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Problem 6.21
For the element in Fig. 6.21a, let x = [1
ξ
η
ξη
] [a
1
a
2
a
3
a
4
]
T
.
(a) Write matrix [A] in the relation
[x
1
x
2
x
3
x
4
]
T
= [A]
[a
1
a
2
a
3
a
4
]
T
.
(b) By inspection of Eqs. 6.23, write [A]
1
in the relation x = [1
ξ
η
] [A]
1
[x
1
x
2
x
3
x
4
]
T
(c) Check answers to parts (a) and (b) by seeing if [A] [A]
1
= [
I
]
Solution:
(a)
Applying the interpolation equation:
x1
ξηξ
η
()
a
1
a
2
a
3
a
4
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
⋅
=
at each node gives:
x
1
x
2
x
3
x
4
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
A
a
1
a
2
a
3
a
4
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
⋅
=
A
1
1
1
1
1
−
1
1
1
−
1
−
1
−
1
1
1
1
−
1
1
−
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:=
(b)
since
a
1
a
2
a
3
a
4
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
A
1
−
x
1
x
2
x
3
x
4
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
⋅
=
then
η
a
1
a
2
a
3
a
4
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
⋅
=
becomes
η
A
1
−
⋅
x
1
x
2
x
3
x
4
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
⋅
=
From Eq. 6.23
N
1
1
4
1
ξ
−
⋅
1
η
−
⋅
=
1
4
1
ξ
−
η
−
ξη
+
⋅
=
1
4
1
η
⋅
11
−
1
−
1
T
⋅
=
N
2
1
4
1
ξ
+
⋅
1
η
−
⋅
=
1
4
1
ξ
+
η
−
ξη
−
⋅
=
1
4
1
η
⋅
11 1
−
1
−
T
⋅
=
N
3
1
4
1
ξ
+
⋅
1
η
+
⋅
=
1
4
1
ξ
+
η
+
ξη
+
⋅
=
1
4
1
η
⋅
1111
T
⋅
=
N
4
1
4
1
ξ
−
⋅
1
η
+
⋅
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This note was uploaded on 01/11/2011 for the course MAE 5020 taught by Professor Folkman during the Fall '10 term at Utah State University.
 Fall '10
 Folkman

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