Civil Engineering Reference
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which is contrary to the classical distribution based on tributary width, where
Rw
= /2. The
shear walls and supporting foundations at grid lines B and C might have been underdesigned
if the analysis were terminated after analyzing the transfer members at grid line 2B and the
loads to lines B/C were assumed to be correct. When compared to significantly larger loads
in real-life examples, 400 lb may seem like an insignificant amount to be concerned about;
however, different diaphragm configurations will cause a different load distribution to line
B/C. The amount of additional load distribution to wall line B/C is controlled by the offset-
to-span ratio. The greater the offset, the greater the load will be.
▲
Example 3.4: Single Notched Diaphragm, Analysis in the Longitudinal Direction, Strut
Offset Out (See Fig. 3.29)
The diaphragm is 60 ft wide by 100 ft deep with a 15 ft outward offset at grid line 2.
A collector would normally be extended the full depth of the cantilever section at grid
line 2, which would simplify the analysis. However, because of the length, an alternate
approach is to extend the collector only a short distance into a transfer diaphragm off the
end of SW1. The offset in the diaphragm creates a discontinuity in the strut/collector at
grid lines 1 and 2. The diaphragm can be broken down into two separate sections. The
main diaphragm lies between grid lines 2 and 3 from C to D. The second diaphragm lies
between grid lines 1 and 3 from A to C. The left end of the upper diaphragm section is
supported by a strut that transfers its reaction
R
1
into the end of the propped cantilever
transfer diaphragm that lies between grid lines B and C. The transfer diaphragm is
tied to the shear wall at grid line 2 by the collector that is embedded into the transfer
diaphragm.
A 200 plf uniform load is applied to the diaphragm. The load is distributed to
each diaphragm section in accordance to its depth. The basic shear diagrams, transfer
diaphragm shears, and loading diagrams can be referenced in Fig. 3.30.
Load distribution into the diaphragm:
Loads between A and C
wD
D
200 40
100
()
AC
AD
-
from
23
to
=
=
=
80
plf
-
from
1to2
=
200 plf
Loads between C and D
wD
D
200 60
100
()
CD
AD
-
from
23
to
=
=
=
120
plf
-
Construction of the basic shear diagrams:
Upper diaphragm 1:
80 60 30
( ()
+
200 15 67 5
75
( (. )
462
0
=
=
4620
lb
=
=
115 5
.plf
R
v
1
1
40
200 15 75
( (.)
+
80 60
(
)(
45
)
=
-
3
180
40
R
=
=
3180
lb
v
=- .plf
79 5
3
75
3
1620
40
VRwx
=- =
4620
-
200 15
()
=
1620
lb
v
=
=
40
.. 5plf
2
1
2
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