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From Eqs. (6.66), it follows that the natural coordinates are defined such that
L 1 =
1 f i falls on point 1 ,
L 1 =
0 f i falls on point 2
L 2 =
1 f i falls on point 2 ,
L 2 =
0 f i falls on point 1
It can be seen that L 1 and L 2 satisfy the properties of Eq. (6.63) and, hence, can be considered
to be forms of Lagrangian interpolation functions. The
ξ
coordinate of any point along the
line can be expressed as
ξ =
L 1
ξ
+
L 2
ξ
(6.68)
1
2
a result which is verified by use of Eq. (6.66). Equations (6.67) and (6.68) placed together
appear as
L 1
L 2
1
ξ
11
ξ
=
ξ
1
2
These relationships, i.e., Eqs. (6.67) and (6.68), can be considered to be the definition of the
natural coordinates, since they lead to
L 1
L 2
ξ 2
1
ξ
1
2 ξ 1 )
1
=
(6.69)
ξ 1
1
which is the same as Eq. (6.66)
A useful property of these natural coordinates is
ξ 2
L 1
L 2
d
1
1
ξ = 12
2
(6.70)
ξ 1
where
2 .
The natural coordinates are useful in representing shape functions for a line divided into
segments. It is convenient to utilize a different node numbering scheme. Designate the left
end point as 0 and the right end point as m , i.e., these points are numbered 0 , 1 , 2 ,
12 is the length from
ξ
1 to
ξ
,m .
Using natural coordinates, each point is identified by its location relative to the two end
points of the line. Define numbers p and q , where p and q are the number of points to the
right and left, respectively, of the point under consideration. Observe Figs. 6.29a, b, and c
wherein the three nodes in Fig. 6.29c are denoted as 20, 11, and 02. Thus, the coordinate
...
ξ
2
ξ
ξ
ξ
of Eq. (6.66) is now
02 and
1 is
20 . The nodal degrees of freedom will now be labeled with
p, q, i.e.,
v
pq or in the case of Fig. 6.29c,
v
20 ,
v
11 and
v
02 . The corresponding shape functions
are N pq , i.e., N 20 ,N 11 , and N 02 .
Shape functions in terms of natural coordinates are given by
N pq
(
L 1 ,L 2
) =
N p
(
L 1
)
N q
(
L 2
)
(6.71a)
where, from the Lagrangian interpolation formula, N p (
L 1 )
and N q (
L 2 )
are defined by
i j = 1
mL k
j
+
1
,
for i
1
N i (
L k ) =
j
k
=
1 , 2 ,
···
(6.71b)
1
for i
=
0
where m is the number of segments.
EXAMPLE 6.8 Three-Node Element
For the element of Fig. 6.29c, the trial function for the displacement can be expressed as
=
v
+
v
+
v
02
u
N 20
N 11
N 02
20
11
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