Shear Walls •Load Distribution to Shear WallsShear Walls 3 9.1.5.2 Deflection calculations shall be based on cracked section properties. Assumed properties shall not exceed half

Shear Walls 1

• Load Distribution to Shear Walls

• Shear wall stiffness

• Diaphragm types

• Types of Masonry Shear Walls

• Maximum Reinforcement Requirements

• Shear Strength

• Example: Single Layer Reinforcing

• Example: Distributed Reinforcing

• Example: Flanged Walls

Shear Walls

Shear Walls 2

_____ 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

Cantilever Shear Wall

Shear Walls 3

9.1.5.2 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 wallℎ = height of wall

𝐴 =shear area; (5/6)𝐴 for a rectangle

𝐸 = 𝐺 = shear modulus (modulus of rigidity); given as 0.4𝐸(4.2.2.2.2)

𝐸 = 𝐸 2 1 𝜈⁄ , where 𝜈 is Poisson’s ratio

𝑡 = thickness of wall

𝐿 = length of wall

Fixed wall (fixed against rotation at top)

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

Shear Walls: Stiffness

𝑘𝐸 𝑡

ℎ𝐿 4

ℎ𝐿 3

𝑘𝐸 𝑡

ℎ𝐿

ℎ𝐿 3

Δ

Δ

Shear Walls 5

Section 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 4 ft.c) Intersecting bond beams at max 4 ft. Reinforcing of at least 0.1in.2 per foot of wall

2-#4’s

Metal 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

T- or L- Shaped Shear Walls

Shear Walls 6

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

Analysis: distinction between 6t and 0.75 times floor-to-floor wall height in compression is not important. Assumed effective width only results in a small shift of neutral axis.

Some people suggest increasing tension flange width by 1.5 for shear capacity design and ductility checks. Reinforcement just outside effective width can be participating.

Effective Flange Width (5.1.1.2.3)

Shear Walls 7

• Rectangular cross-sections

• 𝐼 0.15𝐼

• T-shaped and I-shaped sections

• 𝐼 0.40𝐼

• Shear stiffness

• 𝐴 0.35𝐴

• Partially grouted walls

• 𝐴 total cross-sectional area of face shells plus area of grouted cells

Cracked Moment of Inertia

Seismic Design of Special Reinforced Masonry Shear Walls A Guide for Practicing Engineers

NEHRP Seismic Design Technical Brief No. 9

Shear Walls 8

40in.

112i

n.6𝑡

=48

in.

56in

.

Elevation

Plan

Given: Fully grouted shear wall

Required: Stiffness of wall

Solution: Determine stiffness from basic principles.

Centroid, from outer flange

Example: Flanged Shear Wall

Net area

Net moment of inertia

Shear area

𝐴 7.62𝑖𝑛. 48𝑖𝑛. 7.62𝑖𝑛. 40𝑖𝑛. 671𝑖𝑛.

𝑦7.62 48 3.81 7.62 40 20

67111.16𝑖𝑛.

𝐴 ~𝐴 7.62𝑖𝑛. 40𝑖𝑛. 305𝑖𝑛.

𝐼1

1248 7.62 7.62 48 11.16 3.81

112

7.62 40 7.62 40 11.16 20 86000𝑖𝑛.

Shear Walls 9

Example: Flanged Shear Wall

Uncracked

Cracked

𝑘1

112𝑖𝑛3 1800𝑘𝑠𝑖 0.40 86000𝑖𝑛

112𝑖𝑛305𝑖𝑛 0.35 0.4 1800𝑘𝑠𝑖

111𝑘

𝑖𝑛.

𝑘𝑃Δ

𝑃𝑃ℎ

3𝐸 𝐼𝑃ℎ

𝐴 𝐸

1ℎ

3𝐸 𝐼ℎ

𝐴 𝐸Stiffness

𝑘1

112𝑖𝑛.3 1800𝑘𝑠𝑖 86000𝑖𝑛.

112𝑖𝑛.305𝑖𝑛. 0.4 1800𝑘𝑠𝑖

1

0.00302𝑖𝑛.𝑘 0.00051

𝑖𝑛.𝑘

283𝑘

𝑖𝑛.

Allowable seismic drift: 0.01ℎ = 1.12 in.

Shear Walls 10

__________ shear wall

Coupled Shear Walls





Shear Walls 11

Frame Models



Shear Walls 12

Frame Models



Shear Walls 25

Example: Perforated Shear Wall

Central pier carries about 85% of shear; can design for entire shearGood practice would be to add control joints

10’

18’

5’3’

7’

55’

3’ 8’ 2’ 8’ 18’ 8’ 3’ 3’ 2’

1. Need to provide prescriptive seismic reinforcement everywhere.2. Need drag strut/collector to get shear to solid portion.

Do not skimp on end of wall steel

Con

trol

Joi

nt

Shear Walls 26

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 properly detailed for seismic requirements

3. Most shear walls will have openings

4. Can design only a portion to carry shear load, but need to ca detail rest of structure

5. Can use control joints to isolate piers

Shear Walls: Building Layout

Shear Walls 27

• 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

Diaphragms

Shear Walls 28

Direct Shear:

Torsional Shear:

𝑉 = total shear force on building

𝑘 = relative rigidity of lateral force resisting element i

𝑑 = distance from center of stiffness

𝑒 = eccentricity of load from center of stiffness

Rigid Diaphragms

𝐹 𝑉𝑘

∑ 𝑘

𝐹 𝑉𝑒𝑘 𝑑

∑ 𝑘 𝑑

Total Shear on Wall: 𝐹 𝐹 𝐹

Shear Walls 29

Given: The structure shown is subjected to a 0.2 kip/ft horizontal force. Relative stiffness values are given.

Required: Distribution of force assuming:

• flexible diaphragm

• rigid diaphragm.

k =

4

k =

5

k =

150 ft 50 ft

0.2 kip/ft

PLAN VIEW

Example: Diaphragms

Shear Walls 30

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.

k =

4

k =

5

k =

1

50 ft 50 ft

PLAN VIEW

Example: Flexible Diaphragms

Shear Walls 31

Solution: Rigid diaphragm

k =

4

k =

5

k =

1

50 ft 50 ft

Wal

l 1

Wal

l 2

Wal

l 3

x

Wall x k x(k)

123

Total

Center of stiffness = 350/10 = 35 ft

Wall k d (ft) k(d) k(d2) Fv Ft Ftotal

123

451

-1407565

490011254225

3.912.23.9

Total 10 10250 0.0 20

Example: Rigid Diaphragms

Shear Walls 33

Solution: 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

Example: Diaphragm Design Forces

Shear Walls 34

• 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

Drag Struts and Collectors

Shear Walls 35

Three lateral force resisting systems: Length=20 ft; Height=14 ftW24x68

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 = 3120 ksi, 𝜈 = 0.17, variable thickness, load of 1 kip/ft.

Diaphragm


24 ft 24 ft

Diaphragm Behavior

Shear Walls 36

Diaphragm Behavior

Shear Walls 37

http://skghoshassociates.com/SKGAblog/viewpost.php?id=19

Diaphragm Behavior

Shear Walls 38

Diaphragm Behavior

Shear Walls 39

_____________ (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: Types

Shear Walls 40

_____________ 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: Types

Shear Walls 41

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

Reinforce-ment Ratio

8 in. CMU wall 12 in. CMU wall

As (in.2/ft) Possibilities As (in.2/ft) Possibilities

0.0007 0.064#4@32

#5@56 480.098

#4@24

#5@32

#6@48

0.0010 0.092

#4@24

#5@40

#6@56 48

0.140

#4@16

#5@24

#6@32

#7@48

0.0013 0.119

#4@16

#5@32

#6@40

0.181

#4@8

#5@16

#6@24

#7@40

Shear Walls: Special Reinforced

Max spacing 48

Shear Walls 42

Example: Reinforcement Spacing56’

12’ 12’ 8’ 12’ 12’

9’‐4”

22’

9’‐4”

3’‐4”

Opening Opening

Control Joints

Isolated Piers

Expected Diagonal Cracks in Piers

OpeningOpening

Dis

plac

ed s

hape

Is the reinforcement spacing based on 1/3(3ʹ-4ʺ)?

Shear Walls 43

Minimum shear strength (7.3.2.6.1.1):

• Design shear strength, 𝜙𝑉 , greater than shear corresponding to 1.25 times nominal flexural strength, 𝑀

• Except 𝑉 need not be greater than 2.5𝑉 .

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

Special Walls: Shear Capacity

Shear Walls 44

𝑆Risk Category

I, II, or III IV

𝑆 0.167 A A

0.167 𝑆 0.33 B C

0.33 𝑆 0.50 C D

𝑆 0.50 D D

Seismic Design Category

𝑆Risk Category

I, II, or III IV

𝑆 0.067 A A

0.067 𝑆 0.133 B C

0.133 𝑆 0.20 C D

𝑆 0.20 D D

Seismic Design Category Allowed Shear Walls

A or B

C

D and higher

Allowable seismic drift:Masonry cantilever shear wall structures: 0.010ℎOther masonry shear wall structures: 0.007ℎ

Shear Walls 45

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

Deflection amplification factor, 𝐶 :Deflection under seismic design loads multiplied by deflection amplification factor.

Shear WallBearing Wall System Building Frame System

𝑅 𝐶 𝑅 𝐶

Ordinary plain 1 ½ 1 ¼ 1 ½ 1 ¼

Detailed plain 2 1 ¾ 2 1 ¾

Ordinary reinforced

2 1 ¾ 2 1 ¾

Intermediate reinforced

3 ½ 2 ¼ 4 4

Special reinforced

5 3 ½ 5 ½ 4 ½

Seismic Design Category

Shear Walls 46

Maximum reinforcing

Design with Boundary Elements?No

Is

1 OR 𝑉 3𝐴 𝑓 AND 3

AND

𝑃 0.10𝐴 𝑓 : Geometrically symmetrical walls

𝑃 0.05𝐴 𝑓 : Geometrically unsymmetrical walls

No boundary elements required

Design boundary elements per TMS

402 Section 9.3.6.6.2

Design with TMS 402 Section 9.3.3.2.Area of flexural tensile reinforcement ≤ area required to maintain axial equilibrium under the following conditions

A strain gradient corresponding to 𝜀 in masonry and 𝛼𝜀 in tensile reinforcement

Axial forces from loading combination D + 0.75L + 0.525QE .

Compression reinforcement, with or without lateral restraining reinforcement, can be included.

Is 𝑀 𝑉 𝑑⁄ 1 ?

𝛼 = 1.5Ordinary reinforced walls: 𝛼 = 1.5Intermediate reinforced walls: 𝛼 = 3Special reinforced walls: 𝛼 = 4

Yes No

Yes

Yes No

Shear Walls 48

Maximum reinforcing (9.3.3.5)

𝜌𝐴

𝑡 𝑑

0.64𝑓𝜀

𝜀 𝛼𝜀𝑃

𝑡 𝑑

𝑓 min 𝜀 𝑑𝑑 𝜀 𝛼𝜀 , 𝜀 𝐸

Compression steel with area equal to tension steel

Uniformly distributed reinforcement

𝐴𝑑

0.64𝑏𝑓𝜀

𝜀 𝛼𝜀𝑃𝑑

𝑓𝛼𝜀 𝜀𝜀 𝛼𝜀

Shear Walls 49

Consider a wall with uniformly distributed steel:

Strain

𝛼𝜀

Stress

Steel in tension

Steel in compression

0.8𝑓

𝐴 is total steel𝑑 is actual depth of masonry

Portion of steel in tension

Yielded steel

Elastic steel

Maximum reinforcing

𝜀

𝜀

𝑓

𝐶 𝐶 𝑇 𝑃

𝐶 0.8𝑓 0.8𝜀

𝜀 𝛼𝜀𝑑 𝑡

𝑇 𝑓 𝐴𝛼𝜀

𝜀 𝛼𝜀𝛼𝜀 𝜀

𝛼𝜀12

𝜀𝛼𝜀

𝐶 𝑓 𝐴𝜀

𝜀 𝛼𝜀𝜀 𝜀

𝜀12

𝜀𝜀

𝑓 𝐴𝜀

𝜀 𝛼𝜀𝜀 0.5𝜀

𝜀

𝑓 𝐴𝜀 0.5𝜀𝜀 𝛼𝜀

Shear Walls 50

Maximum reinforcing

𝑇 𝐶 𝐶 𝑃

𝑇 𝐶 𝑓 𝐴𝛼𝜀 0.5𝜀𝜀 𝛼𝜀

𝑓 𝐴𝛼𝜀 0.5𝜀𝜀 𝛼𝜀

𝑓 𝐴𝛼𝜀 𝜀𝜀 𝛼𝜀

𝑇 𝐶 𝑓 𝐴𝛼𝜀 𝜀𝜀 𝛼𝜀

0.8𝑓 0.8𝜀

𝜀 𝛼𝜀𝑑 𝑡 𝑃 𝐶 𝑃

𝐴𝑑

0.64𝑡 𝑓𝜀

𝜀 𝛼𝜀𝑃𝑑

𝑓𝛼𝜀 𝜀𝜀 𝛼𝜀

Shear Walls 51

Maximum reinforcing, εs = 4εy

Shear Walls 52

Maximum reinforcing, εs = 3εy

Shear Walls 53

Maximum reinforcing, εs = 1.5εy

Shear Walls 54

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)

𝑀 𝑉 𝑑⁄ need not be taken > 1.0𝑃 compressive axial load

𝛾 = 0.75 for partially grouted shear walls and 1.0 otherwise

Shear Strength (9.3.4.1.2)

𝑉 6𝐴 𝑓 𝛾 0.25

𝑉 4𝐴 𝑓 𝛾 1.0

Interpolate for 0.25 1.0

𝑉 5 2 𝐴 𝑓 𝛾

𝑉 𝑉 𝑉 𝛾 𝜙 0.8

𝑉 4.0 1.75𝑀

𝑉 𝑑𝐴 𝑓 0.25𝑃

𝑉 0.5𝐴𝑠

𝑓 𝑑

Shear Walls 55

Method𝑉 𝑉⁄

Mean St. Dev.

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

2008 Provisions 0.90 0.26

Multiply shear strengths by 𝐴 𝐴⁄ 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

Partially Grouted Walls

Shear Walls 57

Shear Friction Provisions

𝐴 = area of masonry in compression at nominal moment capacity

𝐴 = reinforcement within net shear area

Coefficient of friction

• 𝜇 = 1.0 for masonry on concrete with unfinished surface, or concrete with a surface that has been intentionally roughened

• UBC (1997) required concrete abutting structural masonry to be roughened to a full amplitude of 1/16 inch.

• 𝜇 = 0.70 for all other conditions

𝑀 𝑉 𝑑 0.5⁄ 𝑉 𝜇 𝐴 𝑓 𝑃

𝑀 𝑉 𝑑⁄ 1.0 𝑉 0.42𝑓 𝐴

Linear interpolation for intermediate values

Shear Walls 58

• Check axial capacity• Check maximum reinforcement• Check shear

Design: Single Layer of ReinforcementCalculate

Is 𝑐 𝑐 ?For CMU, Grade 60 steel𝑐 0.547𝑑

YES

Compression controlsTension controls

NO

𝑎 𝑑 𝑑2 𝑃 𝑑 𝑑 2 𝑀⁄

𝜙 0.8𝑓 𝑡 𝑐

𝑎0.8

𝑐𝜀

𝜀 𝜀𝑑

𝐴 ,0.8𝑓 𝑡 𝑎 𝑃 𝜙⁄

𝜀 𝐸𝑑 𝑐

𝑐

𝐴 ,0.8𝑓 𝑡 𝑎 𝑃 𝜙⁄

𝑓

Shear Walls 59

Given: 2 ft long, 8 ft high CMU pier; Type S masonry cement mortar; Grade 60 steel; fully grouted. 𝑃 = 11 kips, 𝑉 = 7 kips, 𝑀 = 28 k-ft

Required: Required amount of steel

Solution: Choose/determine material properties.

𝑓 = 2000 psi; 𝑓 = 60,000 psi

20 in.

23.625 in.

Example: Single Layer of Reinforcement

𝑎, depth of stress block

𝐴 , , req’darea of steel


𝜙 0.8𝑓 𝑡

𝐴 ,0.8𝑓 𝑡 𝑎 𝑃 𝜙⁄

𝑓

Shear Walls 61

20 in.

23.625 in.

Try #4 bars

Consider second layer of steel:

𝑐 = 2.986 inches; 𝐶 = 29.14kips; 𝑇= 12kips; 𝑇 = 4.92kips


Solve for 𝑐 such that 𝑃 = 11k/0.9 = 12.2k

𝜙𝑀 , design moment

Ignoring the second layer of steel resulted in 𝑐 = 2.482 in., and 𝜙𝑀 = 27.2 k-ft

(2.5% decrease)

𝑃 𝐶 𝑇 𝑇

𝑃 0.8𝑓 0.8𝑐 𝑡 𝐴 𝑓 min 𝜀𝑑 𝑐

𝑐𝐸 , 𝑓 𝐴

1.2𝑘 0.8 2.0𝑘𝑠𝑖 0.8𝑐 7.62𝑖𝑛. 0.20𝑖𝑛. 60𝑘𝑠𝑖 min 0.00254𝑖𝑛. 𝑐

𝑐29000𝑘𝑠𝑖, 60𝑘𝑖𝑠 0.20𝑖𝑛.

𝜙𝑀 0.9 29.14𝑘 12𝑖𝑛.0.8 2.986𝑖𝑛.

212𝑘 20𝑖𝑛. 12𝑖𝑛. 4.92𝑘 4𝑖𝑛. 12𝑖𝑛.

334𝑘 · 𝑖𝑛. 27.9𝑘 · 𝑓𝑡

Shear Walls 62

3. Check axial load

11k < 187k OKApplied load is 6% of axial capacity


Radius of gyration, 𝑟

Slenderness, ratio, ℎ/𝑟

Nominal strength, 𝑃

Design strength, 𝜙𝑃

𝑟 0.289𝑡 0.289 7.62𝑖𝑛 2.20𝑖𝑛.

.

. .43.6 99

𝑃 0.80 0.80𝑓 𝐴 𝐴 𝑓 𝐴 1

0.80 0.80 2.0𝑘𝑠𝑖 7.62𝑖𝑛. 24𝑖𝑛. 0 0 1.

230.6𝑘 0.903 208.2𝑘

𝜙𝑃 0.9 208.2𝑘 187.4𝑘

Shear Walls 63

4. Check maximum reinforcement:

Fully grouted with equal tension and compression reinforcement

Assume axial force is from 0.9D, so 𝑃 for maximum reinforcement is 11k/0.9 = 12.2 kips. For an ordinary wall:


𝜌.

,

𝜌. . .

. . . .

. . .

. ..

. . . , . 0.0245

𝐴 , 𝜌𝑡 𝑑 0.0245 7.62𝑖𝑛. 20𝑖𝑛. 3.74𝑖𝑛. 2 0.20𝑖𝑛. 0.40𝑖𝑛.

OK

Shear Walls 64

Axial Force, Pu As,reqd

Maximum Reinforcement

Ordinary Intermed. Special

0 kips 0.32 in.2 4.35 in.2 1.47 in.2 0.91 in.2

11 kips (6% of axial capacity

0.21 in.2 3.74 in.2 1.15 in.2 0.66 in.2

19 kips (10% of axial capacity)

0.13 in.2 3.30 in.2 0.92 in.2 0.48 in.2

38 kips (20% of axial capacity)

0 2.24 in.2 0.36 in.2 0.06 in.2


Shear Walls 65

5. Check Shear:

Use 𝑀 𝑉 𝑑⁄ = 1.0


𝑀 𝑉 𝑑⁄ratio

Nominal strength, 𝑉

Design strength, 𝜙𝑉

𝑀𝑉 𝑑

28𝑘 · 𝑓𝑡7𝑘 2.0𝑓𝑡

2.00

𝑉 4.0 1.75 𝐴 𝑓 0.25𝑃

4.0 1.75 1.0 7.62𝑖𝑛. 24𝑖𝑛. 2000𝑝𝑠𝑖 0.25 11𝑘18.1𝑘 2.8𝑘 20.9𝑘

𝜙𝑉 0.8 20.9𝑘 16.7𝑘

7k < 16.7k OK

Shear Walls 66

Design: Distributed Reinforcement

Assume tension controls


𝜙 0.8𝑓 𝑡

𝐴 ,0.8𝑓 𝑡 𝑎 𝑃 𝜙⁄

𝑓

𝐴 ,∗ 𝐴 ,

0.65𝑑Approximate required

distributed reinforcement

Approximate 𝑑 0.9𝑑

Spacing of intermediate reinforcing bars often controlled by out-of-plane loading

Shear Walls 67

Example: Distributed Reinforcement

Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type S mortar; 𝑓′𝑚 = 2000 psi; superimposed dead load of 1 kip/ft. In-plane seismic load of 50 kips. 𝑆𝐷𝑆 0.5 (just less than 0.5)

Required: Design the shear wall; ordinary reinforced shear wall

Solution: Check using 0.9D+1.0E load combination.

• 𝑀 50k 10ft 500k ⋅ ft

• Axial load, 𝑃

• Need to know weight of wall to determine 𝑃.

• Need to know reinforcement spacing to determine wall weight

• Estimate wall weight as 45 psf

• Wall weight: 45psf 10ft 16ft 7.2k

• 𝐷 1 k ft⁄ 16ft 7.2k 23.2k

• 𝑃 0.9 0.2𝑆 𝐷 0.80𝐷 0.80 23.2k 18.6k

Shear Walls 68


𝐴 ,0.8𝑓 𝑡 𝑎 𝑃 𝜙⁄

𝑓0.8 2ksi 7.625in. 3.96in. 18.6k 0.9⁄

60ksi0.460in.

𝑎, depth of stress

block

𝐴 , , area of steel

Estimate 𝑑 𝑑 0.9𝑑 0.9 192in. 173in.

𝐴 ,∗ ,

dist. steel𝐴 ,

∗ 𝐴 ,

0.65𝑑0.460in.

0.65 192in.12in.

ft0.044 in. ft⁄

Try #4 @ 48 in. (0.050 in.2/ft)


𝜙 0.8𝑓 𝑡

173in. 173in.2 18.6k 173in. 192in. 2 6000k ⋅ in.⁄

0.9 0.8 2000psi 7.625in.3.96in.

Shear Walls 69

Example: Distributed ReinforcementInteraction diagram: Illustrate with 𝑐 = 54 in., 𝑎 = 0.8(54) = 43.2 in.

8 in.43.2 in.

52 in.92 in.

140 in.

T1 T2 T3 T4 C1C2N.A.

Strain

188 in.

𝜀 0.0025

𝜀. .

.0.0025 0.00620 𝜀

𝜀. .

.0.0025 0.00398 𝜀

𝜀. .

.0.0025 0.00176 𝜀

Shear Walls 70

Example: Distributed ReinforcementInteraction diagram: Illustrate with 𝑐 = 54 in., 𝑎 = 0.8(54) = 43.2 in.

8 in.43.2 in.

52 in.

T1 T2 T3 T4 C1C2N.A.

Strain

𝜀 0.0025

𝜀 0.00620𝑇 60𝑘𝑠𝑖 0.20𝑖𝑛. 12𝑘

𝜀 0.00398𝑇 60𝑘𝑠𝑖 0.20𝑖𝑛. 12𝑘

𝜀 0.00176𝑇 0.00176 29000𝑘𝑠𝑖 0.20𝑖𝑛. 10.2𝑘

𝐶 1.6𝑘𝑠𝑖 8𝑖𝑛. 7.62𝑖𝑛. 97.6𝑘

𝐶 1.6𝑘𝑠𝑖 43.2𝑖𝑛. 8𝑖𝑛. 2.5𝑖𝑛. 140.8𝑘

Shear Walls 71

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

This is for determining interaction diagram; for determining maximum reinforcement could also include compression steel.


𝜙𝑃 𝜙 𝐶 𝐶 𝑇 𝑇 𝑇

0.9 97.6k 140.8k 12.0k 12.0k 10.2k 183.8k

𝜙𝑀 0.9 97.6𝑘192𝑖𝑛.

28𝑖𝑛.

2140.8𝑘

192𝑖𝑛.2

8𝑖𝑛.43.2𝑖𝑛. 8𝑖𝑛.

2

12.0𝑘 188𝑖𝑛.192𝑖𝑛.

212.0𝑘 140𝑖𝑛.

192𝑖𝑛.2

10.2𝑘 92𝑖𝑛.192𝑖𝑛.

2

18430𝑘 · 𝑖𝑛. 1536𝑘 · 𝑓𝑡

Shear Walls 72


At 89% of capacity

Shear Walls 73

Example: Dist. Reinf.; Shear

• Calculate axial force based on 𝑐 = 83.8 in.

• Include compression reinforcement

• 𝜙𝑃 = 323 kips

• Assume a live load of 1 k/ft (often 0; this is live load, not roof live load

• D + 0.75L + 0.525QE = (1k/ft + 0.75(1k/ft))16ft +7.2k = 35.2 kips

Section 9.3.3.2 Maximum ReinforcementSince 𝑀 / 𝑉 𝑑 1, strain gradient is based on 1.5𝜀 .

OK

𝑐 = 0.446(188in.) = 83.8 in.

Strain c/d, CMU c/d, Clay

1.5𝜀 0.446 0.530

3𝜀 0.287 0.360

4𝜀 0.232 0.297

Shear Walls 74

Example: Dist. Reinf.; Shear

𝐴 2 1.25in. 192in. 5 8in. 7.625in. 2.5in. 685in.

OK

𝑀𝑉 𝑑

𝑉 ℎ𝑉 𝑑

ℎ𝑑

120in.192in.

0.625Shear Span:

Max Shear:𝜙𝑉 , 𝜙

43

5 2𝑀

𝑉 𝑑𝐴 𝑓 𝛾

0.843

5 2 0.625 685in. 2000psi 0.75 91.9kip

Masonry Shear:

𝜙𝑉 𝜙 4 1.75𝑀

𝑉𝑑𝐴 𝑓 0.25𝑃 𝛾

0.8 4 1.75 0.625 685in. 2000psi 0.25 18600lb 0.75 56.2kip

OK

Net Shear Area:

Shear Walls 75

Example: Shear Friction Design

Since 0.5 𝑀 𝑉 𝑑⁄ 1.0 use linear interpolation

For 𝑀 𝑉 𝑑 0.5⁄

Specify an unfinished surface: 𝜇 = 1.0

Area of reinforcement crossing shear plane, 𝐴

For 𝑀 𝑉 𝑑 1.0⁄ From interaction diagram spreadsheet, 𝑐 = 7.04 in. @ 𝜙𝑃 = 18.6k

Linear Interpolation

Design Strength OK

𝐴 5 0.20𝑖𝑛. 1.00𝑖𝑛.

𝑉 𝜇 𝐴 𝑓 𝑃 1.0 1.0𝑖𝑛. 60𝑘𝑠𝑖 18.6𝑘 78.6𝑘

𝑉 0.42𝑓 𝐴 0.42 2.0𝑘𝑠𝑖 7.62𝑖𝑛. 7.04𝑖𝑛. 45.1𝑘

𝑉 45.1𝑘1.0 0.625

1.0 0.578.6𝑘 45.1𝑘 70.2𝑘

𝜙𝑉 0.8 70.2𝑘 56.2𝑘 𝑉 50𝑘

Shear Walls 76

Example: Shear Friction Design

If 𝜇 = 0.7For 𝑀 𝑉 𝑑 0.5⁄ 𝑉 55.0𝑘

For 𝑀 𝑉 𝑑 1.0⁄

Linear Interpolation

Design Strength NG

𝑉 45.1𝑘1.0 0.625

1.0 0.555.0𝑘 45.1𝑘 52.5𝑘

𝜙𝑉 0.8 52.5𝑘 42.0𝑘 𝑉 50𝑘

Change dowels to #5 bars (still #4 in wall) 𝐴 5 0.31𝑖𝑛. 1.55𝑖𝑛.

For 𝑀 𝑉 𝑑 0.5⁄ 𝑉 78.1𝑘

𝑐 9.74𝑖𝑛.𝐴 7.62𝑖𝑛. 8𝑖𝑛. 2.5𝑖𝑛. 9.74𝑖𝑛. 8𝑖𝑛. 65.3𝑖𝑛.𝑉 0.42𝑓 𝐴 0.42 2.0𝑘𝑠𝑖 65.3𝑖𝑛. 54.8𝑘

For 𝑀 𝑉 𝑑 1.0 𝑉 45.1𝑘⁄

Linear Interpolation 𝑉 54.8𝑘1.0 0.625

1.0 0.578.1𝑘 54.8𝑘 72.3𝑘

Design Strength OK𝜙𝑉 0.8 72.3𝑘 57.8𝑘 𝑉 50𝑘

Shear Walls 77

Example: Special Reinforced Wall

Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type S mortar; 𝑓′𝑚 = 2000 psi; superimposed dead load of 1 kip/ft. In-plane seismic load of 50 kips. 𝑆𝐷𝑆 0.5

Required: Design the shear wall; special reinforced shear wall

Solution:

• Flexural reinforcement remains the same

• Shear friction strength remains the same

• Shear capacity design: Design shear strength, 𝜙𝑉 , greater than shear corresponding to 1.25𝑀

Shear Walls 78


• For #4 @ 48 in., 𝜙𝑃 𝑃 =18.6 k; 𝜙𝑀 = 552 k-ft; 𝑀 = 613 k-ft

• 1.25𝑀 = 766 k-ft;

• Design for shear of 76.6 kips

• But wait, need to check load combination of 1.2D + 1.0E

• 𝑃 = [1.2 + 0.2(𝑆 )]𝐷 = 1.3𝐷 = 30.1 k, 𝑀 = 709 k-ft ,

• 1.25𝑀 = 886 k-ft


• But wait, Section 7.3.2.6 has maximum spacing requirements:

• min{1/3 length of wall , 1/3 height of wall, 48 in.} = 40 in.

• Decrease spacing to 40 in.

• 𝑀 = 778 k-ft, 1.25𝑀 = 972 k-ft


Shear Walls 79


• Bottom line: any change in wall will change 𝑀 , which will change design requirement

• Often easier to just use 𝑉 = 2.5𝑉 , or 𝜙𝑉 = 𝜙2.5𝑉 = 2.0𝑉 .

• Design for shear of 100 kips

Shear Walls 80


Shear Area:(6 - #4 bars)

𝐴 2 1.25in. 192in. 6 8in. 7.625in. 2.5in. 726in.

Max Shear:

𝜙𝑉 , 𝜙43

5 2𝑀


0.843

5 2 0.625 726in. 2000psi 0.75 97.4kip

Options:

• Design for shear from 1.25𝑀 = 97.2 kips

• Increase 𝑓 to 2100 psi, 𝜙𝑉 , = 99.8 kips

• requires a unit strength of 2250 psi

• Grout more cells

Shear Walls 81


Masonry Shear:

𝑉 4 1.75 𝐴 𝑓 0.25𝑃

4 1.75 0.625 726in. 2000psi 0.25 18600lb99.0kip

RequiredSteel Strength:

𝜙𝑉 𝜙 𝑉 𝑉 𝛾

𝑉 ,𝑉

𝜙𝛾𝑉

100k0.8 0.75

99.0k 67.7k

Determinespacing:Use #5 bars

𝑉 0.5𝐴𝑠

𝑓 𝑑 ⇒ 𝑠0.5𝐴 𝑓 𝑑

𝑉 ,

𝑠0.5 0.31in. 60ksi 192in.

67.7k26.4in.

Use #5 at 24 in. o.c.

Shear Walls 82


Shear Area: 𝐴 7.62in. 192in. 1464in.

Max Shear:𝜙𝑉 , 𝜙

43

5 2𝑀


0.843

5 2 0.625 1464in. 2000psi 1.0 261.9kip

Since horizontal bars are closely spaced, consider fully grouting wall

Masonry Shear:

𝑉 4 1.75 𝐴 𝑓 0.25𝑃

4 1.75 0.625 1464in. 2000psi 0.25 23200lb196.1kip

Wall weight:81psf 10ft 16ft 12.96k𝐷 1 k ft⁄ 16ft 12.96k 28.96k𝑃 0.80𝐷 0.80 28.96k 23.2k

Design Shear:

𝜙𝑉 𝜙 𝛾𝑉 0.8 1.0 196.1𝑘 156.9𝑘 𝑉 100𝑘 OK

Shear Walls 83

Shear Reinforcement

6.1.7.1.1 Except at wall intersections, the end of a horizontal reinforcing bar needed to satisfy shear strength requirements of Section 9.3.4.1.2 shall be bent around the edge vertical reinforcing bar with a 180-degree standard hook.

ASD: 𝑠 ≤ min{𝑑/2, 48 in.} = min{94 in., 48 in.} = 48 in. Code 8.3.5.2.1

In strength design, this provision only applies to beams (9.3.4.2.3 (e))

Suggest that minimum spacing also be applied to shear walls.

Special Reinforced WallsSection 7.3.2.6(d): Shear reinforcement shall be anchored around vertical reinforcing bars with a standard hook.

Shear Walls 84

• Prescriptive Reinforcement Requirements (7.3.2.6)

• 0.0007 in each direction

• 0.002 total

• Vertical: 6(0.20in.2)/1464in.2 = 0.00082 > 0.0007 OK

• Horizontal: 5(0.31in.2)/[120in.(7.625in.)] = 0.00169 > 0.0007 OK

• Total = 0.00082 + 0.00169 = 0.00251 > 0.002 OK

• Note: for fully grouted wall, would need #5 @ 32in. = 0.00127

Example: Special Wall

Shear Walls 85

• Calculate axial force based on 𝑐 = 83.8 in.

• Include compression reinforcement

• 𝜙𝑃 = 297.8 kips

• For 4𝜀 , 𝜙𝑃 = 160.4 kips

• Live load = 1 kip/ft; wall weight = 12 kip

• D + 0.75L + 0.525QE = 16k + 12k + 0.75(16k) = 40 kips

Section 9.3.3.5 Maximum ReinforcementSince 𝑀 𝑉 𝑑⁄ < 1, strain gradient is based on 1.5𝜀 .

OK

𝑐 = 0.446(188in.) = 83.8 in.

Strain c/d, CMU c/d, Clay

1.5𝜀 0.446 0.530

3𝜀 0.287 0.360

4𝜀 0.232 0.297


Shear Walls 86

• Section 9.3.6.5: Maximum reinforcement provisions of 9.3.3.5 do not apply if designed by this section (boundary elements)

• Special boundary elements not required if:

OK

𝑃 0.1𝑓 𝐴 geometrically symmetrical sections𝑃 0.05𝑓 𝐴 geometrically unsymmetrical sections

AND

For our wall, 𝑀 𝑉 𝑑⁄ < 1

𝑃 0.1𝑓 𝐴 0.1 2.0𝑘𝑠𝑖 1464𝑖𝑛. 293 𝑘


1 OR 𝑉 3𝐴 𝑓 AND 3

Shear Walls 87

• Prescriptive Reinforcement Requirements (7.3.2.6)

• 0.0007 in each direction

• 0.002 total

• Spacing Requirements (7.3.2.6)

• Shear Capacity Design (Section 7.3.2.6.1.1)

• 𝜙𝑉 shear corresponding to 1.25𝑀 .

• 𝑉 need not exceed 2.5𝑉

• Maximum Reinforcement Requirements (9.3.3.5; 9.3.6.5)

Special Wall: Summary

Shear Walls 88

Elevation of Shear Wall

Tie

Add 90° corner bar

Strut and Tie ModelConventional Model

Tens

ion

Ste

el

Moment Diagram

Detailing Suggestions

Partially grouted shear walls

• Use two grouted cells at end

• Use 9 gage joint reinforcement every course with high shear

Shear Walls 89

9.1.9.3.2 Maximum specified yield strength of 85,000 psi

• Typical yield strength is 70,000 psi

9.3.3.1 (b) Minimum diameter of 3/16 in. diameter.

9.3.3.3.2.3 Anchor around edge reinforcing bar in the edge cell

• by bar placement between adjacent crosswires, or

• 90° bend in longitudinal wires with 3-in. bend extensions in mortar or grout.

9.3.3.7 Seismic Requirements

• Seismic Design Categories (SDC) A and B

At least two 3/16 in. wires; Maximum spacing of 16 in.

• SDC C, D, E, and F; partially grouted walls

At least two 3/16 in. wires; Maximum spacing of 8 in.

• SDC C, D, E, and F; fully grouted walls

At least four 3/16 in. wires; Maximum spacing of 8 in.

Shear: Joint Reinforcement

Shear Walls 90

Equivalent Joint Reinforcement Options:

Joint ReinforcementEquivalent Bar Reinforcement

Replaces this Reinforcement

2 - 3/16 in. wires at 16 in. 0.0472 in2/ft #4 @ 56 in.; #5 @ 80 in.

2 – 3/16 in. wires at 8 in. 0.0945 in2/ft #4 @ 32 in.; #5 @ 40 in.

4 – 3/16 in. wires at 8 in. 0.189 in2/ft #4 @ 16 in.; #5 @ 24 in.

Bar reinforcement yield stress = 60 ksiJoint reinforcement yield stress = 70 ksi

Shear: Joint Reinforcement

7.3.2.6 Special reinforced masonry shear walls —(b) The maximum spacing of horizontal reinforcement required to resist in-plane shear shall be uniformly distributed, shall be the smaller of one-third the length of the shear wall and one-third the height of the shear wall, and shall be embedded in grout.

Shear Walls 91

Shear Reinforcement

6.1.7.1.1 Except at wall intersections, the end of a horizontal reinforcing bar needed to satisfy shear strength requirements of Section 9.3.4.1.2 shall be bent around the edge vertical reinforcing bar with a 180-degree standard hook.

6.1.7.1.2 At wall intersections, horizontal reinforcing bars needed to satisfy shear strength requirements of Section 9.3.4.1.2 shall be bent around the edge vertical reinforcing bar with a 90-degree standard hook and shall extend horizontally into the intersecting wall a minimum distance at least equal to the development length.

Special Reinforced WallsSection 7.3.2.6(d): Shear reinforcement shall be anchored around vertical reinforcing bars with a standard hook.

Shear Walls 92

Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type S mortar; f’m=2000psi; superimposed dead load of 1 kip/ft. In-plane seismic load (from ASCE 7-10) of 100 kips. SDS = 0.4; intersecting wall on one side.

Required: Design the shear wall; ordinary reinforced shear wall

Solution: Check using 0.9D+1.0E load combination.

1 k/ft

16ft

Effective flange width:

6t = 6(8in.) = 48 in. compression

0.75h = 0.75(120in.) = 90 in. tension

Example: T-Wall

Shear Walls 93

Flange in tension: Approximate as a single layer of reinforcement.Use design procedure for single layer of reinforcement with Pu = 18.4k, Mu = 1000k-ft., b=7.625in., lw=200in. (16 ft + 8 in. flange); d=196in. Required reinforcement: a = 5.9in., As = 0.87in2 or 3 - #5.

Flange in compression: Reinforcement will be approximately the same as for a non-flanged wall. Increase in compression area will only slightly reduce required steel.

Axial load on just the web creates a moment with a small tension in the flange and compression in the web.

Example: T-Wall

Shear Walls 94

Trial Design

Tension: 3 bars total; assume spacing of 48 in. for OOP90 in. flange

Compression: 3 grouted cells within flange

Flange compression area:(48+48+7.62)(2.5) + 3(8)(7.625-2.5) = 382in2

Equivalent thickness of flange = 382in2/7.625in. = 50.1in.

Com

pres

sion

Wid

th

Tens

ion

Wid

th

All reinforcement #5 bars

Example: T-Wall

Shear Walls 95

Check design: conservatively use P =18.4 kips from before(wall weight changes slightly for different loading directions due to different flange widths)

Flange in compression:

Example: T-Wall

Shear Walls 96

Flange in tension:

Example: T-Wall

Shear Walls 97

Check maximum reinforcement with flange in tension:

c/d = 0.446 (α = 1.5)

c = 0.446(196in.) = 87.4 in.

𝜙Pn = 344.3 kips

Pu = D+0.75L

= (12.4+16) +0.75(16)

= 40.4 kips

Perhaps should include steel just outside effective tension flange:

𝜙Pn = 310.8 kips

Example: T-Wall

Shear Walls 98

Shear strength is based on the web, and similar to previous example.

Use #5 @ 24 in.

Example: T-Wall

Shear Walls 99

Check shear at interface. Check using intersecting bond beams.

#[email protected]

in

ft

in

in

in 2

155.01

.12

.24

31.0 2

Shear at interface: V = tension force in reinforcement in flange.

25055.262.7851205.2 ininininininAnv

kipspsiin

fAV

lbkip

mnvnm

6.40200050525.28.0

25.2

100012

If used 𝛾 = 0.75, 𝜙𝑉 30.5 kips

Example: T-Wall

Min. reinf.0.1in2 per foot OK

kipksiinfATVflangeys 8.556031.03 2

Shear area, Anv

5 bond beams

Shear strength NG

Fully grout interface: 2915120625.7 inininAnv

kipspsiinV lbkip

nm 6.73200091525.28.0 100012 OK

Shear Walls 100

Additional cells to be grouted

Summary of intersecting walls

• Little savings in reinforcement

• Interface difficult to design

• Suggest isolating intersecting walls

Example: T-Wall

Documents

Shear Walls •Load Distribution to Shear WallsShear Walls 3 9.1.5.2 Deflection calculations shall be based on cracked section properties. Assumed properties shall not exceed half