FEM: Element Equations
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Transcript of FEM: Element Equations
2nd order DE’s in 1-D
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Introduction to the Finite
Element Method
2nd order DE’s in 1-D
2nd order DE’s in 1-D
Mohammad Tawfik #WikiCourses
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Objectives
• Understand the basic steps of the finite
element analysis
• Apply the finite element method to second
order differential equations in 1-D
2nd order DE’s in 1-D
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The Mathematical Model
• Solve:
• Subject to:
Lx
fcudx
dua
dx
d
0
0
00 ,0 Qdx
duauu
Lx
2nd order DE’s in 1-D
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Step #1: Discretization
• At this step, we divide
the domain into
elements.
• The elements are
connected at nodes.
• All properties of the
domain are defined at
those nodes.
2nd order DE’s in 1-D
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Step #2: Element Equations
• Let’s concentrate our attention to a single element.
• The same DE applies on the element level, hence, we may follow the procedure for weighted residual methods on the element level!
21
0
xxx
fcudx
dua
dx
d
21
2211
21
,
,,
Qdx
duaQ
dx
dua
uxuuxu
xxxx
2nd order DE’s in 1-D
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Polynomial Approximation
• Now, we may propose an approximate
solution for the primary variable, u(x),
within that element.
• The simplest proposition would be a
polynomial!
2nd order DE’s in 1-D
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Polynomial Approximation
• Interpolating the values
of displacement
knowing the nodal
displacements, we may
write:
01 bxbxu
01111 bxbuxu 2
12
11
12
2 uxx
xxu
xx
xxxu
02122 bxbuxu
2nd order DE’s in 1-D
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Polynomial Approximation
euxu
uuu
uxx
xxu
xx
xxxu
2
1
212211
2
12
11
12
2
2nd order DE’s in 1-D
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Step #2: Element Equations
(cont’d)
• Assuming constant
domain properties:
• Applying the
Galerkin method:
21
2
2
0
xxx
fcudx
uda
02
2
Domain
jiijii
j dxfxuxxcudx
xdxa
2nd order DE’s in 1-D
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Step #2: Element Equations
(cont’d)
• Note that:
• And:
ee hdx
xd
hdx
xd 1,
1 21
Domain
ij
x
x
ij
Domain
ij
dxdx
xd
dx
xda
dx
xdxa
dxdx
xdxa
2
1
2
2
2nd order DE’s in 1-D
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Step #2: Element Equations
(cont’d)
• For i=j=1: (and ignoring boundary terms)
• Which gives:
012
1
21
2
2
2
x
x eee
dxh
xxfu
h
xxc
ha
023
1
ee
e
fhu
ch
h
a
2nd order DE’s in 1-D
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Step #2: Element Equations
(cont’d)
• Repeating for all terms:
• The above equation is called the element
equation.
1
1
221
12
611
11
2
1 ee
e
fh
u
uch
h
a
2nd order DE’s in 1-D
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What happens for adjacent
elements?
2nd order DE’s in 1-D
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Objectives
• Learn how the finite element model for the
whole domain is assembled
• Learn how to apply boundary conditions
• Solving the system of linear equations
2nd order DE’s in 1-D
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Recall
• In the previous lecture, we obtained the
element equation that relates the element
degrees of freedom to the externally
applied fields
• Which maybe written:
1
1
221
12
611
11
2
1 ee
e
fh
u
uch
h
a
2
1
2
1
43
21
f
f
u
u
kk
kk
2nd order DE’s in 1-D
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Two–Element example
1
2
1
1
1
2
1
1
1
4
1
3
1
2
1
1
f
f
u
u
kk
kk
2
2
2
1
2
2
2
1
2
4
2
3
2
2
2
1
f
f
u
u
kk
kk
3
2
1
3
2
1
3
2
1
2
4
2
3
2
2
2
1
1
4
1
3
1
2
1
1
0
0
Q
Q
Q
f
f
f
u
u
u
kk
kkkk
kk
2nd order DE’s in 1-D
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Illustration: Bar application
1. Discretization: Divide the bar into N number of elements. The length of each element will be (L/N)
2. Derive the element equation from the differential equation for constant properties an externally applied force:
02
2
xF
x
uEA
02
1
2
x
x
ijij
e
dxfudx
d
dx
d
h
EA
2nd order DE’s in 1-D
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Performing Integration:
1
1
211
11
2
1 e
e
e
e
fh
u
u
h
EA
Note that if the integration is evaluated from 0 to he,
where he is the element length, the same results
will be obtained.
02
1
2
x
x
ijij
e
dxfudx
d
dx
d
h
EA
2nd order DE’s in 1-D
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Two–Element bar example
1
2
1
1
1
2
1
1
11
11
f
f
u
u
h
EA
e
2
2
2
1
2
2
2
1
11
11
f
f
u
u
h
EA
e
0
0
1
2
1
2110
121
011
3
2
1 Rfh
u
u
u
h
EA e
e
2nd order DE’s in 1-D
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Applying Boundary Conditions
2nd order DE’s in 1-D
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Applying BC’s
• For the bar with fixed left side and free
right side, we may force the value of the
left-displacement to be equal to zero:
0
0
1
2
1
2
0
110
121
011
3
2
Rfh
u
uh
EA e
e
2nd order DE’s in 1-D
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Solving
• Removing the first row and column of the
system of equations:
• Solving:
1
2
211
12
3
2 e
e
fh
u
u
h
EA
4
3
2
2
3
2
EA
fh
u
ue
2nd order DE’s in 1-D
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Secondary Variables
• Using the values of the displacements
obtained, we may get the value of the
reaction force:
0
0
1
2
1
2
2
42
30
110
121
011 Rfh
fh
fh e
e
e
2nd order DE’s in 1-D
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Secondary Variables
• Using the first equation, we get:
• Which is the exact value of the reaction
force.
Rfhfh ee 22
3
efhR 2
2nd order DE’s in 1-D
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Summary
• In this lecture, we learned how to
assemble the global matrices of the finite
element model; how to apply the boundary
conditions, and solve the system of
equations obtained.
• And finally, how to obtain the secondary
variables.