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Shear wall Prof Schierle 2
Shear walls
Resist lateral load inshear
Resist load only parallel to wall1 Wood studs with plywood
2 Metal studs with plywood
3 Reinforced Concrete wall
4 Reinforced CMU wall5 Un-reinforced brick wall
(not allowed in seismic areas)
6 Reinforced 2-wythe brick wall
7 Party walls - double studs for 65 STC(STC=Sound TransmissionCoefficient)
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Shear wall Prof Schierle 3
1 X-direction concentric, Y-direction eccentric
2 X-direction eccentric, Y-direction eccentric
3 X-direction concentric, Y-direction concentric
4 X-direction concentric, Y-direction concentric
5 X-direction concentric, Y-direction concentric
6 X-direction concentric, Y-direction concentric
Note: 5 is better than 6 to resist torsion
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Note:eccentricshearwallscausetorsiona
ndshouldbe
avoided
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Note:shearwallsresistlateralload
onlyparalleltowall
1 Shear walls resist only lateral load parallel to wall
2 One-way shear walls collapse @ perpendicular load
3 Eccentric shear walls cause torsion
4 Concentric shear walls resist torsion
Note: Walls in 4 are offset but provide concentric support
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t
Plywood Shear WallPlywood must be nailed to wood framing to resist lateral
shear of wind and seismic forces.
1 Plywood shear wall
2 Plywood shear wall with joint
3 Max. shear wall aspect ratio 1:3.5
(Los Angeles 1:2)4 Plywood nail spacing
A Blocking to transfer shear
B Nail
C Plywood sheathingD Hold-down (essential for short walls)
E Nai l spacing at panel edges (max. 5)
F Nail spacing at other studs (max. 12 )
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Shear wall Prof Schierle 5
Four Town Homes, Beverly Hills Four two-story units over concrete garage 12 concrete slab on columns at 30x30 provides 3-hour fire
separation between garage and residential units above
Concrete slab designed for of 300 psf allows wood framing
anywhere regardless of column locations Double stud party walls for 65 STC sound rating
(STC = Sound Transmission Coefficient = sound rating)
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Rear
Front
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Limitations of:
Height H
Floor Area A
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TerraceH
ousing
Hermos
a
Beac
h
The project size required separation by2-hr f ire walls to comply with area limits
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Terrace Homes Hermosa BeachDesign concept: to minimized grading and retaining walls:
adapt building to site instead of adapting site to building
A 14 x 22 ft module allows shear walls al igned vertically
Each two-story unit has two terraces for outdoor l iving
Terraces provide open space that allowed 33 units at
a lot zoned for only 25 units by conventional planning
Raised rear provides energy-saving cross ventilation
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Height limit 3-story from grade
Rais
edrearprovid
es
crossventilation
Sla
ntedpartyw
allsscreeno
ceanclare
Sunshades,planters,andpartywallsprovi
deprivacy
Length shear wallsWidth shear walls
Communityspace
Terracing
provides
Residentialscale
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Terrace Housing Taipei, ChinaArchitect: G G Schierle
Engineer: China Sincere
200 housing units
Combined shear wall & concrete frame:
Shear walls provide sti ffness
Concrete frames provide ductil ity
for seismic safety
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Rei
nforcedbrickmas
onry
CMU
walls(C
oncreteMasonryUnits)
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Reinforcedconcretewal
l
Rei
nforced
CMUw
all
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Masonry shear reinforcing1 Wall reinforcing for seismic areas
2 Max. bar spacing for required cross-section
areas (0.1% of wall cross-section area)
A Vert ical bars
(max. 4 ft or 6 x wall thickness
B Horizontal bars
(max. 4 ft in seismic areas)C Bars around wall openings, extending min. 24
or 40 bar diameters beyond opening
D Horizontal bars @ wall base and top
E Bars at structural floors and roof
F Spacing of bar sizes # 3 to # 7
G Wall thickness
Rebar diameter Cross-section
Bar # (in) (in2)#3 3/8 0.11
#4 4/8 0.20
#5 5/8 0.31
#6 6/8 0.44
#7 7/8 0.60
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Horizontal Diaphragms
transfer lateral load to shear walls and
other elements two ways1 Flexible diaphragm (wood)
transfers in proportion to tributary area.
Wall reactions are:
R = w (tributary area supported by wall)
w = uniform load
2 Rigid diaphragm (concrete & steel)
transfers in proportion to wall stiffness.Reactions for walls of equal material:
R1 = WL13 / L3 (L3= L13+L23+L33)
R2 = WL23
/ L3
R3 = WL33 / L3
where
L = Lengths of walls
W = Total load supported by all walls
Fl ibl di h / l d ll
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Flexible diaphragm / plywood wallsAssume: DL= 24 psf , Seismic factor CS = 0.15
Flexible floor and roof diaphragms transfer loads proportional to the
tributary area supported by walls. This may be computed as follows:
Unit shear = shear per level / floor area per level Shear per wall = unit shear x tributary area supported by wall
Shear per foot = shear per wall / wall length
Dead load
DL per level: W = 24 psf x 68 x 150/ 1000 W = 235 k
DL at 3 Levels: 3 x 235 k W = 705 kBase shear
V= CS W = 0.15 x705 V = 106 k
Area per level
A= 68 x 150 A = 10,200 f t2
Shear per square foot per level
v0 = V0/A = 106 k x 1000 / 10200 v0 = 10.4 psf
v1 = V1/A = 88 k x 1000 / 10200 v1 = 8.6 psfv2 = V2/A = 53 k x 1000 / 10200 v2 = 5.2 psf
Vertical force distribution
Fx= V wx hx / (w i h i)
Level wx hx = wx hxLevel 2: 235 k x 27 = 6345 k
Level 1: 235 k x 18 = 4230 k
Level 0: 235 k x 9 = 2115 k
w ih I = 12690 k
/ (w i h i) V = Fx Vx = Fx0.50 x 106 = 53 k V2 = 53 k
0.33 x 106 = 35 k + 53 V1 = 88 k
0.17 x 106 = 18 k + 88 V0 = 106 k
V = 106 k
0.50 = 6345 / 12690
0.33 = 4230 / 12690
0.17 = 2115 / 12690
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Wall design (Use Structural I plywood)
Level 0 (v0 = 10.4 psf)
Wall A = 10.4 psf (15) 30/12= 390 plf use 5/16, 6d @ 3 = 390 plf
Wall B = 10.4 psf (19) 30/24= 247 plf use 7/16, 8d @ 6 = 255 pl f
Wall C = 10.4 psf (34) 15/30= 177 plf use 5/16, 6d @ 6 = 200 pl f
Level 1 (v1 = 8.6 psf)
Wall A = 8.6 psf (15) 30/12= 323 plf use 15/32, 10d@6 = 340 plf
Wall B = 8.6 psf (19) 30/24= 204 pl f use 3/8, 8d @ 6 = 230 plf
Wall D = 10.4 psf (34) 30/30= 354 pl f use 3/8, 8d @ 4 = 360 plf
Wall C = 8.6 psf (34) 15/30= 146 plf use 5/16, 6d @ 6 = 200 plf
Wall D = 8.6 psf (34) 30/30= 292 plf use 5/16, 6d @ 4 = 300 plf
Level 2 (v2 = 5.2 psf)
Wall A = 5.2 psf (15) 30/12 =195 plf use 5/16, 6d @ 6 = 200 pl f
Wall B = 5.2 psf (19) 30/24 =124 plf use 5/16, 6d @ 6 = 200 pl fWall C = 5.2 psf (34) 15/30 = 89 plf use 5/16, 6d @ 6 = 200 pl f
Wall D = 5.2 psf (34) 30/30 =177 plf use 5/16, 6d @ 6 = 200 pl f Note:
tosimplify,se
lectonlytwow
allt
ypes
Ri id di h / h ll
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Rigid diaphragm / masonry shear wallsAssume:
Seismic factor CS =0.17
Al lowable masonry shear st ress Fv = 85 ps i
Structural walls DL
Length of walls 12 (30)+14 (12)+8 (24) L = 720
DL = (720) 8(7.625 /12 ) 120 pcf/[(68) 150] DL = 43 psf
Floor/roof (12 slab) 150 psf
Miscellaneous 7 psf
DL
DL = 200 psf
Dead load
DL / level: W = 200 psf x 68 x 150/ 1000 W = 2040 k
DL at 3 Levels: W = 3 x 2040 k W = 6120 k
Base shear
(CS times 1.5 for ASD masonry shear per IBC 2106.5.1)
V=1.5 CS W
V = 1.5 x 0.17 W = 0.26 x 6120 V = 1591 k
Vertical force distributionFx= (V - Ft ) wx hx / (w i h i)
Level wx hx = wx hxLevel 2: 2040 k x 27 = 55080 k
Level 1: 2040 k x 18 = 36720 k
Level 0: 2040 k x 9 = 18360 k
w ih I = 110160 k
/ (w i h i) V = Fx Vx = Fx0.50 x 1591 = 796 k 796 k
0.33 x 1591 = 525 k + 796 = 1321 k
0.17 x 1591 = 270 k + 1321 = 1591 k
V = 1591 k
0.50 = 55080 / 110160
0.33 = 36720 / 110160
0.17 = 18360 / 110160
Ri id di h / h ll
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Rigid diaphragm / masonry shear walls
Assume allowable masonry shear stress Fv = 85 psiRigid diaphragms resist lateral load in proportion to wall stiffness.
For walls of constant height and material, relative stiffness is constant.
In width d irection all walls are equal and, thus, have constant stiffness.In length direction relative wall stiffness is:
R = L x3 / L i
3
B walls R = (12)3 / [(12)3 +(24)3] R= 0.11
C walls R = (24)3 / [(12)3 +(24)3] R= 0.89
Wall cross section areas:
A walls = 12(30)12 (7.625 ) A = 32940 in2
B walls = 14(12)12 (7.625 ) B = 15372 in2
C walls = 8(24)12 (7.625 ) C = 17568 in2
Level 0 (V0 = 1591 k)Wall A = (1591) 1000 / 32940 48 psi < 85
Wall B = (1591) 1000 (0.11) / 15372 19 psi < 85
Wall C = (1591) 1000 (0.89) / 17568 81 psi < 85
Level 1 (V1 = 1321 k)
Wall A = (1321) 1000 / 32940 40 psi < 85Wall B = (1321) 1000 (0.11) / 15372 10 psi < 85
Wall C = (1321) 1000 (0.89) / 17568 67 psi < 85
Level 2 (V2 = 796 k)
Wall A = (796) 1000 / 32940 24 psi < 85
Wall B = (796) 1000 (0.11) / 15372 6 psi < 85Wall C = (796) 1000 (0.89) / 17568 40 psi < 85
From last slide:
Level 2 V2 = 796 k
Level 1 V1 = 1321 k
Level 0 V0 = 1591 k
Base shear V = 1591 k
My projects at Google earth
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My projects at Google earth
Senior Housing San Francisco concrete shear walls
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Senior Housing San Francisco - concrete shear walls
St f d U i it
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Stanford University
Escondido Village
Student Housing
Concrete shear walls
Roxbury Condos Beverley Hills wood shear walls
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Roxbury Condos, Beverley Hills wood shear walls
Terrace Homes Hermosa Beach wood shear walls
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Terrace Homes Hermosa Beach - wood shear walls
Terrace Homes Hermosa Beach wood shear walls
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Terrace Homes Hermosa Beach - wood shear walls
Park City Village 1981 (Olympic Village 2002) wood shear walls
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Park City Village 1981 (Olympic Village 2002) wood shear walls
Park City Village 1981 (Olympic Village 2002) wood shear walls
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Park City Village 1981 (Olympic Village 2002) wood shear walls
2
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Level ski access
ParkC
ityV
illageak
aOlymp
icV
illag
e2002
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Park City Village
Olympic Village2002
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S
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rwa
ll
Stabilize
N
arroww
al
ls
Design stable shear walls
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