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

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

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

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 = 5*wL*4/384*EI*. 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, *q*D, 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 2*A*(*D*/2) , or *AD */2.

∆

=

(7-7)

ii. Shearing Component. The shearing

8 *A*ω G

component of deflection can be derived from the

following equation:

where:

∆w

= ave 61

(7-6)

10

7 -120

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