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4
]
T
,
4, are the
values of the displacement
u
at the nodal points, the shape function of the displacement
u
inside the element will be
If
v
=
[
v
v
v
v
is the nodal displacement vector in which
v
i
,i
=
1
,
2
,
...
1
2
3
v
1
v
2
v
u
=
[
L
1
L
2
L
3
L
4
]
(6.83)
3
v
4
=
N
v
For the case that each side of the tetrahedron contains more than three nodes, the shape
functions for the element can be obtained in a fashion similar to that explained previously
for the two-dimensional case. Divide the tetrahedron into
m
blocks of equal thicknesses
with planes parallel to
S
1
as shown in Fig. 6.33b. Do the same for
S
2
,S
3
,
and
S
4
. Label the
planes parallel to
S
1
as
p
, and those parallel to
S
2
,
S
3
, and
S
4
as
q, r,
and
s
. A point at
the intersection of planes
p, q, r,
and
s
can be designated by four digits
p, q, r,
and
s
. Then
the shape function of Eq. (6.83) can be written as
(
m
+
1
)(
m
+
2
)(
m
+
3
)/
6
u
=
Nv
=
N
pqrs
v
pqrs
(6.84)
with
)
The three-dimensional natural coordinates also have the property that
N
pqrs
=
N
p
(
L
1
)
N
q
(
L
2
)
N
r
(
L
3
)
N
s
(
L
4
a
!
b
!
c
!
d
!
L
1
L
2
L
3
L
4
dV
=
!
6
V
(6.85)
(
+
+
+
+
)
a
b
c
d
3
V
This property is very useful in the construction of stiffness matrices and the loading vectors.
6.5.7 Hermitian Interpolation
Elements used to model a structure undergoing bending very often involve derivatives of
displacements as DOF. Thus, for a beam element the deflection and slope (first derivative
of the deflection) are chosen as nodal DOF. Hermitian interpolation is useful in intro-
ducing derivative variables into the trial functions. Hermitian interpolation differs from
Lagrangian interpolation (where the functions should have
C
0
continuity at the interele-
ment boundaries) in that nodal values of the interpolation functions include derivatives of
the displacements, as well as the displacements themselves. The Hermitian trial functions
are
C
r
continuous at the interelement boundaries.
The shape functions for Hermitian elements are formulated in a fashion similar to that
used in establishing Lagrange elements, but the Hermite polynomial replaces the Lagrange
polynomial.
(
r
≥
1
)
One-Dimensional Case in Cartesian Coordinates
To represent a function
u
that will satisfy conditions on
u
and up to its
m
−
1 derivative on
the end points of an element extending from point
j
=
1 to point
j
=
2, consider the form
N
m
v
(
m
−
1
)
1
N
2
m
v
(
m
−
1
)
2
N
2
v
1
+
N
3
v
1
+···+
N
m
+
2
v
2
+···+
u
=
N
1
v
1
+
+
N
m
+
1
v
2
+
2
m
N
k
(ξ ) v
(
i
−
1
)
=
(6.86)
j
j
=
1
i
=
1
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