WHAT IS DOUBLY REINFORCED BEAM AND HOW TO DESIGN DOUBLY REINFORCED BEAM WITH DESIGN EXAMPLE


WHAT IS DOUBLY REINFORCED BEAM
Concrete Have High Compression Strength And Very Low Tensile Strength For In This Reason Steel Reinforcement Is Provided For Tension Zone Because Steel Reinforcement Have High Tensile Strength.

For Single Reinforced Beam Reinforcement Is Provided Only Tension Zone Only (Bottom Side Of Beam) Other Case Both Tension And Compression Zone Of Beam Is Provided Reinforcement (Top And Bottom Of Beam ) That Is Called Doubly Reinforced Beam.

For Singly Reinforced Beam Minimum Two Numbers Of Bares Provided In Compression zone But It Does Not Consider For Moment Of Resistance Calculation .When We Consider This Reinforced Steel Bas Are Calculation Of Moment Of Resistance At The Time We Have To Provide Additional Tension Steel In Tension Zone Therefore To Reduced The Over Reinforced Section.

For Design Of Doubly Reinforced Beam First Find Out Limiting Moment Of Resistance And Area Of Tension Steel ( Ast ) In Singly Reinforced Section Then To Find Area Of Compression Steel (Asc) And Area Of Addition Tension Steel (Ast) To Resistance The Excess Bending Moment (Mu2)


WHEN WE GO FOR DESIGN OF DOUBLY REINFORCED BEAM.
The Depth Of Beam May Be Restricted For Some Architecture Reason Or Our Requirement. At That Situation Doubly Reinforced Beam  Is designed .If Such Type Of Beam Restricted Depth Are Required To Resist More Moment That Limiting Moment.

When Go Design Moment Of Resistance Is Higher Than Limiting Moment Of Resistance. Design Doubly Reinforced Beam.
DESIGN EXAMPLE
A Simply Supported Beam Have Size 300 mm X 580 mm And 6.5 m Eff Span. It Carries Udl Of 40 KN/M ( Exclusive Of Self Weight ) .M20 Grade Of Concrete Is Used And Fe500 Steel Is Used. To Design The Mid Span Section Of Beam.
SOLUTION
  1. Load On Beam           = 40 KN/M (exclusive self weight)
  2. Self weight of beam = L X B X 1 X unit weight of concrete
                                              = 0.3 x 0.58 x 1 x 25 (note :unit weight of concrete is 25 KN/M)
                                              = 4.35 KN/M
                       Total load W = 40 + 4.35
                                                   = 44.37 KN/M
                                                   = 44.37 x 1.5
                      Design load Wu = 66.53 KN/M               (note: for limit state design factor of safety is 1.5 )

EFF SPAN
Eff span of beam l = 6.5 m
DESIGN BENDING MOMENT
Design BM   Mu = Wul2/8                                      WU = Design load
                      Mu = 66.53 x 6.5 2/8                        L = eff span
                            = 351.36 KN.M          Mu = design  BM
                      Mu = 351.36 X 10 6 N.MM                                     
MOMENT OF RESISTANCE OF SINGLY REINFORCED SECTION
Moment or resistance
Mulim =  Qbd2
Assume eff cover 40mm
Eff depth d = D-40
                    = 580 – 40
               d  = 540 mm
Compression strength of concrete fck  = 25 N/mm2  
tension strength of steel                  fy = 500 n/mm2
Mulim =  Qbd2        ( Q = 3.33 N/MM2 )
                MOMENT OF RESISTANCE FACTOR (FROM SP 16)
CONCRETE GRADE
Q=MUlim/bd2    N/MM2
Fe=250
Fe=415
Fe=500
M15
2.23
2.07
2.00
M20
2.98
2.76
2.66
M25
3.72
3.45
3.33


MUlim = 3.33 x 300 x 5402   
MUlim = 291.31 x 10 6 N.MM
Design bending moment Mu = 351.36 x 106
(condition : Mu < Mulim therefore the beam is designed singly reinforced beam. But there    Mu > Mulim so the beam is designed doubly reinforced beam)
EXTRA BENDING MOMENT
Mu2  = Mu – Mulim
        = 351.36 x 10 6-291.31 x 106
Mu2  = 60.05 x 106 N.mm
AREA OF TENSION STEEL
Total area of tension steel Ast = Ast1+Ast2
AREA OF TENSION STEEL REQUIRED FOR Mulim
Assume the section is balanced section
So ptmax = 0.755 %
CONCRETE GRADE
Ptmax  %
Fe=250
Fe=415
Fe=500
M15
1.316
0.716
0.566
M20
1.758
0.955
0.755
M25
2.197
1.194
0.943


Ast1 = (0.755/100)300x410
      = 930 mm2
AREA OF TENSION STEEL REQUIRED FOR MU2
Ast2 = Mu2/0.87 fy (d-d’)                      (note d’=eff cover (40mm))
Ast2 = (60.05 x 10 6)/((0.87x500(540-40))
Ast2 = 255 mm2
Ast  = Ast1+Ast2
       = 930 + 255
Ast  = 1186 mm2
Provide 6 nos of 16 mm dia bars (Ast=1206 mm2)
AREA OF COMPERESSION STEEL
Asc = Mu2/fsc(d-d’)
fsc = 0.0035 ((Xumax-d’)/ Xumax)
Actual neutral axis Xumax = 0.456d         (for Fe500 steel Xumax=0.456 d from table 13 of design aids SP 16)
Xumax = 0.456x540
Xumax  = 246.24 mm
Fsc = 413.04 +0.0035((246.24-40)/246.24))
Fsc = 413 N/mm2
ASC= (60.05 x 10 2)/413(540-40)
Asc   = 290 mm2
Another method  Asc= 0.87 fy Ast2/fsc
                                   = (0.87 x500 x 255)/413
                                   = 268 mm2
Provide 2 nos of 16 mm dia bars (Ast=402 mm2)

ADDITIONAL POINTS
The compression steel does not exceed 4 % of whole area cross section of beam.(26.5.1.2 of IS 456)
Minimum area of tension steel not less than 0.85 bd/fy
Maximum area of compression steel shall hot more than 0.04 bD (IS 456 – 26.5.1.1.)  
ALSO READ--
HOW TO DESIGN SINGLY REINFORCED BEAM
HOW TO DESIGN SEPTIC TANK FOR RESIDENTIAL COLONY
HOW TO CALCULATE  WATER TANK SIZE AND CAPACITY
HOW TO CALCULATE RENT OF BUILDING
HOW TO FIND UNIT WEIGHT OF STEEL BARS
HOW TO APPROXIMATELY CALCULATE STEEL BAR IN BEAMS,COLUMNS AND SLABS

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