Civil Engineering Reference
In-Depth Information
■
Solution
Per Equation 6.56, the interpolation functions in terms of serendipity or natural coordi-
nates are
1
4
(1
−
r
)(1
−
s
)
N
1
(
r
,
s
)
=
1
4
(1
+
r
)(1
−
s
)
N
2
(
r
,
s
)
=
1
4
(1
+
r
)(1
+
s
)
N
3
(
r
,
s
)
=
1
4
(1
−
r
)(1
+
s
)
N
4
(
r
,
s
)
=
with
r
=
(
x
−
25)
/
15
and
s
=
(
y
−
20)
/
10
. For integration in the natural coordinates,
d
x
=
15 d
r
and d
y
=
10 d
s
. The mass matrix is 8
×
8 and the nonzero terms are defined by
1
1
[
N
]
T
[
N
]
d
V
(
e
)
[
N
]
T
[
N
](15 d
r
)(10 d
s
)
=
t
V
(
e
)
−
1
−
1
1
1
[
N
]
T
[
N
]d
r
d
s
=
150(5)
−
1
−
1
In this solution, we compute a few terms for illustration, then present the overall results.
For example,
1
1
1
1
150(5)
16
N
1
−
r
)
2
(1
−
s
)
2
m
11
=
150(5)
d
r
d
s
=
(1
d
r
d
s
−
1
−
1
−
1
−
1
(1
−
r
)
3
3
1
−
1
=
64
9
(1
−
s
)
3
3
150(5)
16
750
16
4(750)
9
830)(10)
−
6
=
=
(7
.
6(10)
−
3
kg
=
2
.
Similarly,
1
1
1
1
150(5)
16
−
r
2
)(1
−
s
)
2
d
r
d
s
m
12
=
150(5)
N
1
N
2
d
r
d
s
=
(1
−
1
−
1
−
1
−
1
r
−
(1
−
s
)
3
3
(
1)
1
−
1
r
3
3
150(5)
16
=
−
83)(10)
−
6
32
9
150(5)
16
3(10)
−
3
kg
=
(7
.
=
1
.
