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p
η
(ξ
,
η
=
1
)
,
which is applied on the boundary in the
y
direction between nodes 3 and 4.
Then
0
−
p
0
2
ξ
p
η
(ξ
,
η
=
1
)
=−
p
(ξ
,
η
=
1
)
=
(6.39)
p
0
2
ξ
and Eq. (6.38) reduces to
a
1
0
p
i
0
N
T
=
(ξ
,
η
=
1
)
p
(ξ
,
η
=
1
)
d
ξ
(6.40)
where the superscript
i
refers to the
i
th element.
The integral of Eq. (6.40) can be computed as
0
0
0
0
0
0
···
0
0
ξ
0
0
ξ
0
1
−
ξ
0
0
···
−
p
0
2
=
···
···
−
(6.41)
p
0
0
0
2
ξ
0
0
2
ξ
−
ξ
0
ξ
01
2
−
ξ
N
T
(ξ
,
η
=
1
)
p
(ξ
,
η
=
1
)
Upon carrying out the integration of
a
0
N
T
p
d
ξ
,
we find
ap
0
2
u
x
1
u
x
2
u
x
3
u
x
4
u
y
1
u
y
2
u
y
3
u
y
4
p
i
0
T
(
)
=−
(6.42)
[0 000001
/
31
/
6]
Similarly, for an element with the linearly varying distributed load beginning at node
4 with magni
tu
de
p
0
/
2 and terminating at node 3 with magnitude
p
0
,
the loading can be
expressed as
p
η
(ξ
,
η
=
1
)
=−
p
0
(
1
+
ξ)/
2
.
Then
2
0000005
3
ap
0
p
i
0
T
(
)
=−
/
62
/
(6.43)
A more general expression for the loading vector can be derived for the distributed
loading of Fig. 6.13, which acts in the
y
direction between nodes 4 and 3. Then
p
0
ξ
p
(ξ
,
η)
=
=
(6.44)
p
p
η
(ξ
,
η
=
1
)
η
p
3
p
4
4
3
y,h
a
FIGURE 6.13
Applied distributed loading along the element bound-
ary between nodes 4 and 3 of the element of Fig. 6.12.
x,x
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