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This stiffness matrix representing a frame with some of its members rigid will be singular,
i.e., determinant
However, this problem is readily removed by taking advantage
of geometrical conditions resulting from the rigid members. Since bar 2 is rigid in the axial
direction,
U
Xb
must be equal to
U
Xc
(
K
)
=
0
.
.
Also, because bar 3 is rigid,
U
Zc
=
U
Zd
=
0
.
Use these
conditions in (6) to help sort out the dependent rows and columns. Set
U
Zc
0 in (6), delete
the corresponding column, and ignore the corresponding row. Set
U
Xc
equal to
U
Xb
in each
equation of (6). Add the first and fourth equations, and we find
=
Load Load
Case 1 Case 2
.
51872
.
2
−
79412
2040
.
4
6993
.
3
U
Xb
U
Zb
3
.
0
10
√
3
.
−
79412
140994
.
2
−
1963
.
2
−
3141
.
3
20
=
(7)
.
.
−
.
.
.
2040
4
1963
2
11723
8
3141
3
2
.
0
0
b
.
.
−
.
.
.
6993
3
3141
2
3141
2
20269
7
0
0
c
The solutions to this set of equations are
Load Case 1
Load Case 2
.
U
Xb
U
Zb
b
U
Xc
U
Zc
c
0
.
6802 mm
3
.
6497 mm
0
.
3933 mm
2
.
1767 mm
.
10
−
4
10
−
5
1
.
7167
×
rad
−
2
.
5057
×
rad
=
(8)
.
0
.
6802 mm
3
.
6497 mm
0
.
0mm
0
.
0mm
.
−
−
.
×
10
−
4
.
×
10
−
4
2
0006
rad
9
1674
rad
Calculation of Elementary Forces
Consider the forces on bar 1. The force vector
p
1
involves
k
1
v
1
, which we will compute
using the displacements of (8) for each of the loading cases. Because end
a
is fixed,
U
Xa
=
U
Za
=
a
=
Also, due to compatibility, the element displacements referred to the global
coordinates at end
b
are equal to the global displacements at
b
of (8). Thus,
0
.
.
.
.
−
47 210
79 412
−
2040
.
4
0
.
.
.
79 412
−
138 900
−
1178
.
1
0
.
.
.
0
2040
.
4
1178
.
1
2720
.
6
k
1
v
1
=
.
.
.
U
Xb
47 210
−
79 412
2040
.
4
.
.
.
−
U
Zb
79 412
138 900
1178
.
1
.
.
.
2040
.
4
1178
.
1
5441
.
1
b
Load Case 1
Load Case 2
.
−
1
.
2277 kN
0
.
6043 kN
.
−
0
.
8192 kN
−
12
.
4831 kN
m
.
2
.
3183 kN
·
9
.
9429 kN
·
m
=
(9)
.
1
.
2277 kN
−
0
.
6043 kN
.
0
.
8192 kN
12
.
4831 kN
m
.
2
.
7853 kN
·
9
.
8748 kN
·
m
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