Civil Engineering Reference
In-Depth Information
Element III:
5970 2
16
.
V
=
5046 27712
.
+
(
)
=
5970 2
.
lb
v
=
=
3
73 .plf
3
CD
3
CD
77 12
2
()
2
1
6
F
4
=
5046 212
.( )
+
=
4131 2
.
lb
compression
C
l
F
4
=
7182
+
4031 211 313
.
=
,
lb tension
D
Element IV:
6894 2
16
.
V
=
5970 27712
.
+
(
)
=
6894 2
.
lb
v
=
=
4
30 .plf
2
CD
3
CD
77 12
2
()
2
1
6
F
2
=
5970 212
.( )
+
=
4824 2
.
lb
tension
C
l
F
2
=
7182
-
4824 2
.
=
2357 8
.
lb tension
D
Note that the unit shears do not match the shears calculated in the basic shear
diagram at grid line 2. This is so because the basic diaphragm shear was calculated
using a uniform load of 200 plf. The actual load to elements III and IV is 77 plf.
Summing the shear on the elements above and below the opening at grid lines 4, 3,
and 2 will yield the shears previously calculated on the basic shear diagram in Fig. 5.8.
The sum of these shears is a check to verify that the analysis is accurate. The summary
is shown at the bottom of the plan.
Determine the left transfer diaphragm shears, net shears, and transverse collector forces (Fig. 5.10):
Left transfer diaphragm shears:
1
50
3609..
4
=
240 6
.
plf
V
=-
[
4824 216
. ()
+
6441 440
.( )]
=
3609 4
.
lb
v
=
A
A
15
Completing the remaining shear forces on the sheathing element shows that the shear
is negative.
1
16
1992..
2
V
=-
[
6441 410
. ()
+
4824 234
.( )]
=
1992 2
.
lb
v
=
=
132 8
.
plf
D
D
15
Completing the remaining shear forces on the sheathing element shows that the shear
is negative.
2832
15
V
=
1992 2
.
-
4824 2
.
=-
2832
lb
v
=
=
188 8
.plf
BC
BC
Completing the remaining shear forces on the sheathing element shows that the shear
is positive.
The basic diaphragm unit shears are plotted below transfer diaphragm TD1 for a
visual reference.
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