Engineering with Wood - CEMS Home : University of Vermont

```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’
veu
wew=
b=40’
200#/FT
vns
wns = 200#/FT
DIAPHRAGM UNIT SHEARS
vew = wew x b
=
200 x 40
2 x 120
=
33 #/FT
=
200 x 120
2 x 40
=
300 #/FT
2xl
vns = wns x l
2 xb
PANEL LAYOUT AND FASTENER SCHEDULE
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
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 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
Diaphragm layout
Shear Walls

North wall


v=33 #/ft
Nominal nailing required
East and west walls v=300 #/ft
Vns = 300#/ft(40’) = 12000#
20’
40’
T
Nailing pattern
7/16 sheathing
2 x studs
8d nails @ 4” required
A
C
Shear wall elevation
Tiedown Force
ΣMA = 0
0 = (12000 x 20) – (T x 40)
T = 6,000#
WALL DESIGN WITH OPENINGS
DRAG STRUT / COLLECTOR FORCE
OVERTURNING FORCE
```