Civil Engineering Reference
In-Depth Information
The global stiffness matrices of the SDOF column can be written as:
2
4
3
5
12
1+
χ
EI
L
3
K
=
"
#
6
1+
χ
EI
L
2
6
1+
χ
EI
L
2
12
1+
χ
EI
L
3
−
−
K
0
=
2
4
3
5
4+
χ
1+
χ
EI
L
2
−
χ
1+
χ
EI
L
6
1+
χ
EI
L
2
ð4
:
34Þ
−
2
−
χ
1+
χ
EI
L
4+
χ
1+
χ
EI
L
6
1+
χ
EI
L
2
−
K
00
=
6
1+
χ
EI
L
2
6
1+
χ
EI
L
2
12
1+
χ
EI
L
3
−
−
where
χ
=12
EIb
/
GAL
2
. The shear effect is introduced to the stiffness matrices
K
,
K
0
and
K
00
in Eq. (4.34) by the parameter
χ
, and the procedure of obtaining the stiffness matrices can be
found as below. By specifying
E
= 30 GPa and
G
= 11.5 GPa, the governing equation can be
written as:
0.6391
−
0.9586
−
0.9586
0.6391
x
F
m
m
V
−
0.9586
1.9587
0.9171
−
0.9586
−θ″
1
5
×
10
=
ð4
:
35Þ
−
0.9586
0.9171
1.9587
−
0.9586
−θ″
2
0.6391
−
0.9586
−
0.9586
0.6391
−τ″
First, assume that the SDOF column is in the elastic domain, this gives
θ
1
00
=
θ
2
00
= 0 and
τ
00
= 0. Then the restoring force
F
, shear force
V
, moment
m
1
and
m
2
can be calculated
using Eq. (4.35):
<
=
<
=
F
m
1
m
2
V
127
:
8119 kN
−191
:
7178 kN-m
−191
:
7178 kN-m
127
:
8119 kN
=
ð4
:
36Þ
:
;
:
;
Comparing these moment and shear demands in Eq. (4.36) with the cracking and yielding
capacities in Tables 4.1 and 4.2, the SDOF column is in the plastic domain, which means
the assumption that the SDOF column is in the elastic domain is incorrect. Therefore, assume
that the structure is in the plastic domain and is located in the hardening branches, whichmeans
that the RHs and SH begin to deform in the hardening branches. The
k
f
= 2.9×10
5
kN m/rad,
α
f
= 6.6×10
4
kNm/rad,
k
s
= 7.7×10
5
kN/m and
α
s
= 3.8×10
4
kN/m can be computed according
