Título/s: | Argentina document |
Autor/es: | Ortega, Daniel; Hernández Balat, Victorio; Bissio, Juan Francisco |
Institución: | Centro de Investigación de los Reglamentos Nacionales de Seguridad para Obras Civiles. CIRSOC. Buenos Aires, AR Quasdam Ingeniería |
Editor: | INTI-CIRSOC |
Palabras clave: | Estructuras; Cálculo de estructuras; Códigos; Seguridad en la construcción; Flexión; Tensiones; Compresión; Diseño; Elementos estructurales; Cargas |
Idioma: | eng |
Fecha: | 2008 |
Ver+/- 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 Ver+/- | |
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