Civil Engineering Reference
In-Depth Information
for a regular diaphragm (e.g., reduced section, transfer diaphragm, main section, etc.).
The deflection of a notched diaphragm will be greater than that of a rectangular dia-
phragm, due to the change in stiffness of the diaphragm to the left of the notch. Deflec-
tion should always be checked for irregular-shaped diaphragms.
The examples presented in ATC 7 used virtual work to determine the bending and
shear terms of the deflection equation. In the shear deflection term of the equation in
those examples, the shear in the shallower section was adjusted by a ratio of the shal-
lower depth to the total depth b′/b.
∑
(
∆
X
)
C
∆=∆+∆+
0 188
.
Le
+
TL
B
S
n
2
b
where
b
c
mM
EI
mM
EI
∫
∫
∆=
dx
+
dx
B
1
2
a
b
b
c
L
bt
GA
wx
b
b
vV
Gt
∫
∫
=
∫
0
∆=
dx
+
wx dx
dx
S
2
2
2
′
a
b
USDA Forest Service Research Note FPL-0210
8
provided a simplified energy method
for calculation of the shear deflection of beams. The integration is reduced to the multipli-
cation of the area of the unit shear diagram (due to the general loading) and the ordinate
of the shear diagram (due to a unit dummy load) that is applied at the location of interest,
divided by
G
t
. To simplify the overall method of calculating the total deflection, a com-
puter model will be used to determine the bending deflection. The computer results will
be verified by hand, using the moment-area method, for informational purposes only. The
shear deflection will follow the simplified method used in FPL-0210. The overall process is
ideal for irregular-shaped diaphragms because it addresses the different moments of iner-
tia and varying loads and eliminates the need to derivate complex equations.
The diaphragm will be analyzed as a single-span beam with two different stiff-
nesses (moments of inertia). Figure 3.14 shows the anticipated deflected shape. A uni-
formly distributed load of 200 plf was used to determine the chord, collector, and strut
forces for wind loads. However, a seismic loading condition will be used for the deflec-
tion, because it will be more informative. An assumed 200 plf uniform load will be used
for the shallow-depth section, and a 400 plf uniform load will be used for the remainder of
the diaphragm. Also assume that the flanges consists of two 2 × 6, with
E
=×
18 10
6
.
psi.
The moment of inertia is equal to
2
d
II Ad
=+ =
2
2
A
0
2
Reducing the term, assuming
I
0
= 0, the equation becomes
2
A
d
2
2
I
=
2
()
12
=
72
Ad
2
where
d
= depth of the diaphragm, ft and
A
= area of the diaphragm flange, in
2
.
I
1
=
72 16 53514553
(. )(
)
2
=
.
×
10
6
in
4
I
2
=
72 16 550297
(. )(
)
2
=
.
×
10
6
in
4
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