Civil Engineering Reference
In-Depth Information
′
x
F
KK
w
22
×
w
2
6
×
d
21
×
dw
21
×
=
ð6
:
89Þ
′
T
′′
−
Δ
′′
f
KK
w
62
×
w
66
×
61
×
61
×
in which,
,
F
dw
=
x
1
x
2
F
1
w
F
2
w
x
d
=
ð6
:
90Þ
Now the shear wall is first assumed to be linear, this gives
Δ
00
= 0. Then, using the first equa-
tion of Eq. (6.89) gives:
× 10
6
0
:
03
0
:
08
=
4035
:
8 −1569
−1569 985
:
4
F
1
w
F
2
w
ð6
:
91Þ
Solving for the forces in Eq. (6.91) gives:
=
MN
−4
:
446
31
:
762
F
1
w
F
2
w
ð6
:
92Þ
Then substituting Eq. (6.92) back into the second equation of Eq. (6.89) gives:
<
=
<
=
f
1
f
2
f
3
f
4
f
5
f
6
33
:
011
13
:
763
−33
:
011
−13
:
763
27
:
317
31
:
761
=
MN
ð6
:
93Þ
:
;
:
;
Comparing these demands in Eq. (6.93) with the corresponding capacities in Table 6.1,
the shear wall members is in the plastic domain, which means the assumption that the shear
wall members are in the elastic domain is incorrect. Therefore, assume that the shear wall
members are in the plastic domain, which means that the VSH#1,VSH#2, HSH#5 and
HSH#6 begin to deform in the hardening branches. The stiffnesses
K
e
=
1500 MN/m,
K
y
=
15MN/m,
K
cr
=
1050 MN/m,
F
y
=
1.32 MN/m,
K
s
=
2500 MN/m, and
K
scr
= 400
MN/m can be obtained or computed according to the parameters listed in Tables 6.9. Then,
the spring force
f
1
,
f
2
,
f
5
,and
f
6
can be expressed in terms of
δ
1
00
,
δ
2
00
,
δ
5
00
,and
δ
6
00
according to
Eq. (6.17) and Eq. (6.23) as:
<
f
1
= 15
:
15
δ
0
1
+ 1
:
314
f
2
= 15
:
15
δ
0
2
+ 1
:
314
f
5
=476
:
19
δ
0
5
+3
:
00
f
6
=476
:
19
δ
0
6
+3
:
00
ð6
:
94Þ
:
