Introduction: recent masonry code changes Masonry Design ... Convention/2018...Chair, 2016 TMS...

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Masonry Design Example Comparisons Using ASD and SD

Richard Bennett, PhD, PEThe University of Tennessee

Chair, 2016 TMS 402/602 Code Committee2nd Vic Chair, 2022 TMS 402/602 Code Committee

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Outline

Introduction: recent masonry code changes Beams and lintels Non-bearing walls: out-of-plane 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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Reorganization: 2013 TMS 402 Code

Part 1: General

Chapter 1 –General

Requirements

Chapter 2 –Notations & Definitions

Chapter 3 –Quality &

Construction

Part 2: Design Requirements

Chapter 4: General

Analysis & Design

Chapter 5: Structural Elements

Chapter 6: Reinforcement,

Metal Accessories & Anchor Bolts

Chapter 7: Seismic Design Requirements

Part 3: Engineered

Design Methods

Chapter 8: ASD

Chapter 9: SD

Chapter 10: Prestressed

Chapter 11: AAC

Part 4: Prescriptive

Design Methods

Chapter 12: Veneer

Chapter 13: Glass Unit Masonry

Chapter 14: Partition

Walls

Part 5: Appendices &

References

Appendix A – Empirical

Design

Appendix B: Design of

Masonry infill

Appendix C: Limit Design of Masonry

References

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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

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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

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Partially Grouted Shear Wall Factor

Mean St DevFully grouted(Davis et al, 2010) 1.16 0.17

Partially grouted(Minaie et al, 2010) 0.90 0.26

9.3.4.1.2, Equation (9-21)

= 0.75 for partially grouted shear walls= and 1.0 otherwise

0.901.16 0.776

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Beam and Lintel Design

Strength Design (Chapter 9)

• = 0.0035 clay masonry• = 0.0025 concrete masonry• Masonry stress = 0.8• Masonry stress acts over = 0.8• = 0.9 flexure; = 0.8 shear.2.25• Minimum reinf: 1.3

• or 4 3⁄ ,• Maximum reinf: 1.50.8 0.8

Allowable Stress (Chapter 8)

• Allowable masonry stress: 0.45• Allowable steel stress

• 32 ksi, Grade 60 steel⁄ ⁄21 3⁄• No min and max reinforcement• Allowable shear stress2.25

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Beam and Lintel Design

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

stress block

.2. Solve for ,, .3. 0.285 ⁄ for CMU

Grade 60 steel:= 1.5 ksi, 0.00714= 2 ksi, 0.00952

Allowable Stress (Chapter 8)1. Assume value of (or ).

Typically 0.85 < < 0.95.

2. Determine a trial value of , ., /Choose reinforcement.

3. Determine and ; 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 ). Reinforcement is not being used efficiently.

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ASD: Alternate Design MethodCalculate3 2 2 23

Is ?For Grade 60 steel,CMU = 0.312

YES

NO

Compression controls Tension controls

⁄, 21 1

, 1 3,2

Iterate. Use 2 as new guess and repeat.

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Comparison of ASD and SD

8 inch CMUd = 20 in.

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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; ’ = 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• 120 / . 120(7.625 in.)2 / (20 in.) = 349 in. = 29.1 ft

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Allowable Stress Design: Flexure

LoadWeight of fully grouted normal weight: 83 psf

Moment

Determine Assume compression controls

Check if compression controls Compression controls

Calculate modular ratio,

1.5 0.083 2ft 1.5 3.17. . 45.1k ⋅ ft

9.32in.20in. 0.466 0.312900 29000ksi900 2.0ksi 16.1

3 3 . . . ⋅ .. . . 9.32in.

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Allowable Stress Design: Flexure

Find ,Req’d area of steel

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

Use 2 - #9 (As = 2.00 in2)

, 1 1 0.90 9.31in. 7.625in.16.1 0.90ksi 10.466 1 1.94in.

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Strength Design: Flexure

Use 2 - #6 (As = 0.88 in2)= 78.5 k-ft

= 70.6 k-ft

Factored LoadWeight of fully grouted normal weight: 83 psf

Factored Moment

Find Depth of equivalent rectangular stress block

Find ,Req’d area of steel

. . 62.6k ⋅ ft1.2 1.61.2 1.5 0.083 2ft 1.6 1.5 4.40

.20in. 20in. . ⋅. .. . . . 3.78in., . . . . . . . 0.77in.

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Check Min and Max Reinforcement

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

Maximum Reinforcement Check: 0.00952

Section modulus

Cracking moment

Check 1.3

. . . 732in.732in. 160psi117.1k ⋅ in. 9.76k ⋅ ft1.3 1.3 9.76k ⋅ ft 12.7k ⋅ ft78.5k ⋅ ft0.88in.7.625in. 20in. 0.00577

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Summary: Beams, Flexure

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

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)

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Allowable Stress Design: Shear

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

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

Shear at reaction

/2 from face of support

Design shear force

Shear stress

Allowable masonry shear stress

Suggest that be used, not .

. 4in. . 1.17ft. . . . 11.56k

11.56k . .. 9.02k.. . . 59.2psi

2.25 2.25 2000psi 50.3psi- 18 -

Allowable Stress Design: Shear

Use #3 stirrups

Check max shear stress

Req’d steel stress

Determine for a spacing of 8 in.

59.2psiOK

Determine so that no shear reinforcement would be required.

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

2 2 2000psi 89.4psi59.2psi 50.2psi 8.9psi0.5 ⇒ , .. . . . .. . 0.034in.

.. . . 23.5in.

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Strength Design: Shear

Requirement for shear at /2 from face of support

is in ASD, not SD, but assume it applies

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

Shear atreaction

/2 from face of support

Design shear force

Design masonry shear strength

Suggest that be used, not to find

. 4in. . 1.17ft16.00k . .. 12.50k

. . . . . . 16.00k

2.250.8 2.25 7.625in. 20in. 2000psi 12.28k

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Strength Design: Shear

Use #3 stirrups

Check max

Req’d

Determine for a spacing of 8 in.

> 12.50k OK

Determine so that no shear reinforcement would be required.Use inverted bond beam to get slightly greater .

40.9 4 7.625in. 20in. 2000psi 21.82k. . . 0.28k0.5 ⇒ ,.. .. . 0.004in.

. .. . . . 20.4in.

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Non-Bearing Partially Grouted WallsOut-of-Plane

s

b’b

da

t f

As

= effective compressive width per bar = min{ , 6 , 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:ASD: noneSD: Same as for beams

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Non-Bearing Partially Grouted WallsOut-of-Plane

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 0.016 0.030 0.046 0.066

88 0.015 0.027 0.042 0.060

96 0.014 0.025 0.039 0.055

104 0.013 0.023 0.036 0.051

112 0.012 0.021 0.033 0.047

120 0.011 0.020 0.031 0.044

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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:

= 0.45(2000psi) = 900 psi = 1.80 x 106 psi= 32000 psi = 29 x 106 psi= / = 16.1

Moment

Determine Assume compression controls

Check if compression controls

Tension controls

3 3 . . . . . ⋅ . . . 0.346in.. .. . 0.091 0.312

. . 6912 ⋅ . 576 ⋅

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Example: Partially Grouted Wall ASD

Equation / Value Iteration 1 Iteration 2 Iteration 3

(in.) 0.346 0.699 0.709

0.091 0.183 0.183

, (in.2) 0.0584 0.0603 0.0603

, (in.) 0.0784 0.0810 0.0810

2 (in.) 0.699 0.709 0.709

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

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Example: Partially Grouted Wall SD

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:

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

Factored Moment

Find Solve as solid section

Find ,

. 11520 ⋅ . 960 ⋅

, . . . . . . 0.0.0573 .

.3.81in. 3.81in. . ⋅. .. . . 0.179in.

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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

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ASD: Combined Bending and Axial Load Design Method

Calculate

Is ?For Grade 60 steel,CMU = 0.312

YES

Iterate. Use 2 as new guess and repeat.

NO

Compression controls Tension controls

⁄3 2 2 2 2⁄3

, 2 1 1

2 3, ′1 3

,2

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ASD: Combined Bending and Axial Load Design Method

If tension controls; determine from cubic equation.

6 2 2 2 0, 12 1

Determination of . 0.450.45 32ksi29000ksi9000.450.45 3229000900 0.312

For clay masonry, 700 , 0.368

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Strength Design: Combined Bending and Axial Load

Calculate

Is ?For CMU, Grade 60 steel 0.547

YES

Compression controlsTension controls

NO

2 2⁄0.8 0.8

, 0.8 ⁄ , 0.8 ⁄

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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 ( = 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

= 11.8 in

xLoad

= 5.8 in 2.0 in

= 1800 ksi= 16.1

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

Insi

de

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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:

If 0 or , = moment at top

measured down from top of pilaster

2 8 22

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0.6D + 0.6WTop of pilaster

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

Design for = 2.3 k, = 218 k-in.

Location of maximum moment

0.6 9.6k 0.6 8.1k 0.9k0.9k 5.8in. 5.2k ⋅ in.. . ⋅ .. 143.1in.

0.9k 0.6 0.20 143.1in. . 2.3k. ⋅ . . . . . ⋅ .. . . 218k ⋅ in.Maximum

moment

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Assume compression controls; Determine

Determine

3 ⁄3 . . . . . . . . .⁄ ⋅ . . . . . 3.00in.

. .. . 0254 0.312 Tension controls

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Equation / Value Iteration 1 Iteration 2 Iteration 3

(in.) 3.00 3.38 3.40

0.254 0.286 0.288

(k-in.) 15.6 15.3 15.3

, (in.2) 0.585 0.593 0.593

, (in.) 0.678 0.686 0.686

2 (in.) 3.38 3.40 3.40

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

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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

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f’m = 2000 psi

f’m = 1500 psi

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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 ( = 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

= 11.8 in

xLoad

= 5.8 in 2.0 in

Lateral Load= (26psf)(16ft) = 416 lb/ft

Insi

de

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0.9D + 1.0WAt top of pilaster:

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

Find location of maximum moment

0.9 9.6k 1.0 8.1k 0.54k0.54k 5.8in. 3.1k ⋅ in.. . ⋅ .. 143.7in.

,. ⋅ . . . . . ⋅ .. . . 361k ⋅ in.Maximum moment

Design for = 2.7 kips, = 361 k-in.

0.54k 0.9 0.20 143.7in. . 2.69k

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Determine required area of steel (assuming one layer)

Determine

Determine ,

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

⁄.11.8in. 11.8in. . . . . . ⋅ .⁄. . . . . 1.50in., . ⁄. . . . . . . .⁄ 0.57in.

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Example: Pilaster Design

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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

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Design: Bearing Walls – OOP Loads

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

• No second-order analysis required

• No maximum reinforcement limits

• Use previous design procedure

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

• Need to check maximum reinforcement limits

• Need to check deflections

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Strength Design Procedure

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

• 0.05 No height limit

• 0.20 height limited by 30Moment:

factored floor loadfactored wall load

Deflection:

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Solve simultaneous linear equations:

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Deflection LimitCalculated using allowable stress load combinations

Cracking Moment: ⁄ ⁄

Cracked moment of inertia:

Depth to neutral axis: .0.007

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Example: Wind Loads ASD

18 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: ReinforcementSolution:Estimate reinforcement~ . . 0.778Assume = 0.95, = 0.080 in.2/ft

Try #5 @ 48 in. (0.078 in.2/ft)

Cross-sectionof top of wall

Determine eccentricitye = 7.625in/2 – 1.0 in.= 2.81 in.wall weight is 38 psf for

48 in. grout spacing

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Load Comb. (kip/ft) (kip/ft) (k-ft/ft) (k-ft/ft) , (in2)

0.6D+0.6W 0.084 0.364 -0.049 0.753 0.068D+0.6W 0.284 0.751 -0.002 0.777 0.059

Check 0.6D+0.6W

Use #5 @ 48 in. (0.078 in.2/ft)Although close to #5 @ 56 in. (0.066in.2/ft), a wider spacing also reduces wall weight

Find force at top of wall

Find force at midheight

Find moment at top of wall

Find moment at midheight

0.6 0.5 0.6 0.36 0.0840.084 0.6 0.040ksf 2.67ft 9ft 0.3640.084 . ft 0.6 0.032ksf . 0.049 ⋅. . . ⋅ 0.753 ⋅

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Sample Calculations: 0.6D+0.6W1. = 0.312; = 1.19in.2. Assume masonry controls.

Determine .Since 0.478 in. < 1.18 in.

tension controls. 3. Iterate to find , .

Equation / Value Iteration 1 Iteration 2(in.) 0.457 0.7912⁄ 3⁄ (k-ft/ft) 0.1110 0.1076

, (in.2/ft) 0.0658 0.0682

, (in.) 0.10362 (in.) 0.791

3 ⁄3 . . . . . ⋅ . . . 0.457in.

For centered reinforcement, 2 0⁄

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Example: Wind Loads SD

18 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: ReinforcementSolution:Estimate reinforcement~ . 1.30

= 0.24 in., = 0.078 in.2/ftTry #5 @ 48 in. (0.078 in2/ft)

Cross-sectionof top of wall

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

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Summary of Strength Design Load Combination Axial Forces(wall weight is 38 psf for 48 in. grout spacing)

Load Combination

(kip/ft) (kip/ft) (kip/ft)

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

0.9(0.038)(2.67+9) = 0.399 0.489

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

1.2(0.038)(2.67+9) = 0.532 0.972

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

at mid-height of wall between supports (9 ft from bottom)

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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

Cracking moment, :Commentary allows inclusion of axial load (9.3.5.4.4)Use minimum axial load (once wall has cracked, it has cracked)

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

51psi 153psi 68psi

⁄ ⁄ . . . .. .⁄6.97 ⋅ . 0.581 ⋅

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Cracked moment of inertia,

Modular ratio,

Depth to neutral axis,

16.1. . . .. . 0.334in.

16.1 0.0775 . . 3.812in. 0.334in. . . .16.8 .- 54 -


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

= factored moment at top of wall= factored out-of-plane load on parapet

= height of parapet

0.090 . ft . . 0.093 ⋅

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. . ⋅ . ⋅ . . . . .. . . .

1.309 ⋅

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Example: Wind Loads SDCompare to capacity (interaction diagram)

Depth of stress block,

Design moment,

⁄. . . . .. . 0.270in.0.9 . . 0.0775 . 60ksi 3.812in. . .17.19 ⋅ . 1.432 ⋅

,

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OK

Load Combination(kip-ft/ft) (kip-ft/ft)

2nd Order / 1st Order

1.2D+1.6Lr+0.5W 0.799 1.755 0.455 1.074

1.2D+1.0W+0.5Lr 1.416 1.574 0.900 1.097

0.9D+1.0W 1.309 1.432 0.914 1.047

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Check Deflections: Use ASD Load Combinations

Load Combination D+0.6W 0.6D+0.6W(k/ft) 0.5+0.6(-0.36) = 0.284 0.6(0.5)+0.6(-0.36) = 0.084(k/ft) 38(2.67+9) = 0.443 0.6(0.443) = 0.266

(k/ft) 0.727 0.350(in) 0.350 0.325

(in4/ft) 17.48 16.46(k-ft/ft) -0.002 -0.049

(k-ft/ft) 0.805 0.767(in) 0.474 0.426

OKDeflection Limit: 0.007 0.007 18ft 12 . 1.51in.

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Deflections, Sample Calculations: D + 0.6WReplace factored loads with service loads

. . . . . ⋅ . ⋅ . .. .. . . .

0.0393ft 0.474in.- 60 -


Check Maximum Reinforcement: (Section 9.3.3.2)• neutral axis is in face shell• Load combination D + 0.75L +0.525QE (9.3.3.2.1 (d))

• is just dead load = 0.5+0.038(2.67+9) = 0.943k/ft

OK

9.3.3.2 Commentary Equations

.. .. . . .. . . 0.00917

Reinforcement Ratio

. .. . . 0.00169

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Bearing Walls: SD Summary• Factored Loads ,

• Axial load: include wall weight• Moment: include second-order effects

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

• Cross-section properties: , ,• Material properties: , , ,

• Capacity ,• Interaction diagram• Need to only obtain point(s) of interest

• Maximum reinforcement• Shear

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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

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Shear Walls: Shear

Strength Design (Chapter 9)

0.84.0 1.75 0.250.56 0.254 1.0Interpolate for 0.25 1.05 2

Allowable Stress (Chapter 8)

4.0 1.75 0.254.0 1.75 0.25

Special Reinforced Shear Walls0.53 ⁄ 0.252 ⁄ 1.0Interpolate for 0.25 ⁄ 1.05 2

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Shear Walls: Maximum Reinforcement

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

Special reinforced shear walls having• ⁄ 1 and• 0.05 • Provide boundary elements

• Boundary elements not required in certain cases

• OR• Limit reinforcement

• Based on strain condition2

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SD Maximum Reinforcement

Design with Boundary Elements? No

Is1 OR 3 AND 3AND0.10 : Geometrically symmetrical walls0.05 : 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 in masonry and 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 ⁄ 1 ?

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

Yes No

Yes

Yes No

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Shear Walls: Shear Capacity DesignSpecial Reinforced Shear Walls

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

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

• Masonry shear stress reduced for special walls

• Design shear strength, , greater than shear corresponding to 1.25(increases shear at least 1.39 times)

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

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ASD: Distributed Reinforcement

Calculate

Is ?

YESSolve for ,∗

NO

Compression controls

613 3

,∗ 1212 1

Determine from the quadratic equation13 3 66 0,∗ 12 1 112 1

Tension controls∗ = distributed reinforcement (in.2/ft)

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SD: Distributed Reinforcement

Assume tension controls

2 2⁄0.8, 0.8 ⁄

,∗ ,0.65 Approximate required distributed reinforcement

Approximate 0.9

- 69 -

Example: Shear Wall ASD

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

Required: Design the shear wall; ordinary reinforced shear wallSolution: Check using 0.6D+0.7E load combination.• 0.7 50k 10ft 350k ⋅ ft 4200k ⋅ ft• Axial load,

• Need to know weight of wall to determine .• Need to know reinforcement spacing to determine wall weight• Estimate wall weight as 45 psf

• Wall weight: 45psf 10ft 16ft 7.2k• 1 k ft⁄ 16ft 7.2k 23.2k• 0.6 0.7 0.2 0.53 0.53 23.2k 12.3k

- 70 -


613 3 4200k ⋅ in. 12.3k 192in.613 192in. 0.90ksi 7.625in. 12.3k 192in.3 0.0550Calculate ; for preliminary design purposes use full thickness of wall

Since tension controls. Solve quadratic equation.13 3 6 6 013 192in. 32ksi 7.625in.16.11 12.3k 192in.34200k ⋅ in. 12.3k 192in.6 4200k ⋅ in. 12.3k 192in.6 00.147

- 71 -


Calculate required area of reinforcement

,∗ 12 1 112 112 0.147 192in. 7.625in. 32ksi 0.1471 0.147 116.11 12.3k12 1 0.147 32ksi 192in.0.00934 in.in. 0.112 in.ftTry #5 @ 32 in. (0.120in.2/ft)

Due to module; use 40 in. (0.093in.2/ft) for interior bars

32 in. 32 in.40 in. 40 in. 40 in.

- 72 -


Stressed to 93% of allowable

Used equivalent thickness = 3.57 in.(40 in. grout spacing)

- 73 -


Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, ′ =2000psi; #5 at 32in. ends; #5 @ 40in. interior; superimposed dead load of 1 kip/ft. In-plane seismic load of 50 kips.

Required: Design for shearSolution: Net area is shown below

2 1.25in. 192in. 6 8in. 7.625in. 2.5in. 726in.- 74 -


Shear Stress:

OK

120in.192in. 0.625Shear Span:

0.7 50k726in. 48.2psi

Max Shear:, 23 5 223 5 2 0.625 2000psi 0.75 83.8psi

Masonry Shear:

12 4 1.75 0.2512 4 1.75 0.625 2000psi 0.25 12300lb726in. 0.7551.9psiOK

- 75 -

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 of 50 kips. 0.5 (just less than 0.5)

Required: Design the shear wall; ordinary reinforced shear wallSolution: Check using 0.9D+1.0E load combination.• 50k 10ft 500k ⋅ ft• Axial load,

• Need to know weight of wall to determine .• Need to know reinforcement spacing to determine wall weight• Estimate wall weight as 45 psf

• Wall weight: 45psf 10ft 16ft 7.2k• 1 k ft⁄ 16ft 7.2k 23.2k• 0.9 0.2 0.80 0.80 23.2k 18.6k

- 76 -


, 0.8 ⁄0.8 2ksi 7.625in. 3.96in. 18.6k 0.9⁄60ksi 0.460in.

, depth of stress

block

, , area of steel

Estimate d 0.9 0.9 192in. 173in.

,∗ , dist. steel

,∗ ,0.65 0.460in.0.65 192in. 12in.ft 0.044 in. ft⁄Try #4 @ 48 in. (0.050in.2/ft)

2 2⁄0.8173in. 173in. 2 18.6k 173in. 192in. 2 6000k ⋅ in.⁄0.9 0.8 2000psi 7.625in. 3.96in.

- 77 -


-100

0

100

200

300

400

500

600

700

800

0 500 1000 1500 2000 2500

Axi

al (k

ip)

Moment (kip-ft)

Design Strength

Factored Loads Balanced Point

Stressed to 94% of design strength

Used equivalent thickness = 3.39 in.(48 in. grout spacing)

- 78 -

Example: Comparison to ASD

‐100

0

100

200

300

400

500

0 200 400 600 800 1000 1200 1400

Axial (kip)

Moment (k‐ft)

0.7 SD: #4 @ 48 in.

ASD: #5 @ 32,40 in.

SD: 5 - #4

ASD: 6 - #5

- 79 -


Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, ′ =2000 psi; #4 at 48in.; superimposed dead load of 1 kip/ft. In-plane seismic load of 50 kips.

Required: Design for shearSolution: Net area is shown below

2 1.25in. 192in. 5 8in. 7.625in. 2.5in. 685in.- 80 -


OK

120in.192in. 0.625Shear Span:

Max Shear:, 43 5 20.8 43 5 2 0.625 685in. 2000psi 0.75 91.9kip

Masonry Shear:

4 1.75 0.250.8 4 1.75 0.625 685in. 2000psi 0.25 18600lb 0.7556.2kipOK

- 81 -


• Calculate axial force based on = 83.8 in. • Include compression reinforcement • = 323 kips• Assume a live load of 1 k/ft• D + 0.75L + 0.525QE = (1k/ft + 0.75(1k/ft))16ft = 28 kips

Section 9.3.3.2 Maximum ReinforcementSince / 1, strain gradient is based on 1.5 .

OK

= 0.446(188in.) = 83.8 in.

Strain c/d, CMU c/d, Clay1.5 0.446 0.530

3 0.287 0.360

4 0.232 0.297

- 82 -


Shear reinforcement requirements in Strength Design

• 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.

- 83 -


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 of 50 kips. 0.5Required: Design the shear wall; special reinforced shear wallSolution:• Flexural reinforcement remains the same (although ASCE 7 allows a

load factor of 0.9 for ASD and special shear walls)

• Design for 1.5V, or 1.5(0.7)(50 kips) = 52.5 kips (Section 7.3.2.6.1.2)

• 52.5k 726in.⁄ 72.3psi• Maximum 83.8psi OK

- 84 -


Use #5 @ 16 in.

Use #5 bars in bond beams.Determine spacing.

Required steel stress

Masonry Shear:

14 4 1.75 0.2514 4 1.75 0.625 2000psi 0.25 12300lb726in. 36.7psi, 72.3psi0.75 36.7psi 59.7psi

0.5 ⇒ 0.5,0.5 0.31in. 32000psi 192in.59.7psi 726in. 21.9in.

- 85 -


Masonry Shear:

14 4 1.75 0.2514 4 1.75 0.625 2000psi 0.25 15300lb1464in. 35.1psi

Due to closely spaced bond beams, fully grout wall.

Shear Stress:52.5k1464in. 35.9psi7.625in. 192in. 1464in.Shear Area:

81psf 10ft 16ft 13.0kWall weight:

Dead load: 1 k ft⁄ 16ft 13.0k 29.0kAxial load: 0.53 0.53 29.0k 15.3k

- 86 -


Use #4 bars in bond beams.Determine spacing.

Required steel stress

, 35.9psi1.0 35.1psi 0.8psi0.5 ⇒ 0.5,0.5 0.20in. 32000psi 192in.0.8psi 1464in. 524in.

Spacing determined by prescriptive requirements

Maximum Spacing Requirements (7.3.2.6)• minimum{ one-third length, one-third height, 48 in. }min 192 .3 , 120 .3 , 48 . min 64 . , 40 . , 48 . 40 .

- 87 -


Use #5 @ 40 in.

Prescriptive Reinforcement Requirements (7.3.2.6)• 0.0007 in each direction• 0.002

Vertical: ρ 6 0.31 . 2 1464 . 2⁄ 0.00127 OKHorizontal: ρ , max 0.002 0.00127, 0.0007 0.00073Determine bar size for 40 in. spacing, 0.00073 7.625 . 40 . 0.22 .

An alternate is #4 @ 32 in. 0.00082- 88 -


Section 8.3.4.4 Maximum Reinforcement

Requirements only apply to special reinforced shear walls

No need to check maximum reinforcement since only need to check if:

• / 1 and / 0.625• 0.05 ′

• 0.05(2000psi)(1464in2) = 146 kips

• Assume a live load of 1 k/ft

• 1.0 0.7 0.2 1.0 0.7 0.2 0.5 29k 16k 47.0kOK

- 89 -


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

OK

For distributed reinforcement, the reinforcement ratio is obtained as the total area of tension reinforcement divided by . For Assume 5 out of 6 bars in tension.

2 16.1 2ksi2 60ksi 16.1 60ksi2ksi 0.00582

5 0.31in.7.625in. 188in. 0.00108- 90 -


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 of 50 kips. 0.5Required: Design the shear wall; special reinforced shear wallSolution:• Flexural reinforcement remains the same

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

- 91 -


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

• 1.25 = 766 k-ft;

• Design for shear of 76.6 kips

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

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

• 1.25 = 886 k-ft

• Design for shear of 88.6 kips

• 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. • = 778 k-ft, 1.25 = 972 k-ft

• Design for shear of 97.2 kips- 92 -


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

• Often easier to just use = 2.5 , or = 2.5 = 2.0 .

• Design for shear of 100 kips

- 93 -


Shear Area:(6 - #4 bars)

2 1.25in. 192in. 6 8in. 7.625in. 2.5in. 726in.Max Shear:

, 43 5 20.8 43 5 2 0.625 726in. 2000psi 0.75 97.4kipOptions:• Design for shear from 1.25 = 97.2 kips• Increase to 2100 psi, , 99.8 kips

• requires a unit strength of 2250 psi• Grout more cells

- 94 -


Masonry Shear:

4 1.75 0.254 1.75 0.625 726in. 2000psi 0.25 18600lb99.0kipRequiredSteel Strength: , 100k0.8 0.75 99.0k 67.7kDeterminespacing:Use #5 bars

0.5 ⇒ 0.5 ,0.5 0.31in. 60ksi 192in.67.7k 26.4in.Use #5 at 24 in. o.c.

- 95 -


Maximum reinforcement

• = 0.446(188in.) = 83.8 in.

• Calculate axial force based on = 83.8 in. (#4 @ 40 in.)

• Since 1⁄ , = 1.5

• Include compression reinforcement

• = 298 kips

• Assume a live load of 1 k/ft

• D + 0.75L + 0.525QE = 23.2kip + 0.75(16 kip) = 35.2 kips

OK

- 96 -


• 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

0.10 geometrically symmetrical sections0.05 geometrically unsymmetrical sections

AND

OR AND

For our wall, 1⁄ 0.10 0.10 2ksi 1464in. 293k1 33

- 97 -


• Prescriptive Reinforcement Requirements • 0.0007 in each direction• 0.002 total

• Vertical: 6(0.20in2)/1464in2 = 0.00082• Horizontal: 5(0.31in2)/[120in(7.625in)] = 0.00169• Total = 0.00082 + 0.00169 = 0.00251 OK

- 98 -

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 slightly less shear steel

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.

- 99 -

ASD vs SD: Final Outcome

Strength Design

Allowable Stress Design

- 100 -

Thank you

Questions

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