ACI Fall Convention 2008
St. Louis
ACI 314 Committee Meeting
Sunday, November 2
ARGENTINA DOCUMENT
Eng. Daniel Ortega from
INTI -CIRSOC
Eng. Victorio Hernandez Balat
Eng. Juan Francisco Bissio
from Quasdam Ingenieria

INTI CIRSOC
The main purpose of this document is to put forward a proposal for the
development of new design aids based on ACI 318-08.
INTI-CIRSOC, which is the official institution in charge of national safety
and structural codes in Argentina, wants to cooperate with the American
Concrete Institute developing these aids in both US Customary and Metric
Units.
• Flexure - Tension and Compression ~einforce -"jnt (when needed)
. _ Symmetrical
Reinforcement
Unsymmetrical
Reinforcement
Bia"ti?1 Bending
• A Sample Design Aid
• Step by step procedures
• Step by step examples
• The proposed scheme for each set of aids
If this proposal is accepted further discussion will be needed in order to
coordinate details.

RECTANGULAR AND FLANGED SECTIONS WITH TENSION AND
COMPRESSION REINFORCEMENT (IF NECESSARY)
• Sample Design Aid
• Step by step procedures for rectangular and flanged sections
• Step by step examples for rectangular and flanged sections
• Proposed scheme for the whole set of aids

FLEXURE - OPTIMAL DESIGN - fy = 60,000 psi - 'Y = d I h = 0.90 FLE - XX
Ec= 0.003 0.85 fc- -i"~,cl ,COT
d - a /2 d - d'
________________ L_ -l
f'c (psi)
2500 3000 4000 5000 6000 7000 8000 9000
mu, 0.0564 0.0474 0.0359 0.0303 0.0277 0.0257 0.0241 0.0227
P min 0.0030 0.0030 0.0030 0.0032 0.0035 0.0038 0.0040 0.0043
6, 0.850 0.850 0.850 0.801 0.752 0.702 0.653 0.650
0.015 0.015fi.'~~~'),'3
I 0.0100.010 p' = A's I (bw h)
0.05
I
0.10 0.15
I - - I
0.000
0.30 0.35 0.40 0.45
mu = Mu I (bw h2 f'd
0.000
0.00
Developed by Victoria Hernandez Balat, Juan Francisco Bissio
and Daniel Ortega for INTI-CIRSOC, Argentina

FLEXURE - OPTIMAL DESIGN - DESIGN AIDS FLE-XX
PROCEDURES FOR RECTANGULAR AND FLANGED SECTIONS WITH
TENSION AND COMPRESSION REINFORCEMENT (IF NECESSARY)
T
rd' Given: f'e, fy, bw, h, y, MuTTA's Determine: As and A's
~"
h [-d'.'h Step 1: For f'e read mU1 and Pmin1 As- Step 2: Compute mu = Mu 1 (bw h2 f e)bw -oj Ld'
Step 3: ~ If mu ::; mu1 then As = pminbw h (A's = 0)
~ If mu > mu1 then for mu and f'e read p and p' (p' could be zero)
Compute As = P bw h and A's = p' bw h
-I r~d'
d = h - d' = y h
As 1- l..d'
bw-l T
For mu read a/h
Given:
Determine:
f'e, fy, b, hI, bw, h, y, Mu
As and A's
Compute hI 1 hand
mu = Mu 1 (b h2 fe)
important: use b not bw
For mu and f'e read p and p'.
Compute As = p b h and A's =p' b h
Go to step 9
Cnl = 0.85 fe (b - bw) hI
MUI= <l> Cnr{y h - hI 12) = <l> Cnr{d - hI 12) <l> = 0.90
Muw = Mu - MUI
muw = Muw 1 (bw h2 f'e)
Compute As min= pminbw h important: use bw not b
Check As ;:: As min

FLEXURE - OPTIMAL DESIGN - DESIGN AIDS FLE-XX
EXAMPLES FOR RECTANGULAR AND FLANGED SECTIONS WITH
TENSION AND COMPRESSION REINFORCEMENT (IF NECESSARY)
I fd' Given: fc = 4000 psi; fy = 60,000 psiA's bw = 14" ; h = 25" ; Y = d / h == 0.90
~"
h [-d':rh a} Mu = 95 ft-kips = 1140 in-kips1 b} Mu = 205 ft-kips = 2460 in-kipsAs c} Mu = 598 ft-kips = 7175 in-kips-I-- bw ---..j Ld' Determine: As and A's
Step Procedure Case a) Case b) Case c)
1 For f'c read mu1 and Pmin mu1 = 0.0359, Pmin= 0.0030
2 mu = Mu / (b h2 fc) 0.0326 0.0703 0.205
Compare mu:::; mu1 mu> mu1 mu> mu1
For muw and f'c read p and p' or Pmin= 0.0030 P = 0.0062 p=0.01983 o = Dmin if mu:::; mu1 P' = 0 p' = 0.0036
As = P bw h and A's =p' bw h
As = 1.05 in.:.! As = 2.17 in.:.! As = 6.93 in.2
A's = 0.00 in.2 A's = 0.00 in.2 A's = 1.26 in.2

I' b ·1 T~d' Given: fc = 4000 psi ; fy = 60,000 psir;r b = 42" ; hI = 4"A's bw = 14" ; h = 25" ; Y = d / h ::: 0.90
h
~"
d = h - d' = y h a) Mu= 420 ft-kips = 5,040 in-kips1 ~d' 1 b) Mu = 1000 ft-kips = 12,000 in-kipsAs c) Mu = 1170 ft-kips = 14,040 in-kips-I- bw-! T Determine: As and A's
Step Procedure Case a) Case b) Case c)
1 For f'c read mu1 and Pmin mu1 = 0.0359, Pmin= 0.0030
2 hf / h 0.16mu = Mu / (b h£ f c) 0.048 0.1143 0.1337
3 For mu read a / h 0.072 0.185 0.225
4 Compare a / h < hl/ h a / h > hl/ h a / h > hf / hAction go to step 5 go to step 6 qo to step 6
For mu and f'c read p and p' p=0.0041 -------- --------D' = 0
5 As = 4.31 in.LAs=pbh and A's =p' b h A's = 0.00 in.2 -------- --------
Action qo to step 9 -------- --------
Cnf = 0.85 fc (b - bw) hf -------- 380.80 kips 380.80 kips
6 MUI= 0.90 Cnf (yh - hf/ 2) -------- 7025.76 in-kips 7025.76 in-kipsMuw= Mu - MUf -------- 4974.24 in-kips 7014.24 in-kips
muw= Muw/ (bw hL f'c) -------- 0.1421 0.2004
For muwand f'c read p and p' -------- p = 0.0135 p=0.0194
7 D' = 0 D' = 0.0032
Asw= P bw h and A'sw =p' bw h --------- Asw= 4.73 in.L Asw= 6.79 in.L
A'sw = 0.00 in.2 A'sw= 1.12 in.2
8 As = Asw+ Cnf / fy and -------- As = 11.08 in.
2 As= 13.14in.2
A's = A'sw A's = 0.00 in.2 A's=1.12in.2
9 As min= Dminbw h 1.05 in.
2
Check As;O:Asmin Ok Ok Ok

FLEXURE - OPTIMAL DESIGN - DESIGN AIDS FLE-XX
PROPOSED SCHEME FOR THE WHOLE SET OF AIDS
Important: This basic scheme is open to discussion.
d' T~dA's IT IhfI A's
~"
h [dO:'" h
~"
d = h - d' = Y h
1 As 1 As -l dO 1- -Ld' TI-bw --.j I- bw---l
fc (psi):
fy (ksi):
y = d / h:
2500,3000,4000,5000,6000,7000,8000,9000
Grade 60 and 75
0.95,0.90,0.85
fc (MPa):
fy (MPa):
y = d / h:
17,20,25,30,35,40,45,50,55,60
Grade 420 and 520
0.95,0.90,0.85

• Sample Design Aid
• Step by step procedures
• Step by step examples
• Proposed scheme for the whole set of aids

AXIAL LOAD AND UNIAXIAL BENDING
SYMMETRICAL REINFORCEMENT • As = A's AUB-SR-XX
Pu=Pu/(bhf'c) I r rd' f' c = 4,000 psi1.3 - T MuA's fy = 60,000 psiIf"y= 0.80h1.1 1 Ag=bxhAs Ast= pg Ag-0.9 I- b -l Ld' Ast= As + A's
I I I
Ref. Code ACI 318-08
Developed by Victoria Hernandez Balat, Juan Francisco Bissio
and Daniel Ortega for INTI-CIRSOC, Argentina
e I h = Mu I ( Pu h )= ·0.10
\

AXIAL LOAD AND UNIAXIAL BENDING
SYMMETRICAL REINFORCEMENT - As = A's
UNDERSTANDING INTERACTION DIAGRAMS
rd' ee= 0.003I- T ,-A's /// /Pu=Pu/(bhf'cl /I yh / /h /I 1 1 //e=Mu/Pu=0.10h As /I -I I- b ---.j Td' 1 est = ey
Should not be used
10.3.6.1 and 10.3.6.2 2 est = 0.005
I
I
I
I
I
I
I
I CompressionI Axial Forces
I
0.1 I
est < ey }
<1> = 0.65 (ties) 9.3.2.2'!=/0.75 (spiral)
X 2 est = 0.005= var.
0.00-0.1 \ 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55mu = Mu / (b h2 f' cl\
\ Tension
\ Axial Forces
\
\ pg = 0.05
\
\
\
\
e = Mu I Pu = - 0.10 h
For these interaction diagrams: Pu = <1> Pn
Mu = <1> Mn
\
\ Are minimum eccentricities from
\/ R10.3.6 and R10.3.7 valid for tension
axial forces?
\

AXIAL LOAD AND UNIAXIAL BENDING
SYMMETRICAL REINFORCEMENT
DESIGN AIDS AUB-SR-XX
PROCEDURES FOR DESIGN AND VERIFICATION
Given: f c, fv, b, h, 'Y
and qiven Determine Step 1 Step 2 Step 3 Step 4
Compute For mu and Pu ComputeMu, Pu As! mu= Mu / (f'e b h2)?- Pu= Pu / (f'e b h) read Pg As! = pg b h""0
As! c Computeco For Pu and pg ComputeMu .,.?' pg = As! / (b h)
Pu <3 Pu= Pu / (f'e b h) read mu Mu = mu fe b h
2
.•....
As! •.... Compute
Pu E pg = As! / b h For mu and pg Compute:"Q read Pu Pu = Pu fe b hMu co mu = Mu / (f'e b h2)
Q) Compute-co
As! ";:: pg = As! / b h Computea. For pg and eMu, Pu 0 Mu = mu fe b h2•....a. read mu, Pue = Mu / Pu a. Draw line Pu = Pu fe b hco
t5 e = Mu / Pu
As!
Q) Compute Ok if point isQ)
pg = As! / b h Draw point between pg curveMu Safety C/) mu = Mu / (f'e b h2) for mu and Pu and Pu axe or if it isPu Pu = Pu / (f'e b h) on Pn curve

Given: fc = 4000 psi , fy = 60,000 psi; b = 14" , h = 25" , y = (h - 2 d') / h == 0.80
and given Determine Step 1 Step 2 Step 3 Step 4
Mu = 470 ft-kips Compute For mu and Computemu = Mu / (f'c b h2) Puread As! = pg b hMu = 5640 in-kips As! mu= 0.161 pg = 0.026 As! = 9.10 in.2
Pu= 420 kips Pu= Pu / (f'c b h)Pu= 0.300
Compute For Puand Compute
As! = 12.25 in.2 pg = As! / (b h) pg read Mu = mu fc b h2
Mu C/) pg = 0.035 mu= 0.148 Mu = 5180 in-kipsCD
Pu = -168 kips (tension) CD Pu= Pu / (f'c b h)()- pu = - 0.120Q)
"0
As! = 12.25 in.2
"0 Compute For mu and Compute....•0"0 pg = As! / b h pg read Pu = Pufc b h....•
Mu = 470 ft-kips Pu 05' pg = 0.035 Pu= 0.430 Pu = 602 kips-CD mu = Mu / (f' c b h2) Pu = - 105 kipsMu = 5640 in-kips Q) Pu= -0.075Ci m = 0.161-+00
As! = 12.25 in.2
0....• Compute For pg and Compute""'"t! pg = As! / b h e read Mu = mu fc b h2S'
e = Mu / Pu = 0.50 h Mu, Pu ,<-+00 pg = 0.035 mu = 0.18 Mu = 6300 in-kipsQ):J Draw line Pu= 0.36 Pu = Pufc b hQ.
-< e = Moo / P" = 0.5 h Pu = 504 ips
As!= 12.25 in.2 Compute Draw point Point is exterior
Mu = 470 ft-kips pg = As! / b h for mu and Given mu, Pu
Mu = 5640 in-kips pg = 0.035 Pu combination is
Safety mu = Mu / (f'c b h2) UNSAFE
Pu= 665 kips mu= 0.161
Pu= Pu / (f'c b h)
Pu= 0.475

AXIAL LOAD AND UNIAXIAL BENDING
SYMMETRICAL REINFORCEMENT
DESIGN AIDS AUB-SR-XX
PROPOSED SCHEME FOR THE WHOLE SET OF AIDS
2.- Concrete Sections and Reinforcement Arrangements to be Considered
Section R1 Section R2 Section R3
111I h ~I 111I h ~I 111I h ~I
IA AI A AA A 2A 2AA A
I+- yh ---.j I+- yh ---.j I+- yh ---.j
Section R4 Section R5 Section R6
111I h ~I 111I h ~I 111I h ~I
A A • • •
3A 3A • •A A • • •I+- yh ---.j I+- yh ---.j I+- yh ---.j
Section C1 Section A1
111I h ~I 111I h ~I

f c (psi):
fy (ksi):
y:
he/h:
2500,3000,4000,5000,6000,7000,8000,9000
Grade 60 and 75
0.90, 0.80, 0.70
0.10, 0.20, 0.30
fc (MPa):
fy (MPa):
y:
he/h:
17,20,25,30,35,40,45,50,55,60
Grade 420 and 520
0.90, 0.80, 0.70
0.10, 0.20, 0.30
Rectangular Sections: 6 Section Types x 2 Steel Grades x 8 Concrete Strengths x 3 Covers
Rectangular Sections: 288 Aids
Circular Sections:
Circular Sections:
1 Section Type x 2 Steel Grades x 8 Concrete Strengths x 3 Covers
48 Aids
Anular Sections:
Anular Sections:
1 Section Tipe x 2 Steel Grades x 3 he/h Ratios x 8 Concrete Strengths
48 Aids
Rectangular Sections: 6 Section Types x 2 Steel Grades x 10 Concrete Strengths x 3 Covers
Rectangular Sections: 360 Aids
Circular Sections:
Circular Sections:
1 Section Type x 2 Steel Grades x 10 Concrete Strengths x 3 Covers
60 Aids
Annular Sections:
Annular Sections:
1 Section Type x 2 Steel Grades x 3 he/h Ratios x 10 Concrete Strengths
60 Aids

• Sample Design Aid
• Background comments
• Step by step procedures
• Step by step examples
• Proposed scheme for the whole set of aids

AXIAL LOAD AND UNIAXIAL BENDING
UNSYMMETRICAL REINFORCEMENT
~ 1.25
II
"C.
AUB-NSR-XX
- I
e/h=Mu/(Puh)=0.10
I
f'c = 4,000 psi
fy = 60,000 psi
y= 0.80
Ag=bxh
As= pg Ag
A's= pig Ag
I I
Ref. Code ACI 318-08
-A's
As-
Developed by Victoria Hernandez Balat, Juan Francisco Bissio
and Daniel Ortega for INTI-CIRSOC, Argentina
\
\
e / h = Mu / ( Pu h )= - 0.10
\

AXIAL lOAD AND UNIAXIAL BENDING
UNSYMMETRICAL REINFORCEMENT
DESIGN AIDS AUB-NSR-XX
BACKROUND COMMENTS
Symmetric Reinforcement (As = A's)
Non-Symmetric Reinforcement (As #; A's)
Fiqure 1 Fiqure 2 I Fiqure 3 Fiqure 4
ASI AsIa
Symm. As AsIa / 2
A's AstO / 2
Non- ASI AsIa / 2 3 AsIa / 4 3Aslo/4 AsIa / 2As a AstO / 4 AstO / 2 AsIa / 2Symm. A's AsIa / 2 Asto /2 AsIa / 4 a
Figures 1 to 4 represent interaction diagrams developed for a given rectangular section.
Thinner curves represent diagrams corresponding to symmetrical arrangements with
total reinforcement equal to AsIa while thicker curves represent unsymmetrical
arrangements with total reinforcement always smaller than AstQ.
Figures show that, in many cases, unsymmetrical distributions could be much more
economical than symmetrical arrangements.
• Symmetrical distributions are optimal solutions for values of e = mulpu closer to
zero (pure compression or pure tension).
• For any combination (Mu, Pu) it is always possible to find an optimal arrangement
(As, A's) wich minimizes the total amount of reinforcement AstQ.
• These optimal arrangements can be summarized in design aids (graphics)
avoiding trial an error calculations.

AXIAL LOAD AND UNIAXIAL BENDING
UNSYMMETRICAL REINFORCEMENT
DESIGN AIDS AUB-NSR-XX
PROCEDURES FOR DESIGN AND VERIFICATION
Given: fe, fy, b, h, y, Mu and Pu
Determine: As and A's
Step 1: Select appropriate aid for f'e, fy and y
Step 2: Compute
mu = Mu / (f'e b h2)
Pu= Pu / (f'e b h)
Step 3: For mu and Puread pg and p'g
Step 4: Compute
As = pg b h
A's = p'g b h
Given: fe = 4000 psi , fv = 60,000 psi; b = 14" ; h = 25" ; Y = (h - 2 d') / h ::::0.80
and given Determine Step 1 Step 2
Mu = 470 ft-kips
Compute For mu and Compute
mu = Mu / (f'e b h2) puread As = pg b hMu = 5640 in-kips mu=0.161 pg = 0.0055 As = 1.93 in.2L...
Pu= 420 kips (*) E Pu= Pu / (f'e b h) p'g = 0.0105 A's = p'gbh"0 Pu= 0.300 A's = 3.68 in.2'm
Q) ?-- Compute For mu and ComputeMu = 205 ft-kips -As .~ "0 mu = Mu / (f'e b h2) Pu read As = pg b hMu = 2460 in-kips L... cg-ro mu = 0.07 pg = 0.006 As = 2.1 Oin.2
A's c..~ Pu= Put (f'e b h) p'g = 0 A's = p'g b hPu= 0 kips (**) c. uro ••.... Pu= 0 A's = 0 in.2t3 Compute For mu and ComputeMu 598 ft-kips Q)a3 mu = Mu / (f'e b h2) Pu read As = pg b hMu = 7175 in-kips - (f) mu = 0.205 pg = 0.020 As = 7.00 in.2
Pu= 0 kips (***) Pu = Pu / (f'e b h p'g = 0.0035 A's = p'g b hPu= 0 A' - 1.22 in.2s -
(*)
(**)
(***)
Data are the same as for First Example for Axial Load and Uniaxial Bending with
Symmetrical Reinforcement. Results: As = 4.55 in.2 ; A's = 4.55 in.2 (As! = 9.10 in.2)
Data are the same as for Flexure example b. Results: As = 2.17 in.2 ; A's = 0 in.2
Data are the same as for Flexure example c. Results: As = 6.93 in.2 ; A's = 1.26 in.2

Given: fe, fy, b, h, y, As and A's
Plot: Complete Interaction Diagram
1.50 ,0'
."<i' .rll·u..flP.nj.O.lO • [:J ".oo 4,000 pst
'l" ••.~ J ,,", 1 ~ ~,.'"::0000 psi
1.25 ""' ••w:- .,. / h fll ---) p~ ".b.1l
.••.•. '.. "" ..:;.. . .....Pl, .•.•
UIO """'''~ •••, • 4 b _. l d' A·."'p·, ••.•
ReI. Code AC1318-08
Step 1: Select appropriate aids for fe, fy, Y (symmetrical and
unsymmetrical reinforcement aids). For these calculations
unsymmetrical aid will be considered divided into three zones
(see figure)
Step 2: Compute (8V'
pgO = As f (b h)
p'90 = A's f (b h) (7):
PgtO= pgO + p'90 (6).,:",'-
~ = 0.65 (Pgo - p'go)
fyffe
-'.,(3)
,'," (4)
Interaction
Diagram
Step Aid For Point Action Compute
(1 ) If pg ;:: p'g then read muA, PuA mu1= - muAelse determine point from 4 (1) Pu1= PuA
3 UR pg = p'gO (6) Read muB, nuB mu6= - muB(*) p'g = pgO Pu6= PuB
(5) If pg < p'g then read muc, Puc mu5= - mucelse determine point form 4 (5) Pu5= Puc
(1 ) If pg < p'g then read muA, PuA mu1= muAelse determine point from 3 (1) Pu1= PuA
4 UR pg = pgO (4) Read muB, nuB mu4= muB(*) p'g = p'9O Pu4= PuB
(5) If pg ;:: p'g then read muc, Puc mu5= mucelse determine point from 3 (5) Pu5= Puc
5 SR pg = 2 p'9O (2) Read mu, nu mu2= mu(*) fs = 0 Pu2= Pu
6 SR pg = 2 pgO (8) Read mu, nu muB= - mu(*) fs = 0 PuB= Pu
pg = pgO + (3) mu3= muSR Pu3= Pu- /).7 (*) p'go Read mu, nu(7) mu7= - muEt = Cy Pu7= Pu+ /).
(*) SR: Design aid for Symmetrical Reinforcement
UR: Design aid for Unsymmetrical Reinforcement

Given: fc = 4000 psi ; fy = 60,000 psi;
b = 14" ; h = 25" ; 'Y= (h - 2 d') / h ::::0.80
As = 7.00 in.2 ; A's = 3.50 in.2
pgO = As / (b h) = 0.02
p'go = A's / (b h) = 0.01
PgtO = pgO + p'go = 0.03
D. = 0.65 (Pgo - p'go) fy / fc = 0.098
Step Aid For Point Action ComDute
If pgO ;:::p'go then read mu1= - muA= - 0.039
pg = p'9O (1 ) muA= 0.039 Pu1= PuA= 0.84PuA = 0.84pg = 0.01 Read3 UR (6) muB= 0.24 mu6= - muB= - 0.24
p'g = pgO PUB= 0.35 Pu6= PuB= 0.35
p'g = 0.02
(5) If pgO < p'go then read mue, Pueelse determine point form 4 (5) ----------
(1 ) If pgO < p'go then read muA, PuA ----------else determine Doint from 3 (1)
Read muB= 0.24pg = pgO (4 ) muB= 0.24 mu4=4 UR p'g = p'9O PuB= 0.08 Pu4= PuB= 0.08
If pgO ;::: p'9O then read mu5= mue= 0.054(5) mue = 0.054 Pu5= Pue= - 0.41Pue= - 0.41
pg = 2 p'go Read mu2= mu = 0.0885 SR (2) mu = 0.088fs = 0 Pu= 0.53 Pu2= Pu= 0.53
pg = 2 pgO Read mu8 = - mu = 0.1276 SR (8) mu = 0.127fs = 0 Pu= 0.62 PuB= Pu= 0.62
pg = pgO + (3) Read mu3=
mu= 0.186
7 SR p'gO mu=0.186 Pu3= Pu- ~ = 0.153mu7= - mu = - 0.186
£1 = £y (7) Pu= 0.25 Pu+ ~ = 0.348Pu7=

0.6
0.5
0.4
(6)
0.3
0.2
0.1 (4)
3 '().2 0.1 03
'().1 mu
Computer program
Approximate
-0.4

AXIAL LOAD AND UNIAXIAL BENDING
UNSYMMETRICAL REINFORCEMENT
DESIGN AIDS AUB-NSR-XX
PROPOSED SCHEME FOR THE WHOLE SET OF AIDS
Section R1
rd'I~~T
h y h •1~ 1
~ b -I Ld'
fc(psi):
fy (ksi):
y:
2500,3000,4000,5000,6000,7000,8000,9000
Grade 60 and 75
0.90,0.80,0.70
fc (MPa):
fy (MPa):
y:
17,20,25,30,35,40,45,50,55,60
Grade 420 and 520
0.90,0.80,0.70

• Sample Design Aid
• Background comments
• Step by step procedures
• Step by step examples
• Proposed scheme for the whole set of aids

AXIAL LOAD AND BIAXIAL BENDING ABB- XX
~Ub
Pu=Pu/(bhf'e) mux = max (muh, mub)A f' - 4,000 psie-I• • fy = 60,000 psi muh = MUh/ (b h2 f'c) muy = min (muh, mub)T Muh y= 0.80
h 17"Ag=bxh mub = MUb/ (b2 h f'e)1 As = pg Ag• •I- b -I Curves:
pg max= 0.08
~Pg = 0.01 (equidistance) Ref. Code AC1318-08
0.4 0.1 0.5mux 0.1 0.4
0.4 0.4 0.4
0.3 0.3 0.3
Pu = 0.90
0.2 0.2 0.2
0.1 0.1
mux mux
0.5 0.4- 0.3 0.4 0.5
0.1 0.1
0.2 0.2
0.3 0.3
0.4 0.4 0.4
muy O. muy0.4 0.3 0.2 0.1 mux 0.1 0.2 0.3 0.4
DevelopedbyVictoriaHernandezBala!,JuanFranciscoBissioandDanielOrtegaforINTI-CIRSOC,Argentina

AXIAL LOAD AND BIAXIAL BENDING
DESIGN AIDS ABB - XX
BACKROUNDCOMMENTS
Figure 1 shows an interaction surface corresponding to a rectangular section with a symmetrical
arrangement of reinforcement. The surface is expressed in terms of <p Pn, <p Mnhand <p Mnb.
Vertical axis represents axial forces while horizontal axis represent bending moments in two
orthogonal directions. This surface has two vertical planes of symmetry.
Figure 2 shows a horizontal cross section corresponding to a constant axial load. Any cross
section has two axes of symmetry.
0.4 0.3 0.2 0.1 mu. 0.1
B d h· . . Figure 4ase on t IS property It IS
possible to develop rosette type diagrams (Figure 4) which allow direct reading of reinforcement
(in terms of p) for any given set of Pu,muhand mub.Each eight of the diagram represents a cross
sections at a constant value of Pu. Many values of p are plotted on each eight.
~MUb
~
T· •I T MUh
h yh f"1~.. JFigure 3f-b~
If covers are selected as
shown in Figure 3 the
interaction surface in terms
of dimensionless values <p
Pn / (f'e b h), <p Mnh/ (f'e b
h2) and <p Mnb/ (fe b2 h) will
have four vertical planes of
symmetry. In such a case
horizontal cross section
presents four axes of
symmetry.

AXIAL LOAD AND BIAXIAL BENDING
DESIGN AIDS ABB - XX
PROCEDURES FOR DESIGN AND VERIFICATION
~Ub
yb-----I
• • T MUh
h 17"1 L=----=.J. • Given: fe, fy, b, h, y, Muh,MUband PuDetermine: As!
~b4
Step 2: Compute
Pu= Pu / (b h f'c)
muh= MUh/ (b h2 f'c)
mub= MUb/ (b2 h f'c)
mux= max (muh, mub)
muy= min (muh, mub)
Step 3: For the superior immediate value Pu12: Puand for muxand muy read Pg1
Step 4: For the inferior immediate value of Pu2:5Puand for muxand muy read Pg2
Step 5: Interpolate Pg= Pg1+ (P92- P91)x (Pu- Pu1)/ (Pu2- Pu1)
Step 6: Compute As! = Pgb h
Given: fc = 4000 psi
Determine: As!
for
MUh= 470 ft-kips
MUh= 5640 in-kips ~
Q)
(/)
Ste 2 Ste 6
u=Pu/(bhf'c}
Pu= 0.22
m - M / (b h2 f' ) P - P +uh- uh c Read P for Read P for 9 - g1 omputeuh= 0.081 9 9 (P92- P91)X
Ub: MUb/ (b2 h f'c) Pu1= 0.30 u2= 0.15 (Pu- Pu1)/ s! = Pgb hub- 0.220 (Pu2- Pu1)
ux= max (muh,mub) - 0 059 - 0 050 = 18.97 in.2ux= 0.220 P91-. g2-. 9 = 0.054
uy= min (muh,mub)
u = 0.081
Pu= Pu/ (b h f'c)
Pu= 0.30
m - M / (b h2 f' ) P - P +uh- uh c Read P for Read P for 9 - g1 ComputemUh=0.161 9 9 (P92-P91)x
ub= MUb/ (b2 h f'c) = 0 30 = 0 30 (Pu- Pu1)/ sl = Pgb h= 0 Pu1· u2 . P _ P }ub u2 u1
ux= max (muh,mub) 9 10' 2ux= 0.161 g1= 0.026 g2= 0.026 9 = 0.026 sl =. In.
uy= min (muh,mub)
u = 0
MUh= 235 ft-kips
MUh= 2820 in-kips
MUb= 360 ft-kips
MUb= 4320 in-kips
Data are the same as for First Example for Axial Load and Uniaxial Bending with
Symmetrical Reinforcement. Results: As = 4.55 in.2 ; A's = 4.55 in.2 (As!= 9.10 in.2)

These aids can be used to solve many problems ranging from dimensioning to checking safety.
The variety is so wide that could not be fully exposed in this paper.

AXIAL LOAD AND BIAXIAL BENDING
DESIGN AIDS ABB - XX
PROPOSED SCHEME FOR THE WHOLE SET OF AIDS
2.- Concrete Section and Reinforcement Arrangement to be Considered
Seccion R2 Seccion R6 Seccion R7
1-- h ~I 1--
A T •
A A b •A 1 •
---h
• T
yb-i...
fc (psi):
fy (ksi):
y:
2500,3000,4000,5000,6000,7000,8000,9000
Grade 60 and 75
0.90,0.80,0.70
fc (MPa):
fy (MPa):
y:
17,20,25,30,35,40,45,50,55,60
Grade 420 and 520
0.90,0.80,0.70
4.- Estimated of Aids to be Developed
4.1.- U.S. Customary Units Version