Civil Engineering Reference
In-Depth Information
Then, the rotation deformations
θ
1
and
θ
2
at the column ends and the shear deformation
τ
of the
structure can be obtained:
θ
1
=
θ
2
= −0
:
0175 rad,
τ
=0
:
0077m
ð4
:
58Þ
Comparing the demands in Eqs. (4.57) and (4.58) with the corresponding capacities in
Tables 4.1 and 4.2, the structure is in the plastic domain and the RHs and SH behave on
the softening branches, which means that the latest assumption is made correctly and the
solution in Eq. (4.57) are the final results.
4.3.2 Force Analogy Method for Static Multi-Degree-of-Freedom Systems
For a multi-degree-of-freedom (MDOF) system with
n
degrees of freedom (DOFs), the
displacement can be written in vector form as:
<
:
=
;
<
:
=
;
x
0
1
x
0
2
.
x
0
n
x
0
1
x
0
2
.
x
0
n
x
=
x
0
+
x
00
=
+
ð4
:
59Þ
where
x
represents the total displacement vector,
x
0
is the elastic displacement vector, and
x
00
is
the inelastic displacement vector.
Consider a RC frame constructed by flexural members, the
q
f
RHs are located at two
ends of the members to represent the plastic bending behavior and
q
s
SHs are assigned
to the columns to represent the plastic shear behavior. Then the moment vector
m
at
the RHs and shear force vector
V
according to the SHs can be described by the force
vector
f
as:
′
′′
m
1
m
1
′
′′
m
q
f
m
′ ′′
== + =
mm m
q
f
f
+
ð4
:
60Þ
′
′′
V
q
f
+
1
′
V
q
f
+
1
′′
VV V
V
q
f
+
q
s
′
V
q
f
+
q
s
′′
where
m
0
and
V
0
are the vectors of elastic moments and elastic shear forces respectively due to
elastic displacement
x
0
, and
m
00
and
V
00
are the vectors of inelastic moments and shear forces
respectively due to inelastic displacement
x
00
.
Corresponding to the inelastic moment and shear force vectors
m
00
and
V
00
, the inelastic
deformations
θ
00
and
τ
00
can be written in vector form as:
