asd vs. sd a masonry cage fight - seaogseaog.org/asdvssd.pdf · asd vs. sd a masonry cage fight ......
TRANSCRIPT
- 1 -
ASD vs. SDA Masonry Cage Fight
Richard Bennett, PhD, PE
The University of Tennessee
Chair, 2016 TMS 402/602 Code Committee
- 2 -
Outline
Beams and lintels
Non-bearing walls: out-of-plane
Interaction diagrams
Combined bending and axial force: pilasters
Bearing walls: out-of-plane
Shear walls: in-plane
• Partially Grouted Shear Wall
• Special Reinforced Shear Wall
- 3 -
2011 vs 2013 Code
Chapter 1 – General Design Requirements for Masonry
Chapter 2 – Allowable Stress Design of Masonry
Chapter 3 – Strength Design of Masonry
Chapter 4 – Prestressed Masonry
Chapter 5 – Empirical Design of Masonry
Chapter 6 – Veneer
Chapter 7 – Glass Unit Masonry
Chapter 8 – Strength Design of Autoclaved Aerated Concrete Masonry
Appendix B – Design of Masonry Infill
Part 1 – General• Chapter 1 – General Requirements• Chapter 2 – Notation and Definition• Chapter 3 – Quality and Construction
Part 2 – Design Requirements• Chapter 4 – General Analysis and Design Considerations• Chapter 5 – Structural Elements• Chapter 6 – Reinforcement, Metal Accessories, and
Anchor Bolts• Chapter 7 – Seismic Design Requirements
Part 3 – Engineered Design Methods• Chapter 8 – Allowable Stress Design• Chapter 9 – Strength Design of Masonry• Chapter 10 – Prestressed Masonry• Chapter 11 – Strength Design of AAC
Part 4 – Prescriptive Design Methods• Chapter 12 – Veneer• Chapter 13 – Glass Unit Masonry• Chapter 14 – Masonry Partition Walls
Part 5 – Appendices• Appendix A – Empirical Design • Appendix B – Design of Masonry Infill• Appendix C – Limit Design Method
- 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 -
Beam and Lintel Design
Strength Design (Chapter 9)
• mu = 0.0035 clay masonry• mu = 0.0025 concrete masonry• Masonry stress = 0.8f’m• Masonry stress acts over a = 0.8c• = 0.9 flexure; = 0.8 shear
mnvn
m
ysysn
fAV
bf
fAdfAM
25.2
8.02
1
• Minimum reinf: Mn ≥ 1.3Mcr
• or As ≥ (4/3)As,req’d
• Maximum reinf: s ≥ 1.5y
sm
m
y
m
f
f
8.08.0
max
Allowable Stress (Chapter 8)
• Allowable masonry stress: 0.45f’m• Allowable steel stress
• 32 ksi, Grade 60 steel
• No min or max reinforcement requirements
• Allowable shear stress
nnnk 2)( 2
31
kj
jdA
Mf
ss ))((
2
jdkdb
Mfm
mvm fF 25.22
1
- 6 -
Beam and Lintel Design
Strength Design (Chapter 9)
1. Determine a, depth of compressive stress block
2. Solve for As
3. ρ 0.285 ⁄ for CMU
Grade 60 steel:
= 1.5 ksi, ρ 0.00714
= 2 ksi, ρ 0.00952
Allowable Stress (Chapter 8)
1. Assume value of j (or k). Typically 0.85 < j < 0.95.
2. Determine a trial value of As./
Choose reinforcement.
3. Determine k and j; steel stress and masonry stress.
4. Compare calculated stresses to allowable stresses.
5. If masonry stress controls design, consider other options (such as change of member size, or change of f’m). Reinforcement is not being used efficiently.
bf
Mdda
m
n
8.0
22
y
ms f
bafA
8.0
- 7 -
ASD: Alternate Design Method
Calculate
Is k ≥ kbal?For Grade 60 steelkbal = 0.312 n
FF
Fk
sm
mbal
bF
Mddkd
m3
2
223
2
YES
11
2
)(
knF
PbkdF
A
m
m
s
31
kdF
MA
s
s
bF
nFA
s
ss
dkd 222
Iterate. Use (kd)2 as new guess and repeat.
NO
Allowable masonry compression stress
controls
Allowable reinforcement tension stress controls
- 8 -
Comparison of ASD and SD
8 inch CMUd = 20 in.
- 9 -
Example: Beam
Given: 10 ft. opening; dead load of 1.5 kip/ft; live load of 1.5 kip/ft; 24 in. high; Grade 60 steel; Type S masonry cement mortar; 8 in. CMU; f’m = 2000 psiRequired: Design beamSolution:
5.2.1.3: Length of bearing of beams shall be a minimum of 4 in.; typically assumed to be 8 in.
5.2.1.1.1 Span length of members not built integrally with supports shall be taken as the clear span plus depth of member, but need not exceed distance between center of supports.
• Span = 10 ft + 2(4 in.) = 10.67 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• 120b2/d. 120(7.625 in.)2 / (20 in.) = 349 in. = 29.1 ft
- 10 -
Allowable Stress Design: Flexure
ftk
ftwLM ft
k
1.458
67.1017.3
8
22
ftk
ftk
ftk
ftk ftLDw 17.35.12083.05.1 2 Load
Weight of fully grouted normal weight: 83 psf
Moment
Determine kd Assume compression controls
.32.9
625.7900.03
1.452
2
20
2
203
3
2
223
1222
ininksi
ftkinin
bF
Mddkd ft
in
b
312.0466.020
32.9
in
in
d
kdkCheck if compression controls Compression
controls
1.160.2900
29000
900
ksi
ksi
f
E
E
En
m
s
m
sCalculate modular ratio, n
- 11 -
Allowable Stress Design: Flexure
Find AsArea of steel
294.1
1466.01
900.01.16
2625.731.9900.0
112
)(
inksi
ininksi
knF
bkdF
A
b
b
s
Bars placed in bottom U-shaped unit, or knockout bond beam unit.
Use 2 - #9 (As = 2.00 in2)
- 12 -
Strength Design: Flexure
ftk
ftLwM ft
ku
u 6.628
67.1040.4
8
22
Use 2 - #6 (As = 0.88 in2)
ftk
ftk
ftk
ftk
u ftLDw 40.45.16.12083.05.12.16.12.1 2 Factored LoadWeight of fully grouted normal weight: 83 psf
Factored Moment
Find aDepth of equivalent rectangular stress block
Find AsArea of steel
ininksi
ftinftk
ininbf
Mdda
m
n 78.3625.70.28.0
129.0
6.622
20208.0
2 22
277.060
78.3625.70.28.08.0in
ksi
ininksi
f
bafA
y
ms
Mn = 78.5 k-ft
Mn = 70.6 k-ft
- 13 -
Check Min and Max Reinforcement
ftkinkpsiinfSM rncr 76.91.117160732 3
322
7326
24625.7
6in
ininbhSn
ftkMftkftkM ncr 3.757.1276.93.13.1
00577.020625.7
88.0 2
inin
in
bd
As
Minimum Reinforcement Check: fr = 160 psi (parallel to bed joints in running bond; fully grouted)
Maximum Reinforcement Check:
= 2 ksi ρ 0.00952
Section modulus
Cracking moment
Check 1.3Mcr
- 14 -
Summary: Beams, Flexure
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)
ASD: Allowable tension controls for 0.5 k/ft and 1 k/ft.
- 15 -
Allowable Stress Design: Shear
Section 8.3.5.4 allows design for shear at d/2 from face of supports.
Design for DL = 1 k/ft, LL = 1 k/ft
kft
ftftkV 02.9
33.5
17.133.556.11
k
ftftwLV
ftk
ftk
ftk
56.112
67.100.12083.00.1
2
2
ftinin
inft 17.1.4
2
.20.12
1
Shear at reaction
d/2 from face of support
Design shear force
psipsi
fA
Pf
Vd
MF m
nmvm
3.50200025.22
1
25.22
125.075.10.4
2
1
psiinin
k
A
Vf
nvv 2.59
.20.625.7
02.9Shear stress
Allowable masonry shear stress
Suggest that d be used, not dv.
- 16 -
Allowable Stress Design: Shear
2034.0.20320005.0
.8.20.625.79.8
5.05.0
ininpsi
inininpsiA
dF
sAfA
sA
dFAF
v
s
nvsv
n
svvs
psipsifF mv 4.89200022
psipsipsi 9.83.502.59
Check max shear stress
Req’d steel stress
Determine Av for a spacing of 8 in.
> 59.2psi OK
Use #3 stirrups
Determine d so that no shear reinforcement would be required.
.5.233.50.625.7
02.9in
psiin
k
bF
Vd
vm
Use a 32 in. deep beam if possible;will slightly increase dead load.
- 17 -
Strength Design: Shear
Requirement for shear at d/2 from face of support is in ASD, not SD, but assume it applies
Design for DL = 1 k/ft, LL = 1 k/ft
kft
ftftkVu 50.12
33.5
17.133.500.16
k
ftftLwV
ftk
ftk
ftk
uu 00.16
2
67.100.16.12083.00.12.1
2
2
ftinin
inft 17.1.4
2
.20.12
1
Shear at reaction
d/2 from face of support
Design shear force
kpsiinin
fAPfAdV
MV mnvumnv
vu
unm
28.122000.20.625.725.28.0
25.225.075.10.4
Design masonry
shear strength
Suggest that dbe used, not dv to find Anv
- 18 -
Strength Design: Shear
2004.0.20605.0
.828.0
5.05.0
ininksi
inkA
df
sVAdf
s
AV
v
vy
nsvvy
vns
kpsiininfAV mnvn 82.212000.20.625.748.04
kkkVV mu 28.0
8.0
28.1250.12
Check max Vn
Req’d Vns
Determine Av for a spacing of 8 in.
> 12.50k OK
Use #3 stirrups
Determine d so that no shear reinforcement would be required.
.4.20200025.28.0.625.7
50.12
25.2in
psiin
k
fb
Vd
m
u
Use inverted bond beam unit to get slightly greater d.
- 19 -
Non-Bearing Partially Grouted WallsOut-of-Plane
s
b’
b
d
a
t f
As
b = width of effective flange = min{s, 6t, 72 in}
A. Neutral axis in flange:a. Almost always the caseb. Design and analysis for solid
sectionB. Neutral axis in web
a. Design as a T-beam sectionC. Often design based on a 1 ft width
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 0.017 0.030 0.046 0.066
- 20 -
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:
ftftlb
ftinlbft
inftlb ftwh
M 57669128
1612306.0
8
22 2
Fb = 0.45(2000psi) = 900 psi Em = 1.80 x 106 psiFs = 32000 psi Es = 29 x 106 psin = Es/Em = 16.1
Moment
Determine kd Assume compression controls
.346.0
12900.03
576.02
2
81.3
2
81.33
3
2
223
1222
inksi
inin
bF
Mddkd
ftin
ftin
ftftk
b
312.0091.081.3
346.0
in
in
d
kdkCheck if compression
controlsTension controls
- 21 -
Example: Partially Grouted Wall ASD
31
kdF
MA
s
s
bF
nFA
s
ss
dkd 222
Equation / Value Iteration 1 Iteration 2 Iteration 3
kd (in.) 0.346 0.699 0.709
k 0.091 0.183 0.183
(in2) 0.0584 0.0603 0.0603
(in.) 0.0784 0.0810 0.0810
(in.) 0.699 0.709 0.709
Use # 4 @ 40 inches (As=0.060in2/ft)
(close enough for government work)
- 22 -
Example: Partially Grouted Wall SD
Given: 8 in. CMU wall; 16 ft high; Grade 60 steel, = 2000 psi; Wind load of wu = 30 psfRequired: Reinforcing (place in center of wall)Solution:
ft
ftkftinlbft
inftlb
uu
fthwM 96.011520
8
161230
8
22 2
Use # 4 @ 40 inches (As=0.060in2/ft)
inpsi
ininbf
Mdda
ftin
ftinlb
m
n 179.0)12(20008.0
9.0
115202
81.381.38.0
2 22
ft
inftin
y
ms psi
inpsi
f
bafA
2
0573.060000
179.01220008.08.0
Factored Moment
Find aSolve as solid section
Find As
- 23 -
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
- 24 -
Interaction Diagrams
Strength Design (Chapter 9)
• Set masonry strain to εmu
• Vary steel strain• Equivalent rectangular stress block• Nominal axial strength
• h/r ≤ 99
• h/r > 99
• =0.9
Allowable Stress (Chapter 8)
• For k > kbal
• Set masonry strain = Fb/Em; = 0.0005 CMU
• Find steel strain• For k < kbal
• Set steel strain = Fs/Es;= 0.00110 for Grade 60
• Find masonry strain• Allowable axial load
• h/r ≤ 99
• h/r > 99
140
180.08.02
r
hAfAAfP stystnmn
2
h
70r80.080.0
stystnmn AfAAfP
2
7065.025.0
h
rFAAfP sstnma
2
140165.025.0
r
hFAAfP sstnma
- 25 -
Example: 8 in. CMU Bearing Wall ASD
Given: 12 ft high CMU bearing wall, Type S masonry cement mortar; Grade 60 steel in center of wall; #4 @ 48 in.; partial grout; = 2000 psiRequired: Interaction diagram in terms of capacity per foot
Pure Moment: n = 16.1 ρ = 0.00109 nρ = 0.0176
ftftk
ftk-inM 479.0.755
ftink
ftin
sss inksijdFAM 75.5.81.3943.03205.02
ftink
ftinm
m inksi
injdF
kdbM
64.12
81.3943.02
900.081.3171.012
2
171.00176.00176.020176.02)( 22 nnnk
943.03
171.01
31
kj
Find k
Find j
Find Ms
Find Mm
Allowable M
- 26 -
Example: 8 in. CMU Bearing Wall ASD
Pure Axial:
ftkP 3.17
991.5466.2
144
in
in
r
h
NCMA TEK 14-1B Section Properties of Concrete Masonry Wallsr = 2.66 in. An = 40.7in2/ft In = 332.0 in4/ft
ftk
ftin
sstnma
ksi
r
hFAAfP
3.17140
1.54107.400.225.0
140
165.025.0
2
2
2
Find h/r
Find Pa
4.3.3 Radius of gyration Radius of gyration shall be computed using average net cross-sectional area of the member considered.
- 27 -
Example: 8 in. CMU Bearing Wall ASD
Find Cm
Find T
Find P
Find M
Balanced:
ftftk
ftk
M
P
83.1
.844
ftk
ftk
m-TCP 84.460.144.6
ftk
ftin
mm inksibkdfC 44.61219.1900.02
1
2
1
Stress
ftftk
ftink
ftk in
inM
83.10.22
3
19.181.344.6
0.900 ksi32 ksi
Strain0.0005
0.00110
3.82 in
1.19 inkd < 1.25 in.N.A. in face shell
ftk
ftin
ss ksiAfT 6.105.0322
- 28 -
Example: 8 in. CMU Bearing Wall ASD
Find Cm
Find T
Find P
Find M
BelowBalanced:kd = 1.00 in.
ftftk
ftk
M
P
22.1
.622
ftk
ftk
m-TCP 62.260.122.4
ftk
ftin
mm inksibkdfC 22.41200.1703.02
1
2
1
Stress
ftftk
ftink
ftk in
inM
22.17.14
3
00.181.322.4
1800ksi(0.000391) =0.703 ksi
32 ksi
ftk
ftin
ss ksiAfT 6.105.0322
Strain0.000391
0.00110
3.82 in
1.00 in
- 29 -
Example: 8 in. CMU Bearing Wall ASD
AboveBalanced:kd = 2.00 in.
Stress
0.900 ksi
13.1 ksiStrain0.0005
0.00045
3.82 in
2.00 in
Neutral axis is in web
ftk
mm -TCCP 88.8 21
ftftk
ftk
ftk ininininM 58.25.181.325.053.081.328.9
ftk
ftinksiT 66.005.01.13
2
2.00 in
9.28 k/ft
0.66k/ft
1.25+0.75/3=1.5in.
0.338 ksi
1.25 in
ftk
ftin
m inksi
C 28.91225.12
338.0900.01
0.25 k/ft
ftk
ftin
m inksiC 25.0275.0338.02
12
1.5 in
0.53 in
h2
h1
b
x
Centroid
3
2
21
21 b
hh
hhx
ftftk
ftk
M
P
58.2
.888
- 30 -
Example: 8 in. CMU Bearing Wall SD
Given: 12 ft high CMU bearing wall, Type S masonry cement mortar; Grade 60 steel in center of wall; #4 @ 48 in.; partial grout; = 2000 psiRequired: Interaction diagram in terms of capacity per foot
Pure Moment:
Find Mn
Find Mn
ft
ftkftink
ftin
ftin
ftin
m
ysysn
ksi
ksiksi
bf
fAdfAM
934.02.110.2128.0
6005.0
2
1.812in 36005.0
'8.02
1
2
2
ftftk
ftftk
nM 840.0934.00.9
in
ksi
ksi
bf
fAa
ftin
ftin
m
ys 156.00.2128.0
6005.0
'8.0
2
Check to make sure stress block is in face shell
- 31 -
Example: 8 in. CMU Bearing Wall SD
Pure Axial:
991.5466.2
144
in
in
r
h
NCMA TEK 14-1B Section Properties of Concrete Masonry Wallsr = 2.66 in. An = 40.7in2/ft In = 332.0 in4/ft
Find h/r
Find Pn
ftk
ftk
nP 9.393.440.9
ftk
ftin
stystnmn
ksi
r
hAfAAfP
3.44140
1.541007.400.280.08.0
140
180.08.0
2
2
2
- 32 -
Example: 8 in. CMU Bearing Wall SD
Find Cm
Find T
Balanced: Strain
Stress
0.0025
0.002073.82 in
2.09 in
T
Cm
a = 0.8c = 0.8(2.09in.) = 1.67in.
web length = .
.
. 2.0
ftk
ftk
nP 1.200.33.10.240.9
ftk
ftin
webm ininksiC 34.10.225.167.10.280.0,
ftftk
ftk
ftk
n inin
inM
96.5
2
25.167.125.181.33.1
2
25.181.30.249.0
ftk
ftin
shellfacem inksiC 241225.10.280.0,
ftk
ftin
sy ksiAfT 0.305.0602
Find
Find
ftftk
n
ftk
n
M
P
96.5
1.20
0.8f′m = 1.6 ksi
- 33 -
Example: 8 in. CMU Bearing Wall SD
Find Cm
Find T
Find
Find
BelowBalanced:c = 1.25 in.
ftk
ftin
mm inksibafC 2.191200.10.28.08.0
ftk
ftin
sy ksiAfT 0.305.0602
Strain
Stress
0.0025
0.005123.81 in
1.25 in
T
ftk
ftk
mn -TCP 6.140.32.199.0 ft
ftkn
ftk
n
M
P
77.4
6.14
ft
ftkftink
ftk
n
ininM
77.42.57
2
25.18.081.32.199.0
0.8f′m = 1.6 ksi
a = 0.8c1.00 in
- 34 -
Example: 8 in. CMU Bearing Wall SD
AboveBalanced:c = 3.0 in.
Strain0.0025
0.000683.81 in
3.0 inStress
T
Cm
a = 0.8c = 0.8(3.0in.) = 2.4in.
web length = .
.
. 2.0
0.8f′m = 1.6 ksi
Find Cm
Find T
ftk
ftk
nP 0.2499.068.30.240.9
ftk
ftin
webm ininksiC 68.30.225.140.20.280.0,
ftftk
ftk
ftk
n inin
inM
28.6
2
25.140.225.181.368.3
2
25.181.30.249.0
ftk
ftin
shellfacem inksiC 241225.10.280.0,
ftk
ftin
sss ksiAET 99.005.000068.0290002
Find
Find
ftftk
n
ftk
n
M
P
28.6
0.24
- 35 -
Interaction Diagrams
- 36 -
ASD vs SD: Interaction Diagrams
Similar behavior to flexure
• ASD and SD close when allowable tension stress controls
• ASD more conservative when allowable masonry stress controls
SD interaction diagram easier to construct due to equivalent uniform stress vs. linear varying stress
Advantage to SD
- 37 -
Strength Design: Combined Bending and Axial Load Design Method
bf
MtdPdda
m
uu
8.0
2/22
y
ums f
PbafA
/8.0
Calculate
Is c ≥ cbal?For Grade 60 steelcbal = 0.547
YES
Compression controlsTension controls
dcymu
mubal
NO
8.0
ac
ccd
E
PbafA
smu
ums
/8.0
- 38 -
ASD: Combined Bending and Axial Load Design Method
Calculate
Is k ≥ kbal?For Grade 60 steelkbal = 0.312 n
FF
Fk
sm
mbal
bF
MtdPddkd
m3
))2/((2
223
2
YES
11
2
)(
knF
PbkdF
A
m
m
s
32
kdtPM
31
kdF
MMA
s
s
bF
nFAP
s
ss
dkd 222
Iterate. Use (kd)2 as new guess and repeat.
NO
Compression controls
Tension controls
- 39 -
ASD: Combined Bending and Axial Load Design Method If k < kb tension controls. Solve cubic equation for kd.
02226
23
dM
tdPkdM
tdPkd
n
bdFkd
n
bF ss
s
s F
P
kdd
kd
nbkdA
1
2
1
Solve for As.
- 40 -
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 (e=5.8 in.); Pilasters spaced at 16 ft on center; Wall is assumed to span horizontally between pilasters; No ties.Required: Determine required reinforcing using allowable stress design.Solution:
Ver
tical
Spa
nnin
g
d=11.8 in
xLoad
e=5.8 in 2.0 in
Em = 1800ksin = 16.1
Lateral Loadw = 0.6(26psf)(16ft)=250 lb/ft
Insi
de
- 41 -
Example: Pilaster Design ASD
Weight of pilaster:Weight of fully grouted 8 in wall (lightweight units) is 75 psf. Pilaster is like a double thick wall. Weight is 2(75psf)(16in)(1ft/12in) = 200 lb/ft
Usually the load combination with smallest axial load and largest lateral load controls. Try load combination of 0.6D + 0.6W to determine required reinforcement and then check other load combinations.
The location and value of maximum moment can be determined from:
2
22
max282 wh
MwhMM
wh
Mhx
2
If x<0 or x>h, Mmax= M
M = moment at topx measured down from top of pilaster
- 42 -
Example: Pilaster Design ASD
0.6D + 0.6W
Top of pilaster. P = 0.6(9.6)k-0.6(8.1)k = 0.9kips M = 0.9k(5.8in) = 5.2kip-in
inin
inkipin
wh
Mhx
inft
ftk
1.1431288250.0
2.5
2
288
2
inkin
inkinink
wh
MwhMM
ink
ink
3.2182880208.02
)2.5(
8
2880208.0
2
2.5
282 2
22
2
22
max
Find axial force at this point. Include weight of pilaster (200 lb/ft).
kinkPinft
ftk 3.21.14320.06.09.0
.121
Design for P = 2.3 k, M = 218 k-in
Find location of maximum moment
- 43 -
Example: Pilaster Design ASD
in
inksi
inkipininkipinin
bF
MhdPddkd
b
00.36.15900.03
)2182/6.158.113.2(2
2
8.11
2
8.113
3
))2/((2
223
2
2
254.08.11
00.3
in
in
d
kdk
312.0
1.1632
900.0
900.0
ksiksi
ksi
n
FF
Fk
sb
bbal
k < kbal Tension controls; iterate
Assume compression controls;Determine kd
Determine k
Determine kbal
- 44 -
Example: Pilaster Design ASD
Equation / Value Iteration 1 Iteration 2 Iteration 3
kd (in.) 3.00 3.38 3.40
k 0.254 0.286 0.288
(k-in.) 15.6 15.3 15.3
(in2) 0.585 0.593 0.593
(in.) 0.678 0.686 0.686
(in.) 3.38 3.40 3.40
32
kdhPM
31
kdF
MMA
s
s
bF
nFAP
s
ss
dkd 222
Try 2-#5, 4 total, one in each cell
- 45 -
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
- 46 -
Example: Pilaster Design ASD
f’m = 2000 psi
f’m = 1500 psi
- 47 -
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 (e=5.8 in.); Pilasters spaced at 16 ft on center; Wall is assumed to span horizontally between pilasters; No ties.Required: Determine required reinforcing using strength design.Solution:
Ver
tical
Spa
nnin
g
d=11.8 in
xLoad
e=5.8 in 2.0 in
Em = 1800ksin = 16.1
Lateral Loadw = 0.6(26psf)(16ft)=250 lb/ft
Insi
de
- 48 -
Example: Pilaster Design SD
0.9D + 1.0W
At top of pilaster: kkkPu 54.01.80.16.99.0
inkinkePM uu 1.3.8.554.0
inft
inkipin
wh
Mhx
ftk
7.14324416.0
1.3
2
288
2
inkin
inkinink
wh
MwhMM
ink
ink
u
0.3612880347.02
)1.3(
8
2880347.0
2
1.3
282 2
22
2
22
Find axial force at this point. Include weight of pilaster.
kinkPinft
ftk
u 69.27.14320.09.054.0121
Design for Pu = 2.7 kips, Mu = 361 k-in
Find location of maximum moment
- 49 -
Example: Pilaster Design SD
in
inksi
inkininkinin
bf
MhdPdda
m
uu
50.16.150.28.09.0
3612/6.158.117.228.118.11
8.0
2/2
2
2
257.060
9.0/7.250.162.150.28.0/8.0in
ksi
kininksi
f
PbafA
y
ums
Determine a
Required steel = 0.57 in2
Use 2-#5 each face, As = 0.62 in2
Total bars, 4-#5, one in each cell
Determine As
- 50 -
Example: Pilaster Design SD
- 51 -
Example: Pilaster Design
- 52 -
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
- 53 -
Design: Bearing Walls – OOP Loads
Strength Design (Chapter 9)Allowable Stress(Chapter 8)
• No second-order analysis required
• Use previous design procedure
• Second order analysis required• Assumes simple support conditions• Assumes uniform load• Assumes midheight moment is
approximately maximum moment• Valid only for following conditions:
• 0.05 No height limit
• 0.20 height limited by
30
• Need to check maximum reinforcement limits
- 54 -
Strength Design Procedure
uuu
ufu
u Pe
Phw
M 28
2
ufuwu PPP
Puf = Factored floor loadPuw = Factored wall load
Moment: Deflection:
crunm
uu MM
IE
hM
48
5 2
cru
crm
cru
nm
cru MM
IE
hMM
IE
hM
48
5
48
5 22
32
32 bc
cdd
t
f
PAnI sp
y
uscr
bf
PfAc
m
uys
'64.0
Deflection Limit hs 007.0 Calculated using allowable stress load combinations
- 55 -
Strength Design Procedure
48
51
12848
5
48
51
11
48
5
28
2
22
2
22
crm
u
n
crcr
uuf
u
crmu
crm
u
crnm
ucruuf
u
u
IE
hP
I
IM
eP
hw
IEh
IE
hP
IIE
hPMeP
hw
M
nm
u
uuf
u
nmu
nm
u
uuf
u
u
IE
hP
eP
hw
IE
h
IE
hP
eP
hw
M
48
51
2848
5
48
51
28
2
22
2
2
Solve simultaneous linear equations:
Mu > Mcr Mu < Mcr
- 56 -
Example: Wind Loads ASD16
ft
32 p
sfD = 500 lb/ftLr = 400 lb/ftW = -360 lb/ft
2.67
ft
Given: 8 in. CMU shear wall; Grade 60 steel; Type S masonry cement mortar; = 2000 psi; roof forces act on 3 in. wide bearing plate at edge of wall.
Required: Reinforcement
Solution:
Estimate reinforcement
~8
0.6 0.032 168
0.62
Try #4 @ 40 in.
k = 0.185, kd = 0.707 in.
Mm = 1.14 k-ft/ft; Ms = 0.57 k-ft/ft
Lightweight units: wall weight = 40 psf
Cross-sectionof top of wall
Determine eccentricitye = 7.625in/2 – 1.0 in.= 2.81 in.
- 57 -
Example: Wind Loads ASD
Load Comb. Pf (kip/ft) P (kip/ft) Mtop (k-ft/ft) M (k-ft/ft) As (in2)
0.6D+0.6W 0.084 0.340 -0.049 0.590 0.051
D+0.6W 0.284 0.711 -0.002 0.613 0.042
ft
ftkftftk
top ftksfMwhM
590.02
049.0
8
16032.06.0
28
122
Check 0.6D+0.6W
084.036.06.05.06.0 ftk
ftk
ftk
fP
/340.0867.2040.06.0084.0 ftkftftksfP ftk
049.0
2
67.2032.06.0.81.2084.0
2
121
ftftk
inft
ftk
top
ftksfinM
Use #4@ 40 in. (0.06 in2/ft)Although close to #4@48, a wider spacing also reduces wall weight (As = 0.052in2/ft)
Find force at top of wall
Find force at midheight
Find moment at top of wall
Find moment at midheight
- 58 -
Example: Wind Loads ASD
Sample Calculations: 0.6D+0.6W
1. kbal = 0.312; kdbal = 1.19in.2. Assume masonry controls.
Determine kd.Since 0.355 in. < 1.18 in.
tension controls. 3. Iterate to find As.
.355.0
12900.03
590.02
2
.81.3
2
.81.33
3
2
223
.122
2
inksi
inin
bF
Mddkd
ftin
ftin
ftftk
b
3/2/ kdtPM sp
3/1 kdF
MMA
ss
bF
nFAP
s
ss
dkd 222
Equation / Value Iteration 1 Iteration 2
kd (in.) 0.355 0.707
(k-ft/ft) 0.1047 0.1013
(in2/ft) 0.049 0.051
(in.) 0.0804
(in.) 0.707
- 59 -
Example: Wind Loads SD
16 ft
32 p
sf
D = 500 lb/ftLr = 400 lb/ftW = -360 lb/ft
2.67
ft
Given: 8 in. CMU wall; Grade 60 steel; Type S masonry cement mortar; = 2000 psi ; roof forces act on 3 in. wide bearing plate at edge of wall.
Required: Reinforcement
Solution:
Estimate reinforcement
~8
0.032 168
1.02
a = 0.19 in.
As = 0.061 in2/ft
Try #4 @ 40 in.
Cross-sectionof top of wall
Determine eccentricitye = 7.625in/2 – 1.0 in.= 2.81 in.
- 60 -
Example: Wind Loads SD
Summary of Strength Design Load Combination Axial Forces
(wall weight is 40 psf for 40 in. grout spacing)
Load CombinationPuf
(kip/ft)
Puw
(kip/ft)
Pu
(kip/ft)
0.9D+1.0W0.9(0.5)+1.0(-0.36) =
0.0900.9(0.040)(2.67+8) =
0.3840.474
1.2D+1.0W+0.5Lr1.2(0.5)+1.0(-0.36) +0.5(0.4) = 0.440
1.2(0.040)(2.67+8) = 0.512
0.952
Puf = Factored floor load; just eccentrically applied loadPuw = Factored wall load; includes wall and parapet weight, found at
midheight of wall between supports (8 ft from bottom)
- 61 -
Example: Wind Loads SD
Find 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
psicells
cellgroutedpsi
cells
cellsungroutedpsifr 71
5
1153
5
451
Find Mcr, cracking moment:
Commentary allows inclusion of axial load
Use minimum axial load (once wall has cracked, it has cracked)
ftftkip
ftinkip
cr
ftin
ftin
ftlb
nrnucr
M
in
psi
t
IfAPM
606.028.7
81.3
7.336718.42/474
2/
/42
Wall properties determined from NCMA TEK 14-1B Section Properties of Concrete Masonry Walls
- 62 -
Example: Wind Loads SD
Find Icr
in
ksi
ksi
bf
PfAc
ftin
ftk
ftin
m
uys 265.00.264.0
474.06006.0
'64.0 12
2
1.161800
29000
ksi
ksi
E
En
m
s
ftin
ftin
ftk
ftin
sp
y
uscr
ininin
ksi
bccd
d
t
f
PAnI
4
2
8.13
3
265.012265.0812.3
60
474.006.01.16
32
32
32
Find c
Find n
- 63 -
Example: Wind Loads SD
ftftk
ftin
ftk
ftin
ftin
ftk
ftftk
ftftk
crm
u
crnm
ucruuf
u
u
ftin
ksi
ft
ftin
ksi
ftftksf
IEhP
IIEhPMe
Phw
M
009.1
1144
8.13180048
16474.051
1144
8.13
1
337
1180048
16474.0606.05
2
093.0
816032.0
485
1
1148
528
2
22
2
222
2
22
4
44
Pufeu is the moment at the top support of the wall, Mu,top. It includes eccentric axial load and wind load from parapet.
ft
ftkftksfin
hwePM in
ftftkparapetparapetu
uuftopu
093.0
2
67.2032.081.2090.0
2
2
121
2,
,
Find Mu.
- 64 -
Example: Wind Loads SD
ftftkip
ftinkip
ftin
ftk
ysun
ininksi
adfAPM
147.176.13
2
215.0812.36006.09.0/474.09.0
2/
2
in
ksi
ksi
bf
PfAa
ftin
ftk
ftin
m
uys 215.0120.280.0
9.0/474.06006.0
80.0
/2
nftftk
ftftk
u MM 15.101.1 OK
Check other load combinations:
1.2D+1.0W+0.5Lr, Mu = 1.09 k-ft/ft and Mn = 1.29 k-ft/ftOK
Find a
Check area of steel:
Find
Compare
- 65 -
Example: Wind Loads SD
Check Deflections: Use ASD Load Combinations
Load Combination D+0.6W 0.6D+0.6W
Pf (k/ft) 0.5+0.6(-0.36) = 0.284 0.6(0.5)+0.6(-0.36) = 0.084
Pw (k/ft) 40(2.67+8) = 0.427 0.6(0.427) = 0.256
P (k/ft) 0.711 0.340
c (in) 0.280 0.256
Icr (in4/ft) 14.5 13.4
Mtop (k-ft/ft) -0.002 -0.049
δ (in) 0.066 0.044
M (k-ft/ft) 0.617 (cracked) 0.565 (uncracked)
Deflection Limit ininhs 34.1)192(007.0007.0 OK
- 66 -
Example: Wind Loads SD
Deflections, Sample Calculations (D+0.6W):Replace factored loads with service loads
.066.0
1144
5.14180048
16711.051
11728
17.336
5.14606.0
2
002.0
816032.06.0
5.14180048
165
485
1
12848
5
2
22
3
322
2
22
4
4
4
4
in
ftin
ksi
ft
ftinftksf
ksi
ft
IEPh
II
Me
Pwh
IEh
ftin
ftk
ftin
ftin
ftftkft
ftk
ftin
crm
n
crcrf
crm
Check that M > Mcr; M = 0.617 k-ft/ft > 0.606 k-ft/ft = Mcr
- 67 -
Example: Wind Loads SD
Check Maximum Reinforcement:• neutral axis is in face shell• Pu is just dead load = 0.5+0.04(2.67+8) = 0.927k/ft
00918.0
60
81.312
927.0
00207.05.10025.00025.0
0.264.0
64.0
max
ksi
inksi
f
bdP
f
bd
A
ftin
ftk
y
u
ymu
mum
s
00131.081.312
06.02
inbd
A
ftin
ftft
s OK
- 68 -
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
- 69 -
Shear Walls: Shear
Strength Design (Chapter 9)Allowable Stress (Chapter 8)
umnvvu
unm PfA
dV
MV 25.075.10.4
gmnvn fAV 6 25.0/ vuu dVM
vyv
ns dfs
AV
5.0
gnsnmn VVV 8.0
gmnvn fAV 4 0.1/ vuu dVM
gmnvvu
un fA
dV
MV
25
3
40.125.0
vu
u
dV
M
gvsvmv FFF
nm
vvm A
Pf
Vd
MF 25.075.10.4
2
1
sA
dFAF
n
svvs 5.0
gmv fF 3 25.0/ vVdM
gmv fF 2 0.1/ vVdM
nm
vvm A
Pf
Vd
MF 25.075.10.4
4
1
Special Reinforced Shear Walls
Interpolate for 0.25 < (M/Vdv) < 1
= 0.75 partially grouted shear walls; 1.0 otherwise
- 70 -
Shear Walls: Maximum Reinforcement
Strength Design (Chapter 9)Allowable Stress (Chapter 8)
Special reinforced shear walls having• M/(Vd) ≥ 1 and• P > 0.05f′mAn
m
yy
m
f
fnf
fn
2max
• Provide boundary elements, or• Limit reinforcement
Boundary elements not required if:
gA1.0 mu fP symmetrical sections
gA05.0 mu fP unsymmetrical sections
AND
1wu
u
lV
M OR 3wu
u
lV
Mmnu fAV 3
Reinforcement limits: • Maximum stress in steel of αfy• Axial forces D+0.75L+0.525QE
• Compression reinforcement, with or without lateral ties, permitted to be included
= 1.5 ordinary walls; all others = 3 intermediate walls with Mu/(Vudv)≥1 = 4 special walls with Mu/(Vudv)≥1
- 71 -
Shear Walls: Shear Capacity DesignSpecial Reinforced Shear Walls
Strength Design (7.3.2.6.1.2)Allowable Stress (7.3.2.6.1.2)
• Seismic design load required to be increased by 1.5 for shear
• Masonry shear stress reduced for special walls
• Design shear strength, Vn, greater than shear corresponding to 1.25 times nominal flexural strength, Mn
(increases shear at least 1.39 times)
• Except Vn need not be greater than 2.5Vu. (doubles shear)
- 72 -
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 (from ASCE 7-10) of 100 kips. SDS = 0.4
Required: Design the shear wall; ordinary reinforced shear wall
Solution: Check using 0.6D+0.7E load combination.
• Try #6 in each of last 3 cells; #6 @40in.
P = (0.6-0.7(0.2)(SDS))D = (0.6-0.7(0.2)(0.4))(1k/ft(16ft)+7.8k) = 12.9 kips
Weight of wall: [40 psf(12ft)+2(2ft)75psf]10ft = 7800lb
Lightweight units, grout at 40 in. o.c. 40 psf; full grout 75 psf
From interaction diagram OK; stressed to 90% of capacity
- 73 -
Example: Shear Wall ASD
- 74 -
Example: Shear Wall ASD
Calculate net area, Anv, including grouted cells. 2849)5.262.7)(8(91925.2 ininininininAnv
Maximum Fv
625.0192/120// ininVdVhVdM vv
psipsifVd
MF gm
vv 8.8375.02000625.025
3
225
3
2
OK
psiin
lb
A
Vf
nvv 4.82
849
1000007.02
Shear ratio
Shear stress
- 75 -
Example: Shear Wall ASD
in
inpsi
inpsiin
Af
dFAs
sA
dFAF
nvs
sv
n
svvs
9.258494.42
1883200031.05.05.0
5.0
2
psipsipsiFfF vmgvvs 4.425.6775.0/4.82/
Use #5 bars in bond beams.Determine spacing.
Use #5 at 24 in. o.c.
psiin
lbpsi
A
Pf
Vd
MF
nm
vvm
5.67849
870025.02000625.075.10.4
2
1
25.075.10.42
1
2
s ≤ min{d/2, 48 in.} = min{94 in., 48 in.} = 48 in. Code 8.3.5.2.1
Top of wall P = (0.6-0.7(0.2SDS))D = 0.544(1k/ft)(16ft) = 8.70 kips
(critical location for shear):
Determine Fvm
Required steel stress
- 76 -
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 (from ASCE 7-10) of 100 kips. SDS = 0.4
Required: Design the shear wall; special reinforced shear wall
Solution: Check using 0.6D+0.7E load combination.
• Design for 1.5V, or 1.5(70 kips) = 105 kips (Section 7.3.2.6.1.2)
• fv = (105 kips)/(849in2) = 123.7 psi
• But, maximum Fv is 83.8 psi
• Fully grout wall
- 77 -
Example: Shear Wall ASD
psiin
lbpsiFvm 0.34
1464
870025.02000625.075.10.4
4
12
21464.192.625.7 inininAnv
psipsiFff vmvvs 7.370.347.71
in
inpsi
inpsiin
AF
dFAs
nvs
sv 6.1614947.37
1883200031.05.05.0
2
Use #5 @ 16 in.
psi
in
lb
A
Vf
nvv 7.71
1464
1000007.05.12
Calculate net area, Anv, including grouted cells.
Shear stress
Determine Fvm
(special wall)
Use #5 bars in bond beams.Determine spacing.
Required steel stress
- 78 -
Example: Shear Wall ASD
If we needed to check maximum reinforcing, do as follows.
00582.0
200060000
1.16600002
20001.16
2max
psipsi
psi
psi
f
fnf
fn
m
yy
m
Section 8.3.3.4 Maximum ReinforcementNo need to check maximum reinforcement since only need to check if:
• M/(Vdv) ≥ 1 and M/(Vdv) = 0.625
• P > 0.05f′mAn 0.05(2000psi)(1464in2) = 146 kips; P = 12.2 kips
00184.0
.188.625.7
44.06 2
inin
in
bd
As OK
For distributed reinforcement, the reinforcement ratio is obtained as thetotal area of tension reinforcement divided by bd. Assume 6 bars in tension.
- 79 -
Example: Shear Wall ASD
• Prescriptive Reinforcement Requirements (7.3.2.6)
• 0.0007 in each direction
• 0.002 total
• Vertical: 9(0.44in2)/1464in2 = 0.00270
• Horizontal: 7(0.31in2)/[120in(7.625in)] = 0.00237
• Total = 0.00270 + 0.00237 = 0.00507 OK
- 80 -
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 (from ASCE 7-10) of 100 kips. SDS = 0.4
Required: Design the shear wall; ordinary reinforced shear wall
Solution: Check using 0.9D+1.0E load combination.
• Try 2-#5 at end and #5 @ 40 in.
Pu = [0.9 - 0.2(SDS)]D = [0.9 - 0.2(0.4)](1k/ft(16ft)+6.4k) = 18.4 kips
Weight of wall: 40 psf(10ft)(16ft) = 6400 lb
Lightweight units, grout at 40 in. o.c. 40 psf
From interaction diagram OK; stressed to 97% of capacity
- 81 -
Example: Shear Walls SD
- 82 -
Example: Shear Walls SD
2767)5.262.7)(8(71925.2 ininininininAnv
kipspsiin
fAdV
MV gmnv
vu
un
9.10275.02000767625.0253
48.0
253
4
2
OK
Calculate net area, Anv, including grouted cells.
Maximum Vn
- 83 -
Example: Shear Walls SD
in
k
inksiin
V
dfAs
dfs
AV
ns
vyv
vyv
ns
0.287.63
1926031.05.05.0
5.0
kkkVV
V nm
g
uns 7.63
8.0
4.82
75.08.0
100
kipskpsiin
PfAdV
MV
lbkip
umnvvu
unm
4.821.1325.02000767625.075.10.48.0
25.075.10.4
100012
s ≤ min{d/2, 48 in.} = min{94 in., 48 in.} = 48 in. Code 8.3.5.2.1In strength design, this provision only applies to beams (9.3.4.2.3 (e))Suggest that minimum spacing also be applied to shear walls.
Use#5 at 24 in. o.c.
Use #5 bars in bond beams.Determine spacing.
Top of wall Pu = (0.9-0.2SDS)D = 0.82(1k/ft)(16ft) = 13.1 kips
(critical location for shear):
Determine Vnm
Required steel strength
- 84 -
Example: Shear Walls SD
Shear reinforcement requirements (9.3.3.3.2)
• Except at wall intersections, the end of a horizontal reinforcing bar needed to satisfy shear strength requirements shall be bent around the edge vertical reinforcing bar with a 180-degree hook.
• At wall intersections, horizontal reinforcing bars needed to satisfy shear strength requirements shall be bent around the edge vertical reinforcing bar with a 90-degree standard hook and shall extend horizontally into the intersecting wall a minimum distance at least equal to the development length.
- 85 -
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 (from ASCE 7-10) of 100 kips. SDS = 0.4
Required: Design the shear wall; special reinforced shear wall
Solution: Check using 0.9D+1.0E load combination.
• Shear capacity design provisions (Section 7.3.2.6.1.1)
• Vn ≥ shear corresponding to 1.25Mn.
• Minimum increase is 1.25/0.9 = 1.39
• Vn need not exceed 2.5Vu
• Normal design Vn ≥ Vu/ = Vu/0.8 = 1.25Vu
• Increases shear by a factor of 2
• Fully grout wall (Max Vn was 103 kips)
21464
.192.625.7
in
ininAnv
- 86 -
Example: Shear Walls SD
• Previously, Mn = 1028 k-ft; Mn = 1142 k-ft (Pu = 18.4k)
• 1.25Mn = 1428 k-ft; Design for 143 kips
• But wait, wall is fully grouted. Wall weight has increased to 75 psf
• For Pu = 23.0k, fully grouted, Mn = 1178 k-ft, 1.25Mn = 1474 k-ft
• Design for 147 kips
• But wait, need to check load combination of 1.2D + 1.0E
• Pu = [1.2 + 0.2(SDS)]D = 35.8 kips, 1.25Mn = 1601 k-ft
• Design for 160 kips
• Bottom line: any change in wall will change Mn, which will change design requirement; also need to consider all load combinations
• Often easier to just use Vn = 2.5Vu.
- 87 -
Example: Shear Walls SD
in
k
inksiin
V
dfAs
dfs
AV
ns
vyv
vyv
ns
2696.6
1926031.05.05.0
5.0
kkkVV
V nmuns 6.6
8.0
8.1541.160
kipskpsiin
PfAdV
MV
lbkip
umnvvu
unm
8.1541.1325.020001464625.075.10.48.0
25.075.10.4
100012
7.3.2.6 (b) Maximum spacing of reinforcing of 1/3 length, 1/3 height, or 48 in.Use maximum spacing of 1/3(height) = 40 in.
Use #5 at 40 in. o.c.
Use #5 bars in bond beams.Determine spacing.
Determine Vnm
Required steel strength Using shear from 1.25Mn
- 88 -
Example: Shear Walls SD
250k 1005.25.2 kVV un
Use #5 at 32 in. o.c.
Required steel strength
Using shear from Vn = 2.5Vu
kk
kVVV nmnns 4.568.0
8.154250
in
k
inksiin
V
dfAs
ns
vyv 6.314.56
1926031.05.05.0
Use #5 bars in bond beams.Determine spacing.
(Bars at 28, 60, 92 and 116 in. from bottom; top spacing is 24 in.)
- 89 -
Example: Shear Walls SD
• Prescriptive Reinforcement Requirements (7.3.2.6)
• 0.0007 in each direction
• 0.002 total
• Vertical: 7(0.31in2)/1464in2 = 0.00148
• Horizontal: 3(0.31in2)/[120in(7.625in)] = 0.00102
• Total = 0.00148 + 0.00102 = 0.00250 OK
- 90 -
Example: Shear Walls SD
• Calculate axial force based on c = 83.8 in. • Include compression reinforcement • Pn = 726 kips• Assume a live load of 1 k/ft• D + 0.75L + 0.525QE = (1k/ft + 0.75(1k/ft))16ft = 28 kips
Section 9.3.3.5 Maximum ReinforcementSince Mu/(Vudv) < 1, strain gradient is based on 1.5εy.
OK
c = 0.446(188in.) = 83.8 in.
Strain c/d, CMU c/d, Clay
1.5εy 0.446 0.530
3εy 0.287 0.360
4εy 0.232 0.297
- 91 -
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
gA1.0 mu fP geometrically symmetrical sections
gA05.0 mu fP geometrically unsymmetrical sections
AND
1vu
u
dV
MOR 3
vu
u
dV
Mmnu fAV 3 AND
For our wall, Mu/Vudv < 1Pu < 0.1fʹmAg = 0.1(2.0ksi)(1464in2) = 293 kips
- 92 -
Comparison of SD and ASD
- 93 -
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 about same shear steel for ordinary walls; some savings for special walls
Advantage to SD
• But, maximum reinforcement requirements can sometimes be hard to meet; designers then switch to ASD, which requires more steel. This makes no sense.
• SD requires 180° hooks. Some think hard to construct, but not really. Some designers avoid SD solely for this reason.
- 94 -
ASD vs SD: Final Outcome
Strength Design
Allowable Stress Design