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where use has been made of the integral of Eq. (6.78) with, for example,
a
=
1
,b
=
0, and
c
0.
On the element boundaries assume that the tractions
p
have the constant value
p
.On
side 3 of Fig. 6.30,
L
3
=
=
0. Then
p
Side 3
L
1
L
2
0
1
1
0
p
12
ds
p
1
=
=
(6.103)
2
where, as indicated in Eq. (6.70),
12
is the length between nodes 1 and 2. Similar results are
obtained for sides 1 and 2, where
L
1
=
0 and
L
2
=
0, respectively.
Tetrahedral Element
The development of a stiffness matrix for a tetrahedral element can follow the same proce-
dure used for the triangular element. Rather than the displacements
u
x
,u
y
for the triangle,
use
u
x
,u
y
,
and
u
z
for the displacements in the global
x, y,
and
z
directions of the tetrahedron.
Express the displacement vector as
u
x
u
y
u
z
N0
N
0N
v
x
v
y
v
z
N0
N
0N
=
=
v
i
=
u
(6.104)
where
u
x
1
u
x
2
u
x
3
u
x
4
u
y
1
u
y
2
u
y
3
u
y
4
u
z
1
u
z
2
u
z
3
u
z
4
v
x
=
v
y
=
v
z
=
and
N
=
[
L
1
L
2
L
3
L
4
]
=
in which
u
xi
,u
yi
,
and
u
zi
,i
1
,
2
,
3
,
4
,
are the values of
u
x
,u
y
,
and
u
z
at the nodes, and
L
i
,i
=
1
,
2
,
3
,
4 are the shape functions defined in Eq. (6.82).
Then
α
β
γ
δ
1
1
1
1
1
6
V
[1
α
β
γ
δ
2
2
2
2
N
=
[
L
1
L
2
L
3
L
4
]
=
xy
]
(6.105)
α
β
γ
δ
3
3
3
3
α
β
γ
δ
4
4
4
4
A typical expression in the stiffness matrix would be
L
1
L
2
L
3
L
4
N
T
∂
E
ij
∂
x
N
dV
=
∂
E
ij
∂
x
[
L
1
L
2
L
3
L
4
]
dV
x
x
V
V
1
β
β
β
β
β
β
β
1
2
1
3
1
4
2
E
ij
36
V
β
β
β
β
β
β
β
2
1
2
3
2
4
=
(6.106)
2
3
β
β
β
β
β
β
β
3
1
3
2
3
4
2
4
β
β
β
β
β
β
β
4
1
4
2
4
3
For the loading vector, continue to follow the scheme for the triangular element.
Triangular and tetrahedral elements are very useful in structural analysis. They can be
used independently or can be combined to form quadrilateral and hexahedral elements.
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