Civil Engineering Reference
In-Depth Information
Therefore, Eq. (6.89) becomes:
−
4035.8
1569
23
−
679.9
2466.8
−
1569
0.03
0.08
F
F
1
w
−
1569
985.4
404
427
−
583.6
985.4
2
w
′′
23
404
454.4 15.15 175.1
+
427
404
−δ
1.314
1.314
3
3
1
=
−
679.9 427
175.1
185 15.15
+
−
252.9
427
−δ
′′
2
′′
2466.8
−
583.6 427
−
25.29
188.32 476.19
+
−
583.6
−δ
5
−δ
′′
−
1569
985.4
404
427
−
583.6
985.4 476.19
+
6
ð6
:
95Þ
Solving for Eq. (6.95) gives:
<
=
<
=
f
1
f
2
f
5
f
6
3
:
2922
1
:
2574
3
:
1332
2
:
9017
=
MN
ð6
:
96Þ
:
;
:
;
Comparing these spring force demands in Eq. (6.96) with the cracking and yielding capacities in
Table 6.1 shows that the assumption that both the shear wall members are in the plastic domain is
incorrect. Therefore, new trial should be assumed according to the status of the spring forces
in Eq. (6.96). After two trials, assume that the left outer vertical springs of both the shear wall
members and the shear spring of the second floor are in the plastic domain. Then the spring force
f
1
,
f
2
, and
f
6
can be expressed in terms of
δ
1
00
,
δ
2
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
6
= 476
:
2
δ
0
6
+3
:
000
ð6
:
97Þ
:
Therefore, Eq. (6.89) becomes:
4035.8
−
1569 23
−
679.9
−
1569
0.03
F
F
1
w
−
1569
985.4
404
427
985.4
0.08
2
w
′′
23
404
454.39 15.15 175.08
+
404.03
−δ
=
1
.314
1.314
3
1
−
679.9 427
175.08
185.03 15.15 427
+
−δ
′′
2
−
+
−δ
′′
1569
985.4
404.03
427
985.38 476.2
6
ð6
:
98Þ
Solving for Eq. (6.98) gives:
<
=
<
=
;
F
1
w
F
2
w
f
1
f
2
f
6
−2
:
7173
3
:
1707
2
:
3103
1
:
3740
3
:
1707
<
=
<
=
δ
0
1
δ
0
2
δ
0
6
0
:
0657
0
:
0039
0
:
0004
=
m,
=
MN
ð6
:
99Þ
:
;
:
;
:
;
:
