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Bar 2:
α =−
90 (Fig. 5.18b)
.
k bd
...... . ......
k bb
k 2
=
(1b)
.
k db
k dd
Another equivalent representation of k 2 is found by designating d as the beginning node
and b as the end node. Then x would be up and z would be to the left, giving
90 .
α =+
Bar 3:
α =
45 (Fig. 5.18c)
.
k dc
...... . ......
k cd
k dd
k 3
=
(1c)
.
k cc
Since the subscripts in the element stiffness matrices have been identified with the global
nodal numbering scheme, the global stiffness matrix is readily assembled by summation
of submatrices (or stiffness coefficients, if appropriate) using
k i jk
K jk
=
(2)
i
where the summation is taken over all the elements. Thus, the global stiffness matrix would
appear as
K aa
K ab
K ac
K ad
V a
V b
V c
V d
P a
P b
P c
P d
=
K ba
K bb
K bc
K bd
(3)
K ca
K cb
K cc
K cd
K da
K db
K dc
K dd
K
V
=
P
with
3
k i jk =
k jk +
k jk +
k jk
K jk =
(4)
i
=
1
and
U xj
U zj
P x j
P zj
V j =
P j =
P have been established, the boundary
conditions can be introduced. From the constraints shown in Fig. 5.17, it is apparent that
the boundary conditions are
Now that the global stiffness equations KV
=
U Xa =
U Za =
0or V a =
0
U Xb =
U Zb =
0or V b =
0
(5)
U Xc =
U Zc =
0or V c =
0
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