Civil Engineering Reference
In-Depth Information
0210
.( 9929
2
.
)
2
M
max
=
26 357 139 29
.
(
.
) .
−
01530
()(
124 29
.
)
−
=
1917 59
.
ft-k
M
D
1917 59
80
.
F
=
max
=
=
23 97
.
kmaximum chord
force
max
01550
2
.
()
2
M
3
=
25 643 50
.
(
)
−
=
1094 65
.
ft-katgrid line3
M
D
1094 65
80
.
3
F
==
=
13 683
.
kchordforce at grid line3
3
Transverse strut force diagrams at grid line 1 (see Fig. 7.33):
26 357
20
,
v
frame
=
=
1318
plfunitshear alongf
rame
v
=
v
−
v
=
1318
−
329 5
.
=
988 4
.
plfnet
shearalong frame
net
frame
diaph
Starting at grid line B:
F
=−
0 3295 20
.
(
)
=−
659
.
ktension at startofframe
F
=−
65909884 20
.
+
.
()
= +
13 18
.
kcompression at endofframe
F
+
13 18
.
−
0 3295 40
.
()
=
0
katgrid lineA
Therefore the diagram closes.
Transverse strut force diagrams at grid line 4:
25 643
20
,
v
frame
=
= +
1282
plfunitshear along
frame
v
=
v
−
v
=
1282
−
325
=
957 2
.
plfnet s
hear alongframe
net
frame
diaph
Starting at grid line B:
F
=+
0 9572 20
.
(
)
=+
19 14
.
kcompression at endoffframe
F
+
19 14
.
−
0 325 60
.
()
=
0
katgrid lineA
Therefore the diagram closes.
Longitudinal chord force diagrams at grid line B:
VwL
D
−
21 857
,
200 20
80
−
(
)
v
=
2
frame
=
=+
223 2
.plfunitshear at endofOCBF1
frame
Vwx
D
−
21 857
,
−
200 180
80
(
)
v
=
2
=
=−
176
plfun
it shearatstart of OCBF2
frame
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