Civil Engineering Reference
In-Depth Information
etc. in which
k
is a constant. Substituting in Eq. (16.45) for
M
AB
,
M
BA
, etc.
we obtain
M
NS
M
NS
M
NS
M
NS
k
(
M
AS
M
AS
M
AS
M
AS
AB
+
BA
+
CD
+
DC
+
AB
+
BA
+
CD
+
DC
)
+
Ph
=
0
(16.46)
Substituting the calculated values of
M
AS
AB
,
M
AS
AB
, etc. in Eq. (16.46) gives
k
. The actual
sway moments
M
AB
, etc., follow as do the final end moments,
M
AB
(
M
NS
M
AB
),
=
AB
+
etc.
An alternative method of establishing Eq. (16.44) is to consider the equilibrium of
the members AB and DC. Thus, from Fig. 16.46(a) in which we consider the moment
equilibrium of the member AB about B we have
R
A,H
h
−
M
AB
−
M
BA
=
0
which gives
M
AB
+
M
BA
R
A,H
=
h
Similarly, by considering the moment equilibrium of DC about C
M
DC
+
M
CD
R
D,H
=
h
Now, from the horizontal equilibrium of the frame
R
A,H
+
R
D,H
+
P
=
0
so that, substituting for
R
A,H
and
R
D,H
we obtain
M
AB
+
M
BA
+
M
DC
+
M
CD
+
Ph
=
0
which is Eq. (16.44).
E
XAMPLE
16.22
Obtain the bending moment diagram for the portal frame shown
in Fig. 16.47(a). The flexural rigidity of the horizontal member BC is 2
EI
while that
of the vertical members AB and CD is
EI
.
First we shall determine the end moments in the members assuming that the frame
does not sway. The corresponding FEMs are found using the results in Table 16.6 and
