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-1- ASD vs. SD A Masonry Cage Fight Richard Bennett, PhD, PE The University of Tennessee Chair, 2016 TMS 402/602 Code Committee -2- Outline Beams and lintels Non-bearing walls: out-of-plane Interaction diagrams Combined bending and axial force: pilasters Bearing walls: out-of-plane Shear walls: in-plane • Partially Grouted Shear Wall • Special Reinforced Shear Wall

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- 1 -

ASD vs. SDA Masonry Cage Fight

Richard Bennett, PhD, PE

The University of Tennessee

Chair, 2016 TMS 402/602 Code Committee

- 2 -

Outline

Beams and lintels

Non-bearing walls: out-of-plane

Interaction diagrams

Combined bending and axial force: pilasters

Bearing walls: out-of-plane

Shear walls: in-plane

• Partially Grouted Shear Wall

• Special Reinforced Shear Wall

- 3 -

2011 vs 2013 Code

Chapter 1 – General Design Requirements for Masonry

Chapter 2 – Allowable Stress Design of Masonry

Chapter 3 – Strength Design of Masonry

Chapter 4 – Prestressed Masonry

Chapter 5 – Empirical Design of Masonry

Chapter 6 – Veneer

Chapter 7 – Glass Unit Masonry

Chapter 8 – Strength Design of Autoclaved Aerated Concrete Masonry

Appendix B – Design of Masonry Infill

Part 1 – General• Chapter 1 – General Requirements• Chapter 2 – Notation and Definition• Chapter 3 – Quality and Construction

Part 2 – Design Requirements• Chapter 4 – General Analysis and Design Considerations• Chapter 5 – Structural Elements• Chapter 6 – Reinforcement, Metal Accessories, and

Anchor Bolts• Chapter 7 – Seismic Design Requirements

Part 3 – Engineered Design Methods• Chapter 8 – Allowable Stress Design• Chapter 9 – Strength Design of Masonry• Chapter 10 – Prestressed Masonry• Chapter 11 – Strength Design of AAC

Part 4 – Prescriptive Design Methods• Chapter 12 – Veneer• Chapter 13 – Glass Unit Masonry• Chapter 14 – Masonry Partition Walls

Part 5 – Appendices• Appendix A – Empirical Design • Appendix B – Design of Masonry Infill• Appendix C – Limit Design Method

- 4 -

2011 vs 2013 Code

Modulus of rupture values increased by 1/3 for all but fully grouted masonry normal to bed joint

Unit strength tables recalibrated

• Type S mortar, 2000 psi unit strength, f′m = 2000 psi

Shear strength of partially grouted walls: reduction factor of 0.75

- 5 -

Beam and Lintel Design

Strength Design (Chapter 9)

• mu = 0.0035 clay masonry• mu = 0.0025 concrete masonry• Masonry stress = 0.8f’m• Masonry stress acts over a = 0.8c• = 0.9 flexure; = 0.8 shear

mnvn

m

ysysn

fAV

bf

fAdfAM

25.2

8.02

1

• Minimum reinf: Mn ≥ 1.3Mcr

• or As ≥ (4/3)As,req’d

• Maximum reinf: s ≥ 1.5y

sm

m

y

m

f

f

8.08.0

max

Allowable Stress (Chapter 8)

• Allowable masonry stress: 0.45f’m• Allowable steel stress

• 32 ksi, Grade 60 steel

• No min or max reinforcement requirements

• Allowable shear stress

nnnk 2)( 2

31

kj

jdA

Mf

ss ))((

2

jdkdb

Mfm

mvm fF 25.22

1

- 6 -

Beam and Lintel Design

Strength Design (Chapter 9)

1. Determine a, depth of compressive stress block

2. Solve for As

3. ρ 0.285 ⁄ for CMU

Grade 60 steel:

= 1.5 ksi, ρ 0.00714

= 2 ksi, ρ 0.00952

Allowable Stress (Chapter 8)

1. Assume value of j (or k). Typically 0.85 < j < 0.95.

2. Determine a trial value of As./

Choose reinforcement.

3. Determine k and j; steel stress and masonry stress.

4. Compare calculated stresses to allowable stresses.

5. If masonry stress controls design, consider other options (such as change of member size, or change of f’m). Reinforcement is not being used efficiently.

bf

Mdda

m

n

8.0

22

y

ms f

bafA

8.0

- 7 -

ASD: Alternate Design Method

Calculate

Is k ≥ kbal?For Grade 60 steelkbal = 0.312 n

FF

Fk

sm

mbal

bF

Mddkd

m3

2

223

2

YES

11

2

)(

knF

PbkdF

A

m

m

s

31

kdF

MA

s

s

bF

nFA

s

ss

dkd 222

Iterate. Use (kd)2 as new guess and repeat.

NO

Allowable masonry compression stress

controls

Allowable reinforcement tension stress controls

- 8 -

Comparison of ASD and SD

8 inch CMUd = 20 in.

- 9 -

Example: Beam

Given: 10 ft. opening; dead load of 1.5 kip/ft; live load of 1.5 kip/ft; 24 in. high; Grade 60 steel; Type S masonry cement mortar; 8 in. CMU; f’m = 2000 psiRequired: Design beamSolution:

5.2.1.3: Length of bearing of beams shall be a minimum of 4 in.; typically assumed to be 8 in.

5.2.1.1.1 Span length of members not built integrally with supports shall be taken as the clear span plus depth of member, but need not exceed distance between center of supports.

• Span = 10 ft + 2(4 in.) = 10.67 ft

5.2.1.2 Compression face of beams shall be laterally supported at a maximum spacing of:

• 32 multiplied by the beam thickness. 32(7.625 in.) = 244 in. = 20.3 ft• 120b2/d. 120(7.625 in.)2 / (20 in.) = 349 in. = 29.1 ft

- 10 -

Allowable Stress Design: Flexure

ftk

ftwLM ft

k

1.458

67.1017.3

8

22

ftk

ftk

ftk

ftk ftLDw 17.35.12083.05.1 2 Load

Weight of fully grouted normal weight: 83 psf

Moment

Determine kd Assume compression controls

.32.9

625.7900.03

1.452

2

20

2

203

3

2

223

1222

ininksi

ftkinin

bF

Mddkd ft

in

b

312.0466.020

32.9

in

in

d

kdkCheck if compression controls Compression

controls

1.160.2900

29000

900

ksi

ksi

f

E

E

En

m

s

m

sCalculate modular ratio, n

- 11 -

Allowable Stress Design: Flexure

Find AsArea of steel

294.1

1466.01

900.01.16

2625.731.9900.0

112

)(

inksi

ininksi

knF

bkdF

A

b

b

s

Bars placed in bottom U-shaped unit, or knockout bond beam unit.

Use 2 - #9 (As = 2.00 in2)

- 12 -

Strength Design: Flexure

ftk

ftLwM ft

ku

u 6.628

67.1040.4

8

22

Use 2 - #6 (As = 0.88 in2)

ftk

ftk

ftk

ftk

u ftLDw 40.45.16.12083.05.12.16.12.1 2 Factored LoadWeight of fully grouted normal weight: 83 psf

Factored Moment

Find aDepth of equivalent rectangular stress block

Find AsArea of steel

ininksi

ftinftk

ininbf

Mdda

m

n 78.3625.70.28.0

129.0

6.622

20208.0

2 22

277.060

78.3625.70.28.08.0in

ksi

ininksi

f

bafA

y

ms

Mn = 78.5 k-ft

Mn = 70.6 k-ft

- 13 -

Check Min and Max Reinforcement

ftkinkpsiinfSM rncr 76.91.117160732 3

322

7326

24625.7

6in

ininbhSn

ftkMftkftkM ncr 3.757.1276.93.13.1

00577.020625.7

88.0 2

inin

in

bd

As

Minimum Reinforcement Check: fr = 160 psi (parallel to bed joints in running bond; fully grouted)

Maximum Reinforcement Check:

= 2 ksi ρ 0.00952

Section modulus

Cracking moment

Check 1.3Mcr

- 14 -

Summary: Beams, Flexure

Dead Load (k/ft)(superimposed)

Live Load (k/ft)Required As (in2)

ASD SD

0.5 0.50.34

0.34 ( = 1.5 ksi)

0.26

0.26 ( = 1.5 ksi)

1.0 1.00.64

0.65 ( = 1.5 ksi)

0.50

0.52 ( = 1.5 ksi)

1.5 1.51.94

5.09 ( = 1.5 ksi)

0.77

0.80 ( = 1.5 ksi)

ASD: Allowable tension controls for 0.5 k/ft and 1 k/ft.

- 15 -

Allowable Stress Design: Shear

Section 8.3.5.4 allows design for shear at d/2 from face of supports.

Design for DL = 1 k/ft, LL = 1 k/ft

kft

ftftkV 02.9

33.5

17.133.556.11

k

ftftwLV

ftk

ftk

ftk

56.112

67.100.12083.00.1

2

2

ftinin

inft 17.1.4

2

.20.12

1

Shear at reaction

d/2 from face of support

Design shear force

psipsi

fA

Pf

Vd

MF m

nmvm

3.50200025.22

1

25.22

125.075.10.4

2

1

psiinin

k

A

Vf

nvv 2.59

.20.625.7

02.9Shear stress

Allowable masonry shear stress

Suggest that d be used, not dv.

- 16 -

Allowable Stress Design: Shear

2034.0.20320005.0

.8.20.625.79.8

5.05.0

ininpsi

inininpsiA

dF

sAfA

sA

dFAF

v

s

nvsv

n

svvs

psipsifF mv 4.89200022

psipsipsi 9.83.502.59

Check max shear stress

Req’d steel stress

Determine Av for a spacing of 8 in.

> 59.2psi OK

Use #3 stirrups

Determine d so that no shear reinforcement would be required.

.5.233.50.625.7

02.9in

psiin

k

bF

Vd

vm

Use a 32 in. deep beam if possible;will slightly increase dead load.

- 17 -

Strength Design: Shear

Requirement for shear at d/2 from face of support is in ASD, not SD, but assume it applies

Design for DL = 1 k/ft, LL = 1 k/ft

kft

ftftkVu 50.12

33.5

17.133.500.16

k

ftftLwV

ftk

ftk

ftk

uu 00.16

2

67.100.16.12083.00.12.1

2

2

ftinin

inft 17.1.4

2

.20.12

1

Shear at reaction

d/2 from face of support

Design shear force

kpsiinin

fAPfAdV

MV mnvumnv

vu

unm

28.122000.20.625.725.28.0

25.225.075.10.4

Design masonry

shear strength

Suggest that dbe used, not dv to find Anv

- 18 -

Strength Design: Shear

2004.0.20605.0

.828.0

5.05.0

ininksi

inkA

df

sVAdf

s

AV

v

vy

nsvvy

vns

kpsiininfAV mnvn 82.212000.20.625.748.04

kkkVV mu 28.0

8.0

28.1250.12

Check max Vn

Req’d Vns

Determine Av for a spacing of 8 in.

> 12.50k OK

Use #3 stirrups

Determine d so that no shear reinforcement would be required.

.4.20200025.28.0.625.7

50.12

25.2in

psiin

k

fb

Vd

m

u

Use inverted bond beam unit to get slightly greater d.

- 19 -

Non-Bearing Partially Grouted WallsOut-of-Plane

s

b’

b

d

a

t f

As

b = width of effective flange = min{s, 6t, 72 in}

A. Neutral axis in flange:a. Almost always the caseb. Design and analysis for solid

sectionB. Neutral axis in web

a. Design as a T-beam sectionC. Often design based on a 1 ft width

Spacing

(inches)

Steel Area in2/ft

#3 #4 #5 #6

8 0.16 0.30 0.46 0.66

16 0.082 0.15 0.23 0.33

24 0.055 0.10 0.16 0.22

32 0.041 0.075 0.12 0.16

40 0.033 0.060 0.093 0.13

48 0.028 0.050 0.078 0.11

56 0.024 0.043 0.066 0.094

64 0.021 0.038 0.058 0.082

72 0.018 0.033 0.052 0.073

80 0.017 0.030 0.046 0.066

- 20 -

Example: Partially Grouted Wall ASD

Given: 8 in CMU wall; 16 ft high; Grade 60 steel, = 2000 psi; Lateral wind load of 30 psf (factored)Required: Reinforcing (place in center of wall)Solution:

ftftlb

ftinlbft

inftlb ftwh

M 57669128

1612306.0

8

22 2

Fb = 0.45(2000psi) = 900 psi Em = 1.80 x 106 psiFs = 32000 psi Es = 29 x 106 psin = Es/Em = 16.1

Moment

Determine kd Assume compression controls

.346.0

12900.03

576.02

2

81.3

2

81.33

3

2

223

1222

inksi

inin

bF

Mddkd

ftin

ftin

ftftk

b

312.0091.081.3

346.0

in

in

d

kdkCheck if compression

controlsTension controls

- 21 -

Example: Partially Grouted Wall ASD

31

kdF

MA

s

s

bF

nFA

s

ss

dkd 222

Equation / Value Iteration 1 Iteration 2 Iteration 3

kd (in.) 0.346 0.699 0.709

k 0.091 0.183 0.183

(in2) 0.0584 0.0603 0.0603

(in.) 0.0784 0.0810 0.0810

(in.) 0.699 0.709 0.709

Use # 4 @ 40 inches (As=0.060in2/ft)

(close enough for government work)

- 22 -

Example: Partially Grouted Wall SD

Given: 8 in. CMU wall; 16 ft high; Grade 60 steel, = 2000 psi; Wind load of wu = 30 psfRequired: Reinforcing (place in center of wall)Solution:

ft

ftkftinlbft

inftlb

uu

fthwM 96.011520

8

161230

8

22 2

Use # 4 @ 40 inches (As=0.060in2/ft)

inpsi

ininbf

Mdda

ftin

ftinlb

m

n 179.0)12(20008.0

9.0

115202

81.381.38.0

2 22

ft

inftin

y

ms psi

inpsi

f

bafA

2

0573.060000

179.01220008.08.0

Factored Moment

Find aSolve as solid section

Find As

- 23 -

ASD vs SD: Flexural Members

ASD calibrated to SD for wind and seismic loads

When a significant portion of load is dead load, ASD will require more steel than SD

When allowable masonry stress controls in ASD, designs are inefficient

Advantage to SD, which is reason concrete design rapidly switched to SD about 50 years ago

- 24 -

Interaction Diagrams

Strength Design (Chapter 9)

• Set masonry strain to εmu

• Vary steel strain• Equivalent rectangular stress block• Nominal axial strength

• h/r ≤ 99

• h/r > 99

• =0.9

Allowable Stress (Chapter 8)

• For k > kbal

• Set masonry strain = Fb/Em; = 0.0005 CMU

• Find steel strain• For k < kbal

• Set steel strain = Fs/Es;= 0.00110 for Grade 60

• Find masonry strain• Allowable axial load

• h/r ≤ 99

• h/r > 99

140

180.08.02

r

hAfAAfP stystnmn

2

h

70r80.080.0

stystnmn AfAAfP

2

7065.025.0

h

rFAAfP sstnma

2

140165.025.0

r

hFAAfP sstnma

- 25 -

Example: 8 in. CMU Bearing Wall ASD

Given: 12 ft high CMU bearing wall, Type S masonry cement mortar; Grade 60 steel in center of wall; #4 @ 48 in.; partial grout; = 2000 psiRequired: Interaction diagram in terms of capacity per foot

Pure Moment: n = 16.1 ρ = 0.00109 nρ = 0.0176

ftftk

ftk-inM 479.0.755

ftink

ftin

sss inksijdFAM 75.5.81.3943.03205.02

ftink

ftinm

m inksi

injdF

kdbM

64.12

81.3943.02

900.081.3171.012

2

171.00176.00176.020176.02)( 22 nnnk

943.03

171.01

31

kj

Find k

Find j

Find Ms

Find Mm

Allowable M

- 26 -

Example: 8 in. CMU Bearing Wall ASD

Pure Axial:

ftkP 3.17

991.5466.2

144

in

in

r

h

NCMA TEK 14-1B Section Properties of Concrete Masonry Wallsr = 2.66 in. An = 40.7in2/ft In = 332.0 in4/ft

ftk

ftin

sstnma

ksi

r

hFAAfP

3.17140

1.54107.400.225.0

140

165.025.0

2

2

2

Find h/r

Find Pa

4.3.3 Radius of gyration Radius of gyration shall be computed using average net cross-sectional area of the member considered.

- 27 -

Example: 8 in. CMU Bearing Wall ASD

Find Cm

Find T

Find P

Find M

Balanced:

ftftk

ftk

M

P

83.1

.844

ftk

ftk

m-TCP 84.460.144.6

ftk

ftin

mm inksibkdfC 44.61219.1900.02

1

2

1

Stress

ftftk

ftink

ftk in

inM

83.10.22

3

19.181.344.6

0.900 ksi32 ksi

Strain0.0005

0.00110

3.82 in

1.19 inkd < 1.25 in.N.A. in face shell

ftk

ftin

ss ksiAfT 6.105.0322

- 28 -

Example: 8 in. CMU Bearing Wall ASD

Find Cm

Find T

Find P

Find M

BelowBalanced:kd = 1.00 in.

ftftk

ftk

M

P

22.1

.622

ftk

ftk

m-TCP 62.260.122.4

ftk

ftin

mm inksibkdfC 22.41200.1703.02

1

2

1

Stress

ftftk

ftink

ftk in

inM

22.17.14

3

00.181.322.4

1800ksi(0.000391) =0.703 ksi

32 ksi

ftk

ftin

ss ksiAfT 6.105.0322

Strain0.000391

0.00110

3.82 in

1.00 in

- 29 -

Example: 8 in. CMU Bearing Wall ASD

AboveBalanced:kd = 2.00 in.

Stress

0.900 ksi

13.1 ksiStrain0.0005

0.00045

3.82 in

2.00 in

Neutral axis is in web

ftk

mm -TCCP 88.8 21

ftftk

ftk

ftk ininininM 58.25.181.325.053.081.328.9

ftk

ftinksiT 66.005.01.13

2

2.00 in

9.28 k/ft

0.66k/ft

1.25+0.75/3=1.5in.

0.338 ksi

1.25 in

ftk

ftin

m inksi

C 28.91225.12

338.0900.01

0.25 k/ft

ftk

ftin

m inksiC 25.0275.0338.02

12

1.5 in

0.53 in

h2

h1

b

x

Centroid

3

2

21

21 b

hh

hhx

ftftk

ftk

M

P

58.2

.888

- 30 -

Example: 8 in. CMU Bearing Wall SD

Given: 12 ft high CMU bearing wall, Type S masonry cement mortar; Grade 60 steel in center of wall; #4 @ 48 in.; partial grout; = 2000 psiRequired: Interaction diagram in terms of capacity per foot

Pure Moment:

Find Mn

Find Mn

ft

ftkftink

ftin

ftin

ftin

m

ysysn

ksi

ksiksi

bf

fAdfAM

934.02.110.2128.0

6005.0

2

1.812in 36005.0

'8.02

1

2

2

ftftk

ftftk

nM 840.0934.00.9

in

ksi

ksi

bf

fAa

ftin

ftin

m

ys 156.00.2128.0

6005.0

'8.0

2

Check to make sure stress block is in face shell

- 31 -

Example: 8 in. CMU Bearing Wall SD

Pure Axial:

991.5466.2

144

in

in

r

h

NCMA TEK 14-1B Section Properties of Concrete Masonry Wallsr = 2.66 in. An = 40.7in2/ft In = 332.0 in4/ft

Find h/r

Find Pn

ftk

ftk

nP 9.393.440.9

ftk

ftin

stystnmn

ksi

r

hAfAAfP

3.44140

1.541007.400.280.08.0

140

180.08.0

2

2

2

- 32 -

Example: 8 in. CMU Bearing Wall SD

Find Cm

Find T

Balanced: Strain

Stress

0.0025

0.002073.82 in

2.09 in

T

Cm

a = 0.8c = 0.8(2.09in.) = 1.67in.

web length = .

.

. 2.0

ftk

ftk

nP 1.200.33.10.240.9

ftk

ftin

webm ininksiC 34.10.225.167.10.280.0,

ftftk

ftk

ftk

n inin

inM

96.5

2

25.167.125.181.33.1

2

25.181.30.249.0

ftk

ftin

shellfacem inksiC 241225.10.280.0,

ftk

ftin

sy ksiAfT 0.305.0602

Find

Find

ftftk

n

ftk

n

M

P

96.5

1.20

0.8f′m = 1.6 ksi

- 33 -

Example: 8 in. CMU Bearing Wall SD

Find Cm

Find T

Find

Find

BelowBalanced:c = 1.25 in.

ftk

ftin

mm inksibafC 2.191200.10.28.08.0

ftk

ftin

sy ksiAfT 0.305.0602

Strain

Stress

0.0025

0.005123.81 in

1.25 in

T

ftk

ftk

mn -TCP 6.140.32.199.0 ft

ftkn

ftk

n

M

P

77.4

6.14

ft

ftkftink

ftk

n

ininM

77.42.57

2

25.18.081.32.199.0

0.8f′m = 1.6 ksi

a = 0.8c1.00 in

- 34 -

Example: 8 in. CMU Bearing Wall SD

AboveBalanced:c = 3.0 in.

Strain0.0025

0.000683.81 in

3.0 inStress

T

Cm

a = 0.8c = 0.8(3.0in.) = 2.4in.

web length = .

.

. 2.0

0.8f′m = 1.6 ksi

Find Cm

Find T

ftk

ftk

nP 0.2499.068.30.240.9

ftk

ftin

webm ininksiC 68.30.225.140.20.280.0,

ftftk

ftk

ftk

n inin

inM

28.6

2

25.140.225.181.368.3

2

25.181.30.249.0

ftk

ftin

shellfacem inksiC 241225.10.280.0,

ftk

ftin

sss ksiAET 99.005.000068.0290002

Find

Find

ftftk

n

ftk

n

M

P

28.6

0.24

- 35 -

Interaction Diagrams

- 36 -

ASD vs SD: Interaction Diagrams

Similar behavior to flexure

• ASD and SD close when allowable tension stress controls

• ASD more conservative when allowable masonry stress controls

SD interaction diagram easier to construct due to equivalent uniform stress vs. linear varying stress

Advantage to SD

- 37 -

Strength Design: Combined Bending and Axial Load Design Method

bf

MtdPdda

m

uu

8.0

2/22

y

ums f

PbafA

/8.0

Calculate

Is c ≥ cbal?For Grade 60 steelcbal = 0.547

YES

Compression controlsTension controls

dcymu

mubal

NO

8.0

ac

ccd

E

PbafA

smu

ums

/8.0

- 38 -

ASD: Combined Bending and Axial Load Design Method

Calculate

Is k ≥ kbal?For Grade 60 steelkbal = 0.312 n

FF

Fk

sm

mbal

bF

MtdPddkd

m3

))2/((2

223

2

YES

11

2

)(

knF

PbkdF

A

m

m

s

32

kdtPM

31

kdF

MMA

s

s

bF

nFAP

s

ss

dkd 222

Iterate. Use (kd)2 as new guess and repeat.

NO

Compression controls

Tension controls

- 39 -

ASD: Combined Bending and Axial Load Design Method If k < kb tension controls. Solve cubic equation for kd.

02226

23

dM

tdPkdM

tdPkd

n

bdFkd

n

bF ss

s

s F

P

kdd

kd

nbkdA

1

2

1

Solve for As.

- 40 -

Example: Pilaster Design ASD

Given: Nominal 16 in. wide x 16 in. deep CMU pilaster; = 2000 psi; Grade 60 bar in each corner, center of cell; Effective height = 24 ft; Dead load of 9.6 kips and snow load of 9.6 kips act at an eccentricity of 5.8 in. (2 in. inside of face); Factored wind load of 26 psf (pressure and suction) and uplift of 8.1 kips (e=5.8 in.); Pilasters spaced at 16 ft on center; Wall is assumed to span horizontally between pilasters; No ties.Required: Determine required reinforcing using allowable stress design.Solution:

Ver

tical

Spa

nnin

g

d=11.8 in

xLoad

e=5.8 in 2.0 in

Em = 1800ksin = 16.1

Lateral Loadw = 0.6(26psf)(16ft)=250 lb/ft

Insi

de

- 41 -

Example: Pilaster Design ASD

Weight of pilaster:Weight of fully grouted 8 in wall (lightweight units) is 75 psf. Pilaster is like a double thick wall. Weight is 2(75psf)(16in)(1ft/12in) = 200 lb/ft

Usually the load combination with smallest axial load and largest lateral load controls. Try load combination of 0.6D + 0.6W to determine required reinforcement and then check other load combinations.

The location and value of maximum moment can be determined from:

2

22

max282 wh

MwhMM

wh

Mhx

2

If x<0 or x>h, Mmax= M

M = moment at topx measured down from top of pilaster

- 42 -

Example: Pilaster Design ASD

0.6D + 0.6W

Top of pilaster. P = 0.6(9.6)k-0.6(8.1)k = 0.9kips M = 0.9k(5.8in) = 5.2kip-in

inin

inkipin

wh

Mhx

inft

ftk

1.1431288250.0

2.5

2

288

2

inkin

inkinink

wh

MwhMM

ink

ink

3.2182880208.02

)2.5(

8

2880208.0

2

2.5

282 2

22

2

22

max

Find axial force at this point. Include weight of pilaster (200 lb/ft).

kinkPinft

ftk 3.21.14320.06.09.0

.121

Design for P = 2.3 k, M = 218 k-in

Find location of maximum moment

- 43 -

Example: Pilaster Design ASD

in

inksi

inkipininkipinin

bF

MhdPddkd

b

00.36.15900.03

)2182/6.158.113.2(2

2

8.11

2

8.113

3

))2/((2

223

2

2

254.08.11

00.3

in

in

d

kdk

312.0

1.1632

900.0

900.0

ksiksi

ksi

n

FF

Fk

sb

bbal

k < kbal Tension controls; iterate

Assume compression controls;Determine kd

Determine k

Determine kbal

- 44 -

Example: Pilaster Design ASD

Equation / Value Iteration 1 Iteration 2 Iteration 3

kd (in.) 3.00 3.38 3.40

k 0.254 0.286 0.288

(k-in.) 15.6 15.3 15.3

(in2) 0.585 0.593 0.593

(in.) 0.678 0.686 0.686

(in.) 3.38 3.40 3.40

32

kdhPM

31

kdF

MMA

s

s

bF

nFAP

s

ss

dkd 222

Try 2-#5, 4 total, one in each cell

- 45 -

Example: Pilaster Design ASD

0.6D+0.6WP = 2.3kM = 18.2k-ft

D+0.75(0.6W)+0.75SP = 15.3kM = 16.8k-ft

D+SP = 19.2kM = 9.25k-ft

D+0.6WP = 7.1kM = 19.2k-ft

- 46 -

Example: Pilaster Design ASD

f’m = 2000 psi

f’m = 1500 psi

- 47 -

Example: Pilaster Design SD

Given: Nominal 16 in. wide x 16 in. deep CMU pilaster; = 2000 psi; Grade 60 bar in each corner, center of cell; Effective height = 24 ft; Dead load of 9.6 kips and snow load of 9.6 kips act at an eccentricity of 5.8 in. (2 in. inside of face); Factored wind load of 26 psf (pressure and suction) and uplift of 8.1 kips (e=5.8 in.); Pilasters spaced at 16 ft on center; Wall is assumed to span horizontally between pilasters; No ties.Required: Determine required reinforcing using strength design.Solution:

Ver

tical

Spa

nnin

g

d=11.8 in

xLoad

e=5.8 in 2.0 in

Em = 1800ksin = 16.1

Lateral Loadw = 0.6(26psf)(16ft)=250 lb/ft

Insi

de

- 48 -

Example: Pilaster Design SD

0.9D + 1.0W

At top of pilaster: kkkPu 54.01.80.16.99.0

inkinkePM uu 1.3.8.554.0

inft

inkipin

wh

Mhx

ftk

7.14324416.0

1.3

2

288

2

inkin

inkinink

wh

MwhMM

ink

ink

u

0.3612880347.02

)1.3(

8

2880347.0

2

1.3

282 2

22

2

22

Find axial force at this point. Include weight of pilaster.

kinkPinft

ftk

u 69.27.14320.09.054.0121

Design for Pu = 2.7 kips, Mu = 361 k-in

Find location of maximum moment

- 49 -

Example: Pilaster Design SD

in

inksi

inkininkinin

bf

MhdPdda

m

uu

50.16.150.28.09.0

3612/6.158.117.228.118.11

8.0

2/2

2

2

257.060

9.0/7.250.162.150.28.0/8.0in

ksi

kininksi

f

PbafA

y

ums

Determine a

Required steel = 0.57 in2

Use 2-#5 each face, As = 0.62 in2

Total bars, 4-#5, one in each cell

Determine As

- 50 -

Example: Pilaster Design SD

- 51 -

Example: Pilaster Design

- 52 -

ASD vs SD: Pilaster Design

Similar behavior to before

• ASD and SD close when allowable tension stress controls

• ASD more conservative when allowable masonry stress controls

• Less reinforcement required with SD due to small dead load factor

SD design easier as steel has generally yielded

Advantage to SD

- 53 -

Design: Bearing Walls – OOP Loads

Strength Design (Chapter 9)Allowable Stress(Chapter 8)

• No second-order analysis required

• Use previous design procedure

• Second order analysis required• Assumes simple support conditions• Assumes uniform load• Assumes midheight moment is

approximately maximum moment• Valid only for following conditions:

• 0.05 No height limit

• 0.20 height limited by

30

• Need to check maximum reinforcement limits

- 54 -

Strength Design Procedure

uuu

ufu

u Pe

Phw

M 28

2

ufuwu PPP

Puf = Factored floor loadPuw = Factored wall load

Moment: Deflection:

crunm

uu MM

IE

hM

48

5 2

cru

crm

cru

nm

cru MM

IE

hMM

IE

hM

48

5

48

5 22

32

32 bc

cdd

t

f

PAnI sp

y

uscr

bf

PfAc

m

uys

'64.0

Deflection Limit hs 007.0 Calculated using allowable stress load combinations

- 55 -

Strength Design Procedure

48

51

12848

5

48

51

11

48

5

28

2

22

2

22

crm

u

n

crcr

uuf

u

crmu

crm

u

crnm

ucruuf

u

u

IE

hP

I

IM

eP

hw

IEh

IE

hP

IIE

hPMeP

hw

M

nm

u

uuf

u

nmu

nm

u

uuf

u

u

IE

hP

eP

hw

IE

h

IE

hP

eP

hw

M

48

51

2848

5

48

51

28

2

22

2

2

Solve simultaneous linear equations:

Mu > Mcr Mu < Mcr

- 56 -

Example: Wind Loads ASD16

ft

32 p

sfD = 500 lb/ftLr = 400 lb/ftW = -360 lb/ft

2.67

ft

Given: 8 in. CMU shear wall; Grade 60 steel; Type S masonry cement mortar; = 2000 psi; roof forces act on 3 in. wide bearing plate at edge of wall.

Required: Reinforcement

Solution:

Estimate reinforcement

~8

0.6 0.032 168

0.62

Try #4 @ 40 in.

k = 0.185, kd = 0.707 in.

Mm = 1.14 k-ft/ft; Ms = 0.57 k-ft/ft

Lightweight units: wall weight = 40 psf

Cross-sectionof top of wall

Determine eccentricitye = 7.625in/2 – 1.0 in.= 2.81 in.

- 57 -

Example: Wind Loads ASD

Load Comb. Pf (kip/ft) P (kip/ft) Mtop (k-ft/ft) M (k-ft/ft) As (in2)

0.6D+0.6W 0.084 0.340 -0.049 0.590 0.051

D+0.6W 0.284 0.711 -0.002 0.613 0.042

ft

ftkftftk

top ftksfMwhM

590.02

049.0

8

16032.06.0

28

122

Check 0.6D+0.6W

084.036.06.05.06.0 ftk

ftk

ftk

fP

/340.0867.2040.06.0084.0 ftkftftksfP ftk

049.0

2

67.2032.06.0.81.2084.0

2

121

ftftk

inft

ftk

top

ftksfinM

Use #4@ 40 in. (0.06 in2/ft)Although close to #4@48, a wider spacing also reduces wall weight (As = 0.052in2/ft)

Find force at top of wall

Find force at midheight

Find moment at top of wall

Find moment at midheight

- 58 -

Example: Wind Loads ASD

Sample Calculations: 0.6D+0.6W

1. kbal = 0.312; kdbal = 1.19in.2. Assume masonry controls.

Determine kd.Since 0.355 in. < 1.18 in.

tension controls. 3. Iterate to find As.

.355.0

12900.03

590.02

2

.81.3

2

.81.33

3

2

223

.122

2

inksi

inin

bF

Mddkd

ftin

ftin

ftftk

b

3/2/ kdtPM sp

3/1 kdF

MMA

ss

bF

nFAP

s

ss

dkd 222

Equation / Value Iteration 1 Iteration 2

kd (in.) 0.355 0.707

(k-ft/ft) 0.1047 0.1013

(in2/ft) 0.049 0.051

(in.) 0.0804

(in.) 0.707

- 59 -

Example: Wind Loads SD

16 ft

32 p

sf

D = 500 lb/ftLr = 400 lb/ftW = -360 lb/ft

2.67

ft

Given: 8 in. CMU wall; Grade 60 steel; Type S masonry cement mortar; = 2000 psi ; roof forces act on 3 in. wide bearing plate at edge of wall.

Required: Reinforcement

Solution:

Estimate reinforcement

~8

0.032 168

1.02

a = 0.19 in.

As = 0.061 in2/ft

Try #4 @ 40 in.

Cross-sectionof top of wall

Determine eccentricitye = 7.625in/2 – 1.0 in.= 2.81 in.

- 60 -

Example: Wind Loads SD

Summary of Strength Design Load Combination Axial Forces

(wall weight is 40 psf for 40 in. grout spacing)

Load CombinationPuf

(kip/ft)

Puw

(kip/ft)

Pu

(kip/ft)

0.9D+1.0W0.9(0.5)+1.0(-0.36) =

0.0900.9(0.040)(2.67+8) =

0.3840.474

1.2D+1.0W+0.5Lr1.2(0.5)+1.0(-0.36) +0.5(0.4) = 0.440

1.2(0.040)(2.67+8) = 0.512

0.952

Puf = Factored floor load; just eccentrically applied loadPuw = Factored wall load; includes wall and parapet weight, found at

midheight of wall between supports (8 ft from bottom)

- 61 -

Example: Wind Loads SD

Find modulus of rupture; use linear interpolation between no grout and full grout

Ungrouted (Type S masonry cement): 51 psi

Fully grouted (Type S masonry cement): 153 psi

psicells

cellgroutedpsi

cells

cellsungroutedpsifr 71

5

1153

5

451

Find Mcr, cracking moment:

Commentary allows inclusion of axial load

Use minimum axial load (once wall has cracked, it has cracked)

ftftkip

ftinkip

cr

ftin

ftin

ftlb

nrnucr

M

in

psi

t

IfAPM

606.028.7

81.3

7.336718.42/474

2/

/42

Wall properties determined from NCMA TEK 14-1B Section Properties of Concrete Masonry Walls

- 62 -

Example: Wind Loads SD

Find Icr

in

ksi

ksi

bf

PfAc

ftin

ftk

ftin

m

uys 265.00.264.0

474.06006.0

'64.0 12

2

1.161800

29000

ksi

ksi

E

En

m

s

ftin

ftin

ftk

ftin

sp

y

uscr

ininin

ksi

bccd

d

t

f

PAnI

4

2

8.13

3

265.012265.0812.3

60

474.006.01.16

32

32

32

Find c

Find n

- 63 -

Example: Wind Loads SD

ftftk

ftin

ftk

ftin

ftin

ftk

ftftk

ftftk

crm

u

crnm

ucruuf

u

u

ftin

ksi

ft

ftin

ksi

ftftksf

IEhP

IIEhPMe

Phw

M

009.1

1144

8.13180048

16474.051

1144

8.13

1

337

1180048

16474.0606.05

2

093.0

816032.0

485

1

1148

528

2

22

2

222

2

22

4

44

Pufeu is the moment at the top support of the wall, Mu,top. It includes eccentric axial load and wind load from parapet.

ft

ftkftksfin

hwePM in

ftftkparapetparapetu

uuftopu

093.0

2

67.2032.081.2090.0

2

2

121

2,

,

Find Mu.

- 64 -

Example: Wind Loads SD

ftftkip

ftinkip

ftin

ftk

ysun

ininksi

adfAPM

147.176.13

2

215.0812.36006.09.0/474.09.0

2/

2

in

ksi

ksi

bf

PfAa

ftin

ftk

ftin

m

uys 215.0120.280.0

9.0/474.06006.0

80.0

/2

nftftk

ftftk

u MM 15.101.1 OK

Check other load combinations:

1.2D+1.0W+0.5Lr, Mu = 1.09 k-ft/ft and Mn = 1.29 k-ft/ftOK

Find a

Check area of steel:

Find

Compare

- 65 -

Example: Wind Loads SD

Check Deflections: Use ASD Load Combinations

Load Combination D+0.6W 0.6D+0.6W

Pf (k/ft) 0.5+0.6(-0.36) = 0.284 0.6(0.5)+0.6(-0.36) = 0.084

Pw (k/ft) 40(2.67+8) = 0.427 0.6(0.427) = 0.256

P (k/ft) 0.711 0.340

c (in) 0.280 0.256

Icr (in4/ft) 14.5 13.4

Mtop (k-ft/ft) -0.002 -0.049

δ (in) 0.066 0.044

M (k-ft/ft) 0.617 (cracked) 0.565 (uncracked)

Deflection Limit ininhs 34.1)192(007.0007.0 OK

- 66 -

Example: Wind Loads SD

Deflections, Sample Calculations (D+0.6W):Replace factored loads with service loads

.066.0

1144

5.14180048

16711.051

11728

17.336

5.14606.0

2

002.0

816032.06.0

5.14180048

165

485

1

12848

5

2

22

3

322

2

22

4

4

4

4

in

ftin

ksi

ft

ftinftksf

ksi

ft

IEPh

II

Me

Pwh

IEh

ftin

ftk

ftin

ftin

ftftkft

ftk

ftin

crm

n

crcrf

crm

Check that M > Mcr; M = 0.617 k-ft/ft > 0.606 k-ft/ft = Mcr

- 67 -

Example: Wind Loads SD

Check Maximum Reinforcement:• neutral axis is in face shell• Pu is just dead load = 0.5+0.04(2.67+8) = 0.927k/ft

00918.0

60

81.312

927.0

00207.05.10025.00025.0

0.264.0

64.0

max

ksi

inksi

f

bdP

f

bd

A

ftin

ftk

y

u

ymu

mum

s

00131.081.312

06.02

inbd

A

ftin

ftft

s OK

- 68 -

ASD vs SD: Bearing Walls

ASD and SD require approximately the same amount of reinforcement for reasonably lightly loaded walls

SD requires a second-order analysis and a check of deflections

Advantage to ASD

• Interestingly OOP loading is where designers often use SD

- 69 -

Shear Walls: Shear

Strength Design (Chapter 9)Allowable Stress (Chapter 8)

umnvvu

unm PfA

dV

MV 25.075.10.4

gmnvn fAV 6 25.0/ vuu dVM

vyv

ns dfs

AV

5.0

gnsnmn VVV 8.0

gmnvn fAV 4 0.1/ vuu dVM

gmnvvu

un fA

dV

MV

25

3

40.125.0

vu

u

dV

M

gvsvmv FFF

nm

vvm A

Pf

Vd

MF 25.075.10.4

2

1

sA

dFAF

n

svvs 5.0

gmv fF 3 25.0/ vVdM

gmv fF 2 0.1/ vVdM

nm

vvm A

Pf

Vd

MF 25.075.10.4

4

1

Special Reinforced Shear Walls

Interpolate for 0.25 < (M/Vdv) < 1

= 0.75 partially grouted shear walls; 1.0 otherwise

- 70 -

Shear Walls: Maximum Reinforcement

Strength Design (Chapter 9)Allowable Stress (Chapter 8)

Special reinforced shear walls having• M/(Vd) ≥ 1 and• P > 0.05f′mAn

m

yy

m

f

fnf

fn

2max

• Provide boundary elements, or• Limit reinforcement

Boundary elements not required if:

gA1.0 mu fP symmetrical sections

gA05.0 mu fP unsymmetrical sections

AND

1wu

u

lV

M OR 3wu

u

lV

Mmnu fAV 3

Reinforcement limits: • Maximum stress in steel of αfy• Axial forces D+0.75L+0.525QE

• Compression reinforcement, with or without lateral ties, permitted to be included

= 1.5 ordinary walls; all others = 3 intermediate walls with Mu/(Vudv)≥1 = 4 special walls with Mu/(Vudv)≥1

- 71 -

Shear Walls: Shear Capacity DesignSpecial Reinforced Shear Walls

Strength Design (7.3.2.6.1.2)Allowable Stress (7.3.2.6.1.2)

• Seismic design load required to be increased by 1.5 for shear

• Masonry shear stress reduced for special walls

• Design shear strength, Vn, greater than shear corresponding to 1.25 times nominal flexural strength, Mn

(increases shear at least 1.39 times)

• Except Vn need not be greater than 2.5Vu. (doubles shear)

- 72 -

Example: Shear Wall ASD

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

Required: Design the shear wall; ordinary reinforced shear wall

Solution: Check using 0.6D+0.7E load combination.

• Try #6 in each of last 3 cells; #6 @40in.

P = (0.6-0.7(0.2)(SDS))D = (0.6-0.7(0.2)(0.4))(1k/ft(16ft)+7.8k) = 12.9 kips

Weight of wall: [40 psf(12ft)+2(2ft)75psf]10ft = 7800lb

Lightweight units, grout at 40 in. o.c. 40 psf; full grout 75 psf

From interaction diagram OK; stressed to 90% of capacity

- 73 -

Example: Shear Wall ASD

- 74 -

Example: Shear Wall ASD

Calculate net area, Anv, including grouted cells. 2849)5.262.7)(8(91925.2 ininininininAnv

Maximum Fv

625.0192/120// ininVdVhVdM vv

psipsifVd

MF gm

vv 8.8375.02000625.025

3

225

3

2

OK

psiin

lb

A

Vf

nvv 4.82

849

1000007.02

Shear ratio

Shear stress

- 75 -

Example: Shear Wall ASD

in

inpsi

inpsiin

Af

dFAs

sA

dFAF

nvs

sv

n

svvs

9.258494.42

1883200031.05.05.0

5.0

2

psipsipsiFfF vmgvvs 4.425.6775.0/4.82/

Use #5 bars in bond beams.Determine spacing.

Use #5 at 24 in. o.c.

psiin

lbpsi

A

Pf

Vd

MF

nm

vvm

5.67849

870025.02000625.075.10.4

2

1

25.075.10.42

1

2

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

Top of wall P = (0.6-0.7(0.2SDS))D = 0.544(1k/ft)(16ft) = 8.70 kips

(critical location for shear):

Determine Fvm

Required steel stress

- 76 -

Example: Shear Wall ASD

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

Required: Design the shear wall; special reinforced shear wall

Solution: Check using 0.6D+0.7E load combination.

• Design for 1.5V, or 1.5(70 kips) = 105 kips (Section 7.3.2.6.1.2)

• fv = (105 kips)/(849in2) = 123.7 psi

• But, maximum Fv is 83.8 psi

• Fully grout wall

- 77 -

Example: Shear Wall ASD

psiin

lbpsiFvm 0.34

1464

870025.02000625.075.10.4

4

12

21464.192.625.7 inininAnv

psipsiFff vmvvs 7.370.347.71

in

inpsi

inpsiin

AF

dFAs

nvs

sv 6.1614947.37

1883200031.05.05.0

2

Use #5 @ 16 in.

psi

in

lb

A

Vf

nvv 7.71

1464

1000007.05.12

Calculate net area, Anv, including grouted cells.

Shear stress

Determine Fvm

(special wall)

Use #5 bars in bond beams.Determine spacing.

Required steel stress

- 78 -

Example: Shear Wall ASD

If we needed to check maximum reinforcing, do as follows.

00582.0

200060000

1.16600002

20001.16

2max

psipsi

psi

psi

f

fnf

fn

m

yy

m

Section 8.3.3.4 Maximum ReinforcementNo need to check maximum reinforcement since only need to check if:

• M/(Vdv) ≥ 1 and M/(Vdv) = 0.625

• P > 0.05f′mAn 0.05(2000psi)(1464in2) = 146 kips; P = 12.2 kips

00184.0

.188.625.7

44.06 2

inin

in

bd

As OK

For distributed reinforcement, the reinforcement ratio is obtained as thetotal area of tension reinforcement divided by bd. Assume 6 bars in tension.

- 79 -

Example: Shear Wall ASD

• Prescriptive Reinforcement Requirements (7.3.2.6)

• 0.0007 in each direction

• 0.002 total

• Vertical: 9(0.44in2)/1464in2 = 0.00270

• Horizontal: 7(0.31in2)/[120in(7.625in)] = 0.00237

• Total = 0.00270 + 0.00237 = 0.00507 OK

- 80 -

Example: Shear Walls SD

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

Required: Design the shear wall; ordinary reinforced shear wall

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

• Try 2-#5 at end and #5 @ 40 in.

Pu = [0.9 - 0.2(SDS)]D = [0.9 - 0.2(0.4)](1k/ft(16ft)+6.4k) = 18.4 kips

Weight of wall: 40 psf(10ft)(16ft) = 6400 lb

Lightweight units, grout at 40 in. o.c. 40 psf

From interaction diagram OK; stressed to 97% of capacity

- 81 -

Example: Shear Walls SD

- 82 -

Example: Shear Walls SD

2767)5.262.7)(8(71925.2 ininininininAnv

kipspsiin

fAdV

MV gmnv

vu

un

9.10275.02000767625.0253

48.0

253

4

2

OK

Calculate net area, Anv, including grouted cells.

Maximum Vn

- 83 -

Example: Shear Walls SD

in

k

inksiin

V

dfAs

dfs

AV

ns

vyv

vyv

ns

0.287.63

1926031.05.05.0

5.0

kkkVV

V nm

g

uns 7.63

8.0

4.82

75.08.0

100

kipskpsiin

PfAdV

MV

lbkip

umnvvu

unm

4.821.1325.02000767625.075.10.48.0

25.075.10.4

100012

s ≤ min{d/2, 48 in.} = min{94 in., 48 in.} = 48 in. Code 8.3.5.2.1In strength design, this provision only applies to beams (9.3.4.2.3 (e))Suggest that minimum spacing also be applied to shear walls.

Use#5 at 24 in. o.c.

Use #5 bars in bond beams.Determine spacing.

Top of wall Pu = (0.9-0.2SDS)D = 0.82(1k/ft)(16ft) = 13.1 kips

(critical location for shear):

Determine Vnm

Required steel strength

- 84 -

Example: Shear Walls SD

Shear reinforcement requirements (9.3.3.3.2)

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

• At wall intersections, horizontal reinforcing bars needed to satisfy shear strength requirements 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.

- 85 -

Example: Shear Walls SD

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

Required: Design the shear wall; special reinforced shear wall

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

• Shear capacity design provisions (Section 7.3.2.6.1.1)

• Vn ≥ shear corresponding to 1.25Mn.

• Minimum increase is 1.25/0.9 = 1.39

• Vn need not exceed 2.5Vu

• Normal design Vn ≥ Vu/ = Vu/0.8 = 1.25Vu

• Increases shear by a factor of 2

• Fully grout wall (Max Vn was 103 kips)

21464

.192.625.7

in

ininAnv

- 86 -

Example: Shear Walls SD

• Previously, Mn = 1028 k-ft; Mn = 1142 k-ft (Pu = 18.4k)

• 1.25Mn = 1428 k-ft; Design for 143 kips

• But wait, wall is fully grouted. Wall weight has increased to 75 psf

• For Pu = 23.0k, fully grouted, Mn = 1178 k-ft, 1.25Mn = 1474 k-ft

• Design for 147 kips

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

• Pu = [1.2 + 0.2(SDS)]D = 35.8 kips, 1.25Mn = 1601 k-ft

• Design for 160 kips

• Bottom line: any change in wall will change Mn, which will change design requirement; also need to consider all load combinations

• Often easier to just use Vn = 2.5Vu.

- 87 -

Example: Shear Walls SD

in

k

inksiin

V

dfAs

dfs

AV

ns

vyv

vyv

ns

2696.6

1926031.05.05.0

5.0

kkkVV

V nmuns 6.6

8.0

8.1541.160

kipskpsiin

PfAdV

MV

lbkip

umnvvu

unm

8.1541.1325.020001464625.075.10.48.0

25.075.10.4

100012

7.3.2.6 (b) Maximum spacing of reinforcing of 1/3 length, 1/3 height, or 48 in.Use maximum spacing of 1/3(height) = 40 in.

Use #5 at 40 in. o.c.

Use #5 bars in bond beams.Determine spacing.

Determine Vnm

Required steel strength Using shear from 1.25Mn

- 88 -

Example: Shear Walls SD

250k 1005.25.2 kVV un

Use #5 at 32 in. o.c.

Required steel strength

Using shear from Vn = 2.5Vu

kk

kVVV nmnns 4.568.0

8.154250

in

k

inksiin

V

dfAs

ns

vyv 6.314.56

1926031.05.05.0

Use #5 bars in bond beams.Determine spacing.

(Bars at 28, 60, 92 and 116 in. from bottom; top spacing is 24 in.)

- 89 -

Example: Shear Walls SD

• Prescriptive Reinforcement Requirements (7.3.2.6)

• 0.0007 in each direction

• 0.002 total

• Vertical: 7(0.31in2)/1464in2 = 0.00148

• Horizontal: 3(0.31in2)/[120in(7.625in)] = 0.00102

• Total = 0.00148 + 0.00102 = 0.00250 OK

- 90 -

Example: Shear Walls SD

• Calculate axial force based on c = 83.8 in. • Include compression reinforcement • Pn = 726 kips• Assume a live load of 1 k/ft• D + 0.75L + 0.525QE = (1k/ft + 0.75(1k/ft))16ft = 28 kips

Section 9.3.3.5 Maximum ReinforcementSince Mu/(Vudv) < 1, strain gradient is based on 1.5εy.

OK

c = 0.446(188in.) = 83.8 in.

Strain c/d, CMU c/d, Clay

1.5εy 0.446 0.530

3εy 0.287 0.360

4εy 0.232 0.297

- 91 -

Example: Shear Walls SD

• 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

gA1.0 mu fP geometrically symmetrical sections

gA05.0 mu fP geometrically unsymmetrical sections

AND

1vu

u

dV

MOR 3

vu

u

dV

Mmnu fAV 3 AND

For our wall, Mu/Vudv < 1Pu < 0.1fʹmAg = 0.1(2.0ksi)(1464in2) = 293 kips

- 92 -

Comparison of SD and ASD

- 93 -

ASD vs SD: Shear Walls

SD provides significant savings in overturning steel

• Most, if not all, steel in SD is stressed to fy. Stress in steel in ASD varies linearly

SD generally requires about same shear steel for ordinary walls; some savings for special walls

Advantage to SD

• But, maximum reinforcement requirements can sometimes be hard to meet; designers then switch to ASD, which requires more steel. This makes no sense.

• SD requires 180° hooks. Some think hard to construct, but not really. Some designers avoid SD solely for this reason.

- 94 -

ASD vs SD: Final Outcome

Strength Design

Allowable Stress Design