The Reinforced Concrete Design Handbook - ACI 318-19 [PDF]

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An ACI Manual



ACI Reinforced Concrete Design Handbook A Companion to ACI 318-19



Volume 3: Design Aids MNL-17(21)



ACI MNL-17(21)



ACI REINFORCED CONCRETE DESIGN HANDBOOK A Companion to ACI 318-19



VOLUME 2



VOLUME 1 INTRODUCTION



RETAINING WALLS



STRUCTURAL SYSTEMS



SERVICEABILITY



STRUCTURAL ANALYSIS



STRUT-AND-TIE METHOD



DURABILITY



ANCHORING TO CONCRETE



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



ACI MNL-17(21) Volume 3



ACI REINFORCED CONCRETE DESIGN HANDBOOK A Companion to ACI 318-19



First Printing April 2021 ISBN: 978-1-64195-138-8 ACI REINFORCED CONCRETE DESIGN HANDBOOK Volume 3 ~ Tenth 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 non-infringement. 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, through electronic subscription, or reprint, and may be obtained by contacting ACI. ACI codes, specifications, and practices are made available in the ACI Collection of Concrete Codes, Specifications, and Practices. The online subscription to the ACI Collection is always updated, and includes current and historical versions of ACI’s codes and specifications (in both inch-pound and SI units) plus new titles as they are published. The ACI Collection is also available as an eight-volume set of books and a USB drive. American Concrete Institute 38800 Country Club Drive Farmington Hills, MI 48331 Phone: +1.248.848.3700 Fax: +1.248.848.3701 Managing Editor: H. R. Trey Hamilton Staff Engineer: Sureka Sumanasooriya Technical Editor: Carl R. Bischof Director, Publishing Services: Lauren E. Mentz Supervisor, Publishing Services: Ryan M. Jay Lead Production Editor: Kelli R. Slayden Production Editors: Erin N. Azzopardi, Kaitlyn J. Dobberteen, Tiesha Elam, and Hannah E. Genig Graphic Designers: Paul F. Sullivan and Aimee Kahaian Manufacturing: Marie Fuller www.concrete.org



VOLUME 3: CONTENTS APPENDIX A—REFERENCE TABLES



1



APPENDIX B—ANALYSIS TABLES



9



APPENDIX C—SECTIONAL PROPERTIES



23



APPENDIX D––COLUMN INTERACTION DIAGRAMS 25







MNL-17(21)—REFERENCE TABLES



1



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.



Table A-2—Area of bars in a section 1 ft wide Cross section area of bar As (or As′), in.2 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



6.5



7.0



0.19



0.34



0.53



0.75



7.5



0.18



0.32



0.50



0.70



1.03



1.35



1.71



2.18



2.67



3.86



6.86



7.0



0.96



1.26



1.60



2.03



2.50



3.60



6.40



7.5



8.0



0.17



0.30



0.47



8.5



0.16



0.28



0.44



0.66



0.90



1.19



1.50



1.91



2.34



3.38



6.00



8.0



0.62



0.85



1.12



1.41



1.79



2.20



3.18



5.65



9.0



0.15



0.27



8.5



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.



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REF. TABLES



MNL-17(21)—REFERENCE TABLES



2 2



DETAILING MANUALCONCRETE IN ACCORDANCE WITH ACI 318-19—MNL-66(19) ACI REINFORCED DESIGN HANDBOOK—MNL-17(21)



Table A-3—Minimum beam web widths required for two or more bars in one layer for cast-in-place nonprestressed concrete Reference: ACI 318-19 Sections 25.3.2, 25.2.1, 20.5.1.3.1, 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 – 1/2db ≥ 2.0 in. cover required for longitudinal bars and these assumptions are made:



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MNL-66(19)—REFERENCE TABLES TABLES MNL-17(21)—REFERENCE



3



REF. TABLES







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ACI REINFORCED DESIGN HANDBOOK—MNL-17(21) DETAILING MANUALCONCRETE IN ACCORDANCE WITH ACI 318-19—MNL-66(19)



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MNL-17(21)—REFERENCE TABLES TABLES MNL-66(19)—REFERENCE



5



REF. TABLES







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6



ACI REINFORCED CONCRETE DESIGN HANDBOOK—MNL-17(21)



Table A-7—Basic development length ratios of bars in tension Reference: ACI 318-19 Sections 25.4.2.2 and 25.4.2.3 Development length ratios:



ψt ψeψ g f y 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 60,000 psi



fy Bar size



Category α



100,000 psi



80,000 psi



fc′ 3000 4000 5000 6000 8000 10,000 3000 4000 5000 6000 8000 10,000 3000 4000 5000 6000 8000 10,000 psi psi psi psi psi psi psi psi psi psi psi psi psi psi psi psi psi psi



#3 to #6



I



1/25



44



38



34



31



27



24



67



58



52



48



41



37



95



82



74



67



58



52



II



3/50



66



57



51



46



40



36



101



87



78



71



62



55



142



123



110



101



87



78



#7 to #18



I



1/20



55



47



42



39



34



30



84



73



65



59



51



46



119



103



92



84



73



65



II



3/40



82



71



64



58



50



45



126



109



98



89



77



69



178



154



138



126



109



98



Notes: 1. See category chart for Categories I and II. 2. ψt = casting position (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. 4. ψg = 1.0, 1.15, and 1.3 for fy = 60 ksi, 80 ksi, and 100 ksi, respectively



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MNL-17(21)—REFERENCE TABLES TABLES MNL-66(19)—REFERENCE



7



REF. TABLES







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8



ACI REINFORCED CONCRETE DESIGN HANDBOOK—MNL-17(21)



Table A-8—Basic development length ℓdh of standard hooks in tension Reference: ACI 318-19 Sections 25.3.1, 25.4.3.1, 25.4.3.2, and 25.4.10.1.



ψ ψ ψ ψ   dh =  e r o c  db1.5  55λ f c′  This table is calculated with ψe = 1.0, ψt = 1.0, and λ = 1.0. Basic development length ℓdh, in., of standard hooks in tension 60,000 psi



fy Bar size



100,000 psi



80,000 psi



fc′ 3000 4000 5000 6000 8000 db, in. psi psi psi psi psi



10,000 psi



3000 4000 5000 6000 8000 psi psi psi psi psi



10,000 psi



3000 4000 5000 6000 8000 psi psi psi psi psi



10,000 psi



8db, in.



#3



3/8



3.7



3.4



3.3



3.2



2.8



2.5



4.9



4.6



4.1



4.3



3.7



3.3



6.1



5.7



5.5



5.4



4.7



4.2



3



#4



1/2



5.6



5.3



5.1



5.0



4.3



3.9



7.5



7.0



6.3



6.6



5.7



5.1



9.4



8.8



8.5



8.3



7.2



6.4



4



#5



5/8



7.9



7.4



7.1



7.0



6.0



5.4



10.5



9.8



8.8



9.3



8.0



7.2



13.1



12.3



11.9



11.6



10.0



9.0



5



#6



3/4



10.3



9.7



9.4



9.1



7.9



7.1



13.8



12.9



11.6



12.2



10.6



9.4



17.2



16.2



15.6



15.2



13.2



11.8



6



#7



7/8



13.0



12.2



11.8



11.5



10.0



8.9



17.4



16.3



14.6



15.4



13.3



11.9



21.7



20.4



19.6



19.2



16.6



14.9



7



#8



1



15.9



14.9



14.4



14.1



12.2



10.9



21.2



19.9



17.8



18.8



16.3



14.5



26.6



24.9



24.0



23.5



20.3



18.2



8



#9



1.128



19.1



17.9



17.3



16.9



14.6



13.1



25.5



23.9



21.4



22.5



19.5



17.4



31.8



29.8



28.8



28.1



24.4



21.8



9



#10



1.27



22.8



21.4



20.6



20.2



17.5



15.6



30.4



28.5



25.5



26.9



23.3



20.8



38.0



35.7



34.3



33.6



29.1



26.0



10



#11



1.41



26.7



25.0



24.1



23.6



20.4



18.3



35.6



33.4



29.8



31.4



27.2



24.4



44.5



41.7



40.2



39.3



34.0



30.4



11



#14



1.693



35.1



32.9



31.7



31.0



26.9



24.0



46.8



43.9



39.3



41.4



35.8



32.0



58.5



54.9



52.9



51.7



44.8



40.1



14



#18



2.257



54.0



50.7



48.8



47.8



41.4



37.0



72.0



67.6



60.4



63.7



55.1



49.3



90.0



84.5



81.4



79.6



68.9



61.7



18



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. For ψe, ψr, ψo, ψc, and λ values, refer to ACI 318-19 Table 25.4.3.2. 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.



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MNL-17(21)—ANALYSIS TABLES



MNL-17(19)—ANALYSIS TABLES Reproduced with permission form the Canadian Portland Cement Association.



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ACI REINFORCED CONCRETE DESIGN HANDBOOK—MNL-17(21)



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MNL-17(21)—ANALYSIS TABLES



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MNL-17(21)—ANALYSIS TABLES



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MNL-17(21)—ANALYSIS TABLES



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MNL-17(21)—ANALYSIS TABLES



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MNL-17(21)—ANALYSIS TABLES



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ACI REINFORCED CONCRETE DESIGN HANDBOOK—MNL-17(21)



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MNL-17(21)—ANALYSIS TABLES



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ACI REINFORCED CONCRETE DESIGN HANDBOOK—MNL-17(21)



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MNL-17(21)—SECTIONAL PROPERTIES



23



MNL-17(19)—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 I1 =



d4 12



A= = I



d4 I2 = 3



πd 2 = 0.7854d 2 4 vd 4 = 0.0491d 4 64 R=



R1 = 0.2887d



d 4



R2 = 0.57774d A= d 2 y = 0.7071d I1 =



A = 0.8660d 2



d4 12



I = 0.060d 4 R = 0.264d



R = 0.2887d A = bd I1 =



bd 3 12



I2 =



bd 3 3



A = 0.8284d 2 I = 0.055d 4 R = 0.257d



R1 = 0.2887d R2 = 0.57774d



y= I= R=



A=



bd 2



I1 =



bd 3 36



I2 =



bd 3 12



A = bd bd 2



b + d2 bd 6(b 2 + d 2 ) 3



3



bd 6(b 2 + d 2 )



R1 = 0.236d R2 = 0.408d



A = bd y=



b sin ∞ + d cos ∞ 2 I=



R=



b 2 sin 2 (∞ + d 2 cos 2 ∞) 12



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R=



A=



d (b + b ′ ) 2



y=



d (2b + b ′ ) 3(b + b ′ )



y=



d (b + 2b ′ ) 3(b + b ′ )



d 2 (b 2 + 4bb ′ + b ′ 2 ) 36(b + b ′ )



d 2(b 2 + 4bb ′ + b ′ 2 ) 6(b + b ′ )



24



ACI REINFORCED CONCRETE DESIGN HANDBOOK—MNL-17(21)



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 Section of parabola y2 =



A = bt + bc′ y=



d 2b ′ + t 2 (b − b ′ ) 2(bt + bc′ )



y1 = d – y b ′y + by 3 − (b − b ′ )( y − t )3 I= 3



For parabola:



A=



3 1



I A



R=



b2 x d For compliment:



2bd 3



A=



I1 =



8 bd 3 175



I2 =



19 3 bd 480



I1 =



bd 3



37 bd 3 2100



I2 =



1 3 bd 80



A = bt + bc′ y=



d 2b ′ + t 2 (b − b ′ ) 2(bt + bc′ )



A = d2 – a2 I=



y1 = d–y b ′y13 + by 3 − (b − b ′ )( y − t )3 I= 3 I A



R=



A = bt + y=



R=



c(a + b′ ) 2



d 4 − a4 12 d 2 + a2 12



A = bd – ac



3bt 2 + 3b ′c(d + t ) + c(a − b ′ )(c + 3t ) 3 [ 2bt + c(a + b ′ )]



I=



y1 = d – y 4bt 2 + c 3 (3b ′ + a ) I= − A( y − t ) 2 12 I R= A



R=



bd 3 − ac 3 12 bd 3 − ac 3 12(bd − ac)



Ellispe A = 0.7854bd = I1



wbd 3 = 0.0491bd 3 64



A=



π(d 2 − d12 ) = 0.7854(d 2 − d12 ) 4



= I2



wb3 d = 0.0491b3 d 64



I=



π(d 4 − d14 ) = 0.0491(d 4 − d14 ) 64



R1 =



d 4



R2 =



b 4



Parabola Equation: y2 =



b2 4d x



A=



2bd 3



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R = 1 / 4 d 2 + d12



A = 0.8284d 2 − 0.7854d12 = 0.7854(1.055d 2 − d12 ) I = 0.0547 d 4 − 0.0491d14 = 0.0491(1.115d 4 − d14 ) = 0.0491 (1.056) 2 d 4 − d14  R = 1 / 4 1.056d 2 + d12



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 318-19. 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 318-19, 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 = β1c), 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 318-19 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 318-19 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 φ 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 318-19, Section 22.4.2.2). = Pn,max 0.8 0.85 f c′ ( Ag − Ast ) + f y Ast   



(D.1a)



Then, K max = Pn , max / f c′ Ag



(D.1b)



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 ≤ φM n = φ M 2 nx + M 2 ny



(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 0.1 fc′ 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 α



 M nx   M ny   +   M nox   M noy



α



 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 φ, the equation is unchanged. Thus, Mux, Muy, Mox, and Moy, which should correspond to φMnx, φMny, φMnox, and φMnoy, 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 = βM ox and M ny = β 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 β



 M ny + M  noy



  



l og 0.5/ log β



= 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 ox M oy M ox



 M oy   1 − β  M= M ny + M nx    oy  M ox   β 



,



M M= M nx + M ny  ox ox M  oy



,



1− β      β 



(D.2.3e)



(D.2.3f)



For rectangular sections with reinforcement equally distributed on all four faces, the above equations can be approximated by  b 1− β  M= M ny + M nx     oy  h  β 



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 section



10.5



Discussion Calculation Determine column section size. Given: h= 20 in. and b = 16 in. Determine reinforcement ratio ρg Pn = 800 kip using known values of variables on Mn = 5600 in.-kip appropriate interaction diagrams h = 20 in. and compute required cross b = 16 in. section area Ast of longitudinal Ag = b × h = 20 × 16 = 320 in.2 reinforcement. Compute



Kn =



Pn f c ′ Ag



Compute



Rn =



Mn 5600 = Rn = 0.22 f c ′ Ag h (4)(320)(20)



Estimate



γ≈



h−5 h



Determine the appropriate interaction diagrams.



= Kn



γ≈



Design aid



800 = 0.625 (4)(320)



20 − 5 =0.75 20



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 = 2790 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 section



Discussion Determine trial column dimension Pn = 943 kip b corresponding to known values Mn = 3986 in.-kip of variables on appropriate h = 16 in. interaction diagrams. ρg = 0.039 Assume a series of trial column b 24 sizes b, in inches, and compute Ag 384 Ag = b x h, in.2 Compute



Compute



Estimate 10.5



Kn =



Rn =



γ≈



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



Design aid



Calculation



26 416 943 (4)(416)



28 448



= 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 section



10.5



Discussion Calculation Determine reinforcement ration ρg 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



γ≈



Pn



= Kn



f c ’ Ag Mn



= Rn



f c ’ Ag h



h−5 h



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



γ≈



Design aid



943 = 0.73 (4)(324) 3771 = 0.16 (4)(324)(18) 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 S4-60.8. 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 ρg ≤ 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 section



Discussion Calculation 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 2 nx + M 2 ny



Compute Compute



Estimate



Kn =



Rn =



γ≈



29492 + 11832 = 3177 in.-kip



M nr =



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



10.5



Design aid



14



16



18



196



256



324



Pn



297



f c ’ Ag



(5)(196)



Mn f c ’ Ag h



3177 (5)(196)(14)



h−5 h



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



= 0.30



= 0.23



297 (5)(256)



= 0.23



3177 (5)(256)(16)



0.64



0.69



0.058 0.026 Therefore, try h = 15 in. Determine reinforcement ratio ρg Ag = h2 = 152 = 225 in.2 using known values of variables on Pn = 297 kip appropriate interaction diagrams Mn = 3177 in.-kip and compute required cross section area Ast of longitudinal reinforcement. Kn =



Pn f c ’ Ag



= 0.18



3177 (5)(324)(18)



= 0.11



0.72



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



Compute



= 0.16



297 (5)(324)



Kn =



297 (5)(225)



= 0.264



0.012



R5-60.6 R5-60.7 R5-60.8



ACI 318 section



10.5



Discussion Compute



Rn =



Estimate



γ≈



Calculation Mn



= Rn



f c ’ Ag h



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 = ρg Ag and add about 15 percent for skew bending.



γ≈



Design aid



3177 = 0.188 (5)(225)(15) 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 2 Required Ast = 0.037 x 225 in. = 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 section



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 = h 2 , in.2 Compute 113 Compute Estimate



10.5



Rn =



γ≈



Mn f c ’ Ag h



h−5 h



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



Ag =



Compute



h=2



Pn f c ’K n Ag



π



686 (5)(113)(12)



= 0.10



0.58



Design aid



Calculation



16



20



201



314



686 (5)(201)(16)



= 0.04



0.69



686 (5)(314)(20)



= 0.02



0.75



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 C5-60.6 0.9 1.18 1.25 C5-60.7 C5-60.8 0.9 1.17 1.24 298



229



217



19.5



17.1



16.6



Therefore, try 17 in. diameter column. Determine reinforcement ratio ρg using known values of variables on 2 appropriate interaction diagrams  17  = Ag π= 227 in.2   and compute required cross  2 section area Ast of longitudinal reinforcement.



ACI 318 section



Discussion Compute



10.5



Kn =



Compute



Rn =



Estimate



γ≈



Calculation Pn f c ’ Ag Mn



f c ’ Ag h



h−5 h



Determine the appropriate interaction diagrams. Read ρg for Kn and Rn values. Compute required Ast from Ast = ρg Ag.



Kn =



Rn =



γ≈



1343 (5)(227)(17)



1343 (5)(227)(17)



Design aid



= 1.18



= 0.0356



17 − 5 =0.71 17



For a circular column, fc’ = 5 ksi, fy = 60 ksi, use diagram C5-60.7. For Kn = 1.18 and Rn = 0.0356: C5-60.7 γ = 0.70 ρg = 0.040 Required Ast = 0.04 x 227 in.2 = 9.08 in.2



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, Section 6.6.3.1.1,) Second-order analysis of frames for design of slender columns fc′, ksi



3



Ec , ksi



4



5



6



7



8



9



10



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



0.2



901



1008 1104 1192 1275 1352 1425



Notes: 1. Alternatively, the following more refined values of I can be used: For columns:  A I =  0.8 + 25 st  Ag 



 Mu P  − 0.5 u  I g ≤ 0.5 Ig  1 − 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: 𝐼𝐼 = (0.10 + 25𝜌𝜌) �1.2 − 0.2



2. 3. 4. 5.



𝑏𝑏𝑤𝑤 � 𝐼𝐼 ≤ 0.5𝐼𝐼𝑔𝑔 𝑑𝑑 𝑔𝑔



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)



Hinged



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



Stiff



Elastic



Elastic



Hinged



TOP



k



BOTTOM



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



b



γh



h



γh



h



Equal reinforcement on four sides 3 I se = 0.18 ρt b h γ 2



Bars in two end faces 3



I se = 0.25 ρt b h γ 2 b



γh



h



γh



h



Uniformly distributed reinforcement I se = 0.10 ρt h4 γ 2



Bars in two side faces I se = 0.17 ρt b h3γ 2 (3 bars per face) I se = 0.12 ρt b h3γ 2 (6 bars per face)



b = 2h



b



h h = 2b



γh



γh Bars uniformly spaced on all sides I se = 0.22 ρt b h3γ 2



Bars uniformly spaced on all sides I se = 0.13 ρt b h3γ 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.



D.5—Interaction diagrams 2.4 2.2



Ag c /



Kn = Pn / f



ρg = 0.08 0.07



1.8



0.06



1.6



0.05



1.2



γh



fy = 60 ksi



γ = 0.6



2.0



1.4



h



INTERACTION DIAGRAM R3-60.6 f /c = 3 ksi



Kmax e



0.04



Pn



fs/fy = 0



0.03



0.25



0.02



1.0 0.01



0.8



0.50



0.6 0.75



0.4



ε ε t=0 ε t t = 0.0.0035 = 0 04 .0 0 5



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



Rn = Pn e / f / c Ag h



2.4 2.2 2.0 1.8 1.6



Kn = Pn / f



/



c



Ag



1.4 1.2



ρg = 0.08



INTERACTION DIAGRAM R3-60.7 f /c = 3 ksi



0.07



γ = 0.7



fy = 60 ksi



Kmax 0.06



e



fs/fy = 0



0.04



0.03



0.25



0.01



0.50



0.8



0.75



0.6 0.4



ε ε t t = 0 .0 0 ε t = 0.00435 = 0. 005 0



0.2 0.0 0.00



Pn



0.05



0.02



1.0



h



γh



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



2.4 2.2 2.0 1.8 1.6



Kn = Pn / f



/



c



Ag



1.4



ρg = 0.08



h



INTERACTION DIAGRAM R3-60.8 f /c = 3 ksi



γh



fy = 60 ksi



γ = 0.8



0.07



Kmax



0.06



Pn



e 0.05



fs/fy = 0



0.04



0.03



0.25



1.2 0.02



1.0



0.50



0.01



0.8 0.75



0.6 1.0



0.4



ε t = 0.0035 ε t = 0.004 ε 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 0.50 0.55 0.60 Rn = Pn e / f / c Ag h



2.4 2.2 2.0



ρg = 0.08



INTERACTION DIAGRAM R3-60.9 f /c = 3 ksi



0.07



γ = 0.9



h



γh



fy = 60 ksi



Kmax



0.06



1.8



e



0.05



1.6



Pn



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



ε t = 0.0035 ε t = 0.004 ε 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 1.6



h



INTERACTION DIAGRAM R4-60.6 f /c = 4 ksi



ρg = 0.08



γh



fy = 60 ksi



γ = 0.6 0.07



Kmax 0.06



1.4 1.2



0.05 0.04



fs/fy = 0



Kn = Pn / f



/



c



Ag



0.03



1.0



Pn



e



0.02



0.25



0.01



0.8 0.50



0.6 0.75



0.4 ε ε t t = 0 .0 0 = 0. εt 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



Rn = Pn e / f / c Ag h



2.0 1.8



ρg = 0.08



1.4



fy = 60 ksi



Kmax



0.06



e



0.05 0.04



Kn = Pn / f



/



c



Ag



1.2 1.0



γh



γ = 0.7



0.07



1.6



h



INTERACTION DIAGRAM R4-60.7 f /c = 4 ksi



Pn



fs/fy = 0



0.03 0.02



0.25



0.01



0.8



0.50



0.6



0.75



0.4 0.2 0.0 0.00



1.0



ε ε t t = 0 .0 ε t = 0.004035 = 0. 005 0



0.05



0.10



0.15



0.20



Rn = Pn e / f / c Ag h



0.25



0.30



0.35



0.40



2.0 1.8 1.6 1.4



ρg = 0.08



INTERACTION DIAGRAM R4-60.8 f /c = 4 ksi



0.07



γ = 0.8



fy = 60 ksi Kmax



0.06



e



0.05



Ag c /



Kn = Pn / f



1.0



Pn



fs/fy = 0



0.04



1.2



h



γh



0.03



0.25



0.02 0.01



0.50



0.8



0.75



0.6



1.0



0.4



εt= ε t = 0.0035 ε 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



Rn = Pn e / f / c Ag h



2.0 ρg = 0.08



INTERACTION DIAGRAM R4-60.9 f /c = 4 ksi



0.07



γ = 0.9



0.06



Kmax



1.8 1.6 1.4



fy = 60 ksi



e



0.05



Ag c /



Kn = Pn / f



Pn



fs/fy = 0



0.04



1.2



h



γh



0.03



0.25



1.0



0.02 0.01



0.50



0.8



0.75



0.6



1.0



0.4



ε t = 0.0 5 ε t = 0.00403 ε t = 0.0050



0.2 0.0 0.00



0.05



0.10



0.15



0.20



0.25



0.30



Rn = Pn e / f / c Ag h



0.35



0.40



0.45



0.50



1.8 1.6



h



INTERACTION DIAGRAM R5-60.6 f /c = 5 ksi ρg = 0.08



γh



fy = 60 ksi



γ = 0.6



0.07



1.4 1.2



e



Pn



0.05 0.04 0.03



fs/fy = 0



0.02



Kn = Pn / f



/



c



Ag



1.0



Kmax



0.06



0.8



0.01



0.25



0.6



0.50



0.4



0.75



ε ε t t = 0 .0 = 0. 035 0 εt = 0. 040 005



0.2 0.0 0.00



0.05



1.0



0.10



0.15



0.20



0.25



0.30



Rn = Pn e / f / c Ag h



1.8 1.6



h



INTERACTION DIAGRAM R5-60.7 f /c = 5 ksi ρg = 0.08



γh



fy = 60 ksi



γ = 0.7



0.07



1.4



Kmax



0.06



e



0.05



1.2



0.04



fs/fy = 0



0.03 0.02



c



Ag



1.0



0.25



/



Kn = Pn / f



0.8



Pn



0.01



0.50



0.6



0.75



0.4 0.2 0.0 0.00



1.0



εt ε t = = 0.0035 ε t = 0.0040 0.005



0.05



0.10



0.15



0.20



Rn = Pn e / f / c Ag h



0.25



0.30



0.35



1.8 1.6



γh



fy = 60 ksi



ρg = 0.08



γ = 0.8



0.07



1.4



h



INTERACTION DIAGRAM R5-60.8 f /c = 5 ksi



Kmax



0.06



e



0.05



1.2



0.04



Pn



fs/fy = 0



1.0



0.02



0.25



Kn = Pn / f



/



c



Ag



0.03



0.8



0.01



0.50



0.6



0.75



0.4



1.0



ε t = 0.0 5 ε t = 0.00403 ε 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



Rn = Pn e / f / c Ag h



1.8 1.6



ρg = 0.08



γh



fy = 60 ksi



γ = 0.9



0.07



1.4



h



INTERACTION DIAGRAM R5-60.9 f /c = 5 ksi



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 0.8



0.01



0.50



0.6



0.75



1.0



0.4 ε t = 0.0035 ε t = 0.004 ε t = 0.005 0



0.2 0.0 0.00



0.05



0.10



0.15



0.20



0.25



Rn = Pn e / f / c Ag h



0.30



0.35



0.40



0.45



1.6



1.4



h



INTERACTION DIAGRAM R6-60.6 f /c = 6 ksi



ρg = 0.08



γh



fy = 60 ksi



0.07



γ = 0.6



0.06



1.2



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



ε ε t t = 0 .0 = 0. 035 0 εt = 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 Rn = Pn e / f / c Ag h



1.6



1.4



ρg = 0.08 0.07



INTERACTION DIAGRAM R6-60.7 f /c = 6 ksi fy = 60 ksi



γ = 0.7



Kmax



0.06



1.2



h



γh



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



0.2



0.0 0.00



1.0



ε ε t =t = 0.0035 ε t = 0.0040 0.005



0.05



0.10



0.15 Rn = Pn e / f / c Ag h



0.20



0.25



0.30



1.6 ρg = 0.08



1.4



γh



fy = 60 ksi



γ = 0.8



0.07



Kmax



0.06



1.2



h



INTERACTION DIAGRAM R6-60.8 f /c = 6 ksi



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



0.50



0.6



0.75



0.4



0.2



0.0 0.00



1.0



εt ε t = =0 0.0035 .0 ε t = 0 040 .005



0.05



0.10



0.15



0.20



0.25



0.30



γh



ρg = 0.08



INTERACTION DIAGRAM R6-60.9 f /c = 6 ksi



0.07



γ = 0.9



0.06



Kmax



0.35



Rn = Pn e / f / c Ag h



1.6



1.4



1.2



h



fy = 60 ksi



0.05



e



Pn



0.04



1.0



fs/fy = 0



0.03



Kn = Pn / f



/



c



Ag



0.02



0.8



0.25



0.01



0.50



0.6 0.75



0.4



1.0



εt= ε t = 0 0.0035 ε t = 0 .0040 .005



0.2



0.0 0.00



0.05



0.10



0.15



0.20



Rn = Pn e / f / c Ag h



0.25



0.30



0.35



0.40



1.6



1.4



h



INTERACTION DIAGRAM R9-75.6 f /c = 9 ksi



γh



fy = 75 ksi



ρg = 0.08



γ = 0.6



0.07



1.2



0.06



Kmax



0.05



Pn



e



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



ε ε t=0 ε 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



Rn = Pn e / f / c Ag h



1.6



h



INTERACTION DIAGRAM R9-75.7 f /c = 9 ksi



γh



fy = 75 ksi



1.4



γ = 0.7 ρg = 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



ε ε t= ε 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



INTERACTION DIAGRAM R9-75.8 f /c = 9 ksi



γh



fy = 75 ksi



1.4



γ = 0.8 ρg = 0.08



1.2



0.07



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



ε ε t=0 ε t t = 0.0.0038 = 0. 040 005 0



0.2



0.0 0.00



0.05



1.0



0.10



0.15



0.20



0.25



0.30



Rn = Pn e / f / c Ag h



1.6



h



INTERACTION DIAGRAM R9-75.9 f /c = 9 ksi



γh



fy = 75 ksi



1.4 ρg = 0.08



γ = 0.9



0.07



1.2



0.06



Kmax



e



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



0.4



0.75



0.0 0.00



1.0



ε t = 0.0038 ε t = 0.0 0 ε t = 0.00504 0



0.2



0.05



0.10



0.15 Rn = Pn e / f / c Ag h



0.20



0.25



0.30



1.4



1.2



h



INTERACTION DIAGRAM R12-75.6 f /c = 12 ksi



γh



fy = 75 ksi



ρg = 0.08



γ = 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



0.2



0.0 0.000



εt



0.025



ε ε t t = 0 .0 = = 0. 0.0040038 005 0



0.075



0.050



0.75 1.0



0.100



0.125



0.150



0.175



Rn = Pn e / f / c Ag h



1.4



1.2



h



INTERACTION DIAGRAM R12-75.7 f /c = 12 ksi



γh



fy = 75 ksi



ρg = 0.08



γ = 0.7



0.07 0.06 0.05



1.0



Kmax



0.04



Pn



e



0.03 0.02 0.01



Ag



0.8



Kn = Pn / f



/



c



fs/fy = 0



0.6 0.25



0.4



0.50



0.2



0.0 0.000



0.75



ε ε t=0 ε t t = 0 .0 .0 0 3 8 0 = 0. 005 40 0



0.025



0.050



0.075



1.0



0.100



0.125



Rn = Pn e / f / c Ag h



0.150



0.175



0.200



1.4



1.2



h



INTERACTION DIAGRAM R12-75.8 f /c = 12 ksi



γh



fy = 75 ksi



ρg = 0.08



γ = 0.8



0.07 0.06 0.05



1.0



Kmax



0.04



Pn



e



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



ε ε t = 0.0 ε t =t = 0.004038 0.005 0 0



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



Rn = Pn e / f / c Ag h



1.4



1.2



h



INTERACTION DIAGRAM R12-75.9 f /c = 12 ksi ρg = 0.08 0.07



γh



fy = 75 ksi



γ = 0.9



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



0.2



0.75



εt 038 ε t == 0.0 040 ε 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 1.6



Kn = Pn / f



/



c



Ag



1.4



ρg = 0.08



INTERACTION DIAGRAM L3-60.6 f /c = 3 ksi



0.07



γ = 0.6



h



γh



fy = 60 ksi



Kmax 0.06



e



Pn



0.05



0.04



fs/fy = 0 0.03



1.2 0.02



1.0



0.25



0.01



0.8



0.50



0.6 0.75



0.4



ε ε t = 0 .0 ε t t = 0 .0 0 3 5 = 0. 04 005



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 0.55 0.60 Rn = Pn e / f / c Ag h



2.4



INTERACTION DIAGRAM L3-60.7 f /c = 3 ksi



ρg = 0.08



2.2



h



γh



fy = 60 ksi



γ = 0.7



0.07



2.0



Kmax



0.06



1.8



e



Pn



0.05



1.6 0.04



Kn = Pn / f



/



c



Ag



1.4



fs/fy = 0 0.03



1.2 0.02



1.0



0.25



0.01



0.50



0.8 0.6 0.4 0.2 0.0 0.0



0.75



εt= ε t 0.0035 ε t ==0 0.004 .005



0.1



1.0



0.2



0.3



0.4



Rn = Pn e / f / c Ag h



0.5



0.6



0.7



2.4 2.2



γ = 0.8



2.0



Kmax



0.06



1.6



Kn = Pn / f



/



c



Ag



1.4 1.2 1.0



γh



fy = 60 ksi



0.07



1.8



h



INTERACTION DIAGRAM L3-60.8 f /c = 3 ksi



ρg = 0.08



e



0.05



Pn



0.04



fs/fy = 0 0.03



0.02



0.25



0.01



0.50



0.8 0.6



0.75



0.4



1.0



ε t = 0.0035 ε t 0.004 ε t ==0.0 05



0.2 0.0 0.0



0.1



0.2



0.3



0.4



0.5



0.6



0.7



0.8



Rn = Pn e / f / c Ag h



2.4 2.2 2.0 1.8



Kn = Pn / f



/



c



Ag



1.6



h



γh



fy = 60 ksi



0.07



γ = 0.9



0.06



Kmax



0.05



e



Pn



0.04



fs/fy = 0



1.4



0.03



1.2



0.02



1.0



INTERACTION DIAGRAM L3-60.9 f /c = 3 ksi



ρg = 0.08



0.25



0.01



0.50



0.8 0.75



0.6 0.4 0.2 0.0 0.0



1.0



ε t = 0.0035 ε t = 0.004 ε t = 0.005



0.1



0.2



0.3



0.4



0.5



Rn = Pn e / f / c Ag h



0.6



0.7



0.8



0.9



2.0 1.8



INTERACTION DIAGRAM L4-60.6 f /c = 4 ksi



ρg = 0.08



fy = 60 ksi



γ = 0.6



0.07



1.6 1.4



h



γh



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 ε ε tt = 0.00 ε t = 0.00435 = 0. 005



0.2 0.0 0.00



0.05



1.0



0.10



0.15



0.20



0.25



0.30



0.35



0.40



0.45



Rn = Pn e / f / c Ag h



2.0 1.8 1.6 1.4



ρg = 0.08



INTERACTION DIAGRAM L4-60.7 f /c = 4 ksi



0.07



γ = 0.7



h



γh



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 εt ε t == 0.0035 ε 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



Rn = Pn e / f / c Ag h



0.35



0.40



0.45



0.50



0.55



2.0 ρg = 0.08



1.8



1.4



h



γh



fy = 60 ksi



γ = 0.8



0.07



1.6



INTERACTION DIAGRAM L4-60.8 f /c = 4 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



ε t = 0.003 ε t = 0.004 5 ε 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 Rn = Pn e / f / c Ag h



2.0 ρg = 0.08



1.8 1.6 1.4



0.07 0.06



Ag c /



Kn = Pn / f



1.0



γh



fy = 60 ksi



γ = 0.9 Kmax



e



0.05 0.04



1.2



h



INTERACTION DIAGRAM L4-60.9 f /c = 4 ksi



Pn



fs/fy = 0



0.03 0.02



0.25



0.01



0.8



0.50



0.6 0.4 0.2



0.75



1.0



ε t = 0.0035 ε t = 0.004 ε 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



INTERACTION DIAGRAM L5-60.6 f /c = 5 ksi



h



γh



fy = 60 ksi



ρg = 0.08



γ = 0.6



0.07



1.4



Kmax



0.06



e



0.05



1.2



Pn



0.04 0.03



fs/fy = 0



0.02



Kn = Pn / f



/



c



Ag



1.0 0.8



0.01



0.25



0.6



0.50



0.4



0.75



ε ε t = 0 .0 ε t t = 0.00035 = 0. 005 4



0.2 0.0 0.00



0.05



1.0



0.10



0.15



0.20



0.25



0.30



0.35



0.40



Rn = Pn e / f / c Ag h



1.8 1.6



INTERACTION DIAGRAM L5-60.7 f /c = 5 ksi fy = 60 ksi



ρg = 0.08



γ = 0.7



0.07



1.4



h



γh



Kmax



0.06



e



0.05



1.2



0.04 0.03



fs/fy = 0



0.02



Kn = Pn / f



/



c



Ag



1.0



Pn



0.8



0.01



0.25



0.50



0.6



0.75



0.4 εt= ε t = 00.0035 ε t = 0 .004 .005



0.2 0.0 0.00



0.05



0.10



1.0



0.15



0.20



0.25



Rn = Pn e / f / c Ag h



0.30



0.35



0.40



0.45



1.8 1.6



INTERACTION DIAGRAM L5-60.8 f /c = 5 ksi



ρg = 0.08



fy = 60 ksi



γ = 0.8



0.07



1.4



h



γh



Kmax



0.06



e



0.05



1.2



0.04 0.03



fs/fy = 0



0.02



Kn = Pn / f



/



c



Ag



1.0



Pn



0.8



0.25



0.01



0.50



0.6 0.75



0.4 0.2 0.0 0.00



1.0



ε t = 0.0035 ε t = 0.004 ε t = 0.005



0.05



0.10



0.15



0.20



0.25



0.30



0.35



0.40



0.45



0.50



Rn = Pn e / f / c Ag h



1.8 1.6 1.4



INTERACTION DIAGRAM L5-60.9 f /c = 5 ksi



ρg = 0.08



h



γh



fy = 60 ksi



0.07



γ = 0.9



0.06



Kmax e



0.05



1.2



0.04



fs/fy = 0



0.03 0.02



0.25



c



Ag



1.0



0.01



/



Kn = Pn / f



Pn



0.8 0.50



0.6 0.75



0.4 0.2 0.0 0.00



1.0



ε t = 0.0035 ε t = 0.004 ε t = 0.0 05



0.05



0.10



0.15



0.20



0.25



0.30



Rn = Pn e / f / c Ag h



0.35



0.40



0.45



0.50



0.55



1.6



1.4



γh



fy = 60 ksi



ρg = 0.08



γ = 0.6



0.07



Kmax



0.06



1.2



h



INTERACTION DIAGRAM L6-60.6 f /c = 6 ksi



e



0.05



Pn



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



εt ε t = 0 .0 0 ε t = 0 .0 0 3 5 = 0. 005 4



0.2



0.0 0.00



0.05



1.0



0.10



0.15



0.20



0.25



0.30



INTERACTION DIAGRAM L6-60.7 f /c = 6 ksi



γh



Rn = Pn e / f



1.6



1.4



ρg = 0.08



c



0.35



Ag h h



fy = 60 ksi



γ = 0.7



0.07



Kmax



0.06



1.2



/



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



0.2



0.0 0.00



1.0



εt=0 .0035 ε ε t =t 0= 0.004 .005



0.05



0.10



0.15



0.20



Rn = Pn e / f / c Ag h



0.25



0.30



0.35



0.40



1.6



1.4



1.2



ρg = 0.08



INTERACTION DIAGRAM L6-60.8 f /c = 6 ksi



0.07



γ = 0.8



0.06



Kmax



h



γh



fy = 60 ksi



0.05



Pn



e



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



0.2



0.0 0.00



1.0



ε t = 0.0035 ε t = 0.0 ε t = 0.00504



0.05



0.10



0.15



0.20



0.25



0.30



0.35



0.40



0.45



Rn = Pn e / f / c Ag h



1.6



1.4



ρg = 0.08



INTERACTION DIAGRAM L6-60.9 f /c = 6 ksi



0.07



γ = 0.9



0.06



1.2



h



γh



fy = 60 ksi



Kmax



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



0.50



0.6



0.75



0.4



0.2



0.0 0.00



1.0



ε t = 0.0035 ε t = 0.004 ε t = 0.005



0.05



0.10



0.15



0.20



0.25



0.30



Rn = Pn e / f / c Ag h



0.35



0.40



0.45



0.50



1.6



1.4



INTERACTION DIAGRAM L9-75.6 f /c = 9 ksi



h



γh



fy = 75 ksi



γ = 0.6



ρg = 0.08 0.07



1.2



0.06



Kmax



0.05



e



Pn



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



ε ε t = 0.0 ε 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



1.6



INTERACTION DIAGRAM L9-75.7 f /c = 9 ksi fy = 75 ksi



1.4 ρg = 0.08



1.2



h



γh



γ = 0.7



0.07



Kmax



0.06



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



0.0 0.00



0.75



εt ε = 0.0 ε 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



1.4



INTERACTION DIAGRAM L9-75.8 f /c = 9 ksi



h



γh



fy = 75 ksi



γ = 0.8



ρg = 0.08 0.07



1.2



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



0.25



0.6



0.50



0.4 0.75



εt ε = 0.0038 ε t =t 0= 0.004 .005



0.2



0.0 0.00



0.05



1.0



0.10



0.15



0.20



0.25



0.30



0.35



Rn = Pn e / f / c Ag h



1.6



1.4



INTERACTION DIAGRAM L9-75.9 f /c = 9 ksi



h



γh



fy = 75 ksi



γ = 0.9



ρg = 0.08 0.07



1.2



0.06



Kmax



0.05



1.0



e



Pn



0.04 0.03



Kn = Pn / f



/



c



Ag



0.02



0.8



fs/fy = 0



0.01



0.25



0.6 0.50



0.4



0.75



εt ε = 0.00 ε t =t = 0.00438 0.005



0.2



0.0 0.00



0.05



0.10



1.0



0.15



0.20



Rn = Pn e / f / c Ag h



0.25



0.30



0.35



0.40



1.4



1.2



INTERACTION DIAGRAM L12-75.6 f /c = 12 ksi



h



γh



fy = 75 ksi



ρg = 0.08



γ = 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



ε ε t=0 ε 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



Rn = Pn e / f / c Ag h



1.4



1.2



INTERACTION DIAGRAM L12-75.7 f /c = 12 ksi ρg = 0.08 0.07



h



γh



fy = 75 ksi



γ = 0.7



0.06 0.05



1.0



Kmax



0.04



e



Pn



0.03 0.02 0.01



Ag



0.8 Kn = Pn / f



/



c



fs/fy = 0



0.6 0.25



0.4



0.2



0.50



0.75



εt ε t = 0.0038 ε 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



INTERACTION DIAGRAM L12-75.8 f /c = 12 ksi



h



γh



fy = 75 ksi



1.2



ρg = 0.08



γ = 0.8



0.07 0.06



1.0



Kmax



0.05



e



Pn



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



ε t = 0.0038 εt= 04 ε t = 0.00.0 05



0.2



0.0 0.00



0.05



1.0



0.10



0.15



0.20



0.25



0.30



Rn = Pn e / f / c Ag h



1.4



1.2



INTERACTION DIAGRAM L12-75.9 f /c = 12 ksi



h



γh



fy = 75 ksi



ρg = 0.08



γ = 0.9



0.07 0.06



1.0



Kmax



0.05



e



Pn



0.04 0.03 0.02 0.01



fs/fy = 0



Kn = Pn / f



/



c



Ag



0.8



0.6



0.25



0.50



0.4



0.75



ε t = 0.0038 ε t = 0.0 ε t = 0.00504



0.2



0.0 0.00



0.05



0.10



1.0



0.15



0.20



Rn = Pn e / f / c Ag h



0.25



0.30



0.35



2.4



INTERACTION DIAGRAM C3-60.6 f /c = 3 ksi



ρg = 0.08



2.2



h



γh



fy = 60 ksi



0.07



γ = 0.6



2.0



Kmax



0.06



1.8



e



0.05



Pn



1.6 0.04



fs/fy = 0



Kn = Pn / f



/



c



Ag



1.4 0.03



1.2 1.0



0.02



0.25



0.01



0.8



0.50



0.6 0.75



0.4 0.2 0.0 0.00



ε ε t = 0 .0 t= 0



35 0 t = 0 .0 0 4 .0 0 5



ε



0.05



1.0



0.10



0.15



0.20



0.25



0.30



INTERACTION DIAGRAM C3-60.7 f /c = 3 ksi



γh



0.35



Rn = Pn e / f / c Ag h



2.4 2.2 2.0



0.07



1.8



0.06



1.6



0.05



1.4



0.04



Ag c /



Kn = Pn / f



ρg = 0.08



1.2



h



fy = 60 ksi



γ = 0.7 Kmax



e



Pn



fs/fy = 0



0.03



0.25



0.02



1.0 0.50



0.01



0.8 0.75



0.6 0.4 0.2 0.0 0.00



εt ε t = 0 .0 0 ε t = 0 .0 3 5 = 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 ρg = 0.08



INTERACTION DIAGRAM C3-60.8 f /c = 3 ksi



2.0



0.07



γ = 0.8



1.8



0.06



2.2



Ag Kn = Pn / f



/



c



1.2



fy = 60 ksi Kmax



Pn



e 0.05



1.6 1.4



h



γh



fs/fy = 0



0.04



0.25



0.03 0.02



1.0



0.50 0.01



0.8 0.75



0.6 1.0



0.4



εt= ε t = 0.0035 ε t = 0.004 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



Rn = Pn e / f / c Ag h



2.4 2.2 2.0 1.8



ρg = 0.08



INTERACTION DIAGRAM C3-60.9 f /c = 3 ksi



h



γh



fy = 60 ksi



γ = 0.9



0.07



Kmax



0.06



e



fs/fy = 0



0.05



1.6



Pn



0.04



Ag



1.4 Kn = Pn / f



/



c



1.2



0.25 0.03



1.0



0.02



0.50



0.01



0.8



0.75



0.6



1.0



0.4 0.2 0.0 0.00



ε t = 0.0035 ε t = 0.004 ε t = 0.005



0.05



0.10



0.15



0.20



0.25



0.30



Rn = Pn e / f



/



c



0.35



Ag h



0.40



0.45



0.50



0.55



2.0



INTERACTION DIAGRAM C4-60.6 f /c = 4 ksi



ρg = 0.08



1.8 1.6 1.4



h



γh



fy = 60 ksi



0.07



γ = 0.6



0.06



Kmax 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 εt =0 εt . = 0 0035 εt .0 0 =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 0.200 0.225 0.250 0.275 Rn = Pn e / f / c Ag h



2.0 1.8



INTERACTION DIAGRAM C4-60.7 f /c = 4 ksi



h



γh



fy = 60 ksi



ρg = 0.08



γ = 0.7 1.6



0.07



Kmax



0.06



1.4



e



Pn



0.05



fs/fy = 0



1.2



0.04



1.0



0.25



0.02



Kn = Pn / f



/



c



Ag



0.03



0.8



0.01



0.50



0.6



0.75



0.4 0.2 0.0 0.00



1.0



εt = ε t 0 .0 0 3 5 = 0. εt 004 = 0. 005



0.05



0.10



0.15



0.20



Rn = Pn e / f / c Ag h



0.25



0.30



0.35



2.0 1.8



INTERACTION DIAGRAM C4-60.8 f /c = 4 ksi



ρg = 0.08



fy = 60 ksi



γ = 0.8



0.07



1.6 1.4 1.2



Kmax 0.06



e 0.05



Ag c /



Kn = Pn / f



0.8



0.04



0.25



0.02



0.50



0.01



0.75



0.6



1.0



0.4 0.2 0.0 0.00



Pn



fs/fy = 0



0.03



1.0



h



γh



εt= ε t = 00.0035 ε t = 0 .004 .005



0.05



0.10



0.15



0.20



0.25



0.30



0.35



0.40



Rn = Pn e / f / c Ag h



2.0 1.8 1.6 1.4



ρg = 0.08



INTERACTION DIAGRAM C4-60.9 f /c = 4 ksi



0.07



γ = 0.9



h



γh



fy = 60 ksi Kmax



0.06



e



fs/fy = 0



0.05



Pn



0.04



Kn = Pn / f



/



c



Ag



1.2 1.0



0.25



0.03 0.02



0.50 0.01



0.8 0.75



0.6 1.0



0.4 εt=0 .0 ε t = 0 035 .004 εt=0 .005



0.2 0.0 0.00



0.05



0.10



0.15



0.20



Rn = Pn e / f / c Ag h



0.25



0.30



0.35



0.40



1.8 1.6



INTERACTION DIAGRAM C5-60.6 f /c = 5 ksi



ρg = 0.08



fy = 60 ksi



γ = 0.6



0.07



1.4



h



γh



Kmax



0.06



1.2



0.04 0.03



Ag



1.0



fs/fy = 0



0.02



c



0.01



/



Kn = Pn / f



Pn



e



0.05



0.25



0.8 0.6



0.50



0.4



0.75



εt = ε t 0 .0 0 3 5 = 0. 004 εt = 0. 005



0.2 0.0 0.000



0.025



0.050



1.0



0.075



0.100



0.125



0.150



0.175



0.200



0.225



Rn = Pn e / f / c Ag h



1.8 1.6 1.4



1.0



/



Kn = Pn / f



ρg = 0.08



0.8



h



γh



fy = 60 ksi



γ = 0.7



0.07



Kmax 0.06



e



Pn



0.05 0.04



fs/fy = 0



0.03 0.02



c



Ag



1.2



INTERACTION DIAGRAM C5-60.7 f /c = 5 ksi



0.25



0.01



0.50



0.6 0.75



0.4 0.2 0.0 0.00



1.0



ε ε t t== 0.0035 ε t = 0.004 0.005



0.05



0.10



0.15 Rn = Pn e / f



0.20 /



c



Ag h



0.25



0.30



1.8 1.6



INTERACTION DIAGRAM C5-60.8 f /c = 5 ksi ρg = 0.08



h



γh



fy = 60 ksi



γ = 0.8



0.07



1.4 1.2



Kn = Pn / f



/



c



Ag



1.0 0.8



Kmax 0.06



e fs/fy = 0



0.04 0.03



0.25



0.02 0.01



0.50



0.6



0.75



0.4



1.0



εt=0 .0 ε t = 0 035 .004 εt=0 .005



0.2 0.0 0.00



Pn



0.05



0.05



0.10



0.15



0.20



0.25



0.30



Rn = Pn e / f / c Ag h



1.8 1.6



INTERACTION DIAGRAM C5-60.9 f /c = 5 ksi ρg = 0.08



h



γh



fy = 60 ksi



γ = 0.9



0.07



1.4



Kmax



0.06



e



0.05



1.2



0.04 0.03



0.25



0.02



Kn = Pn / f



/



c



Ag



1.0



Pn



fs/fy = 0



0.8



0.01



0.50



0.75



0.6



1.0



0.4 εt=0 .0 ε t = 0 035 .0 ε t = 0 04 .005



0.2 0.0 0.00



0.05



0.10



0.15



0.20



Rn = Pn e / f / c Ag h



0.25



0.30



0.35



1.6 ρg = 0.08



1.4



0.07



INTERACTION DIAGRAM C6-60.6 f /c = 6 ksi fy = 60 ksi



γ = 0.6



0.06



1.2



h



γh



Kmax



0.05



1.0



0.03 0.02



fs/fy = 0



/



c



Ag



0.01



Kn = Pn / f



Pn



e



0.04



0.8 0.25



0.6 0.50



0.4



0.75



εt ε t = 0 .0 03 = εt = 0 0 .0 0 4 5 .0 0 5



0.2



0.0 0.000



0.025



0.050



1.0



0.075



0.100



0.125



0.150



0.175



0.200



0.225



Rn = Pn e / f / c Ag h



1.6 ρg = 0.08



1.4



0.07



INTERACTION DIAGRAM C6-60.7 f /c = 6 ksi fy = 60 ksi



γ = 0.7



0.06



1.2



h



γh



Kmax



0.05



e



0.04



Pn



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



εt = ε t 0 .0 0 3 5 = 0. εt 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



1.4



INTERACTION DIAGRAM C6-60.8 f /c = 6 ksi ρg = 0.08 0.07



1.2



fy = 60 ksi



γ = 0.8 Kmax



0.06



Pn



e



0.05 0.04



1.0



h



γh



fs/fy = 0



0.03



Kn = Pn / f



/



c



Ag



0.02



0.8



0.25



0.01



0.50



0.6 0.75



0.4 1.0



0.2



0.0 0.00



εt=0 .0 ε t = 0 035 .004 εt=0 .005



0.05



0.10



0.15



0.20



0.25



0.30



Rn = Pn e / f / c Ag h



1.6



1.4



ρg = 0.08 0.07



INTERACTION DIAGRAM C6-60.9 f /c = 6 ksi fy = 60 ksi



γ = 0.9



Kmax



0.06



1.2



0.05



e



0.03



Kn = Pn / f



/



c



Ag



0.02



0.8



Pn



fs/fy = 0



0.04



1.0



h



γh



0.25



0.01



0.50



0.6



0.75



1.0



0.4



0.2



0.0 0.00



ε t = 0.0035 ε t = 0.004 ε t = 0.005



0.05



0.10



0.15 Rn = Pn e / f / c Ag h



0.20



0.25



0.30



1.6



INTERACTION DIAGRAM C9-75.6 f /c = 9 ksi



h



γh



fy = 75 ksi



1.4



γ = 0.6 ρg = 0.08



1.2



0.07



Kmax



0.06



Pn



e



0.05



1.0



0.04 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



εt= 0 ε t ε t = 0.0 .0038 0 = 0. 005 4



0.025



0.050



1.0



0.075



0.100



0.125



0.150



Rn = Pn e / f / c Ag h



1.6



INTERACTION DIAGRAM C9-75.7 f /c = 9 ksi



h



γh



fy = 75 ksi



1.4 ρg = 0.08



γ = 0.7



0.07



1.2



Kmax



0.06



e



0.05



Pn



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



0.0 0.000



0.75



εt ε t = 0 .0 0 = 38 εt = 0 0 .0 0 4 .0 0 5



0.2



0.025



0.050



0.075



1.0



0.100



Rn = Pn e / f / c Ag h



0.125



0.150



0.175



1.6



1.4



INTERACTION DIAGRAM C9-75.8 f /c = 9 ksi



h



γh



fy = 75 ksi



γ = 0.8



ρg = 0.08 0.07



1.2



0.06



Kmax



0.05



e



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



εt= 0.00 ε ε t = t = 0.00 38 0.005 4



0.2



0.0 0.000



0.025



0.050



1.0



0.075



0.100



0.125



0.150



0.175



0.200



0.225



Rn = Pn e / f / c Ag h



1.6



INTERACTION DIAGRAM C9-75.9 f /c = 9 ksi



h



γh



fy = 75 ksi



1.4



γ = 0.9 ρg = 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 1.0



εt= ε t = 0.0038 ε t = 0 0.004 .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



1.4



1.2



INTERACTION DIAGRAM C12-75.6 f /c = 12 ksi



h



γh



fy = 75 ksi



γ = 0.6



ρg = 0.08 0.07 0.06



1.0



Kmax



0.05



Pn



e



0.04 0.03 0.02



Kn = Pn / f



/



c



Ag



0.8



0.01



fs/fy = 0



0.6



0.25



0.4 0.50



0.2



0.0 0.000



εt



0.025



ε ε t t = 0. = 0 003 .0 0 8 =0 4 .0 0 5



0.050



0.75 1.0



0.075



0.100



0.125



0.150



Rn = Pn e / f / c Ag h



1.4



1.2



INTERACTION DIAGRAM C12-75.7 f /c = 12 ksi



h



γh



fy = 75 ksi



ρg = 0.08



γ = 0.7



0.07 0.06



1.0



Kmax



0.05



Pn



e



0.04 0.03 0.02



Ag



0.8



0.01



Kn = Pn / f



/



c



fs/fy = 0



0.6 0.25



0.4



0.50



0.0 0.000



0.75



ε ε t t = 0. = 0 003 εt .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



1.2



INTERACTION DIAGRAM C12-75.8 f /c = 12 ksi



h



γh



fy = 75 ksi



γ = 0.8



ρg = 0.08 0.07 0.06



1.0



Kmax



0.05



Pn



e



0.04 0.03 0.02



0.8



0.01



Kn = Pn / f



/



c



Ag



fs/fy = 0



0.6



0.25



0.50



0.4



0.75



0.2



0.0 0.000



εt



0.025



ε ε 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



Rn = Pn e / f / c Ag h



1.4



1.2



INTERACTION DIAGRAM C12-75.9 f /c = 12 ksi



h



γh



fy = 75 ksi



ρg = 0.08



γ = 0.9



0.07 0.06 0.05



1.0



Kmax



0.04



Pn



e



0.03 0.02 0.01



fs/fy = 0



Kn = Pn / f



/



c



Ag



0.8



0.25



0.6



0.50



0.4 0.75



0.2



0.0 0.000



εt



0.025



ε ε t t = 0. = 0 003 = 0 .0 0 4 8 .0 0 5



0.050



1.0



0.075



0.100



0.125



Rn = Pn e / f / c Ag h



0.150



0.175



0.200



2.4



Kn = Pn / f



/



c



Ag



2.2



γh



fy = 60 ksi



ρg = 0.08



γ = 0.6



2.0



0.07



1.8



0.06



1.6



0.05



1.4



0.04



1.2



h



INTERACTION DIAGRAM S3-60.6 f /c = 3 ksi



Kmax e



Pn



fs/fy = 0



0.03



0.02



0.25



1.0 0.01



0.8



0.50



0.6 0.75



0.4



ε εt t=0 ε t = 0 .0 .0 0 3 5 = 0 04 .0 0 0 5



0.2 0.0 0.00



0.05



1.0



0.10



0.15



0.20



0.25



0.30



INTERACTION DIAGRAM S3-60.7 f /c = 3 ksi



γh



0.35



Rn = Pn e / f / c Ag h



2.4 2.2 2.0



0.07



1.8



0.06



1.6



0.05



/



c



Ag



1.4



Kn = Pn / f



ρg = 0.08



1.2



h



fy = 60 ksi



γ = 0.7 Kmax



e



Pn



fs/fy = 0



0.04 0.03



0.25 0.02



1.0



0.01



0.50



0.8 0.75



0.6 0.4 0.2 0.0 0.00



1.0



εt = 0. εt 003 ε t = 0 .0 0 5 4 = 0. 005 0



0.05



0.10



0.15



0.20



Rn = Pn e / f / c Ag h



0.25



0.30



0.35



0.40



2.4 2.2 2.0 1.8 1.6



Kn = Pn / f



/



c



Ag



1.4



ρg = 0.08



INTERACTION DIAGRAM S3-60.8 f /c = 3 ksi



0.07



γ = 0.8



h



γh



fy = 60 ksi



Kmax 0.06



e



Pn



0.05



fs/fy = 0 0.04



0.03



0.25



1.2 0.02



1.0



0.50



0.01



0.8



0.75



0.6 1.0



0.4



εt=0 ε t = 0 .0035 .0 ε 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



Rn = Pn e / f / c Ag h



2.4 2.2 2.0 1.8 1.6



Kn = Pn / f



/



c



Ag



1.4



h



INTERACTION DIAGRAM S3-60.9 f /c = 3 ksi



ρg = 0.08



γh



fy = 60 ksi



γ = 0.9



0.07



Kmax



0.06



e 0.05



Pn



fs/fy = 0



0.04



0.25



0.03



1.2 0.02



1.0



0.50



0.01



0.8



0.75



0.6



1.0



0.4



εt=0 ε t = 0 .0035 .0040 εt=0 .005



0.2 0.0 0.00



0.05



0.10



0.15



0.20



0.25



0.30



Rn = Pn e / f / c Ag h



0.35



0.40



0.45



0.50



0.55



2.0 1.8 1.6



h



INTERACTION DIAGRAM S4-60.6 f /c = 4 ksi ρg = 0.08



γh



fy = 60 ksi



γ = 0.6



0.07



Kmax



0.06



1.4 1.2



e



Pn



0.05 0.04



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 ε εt t=0 . ε t = 0.000035 4 =0 .0 0 0 5



0.2 0.0 0.00



0.05



1.0



0.10



0.15



0.20



0.25



0.30



Rn = Pn e / f / c Ag h



2.0 1.8 1.6



ρg = 0.08



h



INTERACTION DIAGRAM S4-60.7 f /c = 4 ksi



γh



fy = 60 ksi



γ = 0.7 0.07



Kmax



0.06



1.4 1.2



e



Pn



0.05



fs/fy = 0



0.04



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 0.2 0.0 0.00



1.0



ε ε t t = 0 .0 ε t = 0.000435 = 0. 005 0



0.05



0.10



0.15



0.20



Rn = Pn e / f / c Ag h



0.25



0.30



0.35



2.0 1.8



ρg = 0.08



1.4



Kn = Pn / f



/



c



Ag



1.2 1.0



γh



fy = 60 ksi



γ = 0.8



0.07



1.6



h



INTERACTION DIAGRAM S4-60.8 f /c = 4 ksi



Kmax



0.06



e fs/fy = 0 0.04 0.03



0.25



0.02 0.01



0.50



0.8



0.75



0.6



1.0



0.4 εt= ε t = 0 0.0035 .0 ε t = 0 040 .005



0.2 0.0 0.00



Pn



0.05



0.05



0.10



0.15



0.20



0.25



0.30



0.35



0.40



Rn = Pn e / f / c Ag h



2.0 1.8 1.6 1.4



ρg = 0.08



INTERACTION DIAGRAM S4-60.9 f /c = 4 ksi



0.07



γ = 0.9



h



γh



fy = 60 ksi



Kmax



0.06



e



0.05



Pn



fs/fy = 0



0.04



Kn = Pn / f



/



c



Ag



1.2 1.0



0.03



0.25



0.02



0.50



0.01



0.8 0.75



0.6 1.0



0.4



εt=0 ε t = 0 .0035 ε t = 0.0040 .005



0.2 0.0 0.00



0.05



0.10



0.15



0.20



0.25



Rn = Pn e / f / c Ag h



0.30



0.35



0.40



0.45



1.8 1.6



ρg = 0.08



γh



fy = 60 ksi



γ = 0.6



0.07



1.4



h



INTERACTION DIAGRAM S5-60.6 f /c = 5 ksi



Kmax



0.06



1.2



Pn



e



0.05 0.04 0.03



Ag



1.0 c



0.01



/



Kn = Pn / f



fs/fy = 0



0.02



0.25



0.8



0.50



0.6



0.75



0.4 ε ε t t = 0 .0 ε t = 0 .0 0 0 3 5 =0 4 .0 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 0.250 Rn = Pn e / f / c Ag h



1.8 1.6 1.4 1.2



ρg = 0.08 0.07



γ = 0.7



Kmax



0.06



e



Pn



0.05 0.04



fs/fy = 0



0.03 0.02



0.25



/



Kn = Pn / f



0.8



γh



fy = 60 ksi



c



Ag



1.0



h



INTERACTION DIAGRAM S5-60.7 f /c = 5 ksi



0.01



0.50



0.6 0.75



0.4 0.2 0.0 0.00



1.0



ε ε t t = 0 .0 = 0. 035 0 εt = 0. 040 005



0.05



0.10



0.15 Rn = Pn e / f / c Ag h



0.20



0.25



0.30



1.8 1.6



ρg = 0.08



γh



fy = 60 ksi



γ = 0.8



0.07



1.4



h



INTERACTION DIAGRAM S5-60.8 f /c = 5 ksi



Kmax



0.06



e



Pn



0.05



1.2



0.03



0.25



0.02



Kn = Pn / f



/



c



Ag



1.0



fs/fy = 0



0.04



0.8



0.01



0.50



0.6



0.75



0.4



1.0



εt=0 ε t = 0 .0035 ε t = 0 .0040 .005



0.2 0.0 0.00



0.05



0.10



0.15



0.20



0.25



0.30



0.35



Rn = Pn e / f / c Ag h



1.8 1.6



h



INTERACTION DIAGRAM S5-60.9 f /c = 5 ksi ρg = 0.08



γh



fy = 60 ksi



γ = 0.9



0.07



1.4



Kmax



0.06



e



0.05



1.2



fs/fy = 0



0.04 0.03



0.25



0.02



Kn = Pn / f



/



c



Ag



1.0



Pn



0.8



0.01



0.50



0.75



0.6



1.0



0.4 εt= ε t = 0.0035 ε t = 00.0040 .005



0.2 0.0 0.00



0.05



0.10



0.15



0.20



Rn = Pn e / f / c Ag h



0.25



0.30



0.35



1.6



1.4



ρg = 0.08



INTERACTION DIAGRAM S6-60.6 f /c = 6 ksi



0.07



γ = 0.6



fy = 60 ksi



0.06



1.2



h



γh



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



εt



0.2



ε εt t= = 0 0 .0 0 . = 0 004035 .0 0 5



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 ρg = 0.08



1.4



0.07 0.06



1.2



h



INTERACTION DIAGRAM S6-60.7 f /c = 6 ksi



γh



fy = 60 ksi



γ = 0.7



Kmax



0.05



e



0.04



Pn



0.03



1.0



fs/fy = 0



0.02



Kn = Pn / f



/



c



Ag



0.01



0.8



0.25



0.50



0.6



0.75



0.4



0.2



1.0



ε ε t t = 0 .0 = 0 ε t 0 .0 0 4 0 3 5 = 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 0.275 Rn = Pn e / f / c Ag h



1.6



h



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



1.4



0.07



fy = 60 ksi



γ = 0.8



Kmax



0.06



1.2



γh



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



εt=0 ε t = 0 .0035 .0 ε t = 0 040 .005



0.2



0.0 0.00



0.05



0.10



0.15



0.20



0.25



0.30



Rn = Pn e / f / c Ag h



1.6



1.4



ρg = 0.08 0.07



γ = 0.9 Kmax



0.05



e



0.04



1.0 Ag c /



Kn = Pn / f



Pn



fs/fy = 0



0.03 0.02



0.8



γh



fy = 60 ksi



0.06



1.2



h



INTERACTION DIAGRAM S6-60.9 f /c = 6 ksi



0.25



0.01



0.50



0.6



0.75



1.0



0.4 εt=0 ε t = 0 .0035 ε t = 0 .0040 .005



0.2



0.0 0.00



0.05



0.10



0.15



0.20



Rn = Pn e / f / c Ag h



0.25



0.30



0.35



1.6



h



INTERACTION DIAGRAM S9-75.6 f /c = 9 ksi



γh



fy = 75 ksi



1.4



γ = 0.6 ρg = 0.08



1.2



0.07 0.06



Kmax



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



0.0 0.000



εt



0.025



ε εt t=0 = 0 .0 0 = 0 .004038 .0 0 5



0.050



0.075



0.75



1.0



0.100



0.125



0.175



0.150



Rn = Pn e / f / c Ag h



1.6



1.4



h



INTERACTION DIAGRAM S9-75.7 f /c = 9 ksi



γh



fy = 75 ksi



γ = 0.7



ρg = 0.08 0.07



1.2



0.06



Kmax



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



0.50



0.4



0.2



0.0 0.000



0.75



ε ε t t = 0 .0 ε t = = 0.004038 0.005 0



0.025



0.050



0.075



1.0



0.100



0.125



Rn = Pn e / f / c Ag h



0.150



0.175



0.200



1.6



1.4



h



INTERACTION DIAGRAM S9-75.8 f /c = 9 ksi



γh



fy = 75 ksi



γ = 0.8



ρg = 0.08 0.07



1.2



0.06



Kmax



0.05



Pn



e



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



εt ε t = = 0.0038 ε t = 0 0.0040 .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



Rn = Pn e / f / c Ag h



1.6



h



INTERACTION DIAGRAM S9-75.9 f /c = 9 ksi



γh



fy = 75 ksi



1.4



γ = 0.9 ρg = 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



fs/fy = 0



0.01



0.25



0.6 0.50 0.75



0.4



0.2



ε ε t =t =0 0.0038 .0040 εt=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



1.2



INTERACTION DIAGRAM S12-75.6 f /c = 12 ksi



h



γh



fy = 75 ksi



ρg = 0.08



γ = 0.6



0.07 0.06



1.0



Kmax



0.05



e



0.04



Pn



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



εt



0.025



ε ε t t = 0 .0 03 = = 0 . 0 .0 0 4 0 8 005 0



0.050



0.075



0.75 1.0



0.100



0.125



0.150



Rn = Pn e / f / c Ag h



1.4



INTERACTION DIAGRAM S12-75.7 f /c = 12 ksi



h



γh



fy = 75 ksi



1.2



ρg = 0.08



γ = 0.7



0.07 0.06



1.0



Kmax



0.05



e



0.04



Pn



0.03 0.02 0.01



c



Ag



0.8 Kn = Pn / f



/



fs/fy = 0



0.6 0.25



0.4



0.50



0.2



0.0 0.000



0.75



ε ε t t = 0 .0 ε t = 0.004038 = 0. 005 0



0.025



0.050



0.075



1.0



0.100



Rn = Pn e / f / c Ag h



0.125



0.150



0.175



1.4



h



INTERACTION DIAGRAM S12-75.8 f /c = 12 ksi



γh



fy = 75 ksi



1.2



γ = 0.8



ρg = 0.08 0.07 0.06



1.0



Kmax



0.05



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



εt= ε t = 0.0038 ε t = 0.0040 0.005



0.2



0.0 0.000



0.025



0.050



0.075



1.0



0.100



0.125



0.150



0.175



0.200



Rn = Pn e / f / c Ag h



1.4



1.2



h



INTERACTION DIAGRAM S12-75.9 f /c = 12 ksi ρg = 0.08



γh



fy = 75 ksi



γ = 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



0.2



0.0 0.000



1.0



ε ε t =t 0= 0.0038 .0040 εt=0 .005



0.025



0.050



0.075



0.100



0.125



Rn = Pn e / f / c Ag h



0.150



0.175



0.200



0.225



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