Reinforced Concrete Design
Reinforced Concrete Design CIVL 4135
University of Memphis
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MII39 mm mm m T mxmnmi m m mun um I alum In culim I hum m nmmum do 393 A F llll In T CNL 4135 ClassNotes sha h P l hkt 8252008 WE cummgr M 2mmme mm mquot xmnm ma xawm mum Wsnlnln 39 Erwin Sm mm m vm cm 1m INquot CIVL 4135 I 0 ii Chapter 1 Introduction 11 Reading Assignment Chapter 1 Sections 11 through 18 of text 12 Introduction In the design and analysis of reinforced concrete members you are presented with a problem unfamiliar to most of you The mechanics of members consisting of two materials To compound this problem one of the materials concrete behaves differently in tension than in compression and may be considered to be either elastic or inelastic if it is not neglected entirely Although we will encounter some peculiar aspects of behavior of concrete members we will usually be close to a solution for most problems if we can apply the following three basic ideas Geometry of deformation of sections will be consistent under given types of loading ie moment will always cause strain to vary linearly with distance from neutral axis etc Mechanics of materials will allow us to relate stresses to strains Sections will be in equilibrium external moments will be resisted by internal moment external axial load will be equal to the sum of internal axial forces Many new engineers overly impressed speed and apparent accuracy of modern structural analysis computational procedures think less about equilibrium and details We will use some or all of these ideas in solving most of the analysis problems we will have in this course Design of members and structures of reinforced concrete is a problem distinct from but closely related to analysis Strictly speaking it is almost impossible to exactly analyze a concrete structure and to design exactly is no less difficult Fortunately we can make a few fundamental assumptions which make the design of reinforced concrete quite simple if not easy A problem unique to the design of reinforced concrete structures is the need to detail each member throughout Steel structures in general require only the detailed design of connections For concrete structures we must determine not only the area of longitudinal and lateral reinforcement required in each member but also the way to best arrange and connect the reinforcement to insure acceptable structural performance This procedure can be made reasonably simple if not easy Purpose of this course is to establish a firm understanding of behavior of reinforced concrete structures then to develop method used in current practice and to achieve familiarity with codes and specifications governing practical design In this course we will learn to understand the basic performance of concrete and steel as structural materials and the behavior of reinforced concrete members and structures If we understand the basic concepts behind code provisions for design we will be able to 0 Approach the design in a more knowledgeable fashion not like following a black box and 0 Understand and adapt the changes in code provisions better and faster The overall goal is to be able to design reinforced concrete structures that are 0 Safe 0 Economical 0 Efficient Reinforced concrete is one of the principal building materials used in engineered structures because 0 Low cost 0 Weathering and fire resistance 0 Good compressive strength 0 Formability all these criteria make concrete an attractive material for wide range of structural applications such as buildings dams reservoirs tanks etc 13 Design Codes and Specifications Buildings must be designed and constructed according to the provisions of a building code which is a legal document containing requirements related to such things as structural safety fire safety plumbing ventilation and accessibility to the physically disabled A building code has the force of law and is administered by a governmental entity such as a city a county or for some large metropolitan areas a consolidated government Building codes do not give design procedures but specify the design requirements and constraints that must be satisfied Of particular importance to CIVL 4135 Chapter1 Introduction I 2 the structural engineer is the prescription of minimum live loads for buildings While the engineer is encouraged to investigate the actual loading conditions and attempt to determine realistic values the structure must be able to support these speci ed minimum loads Although some large cities write their own building codes many municipalities will adopt a model building code and modify it to suit their particular needs Model codes are written by various nonprofit organizations in a form that is easily adopted by a governmental unit Among the more popular are the BOCA National Building Code the Uniform Building Code the Standard Building Code and Internation Building Code IBC 2003 A related document similar in form to a building code is ASCE 705 Minimum Design Loads for Buildings and other Structures This standard is intended to provide load requirements in a format suitable for adoption by a building code United States does not have a national code governing structural concrete 0 ACI Code American Concrete Institute 0 ACI commentary provides background material rational for code provisions 0 Highway Bridges are designed according to AASHTO which stands for American Association of State Highway and Transportation Officials o AREA stands for American Railway Engineers Association This is manual of railway engineering 14 Loads Loads that act on structures can be divided into three general categories 141 Dead Loads Dead loads are those that are constant in magnitude and fixed in location throughout the lifetime of the structure such as oor fill finish oor and plastered ceiling for buildings and wearing surface sidewalks and curbing for bridges 142 Live Loads Live loads are those that are either fully or partially in place or not present at all may also change in location the minimum live loads for which the oors and roof of a building should be designed are usually specified in building code that governs at the site of construction see Table l Minimum Design Loads for Buildings and Other Structure CIVL 4135 Chapter1 Introduction I 3 143 Environmental Loads Environmental Loads consist of wind earthquake and snow loads such as wind earthquake and snow loads 15 Serviceability Serviceability requires that o De ections be adequately small 0 Cracks if any be kept to a tolerable limits 0 Vibrations be minimized CIVL 4135 Chapter1 Introduction I 4 TABLE M l Un orm psi Cane las Occupancy er use RNm1 kn Aparrmenrs see residenual Access ricer syslcms 39 0 Ice use so 24 2000 89 Computer use 100 479 2000 89 Armories and drill reorns 150 713 Assembly areas and heatcrs Fixed seals fastened to oor 60 287 39 mo 479 Movable 58an 00 479 Platforms assembly 100 479 Siage uors 150 718 Balconies exrerinr 100 479 one and wefamily residences enly and um excccding 100 n 93 ml 50 287 Bowling alleys peelrnerns and similar reerealionnl areas 75 359 Catwalk for mainienanca access 40 192 300 133 Cu 39ders First liner loo 479 mm nears same as occupancy served excepi as indicated Dance halls and ballmoms 100 479 Decks palio and roof Same as area served or or Lhc lypc er neeupnney accommodamd Dining rooms and resrauranrs 100 479 Dwellings sec rcsidcmial Elevator machine rcem graiing on arms of 4 inZ 2580 mm2 l 300 l 33 Finish ighl liner plate mnskru ion on meal of l in 645 mm I 2 00 089 Fire escapes mo 479 n singlerfamlly dwellings only 40 L92 Fixed ladders Sec Semen 44 Garages passengcr vehicles only 40 192 l Noll l ane 2 callnnued CIVL 4135 Chapter 1 Introduction on TABLE 41 conllnuad I mm nu UnWorm as none lo oeeuoaney or Use kNrnz kN Grandsranrls see stadium and arena bleachers Gymnasiums rnain oors and balconies 100 479 More 4 Handrails guardrails and grab bars Scc Section 44 Hospilals Operaring rooms laboralorirs 60 287 1000 445 Private rooms 40 192 I000 445 Wards 40 192 mm 445 Corridors above rs oor 80 383 10000445 Horels see msldenllal Libraries Reading rooms 287 1000 445 Slack rooms 150 718 Note 3 loco 445 Conidors above m oor 80 383 1000 445 Manufacturing Lighl l25 600 2000 890 Heavy 250 l 197 3000 l340 Marquees and canopies 75 359 or ee buildings File lmd compuler moms shall he designed for heavier loads based on anljcipaled occupancy Lobbies and rst oor corridors 100 479 2000 890 Of ces 50 240 2000 890 Corridors above lsl oor 80 383 2000 390 Penal instilulions Cell blocks 40 192 Corn dors 100 479 Residential Dwellings one and twofamily Unlnnabilablc mics wilhnut srorage 10048 Uninhabilzbie attics wllh storage 20 095 Habilable allies and sleeping areas 30 L44 All olher areas except slairs and balconies 40 192 Holels and mullifamily houses Privare moms and corridors serving rhern 40 192 Pu ie rooms and corridors sewing mm 100 479 Reviewing slands grandstands and bleachers loo 479 Note 4 Roofs Sec Sections 43 and 49 continued CIVL 4135 Chapter1 Introduction 6 TABLE 41 continued 1 LOADS Unllurm ps1 cone lhs occupancy or Use kmm2 kN Schools Classrooms 40 192 1000 445 Corridors above rst oor 80 383 1000 445 Firsr hoor corridors 100 479 1000 445 Seurrles slrylighr ribs and accessible ceilings 200 958 Sidewalks vehicular driveways and yards subject to 250 1197 13000 3560 trucking Note 5 Nole 6 Smdiums and arenas Bleachers 100 479 Now 4 Fixed Sears faslened to oor 60 287 ane 4 Stairs and cxhwa s 100 479 Nora 7 Onc and twofamin residences 0111 40 192 Storage areas above ceilings 20 096 Srorage warehouses shall he designed or heavier loads 11 required for anuclpared slorage Lighl 125 600 Heavy 250 1197 Stores Relail Firsr oor 100 479 1000 445 Upper oors 75 359 1000 445 Wholesale all ours 125 600 1000 4 5 Vehicle barriers See Sccuon 44 Walkways and elevaled plartorrns orher 1han exilways l on 287 l L Yards and rename pedeslrians 100479 015 1 live loads o1 Table 41 1 39 1 ad 1 r I nor more rhan nine passengers 3mm 1h 1335 km acung an an area of45 in by 45 in 114 mm py 114 nrni foorprim or a Jack 2 or mechanical parking srnrcrurcs wilhoul step or deck which are used for sroring passenger car only 225011 0 m per wheel 2 Garages aocummoddting rnrelrs and huses shall he designed in accordance wirh an approved rnerhod which conrains provisions or rnrclr and has loadings 3 The loading applies Io srack room iloors rhar suppon nonmphrle douhlelaced library huekslacks 5ubjccl ro rlrc following hnmarions a The nominal hooksrack unir heighr shall nor exceed 90 in 2290 mm r era 1 c l h l 39 ma quot1va wide 4 1h adduron 1n Lhe verrical live loads rhe desrgn shall include horizonral swaying forces applied 1v Each row or he sears as 11pm 24 log linear h a appquot u 39 10 lbs an a 39 39 39 lion 39 row of sears 39 simulmnaously 51 omer uniform loads in accordance win an approved merhpd which conrains provisions for mick loadings shall also bu onsidered where apprnp nw 5 The concenrrared wheel lnad shall be applied on an area 0145 in by 45 in 114 mm hy 114 mm foolpan oi a jack 7 39 load on siair 1 r nan 1 lbs 133 kN CIVL 4135 Chapter1 Introduction 7 16 Safety A structure must be safe against collapse strength of the structure must be adequate for all loads that might act on it If we could build buildings as designed and if the loads and their internal effects can be predicted accurately we do not have to worry about safety But there are uncertainties in 0 Actual loads 0 Forcesloads might be distributed in a manner different from what we assumed o The assumptions in analysis might not be exactly correct 0 Actual behavior might be different from that assumed 0 etc Finally we would like to have the structure safe against brittle failure gradual failure with ample warning permitting remedial measures is preferable to a sudden or brittle failure 17 Design Basis Two philosophies of design have long been prevalent The working stress method focusing on conditions at service load that is when the structure is being used was the principal method used from the early 1900s until the early 1960s Today with few exceptions the strength design method is used focusing on conditions at loads greater than service loads when failure may be immanent The strength design method is deemed conceptually more realistic to establish structural safety 18 Strength Design or Ultimate Strength Design 0 Since 1971 the ACI Code has been totally a strength code with strengt meaning ultimate Select concrete dimensions and reinforcements so that the member strength are adequate to resist forces resulting from certain hypothetical overload stages significantly above loads expected actually to occur in service The design concept is known as strength design Based on strength design the nominal strength of a member must be calculated on the basis of inelastic behavior of material In other words both reinforcing steel and concrete behave inelastically at ultimate strength condition The strength design method may be expressed by the following Strength provide 2 Strength required to carry factored loads ll CIVL 4135 Chapter1 Introduction I 8 where the strength provided such as moment strength is computed in accordance with rules and assumptions of behavior prescribed by a building code and the strength required is that obtained by performing a structural analysis using the factored loads The design procedure is roughly as follows Multiply the working loads by the load factor to obtain the failure loads Determine the crosssectional properties needed to resist failure under these loads A member with these properties is said to have sufficient strength and would be at the verge of failure when subjected to the factored loads Proportion your members that have these properties 19 Allowable Stress Design As an alternate to the strength design method members may be proportioned so that stresses in the steel and concrete resulting from normal service loads are within specified limits an allowable or permissible stress or load These limits known as allowable stresses are only fractions of the failure stresses of the material Allowable stress design is also referred to as working stress design For example in design of steel structures this allowable stress will be in the elastic range of the material and will be less than the yield stress Fy A typical value might be 060Fy The allowable stress is obtained by dividing either the yield stress Fy or the tensile strength F u by a factor of safety This approach to design is also called elastic design or working stress design Working stresses are those due to the working loads which are the applied loads A properly designed member will be stressed to no more than the allowable stress when subjected to working loads The working stress method may be expressed by the following f S allowable stresses faugwaw 12 where f An elastically computed stress such as by using the exure formula f McI for beam CIVL 4135 Chapter 1 Introduction I 9 fallow A limiting stress prescribed by a building code as a percentage of the compressive strength for concrete or of the yield stress fy for the steel reinforcing bars 110 Safety Provisions of the ACI Code Load factors are applied to the loads and a member is selected that will have enough strength to resist the factored loads In addition the theoretical strength of the member is reduced by the application of a resistance factor The criterion that must be satisfied in the selection of a member is F actored Strength 2 F actored Load 13 In this expression the factored load is actually the sum of all working loads to be resisted by the member each multiplied by its own load factor For example dead loads will have load factors that are different from those for live loads The factored strength is the theoretical strength multiplied by a strength reduction factor Equation 13 can therefore be written as Nominal Strength X Strength Reduction Factor 2 Load X Load Factors 14 Since the factored load is a failure load greater than the actual working loads the load factors are usually greater than unity On the other hand the factored strength is a reduced usable strength and the resistance factor is usually less than unity The factored loads are the loads that bring the structure or member to its limit 111 Required Strength Section 92 of the ACI Code The required strength U is expressed in terms of factored loads or related internal moments and forces Factored loads are the loads specified in the general building code multiplied by appropriate factors The factor assigned is in uenced by the degree of accuracy to which the load effect can be determined and the variation which might be expected in the load during the lifetime of the structure Dead loads are assigned a lower load factored than live load because they can be determined more accurately Load factors also account for variability in the structural analysis used to compute moments and shearsThe code gives load factors for specific combinations of loads In assigning factors to combinations of loading some consideration is given to the probability of simultaneous occurrence While most of the usual combinations of loadings are included the designer should not assume that all cases are covered CIVL 4135 Chapter1 Introduction 10 Various load combinations must be considered to determine the most critical design condition This is particularly true when strength is dependent on more than one load effect such as strength for combined exure and axial load or shear strength in members with axial load Since the ACI 318 Building Code is a national code it has to conform to the International Building Code IBC2006 and in turn be consistent with the ASCE7 Standard on Minimum Design Loads for Buildings and Other structures These two standards contain the same probabilistic values for the expected safety resistance factors 47139th where d3 is a strength reduction factor depending on the type of stress being considered in the design such as exure shear or compression etc Table 12 Factored Load Combinations for Determining Required Strength U in ACI Code ACI Section 92 Condition Factored Loads ACI Equation 91 U 14D F ACI Equation 93 U 12D F T 16L H 05Lr orSorR ACI Equation 94 U 12D 16Lr or S orR 10L or 08W ACI Equation 95 U 12D 16W 10L 05Lr orSorR ACI Equation 96 U 12D 10E 10L 02S ACI Equation 97 U 09D 16W 16H ACI Equation 97 U 09D 10E 16H Due Regard is to be given to sign in determining U for combinations of loadings as one type of loading may produce effects of opposite sense to that produced by another type The load combinations with 09D are specifically included for the case where a higher dead load reduces the effects of other loads The loading case may also be critical for tension controlled column sections In such a case a reduction in axial load and an increase in moment may result in critical load combination Except for o The load factor on L in Equation 93 to 95 shall be permitted to be reduced to 05 except for garages areas occupied as places of public assembly and all areas where the live loadL is greater than 100 lbftz Where wind load Whas not been reduced by a directionality factor it shall be permitted to use 13W in place of 16Win Equations 94 and 96 Where earthquake loadE is based on servicelevel seismic forces 14E shall be used in place of 10E in Equations 95 and 97 CIVL 4135 Chapter1 Introduction 11 o The load factor on H shall be equal to zero in Equation 96 and 97 if the structural action due to H counteracts that due to W or E Where lateral earth pressure provides resistance to structural actions from other forces It shall not be included in H but shall be included in the design resistance Read AC1 Sections 922 through 925 112 Reduction in Live Load It is reasonable to assume that the full intensity of live load does not cover the entire oor area Therefore members having an in uence area of 400 ft2 or more can be designed for a reduced live load from the following equation LL0025 15 J 15 xA I where L Reduced design live load per square foot of area supported by the member Lg Unreduced design live load per square foot of area A1 In uence area For other than cantilevered construction A1 is 4 times the tributary area for a column 2 times tributary area for beams or equal area for a twoway slab The reduce design live load cannot be less than 50 of the unit live load Lg for member supporting one floor or less than 40 of the unit live load Lg for members supporting two or more oors For live load of 100 lbft2 or less no reduction can be made for areas used as places of public assembly except that in the case of garages for passenger cars a reduction of up to 20 can be made Live loads in all other cases not stipulated by the code cannot be reduced except as accepted by the jurisdictional authority CIVL 4135 Chapter1 Introduction 12 113 Design Strength Section 93 of the ACI Code The strength of a particular structural unit calculated using the current established procedures is termed nominal strength For example in the case of a beam the resisting moment capacity of the section calculated using the equations of equilibrium and properties of concrete and steel is called the nominal moment capacity Mn of the section The purpose of the strength reduction factor f are MacGregor 1976 and Winter 1979 to allow for understrength members due to variations in material strengths and dimensions to permit for inaccuracies in the design provisions to re ect the degree of ductility and required probability of the member under the load effects being considered to re ect the importance of the member in the structure CIVL 4135 Chapter1 Introduction 13 Table 13 Strength Reduction Factors F of the ACI Code ACI Section 93 Kind of Strength Strength Reduction Factor D 9321 Tension controlled sections as de ned in 1034 9322 Compression controlled sections as de ned in 1093 a Members with spiral reinforcement 070 b Other members 065 For sections in which the net tensile strain in the extreme tension 070 steel is between the limits for compression controlled and tension 16 5 controlled sections may be increased in from that for compression controlled section to 090 as the net tensile strain in the extreme tension steel at nominal strength increases from the compressioncontrolled strain limit 0005 Alternatively when Appendix B is used for members in whichfv does not exceed 60000 psi with symmetrical reinforcement and with h dC djh not less than 07 may be increased linearly to 090 as CDPn decreases from 01 Ag to zero For other reinforced members d3 may be increased linearly to 090 as d3 Pquot decreases from 01 Ag to CDPn whichever is smaller to zero 9323 Shear and torsion 075 9324 Bearing on Concrete 065 9327 Flexure sections without axial load in pretensioned 075 members where strand embedment is less than development length as provided in 12911 935 Plain Concrete 055 CIVL 4135 Chapter1 Introduction 14 114 Example Compute Factored load to be used in the design of a column subjected to the following load effects 9 kips compression from dead load 5 kips compression from roof live load 6 kips compression from snow 7 kips compression from accumulated rain and 8 kips compression from wind Assume live load is greater than 100 lbftz Solution Condition Factored Loads ACI Equation 91 U 14a F U 149 126kz39ps ACI Equation 93 U 12D F T 16L H 05Lr orSorR U 12D 16L 05R U 129 160 057 U 143kz39ps ACI Equation 94 U 12D 16Lr orSorR 10L or 08W U 12D 16R 08W U 129 167 088 U 284 kips ACI Equation 95 U 12D 16W 10L 05Lr orSorR U 12D 16W 05L 05R U 129 168 050 057 U 271 kips ACI Equation 96 U 12D 10E 10L 02S U 129 026 U 12 kips ACI Equation 97 U 09D 16W 16H U 099 168 U 209kz39ps ACI Equation 97 U 09D 10E 16H U 099 U 81 kips Therefore U 284 kips controls CIVL 4135 Chapter1 Introduction 15 115 Ultimate Strength vs Working Stress Design For D andL only 12D16L Rn or 14DS R 16 Working stress design 167D167L Rn 17 Assume a strength reduction factor of d 09 Equation 17 becomes 15D15L 09Rn 1 Therefore the ratio for Ultimate Strength Equation 16 to Working Stress Equation 18 becomes 12D 16L 14D Ratio 0r 19 15D 15L 1216LD or 14 1515LD Ratio 110 Relative Area Weight LKFDI ASD LID Ratio CIVL 4135 Chapter 1 Introduction 0 16 116 References MacGregor J G 1976 Safety and Limit States Design for Reinforced Concrete Canadian Journal of CiVil Engineering 34 December pp 484513 Winter G 1979 Safety and Serviceability Provisions in the ACI Building Code Concrete Design US and European Practices SP59 American Concrete Institute Detroit 1979 pp 3549 Nawy EG 2003 Reinforced Concrete A Fundamental Approach Fifth Edition Prentice Hall CIVL 4135 Chapter1 Introduction 17 Homework Set 1 Draw the shear and moment diagrams for the beams shown below 10 kips 10 kips 20 kips 10 kips 10 kips 11111 8 0n 6 0n 8 0n 7 0n l 10 45quot l 20 kips B I 20 9quot w 25 kft C I 20 9quot 50 kips 50 kips w 10 kft D 1 amp I 20399quot 10 6quot CIVL 4135 Chapter1 Introduction 18 page 151 Chapter 7 Flexural Analysis of NonRectangular Beams 71 Balanced Steel for Beams with NonRectangular Sections In this section we establish a general procedure for the computation of the balanced steel area Asb for a cross section of any shape that is symmetrical with respect to a vertical axis or that is con strained so that under load it de ectsvertically without twisting The resultant C c is not located at a2 because the stress block is not a rectangle passes through the centroid of the stress block areaAc The step by step procedure for computing Ash is detailed below CIVL 4135 Flexure page 152 72 Example Analysis of NonRectangular Sections Find the balanced area Asb for the following section E 0003 O39SSfC b a l c I c Cc 30quot 33 ll b quotd C EAbsFy L y Given fE 5 000 psi fy 60 000 psi Solution Select cd to be right at the borderline of Transition and Tension Controlled g 0375 gt 0375 x 30 1125 inches 0 le 080 X 1128 9 inches CC 085fc39 X Shahed area 085X5ksigtlt16gtlt13gtlt 9 816 340ksz 39 Area of dashed Area of dashed triangle rectangle From Equilibrium 340 kips T C gtAb C gtA 567392 Z Z S fy C Sb fy 60 kipSin2 m CIVL 4135 Flexure 511 Underreinforced Beams Read Sect 3410 of your text We want the reinforced concrete beams to fail in tension because is not a sudden failure Therefore following Figure 53 you have to make sure that you stay in the tension controlled side of the curve In actual practice the upper limit on cd should be somewhat below 0375 for the following reasons 1 material properties are never precisely known 2 strain hardening of the reinforcing steel not accounted for in design may lead to a brittle concrete compression failure E larger than required tending toward over reinforcement the actual steel area provided considering standard rebar sizes will always be equal or Therefore for all members designed according to AC1 318 Code fS fy at failure and the nominal strength is given by a A fy 085 b f0 Mn pfybd21 059f 531 Mn Asfyd g by imposing a strength reduction factor I 09 for bending we get M Mn Asfyd 532 or Mu Mn p fybd21 059 533 C Let us define the strength coe icient of resistance R as R pfy 1 059p fy 534 fc CIVL 4135 103 Flexure Mu Mn bd2R 535 The relationship between 9 andR for various values of f C and fy is shown in the Figure 51 To use the Figure 53 we have to rewrite the Equation 531 in terms of cd As fy As fy d fy As X X X 536 085fc b 085fc b d 085fc bd d fy x x 537 a le 08 p d f p C y 538 X d fc 085 1 We can nd Q to be f0 5 39 p 085 E x l d fy 512 Minimum Steel Area Another mode of failure is when there is not enough reinforcement in a beam If exural strength of the cracked section is less than the moment produced cracking of the previously un cracked section the beam will fail immediately upon formation of the first cracking Aci code im poses a minimum tensile steel area of ACI Eq 10 3 gives gt A min 3 fc bwd 2 de 540 fy fy CIVL 4135 104 Flexure 513 Design of Rectangular Sections in Bending with Tension Reinforcement Only Sineg Reinforced Beams When we design rectangular sections in bending with tension reinforcement only we need to determine b d andAS from the required value oan MuP and the given material properties of f C and fy There are two approaches in determining b d andAS Equation 531 provides the condition of equilibrium Since there are three unknowns but only one equation there are several possible solutions 5131 Case 1 Select the optimum steel ratio Q determine concrete dimensions 1 0 Case 1 Select the optimum steel ratio Q determine concrete dimensions 0 Case 2 Select concrete dimensions b and d then determine the required reinforcement Set the required strengthMu equal to the design strength Mn from equation 531 or 535 MM Mn qbbdZR Using a cd usually 03 we can determine the value of R from Equation 534 Find P from Equation 539 p 085 lfixffiy R pfy 1 0593 4y fc Knowing R we can determine the required bd2 R R Compute AS from bdz AS pbd Select reinforcement and check strength of the section to make sure that Mn 2 MM 541 542 543 5 44 545 546 CIVL 4135 105 Flexure 5132 Case 2 Select concrete dimensions b and d then determine the required reinforce ment This is similar to case 1 except steps taken will be a little different Set the required strengthMu equal to the design strength Mn from equation 531 or 535 MM Mn b d2R 547 Knowing b and d we can determine the strength coefficient of resistance R M M R u n 548 qbbd2 bd2 Knowing R we can determine the reinforcement ratio 9 from Figure 55 or equation 543 1 J1 236Rf c p 549 118fy f 5 Compute AS from AS pbd 5 50 Select re nement and check strength of the section to make sure that Mn 2 M 551 CIVL 4135 106 Flexure H 1 39 6000 T If 40000 I I I f J 3911 fell 5000 quot110 39 F 50000 ii 39 0000 quotquot 10 1400 f quotEU39DU39 100 39 2 03333 It I i I I g ff 40000 e fc39 5000 7 39 00 1 50000 1200 r r 3930 l 39 000 E I V 3000 quotf l 39 I 00000 X J war II n HtTH an n u Ln Bgul D hH GD 0ch Dc an x tar II II h Flah 88 at l I l l u m p31 I Y 1 1 Ha h p 00 a c c c C I m 0 H kgfcm39 I m R MPO 1 gt 03 3000 50 5 2 5V r 50000 600 r 4 400 30 3 I 20 2 I 200 39 39 W r i Mquot H de If h n 0 controls I 10 1 If 01 105 r 0 0010 0020 0030 0040 0050 quot 0 4quot Reinforcement reti0 0 Figure 55 Strength Curves for S ingly Reinforced Rectangular Sections CIVL 4135 107 Flexure 514 Concrete Protection for Reinforcement Section 771 of Code 515 Concrete Proportions To assist the designer further in making choices for beam sizes bar sizes and bar placement the following guideline are suggested These may be regard as role of thumb and are not AC1 Code requirements Undoubtedly situations will arise in which the experienced design will for good and proper reasons make a selection not conforming to the guidelines Use whole inches for overall dimensions except slabs may be in 12 in increments Beam stem widths are most often in multiples of2 or 3 in such as 9 10 1214 15 16 and 18 Minimum speci ed clear cover is measured from outside the stirrup or tie to the face of the concrete Thus beam effective depth d has rarely if ever a dimension to the whole inch An economical rectangular beam proportion is one in which the overall depth to width ratio is between about 15 to 20 unless architectural requirements or construction cost dictates otherwise For T shaped beams typically the ange thickness represents about 20 of overall depth we will talk about treatment of T shaped sections in later sections 516 Reinforcing Bar Selection 1 Maintain bar symmetry about the centroidal axis which lies at right angles to the bending axis ie symmetry about the vertical axis in usual situations Use at least two bars wherever exural reinforcement is required Bars 3 to 11 are more common and larger bars of 14 and 18 are mainly used in columns Use bars 11 and smaller for usual sized beams Use no more than two bar sizes and no more than two standard sizes apart for steel in one face at a given location in the span ie 7 and 9 bars may be acceptable but 9 and 4 bars would not Place bars in one layer if practicable Try to select bar size so that no less than two and no more than ve or six bars are put in one layer Follow requirements of ACI 76l and 762 for clear distance between bars and between lay ers and arrangement between layers When different sizes of bars are used in several layers at a location place the largest bars in the layer nearest the face of beam CIVL 4135 108 Flexure Chapter 8 Flexural Analysis of TBeams 81 Reading Assignments Text Chapter 37 ACI 318 Section 810 82 Occurrence and Con guration of TBeams 39 Common construction type used in conjunction with either on way or two way slabs 39 Sections consists of the ange and web or stem the slab forms the beam ange while the part of the beam projecting below the slab forms is what is called web or stem a one way slab b two way slab 83 Concepts of the effective width Code allowable values In reality the maximum compression stress in T section varies with distance from section Web Real max Longitudinal compression sress Simpli ed equivalent width stress CIVL 4135 156 TiBeam Analysis of TSections in Bending a Two possible locations of equivalent rectangular stress block lead to the following analysis requirements 1 quotaquot less than hf Analyze as for rectangular beam 2 quota greater than hf Special analysis required b 391 l b 1 Fill Emmi 3 l i I 2 I d H 2 1 L mquot39l i L 4 J h LgJ bl a 1 1 b Consider a T section in which entire stress block lies in ange we may assume this to be the case for all sections Statics will tell us if we are correct if a s hf CcTs 085 f39cab A fy therefore Afy pad 1 035113 035 E 0003 inin 035 L 3 a 4 tk V 1 Jigc T din i T W TAlfr LSk ourselves if quotaquot s hf if answer is quotyesquot calculate moment capacity for a rectangular beam if answer is quotnoquot an extensive analysis is required CIVL 4135 157 T Beam Code allows the following maximum effective widths 831 Symmetrical Beam ACI318 Section 8102 lt b hf 1bs f 4 b b 2 2 W S Shf 3 S clear distance between beams k bw gt 832 Flange on one side only Spandrel Beam ACI318 Section 8103 k b D39 3 1b bwsS 2 2b bws6hf clear distance to next web 3bbws k bw gt 833 Isolated TBeam l 2 ACI318 Section 8104 I A 1 b s 4bw CIVL 4135 158 TiBeam 84 Analysis of TBeams a gt hf Consider the total section in two parts 1 Flange overhangs and corresponding steel 2 Stem and corresponding steel 085fc O O O O Asf AsAsf CaseI Case 11 For equilibrium we have 841 Case I Asffy 085fc39hfb bw 81 or 085fc hfb bw 82 Sf f 842 Case 11 AS Asffy 085fc bwa 83 Solve for a A A a S Sf y 84 085fc bW and nominal moment capacity will be M A d hf A A d a 85 n Sffylt 3 s Sfgtfylt 5 CIVL 4135 159 TiBeam 85 Balanced Condition for TBeams See Commentary page 48 of AC1 318 83 old code lt b eu0003 kbwgt From geometry b d wt 86 6 Eu 9 87000 fy CIVL 4135 160 TiBeam 86 Example Analysis of TBeams in Bending 085fc I 40 124quot T 205quot 10quot Find the nominal moment capacity of the beam given above an 0003 AS 688 1112 f6 2 400 psi fy 50 000 psi Solution Check to see if a T beam analysis is required Assume a lt hf Asfy 688X50 a 085fc b 085 x 24 x 40 4 22quot Since 422 in gt 400 in a T beam analysis is required First nd the reinforcement area to balance anges Asf 7 Asf 085b bwhf 085 xx40 10 x4 490 in2 AS Asf 688 490 198 in2 Solve for a 085fc bwa As Asffy Ab8fCffy 08182545210 4 86 i gt 4 O k Assumption is ok CIVL 4135 161 TiBeam c C Note 486 572 i l 085 5 7 279 lt 0375 Tensioncontrolled 20 Find the nominal moment capacity of the beam h M Asffyw 7f fyAs AS d g Mn 49in2 gtlt 50ksi x 205 g 50ksi gtlt 198in2gt x 205 Mn 4530 1790 6320in k This could have been done by statics with TS Asfy CC b bwhf X 085fc abw085fc CIVL 4135 162 TiBeam 87 Example Design of TBeams in Bending Determination of Steel Area for a given Moment A oor system consists of a 3 in concrete slab supported by continuous T beams of 24 ft span 47 in on centers Web dimensions as determined by negative moment requirements at the supports are bW 11 in and d 20 in What tensile steel area is required at midspan to resist a mo ment of 6400 in kips iffy 60000 psi andf C 3000 psi 39 O O O O Asf As Asf Case I Case 11 Solution First determining the effective ange width from Section 831 or ACI 8102 SP0 24 X 12 lt 1 b 4 4 72 m 2bS 16hfbw 16 X3 11 59in 3 b S clear spacing between beams bW center to center spacing between beams 47 in The centerline T beam spacing controls in this case and b 47 inches Assumption Assuming that stress block depth equals to the ange thickness of 3 inches beambe haves like a rectangular shape Mu 6400 2 A 640 87 S fyd a2 09 x 60 x 20 32 m CIVL 4135 163 TiBeam s 7 a Solve for Asfy 640X60 W m 32 m gt hf 30 Assumption 1ncorrect Therefore the beam will act as a T beam and must be designed as a T beam From Case I given above and Section 841 we have 085fc hfb bw 085 x 31m x 3m x 47 11 4 58 M 88 Sf fy 60ksz39 hf 89 qun1 As fyd 7 09 x 458 x 60ksz x 20 32 4570 1n k1ps qan Mu qun1 6400 4570 1830 in kips 810 Find a value by iteration Assume initial a 35 inches Wquot 2 183 2 A A n 186 811 S Sf fyd 22 09 x 60 x 20 352 1quot Find an improve a value a 2 AS Asffy 186 x 60 397 in 812 085fc39 bW 085 X 3 X 11 39 Iterate with the new a 397 in W 2 2 A A n 188 813 S Sf fyd 22 09 x 60 x 20 3972 1quot Find an improve a value A A a S My 2 188 x 60 402 in 814 085fc bW 085 X 3 X 11 W 2 2 A A quot 188 815 S Sf fyd 22 09 x 60 x 20 4022 m CIVL 4135 164 TiBeam CIVL 4135 HOMEWORK SET 7 154 A Determinewhether 39 area meet the AF 1 39 m allowed andthen cal culate the ultimate moment for the section shown below H 16quot gt f C 4 ksi fy 60 ksi 16 n Answer M 2950 in kz39ps 4quot 6 n B a 6quot by 12 hole islocatedin the center of a reinforced concrete cross section to reduce the weight of the beam Determine the nominal exural capacity of the member if the beam is subjected to positive moment fc 35 ksz39 fy 60 ksi 3 10 bars Answer 4049 inkigs C a Determine balanced steelAS for thr symmetrical cross section shown below to be on the borderline of Ten sion and Transition b What is the maximum area of the steel permitted in the cross section by the ACI code c Determine the nominal exural capacity of the member if the beam is subjected to positive moment f C40ksi fy 60 ksi I llt gt Answer 5quot Asb in2 T 5quot l Am 2725quot Mn 2 O n 5 n i T 3 8 CIVL 4135 Flexure