design of shear wall

Shear wall Copyright Prof Schierle 2012 1

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Shear walls• Resist lateral load in shear• Resist load only parallel to wall1 Wood studs with plywood2 Metal studs with plywood3 Reinforced Concrete wall4 Reinforced CMU wall5 Un-reinforced brick wall

(not allowed in seismic areas)6 Reinforced 2-wythe brick wall7 Party walls - double studs for 65 STC

(STC = Sound Transmission Coefficient)


1 X-direction concentric, Y-direction eccentric2 X-direction eccentric, Y-direction eccentric3 X-direction concentric, Y-direction concentric4 X-direction concentric, Y-direction concentric5 X-direction concentric, Y-direction concentric6 X-direction concentric, Y-direction concentricNote: 5 is better than 6 to resist torsion

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1 Shear walls resist only lateral load parallel to wall2 One-way shear walls collapse @ perpendicular load3 Eccentric shear walls cause torsion4 Concentric shear walls resist torsion

Note: Walls in 4 are offset but provide concentric support


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Plywood Shear WallPlywood must be nailed to wood framing to resist lateralshear of wind and seismic forces.

1 Plywood shear wall 2 Plywood shear wall with joint3 Max. shear wall aspect ratio 1:3.5

(Los Angeles 1:2)4 Plywood nail spacing

A Blocking to transfer shearB NailC Plywood sheathingD Hold-down (essential for short walls)E Nail spacing at panel edges (max. 5)F Nail spacing at other studs (max. 12”)


Four Town Homes, Beverly Hills • Four two-story units over concrete garage • 12” concrete slab on columns at 30’x30’ 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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Critical code issuesCheck BEFORE design• Occupancy• Type of Construction• Means of Egress• Fire Resistance• Area Limitations • Height Limitations


Limitations of:• Height H• Floor Area A


Parking• 30’ parking module• 15‘ wall module• 17’ wall module

if required by local ordinance• 12” two-way structural slab

designed for 300 psf allowswood walls at any location

• 12” slab also provides 3 hour fire rating required by code

• Concrete slab requires drop panelsor beams to resist shear stress

• 12” column for one-story parking and up to 3-story wood framing


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The project size required separation by 2-hr fire walls to comply with area limits


Terrace Homes Hermosa BeachDesign concept: to minimized grading and retaining walls: adapt building to site instead of adapting site to buildingA 14 x 22 ft module allows shear walls aligned verticallyEach two-story unit has two terraces for outdoor livingTerraces provide open space that allowed 33 units ata lot zoned for only 25 units by conventional planningRaised rear provides energy-saving cross ventilation


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Terrace Housing Taipei, ChinaArchitect: G G SchierleEngineer: China Sincere

200 housing unitsCombined shear wall & concrete frame:• Shear walls provide stiffness• Concrete frames provide ductility

for seismic safety


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Masonry shear reinforcing1 Wall reinforcing for seismic areas2 Max. bar spacing for required cross-section

areas (0.1% of wall cross-section area)

A Vertical 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 topE Bars at structural floors and roofF Spacing of bar sizes # 3 to # 7G Wall thickness

Rebar diameter Cross-sectionBar # (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


Horizontal Diaphragmstransfer lateral load to shear walls andother elements two ways 1 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

whereL = Lengths of wallsW = Total load supported by all walls


Flexible diaphragm / plywood wallsAssume: DL= 24 psf, Seismic factor CS = 0.15Flexible floor and roof diaphragms transfer loads proportional to thetributary 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 loadDL per level: W = 24 psf x 68’ x 150’/ 1000 W = 235 kDL at 3 Levels: 3 x 235 k W = 705 kBase shearV= CS W = 0.15 x705 V = 106 k

Area per level A= 68 x 150 A = 10,200 ft2

Shear per square foot per levelv0 = V0/A = 106 k x 1000 / 10200 v0 = 10.4 psfv1 = V1/A = 88 k x 1000 / 10200 v1 = 8.6 psfv2 = V2/A = 53 k x 1000 / 10200 v2 = 5.2 psf

Vertical force distributionFx= V wx hx / (wi hi)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’

wihI = 12690 k’

/ (wi hi) V = Fx Vx = Fx0.50 x 106 = 53 k V2 = 53 k 0.33 x 106 = 35 k + 53 V1 = 88 k0.17 x 106 = 18 k + 88 V0 = 106 k

V = 106 k

0.50 = 6345 / 126900.33 = 4230 / 126900.17 = 2115 / 12690


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 plfWall B = 10.4 psf (19’) 30’/24’= 247 plf use 7/16, 8d @ 6” = 255 plfWall C = 10.4 psf (34’) 15’/30’= 177 plf use 5/16, 6d @ 6” = 200 plf

Level 1 (v1 = 8.6 psf)Wall A = 8.6 psf (15’) 30’/12’= 323 plf use 15/32, 10d@6”= 340 plfWall B = 8.6 psf (19’) 30’/24’= 204 plf use 3/8, 8d @ 6” = 230 plf

Wall D = 10.4 psf (34’) 30’/30’= 354 plf use 3/8, 8d @ 4” = 360 plf

Wall C = 8.6 psf (34’) 15’/30’= 146 plf use 5/16, 6d @ 6” = 200 plfWall D = 8.6 psf (34’) 30’/30’= 292 plf use 5/16, 6d @ 4” = 300 plfLevel 2 (v2 = 5.2 psf) Wall A = 5.2 psf (15’) 30’/12’ =195 plf use 5/16, 6d @ 6” = 200 plfWall B = 5.2 psf (19’) 30’/24’ =124 plf use 5/16, 6d @ 6” = 200 plfWall C = 5.2 psf (34’) 15’/30’ = 89 plf use 5/16, 6d @ 6” = 200 plfWall D = 5.2 psf (34’) 30’/30’ =177 plf use 5/16, 6d @ 6” = 200 plf N

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Rigid diaphragm / masonry shear wallsAssume: Seismic factor CS =0.17Allowable masonry shear stress Fv = 85 psi Structural walls DLLength of walls 12 (30’)+14 (12)+8 (24) L = 720’DL = (720’) 8’(7.625”/12”) 120 pcf/[(68) 150] DL = 43 psfFloor/roof (12” slab) 150 psfMiscellaneous 7 psf DL DL = 200 psf

Dead loadDL / level: W = 200 psf x 68’ x 150’/ 1000 W = 2040 kDL 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 / (wi hi)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’

wihI = 110160 k’

/ (wi hi) V = Fx Vx = Fx0.50 x 1591 = 796 k 796 k 0.33 x 1591 = 525 k + 796 = 1321 k0.17 x 1591 = 270 k + 1321 = 1591 k

V = 1591 k

0.50 = 55080 / 1101600.33 = 36720 / 1101600.17 = 18360 / 110160


Rigid diaphragm / masonry shear wallsAssume 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 direction all walls are equal and, thus, have constant stiffness. In length direction relative wall stiffness is:

R = L x3 / L i

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B walls R = (12)3 / [(12)3 +(24)3] R= 0.11C 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 < 85Wall B = (1591) 1000 (0.11) / 15372 19 psi < 85Wall 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 < 85Wall C = (1321) 1000 (0.89) / 17568 67 psi < 85

Level 2 (V2 = 796 k) Wall A = (796) 1000 / 32940 24 psi < 85Wall 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 kLevel 0 V0 = 1591 k Base shear V = 1591 k


My projects at Google earth


Senior Housing, San Francisco - concrete shear walls


Roxbury Condos, Beverley Hills wood shear walls


Terrace Homes Hermosa Beach - wood shear walls


Park City Village 1981 (Olympic Village 2002) wood shear walls


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Park City Village


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design of shear wall

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