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FIGURE 6.29
Shape functions for three-node element.
where, from Eq. (6.71b) with
m
=
2,
N
2
(
L
1
)
=
L
1
(
2
L
1
−
1
)
N
2
(
L
2
)
=
L
2
(
2
L
2
−
1
)
N
1
(
L
1
)
=
2
L
1
N
1
(
L
2
)
=
2
L
2
N
0
(
L
1
)
=
N
0
(
L
2
)
=
1
and from Eq. (6.71a),
=
(
)
(
)
=
(
−
)
N
20
N
2
L
1
N
0
L
2
L
1
2
L
1
1
N
11
=
N
1
(
L
1
)
N
1
(
L
2
)
=
4
L
1
L
2
N
02
=
N
0
(
L
1
)
N
2
(
L
2
)
=
L
2
(
2
L
2
−
1
)
which are plotted in Figs. 6.29d, e, and f. Then the trial displacement function becomes
u
(
L
1
,L
2
)
=
L
1
(
2
L
1
−
1
)v
20
+
4
L
1
L
2
v
11
+
L
2
(
2
L
2
−
1
)v
02
Two-Dimensional Case in Natural Coordinates
The interpolation functions in natural coordinates for a triangle can be established in a
fashion similar to the procedure for one dimension. The natural coordinates for a triangle
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