An ACI Handbook The Reinforced Concrete Design Handbook

An ACI Handbook

The Reinforced Concrete Design Handbook A Companion to ACI 318-14

Volume 3: Design Aids SP-17(14)

ACI SP-17(14)

THE REINFORCED CONCRETE DESIGN HANDBOOK A Companion to ACI 318-14

VOLUME 2

VOLUME 1 BUILDING EXAMPLE

RETAINING WALLS

STRUCTURAL SYSTEMS

SERVICEABILITY

STRUCTURAL ANALYSIS

STRUT-AND-TIE MODEL

DURABILITY

ANCHORING TO CONCRETE

ONE-WAY SLABS TWO-WAY SLABS BEAMS DIAPHRAGMS COLUMNS STRUCTURAL REINFORCED CONCRETE WALLS FOUNDATIONS

ACI SP-17(14) Volume 3

THE REINFORCED CONCRETE DESIGN HANDBOOK A Companion to ACI 318-14

Editors: Andrew Taylor Trey Hamilton III Antonio Nanni

First Printing September 2015

THE REINFORCED CONCRETE DESIGN HANDBOOK Volume 3 ~ Ninth Edition Copyright by the American Concrete Institute, Farmington Hills, MI. All rights reserved. This material may not be reproduced or copied, in whole or part, in any printed, mechanical, electronic, film, or other distribution and storage media, 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 find information 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 errata website at http://concrete.org/Publications/DocumentErrata.aspx. Proper use of this document includes periodically checking 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 and limitations of its content and recommendations and who will accept responsibility for the application of the material it contains. Individuals who use this publication in any way assume all risk and accept total responsibility for the application 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 noninfringement. ACI and its members disclaim liability for damages of any kind, including any special, indirect, incidental, or consequential damages, including without limitation, lost revenues or lost profits, which may result from the use of this publication. It is the responsibility of the user of this document to establish health and safety practices appropriate to the specific circumstances involved with its use. ACI does not make any representations with regard to health and safety issues and the use of this document. The user must determine the applicability of all regulatory limitations before applying the document and must comply with all applicable laws and regulations, including but not limited to, United States Occupational Safety and Health Administration (OSHA) health and safety standards. Participation by governmental representatives in the work of the American Concrete Institute and in the development of Institute standards does not constitute governmental endorsement of ACI or the standards that it develops. 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 together in the annually revised ACI Manual of Concrete Practice (MCP). American Concrete Institute 38800 Country Club Drive Farmington Hills, MI 48331 USA +1.248.848.3700 Managing Editor: Khaled Nahlawi Staff Engineers: Daniel W. Falconer, Matthew R. Senecal, Gregory M. Zeisler, and Jerzy Z. Zemajtis Technical Editors: Shannon B. Banchero, Emily H. Bush, and Cherrie L. Fergusson Manager, Publishing Services: Barry Bergin Lead Production Editor: Carl Bischof Production Editors: Kelli Slayden, Kaitlyn Hinman, Tiesha Elam Graphic Designers: Ryan Jay, Aimee Kahaian Manufacturing: Marie Fuller www.concrete.org

VOLUME 3: CONTENTS APPENDIX A—REFERENCE TABLES

1

APPENDIX B—ANALYSIS TABLES

11

APPENDIX C—SECTIONAL PROPERTIES

25

APPENDIX D––COLUMN INTERACTION DIAGRAMS

27

REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)

1

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.

No. 3

0.11

0.376

0.375

1.18

No. 4

0.20

0.668

0.500

1.57

Nominal perimeter, in.

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.

Table A-2—Area of bars in a section 1 ft wide Bar size

Spacing, in.

#3

#4

#5

#6

#7

#8

#9

#10

#11

#14

#18

Spacing, in.

4.0

0.33

0.60

0.93

1.32

1.80

2.37

3.00

3.81

4.5

0.29

0.53

0.83

1.17

1.60

2.11

2.67

3.39

4.68

—

—

4.0

4.16

6.00

—

5.0

0.26

0.48

0.74

1.06

1.44

1.90

2.40

4.5

3.05

3.74

5.40

9.60

5.0

5.5

0.24

0.44

0.68

0.96

1.31

1.72

6.0

0.22

0.40

0.62

0.88

1.20

1.58

2.18

2.77

3.40

4.91

8.73

5.5

2.00

2.54

3.12

4.50

8.00

6.5

0.20

0.37

0.57

0.81

1.11

6.0

1.46

1.85

2.34

2.88

4.15

7.38

7.0

0.19

0.34

0.53

0.75

6.5

1.03

1.35

1.71

2.18

2.67

3.86

6.86

7.5

0.18

0.32

0.50

7.0

0.70

0.96

1.26

1.60

2.03

2.50

3.60

6.40

8.0

0.17

0.30

7.5

0.47

0.66

0.90

1.19

1.50

1.91

2.34

3.38

6.00

8.5

0.16

8.0

0.28

0.44

0.62

0.85

1.12

1.41

1.79

2.20

3.18

5.65

9.0

8.5

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

14.0

15.0

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

Example: #9 bar spaced 7-1/2 in. apart provides 1.60 in.2/ft of section width.

American Concrete Institute Copyrighted Material—www.concrete.org

REF. TABLES

Cross section area of bar As (or As′ ), in.2

2


Table A-3—Minimum beam web widths required for two or more bars in one layer for cast-in-place nonprestressed 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 + db) where A + B + C - 1/2 db ≥ 2.0 in. cover required for longitudinal bars and these assumptions are made: for ACI 318 B = 0.375 in. for #3 stirrups = 0.500 in. for #4 stirrups D = 1 db ≥ 1 in. ≥ 1-1/3 nominal aggregate size

For both ACI and AASHTO

For AASHTO

A = 1-1/2 in. concrete cover to stirrup B = 0.635 in. for #5 stirrups = 0.750 in. for #6 stirrups

B = 0.375 in. for #3 stirrups (minimum stirrup size for #10 and smaller longitudinal bars) = 0.500 in. for #4 stirrups (minimum stirrup size for #11 and larger longitudinal bars)

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 ≥ 1/2db of longitudinal bars

D = 1-1/2db ≥ 1-1/2 in. ≥ 1-1/2 nominal aggregate size

REF. TABLES

ACI 318-11 3/4-in. aggregate interior exposure #3 stirrups

ACI 318-11 1-in. aggregate interior exposure #3 stirrups

AASHTO requirements cast-in-place concrete 1-in. aggregate exposed to earth or weather

Bar size

Minimum web 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

#5

6 7/8

1 5/8

7 1/4

2

7.37

2.000 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

Notes 2. ACI cover requirements: For exterior 5. Example: Find the minimum web width for a 1. Stirrups: For stirrups larger than those used exposure with use of #6 or larger stirrups, add 1 in. beam reinforced with two #8 bars; a beam for table above, increase web width by the follow- to web width. reinforced with three #8 bars; a beam reinforced ing amounts (in.): 3. AASHTO cover requirements: For interior with three #9 and two #6 bars. exposure, 1/2 in. may be deducted from beam 2 #8 3 #8 3 #9 + 2 #6 Main widths. reinforcement #4 #5 #6 ACI (3/4 in. aggregate) 7 1/4 9 1/4 13 1/4 4. Bars of different sizes: For beams with bars Source size stirrup stirrup stirrup ACI (1 in. aggregate) 7 5/8 10 14 1/2 #4 through #11 3/4 1 1/2 2 1/4 of two or more sizes, determine from table the beam ACI AASHTO 7.75 10.25 15.64 web width required for the given number of largest #14 1/2 1 1/4 2 requirements size bars; then add the indicated increments for each #18 1/4 3/4 1 1/2 smaller bar. #4 through #10 AASHTO #11 through #14 requirements #18

0.75

1.50

2.25

—

0.75

1.50

—

0.49

1.24



3

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)

Columns at left headed 1-5 and 6-10 bars are for 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 outside face(s) of beam. Aggregate size assumed ≤ 3/4 in.

No. of Bar 1 to 5 6 to 10 bars size bars bars A = clear cover of 1-1/2 in. 1 5.5 12.5 B = 3/4 in. diameter of #3 stirrups ACI min. bw, in. 2 7.0 13.5 C = for #11 and smaller bars: twice diameter Size 3 8.0 15.0 #3 of #3 stirrups; for #14 and #18 bars: 1/2 of diameter of bar 4 9.5 16.5 smaller No. of smaller bars D = 1/2 diameter of larger bar 5 11.0 18.0 1 2 3 4 5 bars E = 1/2 spacing for larger bar plus 1/2 spacing for smaller bar (spacing is db for #9 1 5.5 13.0 7.0 8.5 9.5 11.0 12.5 and larger bars, 1 in. for #8 and smaller ACI min. bw, in. 2 7.0 14.5 8.5 9.5 11.0 12.5 14.0 bars) Size 3 8.5 16.0 10.0 11.0 12.5 14.0 15.5 #4 #3 of 4 10.0 17.5 11.5 12.5 14.0 15.5 17.0 smaller No. of smaller bars 5 11.5 19.0 13.0 14.0 15.5 17.0 18.5 1 2 3 4 5 bars 1 5.5 13.5 7.0 8.5 10.0 11.5 13.0 7.0 8.5 9.5 11.0 12.5 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 Size 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 #5 #4 #3 of 4 10.5 18.5 12.0 13.5 15.0 16.5 18.0 11.5 13.0 14.5 16.0 17.0 smaller No. of smaller bars 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 bars 1 5.5 14.0 7.0 9.0 10.5 12.0 13.5 7.0 8.5 10.0 11.5 13.0 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 #6 #5 #4 #3 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 5.5 15.0 7.5 9.0 11.0 12.5 14.5 7.0 9.0 10.5 12.0 13.5 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 #7 9.0 18.5 #6 11.0 12.5 14.5 16.0 18.0 #5 11.0 12.5 14.0 15.5 17.5 #4 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 5.5 15.5 7.5 9.5 11.0 13.0 15.0 7.5 9.0 11.0 12.5 14.5 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 #8 9.5 19.5 #7 11.5 13.0 15.0 17.0 19.0 #6 11.0 13.0 14.5 16.5 18.0 #5 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 5.5 17.0 7.5 9.5 11.5 13.5 15.5 7.5 9.5 11.5 13.0 15.0 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 #9 #8 #7 #6 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 5.5 18.0 8.0 10.0 12.5 14.5 17.0 8.0 10.0 12.0 14.0 16.0 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 #10 #9 #8 #7 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 10.0 12.0 14.0 16.0 8.0 10.5 12.5 15.0 17.0 8.0 1 5.5 19.5 8.0 10.5 13.0 15.5 18.0 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 #8 #11 #10 #9 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 5.5 22.5 8.5 11.5 14.5 17.0 20.0 8.5 11.0 13.5 16.0 18.5 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 #14 12.5 29.5 #11 15.5 18.0 21.0 23.5 26.5 #10 15.0 17.5 20.0 22.5 25.0 #9 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 6.5 29.0 10.0 13.5 16.5 20.0 23.5 9.5 12.5 15.0 18.0 21.0 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 #18 #14 #11 #10 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.


REF. TABLES

ACI min. bw, in.

4


Table A-5—Properties of bundled bars Reference: ACI 318-11 Section 7.6.6.5.

Equivalent diameter, d be =

4 --- A s π

Centroidal distance, x For the bundled bars configuration shown here, the centroidal distance is calculated by the following equation: 5 --- A si d b1 + A s2 ( d b1 + d b2 /2 ) 2 x = --------------------------------------------------------------ΣA si

Bars

Combination of bars Equivalent diameters dbe, 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

REF. TABLES

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

6

4

0.34

0.55

0.47

0.61

0.57

0.67

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

7

0.44

0.69

0.87

8

0.50

0.79

1.00

9

0.56

0.89

1.13

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

1.18

11

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 dbe = 2.26 in., and the centroidal distance x equals 1.13 in.



5

Table A-6—Minimum beam web widths bw for various combinations of bundled bars (interior exposure)

Reference: ACIACI 318-11 SectionsSections 7.2.2, 7.6.6.7.6.6.1,1, 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 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

REF. TABLES

*For beams conforming to AASHTO specifications, add 1 in. to tabulated beam web width. Fore example, for two bundles of three #10, minimum bw = 11.5 in.


6


Table A-7—Basic development length ratios of bars in tension Reference: ACI 318-11 Sections 12.2.2 and 12.2.4

Development length ratios: l ψt ψe fy ----d- = α ------------ --------db λ f′ c

For: ψt = 1.0; ψe = 1.0; and λ = 1.0 (see notes below). Basic development length ratios of bars in tension 40,000 psi

fy Bar size #3 to #6 #7 to #18

fc′ Category

α

80,000 psi

60,000 psi

3000 psi

4000 psi

5000 psi

6000 psi

3000 psi

4000 psi

5000 psi

6000 psi

29

25

23

21

44

38

34

31

8000 10,000 3000 psi psi psi 27

24

58

4000 psi

5000 psi

6000 psi

8000 psi

10,000 psi

51

45

41

36

32

I

1/25

II

3/50

44

38

34

31

66

57

51

46

40

36

88

76

68

62

54

48

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

REF. TABLES

Notes: 1. See category chart for Categories I and II. 2. ψt = 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). ψe = coating factor (1.5 = epoxy-coated reinforcement with cover < 3db or clear spacing < 6db; 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 ld ≥ 12 in.



7

Table A-7—Basic development length ratios of bars in tension (cont.)

CATEGORY CHART

Category I:

 Clear spacing ≥ d b   Clear cover ≥ d b   Code minimum stirrups or ties throughout l d

or  Clear spacing ≥ 2d b   Clear cover ≥ d b

REF. TABLES

Category II: All other cases


8


Table A-8—Basic development length ldh of standard hooks in tension Reference: ACI 318-11 Sections 7.1 and 12.5.1 to 12.5.3 Development length, ldh* = αldh ≥ 8db, less than 6 in., where α represents modifiers from Note 1 below and ldh is basic development length of standard hooks in tension 0.02ψ e f y - db l dh = -------------------λ f c′

This table is calculated with ψe = 1.0 and λ = 1.0. Basic development length ldh, in., of standard hooks in tension 60,000 psi

fy

80,000 psi

fc′ Bar size

db, in.

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

8db , 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

*

REF. TABLES

Note 1: To compute development length ldh for a standard hook in tension, multiply basic development length ldh from table above by applicable modification 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 development length not greater than 3db, where db 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 3db along ldh. α = As required / As provided Note 2: Values of basic development length ldh above the heavy line are less than the minimum development length of 6 in. Development length ldh shall not be less than 8db, nor less than 6 in., whichever is greater.



9

Table A-8—Basic development length ldh 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

ldh, in.

6

7

7

8

9

10

13

14

15

19

25

REF. TABLES

Example: Find minimum embedment depth ldh that will provide 2 in. cover over the tail of a standard 180-degree end hook in a #8 bar. For #8 bar, read ldh = 10 in.



REF. TABLES

10



11

APPENDIX B—ANALYSIS TABLES

ANALYSIS TABLES

Reproduced with permission from the Canadian Portland Cement Association.



ANALYSIS TABLES

12


13

ANALYSIS TABLES




ANALYSIS TABLES

14


15

ANALYSIS TABLES




ANALYSIS TABLES

16


17

ANALYSIS TABLES




ANALYSIS TABLES

18


19

ANALYSIS TABLES




ANALYSIS TABLES

20


21

ANALYSIS TABLES




ANALYSIS TABLES

22


23

ANALYSIS TABLES




ANALYSIS TABLES

24



25

APPENDIX C—SECTIONAL PROPERTIES Table C-1—Properties of sections Dash-and-dot lines are drawn through centers of gravity A = area of section; I = moment of inertia; R = radius of gyration A = d

2 2

πd 2 A = -------- = 0.7854d 4

4

d I 1 = -----12

4

vd 4 I = -------- = 0.0491d 64

4

d I 2 = ----3

d R = --4

R 1 = 0.2887d R 2 = 0.57774d A = d

2

A = 0.8660d

y = 0.7071d

I = 0.060d

4

d I = -----12

2

4

R = 0.264d

R = 0.2887d A = bd 3

bd I 1 = -------12

A = 0.8284d

3

I = 0.055d

bd I 2 = -------3

2

4

R = 0.257d

R 1 = 0.2887d R 2 = 0.5774d A = bd

bd A = -----2

bd y = -------------------2 2 b +d

3

bd I 1 = -------36

3 3

b d I = -----------------------2 2 6(b + d )

3

bd I 2 = -------12

bd R = ----------------------------2 2 6(b + d )

R 1 = 0.236d R 2 = 0.408d d A = --- ( b + b′ ) 2 d ( 2b + b′ ) y = -------------------------3 ( b + b′ )

A = bd bsin∞ + d cos ∞ y = -------------------------------------2

d ( b + 2b′ ) y = -------------------------3 ( b + b′ ) 2

2

2

2

2

2

d 2 2 R = ---------------------- 2 ( b + 4bb′ + b′ ) 6 ( b + b′ )

SECT. PROP.

b sin ( ∞ + d cos ∞ ) R = -------------------------------------------------------12

2

d ( b + 4bb′ + b′ ) I = ----------------------------------------------36 ( b + b′ )


26


Table C-2—Properties of sections Dash-and-dot lines are drawn through centers of gravity A = area of section; I = moment of inertia; R = radius of gyration A = bt + b c′ 2

Section of parabola 2

b 2 y = ----- x d

2

d b′ + t ( b – b′ ) y = --------------------------------------2 ( bt + b c′ )

For parabola:

y1 = d – y 3

3

3

b′y 1 + by – ( b – b′ ) ( y – t ) I = -------------------------------------------------------------------3 I --A

R =

For compliment:

2bd A = --------3

bd A = -----3

8 3 I 1 = --------- bd 175

37 3 I 1 = ------------ bd 2100

19 3 I 2 = --------- b d 480

1 3 I 2 = ------ b d 80

A = bt + b c′ 2

2

d b′ + t ( b – b′ ) y = --------------------------------------2 ( bt + b c′ )

A = d –a

y1 = d – y

d –a I = ---------------12

3

3

3

b′y 1 + by – ( b – b′ ) ( y – t ) I = -------------------------------------------------------------------3 R =

R =

2

2

4

4

2

2

d +a ---------------12

I --A

c ( a + b′ ) A = bt + ---------------------2 2

A = bd – ac

3bt + 3b′c ( d + t ) + c ( a – b′ ) ( c + 3t ) y = -------------------------------------------------------------------------------------------3 [ 2bt + c ( a + b′ ) ]

3

y1 = d – y 2

3

4bt + c ( 3b′ + a ) 2 I = -------------------------------------------- – A ( y – t ) 12

3

bd – ac I = ---------------------12 3

3

bd – ac ----------------------------12 ( bd – ac )

R =

I --A Ellipse

R =

A = 0.7854bd 3

wbd 3 I 1 = ------------- = 0.0491bd 64 3

wb d 3 I 2 = ------------- = 0.0491b d 64 d R 1 = --4

2

2

4

4

π ( d – d1 ) 2 2 A = -----------------------= 0.7854 ( d – d 1 ) 4 π ( d – d1 ) 4 4 I = -----------------------= 0.0491 ( d – d 1 ) 64 2

R = 1 ⁄ 4 d + d1

2

b R 2 = --4 2

2

Parabola Equation: 2

b y = --------x 4d 2

2bd A = --------3

A = 0.8284d – 0.7854d 1 2

2

= 0.7854 ( 1.055d – d 1 ) 4

4

I = 0.0547d – 0.0491d 1 4

4

= 0.0491 ( 1.115d – d 1 ) 4

2 4

= 0.0491 [ ( 1.056 ) d – d 1 ] 2

2

SECT. PROP.

R = 1 ⁄ 4 1.056d + d 1


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 0.85fc′ 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 [Kn = Pn/(fc′Ag)] represents the nondimensional form of the nominal axial load strength of the section. The horizontal coordinate [Rn = Mn/(fc′Agh)] 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 Kn or Rn could lead to

inaccuracies because the e/h lines converge rapidly at the lower ends of the diagrams. Straight lines for the tension steel stress ratios fs/fy have been plotted to assist in designing splices for the reinforcement. Further, the ratio fs/fy = 1.0 represents steel strain εy = fy/Es, 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 Kmax are also provided on each interaction diagram. Here, Kmax refers to the maximum permissible nominal axial load, Pn,max, on a column that is laterally reinforced with ties or hoops (ACI 318-14, Section 22.4.2.1). Pn,max  0.8 0.85 f c  Ag  Ast   f y Ast   

(D.1a)

K max  pmax / fc Ag

(D.1b)

Then,

For columns with spirals, values of Kmax 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.

REFERENCES ACI Committee 340, 1970, “Ultimate Strength Design Handbook,” V. 2, Columns, ACI Special Publication 17A, 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 edition, 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. Everard, 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.

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 M x  M 2ux  M 2uy  M n   M 2nx  M 2ny

(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/Pi) 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 Pn and the nominal biaxial bending moments Mnx and Mny. 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 S1 may be represented by Pn, ex, and ey, as in Fig. D.2.1b, or it may be represented by surface S2 represented by Pn, Mnx, and Mny, as shown in Fig. D.2.1c. Note that S1 is a single curvature surface having no discontinuity at the balance point, whereas S2 has discontinuity. Note that when biaxial bending exists with a nominal axial force smaller than the lesser of Pb or 0.1fc′Ag, it is sufficiently accurate and conservative to ignore the axial force and design the section for bending only.

Fig. D.2.1b—Failure surface S1. Fig. D.2.1c—Failure surface S2. D.2.2 Reciprocal load method—In the reciprocal load method, S1 is inverted by plotting 1/Pn as the vertical axis, giving S3, shown in Fig. D.2.2a. As Figure D.2.2b shows, a true point (1/Pn1, exA, eyB) on this reciprocal failure surface can be approximated by a point (1/Pni, exA, eyB) on a plane S3′ 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 Pny when the load has an eccentricity of exA with ey = 0. Point B represents the nominal axial load strength Pnx when the load has an eccentricity of eyB with ex = 0. Point C is based on the axial strength Po with zero eccentricity. The equation of the plane passing through the three points is 1 1 1 1    Pni Pnx Pny P0

(D.2.2a)

where Pni =

approximation of nominal axial load strength at eccentricities ex and ey

Pnx =

nominal axial load strength for eccentricity ey along the y-axis only (x-axis is axis of bending)

Pny =

nominal axial load strength for eccentricity ex along the x-axis only (y-axis is axis of bending)

P0 =

nominal axial load strength for zero eccentricity

Fig. D.2.2a—Failure surface S3 which is reciprocal of surface S1. Fig. D.2.2b—Graphical representation of reciprocal load method. For design purposes, when φ is constant, 1/Pni in Eq. (D.2.2a) can be used. The variable Kn = Pn /(fc′Ag) can be used directly in the reciprocal equation, as follows (D.2.2b) 1 1 1 1    K ni K nx K ny K 0

(D.2.2b)

where the K values refer to the corresponding Pn values as defined above. Once a preliminary cross section with an estimated steel ratio ρg is selected, calculate Rnx and Rny using the actual bending moments about the cross section x- and y-axes. Obtain the corresponding values of Knx and Kny from the interaction diagrams presented in this chapter as the intersection of Rn value and the assumed steel ratio curve for ρg. Then, obtain the theoretical compression axial load strength K0 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 S2 (Fig. D.2.1c) and works with a load contour defined by a plane at a constant value of Pn (Fig. D.2.3a). The load contour defining the relationship between Mnx and Mny for a constant Pn can be expressed nondimensionally as 

 M nx   M ny      M nox   M noy



   1 

(D.2.3a)

Fig. D.2.3b—Nondimensional load contour at constant Pn. Fig. D.2.3a—Load contour constant Pn on failure surface. For design, when each term is multiplied by , the equation is unchanged. Thus, Mux, Muy, Mox, and Moy, which should correspond to Mnx, Mny, Mnox, and Mnoy, 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 Mnx and Mny are in the same ratio as the uniaxial moment capacities Mox and Moy; thus M nx M ox  M ny M oy

(D.2.3b)

M nx  M ox and M ny  M oy

(D.2.3c)

or

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 Pn/Pog, 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  M nx     M nox 

log 0.5/ log

 M ny  M  noy

  

l og 0.5/ log

1

(D.2.3d)

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

For

M ny M nx

M ny M nx



M oy



M oy

M ox

M ox

,

 M oy   1    M oy  M ny  M nx     M ox    

(D.2.3e)

,

M M ox  M nx  M ny  ox M  oy

(D.2.3f)

 1        

For rectangular sections with reinforcement equally distributed on all four faces, the above equations can be approximated by  b  1   M oy  M ny  M nx      h   

For

M ny M nx



M oy M ox

or

M ny M nx



b h

(D.2.3g)

(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 Mox or Moy. 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.

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 four faces with slenderness ratio below critical value. For a rectangular tied column with bars equally distributed along four faces, find steel area. Given: Loading–– Pu = 560 kip, and Mu = 3920 in.-kip Assume φ = 0.70 Nominal axial load Pn = 560 kip/0.70 = 800 kip Nominal moment Mn = 3920 in.-kip/0.70 = 5600 in.-kip Materials–– Compressive strength of concrete fc′ = 4 ksi Yield strength of reinforcement fy = 60 ksi Normalized maximum size of aggregate is 1 in. Design conditions–– Short column braced against sidesway ACI 318-14 section

10.5

Discussion Determine column section size. Determine reinforcement ratio ρg using known values of variables on appropriate interaction diagrams and compute required cross section area Ast of longitudinal reinforcement.

Calculation Given: h = 20 in. and b = 16 in. Pn = 800 kip Mn = 5600 in.-kip h = 20 in. b = 16 in. Ag = b × h = 20 × 16 = 320 in.2

Compute

Kn 

Pn f c  Ag

Kn 

800  0.625 (4)(320)

Compute

Rn 

Mn f c  Ag h

Rn 

5600  0.22 (4)(320)(20)

Estimate



h5 h



Design aid

20  5  0.75 20

Determine the appropriate interaction diagrams.

For a rectangular tied column with bars along four faces, fc’ = 4 ksi, fy = 60 ksi, and an estimated γ of 0.75, enter diagram R4-60.7 and R4-60.8 with Kn = 0.625 and Rn = 0.22, respectively. Read ρg for Kn and Rn values from Read ρg = 0.041 for γ = 0.7 and ρg = 0.039 for appropriate interaction diagrams. γ = 0.8. Interpolating ρg = 0.040 for γ = 0.75 Compute required Ast from Ast = ρgAg. Required Ast = 0.040 × 320 in.2 = 12.8 in.2

R4-60.7 R4-60.8

D.3.2 Column Example 2—For a specified reinforcement ratio, selection of a column section size for a rectangular tied column with bars on end faces only. For minimum longitudinal reinforcement (ρg= 0.01) and column section dimension h = 16 in., select the column dimension b for a rectangular tied column with bars on end faces only. Given: Loading–– Pu= 660 kip and Mu= 2790 in.-kip Assume φ = 0.70 Nominal axial load Pn = 660 kip/0.70= 943 kip Nominal moment Mn = 4200 in.-kip/0.70= 3986 in.-kip Materials–– Compressive strength of concrete fc’ = 4 ksi Yield strength of reinforcement fy = 60 ksi Nominal maximum size of aggregate is 1 in. Design conditions–– Slenderness effects may be neglected because kℓu/h is known to be below critical value. ACI 318-14 section

Discussion Determine trial column dimension b corresponding to known values of variables on appropriate interaction diagrams. Assume a series of trial column sizes b, in inches, and compute Ag = b x h, in.2

Design aid

Calculation Pn = 943 kip Mn = 3986 in.-kip h = 16 in. ρg = 0.039 b A

24

26

28

384

416

448

g

Compute

Compute

Estimate 10.5

Kn 

Rn 



Pn

943

f c ’ Ag

(4)(384)

 0.61

Mn

3986

f c ’ Ag h

(4)(384)(16)

h5 h

Determine the appropriate interaction diagrams. Read ρg for Kn and Rn values for γ = 0.7, select dimension corresponding to ρg nearest desired value of ρg = 0.01.

0.7

 0.16

943 (4)(416)

 0.57

3986 (4)(416)(16)

 0.15

0.7

943 (4)(448)

 0.53

3986 (4)(448)(16)

 0.14

0.7

For a rectangular tied column with bars along four faces, fc’ = 4 ksi, fy = 60 ksi, and an estimated γ of 0.70, use diagram L4-60.7. 0.018

0.014

Therefore, try a 16 x 28-in. column

0.011

L4-60.7

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–– Pu= 660 kip and Mu= 2640 in.-kip Assume φ = 0.70 Nominal axial load Pn = 660 kip/0.70= 943 kip Nominal moment Mn = 2640 in.-kip/0.70= 3771 in.-kip Materials–– Compressive strength of concrete fc’ = 4 ksi Yield strength of reinforcement fy = 60 ksi Nominal maximum size of aggregate is 1 in. Design conditions–– Column section size h = b = 18 in Slenderness effects may be neglected because kℓu/h is known to be below critical value ACI 318-14 section

10.5

Discussion Calculation Determine reinforcement ration g Pn = 943 kip using known values of variables on Mn = 3771 in.-kip appropriate interaction diagram(s) Ag = b x h = 18 x 18 = 324 in.2 and compute required cross section area Ast of longitudinal reinforcement. Compute

Kn 

Compute

Rn 

Estimate



Pn f c ’ Ag Mn f c ’ Ag h

h5 h

Determine the appropriate interaction diagrams. Read ρg for Kn and Rn.

Kn 

943  0.73 (4)(324)

Rn 

3771  0.16 (4)(324)(18)



Design aid

18  5  0.72 18

For a square spiral column, fc’ = 4 ksi, fy = 60 ksi, and an estimated γ of 0.72, use diagram S4-60.7 and S460.8. For Kn = 0.73 and Rn = 0.16, and for: γ = 0.70 ρg = 0.035 γ = 0.80 ρg = 0.031 γ = 0.72 ρg = 0.034 Ast = 0.034 x 324 in.2 = 11 in.2

S4-60.7 S4-60.8

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  0.04 and bars equally distributed along four faces, subject to biaxial bending. Given: Loading–– Pu= 193 kip, Mux= 1917 in.-kip, and Muy= 769 in.-kip Assume φ = 0.65 Nominal axial load Pn = 193 kip/0.65= 297 kip Nominal moment about x-axis Mnx = 1917 in.-kip/0.65= 2949 in.-kip Nominal moment about y-axis Mny = 769 in.-kip/0.65= 1183 in.-kip Materials–– Compressive strength of concrete fc’ = 5 ksi Yield strength of reinforcement fy = 60 ksi Nominal maximum size of aggregate is 1 in. ACI 318-14 section

Discussion Assume load contour curve at constant Pn is an ellipse, and determine resultant moment Mnx from

For a square column: h = b

M nr  M

M nr  29492  11832  3177 in.-kip

2 nx

M

2 ny

Assume a series of trial column sizes h, in inches. Ag  h2 , in.2 Compute

10.5

Compute

Kn 

Compute

Rn 

Estimate



14

16

18

196

256

324

Pn

297

f c ’ Ag

(5)(196)

Mn f c ’ Ag h

Design aid

Calculation

 0.30

3177

 0.23

(5)(196)(14)

h5 h

297 (5)(256)

 0.23

3177 (5)(256)(16)

0.64

0.69

 0.16

297 (5)(324)

 0.18

3177 (5)(324)(18)

 0.11

0.72

Determine the appropriate interaction diagrams. Read ρg for Kn and Rn values for γ = 0.60, 0.70, and 0.08.

For a square tied column, fc’ = 5 ksi, fy = 60 ksi, use diagram R5-60.6, R5-60.7 and R5-60.8. 0.064 0.030 0.012 0.048 0.026 0.011

Interpolate

0.058 0.026 Therefore, try h = 15 in. Ag = h2 = 152 = 225 in.2 Pn = 297 kip Mn = 3177 in.-kip

Determine reinforcement ratio ρg using known values of variables on appropriate interaction diagrams and compute required cross section area Ast of longitudinal reinforcement. Compute

Kn 

Compute

Rn 

Pn f c ’ Ag Mn f c ’ Ag h

Kn  Rn 

297 (5)(225)

 0.264

3177  0.188 (5)(225)(15)

0.012

R5-60.6 R5-60.7 R5-60.8

ACI 318-14 section

Discussion Estimate

10.5



h5 h

Determine the appropriate interaction diagrams. Read ρg for Kn and Rn values for γ = 0.60 and 0.70.

Compute required Ast from Ast = g Ag and add about 15 percent for skew bending.

Calculation 

Design aid

15  5  0.67 15

For a square tied column, fc’ = 5 ksi, fy = 60 ksi, use diagram R5-60.6 and R5-60.7. For Kn = 0.264 and Rn = 0.188: γ = 0.60 ρg = 0.043 γ = 0.70 ρg = 0.034 γ = 0.37 ρg = 0.037 Required Ast = 0.037 x 225 in.2 = 8.32 in.2 Use Ast = 9.50 in.2

R5-60.6 R5-60.7

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–– Pu = 940 kip and Mu = 480 in.-kip Assume φ = 0.70 Nominal axial load Pn = 940 kip/0.70 = 1343 kip Nominal moment Mn = 480 in.-kip/0.70 =686 in.-kip Materials–– Compressive strength of concrete f c’ = 5 ksi Yield strength of reinforcement fy = 60 ksi Nominal maximum size of aggregate is 1 in. Design condition–– Slenderness effects may be neglected because kℓu/h is known to be below critical value ACI 318-14 section

10.5

Discussion Determine trial column dimension Pn = 1343 kip b corresponding to known values Mn = 686 in.-kip of variables on appropriate ρg = 0.04 interaction diagrams. Assume a series of trial column 12 sizes b, in inches. Ag  h2 , in.2 Compute 113 Compute

Rn 

Estimate



Mn

686

f c ’ Ag h

(5)(113)(12)

h5 h

Determine the appropriate interaction diagrams. Read Kn and Rn and ρg values for γ = 0.60, 0.70, and 0.08.

Ag 

Compute

h2

16

20

201

314

686 (5)(201)(16)

0.58

 0.04

0.69

Pn f c ’K n Ag



Compute

Kn 

Pn f c ’ Ag

(5)(314)(20)

1.17

1.24

298

229

217

19.5

17.1

16.6

2

 17  Ag      227 in.2  2

Kn 

1343 (5)(227)(17)

 1.18

 0.02

0.75

0.9

Therefore, try 17 in. diameter column. Determine reinforcement ratio ρg using known values of variables on appropriate interaction diagrams and compute required cross section area Ast of longitudinal reinforcement.

686

For a circular column, fc’ = 5 ksi, fy = 60 ksi, use diagram C5-60.6, C5-60.7 and C5-60.8. No chart 1.08 1.23 0.9 1.18 1.25

Interpolate Compute

 0.10

Design aid

Calculation

C5-60.6 C5-60.7 C5-60.8

ACI 318-14 section

10.5

Discussion Compute

Rn 

Estimate



Calculation Mn f c ’ Ag h

h5 h

Rn 



1343 (5)(227)(17)

Design aid

 0.0356

17  5  0.71 17

Determine the appropriate interaction diagrams. Read ρg for Kn and Rn values.

For a circular column, fc’ = 5 ksi, fy = 60 ksi, use diagram C5-60.7. For Kn = 1.18 and Rn = 0.0356: γ = 0.70 ρg = 0.040

Compute required Ast from Ast = g Ag.

Required Ast = 0.04 x 227 in.2 = 9.08 in.2

C5-60.7

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)

Table D.4.2—Effective Length Factor, Jackson and Moreland alignment chart for columns in unbraced (sway) frames (Column Research Council 1966)

Table D.4.3—Recommended flexural rigidities (EI) for use in first-order and second order analyses of frames for design of slender columns (ACI 318-14, Section 6.6.3.1.1,) Second-order analysis of frames for design of slender columns fc′, ksi

3

4

5

6

7

8

9

10

Ec , ksi

3120

3605

4031

4415

4769

5098

5407

5700

Ec I / Ig , ksi

I/Ig

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 slabs

780

901

1008

1104

1192

1275

1352

1425

0.2

Notes: 1. Alternatively, the following more refined values of I can be used: For columns:  A I   0.8  25 st  Ag 

 M P   1  u  0.5 u  I g  0.5Ig Po    Pu h

Where Pu and Mu are for the particular load combinations under consideration, or the combination of Pu and Mu that leads to the smallest value of I, I need not be less than 0.35Ig. For beams: 𝐼 = (0.10 + 25𝜌) (1.2 − 0.2

2.

3. 4. 5.

𝑏𝑤 ) 𝐼 ≤ 0.5𝐼𝑔 𝑑 𝑔

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 0.25Ig. When sustained lateral loads are applied, I for columns should be divided by (1+βds), where βds is the ratio of maximum factored sustained shear within a story to the maximum factored story shear for the same load combination. βds shall not be taken greater than 1.0. The above values are applicable to normal-density concrete with wc between 90 and 155 lb/ft3. The moment of inertia of a T-beam should be based on the effective flange width as shown in Flexure 6. It is generally accurate to take Ig of a T-beam as two times the Ig for the web. Member area will not be reduced for analysis.

Table D.4.4—Effective length factor k for columns in braced frames (Concrete Design Handbook 2005)

0.81

0.91

0.95

1.00

Elastic

0.77

0.86

0.90

0.95

Elastic

0.74

0.83

0.86

0.91

Stiff

0.67

0.74

0.77

0.81

Elastic

Elastic

Hinged

TOP

Hinged

Stiff

k

BOTTOM

Table D.4.5—Moment of inertia of reinforcement about sectional centroid (Based on Table 12-1, MacGregor 1997) b

b

h

h

h

h

Bars in two end faces

Equal reinforcement on four sides

I se = 0.25 t b h3 2

I se = 0.18 t b h  2 3

b

h

h

h

h

Uniformly distributed reinforcement I se = 0.10 t h4  2

Bars in two side faces I se = 0.17 t b h3 2 (3 bars per face) I se = 0.12 t b h3 2 (6 bars per face)

b = 2h

b

h

h h

h = 2b

Bars uniformly spaced on all sides I se = 0.22 t b h3 2

Bars uniformly spaced on all sides I se = 0.13 t b h3 2

γ is the ratio of the distance between the centers of the outermost bars to the column dimension perpendicular to the axis of bending.

COLUMNS 3.1.1 - Nominal load-moment strength interaction diagram, R3-60.6

D.5—Interaction diagrams 2.4 2.2 2.0

h

INTERACTION DIAGRAM R3-60.6 f /c = 3 ksi

g = 0.08

h

fy = 60 ksi

 = 0.6

0.07

Kmax

1.8

0.06

1.6

0.05

Kn = Pn / f

/

c

Ag

1.4

e

Pn

0.04

fs/fy = 0 0.03

1.2 0.25

0.02

1.0 0.01

0.8

0.50

0.6 0.75

0.4

  t = 0 t t = 0.0.0035 = 0 04 .0 0 5

0.2

1.0

0.0 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 COLUMNS 3.1.2 - Nominal load-moment strength interaction diagram, R3-60.7 Rn = Pn e / f / c Ag h

2.4 2.2 2.0

g = 0.08


0.07

 = 0.7

h

h

fy = 60 ksi

Kmax

1.8 1.6

0.06

e

Pn

0.05

fs/fy = 0

0.04

K n = Pn / f

/

c

Ag

1.4 0.03

1.2

0.25 0.02

1.0 0.01

0.50

0.8 0.75

0.6 0.4

 t t = 0.00 t = 0.00435 = 0. 005 0

0.2 0.0 0.00

0.05

0.10

0.15

1.0

0.20

0.25

0.30

Rn = Pn e / f / c Ag h

0.35

0.40

0.45

0.50

COLUMNS 3.1.3 - Nominal load-moment strength interaction diagram, R3-60.8 2.4 2.2

g = 0.08

h


h

fy = 60 ksi

 = 0.8

0.07

2.0 Kmax 0.06

1.8

Pn

e 0.05

1.6 fs/fy = 0

0.04

Kn = Pn / f

/

c

Ag

1.4 0.03

0.25

1.2 0.02

1.0 0.50

0.01

0.8 0.75

0.6 1.0

0.4

t = 0.0035 t = 0.004 0 t = 0.005

0.2

0.0 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 COLUMNS 3.1.4 - Nominal load-moment strength interaction diagram, R3-60.9 Rn = Pn e / f / c Ag h

2.4 g = 0.08


0.07

 = 0.9

2.2

h

h

fy = 60 ksi

2.0 Kmax

0.06

1.8

e

Pn

0.05

1.6

fs/fy = 0 0.04

Kn = Pn / f

/

c

Ag

1.4 0.03

0.25

1.2 0.02

1.0

0.50 0.01

0.8

0.75

0.6 1.0

0.4 0.2

t = 0.0035 t = 0.004 t = 0.0050

0.0 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 0.65 Rn = Pn e / f / c Ag h

2.0

1.8

h


g = 0.08

h

fy = 60 ksi

 = 0.6 0.07

1.6

Kmax 0.06

Pn

e

1.4

0.05 0.04

1.2

fs/fy = 0

Kn = P n / f

/

c

Ag

0.03

1.0

0.02

0.25 0.01

0.8 0.50

0.6 0.75

0.4  t t = 0.00 = 0. t 0 35 = 0. 040 005

0.2

0.0 0.00

0.05

1.0

0.10

0.15

0.20

0.25

0.30

0.35

/ COLUMNS 3.2.2 - Nominal load-moment interaction diagram, R4-60.7 Rn = Pn e / fstrength c Ag h

2.0

1.8

h


g = 0.08

h

fy = 60 ksi

 = 0.7

0.07

1.6

Kmax

0.06

e

1.4

Pn

0.05

0.04

fs/fy = 0

1.2

Kn = P n / f

/

c

Ag

0.03

1.0

0.02

0.25

0.01

0.8

0.50

0.6 0.75

0.4

 t t = 0.0 t = 0.004035 = 0. 005 0

0.2

0.0 0.00

0.05

0.10

1.0

0.15

0.20

Rn = Pn e / f

0.25

/ c

Ag h

0.30

0.35

0.40

COLUMNS 3.2.3 - Nominal load-moment strength interaction diagram, R4-60.8 2.0

1.8

g = 0.08

h

fy = 60 ksi

 = 0.8

0.07

1.6

h


Kmax

0.06

e

1.4

Pn

0.05

fs/fy = 0

0.04

1.2

Kn = Pn / f

/

c

Ag

0.03

1.0

0.25

0.02

0.01

0.50

0.8

0.75

0.6

1.0

0.4

t = t = 0.0035 t = 00.0040 .005

0.2

0.0 0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

COLUMNS 3.2.4 - Nominal load-moment strength interaction diagram, R4-60.9 / Rn = Pn e / f

c

Ag h

2.0 g = 0.08


0.07

 = 0.9

0.06

Kmax

1.8

1.6

1.4

fy = 60 ksi

e

0.05

Ag c /

Kn = P n / f

Pn

fs/fy = 0

0.04

1.2

h

h

0.03

0.25

1.0

0.02

0.01

0.50

0.8 0.75

0.6 1.0

0.4 t = 0.003 t = 0.00405 t = 0.005

0.2

0.0 0.00

0.05

0.10

0.15

0.20

0.25

0.30


0.35

0.40

0.45

0.50

1.8

h


1.6

g = 0.08

h

fy = 60 ksi

 = 0.6

0.07

Kmax

1.4 0.06

e

Pn

0.05

1.2 0.04 0.03

fs/fy = 0

Ag

1.0 Kn = P n / f

/

c

0.02 0.01

0.8

0.25

0.6

0.50

0.75

0.4  t t = 0.0 = 0. 035 0 t = 0. 040 005

0.2

1.0

0.0 0.00 0.05 0.10 0.15 0.20 0.25 0.30 COLUMNS 3.3.2 - Nominal load-moment strength interaction diagram, R5-60.7 / Rn = Pn e / f

c

Ag h

1.8

h


1.6

g = 0.08

h

fy = 60 ksi

 = 0.7

0.07

1.4

Kmax

0.06

e

Pn

0.05

1.2 0.04

fs/fy = 0

0.03

1.0 Kn = P n / f

/

c

Ag

0.02

0.25

0.01

0.8 0.50

0.6 0.75

0.4 t t = = 0.0035 t = 0.0040 0.005

0.2

0.0 0.00

0.05

0.10

1.0

0.15

0.20


0.25

0.30

0.35

1.8

h


1.6

g = 0.08

h

fy = 60 ksi

 = 0.8 0.07

1.4

Kmax

0.06

e

Pn

0.05

1.2

0.04

fs/fy = 0

1.0 0.02

Kn = Pn / f

/

c

Ag

0.03

0.25 0.01

0.8 0.50

0.6 0.75

0.4

1.0

t = 0.0 t = 0.004035 t = 0.005 0

0.2

0.0 0.00

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 COLUMNS 3.3.4 - Nominal load-moment strength interaction diagram, R5-60.9 / Rn = Pn e / f

c

Ag h

1.8

1.6

g = 0.08

h

fy = 60 ksi

 = 0.9

0.07

1.4

h


Kmax

0.06

e

0.05

1.2

0.04

Pn

fs/fy = 0

0.03 0.02

0.25

Kn = Pn / f

/

c

Ag

1.0

0.01

0.8

0.50

0.6

0.75

1.0

0.4 t = 0.0035 t = 0.004 t = 0.005 0

0.2

0.0 0.00

0.05

0.10

0.15

0.20

0.25


0.30

0.35

0.40

0.45

1.6

1.4

h


g = 0.08

h

fy = 60 ksi

0.07

 = 0.6

0.06

Kmax 0.05

1.2

e

0.04

Pn

0.03

1.0

0.02

fs/fy = 0

Kn = P n / f

/

c

Ag

0.01

0.8 0.25

0.6 0.50

0.4

0.75

 t t = 0.0 = 0. 035 0 t = 0. 040 005

0.2

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 0.275 COLUMNS 3.4.2 - Nominal load-moment strength interaction diagram, R6-60.7 Rn = Pn e / f / c Ag h

1.6 g = 0.08

1.4 0.07

h

fy = 60 ksi

 = 0.7

Kmax

0.06

1.2

h


0.05

e

Pn

0.04

1.0

0.03

fs/fy = 0

Kn = Pn / f

/

c

Ag

0.02

0.8

0.01

0.25

0.6

0.50

0.75

0.4  t =t = 0.0035 t = 0.0040 0.005

0.2

0.0 0.00

0.05

0.10

1.0

0.15 Rn = Pn e / f / c Ag h

0.20

0.25

0.30

COLUMNS 3.4.3 - Nominal load-moment strength interaction diagram, R6-60.8 1.6 g = 0.08

1.4

h

fy = 60 ksi

 = 0.8

0.07

Kmax

0.06

1.2

h


0.05

e

Pn

0.04

1.0

0.03

fs/fy = 0

Kn = P n / f

/

c

Ag

0.02

0.8

0.01

0.25

0.50

0.6 0.75

0.4 1.0

t t = =0 0.0035 .0 t = 0 040 .005

0.2

0.0 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 COLUMNS 3.4.4 - Nominal load-moment strength interaction diagram, R6-60.9 Rn = Pn e / f / c Ag h

1.6

1.4

1.2

g = 0.08


0.07

 = 0.9

0.06

Kmax

h

h

fy = 60 ksi

0.05

e

Pn

0.04

1.0

fs/fy = 0

0.03

K n = Pn / f

/

c

Ag

0.02

0.8

0.25

0.01

0.50

0.6 0.75

0.4

1.0

t = t = 0 0.0035 t = 0 .0040 .005

0.2

0.0 0.00

0.05

0.10

0.15

0.20


0.25

0.30

0.35

0.40

1.6

h


1.4

h

fy = 75 ksi

g = 0.08

 = 0.6

0.07

1.2

0.06

Kmax

0.05

Pn

e

0.04

1.0

0.03 0.02

Kn = Pn / f

/

c

Ag

0.01

0.8 fs/fy = 0

0.6 0.25

0.4

0.50

  t = 0 t t = 0.00.0038 = 0. 005 40

0.2

0.75

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 COLUMNS 3.5.2 - Nominal load-moment strength interaction diagram, R9-75.7 Rn = Pn e / f / c Ag h

1.6

h


h

fy = 75 ksi

1.4

 = 0.7 g = 0.08

1.2

0.07 0.06

Kmax

e

Pn

0.05

1.0

0.04 0.03

Kn = Pn / f

/

c

Ag

0.02

0.8

0.01

fs/fy = 0

0.6

0.25

0.50

0.4

0.2

0.75

  t= t t = 0.00.0038 = 0. 005 040 0

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 Rn = Pn e / f / c Ag h

1.6

h


h

fy = 75 ksi

1.4

 = 0.8 g = 0.08 0.07

1.2

Kmax

0.06

Pn

e

0.05

1.0

0.04 0.03

Kn = Pn / f

/

c

Ag

0.02

0.8

0.01

fs/fy = 0

0.25

0.6

0.50

0.4 0.75

  t = 0 t t = 0.0.0038 = 0. 040 005 0

0.2

1.0

0.0 0.00 0.05 0.10 0.15 0.20 0.25 0.30 COLUMNS 3.5.4 - Nominal load-moment strength interaction diagram, R9-75.9 Rn = Pn e / f / c Ag h

1.6

h


h

fy = 75 ksi

1.4 g = 0.08

 = 0.9

0.07

1.2

0.06

Kmax

e

Pn

0.05 0.04

1.0 0.03

Kn = Pn / f

/

c

Ag

0.02

0.8

fs/fy = 0

0.01

0.25

0.6 0.50

0.4

0.75

t = 0.0038 t = 0.0 0 t = 0.00504 0

0.2

0.0 0.00

0.05

0.10

1.0

0.15 Rn = Pn e / f / c Ag h

0.20

0.25

0.30

1.4

h


1.2

h

fy = 75 ksi

g = 0.08

 = 0.6

0.07 0.06 0.05

1.0

Kmax

Pn

e

0.04 0.03 0.02 0.01

K n = Pn / f

/

c

Ag

0.8

fs/fy = 0

0.6

0.25

0.4 0.50

  t = t t = 0.00.0038 = 0. 005 040 0

0.2

0.75

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 COLUMNS 3.6.2 - Nominal load-moment strength interaction diagram, R12-75.7 / Rn = Pn e / f

c

Ag h

1.4 h


1.2

h

fy = 75 ksi

g = 0.08

 = 0.7

0.07 0.06 0.05

1.0

Kmax

Pn

e

0.04 0.03 0.02 0.01

c

Ag

0.8

Kn = Pn / f

/

fs/fy = 0

0.6 0.25

0.4

0.50

t

0.2

0.0 0.000

0.025

0.050

0.75

 t t = 0.0 = 03 = 0 . 0 .0 0 4 0 8 005 0

0.075

1.0

0.100

0.125


0.150

0.175

0.200

1.4

h


1.2

g = 0.08

h

fy = 75 ksi

 = 0.8

0.07 0.06 0.05

1.0

Kmax

0.04

e

Pn

0.03 0.02 0.01

Ag

0.8

Kn = P n / f

/

c

fs/fy = 0

0.6

0.25

0.50

0.4 0.75

  t = 0.0 t =t = 0.004038 0.005 0 0

0.2

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 COLUMNS 3.6.4 - Nominal load-moment strength interaction diagram, R12-75.9 Rn = Pn e / f / c Ag h

1.4

h


1.2

g = 0.08

h

fy = 75 ksi

 = 0.9

0.07 0.06 0.05

1.0

0.04

Kmax

e

Pn

0.03 0.02 0.01

0.8

Kn = Pn / f

/

c

Ag

fs/fy = 0

0.6

0.25

0.50

0.4 0.75

0.2

t 038 t == 0.0 040 t = 0.00.0 050

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 Rn = Pn e / f / c Ag h

2.4 2.2 2.0 1.8

g = 0.08

INTERACTION DIAGRAM L3-60.6 f /c = 3 ksi

0.07

 = 0.6

h

h

fy = 60 ksi

Kmax 0.06

e

Pn

0.05

1.6 0.04

Ag

1.4

fs/fy = 0

Kn = P n / f

/

c

0.03

1.2 0.02

0.25

1.0 0.01

0.8

0.50

0.6 0.75

0.4

   t = 0 .0 t t = 0.0 035 = 0. 04 005

0.2

1.0

0.0 0.00 0.05 3.7.2 0.10 - 0.15 0.20 0.25 0.30strength 0.35 interaction 0.40 0.45 diagram, 0.50 0.55 0.60 COLUMNS Nominal load-moment L3-60.7 Rn = Pn e / f / c Ag h

2.4


g = 0.08

2.2

h

h

fy = 60 ksi

 = 0.7

0.07

2.0 Kmax

0.06

1.8

e

Pn

0.05

1.6 0.04

1.4 Ag c /

Kn = Pn / f

fs/fy = 0 0.03

1.2 0.02

0.25

1.0 0.01

0.50

0.8 0.6 0.4 0.2 0.0 0.0

0.75

t = t 0.0035 t ==0 0.004 .005

0.1

1.0

0.2

0.3

0.4


0.5

0.6

0.7

2.4 g = 0.08


0.07

 = 0.8

2.2

h

h

fy = 60 ksi

2.0 Kmax

0.06

1.8

e

0.05

Pn

1.6 0.04

fs/fy = 0

1.4

Kn = P n / f

/

c

Ag

0.03

1.2 0.02

0.25

1.0 0.01

0.50

0.8 0.6

0.75

0.4

t = 0.0035 t 0.004 t ==0.0 05

0.2

1.0

0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 COLUMNS 3.7.4 - Nominal load-moment strength interaction diagram, L3-60.9 Rn = Pn e / f / c Ag h

2.4

g = 0.08

h


2.2

h

fy = 60 ksi

0.07

 = 0.9

2.0 0.06

1.8 1.6

Kmax

0.05

e

Pn

0.04

Kn = Pn / f

/

c

Ag

fs/fy = 0

1.4

0.03

1.2

0.02

1.0

0.25

0.01

0.50

0.8 0.75

0.6 0.4 0.2 0.0 0.0

1.0

t = 0.0035 t = 0.004 t = 0.005

0.1

0.2

0.3

0.4

0.5


0.6

0.7

0.8

0.9

2.0

1.8

h


g = 0.08

h

fy = 60 ksi

 = 0.6

0.07

1.6

Kmax 0.06

e

1.4

Pn

0.05

0.04

1.2

fs/fy = 0

Kn = Pn / f

/

c

Ag

0.03

1.0

0.02

0.25 0.01

0.8 0.50

0.6 0.75

0.4  tt = 0.00 t = 0.00435 = 0. 0 05

0.2

1.0

0.0 0.00

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 COLUMNS 3.8.2 - Nominal load-moment strength interaction diagram, L4-60.7 Rn = Pn e / f / c Ag h

2.0 g = 0.08


0.07

 = 0.7

1.8

1.6

1.4

h

h

fy = 60 ksi

Kmax

0.06

e

0.05

Pn

0.04

1.2

fs/fy = 0

Kn = Pn / f

/

c

Ag

0.03

1.0

0.02

0.25 0.01

0.8 0.50

0.6 0.75

0.4 t t == 0.0035 t = 0 0.004 .005

0.2

0.0 0.00

0.05

0.10

0.15

1.0

0.20

0.25

0.30


0.35

0.40

0.45

0.50

0.55

2.0 g = 0.08


0.07

 = 0.8

1.8

1.6

1.4

h

h

fy = 60 ksi

Kmax

0.06

e

0.05

Pn

0.04

Kn = Pn / f

/

c

Ag

1.2

1.0

fs/fy = 0 0.03

0.02

0.25 0.01

0.8

0.50

0.6

0.75

0.4 1.0

t = 0.003 t = 0.004 5 t = 0.005

0.2

0.0 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 COLUMNS 3.8.4 - Nominal load-moment strength interaction diagram, L4-60.9 Rn = Pn e / f / c Ag h

2.0 g = 0.08

1.8 0.07

1.6

0.06

h


h

fy = 60 ksi

 = 0.9 Kmax

0.05

e

Pn

1.4 0.04

fs/fy = 0

K n = Pn / f

/

c

Ag

1.2

1.0

0.03

0.02

0.25

0.01

0.8

0.50

0.6

0.4

0.2

0.75

1.0

t = 0.0035 t = 0.004 t = 0.005

0.0 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 0.65 0.70 Rn = Pn e / f / c Ag h

1.8

1.6

h

INTERACTION DIAGRAM L5-60.6 f /c = 5 ksi g = 0.08

h

fy = 60 ksi

 = 0.6

0.07

1.4

Kmax

0.06

e

Pn

0.05

1.2

0.04 0.03

fs/fy = 0 0.02

Kn = Pn / f

/

c

Ag

1.0

0.01

0.8

0.25

0.6

0.50

0.4

0.75

   t = 0 .0 t t = 0.00035 = 0. 005 4

0.2

1.0

0.0 0.00

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 COLUMNS 3.9.2 - Nominal load-moment strength interaction diagram, L5-60.7 Rn = Pn e / f / c Ag h

1.8

1.6

h


g = 0.08

h

fy = 60 ksi

 = 0.7 0.07

1.4

Kmax

0.06

e

0.05

1.2

0.04 0.03

fs/fy = 0

0.02

K n = Pn / f

/

c

Ag

1.0

Pn

0.01

0.25

0.8

0.50

0.6 0.75

0.4 t = t = 00.0035 t = 0 .004 .005

0.2

0.0 0.00

0.05

0.10

1.0

0.15

0.20

0.25


0.30

0.35

0.40

0.45

1.8

1.6

g = 0.08

h

fy = 60 ksi

 = 0.8

0.07

1.4

h


Kmax

0.06

e

0.05

1.2

0.04 0.03

fs/fy = 0

0.02

K n = Pn / f

/

c

Ag

1.0

Pn

0.25

0.01

0.8 0.50

0.6 0.75

0.4 t = 0.0035 t = 0.004 t = 0.005

0.2

1.0

0.0 0.00

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 COLUMNS 3.9.4 - Nominal load-moment strength interaction diagram, L5-60.9 Rn = Pn e / f / c Ag h

1.8

1.6

g = 0.08

h

fy = 60 ksi

 = 0.9

0.07

1.4

h


0.06

Kmax e

0.05

1.2

0.04

fs/fy = 0

0.03 0.02

c

Ag

1.0

0.25

/

K n = Pn / f

Pn

0.01

0.8 0.50

0.6 0.75

0.4 1.0

t = 0.0035 t = 0.004 t = 0.0 05

0.2

0.0 0.00

0.05

0.10

0.15

0.20

0.25

0.30


0.35

0.40

0.45

0.50

0.55

1.6

h


1.4

h

fy = 60 ksi

 = 0.6

0.07

Kmax

0.06

1.2

e

0.05

Pn

0.04

1.0

0.03 0.02

Kn = P n / f

/

c

Ag

fs/fy = 0 0.01

0.8 0.25

0.6 0.50

0.4

0.75

t t = 0.00 t = 0.00 35 = 0. 005 4

0.2

1.0


1.6 g = 0.08

h

fy = 60 ksi

1.4

 = 0.7

0.07

Kmax

0.06

1.2

h


0.05

e

Pn

0.04

1.0

0.03

fs/fy = 0

Kn = Pn / f

/

c

Ag

0.02

0.8

0.01

0.25

0.6

0.50

0.75

0.4 t =  0.0035 t =t 0= 0.004 .005

0.2

0.0 0.00

0.05

0.10

1.0

0.15

0.20


0.25

0.30

0.35

0.40

1.6

1.4

1.2

g = 0.08


0.07

 = 0.8

0.06

Kmax

h

h

fy = 60 ksi

0.05

e

Pn

0.04

1.0

0.03

fs/fy = 0

K n = Pn / f

/

c

Ag

0.02 0.01

0.8

0.25

0.6

0.50

0.75

0.4 1.0

t = 0.0035 t = 0.0 t = 0.00504

0.2

0.0 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 COLUMNS 3.10.4 - Nominal load-moment strength interaction diagram, L6-60.9 Rn = Pn e / f / c Ag h

1.6

1.4

g = 0.08


0.07

 = 0.9

fy = 60 ksi

0.06

1.2

h

h

Kmax

0.05

e

Pn

0.04

1.0

0.03

fs/fy = 0

K n = Pn / f

/

c

Ag

0.02 0.01

0.8

0.25

0.50

0.6

0.75

0.4

0.2

0.0 0.00

1.0

t = 0.0035 t = 0.004 t = 0.005

0.05

0.10

0.15

0.20

0.25

0.30


0.35

0.40

0.45

0.50

1.6

1.4

h


h

fy = 75 ksi

 = 0.6

g = 0.08 0.07 0.06

1.2

Kmax

0.05

e

Pn

0.04

1.0

0.03 0.02

K n = Pn / f

/

c

Ag

0.01

0.8 fs/fy = 0

0.6 0.25

0.4 0.50

  t = 0.0 t =t = 0.000438 0.005

0.2

0.75

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 Rn = Pn e / f / c Ag h

COLUMNS 3.11.2 - Nominal load-moment strength interaction diagram, L9-75.7 1.6

h


h

fy = 75 ksi

1.4

 = 0.7 g = 0.08 0.07

1.2

Kmax

0.06

e

Pn

0.05 0.04

1.0

0.03

K n = Pn / f

/

c

Ag

0.02

0.8

0.01

fs/fy = 0

0.6

0.25

0.50

0.4

0.2

0.0 0.00

0.75

t  = 0.0 t =t = 0.000438 0.005

0.05

0.10

1.0

0.15 Rn = Pn e / f / c Ag h

0.20

0.25

0.30

1.6

h


h

fy = 75 ksi

1.4

 = 0.8

g = 0.08 0.07

1.2

0.06

Kmax

e

Pn

0.05 0.04

1.0

0.03

K n = Pn / f

/

c

Ag

0.02

0.8

0.01

fs/fy = 0

0.25

0.6 0.50

0.4 0.75

t  = 0.0038 t =t 0= 0.004 .005

0.2

1.0


1.6

1.4

h


h

fy = 75 ksi

 = 0.9

g = 0.08 0.07

1.2

0.06

Kmax

e

Pn

0.05 0.04

1.0

0.03

K n = Pn / f

/

c

Ag

0.02

0.8

fs/fy = 0

0.01

0.25

0.6 0.50

0.4

0.75

t  = 0.00 t =t = 0.00438 0.005

0.2

0.0 0.00

0.05

0.10

1.0

0.15

0.20


0.25

0.30

0.35

0.40

1.4

1.2

h


h

fy = 75 ksi

 = 0.6

0.07 0.06 0.05

1.0

Kmax

0.04

Pn

e

0.03 0.02 0.01

Kn = Pn / f

/

c

Ag

0.8

fs/fy = 0

0.6

0.25

0.4 0.50

  t = 0 t t = 0.0.0038 = 0. 0 005 4

0.2

0.75

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 COLUMNS 3.12.2 - Nominal load-moment strength interaction diagram, L12-75.7 Rn = Pn e / f / c Ag h

1.4

1.2

h


h

fy = 75 ksi

 = 0.7

0.07 0.06 0.05

1.0

Kmax

0.04

e

Pn

0.03 0.02 0.01

c

Ag

0.8

Kn = Pn / f

/

fs/fy = 0

0.6 0.25

0.4

0.2

0.50

0.75

t t = 0.0038 t = 0= 0.004 .005

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 Rn = Pn e / f / c Ag h

1.4

h


h

fy = 75 ksi

1.2

g = 0.08

 = 0.8

0.07 0.06

1.0

Kmax

0.05

e

Pn

0.04 0.03 0.02 0.01

Ag

0.8

K n = Pn / f

/

c

fs/fy = 0

0.6 0.25

0.50

0.4 0.75

t = 0.0038 t = 04 t = 0.00.0 05

0.2

1.0

0.0 0.00 0.05 0.10 0.15 0.20 0.25 0.30 COLUMNS 3.12.4 - Nominal load-moment strength interaction diagram, L12-75.9 Rn = Pn e / f / c Ag h

1.4

h


h

fy = 75 ksi

1.2

g = 0.08

 = 0.9

0.07 0.06

1.0

Kmax

0.05

e

Pn

0.04 0.03 0.02

0.8

0.01

K n = Pn / f

/

c

Ag

fs/fy = 0

0.6

0.25

0.50

0.4 0.75

t = 0.0038 t = 0.004 t = 0.005

0.2

0.0 0.00

0.05

0.10

1.0

0.15

0.20


0.25

0.30

0.35

2.4

h

INTERACTION DIAGRAM C3-60.6 f /c = 3 ksi

g = 0.08

2.2

h

fy = 60 ksi

0.07

 = 0.6

2.0

Kmax

0.06

1.8

e

0.05

Pn

1.6 0.04

fs/fy = 0

1.4

Kn = Pn / f

/

c

Ag

0.03

1.2 0.02

1.0

0.25

0.01

0.8

0.50

0.6 0.75

0.4 0.2

  t = 0.0 t = 0 03  .00 5 t =0

.00

1.0

4

5

0.0 0.00 0.05 0.10 load-moment 0.15 0.20 interaction 0.25 diagram, 0.30 C3-60.7 0.35 COLUMNS 3.13.2 - Nominal strength Rn = Pn e / f / c Ag h

2.4 2.2 2.0

h


g = 0.08

h

fy = 60 ksi

 = 0.7 0.07

Kmax

1.8

0.06

1.6

0.05

e

Pn

fs/fy = 0

Ag

1.4

0.03

c /

Kn = P n / f

0.04

1.2

0.25 0.02

1.0 0.50

0.01

0.8 0.75

0.6 0.4 0.2 0.0 0.00

t t = 0.00 t = 0.0 35 = 0 04 .0 0 5

0.05

0.10

1.0

0.15

0.20 Rn = Pn e / f

0.25 / c

Ag h

0.30

0.35

0.40

2.4 g = 0.08


0.07

 = 0.8

2.2 2.0

h

h

fy = 60 ksi Kmax

0.06

1.8

Pn

e 0.05

1.6

fs/fy = 0

0.04

Kn = P n / f

/

c

Ag

1.4 0.25

0.03

1.2 0.02

1.0

0.50 0.01

0.8 0.75

0.6 1.0

0.4 t = t = 0.0035 t = 0.004 0.005

0.2

0.0 0.00 0.05 0.15 0.20 strength 0.25 interaction 0.30 0.35 0.40 0.45 COLUMNS 3.13.4 0.10 - Nominal load-moment diagram, C3-60.9 Rn = Pn e / f / c Ag h

2.4 2.2

g = 0.08


0.07

 = 0.9

h

h

fy = 60 ksi

2.0

Kmax 0.06

1.8

e

Pn

fs/fy = 0

0.05

1.6 0.04

Ag

1.4

0.25

Kn = P n / f

/

c

0.03

1.2 0.02

0.50

1.0 0.01

0.75

0.8 0.6

1.0

0.4 0.2 0.0 0.00

t = 0.0035 t = 0.004 t = 0.005

0.05

0.10

0.15

0.20

0.25

0.30

0.35


0.40

0.45

0.50

0.55

2.0 g = 0.08


0.07

 = 0.6

0.06

Kmax

1.8

1.6

h

h

fy = 60 ksi

e

0.05

Pn

1.4 0.04

Kn = Pn / f

/

c

Ag

1.2

fs/fy = 0 0.03 0.02

1.0 0.25

0.01

0.8 0.50

0.6 0.75

0.4 t =0 t .0 0 35 = t = 0 0.004 .00 5

0.2

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 0.275

COLUMNS 3.14.2 - Nominal load-moment strength interaction diagram, C4-60.7 / Rn = Pn e / f

2.0

1.8

c

Ag h

h

INTERACTION DIAGRAM C4-60.7 f /c = 4 ksi g = 0.08

h

fy = 60 ksi

 = 0.7 0.07

Kmax

1.6 0.06

e

1.4

Pn

0.05

fs/fy = 0

1.2

0.04

1.0

0.25 0.02

Kn = Pn / f

/

c

Ag

0.03

0.01

0.8

0.50

0.6 0.75

0.4

0.2

0.0 0.00

1.0

t = t 0.0035 = 0. t 004 = 0. 005

0.05

0.10

0.15

0.20


0.25

0.30

0.35

2.0

1.8

h


g = 0.08

h

fy = 60 ksi

 = 0.8

0.07

1.6

Kmax 0.06

e

1.4

0.05

Pn

fs/fy = 0

0.04

1.2

Kn = Pn / f

/

c

Ag

0.03

1.0

0.25

0.02

0.50

0.01

0.8 0.75

0.6 1.0

0.4

0.2

t = t = 00.0035 t = 0 .004 .005

0.0 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 COLUMNS 3.14.4 - Nominal load-moment strength interaction diagram, C4-60.9 Rn = Pn e / f / c Ag h

2.0

1.8

g = 0.08


0.07

 = 0.9

h

h

fy = 60 ksi

Kmax

1.6 0.06

1.4

e

fs/fy = 0

0.05

Pn

0.04

1.2

0.25

Kn = Pn / f

/

c

Ag

0.03

1.0

0.02

0.50 0.01

0.8 0.75

0.6 1.0

0.4 t = 0 .0 t = 0 035 .004 t = 0 .005

0.2

0.0 0.00

0.05

0.10

0.15

0.20


0.25

0.30

0.35

0.40

1.8

1.6

g = 0.08

h

fy = 60 ksi

 = 0.6

0.07

1.4

h


Kmax

0.06

e

0.05

1.2

Pn

0.04 0.03

fs/fy = 0

Ag

1.0

0.02

Kn = Pn / f

/

c

0.01

0.25

0.8

0.6

0.50

0.4

0.75

t = t 0.0035 = 0. 004 t = 0. 005

0.2

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.175 0.200 0.225 COLUMNS 3.15.2 - Nominal load-moment strength0.150 interaction diagram, C5-60.7 Rn = Pn e / f / c Ag h

1.8

h


1.6

g = 0.08

h

fy = 60 ksi

 = 0.7

0.07

Kmax

1.4 0.06

e

Pn

0.05

1.2 0.04

fs/fy = 0

0.02

Kn = Pn / f

/

c

Ag

0.03

1.0 0.25

0.8

0.01

0.50

0.6 0.75

0.4

0.2

0.0 0.00

1.0

 t t== 0.0035 t = 0.004 0.005

0.05

0.10

0.15 Rn = Pn e / f

0.20 / c

Ag h

0.25

0.30

COLUMNS 3.15.3 - Nominal load-moment strength interaction diagram, C5-60.8 1.8

h


1.6

g = 0.08

h

fy = 60 ksi

 = 0.8

0.07

Kmax

1.4 0.06

e

Pn

0.05

1.2

fs/fy = 0 0.04 0.03

1.0

0.25

K n = Pn / f

/

c

Ag

0.02

0.8

0.01

0.50

0.6 0.75

0.4

1.0

t = 0 .0 t = 0 035 .004 t = 0 .005

0.2

0.0 0.00

0.05

0.10

0.15

0.20

0.25

0.30

COLUMNS 3.15.4 - Nominal load-moment strength interaction diagram, C5-60.9 Rn = Pn e / f / c Ag h

1.8

h


1.6

g = 0.08

h

fy = 60 ksi

 = 0.9 0.07

1.4

Kmax

0.06

e 0.05

Pn

fs/fy = 0

1.2 0.04 0.03

0.25

Ag

1.0 Kn = Pn / f

/

c

0.02 0.01

0.50

0.8

0.75

0.6

1.0

0.4 t = 0 .0 t = 0 035 .0 t = 0 04 .005

0.2

0.0 0.00

0.05

0.10

0.15

0.20


0.25

0.30

0.35

1.6 g = 0.08

1.4

h

fy = 60 ksi

0.07

 = 0.6

0.06

1.2

h


Kmax

0.05

e

Pn

0.04 0.03

1.0

0.02

fs/fy = 0

Kn = P n / f

/

c

Ag

0.01

0.8 0.25

0.6 0.50

0.4 0.75

t t = 0.0 03 = t = 0 0 .0 0 4 5 .0 0 5

0.2

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 COLUMNS 3.16.2 - Nominal load-moment strength interaction diagram, C6-60.7 Rn = Pn e / f / c Ag h

1.6 g = 0.08

1.4

0.07

h

fy = 60 ksi

 = 0.7

0.06

1.2

h


Kmax

0.05

e

Pn

0.04 0.03

1.0

fs/fy = 0

0.02

Kn = Pn / f

/

c

Ag

0.01

0.8

0.25

0.6

0.50

0.75

0.4

0.2

1.0

t = t 0.0035 = 0. t 004 = 0. 005

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 Rn = Pn e / f / c Ag h

1.6

h


1.4

g = 0.08 0.07

h

fy = 60 ksi

 = 0.8 Kmax

0.06

1.2 0.04

1.0

Pn

e

0.05

fs/fy = 0

0.03

Kn = P n / f

/

c

Ag

0.02

0.8

0.25

0.01

0.50

0.6 0.75

0.4 1.0

0.2

t = 0 .0 t = 0 035 .004 t = 0 .005

0.0 0.30 0.25 0.20 0.15 0.10 0.05 0.00 COLUMNS 3.16.4 - Nominal load-moment strength interaction diagram, C6-60.9 Rn = Pn e / f / c Ag h

1.6 g = 0.08

1.4 0.07

fy = 60 ksi Kmax

0.05

e

0.03

Kn = Pn / f

/

c

Ag

0.02

0.8

Pn

fs/fy = 0

0.04

1.0

h

 = 0.9

0.06

1.2

h


0.25

0.01

0.50

0.6 0.75

1.0

0.4

0.2

0.0 0.00

t = 0.0035 t = 0.004 t = 0.005

0.05

0.10

0.15 Rn = Pn e / f / c Ag h

0.20

0.25

0.30

1.6

h


h

fy = 75 ksi

1.4

 = 0.6 g = 0.08 0.07

1.2

Kmax

0.06

Pn

e

0.05 0.04

1.0

0.03

Kn = Pn / f

/

c

Ag

0.02 0.01

0.8 fs/fy = 0

0.6 0.25

0.4 0.50

0.2

t = 0 t t = 0.0 .0038 0 = 0. 005 4

0.75

1.0


1.6

h


h

fy = 75 ksi

1.4 g = 0.08

 = 0.7

0.07

1.2

Kmax

0.06

e

Pn

0.05 0.04

1.0

0.03

Kn = Pn / f

/

c

Ag

0.02 0.01

0.8

fs/fy = 0

0.6

0.25

0.4

0.50

0.2

0.0 0.000

0.75

t t = 0.00 = 38 t = 0 0.004 .00 5

0.025

0.050

0.075

1.0

0.100


0.125

0.150

0.175

1.6

1.4

h


h

fy = 75 ksi

 = 0.8

g = 0.08 0.07

1.2

0.06

Kmax

e

Pn

0.05 0.04

1.0

0.03

Kn = Pn / f

/

c

Ag

0.02 0.01

fs/fy = 0

0.8

0.25

0.6

0.50

0.4 0.75

0.2

t = 0.00  t =t = 0.00 38 0.005 4

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 COLUMNS 3.17.4 - Nominal load-moment strength interaction diagram, C9-75.9 Rn = Pn e / f / c Ag h

1.6

h


h

fy = 75 ksi

1.4

 = 0.9 g = 0.08

1.2

0.07 0.06

Kmax

e

Pn

0.05

1.0

0.04

Kn = Pn / f

/

c

Ag

0.03

fs/fy = 0

0.02

0.8

0.01

0.25

0.6 0.50

0.4

0.2

0.75

t = 0 t = .0038 t = 0 0.004 .005

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 Rn = Pn e / f / c Ag h

1.4

h


h

fy = 75 ksi

1.2

 = 0.6

g = 0.08 0.07 0.06

1.0

Kmax

0.05

Pn

e

0.04 0.03 0.02 0.01

Kn = P n / f

/

c

Ag

0.8

fs/fy = 0

0.6

0.25

0.4 0.50

0.2 t

 t t = 0. = 0 003 .0 0 8 =0 4 .0 0 5

0.75

1.0


1.4 INTERACTION DIAGRAM C12-75.7 f /c = 12 ksi

h

h

fy = 75 ksi

1.2

g = 0.08

 = 0.7

0.07 0.06

Kmax

0.05

1.0

Pn

e

0.04 0.03 0.02 0.01

c

Ag

0.8

Kn = P n / f

/

fs/fy = 0

0.6 0.25

0.4 0.50

0.0 0.000

0.75

 t t = 0. = 0 003 t .0 0 8 =0 4 .0 0 5

0.2

0.025

0.050

1.0

0.075 Rn = Pn e / f / c Ag h

0.100

0.125

0.150

1.4

h


h

fy = 75 ksi

1.2

 = 0.8

g = 0.08 0.07 0.06

Kmax

0.05

1.0

Pn

e

0.04 0.03 0.02 0.01

0.8

Kn = P n / f

/

c

Ag

fs/fy = 0

0.6 0.25

0.50

0.4

0.75

 t t = 0. = 0 003 t .0 0 8 =0 4 .0 0 5

0.2

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 COLUMNS 3.18.4 - Nominal load-moment strength interaction diagram, C12-75.9 Rn = Pn e / f / c Ag h

1.4

h


1.2

h

fy = 75 ksi

g = 0.08

 = 0.9

0.07 0.06 0.05

1.0

Kmax

Pn

e

0.04 0.03 0.02 0.01

0.8

Kn = Pn / f

/

c

Ag

fs/fy = 0

0.25

0.6

0.50

0.4 0.75

t

0.2

0.0 0.000

0.025

 t t = 0. = 0 003 .0 0 8 =0 4 .0 0 5

0.050

1.0

0.075

0.100

0.125


0.150

0.175

0.200

2.4 2.2

g = 0.08

h

INTERACTION DIAGRAM S3-60.6 f /c = 3 ksi

h

fy = 60 ksi

 = 0.6

Kn = Pn / f

/

c

Ag

2.0

0.07

Kmax

1.8

0.06

1.6

0.05

1.4

0.04

e

Pn

fs/fy = 0

0.03

1.2 0.02

0.25

1.0 0.01

0.8

0.50

0.6 0.75

0.4

 t t = 0 t = 0.0.0035 = 0 04 .0 0 0 5

0.2

1.0

0.0 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 COLUMNS 3.19.2 - Nominal load-moment strength interaction diagram, S3-60.7 Rn = Pn e / f / c Ag h

2.4 2.2

h


g = 0.08

h

fy = 60 ksi

 = 0.7 2.0

0.07

Kmax

1.8

0.06

1.6

0.05

Kn = P n / f

/

c

Ag

1.4

e

Pn

fs/fy = 0

0.04

0.03

1.2

0.25 0.02

1.0 0.01

0.50

0.8 0.75

0.6 0.4

t = 0. t 003 t = 0.00 5 4 = 0. 005 0

0.2 0.0 0.00

0.05

0.10

1.0

0.15

0.20


0.25

0.30

0.35

0.40

2.4 2.2 2.0

g = 0.08


0.07

 = 0.8

h

h

fy = 60 ksi

Kmax 0.06

1.8

e

Pn

0.05

1.6

fs/fy = 0 0.04

Kn = P n / f

/

c

Ag

1.4 0.03

0.25

1.2 0.02

1.0

0.50 0.01

0.8 0.75

0.6 1.0

0.4

t = 0 t = 0 .0035 .0 t = 0 040 .005

0.2

0.0 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 COLUMNS 3.19.4 - Nominal load-moment strength interaction diagram, S3-60.9 Rn = Pn e / f / c Ag h

2.4 2.2 2.0

h


g = 0.08

h

fy = 60 ksi

 = 0.9

0.07

Kmax 0.06

1.8

e 0.05

Pn

fs/fy = 0

1.6 0.04

Kn = Pn / f

/

c

Ag

1.4 0.25

0.03

1.2 0.02

0.50

1.0 0.01

0.8

0.75

0.6

1.0

0.4

t = 0 t = 0 .0035 .0040 t = 0 .005

0.2 0.0 0.00

0.05

0.10

0.15

0.20

0.25

0.30


0.35

0.40

0.45

0.50

0.55

2.0

h


1.8

1.6

g = 0.08

h

fy = 60 ksi

 = 0.6

0.07

Kmax

0.06

1.4

e

Pn

0.05 0.04

1.2

fs/fy = 0

Kn = P n / f

/

c

Ag

0.03

1.0

0.02

0.25 0.01

0.8 0.50

0.6 0.75

0.4  t t = 0 . t = 0.000035 4 =0 .0 0 0 5

0.2

1.0

0.0 0.00 0.05 0.10 0.15 0.20 0.25 0.30 COLUMNS 3.20.2 - Nominal load-moment strength interaction diagram, S4-60.7 Rn = Pn e / f / c Ag h

2.0

1.8

g = 0.08

h


h

fy = 60 ksi

 = 0.7 0.07

Kmax

1.6 0.06

e

1.4

Pn

0.05

fs/fy = 0

0.04

1.2

Kn = Pn / f

/

c

Ag

0.03

1.0

0.25

0.02 0.01

0.8

0.50

0.6

0.75

0.4 t

0.2

0.0 0.00

0.05

 t t = 0.0 = 0. 035 = 0. 0040 005

0.10

1.0

0.15

0.20


0.25

0.30

0.35

2.0

1.8

g = 0.08

h


h

fy = 60 ksi

 = 0.8

0.07

1.6

Kmax 0.06

e

1.4

Pn

0.05

fs/fy = 0 0.04

1.2

Kn = Pn / f

/

c

Ag

0.03

0.25

1.0

0.02 0.01

0.50

0.8 0.75

0.6 1.0

0.4 t = t = 0 0.0035 .0 t = 0 040 .005

0.2

0.0 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 COLUMNS 3.20.4 - Nominal load-moment strength interaction diagram, S4-60.9 Rn = Pn e / f / c Ag h

2.0

1.8

g = 0.08


0.07

 = 0.9

h

h

fy = 60 ksi

1.6

Kmax

0.06

e

1.4

0.05

Pn

fs/fy = 0

0.04

1.2

Kn = Pn / f

/

c

Ag

0.03

1.0

0.25

0.02

0.50

0.01

0.8 0.75

0.6 1.0

0.4

t = 0 t = 0 .0035 t = 0.0040 .005

0.2

0.0 0.00

0.05

0.10

0.15

0.20

0.25


0.30

0.35

0.40

0.45

1.8

1.6

g = 0.08

h

fy = 60 ksi

 = 0.6

0.07

1.4

h


Kmax

0.06

1.2

Pn

e

0.05 0.04 0.03

Ag

1.0

fs/fy = 0

0.02

Kn = Pn / f

/

c

0.01

0.25

0.8

0.50

0.6

0.75

0.4  t t = 0.0 t = 0.00 035 =0 4 .0 0 0 5

0.2

1.0


1.8

h


1.6

g = 0.08 0.07

1.4

h

fy = 60 ksi

 = 0.7

Kmax

0.06

e

Pn

0.05

1.2 0.04

fs/fy = 0 0.03

Ag

1.0 c

0.02

Kn = P n / f

/

0.25 0.01

0.8 0.50

0.6 0.75

0.4

0.2

0.0 0.00

1.0

 t t = 0.0 = 0. 035 0 t = 0. 040 005

0.05

0.10

0.15 Rn = Pn e / f / c Ag h

0.20

0.25

0.30

1.8

h


1.6

g = 0.08

h

fy = 60 ksi

 = 0.8

0.07

1.4

Kmax 0.06

e

Pn

0.05

1.2

fs/fy = 0

0.04 0.03

Ag

1.0 0.25

Kn = Pn / f

/

c

0.02 0.01

0.8 0.50

0.6

0.75

0.4

1.0

t = 0 t = 0 .0035 t = 0 .0040 .005

0.2


1.8

h


1.6

g = 0.08

h

fy = 60 ksi

 = 0.9

0.07

1.4

Kmax 0.06

e

Pn

0.05

1.2

fs/fy = 0 0.04 0.03

1.0

0.25

Kn = Pn / f

/

c

Ag

0.02 0.01

0.50

0.8

0.75

0.6

1.0

0.4 t = t = 0.0035 t = 00.0040 .005

0.2

0.0 0.00

0.05

0.10

0.15

0.20


0.25

0.30

0.35

1.6

1.4

g = 0.08


0.07

 = 0.6

fy = 60 ksi

0.06

1.2

h

h

Kmax

0.05

e

0.04

Pn

0.03

1.0

0.02

fs/fy = 0

Kn = Pn / f

/

c

Ag

0.01

0.8 0.25

0.6 0.50

0.4

0.75

t

0.2

 t t = = 0 0 .0 0 . = 0 004035 .0 0 5

1.0


1.6 g = 0.08

1.4

0.07 0.06

1.2

h


h

fy = 60 ksi

 = 0.7

Kmax

0.05

e

0.04

Pn

0.03

1.0

fs/fy = 0

0.02

Kn = P n / f

/

c

Ag

0.01

0.8

0.25

0.50

0.6

0.75

0.4

0.2

 t t = 0.0 = 0 t 0.0040 35 = 0. 005

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 0.275 Rn = Pn e / f / c Ag h

1.6

h

INTERACTION DIAGRAM S6-60.8 f /c = 6 ksi g = 0.08

1.4

h

fy = 60 ksi

 = 0.8

0.07

Kmax

0.06

1.2

0.05

e

Pn

0.04

1.0

fs/fy = 0

0.03

Kn = Pn / f

/

c

Ag

0.02

0.25

0.01

0.8

0.50

0.6 0.75

0.4

1.0

t = 0 t = 0 .0035 .0 t = 0 040 .005

0.2

0.0 0.00 0.05 0.10 0.15 0.20 0.25 0.30 COLUMNS 3.22.4 - Nominal load-moment strength interaction diagram, S6-60.9 Rn = Pn e / f / c Ag h

1.6 g = 0.08

1.4 0.07

h

fy = 60 ksi

 = 0.9 Kmax

0.06

1.2

h


0.05

e

Pn

0.04

fs/fy = 0

1.0

0.03 0.02

K n = Pn / f

/

c

Ag

0.25

0.8

0.01

0.50

0.6

0.75

1.0

0.4 t = 0 t = 0 .0035 t = 0 .0040 .005

0.2

0.0 0.00

0.05

0.10

0.15

0.20


0.25

0.30

0.35

1.6

h


h

fy = 75 ksi

1.4

 = 0.6 g = 0.08 0.07

1.2

0.06

Kmax

Pn

e

0.05 0.04

1.0

0.03 0.02

Kn = Pn / f

/

c

Ag

0.01

0.8 fs/fy = 0

0.6 0.25

0.4

0.50

t

0.2

 t t = 0 = 0 .00 = 0 .004038 .0 0 5

0.75

1.0


1.6

h


h

fy = 75 ksi

1.4

 = 0.7

g = 0.08 0.07

1.2

0.06

Kmax

Pn

e

0.05 0.04

1.0

0.03 0.02

Kn = Pn / f

/

c

Ag

0.01

0.8

fs/fy = 0

0.6

0.25

0.50

0.4

0.2

0.0 0.000

0.75

 t t = 0.0 t = = 0.004038 0.005 0

0.025

0.050

0.075

1.0

0.100

0.125


0.150

0.175

0.200

1.6

h


1.4

h

fy = 75 ksi

 = 0.8

g = 0.08 0.07

1.2

0.06

Kmax

e

0.05

Pn

0.04

1.0

0.03

Kn = Pn / f

/

c

Ag

0.02 0.01

fs/fy = 0

0.8

0.25

0.6 0.50

0.4 0.75

t t = = 0.0038 t = 0 0.0040 .005

0.2

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 COLUMNS 3.23.4 - Nominal load-moment strength interaction diagram, S9-75.9 Rn = Pn e / f / c Ag h

1.6

h


h

fy = 75 ksi

1.4

 = 0.9 g = 0.08

1.2

0.07

Kmax

e

0.06

Pn

0.05

1.0

0.04 0.03

Kn = Pn / f

/

c

Ag

0.02

0.8

fs/fy = 0

0.01

0.25

0.6 0.50

0.75

0.4

0.2

 t =t =0 0.0038 .0040 t = 0 .005

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 Rn = Pn e / f / c Ag h

1.4 INTERACTION DIAGRAM S12-75.6 f /c = 12 ksi

h

h

fy = 75 ksi

1.2

g = 0.08

 = 0.6

0.07 0.06

Kmax

0.05

1.0

e

Pn

0.04 0.03 0.02 0.01

Kn = Pn / f

/

c

Ag

0.8

fs/fy = 0

0.6 0.25

0.4 0.50

t

0.2

 t t = 0.0 03 = = 0. 0.0040 8 0 05 0

0.75

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 COLUMNS 3.24.2 - Nominal load-moment strength interaction diagram, S12-75.7 / Rn = Pn e / f

c

Ag h

1.4 INTERACTION DIAGRAM S12-75.7 f /c = 12 ksi

h

h

fy = 75 ksi

1.2

g = 0.08

 = 0.7

0.07 0.06

Kmax

0.05

1.0

e

Pn

0.04 0.03 0.02 0.01

Kn = Pn / f

/

c

Ag

0.8 fs/fy = 0

0.6 0.25

0.4

0.50

0.2

0.0 0.000

0.75

 t t = 0.0 t = 0.004038 = 0. 0 05 0

0.025

0.050

0.075

1.0

0.100


0.125

0.150

0.175

1.4

h


h

fy = 75 ksi

1.2

g = 0.08

 = 0.8

0.07 0.06

Kmax

0.05

1.0

Pn

e

0.04 0.03 0.02 0.01

Ag

0.8

Kn = Pn / f

/

c

fs/fy = 0

0.6

0.25

0.50

0.4 0.75

t = t = 0.0038 t = 0.0040 0.005

0.2

1.0

0.0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 COLUMNS 3.24.4 - Nominal load-moment strength interaction diagram, S12-75.9 Rn = Pn e / f / c Ag h

1.4 h


1.2

g = 0.08

h

fy = 75 ksi

 = 0.9

0.07 0.06 0.05

1.0

Kmax

e

0.04

Pn

0.03 0.02 0.01

0.8

Kn = Pn / f

/

c

Ag

fs/fy = 0

0.25

0.6

0.50

0.4

0.75

 t =t 0= 0.0038 .0040 t = 0 .005

0.2

0.0 0.000

0.025

0.050

0.075

0.100

1.0

0.125


0.150

0.175

0.200

0.225

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