Presentation on theme: "Horizontal Diaphragms by Bart Quimby, P.E., Ph.D UAA Civil Engineering CE 434 - Timber Design."— Presentation transcript:
Horizontal Diaphragms by Bart Quimby, P.E., Ph.D UAA Civil Engineering CE 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.
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”, 4 th 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”, 4 th edition
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(L 2 ) 2 /8 C = M / L 1 T = M / L 1 v = w(L 2 )/(2L 1 )
Diaphragm Forces in Dir. 2 The diaphragm must be analyzed and designed to handle the forces in both principle directions. v = w(L 2 )/(2L 1 ) MC = M / L 2 T = M / L 2
Maximum Diaphragm Ratios 2003 IBC IBC Table (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 (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 A, 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 states: “The allowable shear capacities in Table 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 )
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.
Roof Diaphragm: Direction 1 Parameters: – ½” C-DX plywood –2x Hem Fir Framing –V max = 150 plf –Case I layup Design nailing for the diaphragm (IBC) –Unblocked, 8d nails V allow = 1.4*240 *(1-(.5-.43)) V allow = 313 plf > V max
Roof Diaphragm: Direction 2 Parameters: – ½” C-DX plywood –2x Hem Fir Framing –V max = 43.3 plf –Case 3 layup Design nailing for the diaphragm –Unblocked, 8d nails V allow = 1.4*180*(1-(.5-.43)) V allow = 234 plf > V max
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 6” O.C. Edge and Boundary nailing 12” O.C. Field nailing
Roof Diaphragm Chords: Direction 1 Moment = 90 ft-k Depth = 40 ft Chord Force = k
Roof Diaphragm Chords: Direction 2 Moment = 82.7 ft-k Depth = 60 ft Chord Force = k
Chord Design Hem Fir #2 Try (1) 2x4 Check Tension: –F’ t = (525 psi)(1.6)(1.5) –F’ t = 1260 psi –f t = 2250 # / 5.25 in 2 –f t = 429 psi < F’ t Try (1) 2x4 Check Compression: –F’ c = (1300 psi)(1.6)(1.15) –F’ c = 2392 psi –f c = 2250 # / 5.25 in 2 –f c = 429 psi < F’ c (1) 2x4 is adequate in both directions