are given simply as:

xx = @2w@x2 ; yy =

@2w@y2 and xy = 2

@2w@x@y :

Therefore we have: "xx= z xx"yy =z yy"xy =z xy

:

6.1.3 Constitutive relationships

Plane stress state can be used (Assumption 3), and then:8


68/110


(N.m/m N) are denoted Mxx; Myyand Mxywith Mxx = Z t=2

t=2

z xxdz

Myy =Z t=2t=2

z yydz

Mxy =

Z t=2t=2

z xydz

Mxy = Myx because xy =yx

:

Finally,

8


69/110


Figure 6.4: Dierential plate element with dierential moments and forces.

6.1.4 Equilibrium equations

Note that Q andM follow the same rules as stresses: a positive stress actson a positive face in the positive direction (Fig. 6.5).

Figure 6.5: Positive normal and shear stresses.

From Fig. 6.6, the equilibrium in the z-direction yields:@Qx

@x dxdy+ @Qy

@y dxdy+ qdxdy= 0Equilibrium of moments about thex-axis yields:( @Mxy@x dx)dy (

@Myy@y dy)dx + Qydxdy= 0


70/110


Equilibrium of moments about they-axis yields:

(

@Myx

@y dy)dx + (

@Mxx

@x dx)dy Qxdxdy= 0

Fig. 6.8: Equilibirum of the elemental plate.

Then, @Qx@x

+ @Qy@y

+ q= 0@Mxx

@x + @Mxy

@y Qx = 0@Myy

@y + @Mxy

@x Qy = 0

; therefore Qx= D @@x

( @2w

@x2 + @

2w@y2

)

Qy = D @@y (

@2w@x2 +

@2w@y2)

;

and nally,D( @4w

@x4 + 2

@4w

@x2

@y2 +

@4w

@y4) =q:

In concise form,r4w= q=D: These are the governing equations of platedeection.

Note: the theory of thin plates neglects the eects on bending of zz ; aswell as "xz =

xzG and"yz =

yzG . However, the shearing forces Qx andQy

resulting from xz andyz are not negligible. In fact, they are of the sameorder of magnitude as the lateral loads and moments.

6.2 Finite element formulation (rectangular

element)

6.2.1 Element type

We will consider the 12-degree of freedom (d.o.f) at-plate bending elementshown in Fig. 6.7:


71/110


Figure 6.7: Rectangular shell element.

Each node has 3 d.o.f.: a transverse displacementw in thez-direction, arotationxabout thex-axis and a rotationyabout they-axis. The rotationsare related to the transverse displacement by x =

@w@y

and y = @w@x

(anegative w displacement is required to produce a positive rotation about the

y-axis). The displacement vector for the element is:fUegT = [wi xi yi wj xj yj wk xk yk wl xl yl] :

6.2.2 Displacement function

Because there are 12 d.o.f., we select a 12-term polynomial in x and y asfollows:

w= c1+ c2x + c3y+ c4x2 + c5xy+ c6y

2 + c7x3 + c8x

2y+ c9xy2 + c10y

3 +c11x

3y+ c12xy3:

The function allows for rigid-body motion and constant strain, but con-tinuity of slopes between elements is not ensured. Along side i j (x-axis),

w= c1+ c2x + c4x2 + c7x3@w@x

=c2+ 2c4x + 3c7x2

@w@y =c3+ c5x + c8x

2 + c11x3

:

The displacement w is a cubic as used for the beam element, and @w@x

isthe same as in beam bending. The four constantsc1; c2; c4; c7 can be denedby invoking the end point conditions of (wi; wj ; yi; yj ). Therefore, w and@w@x are completely dened along side i j: On the other hand, the normalslope @w

@yis a cubic in x;with four constants, and only two d.of. left (xi; xj).

Therefore, this slope is not uniquely dened, and function w is said to benonconforming. However, this element has proven to give acceptable results,

and to converge.Constants c1 to c12 can be determined by expressing 12 simultaneous

equations linking them to the values ofw and its slopes at the nodes:


72/110


73/110


Focusing on the left-hand side,

RVe

f"gT fg dV = RVe

8


74/110


ffeb g =

RVe[N]

T fbg dV

Surface tractions:

ffes g =R

Ae[N]T qdA

For a uniform load qacting over the surface of an element of dimensions2b 2c, the force and moments at nodei are (similar expressions for j; k;l):8


75/110

Chapter 7

Method of weighted residuals

So far, we have always used the principle of virtual displacements (or, equiva-lently, the minimum of potential energy) towards nite element (FE) formula-tion. What happens when such principles are not available? FE formulationcan in fact be generalized to virtually all problems describable by ordinaryor partial dierential equations (ODE or PDE). For simplicity, let us limitourselves here to linear statics, with [K]fUg = ffg being the end point ofthe formulation.

7.1 The method of weighted residuals

7.1.1 General concepts

Consider a governing equation involving one independent variable:f [T(x)] = 0in domain where T represents the function sought (e.g. temperature), which is a

function ofx only. In addition, let us specify the boundary conditions (BC):g1[T(x)] = 0on1g2[T(x)] = 0on2

where 1 and 2 include only those parts of that are on the bound-ary. Let us approximate the solution with an approximate functionbT givenbybT =bT(x; a1; ; an) = Xni=1 ai Ni(x) which has n unknown constantparametersai and satises the boundary conditions exactly. TheNi(x) are

referred to as trial functions. If the approximate solutionbTis substituted inthe governing equation, some residual error R(x; a1; ; an)appears:

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CHAPTER 7. METHOD OF WEIGHTED RESIDUALS 70

f [

bT(x; a1; ; an)] =R(x; a1; ; an):

The method of weighted residuals requires that parameters ai be deter-mined by satisfying:Z

wi(x)R(x; a1; ; an)dx= 0 ; 1 i n

where thewi(x)aren arbitrary weighting functions. Many functions canbe used; the most popular methods are described below.

7.1.2 Point collocation

wi(x) =(x xi)(Dirac function) such that

Z ba

(x xi)dx= 1 for x = xi

Z ba

(x xi)dx= 0 for x 6=xi.

Thexi are the collocation points and are selected arbitrarily by the ana-lyst.

If

Z

(x xi)R(x; a1; ; an)dx= 0is evaluated atn collocation points,

nalgebraic equations in n unknowns (ai) result:R(x1; a1; ; an) = 0R(x2; a1; ; an) = 0

...R(x

n; a

1; ; a

n) = 0

Once theaiare determined, the approximate solutionbTis found, and theproblem is solved.

Example 14 solve the ODE: d2T

dx2 + 1000x2 = 0 for0 x 1 withT(0) =

T(1) = 0; usingN1(x) =x(1 x2) (satises BC)bT =a1N1(x) =a1x(1 x2)

R(x; a1) = d2bT

dx2 + 1000x2 = 6a1x + 1000x

2

Collocation point (chosen): x1 = 12

Z 1

0

(x 12)R(x; a1)dx= 0 , Z 1

0

(x 12) (6a1x + 1000x2)dx= 0

, R(x1; a1) = 0, 6a1(

12

) + 1000(12

)2 = 0, a1 =

100012

Finally,bT(x) = 100012 x(1 x2) (dashed line in gure below).


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7.1.3 Subdomain collocation

The domain is subdivided into n subdomains over which wi(x) is unityand0 elsewhere.

w1(x) = 1for x in 1

0for x not in1...

wn(x) = 1for x in n

0for x not inn

Figure 7.1: Subdomain collocation.

IfZ

wi(x) R(x; a1; ; an)dx = 0 is evaluated for 1 i n; n integralequations result:Z

1

R(x; a1; ; an)dx= 0

...Zn

R(x; a1; ; an)dx= 0

These equations can be solved for ai; andbT determined.Example 15 same problem as above: solve the ODE: d

2Tdx2

+ 1000x2 = 0 for

0 x 1 withT(0) =T(1) = 0; usingN1(x) =x(1 x2)bT =a1N1(x) =a1x(1 x2)Only one integral of the residual needs to be evaluated (because one

parameter,a1):


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R(x; a1) = d2bT

dx2 + 1000x2 = 6a1x + 1000x

2

Z 10

R(x; a1)dx= 0 , a1 = 10009

Finally,bT(x) = 10009 x(1 x2) (thin solid line in Fig. 7.2).Note: this method is equivalent to the Ritz (variational) method.

7.1.4 Least squares

The method of least squares requires that the integral Iof the square of theresidualR be minimized:

I=

Z

[R(x; a1; ; an)]2dx:

In other words, parameters ai need to be determined so that I is mini-mized. This is done by writing:@I

@a1= 0 = @

@a1

Z

[R(x; a1; ; an)]2dx= 0

...

@I@an

= 0 = @@an

Z

[R(x; a1; ; an)]2dx= 0

Because the limits on the integral are not functions ofai; the order ofintegration and dierentiation may be interchanged to give:

Z@

@a1[R(x; a1; ; an)]

2dx= 0 etc... , or

ZR @R

@a1dx= 0 etc...

Finally, we observe that in this method, wi(x) = @R@ai ; 1 i n:Then equations obtained will be used to solve for the n parametersai:

Example 16 from above: R(x; a1) = d2bT

dx2 + 1000x2 = 6a1x + 1000x

2

@R@a1

= 6xZ 10

R @R@a1

dx= 0 , a1 = 1000

8

Finally,

bT(x) = 10008 x(1 x

2) (dotted line in Fig. 7.2).

7.1.5 GalerkinIn the Galerkin method, the trial functionsNi(x)are used as weighting func-tions, that is wi(x) =Ni(x):


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IfZNi(x) R(x; a1; ; an)dx= 0 is evaluated for 1 i n; n integralequations result and can be solved for n parametersai:Example 17 N1(x) =x(1 x

2) from aboveZ 10

x(1 x2)(6a1x + 1000x2)dx= 0 , a1=

500048

Finally,bT(x) = 500048

x(1 x2) (medium solid line in Fig. 7.2).Note: this method is equivalent to the Rayleigh-Ritz (variational) method.

7.1.6 Comparison with exact solution

The exact solution to the above problem is T(x) = 100012 x(1 x3)(thick solid

line in Fig. 7.2), while our approximate solution was assumed to be of the

formbT(x) =a1x(1 x2): None of the methods seems to do an outstandingjob, but the trial functions have not been worked on. In what follows, wewill use the Galerkin method in combination with trial functions that are nolonger applied to the entire domain, but applied locally over each element.This is a very successful method.

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00

10

20

30

40

50

x

y

Figure 7.2: Comparison between solutions.


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7.2 The Galerkin nite element method

7.2.1 Formulation

Piecewise continuous trial functions (or shape functions) will be employedfor each element. For a 1-D problem, as in Fig. 7.3:

Figure 7.3: Piecewise formulation.

The problem domain a x b is divid

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