Sample Beam Design
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DESIGN OF REINFORCED RECTANGULAR BEAM KNOWN BEAM DIMENSION Ref. Code: NSCP 2001, 5th Edition BEAM DIMENSIONS b 300 mm width of beam h 660 mm height of beam d 600 mm effective depth of tension bars d' 60 mm effective depth of compression bars DESIGN PARAMETERS fc' 30 Mpa compressive strength of concrete fy 414 Mpa yield strength of reinforcement Mu 650000000 N.mm ultimate moment φ 0.9 strength reduction factor defined in section 409.4 (factor defined in section 410.3.7.3) β = 0.85 β = (for fc'>30) Design β = 0.85 (reinforcement ratio producing balanced strain conditions) = = 0.0309793328 (maximum steel ratio) = = 0.0232344996 (reinforcement index) = 0.0209110497 ω = ρfy/fc' = 0.2885724852 = = 775788122.24 N.mm = = 698209310.01 N.mm COMPRESSION REINFORCEMENT IS NOT NEEDED PROCEED TO STEP 2 !!!!! STEP 2 (SINGLY REINFORCED BEAM) (proceed only if compression ω = (1-(1-2.36(Mu/φfc'bd2))1/2)/1.18 = 0.2640417359 Computation of β: (for fc'≤30) 0.85-0.05/7(fc'-30) ≥0.6 Computation of ρbal: ρbal 0.85fc'β600/fy(600+fy) Computation of ρmax: ρmax 0.75ρbal Computation of ω: assume ρ = 90% of ρmax Computation of Mu1: Mn1 fc'bd 2 ω(1-0.59ω) Mu1 φMn1 Determine if compression reinforcement is necessary: (if Mu1 < Mu) Solve for ω:
Transcript of Sample Beam Design
DESIGN OF REINFORCED RECTANGULAR BEAMKNOWN BEAM DIMENSION
Ref. Code: NSCP 2001, 5th Edition
BEAM DIMENSIONSb 300 mm width of beamh 660 mm height of beamd 600 mm effective depth of tension barsd' 60 mm effective depth of compression bars
DESIGN PARAMETERSfc' 30 Mpa compressive strength of concretefy 414 Mpa yield strength of reinforcementMu 650000000 N.mm ultimate momentφ 0.9 strength reduction factor defined in section 409.4
(factor defined in section 410.3.7.3)β = 0.85β = (for fc'>30)Design β = 0.85
(reinforcement ratio producing balanced strain conditions)
= = 0.0309793328
(maximum steel ratio)
= = 0.0232344996
(reinforcement index)
= 0.0209110497ω = ρfy/fc' = 0.2885724852
= = 775788122.24 N.mm
= = 698209310.01 N.mm
COMPRESSION REINFORCEMENT IS NOT NEEDED PROCEED TO STEP 2 !!!!!
STEP 2 (SINGLY REINFORCED BEAM) (proceed only if compression reinforcement is not needed)
ω = (1-(1-2.36(Mu/φfc'bd2))1/2)/1.18 = 0.2640417359
ρ = ωfc'/fy = 0.0191334591
Computation of β:(for fc'≤30)
0.85-0.05/7(fc'-30) ≥0.65
Computation of ρbal:
ρbal 0.85fc'β600/fy(600+fy)
Computation of ρmax:
ρmax 0.75ρbal
Computation of ω:
assume ρ = 90% of ρmax
Computation of Mu1:
Mn1 fc'bd2ω(1-0.59ω)
Mu1 φMn1
Determine if compression reinforcement is necessary: (if Mu1 < Mu)
Solve for ω:
Solve for ρ:
Solve for As:
As = ρbd = 3444.0226
Bars Selection:try diam. 28 mm
= = 615.7521601
= 5.5931961994say 6 pcs
try diam. 25 mm
= = 490.87385212
= 7.0161053126say 8 pcs
Check ductility:
= 1.4/fy = 0.0033816425 (minimun steel ratio)
As = = 3694.5129606 (actual steel area)ρ = As/bd = 0.020525072 (actual ratio of tension reinforcement)
OK!
OK!
STEP 3 (DOUBLY REINFORCED BEAM) (proceed only if compression reinforcement is needed)
= assumed ρ * bd =
= = N.mm
Check if compression steel yields:
a = = mmc = = mmfs' = 600*(c-d')/c =COMPRESSION STEEL DOES NOT YIELD !!!!
= = 0.0000
As' = = 0.0000 (compression steel area if compression steel yields)
As' = = (compression steel area if compression steel does not yield)
As = = 0.0000 (tension steel area)
mm2
A28 π*d2/4 mm2 (area of 28mmφ bar)
N = As/A28
A25 π*d2/4 mm2 (area of 25mmφ bar)
N = As/A25
Use 6-28mmφ bars
ρmin
6A28
ρmin < ρ
ρ < ρmax
Solve for As1:
As1 mm2
Solve for Mu2:
Mu2 Mu - Mu1
As1fy/0.85fc'ba/β
if compression steel yields use As' = As2
if compression steel does not yield use As' = As2fy/fs'
Solve for As2 and As':
As2 Mu2/φfy(d-d') mm2
As2 mm2
As2fy/fs' mm2
Solve for As:
As1 + As2 mm2
Bars Selection (compression):try diam. 20 mm
= = 314.15926536
= 0say pcs
try diam. 25 mm
= = 490.87385212
= 0say pcs
Bars Selection (tension):try diam. 20 mm
= = 314.15926536
= 0say pcs
try diam. 25 mm
= = 490.87385212
= 0say pcs
A20 π*d2/4 mm2 (area of 20mmφ bar)
N = As'/A28
A25 π*d2/4 mm2 (area of 25mmφ bar)
N = As'/A25
A20 π*d2/4 mm2 (area of 20mmφ bar)
N = As/A28
A25 π*d2/4 mm2 (area of 25mmφ bar)
N = As/A25
(proceed only if compression reinforcement is not needed)
(actual ratio of tension reinforcement)
(proceed only if compression reinforcement is needed)
(compression steel area if compression steel yields)
(compression steel area if compression steel does not yield)
(tension steel area)
(area of 28mmφ bar)
(area of 25mmφ bar)
(area of 20mmφ bar)
(area of 25mmφ bar)
(area of 20mmφ bar)
(area of 25mmφ bar)
DESIGN OF REINFORCED RECTANGULAR BEAMKNOWN BEAM DIMENSION
Ref. Code: NSCP 2001, 5th Edition
BEAM DIMENSIONSb 250 mm width of beamh 460 mm height of beamd 400 mm effective depth of tension barsd' 60 mm effective depth of compression bars
DESIGN PARAMETERSfc' 20.7 Mpa compressive strength of concretefy 344.7 Mpa yield strength of reinforcementMu 272000000 N.mm ultimate momentφ 0.9 strength reduction factor defined in section 409.4
(factor defined in section 410.3.7.3)β = 0.85β = (for fc'>30)Design β = 0.85
(reinforcement ratio producing balanced strain conditions)
= = 0.0275565122
(maximum steel ratio)
= = 0.0206673842
(reinforcement index)
= 0.0186006457ω = ρfy/fc' = 0.3097411877
= = 209597288.37 N.mm
= = 188637559.53 N.mm
COMPRESSION REINFORCEMENT IS NEEDED PROCEED TO STEP 3 !!!!!
STEP 2 (SINGLY REINFORCED BEAM) (proceed only if compression reinforcement is not needed)
ω = (1-(1-2.36(Mu/φfc'bd2))1/2)/1.18 =
ρ = ωfc'/fy =
Computation of β:(for fc'≤30)
0.85-0.05/7(fc'-30) ≥0.65
Computation of ρbal:
ρbal 0.85fc'β600/fy(600+fy)
Computation of ρmax:
ρmax 0.75ρbal
Computation of ω:
assume ρ = 90% of ρmax
Computation of Mu1:
Mn1 fc'bd2ω(1-0.59ω)
Mu1 φMn1
Determine if compression reinforcement is necessary: (if Mu1 < Mu)
Solve for ω:
Solve for ρ:
Solve for As:
As = ρbd =
Bars Selection:try diam. 28 mm
= = 615.7521601
= 0say pcs
try diam. 25 mm
= = 490.87385212
= 0say pcs
Check ductility:
= 1.4/fy = (minimun steel ratio)
As = = (actual steel area)ρ = As/bd = (actual ratio of tension reinforcement)
REDESIGN!
REDESIGN!
STEP 3 (DOUBLY REINFORCED BEAM) (proceed only if compression reinforcement is needed)
= assumed ρ * bd = 1860.0646
= = 83362440.465 N.mm
Check if compression steel yields:
a = = 145.76055891 mmc = = 171.48301048 mmfs' = 600*(c-d')/c = 390.06666667COMPRESSION STEEL YIELDS !!!!
= = 790.3286
As' = = 790.3286 (compression steel area if compression steel yields)
As' = = (compression steel area if compression steel does not yield)
As = = 2650.3932 (tension steel area)
mm2
A28 π*d2/4 mm2 (area of 28mmφ bar)
N = As/A28
A25 π*d2/4 mm2 (area of 25mmφ bar)
N = As/A25
ρmin
6A28
ρmin < ρ
ρ < ρmax
Solve for As1:
As1 mm2
Solve for Mu2:
Mu2 Mu - Mu1
As1fy/0.85fc'ba/β
if compression steel yields use As' = As2
if compression steel does not yield use As' = As2fy/fs'
Solve for As2 and As':
As2 Mu2/φfy(d-d') mm2
As2 mm2
As2fy/fs' mm2
Solve for As:
As1 + As2 mm2
Bars Selection (compression):try diam. 20 mm
= = 314.15926536
= 2.5156941374say 3 pcs
try diam. 25 mm
= = 490.87385212
= 1.6100442479say 2 pcs
Bars Selection (tension):try diam. 20 mm
= = 314.15926536
= 8.4364635644say 9 pcs
try diam. 25 mm
= = 490.87385212
= 5.3993366812say 6 pcs
A20 π*d2/4 mm2 (area of 20mmφ bar)
N = As'/A28
A25 π*d2/4 mm2 (area of 25mmφ bar)
N = As'/A25
Use 3-20mmφ bars or 2-25mmφ bars for compression
A20 π*d2/4 mm2 (area of 20mmφ bar)
N = As/A28
A25 π*d2/4 mm2 (area of 25mmφ bar)
N = As/A25
Use 6-25mmφ bars for tension
(proceed only if compression reinforcement is not needed)
(actual ratio of tension reinforcement)
(proceed only if compression reinforcement is needed)
(compression steel area if compression steel yields)
(compression steel area if compression steel does not yield)
(tension steel area)
(area of 28mmφ bar)
(area of 25mmφ bar)
(area of 20mmφ bar)
(area of 25mmφ bar)
bars for compression
(area of 20mmφ bar)
(area of 25mmφ bar)
