Dhaka University of Engineering & Technology, Gazipur-1700

DEPARTMENT OF CIVIL ENGINEERINGCE-2105DESIGN OF CONCRETE STRUCTURE-IDESIGN OF BEAM(AS PER ACI CODE)Course Teacher- Presented By-Dr. Md. Rezaul Karim Md. Jahidur RahmanAssociate Professor S.ID- 121041

Dhaka University of Engineering & Technology, Gazipur-1700

11Content2USD(Ultimate Strength Design)Classification of beam with respect to design systemAssumptionsEvolution of design parameters Moment Factors Kn, Balanced Reinforcement Ratio bCalculating Strength Reduction Factor CalculatingDesign procedure for Singly Reinforced BeamDesign procedure for Doubly Reinforced BeamDesign procedure for T-BeamAppendixJahidur Rahman

Ultimate Strength Design(USD)

3Assuming tensile failure conditionAdditional strength of steel after yieldingACI code emphasizes this methodJahidur RahmanEmphasizes= 3

Classification of beam with respect to design system

Rectangular beam (reinforced at tension zone only)Doubly reinforced beam (reinforced at both tension and compression zone)T section beam (both beam and slab are designed together)

ASSUMPTIONS

5Plane sections before bending remain plane and perpendicular to the N.A. after bendingStrain distribution is linear both in concrete & steel and is directly proportional to the distance from N.A.Strain in the steel & surrounding concrete is the same prior to cracking of concrete or yielding of steelConcrete in the tension zone is neglected in the flexural analysis & design computation

c=0.003s = fy / Eshdc0.85fcaa/2d-a/2bCTJahidur Rahman6Concrete stress of 0.85fc is uniformly distributed over an equivalent compressive zone. fc = Specified compressive strength of concrete in psi.Maximum allowable strain of 0.003 is adopted as safe limiting value in concrete.The tensile strain for the balanced section is fy/EsMoment redistribution is limited to tensile strain of at least 0.0075syfyfsIdealizedActualEs1Jahidur RahmanAdopted= 6

EVALUATION OF DESIGN PARAMETERS

Total compressive force -C = 0.85fc ba(Refer stress diagram)

Total Tensile force-T = As fyC = T0.85fc ba = As fya = As fy / (0.85fc b) = d fy / (0.85 fc)[ = As / bd]Moment of Resistance/Nominal Moment- Mn = 0.85fc ba (d a/2) or, Mn = As fy (d a/2) = bd fy [ d (dfyb / 1.7fc) ] = fc [ 1 0.59 ] bd2 = fy / fc Mn = Kn bd2 Kn = fc [ 1 0.59 ] Ultimate Moment- Mu = Mn = Kn bd2 ( = Strength Reduction Factor)7Jahidur Rahman7Balaced Reinforcement Ratio ( b)From strain diagram, similar trianglescb / d = 0.003 / (0.003 + fy / Es); Es = 29x106 psi cb / d = 87,000 / (87,000+fy)

b = Asb / bd = 0.85fc ab / (fy. d) = 1 ( 0.85 fc / fy) [ 87,000 / (87,000+fy)]

Relationship b / n the depth `a of the equivalent rectangular stress block & depth `c of the N.A. is a = 1c1= 0.85 ; fc 4000 psi 1= 0.85 - 0.05(fc 4000) / 1000; 4000 < fc 80001= 0.65 ; fc> 8000 psi 8Jahidur RahmanFor beams the ACI code limits the max. amount of steel to 75% of that required for balanced section. 0.75 b

Min. reinforcement is greater of the following:Asmin = 3fc x bwd / fy or 200 bwd / fy min = 3fc / fy or 200 / fy

For statically determinate member, when the flange is in tension, the bw is replaced with 2bw or bf whichever is smaller

The above min steel requirement need not be applied, if at every section, Ast provided is at least 1/3 greater than the analysis

9Jahidur Rahman

YesNo

YesNoCalculating

Calculating

Jahidur Rahman12SINGLY REINFORCED BEAMBeam isreinforcednear the tensile faceReinforcement resists the tension.Concrete resists the compression.

12

DESIGN PROCEDURE FOR SINGLY REINFORCED BEAM131. Determine the service loads3. Calculate d= h Effective cover2. Assume `h` as per the support conditions [As per ACI code in table 9.5(a)] 4. Assume the value of `b` by the rule of thumb5. Estimate self weight6. Primary elastic analysis and derive B.M (M), Shear force (V) values7. Compute min and b8.Choose between min and b13Note Below:Design the steel reinforcement arrangement with appropriate cover and spacing stipulated in code. Bar size and corresponding no. of bars based on the bar size #n.Check crack width as per codal provisions.

14Jahidur Rahman9. Calculate , Kn10. From Kn & M calculate `d required11. Check the required `d with assumed `d

12. With the final values of , b, d determine the Total As requiredOK

DESIGN PROCEDURE FOR SINGLY REINFORCED BEAM BY FLOWCHART15Actual steel ratio

Maximum steel ratio

Over reinforced beamYesUnder reinforced beamNo

Ultimate moment

15Jahidur Rahman16DOUBLY REINFORCED BEAMBeam isfixed for Architectural purposes.Reinforcement are provided both in tension and compression zone.Concrete has limitation to resist the total compression so extra reinforcement is required.

Actual steel ratio

Maximum steel ratio

rectangular beamYesDoubly reinforced beamNoDESIGN PROCEDURE FOR DOUBLY REINFORCED BEAMConcrete section, Area of steel are known

YesNoCompression and tension bar reach in yieldingCompression bar does not reach in yielding

DESIGN PROCEDURE FOR DOUBLY REINFORCED BEAMLoad or ultimate moment is givenMaximum steel ratio

Reinforcement required in comp. zone too

rectangular beamYesDoubly reinforced beamNo

Jahidur Rahman19T- REINFORCED BEAMA part of slab acts as the upper part of beam.Resulting cross section is T shaped.The slab portion of the beam is flange.The beam projecting bellow is web or stem.

DESIGN PROCEDURE FOR T-REINFORCED BEAMAre of concrete section, Area of steel are knownAdditional moment

Nominal moment calculation

Rectangular beam analysisYesT - beam analysisNo

modification

Total moment

DESIGN PROCEDURE FOR T-REINFORCED BEAMMoment or loading is givenTotal area of steel

Nominal moment calculation

Rectangular beam analysisYesT - beam analysisNoArea of steel in flange

Area of steel in web

Additional moment

T section:

L section:

Isolated beam:

Check for

Jahidur Rahman22APPENDIXAS PER TABLE 9.5 (a)

Values given shall be used directly for members with normal weight concrete (Wc = 145 lb/ft3) and Grade 60 reinforcement

For structural light weight concrete having unit wt. In range 90-120 lb/ft3 the values shall be multiplied by (1.65 0.005Wc) but not less than 1.09

For fy other than 60,000 psi the values shall be multiplied by (0.4 + fy/100,000)

`h` should be rounded to the nearest whole number23Jahidur RahmanRULE OF THUMBd/b = 1.5 to 2.0 for beam spans of 15 to 25 ft.d/b = 3.0 to 4.0 for beam spans > 25 ft.`b` is taken as an even numberLarger the d/b, the more efficient is the section due to less deflectionCLEAR COVER

Not less than 1.5 in. when there is no exposure to weather or contact with the groundFor exposure to aggressive weather 2 in.Clear distance between parallel bars in a layer must not be less than the bar diameter or 1 in.24Jahidur RahmanBAR SIZE#n = n/8 in. diameter for n 8.Ex. #1 = 1/8 in.. #8 = 8/8 i.e., I in.

25Jahidur Rahman26

Jahidur Rahman26Sheet1Simply SupportedOne End ContinuousBoth End ContinuousCantileverL / 16L / 18.5L / 21L/8

Sheet2

Sheet3

Sheet1DESIGN OF SINGLY REINFORCED BEAM (AS PER ACI)SpanL =12ftEnd ConditionOne End Cont.Nominal Bending momentMn =in.lbNominal Shear forceVn =lbStrength of Concretefc' =4000psiStrength of Steelfy =60000psiStrength reduction factorf =0.9Mu ==0in.lb0.85As per codal conditionsBalanced Reinforcement ratio=0.029Recom. reinforcement ratio=0.014=0.214Kn ==747.33Assume breadth of beamb =0.5 dMu ===Effective depth As per designd ==0.000in.Effective depth As per deflectiond =8.0inFinal Effective depthd =8.0inb =4.0inAst =0.456294.263

Sheet2Weight, Area and Perimeter of individual barsBar NoWt.per Foot (lb)Stamdard Nominal DimensionsPerimeter (in.)inchmm30.3760.37590.111.17840.6680.500130.201.57151.0430.625160.311.96361.5020.750190.442.35672.0440.875220.602.74982.6701.000250.793.14293.4001.128281.003.544104.3031.270311.273.990115.3131.410331.564.430147.6501.693432.255.3191813.6002.257564.007.091

Sheet3