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ModelEqn 1D - 1JN Reddy

FINITE ELEMENT ANALYSIS OF A 1D MODEL PROBLEM

WITH A SINGLE VARIABLE

Finite element model development of a linear 1D model differential equation involving a singledependent unknown (governing equations, FE model development weak form).


GOVERNING EQUATIONGOVERNING EQUATION

pointboundary aat)(

)(in)()()(

Puubdxdua

L,xfuxcdxduxa

dxd

=−+

=Ω=+⎟⎠⎞

⎜⎝⎛−

0

0

E, A

b=k L

u(L)

u0

P

x, u

a = EAk, A

L

u(L)u0

P

x, u

a = kA

Elastic deformation of a bar Heat transfer in a bar

uninsulated bar

0=−+

− fcu

dxdua

dxd


FINITE ELEMENT APPROXIMATIONFINITE ELEMENT APPROXIMATION(NEED FOR SEEKING SOLUTION ON SUB(NEED FOR SEEKING SOLUTION ON SUB--INTERVALS)INTERVALS)

Approximation of the actual solutionover the entire domain requires higher-order approximation1.

Actual solution may be defined by Sub-intervals because of discontiuityof the data a(x).2.

Approximation over sub-intervals subintervals allows lower-order Approximation of the actual solution


FINITE ELEMENT DISCRETIZATIONFINITE ELEMENT DISCRETIZATION

Approximation over sub-intervals subintervals allows lower-order Approximation of the actual solution

x

x xaQ bQ

axx = bxx =

A typical element (geometry and ‘forces’)

lengthelement heatsor forcesend

b =−= a

ba

xxhQ,Q


bba

x

x

xb

xa

x

x

x

x

x

x

x

x

QxwQxwdxwfcwudxdu

dxdwa

dxduaxw

dxduaxwdxwfcwu

dxdu

dxdwa

dxduawdxwfcwu

dxdu

dxdwa

dxxfuxcdxduxa

dxdw

b

a

ba

b

a

b

a

b

a

b

a

⋅−−⎥⎦⎤

⎢⎣⎡ −+=

⎟⎠⎞

⎜⎝⎛⋅−⎟

⎠⎞

⎜⎝⎛−⋅−⎥⎦

⎤⎢⎣⎡ −+=

⎥⎦⎤

⎢⎣⎡ ⋅−⎥⎦

⎤⎢⎣⎡ −+=

⎥⎦

⎤⎢⎣

⎡−+⎟

⎠⎞

⎜⎝⎛−=

∫

∫

∫

∫

)()(

)()(

)()()(

a

0

WEAK FORM OVER AN ELEMENTWEAK FORM OVER AN ELEMENT

Heat/Force input

bx

Qdxdua

b

≡⎟⎠⎞

⎜⎝⎛

ax

Qdxdua

a

≡⎟⎠⎞

⎜⎝⎛−

he

21

Heat input

Force out


LINEAR AND BILINEAR FORMSLINEAR AND BILINEAR FORMSAND THE VARIATIONAL PROBLEMAND THE VARIATIONAL PROBLEM

wwlu,wB allfor holds)()( =

⎥⎦⎤

⎢⎣⎡ ⋅++=⎥⎦

⎤⎢⎣⎡ += ∫∫ bba

x

x

x

xQxwQxwdxwfwl,dxcwu

dxdu

dxdwau,wB

a

b

b

a

)()()()( a

Bilinear Form and Linear Form

Weak Form

)()(

)()(

)()(

a

a

wlu,wB

QxwQxwdxwfdxcwudxdu

dxdwa

QxwQxwdxwfcwudxdu

dxdwa

bba

x

x

x

x

bba

x

x

a

b

b

a

b

a

−=

⎥⎦⎤

⎢⎣⎡ ⋅++−⎥⎦

⎤⎢⎣⎡ +=

⋅−−⎥⎦⎤

⎢⎣⎡ −+=

∫∫

∫0

Variational Problem: Find u such that


EQUIVALENCE BETWEEN MINIMUM OF A EQUIVALENCE BETWEEN MINIMUM OF A QUADRATIC FUNCTIONAL AND WEAK FORMQUADRATIC FUNCTIONAL AND WEAK FORM

δuulu,δuBδI allfor )()( 00 =δ−⇒=

which is the same as the weak form or the variationalproblem with δu = w

Quadratic Functional:

Variational Problem: Find u such that I(u) is a minimum:

⎥⎦⎤

⎢⎣⎡ ⋅++−

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛=

≡

∫∫ bba

x

x

x

xQxuQxudxufdxcu

dxdua

ulu,uBuI

a

b

b

a

)()(

)(-)()(

a2

2

21

21

Strain energyWork done by applied forces


FINITE ELEMENT MODELFINITE ELEMENT MODEL

)()()(1

xψuxUxu ej

n

j

ej

e ∑=

=≈

Finite element model

bbiaai

x

xi

ei

x

xji

jieij

eee

QxψQxψdxfψF

,dxψcψdx

dψdx

dψaK

FuK

b

a

b

a

)( )(

][

++=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

=

∫

∫

Finite element approximation (to be derived later)


APPROXIMATION FUNCTIONS FORAPPROXIMATION FUNCTIONS FORLINEAR ELEMENTLINEAR ELEMENT

he

21

xccxU 21 +=)(

1uuxU aa =≡)(2uuxU bb =≡)(

xccxUxu 21 +=≈ )()(

2u

bb

aa

xccuxUxccuxU

212

211

+==+==

)()(

abab

ab

xxuuc,

xxxuxu

c−−

=−−

=→ 122

211

he

21

ab

b

xxxx

xψ−−

≡)(1ab

a

xxxx

xψ−−

≡)(2

xaxx = bxx =

2211

1221

21

uxψuxψ

xxxuu

xxxuxu

xccxUxu

abab

ab

)()(

)()(

+=−−

+−−

=

+=≈

1u


ALTERNATE DERIVATION OF ALTERNATE DERIVATION OF APPROXIMATION FUNCTIONSAPPROXIMATION FUNCTIONS

(Linear Element)(Linear Element)

2u

he

21

ab

b

xxxx

xψ−−

≡)(1ab

a

xxxx

xψ−−

≡)(2

xaxx = bxx =

1u

2211 uxψuxψxU )()()( +=

⎩⎨⎧

=≠

=≡ji,ji,

δxψ ijji ifif

)(10

ba xx,xx ≡≡ 21

1-

1-

)(1)(and

)()()(1)(and

)()(

abb

a

aba

b

xxαxψ

xxαxψxxαxψ

xxαxψ

−=→=

−=−=→=

−=

22

22

11

11

(interpolation functions)

Alternate derivation usingthe interpolation property


ALTERNATE DERIVATION OF ALTERNATE DERIVATION OF APPROXIMATION FUNCTIONSAPPROXIMATION FUNCTIONS

(Quadratic Element)(Quadratic Element)

he

21

xaxx = ea hxx +=

1u

332211 uxψuxψuxψxU )()()()( ++=

233

33

222

22

211

11

250

4

250

hαhxψ

xh.xxxαxψh

αhxψ

xhxxxαxψh

αxψ

xh.xxhxαxψ

a

aa

a

aa

a

aa

−=→=+

−+−=

=→=+

−+−=

=→=

−+−+=

1)(and

)()()(

1)0.5(and

)()()(

1)(and

))(()(

Alternate derivation usingthe interpolation property

ea h.xx 50+=

3

)(xψ1 )(xψ3)(xψ2

3u

1.0 1.01.0(b)

1 2 3

)(1 xeψ )(2 x

eψ )(3 xeψ

1 2 3xhe

(a)

ex3ex2ex1

x

)(xeiψ

x


NUMERICAL EVALUATION OF COEFFICIENTSNUMERICAL EVALUATION OF COEFFICIENTSfor elementfor element--wise constant datawise constant data

i

x

xi

ei

x

xji

jieij

eee

QdxfψF,dxψcψdx

dψdx

dψaK

ff,cc,aa:data

b

a

b

a

constantFor

+=⎟⎟⎠

⎞⎜⎜⎝

⎛+=

===

∫∫

⎭⎬⎫

⎩⎨⎧

+⎭⎬⎫

⎩⎨⎧

=⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−

−=

2

1

11

22112

61111

QQhf

F,hc

ha

K eeeee

e

ee ][

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧+

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−=

3

2

1

141

24212162124

307818168187

3QQQ

hfF,

hcha

K eeeee

e

ee ][

Linear element:

Quadratic element:

eh

eh

1 2 3

1 2


NUMERICAL EXAMPLE NUMERICAL EXAMPLE -- 11

elementeach in0=⎟⎠⎞

⎜⎝⎛−

dxduEA

dxd

lb3000

,psi1010psi,1014

psi,1030 in., 0.1

in., 0.875 in., 0.5

ft.,6ft,4 ft.,8

66

6

321

=

×=×=

×==

==

===

P

EE

Ed

dd

hhh

ab

sa

bs

Rigid member (constrained to move horizontally)

h22P

Steel (Es , As)

Aluminum (Ea , Aa)

Brass (Eb , Ab)

h3

h1

Problem: Wish to determine the deformation and stresses in the three members of the structure.

Solution: (1) Note that the governing equation is


NUMERICAL EXAMPLE NUMERICAL EXAMPLE –– 1 (continued)1 (continued)

(2) We use linear elements to represent the members of thestructure and label the elements, element nodes and globalnodes.

(3) The element equations for a typical element are

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧⎥⎦

⎤⎢⎣

⎡−

−e

e

e

e

e

ee

QQ

uu

hAE

2

1

2

1

1111

)( 00 === eeeee f,c,AEa

Rigid member (constrained to move horizontally)

h22P

Steel (Es , As)

Aluminum (Ea , Aa)

Brass (Eb , Ab)

1

2

3

1

3

4Global node numberS

1 2

21

1 2h33 3

h1

Element node numbers

2



(4) The displacement boundary conditions expressed in terms ofthe global displacements are

000 421 === U,U,U

(5) The force equilibrium conditions are obtained by looking at the equilibrium of forces on the rigid bar. We have

3)(1

1Q )(12Q1

2)(21Q )(2

2Q)(3

1Q )(32Q

)(12Q

)(22Q

)(31QP2

PQQQQQPQ 202 31

22

12

31

22

12 =++→=−−+− )()()()()()(



(6) The equilibrium conditions suggest that we must add thesecond equation of element 1, the second equation of element2, and the first equation of element 3 so that we can replacethe sum of the Q’s with 2P. The element equations in termsof the global displacements are

Using the fact that U1=U2=U4=0, we obtain

PQQQUh

AEh

AEhAE 23

122

123

3

33

2

22

1

11 =++=⎟⎟⎠

⎞⎜⎜⎝

⎛++

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

==

⎥⎦

⎤⎢⎣

⎡−

−2

2

32

22

2

22

2

1

2

1

1111

QQ

UuUu

hAE

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

==

⎥⎦

⎤⎢⎣

⎡−

−1

1

31

11

1

11

2

1

2

1

1111

QQ

UuUu

hAE

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

==

⎥⎦

⎤⎢⎣

⎡−

−3

3

43

33

3

33

2

1

2

1

1111

QQ

UuUu

hAE

0 0

0



Thus we can compute U3 , which represents the elongation inelements 1 and 2 and compression in element 3.

This completes Example 1 (one can substitute the given data to obtain numerical values of the displacements, forces, and stresses in the members).

)(eQ2

33

33323

2

22223

1

1112 U

hAE

Q,Uh

AEQ,UhAEQ −=== )()()(

(7) We can then determine the member forces using

33

3

3

323

32

2

2

222

31

1

1

121 U

hE

AQσ,U

hE

AQσ,U

hE

AQσ −=====

)()(

)()(

)()(

The member stresses are then can be computed from


NUMERICAL EXAMPLE NUMERICAL EXAMPLE –– 22

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

0100

1022

2

==

<<−=−−

)()(

in

u,u

xxudx

ud

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

dxψxf,xf,c,a i

x

x

ei

b

a∫−=−=−== 2211

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

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+−−−−−

−=⎭⎬⎫

⎩⎨⎧⎟⎟⎠

⎞⎜⎜⎝

⎛⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡−

−e

e

ababx

ababx

ee

ee

e QQ

xxxxxxxx

huuh

h a

b

2

14433

3

44333

2

1 12112

611111

)()()()(

41

41


NUMERICAL EXAMPLE NUMERICAL EXAMPLE –– 2 (continued)2 (continued)(2) We consider a mesh of 4 linear elements (h = 0.25).

The element equations are

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+⎭⎬⎫

⎩⎨⎧

−=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧⎥⎦

⎤⎢⎣

⎡−

−1

1

1

1

241

2

1

2

1

003910001300

94979794

QQ

.

.uu

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+⎭⎬⎫

⎩⎨⎧

−=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧⎥⎦

⎤⎢⎣

⎡−

−2

2

2

2

241

2

1

2

1

022320014320

94979794

QQ

.

.uu

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+⎭⎬⎫

⎩⎨⎧

−=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧⎥⎦

⎤⎢⎣

⎡−

−3

3

3

3

241

2

1

2

1

055990042970

94979794

QQ

.

.uu

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+⎭⎬⎫

⎩⎨⎧

−=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧⎥⎦

⎤⎢⎣

⎡−

−4

4

4

4

241

2

1

2

1

105470087240

94979794

QQ

.

.uu

1 2 3 4 5

1 2 3 4

1 21

1 22

1 23

1 24

)(11Q )(1

2Q

)(21Q )(2

2Q

)(31Q )(3

2Q

)(41Q )(4

2Q



(3) The boundary conditions are

1 2 3 4 5

1 2 3 4

1 2

1)(1

1Q )(12Q )(2

1Q )(22Q

2 1

2

0000 41

42

31

32

31

22

21

12 =+=+=+=+ )()()()()()()()( QQ,QQ,QQ,QQ

00 51 == U,U

The equilibrium conditions are


NUMERICAL EXAMPLE NUMERICAL EXAMPLE –– 2 (continued)2 (continued)(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

0

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎦

⎤⎢⎣

⎡−−

=⎭⎬⎫

⎩⎨⎧

0.26386 0.09520

0.105470.001302

3.9167 4.0417004.0417 3.9167

4

3

2

42

11

UUU

QQ

00



0.0 0.2 0.4 0.6 0.8 1.0

Coordinate, x

-0.05

-0.04

-0.03

-0.02

-0.01

0.00

Sol

uti

on, u(x)

Analytical

4L2Q4Q

FEM

Plot of the solution

0.0 0.2 0.4 0.6 0.8 1.0

Coordinate, x

-0.15

-0.10

-0.05

0.00

0.05

0.10

0.15

0.20

0.25

0.30

Sol

uti

on, du/dx

Analytical

4L2Q4Q

FEM

& "

"

"

"

"

*

**

*

*

&

& &

& &

& &

Plot of the derivative of the solution

finite element analysis of - university of … · jn reddy modeleqn 1d - 1 finite element analysis...

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