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Horizontal Diaphragms

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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Horizontal Diaphragms

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  1. Horizontal Diaphragms by Bart Quimby, P.E., Ph.D UAA Civil Engineering CE 434 - Timber Design

  2. 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.

  3. The Building

  4. Tributary Areas

  5. 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.

  6. 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.

  7. Elements for Direction 1

  8. 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.

  9. Shear Wall Free Body Diagram

  10. Elements for Direction 2

  11. 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.

  12. Shear Wall Free Body Diagram

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

  14. 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”.

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

  16. 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

  17. 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

  18. 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.

  19. 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

  20. 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]

  21. 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.

  22. 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”

  23. 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)

  24. Nailing Definitions

  25. 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.

  26. Typical Chord • Roof Chord Member

  27. 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.

  28. 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

  29. 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

  30. 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

  31. 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

  32. Roof Diaphragm Chords: Direction 1 • Moment = 90 ft-k • Depth = 40 ft • Chord Force = + 2.25 k

  33. Roof Diaphragm Chords: Direction 2 • Moment = 82.7 ft-k • Depth = 60 ft • Chord Force = + 1.38 k

  34. 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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