Masonry Design Example for building with brick materials.

- 1 -
Masonry Design Example
Comparisons Using ASD and SD
Richard Bennett, PhD, PE
The University of Tennessee
Chair, 2016 TMS 402/602 Code Committee
2nd Vic Chair, 2022 TMS 402/602 Code Committee
- 2 -
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
- 3 -
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
- 4 -
2011 vs 2013 Code
 Modulus of rupture values increased by 1/3 for all but fully
grouted masonry normal to bed joint
 Unit strength tables recalibrated
• Type S mortar, 2000 psi unit strength, f′m = 2000 psi
 Shear strength of partially grouted walls: reduction factor of
0.75

- 5 -
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
- 6 -
Partially Grouted Shear Wall Factor
Mean St Dev
Fully 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
0.90
1.16
0.776
- 7 -
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: 1.5
0.8 0.8
Allowable Stress (Chapter 8)
• Allowable masonry stress: 0.45
• Allowable steel stress
• 32 ksi, Grade 60 steel
⁄ ⁄
2
1 3
⁄
• No min and max reinforcement
• Allowable shear stress
2.25
- 8 -
Beam and Lintel Design
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
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.

- 9 -
ASD: Alternate Design Method
Calculate
3
2 2
2
3
Is ?
For Grade 60 steel,
CMU = 0.312
YES
NO
Compression controls Tension controls
⁄
,
2
1
1
,
1
3
,
2
Iterate. Use 2 as
new guess and repeat.
- 10 -
Comparison of ASD and SD
8 inch CMU
d = 20 in.
- 11 -
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 psi
Required: Design beam
Solution:
5.2.1.3: Length of bearing of beams shall be a minimum of 4 in.; typically
assumed to be 8 in.
5.2.1.1.1 Span length of members not built integrally with supports shall be taken
as the clear span plus depth of member, but need not exceed distance between
center of supports.
• Span = 10 ft + 2(4 in.) = 10.67 ft
5.2.1.2 Compression face of beams shall be laterally supported at a maximum
spacing of:
• 32 multiplied by the beam thickness. 32(7.625 in.) = 244 in. = 20.3 ft
• 120 / . 120(7.625 in.)2 / (20 in.) = 349 in. = 29.1 ft
- 12 -
Allowable Stress Design: Flexure
Load
Weight 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 0.312
900
29000ksi
900 2.0ksi
16.1
3 3
. . . ⋅
.
. . .
9.32in.

- 13 -
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
1
0.466
1
1.94in.
- 14 -
Strength Design: Flexure
Use 2 - #6 (As = 0.88 in2)
= 78.5 k-ft
= 70.6 k-ft
Factored Load
Weight of fully grouted
normal weight: 83 psf
Factored Moment
Find
Depth of equivalent
rectangular stress block
Find ,
Req’d area of steel
. .
62.6k ⋅ ft
1.2 1.6
1.2 1.5 0.083 2ft 1.6 1.5 4.40
.
20in. 20in.
. ⋅
.
.
. . . .
3.78in.
,
. . . . . . .
0.77in.
- 15 -
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. 160psi
117.1k ⋅ in. 9.76k ⋅ ft
1.3 1.3 9.76k ⋅ ft 12.7k ⋅ ft
78.5k ⋅ ft
0.88in.
7.625in. 20in.
0.00577
- 16 -
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 0.5
0.34
0.34 ( = 1.5 ksi)
0.26
0.26 ( = 1.5 ksi)
1.0 1.0
0.64
0.65 ( = 1.5 ksi)
0.50
0.52 ( = 1.5 ksi)
1.5 1.5
1.94
5.09 ( = 1.5 ksi)
0.77
0.80 ( = 1.5 ksi)

- 17 -
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.
59.2psi
OK
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 89.4psi
59.2psi 50.2psi 8.9psi
0.5 ⇒
,
.
. . . . .
. .
0.034in.
.
. . .
23.5in.
- 19 -
Strength Design: Shear
Requirement for shear at
is in ASD, not SD, but
assume it applies
Design for DL = 1 k/ft, LL = 1 k/ft
Shear at
reaction
Design shear force
Design masonry
shear strength
Suggest that be used,
not to find
.
4in.
.
1.17ft
16.00k
. .
.
12.50k
. . . . . .
16.00k
2.25
0.8 2.25 7.625in. 20in. 2000psi 12.28k
- 20 -
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 .
4
0.9 4 7.625in. 20in. 2000psi 21.82k
. .
.
0.28k
0.5 ⇒
,
.
. .
. .
0.004in.
.
.
. . . .
20.4in.

- 21 -
Non-Bearing Partially Grouted Walls
Out-of-Plane
s
b’
b
d
a
t
f
As
= effective compressive width per bar = min{ , 6 , 72 in} (5.1.2)
A. Neutral axis in flange:
a. Almost always the case
b. Design for solid section
B. Neutral axis in web
a. Design as a T-beam section
C. Often design based on a 1 ft width
Minimum reinforcement:
No requirements
Maximum reinforcement:
ASD: none
SD: Same as for beams
- 22 -
Non-Bearing Partially Grouted Walls
Out-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
- 23 -
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
⋅
- 24 -
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)

- 25 -
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.
- 26 -
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
- 27 -
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
- 28 -
ASD: Combined Bending and
Axial Load Design Method
If tension controls; determine from cubic equation.
6 2 2 2
0
,
1
2
1
Determination of .
0.45
0.45
32ksi
29000ksi
900
0.45
0.45
32
29000
900
0.312
For clay masonry, 700 , 0.368

- 29 -
Strength Design: Combined
Bending and Axial Load
Calculate
Is ?
For CMU, Grade 60 steel
0.547
YES
Compression controls
Tension controls
NO
2 2
⁄
0.8 0.8
,
0.8 ⁄
,
0.8 ⁄
- 30 -
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:
Vertical
Spanning
= 11.8 in
x
Load
= 5.8 in 2.0 in
= 1800 ksi
= 16.1
Lateral Load
= 0.6(26psf)(16ft) = 250 lb/ft
Inside
- 31 -
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 2
2
- 32 -
0.6D + 0.6W
Top 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 0.9k
0.9k 5.8in. 5.2k ⋅ in.
. . ⋅ .
.
143.1in.
0.9k 0.6 0.20 143.1in.
.
2.3k
. ⋅ . .
.
. . ⋅ .
.
.
.
218k ⋅ in.
Maximum
moment

- 33 -
Assume compression controls; Determine
Determine
3
⁄
3
. . . . . . . . .⁄ ⋅ .
.
. . .
3.00in.
. .
. .
0254 0.312 Tension controls
- 34 -
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 -
0.6D+0.6W
P = 2.3k
M = 18.2k-ft
D+0.75(0.6W)+0.75S
P = 15.3k
M = 16.8k-ft
D+S
P = 19.2k
M = 9.25k-ft
D+0.6W
P = 7.1k
M = 19.2k-ft
- 36 -
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:
Vertical
Spanning
= 11.8 in
x
Load
= 5.8 in 2.0 in
Lateral Load
= (26psf)(16ft) = 416 lb/ft
Inside
- 38 -
0.9D + 1.0W
At 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 0.54k
0.54k 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 -
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 -

- 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
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 30
Moment:
factored floor load
factored wall load
Deflection:

- 45 -
Solve simultaneous linear equations:
- 46 -
Deflection Limit
Calculated 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
psf
D = 500 lb/ft
Lr = 400 lb/ft
W = -360 lb/ft
2.67
ft
Given: 8 in. CMU wall; Grade 60 steel; Type S masonry
cement mortar; = 2000 psi; roof forces act on 3 in.
wide bearing plate at edge of wall.
Required: Reinforcement
Solution:
Estimate reinforcement
~
. .
0.778
Assume = 0.95
, = 0.080 in.2/ft
Try #5 @ 48 in. (0.078 in.2/ft)
Cross-section
of top of wall
Determine eccentricity
e = 7.625in/2 – 1.0 in.
= 2.81 in.
wall weight is 38 psf for
48 in. grout spacing
- 48 -
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 0.068
D+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 0.084
0.084 0.6 0.040ksf 2.67ft 9ft 0.364
0.084
.
ft 0.6 0.032ksf
.
0.049
⋅
. . .
⋅
0.753
⋅

- 49 -
Sample Calculations: 0.6D+0.6W
1. = 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 0.791
2
⁄ 3
⁄ (k-ft/ft) 0.1110 0.1076
, (in.2/ft) 0.0658 0.0682
,
(in.) 0.1036
2 (in.) 0.791
3
⁄
3
. . . . .
⋅ .
.
. 0.457in.
For centered reinforcement, 2 0
⁄
- 50 -
Example: Wind Loads SD
18
ft
32
psf
D = 500 lb/ft
Lr = 400 lb/ft
W = -360 lb/ft
2.67
ft
Given: 8 in. CMU wall; Grade 60 steel; Type S masonry
cement mortar; = 2000 psi; roof forces act on 3 in.
wide bearing plate at edge of wall.
Required: Reinforcement
Solution:
Estimate reinforcement
~
.
1.30
= 0.24 in.
, = 0.078 in.2/ft
Try #5 @ 48 in. (0.078 in2/ft)
Cross-section
of top of wall
Determine eccentricity
e = 7.625in/2 – 1.0 in.
= 2.81 in.
- 51 -
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+0.5Lr
1.2(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 -
Modulus of rupture: use linear interpolation between no grout and full grout
Ungrouted (Type S masonry cement): 51 psi
Fully grouted (Type S masonry cement): 153 psi
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 -
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
⋅
- 55 -
. .
⋅ .
⋅
.
.
.
.
.
.
.
. .
1.309
⋅
- 56 -
Compare to capacity (interaction diagram)
Depth of stress
block,
Design
moment,
⁄
.
.
.
. .
.
. 0.270in.
0.9
.
.
0.0775
.
60ksi 3.812in.
. .
17.19
⋅ .
1.432
⋅
,

- 57 -
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 -
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
OK
Deflection Limit: 0.007 0.007 18ft 12
.
1.51in.
- 59 -
Deflections, Sample Calculations: D + 0.6W
Replace 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

- 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
0.8
4.0 1.75 0.25
0.5
6 0.25
4 1.0
Interpolate for 0.25 1.0
5 2
4.0 1.75 0.25
4.0 1.75 0.25
Special Reinforced Shear Walls
0.5
3 ⁄ 0.25
2 ⁄ 1.0
Interpolate for 0.25 ⁄ 1.0
5 2
- 64 -
Shear Walls: Maximum Reinforcement
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 condition
2

- 65 -
SD Maximum Reinforcement
Design with Boundary Elements?
No
Is
1 OR 3 AND 3
AND
0.10 : Geometrically symmetrical walls
0.05 : 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 ?
= 1.5
Ordinary reinforced walls: = 1.5
Intermediate reinforced walls: = 3
Special reinforced walls: = 4
Yes No
Yes
Yes No
- 66 -
Shear Walls: Shear Capacity Design
Special Reinforced Shear Walls
• 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 ?
YES
Solve for ,
∗
NO
Compression controls
6
1
3 3
,
∗
1
2
1
2
1
Determine from the quadratic equation
1
3 3 6
6
0
,
∗
1
2 1
1
1
2
1
Tension controls
∗
= distributed reinforcement (in.2/ft)
- 68 -
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 wall
Solution: 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 -
6
1
3 3
4200k ⋅ in. 12.3k
192in.
6
1
3
192in. 0.90ksi 7.625in. 12.3k
192in.
3
0.0550
Calculate ; for preliminary design purposes use full thickness of wall
Since tension controls. Solve quadratic equation.
1
3 3 6 6
0
1
3
192in. 32ksi
7.625in.
16.11
12.3k
192in.
3
4200k ⋅ in. 12.3k
192in.
6
4200k ⋅ in. 12.3k
192in.
6
0
0.147
- 71 -
Calculate required area of reinforcement
,
∗
1
2 1
1
1
2
1
1
2
0.147 192in. 7.625in. 32ksi
0.147
1 0.147
1
16.11
12.3k
1
2
1 0.147 32ksi 192in.
0.00934
in.
in.
0.112
in.
ft
Try #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 shear
Solution: Net area is shown below
2 1.25in. 192in. 6 8in. 7.625in. 2.5in. 726in.
- 74 -
Shear Stress:
OK
120in.
192in.
0.625
Shear Span:
0.7 50k
726in.
48.2psi
Max Shear:
,
2
3
5 2
2
3
5 2 0.625 2000psi 0.75 83.8psi
Masonry Shear:
1
2
4 1.75 0.25
1
2
4 1.75 0.625 2000psi 0.25
12300lb
726in.
0.75
51.9psi
OK
- 75 -
Example: Shear Walls SD
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 wall
Solution: 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
⁄
0.8
173in. 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
Axial
(kip)
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 shear
Solution: Net area is shown below
2 1.25in. 192in. 5 8in. 7.625in. 2.5in. 685in.
- 80 -
OK
120in.
192in.
0.625
Shear Span:
Max Shear:
,
4
3
5 2
0.8
4
3
5 2 0.625 685in. 2000psi 0.75 91.9kip
Masonry
Shear:
4 1.75 0.25
0.8 4 1.75 0.625 685in. 2000psi 0.25 18600lb 0.75
56.2kip
OK

- 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 Reinforcement
Since / 1, strain gradient is based on 1.5 .
OK
= 0.446(188in.) = 83.8 in.
Strain c/d, CMU c/d, Clay
1.5 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 -
seismic load of 50 kips. 0.5
Required: Design the shear wall; special reinforced shear wall
Solution:
• 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:
1
4
4 1.75 0.25
1
4
4 1.75 0.625 2000psi 0.25
12300lb
726in.
36.7psi
,
72.3psi
0.75
36.7psi 59.7psi
0.5 ⇒
0.5
,
0.5 0.31in. 32000psi 192in.
59.7psi 726in.
21.9in.

- 85 -
Masonry Shear:
1
4
4 1.75 0.25
1
4
4 1.75 0.625 2000psi 0.25
15300lb
1464in.
35.1psi
Due to closely spaced bond beams, fully grout wall.
Shear Stress:
52.5k
1464in.
35.9psi
7.625in. 192in. 1464in.
Shear Area:
81psf 10ft 16ft 13.0k
Wall weight:
Dead load: 1 k ft
⁄ 16ft 13.0k 29.0k
Axial load: 0.53 0.53 29.0k 15.3k
- 86 -
Use #4 bars in
bond beams.
Determine
spacing.
Required steel
stress
,
35.9psi
1.0
35.1psi 0.8psi
0.5 ⇒
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 OK
Horizontal: ρ , max 0.002 0.00127, 0.0007 0.00073
Determine 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
• 1.0 0.7 0.2 1.0 0.7 0.2 0.5 29k 16k 47.0k
OK

- 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 2ksi
2 60ksi 16.1
60ksi
2ksi
0.00582
5 0.31in.
7.625in. 188in.
0.00108
- 90 -
seismic load of 50 kips. 0.5
Required: Design the shear wall; special reinforced shear wall
Solution:
• 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
• 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
- 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:
,
4
3
5 2
0.8
4
3
5 2 0.625 726in. 2000psi 0.75 97.4kip
Options:
• 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 0.25
4 1.75 0.625 726in. 2000psi 0.25 18600lb
99.0kip
Required
Steel
Strength: ,
100k
0.8 0.75
99.0k 67.7k
Determine
spacing:
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
• 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 sections
0.05 geometrically unsymmetrical sections
AND
OR AND
For our wall, 1
⁄
0.10 0.10 2ksi 1464in. 293k
1 3
3

- 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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