Civil Engineering Reference
In-Depth Information
x
3
E
,
I
b
,
L
b
#5
#6
x
2
3
x
1
#2
#4
1
2
E
,
I
c
,
L
c
E
,
I
c
,
L
c
PH
L #1
#3
Figure 2.6
One-story one-bay moment-resisting frame
2
2
2
2
6
EI
L
6
EI
L
6
EI
L
6
EI
L
0
0
←
x
c
c
c
c
c
c
c
c
1
′
K
=
2
EI
L
4
EI
L
0
0
4
EI
L
2
EI
L
←
x
ð2
:
54bÞ
c
c
c
c
b
b
b
b
2
0
0
2
EI
L
4
EI
L
2
EI
L
4
EI
L
←
x
c
c
c
c
b
b
b
b
3
θ
′
4
EI
L
2
EI
L
0
0
0
0
←
c
c
c
c
1
θ
′
2
EI
L
4
EI
L
0
0
0
0
←
c
c
c
c
2
θ
′
0
0
4
EI
L
2
EI
L
0
0
←
c
c
c
c
3
K
′
=
ð2
:
54cÞ
θ
′
0
0
2
EI
L
4
EI
L
0
0
←
c
c
c
c
4
θ
′
0
0
0
0
4
EI
L
2
EI
L
←
b
b
b
b
5
θ
′
0
0
0
0
2
EI
L
4
EI
L
←
b
b
b
b
6
2.4 Force Analogy Method for Static
Multi-Degree-of-Freedom Systems
At this stage, similar to the SDOF system, it can be seen that the FAM has divided the nonlinear
MDOF structural analysis problem into (1) the traditional stiffness method of analysis of the
elastic portion, and (2) the traditional stiffness method of analysis of the inelastic portion. For
the elastic portion, the applied forces
F
a
cause elastic displacements
x
0
(see Eq. (2.51)) and
elastic moments
m
0
(see Eq. (2.52)) of the form:
F
a
=
Kx
0
,
m
0
=
K
0
T
x
0
ð2
:
55Þ
For the inelastic portion, the plastic rotations
Θ
00
cause inelastic displacements
x
00
(see
Eq. (2.45)) and inelastic moments
m
00
(see Eq. (2.50)) of the form:
x
00
=
K
−1
K
0
Θ
00
,
m
00
= −
K
00
−
K
0
T
K
−1
K
0
Θ
00
ð2
:
56Þ
The objective now is to combine Eqs. (2.55) and (2.56) and represent the analytical solution
using the total displacements
x
and the total moments
m
, where according to Eqs. (2.35) and
(2.36), they are
