Civil Engineering Reference
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do not contain an opening. Wind loads of 123 and 77 plf are applied to the windward and
leeward sides of the diaphragm, respectively, at the area of the opening. It is important
during the analysis to pay attention to the magnitude of the chord and collector forces
and where they occur. A better sense of the critical nature of these elements can be
obtained by this exercise.
Construction of the basic shear diagram (see Fig. 8.7):
The diaphragm unit shears must be
determined at each strut, collector, and change in diaphragm depth locations.
Diaphragm 1:
RR
wL
1
200 100
2
(
)
== =
=
10 000
,
lb
6
2
R
D
10 000
40
,
1
v
==
=+
250
plfdiaphragm unit shearattgrid line1
1
VRwx
2
=− =
10 000
,
−
200 30
(
)
=
4000
lbshear at grid line2
1
V
D
4000
40
2
v
== =+
100
plfdiaphragm unit shea
rr eftsideofgrid line
,
2
2
L
V
D
4000
80
2
v
== =+
50
plfdiaphragm unit shear
rightsideofgrid line2
2
R
VVwx
3
=− =
4000
−
200 20
()
=
0
lb shearatgrid line
3
2
v
3
=
0
plf
lb shearatgrid line4
VVwx
4
=− =−
0
200 10
()
=−
2000
3
V
D
==
−
2000
50
4
v
=−
40
plfdiaphragm unit shea
r, left side of grid line4
4
L
V
D
==
−
2000
60
4
v
=− .
33 33
plfdiaphragm unit s
hear
,
rightsideofgrid line
4
4
R
VVwx
5
=− =−
2000
−
200 15
()
= −
5000
lb shearatgridline5
4
V
D
==
−
5000
60
5
v
=− .
83 33
plfdiaphragm unit s
hear
,
left side of grid line
5
5
L
V
D
==
−
5000
90
5
v
=− .
55 56
plfdiaphragm unit s
hear
,
rightsideofgrid line
5
5
R
V
6
=− ,
10 000
lbshear at grid line
6
V
D
==
−
10 000
90
,
6
v
=−
111 11
.
lf diaphragmunitshear at grid line6
6
Diaphragm 2:
wL
200 90
2
()
RR V
====
=
9000
lb
6
10
6
2
V
D
9000
90
v
== =+
6
100
plfdiaphragm unit shearatgrid line6
6
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