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
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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
REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
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.
Minimum
web width for
2 bars, in.
Increment for
each added bar,
in.
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
American Concrete Institute Copyrighted Material—www.concrete.org
REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
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.
American Concrete Institute Copyrighted Material—www.concrete.org
REF.
TABLES
ACI min. bw, in.
4
REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
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.
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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.
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REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
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.
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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
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REINFORCED CONCRETE DESIGN HANDBOOK IN ACCORDANCE WITH ACI 318-14—SP-17(14)
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.
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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.
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REF.
TABLES
10
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11
APPENDIX B—ANALYSIS TABLES
ANALYSIS
TABLES
Reproduced with permission from the Canadian Portland Cement Association.
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ANALYSIS
TABLES
12
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13
ANALYSIS
TABLES
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ANALYSIS
TABLES
14
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15
ANALYSIS
TABLES
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ANALYSIS
TABLES
16
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17
ANALYSIS
TABLES
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ANALYSIS
TABLES
18
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19
ANALYSIS
TABLES
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ANALYSIS
TABLES
20
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21
ANALYSIS
TABLES
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ANALYSIS
TABLES
22
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23
ANALYSIS
TABLES
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ANALYSIS
TABLES
24
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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′ )
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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
American Concrete Institute Copyrighted Material—www.concrete.org
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 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 318‐14, Section
22.4.2.1).
Pn,max  0.8 0.85 f c  Ag  Ast   f y Ast 


(D.1a)
K max  pmax / f c 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 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

 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 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   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 ox
M oy
M ox
 M oy   1   
M oy  M ny  M nx 


 M ox    
,
M
M ox  M nx  M ny  ox
M
 oy
,
 1  
 

  
(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  
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
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.
b = 16 in.
and compute required cross
section area Ast of longitudinal
Ag = b × h = 20 × 16 = 320 in.2
reinforcement.
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 Pn = 943 kip
b corresponding to known values Mn = 3986 in.‐kip
h = 16 in.
of variables on appropriate
ρg = 0.039
interaction diagrams.
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 

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 318‐14
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 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 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
Calculation
Assume load contour curve at
constant Pn is an ellipse, and
determine resultant moment Mnx
For a square column: h = b
from
M nr  29492  11832  3177 in.-kip
M nr  M 2 nx  M 2 ny
Assume a series of trial column
sizes h, in inches.
Ag  h 2 , in.2
Compute
10.5
Design
aid
Compute
Kn 
Compute
Rn 
Estimate

14
16
18
196
256
324
Pn
297
f c ’ Ag
(5)(196)
Mn
3177
f c ’ Ag h
(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
0.64
297
(5)(256)
 0.23
3177
(5)(256)(16)
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 318‐14
section
10.5
Discussion
Compute
Rn 
Estimate

Calculation
Mn
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 = g Ag and add about 15
percent for skew bending.
Rn 

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 318‐14
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  h 2 , in.2
113
Compute
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.
Interpolate
Compute
Ag 
Compute
h2
Pn
f c ’K n
Ag

 0.10
Design
aid
Calculation
16
20
201
314
686
(5)(201)(16)
0.58
 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      227 in.2
and compute required cross
 2
section area Ast of longitudinal
reinforcement.
ACI 318‐14
section
Discussion
Compute
10.5
Kn 
Compute
Rn 
Estimate

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 = g Ag.
Kn 
Rn 

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 318‐14, 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:
0.10
2.
3.
4.
5.
25
1.2
0.2
0.5
For continuous beams, I can be taken as the average of values for the critical positive and negative moment
sections. I need not be less than 0.25Ig.
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.
Hinged
Elastic
Elastic
Stiff
TOP
Table D.4.4—Effective length factor k for columns in braced frames (Concrete Design Handbook 2005)
Table D.4.5—Moment of inertia of reinforcement about sectional centroid (Based on Table 12‐1, MacGregor 1997)
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
1.8
0.06
1.6
0.05
Ag
c
Kn = Pn / f
1.0
h
fy = 60 ksi
 = 0.6
0.07
1.2
/
g = 0.08
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
0.01
0.8
0.50
0.6
0.75
0.4

 t = 0
t t = 0.0.0035
= 0 04
. 00
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
g = 0.08
INTERACTION DIAGRAM R3-60.7
f /c = 3 ksi
0.07
 = 0.7
h
h
fy = 60 ksi
Kmax
0.06
e
Pn
0.05
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 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
2.4
2.2
2.0
1.8
g = 0.08
h
INTERACTION DIAGRAM R3-60.8
f /c = 3 ksi
h
fy = 60 ksi
 = 0.8
0.07
Kmax
0.06
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 = 0.0
t = 0.0 0305
t = 0.0004
5
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
g = 0.08
INTERACTION DIAGRAM R3-60.9
/
f c = 3 ksi
0.07
 = 0.9
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
INTERACTION DIAGRAM R4-60.6
f /c = 4 ksi
g = 0.08
h
fy = 60 ksi
 = 0.6
0.07
1.6
Kmax
0.06
Pn
e
1.4
1.2
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

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
Rn = Pn e / f / c Ag h
2.0
1.8
g = 0.08
h
fy = 60 ksi
 = 0.7
0.07
1.6
h
INTERACTION DIAGRAM R4-60.7
/
f c = 4 ksi
Kmax
0.06
e
1.4
Pn
0.05
0.04
fs/fy = 0
1.2
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 = 0.0
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
2.0
1.8
1.6
1.4
g = 0.08
INTERACTION DIAGRAM R4-60.8
f /c = 4 ksi
0.07
 = 0.8
h
h
fy = 60 ksi
Kmax
0.06
e
0.05
Pn
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
Rn = Pn e / f / c Ag h
2.0
g = 0.08
INTERACTION DIAGRAM R4-60.9
f /c = 4 ksi
0.07
 = 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
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
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
g = 0.08
h
fy = 60 ksi
 = 0.6
0.07
1.4
Kmax
0.06
e
Pn
0.05
1.2
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 = 0.0
= 0. 035
0
t
= 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
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
0.02
c
Ag
1.0
Kn = Pn / 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 = = 0.0035
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
h
INTERACTION DIAGRAM R5-60.8
f /c = 5 ksi
h
fy = 60 ksi
g = 0.08
 = 0.8
0.07
1.4
Kmax
0.06
e
0.05
1.2
Pn
0.04
fs/fy = 0
1.0
c
Ag
0.03
0.02
Kn = Pn / f
/
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
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
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
0.01
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
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
g = 0.08
h
fy = 60 ksi
0.07
 = 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

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
Rn = Pn e / f / c Ag h
1.6
1.4
g = 0.08
0.07
h
fy = 60 ksi
 = 0.7
Kmax
0.06
1.2
h
INTERACTION DIAGRAM R6-60.7
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
0.6
0.50
0.75
0.4
0.2
0.0
0.00
1.0

t =t = 0.0035
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
g = 0.08
1.4
h
fy = 60 ksi
 = 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
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
h
g = 0.08
INTERACTION DIAGRAM R6-60.9
f /c = 6 ksi
0.07
 = 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
 t =
t = 0 0.0035
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
h
fy = 75 ksi
 = 0.6
g = 0.08
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
Rn = Pn e / f / c Ag h
1.6
h
INTERACTION DIAGRAM R9-75.7
f /c = 9 ksi
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
INTERACTION DIAGRAM R9-75.8
f /c = 9 ksi
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
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
h
fy = 75 ksi
1.4
g = 0.08
 = 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
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
0.20
/
c
Ag h
0.25
0.30
1.4
1.2
h
INTERACTION DIAGRAM R12-75.6
f /c = 12 ksi
h
fy = 75 ksi
g = 0.08
 = 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 =
t t = 0.00.0038
= 0.
005 040
0
0.2
0.0
0.000
0.025
0.050
0.75
1.0
0.075
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
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
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

 t = 0
t t = 0.0 .0038
0
= 0.
005 40
0
0.025
0.050
0.075
1.0
0.100
Rn = Pn e / f
0.125
/
c
Ag h
0.150
0.175
0.200
1.4
1.2
h
INTERACTION DIAGRAM R12-75.8
f /c = 12 ksi
h
fy = 75 ksi
g = 0.08
 = 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

 t = 0.0
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
g = 0.08
0.07
h
fy = 75 ksi
 = 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
0.75
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
c
/
Kn = Pn / f
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 = 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 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
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
Kn = Pn / f
/
c
Ag
1.4
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
Rn = Pn e / f / c Ag h
0.5
0.6
0.7
2.4
h
INTERACTION DIAGRAM L3-60.8
f /c = 3 ksi
g = 0.08
2.2
h
fy = 60 ksi
0.07
 = 0.8
2.0
Kmax
0.06
1.8
e
0.05
Pn
1.6
0.04
Kn = Pn / f
/
c
Ag
1.4
1.2
1.0
fs/fy = 0
0.03
0.02
0.25
0.01
0.50
0.8
0.6
0.75
0.4
1.0
t = 0.0035
t 0.004
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
fy = 60 ksi
Kmax
0.05
e
Pn
0.04
c
fs/fy = 0
0.03
1.2
0.02
1.0
Ag
h
h
 = 0.9
0.06
1.4
/
INTERACTION DIAGRAM L3-60.9
f /c = 3 ksi
0.07
Kn = Pn / f
1.6
g = 0.08
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
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
g = 0.08
fy = 60 ksi
 = 0.6
0.07
1.6
1.4
h
h
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

tt = 0.00
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
g = 0.08
INTERACTION DIAGRAM L4-60.7
f /c = 4 ksi
0.07
 = 0.7
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
 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
Rn = Pn e / f / c Ag h
0.35
0.40
0.45
0.50
0.55
2.0
g = 0.08
INTERACTION DIAGRAM L4-60.8
f /c = 4 ksi
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
Rn = Pn e / f / c Ag h
2.0
g = 0.08
1.8
0.07
1.6
0.06
h
INTERACTION DIAGRAM L4-60.9
f /c = 4 ksi
h
fy = 60 ksi
 = 0.9
Kmax
e
0.05
Pn
1.4
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
0.4
0.2
0.75
1.0
t = 0.0035
t = 0.00
t = 0.0054
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
h
fy = 60 ksi
g = 0.08
 = 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
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
g = 0.08
h
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
Kn = Pn / f
/
c
Ag
1.0
Pn
0.01
0.25
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
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
g = 0.08
fy = 60 ksi
 = 0.8
0.07
1.4
h
h
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
0.25
0.01
0.8
0.50
0.6
0.75
0.4
t = 0.0035
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
0.40
0.45
0.50
Rn = Pn e / f / c Ag h
1.8
1.6
INTERACTION DIAGRAM L5-60.9
f /c = 5 ksi
g = 0.08
fy = 60 ksi
 = 0.9
0.07
1.4
h
h
0.06
Kmax
e
0.05
1.2
0.04
fs/fy = 0
0.03
0.02
0.25
Kn = Pn / f
/
c
Ag
1.0
Pn
0.01
0.8
0.50
0.6
0.75
0.4
0.2
0.0
0.00
1.0
t = 0.0035
t = 0.004
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
h
INTERACTION DIAGRAM L6-60.6
f /c = 6 ksi
h
fy = 60 ksi
g = 0.08
 = 0.6
0.07
Kmax
0.06
1.2
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
t
t = 0.00
t = 0.00 35
= 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
h
Rn = Pn e / f
1.6
g = 0.08
1.4
c
0.35
Ag h
h
fy = 60 ksi
 = 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
t =
 0.0035
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
g = 0.08
INTERACTION DIAGRAM L6-60.8
f /c = 6 ksi
0.07
 = 0.8
0.06
Kmax
h
h
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
t = 0.0035
t = 0.0
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
g = 0.08
1.4
h
h
fy = 60 ksi
 = 0.9
0.07
0.06
1.2
INTERACTION DIAGRAM L6-60.9
f /c = 6 ksi
Kmax
0.05
e
Pn
0.04
1.0
0.03
fs/fy = 0
Kn = 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
Rn = Pn e / f
0.30
/
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
h
fy = 75 ksi
 = 0.6
g = 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

 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
1.6
INTERACTION DIAGRAM L9-75.7
f /c = 9 ksi
h
h
fy = 75 ksi
1.4
 = 0.7
g = 0.08
0.07
1.2
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
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
INTERACTION DIAGRAM L9-75.8
f /c = 9 ksi
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
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
 = 0.0038
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
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
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
 = 0.00
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
h
fy = 75 ksi
g = 0.08
 = 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
Rn = Pn e / f / c Ag h
1.4
1.2
INTERACTION DIAGRAM L12-75.7
/
f c = 12 ksi
g = 0.08
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
INTERACTION DIAGRAM L12-75.8
f /c = 12 ksi
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
Kn = Pn / f
/
c
fs/fy = 0
0.6
0.25
0.50
0.4
0.75
t = 0.0038
t = 0.004
t = 0.00
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
1.4
INTERACTION DIAGRAM L12-75.9
f /c = 12 ksi
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
Kn = 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
Rn = Pn e / f / c Ag h
0.25
0.30
0.35
2.4
INTERACTION DIAGRAM C3-60.6
/
f c = 3 ksi
g = 0.08
2.2
h
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
1.0
0.02
0.25
0.01
0.8
0.50
0.6
0.75
0.4
0.2
0.0
0.00


 t = 0 .0
t = 0 03
5
t = 0 .0 0 4
.0 0
5
0.05
1.0
0.10
0.15
0.20
0.25
0.30
INTERACTION DIAGRAM C3-60.7
/
f c = 3 ksi
h
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
g = 0.08
1.2
h
fy = 60 ksi
 = 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
t
t = 0.00
t = 0.0 35
= 0 04
.00
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
INTERACTION DIAGRAM C3-60.8
f /c = 3 ksi
0.07
 = 0.8
2.2
2.0
h
h
fy = 60 ksi
Kmax
0.06
1.8
0.05
1.6
c
1.0
/
fs/fy = 0
0.04
0.25
0.03
1.2
Kn = Pn / f
Ag
1.4
Pn
e
0.02
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
0.10
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
g = 0.08
INTERACTION DIAGRAM C3-60.9
f /c = 3 ksi
0.07
 = 0.9
h
h
fy = 60 ksi
Kmax
0.06
1.8
e
Pn
fs/fy = 0
0.05
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
0.75
0.8
0.6
1.0
0.4
0.2
0.0
0.00
t = 0.0
t = 0.00035
t = 0.0054
0.05
0.10
0.15
0.20
0.25
0.30
Rn = Pn e / f
/
0.35
c Ag h
0.40
0.45
0.50
0.55
2.0
INTERACTION DIAGRAM C4-60.6
f /c = 4 ksi
g = 0.08
1.8
1.6
1.4
h
h
fy = 60 ksi
0.07
 = 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
t
=0
t
.00
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
Rn = Pn e / f / c Ag h
2.0
1.8
INTERACTION DIAGRAM C4-60.7
f /c = 4 ksi
h
h
fy = 60 ksi
g = 0.08
 = 0.7
1.6
0.07
Kmax
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.
0 05
0.05
0.10
0.15
Rn = Pn e / f
0.20
/
c
Ag h
0.25
0.30
0.35
2.0
1.8
INTERACTION DIAGRAM C4-60.8
f /c = 4 ksi
g = 0.08
h
h
fy = 60 ksi
 = 0.8
0.07
1.6
Kmax
0.06
1.4
e
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
0.0
0.00
 t =
t = 00.0035
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
g = 0.08
INTERACTION DIAGRAM C4-60.9
f /c = 4 ksi
0.07
 = 0.9
h
h
fy = 60 ksi
Kmax
0.06
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
Rn = Pn e / f
0.25
/
c
Ag h
0.30
0.35
0.40
1.8
1.6
INTERACTION DIAGRAM C5-60.6
f /c = 5 ksi
g = 0.08
fy = 60 ksi
 = 0.6
0.07
1.4
h
h
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
t
=
t 0.0035
= 0.
004
t
= 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
INTERACTION DIAGRAM C5-60.7
f /c = 5 ksi
g = 0.08
h
h
fy = 60 ksi
 = 0.7
0.07
1.4
Kmax
0.06
1.2
e
0.05
/
Kn = Pn / f
fs/fy = 0
0.03
0.02
c
Ag
0.04
1.0
Pn
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
1.8
1.6
INTERACTION DIAGRAM C5-60.8
f /c = 5 ksi
g = 0.08
h
h
fy = 60 ksi
 = 0.8
0.07
1.4
Kmax
0.06
e
Pn
0.05
1.2
fs/fy = 0
0.04
1.0
0.03
0.25
Kn = 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 0 3 5
.004
 t = 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
1.8
1.6
INTERACTION DIAGRAM C5-60.9
f /c = 5 ksi
h
h
fy = 60 ksi
g = 0.08
 = 0.9
0.07
1.4
Kmax
0.06
e
0.05
1.2
Pn
fs/fy = 0
0.04
0.03
1.0
0.25
Kn = Pn / f
/
c
Ag
0.02
0.8
0.01
0.50
0.75
0.6
1.0
0.4
 t = 0
.0
 t = 0 0 3 5
.0
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
g = 0.08
1.4
INTERACTION DIAGRAM C6-60.6
f /c = 6 ksi
fy = 60 ksi
 = 0.6
0.07
0.06
1.2
h
h
Kmax
0.05
Pn
e
0.04
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
t = 0.0
03
=
t
= 0 0.004 5
.00
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
g = 0.08
1.4
0.07
INTERACTION DIAGRAM C6-60.7
/
f c = 6 ksi
fy = 60 ksi
 = 0.7
0.06
1.2
h
h
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
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
1.4
INTERACTION DIAGRAM C6-60.8
f /c = 6 ksi
g = 0.08
0.07
fy = 60 ksi
 = 0.8
Kmax
0.06
1.2
Pn
e
0.05
0.04
1.0
h
h
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
 t = 0
.0
 t = 0 0 3 5
.004
 t = 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
g = 0.08
0.07
INTERACTION DIAGRAM C6-60.9
f /c = 6 ksi
fy = 60 ksi
 = 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
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
INTERACTION DIAGRAM C9-75.6
f /c = 9 ksi
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
0.0
0.000
0.75
t =
0
t 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
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
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
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 =
0.00

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
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
1.0
t = 0
t = .0038
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
INTERACTION DIAGRAM C12-75.6
f /c = 12 ksi
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 = Pn / f
/
c
Ag
0.8
fs/fy = 0
0.6
0.25
0.4
0.50
0.2
t
0.0
0.000
0.025

t t = 0.
= 0 0 03
.00 8
=0
4
. 00
5
0.050
0.75
1.0
0.075
0.100
0.125
0.150
Rn = Pn e / f / c Ag h
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
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
0.4
0.50
0.0
0.000
0.75

t t = 0.
= 0 003
t
. 00 8
=0
4
. 00
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
INTERACTION DIAGRAM C12-75.8
f /c = 12 ksi
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 = Pn / f
/
c
Ag
fs/fy = 0
0.6
0.25
0.50
0.4
0.75

t t = 0.
= 0 003
t
.00 8
=0
4
. 00
5
0.2
0.0
0.000
0.025
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
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
.00 8
=0
4
.00
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
2.2
h
fy = 60 ksi
 = 0.6
0.07
Kmax
0.06
1.6
0.05
1.4
0.04
e
Pn
fs/fy = 0
c
0.03
1.2
1.0
/
g = 0.08
1.8
Kn = Pn / f
Ag
2.0
h
INTERACTION DIAGRAM S3-60.6
/
f c = 3 ksi
0.02
0.25
0.01
0.8
0.50
0.6
0.75
0.4

t t = 0
t = 0.0.0035
= 0 04
. 00 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
h
0.35
Rn = Pn e / f / c Ag h
2.4
2.2
2.0
g = 0.08
h
fy = 60 ksi
 = 0.7
0.07
Kmax
1.8
0.06
1.6
0.05
Kn = Pn / f
/
c
Ag
1.4
1.2
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
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
Rn = Pn e / f / c Ag h
0.25
0.30
0.35
0.40
2.4
2.2
2.0
g = 0.08
INTERACTION DIAGRAM S3-60.8
/
f c = 3 ksi
0.07
 = 0.8
h
h
fy = 60 ksi
Kmax
1.8
1.6
0.06
e
Pn
0.05
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
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
Rn = Pn e / f / c Ag h
2.4
2.2
2.0
1.8
h
INTERACTION DIAGRAM S3-60.9
f /c = 3 ksi
g = 0.08
h
fy = 60 ksi
 = 0.9
0.07
Kmax
0.06
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
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
g = 0.08
h
fy = 60 ksi
 = 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

t t = 0
.
t = 0.000035
4
=0
.00 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
g = 0.08
h
INTERACTION DIAGRAM S4-60.7
f /c = 4 ksi
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
1.0

t t = 0.0
= 0. 035
= 0. 0040
005
0.10
0.15
0.20
Rn = Pn e / f / c Ag h
0.25
0.30
0.35
2.0
1.8
h
INTERACTION DIAGRAM S4-60.8
f /c = 4 ksi
g = 0.08
h
fy = 60 ksi
 = 0.8
0.07
1.6
Kmax
0.06
1.4
e
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 0 4 0
.005
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
2.0
1.8
1.6
1.4
g = 0.08
INTERACTION DIAGRAM S4-60.9
f /c = 4 ksi
0.07
 = 0.9
h
h
fy = 60 ksi
Kmax
0.06
e
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
Rn = Pn e / f / c Ag h
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
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
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
. 00 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
g = 0.08
0.07
1.4
h
INTERACTION DIAGRAM S5-60.7
f /c = 5 ksi
h
fy = 60 ksi
 = 0.7
Kmax
0.06
e
Pn
0.05
1.2
0.04
fs/fy = 0
0.03
1.0
Ag
0.02
Kn = Pn / f
/
c
0.25
0.8
0.01
0.50
0.6
0.75
0.4
0.2
0.0
0.00
1.0

t t = 0.00
= 0.
0 35
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
1.6
h
INTERACTION DIAGRAM S5-60.8
f /c = 5 ksi
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
1.0
0.25
Kn = Pn / f
/
c
Ag
0.02
0.8
0.01
0.50
0.6
0.75
0.4
1.0
 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
Rn = Pn e / f / c Ag h
1.8
1.6
h
INTERACTION DIAGRAM S5-60.9
f /c = 5 ksi
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
0.25
0.02
Kn = Pn / f
/
c
Ag
1.0
0.01
0.8
0.50
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
Rn = Pn e / f / c Ag h
0.25
0.30
0.35
1.6
1.4
g = 0.08
INTERACTION DIAGRAM S6-60.6
f /c = 6 ksi
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. 00
.
= 0 004035
.00
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
g = 0.08
1.4
0.07
0.06
1.2
h
INTERACTION DIAGRAM S6-60.7
f /c = 6 ksi
h
fy = 60 ksi
 = 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

t t = 0.0
=
0
t 0.0040 35
= 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
g = 0.08
1.4
0.07
fy = 60 ksi
 = 0.8
Kmax
0.06
1.2
h
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 0 4 0
.005
0.2
0.0
0.00
0.05
0.10
0.15
Rn = Pn e / f
1.6
1.4
g = 0.08
0.07
c
0.30
Ag h
h
h
fy = 60 ksi
 = 0.9
Kmax
0.05
e
0.04
1.0
/
0.25
INTERACTION DIAGRAM S6-60.9
f /c = 6 ksi
0.06
1.2
0.20
Pn
fs/fy = 0
0.03
0.02
Kn = 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
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
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
0.0
0.000
0.025

t t = 0
= 0 .00
= 0 .004038
.00
5
0.050
0.075
0.75
1.0
0.100
0.125
0.150
0.175
Rn = Pn e / f / c Ag h
1.6
h
INTERACTION DIAGRAM S9-75.7
f /c = 9 ksi
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
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
h
fy = 75 ksi
 = 0.8
g = 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
t
t = = 0.0038
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
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
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
1.0
Kmax
0.05
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
t
0.025

t t = 0.00
=
3
= 0. 0. 0040 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
h
fy = 75 ksi
1.2
g = 0.08
 = 0.7
0.07
0.06
1.0
Kmax
0.05
e
Pn
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
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
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
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
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
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
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
Rn = Pn e / f / c Ag h
0.150
0.175
0.200
0.225
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