Shear Walls

• Types of Shear Walls• Shear Wall Stiffness• Maximum Reinforcement Requirements• Shear Strength• Example: simple buildingExample: simple building• Shear walls with openings

Shear Walls 1

Shear Walls

_____________ (unreinforced) shear wall (1.17.3.2.2): Unreinforced wall

_____________ (unreinforced) shear wall (1.17.3.2.3): Unreinforced wall ith i ti i f twith prescriptive reinforcement.

Within 16 in. of top of wall Structurally connected floor and roof levels

40db or 24 in.

≤ 16 i≤ 8 in.

≤ 8 i ≤ 16 in.

Control joint

≤ 8 in.

Corners and end of walls

Joint reinforcement at 16 in. o.c. or bond beams at 10 ft.

Reinforcement not required at openings smaller than 16 in in either vertical or horizontal direction

≤ 10 ft.

Shear Walls 2

Reinforcement not required at openings smaller than 16 in. in either vertical or horizontal direction

Reinforcement of at least 0.2 in2

Shear Walls

_____________ reinforced shear wall (1.17.3.2.4): Reinforced wall with prescriptive reinforcement of detailed plain shear wall.

______________ reinforced shear wall (1.17.3.2.5): Reinforced wall with prescriptive reinforcement of detailed plain shear wall. Spacing of vertical reinforcement reduced to 48 inches.

___________ reinforced shear wall (1.17.3.2.6):1. Maximum spacing of vertical and horizontal reinforcement is min{1/3 length of

wall, 1/3 height of wall, 48 in. [24 in. for masonry in other than running bond]}., g , [ y g ]}2. Minimum area of vertical reinforcement is 1/3 area of shear reinforcement3. Shear reinforcement anchored around vertical reinforcing with standard hook4. Sum of area of vertical and horizontal reinforcement shall be 0.002 times gross

cross sectional area of wallcross-sectional area of wall5. Minimum area of reinforcement in either direction shall be 0.0007 times gross

cross-sectional area of wall [0.0015 for horizontal reinforcement for masonry in other than running bond].

Shear Walls 3

Minimum Reinforcement of Special Shear W llWalls

Reinforce- 8 in. CMU wall 12 in. CMU wallReinforce-ment Ratio

8 in. CMU wall 12 in. CMU wall

As (in2/ft) Possibilities As (in2/ft) Possibilities

#4@32#4@24

0.0007 0.064#4@32

#5@560.098 #5@32

#6@48

#4@24 #4@16

0.0010 0.092

@

#5@32

#6@40

0.140

@

#5@24

#6@32

#4@16

0.0013 0.119

#4@16

#5@32

#6@40

0.181#5@16

#6@24

Shear Walls 4

Use specified dimensions, e.g. 7.625 in. for 8 in. CMU walls.

Special Shear Walls: Design

Minimum strength, strength design (1.17.3.2.6.1.1): Design shear strength, Vn, greater than 1.25 times nominal flexural strength M except V need not benominal flexural strength, Mn, except Vn need not be greater than 2.5Vu.

Shear Walls 5

Seismic Design Category

Seismic Design Category Allowed Shear Walls

A or B

C

D and higher

Response modification factor:Seismic design force divided by response modification factor, which accounts for

Shear Wall R

Ordinary plain 1.5

Detailed plain 2ductility and energy absorption. Detailed plain 2

Ordinary reinforced 2

Intermediate reinforced 3.5Knoxville, Tennessee

Special reinforced 5Knoxville, TennesseeCategory C for Use Group I and IICategory D for Use Group III (essential

facilities

Shear Walls 6

Shear Walls: Stiffness

d

hh/d < 0.25 0.25 < h/d < 4.0 h/d >4.0

_____ stiffness predominates

_______ stiffness predominates

Both shear and bending stiffness

are important

Shear Walls 7

Coupled Shear Walls

Sh W ll Shear Wall

Shear Walls 8

________ Shear Wall ___________ Shear Wall

Shear Wall Stiffness3.1.5.1 Deflection calculations shall be based on cracked section properties. Assumed properties shall not exceed half of gross section properties, unless a cracked-section analysis is performed.

Cantilever wall

h = height of wallA =shear area; 5/6A for a

2

hh

Etkcant

cantAv shear area; 5/6A for a

rectangleG = shear modulus; given

as 0.4E (1.8.2.2.2)hi k f ll

34

Lh

Lh

Fi d ll t = thickness of wallL = length of wall

Fixed wall (fixed against rotation at top)

fixed

2

hh

Etk fixed

3

Lh

Lh

Real wall is probably between two cases; diaphragm provides some

Shear Walls 9

Real wall is probably between two cases; diaphragm provides some rotational restraint, but not full fixity.

T- or L- Shaped Shear WallsSection 1.9.4 Wall intersections designed either to:a) ____________________: b) ____________________

Connection that transfers shear: (must be in running bond)a) Fifty percent of masonry units interlockb) Steel connectors at max 4ft.c) Intersecting bond beams at max 4 ft Reinforcing of at least 0 1in2 per foot of wallc) Intersecting bond beams at max 4 ft. Reinforcing of at least 0.1in per foot of wall

Metal lath or wire

2-#4’sscreen to support grout

1/4in. x 11/2in. x 28in. with 2in. long 90 deg bends at each end to form U or Z shape

Shear Walls 11

Effective Flange Width (1.9.4.2.3)

Effective flange width on either side of web shall be smaller of actual flange width, distance to a movement joint, or:

• Flange in compression: 6t• Flange in tension:Flange in tension:

• Unreinforced masonry: 6t• Reinforced masonry: 0.75 times floor-to-floor wall

heightheight

Shear Walls 12

Maximum reinforcing (3.3.3.5)

No limits on maximum reinforcing for following case (3.3.3.5.4):

5.1 and 1 RdV

MuSquat walls, not designed for ductility

dV vu

In other cases, can design by either providing boundary elements or limiting reinforcement.

Boundary element design (3.3.6.5):More difficult with masonry than concreteBoundary elements not required if:

gA1.0 mu fP geometrically symmetrical sections

gA05.0 mu fP geometrically unsymmetrical sections

ANDAND

1u

lV

MOR 3u

lV

Mmnu fAV 3 AND

Shear Walls 13

wulV wulV

Maximum reinforcing (3.3.3.5)

Reinforcement limits: Calculated usingMaximum stress in steel of fyAxial forces taken from load combination D+0.75L+0.525QEAxial forces taken from load combination D 0.75L 0.525QE

Compression reinforcement, with or without lateral ties, permitted to beincluded for calculation of maximum flexural tensile reinforcement

Uniformly distributed reinforcement

muy

vymu

mum

v

s

f

dP

bf

d

A

64.0

ymu

yf

P

= 1.5 ordinary walls = 3 intermediate walls = 4 special walls

Compression steel with area equal to tension steel syymumuy

ymu

mum

s

Edd

f

bdP

bf

bd

A

,min

64.0

Shear Walls 14

d

Maximum reinforcing

Consider a wall with uniformly distributed steel:

y PTCC ssm

Strain mu

bdfC vymu

mumm

8.08.0

1

y

mu

fy 0.8f’m

y

y

y

yy

ymu

ysys AfT

2

1

Portion of steel Yielded Elastic

Steel in tension

in tension steel steel

mu

y

mu

ymu

ymu

musys AfC

2

1

Stress Steel in compression

A taken as total steel

mu

ymu

ymu

musy Af

5.0

Shear Walls 15

As taken as total steeldv is actual depth of masonry

ymu

ymusy Af

5.0

Maximum reinforcingPCCT mss

ymu

muysy

ymu

ymusy

ymu

yysyss AfAfAfCT

5.05.0

PCPbdfAfCT mvmu

mmuy

syss

8.08.0ymuymu

mu P

fb

640

muy

y

vymum

v

s

f

dfb

d

A

64.0

Shear Walls 16

ymu

y

Maximum reinforcingMaximum steel on a per foot length of wall basis given for 8 in. wall

0.025

0.03

0.035

2.5

3s=4y

0.05

0.06

4

4.5

5Clay f'm=4ksi

Clay f'm=2ksi

s=2y

0.01

0.015

0.02

0.025

As/(

bd

v)

1

1.5

2

As (

in2/f

t)

Clay f'm=4ksi

Clay f'm=2ksiCMU f'm=2ksi

CMU f'm=1.5ksi0.02

0.03

0.04

As/(

bd

v)

1 5

2

2.5

3

3.5

As (

in2/f

t)

CMU f'm=2ksi

0

0.005

0.01

0 0.05 0.1 0.15 0.2

0

0.5

0

0.01

0 0.05 0.1 0.15 0.2 0.25 0.3

0

0.5

1

1.5

CMU f'm=1.5ksi

[P/(bd)]/f'm [P/(bd)]/f'm

Shear Walls 17

Shear Strength (3.3.4.1.2)

M

_______nV 8.0

M /V d d t b t k 1 0umn

vu

um PfA

dV

MV 25.075.10.4

Mu/Vudv need not be taken > 1.0

Pu = axial load

vyv

s dfs

AV

5.0

Vertical reinforcement shall not be less than one-third horizontal reinforcement; reinforcement shall be uniformly distributed, max spacing of 8 ft (3.3.6.2)

Maximum Vn is:mnn fAV 6 25.0)/( vuu dVM

mnn fAV 4 0.1)/( vuu dVM

Interpolate for values of M /V d between 0 25 and 1 0

Shear Walls 18

Interpolate for values of Mu/Vudv between 0.25 and 1.0

mnvu

un fA

dV

MV

25

3

4

ExampleGiven: 10ft. high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, f’m=1500psi; 2-#5 each end; #5 at 48in. (one space will actually be 40 in.); superimposed dead load of 1kip/ft.R i d M i i d l d b d fl l it h t lRequired: Maximum wind load based on flexural capacity; shear steel required to achieve this capacity. Solution: Check capacity in flexure using 0.9D+1.0W load combination.

P 0 9(1k/ft 0 38k/ft) 1 24k/ft T t l 1 24k/ft(16ft) 19 88ki

Weight of wall: 38 psf(10ft) = 0.38 k/ft(Wall weight from lightweight units, grout at 48 in. o.c.)

Pu=0.9(1k/ft+0.38k/ft)=1.24k/ft Total: 1.24k/ft(16ft) = 19.88kips

Shear Walls 19

ExampleSolve for stresses, forces and moment in terms of c, depth to neutral axis.

T1 T2 T3 T4 C1C2

8 inc

52 in92 in

140 in188 in

Shear Walls 20

ExampleSolve for stresses, forces and moment in terms of c, depth to neutral axis.

T1 T2 T3 T4 C1C2

8 inc

52 in92 in

140 in188 in

• Use trial and error to find c such that Pn = 19.88 kips/0.9 = 22.09 kips

• Pn = C1 + C2 – T1 – T2 – T3 – T4

• c = 28.2 inches

• Sum moments about middle of wall (8 ft from end) to find Mn

• M = 1123 kip-ft; M = 0 9(1123 kip-ft) = 1010 kip-ft

Shear Walls 22

Mn 1123 kip ft; Mu 0.9(1123 kip ft) 1010 kip ft

• Wind = Mu/h = (1010 kip-ft)/(10ft) = 101 kips

ExampleCalculate net area, An, including grouted cells. Net area

Net area:

vuu dVM /

25.0/ vuu dVMmnn fAV 4Maximum Vn 0.1/ vuu dVMmnn fAV 6

vuu dV/

mnvuu

n fAdVM

V25.00.1

25.0/26

kipskipsVV nu 1.1066.1328.08.0 Maximum Vu

Shear Walls 23

Example

umnvu

um PfA

dV

MV 25.075.10.4

Pu = 0.9(1k/ft) = 0.9k/ftTotal: 0.9k/ft(16ft) = 14.4 kips

vu

kkkVVV mns 6.457.808.0/101

Use #5 bars in bond beams Determine spacing

vyv

vyv

s V

dfAsdf

s

AV

5.0 5.0

Use #5 bars in bond beams. Determine spacing.

sVs

Use 4 bond beams, spaced at 32 in. o.c. vertically.

Shear Walls 25

Actual construction has 15 courses: Bond beam in course 4, 8, 12, and 15 from bottom

Example: Maximum Reinforcing

3.3.3.5 No limit to reinforcement since Mu/(Vudv) < 1.

If we needed to check maximum reinforcing, there are at least two ways.g, y

Check on axial load:a) Set steel strain to limit and

find cSteel strain = 1.5y = 0.00310c 0 0025/(0 0025+0 00310)*192 85 7 infind c

b) Find axial load for this cc) If applied axial load is less

than axial load, OK

c = 0.0025/(0.0025+0.00310)*192 = 85.7 in.From spreadsheet, Pu = 181 kips OK

Find steel strain for given axial load:) Th h t i l d ( l ) P = 22 08 kips (Dead load)a) Through trial and error (or solver),

find c for given axial loadb) Determine steel strainc) If steel strain greater than limit, OK

P = 22.08 kips (Dead load)c = 28.46 in.s = (192-28.5)/28.5*0.0025 = 0.0143

= 6.93y OK

Shear Walls 27

) g , yNote: did not include compression steel which is allowed when checking max reinforcing

Example: Issues with Strength Design

If this were a special shear wall:

• Vn ≥ shear corresponding to 1.25Mn (1.17.3.2.6.1.1)

• 1.25Mn = 1.25(1123) = 1404k-ft

• Vn = Mn/h = 1404k-ft/10ft = 140.4kips

• Vn = 140.4k/0.8 = 175.4kipsVn 140.4k/0.8 175.4kips

• But, maximum Vn is 132.6 kips (76% of required).

Cannot build wall. Need to take out some flexural steel or fully grout to get greater shear capacity.

Shear Walls 28

Partially Grouted Shear Walls

Recent research has shown that TMS Code equations overestimate capacity of partially grouted shear walls.

T ibl difi tiTwo possible modifications:

• Only use face shell area, 2.5in(192in) = 480in2 (30% reduction)

• Multiply An by An/Ag, an empirical reduction (53% reduction)n n g

Shear Walls 29

ExampleGiven: Wall system constructed with 8 in. grouted CMU (Type S mortar).Required: Determine the shear distribution to each wall

S l ti U

12in

Solution: Use gross properties (will not make any difference for shear distribution)1 for shear distribution).

40in 40in64in

Elevation

40in 40in

6t=

48in

56in

ba c

Shear Walls 30

Plan

ExamplePier b

inE

inin

inin

inE

Lh

Lh

Etkb 286.0

364

1124

64112

62.7

3422

ininLL 6464

Piers a and c Determine stiffness from basic principles.

A

Find centroid from outer flange surface

y

I

bAA

Shear Walls 31

webv AA

Example

inE

Einin

inEin

GAPh

EIPh

PPk

v

ca 157.0

4.0305112

860003112

1

3 24

33,

v

Portion of shear load = stiffness of pier over the sum of the stiffnesses.

VVinE

Vk

V a 260157.0

VVinEinEinE

Vk

V aa 26.0

157.0286.0157.0

Vb = 0.48V

Deflection calculations:k = (0.600in)E = (0.600in)(1350ksi) = 810kip/in

Reduce by a factor of 2 to account for cracking: k = 405kip/inReduce by a factor of 2 to account for cracking: k = 405kip/in

Shear Walls 33

Example

Required: Design the system for a horizontal earthquake load of 35 kips, a dead load of 1200 lb/ft length of building, a live load of 1000 lb/ft length of building, and a vertical earthquake force of 0.2 times the dead load. Use a special reinforced shear wall (R>1.5).

Solution: Use strength design. Check deflection.

inink

k

k

P086.0

/405

35

000768.0112

086.0

in

in

hor

1302

h

112inh 1302

Shear Walls 34

Example

Wall b: Vub = 0.48Vu = 0.48(35k) = 16.8k

D = 1.2k/ft(_____________)(1ft/12in) + 0.075ksf(112in)(64in)(1ft2/144in2) = 14.1kips

Tributary width

Use load combination 0.9D+EMu = 16.8k(112in)(1ft/12in) = 156.8k-ft Pu = 0.9(14.1k) - 0.2(14.1k) = 9.87k

Tributary width

Minimum prescriptive reinforcement for special shear walls:Sum of area of vertical and horizontal reinforcement shall be 0.002 times

gross cross-sectional area of wallMinimum area of reinforcement in either direction shall be 0.0007 times gross

cross-sectional area of wall

Use prescriptive reinforcement of (0.002-0.0007)Ag = 0.0013Ag in vertical directionAs = 0.0013(64in)(7.625in) = 0.63in2

Shear Walls 35

Example: Flexural Steel

Use load combination 0.9D+E Mu = 156.8k-ft Pu = 9.87k

Try 2-#4 each end; 1-#4 middle: As=1.00in2y ; s

Middle bar is at 28 in. (8 cells, cannot place bar right in center)

From spreadsheet: Mn=150.5k-ft NG

N t d l t fi d lNote: used solver to find value

Try 1-#6 each end; 1-#6 middle: As=1.32in2 Mn=183.4k-ft OK

Try 2-#5 each end; 1-#5 middle: As=1.55in2 Mn=214.1k-ft OK

Use 2-#5 each end; 1-#5 middle: As=1.59in2

Shear Walls 36

Example: Maximum Reinforcing

Check maximum reinforcing:

Axial force from load combination D+0.75L+0.525QE

L 1k/ft(104in)(1ft/12in) 8 67kipL=1k/ft(104in)(1ft/12in) = 8.67kip

D+0.75L+0.525QE = 14.1k+0.75(8.67k)+0.525(0) = 20.60kip

For maximum reinforcing calculations, compression steel can be included even if t l t ll t dif not laterally supported

For #5 bars and P = 20.60kip, c = 7.77in. (from solver seek on spreadsheet)

ininkdd 77760yymus in

inine

kd

kdd 412.80168.00025.077.7

77.760

OK

Shear Walls 37

OK

Example: Shear 175.164

112

inV

inV

dV

M

u

u

vu

u Use Mu/Vudv = 1.0

Find Pu neglecting wall weightPu = 0.9(10.4k) - 0.2(10.4k) = 7.28k

PfAdVMV umnvuunm 25.0/75.10.4

kipsklbkpsiinin 3.4428.725.01000/11500)64)(625.7(25.2

kipsVn 3.44Ignore any shear steel: pnIgnore any shear steel:

kipskipsVn 5.353.448.0 OK

Shear Walls 38

Example: Shear

Mn = Mn/ = 214.1k-ft/0.9 = 237.9k-ft

V corresponding to Mn: [237.9k-ft/(112in)](12in/ft) = 25.5kips

1.25 times V: 1.25(25.5k) = 31.9kips < Vn=35.5kips OK

Minimum prescriptive horizontal reinforcement: 0.0007Ag

As = 0.0007(64in)(7.625in) = 0.34in2

Maximum spacing is 48 inMaximum spacing is 48 in.

Use 3 bond beams, 40in. o.c. (top spacing is only 32in.)

Wall b: Vertical bars: 2-#5 each end; 1-#5 middleHorizontal bars: #4 in bond beam at 40 in. o.c. (Total 3-#4)

Shear Walls 39

Example

Wall a: Vua = 0.26Vu = 0.26(35k) = 9.1kD = 1.2k/ft(20in+40in)(1ft/12in) + 0.075ksf(112in)(40in+48in)(1ft2/144in2) = 11.1kips

Use load combination 0.9D+EMu = 9.1k(112in)(1ft/12in) = 84.9k-ft Pu = 0.9(11.1k) - 0.2(11.1k) = 7.77k

Minimum prescriptive reinforcement: 0.0013Ag

As = 0.0013(88in)(7.625in) = 0.87in2

Design issues:Flange in tension: easy to get steel; problems with max

Flange in compression: hard to get steel; max is easySteel in both flange and web affect max

Need to also consider loads in perpendicular directionNeed to check shear at interface of flange and web

Shear Walls 40

ExampleUse load combination 0.9D+E Mu = 84.9k-ft Pu = 7.77k

Axial force for max from load combination D+0.75L+0.525QEL=1k/ft(60in)(1ft/12in) = 5 00kip

To be consistent with other wall try #5 bars

L=1k/ft(60in)(1ft/12in) = 5.00kipD+0.75L+0.525QE = 11.1k+0.75(5.00k)+0.525(0) = 14.85kip

To be consistent with other wall, try #5 barsTry 3 in the flange (either distributed as shown, or one in

the corner, and two in the jambTry 2 in the web jamb

Flange in tension: Mn=146.4k-ftMax reinforcing check: s=0.0101=4.86y

Note: with 2 #5 in flange M =104 8k ft but might need 3 bars for

OK

Note: with 2-#5 in flange, Mn=104.8k-ft, but might need 3 bars for perpendicular direction

Flange in compression: Mn=124.3k-ft OK

Shear Walls 41

Max reinforcing check: s=0.0417=20.2y

Example: Shear

18.240

112

inV

inV

dV

M

u

u

vu

u Use Mu/Vudv = 1.0

kipsklbkpsiininVm 6.272.425.01000/11500)40)(625.7(25.2

kipsVn 6.27Ignore any shear steel: kipskipsVn 1.226.278.0 OK

Mn = Mn/ = 146.4k-ft/0.9 = 162.7k-ftV corresponding to Mn: [162.7k-ft/(112in)](12in/ft) = 17.4kips1 25 times V: 1 25(17 4k) = 21 8kips < V =22 1kips OK1.25 times V: 1.25(17.4k) = 21.8kips < Vn=22.1kips

Minimum prescriptive horizontal reinforcement: 0.0007Ag

Use 3 bond beams 40in o c (top spacing is only 32in )

OK

Walls a,c: Vertical bars: 5-#5 as shown in previous slideHori ontal bars #4 in bond beam at 40 in o c (Total 3 #4)

Use 3 bond beams, 40in. o.c. (top spacing is only 32in.)

Shear Walls 42

Horizontal bars: #4 in bond beam at 40 in. o.c. (Total 3-#4)

ExampleSection 1.9.4.2.4 requires shear to be checked at interface. Check intersection of flange and web in walls a and c.

kipsksiinfAV ysu 8.556031.03 2 Approximate shear force is tension force in flexural steel.

Conservatively take Mu/Vudv=1.0, and Pu=0.

kipslbkpsiinin

PfAVdMV umnvm

4.741000/11500)112)(625.7(0.175.10.4

25.0/75.10.4

Conservatively take Mu/Vudv 1.0, and Pu 0.

kipskipsVn 5.594.748.0 OK

pp))((

Shear Walls 43

Example: Drag Struts

Distributed shear force: ftkft

in

in

k

L

Vu /875.11

12

224

35

Look at top of wall:

1.88k/ft 1.88k/ft 1.88k/ft 1.88k/ft1.88k/ft

3 33f 3 33f33f3 33f 3 33f

Drag strut Drag strut9.1k 9.1k16.8k

1.88k/ft 88 / 88 / 88 /88 /

2.8k 3.4k 2.8k3.4k

3.33ft 3.33ft5.33ft3.33ft 3.33ft

Shear Walls 44

Shear Walls: Building Layout

1. All elements either need to be isolated, or will participate in carrying the load

2. Elements that participate in carrying the load need to be properly2. Elements that participate in carrying the load need to be properly detailed for seismic requirements

3. Most shear walls will have openings4. Can design only a portion to carry shear load, but need to detail rest of g y p y

structure

Shear Walls 45

Shear Walls with Openings

Shear walls with openings will be composed of solid wall portions, and portions with piers between openings.

Pi• Requirements only in strength design• Length ≥ 3 x thickness; width ≤ 6 x thickness • Height < 5 x length

Piers:

• Height < 5 x length • Factored axial compression ≤ 0.3Anf’m (3.3.4.3.1)

• One bar in each end cell (3.3.4.3.2)

• Minimum reinforcement is 0 0007bd (3 3 4 3 2)Minimum reinforcement is 0.0007bd (3.3.4.3.2)

Shear Walls 46

Shear Walls with Openings1. Divide wall into piers.2. Find flexibility of each pier3. Stiffness is reciprocal of flexibility4 Di t ib t l d di t tiff

Top spandrelh t

tt h

hk

h

4. Distribute load according to stiffness

1

5

12

2

322

6

3

EA

h

kkkk

kkkk

EI

hf

bbtt

bbttBottom spandrelh b

bb h

hk

bbtt

bbtt

btt kkkk

kkVhM

2

1If hb = 0

3

bbtt

tbb kkkk

kkVhM

2

1

1

5

12

12

2

6

3

EA

h

k

k

EI

hf

t

t

k 1 k

12 t

tt k

kVhM

12

1

t

tb k

kVhMAs the top spandrel decreases in height,

the top approaches a fixed condition against rotation.

Shear Walls 47

Qamaruddin, M., Al-Oraimi, S., and Hago, A.W. (1996). “Mathematical model for lateral stiffness of shear walls with openings.” Proceedings, Seventh North American Masonry Conference, 605-617.

Example - Shear Walls with Openings

ft

Given: Required:A. Stiffness of wallB. Forces in each pier under

4ft

4f 10 kip horizontal load

4

A B C

8ft

2ft 2ft 2ft4ft 6ft

Shear Walls 48

Example - Shear Walls with OpeningsSample calculations: Pier Bh = 4ft ht = 4ft hb = 8ft kt = h/ht = 4ft/4ft = .0 kb = h/hb = 4ft/8ft = 0.5

3tL 12

tLI tfttLA 2

123223 hkkkkh

1

5

12

2

322

6

3

EA

h

kkkk

kkkk

EI

hf

bbtt

bbtt

EtEtf

k 0210.0/6.47

11

Shear Walls 49

Example - Shear Walls with Openings

Pier h (ft) ht (ft) hb (ft) kt kb I (ft3) A (ft) f s f k

A 12 4 0 3 inf 0.667t 2t 308.4 18 326.4/Et 0.0031Et

B 4 4 8 1 0.5 0.667t 2t 41.6 6 47.6/Et 0.0210Et

C 4 4 8 1 0.5 0.667t 2t 41.6 6 47.6/Et 0.0210Et

f is flexural deformation; s is shear deformation

Total stiffness is 0.0451EtAverage stiffness from finite element analysis 0.0440Et Solid wall stiffness 0 1428Et (free at top); 0 25Et (fixed at top)Solid wall stiffness 0.1428Et (free at top); 0.25Et (fixed at top)

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Example - Shear Walls with OpeningsSample calculations: Pier B kipsk

Et

EtV

k

kV b

b 66.4100451.0

0210.0

kk 1

bbtt

btt kkkk

kkVhM

2

1

tb kk

VhM1

bbtt

tbb kkkk

VhM2

Pier V (kips) Mt (k-ft) Mb (k-ft)

A 0.68 3.50 4.66

B 4.66 11.19 7.46

C 4.66 11.19 7.46

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Example - Shear Walls with Openings

To find axial force in piers, look at FBD of top 4ft of wall.

10k

Moment direction on piers

4ft

4.66k 4.66k0.68k

2ft 2ft 2ft4ft 6ft

3.50k-ft 11.19k-ft 11.19k-ft

This 65.9k-ft moment is carried by axial forces in the piers. Axial forces are d t i d f M /I h h i i id d t t ddetermined from Mc/I, where each pier is considered a concentrated area.Find centroid from left end.

ii Ayy

Shear Walls 54

iA

y

Example - Shear Walls with Openings

The axial force in pier A is:

FBD’s of piers:11.19k-ft

4.90k11.19k-ft

0.45k3.50k-ft

4.45k

FBD s of piers:

B C

4.66k4.66k0.68k

A 4.66k7.46k-ft0.45k

4.66k7.46k-ft4.90k

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0.68k4.66k-ft4.45k

Example - Shear Walls with OpeningsFBD’s of entire wall to obtain forces in bottom right spandrel:

10kFxkkFx 068.010

kFx

x

32.9

FkF 0454

A B C

kFy

FykFy

45.4

045.4

x

y

ftkMz

Mzftkftk

ftkftkM z

8110

0)11(45.4)1(45.4

)66.4()16(10

0.68k4 66k ft

Fx=9.32kM 110 8k ft

ftkMz 8.110

Z

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4.66k-ft4.45k Mz=110.8k-ftFy=4.45k