Civil Engineering Reference
In-Depth Information
Initially, the FEMs produced by the applied loads are calculated. Thus, fromTable 16.6
40
×
6
M
BA
=−
M
BA
=−
=−
30 kNm
8
6
2
20
×
M
BC
=−
M
CB
=−
=−
60 kNm
12
M
CD
=
M
DC
=
0
=
T
he DFs a
re calculated as before. Note that the length of the member CD
√
6
2
+
4
.
5
2
=
7
.
5m.
K
BA
K
BA
+
4
EI
/
6
4
EI
/
6
DF
BA
=
K
BC
=
4
EI
/
6
=
0
.
5
+
Hence
DF
BC
=
1
−
0
.
5
=
0
.
5
K
CB
K
CB
+
4
EI
/
6
4
EI
/
6
DF
CB
=
K
CD
=
3
EI
/
7
.
5
=
0
.
625
+
Therefore
DF
CD
=
1
−
0
.
625
=
0
.
375
No-sway case
A
B
C
D
DFs
-
0.5
0.5
0.625
0.375
1.0
FEMs
30.0
30.0
60.0
60.0
0
0
Balance
15.0
15.0
37.5
22.5
Carry over
7.5
18.8
7.5
Balance
9.4
9.4
4.7
2.8
Carry over
4.7
2.4
4.7
Balance
1.2
1.2
2.9
1.8
Carry over
0.6
1.5
0.6
Balance
0.75
0.75
0.38
0.22
Final moments(
M
NS
)
17.2
56.35
56.35
27.32
27.32
0
Unlike the frame in Ex. 16.22 the frame itself in this case is unsymmetrical. There-
fore the geometry of the frame, after an imposed arbitrary sway, will not have the
simple form shown in Fig. 16.47(b). Furthermore, since the member CD is inclined,
an arbitrary sway will cause a displacement of the joint C relative to the joint B. This
also means that in the application of the principle of virtual work a virtual rotation of
the member AB will result in a rotation of the member BC, so that the end moments
M
BC
and
M
CB
will do work; Eq. (16.46) cannot, therefore, be used in its existing form.
In this situation we can make use of the geometry of the frame after an arbitrary
