Civil Engineering Reference
In-Depth Information
P
P
E
,
I
b
,
L
b
x
3
#5
#6
F
0
x
1
3
#2
x
2
#4
1
2
E
,
I
c
,
L
c
E
,
I
c
,
L
c
#3
PHL #1
Figure 7.6
One-story one-bay moment-resisting frame.
PHLs (
q
= 6). For simplicity, let
I
c
=
I
b
=
I
and
L
c
=
L
b
=
L
. Assume that only a lateral force of
F
o
is applied at the horizontal degree of freedom
x
1
, this gives
F
1
=
F
o
and
F
2
=
F
3
=0.
Since the axial compressive force in Member 1 (denote as
P
1
) can be different from that of
Member 2 (denote as
P
2
) due to overturning caused the lateral applied load
F
o
, the resulting
stability coefficients can be different as well. The axial compressive force in Member 3 is
assumed to be negligible (i.e.
P
3
≈
0) due to the presence of a slab, even though it is not shown
in the figure. Therefore, let
s
s
P
1
EI
P
2
EI
λ
1
=
×
L
,
λ
2
=
×
L
ð7
:
51Þ
It then follows that the stiffness matrices of the FAM for this one-story frame become similar to
those in Eq. (2.54), i.e.
2
3
s
0
1
EI
=
L
3
+
s
0
2
EI
=
L
3
s
1
EI
=
L
2
s
2
EI
=
L
2
x
1
x
2
x
3
4
5
K
=
s
1
EI
=
L
2
ð7
:
52aÞ
s
1
EI
=
L
+4
EI
=
L
2
EI
=
L
s
2
EI
=
L
2
2
EI
=
L
s
2
EI
=
L
+4
EI
=
L
2
2
2
2
s
EI
L
s
EI
L
s
EI
L
s
EI
L
0
0
←
x
1
1
2
2
1
′
K
=
s
c
EI
L
s
EI
L
0
0
4
EI
L
2
EI
L
←
x
ð7
:
52bÞ
1
1
1
2
0
0
s
c
EI
L
s
EI
L
2
EI
L
4
EI
L
←
x
2
2
2
3
s
EI
L
s
c
EI
L
0
0
0
0
←
θ
1
1
1
1
s
c
EI
L
s
EI
L
0
0
0
0
←
θ
1
1
1
2
0
0
s
EI
L
s
c
EI
L
0
0
←
θ
2
2
2
3
′
K
=
ð7
:
52cÞ
0
0
s
c
EI
L
s
EI
L
0
0
←
θ
2
2
2
4
0
0
0
0
4
EI
L
2
EI
L
←
θ
5
0
0
0
0
2
EI
L
4
EI
L
←
θ
6
