Civil Engineering Reference
In-Depth Information
The application of the dead loads affects the unit shears as follows:
230 8
2
()
2
1
8
T
SW
=
2000 8
()
-
=
1080
lb
1
1080
8
v
left
=
= +
135
plfunitshear
,
chordforc
eedivided by wall height
230 8
2
()
2
1
8
C
SW
=
2000 8
()
+
=
2920
lb
1
V
h
2920
8
right
v
=
=
=+
365
plfunitshea
rrchord forcedivided by wall height
,
right
wall
135
+
365
v
average
=
=+
250
plfsameascalcula
,
tedfor basicwallshear
8
The results show that the calculated shear at the left chord member is less than
the basic wall shear
V
/
L
and the shears at the right chord exceed the basic wall shear.
However, the average of these two shears is equal to the basic wall shear. None of the
APA tests
3
on diaphragms or shear walls were set up to directly measure the variance in
shear across the wall section. It has been assumed and widely accepted that the shears will
be uniformly distributed across the wall. The calculations show that this is not the case,
especially when the wall is being used as a transfer diaphragm or a complex shear wall.
2000
+
3000
atfirst-floor wall
v
SW
=
=
416 67
.
plfbasic wall shear
2
8
Overturning forces at first-floor (see Figs. 11.8 and 11.9):
1
12
[
( (
]]
=
T
SW2
=
2000 17
()
+
3000 9
()
-
230 84 330 12 6
()()
-
2490 lb
1
12
[
]
=
C
SW2
=
2000 17
()
+
3000 9
()
+
230 88 330 12 6
()()
+
( () (
+
450
+
960 12
)()
9700
lb
Since the 2000 lb and 3000 lb shear forces must be translated to the top of SW2, the
associated overturning force of the rim joist applied to the top of the wall is equal to
∑
Vd
5000 1
12
()
flr
TC
=
=
=
=
416 67
.
lb
rimrim
L
SW
1
Transfer
diaphragm
shear includes the floor dead load with the overturning force from the second-
floor wall, SW1.
See the free-body diagrams on the right of Fig. 11.9.
2
()
330 12
2
1
12
V
left
=
-
1080 8
()
=
1260
lb
V
h
1260
8
v
=
left
wall
=
=-
157 7
.
plffromshea
rs acting on shtg element
.
left
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