Civil Engineering Reference
In-Depth Information
4kN
B
C
B
C
2kN
2
EI
2
EI
10 m
EI
EI
EI
EI
A
D
A
D
5m
10 m
F
IGURE
16.47
Portal frame of
Ex. 16.22
(a)
(b)
are as follows:
M
AB
=
M
BA
=
M
CD
=
M
DC
=
0
0
10
2
4
×
5
×
M
BC
=−
=−
8
.
89 kNm
15
2
5
2
4
×
10
×
M
CB
=+
=+
4
.
44 kNm
15
2
The DFs are
K
BA
K
BA
+
4
EI
/
10
DF
BA
=
K
BC
=
2
EI
/
15
=
0
.
43
4
EI
/
10
+
4
×
Hence
DF
BC
=
1
−
0
.
43
=
0
.
57
From the symmetry of the frame, DF
CB
=
0
.
57 and DF
CD
=
0.43.
The no-sway moments are determined in the table overleaf. We now assume that the
frame sways by an arbitrary amount,
, as shown in Fig. 16.47(b). Since we are ignoring
the effect of axial strains, the horizontal movements of B and C are both
δ
δ
. The FEMs
corresponding to this sway are then (see Table 16.6)
6
EI
δ
10
2
M
AB
=
M
BA
=−
M
DC
=
M
CD
=
M
BC
=
M
CB
=
0
10
2
/6
EI
. Then
Suppose that
δ =
100
×
M
AB
=
M
BA
=
M
DC
=
M
CD
=−
100 kNm (a convenient value)
TheDFs for themembers are the same as those in the no-sway case since they are func-
tions of themember stiffness. We now obtain themember endmoments corresponding
to the arbitrary sway.
