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2 CECW-ED Engineer Manual Department of the Army U.S. Army Corps of Engineers Washington, DC Engineering an Design STRENGTH DESIGN FOR REINFORCED CONCRETE HYDRAULIC STRUCTURES Distribution Restriction Statement Approve for public release; istribution is unlimite. EM June 1992

3 30 June 1992 US Army Corps of Engineers ENGINEERING AND DESIGN Strength Design for Reinforce-Concrete Hyraulic Structures ENGINEER MANUAL

4 DEPARTMENT OF THE ARMY EM US Army Corps of Engineers CECW-ED Washington, DC Engineer Manual No June 1992 Engineering an Design STRENGTH DESIGN FOR REINFORCED-CONCRETE HYDRAULIC STRUCTURES 1. Purpose. This manual provies guiance for esigning reinforce concrete hyraulic structures by the strength-esign metho. Plain concrete an prestresse concrete are not covere in this manual. 2. Applicability. This manual applies to all HQUSACE/OCE elements, major suborinate commans, istricts, laboratories, an fiel operating activities having civil works responsibilities. FOR THE COMMANDER: Colonel, Corps of Engineers Chief of Staff This manual supersees ETL , Strength Design Criteria for Reinforce Concrete Hyraulic Structures, ate 10 March 1988 an EM , Details of Reinforcement-Hyraulic Structures, ate 21 May 1971.

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7 CHAPTER 1 INTRODUCTION 1-1. Purpose This manual provies guiance for esigning reinforce-concrete hyraulic structures by the strength-esign metho Applicability This manual applies to all HQUSACE/OCE elements, major suborinate commans, istricts, laboratories, an fiel operating activities having civil works responsibilities References a. EM , Working Stresses for Structural Design. b. EM , Conuits, Culverts, an Pipes. c. CW-03210, Civil Works Construction Guie Specification for Steel Bars, Wele Wire Fabric, an Accessories for Concrete Reinforcement.. American Concrete Institute, "Builing Coe Requirements an Commentary for Reinforce Concrete," ACI 318, Box 19150, Refor Station, Detroit, MI e. American Concrete Institute, "Environmental Engineering Concrete Structures," ACI 350R, Box 19150, Refor Station, Detroit, MI f. American Society for Testing an Materials, "Stanar Specification for Deforme an Plain Billet-Steel Bars for Concrete Reinforcement," ASTM A , 1916 Race St., Philaelphia, PA g. American Weling Society, "Structural Weling Coe-Reinforcing Steel," AWS D , 550 NW Le Jeune R., P.O. Box , Miami, FL h. Liu, Tony C (Jul). "Strength Design of Reinforce Concrete Hyraulic Structures, Report 1: Preliminary Strength Design Criteria," Technical Report SL-80-4, US Army Engineer Waterways Experiment Station, 3909 Halls Ferry Roa, Vicksburg, MS i. Liu, Tony C., an Gleason, Scott (Sep). "Strength Design of Reinforce Concrete Hyraulic Structures, Report 2: Design Ais for Use in the Design an Analysis of Reinforce Concrete Hyraulic Structural Members Subjecte to Combine Flexural an Axial Loas," Technical Report SL-80-4, US Army Engineer Waterways Experiment Station, 3909 Halls Ferry Roa, Vicksburg, MS

8 j. Liu, Tony C (Sep). "Strength Design of Reinforce Concrete Hyraulic Structures, Report 3: T-Wall Design," Technical Report SL-80-4, US Army Engineer Waterways Experiment Station, 3909 Halls Ferry Roa, Vicksburg, MS Backgroun a. A reinforce concrete hyraulic structure is one that will be subjecte to one or more of the following: submergence, wave action, spray, chemically contaminate atmosphere, an severe climatic conitions. Typical hyraulic structures are stilling-basin slabs an walls, concrete-line channels, portions of powerhouses, spillway piers, spray walls an training walls, floowalls, intake an outlet structures below maximum high water an wave action, lock walls, guie an guar walls, an retaining walls subject to contact with water. b. In general, existing reinforce-concrete hyraulic structures esigne by the Corps, using the working stress metho of EM , have hel up extremely well. The Corps began using strength esign methos in 1981 (Liu 1980, 1981 an Liu an Gleason 1981) to stay in step with inustry, universities, an other engineering organizations. ETL , "Strength Design Criteria for Reinforce Concrete Hyraulic Structures," ate 15 September 1981, was the first ocument proviing guiance issue by the Corps concerning the use of strength esign methos for hyraulic structures. The labor-intensive requirements of this ETL regaring the application of multiple loa factors, as well as the fact that some loa-factor combination conitions resulte in a less conservative esign than if working stress methos were use, resulte in the evelopment of ETL , "Strength Design Criteria for Reinforce Concrete Hyraulic Structures," ate 10 March c. The revise loa factors in ETL were intene to ensure that the resulting esign was as conservative as if working stress methos were use. Also, the single loa factor concept was introuce. The guiance in this ETL iffere from ACI 318 Builing Coe Requirements an Commentary for Reinforce Concrete primarily in the loa factors, the concrete stressstrain relationship, an the yiel strength of Grae 60 reinforcement. ETL guiance was intene to result in esigns equivalent to those resulting when working stress methos were use.. Earlier Corps strength esign methos eviate from ACI guiance because ACI 318 inclues no provisions for the serviceability nees of hyraulic structures. Strength an stability are require, but serviceability in terms of eflections, cracking, an urability eman equal consieration. The importance of the Corps hyraulic structures has cause the Corps to move cautiously, but eliberately, towar exclusive use of strength esign methos. e. This manual moifies an expans the guiance in ETL with an approach similar to that of ACI 350R-89. The concrete stress-strain relationship an the yiel strength of Grae 60 reinforcement given in ACI 318 are aopte. Also, the loa factors bear a closer resemblance to ACI 318 an 1-2

9 are moifie by a hyraulic factor, H f, to account for the serviceability nees of hyraulic structures. f. As in ETL , this manual allows the use of a single loa factor for both ea an live loas. In aition, the single loa factor metho is require when the loas on the structural component inclue reactions from a soil-structure stability analysis General Requirements Reinforce-concrete hyraulic structures shoul be esigne with the strength esign metho in accorance with the current ACI 318, except as hereinafter specifie. The notations use are the same as those use in the ACI 318 Coe an Commentary, except those efine herein Scope a. This manual is written in sufficient etail to not only provie the esigner with esign proceures, but to also provie examples of their application. Also, erivations of the combine flexural an axial loa equations are given to increase the esigner s confience an unerstaning. b. General etailing requirements are presente in Chapter 2. Chapter 3 presents strength an serviceability requirements, incluing loa factors an limits on flexural reinforcement. Design equations for members subjecte to flexural an/or axial loas (incluing biaxial bening) are given in Chapter 4. Chapter 5 presents guiance for esign for shear, incluing provisions for curve members an special straight members. The appenices inclue notation, equation erivations, an examples. The examples emonstrate: loa-factor application, esign of members subjecte to combine flexural an axial loas, esign for shear, evelopment of an interaction iagram, an esign of members subjecte to biaxial bening Computer Programs Copies of computer programs, with ocumentation, for the analysis an esign of reinforce-concrete hyraulic structures are available an may be obtaine from the Engineering Computer Programs Library, US Army Engineer Waterways Experiment Station, 3909 Halls Ferry Roa, Vicksburg, Mississippi For esign to account for combine flexural an axial loas, any proceure that is consistent with ACI 318 guiance is acceptable, as long as the loa factor an reinforcement percentage guiance given in this manual is followe Recission Corps library computer program CSTR (X0066), base on ETL , is replace by computer program CASTR (X0067). Program CASTR is base on this new engineer manual. 1-3

10 30 Apr 92 CHAPTER 2 DETAILS OF REINFORCEMENT 2-1. General This chapter presents guiance for furnishing an placing steel reinforcement in various concrete members of hyraulic structures Quality The type an grae of reinforcing steel shoul be limite to ASTM A 615 (Billet Steel), Grae 60. Grae 40 reinforcement shoul be avoie since its availability is limite an esigns base on Grae 40 reinforcement, utilizing the proceures containe herein, woul be overly conservative. Reinforcement of other graes an types permitte by ACI 318 may be permitte for special applications subject to the approval of higher authority Anchorage, Bar Development, an Splices The anchorage, bar evelopment, an splice requirements shoul conform to ACI 318 an to the requirements presente below. Since the evelopment length is epenent on a number of factors such as concrete strength an bar position, function, size, type, spacing, an cover, the esigner must inicate the length of embement require for bar evelopment on the contract rawings. For similar reasons, the rawings shoul show the splice lengths an special requirements such as staggering of splices, etc. The construction specifications shoul be carefully eite to assure that they agree with reinforcement etails shown on the rawings Hooks an Bens Hooks an bens shoul be in accorance with ACI Bar Spacing a. Minimum. The clear istance between parallel bars shoul not be less than 1-1/2 times the nominal iameter of the bars nor less than 1-1/2 times the maximum size of coarse aggregate. No. 14 an No. 18 bars shoul not be space closer than 6 an 8 inches, respectively, center to center. When parallel reinforcement is place in two or more layers, the clear istance between layers shoul not be less than 6 inches. In horizontal layers, the bars in the upper layers shoul be place irectly over the bars in the lower layers. In vertical layers, a similar orientation shoul be use. In construction of massive reinforce concrete structures, bars in a layer shoul be space 12 inches center-to-center wherever possible to facilitate construction. b. Maximum. The maximum center-to-center spacing of both primary an seconary reinforcement shoul not excee 18 inches. 2-1

11 30 Apr Concrete Protection for Reinforcement The minimum cover for reinforcement shoul conform to the imensions shown below for the various concrete sections. The imensions inicate the clear istance from the ege of the reinforcement to the surface of the concrete. CONCRETE SECTION MINIMUM CLEAR COVER OF REINFORCEMENT, INCHES Unforme surfaces in contact with founation 4 Forme or screee surfaces subject to cavitation or abrasion erosion, such as baffle blocks an stilling basin slabs 6 Forme an screee surfaces such as stilling basin walls, chute spillway slabs, an channel lining slabs on grae: Equal to or greater than 24 inches in thickness 4 Greater than 12 inches an less than 24 inches in thickness 3 Equal to or less than 12 inches in thickness will be in accorance with ACI Coe 318. NOTE. In no case shall the cover be less than: 1.5 times the nominal maximum size of aggregate, or 2.5 times the maximum iameter of reinforcement Splicing a. General. Bars shall be splice only as require an splices shall be inicate on contract rawings. Splices at points of maximum tensile stress shoul be avoie. Where such splices must be mae they shoul be staggere. Splices may be mae by lapping of bars or butt splicing. b. Lappe Splices. Bars larger than No. 11 shall not be lap-splice. Tension splices shoul be staggere longituinally so that no more than half of the bars are lap-splice at any section within the require lap length. If staggering of splices is impractical, applicable provisions of ACI 318 shoul be followe. c. Butt Splices (1) General. Bars larger than No. 11 shall be butt-splice. Bars No. 11 or smaller shoul not be butt-splice unless clearly justifie by esign etails or economics. Due to the high costs associate with butt splicing of bars larger than No. 11, especially No. 18 bars, careful 2-2

12 30 Apr 92 consieration shoul be given to alternative esigns utilizing smaller bars. Butt splices shoul be mae by either the thermit weling process or an approve mechanical butt-splicing metho in accorance with the provisions containe in the following paragraphs. Normally, arc-wele splices shoul not be permitte ue to the inherent uncertainties associate with weling reinforcement. However, if arc weling is necessary, it shoul be one in accorance with AWS D1.4, Structural Weling Coe-Reinforcing Steel. Butt splices shoul evelop in tension at least 125 percent of the specifie yiel strength, f y, of the bar. Tension splices shoul be staggere longituinally at least 5 feet for bars larger than No. 11 an a istance equal to the require lap length for No. 11 bars or smaller so that no more than half of the bars are splice at any section. Tension splices of bars smaller than No. 14 shoul be staggere longituinally a istance equal to the require lap length. Bars Nos. 14 an 18 shall be staggere longituinally, a minimum of 5 feet so that no more than half of the bars are splice at any one section. (2) Thermit Weling. Thermit weling shoul be restricte to bars conforming to ASTM A 615 (billet steel) with a sulfur content not exceeing 0.05 percent base on lale analysis. The thermit weling process shoul be in accorance with the provisions of Guie Specification CW (3) Mechanical Butt Splicing. Mechanical butt splicing shall be mae by an approve exothermic, threae coupling, swage sleeve, or other positive connecting type in accorance with the provisions of Guie Specification CW The esigner shoul be aware of the potential for slippage in mechanical splices an insist that the testing provisions containe in this guie specification be inclue in the contract ocuments an utilize in the construction work Temperature an Shrinkage Reinforcement a. In the esign of structural members for temperature an shrinkage stresses, the area of reinforcement shoul be times the gross crosssectional area, half in each face, with a maximum area equivalent to No. 9 bars at 12 inches in each face. Generally, temperature an shrinkage reinforcement for thin sections will be no less than No. 4 bars at 12 inches in each face. b. Experience an/or analyses may inicate the nee for an amount of reinforcement greater than inicate in paragraph 2-8a if the reinforcement is to be use for istribution of stresses as well as for temperature an shrinkage. c. In general, aitional reinforcement for temperature an shrinkage will not be neee in the irection an plane of the primary tensile reinforcement when restraint is accounte for in the analyses. However, the primary reinforcement shoul not be less than that require for shrinkage an temperature as etermine above. 2-3

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14 Change 1 20 Aug Require Strength a. General. Reinforce concrete hyraulic structures an hyraulic structural members shall be esigne to have a require strength, U h, to resist ea an live loas in accorance with the following provisions. The hyraulic factor is to be applie in the etermination of the require nominal strength for all combinations of axial loa, moments an shear (iagonal tension). In particular, the shear reinforcement shoul be esigne for the excess shear, the ifference between the hyraulic factore ultimate shear force,, an the shear strength provie by the concrete, φv c, where φ is the concrete resistance factor for shear esign. Therefore, the esign shear for the reinforcement, V s, is given by V uh V s V uh 1.3 φ φv c (3.1) b. Single Loa Factor Metho. In the single loa factor metho, both the ea an live loas are multiplie by the same loa factor. where where ( D L) U = factore loas for a nonhyraulic structure D = internal forces an moments from ea loas L = internal forces an moments from live loas U h H f = factore loas for a hyraulic structure = hyraulic factor. U = (3.2) [. ( D L) ] U h = H f (3.3) For hyraulic structures the basic loa factor, 1.7, is multiplie by a hyraulic factor, H f, where H =1.3, except for members in irect tension. For members in irect tension, H = f Other values may be use subject to consultation with an approval from CECW-ED. An exception to the above occurs when resistance to the effects of unusual or extreme loas such as win, earthquake or other forces of short uration an low probability of occurrence are inclue in the esign. For those cases, one of the following loaing combinations shoul be use: f 3-2

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28 CHAPTER 4 FLEXURAL AND AXIAL LOADS 4-1. Design Assumptions an General Requirements a. The assume maximum usable strain ε c at the extreme concrete compression fiber shoul be equal to in accorance with ACI 318. b. Balance conitions for hyraulic structures exist at a cross section when the tension reinforcement ρ b reaches the strain corresponing to its specifie yiel strength f y just as the concrete in compression reaches its esign strain ε c. c. Concrete stress of 0.85f c shoul be assume uniformly istribute over an equivalent compression zone boune by eges of the cross section an a straight line locate parallel to the neutral axis at a istance a = β 1 c from the fiber of maximum compressive strain.. Factor β 1 shoul be taken as specifie in ACI 318. e. The eccentricity ratio e / shoul be efine as e M u /P u h/2 (4-11)* where e = eccentricity of axial loa measure from the centroi of the tension reinforcement 4-2. Flexural an Compressive Capacity - Tension Reinforcement Only a. The esign axial loa strength φp n at the centroi of compression members shoul not be taken greater than the following: φp n(max) 0.8φ 0.85f c(a g ρb) f y ρb (4-12) b. The strength of a cross section is controlle by compression if the loa has an eccentricity ratio e / no greater than that given by Equation 4-3 an by tension if e / excees this value. * P u is consiere positive for compression an negative for tension. 4-1

29 e b 2k b k 2 b ρf 2k y b 0.425f c (4-13) where k b β 1 E s ε c E s ε c f y (4-14) c. Sections controlle by tension shoul be esigne so φp n φ 0.85f ck u ρf y b (4-15) an φm n φ 0.85f ck u ρf y e 1 h 2 b 2 (4-16) where k u shoul be etermine from the following equation: k u e 1 2 ρf y 0.425f c e e 1 (4-17). Sections controlle by compression shoul be esigne so φp n φ 0.85f ck u ρf s b (4-18) an φm n φ 0.85f ck u ρf s e 1 h 2 b 2 (4-19) 4-2

30 where f s E s ε c β 1 k u k u f y (4-20) an k u shoul be etermine from the following equation by irect or iterative metho: k 3 u 2 e 1 k 2 u E s ε c ρe 0.425f c k u β 1 E s ε c ρe 0.425f c 0 (4-21) e. The balance loa an moment can be compute using either Equations 4-5 an 4-6 or Equations 4-8 an 4-9 with k u = k b an e b e =. The values of e b/ an k b are given by Equations 4-3 an 4-4, respectively Flexural an Compressive Capacity - Tension an Compression Reinforcement a. The esign axial loa strength φp n of compression members shoul not be taken greater than the following: φp n(max) 0.8φ 0.85f c A g ρ ρ b (4-22) f y ρ ρ b b. The strength of a cross section is controlle by compression if the loa has an eccentricity ratio e / no greater than that given by Equation 4-13 an by tension if e / excees this value. e b 2k b k 2 b 2k b ρf y 0.425f c ρ f s f c ρ f s 0.425f c (4-23) The value k b is given in Equation 4-4 an f s is given in Equation 4-16 with k u = k b. c. Sections controlle by tension shoul be esigne so 4-3

31 φp n φ 0.85f ck u ρ f s ρf y b (4-24) an φm n φ 0.85f ck u ρ f s ρf y e 1 h 2 b 2 (4-25) where f s k u β 1 β 1 k u E s ε y f y (4-26) an k u shoul be etermine from the following equation by irect or iterative methos: k 3 u 2 e 1 β 1 k 2 u f y 0.425f c ρ e 1 ρe (4-27) 2 β 1 e f 1 k y β 1 u ρ 0.425f c e 1 ρe 0. Sections controlle by compression shoul be esigne so φp n φ 0.85f ck u ρ f s ρf s b (4-28) 4-4

32 an φm n φ 0.85f ck u ρ f s ρf s e 1 h 2 b 2 (4-29) where f s E s ε c β 1 k u k u f y (4-30) an f s E s ε c k u β 1 k u f y (4-31) an k u shoul be etermine from the following equation by irect or iterative methos: k 3 u 2 e 1 k 2 u E s ε c 0.425f c ρ ρ e ρ 1 k u β 1 E s ε c 0.425f c ρ e (4-32) 1 ρ e 0 Design for flexure utilizing compression reinforcement is iscourage. However, if compression reinforcement is use in members controlle by compression, lateral reinforcement shall be provie in accorance with the ACI Builing Coe. e. The balance loa an moment shoul be compute using Equations 4-14, 4-15, 4-16, an 4-17 with k u = k b an e =. The values of e b/ an k b are given by Equations 4-13 an 4-4, respectively Flexural an Tensile Capacity a. The esign axial strength φp n of tensile members shoul not be taken greater than the following: e b 4-5

33 φp n(max) 0.8φ ρ ρ f y b (4-33) b. Tensile reinforcement shoul be provie in both faces of the member if the loa has an eccentricity ratio e / in the following range: 1 h 2 e 0 The section shoul be esigne so φp n φ ρf y ρ f s b (4-24) an φm n φ ρf y ρ f s 1 h 2 e b 2 (4-25) with k u s f y k u 1 f f y (4-26) an k u shoul be etermine from the following equation: ρ k u ρ e 1 ρ 1 e ρ e e (4-27) c. Sections subjecte to a tensile loa with an eccentricity ratio e / < 0 shoul be esigne using Equations 4-5 an 4-6. The value of k u is k u e 1 e 1 2 ρf y 0.425f c e (4-28) 4-6

34 . Sections subject to a tensile loa with an eccentricity ratio e / < 0 shoul be esigne using Equations 4-14, 4-15, 4-16, an 4-17 if A s > 0 an c > Biaxial Bening an Axial Loa a. Provisions of paragraph 4-5 shall apply to reinforce concrete members subjecte to biaxial bening. b. For a given nominal axial loa P n = P u /φ, the following nonimensional equation shall be satisfie: (M nx /M ox ) K (M ny /M oy ) K 1.0 (4-29) where M nx, M ny = nominal biaxial bening moments with respect to the x an y axes, respectively M ox, M oy = uniaxial nominal bening strength at P n about the x an y axes, respectively K = 1.5 for rectangular members = 1.75 for square or circular members = 1.0 for any member subjecte to axial tension c. M ox an M oy shall be etermine in accorance with paragraphs 4-1 through

35 30 Apr 92 CHAPTER 5 SHEAR 5-1. Shear Strength The shear strength V c provie by concrete shall be compute in accorance with ACI 318 except in the cases escribe in paragraphs 5-2 an Shear Strength for Special Straight Members The provisions of this paragraph shall apply only to straight members of box culvert sections or similar structures that satisfy the requirements of 5-2.a an 5-2.b. The stiffening effects of wie supports an haunches shall be inclue in etermining moments, shears, an member properties. The ultimate shear strength of the member is consiere to be the loa capacity that causes formation of the first incline crack. a. Members that are subjecte to uniformly (or approximately uniformly) istribute loas that result in internal shear, flexure, an axial compression (but not axial tension). b. Members having all of the following properties an construction etails. (1) Rectangular cross-sectional shapes. (2) n/ between 1.25 an 9, where n is the clear span. (3) f c not more than 6,000 psi. (4) Rigi, continuous joints or corner connections. (5) Straight, full-length reinforcement. Flexural reinforcement shall not be terminate even though it is no longer a theoretical requirement. (6) Extension of the exterior face reinforcement aroun corners such that a vertical lap splice occurs in a region of compression stress. (7) Extension of the interior face reinforcement into an through the supports. c. The shear strength provie the concrete shall be compute as n V c 11.5 f c N 1 u /A g b 5 f c (5-1) 5-1

36 30 Apr 92 at a istance of 0.15 n from the face of the support.. The shear strength provie by the concrete shall not be taken greater than V c 2 n 12 f c b (5-2) an shall not excee 10 f c b Shear Strength for Curve Members At points of maximum shear, for uniformly loae curve cast-in-place members with R/ > 2.25 where R is the raius curvature to the centerline of the member: V c 4 f c N 1 u /A g b 4 f c (5-3) The shear strength shall not excee 10 f c b Empirical Approach Shear strength base on the results of etaile laboratory or fiel tests conucte in consultation with an approve by CECW-ED shall be consiere a vali extension of the provisions in paragraphs 5-2 an

37 APPENDIX A NOTATION a e e b Depth of stress block at limiting value of balance conition (Appenix D) Minimum effective epth that a singly reinforce member may have an maintain steel ratio requirements (Appenix D) Eccentricity of axial loa measure from the centroi of the tension reinforcement Eccentricity of nominal axial loa strength, at balance strain conitions, measure from the centroi of the tension reinforcement H f Hyraulic structural factor equal to 1.3 k b k u Ratio of stress block epth (a) to the effective epth () at balance strain conitions Ratio of stress block epth (a) to the effective epth () K Exponent, equal to 1.0 for any member subject to axial tension, 1.5 for rectangular members an 1.75 for square or circular members, use in nonimensional biaxial bening expression n M DS M nx, M ny Clear span between supports Bening moment capacity at limiting value of balance conition (Appenix D) Nominal biaxial bening moments with respect to the x an y axes, respectively M ox, M oy Uniaxial nominal bening strength at P n about the x an y axes, respectively R Raius of curvature to centerline of curve member A-1

38 APPENDIX B DERIVATION OF EQUATIONS FOR FLEXURAL AND AXIAL LOADS B-1. General Derivations of the esign equations given in paragraphs 4-2 through 4-4 are presente below. The esign equations provie a general proceure that may be use to esign members for combine flexural an axial loa. B-2. Axial Compression an Flexure a. Balance Conition From Figure B-1, the balance conition, Equations 4-3 an 4-4 can be erive as follows: From equilibrium, P u φ 0.85 f c bk u A s f s (B-1) let j u a 2 k u 2 (B-2) from moment equilibrium, P u e φ 0.85 f c bk u j u (B-3) Rewrite Equation B-3 as: P u e φ 0.85 f c bk u k u 2 k 2 u 0.85 f c b 2 k u f c 2k u k 2 u b 2 (B-4) B-1

39 From the strain iagram at balance conition (Figure B-1): c b ε c ε c ε y k b β 1 ε c ε c ε y (B-5) since ε y f y = E s k b β 1 E s ε c E s ε c f y (B-6, Eq. 4-4) since e b P b e P b (B-7) e b is obtaine by substituting Equations B-4 an B-1 into Equation B-7 with k u = k b, f s = f y an P u = P b. e b 0.425f c 2k b k 2 b b f c k b b f y ρb (B-8) Therefore e b 2k b k 2 b f 2k y ρ b 0.425f c (B-9, Eq. 4-3) B-2

40 b. Sections Controlle by Tension (Figure B-1). φp n is obtaine from Equation B 1 with f s f y as: φp n φ 0.85 f c bk u A s f y (B-10, Eq. 4-5) φp n φ 0.85 f c k u ρf y b The esign moment φm n is expresse as: φm n φp n e φm n φp n e 1 h 2 (B-11) Therefore, φm n φ 0.85 f c k u f y ρ e 1 h b 2 2 (B-12, Eq. 4-6) Substituting Equation B-1 with f s = f y into Equation B-4 gives 0.85 f c k u b f y ρb e 0.425f c 2k u k 2 u b 2 (B-13) which reuces to k 2 u 2 e 1 k u f y ρe 0.425f c 0 (B-14) Solving by the quaratic equation: k u e 1 2 ρf y 0.425f c e e 1 (B-15, Eq. 4-7) B-3

41 c. Sections Controlle by Compression (Figure B-1) φp n is obtaine from Equation B-1 φp n φ 0.85 f c k u ρf s b (B-16, Eq. 4-8) an φm n is obtaine by multiplying Equation B-16 by e. φm n φ 0.85 f c k u ρf s e 1 h 2 b 2 (B-17) The steel stress, f s, is expresse as f s = E s ε s. From Figure B-1. c ε c ε c ε s or k u β 1 ε c ε c ε s Therefore, f s E s ε c β 1 k u k u (B-18, Eq. 4-10) Substituting Equations B-1 an B-18 into B-4 gives 0.85 f c k u be E s ε c β 1 k u ρbe k u f c 2k u k 2 u b 2 (B-19) B-4

42 which can be arrange as k 3 u 2 e 1 k 2 u E s ε c ρe 0.425f c k u β 1 E s ε c ρe 0.425f c 0 (B-20, Eq. 4-11) B-3. Flexural an Compressive Capacity-Tension an Compression Reinforcement (Figure B-2) a. Balance Conition Using Figure B-2, the balance conition, Equation 4-13 can be erive as follows: From equilibrium, P u φ 0.85 f c k u b f s ρ b f s ρb (B-21) In a manner similar to the erivation of Equation B-4, moment equilibrium results in P u e φ f c 2k u k 2 u b 2 f s ρ b( ) (B-22) As in Equation B-6, k b β 1 E s ε c E s ε c f y (B-23) since e b P b e P s (B-24) an using Equations B-21 an B-22: e b f c (2k b k 2 b)b 2 f s ρ b( ) 0.85 f c k b b f s ρ b f s ρb (B-25) B-5

43 which can be rewritten as e b 2k b k 2 b 2k b f s ρ 0.425f c f sρ ( ) 0.425f c f y ρ 0.425f c or e b 2k b 2k b k 2 b f y ρ 0.425f c f sρ f c f sρ 0.425f c (B-26, Eq. 4-13) b. Sections Controlle by Tension (Figure B-2) φp n is obtaine as Equation B-21 with f s = f y. φp n φ 0.85 f c k u ρ f s ρf y b (B-27, Eq. 4-14) Using Equations B-11 an B-27, φm n φ 0.85 f c k u ρ f s ρf y e 1 h 2 b 2 (B-28, Eq. 4-15) From Figure B-2 ε s c ε y c ; f s E s ε s ; c k u β 1 B-6

44 Therefore, f s k u E s β 1 ε y k u β 1 or f s k u β 1 β 1 k u E s ε y (B-29, Eq. 4-16) Substituting Equation B-21 with f s = f y into Equation B-22 gives, 0.85 f c k u b f sρ b f y ρb e f c 2k u k 2 u b 2 f sρ b( ) (B-30) Using Equation B-29, Equation B-30 can be written as: k 3 u 2 e 1 β 1 k 2 u f y 0.425f c ρ e 1 ρe (B-31, Eq. 4-17) 2 β 1 e f 1 k y β 1 u 0.425f c ρ e 1 ρe 0 c. Sections Controlle by Compression (Figures B-2) φp n is obtaine from equilibrium φp n φ 0.85 f c k u ρ f s ρf s b (B-32, Eq. 4-18) B-7

45 Using Equations B-11 an B-32, φm n φ 0.85 f c k u ρ f s ρf s e 1 h 2 b 2 (B-33, Eq. 4-19) From Figure B-2 ε s c ε c c ; f s E s ε s ; c k u β 1 which can be written as f s E s ε c β 1 k u k u (B-34, Eq. 4-20) Also, ε s c ε c c which can be rewritten as f s E s ε c k u β 1 k u (B-35, Eq. 4-21) From Equations B-21 an B f c k u b f sρ b f s ρb e f c 2k u k 2 u b 2 f s ρ b( ) (B-36) B-8

46 Substituting Equations B-34 an B-35 with k b = k u into Equation B-36 gives k 3 u 2 e β 1 E s ε c 0.425f c 1 k 2 u ρ E s ε c 0.425f c e (ρ ρ ) e 1 ρ e ρ 1 0 k u (B-37, Eq. 4-22) B-4. Flexural an Tensile Capacity a. Pure Tension (Figure B-3) From equilibrium (ouble reinforcement) φp n φ A s A s f y (B-38) For esign, the axial loa strength of tension members is limite to 80 percent of the esign axial loa strength at zero eccentricity. Therefore, φp n(max) 0.8φ (ρ ρ )f y b (B-39, Eq. 4-23) P u φ b. For the case where 1 - h e 0, the applie tensile resultant 2 lies between the two layers of steel. From equilibrium φp n φ A s f y A sf s or φp n φ ρf y ρ f s b (B-40, Eq. 4-24) an φm n φ P n 1 h e 2 B-9

47 or φm n φ ρf y ρ f s 1 h 2 e b 2 (B-41, Eq. 4-25) From Figure B-3, ε s a ε y a which can be rewritten as k u s f y k u 1 f (B-42, Eq. 4-26) From Figure B-3 equilibrium requires: A s f s e A sf s ( e ) (B-43) Substituting Equation B-42 an f s = f y into Equation B-43 results in k u ρ ρ e 1 ρ 1 e ρ e e (B-44, Eq. 4-27) c. The case where (e /) < 0 is similar to the combine flexural an compression case. Therefore, k u is erive in a manner similar to the erivation of Equation B-15 an is given as k u e 1 e 1 2 ρf y 0.425f c e (B-45, Eq. 4-28) B-10

48 Figure B-1. Axial compression an flexure, single reinforcement B-11

49 Figure B-2. Axial compression an flexure, ouble reinforcement B-12

50 Figure B-3. Axial tension an flexure, ouble reinforcement B-13

51 APPENDIX C INVESTIGATION EXAMPLES C-1. General For the esigner s convenience an reference, the following examples are provie to illustrate how to etermine the flexural capacity of existing concrete sections in accorance with this Engineer Manual an ACI 318. C-2. Analysis of a Singly Reinforce Beam Given: f c = 3 ksi β 1 = 0.85 f y = 60 ksi E s = 29,000 ksi A s = 1.58 in. 2 Solution: 1. Check steel ratio ρ act A s b (20.5) C-1

52 f c ρ b 0.85β 87,000 1 f y 87,000 f y 0.85 (0.85) ,000 87,000 60,000 in accorance with Paragraph 3-5 check: 0.25ρ b ρ b ρ act ρ b < ρ act < 0.375ρ b ρ act is greater than the recommene limit, but less than the maximum permitte upper limit not requiring special stuy or investigation. Therefore, no special consieration for serviceability, constructibility, an economy is require. This reinforce section is satisfactory. 2. Assume the steel yiels an compute the internal forces: T A s f y 1.58 (60) 94.8 kips C 0.85 f c ba C 0.85 (3)(12)a 30.6a 3. From equilibrium set T = C an solve for a: a a 3.10 in. Then, a β 1 c c in. 4. Check ε s to emonstrate steel yiels prior to crushing of the concrete: C-2

53 ε s 20.5 c c ε s ε y f y E s 60 29, ε s > ε y Ok, steel yiels 5. Compute the flexural capacity: φm n φ A s f y ( a/2) 0.90 (94.8) in. k ft k C-3. Analysis of an Existing Beam - Reinforcement in Both Faces C-3

54 Given: f c = 3,000 psi ε c = f y = 60,000 psi β 1 = 0.85 A s = 8.00 in. 2 E s = 29,000,000 psi A s = 4.00 in. 2 Solution: 1. First analyze consiering steel in tension face only ρ A s b 8 (60)(12) ρ bal 0.85 β 1f c f y 87,000 87,000 f y ρ ρ bal 0.51ρ b Note: ρ excees maximum permitte upper limit not requiring special stuy or investigation ρ b. See Chapter 3. T A s f y T 8(60) 480 kips then C c 0.85f c ba 30.6a T C c a 15.7 in. an c in. By similar triangles, emonstrate that steel yiels ε c ε s(2) 54 c ε s(2) > ε y ok; both layers of steel yiel. Moment capacity = 480 kips ( - a/2) = 480 kips (52.15 in.) M = 25,032 in.-k C-4

55 2. Next analyze consiering steel in compression face ρ (60) ρ ρ β 1f c f y 87, ,000 f y ρ - ρ compression steel oes not yiel, must o general analysis using σ : ε compatability Locate neutral axis T 480 kips C c 0.85f c ba 30.6a C s A s (f s 0.85f c) 4(f s 2.55) By similar triangles ε s c c Substitute c Then ε s a a Since f s E ε s f s a a ksi C-5

56 Then C s a T C c C s 480 kips kips Substitute for C c an C s an solve for a 30.6a a a 58 0 Then a 10.3 in. an c 12.1 in a 480 Check ε s > ε y By similar triangles ε s 12.1 ε s > C c C s 30.6a 315 kips 4(41.37) 165 kips C c C s 480 kips T (165)(6) Resultant of C c an C 2 s 480 Internal Moment Arm in. M 480(54.6) 26,208 in. k 5.4 in. Comparison Tension Steel Compression Only Steel a 15.7 in in. c in in. Arm in in. M 25,032 in.-k 26,208 in.-k 4.7 percent increase C-6

57 APPENDIX D DESIGN EXAMPLES D-1. Design Proceure For convenience, a summary of the steps use in the esign of the examples in this appenix is provie below. This proceure may be use to esign flexural members subjecte to pure flexure or flexure combine with axial loa. The axial loa may be tension or compression. Step 1 - Compute the require nominal strength M n, P n where M u an P u are etermine in accorance with paragraph 4-1. M n M u φ P n P u φ Note: Step 2 below provies a convenient an quick check to ensure that members are size properly to meet steel ratio limits. The expressions in Step 2a are aequate for flexure an small axial loa. For members with significant axial loas the somewhat more lengthy proceures of Step 2b shoul be use. Step 2a - Compute from Table D-1. The term is the minimum effective epth a member may have an meet the limiting requirements on steel ratio. If the member is of aequate epth to meet steel ratio requirements an A s is etermine using Step 3. Step 2b - When significant axial loa is present, the expressions for become cumbersome an it becomes easier to check the member size by etermining M DS. M DS is the maximum bening moment a member may carry an remain within the specifie steel ratio limits. M DS 0.85f c a b a /2 h/2 P n (D-1) where a K (D-2) an K is foun from Table D-1. Step 3 - Singly Reinforce - When (or M n M DS ) the following equations are use to compute A s. D-1

58 K u 1 1 M n P n h/ f cb 2 (D-3) A s 0.85f c K u b P n f y (D-4) Table D-1 Minimum Effective Depth f c (psi) f y (psi) ρ ρ b K (in.) M n b M n b M n b * See Section 3-5. Maximum Tension Reinforcement ** M n units are inch-kips. D-2

59 where ρ β ρ 1 ε c K b f ε y c E s M n 0.85f ck b 1 k 2 D-2. Singly Reinforce Example The following example emonstrates the use of the esign proceure outline in paragraph D-1 for a Singly Reinforce Beam with the recommene steel ratio of 0.25 ρ b. The require area of steel is compute to carry the moment at the base of a retaining wall stem. Given: M = k-ft (where M = moment from unfactore ea an live loas) f c = 3.0 ksi f y = 60 ksi = 20 in. First compute the require strength, M u. M u 1.7 H f D L M u (1.7)(1.3)(41.65) k ft Step 1. M n = M u /φ = /0.90 = k-ft D-3

60 Step M n b in. > therefore member size is aequate (Table D-1) Step 3. K u 1 1 K u 1 1 M n P n h/ f cb 2 ( )(12) (0.425)(3.0)(12)(20) (D-3) 0.85f c K A u b (0.85)(3.0)( )(12)(20) s f y 60 A s 1.08 sq in. (D-4) D-3. Combine Flexure Plus Axial Loa Example The following example emonstrates the use of the esign proceure outline in paragraph D-1 for a beam subjecte to flexure plus small axial compressive loa. The amount of tensile steel require to carry the moment an axial loa at the base of a retaining wall stem is foun. Given: M = k-ft P = 5 kips (weight of stem) where M an P are the moment an axial loa from an unfactore analysis. f c f y = 3.0 ksi = 60 ksi = 20 in. h = 24 in. D-4

61 First compute the require strength, M u, P u M u 1.7 H f (D L) M u (1.7)(1.3)(41.65) k ft P u 1.7 H f (D L) P u (1.7)(1.3)(5.0) kips Since axial loa is present a value must be foun for φ. For small axial loa φ = [(0.20 P u )/(0.10f c Ag )] φ = 0.88 Step 1. M n = M u /φ = /0.88 = k-ft P n = P u /φ = 11.05/0.88 = kips Step 2. a = K (D-2) a = ( )(20) = M DS = 0.85f c a b( - a /2.0) - ( - h/2.0)p n (D-1) M DS = (0.85)(3.0)(2.515)(12)( ) - (20-12)(12.56) M DS = k-in. or k-ft M DS > M n therefore member size is aequate D-5

62 Step 3. K u 1 1 M n P n ( h/2) 0.425f cb 2 K u 1 1 (12) (20 12) (0.425)(3.0)(12)(20) 2 K u f A ck u b s f y P n A s (0.85)(3.0)( )(12)(20) (D-4) A s 0.99 sq in. D-4. Derivation of Design Equations The following paragraphs provie erivations of the esign equations presente in paragraph D-1. (1) Derivation of Design Equations for Singly Reinforce Members. The figure below shows the conitions of stress on a singly reinforce member subjecte to a moment M n an loa P n. Equations for esign may be evelope by satisfying conitions of equilibrium on the section. By requiring the ΣM about the tensile steel to equal zero M n 0.85f c ab( a/2) P n ( h/2) (D-5) D-6

63 By requiring the ΣH to equal zero A s f y 0.85f c ab P n (D-6) Expaning Equation D-5 yiels M n 0.85f c ab 0.425f ca 2 b P n ( h/2) Let a = K u then M n 0.85f ck u b f ck 2 u 2 b P n ( h/2) The above equation may be solve for K u equation using the solution for a quaratic K u 1 1 M n P n ( h/2) 0.425f cb 2 (D-3) Substituting K u for a in Equation D-6 then yiels A s 0.85f c K u b P n f y (2) Derivation of Design Equations for Doubly Reinforce Members. The figure below shows the conitions of stress an strain on a oubly reinforce member subjecte to a moment M n an loa P n. Equations for esign are evelope in a manner ientical to that shown previously for singly reinforce beams. D-7

64 Requiring ΣH to equal zero yiels 0.85f A ck b P n A sf s s f y (D-7) By setting a = β 1 c an using the similar triangles from the strain iagram above, ε s an f s may be foun: f s a β 1 ε c E s a An expression for the moment carrie by the concrete (M DS ) may be foun by summing moments about the tensile steel of the concrete contribution. M DS 0.85f ca b a /2 h/2 P n (D-1) Finally, an expression for A s may be foun by requiring the compression steel to carry any moment above that which the concrete can carry (M n - M DS ). A s M n M DS f s (D-8) (3) Derivation of Expression of. The expression for is foun by substituting a = k in the equation shown above for M DS an solving the resulting quaratic expression for. M DS 0.85f ck b 1 K /2 (D-9) D-5. Shear Strength Example for Special Straight Members Paragraph 5.2 escribes the conitions for which a special shear strength criterion shall apply for straight members. The following example emonstrates the application of Equation 5-1. Figure D-1 shows a rectangular conuit with factore loas, 1.7 H f (ea loa + live Loa). The following parameters are given or compute for the roof slab of the conuit. D-8

65 f c n = 4,000 psi = 10.0 ft = 120 in. = 2.0 ft = 24 in. b = 1.0 ft (unit with) = 12 in. N u A g = 6.33(5) = 31.7 kips = 2.33 sq ft = 336 sq in. V c in. 24 in. 4, ,700 lb 336 sq in. 5 4,000 (12 in.)(24 in.) (D-10, Eq. 5-1) V c 134,906 lb kips Check limit V c 10 f c b 10 4,000 (12 in.)(24 in.) 182,147 lb Compare shear strength with applie shear. φv c 0.85(134.9 kips) kips V u at 0.15( n ) from face of the support is V u w n n 15.0 kips/ft 10 ft 2 (0.15)(10 ft) 52.5 kips < φv c ; shear strength aequate D-6. Shear Strength Example for Curve Members Paragraph 5-3 escribes the conitions for which Equation 5-3 shall apply. The following example applies Equation 5-3 to the circular conuit presente in Figure D-2. Factore loas are shown, an the following values are given or compute: D-9

66 f c = 4,000 psi b = 12 in. = 43.5 in. A g N u V u = 576 sq in. = kips = 81.3 kips at a section 45 egrees from the crown V c 4 4, ,500 lb 576 sq in. 4 4,000 (12 in.)(43.5 in.) V c 192,058 lb kips Check limit V c 10 f c b 10 4,000 (12 in.)(43.5 in.) 330,142 lb Compare shear strength with applie shear φv c = 0.85(192.1 kips) = kips V u < φv c ; shear strength aequate D-10

67 Figure D-1. Rectangular conuit Figure D-2. Circular conuit D-11

68 APPENDIX E INTERACTION DIAGRAM E-1. Introuction A complete iscussion on the construction of interaction iagrams is beyon the scope of this manual; however, in orer to emonstrate how the equations presente in Chapter 4 may be use to construct a iagram a few basic points will be compute. Note that the effects of φ, the strength reuction factor, have not been consiere. Using the example cross section shown below compute the points efine by 1, 2, 3 notations shown in Figure E-1. Given: f c = 3.0 ksi f y = 60 ksi A s = 2.0 sq in. = 22 in. h = 24 in. b = 12 in. Figure E-1. Interaction iagram E-1

69 E-2. Determination of Point 1, Pure Flexure φ M n φ 0.85 f c ab( a/2) a A s f y 0.85 f c b (2.0)(60.0) (0.85)(3.0)(12) M n (0.85)(3.0)(3.922)(12)( ) in. (D-5) M n M n k in k ft E-3. Determination of Point 2, Maximum Axial Capacity E-2

70 φp n(max) φ 0.80 P o φ P n(max) φ f c A g ρb f y ρb P n(max) 0.80 (0.85)(3.0)( ) (60.0)(2.0) (4-2) P n(max) 0.80(849.3) kips E-4. Determination of Point 3, Balance Point β (1) Fin k 1 E s ε c b E s ε c f y k b (0.85)(29,000)(0.003) (29,000)(0.003) (4-4) (2) Fin e b e b 2k u k 2 u pf 2k y u 0.425f c (2)(0.5031) (0.5031) 2 (2)(0.5031) ( )(60) (0.425)(3.0) (4-3) E-3

71 (3) Fin φ P b φ 0.85 f c k b ρf y b P b [(0.85)(3.0)(0.5031) ( )(60.0)](12)(22.0) P b kips (4) Fin φm b φ 0.85f c k b ρf y e 1 h 2 b 2 M b [(0.85)(3.0)(0.5031) ( )(60)] [ (1 24.0/44.0)](12)(22.0) 2 M b k in. (4-6) M b k ft E-4

72 Figure E-2. Interaction iagram solution E-5

73 APPENDIX F AXIAL LOAD WITH BIAXIAL BENDING - EXAMPLE F-1. In accorance with paragraph 4-5, esign an 18- by 18-inch reinforce concrete column for the following conitions: f c f y = 3,000 psi = 60,000 psi P u = 100 kips, P n = P u /0.7 = kips M ux = 94 ft-kips, M nx = M ux /0.7 = ft-kips M uy = 30 ft-kips, M ny = M uy /0.7 = 42.8 ft-kips Let concrete cover plus one-half a bar iameter equal 2.5 in. F-2. Using uniaxial esign proceures (Appenix E), select reinforcement for P n an bening about the x-axis since M nx > M ny. The resulting cross-section is given below. F-3. Figures F-1 an F-2 present the nominal strength interaction iagrams about x an y axes. It is seen from Figure F-2 that the member is aequate for uniaxial bening about the y-axis with P n = kips an M ny = 42.8 ftkips. From Figures F-1 an F-2 at P n = kips: M ox M oy = ft-kips = ft-kips F-1

74 For a square column, must satisfy: (M nx /M ox ) (M ny /M oy ) (134.3/146.1) (42.8/145.9) 1.75 = 0.98 < 1.0 If a value greater than 1.0 is obtaine, increase reinforcement an/or increase member imensions. Figure F-1. Nominal strength about the X-axis F-2

75 Figure F-2. Nominal strength about the Y-axis F-3