FEM: Bars and Trusses
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Introduction to the Finite
Element Method
Bars and Trusses
FEM: Bars and Trusses
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Bar Example (Ex. 4.5.2, p. 187)
• Consider the bar shown in the above figure.
• It is composed of two different parts. One steel tapered part, and uniform Aluminum part.
• Calculate the displacement field using finite element method.
FEM: Bars and Trusses
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Bar Example
• The bar may be represented by two
elements.
• The stiffness matrices of the two elements
may be obtained using the following
integration:
2
1
2
122
2221
2
1
11
11x
x
ee
ee
x
x
e dx
hh
hhxEAdx
dx
d
dx
d
dx
ddx
d
xEAK
FEM: Bars and Trusses
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Bar Example
• For the Aluminum bar: E=107 psi, and A=1
in2. we get:
• For the Steel bar: E=38107 psi, and
A=(1.5-0.5x/96) in2. we get:
11
11
120
10
11
11
120
10 7
2
7 2
1
x
x
Al dxK
11
11
96
10.75.4
11
11
96
5.05.1
96
10.3 7
2
7 2
1
x
x
Fe dxx
K
FEM: Bars and Trusses
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Bar Example
• Assembling the Stiffness matrix and
utilizing the external forces, we get:
• The boundary conditions may be applied
and the system of equations solved.
0
0
10
10.2
0
33.833.80
33.88.575.49
05.495.49
105
5
3
2
1
4
R
u
u
u
FEM: Bars and Trusses
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Bar Example
• Solving, we get:
• For the secondary
variables:
inu
u
181.0
061.0
3
2
lbR 30000
FEM: Bars and Trusses
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Trusses
• A truss is a set of bars that are connected
at frictionless joints.
• The Truss bars are generally oriented in
the plain.
FEM: Bars and Trusses
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Trusses
• Now, the problem lies in the
transformation of the local displacements
of the bar, which are always in the
direction of the bar, to the global degrees
of freedom that are generally oriented in
the plain.
FEM: Bars and Trusses
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Equation of Motion
0
0
0000
0101
0000
0101
2
1
2
2
1
1
F
F
v
u
v
u
h
EA
FEM: Bars and Trusses
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Transformation Matrix
DOF
dTransformeDOFLocal
v
u
v
u
CosSin
SinCos
CosSin
SinCos
v
u
v
u
2
2
1
1
2
2
1
1
00
00
00
00
DOF
dTransformeDOFLocal T
FEM: Bars and Trusses
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The Equation of Motion Becomes
• Substituting into the
FEM:
• Transforming the
forces:
• Finally:
FTK
FTTKTTT
FK
FEM: Bars and Trusses
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Recall
TKTKT
CosSin
SinCos
CosSin
SinCos
T
00
00
00
00
Where:
0000
0101
0000
0101
h
EAK
FEM: Bars and Trusses
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Element Stiffness Matrix in Global
Coordinates
CosSin
SinCos
CosSin
SinCos
CosSin
SinCos
CosSin
SinCos
h
EAK
00
00
00
00
0000
0101
0000
0101
00
00
00
00
FEM: Bars and Trusses
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Element Stiffness Matrix in Global
Coordinates
22
22
22
22
22
12
2
1
22
12
2
1
22
12
2
1
22
12
2
1
SinSinSinSin
SinCosSinCos
SinSinSinSin
SinCosSinCos
h
EAK
FEM: Bars and Trusses
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Example: 4.6.1 pp. 196-201
• Use the finite element analysis to find the
displacements of node C.
FEM: Bars and Trusses
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Element Equations
0000
0101
0000
0101
1
L
EAK
1010
0000
1010
0000
2
L
EAK
3536.03536.03536.03536.0
3536.03536.03536.03536.0
3536.03536.03536.03536.0
3536.03536.03536.03536.0
3
L
EAK
FEM: Bars and Trusses
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Assembly Procedure
3536.13536.0103536.03536.0
3536.03536.0003536.03536.0
101000
000101
3536.03536.0003536.03536.0
3536.03536.0013536.03536.1
L
EAK
FEM: Bars and Trusses
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Global Force Vector
P
P
F
F
F
F
F
F
F
F
F
F
Fy
x
y
x
y
x
y
x
y
x
2
2
2
1
1
3
3
2
2
1
1
Remember!
NO distributed load
is applied to a truss
FEM: Bars and Trusses
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Boundary Conditions
02211 VUVU
Remove the corresponding rows and columns
P
P
V
U
L
EA
23536.13536.0
3536.03536.0
3
3
Continue! (as before)
FEM: Bars and Trusses
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Results
EA
PLV
EA
PLU
3 ,828.5 33
PFF
PFPF
yx
yx
3 ,0
, ,
22
11
FEM: Bars and Trusses
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Postcomputation
e
e
e
e
e
A
P
A
P 21
e
e
eee
e
u
u
L
EA
P
P
2
1
2
1
11
11
2
2
1
1
2
2
1
1
00
00
00
00
v
u
v
u
CosSin
SinCos
CosSin
SinCos
v
u
v
u
FEM: Bars and Trusses
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Postcomputation
A
P
A
P2 ,
3 ,0 )3()2()1(
FEM: Bars and Trusses
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Summary
• In this lecture we learned how to apply the
finite element modeling technique to bar
problems with general orientation in a
plain.
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