More Details on

Variational Formulation

Axially Loaded Member

Axially Loaded Member

Let us remember the initial form of the problem

Integrating by parts

The Strong Form

dxqzupuvquLv 10

])''([)][,(

.0)1()0(

10),()())((][

uu

xxquxzdxdu

xpdxd

uL

0')''(])''([ 101

0

1

0

vpudxvqvzupuvdxqzupuv

The Weak Form

The problem can be rewritten as

where

The integration by parts eliminated the second derivatives from the problem making it possible less continouity than the previous form. This is why this form is called weak form of the problem. A(v,u) is called Strain Energy.

0),(),( qvuvA

dxvzupuvuvA )''(),(1

0

Let us take the initial value problem with constant coefficients

As a first step let us divide the domain in n subintervals with the following mesh

Each subinterval is called finite element.

Finite Element Discretization

.0)1()0(0,

10),(''

uu

zpxxqzupu

njxx jj :1),,( 1

1...0 10 nxxx

The problem at this point can be easily solved using the previously derived Galerkins Method

A little more work is needed to convert this problem into matrix notation

The Trial Basis

N

kjkjk njqNNNAu

1,...,2,1,),(),(

Restricting U over the typical finite element we can write

Which in turn can be written as

in the same way

Matrix Form of the Problem

],[)]()([)()(][)( 11111 jj

j

jjj

j

jjj xxxu

uxNxN

xNxN

uuxU

],[)()()( 111 jjjjjj xxxxNuxNuxU

],[)]()([)()(][)( 11111 jj

j

jjj

j

jjj xxxd

dxNxN

xNxN

ddxV

Taking the derivative

Derivative of V is analogus

Matrix Form of the Problem

1

11

1 ],[]/1/1[/1/1][)('

jjj

jjj

jjj

j

jjj

xxh

xxxu

uhh

hh

uuxU

],[]/1/1[/1/1][)(' 111 jj

j

jjj

j

jjj xxxd

dhh

hh

ddxV

The variational formula can be elementwise defined as follows:

Matrix Form of the Problem

j

j

j

j

j

j

x

x

x

x

Mj

x

x

Sj

Mj

Sjj

N

jjj

dxVqqV

dxzVUUVA

dxUpVUVA

UVAUVAUVA

qVUVA

1

1

1

),(

),(

''),(),(),(),(

0]),(),([1

Substituting U,V,U and V into these formulae we obtain The Element Stiffnes Matrix

1111

][1111][),(

]/1/1[/1/1][),(

11

121

11

1

1

jj

j

jjjj

j

jx

x jjj

Sj

j

jx

xjj

j

jjj

Sj

hpK

u

uKdd

u

udx

hpddUVA

u

udxhh

hh

pddUVA

j

j

j

j

The Element Mass Matrix

2112

6

][),(

][][),(

11

11

11

1

jj

j

jjjj

Sj

j

jx

xjj

j

jjj

Mj

zhM

u

uMddUVA

u

udxNN

NN

zddUVA jj

The same way

The external work integral cannot be evaluated for every function q(x)

We can consider a linear interpolant of q(x) for simplicity.

Substituting and evaluating the integral

External Work Integral

jjxx dxVqqV 1),(],[),()()( 111 jjjjjj xxxxNqxNqxq

Element load vector

jj

jjjj

jjjj

jjj

x

x j

jjjj

qqqqh

l

ldddxq

qNN

NN

ddqV jj

22

6

],[],[],[),(

1

1

11

11

11

Now the task is to assemble the elements into the whole system in fact we have to sum each integral over all the elements

For doing so we can extend the dimension of each element matrix to n and then put the 2x2 matrix at the appropriate position inside it

With all matrices and vectors having the same dimension the summation looks like

Assembling

Assembling

N

j

SjA

1uKdT

11121

.........

121121

11

hpK

1

2

1

1

2

1

......

nn d

dd

d

u

u

u

u

Doing the same for the Mass Matrix Assembling

u1

N

j

MjA Md

T

21141

.........

141141

12

6zhM

Doing the same for the Load Vector

Assembling

forces nodal applied

1

2

1

0

1

12

321

210

10

...

24...

44

2

2

n

n

nn

nnn

FF

FFF

qqqqq

qqqqqq

qq

hf

N

jj fqV

1),( Td