Masonry Design Example for building with brick materials.

Masonry Design Example
Comparisons Using ASD and SD

1. -1-
Masonry Design Example
Comparisons Using ASD and SD
Richard Bennett, PhD, PE
The University of Tennessee
Chair, 2016 TMS 402/602 Code Committee
2
nd
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

2. -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
1,9001,900 2,350
2,0002,000 2,650
2,2502,600 3,400
2,5003,250 4,350
2,7503,900 ----
TMS 602 Table 2
-6-
Partially Grouted Shear Wall Factor
8
????????????
8
???????
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)
?
C
= 0.75 for partially grouted shear walls
= and 1.0 otherwise
0.90
1.16
L 0.776
-7-
Beam and Lintel Design
Strength Design (Chapter 9)
•? ??
= 0.0035 clay masonry
•?
??
= 0.0025 concrete masonry
• Masonry stress = 0.8B
?
?
• Masonry stress acts over == 0.8?
•?= 0.9 flexure; ?= 0.8 shear
/
?
L#
?
B
?
@F
5
6
?
?
?
?
4.<?
?
?
?
8
?
L2.25#
??
B
?
?
• Minimum reinf: /
?
R1.3/
??
•or #
?
R
43⁄#
?,????
• Maximum reinf: ?
?
R1.5?
?
?
???
L
0.8
0.8 B
?
?
B
?
?
?
?
?
E?
?
Allowable Stress (Chapter 8)
• Allowable masonry stress: 0.45B
?
?
• Allowable steel stress
• 32 ksi, Grade 60 steel
JL'
?
'
?
⁄ ? L #
?
>@ ⁄
GL
J?
6
E2J?FJ?
FL1FG 3⁄
B
?
L
?
? ?
??
B
?
L
6?
?
??
??
• No min and max reinforcement
• Allowable shear stress
(
?
L
5
6
2.25
B
?
?
-8-
Beam and Lintel Design
Strength Design (Chapter 9)
1. Determine =, depth of compressive
stress block
=L@F
@
6
F
6?
?
4.<?
?
?
?
2. Solve for #
?,????
#
?,????
L
4.<?
?
?
??
?
?
3.
?
???
L 0.285
B
?
?
B
?
⁄
for CMU
Grade 60 steel:
B
?
?
= 1.5 ksi, ?
???
L 0.00714
B
?
?
= 2 ksi, ?
???
L 0.00952
Allowable Stress (Chapter 8)
1. Assume value of F(or G).
Typically 0.85 < F< 0.95.
2. Determine a trial value of #
?,????
.
#
?,????
L//
(
?
F@
Choose reinforcement.
3. Determine Gand F; 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 B
?
?
). Reinforcement is
not being used efficiently.

3. -9-
ASD: Alternate Design Method
Calculate
G@ L 3
@
2
F
@
2
6
F
2/
3(
?
P
??

Is G R G
>=H
?
For Grade 60 steel,
CMU G
>=H
= 0.312
YES
NO
Compression controlsTension controls
G
???
L
(
?
(
?
E(
?
J⁄
#
?,????
L
(
?
G@ P
??
2
J(
?
1
G
F1
#
?,????
L
/
(
?
@
1F
G
3
?L
#
?,????
(
?
J
(
?
P
??
G@
6
L
?
6
E2?@F?
Iterate. Use :G@;
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; B’
I
=
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.
5.2.1.1.1Span 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
5.2.1.2Compression 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>
6
/@. 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 G@
Assume compression controls
Check if compression controls
Compression
controls
Calculate modular ratio, J
SL&E.L
1.5
i
dr
E0.083
i
dr
.
2ft E 1.5
i
dr
L3.17
i
dr
/L
??
.
<
L
7.5;
a
\j
54.:;dr
.
<
L45.1k⋅ft
GL
G@
@
L
9.32in.
20in.
L 0.466 P 0.312
JL
'
?
'
?
L
'
?
900B
?
?
L
29000ksi
900
2.0ksi
L16.1
G@ L 3
?
6
F
?
6
6
F
6?
7?
?
?
L3
64gl.
6
F
64gl.
6
6
F
6:89.6i⋅dr;
-._d.
\j
7:4.=4iqg;:;.:69gl.;
L 9.32in.

4. -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 (A
s
= 2.00 in
2
)
#
?,????
L
(
?
G@ >
J(
?
1
G
F1
L
0.90GOE
9.31in. 7.625in.
16.1:0.90ksi;
1
0.466
F1
L1.94in.
6
-14-
Strength Design: Flexure
Use 2 - #6 (A
s
= 0.88 in
2
)
/
?
= 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
/
?
L
?
?
?
.
<
L
8.84
a
\j
54.:;dr
.
<
L62.6k⋅ft
SL1.2&E1.6.
L1.2
1.5
i
dr
E 0.083
i
dr
.
2ft E 1.6
1.5
i
dr
L4.40
i
dr
=L@F
@
6
F
6?
?
4.<?
?
?
?
L20in.F
20in.
6
F
6
2..5a⋅\j
,.5
-._d.
\j
4.<
6.4iqg
;.:69gl.
L3.78in.
#
?,????
L
4.<?
?
?
??
?
?
L
4.<:6.4iqg;:;.:69gl.;:7.;<gl.;
:4iqg
L0.77in.
6
-15-
Check Min and Max Reinforcement
Minimum Reinforcement Check:B
?
= 160 psi
(parallel to bed joints
in running bond; fully grouted)
Maximum Reinforcement Check: ?
???
L 0.00952
Section modulus
Cracking moment
Check 1.3/
??
5
?
L
??
.
:
L
:;.:69gl.;
68gl.
.
:
L732in.
7
/
??
L5
?
B
?
L732in.
7
160psi
L 117.1k ⋅ in. L 9.76k ⋅ ft
1.3/
??
L1.3
9.76k ⋅ ft L 12.7k ⋅ ft
Q/
?
L78.5k⋅ft
?L
#
?
>@
L
0.88in.
6
:7.625in. ;:20in. ;
L 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 A
s
(in
2
)
ASD SD
0.5 0.5
0.34
0.34 (B
?
?
= 1.5 ksi)
0.26
0.26 (B
?
?
= 1.5 ksi)
1.0 1.0
0.64
0.65 (B
?
?
= 1.5 ksi)
0.50
0.52 (B
?
?
= 1.5 ksi)
1.5 1.5
1.94
5.09 (B
?
?
= 1.5 ksi)
0.77
0.80 (B
?
?
= 1.5 ksi)

5. -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 @
?
.
64gl.
6
E4in.
5dr
56gl.
L1.17ft
8L
??
6
L
5.4
a
\j
>4.4<7
a
\j
.
6dr >5.4
a
\j
54.:;dr
6
L11.56k
8L11.56k
9.77dr?5.5;dr
9.77dr
L9.02k
B
?
L
?
? ??
L
=.46i
;.:69gl.:64gl.;
L59.2psi
(
??
L
5
6
2.25
B
?
?
L
5
6
2.25
2000psi L 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.
P 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.
(
?
Q2
B
?
?
L2
2000psi L 89.4psi
59.2psi F 50.2psi L 8.9psi
(
??
L0.5
?
?
?
?
?
?
?
?
⇒ #
?
L
?
??,????
?
?
?
4.9?
?
?
#
?
L
<.=nqg:;.:69gl.;:64gl.;:<gl.;
4.9:76444nqg;:64gl.;
L0.034in.
6
@L
?
??
??
L
=.46i
;.:69gl.:94.7nqg;
L23.5in.
-19-
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 at
reaction
@/2 from face of support
Design shear force
Design masonry
shear strength
Suggest that @be used,
not @
?
to find #
??
64gl.
6
E4in.
5dr
56gl.
L1.17ft
8
?
L16.00k
9.77dr?5.5;dr
9.77dr
L12.50k
8
?
L
?
?
?
6
L
5.6
5.4
a
\j
>4.4<7
a
\j
.
6dr >5.:
5.4
a
\j
54.:;dr
6
L16.00k
?8
??
L?2.25#
??
B
?
?
L 0.8:2.25;:7.625in. ;:20in. ;
2000psi L 12.28k
-20-
Strength Design: Shear
Use #3 stirrups
Check max 8
?
Req’d8
??
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 @.
?8
?
Q?4#
??
B
?
?
L 0.9:4;:7.625in. ;:20in. ;
2000psi L 21.82k
?
?
?%?
??
%
L
56.94i?56.6<i

4.<
L0.28k
8
??
L0.5
?
?
?
B
?
@⇒#
?
L
?
??,????
?
4.9?
?
?
?
#
?
L
4.6<i:<gl.;
4.9::4iqg;:64gl.;
L0.004in.
6
@L
?
?
?
%6.69
?
?
?
L
56.94i
;.:69gl.:4.<:6.69;
6444nqg;
L20.4in.

6. -21-
Non-Bearing Partially Grouted Walls Out-of-Plane
s
b’
b
d
a
t
f
A
s
>= effective compressive width per bar = min{ O, 6P, 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 in
2
/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, B
?
?
= 2000 psi; Lateral wind load
of S
?
= 30 psf (factored)
Required: Reinforcing (place in center of wall)
Solution:
(
?
= 0.45(2000psi) = 900 psi'
?
= 1.80 x 10
6
psi
(
?
= 32000 psi'
?
= 29 x 10
6
psi
J= '
?
/'
?
= 16.1
Moment
Determine G@
Assume compression controls
Check if compression
controls
Tension controls
G@ L 3
?
6
F
?
6
6
F
6?
7?
?
?
L3
7.<5gl.
6
F
7.<5gl.
6
6
F
6
4.9;:
a⋅\j
\j
-._d.
\j

7:4.=4iqg;
-._d.
\j
L0.346in.
GL
??
?
L
4.78:gl.
7.<5gl.
L 0.091 O 0.312
/L
??
.
<
L
4.:
74
bX
\j
.
56
_d.
\j
5:dr
.
<
L 6912
j`⋅gl.
dr
L576
j`⋅dr
dr
-24-
Example: Partially Grouted Wall ASD
Equation / Value Iteration 1 Iteration 2 Iteration 3
G@(in.) 0.346 0.699 0.709
G0.091 0.183 0.183
#
?,????
L
?
?
?
?
5?
?
/
(in.
2
)
0.0584 0.0603 0.0603
?L
?
?,????
?
?
?
?
?
?
??
(in.)0.0784 0.0810 0.0810
G@
6
L
?
6
E2?@F?(in.)0.699 0.709 0.709 Use # 4 @ 40 inches (A
s
= 0.060in
2
/ft)
(close enough)

7. -25-
Example: Partially Grouted Wall SD
Given: 8 in CMU wall; 16 ft high; Grade 60 steel, B
?
?
= 2000 psi; Lateral wind
load of S
?
= 30 psf (factored)
Required: Reinforcing (place in center of wall)
Solution:
Use #4 @ 40 inches (A
s
= 0.060in
2
/ft)
Factored Moment
Find = Solve as solid
section Find
#
?,????
/
?
L
?
?
?
.
<
L
74
bX
\j
.
56
_d.
\j
5:dr
.
<
L 11520
j`⋅gl.
dr
L960
j`⋅dr
dr
#
?,????
L
4.<?
?
?
??
?
?
L
4.<
6.4iqg
56
_d.
\j
:4?5;=gl?;
:4iqg
L 0.0.0573
gl.
.
dr
=L@F
@
6
F
6?
?
4.<?
?
?
?
L3.81in.F
3.81in.
6
F
6
,.52,
a⋅\j
\j
,.5
-._d.
??
4.<
6.4iqg
-.??.
\j
L 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 G R G
>=H
?
For Grade 60 steel,
CMU G
>=H
= 0.312
YES
Iterate. Use :G@;
2
as new
guess and repeat.
NO
Compression controlsTension controls
G
???
L
(
?
(
?
E(
?
J⁄
G@ L 3
@
2
F
@
2
6
F
2:2:@ F @
?
2; E /; ⁄
3(
?
P
??

#
?,????
L
(
?
G@ P
??
2
F2
J(
?
1
G
F1
/
?
L2
@
?
2
F
G@
3
#
?,????
L
/F/′
(
?
@
1F
G
3
?L
2E#
?,????
(
?
J
(
?
P
??
G@
6
L
?
6
E2?@F?
-28-
ASD: Combined Bending and
Axial Load Design Method
If G O G
>=H
tension controls; determine G@from cubic equation.
P
??
(
?
6J
G@
7
F
P
??
@(
?
2J
G@
6
F
2
@F
@
?
2
E/
G@ E
2
@F
@
?
2
E/ @L0
#
?,????
L
1
2
G@ P
??
1
J
G@
@FG@
F
2
(
?
Determination of G
>=H
.
G
???
L
(
?
(
?
E
(
?
J
L
(
?
(
?
E
(
?
'
?
'
?
L
r?vwB
?
?
r?vwB
?
?
E
32ksi
29000ksi
900B
?
?
L
0.45
0.45 E
32
29000
900
L0.312
For clay masonry, '
?
L 700B
?
?
, G
???
L 0.368

8. -29-
Strength Design: Combined
Bending and Axial Load
Calculate
Is ? R ?
???
?
For CMU, Grade 60 steel
?
???
L 0.547@
YES
Compression controls
Tension controls
NO
=L@F
@
6
F
2>2
?
:@ F @
?
2; E /
?
? ⁄
?
0.8B
?
?
P
??
? L
=
0.8
?
???
L
?
??
?
??
E?
?@
#
?,????
L
0.8B
?
?
P
??
=F2
?
?⁄
?
??
'
?
@F?
?
#
?,????
L
0.8B
?
?
P
??
=F2
?
?⁄
B
?
-30-
Example: Pilaster Design ASD
Given: Nominal 16 in. wide x 16 in. deep CMU pilaster; B
?
?
= 2000 psi; Grade 60
bar in each corner, center of cell; Effe ctive 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 ( A= 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
A= 5.8 in
2.0 in
'
?
= 1800 ksi
J= 16.1
Lateral Load
S= 0.6(26psf)(16ft) = 250 lb/ft
Inside
-31-
Example: Pilaster Design ASD
Weight of pilaster:
Weight of fully grouted 8 in wall (light weight 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 TO0or TPD, /
???
L /
?
/
?
= moment at top
Tmeasured down from top of pilaster
/
???
L
/
?
2
E
SD
6
8
E
/
?
6
2SD
6
TL
D
2
F
/
?
SD
-32-
Example: Pilaster Design ASD
0.6D + 0.6W
Top of pilaster Find axial force at this point. Include weight of pilaster (200 lb/ft).
Design for 2= 2.3 k, /= 218 k-in.
Location of
maximum moment
2
?
L0.6
9.6k F 0.6
8.1k L 0.9k
/
?
L0.9k
5.8in. L 5.2k ⋅ in.
TL
?
6
F
?
?
??
L
6<<gl.
6
F
9.6i⋅gl.
4.694
a
\j
68dr
L143.1in.
2L0.9kE0.60.20
i
dr
143.1in.
5dr
56gl.
L2.3k
/
???
L
?
?
6
E
??
.
<
E
?
?
.
6??
.
L
9.6i⋅gl.
6
E
4.464<
a
_d.
:6<<gl.;
.
<
E
9.6i⋅gl.
.
6
4.464<
a
_d.
:6<<gl.;
.
L 218k ⋅ in.
Maximum
moment

9. -33-
Example: Pilaster Design ASD Assume compression controls; Determine G@
Determine G
G@ L 3
?
6
F
?
6
6
F
6:?:???
?
6;>?; ⁄
7?
?
?
??
L3
55.<gl.
6
F
55.<gl.
6
6
F
6
6.7i
55.<gl.?59.:gl. 6 ⁄>65<i⋅gl.
-._d.
\j

7:4.=4iqg;
59.:gl.
L3.00in.
GL
??
?
L
7.44gl.
55.<5gl.
L 0254 O 0.312
Tension controls
-34-
Example: Pilaster Design ASD
Equation / Value Iteration 1 Iteration 2 Iteration 3
G@(in.) 3.00 3.38 3.40
G0.254 0.286 0.288
/
?
L2
?
?
6
F
??
7
(k-in.) 15.6 15.3 15.3
#
?,????
L
????
?
?
?
5?
?
/
(in.
2
)
0.585 0.593 0.593
?L
?>?
?,????
?
?
?
?
?
?
??
(in.)0.678 0.686 0.686
G@
6
L
?
6
E2?@F?(in.)3.38 3.40 3.40
Try 2 - #5, 4 total, one in each cell
-35-
Example: Pilaster Design ASD
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-
Example: Pilaster Design ASD
f’
m
= 2000 psi
f’
m
= 1500 psi

10. -37-
Example: Pilaster Design SD
Given: Nominal 16 in. wide x 16 in. deep CMU pilaster; B
?
?
= 2000 psi; Grade 60
bar in each corner, center of cell; Effe ctive 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 ( A= 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
A= 5.8 in
2.0 in
Lateral Load
S
?
= (26psf)(16ft) = 416 lb/ft
Inside
-38-
Example: Pilaster Design SD
0.9D + 1.0W At top of pilaster: Find axial force at this point. Include weight of pilaster. Find location of
maximum moment
2
??
L0.9
9.6k F 1.0
8.1k L 0.54k
/
??
L0.54k
5.8in. L 3.1k ⋅ in.
TL
?
6
F
?
??
?
?
?
L
6<<gl.
6
F
7.5i⋅gl.
4.85:
a
\j
68dr
L143.7in.
/
?,???
L
?
?? 6
E
?
?
?
.
<
E
?
??
.
6?
?
?
.
L
7.5i⋅gl.
6
E
4.478;
a
_d.
:6<<gl.;
.
<
E
7.5i⋅gl.
.
6
4.478;
a
_d.
:6<<gl.;
.
L361k⋅in.
Maximum
moment
Design for 2
?
= 2.7 kips, /
?
= 361 k-in.
2
?
L 0.54k E 0.9
0.20
i
dr
143.7in.
5dr
56gl.
L2.69k
-39-
Example: Pilaster Design SD
Determine required area of steel (assuming one layer)
Determine =
Determine #
O
,
NAM@
Try 2 - #5, 4 total, one in each cell
=L@F
@
6
F
6>?
?
:???
?
6;>?
?
? ⁄
%
4.<?
?
?
?
??
L11.8in.F
:11.8in. ;
6
F
6>6.;i:55.<gl.?59.:gl. 6;>7:5i⋅gl.? ⁄
4.=
4.<
6.4iqg
59.:gl.
L1.50in.
#
?,????
L
4.<?
?
?
?
??
???
?
%⁄
?
?
L
4.<:6.4iqg;:59.:gl.;:5.94gl.;?6.;i 4.= ⁄
:4iqg
L0.57in.
6
-40-
Example: Pilaster Design SD

11. -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:
•
?
?
?
?
Qr?rwB
?
?
No height limit
•
?
?
?
?
Qr?trB
?
?
height limited by
?
?
Q30
Moment: /
?
L
?
?
?
.
<
E2
??

? 6E2
?
?
?
2
?
L2
??
E2
??
2
??
Lfactored floor load
2
??
Lfactored wall load
Deflection:
/
?
Q/
??
?
?
L
9?
?
?
.
8<?
?
?
?
/
?
P/
??
?
?
L
9?
??
?
.
8<?
?
?
?
E
9
?
?
??
??
?
.
8<?
?
?
??

12. -45-
Strength Design Procedure
Solve simultaneous linear equations:
y
?
Py
??
/
?
L
???
.
4
>?
??
??
.
>
1??????
.
04??
-
??
?
-
???
5?
1???
.
04?????
?
?
L
1?
.
04?????
???
.
4
>?
??
??
.
>?
??
???
??
?5
5?
1???
.
04?????
y
?
Qy
??
/
?
L
???
.
4
>?
??
??
.
5?
1???
.
04????
?
?
L
1?
.
04????
???
.
4
>?
??
??
.
5?
1???
.
04????
-46-
Strength Design Procedure
Deflection Limit
Calculated using allowable stress load combinations
Cracking Moment:
/
??
L
?
?
?
?
⁄>?
?
?
?
?
??
6⁄
Cracked moment of inertia:
+
??
LJ
#
?
E
?
?
?
?
?
??
6?
@F?
6
E
??
/
7
Depth to neutral axis:
?L
?
?
?
?
>?
?
4.:8?
?
?
?
?
?
Q0.007D
-47-
Example: Wind Loads ASD
18 ft
32 psfD = 500 lb/ft
L
r
= 400 lb/ft
W = -360 lb/ft
2.67 ft
Given: 8 in. CMU wall; Grade 60 steel; Type S masonry cement mortar; B
?
?
= 2000 psi; roof forces act on 3 in.
wide bearing plate at edge of wall.
Required: Reinforcement
Solution: Estimate reinforcement
/~
??
.
<
L
4.:
4.476iqd
5<dr
.
<
L0.778
i?dr
dr
Assume F= 0.95
#
?,????
L
?
? ?
??
= 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-
Example: Wind Loads ASD
Load Comb.2
?
(kip/ft)2(kip/ft)/
?
(k-ft/ft)/(k-ft/ft)#
?,????
(in
2
)
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
2
?
L0.6
0.5
i
dr
E0.6
F0.36
i
dr
L 0.084
i
dr
2L0.084
i
dr
E0.6
0.040ksf
2.67ft E 9ft L 0.364
i
dr
/
?
L 0.084
i
dr
6.<5
56
ft F 0.6
0.032ksf
6.:;dr
.
6
LF0.049
i⋅dr
dr
/L
??
.
<
E
?
?
6
L
4.:
4.476iqd
5<dr
.
<
E
?4.48=
a⋅\j
\j 6
L 0.753
i⋅dr
dr

13. -49-
Example: Wind Loads ASD
Sample Calculations: 0.6D+0.6W 1.
G
???
= 0.312; G@
???
= 1.19in.
2. Assume masonry controls
.
Determine G@.
Since 0.478 in. < 1.18 in.
tension controls.
3. Iterate to find #
?,????
.
Equation / Value Iteration 1 Iteration 2
G@(in.) 0.457 0.791
/
?
L2
@
?
2⁄FG@ 3⁄(k-ft/ft)
0.1110 0.1076
#
?,????
L
????
?
?
?
5?
?
/
(in.
2
/ft)
0.0658 0.0682
?L
?>?
?,????
?
?
?
?
?
?
??
(in.)
0.1036
G@
6
L
?
6
E2?@F?(in.)
0.791
G@ L 3
?
6
F
?
6
6
F
6:?:???
?
6;>?; ⁄
7?
?
?
??
L3
7.<5gl.
6
F
7.<5gl.
6
6
F
6
4.;97
a⋅\j
\j

-._d.
\j

7:4.=4iqg;
-._d.
\j
L 0.457in.
For centered reinforcement, :@ F @
?
2; L 0 ⁄
-50-
Example: Wind Loads SD
18 ft
32 psfD = 500 lb/ft
L
r
= 400 lb/ft
W = -360 lb/ft
2.67 ft
Given: 8 in. CMU wall; Grade 60 steel; Type S masonry cement mortar; B
?
?
= 2000 psi; roof forces act on 3 in.
wide bearing plate at edge of wall.
Required: Reinforcement
Solution: Estimate reinforcement
/
?
~
??
.
<
L
4.476iqd
5<dr
.
<
L1.30
i?dr
dr
== 0.24 in.
#
?,????
= 0.078 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.
-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
2
??

(kip/ft)
2
??
(kip/ft)
2
?
(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+0.5L
r
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
2
??
= Factored floor load; just eccentrically applied load
2
??
= 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 inte rpolation 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
B
?
L51psi
9 slepmsrcb acjjq
: acjjq
E 153psi
5 epmsrcb acjjq
: acjjq
L68psi
/
??
L
?
?
?
?
⁄>?
?
?
?
?
??
6⁄
L
8<=
bX
\j
84.;
_d.
.
\j
W >:<nqg 776.4
_d.
0
\j
;.:6gl. 6 ⁄
L6.97
i⋅gl.
dr
L 0.581
i⋅dr
dr

14. -53-
Example: Wind Loads SD Cracked moment of inertia, +
??
Modular ratio, J
Depth to
neutral axis, ?
JL
?
?
?
?
L
6=444iqg
5<44iqg
L16.1
?L
?
?
?
?
>?
?
4.:8?
?
?
?
L
4.4;;9
_d.
.
\j
:4iqg >4.8<=
a
\j
4.:8
6iqg 56
_d.
\j
L0.334in.
+
??
LJ
#
?
E
?
?
?
?
@F?
6
E
??
/
7
L16.1
0.0775
gl.
.
dr
E
4.48<=
a
\j
:4iqg
3.812in. F0.334in.
6
E
56
_d.
\j
4.778gl.
/
7
L16.8
gl.
0
dr
-54-
Example: Wind Loads SD
Find /
?
2
??
A
?
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
S
??
= factored out-of-plane load on parapet
D
?
= height of parapet
/
??
L2
??
A
?
F
?
??
?
?
.
6
L0.090
i
dr
6.<5
56
ft F
4.476iqd
6.:;dr
.
6
LF0.093
i⋅dr
dr
-55-
Example: Wind Loads SD
/
?
L
???
.
4
>
?
??
.
>
1??????
.
04??
-
??
?
-
???
5?
1???
.
04?????
L
,.,/.ai\
-4\j
.
4
>
7,.,5/
a⋅\j
\j
.
>
1
,.14-
a⋅\j
\j
,.045
a
\j
-4\j
.
04
-4,,ai_
-
//.
_d.
0
\j
?
-
-2.4
_d.
0
\j
-00_d.
.
-\j
.
5?
1
,.045
a
\j
-4\j
.
04
-4,,ai_
-2.4
_d.
0
\j
-00_d.
.
-\j
.

L1.309
i⋅dr
dr
-56-
Example: Wind Loads SD Compare to capacity (interaction diagram) Depth of stress
block, =
Design
moment, ?/
?
=L
?
?
?
?
>?
?
%⁄
4.<?
?
?
?
L
4.4;;9
_d.
.
\j
>4.8<=
a
\j
4.= W
4.<
6iqg
56
_d.
\j
L0.270in.
?/
?
L?
?
?
%
E#
?
B
?
@F
?
6
L0.9
4.8<=
a
\j
4.=
E 0.0775
gl.
.
dr
60ksi
3.812in. F
4.6;4gl.
6
L17.19
i⋅gl.
dr
L1.432
i⋅dr
dr
?/
?
?2
?
/
?
,2
?

15. -57-
Example: Wind Loads SD
OK
Load Combination/
?
(kip-ft/ft)
?/
?
(kip-ft/ft)
/
?
?/
?
2
nd
Order /
1
st
Order
1.2D+1.6L
r+0.5W 0.799 1.755 0.455 1.074
1.2D+1.0W+0.5L
r
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
2(k/ft) 0.5+0.6(-0.36) = 0.284 0.6(0.5)+0.6(-0.36) = 0.084
2
?
(k/ft)38(2.67+9) = 0.4430.6(0.443) = 0.266
2(k/ft)0.7270.350
?(in)0.3500.325
+
??
(in
4
/ft)17.4816.46
/
?
(k-ft/ft)-0.002-0.049
/(k-ft/ft)0.8050.767
?(in)0.4740.426
OK
Deflection Limit: ?
?
Q 0.007D L 0.007
18ft
12
gl.
dr
L1.51in.
-59-
Example: Wind Loads SD
Deflections, Sample Calculations: D + 0.6W
Replace factored loads with service loads
?L
1?
.
04?????
??
.
4
>
?
?
.
>?
??
???
??
?5
5?
1??
.
04?????
L
1:-4\j;
.
04-4,,ai_
-3.1
_d.
0
\j
,.2:,.,/.ai\;:-4\j;
.
4
>
7,.,,.
a⋅\j
\j
.
>4.9<5
a⋅\j
\j
-3.1
_d.
0
\j
//.
_d.
0
\j
?5
-00_d.
.
\j
.
5?
1:,.3.3
a
\j
;:-4\j;
.
04
-4,,ai_
-3.1
_d.
0
\j
-00_d.
.
\j
.
L 0.0393ft L 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+0.525Q
E
(9.3.3.2.1 (d))
•2
?
is just dead load = 0.5+0.038(2.67+9) = 0.943k/ft
OK
9.3.3.2
Commentary
Equations
?
???
L
4.:8?
?
?
??
??6?
?
??
??
?
?
L
4.:8:6iqg;
,.,,.1
,.,,.16-.1:,.,,.,3;
?
,.50/
a
\j
-.
_d.
\j
:/.4-_d.;
:4iqg
L 0.00917
Reinforcement
Ratio
?L
?
?
??
L
4.4;;9
_d.
.
\j
56
_d.
\j
:7.<5gl.;
L0.00169

16. -61-
Bearing Walls: SD Summary
• Factored Loads /
?
?2
?
• Axial load: include wall weight
• Moment: include second-order :2 F ?;effects
• Procedures
• Slender wall
• Moment magnification
• Second-order analysis
• Cross-section properties: +
?
?+
??
?#
?
• Material properties: B
?
?/
??
?'
?
?J
• Capacity ?/
?
,?2
?
• 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)
8
?
L
8
??
E8
??
?
?
?L0.8
8
??
L
4.0 F 1.75
/
?
8
?
@
?
#
??
B
?
?
E 0.252
?
8
??
L0.5
?
?
?
B
?
@
?
8
?
Q
6#
??
B
?
?
?
?

?
?
?
?
?
?
Q 0.25
8
?
Q
4#
??
B
?
?
?
?

?
?
?
?
?
?
R1.0
Interpolate for 0.25 O?
?
?
?
?
?
O1.0
8
?
Q
8
7
5F2
?
?
?
?
?
?
#
??
B
?
?
?
?
Allowable Stress (Chapter 8)
(
?
L
(
??
E(
??
?
?
(
??
L
5
6
4.0 F 1.75
?
??
?
B
?
?
E0.25
?
?
?
(
??
L
5 6
4.0 F 1.75
?
??
?
B
?
?
E0.25
?
?
?
Special Reinforced Shear Walls
(
??
L0.5
?
?
?
?
?
?
?
??
?
(
?
Q
3
B
?
?
?
?

/8@
?
⁄ Q 0.25
(
?
Q
2
B
?
?
?
?

/8@ ⁄
?
R 1.0
Interpolate for 0.25 O
/8@ ⁄
?
O1.0
(
?
Q
6
7
5F2
?
??
?
B
?
?
?
?
-64-
Shear Walls: Maximum Reinforcement
Strength Design (Chapter 9)
Allowable Stress (Chapter 8)
Special reinforced shear walls having
•/8@
?
⁄R1and
•2P0.05B
?
?
#
?
• Provide boundary elements
• Boundary elements not
required in certain cases
•OR
• Limit reinforcement
• Based on strain condition
?
???
L
JB
?
?
2B
?
JE
B
?
B
?
I

17. -65-
SD Maximum Reinforcement
Design with Boundary Elements?
No
Is
?
?
?
?
?
?
Q1OR
8
?
Q3#
??
B
?
?
AND
?
?
?
?
?
?Q3
AND
2
?
Q 0.10#
?
B
?
?
:Geometrically symmetrical walls
2
?
Q 0.05#
?
B
?
?
: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 + 0.525Q
E
.
Compression reinforcement, with or
without lateral restraining reinforcement,
can be included.
Is /
?
8
?
@
?
⁄R1?
?= 1.5
Ordinary reinforced walls: ?= 1.5
Intermediate reinforced walls: ?= 3
Special reinforced walls: ?= 4
Yes
No
Yes
YesNo
-66-
Shear Walls: Shear Capacity Design Special 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, ?8
?
,
greater than shear
corresponding to s?tw/
?
(increases shear at least
1.39 times)
• Except 8
?
need not be
greater than t?w8
?
. (doubles
shear)
-67-
ASD: Distributed Reinforcement
Calculate
Is G R G
>=H
?
YES
Solve for #
?,????
∗
NO
Compression controls
GL
/E2
@
?
6
1
3
@
?
6
(
?
P
??
F2
@
?
3
#
?,????
∗
L
1
2
G@
?
(
?
P
??
F2
1
2
1FG
6
G
@
?
J(
?
Determine G@from the quadratic equation
1
3
@
?
6
(
?
P
?? J
E2
@
?
3
G
6
E
/F2
@
?
6
G
F
/E2
@
?
6
L0
#
?,????
∗
L
1
2
G@
?
P
??
(
?
G
1FG
1
J
F2
1
2
1FG (
?
@
?
Tension controls
#
?
∗
= distributed reinforcement (in.
2
/ft)
-68-
SD: Distributed Reinforcement
Assume tension controls
=L@F
@
6
F
2>2
?
:@ F @
?
2; E /
?
? ⁄
?
r?zB
?
?
P
??
#
?,????
L
r?zB
?
?
P
??
=F2
?
?⁄
B
?
#
?,????
∗
L
#
?,????
r?xw@
?
Approximate required
distributed reinforcement
Approximate @L0.9@
?

18. -69-
Example: Shear Wall ASD
Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type S mortar; B′
I
= 2000psi; superimposed dead load of 1 kip/ft. In-plane
seismic load of 50 kips. 5
&5
L 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. •/L0.7
50k
10ft L 350k ⋅ ft L 4200k ⋅ ft
• Axial load, 2
• Need to know weight of wall to determine 2.
• Need to know reinforcement spacing to determine wall weight
• Estimate wall weight as 45 psf
• Wall weight: 45psf
10ft
16ft L 7.2k
•&L1k ft⁄
16ft E 7.2k L 23.2k
•2L
0.6 F 0.7
0.2 5
??
& L 0.53& L 0.53
23.2k L 12.3k
-70-
Example: Shear Wall ASD
GL
/E2
@
?
6
1
3
@
?
6
(
?
P
??
F2
@
?
3L
4200k ⋅ in. E12.3k
192in.
6
1 3
192in.
6
:0.90ksi;: 7.625in. ; F 12.3k
192in.
3
L 0.0550
Calculate G; for preliminary design purposes use full thickness of wall
Since GOG
???
tension controls. Solve quadratic equation.
1
3
@
?
6
(
?
P
?? J
E2
@
?
3
G
6
E
/F2
@
?
6
GF
/E2
@
?
6
L0
1
3
192in.
6
:32ksi;
7.625in.
16.11
E12.3k
192in.
3
G
6
E
4200k ⋅ in. F12.3k
192in.
6
GF
4200k ⋅ in. E12.3k
192in.
6
L0
GL0.147
-71-
Example: Shear Wall ASD
Calculate required area of reinforcement
#
?,????
∗
L
1
2
G@
?
P
??
(
?
G
1FG
1
J
F2
1
2
1FG (
?
@
?
L
1
2
:0.147;:192in. ;:7.625in. ;:32ksi;
0.147
1F0.147
1
16.11
F12.3k
1
2
1 F 0.147 :32ksi;:192in. ;
L 0.00934
in.
6
in.
L0.112
in.
6
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-
Example: Shear Wall ASD
Stressed to 93% of
allowable
Used equivalent thickness = 3.57 in.
(40 in. grout spacing)

19. -73-
Example: Shear Wall ASD
Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, B′
I
=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
#
??
L2
1.25in.
192in. E 6
8in.
7.625in. F2.5in. L 726in.
6
-74-
Example: Shear Wall ASD Shear Stress:
OK
/
8@
?
L
8D
8@
?
L
D
@
L
120in.
192in.
L0.625
Shear Span:
B
?
L
8
#
??
L
0.7:50k;
726in.
6
L48.2psi
Max Shear:
(
?,???
L
2 3
5F2
/
8@
?
B
?
?
?
?
L
2
3
5F2
0.625
2000psi 0.75 L 83.8psi
Masonry Shear:
(
?
L:(
??
;?
?
L
1 2
4F1.75
/
8@
?
B
?
?
E0.25
2
#
?
?
?
L
1 2
4F1.75
0.625
2000psi E 0.25
12300lb
726in.
6
0.75
L51.9psi
OK
-75-
Example: Shear Walls SD
Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type S mortar; B′
I
= 2000 psi; superimposed dead load of 1 kip/ft. In-plane
seismic load of 50 kips. 5
&5
L 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. •/
?
L
50k
10ft L 500k ⋅ ft
• Axial load, 2
?
• Need to know weight of wall to determine 2.
• Need to know reinforcement spacing to determine wall weight
• Estimate wall weight as 45 psf
• Wall weight: 45psf
10ft
16ft L 7.2k
•&L1k ft⁄
16ft E 7.2k L 23.2k
•2
?
L
0.9 F 0.25
??
& L 0.80& L 0.80
23.2k L 18.6k
-76-
Example: Shear Walls SD
#
?,????
L
0.8B
?
?
P
??
=F2
?
?⁄
B
?
L
0.8:2ksi;:7.625in. ;:3.96in. ; F 18.6k 0.9 ⁄
60ksi
L0.460in.
6
=, depth
of stress
block
#
?,????
,
area of
steel
Estimate d
@ L 0.9@
?
L0.9
192in. L 173in.
#
?,????
∗
,
dist. steel
#
?,????
∗
L
#
?,????
0.65@
?L
0.460in.
6
0.65
192in.
12in.
ft
L0.044in.
6
ft⁄
Try #4 @ 48 in. (0.050in.
2
/ft)
=L@F
@
6
F
2>2
?
:@ F @
?
2; E /
?
⁄?
?
0.8B
?
?
P
??
L173in.F
173in.
6
F
2>18.6k:173in. F 192in. 2; E 6000k ⋅ in. ⁄?
0.9
0.8
2000psi
7.625in.
L 3.96in.

20. -77-
Example: Shear Walls SD
-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-
Example: Shear Walls SD
Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, B′
I
=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
#
??
L2
1.25in.
192in. E 5
8in.
7.625in. F2.5in. L 685in.
6
-80-
Example: Shear Walls SD
OK
/
?
8
?
@
?
L
8
?
D
8
?
@
?
L
D
@
L
120in.
192in.
L0.625
Shear Span:
Max Shear:
?8
?,???
L?
4
3
5F2
/
?
8
?
@
?
#
??
B
?
?
?
?
L0.8
4
3
5F2
0.625
685in.
6
2000psi 0.75 L 91.9kip
Masonry
Shear:
?8
??
L?
4F1.75
/
8@
?
#
??
B
?
?
E0.252
?
?
?
L0.8
4F1.75
0.625
685in.
6
2000psi E 0.25
18600lb 0.75
L56.2kip
OK

21. -81-
Example: Shear Walls SD
• Calculate axial force based on ?= 83.8 in.
• Include compression reinforcement
•?2
?
= 323 kips
• Assume a live load of 1 k/ft
•D + 0.75L + 0.525Q
E
= (1k/ft + 0.75(1k/ft))16ft = 28 kips
Section 9.3.3.2 Maximum Reinforcement
Since /
?
/:8
?
@
?
; O 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-
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.
-83-
Example: Shear Wall ASD
Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type S mortar; B′
I
= 2000 psi; superimposed dead load of 1 kip/ft. In-plane
seismic load of 50 kips. 5
&5
L0.5
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)
•B
?
L 52.5k 726in.
6
⁄L72.3psi
• Maximum (
?
L83.8psiOK
-84-
Example: Shear Wall ASD
Use #5 @ 16 in.
Use #5 bars in
bond beams.
Determine
spacing.
Required steel
stress
Masonry Shear:
(
??
L
1
4
4F1.75
/
8@
?
B
?
?
E0.25
2
#
?
L
1
4
4F1.75
0.625
2000psi E 0.25
12300lb
726in.
6
L36.7psi
(
??,????
L
B
?
?
?
F(
??
L
72.3psi
0.75
F 36.7psi L 59.7psi
(
??
L0.5
#
?
(
?
@
?
#
??
O
⇒ O L
0.5#
?
(
?
@
?
(
??,????
#
??
OL
0.5
0.31in.
6
32000psi
192in.
59.7psi
726in.
6
L21.9in.

22. -85-
Example: Shear Wall ASD
Masonry Shear:
(
??
L
1
4
4F1.75
/
8@
?
B
?
?
E0.25
2
#
?
L
1
4
4F1.75
0.625
2000psi E 0.25
15300lb
1464in.
6
L35.1psi
Due to closely spaced bond beams, fully grout wall.
Shear Stress:
B
?
L
8
#
??
L
52.5k
1464in.
6
L35.9psi
#
??
L7.625in.
192in. L 1464in.
6
Shear Area:
81psf
10ft
16ft L 13.0k
Wall weight:
Dead load:
&L1k ft⁄
16ft E 13.0k L 29.0k
Axial load:
2 L 0.53& L 0.53
29.0k L 15.3k
-86-
Example: Shear Wall ASD
Use #4 bars in
bond beams.
Determine
spacing.
Required steel
stress
(
??,????
L
B
?
?
?
F(
??
L
35.9psi
1.0
F35.1psiL0.8psi
(
??
L0.5
#
?
(
?
@
?
#
??
O
⇒ O L
0.5#
?
(
?
@
?
(
??,????
#
??
OL
0.5
0.20in.
6
32000psi
192in.
0.8psi
1464in.
6
L524in.
Spacing determined by prescriptive requirements Maximum Spacing Requirements (7.3.2.6)
• minimum{ one-third length, one-third height, 48 in. }
O
???
Lmin
192EJ.
3
,
120EJ.
3
,48EJ. Lmin
64EJ. , 40EJ. , 48EJ. L 40EJ.
-87-
Example: Shear Wall ASD
Use #5 @ 40 in.
Prescriptive Reinforcement Requirements (7.3.2.6)
•?R0.0007in each direction
•?
?
E?
?
R0.002
Vertical: ρ
?
L 6:0.31EJ.
2
; 1464EJ.
2
⁄L 0.00127OK
Horizontal: ρ
?,????
Lmax
0.002 F 0.00127, 0.0007 L 0.00073
Determine bar size for 40 in. spacing
#
?,????
L?
?
P
??
OL0.00073
7.625EJ.
40EJ. L 0.22EJ.
6
An alternate is #4 @ 32 in. :?
?
L0.00082;
-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:
•//:8@
R
; R 1and //:8@
R
; L 0.625
•2 P 0.05B′
I
#
J
• 0.05(2000psi)(1464in
2
) = 146 kips
• Assume a live load of 1 k/ft
•2L
1.0 E 0.7
0.2 5
??
E.L
1.0 E 0.7
0.2 0.5 29k E 16k L 47.0k
OK

23. -89-
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.
?
???
L
JB
?
?
2B
?
JE
B
?
B
?
I
L
16.1:2ksi;
2:60ksi;
16.1 E
60ksi
2ksi
L 0.00582
?L
#
?
>@
L
5
0.31in.
6
7.625in. :188in. ;
L0.00108
-90-
Example: Shear Walls SD
Given: 10 ft high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, Type S mortar; B′
I
= 2000 psi; superimposed dead load of 1 kip/ft. In-plane
seismic load of 50 kips. 5
&5
L0.5
Required: Design the shear wall; special reinforced shear wall
Solution: • Flexural reinforcement remains the same
• Shear capacity design: Design shear strength, ?8
?
, greater than shear
corresponding to 1.25/
?
-91-
Example: Shear Walls SD
• For #4 @ 48 in., ?2
?
L2
?
=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
•2
?
= [1.2 + 0.2(5
??
)]&= 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 8
?
= 2.58
?
, or ?8
?
= ?2.58
?
= 2.08
?
.
•Design for shear of 100 kips

24. -93-
Example: Shear Walls SD Shear Area:
(6 - #4 bars)
#
??
L2
1.25in.
192in. E 6
8in.
7.625in. F2.5in. L 726in.
6
Max Shear:
?8
?,???
L?
4
3
5F2
/
?
8
?
@
?
#
??
B
?
?
?
?
L0.8
4
3
5F2
0.625
726in.
6
2000psi 0.75 L 97.4kip
Options:
• Design for shear from 1.25/
?
= 97.2 kips
• Increase B
?
?
to 2100 psi, ?8
?,???
L99.8kips
• requires a unit strength of 2250 psi
• Grout more cells
-94-
Example: Shear Walls SD Masonry
Shear:
8
??
L
4F1.75
?
??
?
#
??
B
?
?
E0.252
?
L
4F1.75
0.625
726in.
6
2000psi E 0.25
18600lb
L99.0kip
Required
Steel
Strength:
?8
?
L?
8
??
E8
??
?
?
8
??,????
L
8
?
??
?F8
??
L
100k
0.8
0.75
F 99.0k L 67.7k
Determine
spacing:
Use #5 bars
8
??
L0.5
#
?
O
B
?
@
?
⇒ O L
0.5#
?
B
?
@
?
8
??,????
OL
0.5
0.31in.
6
60ksi
192in.
67.7k
L26.4in.
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 /
?
8
?
@
?
Q1 ⁄, ?= 1.5
• Include compression reinforcement
•?2
?
= 298 kips
• Assume a live load of 1 k/ft
•D + 0.75L + 0.525Q
E
= 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
2
?
Q0.10B
?
?
#
?
geometrically symmetrical sections
2
?
Q0.05B
?
?
#
?
geometrically unsymmetrical sections
AND
OR
AND
For our wall, /
?
8
?
@
?
O1 ⁄
2
?
Q0.10B
?
?
#
?
L0.10
2ksi
1464in.
6
L293k
/
?
8
?
@
?
Q1
/
?
8
?
@
? Q3 8
?
Q3#
?
B
?
?

25. -97-
Example: Shear Walls SD
• Prescriptive Reinforcement Requirements
• 0.0007 in each direction
• 0.002 total
• Vertical: 6(0.20in
2
)/1464in
2
= 0.00082
• Horizontal: 5(0.31in
2
)/[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 f
y
. 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