Civil Engineering Reference
In-Depth Information
hence, the element stiffness matrix is given, formally, by
k
(
e
)
=
[
B
T
] [
D
][
B
] d
V
(
e
)
V
(
e
)
s
−
1
r
−
1
0
4
a
4
b
1
−
s
4
a
1
+
r
4
b
0
−
1
+
s
4
a
1
+
r
4
b
0
1
−
0
1
+
s
4
a
1
−
r
4
b
1
1
−
0
Etab
(1
+
)(1
−
2
)
1
−
0
=
r
−
1
s
−
1
(1
+
)(1
−
2
)
2(1
0
0
0
−
1
−
1
+
)
4
b
4
a
1
+
r
4
b
1
−
s
4
a
−
0
1
+
r
4
b
1
+
s
4
a
0
1
−
r
4
b
1
+
s
4
a
0
−
s
−
1
4
a
1
−
s
4
a
1
+
s
4
a
1
+
s
4
a
−
0
0
0
0
r
−
1
4
b
1
+
r
4
b
1
+
r
4
b
1
−
r
4
b
0
0
0
0
−
(9.64)
×
d
r
d
s
r
−
1
4
b
1
+
r
4
b
1
+
r
4
b
1
−
r
4
b
s
−
1
4
a
1
−
s
4
a
1
+
s
4
a
1
+
s
4
a
−
−
The element stiffness matrix as defined by Equation 9.64 is an
8
×
8
symmetric
matrix, which therefore, contains 36 independent terms. Hence, 36 integrations
are required to obtain the complete stiffness matrix. The integrations are straight-
forward but algebraic tedious. Here, we develop only a single term of the stiff-
ness matrix in detail, then discuss the more-efficient numerical methods used in
finite element software packages.
If we carry out the matrix multiplications just indicated, the first diagonal
term of the stiffness matrix is found (after a bit of algebra) to be
1
1
1
1
Etb
16
a
(1
Eta
32
b
(1
k
(
e
)
11
1)
2
d
r
d
s
1)
2
d
r
d
s
=
(
s
−
+
(
r
−
+
2
)
+
)
−
1
−
1
−
1
−
1
(9.65)
and this term evaluates to
1
−
1
+
1
1)
3
1)
3
Etb
16
a
(1
2(
s
−
Eta
32
b
(1
2(
r
−
k
(
e
)
11
=
+
2
)
3
+
)
3
−
1
16
3
16
3
Etb
16
a
(1
Eta
32
b
(1
=
+
(9.66)
+
2
)
+
)
