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The forces at nodes
a
,
b
, and
c
are the unknown reactions. Thus, the equilibrium equations
of (3) become
.
K
aa
K
ab
K
ac
K
ad
0
0
0
...
V
d
P
a
P
b
P
c
...
P
d
?
?
?
...
P
d
.
K
ba
K
bb
K
bc
K
bd
.
=
(6)
K
ca
K
cb
K
cc
K
cd
...
...
...
.
...
.
K
da
K
db
K
dc
K
dd
If the rows corresponding to the unknown reactions are ignored for the moment, then (6)
reduces to
K
dd
V
d
=
P
d
(7)
where
3
k
i
dd
=
k
dd
+
k
dd
+
k
dd
K
dd
=
(8)
i
=
1
P
Xd
P
Zd
P
X
P
Z
P
d
=
=
(9)
To complete the calculations, it is apparent that it is necessary to find only the element
stiffness matrices
k
i
dd
. Suppose the nodes are located at the
XZ
coordinates: Node
a
: (0, 0),
Node
b
: (1, 0), Node
c
: (2, 0), Node
d
: (1, 1). Then, if
EA
is assigned a unit magnitude,
k
i
dd
are given by [Eqs. (5.87), (5.86), and (5.85)]
Bar 1:
√
2
,
√
2
=
E
A
/
=
1
/
√
2
√
2
135
◦
,
α
=
sin
α
=
/
2
,
cos
α
=−
/
2
.
1
··· · ···
1
√
2
k
dd
=
/
4
(10a)
.
1
1
Bar 2:
=
1
,
EA
/
=
1
90
◦
,
α
=−
sin
α
=−
1
,
cos
α
=
0
.
0
··· · ···
0
k
dd
=
(10b)
.
0
1
Bar 3:
=
√
2
,
/
√
2
EA
/
=
1
α
=
√
2
45
◦
,
α
=
sin
α
=
cos
/
2
.
−
1
1
√
2
k
dd
=
···
·
···
/
4
(10c)
.
−
1
1
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