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Horizontal Diaphragms

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Lateral Forces

Lateral forces result from either wind

loading or seismic motion.

In either case, the diaphragms are generally

loaded with distributed loads.

The example here is more closely

associated with wind loading.

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The Building

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Tributary Areas

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Loadings for Roof Diaphragm

The upper beamdiagram is for loading in

the 2 direction.

The lower beam

diagram is for loading inthe 1 direction.

The distributed loads

equal the pressure times

the tributary height ofthe exposed area.

The unit shears equal the

beam reaction divided

by the length of the

edge.

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Loadings for Floor Diaphragm

Note that the unit

shears at the ends of

the diaphragm are the

result of the

interaction with the

shear walls that are

providing lateral

support for the

diaphragm. These forces are

transferred to the

shear walls.

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Elements for Direction 1

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Idealized Diagram for Dir. 1

Green arrows are unitshears at edge of roof

diaphragm.

Yellow arrows are unit

shears at edge of floordiaphragm.

Shear in upper part of

shear wall is from roof

diaphragm only. Shear (red arrows )in

lower part of shear wall

includes both horizontal

diaphragms.

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Shear Wall Free Body Diagram

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Elements for Direction 2

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Idealized Diagram for Dir. 2

Green arrows are unitshears at edge of roof

diaphragm.

Yellow arrows are unit

shears at edge of floor

diaphragm.

Shear in upper part of

shear wall is from roof

diaphragm only.

Shear (red arrows )in

lower part of shear

wall includes both

horizontal diaphragms.

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Shear Wall Free Body Diagram

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Another View

Amrhein, James E

Reinforced Masonry Engineering

Handbook, 4th edition

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Diaphragms are Beams

Like beams, diaphragms carry loads in bending. Wood diaphragms are considered to be simply

supported.

This results in both internal bending moment and

shear.

The diaphragm can be considered to be similar to

a wide flange beam where the flanges (diaphragm

chords) take all the bending and the web (theplywood sheathing) takes all the shear.

In diaphragms, the shear force is expressed in

terms of unit shear.

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Beam Behavior of Diaphragms

Amrhein, James E

Reinforced Masonry Engineering

Handbook, 4th edition

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Diaphragm Forces in Dir. 1

Unit shear, v, equals the shear force, V, at

a location along the span divided by the

depth of the diaphragm at that location. Moment is taken by chord forces whose

magnitudes equal the Moment at a

particular location divided by the

diaphragm depth at the same location.

M = w(L2)2/8

C = M / L1

T = M / L1

v = w(L2)/(2L1)

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Diaphragm Forces in Dir. 2

The diaphragm must be

analyzed and designed to

handle the forces in both

principle directions.

v = w(L2)/(2L1)

M C = M / L2T = M / L2

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Maximum Diaphragm Ratios

2003 IBC

IBC Table 2305.2.3 (text pg C.42) - Rules of

Thumb used to control diaphragm deflections.

If the span to width ratios are too large, then thediaphragm is not stiff enough to transfer the forces

without significant deflection.

Deflection is a function of beam bending, shear

deflection, nail slip in diaphragm and slip in chord

connections.

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Shear Capacity of Horizontal

Wood Diaphragms2003 IBC

UBC Table 2306.3.1 (pgs C.45-C.47)

Also see Special Design Provisions for Wind & Seismic Table A.4.2A

Shear capacity depends on the following design variables:

supporting member species

plywood grade

nail size (and penetration)

plywood thickness (normally selected for vert. loads) support widths

nail spacing

blocking

layup

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Footnote a

Use of supporting lumber species other than

Douglas Fir or Southern Pine

(1) find specific gravity of supporting framing(see NDS Table 11.3.2A, NDS pg 74)

For Staples: Use Structural I values multiplied

by either 0.82 or 0.65 depending on specific

gravity of supporting members.

For Nails: Use values from table for actual grade

of plywood used multiplied by min[(.5+S.G),1]

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Footnote b

Field nailing requirement

Spacing of fasteners along intermediate

framing to be 12 O.C. unless supportingmember spacing equals 48 or more, then

use 6 O.C. nail spacing.

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Use With Wind Loads

IBC-03 2306.3.1 states:

The allowable shear capacities in Table

2306.3.1 for horizontal wood structuralpanel diaphragms shall be increased 40

percent for wind design

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Some Definitions

Nailing:

Boundary nailing: Nailing at all intersections

with shear walls. (parallel to direction of force.)Edge nailing: nailing along any other

supported plywood edge.

Field nailing: nailing along supports but not ata plywood edge.

Layup cases (See IBC Table 2306.3.1)

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Nailing Definitions

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Chord Design The chords are axial force

members that generally have

full lateral support in both

principle directions.

The top plates of the supporting

walls are generally used as the

chord members.

Due to the reversing nature of

the loads being resisted, the

chord forces are considered tobe both tension and

compression.

Design as an axial force

member.

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Typical Chord

Roof Chord Member

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Example

Consider the building introduced in the

lecture on structural behavior:

We spent some time

determining forces in thehorizontal and vertical

diaphragms (shear walls)

in an earlier lecture.

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Applied Forces: Wind

Direction #1Roof = 12,000 # = 200 plf

2nd flr = 6,300 # = 105 plf

Direction #2Roof = 5,200 # = 60 plf to 200 plf

2nd flr = 4,200 # = 105 plf

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Roof Diaphragm:

Direction 1 Parameters:

C-DX plywood

2x Hem Fir Framing Vmax = 150 plf

Case I layup

Design nailing for the

diaphragm (IBC) Unblocked, 8d nails

Vallow = 1.4*240 *(1-(.5-.43))

Vallow = 313 plf > Vmax

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Roof Diaphragm: Direction 2

Parameters: C-DX plywood

2x Hem Fir Framing

Vmax = 43.3 plf

Case 3 layup

Design nailing for thediaphragm

Unblocked, 8d nails

Vallow = 1.4*180*(1-(.5-.43))

Vallow = 234 plf > Vmax

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Roof Diaphragm Sheathing

Summary After determining

the needs in each

direction the designof the roof can be

specified.

Result:

C-DX plywood

Unblocked

8d @ 6 O.C. Edge

and Boundary

nailing 8d @ 12 O.C.

Field nailing

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Roof Diaphragm Chords:

Direction 1 Moment = 90 ft-k

Depth = 40 ft

Chord Force = + 2.25 k

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Roof Diaphragm Chords:

Direction 2 Moment = 82.7 ft-k

Depth = 60 ft

Chord Force = + 1.38 k

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Chord DesignHem Fir #2

Try (1) 2x4

Check Tension: Ft = (525 psi)(1.6)(1.5)

Ft = 1260 psi

ft = 2250 # / 5.25 in2

ft= 429 psi < Ft

Try (1) 2x4

Check Compression: Fc = (1300 psi)(1.6)(1.15)

Fc = 2392 psi

fc = 2250 # / 5.25 in2

fc= 429 psi < Fc

(1) 2x4 is adequate in both directions