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
Bia"ti?1 Bending
Unsymmetrical Reinforcement
• 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
,COT Ec= 0.003
0.85 fc
- -i"~,cl
-l d - a /2 d - d'
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ L\_
mu,
P min
6,
2500 0.0564 0.0030 0.850
3000 0.0474 0.0030 0.850
4000 0.0359 0.0030 0.850
f'c (psi)
5000 0.0303
6000 0.0277
0.0032
0.0035
0.801
0.752
7000 0.0257 0.0038 0.702
8000 0.0241 0.0040 0.653
9000 0.0227 0.0043 0.650
0.015 0.010
fi.' ~~~
'),'3
I
p' = A's I (bw h)
0.015 0.010
0.000
0.00
0.05
0.10
0.15
I
I
-- I
0.000
0.30
0.35
0.40
0.45
mu = Mu I (bw h2 f'd
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)
rd'
~"
T
h
A's [T-Td'.'h
- 1 As
Given: Determine:
Step 1:
Step 2:
f'e, fy, bw, h, y, Mu As and A's
For f'e read mU1 and Pmin
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'
As
1 d = h - d' = y h
-
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)
= Muw Mu - MUI = muw Muw 1 (bw h2 f'e)
<l> = 0.90
Compute As min= pminbw h Check As ;:: As min
important: use bw not b
FLEXURE - OPTIMAL DESIGN - DESIGN AIDS FLE-XX EXAMPLES FOR RECTANGULAR AND FLANGED SECTIONS WITH TENSION AND COMPRESSION REINFORCEMENT (IF NECESSARY)
fd' A's
I [-d':rh h
Given: a}
fc = 4000 psi; fy = 60,000 psi bw = 14" ; h = 25" ; Y = d / h == 0.90
Mu = 95 ft-kips = 1140 in-kips
1 - ~" As
b}
Mu = 205 ft-kips = 2460 in-kips
c} Mu = 598 ft-kips = 7175 in-kips
I-- bw ---..j Ld'
Determine: As and A's
Step 1 2
3
Procedure For f'c read mu1 and Pmin
mu = Mu / (b h2 fc) Compare
For muw and f'c read p and p' or o = Dmin if mu:::; mu1
As = P bw h and A's =p' bw h
Case a)
Case b)
Case c)
mu1 = 0.0359, Pmin= 0.0030
0.0326
0.0703
0.205
mu:::; mu1
Pmin= 0.0030 As = 1.05 in.:.! A's = 0.00 in.2
mu> mu1 P = 0.0062
P' = 0 As = 2.17 in.:.! A's = 0.00 in.2
mu> mu1 p=0.0198
p' = 0.0036 As = 6.93 in.2 A's = 1.26 in.2
r;r I'
h
1 ~"
b A's
·1 T~d'
d = h - d' = y h
- ~d' 1 As
I- bw-! T
Given:
fc = 4000 psi ; fy = 60,000 psi b = 42" ; hI = 4" bw = 14" ; h = 25" ; Y = d / h ::: 0.90
a)
Mu= 420 ft-kips = 5,040 in-kips
b)
Mu = 1000 ft-kips = 12,000 in-kips
c)
Mu = 1170 ft-kips = 14,040 in-kips
Determine: As and A's
Step 1 2
3 4
Procedure
For f'c read mu1 and Pmin hf / h
mu = Mu / (b h£ f c) For mu read a / h
Compare Action
For mu and f'c read p and p'
5
As=pbh
and A's =p' b h
Action
Cnf = 0.85 fc (b - bw) hf
6
MUI= 0.90 Cnf (yh - hf/ 2)
Muw= Mu - MUf
muw= Muw/ (bw hL f'c)
For muwand f'c read p and p'
7 Asw = P bw h and A'sw =p' bw h
8
As = Asw + Cnf / fy and
A's = A'sw
9
As min= Dminbw h
Check As;O:Asmin
Case a)
Case b)
Case c)
0.048
mu1 = 0.0359, Pmin= 0.0030
0.16
0.1143
0.1337
0.072
0.185
0.225
a / h < hl/ h go to step 5
a / h > hl/ h go to step 6
a / h > hf / h qo to step 6
p=0.0041
D' = 0 As = 4.31 in.L A's = 0.00 in.2 qo to step 9
--------
---------------
----------------------
---------------
380.80 kips 7025.76 in-kips
380.80 kips 7025.76 in-kips
--------
--------
4974.24 in-kips 0.1421
7014.24 in-kips 0.2004
----------------
--------
p = 0.0135
D' = 0 Asw = 4.73 in.L A'sw = 0.00 in.2 As = 11.08 in.2 A's = 0.00 in.2
1.05 in.2
p=0.0194
D' = 0.0032
Asw = 6.79 in.L A'sw= 1.12 in.2
As= 13.14in.2 A's=1.12in.2
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'
IT A's
h
[dO:'"
1 - ~" As
I-bw --.j Ld'
IhfI
h
1 ~"
A's
T~d
d = h - d' = Y h
- dO 1 As
-l
I- bw---l T
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
1.3
1.1
0.9
rd' f' c = 4,000 psi
r - T A's
Mu
fy = 60,000 psi
y= 0.80
h
Ag=bxh
1- If" As
Ast= pg Ag
I- b -l Ld'
Ast= As + A's
III
Ref. Code ACI 318-08
e I h = Mu I ( Pu h )= ·0.10 \\
Developed by Victoria Hernandez Balat, Juan Francisco Bissio and Daniel Ortega for INTI-CIRSOC, Argentina
AXIAL LOAD AND UNIAXIAL BENDING
= SYMMETRICAL REINFORCEMENT - As A's
UNDERSTANDING INTERACTION DIAGRAMS
Pu=Pu/(bhf'cl
I
I
e=Mu/Pu=0.10h
I
I
Should not be used
10.3.6.1 and 10.3.6.2
I
I- rd' T A's
1 - 1 h
yh
As
I- b ---.j T d'
,- ee= 0.003
// //
/
/
/
/ /
/
/
1 est = ey 2 est = 0.005
I I I I I I
I Compression I Axial Forces I 0.1 I
est < ey
}
<1> = 0.65 (ties)
'!=/0.75 (spiral)
9.3.2.2
X = var.
= 2 est 0.005
0.00 -0.1 \\
\\
\\ \\
Tension Axial Forces
0.20
0.25
0.30
0.35
0.40 0.45 0.50 0.55
mu = Mu / (b h2 f' cl
\\
\\ pg = 0.05
\\
For these interaction
\\
\\
\\
e = Mu I Pu = - 0.10 h
diagrams:
\\
\\ \\/
\\
Are minimum eccentricities from R10.3.6 and R10.3.7 valid for tension axial forces?
Pu = <1> Pn Mu = <1> Mn
AXIAL LOAD AND UNIAXIAL BENDING SYMMETRICAL REINFORCEMENT DESIGN AIDS AUB-SR-XX
PROCEDURES FOR DESIGN AND VERIFICATION
and qiven Mu, Pu
Determine
As!
As!
Mu Pu
As!
Pu Mu
As!
e = Mu / Pu
Mu, Pu
As!
Mu
Safety
Pu
Given: f c, fv, b, h, 'Y
Step 1
Step 2
Step 3
Compute
?""0
mu= Mu / (f'e b h2) Pu= Pu / (f'e b h)
For mu and Pu read Pg
cco .,.?' .•...<.3
Compute pg = As! / (b h) Pu= Pu / (f'e b h)
For Pu and pg read mu
•....
Compute
E
:"Q
-co
Q)
co
pg = As! / b h mu = Mu / (f'e b h2)
Compute
For mu and pg read Pu
"a;::. •a0..... a. co
pg = As! / b h Draw line
For pg and e read mu, Pu
t5
e = Mu / Pu
Q)
Compute
Q)
C/)
pg = As! / b h
Draw point
mu = Mu / (f'e b h2) for mu and Pu
Pu = Pu / (f'e b h)
Step 4 Compute As! = pg b h
Compute Mu = mu fe b h2
Compute Pu = Pu fe b h
Compute Mu = mu fe b h2 Pu = Pu fe b h
Ok if point is between pg curve and Pu axe or if it is 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
Mu = 5640 in-kips
As!
Pu= 420 kips
Compute
= mu Mu / (f'c b h2)
mu= 0.161 Pu= Pu / (f'c b h) Pu= 0.300
For mu and Pu read pg = 0.026
Compute
As! = pg b h As! = 9.10 in.2
As! = 12.25 in.2 Pu = -168 kips (tension)
As! = 12.25 in.2 Mu = 470 ft-kips Mu = 5640 in-kips As! = 12.25 in.2 e = Mu / Pu = 0.50 h
As!= 12.25 in.2 Mu = 470 ft-kips Mu = 5640 in-kips
Pu= 665 kips
Mu Pu Mu, Pu Safety
Compute
For Pu and Compute
pg = As! / (b h)
pg read
Mu = mu fc b h2
C/) CD
pg = 0.035
-CD
()
Pu= Pu / (f'c b h) pu = - 0.120
Q)
mu= 0.148 Mu = 5180 in-kips
"0
".0...•
Compute
0
For mu and Compute
-".0...•
05'
CD
pg = As! / b h pg = 0.035
pg read Pu= 0.430
Pu = Pufc b h Pu = 602 kips
Q)
mu = Mu / (f' c b h2)
Pu= -0.075
Pu = - 105 kips
Ci
-+00
m = 0.161
.0...• Compute
For pg and Compute
""'"t!
S'
pg = As! / b h
e read
Mu = mu fc b h2
,<-+00
pg = 0.035
Q)
mu = 0.18
Mu = 6300 in-kips
:J
Q.
Draw line
Pu= 0.36
Pu = Pufc b h
-<
e = Moo / P" = 0.5 h
Pu = 504 ips
Compute
pg = As! / b h pg = 0.035 mu = Mu / (f' c b h2) mu= 0.161 Pu= Pu / (f'c b h) Pu= 0.475
Draw point for mu and Pu
Point is exterior Given mu, Pu combination is UNSAFE
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
111I
h
~I
IA AI
I+- yh ---.j
Section R2
111I
h
~I
A
A
A
A
I+- yh ---.j
Section R3
111I
h
~I
A
2A
2A
A
I+- yh ---.j
Section R4
111I
h
~I
A
3A
3A
A
I+- yh ---.j
Section R5
111I
h
~I
A
A
I+- yh ---.j
Section R6
111I
h
~I
• •• •• • ••
I+- yh ---.j
Section C1
111I
h
~I
Section A1
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: 1 Section Type x 2 Steel Grades x 3 he/h Ratios x 10 Concrete Strengths Annular Sections: 60 Aids
• Sample Design Aid • Background comments • Step by step procedures • Step by step examples • Proposed scheme for the whole set of aids
~ 1.25
II
C".
AXIAL LOAD AND UNIAXIAL BENDING UNSYMMETRICAL REINFORCEMENT
-
I
e/h=Mu/(Puh)=0.10
I
-A's
-As
AUB-NSR-XX
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
\\
\\ e / h = Mu / ( Pu h )= - 0.10
\\
Developed by Victoria Hernandez Balat, Juan Francisco Bissio and Daniel Ortega for INTI-CIRSOC, Argentina
AXIAL lOAD AND UNIAXIAL BENDING UNSYMMETRICAL REINFORCEMENT
DESIGN AIDS AUB-NSR-XX BACKROUND COMMENTS
ASI
Symm.
As
A's
Non-
ASI
Symm.
As
A's
Fiqure 1
AsIa / 2
a
AsIa / 2
= Symmetric Reinforcement (As A's)
Non-Symmetric Reinforcement (As #; A's)
Fiqure 2
3 AsIa / 4 AstO / 4 Asto /2
I
AsIa AsIa / 2 AstO / 2
Fiqure 3
3Aslo/4 AstO / 2 AsIa / 4
Fiqure 4
AsIa / 2 AsIa / 2
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 Pu read 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;
and given
Determine
Step 1
Mu = 470 ft-kips Mu = 5640 in-kips
Pu= 420 kips (*)
Mu = 205 ft-kips Mu = 2460 in-kips
As
Pu= 0 kips (**)
A's
Mu 598 ft-kips Mu = 7175 in-kips -
L...
E
"0
'mQ) ?--
- c .~ "0 L... cg-r.o.~ cro. ••.u...
t3
Q)
a3
(f)
Pu= 0 kips (***)
b = 14" ; h = 25" ; Y = (h - 2 d') / h ::::0.80
Step 2
Compute mu = Mu / (f'e b h2) mu=0.161
For mu and puread pg = 0.0055
Compute
As = pg b h As = 1.93 in.2
Pu= Pu / (f'e b h) Pu= 0.300
p'g = 0.0105 A's = p'gbh A's = 3.68 in.2
Compute
For mu and
mu = Mu / (f'e b h2) Pu read
mu = 0.07
pg = 0.006
Compute
As = pg b h As = 2.1 Oin.2
Pu= Put (f'e b h) p'g = 0
A's = p'g b h
Pu= 0
A's = 0 in.2
Compute
For mu and
mu = Mu / (f'e b h2) Pu read
mu = 0.205
pg = 0.020
Compute
As = pg b h As = 7.00 in.2
Pu = Pu / (f'e b h p'g = 0.0035 A's = p'g b h
Pu= 0
A' s -- 1.22 in.2
(*)
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
[:J Plot: Complete Interaction Diagram
1.50
."<i' 'l" ••~.
.rll·u..flP.nj.O.lO J
,0'
• ,,", 1 ~
~".,o.o'"::040,00000
pst psi
Step
1: Select
1.25
""' ••w:- .,. /
.••.•. '..""
h
fll ---) p~ ".b.1l
..:;.. .
.....Pl, .•.•
unsymmetrical
appropriate aids for fe, fy, Y (symmetrical and reinforcement aids). For these calculations
UIO
"""'''~ •••,
• 4 b \_. l d'
A·."'p·, ••.•
unsymmetrical aid will be considered divided into three zones
ReI. Code AC1318-08
(see figure)
Step 2: Compute
pgO = As f (b h) p'90 = A's f (b h)
PgtO = pgO + p'90
~ = 0.65 (Pgo - p'go)
fyffe
(8V'
(7): (6).,:",'-
Interaction Diagram
-'.,(3) ,'," (4)
Step Aid
UR
3
(*)
4
UR (*)
SR
5
(*)
SR
6
(*)
7
SR
(*)
For
pg = p'gO p'g = pgO
pg = pgO p'g = p'9O
pg = 2 p'9O fs = 0
pg = 2 pgO fs = 0
pg = pgO +
p'go
= Et Cy
Point (1 ) (6) (5) (1 ) (4) (5) (2) (8) (3) (7)
Action If pg ;:: p'g then read muA, PuA else determine point from 4 (1) Read muB, nuB If pg < p'g then read muc, Puc else determine point form 4 (5) If pg < p'g then read muA, PuA else determine point from 3 (1) Read muB, nuB If pg ;:: p'g then read muc, Puc else determine point from 3 (5) Read mu, nu
Read mu, nu
Read mu, nu
(*)
SR: Design aid for Symmetrical Reinforcement
UR: Design aid for Unsymmetrical Reinforcement
Compute
mu1 = - muA = Pu1 PuA mu6 = - muB = Pu6 PuB mu5 = - muc = Pu5 Puc = mu1 muA = Pu1 PuA = mu4 muB = Pu4 PuB = mu5 muc = Pu5 Puc mu2 = mu Pu2= Pu muB= - mu PuB= Pu mu3 = mu = Pu3 Pu - /). mu7 = - mu = Pu7 Pu + /).
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
3
UR
4
UR
5
SR
6
SR
7
SR
For
pg = p'9O pg = 0.01 p'g = pgO p'g = 0.02
pg = pgO p'g = p'9O
pg = 2 p'go fs = 0
pg = 2 pgO fs = 0
pg = pgO + p'gO
£1 = £y
Point (1 ) (6) (5) (1 ) (4 ) (5) (2) (8) (3) (7)
Action
If pgO ;:::p'go then read muA = 0.039 PuA = 0.84 Read muB = 0.24 PUB= 0.35
If pgO < p'go then read mue, Pue else determine point form 4 (5)
If pgO < p'go then read muA, PuA else determine Doint from 3 (1) Read muB = 0.24 PuB= 0.08 If pgO ;::: p'9O then read mue = 0.054 Pue= - 0.41 Read mu = 0.088 Pu = 0.53 Read mu = 0.127 Pu = 0.62
Read mu=0.186 Pu = 0.25
ComDute mu1 = - muA= - 0.039
Pu1= PuA= 0.84
mu6 = - muB= - 0.24 Pu6= PuB= 0.35
----------
----------
mu4 = muB = 0.24 Pu4= PuB= 0.08
mu5 = mue= 0.054 Pu5= Pue= - 0.41
mu2 = mu = 0.088 Pu2= Pu = 0.53
mu8 = - mu = 0.127 PuB= Pu = 0.62
mu3 = mu= 0.186 Pu3= Pu- ~ = 0.153 mu7 = - mu = - 0.186 Pu7= Pu + ~ = 0.348
0.6
0.5
0.4
(6)
0.3
0.2
0.1
3
'().2
0.1
'().1
Computer program
Approximate
-0.4
(4)
03
mu
AXIAL LOAD AND UNIAXIAL BENDING UNSYMMETRICAL REINFORCEMENT
DESIGN AIDS AUB-NSR-XX PROPOSED SCHEME FOR THE WHOLE SET OF AIDS
Section R1
I~~Trd'
h
yh •
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
f' e-- 4,000 psi
A fy = 60,000 psi • • y= 0.80
T Muh
I Ag=bxh
h
1 • • 17"As = pg Ag
Pu=Pu/(bhf'e) muh = MUh/ (b h2 f'c) mub = MUb/ (b2 h f'e)
mux = max (muh, mub) muy = min (muh, mub)
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
mux 0.5
0.1
0.4-
0.3
0.1
mux
0.4
0.5
0.1
0.2
0.2
0.3
0.3
0.4
0.4
0.4
muy
0.4
0.3
0.2
O.
0.1
mux
0.1
0.2
0.3
0.4
muy
DevelopebdyVictoriaHernandeBzala!,JuanFranciscoBissio andDanieOl rtegaforINTI-CIRSOCA,rgentina
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.
~MUb
~
T TI · •
MUh
f" h
yh
1.~. J
f-b~
Figure 3
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.
0.4 0.3
0.2
0.1
mu. 0.1
Based
on
t h·IS
property
I.t
.
IS
Figure 4
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.
AXIAL LOAD AND BIAXIAL BENDING DESIGN AIDS ABB - XX
PROCEDURES FOR DESIGN AND VERIFICATION
~Ub
yb -----I
•
1h L=----=.J.
• T MUh
Given: fe, fy, b, h, y, Muh,MUband Pu
17"Determine: 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 mux and 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! = Pg b h
Given: fc = 4000 psi Determine: As!
for
MUh= 235 ft-kips MUh= 2820 in-kips MUb= 360 ft-kips MUb= 4320 in-kips
MUh= 470 ft-kips MUh= 5640 in-kips ~
Q)
(/)
Ste 2
Ste 6
u=Pu/(bhf'c}
Pu = 0.22
muh-- Muh/ (b h2 f' c) Read P for Read P for P9 -- Pg1+
uh= 0.081
9
9 (P92- P91)X
Ub:
MUb/
2 (b
h
f'c)
Pu1= 0.30
u2= 0.15 (Pu - Pu1)/
ub- 0.220
(Pu2- Pu1)
ompute s! = Pg b h
ux= max (muh,mub) - 0 059 - 0 050
= 18.97 in.2
ux= 0.220
P91-.
g2-.
9 = 0.054
uy= min (muh,mub)
u = 0.081
Pu = Pu/ (b h f'c)
Pu = 0.30
m uh-- M uh/ (b h2 f' c) Read P for Read P for P9 -- Pg1+ Compute
mUh=0.161
9
9 (P92-P91)x
uubb== 0MUb/
2 (b h f'c)
Pu1=·
0 30
u2= 0 .30 (PPuu2-\_PPuu11)}/
sl = Pg b h
ux= max (muh,mub)
9 10' 2
ux= 0.161
g1= 0.026 g2= 0.026 9 = 0.026 sl =. In.
uy= min (muh,mub) u =0
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-- --h
T A
A
A
•• b•
A
1•
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
Ver+/-
