outline - umn ccaps · · 2018-02-12ncma tek notes 10-01a, 10-02c, 10-03 expands with time:...

1

S. K. Ghosh Associates Inc.

www.skghoshassociates.com

- 1 -

ASD vs. SDA Masonry Cage Fight

Richard Bennett, PhD, PEThe University of Tennessee

Chair, 2016 TMS 402/602 Code Committee

- 2 -

Outline

q Introduction: masonry basics and code changesq Beams and lintelsq Non-bearing walls: out-of-planeq Combined bending and axial force: pilastersq Bearing walls: out-of-planeq Shear walls: in-plane

• Partially Grouted Shear Wall• Special Reinforced Shear Wall

- 3 -

OUTLINE

Introduction

• Masonry Basics

• Recent Code Changes

- 4 -

Basic Masonry Terminology

Course

Bed Joint

Less than ¼ unit overlap

Head Joint

Running Bond Not Laid in Running Bond

2



- 5 -


Bond Beams : Continuously reinforced horizontal element

- 6 -


Pilaster: wall portion, generally projecting from either or both wall faces, serving as vertical column and/or beam

http://theconstructor.org/construction/masonry-pilaster-wall-design-details/14424/

- 7 -

Masonry Units

- 8 -

Concrete Masonry Units (CMU)

n specified by ASTM C90n minimum specified compressive

strength (net area) of 2000 psi (average)n Changed from 1900 psi in 2014!

n net area is about 55% of gross arean Changed from minimum web thickness

to normalized web area in 2012

n Type I and Type II designations no longer exist

http://www.oneontablock.com/

3



- 9 -

Clay Masonry Units

n Typically specified by ASTM C216 (facing brick, solid units), or C652 (hollow brick)

n If cores occupy £ 25% of net area, units can be considered 100% solid

n ASTM C652: Hollow unitsn Used for reinforced masonry, similar to cmun Increasingly veneers are C652 brick

- 10 -

CMU vs. Clay Units

Clay and concrete masonry behave quite differently, especially when exposed to moisture

Concrete ClayCementious: shrinks with

timeUse control joints: See

NCMA TEK Notes 10-01A, 10-02C, 10-03

Expands with time: Irreversible moisture expansion

Use expansion joints: SeeBIA Tech Note 18A

Need both vertical and horizontal expansion joints

- 11 -

CMU vs. Clay Units

Before leaving units…Block Shrinkage Block Shrinkage

Brick Expansion Brick Expansion

- 12 -

Masonry Mortar

M a S o N w O r Kstrongest weakest

Typical uses:Type S: Most structural masonryType N: Partitions and veneerType O: Tuckpointing

Cementious SystemsPortland cement - limeMasonry cementMortar cement

ASTM C270 Specification for Mortar for Unit MasonryASTM C780 Method for Preconstruction and Construction Evaluation of

Mortars for Plain and Reinforced Unit MasonryASTM C1586 Standard Guide for Quality Assurance of Mortars

4



- 13 -

Masonry Mortar: Specifications

Proportion: compliance is verified only by verifying volume proportions. n Type S cement - lime:

n 1 part cementn ~ 0.5 parts hydrated

mason’s limen ~ 4.5 parts sand

n Type N masonry cement:n 1 part Type N masonry

cement (single – bag)n ~ 3 parts sand

Property:n In laboratory, establish

proportions that will meet the required compressive strength, air content and retentivity (ability to retain water)

n verify in field that volume proportions meet proportion limits.

Proportion specification is the default- 14 -

Masonry Grout

n Two kinds of groutn fine grout (cement, sand, water)n coarse grout (cement, sand, gravel,

water)n Grout specified by

n proportionn compressive strength (required if f′m >

2000 psi)n Slump between 8 and 11 inches

n Masonry units absorb water from the grout, decreasing its water – cement ratio and increasing its compressive strength.

- 15 -

Masonry Compressive Strength, 𝒇𝒎#

n Unit strength method (TMS 602 1.4 B 2)n Estimates a conservative assemblage strength

based on mortar type and unit strength (from manufacturer)

n Prism test method (TMS 602 1.4 B 3)n Pro: can permit optimization of materialsn Con: requires testing, qualified testing lab, and

procedures in case of non - complying resultsn Must comply with ASTM C1314

- 16 -

CMU Unit Strength Table

Net area compressive

strength of concrete masonry, psi

Net area compressive strength of ASTM C90 concrete masonry units, psi (MPa)

Type M or S Mortar Type N Mortar

1,700 --- 1,900

1,900 1,900 2,350

2,000 2,000 2,650

2,250 2,600 3,400

2,500 3,250 4,350

2,750 3,900 ----

TMS 602 Table 2

5



- 17 -

Masonry Material Properties

Property CMU ClayModulus of Elasticity 900𝑓'# 700𝑓'#

Modulus of rigidity (shear modulus) 0.4𝐸'

Coefficient of thermal expansion 4.5 x 10-6 in./in./°F 4.0 x 10-6 in./in./°F

Coefficient of moistureexpansion N/A 3 x 10-4 in./in.

Coefficient of shrinkage 𝑘' = 0.5𝑠0C90 limits shrinkage

to 0.065%N/A

Coefficient of creep 2.5 x 10-7 /psi 0.7 x 10-7 /psi

- 18 -

Brief History of TMS 402/602

1988: First edition1995: Seismic requirements moved from Appendix to main body of code;

chapter on veneers added; chapter on glass block added1998: Major reorganization of code; prestressed masonry chapter added2002: Strength design chapter added; definitions of shear walls added to

correspond to IBC definitions; code moved to a three year revision cycle2005: Changed lap splice requirements, requiring much longer lap lengths2008: Major reorganization of seismic requirements; added AAC masonry in

Appendix2011: Eliminated one-third stress increase and recalibrated allowable stresses;

added infill provisions in Appendix2013: Changes for partially grouted shear walls; updated unit strength table;

limit states appendix2016: Add shear friction provisions; increase cavity width; update anchor bolts2022: Go to a six-year cycle

- 19 -

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

q Part 1 – General• Chapter 1 – General Requirements• Chapter 2 – Notation and Definition• Chapter 3 – Quality and Constructionq 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 Requirementsq 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 AACq Part 4 – Prescriptive Design Methods• Chapter 12 – Veneer• Chapter 13 – Glass Unit Masonry• Chapter 14 – Masonry Partition Wallsq Part 5 – Appendices• Appendix A – Empirical Design • Appendix B – Design of Masonry Infill• Appendix C – Limit Design Method

- 20 -

2011 vs 2013 Code

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

q Unit strength tables recalibrated

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

q Shear strength of partially grouted walls: reduction factor of 0.75

6



- 21 -

OUTLINE

Flexural Members

• Beams and Lintels

• Non-Bearing Walls

- 22 -

Beam and Lintel Design

Strength Design (Chapter 9)

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

mnvn

m

ysysn

fAV

bffA

dfAM

¢=

÷÷ø

öççè

æ¢

-=

25.2

8.021

• Minimum reinf: Mn ≥ 1.3Mcr• or As ≥ (4/3)As,req’d

• Maximum reinf: es ≥ 1.5ey

( )÷÷ø

öççè

æ+

¢=

sm

m

y

m

ff

eeer 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

rrr nnnk -+= 2)( 2

31 kj -=

jdAMfs

s = ))((2jdkdb

Mfm =

mvm fF ¢= 25.221

- 23 -

Beam and Lintel Design

Strength Design (Chapter 9)1. Determine a, depth of compressive

stress block

2. Solve for As

3. ρ345 = 0.285 𝑓'# 𝑓8⁄ for CMUGrade 60 steel:𝑓'# = 1.5 ksi, ρ345 = 0.00714𝑓'# = 2 ksi, ρ345 = 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.𝐴< = 𝑀/ 𝐹<𝑗𝑑

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.

bfMddam

n

¢--=8.022

y

ms f

bafA¢

=8.0

- 24 -

ASD: Alternate Design MethodCalculate

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

FF

Fks

m

mbal

+=

( )úúû

ù

êêë

é-÷

øö

çèæ-=

bFMddkdm3

222

32

YES

÷øö

çèæ -

-=

112)(

knF

PbkdF

Am

m

s

÷øö

çèæ -

=

31 kdF

MAs

s

bFnFA

s

ss=z

( ) zzz -+= dkd 222

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

NO

Allowable masonry compression stress

controls

Allowable reinforcement tension stress controls

7



- 25 -

Comparison of ASD and SD

8 inch CMUd = 20 in.

- 26 -

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 ft5.2.1.2 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

- 27 -

Allowable Stress Design: Flexure

( )( )ftk

ftwLM ftk

-=== 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

625.7900.031.452

220

2203

32

223

1222

ininksi

ftkininbFMddkd ft

in

b

=úúû

ù

êêë

é --÷

øö

çèæ-=

úúû

ù

êêë

é-÷

øö

çèæ-=

312.0466.02032.9

>===inin

dkdkCheck if compression controls Compression

controls

( ) 1.160.2900

29000900

==¢

==ksiksi

fE

EEn

m

s

m

sCalculate modular ratio, n

- 28 -

Allowable Stress Design: Flexure

Find AsArea of steel

( )( )

( )294.1

1466.01900.01.16

2625.731.9900.0

112)(

inksi

ininksi

knF

bkdF

Ab

b

s =÷øö

çèæ -

=÷øö

çèæ -

=

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

Use 2 - #9 (As = 2.00 in2)

8



- 29 -

Strength Design: Flexure

( )( )ftk

ftLwM ftk

uu -=== 6.62

8

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

( ) ( )( ) ininksiftinftk

ininbf

Mddam

n 78.3625.70.28.0

129.0

6.6222020

8.02 22 =

÷÷ø

öççè

æ÷øö

çèæ -

--=¢

--=

( )( )( ) 277.060

78.3625.70.28.08.0 inksi

ininksifbafAy

ms ==

¢=

Mn = 78.5 k-ft𝜙Mn = 70.6 k-ft

- 30 -

Check Min and Max Reinforcement

( )( ) ftkinkpsiinfSM rncr -=-=== 76.91.117160732 3

( )( ) 322

732624625.7

6inininbhSn ===

( ) ftkMftkftkM ncr -=£-=-= 3.757.1276.93.13.1

( )( ) 00577.020625.7

88.0 2

===inin

inbdAsr

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

Maximum Reinforcement Check:𝑓'# = 2 ksi ρ345 = 0.00952

Section modulus

Cracking moment

Check 1.3Mcr

- 31 -

Modulus of Rupture (Table 9.1.9.2)Masonry Type Mortar Type

Portland cement/lime or mortar cement

Masonry Cement

M or S N M or S N

Normal to Bed JointsSolid UnitsHollow Units*

UngroutedFully Grouted

133

84163

100

64158

80

51153

51

31145

Parallel to bed joints in running bondSolid UnitsHollow Units

Ungrouted and partially groutedFully grouted

267

167267

200

127200

160

100160

100

64100

Parallel to bed joints not laid in running bondContinuous grout section parallel to

bed jointsOther

335

0

335

0

335

0

335

0- 32 -

Summary: Beams, Flexure

Dead Load (k/ft)(superimposed) Live Load (k/ft)

Required As (in2)ASD SD

0.5 0.50.34

0.34 (𝑓'# = 1.5 ksi)0.26

0.26 (𝑓'# = 1.5 ksi)

1.0 1.00.64

0.65 (𝑓'# = 1.5 ksi)0.50

0.52 (𝑓'# = 1.5 ksi)

1.5 1.51.94

5.09 (𝑓'# = 1.5 ksi)0.77

0.80 (𝑓'# = 1.5 ksi)

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

9



- 33 -

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 02.933.5

17.133.556.11 =÷÷ø

öççè

æ -=

( )( )( )k

ftftwLV ftk

ftk

ftk

56.112

67.100.12083.00.1

2

2

=++

==

ftinininft 17.1.4

2.20

.121 =÷

øö

çèæ +

Shear at reaction

d/2 from face of support

Design shear force

[ ]

[ ] psipsi

fAPf

VdMF m

nmvm

3.50200025.221

25.22125.075.10.4

21

==

¢=+úû

ùêë

é¢÷÷

ø

öççè

æ÷øö

çèæ-=

( ) psiinin

kAVfnv

v 2.59.20.625.7

02.9===Shear stress

Allowable masonry shear stress

Suggest that d be used, not dv.

- 34 -

Allowable Stress Design: Shear

( )( )( )( )( )

2034.0.20320005.0.8.20.625.79.8

5.05.0

ininpsiinininpsiA

dFsAfA

sAdFAF

v

s

nvsv

n

svvs

==

=Þ÷÷ø

öççè

æ=

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

02.9 inpsiin

kbFVdvm

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

- 35 -

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 50.1233.5

17.133.500.16 =÷÷ø

öççè

æ -=

( )( ) ( )( )( )k

ftftLwV ftk

ftk

ftk

uu 00.16

2

67.100.16.12083.00.12.1

2

2

=++

==

ftinininft 17.1.4

2.20

.121 =÷

øö

çèæ +

Shear at reaction

d/2 from face of support

Design shear force

( )( )( ) kpsiinin

fAPfAdVMV mnvumnv

vu

unm

28.122000.20.625.725.28.0

25.225.075.10.4

==

¢=+¢úû

ùêë

é÷÷ø

öççè

æ-= fffDesign masonry

shear strengthSuggest that dbe used, not dv to find Anv

- 36 -

Strength Design: Shear

( )( )( )

2004.0.20605.0

.828.0

5.05.0

ininksi

inkA

dfsVAdf

sAV

v

vy

nsvvy

vns

==

=Þ÷øö

çèæ=

( )( )( ) kpsiininfAV mnvn 82.212000.20.625.748.04 ==¢£ ff

kkkVV mu 28.08.028.1250.12

=-

=-ff

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

50.1225.2

inpsiin

kfb

Vdm

u ==¢

=f

Use inverted bond beam unit to get slightly greater d.

10



- 37 -

Non-Bearing Partially Grouted WallsOut-of-Plane

s

b’b

da

t f

As

b = effective compressive width per bar = min{s, 6t, 72 in} (5.1.2)

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

B. Neutral axis in weba. Design as a T-beam section

C. Often design based on a 1 ft width

Minimum reinforcement:No requirements

Maximum reinforcement:Same requirement as beams

- 38 -

Design AidSpacing(inches)

Steel Area in2/ft#3 #4 #5 #6

8 0.16 0.30 0.46 0.6616 0.082 0.15 0.23 0.33

24 0.055 0.10 0.16 0.2232 0.041 0.075 0.12 0.1640 0.033 0.060 0.093 0.1348 0.028 0.050 0.078 0.1156 0.024 0.043 0.066 0.094

64 0.021 0.038 0.058 0.08272 0.018 0.033 0.052 0.07380 0.016 0.030 0.046 0.06688 0.015 0.027 0.042 0.06096 0.014 0.025 0.039 0.055104 0.013 0.023 0.036 0.051112 0.012 0.021 0.033 0.047120 0.011 0.020 0.031 0.044

- 39 -

Example: Partially Grouted Wall ASD

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

ftftlb

ftinlbft

inftlb ftwhM -- ==== 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

12900.03576.02

281.3

281.33

32

223

1222

inksi

ininbFMddkd

ftinftin

ftftk

b

=úúû

ù

êêë

é-÷

øö

çèæ-=

úúû

ù

êêë

é-÷

øö

çèæ-=

-

312.0091.081.3346.0

<===inin

dkdkCheck if compression

controls Tension controls

- 40 -

Example: Partially Grouted Wall ASD

÷øö

çèæ -

=

31 kdF

MAs

s

( )bFnFA

s

ss=z

( ) zzz -+= 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)

11



- 41 -

Example: Partially Grouted Wall SDGiven: 8 in. CMU wall; 16 ft high; Grade 60 steel, 𝑓'# = 2000 psi; Wind load of wu = 30 psfRequired: Reinforcing (place in center of wall)Solution:

( )( )ftftk

ftinlbft

inftlb

uu

fthwM -- ==== 96.0115208

161230

8

22 2

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

( ) ( ) inpsi

ininbf

Mddaftin

ftinlb

m

n 179.0)12(20008.0

9.011520

281.381.3

8.02 22 =

÷÷ø

öççè

æ

--=¢

--=

-

( )( )( )ftinft

in

y

ms psi

inpsifbafA 20573.0

60000179.01220008.08.0

==¢

=

Factored Moment

Find aSolve as solid section

Find As

- 42 -

ASD vs SD: Flexural Members

q ASD calibrated to SD for wind and seismic loads

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

q When allowable masonry stress controls in ASD, designs are inefficient

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

- 43 -

OUTLINE

Combined Bending and Axial Loads

• Interaction Diagrams

• Pilaster

- 44 -

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

• f=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

úúû

ù

êêë

é÷øö

çèæ-+-¢=

rhAfAAfP stystnmn

( )[ ]2

h70r80.080.0 ÷

øö

çèæ+-¢= stystnmn AfAAfP

( )27065.025.0 ÷øö

çèæ+¢=hrFAAfP sstnma

( )úúû

ù

êêë

é÷øö

çèæ-+¢=

2

140165.025.0

rhFAAfP sstnma

12



- 45 -

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; 𝑓'# = 2000 psiRequired: Interaction diagram in terms of capacity per footPure Moment: n = 16.1 ρ = 0.00109 nρ = 0.0176

ftftk

ftk-inM -== 479.0.755

( )( )( ) ftink

ftin

sss inksijdFAM -=== 75.5.81.3943.03205.0 2

( ) ( ) ( ) ( )[ ] ( )[ ]

ftink

ftinm

m inksiinjdFkdbM

-=

==

64.12

81.3943.02900.081.3171.012

2

( ) ( ) 171.00176.00176.020176.02)( 22 =-+=-+= rrr nnnk

943.03171.01

31 =-=-=kj

Find k

Find j

Find Ms

Find Mm

Allowable M

- 46 -


Pure Axial:

ftkP 3.17 =

991.5466.2144

£==inin

rh

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

rhFAAfP

3.17140

1.54107.400.225.0

140

165.025.0

2

2

2 =úúû

ù

êêë

é÷øö

çèæ-+=

úúû

ù

êêë

é÷øö

çèæ-+¢=

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.

- 47 -


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 44.61219.1900.021

21

===

Stress

ftftk

ftink

ftk ininM -- ==÷

øö

çèæ -= 83.10.22

319.181.344.6

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

===

- 48 -


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 22.41200.1703.021

21

===

Stress

ftftk

ftink

ftk ininM -- ==÷

øö

çèæ -= 22.17.14

300.181.322.4

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

13



- 49 -


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 66.005.01.132

==

2.00 in

9.28 k/ft0.66k/ft

1.25+0.75/3=1.5in.

0.338 ksi

1.25 in

( ) ( )( ) ftk

ftin

m inksiC 28.91225.12338.0900.0

1 =+

=0.25 k/ft

( )( ) ftk

ftin

m inksiC 25.0275.0338.021

2 ==

1.5 in

0.53 in

h2h1

bx

Centroid

32

21

21 bhhhhx

++

=

ftftk

ftk

M

P-=

=

58.2

.888

- 50 -

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; 𝑓'# = 2000 psiRequired: Interaction diagram in terms of capacity per footPure Moment:

Find Mn

Find 𝜙Mn

( ) ( )( )( ) ft

ftkftink

ftinftin

ftin

m

ysysn

ksiksi

ksi

bffA

dfAM

-- ==÷÷ø

öççè

æ-=

÷÷ø

öççè

æ-=

934.02.110.2128.0

6005.021.812in 36005.0

'8.021

2

2

( ) ftftk

ftftk

nM-- == 840.0934.00.9 f

( )( )( ) in

ksiksi

bffA

aftinftin

m

ys 156.00.2128.0

6005.0'8.0

2

===Check to make sure stress block is in face shell

- 51 -


Pure Axial:

991.5466.2144

£==inin

rh

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 ==f

( )[ ]

( )( )[ ] ftk

ftin

stystnmn

ksi

rhAfAAfP

3.44140

1.541007.400.280.08.0

140

180.08.0

2

2

2 =úúû

ù

êêë

é÷øö

çèæ-+-=

úúû

ù

êêë

é÷øö

çèæ-+-¢=

- 52 -


Find Cm

Find T

Balanced: Strain

Stress

0.0025

0.002073.82 in

2.09 in

TCm

a = 0.8c = 0.8(2.09in.) = 1.67in.web length = EFG.

HEFG.IJFG.KL

= 2.0 FGKL

( ) ftk

ftk

nP 1.200.33.10.240.9 =-+=f

( )[ ]( )( ) ftk

ftin

webm ininksiC 34.10.225.167.10.280.0, =-=

ftftk

ftk

ftk

n inininM -=úû

ùêë

é÷øö

çèæ -

--+÷øö

çèæ -= 96.5

225.167.125.181.33.1

225.181.30.249.0f

( )[ ]( )( ) ftk

ftin

shellfacem inksiC 241225.10.280.0, ==

( )( ) ftk

ftin

sy ksiAfT 0.305.0602

===

Find 𝜙𝑃G

Find 𝜙𝑀G

ftftk

n

ftk

n

M

P-=

=

96.5

1.20

f

f

0.8f′m = 1.6 ksi

14



- 53 -


Find Cm

Find T

Find 𝜙𝑃G

Find 𝜙𝑀G

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 =-==ffftftk

n

ftk

n

M

P-=

=

77.4

6.14

f

f

( )ftftk

ftink

ftk

nininM -- ==ú

û

ùêë

é÷øö

çèæ -= 77.42.57

225.18.081.32.199.0f

0.8f′m = 1.6 ksi

a = 0.8c1.00 in

- 54 -


AboveBalanced:c = 3.0 in.

Strain0.0025

0.000683.81 in

3.0 inStress

TCm

a = 0.8c = 0.8(3.0in.) = 2.4in.web length = EFG.

HEFG.IJFG.KL

= 2.0 FGKL

0.8f′m = 1.6 ksi

Find Cm

Find T

( ) ftk

ftk

nP 0.2499.068.30.240.9 =-+=f

( )[ ]( )( ) ftk

ftin

webm ininksiC 68.30.225.140.20.280.0, =-=

ftftk

ftk

ftk

n inininM -=úû

ùêë

é÷øö

çèæ -

--+÷øö

çèæ -= 28.6

225.140.225.181.368.3

225.181.30.249.0f

( )[ ]( )( ) ftk

ftin

shellfacem inksiC 241225.10.280.0, ==

( )( ) ftk

ftin

sss ksiAET 99.005.000068.0290002

=== e

Find 𝜙𝑃G

Find 𝜙𝑀G

ftftk

n

ftk

n

M

P-=

=

28.6

0.24

f

f

- 55 -

Interaction Diagrams

- 56 -

ASD vs SD: Interaction Diagrams

q Similar behavior to flexure• ASD and SD close when allowable tension stress controls

• ASD more conservative when allowable masonry stress controls

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

q Advantage to SD

15



- 57 -

ASD: Combined Bending and Axial Load Design Method

Calculate

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

FF

Fks

m

mbal

+=

úú

û

ù

êê

ë

é +--÷

øö

çèæ-=

bFMtdPddkd

m3))2/((2

223

2

YES

÷øö

çèæ -

-=

112)(

knF

PbkdF

Am

m

s

÷øö

çèæ -=¢

32kdtPM

÷øö

çèæ -

¢-=

31 kdF

MMAs

s

( )bF

nFAP

s

ss+=z

( ) zzz -+= dkd 222

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

NO

Compression controlsTension controls

- 58 -

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

[ ] [ ] [ ] 02226

23 =÷÷ø

öççè

æ+÷øö

çèæ -+÷÷

ø

öççè

æ+÷øö

çèæ --- dMtdPkdMtdPkd

nbdFkd

nbF ss

( )( )s

s FP

kddkd

nbkdA -÷

øö

çèæ

-=

121

Solve for As.

- 59 -

Strength Design: Combined Bending and Axial Load Design Method

( )[ ]( )bf

MtdPddam

uu

¢+-

--=8.02/22

f

y

ums f

PbafA f/8.0 -¢=

Calculate

Is c ≥ cbal?For Grade 60 steelcbal = 0.547

YES

Compression controlsTension controls

dcymu

mubal ee

e+

=

NO

8.0ac =

÷øö

çèæ --¢

=

ccdE

PbafAsmu

ums

e

f/8.0

- 60 -

Example: Pilaster Design ASDGiven: Nominal 16 in. wide x 16 in. deep CMU pilaster; 𝑓'# = 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:

Verti

cal S

pann

ing

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

16



- 61 -

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

max 282 whMwhMM ++=

whMhx -=

2

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

M = moment at topx measured down from top of pilaster

- 62 -


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

( )( ) inininkipin

whMhx

inft

ftk

1.1431288250.0

2.52

2882

=-

-=-=

( )( )( )

inkin

inkininkwhMwhMM

ink

ink

-=-

++-

=++= 3.2182880208.02)2.5(

82880208.0

22.5

282 2

22

2

22

max

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

( )( ) kinkP inft

ftk 3.21.14320.06.09.0 .12

1 =+=

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

Find location of maximum moment

- 63 -


( )( )( ) in

inksiinkipininkipinin

bFMhdPddkd

b

00.36.15900.03

)2182/6.158.113.2(228.11

28.113

3))2/((2

223

2

2

=úú

û

ù

êê

ë

é -+--÷

øö

çèæ-=

úú

û

ù

êê

ë

é +--÷

øö

çèæ-=

254.08.1100.3

===inin

dkdk

312.0

1.1632900.0

900.0=

+=

+= ksiksi

ksi

nFF

Fks

b

bbal

k < kbal Tension controls; iterate

Assume compression controls;Determine kd

Determine k

Determine kbal

- 64 -


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

÷øö

çèæ -=¢

32kdhPM

÷øö

çèæ -

¢-=

31 kdF

MMAs

s

( )bF

nFAP

s

ss+=z

( ) zzz -+= dkd 222

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

17



- 65 -


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

- 66 -


f’m = 2000 psi

f’m = 1500 psi

- 67 -

Example: Pilaster Design SDGiven: Nominal 16 in. wide x 16 in. deep CMU pilaster; 𝑓'# = 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:

Verti

cal S

pann

ing

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

- 68 -

Example: Pilaster Design SD

0.9D + 1.0WAt top of pilaster:

( ) ( ) kkkPu 54.01.80.16.99.0 =-+=

( ) inkinkePM uu -=== 1.3.8.554.0

( ) inftinkipin

whMhx

ftk 7.14324416.0

1.32

2882

=-

-=-=

( )( )( )

inkin

inkininkwhMwhMM

ink

ink

u -=-

++-

=++= 0.3612880347.02)1.3(

82880347.0

21.3

282 2

22

2

22

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

( )( ) kinkP inft

ftk

u 69.27.14320.09.054.0 121 =+=

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

Find location of maximum moment

18



- 69 -


( )[ ]( )

( ) ( )[ ]( )( )( ) in

inksiinkininkinin

bfMhdPdda

m

uu

50.16.150.28.09.03612/6.158.117.228.118.11

8.02/2

2

2

=-+-

--=

¢+-

--=f

( )( )( ) 257.060

9.0/7.250.162.150.28.0/8.0 inksi

kininksifPbafA

y

ums =

-=

-¢=

f

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

- 70 -


- 71 -

Example: Pilaster Design

- 72 -

ASD vs SD: Pilaster Design

q 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

q SD design easier as steel has generally yielded

q Advantage to SD

19



- 73 -

OUTLINE

Bearing Walls: Out-of-Plane Loads

- 74 -

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• Slender wall procedure• Moment magnification• Second-order analysis

• Need to check maximum reinforcement limits

- 75 -

Slender Wall Procedure• Assumes simple support conditions.• Assumes midheight moment is approximately maximum moment• Assumes uniform load over entire height• Valid only for the following conditions:

• NOPQ≤ 0.05𝑓'# No height limit

• NOPS≤ 0.20𝑓'# height limited by T

L≤ 30

uuu

ufu

u PePhwM d++=28

2

ufuwu PPP +=

Puf = Factored floor loadPuw = Factored wall load

Moment: Deflection:

crunm

uu MM

IEhM

<= 485 2

d

( )cru

crm

cru

nm

cru MM

IEhMM

IEhM

>-

+= 48

5 485 22

d

- 76 -

Slender Wall Procedure

4851

12848

5

4851

1148

528

2

22

2

22

crm

u

n

crcr

uuf

u

crmu

crm

u

crnm

ucruuf

u

u

IEhP

IIMePhw

IEh

IEhP

IIEhPMePhw

M

-

úúû

ù

êêë

é÷÷ø

öççè

æ-++

=

-

÷÷ø

öççè

æ-++

=

d

nm

u

uuf

u

nmu

nm

u

uuf

u

u

IEhP

ePhwIE

h

IEhP

ePhw

M

4851

28485

4851

28

2

22

2

2

-

úû

ùêë

é+

=

-

+=

d

Solve simultaneous linear equations:

Mu > Mcr Mu < Mcr

20



- 77 -

Slender Wall Procedure

( )32

32 bccd

dt

fPAnI sp

y

uscr +-÷

÷ø

öççè

æ+=

bfPfA

cm

uys

'64.0+

=

For centered bars: ( )3

32 bccd

fPAnIy

uscr +-÷

÷ø

öççè

æ+=

( )2/

/

sp

nrnucr t

IfAPM += What axial load, Pu, should be used?

Suggest using smallest value of Pu.

Deflection Limit hs 007.0£d Calculated using allowable stress load combinations

- 78 -

Moment Magnification Procedure

Mu < Mcr: Ieff = 0.75InMu ≥ Mcr: Ieff = Icr

4851

1148

528

2

22

crm

u

crnm

ucruuf

u

u

IEhP

IIEhPMePhw

M-

÷÷ø

öççè

æ-++

=

First Order Moment

Always Negative

548

= 0.104~1𝜋J

= 0.101

Complementary Moment

2

2

0,

1

1

hIE

P

PP

MM

effme

e

u

uu

p

y

y

=

-=

= Equation 9-27

Equation 9-28

Equation 9-29

- 79 -

Example: Wind Loads ASD

16 ft

32 p

sf

D = 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; 𝑓'# = 2000 psi; roof forces act on 3 in. wide bearing plate at edge of wall.Required: ReinforcementSolution:Estimate reinforcement

𝑀~𝑤ℎJ

8=0.6 0.032𝑘𝑠𝑓 16𝑓𝑡 J

8= 0.62

𝑘 − 𝑓𝑡𝑓𝑡

Try #4 @ 40 in.k = 0.185, kd = 0.707 in.Mm = 1.14 k-ft/ft; Ms = 0.57 k-ft/ftLightweight units: wall weight = 40 psf

Cross-sectionof top of wall

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

- 80 -


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 0.051D+0.6W 0.284 0.711 -0.002 0.613 0.042

( )( )ftftkft

ftktop ftksfMwhM -

-

=-

+=+= 590.02

049.08

16032.06.0 28

122

Check 0.6D+0.6W( ) ( ) 084.036.06.05.06.0 ft

kftk

ftk

fP =-+=

( )( ) /340.0867.2040.06.0084.0 ftkftftksfP ftk =++=

( ) ( ) ( ) 049.02

67.2032.06.0.81.2084.02

121

ftftk

inft

ftk

topftksfinM --=-=

Use #4@ 40 in. (0.06 in2/ft)Although close to #4@48 (0.05in2/ft), 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

21



- 81 -


Sample Calculations: 0.6D+0.6W1. 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

12900.03590.02

2.81.3

2.81.33

32

223

.122

2

inksi

inin

bFMddkd

ftinftin

ftftk

b

=úúû

ù

êêë

é-÷

øö

çèæ-=

úúû

ù

êêë

é-÷

øö

çèæ-=

-

( )3/2/ kdtPM sp -=¢

( )3/1 kdFMMA

ss -

¢-=

( )bF

nFAP

s

ss+=z

( ) zzz -+= dkd 222

Equation / Value Iteration 1 Iteration 2kd (in.) 0.355 0.707

(k-ft/ft) 0.1047 0.1013

(in2/ft) 0.049 0.051

(in.) 0.0804

(in.) 0.707

- 82 -

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; 𝑓'# = 2000 psi ; roof forces act on 3 in. wide bearing plate at edge of wall.Required: ReinforcementSolution:Estimate reinforcement

𝑀~𝑤ℎJ

8=

0.032𝑘𝑠𝑓 16𝑓𝑡 J

8= 1.02

𝑘 − 𝑓𝑡𝑓𝑡

a = 0.19 in.As = 0.061 in2/ftTry #4 @ 40 in.

Cross-sectionof top of wall

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

- 83 -


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+1.0W 0.9(0.5)+1.0(-0.36) = 0.090

0.9(0.040)(2.67+8) = 0.384 0.474

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)

- 84 -


Find modulus of rupture; use linear interpolation between no grout and full groutUngrouted (Type S masonry cement): 51 psiFully grouted (Type S masonry cement): 153 psi

( ) psicells

cellgroutedpsicells

cellsungroutedpsifr 71 5

11535

451 =÷øö

çèæ+÷÷

ø

öççè

æ=

Find Mcr, cracking moment:Commentary allows inclusion of axial loadUse minimum axial load (once wall has cracked, it has cracked)

( ) ( ) ( )( )( )

ftftkip

ftinkip

cr

ftin

ftin

ftlb

nrnucr

Minpsi

tIfAPM

-- ==

+=

+=

606.028.781.3

7.336718.42/4742/

/42

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

22



- 85 -


Find Icr

( )( ) in

ksiksi

bfPfA

cftin

ftk

ftin

m

uys 265.00.264.0

474.06006.0'64.0 12

2

=+

=+

=

1.16180029000

===ksiksi

EEnm

s

( )

( ) ( )

ftin

ftin

ftk

ftin

sp

y

uscr

ininin

ksi

bccddt

fPAnI

4

2

8.13

3265.012

265.0812.360474.0

06.01.16

323

2

32

=

+-÷÷ø

öççè

æ+=

+-÷÷ø

öççè

æ+=

Find c

Find n

- 86 -


( ) ( )( )( )( )

( )( )( )( )

ftftk

ftin

ftk

ftin

ftin

ftk

ftftk

ftftk

crm

u

crnm

ucruuf

u

u

ftin

ksift

ftin

ksiftftksf

IEhP

IIEhPMePhw

M

-

--

=

÷÷ø

öççè

æ-

÷÷ø

öççè

æ÷÷ø

öççè

æ-+

-+

=

-

÷÷ø

öççè

æ-++

=

009.1

1144

8.1318004816474.05

1

1144

8.131

3371

18004816474.0606.05

2093.0

816032.0

4851

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.

( ) ( )ftftkftksfin

hwePM in

ftftkparapetparapetu

uuftopu-

-=-=-= 093.0267.2032.081.2090.0

2

2

121

2,

,

Find Mu.

- 87 -


( )

( )( )

ftftkip

ftinkip

ftin

ftk

ysun

ininksi

adfAPM

-- ==

÷øö

çèæ -+=

÷øö

çèæ -+=

147.176.132215.0812.36006.09.0/474.09.0

2/

2

fff

( )( )( ) in

ksiksi

bfPfA

aftinftk

ftin

m

uys 215.0120.280.0

9.0/474.06006.080.0

/ 2

=+

=¢

+=

f

nftftk

ftftk

u MM f=£= -- 15.101.1 OK

Check other load combinations:1.2D+1.0W+0.5Lr, Mu = 1.09 k-ft/ft and 𝜙Mn = 1.29 k-ft/ft OK

Find a

Check area of steel:

Find 𝜙𝑀G

Compare

- 88 -


Check Deflections: Do a quick check with SD Load Combinations

Deflection Limit ininhs 34.1)192(007.0007.0 ==£d OK

( )

.791.00659.0

474.0093.0816032.0009.1

282

2

inft

ftksf

PePhwM

u

uftk

ftftk

ftftk

uuu

ufu

u

==

+-=

++=

--

d

d

d

0.9D+1.0W

1.2D+1.0W+0.5Lr .854.00711.0 inftu ==d

23



- 89 -


Check Deflections: Use ASD Load Combinations

Load Combination D+0.6W 0.6D+0.6WPf (k/ft) 0.5+0.6(-0.36) = 0.284 0.6(0.5)+0.6(-0.36) = 0.084Pw (k/ft) 40(2.67+8) = 0.427 0.6(0.427) = 0.256P (k/ft) 0.711 0.340c (in) 0.280 0.256Icr (in4/ft) 14.5 13.4Mtop (k-ft/ft) -0.002 -0.049δ (in) 0.066 0.044M (k-ft/ft) 0.617 (cracked) 0.565 (uncracked)

Deflection Limit ininhs 34.1)192(007.0007.0 ==£d OK

- 90 -


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

( )( )( )

( )( )( )

( )( )( )( )

.066.0

1144

5.1418004816711.05

1

117281

7.3365.14

606.02

002.08

16032.06.05.14180048

165

4851

12848

5

2

22

3

322

2

22

4

4

4

4

in

ftin

ksift

ftinftksf

ksift

IEPh

IIMePwh

IEh

ftin

ftk

ftinftin

ftftkft

ftk

ftin

crm

n

crcrf

crm

=

-

úúû

ù

êêë

é÷÷ø

öççè

æ-+

-+

=

-

úû

ùêë

é÷÷ø

öççè

æ-++

=

--

d

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

- 91 -


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

6081.312

927.000207.05.10025.0

0025.00.264.0

64.0

max

=

-÷÷ø

öççè

æ+

-÷÷ø

öççè

æ

+¢

==

ksiin

ksi

fbdPf

bdA

ftin

ftk

y

u

ymu

mum

saee

e

r

( ) 00131.081.312

06.02

===inbd

A

ftin

ftft

sr OK

- 92 -

ASD vs SD: Bearing Walls

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

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

q Advantage to ASD• Interestingly OOP loading is where designers often use SD

24



- 93 -

OUTLINE

Shear Walls: In-Plane Loads

- 94 -

Shear Walls: Shear

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

umnvvu

unm PfA

dVMV 25.075.10.4 +¢

úúû

ù

êêë

é÷÷ø

öççè

æ-=

( ) gmnvn fAV g¢£ 6 ( ) 25.0/ £vuu dVM

vyv

ns dfsAV ÷øö

çèæ= 5.0

( ) gnsnmn VVV g+= 8.0=f

( ) gmnvn fAV g¢£ 4 ( ) 0.1/ ³vuu dVM

gmnvvu

un fA

dVMV g÷÷

ø

öççè

æ¢÷÷

ø

öççè

æ-= 25

34

0.125.0 <<vu

u

dVM

( ) gvsvmv FFF g+=

nm

vvm A

PfVdMF 25.075.10.4

21

+úúû

ù

êêë

é¢÷

÷ø

öççè

æ÷÷ø

öççè

æ-=

÷÷ø

öççè

æ=

sAdFAF

n

svvs 5.0

( ) gmv fF g¢£ 3 ( ) 25.0/ £vVdM

( ) gmv fF g¢£ 2 ( ) 0.1/ ³vVdM

nm

vvm A

PfVdMF 25.075.10.4

41

+úúû

ù

êêë

é¢÷

÷ø

öççè

æ÷÷ø

öççè

æ-=

Special Reinforced Shear Walls

Interpolate for 0.25 < (M/Vdv) < 1

𝛾 = 0.75 partially grouted shear walls; 1.0 otherwise

- 95 -

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

÷÷ø

öççè

æ¢

+

¢=

m

yy

m

ff

nf

fn

2maxr

• Provide boundary elements, or• Limit reinforcement

Boundary elements not required if:gA1.0 mu fP ¢£ symmetrical sections

gA05.0 mu fP ¢£ unsymmetrical sections

AND

1£wu

u

lVM OR 3£

wu

u

lVM

mnu 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

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

- 96 -

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, fVn, 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)

25



- 97 -

Example: Shear Wall ASD

Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type S mortar; 𝑓'# = 2000 psi; superimposed dead load of 1 kip/ft. In-plane seismic load of 100 kips. SDS = 0.4Required: Design the shear wall; ordinary reinforced shear wallSolution: 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 = 7800lbLightweight units, grout at 40 in. o.c. 40 psf; full grout 75 psf

From interaction diagram OK; stressed to 90% of capacity

- 98 -


- 99 -


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

Maximum Fv

( ) ( ) ( ) 625.0192/120// === ininVdVhVdM vv

( )( ) psipsifVdMF gmv

v 8.8375.02000625.0253225

32

=÷øö

çèæ -=÷

÷ø

öççè

æ¢÷÷

ø

öççè

æ-= g

OK( ) psiin

lbAVfnv

v 4.828491000007.0

2 ===

Shear ratio

Shear stress

- 100 -


( )( )( )( ) in

inpsiinpsiin

AfdFAs

sAdFAF

nvs

sv

n

svvs

9.258494.42

1883200031.05.05.0

5.0

2 ===

Þ÷÷ø

öççè

æ=

psipsipsiFfF vmgvvs 4.425.6775.0/4.82/ =-=-= g

Use #5 bars in bond beams.Determine spacing.

Use #5 at 24 in. o.c.

( )( )[ ] psiinlbpsi

APf

VdMF

nm

vvm

5.67849870025.02000625.075.10.4

21

25.075.10.421

2 =+-=

+úúû

ù

êêë

é¢÷

÷ø

öççè

æ÷÷ø

öççè

æ-=

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

26



- 101 -


Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type S mortar; 𝑓'# = 2000 psi; superimposed dead load of 1 kip/ft. In-plane seismic load of 100 kips. SDS = 0.4Required: Design the shear wall; special reinforced shear wallSolution: 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

- 102 -


( )( )[ ] psiinlbpsiFvm 0.34

1464870025.02000625.075.10.4

41

2 =+-=

( ) 21464.192.625.7 inininAnv ==

psipsiFff vmvvs 7.370.347.71 =-=-=

( )( )( )( ) in

inpsiinpsiin

AFdFAs

nvs

sv 6.1614947.37

1883200031.05.05.0 2 ===

Use #5 @ 16 in.

( )[ ] psiin

lbAVfnv

v 7.7114641000007.05.1

2 ===

Calculate net area, Anv, including grouted cells.

Shear stress

Determine Fvm

(special wall)


Required steel stress

- 103 -


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

( )

( )00582.0

2000600001.16600002

20001.16

2max =

÷÷ø

öççè

æ+

=

÷÷ø

öççè

æ¢

+

¢=

psipsipsi

psi

ff

nf

fn

m

yy

mr

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

.188.625.744.06 2

===inin

inbdAsr OK

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

- 104 -


• 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

27



- 105 -

Example: Shear Walls SD

Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type S mortar; 𝑓'# = 2000 psi; superimposed dead load of 1 kip/ft. In-plane seismic load of 100 kips. SDS = 0.4Required: Design the shear wall; ordinary reinforced shear wallSolution: 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 lbLightweight units, grout at 40 in. o.c. 40 psf

From interaction diagram OK; stressed to 97% of capacity

- 106 -


- 107 -


( ) 2767)5.262.7)(8(71925.2 ininininininAnv =-+=

( )( )( ) kipspsiin

fAdVMV gmnv

vu

un

9.10275.02000767625.025348.0

2534

2 =úûù

êëé -

=úû

ùêë

é¢÷÷

ø

öççè

æ-= gff OK

Calculate net area, Anv, including grouted cells.

Maximum Vn

- 108 -


( )( )( ) ink

inksiinV

dfAs

dfsAV

ns

vyv

vyv

ns

0.287.63

1926031.05.05.0

5.0

===

Þ÷øö

çèæ=

( ) kkkVVV nm

g

uns 7.63

8.04.82

75.08.0100

=-=-=f

ffg

( )( )( ) ( )[ ] kipskpsiin

PfAdVMV

lbkip

umnvvu

unm

4.821.1325.02000767625.075.10.48.0

25.075.10.4

100012 =+-=

úúû

ù

êêë

é+¢÷

÷ø

öççè

æ÷÷ø

öççè

æ-=ff

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.


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

Determine 𝜙Vnm

Required steel strength

28



- 109 -


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.

- 110 -


Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type S mortar; 𝑓'# = 2000 psi; superimposed dead load of 1 kip/ft. In-plane seismic load (from ASCE 7-10) of 100 kips. SDS = 0.4Required: Design the shear wall; special reinforced shear wallSolution: Check using 0.9D+1.0E load combination.• Shear capacity design provisions (Section 7.3.2.6.1.1)

• 𝜙Vn ≥ shear corresponding to 1.25Mn.

• Minimum increase is 1.25/0.9 = 1.39

• Vn need not exceed 2.5Vu

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

• Increases shear by a factor of 2

• Fully grout wall (Max 𝜙Vn was 103 kips)

( )21464

.192.625.7

in

ininAnv=

=

- 111 -


• Previously, 𝜙Mn = 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.- 112 -


( )( )( ) ink

inksiinV

dfAs

dfsAV

ns

vyv

vyv

ns

2696.6

1926031.05.05.0

5.0

===

Þ÷øö

çèæ=

kkkVVV nmuns 6.6

8.08.1541.160

=-

=-

=ff

( )( )( ) ( )[ ] kipskpsiin

PfAdVMV

lbkip

umnvvu

unm

8.1541.1325.020001464625.075.10.48.0

25.075.10.4

100012 =+-=

úúû

ù

êêë

é+¢÷

÷ø

öççè

æ÷÷ø

öççè

æ-=ff

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.



Determine 𝜙Vnm

Required steel strength Using shear from 1.25Mn

29



- 113 -


( ) 250k 1005.25.2 === kVV un


Required steel strength

Using shear from Vn = 2.5Vu

kkkVVV nmnns 4.568.08.154250 =-=-=

( )( )( ) ink

inksiinV

dfAs

ns

vyv 6.314.56

1926031.05.05.0===


- 114 -


• 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

- 115 -


• Calculate axial force based on c = 83.8 in. • Include compression reinforcement • 𝜙Pn = 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, Clay1.5εy 0.446 0.5303εy 0.287 0.3604εy 0.232 0.297

- 116 -


• 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

1£vu

u

dVM OR 3£

vu

u

dVM

mnu fAV ¢£ 3 AND

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

30



- 117 -

Comparison of SD and ASD

- 118 -

ASD vs SD: Shear Walls

q 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

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

q 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.

- 119 -

ASD vs SD: Final Outcome

Strength Design

Allowable Stress Design

outline - umn ccaps · · 2018-02-12ncma tek notes 10-01a, 10-02c, 10-03 expands with time:...

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