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DESIGN OF REINFORCED RECTANGULAR BEAM KNOWN BEAM DIMENSION
Ref. Code: NSCP 2001, 5th Edition
b h d d'
300 660 600 60
fc' 30 fy 414 Mu 650000000 φ 0.9
mm mm mm mm
Mpa Mpa N.mm
BEAM DIMENSIONS width of beam height of beam effective depth of tension bars effective depth of compression bars DESIGN PARAMETERS compressive strength of concrete yield strength of reinforcement ultimate moment strength reduction factor defined in section 409.4
Computation of β: (factor defined in section 410.3.7.3) (for fc'≤30) β = 0.85 0.85-0.05/7(fc'-30) ≥0.65 β = (for fc'>30) Design β = 0.85 Computation of ρbal: (reinforcement ratio producing balanced strain conditions) 0.85fc'β600/fy(600+fy) ρbal = = 0.030979333 Computation of ρmax: ρmax 0.75ρbal =
(maximum steel ratio)
Computation of ω: assume ρ = 90% of ρmax
(reinforcement index)
=
0.0232345
= =
0.02091105 0.288572485
Computation of Mu1: fc'bd2ω(1-0.59ω) Mn1 =
=
775788122.2 N.mm
Mu1
=
ω
=
=
ρfy/fc'
φMn1
698209310
N.mm
Determine if compression reinforcement is necessary: (if Mu1 < Mu) COMPRESSION REINFORCEMENT IS NOT NEEDED PROCEED TO STEP 2 !!!!! (proceed only if compression reinforcement is not needed)
STEP 2 (SINGLY REINFORCED BEAM) Solve for ω: ω= (1-(1-2.36(Mu/φfc'bd2))1/2)/1.18
=
Solve for ρ: ρ =
=
ωfc'/fy
0.264041736
0.019133459
Solve for As: As
=
Bars Selection: try diam. A28 =
ρbd
=
28 π*d2/4
= = say
25 π*d2/4
(area of 28mm φ bar )
5.593196199 6 pcs
=
2 490.8738521 mm
(area of 25mm φ bar )
= 7.016105313 say 8 pcs Use 6-28mmφ bars
Check ductility: ρmin = = =
ρ < ρmax
1.4/fy
=
0.003381643 (minimun steel ratio)
6A28
= =
3694.512961 (actual steel area) 0.020525072 (actual ratio of tension reinforcement)
As/bd OK! OK!
(proceed only if compression reinforcement is needed)
STEP 3 (DOUBLY REINFORCED BEAM) Solve for As1: As1
2 615.7521601 mm
mm
N = As/A25
As ρ ρmin < ρ
mm2
mm
N = As/A28 try diam. A25 =
3444.0226
=
Solve for Mu2: Mu2 =
assumed ρ * bd
Mu - Mu1
Check if compression steel yields: As1fy/0.85fc'b a = a/β c = fs' = 600*(c-d')/c COMPRESSION STEEL DOES NOT YIELD !!!!
2
=
mm
=
N.mm
= = =
mm mm
if compression steel yields use As' = As2 if compression steel does not yield use As' = As2fy/fs' Solve for As2 and As': As2
=
As'
=
As'
=
Mu2/φfy(d-d') As2 As2fy/fs'
=
0.0000
mm2
=
0.0000
mm2
=
mm
2
(compression steel area if compressio
(compression steel area if compressio
Solve for As: As
=
As1 + As2
=
Bars Selection (compression): try diam. 20 mm 2 π*d /4 A20 = N = As'/A28 try diam. A25 =
ession steel area if compression steel does not yield)
n steel area)
f 20mm φ bar )
f 25mm φ bar )
f 20mm φ bar )
f 25mm φ bar )
DESIGN OF REINFORCED RECTANGULAR BEAM KNOWN BEAM DIMENSION
Ref. Code: NSCP 2001, 5th Edition
b h d d'
250 460 400 60
fc' 20.7 fy 344.7 Mu 272000000 φ 0.9
mm mm mm mm
Mpa Mpa N.mm
BEAM DIMENSIONS width of beam height of beam effective depth of tension bars effective depth of compression bars DESIGN PARAMETERS compressive strength of concrete yield strength of reinforcement ultimate moment strength reduction factor defined in section 409.4
Computation of β: (factor defined in section 410.3.7.3) (for fc'≤30) β = 0.85 0.85-0.05/7(fc'-30) ≥0.65 β = (for fc'>30) Design β = 0.85 Computation of ρbal: (reinforcement ratio producing balanced strain conditions) 0.85fc'β600/fy(600+fy) ρbal = = 0.027556512 Computation of ρmax: ρmax 0.75ρbal =
(maximum steel ratio)
Computation of ω: assume ρ = 90% of ρmax
(reinforcement index)
=
0.020667384
= =
0.018600646 0.309741188
Computation of Mu1: fc'bd2ω(1-0.59ω) Mn1 =
=
209597288.4 N.mm
Mu1
=
188637559.5 N.mm
ω
=
=
ρfy/fc'
φMn1
Determine if compression reinforcement is necessary: (if Mu1 < Mu) COMPRESSION REINFORCEMENT IS NEEDED PROCEED TO STEP 3 !!!!! (proceed only if compression reinforcement is not needed)
STEP 2 (SINGLY REINFORCED BEAM) Solve for ω: ω= (1-(1-2.36(Mu/φfc'bd2))1/2)/1.18
=
Solve for ρ: ρ =
=
ωfc'/fy
Solve for As: As
=
Bars Selection: try diam. A28 =
ρbd
28 π*d2/4
mm2
=
2 615.7521601 mm
mm
N = As/A28
= say
try diam. A25 =
25 π*d2/4
= = say
Check ductility: ρmin = = =
ρ < ρmax
0 pcs 2 490.8738521 mm
(area of 25mm φ bar )
0 pcs
1.4/fy
=
(minimun steel ratio)
6A28
= =
(actual steel area) (actual ratio of tension reinforcement)
As/bd REDESIGN! REDESIGN!
(proceed only if compression reinforcement is needed)
STEP 3 (DOUBLY REINFORCED BEAM) Solve for As1: As1
(area of 28mm φ bar )
mm
N = As/A25
As ρ ρmin < ρ
=
=
Solve for Mu2: Mu2 =
assumed ρ * bd
Mu - Mu1
Check if compression steel yields: As1fy/0.85fc'b a = a/β c = fs' = 600*(c-d')/c COMPRESSION STEEL YIELDS !!!!
=
1860.0646
mm
2
=
83362440.47 N.mm
= = =
145.7605589 mm 171.4830105 mm 390.0666667
if compression steel yields use As' = As2 if compression steel does not yield use As' = As2fy/fs' Solve for As2 and As': As2
=
As'
=
As'
=
Mu2/φfy(d-d') As2 As2fy/fs'
=
790.3286
mm2
=
790.3286
mm2
=
mm
2
(compression steel area if compressio
(compression steel area if compressio
Solve for As: As
=
As1 + As2
Bars Selection (compression): try diam. 20 mm 2 π*d /4 A20 = N = As'/A28 try diam. A25 =
25 π*d2/4
=
314.1592654 mm
2
(area of 20mm φ bar )
= say
2.515694137 3 pcs 2 490.8738521 mm
(area of 25mm φ bar )
(tension steel area)
=
= 1.610044248 say 2 pcs Use 3-20mmφ bars or 2-25mmφ bars for compression