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Element 2:
Here the local and global coordinates coincide so that
k
2
=
k
2
. Substitute
=
3
.
0m,
m
2
,
and
EA
EI
=
4
.
712 MN
·
=
640 MN in Eq. (5.94) giving
k
bb
k
bc
k
2
=
k
cb
k
cc
.
213
.
33
0
0
−
213
.
33
0
0
0
2
.
0942
−
3
.
1413
0
−
2
.
0942
−
3
.
1413
1413
··························· · ···························
−
0
−
3
.
1413
6
.
2827
0
3
.
1413
3
.
=
(4)
.
213
.
33
0
0
213
.
33
0
0
0
−
2
.
0942
3
.
1413
0
2
.
0942
3
.
1413
0
−
3
.
1413
3
.
1413
0
3
.
1413
6
.
2827
Element 3:
Use
=
.
=
.
·
=
3
0m,
EI
10
49 MN
m
,
and
EA
1320 MN in Eq. (5.94). Also, for a global
α
=−
90
◦
. This leads to
coordinate system
XZ
placed at point
c
,
k
cc
k
cd
k
3
T
3
k
3
T
3
T
=
=
k
dc
k
dd
.
4
.
6622
0
6
.
9933
−
4
.
6622
0
6
.
9933
0
440
0
0
−
440
0
9933
··························· · ···························
−
6
.
9933
0
13
.
987
−
6
.
9933
0
6
.
=
(5)
.
4
.
6622
0
−
6
.
9933
4
.
6622
0
−
6
.
9933
0
−
440
0
0
440
0
6
.
9933
0
6
.
9933
−
6
.
9933
0
13
.
987
The global stiffness matrix is assembled by superimposing the element stiffness matrices
using
M
k
i
jk
K
jk
=
(6)
i
=
1
. By observation of
k
i
jk
of (3), (4),
where the summation is taken over all beam elements
(
M
)
and (5), we see that
k
i
jk
K
jk
=
i
=
1
,
2
,
or 3
(7)
with the exception of
K
bb
and
K
cc
which are given by
K
bb
=
k
bb
+
k
bb
(8)
k
cc
Thus, assembly leads to a global stiffness matrix that appears as
k
cc
+
K
cc
=
k
1
K
=
k
2
(9)
k
3
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