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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
- 8 -
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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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:
If 0 or , = moment at top
measured down from top of pilaster
2 8 22
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Example: Pilaster Design ASD
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
- 33 -
Example: Pilaster Design ASD
Assume compression controls; Determine
Determine
3 ⁄3 . . . . . . . . .⁄ ⋅ . . . . . 3.00in.
. .. . 0254 0.312 Tension controls
- 34 -
Example: Pilaster Design ASD
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
- 35 -
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
- 36 -
Example: Pilaster Design ASD
f’m = 2000 psi
f’m = 1500 psi
- 37 -
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
- 38 -
Example: Pilaster Design SD
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
- 39 -
Example: Pilaster Design SD
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.
- 40 -
Example: Pilaster Design SD
- 41 -
Example: Pilaster Design
- 42 -
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
- 43 -
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
- 44 -
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:
- 45 -
Strength Design Procedure
Solve simultaneous linear equations:
- 46 -
Strength Design Procedure
Deflection LimitCalculated using allowable stress load combinations
Cracking Moment: ⁄ ⁄
Cracked moment of inertia:
Depth to neutral axis: .0.007
- 47 -
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
- 48 -
Example: Wind Loads ASD
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 ⋅
- 49 -
Example: Wind Loads ASD
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⁄
- 50 -
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.
- 51 -
Example: Wind Loads SD
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)
- 52 -
Example: Wind Loads SD
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 ⋅
- 53 -
Example: Wind Loads SD
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 -
Example: Wind Loads SD
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 ⋅
- 55 -
Example: Wind Loads SD
. . ⋅ . ⋅ . . . . .. . . .
1.309 ⋅
- 56 -
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 ⋅
,
- 57 -
Example: Wind Loads SD
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
- 58 -
Example: Wind Loads SD
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.
- 59 -
Example: Wind Loads SD
Deflections, Sample Calculations: D + 0.6WReplace factored loads with service loads
. . . . . ⋅ . ⋅ . .. .. . . .
0.0393ft 0.474in.- 60 -
Example: Wind Loads SD
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
- 61 -
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
- 62 -
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
- 63 -
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
- 64 -
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
- 65 -
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
- 66 -
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)
- 67 -
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
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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 -
Example: Shear Wall ASD
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 -
Example: Shear Wall ASD
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 -
Example: Shear Wall ASD
Stressed to 93% of allowable
Used equivalent thickness = 3.57 in.(40 in. grout spacing)
- 73 -
Example: Shear Wall ASD
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 -
Example: Shear Wall ASD
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
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Example: Shear Walls SD
, 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.
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Example: Shear Walls SD
-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)
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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
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Example: Shear Walls SD
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 -
Example: Shear Walls SD
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
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Example: Shear Walls SD
• 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
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Example: Shear Walls SD
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.
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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 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
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Example: Shear Wall ASD
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.
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Example: Shear Wall ASD
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 -
Example: Shear Wall ASD
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 -
Example: Shear Wall ASD
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 -
Example: Shear Wall ASD
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
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Example: Shear Wall ASD
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 -
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. 0.5Required: 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
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Example: Shear Walls SD
• 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 -
Example: Shear Walls SD
• 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
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Example: Shear Walls SD
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 -
Example: Shear Walls SD
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 -
Example: Shear Walls SD
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 -
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
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 -
Example: Shear Walls SD
• 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.
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ASD vs SD: Final Outcome
Strength Design
Allowable Stress Design
- 100 -
Thank you
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