shear walls - engineering fundamentals...

Shear Walls 1

Shear Walls

• Load Distribution to Shear Walls• Shear wall stiffness• Shear walls with openings• Diaphragm types

• Types of Masonry Shear Walls• Maximum Reinforcement Requirements• Shear Strength• Example: simple building

Shear Walls 2

Shear Walls: Stiffness

_____ stiffness predominates

_______ stiffness predominates

Both shear and bending stiffness

are important

d

h

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

__________

___________

Lateral Force Resisting System

Shear Walls 3

Shear Wall Stiffness9.1.5.2 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

cant

h = height of wallAv =shear area; (5/6)A for

a rectangleEv = G = shear modulus

(modulus of rigidity); given as 0.4Em(4.2.2.2.2)

Ev = ___________, where ν is Poisson’s ratio

t = thickness of wallL = length of wall

342

Lh

Lh

tEk mcant

Fixed wall (fixed against rotation at top)

fixed

32

Lh

Lh

tEk mfixed

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

Shear Walls 5

T- or L- Shaped Shear WallsSection 5.1.1 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 wall

2-#4’sMetal lath or wire screen 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 6

Effective Flange Width (5.1.1.2.3)

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:

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

height

Shear Walls 7

Example: Flanged Shear Wall

40in.

112i

n.6t

=48i

n.56

in.

Elevation

Plan

Given: Fully grouted shear wallRequired: Stiffness of wallSolution: Determine stiffness from basic principles.

A

Find centroid from outer flange surface in

inininininininy 16.11

671204062.781.34862.7

2

I

vA

Shear Walls 9

Example: Flanged Shear Wall

m

mmvvnm

Ein

Einin

inEin

EAPh

IEPh

PPk 157.0

4.0305112

860003112

1

3 2433

Shear Walls 10

Shear Walls with Openings

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.

1. Divide wall into piers.2. Find flexibility of each pier3. Stiffness is reciprocal of flexibility4. Distribute load according to stiffness

1

512

2322

6

3

EAh

kkkkkkkk

EIhf

bbtt

bbtt

bbtt

btt kkkk

kkVhM21

bbtt

tbb kkkk

kkVhM21

Bottom spandrel

Top spandrel

h bh t

tt h

hk

bb h

hk

If hb = 0

1

512

122

6

3

EAh

kk

EIhf

t

t

12 t

tt k

kVhM

121

t

tb k

kVhMAs the top spandrel decreases in height, the top approaches a fixed condition against rotation.

h

Shear Walls 11

Example - Shear Walls with Openings

2ft 2ft 2ft4ft 6ft

4ft

4ft

8ft

Given:

A B C

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

10 kip horizontal load

Shear Walls 12

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

12

3tLI tfttLA 2

1

512

2322

6

3

EAh

kkkkkkkk

EIhf

bbtt

bbtt

EtEtf

k 0210.0/6.47

11

Shear Walls 14

Example - Shear Walls with OpeningsPier 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 0.0031EtB 4 4 8 1 0.5 0.667t 2t 41.6 6 47.6/Et 0.0210EtC 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)

Shear Walls 15

Example - Shear Walls with OpeningsSample calculations: Pier B kipsk

EtEtV

kkV bb 66.4100451.0

0210.0

bbtt

btt kkkk

kkVhM21

bbtt

tbb kkkk

kkVhM21

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

A 0.68 3.50 4.66B 4.66 11.19 7.46C 4.66 11.19 7.46

Shear Walls 17


2ft 2ft 2ft4ft 6ft

4ft

To find axial force in piers, look at FBD of top 4ft of wall.10k

4.66k 4.66k0.68k3.50k-ft 11.19k-ft 11.19k-ft

Moment direction on piers

This 65.9k-ft moment is carried by axial forces in the piers. Axial forces are determined from Mc/I, where each pier is considered a concentrated area.Find centroid from left end.

i

ii

AAy

y

Shear Walls 19


The axial force in pier A is:

FBD’s of piers:

A

B C

0.68k4.66k-ft4.45k

4.66k7.46k-ft0.45k

4.66k7.46k-ft4.90k

4.66k11.19k-ft

4.90k

4.66k11.19k-ft

0.45k

0.68k3.50k-ft

4.45k

Shear Walls 21

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

0.68k4.66k-ft4.45k

A B C

10k

Fx=9.32kMz=110.8k-ft

Fy=4.45k

xy

kFxFxkkFx

32.9068.010

kFyFykFy

45.4

045.4

ftkMzMzftkftk

ftkftkM z

8.1100)11(45.4)1(45.4)66.4()16(10

Z

Shear Walls 22

Shear Walls with Openings

• Requirements only in strength design• Length ≥ 3 x thickness; width ≤ 6 x thickness • Height < 5 x length • Factored axial compression ≤ 0.3Anf’m (9.3.4.3.1)• One bar in each end cell (9.3.4.3.2)• Minimum reinforcement is 0.0007bd (9.3.4.3.2)

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

Piers:

Shear Walls 23

Shear Walls: Building Layout1. All elements either need to be isolated, or will participate in carrying

the load2. 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

structure

Shear Walls 24

Coupled Shear Walls

________ Shear Wall ___________ Shear Wall

Shear Walls 25

Diaphragms

• Diaphragm: _____________ system that transmits ____________ forces to the vertical elements of the lateral load resisting system.

• Diaphragm classification:• ______________: distribution of shear force is based on

tributary ________ (wind) or tributary _______ (earthquake)• ____________: distribution of shear force is based on

relative ______________.

__________

___________


Typical classifications:__________: Precast planks without topping, metal deck without concrete, plywood sheathing_________: Cast-in-place concrete, precast concrete with concrete topping, metal deck with concrete

Shear Walls 26

Rigid Diaphragms

Direct Shear:

i

iv RR

RRVF

Torsional Shear: 2ii

iitt dRR

dRRVeF

V = total shear forceRRi = relative rigidity of lateral force resisting element idi = distance from center of stiffnesse = eccentricity of load from center of stiffness

Shear Walls 27

Diaphragms: Example

Given: The structure shown is subjected to a 0.2 kip/ft horizontal force. Relative rigidities are given, where the relative rigidity is a normalized stiffness.Required: Distribution of force assuming:• flexible diaphragm• rigid diaphragm.

RR

= 4

RR

= 5

RR

= 1

50 ft 50 ft

0.2 kip/ft

PLAN VIEW

Shear Walls 28

Flexible Diaphragms: Example

Solution: Flexible diaphragm – windDistribute based on tributary area

For seismic, the diaphragm load would be distributed the same (assuming a uniform mass distribution), but when wall weights were added in, the forces could be different.

RR

= 4

RR

= 5

RR

= 1

50 ft 50 ft

PLAN VIEW

Shear Walls 29

Rigid Diaphragms: Example

Solution: Rigid diaphragm

RR

= 4

RR

= 5

RR

= 1

50 ft 50 ft

Wal

l 1

Wal

l 2

Wal

l 3

x

Wall x RR x(RR)123

050

100

451

0250100

Total 10 350

Center of stiffness = 350/10 = 35 ft

Wall RR d (ft) RR(d) RR(d2) Fv Ft Ftotal123

451

-351565

-1407565

490011254225

8102

-4.12.21.9

3.912.23.9

Total 10 350 10250 20 0.0 20

Shear Walls 30

Diaphragms Design ForcesSolution: Forces are shown for a rigid diaphragm.

The moment is generally taken through chord forces, which are simply the moment divided by the width of the diaphragm. In masonry structures, the chord forces are often take by bond beams.

50 ft 50 ft

0.2 kip/ft

3.9 k 12.2 k 3.9 k

V (k)

M (k-ft)

3.9

-3.9

6.1

-6.1

38

19.5 ft

38

-55

Shear Walls 31

Drag Struts and Collectors• Shear forces: generally considered

to be uniformly distributed across the width of the diaphragm.

• Drag struts and collectors: transfer load from the diaphragm to the lateral force resisting system.

w

PLAN VIEW

v

vL/2 vL/2

WEST WALL ELEVATION

v

vL

EAST WALL ELEVATION

L/3 L/3 L/2

Shear Walls 32

Diaphragm BehaviorThree lateral force resisting systems: Length=20 ft; Height=14 ft

W24x68

W14

x68

Moment Resisting Framek = 83.2 kip/in

W16x40

W14

x68

Braced Framek = 296 kip/in

L 4x4x5/16

W16x40

Masonry Shear Wallk = 1470 kip/in

L 4x4x5/16

E = 1800 ksiFace shell bedding

End cells fully grouted

Lateral force resisting systems at 24 ft o.c. Diaphragm assumed to be concrete slab, E=3120ksi, ν=0.17, variable thickness, load of 1 kip/ft.

Diaphragm


24 ft 24 ft

Shear Walls 33

Diaphragm Behavior

Shear Walls 34

Shear Walls_____________ (unreinforced) shear wall (7.3.2.2): Unreinforced wall

_____________ (unreinforced) shear wall (7.3.2.3): Unreinforced wall with prescriptive reinforcement.

40db or 24 in.

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

≤ 16 in.

Control joint

≤ 8 in.

≤ 10 ft.

≤ 8 in.Corners and end of walls

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

Reinforcement of at least 0.2 in2

Shear Walls 35

Shear Walls_____________ reinforced shear wall (7.3.2.4): Reinforced wall with prescriptive reinforcement of detailed plain shear wall.

______________ reinforced shear wall (7.3.2.5): Reinforced wall with prescriptive reinforcement of detailed plain shear wall. Spacing of vertical reinforcement reduced to 48 inches.

___________ reinforced shear wall (7.3.2.6):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]}.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 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 36

Minimum Reinforcement of Special Shear Walls

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

Reinforce-ment Ratio

8 in. CMU wall 12 in. CMU wallAs (in2/ft) Possibilities As (in2/ft) Possibilities

0.0007 0.064#4@32#5@56

0.098#4@24#5@32#6@48

0.0010 0.092#4@24#5@32#6@40

0.140#4@16#5@24#6@32

0.0013 0.119#4@16#5@32#6@40

0.181#5@16#6@24

Shear Walls 37

Special Shear Walls: Shear Capacity

Minimum shear strength (7.3.2.6.1.1):• Design shear strength, Vn, greater than shear

corresponding to 1.25 times nominal flexural strength, Mn

• Except Vn need not be greater than 2.5Vu.

Normal design: φVn has to be greater than Vu. Thus, Vn has to be greater than Vu/φ = Vu/0.8 = 1.25Vu. Thus, the requirement approximately doubles the shear.

Shear Walls 38

Seismic Design CategorySeismic Design Category Allowed Shear Walls

A or BC

D and higher

Response modification factor:Seismic design force divided by response modification factor, which accounts for ductility and energy absorption.

Shear Wall ROrdinary plain 1.5Detailed plain 2Ordinary reinforced 2Intermediate reinforced 3.5Special reinforced 5

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

facilities

Shear Walls 39

Maximum reinforcing (9.3.3.5)No limits on maximum reinforcing for following case (9.3.3.5.4):

5.1 and 1 RdV

M

vu

u Squat walls, not designed for ductility

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

Boundary element design (9.3.6.5):More difficult with masonry than concreteBoundary elements not required if:

gA1.0 mu fP geometrically symmetrical sections

gA05.0 mu fP geometrically unsymmetrical sections

AND

1wu

u

lVM OR 3

wu

u

lVM

mnu fAV 3 AND

Shear Walls 40

Maximum reinforcing (9.3.3.5)

Reinforcement limits: Calculated usingMaximum stress in steel of fyAxial forces taken from load combination D+0.75L+0.525QECompression reinforcement, with or without lateral ties, permitted to be

included for calculation of maximum flexural tensile reinforcement

Uniformly distributed reinforcement

Compression steel with area equal to tension steel

ymu

muyy

vymu

mum

v

s

f

dPbf

dA

64.0

syymumuyymu

mum

s

Eddf

bdPbf

bdA

,min

64.0

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

Shear Walls 41

Maximum reinforcingConsider a wall with uniformly distributed steel:

Strain mu

y

Stress

Steel in tension

fy

Steel in compression

0.8f’m

PTCC ssm

bdfC vymu

mumm

8.08.0

y

y

y

yy

ymu

ysys AfT

21

y

As taken as total steeldv is actual depth of masonry

Portion of steel in tension

Yielded steel

Elastic steel

ymu

ymusy

mu

ymu

ymu

musy

mu

y

mu

ymu

ymu

musys

Af

Af

AfC

5.0

5.0

21

Shear Walls 42

Maximum reinforcingPCCT mss

ymu

muysy

ymu

ymusy

ymu

yysyss AfAfAfCT

5.05.0

PCPbdfAfCT mvymu

mum

ymu

muysyss

8.08.0

ymu

muyy

vymu

mum

v

s

f

dPfb

dA

64.0

Shear Walls 43

Maximum reinforcing, εs = 4εy

Shear Walls 44

Maximum reinforcing, εs = 3εy

Shear Walls 45

Maximum reinforcing, εs = 1.5εy

Shear Walls 46

Shear Strength (9.3.4.1.2)

umnvvu

um PfAdV

MV 25.075.10.4

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

vyv

s dfsAV

5.0

Vertical reinforcement shall not be less than one-third horizontal reinforcement; reinforcement shall be uniformly distributed, max spacing of 8 ft (9.3.6.2)

gnsnmn VVV 8.0

Mu/Vudv need not be taken > 1.0Pu = axial load

gmnvn fAV 4 0.1)/( vuu dVMInterpolate for values of Mu/Vudv between 0.25 and 1.0

gmnvvu

un fAdV

MV

25

34

γg = 0.75 for partially grouted shear walls and 1.0 otherwise

2013 Draft Code Changes 47

Partially Grouted Walls

MethodVexp / Vn

Mean St. Dev.

Partially Grouted Walls (Minaie et al, 2010; 60 tests)

2008 Provisions 0.90 0.26Multiply shear strengths by Ag/An 1.53 0.43

Using just face shells 1.77 0.78

Fully Grouted Walls (Davis et al, 2010; 56 tests)

2008 Provisions 1.16 0.17

0.90/1.16 = 0.776; rounded to 0.75

Strength Design 48

ExampleGiven: 10ft. high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, f’m=2000psi; 2-#5 each end; #5 at 48in. (one space will actually be 40 in.); superimposed dead load of 1 kip/ft. SDS = 0.4Required: Maximum seismic load based on flexural capacity; shear steel required to achieve this capacity. Solution: Check capacity in flexure using 0.9D+1.0E load combination.

Pu = (0.9-0.2SDS)D = (0.9-0.2*0.4)(1k/ft+0.38k/ft) = 0.82(1.38k/ft) = 1.13 k/ft

Total = 1.13k/ft(16ft) = 18.1 kips

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

Shear Walls 49

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

8 inc

52 in92 in

140 in188 in

T1 T2 T3 T4 C1C2

Strength Design 51

Example• Use trial and error to find c such that Pn = 18.1 kips/0.9 = 20.1 kips

• Pn = C1 + C2 – T1 – T2 – T3 – T4• c = 14.85 inches

• C1 = 97.6 kips; C2 = 15.5 kips

• T1 = 37.2 kips; T2 = 18.6 kips; T3 = 18.6 kips T4 = 18.6 kips (all yielded)

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

• Mn = 37.2(188-96) + 18.6(140-96) + 18.6(92-96) + 18.6(52-96) + 97.6[96-8/2] + 15.5[96-8-(0.8*14.85-8)/2]

• Mn = 13660 kip-in = 1140 kip-ft; Mu = 0.9(1140 kip-ft) = 1025 kip-ft

• Seismic = Mu/h = (1025 kip-ft)/(10ft) = 102.5 kips

Strength Design 52

ExampleCalculate net shear area, Anv, including grouted cells. Net area

Net area: 2685)5.262.7)(8(51925.2 ininininininAnv

Maximum Vn

625.016/10// ftftdVhVdVM vuuvuu

kipskipsVV nu 9.919.1148.08.0 Maximum Vu

Maximum shear controls; seismic = 91.9 kips

Strength Design 54

Example

kkVVVg

nmgnns 2.6475.0

0.8975.09.114

kipskipspsiinVnm 0.891.1325.02000685625.075.10.4 2

Top of wall (critical location for shear):Pu = (0.9-0.2SDS)D = 0.82(1k/ft) = 0.82 k/ftTotal: 0.82k/ft(16ft) = 13.1 kips

ns

vyvvy

vns V

dfAsdf

sAV

5.0 5.0

Use #5 bars in bond beams. Determine spacing.

Use __ bond beams, spaced at ____ in. o.c. vertically.

Example: Maximum ReinforcingCheck with boundary elements:

• Geometrically symmetrical section:

• 0.1fʹmAn = 0.1(2.0ksi)(685in2) = 137 kips

• Check with Pu = (1.2+0.2SDS)D = 1.28(1.38k/ft) = 1.77 k/ft

• Total = 1.77k/ft(16ft) = 28.3 kips OK

• Since Mu(Vudv) = 0.625 ≤ 1 No boundary elements required

• Maximum reinforcement provisions satisfied

If we needed to check maximum reinforcing, there are at least two ways. These are shown on the following slides.A live load of 1 k/ft is assumed for maximum reinforcement calculations.

Strength Design 56

Strength Design 57

Example: Maximum Reinforcing

Check on axial load:a) Set steel strain to limit and find cb) Find axial load for this cc) If applied axial load is less than axial load, OK

Steel strain = 1.5y = 0.00310c = 0.0025/(0.0025+0.00310)*192 = 85.7 in.From spreadsheet, Pu = 258 kips

Pu = D+0.75L+0.525QE = (1.38k/ft + 0.75(1k/ft))16ft = 34.1 kips OK

Note: Did not include compression steel which is allowed when checking maximum reinforcement.If compression steel had been included, Pu = 299 kips

Strength Design 58

Example: Maximum Reinforcing

Find steel strain for given axial 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 = 34.1 kips (from D+0.75L+0.525QE)c = 20.4 in. (from spreadsheet and solver)s = (192-20.4)/20.4*0.0025 = 0.0210 = 10.2y OK

Note: Did not include compression steel which is allowed when checking maximum reinforcing. If compression steel had been included: c = 11.3 in. s = 0.0400 = 19.3y

Strength Design 59

Example: Special Shear Wall

If this were a special shear wall:• Vn ≥ shear corresponding to 1.25Mn (7.3.2.6.1.1)

• 1.25Mn = 1.25(1140 k-ft) = 1425 k-ft

• Vn = Mn/h = 1425k-ft/10ft = 142.5 kips

• Vn = 142.5k/0.8 = 178.1kips

• But, maximum Vn is 114.9 kips (64% of required).

Cannot build wall. Need to fully grout to get greater shear capacity.

Strength Design 60

Example: Special Shear WallFully grouted:

• Use spreadsheet to find Mn corresponding to Pn = 20.1 kips

• Mn = 1156 kip-ft; Mu = 1041 kip-ft a 1.4% increase

• Seismic = 104.1 kip-ft

• 1.25Mn = 1.25(1156 k-ft) = 1445 k-ft

• Vn = Mn/h = 1445k-ft/10ft = 144.5 kips

• Vn = 144.5k/0.8 = 180.6kips

21494.196.625.7 inininAnv

kipskipspsiinVnm 2.1941.1325.020001494625.075.10.4 2 No shear steel required; choose horizontal reinforcement to satisfy prescriptive requirements.

Strength Design 61


Prescriptive reinforcement:• Minimum reinforcement in either direction: 0.0007A

• Vertical: 0.0007(1494 in2) = 1.05 in2

• 7-#5’s provided: 7(0.31 in2) = 2.17 in2 OK• Spacing is min{1/3 length of wall, 1/3 height of wall, 48 in.} = {192

in./3, 120 in./3, 48 in.} = min{64, 40, 48 in.}• Need to decrease spacing of vertical reinforcement to 40 inches

• Use spacings of 40, 32, 40, 32, 40 inch to match wall coursing.• This will have a small impact on moment capacity, increasing

Mu from 1041 kip-ft to 1166 kip-ft, a 12% increase• This increases the required nominal shear strength to 202.4

kips. This is slightly greater than the nominal shear strength due to the masonry (194.2 kips), but with the prescriptive reinforcement, everything will be OK.

Strength Design 62


Prescriptive reinforcement:• Minimum reinforcement in either direction: 0.0007A

• Horizontal: 0.0007(120 in)(7.625 in) = 0.64 in2

• Use #5’s at 40 in (required spacing); total is 3 bars• 3(0.31 in2) = 0.92 in2 OK

• Total reinforcement: 0.002A• Vertical: 2.79in2/1494in2 = 0.00187A• Horizontal: 0.92in2/915in2 = 0.00100A• Total = 0.00187A + 0.00100A = 0.00287A OK

Shear Walls 63

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

40in 40in64in

112i

n

Elevation

Plan

6t=4

8in

56in

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

ba c

40in 40in

Shear Walls 64

ExamplePier b inE

inin

inin

inE

Lh

Lh

Etkb 286.03

641124

64112

62.7

3422

Piers a and c Previously determined stiffness from basic principles.

inEk ca 157.0,

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

VVinEinEinE

inEVk

kV aa 26.0157.0286.0157.0157.0

Vb = 0.48V

Deflection calculations:k = (0.600in.)E = (0.600in.)(1800ksi) = 1080 kip/inReduce by a factor of 2 to account for cracking: k = 540 kip/in

Shear Walls 65

ExampleRequired: 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

kkP 065.0

/54035

000579.0112065.0

inin

hor

1728h

Shear Walls 66

Example

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

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

Wall b: Vub = 0.48Vu = 0.48(35k) = 16.8kD = 1.2k/ft(_____________)(1ft/12in) + 0.075ksf(112in)(64in)(1ft2/144in2) = 14.1kips

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

Shear Walls 67

Example: Flexural SteelUse load combination 0.9D+E Mu = 156.8k-ft Pu = 9.87k

Try 2-#4 each end; 1-#4 middle: As=1.00in2

Middle bar is at 28 in. (8 cells, cannot place bar right in center)From spreadsheet: Mn=152.7k-ft NG

Note: used solver to find value

Try 1-#6 each end; 1-#6 middle: As=1.32in2 Mn=194.1k-ft OKTry 2-#5 each end; 1-#5 middle: As=1.55in2 Mn=219.1k-ft OK

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

Shear Walls 68

Example: Maximum ReinforcingCheck maximum reinforcing:Axial force from load combination D+0.75L+0.525QE

L=1k/ft(104in)(1ft/12in) = 8.67kipD+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 not laterally supportedFor #5 bars and P = 20.60kip, c = 6.35in.. (from solver on spreadsheet)

yymus ininine

kdkdd 421.100211.00025.0

35.635.660

OK

Shear Walls 69

Example: Shear

kipsklbkpsiinin

PfAdVMV umnvuunm9.5028.725.01000/12000)64)(625.7(25.2

25.0/75.10.4

175.164112

inVinV

dVM

u

u

vu

u Use Mu/Vudv = 1.0

kipsVn 9.50Ignore any shear steel:

kipskipskipsVn 8.167.409.508.0 OK

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

Shear Walls 70

Example: Shear

Mn = Mn/ = 219.1k-ft/0.9 = 243.4k-ftV corresponding to Mn: [243.3k-ft/(112in)](12in/ft) = 26.1kips1.25 times V: 1.25(26.1k) = 32.6kips < Vn=40.7kips OK

Wall b: Vertical bars: 2-#5 each end; 1-#5 middleHorizontal bars: #4 at 24 in. o.c. (Total 5-#4)

Minimum prescriptive horizontal reinforcement: 0.0007AgAs = 0.0007(64in)(7.625in) = 0.34in2

Maximum spacing is {64in./3, 48in.} = 21.3 in.Use bond beams at 16 in. o.c. (I would be fine with 24 in.)

Shear Walls 71

Example

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: 0.0013AgAs = 0.0013(88in)(7.625in) = 0.87in2

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

Design issues:Flange in tension: easy to get steel; problems with maxFlange in compression: hard to get steel; max is easySteel in both flange and web affect maxNeed to also consider loads in perpendicular directionNeed to check shear at interface of flange and web

Shear Walls 72

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

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=150.8k-ftMax reinforcing check: s=0.0127=6.14y

Note: with 2-#5 in flange, Mn=106.9k-ft, but might need 3 bars for perpendicular direction or out-of-plane

Axial force for max from load combination D+0.75L+0.525QEL=1k/ft(60in)(1ft/12in) = 5.00kipD+0.75L+0.525QE = 11.1k+0.75(5.00k)+0.525(0) = 14.85kip

Flange in compression: Mn=125.8k-ftMax reinforcing check: s=0.0564=27.3y

OK

OK

Shear Walls 73

Example: Shear

Mn = Mn/ = 150.8k-ft/0.9 = 167.6k-ftV corresponding to Mn: [167.6k-ft/(112in)](12in/ft) = 18.0kips1.25 times V: 1.25(18.0k) = 22.4kips < Vn=25.4kips

Walls a,c: Vertical bars: 5-#5 as shown in previous slideHorizontal bars: #4 at 24 in. o.c. (Total 5-#4)

Minimum prescriptive horizontal reinforcement: 0.0007AgUse 5 bond beams, 24in. o.c. (top spacing is only 32in.)

kipsklbkpsiininVm 7.312.425.01000/12000)40)(625.7(25.2

18.240112

inVinV

dVM

u

u

vu

u Use Mu/Vudv = 1.0

kipsVn 7.31Ignore any shear steel: kipskipsVn 4.257.318.0 OK

OK

Shear Walls 74

ExampleSection 5.1.1.2.4 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.

kipskipsVn 7.689.858.0 OK

kipslbkpsiinin

PfAVdMV umnvm9.851000/12000)112)(625.7(0.175.10.4

25.0/75.10.4

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

Shear Walls 75

Example: Drag StrutsDistributed shear force: ftk

ftin

ink

LVu /875.1

112

22435

Look at top of wall:

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

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

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

2.8k 3.4k 2.8k3.4k

shear walls - engineering fundamentals...

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