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Engineering with Wood - PowerPoint PPT Presentation

Engineering with Wood. Shear Walls and Diaphragms Why Buildings Don’t Fall Over. Presenters: David W. Boehm, P.E. Gary Sweeny, P.E. Plan View. Assume seismic load is also 200 plf. l = 120’. 200 x 40 2 x 120. v ew = w ew x b. =. =. 33 #/FT. 2 x l. 200 x 120 2 x 40.

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Engineering with Wood

Shear Walls and Diaphragms

Why Buildings Don’t Fall Over

Presenters: David W. Boehm, P.E.

Gary Sweeny, P.E.

l = 120’

200 x 402 x 120

vew = wew x b

=

=

33 #/FT

2 x l

200 x 120 2 x 40

vns = wns x l

=

=

300 #/FT

veu

wew=

200#/FT

b=40’

vns

wns = 200#/FT

DIAPHRAGM UNIT SHEARS

2 x b

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

Case 3 v = 33 #/FT

Unblocked 6” max spacing at panel edges

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 ft

Assume allowable ft = 1150 psi

Area required = 9,000# = 7.8 in2

1150 psi

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

40’

T

C

Shear Walls

• North wall v=33 #/ft

• Nominal nailing required

• East and west walls v=300 #/ft

Vns = 300#/ft(40’) = 12000#

20’

Nailing pattern

7/16 sheathing2 x studs8d nails @ 4” required

Shear wall elevation

ΣMA = 0

0 = (12000 x 20) – (T x 40)

T = 6,000#