Civil Engineering Reference
In-Depth Information
Load case D
+
W:
∑
V
3
=
2 875
.
+
2 863
.
+
2 923
.
=
8 661
.
,ksum of applieddloads at grid line3
∑
M
2
=
0
F
3
=−
[.
9 696 17
() .
−
0 234 6018 8168 686
( )
+
.
()
+
.
()
+
.
61 12
() .
+
0 935 18 5
(
.)
1
12
+
8 768 14
.
(
)]
=
74
.
39 kcompression
∑
M
3
=
0
F
2
=+
[.
9 696 17
() .
−
0 234 6018 42424 12
( )
+
.
()
+
.
(
)
+
1
68 60935 18 5
()
−
.
(
.)
1
12
−
8 768 14
.
(
)]
=
52
.
72 kcompression
V
∑
0
V
=
5 272
.
+
7 439
.
+
0 234
.
−
2 424
.
−
018168
.
−
.
−
2 875
.
−
2
863
−
2 923
.
=
0
Therefore, OK.
Load case W
+
0.6D:
∑
M
3
=
0
hold-downconnection
F
2
=+
[.
9 696 17
() .
−
0 234 606018
( )
+
. (. )( )
42424
+
.
(12
)
+
061686
. (. )( )
−
0 935 18 5
.
(. )
1
12
−
8 768 1
.
(
4
)]
=
4 912
.
kcompression
The diaphragm nailing should be based on enveloping the maximum unit shear
calculated and shown in Fig. 7.15. Consideration should be given to stiffening up the
clerestory wall, which will reduce the deflection taking place along the vertical offset. A
possibility of modifying the clerestory wall is shown in Fig. 7.20.
▲
F
i g u r e
7.20 Alternate stiffened wall at offset.
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