150956631 Engineering Guide Wood for Wood Frame Construction
Engineering with Wood
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Transcript of Engineering with Wood
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Engineering with Engineering with WoodWood
Presenters: David W. Boehm, P.E.Presenters: David W. Boehm, P.E.
Gary Sweeny, P.E.Gary Sweeny, P.E.
Shear Walls and DiaphragmsWhy Buildings Don’t Fall Over
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Plan View
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Assume seismic load is also 200 plf
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veu
l = 120’
DIAPHRAGM UNIT SHEARS
wew=
200#/FT
b=40’
wns = 200#/FT
vns = wns x l 200 x 120 2 x 40
300 #/FT= =
vn
s
vew = wew x b 200 x 402 x 120
33 #/FT= =
2 x l
2 x b
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PANEL LAYOUT AND FASTENER SCHEDULE
North-South Loading
Case 1 v = 300 #/FT
Assume 8 d nails
15/32 plywood
2” nominal framing
Choose: Blocked Diaphragm
8 d nails @ 4” panel edges
8 d nails @ 6” interior
East-West Loading
Case 3 v = 33 #/FT
Unblocked 6” max spacing at panel edges
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DIAPHRAGM CHORD SIZE
Moment due to N-S wind
m = wl 2 = 200 1202 = 360,000 FT-LBS 8 8
Axial load in chords = C = T = M = 360,000 ft-lbs = 9,000 LBS b 40 ftAssume allowable ft = 1150 psi
Area required = 9,000# = 7.8 in2
Assume 2 x 8 wall plate, bolted
Area of 2 x 8 with bolt hole
A = 1.5 x (7.25 - .875) = 9.56 in2
Use double 2 x 8 top plate / chord to allow for splice
1150 psi
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Diaphragm layout
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Shear WallsShear Walls North wallNorth wall v=33 #/ftv=33 #/ft
Nominal nailing requiredNominal nailing required East and west walls v=300 #/ftEast and west walls v=300 #/ft
20’
Shear wall elevation
Nailing pattern
7/16 sheathing2 x studs8d nails @ 4” required
A40’
T C
Vns = 300#/ft(40’) = 12000#
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Tiedown ForceTiedown Force
ΣΣMMA A = 0= 0
0 = (12000 x 20) – (T x 40)0 = (12000 x 20) – (T x 40)
T = 6,000#T = 6,000#
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WALL DESIGN WITH OPENINGS
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DRAG STRUT / COLLECTOR FORCE
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OVERTURNING FORCE
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