horizdiap
TRANSCRIPT
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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