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Horizontal Diaphragms. by Bart Quimby, P.E., Ph.D UAA Civil Engineering CE 434 - Timber Design. Lateral Forces. Lateral forces result from either wind loading or seismic motion. In either case, the diaphragms are generally loaded with distributed loads.

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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.

Loadings for Roof Diaphragm

- The upper “beam” diagram is for loading in the “2” direction.
- The lower “beam” diagram is for loading in the “1” direction.
- The distributed loads equal the pressure times the tributary height of the exposed area.
- The unit shears equal the “beam” reaction divided by the length of the edge.

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.

Idealized Diagram for Dir. 1

- Green arrows are unit shears 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.

Idealized Diagram for Dir. 2

- Green arrows are unit shears 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.

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 (the plywood sheathing) takes all the shear.
- In diaphragms, the shear force is expressed in terms of “unit shear”.

Diaphragm Forces in Dir. 1

C = M / L1

- 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

v = w(L2)/(2L1)

T = M / L1

Diaphragm Forces in Dir. 2

- The diaphragm must be analyzed and designed to handle the forces in both principle directions.

v = w(L2)/(2L1)

T = M / L2

M

C = M / L2

Maximum Diaphragm Ratios2003 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 the diaphragm 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.

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

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]

Footnote “b”

- Field nailing requirement
- Spacing of fasteners along intermediate framing to be 12” O.C. unless supporting member spacing equals 48” or more, then use 6” O.C. nail spacing.

Use With Wind Loads

- IBC-03 2306.3.1 states:
“The allowable shear capacities in Table 2306.3.1 for horizontal wood structural panel diaphragms shall be increased 40 percent for wind design”

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 at a plywood edge.

- Layup cases (See IBC Table 2306.3.1)

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 to be both tension and compression.
- Design as an axial force member.

Typical Chord

- Roof Chord Member

Example

- Consider the building introduced in the lecture on structural behavior:

We spent some time determining forces in the horizontal and vertical diaphragms (shear walls) in an earlier lecture.

Applied Forces: Wind

Direction #1

Roof = 12,000 # = 200 plf

2nd flr = 6,300 # = 105 plf

Direction #2

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

2nd flr = 4,200 # = 105 plf

Roof Diaphragm: Direction 1

- Parameters:
- ½” 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

- Unblocked, 8d nails

Roof Diaphragm: Direction 2

- Parameters:
- ½” C-DX plywood
- 2x Hem Fir Framing
- Vmax = 43.3 plf
- Case 3 layup

- Design nailing for the diaphragm
- Unblocked, 8d nails
- Vallow = 1.4*180*(1-(.5-.43))
- Vallow = 234 plf > Vmax

- Unblocked, 8d nails

Roof Diaphragm Sheathing Summary

- After determining the needs in each direction the design of the roof can be specified.

- Result:
- ½” C-DX plywood
- Unblocked
- 8d @ 6” O.C. Edge and Boundary nailing
- 8d @ 12” O.C. Field nailing

Roof Diaphragm Chords: Direction 1

- Moment = 90 ft-k
- Depth = 40 ft
- Chord Force = + 2.25 k

Roof Diaphragm Chords: Direction 2

- Moment = 82.7 ft-k
- Depth = 60 ft
- Chord Force = + 1.38 k

Chord DesignHem Fir #2

- Try (1) 2x4
- Check Tension:
- F’t = (525 psi)(1.6)(1.5)
- F’t = 1260 psi
- ft = 2250 # / 5.25 in2
- ft = 429 psi < F’t

- Try (1) 2x4
- Check Compression:
- F’c = (1300 psi)(1.6)(1.15)
- F’c = 2392 psi
- fc = 2250 # / 5.25 in2
- fc = 429 psi < F’c

(1) 2x4 is adequate in both directions

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