DESIGN OF BEAM(AS PER ACI CODE)

CONTENTASSUMPTIONSEVALUATION OF DESIGN PARAMETERSMOMENT FACTORS Kn, STRENGTH REDUCTION FACTOR BALANCED REINFORCEMENT RATIO b

DESIGN PROCEDURE FOR SINGLY REINFORCED BEAMCHECK FOR CRACK WIDTH

DESIGN PROCEDURE FOR DOUBLY REINFORCED BEAMFLANGED BEAMST BEAMSL - BEAMS

ASSUMPTIONSPlane 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 computationc=0.003s = fy / Eshdc0.85fcaa/2d-a/2bCTTO SLIDE-5

Concrete 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 0.0075syfyfsIdealizedActualEs1

Total compressive force C = 0.85fc ba(Refer stress diagram)Total Tensile forceT = As fyC = T0.85fc ba = As fya = As fy / (0.85fc b) = d fy / (0.85 fc) = As / bdMoment of Resistance, 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 ] Mu = Mn = Kn bd2 = Strength Reduction Factor

EVALUATION OF DESIGN PARAMETERSTO SLIDE-7

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

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

In case of statically determinate structure ductile failure is essential for proper moment redistribution. Hence, for beams the ACI code limits the max. amount of steel to 75% of that required for balanced section. For practical purposes, however the reinforcement ratio ( = As / bd) should not normally exceed 50% to avoid congestion of reinforcement & proper placing of concrete. 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

DESIGN PROCEDURE FOR SINGLY REINFORCED BEAMDetermine the service loadsAssume `h` as per the support conditions according to Table 9.5 (a) in the codeCalculate d = h Effective coverAssume the value of `b` by the rule of thumb. Estimate self weightPerform preliminary elastic analysis and derive B.M (M), Shear force (V) valuesCompute min and bChoose between min and bCalculate , KnFrom Kn & M calculate `d required (Substitute b interms of d)Check the required `d with assumed `dRevise & repeat the steps, if necessary BACK

With the final values of , b, d determine the Total As requiredDesign 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 widths as per codal provisionsEXAMPLE

DESIGN PROCEDURE FOR DOUBLY REINFORCED BEAMMoment of resistance of the sectionMu = Mu1 + Mu2Mu1= M.R. of Singly reinforced section= As1 fy (d a/2) ;As1 = Mu1 / [ fy (d a/2) ]Mu2= As2 fy (d d);As2 = Mu2 / [ fy (d d) ]Mu= As1 fy (d a/2) + As2 fy (d d)If Compression steel yields, fy / EsI.e.,0.003 [ 1 (0.85 fc 1 d) / ((- ) fyd) ] fy / Es If compression steel does not yield,fs = Es x 0.003 [ 1 (0.85 fc 1 d) / ((- ) fyd) ] Balanced section for doubly reinforced section is b = b1 + (fs / fy) b1 = Balanced reinforcement ratio for S.R. sectionEND

Mu = MnThe design strength of a member refers to the nominal strength calculated in accordance with the requirements stipulated in the code multiplied by a Strength Reduction Factor , which is always less than 1.DESIGN STRENGTHWhy ?To allow for the probability of understrength members due to variation in material strengths and dimensionsTo allow for inaccuracies in the design equationsTo reflect the degree of ductility and required reliability of the member under the load effects being considered.To reflect the importance of the member in the structureRECOMMENDED VALUEBeams in Flexure...0.90Beams in Shear & Torsion 0.85 BACK

AS PER TABLE 9.5 (a)

BACKValues 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 number

Sheet1

Simply SupportedOne End ContinuousBoth End ContinuousCantilever

L / 16L / 18.5L / 21L/8

Sheet2

Sheet3

RULE 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 deflection BACKCLEAR 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.

BAR SIZE#n = n/8 in. diameter for n 8.Ex. #1 = 1/8 in.. #8 = 8/8 i.e., I in.

BACK

Sheet1

DESIGN OF SINGLY REINFORCED BEAM (AS PER ACI)

SpanL =12ft

End ConditionOne End Cont.

Nominal Bending momentMn =in.lb

Nominal Shear forceVn =lb

Strength of Concretefc' =4000psi

Strength of Steelfy =60000psi

Strength reduction factorf =0.9

Mu =

=0in.lb

0.85As per codal conditions

Balanced Reinforcement ratio

=0.029

Recom. reinforcement ratio

=0.014

=0.214

Kn =

=747.33

Assume breadth of beamb =0.5 d

Mu =

=

=

Effective depth As per designd =

=0.000in.

Effective depth As per deflectiond =8.0in

Final Effective depthd =8.0in

b =4.0in

Ast =0.456294.263

Sheet2

Weight, Area and Perimeter of individual bars

Bar NoWt.per Foot (lb)Stamdard Nominal Dimensions

Perimeter (in.)

inchmm

30.3760.37590.111.178

40.6680.500130.201.571

51.0430.625160.311.963

61.5020.750190.442.356

72.0440.875220.602.749

82.6701.000250.793.142

93.4001.128281.003.544

104.3031.270311.273.990

115.3131.410331.564.430

147.6501.693432.255.319

1813.6002.257564.007.091

Sheet3

CRACK WIDTHw=0.000091.fs.3(dc.A)Where,w=Crack width=0.016 in. for an interior exposure condition=0.013 in. for an exterior exposure conditionfs=0.6 fy, kipsdc=Distance from tension face to center of the row of bars closest to the outside surfaceA=Effective tension area of concrete divided by the number of reinforcing bars=Aeff / NAeff=Product of web width and a height of web equal to twice the distance from the centroid of the steel and tension surfaceN=Total area of steel As / Area of larger bar BACK

dcTension facedbwAeff = bw x 2d BACK

FLANGED BEAMSEFFECTIVE OVERHANG, r r T BEAMr 8 hfr lnr LL BEAMr 6 hfr lnr 1/12 L

Case-1: Depth of N.A `c < hf0.85fc b a = As fya = As fy / [ 0.85fc b] Mn = As fy (d a/2)

c=0.003s = fy / EsdcaStrain DiagramCase-2: Depth of N.A `c > hfi) a < hf

0.85fc b a = As fya = As fy / [ 0.85fc b] Mn = As fy (d a/2) b

c=0.003s = fy / EsdcaStrain DiagramCase-2: Depth of N.A `c > hfii) a > hf

Part-10.85fc bw a = As1 fyPart-2 0.85fc (b-bw) hf = As2 fy0.85fc bw a + 0.85fc (b-bw) hf = As fy a = [As fy - 0.85fc (b-bw) hf ] / [ 0.85fc bw]b

Moment of resistance of the sectionMn = Mn1 + Mn2Mn1= As1 fy (d a / 2) Mn2= As2 fy (d hf / 2) Moment RedistributionFor continuous beam members,Code permits Max of 20%when et 0.0075 at that section

Balaced Reinforcement Ratio ( b)b = (bw / b) [b + f ]b = Asb / bwd = 0.85fc ab / (fy. d)= 1 ( 0.85 fc / fy) [ 87,000 / (87,000+fy)]f = 0.85fc (b-bw) hf / (fy bw d) 0.75 b

Min. reinforcement is greater of the following: w = 3fc / fy or 200 / fy; for +ve Reinf. min = 6fc / fy or 200 / fy; for -ve Reinf.