computed
in
the
following
paragraphs.
where:
2.
Deflection calculations.
The total
computed deflection of diaphragms (∆ d) under the
∆w =
web component of diaphragm
prescribed static seismic forces consists of the sum of
deflection, in. (mm).
two components: the first component is the flexural
deflection (∆ f); the second component is the shearing
qave = average shear in diaphragm, lbs. /ft.
deflection (∆ w). When most beams are designed, the
(N/m).
flexural component is usually all that is calculated,
but for diaphragms, which are like deep beams, the
L1 =
distance from adjacent vertical
shearing component must be added to the flexural
resisting element (i.e., such as a shear wall) and the
component.
point to which the deflection is to be determined, ft.
(m).
i.
Flexural component.
This is
calculated in the same way as for any beam. For
F = flexibility factor, micro inches per foot
example, for a simple beam with uniform load, the
of span stressed with a shear of one pound per foot
flexural component is obtained from the familiar
(micro millimeters per meter of span stressed with a
formula ∆ f = 5wL4/384EI. The only question is the
shear of one Newton per meter of span).
value of the moment of inertia, I. For diaphragms
whose webs have uniform properties in both
Values of the flexibility factor, F, and the allowable
directions (concrete or a flat steel plate), the moment
shear per foot, qD, for steel decking are given in
of inertia is simply that of the diaphragm cross-
manufacturers' catalogs, as well as the Diaphragm
section.
For diaphragms of fluted steel deck, or
Design Manual of the Steel Deck Institute (SDI).
diaphragms of wood, whose stiffness is influenced by
Deflection calculations for concrete diaphragms are
nail slip and chord-joint slip, the flexural resistance
seldom required, but the deflection can be calculated
of the diaphragm web is generally negligible, and the
by the conventional beam theory. For example, for a
moment of inertia is based on the properties of the
diaphragm with a single span of length, L, with a
diaphragm chords. For a diaphragm of depth D with
total load, W, that is uniformly distributed, the
chord members each having area A, the moment of
maximum shearing deflection is:
2
2
inertia, I, equals 2A(D/2) , or AD /2.
α WL
∆
=
(7-7)
ii. Shearing Component. The shearing
w
8 Aω G
component of deflection can be derived from the
following equation:
where:
q LF
" = a form factor (L/D for prismatic webs)
∆w
= ave 61
(7-6)
10
Aw = area of the web
7 -120
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