the finite element method - tamu...

JN Reddy

The Finite Element Method

Read: Chs. 3 and 4

1D Problems Governed by Second-Order Equation

CONTENTS• Model differential equation • Finite element approximation • Finite element discretization• Development of weak form and

the definition of primary and secondary variables (duality)

• Essential and natural BCs• Linear and bilinear forms • Finite element model • Interpolation functions• Assembly of elements • Numerical examples

JN Reddy

When you are analyzing an engineering problem using a FEM program, you should know: (1) the mathematical model (governing

equations) that the program is solving, (2) the duality pairs for specifying the boundary

conditions, and (3) know the restrictions of the mathematical and

computational models.

Important Comments• Knowing is power. Your confidence in what you

do goes up.

• Mathematics is the language of an engineer.

• “Mathematics is the language with which God has written the universe.” −Galileo Galilei

JN Reddy

Major Steps of Finite Element Formulation

Finite Element Model Development

Governing (Differential) Equations

Engineering ProblemFormulation

Virtual work statements

Weak Form Development

Basic Concepts: 3

Solid mechanicsAll other fields

JN Reddy Basic Concepts: 4

Model Differential Equation Boundary Conditions and data

0ˆ, ( ), ( ), ( ), , , ,L a x c x f x b u u P

Data (i.e., information you need to solve the problem)

0 0( ) ( ) ( ) in ( , )d dua x c x u f x Ldx dx

− + − = Ω =

Governing equation

0ˆ or ( ) at a boundary pointduu u a b u u Pdx

= + − =

Boundary conditions

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ENGINEERING EXAMPLES OF THE MODEL PROBLEM IN 1-D

Basic Concepts: 5

Elastic deformation of a bar

( )f xa EA=0u

P( )u L

b k=

x L

c


− + − = Ω =

0 0( )u =

( ) axial displacement( ) body force( ) modulus of elasticity( ) area of cross section

spring elastic (linear) stiffnesssurface shear resistance (linear)axial point load at

u xf xE xA xkcP x L

====

=== =

0( )x L

duEA k u u Pdx =

+ − =

JN Reddy Introduction: 6


1D Heat flow through a fin


− + − = Ω =

( ) ( ) , absolutetemperature( ) internal heat generation( ) thermal conductivity( ) area of cross section

heat transfer coefficientperimeterheat input at ambient temperature

u x T x Tf xk xA x

PQ x LT

β

∞

∞

= −===

=== ==

uninsulated bara kAc Pβ== ( )f x

xL

u T T¥= -

0T

ˆx L

dukA u Qdx

b=

é ùê ú+ =ê úë û00( ) ,u T T¥= -

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Flow of a viscous fluid through a channel

Basic Concepts: 7

0 0in ( , )d dua f bdy dy

− − = Ω =

00

( ) , axial velocity

, pressure gradientviscosity of the fluidshear stress at

, Poiseuille flow, Couette flow

x

L

L

L

u y vdpfdx

aQ x Luu

μτ

=

= −

= == = ==≠

00( ) , ( ) Lu u u b u= =

Luy

b

x2b

0 0u =

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Finite Element Approximation(Need for Seeking Solution on Sub-intervals)

Approximation of the actual solutionover the entire domain requires higher-order (or degree) polynomials.1.

Actual solution may be defined by piecewise continuous functions because of discontinuity of the data.2.

A1 A2 A3A = Area of cross section

x

u(x)

Material 1 Material 2 Material 3x

u(x)Three segments of the solution

Basic Concepts: 8

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FINITE ELEMENT DISCRETIZATION

Approximation over subintervals (or elements) allows lower-order approximation of the actual solutionx

A “free-body-diagram” of typical element (geometry and ‘forces’)

b

, end forces or heats (scalar)

element lengtha b

a

Q Q

h x x= - =

Basic Concepts: 9

x xaQ bQ

axx = bxx =

( )f x

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{ }

0

0

( ) ( ) ( ) ( )

We seek to make ( ) zero in a weighted-residual sense:

( ) , is the weight function from a set of weight functions

( )

b

a

hh

x

i ixi

hi

dud a x c x u f x R xdx dx

R x

w R x dx ww

dudw a xdx dx

é ùæ ö÷çê ú- + - = ¹÷ç ÷çê úè øë û

=

æ öç= - ççè

ò

( ) ( )b

a

x

hxc x u f x dx

é ù÷ê ú+ -÷÷ê úøë ûò

Approximate Solution and Residual of the Approximation

Approximate solution: ( ) ( )hu x u x»

Basic Concepts: 10

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Product rule of differentiation (an identity) and integration-by-parts

( ) ( ) ( )

( ) ( ) ( )

h i h hi i

h h i hi i

du dw du dud dw a x a x w a xdx dx dx dx dx dx

du du dw dud dw a x w a x a xdx dx dx dx dx dx

æ ö æ ö÷ ÷ç ç= +÷ ÷ç ç÷ ÷ç çè ø è ø

æ ö æ ö÷ ÷ç ç- =- +÷ ÷ç ç÷ ÷ç çè ø è ø

Trading of Differentiation between the weight function and the variable

( ) ( ) ( )

( ) ( )

b b

a a

bb

aa

x xh h i h

i ix x

xx

i h hix

x

du du dw dud dw a x dx w a x a x dxdx dx dx dx dx dx

dw du dua x dx w a xdx dx dx

é ùæ ö æ ö÷ ÷ç çê ú- = - +÷ ÷ç ç÷ ÷ç çê úè ø è øë û

æ ö÷ç= - ÷ç ÷çè ø

ò ò

ò

Basic Concepts: 11

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Examine the boundary term(s) obtained during integration-by-parts:

The expression always contains the weight function w and acoefficient that depends on the dependent unknown. In thiscase the coefficient is a(duh/dx). We will term the coefficient asecondary variable (a name we choose to give).

Identification of the Primary and Secondary Variables

b

a

xh

ix

duw adx

⋅

Secondary variable

The weight function wi in the boundary term when replacedwith the dependent variable uh of the problem is termed theprimary variable.

b

a

xh

ix

duw adx

⋅

hu Primary variable

Basic Concepts: 12

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0b

a

bb

aa

b

aa

xh

i hx

xx

i h hi h i ix

x

xi h h h

i h i i a i bxx

dudw a x c x u f x dxdx dx

dw du dua cw u w f dx w adx dx dx

dw du du dua cw u w f dx w x a w x adx dx dx dx

é ùæ ö÷çê ú= - + -÷ç ÷çê úè øë û

é ù é ùê ú ê ú= + - - ⋅ê ú ê úë û ë û

é ù æ ö æ ö÷ ÷ç çê ú= + - - ⋅ - - ⋅÷ ÷ç ç÷ç çê ú è ø è øë û

ò

ò

ò

( ) ( ) ( )

( ) ( )b

b

a

x

xi h

i h i i a i b bx

dw dua cw u w f dx w x Q w x Qdx dx

÷

é ùê ú= + - - - ⋅ê úë û

ò a( ) ( )

WEAK FORM DEVELOPMENT OVER AN ELEMENT

weight functionw −

secondary variable

Basic Concepts: 13

0b

a

xi h

i h i i a i b bx


é ùê ú= + - - - ⋅ê úë û

ò a( ) ( )

Final Weak Form

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Interpretation of the secondary variables

Force

b

hb

x

dua Qdx

æ ö÷ç º÷ç ÷çè øa

ha

x

dua Qdx

æ ö÷ç- º÷ç ÷çè ø he 21

ForceDispl. Displ.

Axial deformation of a bar

Axial heat flow

Heat input

b

hb

x

dua Qdx

æ ö÷ç º÷ç ÷çè øa

ha

x

dua Qdx

æ ö÷ç- º÷ç ÷çè ø he 21

Heat inputTemp. Temp.

Basic Concepts: 13

JN Reddy

Primary variables and secondary variables alwaysappear in pairs. They are like `cause’ and `effect’ (i.e.,one is the result of the other). For example, when uh isthe temperature, a(duh/dx) is heat (and heat causestemperature). When uh is the displacement, a(duh/dx)is the force. This duality exists in every engineeringproblem.

Primary and Secondary Variables(Some Remarks)

Essential Boundary Conditions: Specifying a primaryvariable at a boundary point of the domain is called anessential (or Dirichlet) boundary condition.

Essential and Natural Boundary Conditions

Natural Boundary Conditions: Specifying a secondaryvariable at a boundary point of the domain is called anatural (or Neumann) boundary condition.

Basic Concepts: 15

JN Reddy

Recall that the primary and secondary variables appearas pairs. One may specify only one element (variable)of each pair at a boundary point. Thus, whenis the pair, only u or a(du/dx), and never, both may bespecified at a boundary point.

)]/(,[ dxduau

Essential and Natural Boundary Conditions

0

0

00

0 00

1 0

2 0

3

4

. ( ) , ( )

. ( ) ,

. , ( )

. , ( )

L

Lx L

Lx

L Lx x L

u u u L u

duu u a Pdx

dua P u L udx

du dua P a P P Pdx dx

=

=

= =

= =

æ ö÷ç= =÷ç ÷çè ø

æ ö÷ç = =÷ç ÷çè ø

æ ö æ ö÷ ÷ç ç= = =÷ ÷ç ç÷ ÷ç çè ø è ø

Thus, for a problem with two boundary points, there arefour combinations of boundary conditions:

0P LP

LP

0P

Basic Concepts: 16

JN Reddy

Linear, Bilinear Forms, and the Variational Problem

( , ) ( ) holds for all i h i iB w u l w w=

a( , ) , ( ) ( ) ( )b a

a b

x xi hi h i h i i i a i b bx x

dw duB w u a cw u dx l w w f dx w x Q w x Qdx dx

= + = + + ⋅

Bilinear Form and Linear Form

Weak Form

a

a

0 ( ) ( )

( ) ( )

( , ) ( )

b

a

b a

a b

x i hi h i i a i b bx

x xi hi h i i a i b bx x

i h i


dw dua cw u dx w f dx w x Q w x Qdx dx

B w u l w

= + − − − ⋅

= + − + + ⋅

= −

Variational Problem: Find u such that

Basic Concepts: 17

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Equivalence Between Minimum of a the Total Potential Energy and Weak Form

Total potential energy (of uniaxial members):

221

21

b b

a a

E

nx x e ef i ix x i

U W

duEA c u dx uf dx u Qdx =

= +

é ù æ öæ ö ÷ê ú÷ çç= + - + ÷÷ çç ÷ê ú÷ç ÷çè ø è øê úë ûåò ò

Π

1

b b

a a

nx x e ef i ix x i

d u duEA c uu dx u f dx u Qdx dxd

d d d d=

é ù æ öæ öæ ö ÷÷ ÷ çç çê ú= + - + ÷÷ ÷ çç ç ÷÷ ÷ç ç ÷çê úè øè ø è øë ûåò òΠ

0Now let . Then gives the weak form:iu wd d= =Π

Basic Concepts: 18

1

0 ( )b b

a a

nx x eif i i i j jx x j

dw duEA c w u dx w f dx w x Qdx dx =

æ öé ùæ öæ ö ÷ç÷ ÷ç çê ú ÷= + - +÷ ç÷ç ç ÷÷ ÷ ççê ú÷ç ÷çè øè ø è øë ûåò ò

JN Reddy

22

0

12

dd d d d

dd d d d

d d

d

é ùê ú= + - - -ê úë û

é ù é ùê ú= + - + +ê úê ú ê úë ûë û

é ùæ ö é ùê ú÷ç= + - + +÷ ê úçê ú÷ç ê úè ø ë ûê úë û

=

ò

ò ò

ò ò

a

a

a

( ) ( )

( ) ( )

( ) ( )

b

a

b a

a b

b a

a b

x

a b bx

x x

a b bx x

x x

a b bx x

d u dua c uu u f dx u x Q u x Qdx dx

d u dua c uu dx u f dx u x Q u x Qdx dx

dua cu dx u f dx u x Q u x Qdx

[ ]12

12

d- =

= -

( , ) ( ) ( )

or

( ) ( , ) ( )

B u u l u I u

I u B u u l u

Equivalence Between Minimum of a Quadratic Functional and Weak Form

Replace w with δu

Basic Concepts: 19

[when B(w,u) is bilinear and symmetric]

JN Reddy

FINITE ELEMENT MODEL

1( ) ( ) ( )

ne e eh j j

ju x u x u x

=

≈ = yFinite element approximation

11

0 a( ) ( )

( ) ( )

b a

a b

b a

a b

x xi h

i h i i a i b bx x

n x xj e eij i j i i a i b n

x xj

dw dua cw u dx w f dx w x Q w x Qdx dx

ddu a c dx f dx x Q x Qdx dx

yyy y y y y

=

é ùæ öç ÷ ê ú= + - + + ⋅ç ÷ ê úç ÷ç ÷è ø ë ûé ù é ùê ú ê ú= + - + +ê ú ê úê ú ë ûë û

ò ò

å ò ò

he

211eQ e

nQsF cu=

( )f x

( )= ( )ei iw x xy

Basic Concepts: 20

(is a set of algebraic relations between the primaryand the secondary variables at the nodes)

1 2 2

1

1

[ ]{ } {

(

}

( , ) ,

( ) ( ) ( ))

b

a

b

a

ne e e e e eij j i

j

eex je e eiij i j e e

e e

i jx

xe e e e ei i e i i ii n nx

K u F K u F

ddK B a c dxdx dx

F l f dx x Q x Qx Q

yyy y y

y y

y

y y y

=

= =

= = +

= = + ++ +

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Approximation (interpolation or shape) Functions for Linear Element

1 1 2 2 1 2( ) ; ( )h a a h b bu x u c c x u x u c c x= = + = = +

1 2 2 11 2,b a

b a b a

u x u x u uc cx x x x

- - = =

- -

1 2

1 2 2 1

1 1 2 2

( ) ( )

( ) ( )

h

b a

b a b a

u x u x c c x

u x u x u u xx x x x

x u x u

≈ = +

− −= +− −

= +y y

1 2( ) ( )hu x u x c c x≈ = +

he

21

2u1u

he

21

1( ) b

b a

x xxx x

y−

≡− 2( ) a

b a

x xxx x

y−

≡−

xaxx = bxx =

1.0

1 2( )hu x c c x= +

1( )h a au x u u≡ =2( )h b bu x u u≡ =1 2 1 2Rewrite and in terms of andc c u u

Basic Concepts: 21

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Interpolation Propertiesof the Approximation Functions

2u

he

21

1 1( )u xy 2 2( )u xy

xaxx = bxx =

1u

1 1 2 2( ) ( ) ( )hu x x u x u= +y y

0,( )

1,i j ij

if i jx

if i j≠

≡ = =y d ba xx,xx ≡≡ 21

Interpolation property

1( ) 1

n

jj

x=

=y

Partition of unity

Basic Concepts: 22

JN Reddy

Derivation of Approximation Functions(Quadratic Element)

1 1 1 1 2

2 2 2 2 2

3 3 3 3 2

20 5 0 1

40 0 5 1

20 0 5 1

( ) ( )( . ), ( )

( ) ( )( ), ( . )

( ) ( )( . ), ( )

x h x h xh

x h x x hh

x x h x hh

= − − = → =

= − − = → =

= − − = → = −

y a y a

y a y a

y a y a

Derivation using the interpolation property

he

21

xaxx = ea hxx +=

1u

1 1 2 2 3 3( ) ( ) ( ) ( )hu x x u x u x u= + +y y y

ea h.xx 50+=

3

)(xψ1 )(xψ3)(xψ2

3u

1.0 1.01.0(b)

1 2 3

1 ( )e xy2 ( )e xy

3 ( )e xy

x

)(xeiψ

1 23

x he

(a)

ex3ex2ex1

x

21 2 3( ) ( )hu x u x c c x c x» = + +

1 2

3

21 1 4 1

21

( ) , ( )

( )

x x x xx xh h h h

x xxh h

= − − = −

= − −

y y

y

Basic Concepts: 23

JN Reddy

Derivation of Approximation Functions(Cubic Element)

1 1

1 1 3

1

23 3

90 12

3 31 1 12

( ) ( )

( )

( )

h hx x x x h

h

x x xxh h h

æ öæ ö÷ ÷ç ç= - - -÷ ÷ç ç÷ ÷ç çè øè ø

= =-

æ öæ öæ ö÷ ÷ ÷ç ç ç= - - -÷ ÷ ÷ç ç ç÷ ÷ ÷ç ç çè øè øè ø

y a

y a

y

Derivation using the interpolation property

2 31 2 3 4( ) ( )hu x u x c c x c x c x» = + + +1

2 3

hex

4

Basic Concepts: 24


Approximation Functions in terms of the local coordinates

1 3xhe

3ex

1ex

x

x

0x = ex h=1x =- 1x =+ 1

1 11 22 2

1 2

1 1

2 2

( ) ( )

e

e e

e ee

x x x

x x x

hx x

x x

x

= +

= - + +

+= +

Relations between various coordinates

1 2

1 11 22 2

1 0

1 1 1 1

( ) , ( ) ,

( ) ( ), ( ) ( ),

ee e

x xx x x hh h

x

y y

y x x y x x

= - = < <

= - = + - < <

1 2

2 1

1

1

( ) ( ),

( ) ( )

e

e

e

e

x x xh

x x xh

y

y

üïï= - ïïïïýïïï= - ïïïþ

1 2e ex x x< <


NUMERICAL EVALUATION OF COEFFICIENTSin various coordinates

2

1

0

0

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

e

e

e

b e

a

x je iij i jx

h jii j

x hei i i i ix

ddK a x c x x x dxdx dx

dda x c x x x dxdx dx

F f x x dx Q f x x dx Q

yyy y

yyy y

y y

æ ö÷ç ÷= +ç ÷ç ÷çè ø

æ ö÷ç ÷= +ç ÷ç ÷çè ø

= + = +

ò

ò

ò ò1

1

2

( ) ( ) ( ) ( ) b

a

xei i i i ix

e

F f x x dx Q f Jd Q

hdxJd

y x y x x

x

+

-= + = +

= =

ò ò


NUMERICAL EVALUATION OF COEFFICIENTSfor element-wise constant data

For constant : , ,

, b b

a a

e e e

x xje eiij i j i i ix x

data a a c c f f

ddK a c dx F f dx Qdx dx

= = =

æ ö÷ç ÷= + = +ç ÷ç ÷çè øò òyy

yy y

1

2

1 1 2 1 11 1 1 2 16 2

[ ] , { }e ee e e e e

e

Qa c h f hK FQh

− = + = + −

Linear element:eh

1 2 2 2,u Q1 1,u Q

JN Reddy

NUMERICAL EVALUATION OF COEFFICIENTSfor element-wise constant data

known

Basic Concepts: 28

{ }1

2

3

7 8 1 4 2 1 1[ ] 8 16 8 2 16 2 4

3 30 61 8 7 1 2 4 1

e ee e e e e

e

Qa c h f hK , F Qh

Q

− − = − − + = + − −

Quadratic element: eh

1 2 31 1,u Q 3 3,u Q

2 2,u Q

For constant : , ,

, b b

a a

e e e

x xje eiij i j i i ix x

data a a c c f f

ddK a c dx F f dx Qdx dx

= = =

æ ö÷ç ÷= + = +ç ÷ç ÷çè øò òyy

yy y


Other (Discrete) Elements

Td, GB

LSteel •

32

4dJ π=θ

e1θ,T e

1e2θ,T e

2

eh

Torsion element:

1

2

1 11 1

[ ] , { }e

e ee ee

e

TG JK Fh T

ì üé ù ï ï- ï ïê ú= = í ýê ú ï ï-ë û ï ïî þ

1

2

1 11 1

[ ] , { }e

e ee e

FK k FF


Spring element: 1 2ek11 ,e eu F

22 ,e eu F


1

2

1 11 1

[ ] , { }e

e ee e

IK k FI


Electrical element:1

ee

kR

= 1 2

Electrical resistance

eR =

eR

2 2,e eV I1 1,e eV I

Pipe flow element:

1

2

1 11 1

[ ] , { }e

e ee e

QK k FQì üé ù ï ï- ï ïê ú= = í ýê ú ï ï-ë û ï ïî þ

4

128ee

dkh

pm

=

d 1 1,e eP Q

eh

2 2,e eP Q

Other (Discrete) Elements (continued)

JN Reddy

Representation of Point Sourcesat points other than nodes

1 41 0 1 0( )f F sy=

0s3

2 0 2 0( )f F sy=

3 0 3 0( )f F sy=

4 0 4 0( )f F sy=

0( ) ( )

hei if f s s dsy= ò

eh

1 4

( )f s

s 3

eh

1 4

0F

s0s 3

0 00

0 0 0 00

( ) ( ), ( ) ( )

( ) ( ) ( )

hei i

hei i i

f s F s s f f s s ds

f F s s s ds F s

d y

d y y

= - =

= - =

ò

ò

Basic Concepts: 31

JN Reddy Basic Concepts: 32

Equivalence Between Minimum of a the Total Potential Energy and Weak Form

321 1 1

2 2 21

2 21 12 2

,( )

e e e

b

ea

ij ij xx xx xxV V Vi j

x

V x

U dv dv E dv

du duE dv EA dxdx dx

s e s e e=

= = =

æ ö æ ö÷ ÷ç ç= =÷ ÷ç ç÷ ÷ç çè ø è ø

å ò ò ò

ò ò

Strain energy:

enQ1

eQ 21 n

2eQ

1-enQ

enu1

eu 21 n

2eu

1-enu

Work done by external forces:

1

b

a

nx e eE i ix i

W uf dx u Q=

æ ö÷ç=- + ÷ç ÷ç ÷è øåò

JN Reddy

ASSEMBLY OF ELEMENTS

Assembly of elements is based on two requirements:

• Continuity of the primary variables across theelement boundaries.

• Balance of the secondary variables between theelement boundaries.

(1) 12 3 … n

e1

2 3 … mf

( ) ( )1

e fnu u=

(2) ( ) ( )1 0 or equal to externally applied sourcee f

nQ Q+ =

Basic Concepts: 33

JN Reddy

POST-COMPUTATION OF VARIABLES(1) (1) (1) (1)

11

(2) (2) (2) (2)1

1

( ) ( ) ( ) ( )1

1

( ),

( ),

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

( ),

n

j j nj

n

j j nejh

nN N N Nj j n

j

u x x x x

u x x x xu

u x x x x

y

y

y

=

=

=

ìïï £ £ïïïïïïï £ £ïï= íïïïïïïï £ £ïïïïî

å

å

åx

aQbQ

1e

ax x x= = eb nx x x= =

• • •• •1 2 n

x

eth element

•

(1)(1) (1) (1)

11

(2)(2) (2) (2)

11

(1)( ) ( ) ( )

11

,

,

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

,

nj

j nj

nje

j nhj

njN N N

j nj

du x x xdx

du x x xdudx

dx

du x x x

dx

y

y

y

=

=

=

ìïï £ £ïïïïïïïï £ £ïï= íïïïïïïïï £ £ïïïïî

å

å

åBasic Concepts: 34

JN Reddy

A DIFFERENTIAL EQUATION

FE Solution: We have the following correspondence compared to the model equation:

2 21 1= =- =- =-òb

a

xei ix

a , c , f x , f x dxy

(1) We wish to use a mesh of linear elements to solve the problem. The equations of a typical element are

1

2

1 1

2 2

1 1 2 1161 1 1 2

ì üï ïì ü ì üæ öé ù é ù ï ï ï ï- ï ï÷ï ï ï ï ï ïç ê ú ê ú÷- = +ç í ý í ý í ý÷ç ê ú ê ú÷ï ï ï ï ï ïç -è øë û ë û ï ï ï ï ï ïî þ î þ ï ïî þ

ee ee

e e ee

Qu fhh u f Q

Basic Concepts: 35

Problem: Wish to determine the numerical solution of the differential equation

22

2 0 1

0 0 1 0

in

( ) , ( )

d u u x xdx

u u

- - =- < <

= =

1


( )22

0y y y= =- =- +ò ò ò( ) ( ) ( )b b e

a a

x x hei i i a ix x

f f x x dx x dx x x x dx

Details of the computation of the nodal source vector due to the distributed source

● ● ●●●

x xax bx

eh

= + ax x x

1 21( ) , ( )e e

x xx xh h

y y= - =

( ) ( )

( )

( )

2 2 2

0 0

2 21 0

2 22 0

2

2 1

2

y y=- + =- + +

æ ö÷ç ÷=- + + -ç ÷ç ÷çè ø

æ ö÷ç ÷=- + + ç ÷ç ÷çè ø

ò ò

ò

ò

( ) ( )e e

e

e

h hei a i a a i

hea a

e

hea a

e

f x x x dx x x x x x dx

xf x x x x dxh

xf x x x x dxh


Details of the computation of the nodal source vector due to the distributed source

( )

( )

2 21 0

3 4 3 22 2 2

0 0

3 4 3 22 2 2

42 2

2 0

2 1

1 23 4 3 2

213 4 3 2

12

æ ö÷ç ÷=- + + -ç ÷ç ÷çè ø

é ù é ùê ú ê ú=- + + + + +ê ú ê úë û ë ûé ù é ùê ú ê ú=- + + + + +ê ú ê úë û ë û

æ ö÷ç ÷=- + + =-ç ÷ç ÷çè ø

ò

ò

e

e e

e

hea a

e

h h

a a a ae

e e e ee a e a a a

e

he ea a

e e

xf x x x x dxh

x x x xx x x x x xh

h h h hh x h x x xh

hxf x x x x dxh h

3 22

1

1 2 1 2 3

24 3 2

0

é ùê ú+ +ê úë û

= =

= + = + +

Note that for Element 1, for Element 2,for Element 3, and for Element 4

e ea a

a a

a a

h hx x

x x hx h h x h h h

JN Reddy

(2) We consider a mesh of 4 linear elements (h =0.25). The element equations are 1 2 3 4 5

1 2 3 4

1

+

−=

−

−1

1

1

1

241

2

1

2

1

003910001300

94979794

QQ

.

.uu

1 2)(1

1Q )(12Q

+

−=

−

−2

2

2

2

241

2

1

2

1

022320014320

94979794

QQ

.

.uu

1 22)(2

1Q )(22Q

+

−=

−

−3

3

3

3

241

2

1

2

1

055990042970

94979794

QQ

.

.uu

1 23)(3

1Q )(32Q

+

−=

−

−4

4

4

4

241

2

1

2

1

105470087240

94979794

QQ

.

.uu

1 24)(4

1Q )(42Q

Basic Concepts: 38

A DIFFERENTIAL EQUATION (cont.)

JN Reddy

(3) The boundary conditions on the primary variables are

(1) (2) (2) (3) (3) (4)2 1 2 1 2 10 0 0Q Q , Q Q , Q Q+ = + = + =

00 51 == U,U

1 2 3 4 51 2 3 4

1 2

1)(1

1Q )(12Q )(2

1Q )(22Q

2 1

2

The equilibrium conditions are

Basic Concepts: 39


JN Reddy

(4) The assembled equations are

+++

+

−=

−−−

−−−−

−

42

41

32

31

22

21

12

11

5

4

3

2

1

94970009718897000971889700097188970009794

241

QQQQQQQ

Q

UUUUU

0.1054700.1432300.0651040.0182290.001302

−=

−−−

−

0.143230.065100.01823

7.83334.041704.0417 7.83334.0417

4.04177.8333

4

3

20

UUU

(5) The condensed equations for the unknown U’s and Q’s are

211

342

4

-4.0417 0 0 0.00130 0.095200 0 4.0417 0.10547 0.26386

UQ

UQ

U

= + = −

000

0

0

Basic Concepts: 40


JN Reddy

0.0 0.2 0.4 0.6 0.8 1.0Coordinate, x

-0.05

-0.04

-0.03

-0.02

-0.01

0.00So

luti

on, u

(x)

Analytical

4L2Q4Q

Plot of the solution,

Basic Concepts: 41

andh exactu u

FEM

Uniform mesh of 4 linear elements

2 quadratic elements

Nonuniform mesh of 4 linear elements


JN Reddy

0.0 0.2 0.4 0.6 0.8 1.0Coordinate, x

-0.15

-0.10

-0.05

0.00

0.05

0.10

0.15

0.20

0.25

0.30

Solu

tion

, du/

dx

Analytical

4L2Q4Q

FEM

Plot of the derivative of the solution

Basic Concepts: 42

111

1 1

e ee en nj je eh h

j jj j

d ddu duu udx dx dx dx

y y +++

= =

= ¹ =å å

Same at the nodes common to elements

Not the same at the nodes common to elements


JN ReddyBasic Concepts: 43

Rigid bar

211

°°°°

2

3

4

2

21

2

Global nodes

Element nodes

(2)2 3u U≡

2k

3

(1)1 1u U≡

P

3k

1k

°°°

°

° 1

(3)2 4u U≡

(1) (2) (3)2 1 1 3u u u U= = ≡

2

2

1

ek1 2

e1δ e

2δ

1eF eF2

1

2

1 11 1

[ ] , { }e

e ee e

FK k F

F

− = = −

Typical spring element and its force-displacement relations

A network of linear elasticsprings with finite element representation

k1°°

11F

12F

31F

P

°°°

°°°°

°

°

21F

°°2

2F

°°3

2F

031

21

12 =−−− FFFP

2k

3kIndividual elements and their forces

Balance of forces

1 22 32 4

[B] =

A Network of Springs (No Des)

2

JN Reddy

Rigid bar

211

°°°°

2

3

4

2

21

2

Global nodes

Element nodes

2U3U

4U

2k

3

1U

P

3k

1k

°°°

°

°1

1 °

°

°B

é ùê úê ú= ê úê úë û

1 2

2 3

2 4

K

K K K K

K K K K

K K K K

K K K K

é ùê úê úê ú= ê úê úê úë û

11 12 13 14

21 22 23 24

31 32 33 34

41 42 43 44

( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( )

, ,,

, ,, .

K k K k

K k k k

K k K k

K k K k

= =

= + +

= =

= =

1 111 11 12 12

1 2 322 22 11 11

2 333 22 44 22

2 323 12 24 12

Basic Concepts: 44

Network of Springs (cont.)

JN Reddy

0

00

P

1 11 1

1 1 12 2

1 11 1

( ) ( )

( ) ( )

u Fk

u F

− = −

2 21 1

2 2 22 2

1 11 1

( ) ( )

( ) ( )

u Fk

u F

− = −

3 31 1

3 3 32 2

1 11 1

( ) ( )

( ) ( )

u Fk

u F

− = −

( ) ( ) ( )P F F F- - - =1 2 32 1 1 0------- (1)

---- (2)------- (3)

------- (4)

( )

( ) ( ) ( )

( )

( )

k k U F

k k k k k k U F F F

k k U F

k k U F

ì üé ù ì ü- ï ïï ï ï ïï ïê ú ï ïï ïê ú ï ïï ï- + + - - + +ï ïï ï ï ïê ú =í ý í ýê ú ï ï ï ï- ï ï ï ïê ú ï ï ï ïê ú ï ï ï ï- ï ï ï ïë û î þ ï ïî þ

11 1 1 1

1 2 31 1 2 3 2 3 2 2 1 1

22 2 3 2

33 3 4 2

0 0

0 0

0 0

PU

k k k=

+ +21 2 3

( ) ( ) ( ), ,F kU F kU F kU=- =- =-1 2 31 1 2 2 2 2 2 3 2

Basic Concepts: 45

Network of Springs (Cont.)

JN Reddy

Examples of Uniaxially-Loaded Members

2 1 2,PL EAu Q u PEA L

= =- =-

P

,E A

Problem

L

1 1

2 2

1 1 01 1 0

u QAEu QLì ü ì üé ù ì üï ï ï ï- ï ïï ï ï ï ï ïê ú = +í ý í ý í ýê ú ï ï ï ï ï ï- ï ïë û î þï ï ï ïî þ î þ

• •1 2

1 2

Finite element mesh

L

1 1,u Q 2 2,u Q

0P

P

,E A

ProblemL

k

Spring force

2ku •

Basic Concepts: 46

2 2Q P ku= -

Solution:

3

4

0d duEAdx dx

− =

GDE

JN Reddy

UNIAXIALLY-LOADED MEMBERS (cont.)

P

1 1,E A 2 2,E A

h1 h2

Problem P

h1 h2

• • •1 2 1 21 2 3

Finite element mesh

Inter-element compatibility

h1 h2

• • •1 2 1 21 2 32

•1

1( )F 1

2( )F 2

1( )F 2

2( )F

11( )u 1

2( )u 2

1( )u 2

2( )u

1 22 1( ) ( )F F P+ =

11 11 2

2 1 22

2 3

( )

( ) ( )

( )

u Uu u Uu U

=

= =

=

Basic Concepts: 47

5 0d duEAdx dx

− =

GDE

JN Reddy

P

h1 h2

• • •1 2 1 21 2 3

Finite element mesh

212 2 2 2

2 22

2 2 2 2 2

1111 1 1 1

1 11

2 21 1 1 1

1 1

3

2 2

0

0

( )

( )

( )

( ) FE A E Ah h

E A E A Fh

FUE A E Ah h

U FE A E Ah h

Uh

ì üì ü ï ïé ù ï ï ï ïï ïê ú ï ï- ï ï ï ïï ïê ú ï ïï ï ï ïê ú ï ï ï ïï ïê ú ï ïï ï ï ïï ïê ú- + =í ý í ýê ú ï ï ï ïï ï ï ïê ú ï ï ï ïê ú ï ï ï ïï ï ï ïê ú ï ï ï ïï ï ï ïê ú ï ï ï ïê úë û ï ï ï ïî

+-

þ î-

þ

Boundary conditions: 1 3 0U U= = 1 22 1( ) ( )F F P+ =

Solution: 21 21

1

2

2

PU E Ah

E Ah

=+

Post-computation: 1 21 211

2

22 2 2

1

( ) ( ),E A E AF U F Uh h

=- =-

Basic Concepts: 48

UNIAXIALLY-LOADED MEMBERS (cont.)

JN ReddyIntroduction: 49

APPLICATIONS TO HEAT TRANSFER

ee

e

kSh

≡ 1

2

1 1 121 1 1

2 16 1 2

[ ] { },e e e eee

e Qf hK c hs FQ

− = = + −

+

50

1 1

2 2

3 32

70 W/(m C), 2 cm 40 W/(m C), 2.5 cm 20 W/(m C), 4 cm

C, 10 W/(m C)

k hk hk hT β∞

= ⋅ =

= ⋅ =

= ⋅ =

= = ⋅

Material 1, k1

Material 2, k2

Material 3, k3

h3h1 h2

Surface area,A (=1)

T∞= 50ºC

T0= 200ºC

● ● ● ●

11 1 1 1

1 21 1 2 2 2 2 1

2 32 2 3 3 3 2 1

33 3 4 2

0 0 00 0

0 00 0 0

( )

( ) ( )

( ) ( )

( )

s s U Qs s s s U Q Q

s s s s U Q Qs s U Q

− − + − + = + − + − + −

2000

0

4( )T Tβ ∞− −

6 1D Heat flow through a composite wall

0 1

0

,

( )

d dTk Adx dxdTk T Tdx

β ∞

− = =

+ − =GDE

JN Reddy

0( )d dTkA P T Tdx dx

d dTkA P T P Tdx dx

β

β β

∞

∞

− + − = − + =

2

2

1

2

1

1

2 11 1 121 1 1 2 16

[ ] { },e ee e

e

s ss

QP T hAkKQs

P Fh

hβ β ∞− = =

+ +

≡

−

f

111 2 1

1 222 1 2 2 1

2 332 1 2 2 1

3 442 1 2 2 1

452 1 2

0 0 02 0 0

0 2 00 0 20 0 0

Us s f QUs s s f f Q QUs s s f f Q QUs s s f f Q QUs s f Q

+ + = + + +

+ +

000

42

( )

dTQ kdx

T Tβ ∞

=

= − −

known

7 1D Heat flow through a fin with convection

Fin’s lateral surface as well as the end x = L are exposed to ambient temperature, T∞

T(0) = T0

100 mm

5 mm

1 mmx

Problem: Find the temperature distribution


EXERCISE PROBLEM FOR DISCUSSION

1

2

●

●

●

1

2

3

Problem set up and FEA10 cmp

20 cm

Steel (Es = 200 GPa, As= 60 mm2 )

Aluminum (Ea = 70 GPa,Aa= 600 mm2 )

A

C

B

xp

510 NP Dpp= =

D

Problem: Find stresses in each member

0d duEAdx dx

− =

GDE



Problem set up and FEA

Problem: Find stresses in each member

A2 =0.25 in.2

F1 =2000 lb

Rigid member

°

°

°

°°

A20 in.

B

C

30 in.

30 in. 25 in.E

A1 =0.4 in.2 E =30×106 psi

D°

F2 = 6000 lb

1

2

●

●

●

1

2

30d duEAdx dx

− =

GDE



2

2

1

1

( )

, ,

,r

r

d trt tr dr r

du u u du Ec c cdr r r d

r

r

q

q

ss rw

s n s nn

- +

æ ö æ ö÷ ÷ç ç= + = + =÷ ÷ç ç÷ ÷ç çè ø è ø -

=GDE:

Problem: Find the hoop stress in the rotating circular disc

●( , )r q

rq

w

2

00

0

10

2

( )

( ) ( )

b

a

b

a

r

rr

r

r a a b br

d tw tr f rdrdr dr r

dwtr wt wf r dr Q w r Q w rdr

pq

q

ss q

p s s

é ùê ú= - + -ê úë û

æ ö÷ç= + - - -÷ç ÷çè ø

ò ò

ò

Weak Form:2

0 2 2, ( ) , ( )a r a b r bf t r Q tr Q trrw p s p s= º - º


0

12

2

[ ]{ } { } or

( ) ( )

b

a

b

a

e e e e e e

r j je iij j i jr

rei i a i a b i br

K u F

d ddK ct r drdr dr r r dr

F f rdr Q r Q r

y yy np y y y n

p y y y

= =

é ùæ ö æ ö÷ ÷ç çê ú÷ ÷= + + +ç ç÷ ÷ê úç ç÷ ÷ç çè ø è øë û

= + +

ò

ò

K u F

20 2 2, ( ) , ( )a r a b r bf t r Q tr Q trrw p s p s= º - º

EXERCISE PROBLEM FOR DISCUSSION: spinning disc problem

(continued)

Finite Element Model:

JN Reddy

SUMMARY

Beginning with a model second-order differentialequation that arises, for example, in connection withaxial deformation of bars, 1-D heat transfer in fins ofa heat exchanger, or 1-D flow through pipes andchannels, the following steps are used to in the finiteelement analysis of the problem:

1. Divided the domain into subdomains, called finiteelements.

2. Over each element, an integral statement, calledweakform, is developed. The weak form is equivalent tothe differential equation as well as specified naturalboundary conditions on the boundary of the element.

Basic Concepts: 55

JN Reddy

SUMMARY (continued)

3. Using polynomial approximation of the variables, asystem of algebraic equations, called finite elementmodel, is developed. The model relates the nodal valuesof the PVs and the SVs.

4. The element equations are then assembled to eliminateexcessive unknown SVs by requiring continuity of PVsand balance of SVs at the nodes.

5. The assembled system of equations are then solved forthe unknown PVs at the nodes by using the knownboundary conditions.

6. Post-computation may be used to compute SVs and PVs atpoints other than nodes. The SVs are discontinuousbetween elements.

Basic Concepts: 56

the finite element method - tamu...

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