sp-17-14_da
TRANSCRIPT
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 1/104
An ACI Handbook
The Reinforced Concrete Design Handbook
A Companion to ACI 318-14
Volume 3: Design Aids
SP-17(14)
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 2/104
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 3/104
ACI SP-17(14)
THE REINFORCED CONCRETE DESIGN HANDBOOK
A Companion to ACI 318-14
VOLUME 1
BUILDING EXAMPLE
STRUCTURAL SYSTEMS
STRUCTURAL ANALYSIS
DURABILITY
ONE-WAY SLABS
TWO-WAY SLABS
BEAMS
DIAPHRAGMS
COLUMNS
STRUCTURAL REINFORCED CONCRETE WALLS
FOUNDATIONS
VOLUME 2
RETAINING WALLS
SERVICEABILITY
STRUT-AND-TIE MODEL
ANCHORING TO CONCRETE
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 4/104
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 5/104
ACI SP-17(14)
Volume 3
THE REINFORCED CONCRETE
DESIGN HANDBOOK
A Companion to ACI 318-14
Editors:
Andrew TaylorTrey Hamilton III
Antonio Nanni
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 6/104
First PrintingSeptember 2015
THE REINFORCED CONCRETE DESIGN HANDBOOK
Volume 3 ~ Ninth EditionCopyright by the American Concrete Institute, Farmington Hills, MI. All rights reserved. This material may not bereproduced or copied, in whole or part, in any printed, mechanical, electronic, film, or other distribution and storagemedia, without the written consent of ACI.
The technical committees responsible for ACI committee reports and standards strive to avoid ambiguities,omissions, and errors in these documents. In spite of these efforts, the users of ACI documents occasionally findinformation or requirements that may be subject to more than one interpretation or may be incomplete or incorrect.Users who have suggestions for the improvement of ACI documents are requested to contact ACI via the erratawebsite at http://concrete.org/Publications/DocumentErrata.aspx. Proper use of this document includes periodicallychecking for errata for the most up-to-date revisions.
ACI committee documents are intended for the use of individuals who are competent to evaluate the significance andlimitations of its content and recommendations and who will accept responsibility for the application of the materialit contains. Individuals who use this publication in any way assume all risk and accept total responsibility for theapplication and use of this information.
All information in this publication is provided “as is” without warranty of any kind, either express or implied,including but not limited to, the implied warranties of merchantability, fitness for a particular purpose or non-infringement.
ACI and its members disclaim liability for damages of any kind, including any special, indirect, incidental, orconsequential damages, including without limitation, lost revenues or lost profits, which may result from the use ofthis publication.
It is the responsibility of the user of this document to establish health and safety practices appropriate to the specificcircumstances involved with its use. ACI does not make any representations with regard to health and safety issuesand the use of this document. The user must determine the applicability of all regulatory limitations before applyingthe document and must comply with all applicable laws and regulations, including but not limited to, United StatesOccupational Safety and Health Administration (OSHA) health and safety standards.
Participation by governmental representatives in the work of the American Concrete Institute and in thedevelopment of Institute standards does not constitute governmental endorsement of ACI or the standards that itdevelops.
Order information: ACI documents are available in print, by download, on CD-ROM, through electronic subscription,or reprint and may be obtained by contacting ACI. Most ACI standards and committee reports are gathered togetherin the annually revised ACI Manual of Concrete Practice (MCP).
American Concrete Institute38800 Country Club DriveFarmington Hills, MI 48331 USA+1.248.848.3700
Managing Editor: Khaled NahlawiStaff Engineers: Daniel W. Falconer, Matthew R. Senecal, Gregory M. Zeisler, and Jerzy Z. ZemajtisTechnical Editors: Shannon B. Banchero, Emily H. Bush, and Cherrie L. FergussonManager, Publishing Services: Barry BerginLead Production Editor: Carl Bischof Production Editors: Kelli Slayden, Kaitlyn Hinman, Tiesha ElamGraphic Designers: Ryan Jay, Aimee KahaianManufacturing: Marie Fuller
www.concrete.org
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 7/104
VOLUME 3: CONTENTS
APPENDIX A—REFERENCE TABLES 1
APPENDIX B—ANALYSIS TABLES 11
APPENDIX C—SECTIONAL PROPERTIES 25
APPENDIX D––COLUMN INTERACTIONDIAGRAMS 27
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 8/104
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 9/104
REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14) 1
American Concrete Institute Copyrighted Material—www.concrete.org
APPENDIX A—REFERENCE TABLES
Table A-1—Nominal cross section area, weight, and nominal diameter of ASTM standard reinforcing bars
Bar size designation Nominal cross section area, in.2 Weight, lb/ft Nominal diameter, in. Nominal perimeter, in.
No. 3 0.11 0.376 0.375 1.18
No. 4 0.20 0.668 0.500 1.57
No. 5 0.31 1.043 0.625 1.96
No. 6 0.44 1.502 0.750 2.36
No. 7 0.60 2.044 0.875 2.75
No. 8 0.79 2.670 1.000 3.14
No. 9 1.00 3.400 1.128 3.54
No. 10 1.27 4.303 1.270 3.99
No. 11 1.56 5.313 1.410 4.43
No. 14 2.25 7.650 1.693 5.32
No. 18 4.00 13.600 2.257 7.09
Note: The nominal dimensions of a deformed bar are equivalent to those of a plain bar having the same mass per foot as the deformed bars.
Example: #9 bar spaced 7-1/2 in. apart provides 1.60 in.2/ft of section width.
Table A-2—Area of bars in a section 1 ft wide
Spacing,in.
Cross section area of bar As (or As′ ), in.2
Spacing,in.
Bar size
#3 #4 #5 #6 #7 #8 #9 #10 #11 #14 #18
4.0 0.33 0.60 0.93 1.32 1.80 2.37 3.00 3.81 4.68 — — 4.0
4.5 0.29 0.53 0.83 1.17 1.60 2.11 2.67 3.39 4.16 6.00 — 4.5
5.0 0.26 0.48 0.74 1.06 1.44 1.90 2.40 3.05 3.74 5.40 9.60 5.0
5.5 0.24 0.44 0.68 0.96 1.31 1.72 2.18 2.77 3.40 4.91 8.73 5.5
6.0 0.22 0.40 0.62 0.88 1.20 1.58 2.00 2.54 3.12 4.50 8.00 6.0
6.5 0.20 0.37 0.57 0.81 1.11 1.46 1.85 2.34 2.88 4.15 7.38 6.5
7.0 0.19 0.34 0.53 0.75 1.03 1.35 1.71 2.18 2.67 3.86 6.86 7.0
7.5 0.18 0.32 0.50 0.70 0.96 1.26 1.60 2.03 2.50 3.60 6.40 7.5
8.0 0.17 0.30 0.47 0.66 0.90 1.19 1.50 1.91 2.34 3.38 6.00 8.0
8.5 0.16 0.28 0.44 0.62 0.85 1.12 1.41 1.79 2.20 3.18 5.65 8.5
9.0 0.15 0.27 0.41 0.59 0.80 1.05 1.33 1.69 2.08 3.00 5.33 9.0
9.5 0.14 0.25 0.39 0.56 0.76 1.00 1.26 1.60 1.97 2.84 5.05 9.5
10.0 0.13 0.24 0.37 0.53 0.72 0.95 1.20 1.52 1.87 2.70 4.80 10.0
10.5 0.13 0.23 0.35 0.50 0.69 0.90 1.14 1.45 1.78 2.57 4.57 10.5
11.0 0.12 0.22 0.34 0.48 0.65 0.86 1.09 1.39 1.70 2.45 4.36 11.0
11.5 0.11 0.21 0.32 0.46 0.63 0.82 1.04 1.33 1.63 2.35 4.17 11.5
12.0 0.11 0.20 0.31 0.44 0.60 0.79 1.00 1.27 1.56 2.25 4.00 12.0
13.0 0.10 0.18 0.29 0.41 0.55 0.73 0.92 1.17 1.44 2.08 3.69 13.0
14.0 0.09 0.17 0.27 0.38 0.51 0.68 0.86 1.09 1.34 1.93 3.43 14.015.0 0.09 0.16 0.25 0.35 0.48 0.63 0.80 1.02 1.25 1.80 3.20 15.0
16.0 0.08 0.15 0.23 0.33 0.45 0.59 0.75 0.95 1.17 1.69 3.00 16.0
17.0 0.08 0.14 0.22 0.31 0.42 0.56 0.71 0.90 1.10 1.59 2.82 17.0
18.0 0.07 0.13 0.21 0.29 0.40 0.53 0.67 0.85 1.04 1.50 2.67 18.0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 10/104
2 REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
American Concrete Institute Copyrighted Material—www.concrete.org
Table A-3—Minimum beam web widths required for two or more bars in one layer for cast-in-place non-
prestressed concrete
Reference: ACI 318-11, Sections 7.2.2, 7.6.1, 7.7.1(c), and AASHTO Standard Specifications for Highway Bridges (17th
edition, 2002) Division I, Sections 8.17.3.1, 8.21.1, 8.22.1, 8.23.2.2, and Table 8.23.2.1. For use of this Design Aid, see Flexure
Example 1.
Minimum beam width = 2(A + B + C) + (n – 1)(D + d b) where A + B + C - 1/2 d b ≥ 2.0 in. cover required for longitudinal bars
and these assumptions are made:
for ACI 318
For both ACI and AASHTO For AASHTO
Bar size
ACI 318-113/4-in. aggregateinterior exposure
#3 stirrups
ACI 318-111-in. aggregate
interior exposure#3 stirrups
AASHTO requirementscast-in-place concrete
1-in. aggregateexposed to earth or weather
Minimumweb width for
2 bars, in.
Increment for each added bar,
in.
Minimumweb width for
2 bars, in.
Increment for each added bar,
in.
Minimumweb width for
2 bars, in.
Increment for each added bar,
in.
#4 6 3/4 1 1/2 7 1/8 1 7/8 7.25 2.000
#5 6 7/8 1 5/8 7 1/4 2 7.37 2.125
#6 7 1 3/4 7 3/8 2 1/8 7.50 2.250
#7 7 1/8 1 7/8 7 1/2 2 1/4 7.62 2.375
#8 7 1/4 2 7 5/8 2 3/8 7.75 2.500
#9 7 1/2 2 1/4 7 3/4 2 1/2 8.32 2.820
#10 7 7/8 2 1/2 7 7/8 2 5/8 8.68 3.175
#11 8 1/8 2 7/8 8 1/8 2 7/8 9.52 3.525
#14 8 7/8 3 3/8 8 7/8 3 3/8 10.23 4.232
#18 10 1/2 4 1/2 10 1/2 4 1/2 11.90 5.642
B = 0.375 in. for #3 stirrups = 0.500 in. for #4 stirrups
D = 1 d b
≥ 1 in.≥ 1-1/3 nominal aggregate size
B = 0.375 in. for #3 stirrups (minimumstirrup size for #10 andsmaller longitudinal bars)
= 0.500 in. for #4 stirrups (minimumstirrup size for #11 and largerlongitudinal bars)
D = 1-1/2d b≥ 1-1/2 in.≥ 1-1/2 nominal aggregate size
A = 1-1/2 in. concrete cover to stirrupB = 0.635 in. for #5 stirrups
= 0.750 in. for #6 stirrups
C = stirrup bend radius of 2 stirrup
bar diameter for #5 and smaller
stirrups
= stirrup bend radius of 3 stirrup
bar diameters for #6 stirrups
≥ 1/2d b of longitudinal bars
Notes
1. Stirrups: For stirrups larger than those used
for table above, increase web width by the follow-
ing amounts (in.):
2. ACI cover requirements: For exterior
exposure with use of #6 or larger stirrups, add 1 in.
to web width.
3. AASHTO cover requirements: For interior
exposure, 1/2 in. may be deducted from beam
widths.
4. Bars of different sizes: For beams with bars
of two or more sizes, determine from table the beam
web width required for the given number of largest
size bars; then add the indicated increments for each
smaller bar.
5. Example: Find the minimum web width for a
beam reinforced with two #8 bars; a beam
reinforced with three #8 bars; a beam reinforced
with three #9 and two #6 bars.
Source
Mainreinforcement
size#4
stirrup#5
stirrup#6
stirrup
ACIrequirements
#4 through #11 3/4 1 1/2 2 1/4
#14 1/2 1 1/4 2
#18 1/4 3/4 1 1/2
AASHTOrequirements
#4 through #10 0.75 1.50 2.25
#11 through #14 — 0.75 1.50
#18 — 0.49 1.24
2 #8 3 #8 3 #9 + 2 #6
ACI (3/4 in. aggregate) 7 1/4 9 1/4 13 1/4
ACI (1 in. aggregate) 7 5/8 10 14 1/2
AASHTO 7.75 10.25 15.64
T A B L E S
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 11/104
REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14) 3
American Concrete Institute Copyrighted Material—www.concrete.org
Table A-4—Minimum beam web widths for various bar combinations (interior exposure)Reference: ACI 318-11 Sections 7.2.2, 7.6.1, and 7.7.1(c)
No. of bars
ACI min. bw, in. Columns at left headed 1-5 and 6-10 bars arefor minimum web width bw of beam having bars
of one size only. Remaining columns are for combinations of 1 to 5 bars of each of two sizes. Calculated values of beam web width bw
rounded upward to nearest half inch. Where bars of two sizes are used, larger
bar(s) assumed to be placed along outsideface(s) of beam. Aggregate size assumed ≤ 3/4 in.
Barsize
1 to 5 bars
6 to 10 bars
1
#3
5.5 12.5
Sizeof
smaller bars
ACI min. bw, in.
A =B =C =
D =E =
clear cover of 1-1/2 in.3/4 in. diameter of #3 stirrupsfor #11 and smaller bars: twice diameter of #3 stirrups; for #14 and #18 bars: 1/2diameter of bar 1/2 diameter of larger bar 1/2 spacing for larger bar plus 1/2 spac-ing for smaller bar (spacing is d b for #9
and larger bars, 1 in. for #8 and smaller bars)
2 7.0 13.5
3 8.0 15.0
No. of smaller bars4 9.5 16.5
5 11.0 18.0 1 2 3 4 5
1
#4
5.5 13.0
#3
7.0 8.5 9.5 11.0 12.5
Sizeof
smaller bars
ACI min. bw, in.2 7.0 14.5 8.5 9.5 11.0 12.5 14.0
3 8.5 16.0 10.0 11.0 12.5 14.0 15.5
No. of smaller bars4 10.0 17.5 11.5 12.5 14.0 15.5 17.0
5 11.5 19.0 13.0 14.0 15.5 17.0 18.5 1 2 3 4 5
1
#5
5.5 13.5
#4
7.0 8.5 10.0 11.5 13.0
#3
7.0 8.5 9.5 11.0 12.5
Sizeofsmaller bars
ACI min. bw, in.2 7.0 15.0 8.5 10.0 11.5 13.0 14.5 8.5 10.0 11.0 12.5 14.0
3 8.5 17.0 10.0 11.5 13.0 14.5 16.0 10.0 11.5 13.0 14.0 15.5 No. of smaller bars4 10.5 18.5 12.0 13.5 15.0 16.5 18.0 11.5 13.0 14.5 16.0 17.0
5 12.0 20.0 13.5 15.0 16.5 18.0 19.5 13.5 14.5 16.5 17.5 19.0 1 2 3 4 5
1
#6
5.5 14.0
#5
7.0 9.0 10.5 12.0 13.5
#4
7.0 8.5 10.0 11.5 13.0
#3
7.0 8.5 10.0 11.0 12.5
2 7.0 16.0 9.0 10.5 12.0 13.5 15.5 8.5 10.0 11.5 13.0 14.5 8.5 10.0 11.5 12.5 14.0
3 9.0 17.5 10.5 12.0 14.0 15.5 17.0 10.5 12.0 13.5 15.0 16.5 10.5 11.5 13.0 14.5 16.0
4 10.5 19.5 12.5 14.0 15.5 17.0 19.0 12.0 13.5 15.0 16.5 18.0 12.0 13.5 15.0 16.0 17.5
5 12.5 21.0 14.0 15.5 17.5 19.0 20.5 14.0 15.5 17.0 18.5 20.0 14.0 15.0 16.5 18.0 19.5
1
#7
5.5 15.0
#6
7.5 9.0 11.0 12.5 14.5
#5
7.0 9.0 10.5 12.0 13.5
#4
7.0 8.5 10.0 11.5 13.0
2 7.5 16.5 9.0 11.0 12.5 14.5 16.0 9.0 10.5 12.0 14.0 15.5 9.0 10.5 12.0 13.5 15.0
3 9.0 18.5 11.0 12.5 14.5 16.0 18.0 11.0 12.5 14.0 15.5 17.5 10.5 12.0 13.5 15.0 16.5
4 11.0 20.5 13.0 14.5 16.5 18.0 20.0 12.5 14.5 16.0 17.5 19.0 12.5 14.0 15.5 17.0 18.5
5 13.0 22.5 14.5 16.5 18.0 20.0 21.5 14.5 16.0 18.0 19.5 21.0 14.5 16.0 17.5 19.0 20.5
1
#8
5.5 15.5
#7
7.5 9.5 11.0 13.0 15.0
#6
7.5 9.0 11.0 12.5 14.5
#5
7.5 9.0 10.5 12.0 14.0
2 7.5 17.5 9.5 11.0 13.0 15.0 17.0 9.0 11.0 12.5 14.5 16.0 9.0 10.5 12.5 14.0 15.5
3 9.5 19.5 11.5 13.0 15.0 17.0 19.0 11.0 13.0 14.5 16.5 18.0 11.0 12.5 14.5 16.0 17.5
4 11.5 21.5 13.5 15.0 17.0 19.0 21.0 13.0 15.0 16.5 18.5 20.0 13.0 14.5 16.5 18.0 19.5
5 13.5 23.5 15.5 17.0 19.0 21.0 23.0 15.0 17.0 18.5 20.5 22.0 15.0 16.5 18.5 20.0 21.5
1
#9
5.5 17.0
#8
7.5 9.5 11.5 13.5 15.5
#7
7.5 9.5 11.5 13.0 15.0
#6
7.5 9.0 11.0 12.5 14.5
2 8.0 19.0 10.0 12.0 14.0 16.0 18.0 9.5 11.5 13.5 15.5 17.0 9.0 11.5 13.0 15.0 16.5
3 10.0 21.5 12.0 14.0 16.0 18.0 20.0 12.0 14.0 15.5 17.5 19.5 12.0 13.5 15.5 17.0 19.0
4 12.5 23.5 14.5 16.5 18.5 20.5 22.5 14.0 16.0 18.0 20.0 21.5 14.0 16.0 17.5 19.5 21.0
5 14.5 26.0 16.5 18.5 20.5 22.5 24.5 16.5 18.5 20.0 22.0 24.0 16.5 18.0 20.0 21.5 23.5
1
#10
5.5 18.0
#9
8.0 10.0 12.5 14.5 17.0
#8
8.0 10.0 12.0 14.0 16.0
#7
7.5 9.5 11.5 13.5 15.0
2 8.0 20.5 10.5 12.5 15.0 17.0 19.5 10.0 12.0 14.0 16.0 18.0 10.0 12.0 13.5 15.5 17.5
3 10.5 23.5 13.0 15.0 17.5 19.5 22.0 12.5 14.5 16.5 18.5 20.5 12.5 14.5 16.0 18.0 20.0
4 13.0 26.0 15.5 17.5 20.0 22.0 24.5 15.0 17.0 19.0 21.0 23.0 15.0 17.0 18.5 20.5 22.5
5 15.5 28.5 18.0 20.0 22.5 24.5 27.0 17.5 19.5 21.5 23.5 25.5 17.5 19.5 21.5 23.0 25.0
1
#11
5.5 19.5
#10
8.0 10.5 13.0 15.5 18.0
#9
8.0 10.5 12.5 15.0 17.0
#8
8.0 10.0 12.0 14.0 16.0
2 8.5 22.5 11.0 13.5 16.0 18.5 21.0 10.5 13.0 15.0 17.5 19.5 10.5 12.5 14.5 16.5 18.5
3 11.0 25.0 13.5 16.0 19.0 21.5 24.0 13.5 15.5 18.0 20.0 22.5 13.0 15.0 17.0 19.0 21.0
4 14.0 28.0 16.5 19.0 21.5 24.0 26.5 16.0 18.5 20.5 23.0 25.0 16.0 18.0 20.0 22.0 24.0
5 17.0 31.0 19.5 22.0 24.5 27.0 29.5 19.0 21.5 23.5 26.0 28.0 19.0 21.0 23.0 25.0 27.0
1
#14
5.5 22.5
#11
8.5 11.5 14.5 17.0 20.0
#10
8.5 11.0 13.5 16.0 18.5
#9
8.5 10.5 13.0 15.0 17.5
2 9.0 26.0 12.0 14.5 17.5 20.5 23.0 11.5 14.0 16.5 19.0 22.0 11.5 13.5 16.0 18.0 20.5
3 12.5 29.5 15.5 18.0 21.0 23.5 26.5 15.0 17.5 20.0 22.5 25.0 14.5 17.0 19.0 21.5 23.5
4 16.0 33.0 18.5 21.5 24.5 27.0 30.0 18.5 21.0 23.5 26.0 28.5 18.0 20.5 22.5 25.0 27.0
5 19.0 36.0 22.0 25.0 27.5 30.5 33.5 22.0 24.5 27.0 29.5 32.0 21.5 23.5 26.0 28.5 30.5
1
#18
6.5 29.0
#14
10.0 13.5 16.5 20.0 23.5
#11
9.5 12.5 15.0 18.0 21.0
#10
9.5 12.0 14.5 17.0 19.5
2 11.0 33.5 14.0 17.5 21.0 24.5 27.5 13.5 16.5 19.0 22.0 25.0 13.5 16.0 18.5 21.0 23.5
3 15.5 38.0 18.5 22.0 25.5 29.0 32.0 18.0 21.0 23.5 26.5 29.5 18.0 20.5 23.0 25.5 28.0
4 20.0 42.5 23.0 26.5 30.0 33.5 36.5 22.5 25.5 28.5 31.0 34.0 22.5 25.0 27.5 30.0 32.5
5 24.5 47.0 27.5 31.0 34.5 38.0 41.0 27.0 30.0 33.0 35.5 38.5 27.0 29.5 32.0 34.5 37.0
Examples: For 2 #6 bars, minimum bw = 7.0 in. For 8 #6 bars, minimum bw =17.5 in. For 2 #7 bars plus 3 #6 bars, minimum bw = 12.5 in.
For 3 #6 bars plus 5 #4 bars, minimum bw = 16.5 in.
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 12/104
4 REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
American Concrete Institute Copyrighted Material—www.concrete.org
Table A-5—Properties of bundled bars
Reference: ACI 318-11 Section 7.6.6.5.
Equivalent diameter,
Centroidal distance, x
For the bundled bars configurationshown here, the centroidal distance iscalculated by the following equation:
Bars Combination of bars
Equivalent diameters d be, in.
#8 1.42 1.74 2.01 #7 #6 1.15 1.37 1.45 1.56 1.63 1.69
9 1.60 1.95 2.26 7 5 1.08 1.25 1.39 1.40 1.52 1.64
10 1.80 2.20 2.54 8 7 1.33 1.59 1.67 1.82 1.88 1.94
11 1.99 2.44 2.82 8 6 1.25 1.46 1.60 1.64 1.77 1.89
9 8 1.51 1.81 1.88 2.07 2.13 2.20
9 7 1.43 1.67 1.82 1.89 2.02 2.14
10 9 1.70 2.04 2.12 2.33 2.40 2.47
10 8 1.62 1.90 2.06 2.15 2.29 2.42
11 10 1.90 2.28 2.36 2.61 2.68 2.75
11 9 1.81 2.13 2.29 2.41 2.55 2.69
Centroidal distance x , from bottom of bundle, in.
#4 0.25 0.39 0.50 #4 #3 0.23 0.38 0.33 0.43 0.40 0.46
5 0.31 0.49 0.62 5 4 0.29 0.47 0.43 0.55 0.53 0.58
6 0.37 0.59 0.75 5 3 0.28 0.46 0.37 0.49 0.44 0.55
6 5 0.35 0.57 0.53 0.67 0.66 0.71
7 0.44 0.69 0.87 6 4 0.34 0.55 0.47 0.61 0.57 0.67
8 0.50 0.79 1.00
9 0.56 0.89 1.13 7 6 0.41 0.67 0.62 0.80 0.78 0.83
7 5 0.39 0.65 0.56 0.73 0.69 0.79
10 0.63 1.00 1.27 8 7 0.47 0.77 0.72 0.93 0.90 0.95
11 0.70 1.11 1.41 8 6 0.46 0.75 0.66 0.86 0.81 0.92
9 8 0.54 0.86 0.82 1.05 1.03 1.08
9 7 0.52 0.85 0.75 0.98 0.94 1.04
10 9 0.60 0.98 0.92 1.19 1.16 1.22
10 8 0.58 0.95 0.86 1.11 1.07 1.1811 10 0.67 1.08 1.03 1.32 1.31 1.36
11 9 0.65 1.06 0.96 1.24 1.20 1.31
Example: Find the equivalent diameter of a single bar for 4 #9 bars. For 4 #9 bars, read d be = 2.26 in.,
and the centroidal distance x equals 1.13 in.
d be
4
π--- As=
x
5
2--- Asi d b1 As2 d b1 d b2/2+( )+
Σ Asi
---------------------------------------------------------------=
T A B L E S
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 13/104
5
American Concrete Institute Copyrighted Material—www.concrete.org
Calculated values of beam width bw rounded
upward to nearest half-inch.
Assumptions:Aggregate size: ≤ 3/4 in.
Clear cover of 1-1/2 in.
No. 3 stirrups
Minimum beam web width bw, in.*
Bar size
Two bundles
#8 10.0 10.0 10.5
#9 10.5 11.0 11.0#10 11.0 11.5 12.0
#11 11.5 12.0 12.5
Three bundles
#8 13.5 14.0 14.5
#9 14.5 15.0 15.5
#10 15.5 16.0 17.0
#11 16.5 17.5 18.0
Four bundles
#8 17.0 17.5 18.5
#9 18.0 19.0 20.0
#10 20.0 21.0 22.0
#11 21.5 22.5 24.0
*For beams conforming to AASHTO specifications, add 1 in. to tabulated beam webwidth. Fore example, for two bundles of three #10, minimum bw = 11.5 in.
REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
Reference:Reference: ACIACI 318-11318-11 SectionsSections 7.2.2,7.2.2, 7.6.6.7.6.6.1,1, 7.6.6.2,7.6.6.2, 7.6.6.3,7.6.6.3, 7.6.6.7.6.6.5,5, andand 77.7.1.7.1
Table A-6—Minimum beam web widths b w for various combinations of bundled bars (interior exposure)
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 14/104
6 REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
American Concrete Institute Copyrighted Material—www.concrete.org
For: ψ t = 1.0; ψ e = 1.0; and λ = 1.0 (see notes below).
Basic development length ratios of bars in tension
Bar
size Category
f y 40,000 psi 60,000 psi 80,000 psi
f c′3000
psi
4000
psi
5000
psi
6000
psi
3000
psi
4000
psi
5000
psi
6000
psi
8000
psi
10,000
psi
3000
psi
4000
psi
5000
psi
6000
psi
8000
psi
10,000
psiα
#3 to#6
I 1/25 29 25 23 21 44 38 34 31 27 24 58 51 45 41 36 32
II 3/50 44 38 34 31 66 57 51 46 40 36 88 76 68 62 54 48
#7 to
#18
I 1/20 37 32 28 26 55 47 42 39 34 30 73 63 57 52 45 40
II 3/40 55 47 42 39 82 71 64 58 50 45 110 95 85 77 67 60
Notes:
1. See category chart for Categories I and II.
2. ψ t = bar location factor (1.3 for bars placed such that more than 12 in. of fresh concrete is cast below the development length or splice; 1.0 for other bars).
ψ e = coating factor (1.5 = epoxy-coated reinforcement with cover < 3d b or clear spacing < 6d b; 1.2 = all other epoxy-coated reinforcement; and 1.0 =
uncoated and zinc-coated [galvanized] reinforcement).
λ = lightweight-aggregate concrete factor (0.75 for lightweight concrete, and 1.0 for normalweight concrete).
3. Minimum development length l d ≥ 12 in.
Development length ratios:
l d
d b----- α
ψ t ψ eλ
------------ f y
f c′---------=
Table A-7—Basic development length ratios of bars in tension
Reference: ACI 318-11 Sections 12.2.2 and 12.2.4
T A B L E S
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 15/104
REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14) 7
American Concrete Institute Copyrighted Material—www.concrete.org
Table A-7—Basic development length ratios of bars in tension (cont.)
CATEGORY CHART
or
Category II: All other cases
Category I:
Clear spacing d b≥
Clear cover d b≥
Code minimum stirrups or ties throughout l d
Clear spacing 2d b≥
Clear cover d b≥
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 16/104
8 REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
American Concrete Institute Copyrighted Material—www.concrete.org
Table A-8—Basic development length l dh of standard hooks in tension
Reference: ACI 318-11 Sections 7.1 and 12.5.1 to 12.5.3
This table is calculated with ψ e = 1.0 and λ = 1.0.
Basic development length l dh, in., of standard hooks in tension
Bar size
f y 60,000 psi 80,000 psi
f c′
3000 psi 4000 psi 5000 psi 6000 psi 8000 psi 10,000 psi 3000 psi 4000 psi 5000 psi 6000 psi 8000 psi 10,000 psi 8d b , in.d b , in.
#3 0.375 8.2 7.1 6.4 5.8 5.0 4.5 11.0 9.5 8.5 7.7 6.7 6.0 3
#4 0.5 11.0 9.5 8.5 7.7 6.7 6.0 14.6 12.6 11.3 10.3 8.9 8.0 4
#5 0.625 13.7 11.9 10.6 9.7 8.4 7.5 18.3 15.8 14.1 12.9 11.2 10.0 5
#6 0.75 16.4 14.2 12.7 11.6 10.1 9.0 21.9 19.0 17.0 15.5 13.4 12.0 6
#7 0.875 19.2 16.6 14.8 13.6 11.7 10.5 25.6 22.1 19.8 18.1 15.7 14.0 7
#8 1 21.9 19.0 17.0 15.5 13.4 12.0 29.2 25.3 22.6 20.7 17.9 16.0 8
#9 1.128 24.7 21.4 19.1 17.5 15.1 13.5 33.0 28.5 25.5 23.3 20.2 18.0 9
#10 1.27 27.8 24.1 21.6 19.7 17.0 15.2 37.1 32.1 28.7 26.2 22.7 20.3 10
#11 1.41 30.9 26.8 23.9 21.8 18.9 16.9 41.2 35.7 31.9 29.1 25.2 22.6 11
#14 1.693 37.1 32.1 28.7 26.2 22.7 20.3 49.5 42.8 38.3 35.0 30.3 27.1 14
#18 2.257 49.5 42.8 38.3 35.0 30.3 27.1 65.9 57.1 51.1 46.6 40.4 36.1 18
Note 1: To compute development length l dh* for a standard hook in tension, multiply basic development length l dh from table above by applicable modifica-
tion factors:
α = 0.7 for #11 and smaller bars with side cover normal to plane of hook not less than 2-1/2 in.; and for 90-degree hook, cover on bar extension
beyond hook not less than 2 in.
α = 0.8 for #11 and smaller bars with a 90-degree hook enclosed vertically or horizontally within ties or stirrup ties spaced along the full develop-
ment length not greater than 3d b, where d b is diameter of hooked bar.
α = 0.8 for #11 and smaller bars with 180-degree hook that are enclosed within ties or stirrups perpendicular to the bar being developed, spaced
not greater than 3d b along l dh.
α = As required / As provided
Note 2: Values of basic development length l dh above the heavy line are less than the minimum development length of 6 in. Development length l dh shall not
be less than 8d b, nor less than 6 in., whichever is greater.
Development length, l dh* = αl dh ≥ 8db,
less than 6 in., where α represents
modifiers from Note 1 below and l dh is
basic development length of standard
hooks in tension
l dh
0.02ψ e f y
λ f c′---------------------d b=
T A B L E S
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 17/104
9
American Concrete Institute Copyrighted Material—www.concrete.org
REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
Table A-8—Basic development length l dh of standard hooks in tension (cont.)
Reference: ACI 318-11 Sections 7.1 and 7.21
Bar size #3 #4 #5 #6 #7 #8 #9 #10 #11 #14 #18
l dh, in. 6 7 7 8 9 10 13 14 15 19 25
Example: Find minimum embedment depth l dh that will provide 2 in. cover over the tail of a standard 180-degree end hook
in a #8 bar. For #8 bar, readl dh = 10 in.
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 18/104
10 REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
American Concrete Institute Copyrighted Material—www.concrete.org
T A B L E S
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 19/104
REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14) 11
American Concrete Institute Copyrighted Material—www.concrete.org
APPENDIX B—ANALYSIS TABLES Reproduced with permission from the Canadian Portland Cement Association.
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 20/104
12 REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
American Concrete Institute Copyrighted Material—www.concrete.org
T A B L E S
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 21/104
REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14) 13
American Concrete Institute Copyrighted Material—www.concrete.org
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 22/104
14 REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
American Concrete Institute Copyrighted Material—www.concrete.org
T A B L E S
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 23/104
REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14) 15
American Concrete Institute Copyrighted Material—www.concrete.org
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 24/104
16 REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
American Concrete Institute Copyrighted Material—www.concrete.org
T A B L E S
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 25/104
REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14) 17
American Concrete Institute Copyrighted Material—www.concrete.org
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 26/104
18 REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
American Concrete Institute Copyrighted Material—www.concrete.org
T A B L E S
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 27/104
REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14) 19
American Concrete Institute Copyrighted Material—www.concrete.org
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 28/104
20 REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
American Concrete Institute Copyrighted Material—www.concrete.org
T A B L E S
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 29/104
REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14) 21
American Concrete Institute Copyrighted Material—www.concrete.org
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 30/104
22 REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
American Concrete Institute Copyrighted Material—www.concrete.org
T A B L E S
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 31/104
REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14) 23
American Concrete Institute Copyrighted Material—www.concrete.org
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 32/104
24 REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
American Concrete Institute Copyrighted Material—www.concrete.org
T A B L E S
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 33/104
REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14) 25
American Concrete Institute Copyrighted Material—www.concrete.org
APPENDIX C—SECTIONAL PROPERTIES
Dash-and-dot lines are drawn through centers of gravity
A = area of section; I = moment of inertia; R = radius of gyration
A = bd
A d 2
=
I 1d
4
12------=
I 2d
4
3-----=
R1 0.2887d =
R2 0.57774d =
A πd
2
4-------- 0.7854d
2= =
I vd
4
64-------- 0.0491d
4= =
Rd
4---=
A d 2
=
y 0.7071d =
I d
4
12------=
R 0.2887d =
A 0.8660d 2
=
I 0.060d 4
=
R 0.264d =
I 1bd
3
12--------=
I 2bd
3
3--------=
R1 0.2887d =
R2 0.5774d =
A 0.8284d 2
=
I 0.055d 4
=
R 0.257d =
A bd =
ybd
b2
d 2
+
--------------------=
I b
3d
3
6 b2
d 2
+( )------------------------=
Rbd
6 b2
d 2
+( )-----------------------------=
Abd
2------=
I 1 bd
3
36--------=
I 2bd
3
12--------=
R1 0.236d =
R2 0.408d =
A bd =
Ad
2--- b b′+( )=
yd 2b b′+( )3 b b′+( )
--------------------------=
yd b 2b′+( )3 b b′+( )
--------------------------=
I d
2b
24bb ′ b′
2+ +( )
36 b b′+( )-----------------------------------------------=
Rd
6 b b ′+( )---------------------- 2 b
24bb ′ b′
2+ +( )=
Rb
2 sin
2∞ d
2cos
2∞+( )
12--------------------------------------------------------=
ybsin∞ d ∞cos+
2--------------------------------------=
Table C-1—Properties of sections
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 34/104
26 REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
American Concrete Institute Copyrighted Material—www.concrete.org
Dash-and-dot lines are drawn through centers of gravity
A = area of section; I = moment of inertia; R = radius of gyration
Section of parabola
For parabola: For compliment:
Ellipse
Parabola
Equation:
A bt bc′+=
yd
2b′ t
2b b′ – ( )+
2 bt bc′+( )
---------------------------------------=
y1 d y – =
I b′ y1
3by
3b b′ – ( ) y t – ( )
3 – +
3---------------------------------------------------------------------=
R I
A---=
y2 b
2
d ----- x =
A2bd
3---------= A
bd
3------=
I 18
175---------bd
3= I 1
37
2100------------bd
3=
I 219
480---------b
3d = I 2
1
80------b
3d =
A bt bc′+=
yd
2b′ t
2b b′ – ( )+
2 bt bc′+( )---------------------------------------=
y1 d y – =
I
b′ y1
3by
3b b′ – ( ) y t – ( )
3 – +
3---------------------------------------------------------------------=
R I
A---=
A d 2
a2
– =
I d
4a
4 –
12----------------=
R d 2
a2
+12
----------------=
A bt c a b′+( )
2----------------------+=
y3bt
23b′c d t +( ) c a b′ – ( ) c 3t +( )+ +
3 2bt c a b′+( )+[ ]--------------------------------------------------------------------------------------------=
y1 d y – =
I 4bt
2c
33b′ a+( )+
12-------------------------------------------- A y t – ( )
2 – =
R I
A---=
A bd ac – =
I bd
3ac
3 –
12----------------------=
Rbd
3ac
3 –
12 bd ac – ( )-----------------------------=
A 0.7854bd =
I 1wbd
3
64------------- 0.0491bd
3= =
I 2wb
3d
64------------- 0.0491b
3d = =
R1
d
4---=
R2
b
4---=
A π d
2d 1
2 – ( )
4------------------------ 0.7854 d
2d 1
2 – ( )= =
I π d
4d 1
4 – ( )
64------------------------ 0.0491 d
4d 1
4 – ( )= =
R 1 4 ⁄ d 2
d 12
+=
y2 b
2
4d x
--------=
A2bd
3---------=
A 0.8284d 2
0.7854d 12
– =
0.7854 1.055d 2
d 1
2 –
( )=
I 0.0547d 4
0.0491d 14
– =
0.0491 1.115d 4
d 14
– ( )=
0.0491 1.056( )2d
4d 1
4 – [ ]=
R 1 4 ⁄ 1.056d 2
d 12
+=
Table C-2—Properties of sections
P R O P .
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 35/104
APPENDIX D –– COLUMN INTERACTION DIAGRAMS (DESIGN AIDS)
D.1 — Column interaction diagrams
The column axial load-bending moment interaction diagrams illustrated in D.5 conform to ACI 318-14. The
equations used to generate data for plotting the interaction diagrams were originally developed for ACI SP-7 (Everard
and Cohen 1964). In addition, complete derivations of the equations for square and circular columns having the steel
arranged in a circle have been published in the ACI Structural Journal (Everard 1997). The interaction diagrams
contained in SP-7 were subsequently published in SP-17A (ACI Committee 340 1970). The related equations were
derived considering the following:
(a) For rectangular and square columns having steel bars placed on the end faces only, reinforcement was assumed
to consist of two equal thin strips parallel to the compression face of the section;
(b) For rectangular and square columns having steel bars equally distributed along all four faces, the reinforcement
was considered to consist of a thin rectangular or square tube; and
(c) For square and circular sections having steel bars arranged in a circle, the reinforcement was considered to consist
of a thin circular tube.
The interaction diagrams were developed using the rectangular stress block (ACI 318-14, Section 22.2.2.4). In
all cases, for reinforcement within the compressed portion of the depth perpendicular to the compression face of the
concrete (a =
1c), the compression stress in the steel was reduced by 0.85 f c′ to account for the concrete area that is
displaced by the reinforcing bars within the compression stress block.
The interaction diagrams were plotted in nondimensional form. The vertical coordinate [ K n = P n/( f c′ A g )] represents
the nondimensional form of the nominal axial load strength of the section. The horizontal coordinate [ Rn = M n/( f c′ A g h)]
represents the nondimensional nominal bending moment strength of the section. The nondimensional forms were
used so that the interaction diagrams could be used with any system of units (SI or in.-lb units). Because ACI 318-14
contains different φ factors in Chapter 21, the strength reduction factor φ was considered as 1.0 so that the nominal
values in the interaction diagrams could be used with any set of φ factors.
Note that the
φ factors provided in Chapter 21 of ACI 318-14 are based on strain values in the tension reinforcement
farthest from the compression face of a member, or at the centroid of the tension reinforcement.
Also note the eccentricity ratios (e/h = M / P ), sometimes included as diagonal lines on interaction diagrams, are
not included in the interaction diagrams. Using the eccentricity ratio as a coordinate with K n or Rn could lead to
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 36/104
inaccuracies because the e/h lines converge rapidly at the lower ends of the diagrams. Straight lines for the tension
steel stress ratios f s/ f y have been plotted to assist in designing splices for the reinforcement. Further, the ratio
f s/ f y = 1.0 represents steel strain ε y = f y/ E s, which is the boundary point for the compression control φ factor, and the
beginning of the transition zone for linear increase of the φ factor to that for tension control.
To provide interpolation for the factor, other strain lines were plotted. The strain line for εt = 0.005, the
beginning of the zone for tension control, has been plotted on all diagrams. The intermediate strain line for
εt
= 0.0035
has been plotted for steel yield strength 60.0 ksi. The intermediate strain line for εt = 0.0038 has been plotted for steel
yield strength 75.0 ksi. Note that all strains refer to the reinforcing bar or bars farthest from the compression face of
the section. Discussions and tables related to the strength reduction factors are contained in two publications in
Concrete International (Everard 2002a,b).
Straight lines for K max are also provided on each interaction diagram. Here, K max refers to the maximum
permissible nominal axial load, P n,max, on a column that is laterally reinforced with ties or hoops (ACI 318-14, Section
22.4.2.1).
0.8 0.85n,max c g st y st
P f A A f A
(D.1a)
Then,
/max max c g K p f A
(D.1b)
For columns with spirals, values of K max from the interaction diagrams are multiplied by 0.85/0.80 ratio.
The number of longitudinal reinforcing bars is not limited to the number shown on the interaction diagrams. The
diagrams only illustrate the type of reinforcement patterns. However, for circular and square columns with steel
arranged in a circle, and for rectangular or square columns with steel equally distributed along all four faces, 12 bars
are preferred; using at least 8 bars is good practice. Although side steel was assumed to be 50 percent of the total steel
for columns having longitudinal steel equally distributed along all four faces, accurate and conservative designs can
result when side steel is only 30 percent of the total steel. The maximum number of bars used in any column cross
section is limited by the maximum allowable steel ratio of 0.08, the cover, and spacing between bars.
Tension axial loads are not included in the interaction diagrams.
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 37/104
REFERENCES
ACI Committee 340, 1970, “Ultimate Strength Design Handbook,” V. 2, Columns, ACI Special Publication 17 -
A, American Concrete Institute, Farmington Hills, MI, 226 pp.
Bresler, B., 1960, “Design Criteria for Reinforced Concrete Column under Axial Load and Biazial Bending,” ACI
JOURNAL Proceedings, V. 57, No. 5, Nov. pp. 481-490.
Column Research Council, 1966, “Guide to Design Criteria for Metal Compression Members,” second ed ition,
Fritz Engineering Laboratory, Lehigh University, Bethlehem, PA, 217 pp.
“Concrete Design Handbook,” 2005, third edition, Cement Association of Canada, Ottawa, ON.
Everard, N.J., 1997, “Axial Load-Moment Interaction for Cross-Sections Having Longitudinal Reinforcement
Arranged in a Circle”, ACI Structural Journal , V. 94, No. 6, Nov.-Dec., pp. 695-699.
Everard, N. J., 2002a, “Designing With ACI 318-02 Strength Reduction Factors,” Concrete International , V. 24,
No. 7, July, pp. 71-74.
Ever ard, N. J., 2002b, “Strain-Related Strength Reduction Factors (Φ) According to ACI 318-02, Concrete
International, Aug, V. 34, No. 8, pp. 91-93.
Everard, N. J., and Cohen, E., 1964, “Ultimate Strength Design of Reinforced Concrete Columns,” SP -7,
American Concrete Institute, Farmington Hills, MI, pp. 152-182.
MacGregor, J. G., 1997, Reinforced Concrete: Mechanics and Design, third edition, Prentice Hall, Englewood
Cliffs, New Jersey, 939 pp.
Parma, A. L.; Nieves, J. M.; and Gouwens, A., 1966, “Capacity of Reinforced Rectangular Columns Subject to
Biaxial Bending,” ACI JOURNAL Proceedings, V. 63, No. 9, Sept., pp. 911-923.
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 38/104
D.2 — Columns subject to biaxial bending
D.2.1 General — Most columns are subjected to significant bending in one direction while subjected to
relatively small bending moments in the orthogonal direction. These columns are designed using the interaction
diagrams discussed in the preceding section for uniaxial bending and, when required, checked for strength in the
orthogonal direction. Other columns, such as corner columns, are subjected to equally significant bending moments
in two orthogonal directions and, therefore, might have to be designed for biaxial bending.
A circular column subjected to moments about two axes may be designed as a uniaxial column acted upon by
the resultant moment
2 2 2 2
x ux uy n nx ny M M M M M (D.2.1)
For the design of rectangular columns subjected to moments about two axes, this Handbook provides design
aids for two methods: The reciprocal load (1/ P i) method (Bresler 1960) and the load contour method (Parme et al.
1966). The reciprocal load method is convenient for making a trial section analysis. The load contour method is
suitable for selecting a column cross section. Both methods use a failure surface to reflect the interaction of three
variables: the nominal axial load P n and the nominal biaxial bending moments M nx and M ny. In combination, these
variables cause failure strain at the extreme compression fiber; that is, the failure surface reflects the strength of
short compression members subject to biaxial bending and compression. Figure D.2.1a illustrates the bending axes,
eccentricities, and biaxial moments.
Fig. D.2.1a — Notation used for column sections subjected to biaxial bending.
A failure surface S 1 may be represented by P n, e x, and e y, as in Fig. D.2.1b, or it may be represented by surface
S 2 represented by P n, M nx, and M ny, as shown in Fig. D.2.1c. Note that S 1 is a single curvature surface having no
discontinuity at the balance point, whereas S 2 has discontinuity.
Note that when biaxial bending exists with a nominal axial force smaller than the lesser of P b or 0.1 f c′ A g , it is
sufficiently accurate and conservative to ignore the axial force and design the section for bending only.
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 39/104
Fig. D.2.1b — Failure surface S 1.
Fig. D.2.1c — Failure surface S 2.
D.2.2 Reciprocal load method — In the reciprocal load method, S 1 is inverted by plotting 1/ P n as the vertical
axis, giving S 3, shown in Fig. D.2.2a. As Figure D.2.2b shows, a true point (1/ P n1, e xA, e yB) on this reciprocal failure
surface can be approximated by a point (1/ P ni, e xA, e yB) on a plane S 3′ passing through points A, B, and C. Each
point is approximated by a different plane, that is, the entire failure surface is defined by an infinite number of
planes.
Point A represents the nominal axial load strength P ny when the load has an eccentricity of e xA with e y = 0. Point
B represents the nominal axial load strength P nx when the load has an eccentricity of e yB with e x = 0. Point C is
based on the axial strength P o with zero eccentricity. The equation of the plane passing through the three points is
0
1 1 1 1
ni nx ny P P P P (D.2.2a)
where
P ni = approximation of nominal axial load strength at eccentricities ex and ey
P nx = nominal axial load strength for eccentricity ey along the y-axis only (x-axis is axis of bending)
P ny = nominal axial load strength for eccentricity ex along the x-axis only (y-axis is axis of bending)
P 0 = nominal axial load strength for zero eccentricity
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 40/104
Fig. D.2.2a — Failure surface S 3 which is reciprocal
of surface S 1.
Fig. D.2.2b — Graphical representation of reciprocal
load method.
For design purposes, when φ is constant, 1/ P ni in Eq. (D.2.2a) can be used. The variable K n = P n /( f c′A g ) can be
used directly in the reciprocal equation, as follows (D.2.2b)
0
1 1 1 1
ni nx ny K K K K (D.2.2b)
where the K values refer to the corresponding P n values as defined above.
Once a preliminary cross section with an estimated steel ratio ρ g is selected, calculate Rnx and Rny using the
actual bending moments about the cross section x- and y-axes. Obtain the corresponding values of K nx and K ny from
the interaction diagrams presented in this chapter as the intersection of Rn value and the assumed steel ratio curve
for ρ g . Then, obtain the theoretical compression axial load strength K 0 at the intersection of the steel ratio curve and
the vertical axis for zero Rn.
D.2.3 Load contour method — The load contour method uses the failure surface S 2 (Fig. D.2.1c) and works with
a load contour defined by a plane at a constant value of P n (Fig. D.2.3a). The load contour defining the relationship
between M nx and M ny for a constant P n can be expressed nondimensionally as
1nynx
nox noy
M M
M M
(D.2.3a)
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 41/104
Fig. D.2.3a — Load contour constant Pn on failure
surface.
Fig. D.2.3b — Nondimensional load contour at constant
Pn.
For design, when each term is multiplied by , the equation is unchanged. Thus, M ux, M uy, M ox, and M oy, which
should correspond to M nx, M ny, M nox, and M noy, are used in the remainder of this section. To simplify the equation
for application, a point on the nondimensional diagram (Fig. D.2.3b) is defined such that the biaxial moment
capacities M nx and M ny are in the same ratio as the uniaxial moment capacities M ox and M oy; thus
nx ox
ny oy
M M
M M (D.2.3b)
or
nx ox ny oy M and M M (D.2.3c)
In a physical sense, the ratio β is the constant portion of the uniaxial moment capacities that may act
simultaneously on the column section. The actual value of β depends on the ratio P n/ P og , as well as properties of the
material and cross section. The usual range is 0.55 to 0.70 and an average value of 0.65 is suggested for design.
The load contour equation (Eq. (D.2.3a)) may be written in terms of β, as shown
l 0.5 /0.5/
1
og log log log
nynx
nox noy
M M
M M
(D.2.3d)
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 42/104
Figure D.2.3b illustrates the relationship using β. The true relationship between Points A, B, and C is a curve, but
for design purposes, it may be approximated by straight lines. The load contour equations as straight line
approximation are
For ny oy
nx ox
M M
M M ,
1oy
oy ny nx
ox
M M M
M
(D.2.3e)
For ny oy
nx ox
M M
M M ,
1ox
ox nx ny
oy
M M M
M
(D.2.3f)
For rectangular sections with reinforcement equally distributed on all four faces, the above equations can be
approximated by
1oy ny nx
b M M
h
(D.2.3g)
For ny oy
nx ox
M M
M M or ny
nx
M b
M h (D.2.3f)
where b and h are dimensions of the rectangular column section parallel to x- and y-axes, respectively. Using the
straight line approximation equations, the design problem can be solved by converting nominal moments into
equivalent uniaxial moment capacities M ox or M oy. This is accomplished by:
(a) Assuming a value for b/h
(b) Estimating β as 0.65
(c) Calculating the approximate equivalent uniaxial bending moment using Eq. (D.2.3e) or (D.2.3f)
(d) Choosing the trial section and reinforcement using the methods for uniaxial bending and axial load.
The trial section should be verified using the load contour method or the reciprocal load method.
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 43/104
D.3 — Column examples using interaction diagrams (D.5)
D.3.1 Column Example 1 — Determination of required area of steel for a rectangular tied column with bars on fourfaces with slenderness ratio below critical value.
For a rectangular tied column with bars equally distributed along fourfaces, find steel area.
Given:
Loading ––
P u = 560 kip, and M u = 3920 in.-kipAssume
φ = 0.70
Nominal axial load P n = 560 kip/0.70 = 800 kip
Nominal moment M n = 3920 in.-kip/0.70 = 5600 in.-kip
Materials –– Compressive strength of concrete f c′ = 4 ksiYield strength of reinforcement f y = 60 ksi
Normalized maximum size of aggregate is 1 in.
Design conditions –– Short column braced against sidesway
ACI 318-14
section Discussion Calculation Design aid
Determine column section size. Given: h = 20 in. and b = 16 in.
Determine reinforcement ratio ρ g using known values of variables
on appropriate interactiondiagrams and compute required
cross section area A st oflongitudinal reinforcement.
P n = 800 kip M n = 5600 in.-kip
h = 20 in.b = 16 in.
A g = b × h = 20 × 16 = 320 in.2
Computen
n
c g
f
P K
A
8000.625
(4)(320)n K
Computen
n
c g f
M R
A h
56000.22
(4)(320)(20)n R
Estimate5
h
20 50.75
20
10.5 Determine the appropriateinteraction diagrams.
For a rectangular tied column with bars along fourfaces, f c’ = 4 ksi, f y = 60 ksi, and an estimated γ of0.75, enter diagram R4-60.7 and R4-60.8 with
K n = 0.625 and Rn = 0.22, respectively.
Read ρ g for K n and Rn values from
appropriate interaction diagrams.
Read ρ g = 0.041 for γ = 0.7 and ρ g = 0.039 for
γ = 0.8. Interpolating ρ g = 0.040 for γ = 0.75
R4-60.7
R4-60.8
Compute required A st from
A st = ρ g A g . Required A st = 0.040 × 320 in.2 = 12.8 in.2
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 44/104
D.3.2 Column Example 2 — For a specified reinforcement ratio, selection of a column section size for a rectangulartied column with bars on end faces only.
For minimum longitudinal reinforcement (ρg= 0.01) and column sectiondimension h = 16 in., select the column dimension b for a rectangular tied
column with bars on end faces only.
Given:
Loading –– P u= 660 kip and M u= 2790 in.-kip
Assume φ = 0.70 Nominal axial load P n = 660 kip/0.70= 943 kip
Nominal moment M n = 4200 in.-kip/0.70= 3986 in.-kip
Materials –– Compressive strength of concrete f c’ = 4 ksi
Yield strength of reinforcement f y = 60 ksi Nominal maximum size of aggregate is 1 in.
Design conditions –– Slenderness effects may be neglected because k ℓ u/h is known to be below
critical value.
ACI 318-14
section Discussion Calculation Design
aid
Determine trial column dimensionb corresponding to known valuesof variables on appropriateinteraction diagrams.
P n = 943 kip M n = 3986 in.-kiph = 16 in.ρ g = 0.039
Assume a series of trial columnsizes b, in inches, and compute
A g = b x h, in.2
b 24 26 28
A
g 384 416 448
Compute’
c
n
g
n
f P K
A 943 0.61
(4)(384) 943 0.57
(4)(416) 943 0.53
(4)(448)
Compute’
c
n
g
n
f
M R
A h
39860.16
(4)(384)(16)
39860.15
(4)(416)(16)
39860.14
(4)(448)(16)
Estimate5
h
0.7 0.7 0.7
10.5 Determine the appropriate
interaction diagrams.
For a rectangular tied column with bars along four
faces, f c’ = 4 ksi, f y = 60 ksi, and an estimated γ of 0.70,
use diagram L4-60.7.Read ρ g for K n and Rn values forγ = 0.7, select dimension
corresponding to ρ g nearestdesired value of ρ g = 0.01.
0.018 0.014 0.011 L4-60.7
Therefore, try a 16 x 28-in. column
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 45/104
D.3.3 Example 3 — Selection of reinforcement for a square spiral column (slenderness ratio is below critical value).
For the square spiral column section shown, select reinforcement.
Given:
Loading –– P u= 660 kip and M u= 2640 in.-kip
Assume φ = 0.70
Nominal axial load P n = 660 kip/0.70= 943 kip Nominal moment M n = 2640 in.-kip/0.70= 3771 in.-kip
Materials ––
Compressive strength of concrete f c’ = 4 ksiYield strength of reinforcement f y = 60 ksi
Nominal maximum size of aggregate is 1 in.
Design conditions –– Column section size h = b = 18 inSlenderness effects may be neglected because k ℓ u/h is known to be
below critical value
ACI 318-14
section Discussion
Calculation
Design
aid Determine reinforcement ration g using known values of variables onappropriate interaction diagram(s)
and compute required cross sectionarea A st of longitudinal
reinforcement.
P n = 943 kip M n = 3771 in.-kip
A g = b x h = 18 x 18 = 324 in.2
Compute’
c
n
g
n
f
P K
A
9430.73
(4)(324)n K
Compute’
c
n
g
n
f
M R
A h
37710.16
(4)(324)(18)n
R
Estimate 5h 18 5 0.72
18
10.5 Determine the appropriate
interaction diagrams.
For a square spiral column, f c’ = 4 ksi, f y = 60 ksi, and
an estimated γ of 0.72, use diagram S4-60.7 and S4-60.8.
Read ρ g for K n and Rn. For K n = 0.73 and Rn = 0.16, and for:γ = 0.70 ρ g = 0.035
γ = 0.80 ρ g = 0.031γ = 0.72 ρ g = 0.034
A st = 0.034 x 324 in.2 = 11 in.2
S4-60.7S4-60.8
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 46/104
D.3.4 Example 4 — Design of square column section subject to biaxial bending using resultant moment
Select column section size and reinforcement for a square column with g 0.04and bars equally distributed along four faces, subject to biaxial bending.
Given:
Loading –– P u= 193 kip, M ux= 1917 in.-kip, and M uy= 769 in.-kip
Assume φ = 0.65
Nominal axial load P n = 193 kip/0.65= 297 kip
Nominal moment about x-axis M nx = 1917 in.-kip/0.65= 2949 in.-kip Nominal moment about y-axis M ny = 769 in.-kip/0.65= 1183 in.-kip
Materials ––
Compressive strength of concrete f c’ = 5 ksiYield strength of reinforcement f y = 60 ksi
Nominal maximum size of aggregate is 1 in.
ACI 318-14
section Discussion Calculation
Design
aid
Assume load contour curve atconstant P n is an ellipse, and
determine resultant moment M nx from
2 2
nr nx ny M M
For a square column: h = b 2 2
2949 1183 3177 in.-kipnr
Assume a series of trial columnsizes h, in inches.
14 16 18
Compute , in. g A h 196 256 324
Compute’
c
n
g
n
f
P K
A
2970.30
(5)(196)
2970.23
(5)(256)
2970.18
(5)(324)
Compute ’c
n
g
n
f
M
R A h
0.23
(5)(196)(14)
3177
0.16(5)(256)(16)
3177
0.11(5)(324)(18)
Estimate5
h
0.64 0.69 0.72
10.5 Determine the appropriateinteraction diagrams.
For a square tied column, f c’ = 5 ksi, f y = 60 ksi, usediagram R5-60.6, R5-60.7 and R5-60.8.
Read ρ g for K n and Rn values forγ = 0.60, 0.70, and 0.08.
Interpolate
0.064 0.030 0.0120.048 0.026 0.011
0.058 0.026 0.012
R5-60.6R5-60.7R5-60.8
Therefore, try h = 15 in.
Determine reinforcement ratio ρ g using known values of variables
on appropriate interactiondiagrams and compute required
cross section area A st oflongitudinal reinforcement.
A g = h2 = 152 = 225 in.2
P n = 297 kip
M n = 3177 in.-kip
Compute’
c
n
g
n
f
P K
A
2970.264
(5)(225)n K
Compute’
c
n
g
n
f
M R
A h
31770.188
(5)(225)(15)n
R
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 47/104
ACI 318-14
section Discussion Calculation
Design
aid
Estimate5
h
15 50.67
15
10.5 Determine the appropriateinteraction diagrams.
For a square tied column, f c’ = 5 ksi, f y = 60 ksi, usediagram R5-60.6 and R5-60.7.
Read ρ g for K n and Rn values forγ = 0.60 and 0.70.
For K n = 0.264 and Rn = 0.188:γ = 0.60 ρ g = 0.043γ = 0.70 ρ g = 0.034
γ = 0.37 ρ g = 0.037
R5-60.6R5-60.7
Compute required A st from A st = g A g and add about 15 percent for skew bending.
Required A st = 0.037 x 225 in.2 = 8.32 in.2
Use A st = 9.50 in.2
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 48/104
D.3.5 Example 5 — Design of circular spiral column section subject to very small design moment
For a circular spiral column, select column section diameter h and choose
reinforcement. Use relatively high proportion of longitudinal steel.
Given:
Loading –– P u = 940 kip and M u = 480 in.-kip
Assume
φ = 0.70
Nominal axial load P n = 940 kip/0.70 = 1343 kip
Nominal moment M n = 480 in.-kip/0.70 =686 in.-kip
Materials ––
Compressive strength of concrete c’ = 5 ksi
Yield strength of reinforcement f y = 60 ksi
Nominal maximum size of aggregate is 1 in.
Design condition –– Slenderness effects may be neglected because kℓ u/h is known to be below critical value
ACI 318-14
section Discussion Calculation Design
aid
Determine trial column dimensionb corresponding to known values
of variables on appropriateinteraction diagrams.
P n = 1343 kip M n = 686 in.-kip
ρ g = 0.04
Assume a series of trial columnsizes b, in inches.
12 16 20
Compute , in. g
A h 113 201 314
Compute’
c
n
g
n
f
M R
A h
6860.10
(5)(113)(12)
6860.04
(5)(201)(16)
6860.02
(5)(314)(20)
Estimate 5hh 0.58 0.69 0.75
10.5 Determine the appropriateinteraction diagrams.
For a circular column, f c’ = 5 ksi, f y = 60 ksi, usediagram C5-60.6, C5-60.7 and C5-60.8.
Read K n and Rn and ρ g values forγ = 0.60, 0.70, and 0.08.
Interpolate
No chart 1.08 1.230.9 1.18 1.25
0.9 1.17 1.24
C5-60.6C5-60.7C5-60.8
Compute’
c
n
g
n f K
P A 298 229 217
Compute 2 g A
h
19.5 17.1 16.6
Therefore, try 17 in. diameter column.
Determine reinforcement ratio ρ g using known values of variables
on appropriate interactiondiagrams and compute required
cross section area A st oflongitudinal reinforcement.
2
217227 in.
2 g A
Compute ’c
n
g
n
f
P K
A
1343
(5)(227)(17)1.18
n K
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 49/104
ACI 318-14
section Discussion Calculation
Design
aid
Compute’
c
n
g
n
f
M R
A h
1343
(5)(227)(17)0.0356
n R
Estimate5
h
17 50.71
17
10.5 Determine the appropriateinteraction diagrams.
For a circular column, f c’ = 5 ksi, f y = 60 ksi, usediagram C5-60.7.
Read ρ g for K n and Rn values. For K n = 1.18 and Rn = 0.0356:γ = 0.70 ρ g = 0.040
C5-60.7
Compute required A st from
A st = g A g . Required A st = 0.04 x 227 in.2 = 9.08 in.2
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 50/104
D.4 — Column design aids
Table D.4.1 — Effective Length Factor, Jackson and Moreland alignment chart for columns in braced (nonsway)
frames (Column Research Council 1966)
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 51/104
Table D.4.2 — Effective Length Factor, Jackson and Moreland alignment chart for columns in unbraced (sway)frames (Column Research Council 1966)
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 52/104
Table D.4.3 — Recommended flexural rigidities (EI) for use in first-order and second order analyses of frames fordesign of slender columns (ACI 318-14, Section 6.6.3.1.1,)
Second-order analysis of frames for design of slender columns
f c ′, ksi 3 4 5 6 7 8 9 10
I /I g
E c , ksi 3120 3605 4031 4415 4769 5098 5407 5700
E c I / I g , ksi
Beams 1092 1262 1411 1545 1669 1784 1892 1995 0.3
Columns 2184 2524 2822 3091 3338 3569 3785 3990 0.7
Walls (uncracked) 2184 2524 2822 3091 3338 3569 3785 3990 0.7
Walls (cracked) 1092 1262 1411 1545 1669 1784 1892 1995 0.3
Flat plates
Flat slabs780 901 1008 1104 1192 1275 1352 1425 0.2
Notes:
1. Alternatively, the following more refined values of I can be used:
For columns:
0.8 25 1 0.5 0.5 st u u
g
g u o
A M P I I Ig
A P h P
Where P u and M u are for the particular load combinations under consideration, or the combination of P u and M u that leads to the smallest value of I , I need not be less than 0.35 I g .
For beams:
= 0.10+25 (1.2−0.2
) ≤ 0.5
For continuous beams, I can be taken as the average of values for the critical positive and negative moment
sections. I need not be less than 0.25 I g .2. When sustained lateral loads are applied, I for columns should be divided by (1+ β ds), where β ds is the ratio of
maximum factored sustained shear within a story to the maximum factored story shear for the same loadcombination. β ds shall not be taken greater than 1.0.
3.
The above values are applicable to normal-density concrete with wc between 90 and 155 lb/ft3.
4. The moment of inertia of a T-beam should be based on the effective flange width as shown in Flexure 6. It isgenerally accurate to take I g of a T-beam as two times the I g for the web.
5. Member area will not be reduced for analysis.
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 53/104
Table D.4.4 — Effective length factor k for columns in braced frames (Concrete Design Handbook 2005)
T O P
Hinged
H i n g e d
Elastic
E l a s t i c
Elastic
E l a s t i c
Stiff
S t i f f
BOTTOM
0.81 0.91 0.95 1.00
0.77 0.86 0.90 0.95
0.67 0.74 0.77 0.81
0.74 0.83 0.86 0.91
k
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 54/104
Table D.4.5 — Moment of inertia of reinforcement about sectional centroid (Based on Table 12-1, MacGregor1997)
I = 0.25 b h se3 2
Bars in two end facesEqual reinforcement on four
sides
t I = 0.18 b h se3 2
t
Bars in two side faces
I = 0.17 b h se3 2
t (3 bars per face)
I = 0.12 b h se3 2
t(6 bars per face)
I = 0.10 h se4 2
t
Uniformly distributed
reinforcement
I = 0.13 b h se3 2
t
Bars uniformly spaced on all sides
Bars uniformly spaced on all sides
I = 0.22 b h se3 2
t
b b
b
b b = 2h
h h
h
h
h h
h
h
h
hh
h = 2b
γ is the ratio of the distance between the centers of the outermost bars to the column dimension perpendicular to theaxis of bending.
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 55/104
D.5 — Interaction diagrams
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.450.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
COLUMNS 3.1.1 - Nominal load-moment strength interaction diagram, R3-60.6
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R3-60.6
Pne
h
h
= 0.6
f /
c = 3 ksi
f y = 60 ksi
t = 0 .0 0 3 5 t = 0 .0 0 5
t = 0 .0 0 4
COLUMNS 3.1.2 - Nominal load-moment strength interaction diagram, R3-60.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
K n
=
P n
/ f /
c
A g
Rn
= Pn e / f
/
c A
gh
INTERACTION DIAGRAM R3-60.7
Pne
h
h
= 0.7
f /
c = 3 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 56/104
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.600.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n =
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R3-60.8
Pne
h
h
= 0.8
f /
c = 3 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.650.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
COLUMNS 3.1.4 - Nominal load-moment strength interaction diagram, R3-60.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R3-60.9
Pne
h
h
= 0.9
f /
c = 3 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 57/104
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.350.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R4-60.6
Pne
h
h
= 0.6
f /
c = 4 ksi
f y = 60 ksi
t = 0 .0 0 5
t = 0
.0 0 3 5
t = 0 .0 0 4 0
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.400.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
COLUMNS 3.2.2 - Nominal load-moment strength interaction diagram, R4-60.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n = P
n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R4-60.7
Pne
h
h
= 0.7
f /
c = 4 ksi
f y = 60 ksi
t = 0 .0 0 3 5 t = 0 .0 0 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 58/104
COLUMNS 3.2.3 - Nominal load-moment strength interaction diagram, R4-60.8
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.450.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R4-60.8
Pne
h
h
= 0.8
f /
c = 4 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
COLUMNS 3.2.4 - Nominal load-moment strength interaction diagram, R4-60.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
K
n =
P n
/ f / c
A g
Rn = Pn e / f/
c Ag h
INTERACTION DIAGRAM R4-60.9
Pne
h
h
= 0.9
f /
c = 4 ksi
f y = 60 ksi
t = 0 .0 0 5
t = 0 .0 0 3 5 t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 59/104
0.00 0.05 0.10 0.15 0.20 0.25 0.300.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n =
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R5-60.6
Pne
h
h
= 0.6
f /
c = 5 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.350.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
COLUMNS 3.3.2 - Nominal load-moment strength interaction diagram, R5-60.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R5-60.7
Pne
h
h
= 0.7
f /
c = 5 ksi
f y = 60 ksi
t = 0 .0 0 5
t = 0 .0 0 3 5 t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 60/104
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.400.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
f s/f y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
Rn = P
n e / f
/
c A
gh
K n =
P n
/ f / c
A g
INTERACTION DIAGRAM R5-60.8
Pne
h
h
= 0.8
f /
c = 5 ksi
f y = 60 ksi
t = 0 .0 0 5
t = 0 .0 0 3 5 t = 0 .0 0 4 0
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.450.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
COLUMNS 3.3.4 - Nominal load-moment strength interaction diagram, R5-60.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n =
P n
/ f /
c
A g
Rn
= Pn e / f
/
c A
gh
INTERACTION DIAGRAM R5-60.9
Pne
h
h
= 0.9
f /
c = 5 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 61/104
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 0.2750.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R6-60.6
Pne
h
h
= 0.6
f /
c = 6 ksi
f y = 60 ksi
t = 0 .0 0 5
t = 0 .0 0 3 5
t = 0 .0 0 4 0
COLUMNS 3.4.2 - Nominal load-moment strength interaction diagram, R6-60.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.00 0.05 0.10 0.15 0.20 0.25 0.300.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R6-60.7
Pne
h
h
= 0.7
f /
c = 6 ksi
f y = 60 ksi
t = 0 .0 0 3 5 t = 0 .0 0 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 62/104
COLUMNS 3.4.3 - Nominal load-moment strength interaction diagram, R6-60.8
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.350.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R6-60.8
Pne
h
h
= 0.8
f /
c = 6 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.400.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
COLUMNS 3.4.4 - Nominal load-moment strength interaction diagram, R6-60.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R6-60.9
Pne
h
h
= 0.9
f /
c = 6 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 63/104
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.2000.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R9-75.6
Pne
h
h
= 0.6
f /
c = 9 ksi
f y = 75 ksi
t = 0 .0 0 3 8 t = 0 .0 0 5
t = 0 .0 0 4 0
COLUMNS 3.5.2 - Nominal load-moment strength interaction diagram, R9-75.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.2500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R9-75.7
Pne
h
h
= 0.7
f /
c = 9 ksi
f y = 75 ksi
t = 0 .0 0 5 0
t = 0 .0 0 3 8
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 64/104
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.300.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R9-75.8
Pne
h
h
= 0.8
f /
c = 9 ksi
f y = 75 ksi
t = 0 .0 0 5 0
t = 0 .0 0 3 8
t = 0 .0 0 4 0
0.00 0.05 0.10 0.15 0.20 0.25 0.300.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
COLUMNS 3.5.4 - Nominal load-moment strength interaction diagram, R9-75.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n =
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R9-75.9
Pne
h
h
= 0.9
f /
c = 9 ksi
f y = 75 ksi
t = 0 .0 0 3 8
t = 0 .0 0 5 0
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 65/104
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.1750.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R12-75.6
Pne
h
h
= 0.6
f /
c = 12 ksi
f y = 75 ksi
t = 0 .0 0 3 8 t = 0 .0 0 5 0
t = 0 .0 0 4 0
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.2000.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
COLUMNS 3.6.2 - Nominal load-moment strength interaction diagram, R12-75.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R12-75.7
Pne
h
h
= 0.7
f /
c = 12 ksi
f y = 75 ksi
t = 0 .0 0 5 0
t = 0 .0 0 3 8
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 66/104
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.2250.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R12-75.8
Pne
h
h
= 0.8
f /
c = 12 ksi
f y = 75 ksi
t = 0 .0 0 3 8 t = 0 .0
0 5 0
t = 0 .0 0 4 0
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.2500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
COLUMNS 3.6.4 - Nominal load-moment strength interaction diagram, R12-75.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM R12-75.9
Pne
h
h
= 0.9
f /
c = 12 ksi
f y = 75 ksi
t = 0 .0 0 3 8
t = 0 .0 0 5 0
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 67/104
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.600.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L3-60.6
Pne
h
h
= 0.6
f /
c = 3 ksi
f y = 60 ksi
t = 0 .0 0 3 5 t = 0 .0 0 5
t = 0 .0 0 4
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.70.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
COLUMNS 3.7.2 - Nominal load-moment strength interaction diagram, L3-60.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L3-60.7
Pne
h
h
= 0.7
f /
c = 3 ksi
f y = 60 ksi
t = 0 .0 0 3 5 t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 68/104
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
f s/f y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n =
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L3-60.8
Pne
h
h
= 0.8
f /
c = 3 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
COLUMNS 3.7.4 - Nominal load-moment strength interaction diagram, L3-60.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L3-60.9
Pne
h
h
= 0.9
f /
c = 3 ksi
f y = 60 ksi
t = 0.0035
t = 0.005 t = 0.004
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 69/104
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.450.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L4-60.6
Pne
h
h
= 0.6
f /
c = 4 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
COLUMNS 3.8.2 - Nominal load-moment strength interaction diagram, L4-60.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.550.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L4-60.7
Pne
h
h
= 0.7
f /
c = 4 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 70/104
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.600.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L4-60.8
Pne
h
h
= 0.8
f /
c = 4 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
COLUMNS 3.8.4 - Nominal load-moment strength interaction diagram, L4-60.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.700.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L4-60.9
Pne
h
h
= 0.9
f /
c = 4 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 71/104
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.400.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L5-60.6
Pne
h
h
= 0.6
f /
c = 5 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
COLUMNS 3.9.2 - Nominal load-moment strength interaction diagram, L5-60.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.450.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L5-60.7
Pne
h
h
= 0.7
f /
c = 5 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 72/104
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
f s/f y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L5-60.8
Pne
h
h
= 0.8
f /
c = 5 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
COLUMNS 3.9.4 - Nominal load-moment strength interaction diagram, L5-60.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.550.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
K n =
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L5-60.9
Pne
h
h
= 0.9
f /
c = 5 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 73/104
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.350.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L6-60.6
Pne
h
h
= 0.6
f /
c = 6 ksi
f y = 60 ksi
t = 0 .0 0 3 5 t = 0 .0 0 5
t = 0 .0 0 4
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.400.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
COLUMNS 3.10.2 - Nominal load-moment strength interaction diagram, L6-60.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n
=
P n
/ f /
c
A g
Rn = P
n e / f /
c A
gh
INTERACTION DIAGRAM L6-60.7
Pne
h
h
= 0.7
f /
c = 6 ksi
f y = 60 ksi
t = 0 .0 0 3 5 t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 74/104
f s/f y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.450.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L6-60.8
Pne
h
h
= 0.8
f /
c = 6 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
COLUMNS 3.10.4 - Nominal load-moment strength interaction diagram, L6-60.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
g = 0.08
0.01
0.02
0.03
0.04
0.05
0.06
0.07
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L6-60.9
Pne
h
h
= 0.9
f /
c = 6 ksi
f y = 60 ksi
t = 0 .0 0 5
t = 0 .0 0 3 5 t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 75/104
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.2500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L9-75.6
Pne
h
h
= 0.6
f /
c = 9 ksi
f y = 75 ksi
t = 0 .0 0 5
t = 0 .0 0 3 8
t = 0 .0 0 4
0.00 0.05 0.10 0.15 0.20 0.25 0.300.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
COLUMNS 3.11.2 - Nominal load-moment strength interaction diagram, L9-75.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L9-75.7
Pne
h
h
= 0.7
f /
c = 9 ksi
f y = 75 ksi
t = 0 .0 0 5
t = 0 .0 0 3 8 t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 76/104
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.350.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n =
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L9-75.8
Pne
h
h
= 0.8
f /
c = 9 ksi
f y = 75 ksi
t = 0 .0 0 5
t = 0 .0 0 3 8 t = 0 .0 0 4
COLUMNS 3.11.4 - Nominal load-moment strength interaction diagram, L9-75.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.400.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L9-75.9
Pne
h
h
= 0.9
f /
c = 9 ksi
f y = 75 ksi
t = 0 .0 0 5
t = 0 .0 0 3 8 t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 77/104
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.2250.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L12-75.6
Pne
h
h
= 0.6
f /
c = 12 ksi
f y = 75 ksi
t = 0 .0 0 3
8
t = 0 .0 0 5
t = 0 .0 0 4
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.2500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
COLUMNS 3.12.2 - Nominal load-moment strength interaction diagram, L12-75.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L12-75.7
Pne
h
h
= 0.7
f /
c = 12 ksi
f y = 75 ksi
t = 0 .0 0 3 8 t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 78/104
0.01
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
g = 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.300.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L12-75.8
Pne
h
h
= 0.8
f /
c = 12 ksi
f y = 75 ksi
t = 0 .0 0 3 8
t = 0 .0 0 5
t = 0 .0 0 4
COLUMNS 3.12.4 - Nominal load-moment strength interaction diagram, L12-75.9
0.01
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
g = 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.350.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM L12-75.9
Pne
h
h
= 0.9
f /
c = 12 ksi
f y = 75 ksi
t = 0 .0 0 3 8
t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 79/104
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.350.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
f s/f y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
= 0.6
INTERACTION DIAGRAM C3-60.6
f /
c = 3 ksi
f y = 60 ksi
Pne
h
h
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
t = 0 .0 0 3 5 t = 0 .0 0 5
t = 0 .0 0 4
COLUMNS 3.13.2 - Nominal load-moment strength interaction diagram, C3-60.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.400.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
g = 0.08
K n =
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
= 0.7
INTERACTION DIAGRAM C3-60.7
f /
c = 3 ksi
f y = 60 ksi
Pne
h
h
t = 0 .0 0 3 5 t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 80/104
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g= 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.450.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
= 0.8
INTERACTION DIAGRAM C3-60.8
f /
c = 3 ksi
f y = 60 ksi
Pne
h
h
t = 0 .0 0 3 5 t = 0 .0 0 5
t = 0 .0 0 4
COLUMNS 3.13.4 - Nominal load-moment strength interaction diagram, C3-60.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.550.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
K n =
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
g = 0.08
= 0.9
INTERACTION DIAGRAM C3-60.9
f /
c = 3 ksi
f y = 60 ksi
Pne
h
h
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 81/104
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 0.2750.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
= 0.6
INTERACTION DIAGRAM C4-60.6
f /
c = 4 ksi
f y = 60 ksi
Pne
h
h
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
COLUMNS 3.14.2 - Nominal load-moment strength interaction diagram, C4-60.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.350.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
g = 0.08
= 0.7
INTERACTION DIAGRAM C4-60.7
f /
c = 4 ksi
f y = 60 ksi
Pne
h
h
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 82/104
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.400.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
g = 0.08
= 0.8
INTERACTION DIAGRAM C4-60.8
f /
c = 4 ksi
f y = 60 ksi
Pne
h
h
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.400.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
COLUMNS 3.14.4 - Nominal load-moment strength interaction diagram, C4-60.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
g = 0.08
= 0.9
INTERACTION DIAGRAM C4-60.9
f /
c = 4 ksi
f y = 60 ksi
Pne
h
h
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 83/104
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.2250.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
f s/f y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
= 0.6
INTERACTION DIAGRAM C5-60.6
f /
c = 5 ksi
f y = 60 ksi
Pne
h
h
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
0.00 0.05 0.10 0.15 0.20 0.25 0.300.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
COLUMNS 3.15.2 - Nominal load-moment strength interaction diagram, C5-60.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g= 0.08
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
= 0.7
INTERACTION DIAGRAM C5-60.7
f /
c = 5 ksi
f y = 60 ksi
Pne
h
h
t = 0 .0 0 3 5 t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 84/104
0.00 0.05 0.10 0.15 0.20 0.25 0.300.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
. . , .
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
= 0.8
INTERACTION DIAGRAM C5-60.8
f /
c = 5 ksi
f y = 60 ksi
Pne
h
h
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
COLUMNS 3.15.4 - Nominal load-moment strength interaction diagram, C5-60.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.350.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
= 0.9
INTERACTION DIAGRAM C5-60.9
f /
c = 5 ksi
f y = 60 ksi
Pne
h
h
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 85/104
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.2250.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
= 0.6
INTERACTION DIAGRAM C6-60.6
f /
c = 6 ksi
f y = 60 ksi
Pne
h
h
t = 0 .0 0 3 5 t = 0 .0 0 5
t = 0 .0 0 4
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.2500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
COLUMNS 3.16.2 - Nominal load-moment strength interaction diagram, C6-60.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
= 0.7
INTERACTION DIAGRAM C6-60.7
f /
c = 6 ksi
f y = 60 ksi
Pne
h
h
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 86/104
0.00 0.05 0.10 0.15 0.20 0.25 0.300.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
= 0.8
INTERACTION DIAGRAM C6-60.8
f /
c = 6 ksi
f y = 60 ksi
Pne
h
h
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
0.00 0.05 0.10 0.15 0.20 0.25 0.300.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
COLUMNS 3.16.4 - Nominal load-moment strength interaction diagram, C6-60.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
= 0.9
INTERACTION DIAGRAM C6-60.9
f /
c = 6 ksi
f y = 60 ksi
Pne
h
h
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 87/104
0.000 0.025 0.050 0.075 0.100 0.125 0.1500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
= 0.6
INTERACTION DIAGRAM C9-75.6
f /
c = 9 ksi
f y = 75 ksi
Pne
h
h
t = 0 .0 0 3 8 t = 0 .0 0 4
t = 0 .0 0 5
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.1750.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
COLUMNS 3.17.2 - Nominal load-moment strength interaction diagram, C9-75.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
= 0.7
INTERACTION DIAGRAM C9-75.7
f /
c = 9 ksi
f y = 75 ksi
Pne
h
h
t = 0 .0 0 3 8 t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 88/104
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
g = 0.08
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.2250.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
= 0.8
INTERACTION DIAGRAM C9-75.8
f /
c = 9 ksi
f y = 75 ksi
Pne
h
h
t = 0 .0 0 3 8
t = 0 .0 0 5
t = 0 .0 0 4
0.03
0.02
0.01
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.2500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
COLUMNS 3.17.4 - Nominal load-moment strength interaction diagram, C9-75.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
= 0.9
INTERACTION DIAGRAM C9-75.9
f /
c = 9 ksi
f y = 75 ksi
Pne
h
h
t = 0 .0 0 5
t = 0 .0 0 3 8 t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 89/104
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0.000 0.025 0.050 0.075 0.100 0.125 0.1500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
g = 0.08 = 0.6
INTERACTION DIAGRAM C12-75.6
f /
c = 12 ksi
f y = 75 ksi
Pne
h
h
t = 0 .0 0 3 8 t = 0 .0 0 5
t = 0 .0 0 4
COLUMNS 3.18.2 - Nominal load-moment strength interaction diagram, C12-75.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.070.06
0.05
0.04
0.03
0.02
0.01
0.000 0.025 0.050 0.075 0.100 0.125 0.1500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
g = 0.08
= 0.7
INTERACTION DIAGRAM C12-75.7
f /
c = 12 ksi
f y = 75 ksi
Pne
h
h
t = 0 .0 0 3 8 t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 90/104
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.1750.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
g = 0.08
= 0.8
INTERACTION DIAGRAM C12-75.8
f /
c = 12 ksi
f y = 75 ksi
Pne
h
h
t = 0 .0 0 3 8
t = 0 .0 0 5
t = 0 .0 0 4
COLUMNS 3.18.4 - Nominal load-moment strength interaction diagram, C12-75.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.2000.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
= 0.9
INTERACTION DIAGRAM C12-75.9
f /
c = 12 ksi
f y = 75 ksi
Pne
h
h
t = 0 .0 0 3 8 t = 0 .0 0 5
t = 0 .0 0 4
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 91/104
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.350.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S3-60.6
Pne
h
h
= 0.6
f /
c = 3 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
COLUMNS 3.19.2 - Nominal load-moment strength interaction diagram, S3-60.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.400.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S3-60.7
Pne
h
h
= 0.7
f /
c = 3 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 92/104
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
K max
f s/f
y = 0
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S3-60.8
Pne
h
h
= 0.8
f /
c = 3 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
COLUMNS 3.19.4 - Nominal load-moment strength interaction diagram, S3-60.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.550.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S3-60.9
Pne
h
h
= 0.9
f /
c = 3 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 93/104
f s/f y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.300.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S4-60.6
Pne
h
h
= 0.6
f /
c = 4 ksi
f y = 60 ksi
t =
0 .0 0 3 5 t = 0 .0 0 5
t = 0 .0 0 4 0
COLUMNS 3.20.2 - Nominal load-moment strength interaction diagram, S4-60.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.350.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S4-60.7
Pne
h
h
= 0.7
f /
c = 4 ksi
f y = 60 ksi
t = 0 .0 0 3 5 t = 0 .0 0 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 94/104
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.400.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
K n =
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S4-60.8
Pne
h
h
= 0.8
f /
c = 4 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
COLUMNS 3.20.4 - Nominal load-moment strength interaction diagram, S4-60.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.450.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S4-60.9
Pne
h
h
= 0.9
f /
c = 4 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 95/104
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.2500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S5-60.6
Pne
h
h
= 0.6
f /
c = 5 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
COLUMNS 3.21.2 - Nominal load-moment strength interaction diagram, S5-60.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.300.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S5-60.7
Pne
h
h
= 0.7
f /
c = 5 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 96/104
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.350.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S5-60.8
Pne
h
h
= 0.8
f /
c = 5 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
COLUMNS 3.21.4 - Nominal load-moment strength interaction diagram, S5-60.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.350.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S5-60.9
Pne
h
h
= 0.9
f /
c = 5 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 97/104
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.2500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S6-60.6
Pne
h
h
= 0.6
f /
c = 6 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
COLUMNS 3.22.2 - Nominal load-moment strength interaction diagram, S6-60.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 0.2750.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S6-60.7
Pne
h
h
= 0.7
f /
c = 6 ksi
f y = 60 ksi
t = 0 .0 0 5
t = 0 .0 0 3 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 98/104
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.300.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S6-60.8
Pne
h
h
= 0.8
f /
c = 6 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
COLUMNS 3.22.4 - Nominal load-moment strength interaction diagram, S6-60.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.350.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S6-60.9
Pne
h
h
= 0.9
f /
c = 6 ksi
f y = 60 ksi
t = 0 .0 0 3 5
t = 0 .0 0 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 99/104
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.1750.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S9-75.6
Pne
h
h
= 0.6
f /
c = 9 ksi
f y = 75 ksi
t = 0 .0 0 3 8 t = 0 .0 0 5
t =
0 .0 0 4 0
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.2000.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
COLUMNS 3.23.2 - Nominal load-moment strength interaction diagram, S9-75.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S9-75.7
Pne
h
h
= 0.7
f /
c = 9 ksi
f y = 75 ksi
t = 0 .0 0 3 8 t = 0 .0 0 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 100/104
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.2250.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S9-75.8
Pne
h
h
= 0.8
f /
c = 9 ksi
f y = 75 ksi
t = 0 .0 0 3 8
t = 0 .0 0 5
t = 0 .0 0 4 0
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.2500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
COLUMNS 3.23.4 - Nominal load-moment strength interaction diagram, S9-75.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S9-75.9
Pne
h
h
= 0.9
f /
c = 9 ksi
f y = 75 ksi
t = 0 .0 0 3 8
t = 0 .0 0 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 101/104
0.000 0.025 0.050 0.075 0.100 0.125 0.1500.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S12-75.6
Pne
h
h
= 0.6
f /
c = 12 ksi
f y = 75 ksi
t = 0 .0 0 3 8 t = 0 .0 0 5 0
t = 0 .0 0 4 0
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.1750.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
COLUMNS 3.24.2 - Nominal load-moment strength interaction diagram, S12-75.7
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S12-75.7
Pne
h
h
= 0.7
f /
c = 12 ksi
f y = 75 ksi
t = 0 .0 0 3 8 t = 0 .0 0 5
t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 102/104
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.2000.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n
=
P n
/ f /
c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S12-75.8
Pne
h
h
= 0.8
f /
c = 12 ksi
f y = 75 ksi
t = 0 .0 0 5
t = 0 .0 0 3 8 t = 0 .0 0 4 0
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.2250.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
COLUMNS 3.24.4 - Nominal load-moment strength interaction diagram, S12-75.9
f s/f
y = 0
K max
0.25
0.50
0.75
1.0
0.07
0.06
0.05
0.04
0.03
0.02
0.01
g = 0.08
K n =
P n
/ f / c
A g
Rn = P
n e / f
/
c A
gh
INTERACTION DIAGRAM S12-75.9
Pne
h
h
= 0.9
f /
c = 12 ksi
f y = 75 ksi
t = 0 .0 0 5
t = 0 .0 0 3 8 t = 0 .0 0 4 0
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 103/104
7/25/2019 SP-17-14_DA
http://slidepdf.com/reader/full/sp-17-14da 104/104
38800 Country Club Drive
Farmington Hills, MI 48331 USA
+1.248.848.3700
www.concrete.org