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. The example here is more closely associated with wind loading.

The Building

Tributary Areas

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.

Elements for Direction 1

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.

Shear Wall Free Body Diagram

Elements for Direction 2

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.

Shear Wall Free Body Diagram

Another View Amrhein, James E “Reinforced Masonry Engineering Handbook”, 4th edition

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”.

Beam Behavior of Diaphragms Amrhein, James E “Reinforced Masonry Engineering Handbook”, 4th edition

Diaphragm Forces in Dir. 1 C = M / L1 M = w(L2)2/8 v = w(L2)/(2L1) T = 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.

Diaphragm Forces in Dir. 2 v = w(L2)/(2L1) The diaphragm must be analyzed and designed to handle the forces in both principle directions. T = M / L2 M C = M / L2

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 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 Diaphragms 2003 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: Layup cases (See IBC Table 2306.3.1) 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)

Nailing Definitions

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

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

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 Design Hem Fir #2 Try (1) 2x4 Check Tension: Try (1) 2x4 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