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

Plan View

Assume seismic load is also 200 plf

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

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

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

Diaphragm layout

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#

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#

WALL DESIGN WITH OPENINGS

DRAG STRUT / COLLECTOR FORCE

OVERTURNING FORCE