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Engineering with Wood Presenters: David W. Boehm, P.E. Gary Sweeny, P.E. Gary Sweeny, P.E. Shear Walls and Diaphragms Why Buildings Don’t Fall Over.

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Presentation on theme: "Engineering with Wood Presenters: David W. Boehm, P.E. Gary Sweeny, P.E. Gary Sweeny, P.E. Shear Walls and Diaphragms Why Buildings Don’t Fall Over."— Presentation transcript:

1 Engineering with Wood Presenters: David W. Boehm, P.E. Gary Sweeny, P.E. Gary Sweeny, P.E. Shear Walls and Diaphragms Why Buildings Don’t Fall Over

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17 Plan View

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19 Assume seismic load is also 200 plf

20 v eu l = 120’ DIAPHRAGM UNIT SHEARS w ew = 200#/FT b=40’ w ns = 200#/FT v ns = w ns x l 200 x 120 2 x 40 300 #/FT== v ns v ew = w ew x b 200 x 40 2 x 120 33 #/FT== 2 x l 2 x b

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22 PANEL LAYOUT AND FASTENER SCHEDULE North-South Loading Case 1v = 300 #/FT Assume8 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 3v = 33 #/FT Unblocked 6” max spacing at panel edges

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24 DIAPHRAGM CHORD SIZE Moment due to N-S wind m = w l 2 = 200 120 2 = 360,000 FT-LBS 8 8 Axial load in chords = C = T = M = 360,000 ft-lbs = 9,000 LBS b 40 ft Assume allowable ft = 1150 psi Area required = 9,000# = 7.8 in 2 Assume 2 x 8 wall plate, bolted Area of 2 x 8 with bolt hole A = 1.5 x (7.25 -.875) = 9.56 in 2 Use double 2 x 8 top plate / chord to allow for splice 1150 psi

25 Diaphragm layout

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27 Shear Walls North wallv=33 #/ft North wallv=33 #/ft Nominal nailing required Nominal nailing required East and west walls v=300 #/ft East and west walls v=300 #/ft 20’ Shear wall elevationNailing pattern 7/16 sheathing 2 x studs 8d nails @ 4” required A 40’ TC V ns = 300#/ft(40’) = 12000#

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29 Tiedown Force ΣM A = 0 0 = (12000 x 20) – (T x 40) T = 6,000#

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31 WALL DESIGN WITH OPENINGS

32 DRAG STRUT / COLLECTOR FORCE

33 OVERTURNING FORCE

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