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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
hf
N
jj fqV
1),( Td