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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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1 X-direction concentric, Y-direction eccentric

2 X-direction eccentric, Y-direction eccentric

3 X-direction concentric, Y-direction concentric




Note: 5 is better than 6 to resist torsion

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Note:eccentricshearwallscausetorsiona

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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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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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6/33Shear wall Prof Schierle 6

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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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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Stabilize

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Design stable shear walls

12 shear wall

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