BEAM Sample
BEAM Sample
BEAM Sample
d= 387.5 mm
ρbal = 0.85*28*0.85*600
420(600+420)
ρbal = 0.0283333333
ρmax = 0.75ρbal
ρmax = 0.75*0.03
ρmax = 0.02125
use ρ= 0.02125
Mu = (ϕ)(fc')(ω)(b)(d2)(1 - 0.59ω)
Mu = (0.9)(28)(0.31875)(350)(387.5^2)(1 - 0.59*0.31875)
Mu = 342.75578411 kN-m
Mn = Mu/ϕ
Mn = 342.76/0.9
Mn = 380.83976012 kN-m
380.8398>0.0000
Therefore, design as singly reinforced
IF DOUBLY IF SINGLY
Check if Compression Steel Yields, when f's >fy Solve for Coefficient of Resista
As1 = ρmaxbd *rho max is used not rho M u=
As1 = 0.02125*350*387.5 Rn =
As1 = 2882.03125 mm2 Rn =
Mn1 = Mn Rn=
Mn1 = 380.83976012 kN-m
Solve for Steel Ratio
Mn2 = Mn - Mn1 ρ=
Mn2 = 0.0000-380.8398 ρ=
Mn2 = -380.8398 kN-m ρ=
ρmax =
Mn2 = As2*fy*(d-d')
use ρ=
As2 = Mn2 = -380.84
fy*(d-d') 420*(387.5-50) Solve for Tension Reinforcem
A s=
As2 = -2686.700248 mm2 A s=
A s=
0.85*f'c*a*b = As1*fy
a= As1*fy = 2882.03125*420 No. of Bars = As/Ab
0.85*f'c*b 0.85*28*350 As/Ab=
N=
a= 145.3125 mm
c= a/β1
c= 145.3125/0.85 Check for Spacing
c= 170.95588235 mm S=
f's= 600(c-d')/c
f's= 600(170.956-50)/170.956 S=
f's= 424.51612903
S=
424.5161>420
Therefore, compression reinforcement yields The proponents reco
reinforcement
Solve Area of Reinforcement
A's= As2
A's= -2686.700248
Vc= 0.17λ(√f'c)bwd λ= 1
Vc= 0.17*1*(√28)*350*387.5
Vc= 122.00220733 kN
ϕVc = 91.501655499 kN
Vu= 234.799 kN
Vs= Vn - Vc
Vs= 313.07 - 122.00 s=
Vs= 191.063126 kN
use s=
0.66(√f'c)bwd= 0.66(√28)*350*387.5
= 473.656 kN
Vs<0.66(√f'c)bd; Okay!
3
of bars between layers assumed to be equal to size of bar
or Coefficient of Resistance
ϕRnbd2
Mu/ϕbd2
0.00/0.9*350*387.5*^2
0 Mpa
or Steel Ratio
0.85(f'c/fy)(1-√(1-2Rn/0.85f'c))
0.85(28/420)(1-√(1-2*(0.000)/0.85*28))
0
0.0033333333
or Tension Reinforcement
ρbd Ab= πdb2/4
0.0033*350*387.5 Ab= π*25^2/4
452.08333333 mm2 Ab= 490.8739 mm2
for Spacing
b - 2cc - 2dbs - N(db)
N-1
530 mm
Sample Beam
Given
Mu= 999.06
Vu= 234.799
Concrete Strength f'c= 28 Mpa
Yield Strength of Steel fy= 420 Mpa
Reduction Factor ϕ= 0.9 (Flexure)
ϕ= 0.75 (Shear)
β1 = 0.85
Es= 200000 Mpa
Total Depth h= 450 mm
Beam Width b= 350 mm
for
Distance of extreme
doubly
fiber to centroid of d'= 50 mm reinforce
compresion rebar
d only
Clear Cover cc= 50 mm
Diameter of Bar db= 28 mm ∅ Rebar
db= 25 mm ∅ Rebar
Using 10 mm ∅ stirrups
Assumed no of layer/s n= 2 number of layer/s of bars
of tension bars
Assumed no of layer/s n= 1 number of layer/s of bars
of compression bars
d= 358 mm
ρbal = 0.85f'cβ1*600
fy(600+fy)
ρbal = 0.85*28*0.85*600
420(600+420)
ρbal = 0.0283333333
ρmax = 0.75ρbal
ρmax = 0.75*0.03
ρmax = 0.02125
use ρ= 0.02125
Mu = (ϕ)(fc')(ω)(b)(d2)(1 - 0.59ω)
Mu = (0.9)(28)(0.31875)(350)(358^2)(1 - 0.59*0.31875)
Mu = 292.55493737 kN-m
Mn = Mu/ϕ
Mn = 292.56/0.9
Mn = 325.06104152 kN-m
325.0610<1,110.0667
Therefore, design as doubly reinforced
IF DOUBLY
Check if Compression Steel Yields, when f's >fy
As1 = ρmaxbd *rho max is used not rho
As1 = 0.02125*350*358
As1 = 2662.625 mm2
Mn1 = Mn
Mn1 = 325.06104152 kN-m
Mn2 = Mn - Mn1
Mn2 = 1,110.0667-325.0610
Mn2 = 785.0056 kN-m
Mn2 = As2*fy*(d-d')
0.85*f'c*a*b = As1*fy
a= As1*fy = 2662.625*420
0.85*f'c*b 0.85*28*350
a= 134.25 mm
c= a/β1
c= 134.25/0.85
c= 157.94117647 mm
f's= 600(c-d')/c
f's= 600(157.941-50)/157.941
f's= 410.05586592
410.0559<420
Therefore, compression reinforcement does not yields
Vc= 0.17λ(√f'c)bwd λ= 1
Vc= 0.17*1*(√28)*350*358
Vc= 112.71429735 kN
ϕVc = 84.535723015 kN
Vu= 234.799 kN
Vn= Vu/ϕ
Vn= 234.799/0.75
Vn= 313.06533333 kN
Vs= Vn - V c
Vs= 313.07 - 112.71
Vs= 200.35103598 kN
0.66(√f'c)bwd= 0.66(√28)*350*358
= 437.597 kN
Vs<0.66(√f'c)bd; Okay!
smax= d/2
smax= 358/2
smax= 179 180
4
tension
compresion
use ρ= Err:502
S= Err:502
Err:502
S= Err:502 Err:502
Err:502
if avmin reqd
Solve for Spacing of Stirrups
Av= 2Ab = 2(π/4)*10^2
Av= 157.07963268
d= 387.5 mm
ρbal = 0.85f'cβ1*600
fy(600+fy)
ρbal = 0.85*28*0.85*600
420(600+420)
ρbal = 0.0283333333
ρmax = 0.75ρbal
ρmax = 0.75*0.03
ρmax = 0.02125
use ρ= 0.02125
Mu = (ϕ)(fc')(ω)(b)(d2)(1 - 0.59ω)
Mu = (0.9)(28)(0.31875)(350)(387.5^2)(1 - 0.59*0.31875)
Mu = 342.75578411 kN-m
Mn = Mu/ϕ
Mn = 342.76/0.9
Mn = 380.83976012 kN-m
380.8398<543.5589
Therefore, design as doubly reinforced
IF DOUBLY
Check if Compression Steel Yields, when f's >fy
As1 = ρmaxbd *rho max is used not rho
As1 = 0.02125*350*387.5
As1 = 2882.03125 mm2
Mn1 = Mn
Mn1 = 380.83976012 kN-m
Mn2 = Mn - Mn1
Mn2 = 543.5589-380.8398
Mn2 = 162.7191 kN-m
Mn2 = As2*fy*(d-d')
0.85*f'c*a*b = As1*fy
a= As1*fy = 2882.03125*420
0.85*f'c*b 0.85*28*350
a= 145.3125 mm
c= a/β1
c= 145.3125/0.85
c= 170.95588235 mm
f's= 600(c-d')/c
f's= 600(170.956-50)/170.956
f's= 424.51612903
424.5161>420
Therefore, compression reinforcement yields
Vc= 0.17λ(√f'c)bwd λ= 1
Vc= 0.17*1*(√28)*350*387.5
Vc= 122.00220733 kN
ϕVc = 91.501655499 kN
Vu= 0 kN
Vu<ϕVc/2
Therefore, beam does not need shear
reinforcements
If stirrups are reqd
Solve for the Shear Strength Provided by the Stirrup
Vn= Vu/ϕ
Vn= 0/0.75
Vn= 0 kN
Vs= Vn - Vc
Vs= 0.00 - 122.00
Vs= -122.0022073 kN
0.66(√f'c)bwd= 0.66(√28)*350*387.5
= 473.656 kN
Vs<0.66(√f'c)bd; Okay!
smax= d/2
smax= 387.5/2
smax= 193.75 190
-2
tension
compresion
use ρ= 0.02125
d= 389 mm
ρbal = 0.85*28*0.85*600
420(600+420)
ρbal = 0.0283333333
ρmax = 0.75ρbal
ρmax = 0.75*0.03
ρmax = 0.02125
use ρ= 0.02125
Mu = (ϕ)(fc')(ω)(b)(d2)(1 - 0.59ω)
Mu = (0.9)(28)(0.31875)(350)(389^2)(1 - 0.59*0.31875)
Mu = 345.41451326 kN-m
Mn = Mu/ϕ
Mn = 345.42/0.9
Mn = 383.79390362 kN-m
383.7939>0.0000
Therefore, design as singly reinforced
IF DOUBLY IF SINGLY
Check if Compression Steel Yields, when f's >fy Solve for Coefficient of Resista
As1 = ρmaxbd *rho max is used not rho M u=
As1 = 0.02125*350*389 Rn =
As1 = 2893.1875 mm2 Rn =
Mn1 = Mn Rn=
Mn1 = 383.79390362 kN-m
Solve for Steel Ratio
Mn2 = Mn - Mn1 ρ=
Mn2 = 0.0000-383.7939 ρ=
Mn2 = -383.7939 kN-m ρ=
ρmax =
Mn2 = As2*fy*(d-d')
use ρ=
As2 = Mn2 = -383.794
fy*(d-d') 420*(389-50) Solve for Tension Reinforcem
A s=
As2 = -2695.560497 mm2 A s=
A s=
0.85*f'c*a*b = As1*fy
a= As1*fy = 2893.1875*420 No. of Bars = As/Ab
0.85*f'c*b 0.85*28*350 As/Ab=
N=
a= 145.875 mm
c= a/β1
c= 145.875/0.85 Check for Spacing
c= 171.61764706 mm S=
f's= 600(c-d')/c
f's= 600(171.618-50)/171.618 S=
f's= 425.19280206
S=
425.1928>420
Therefore, compression reinforcement yields The proponents reco
reinforcement
Solve Area of Reinforcement
A's= As2
A's= -2695.560497
Vc= 0.17λ(√f'c)bwd λ= 1
Vc= 0.17*1*(√28)*350*389
Vc= 122.47447394 kN
ϕVc = 91.855855455 kN
Vu= 0 kN
Vs= Vn - Vc
Vs= 0.00 - 122.47 s=
Vs= -122.4744739 kN
use s=
0.66(√f'c)bwd= 0.66(√28)*350*389
= 475.489 kN
Vs<0.66(√f'c)bd; Okay!
-2
of bars between layers assumed to be equal to size of bar
or Coefficient of Resistance
ϕRnbd2
Mu/ϕbd2
0.00/0.9*350*389*^2
0 Mpa
or Steel Ratio
0.85(f'c/fy)(1-√(1-2Rn/0.85f'c))
0.85(28/420)(1-√(1-2*(0.000)/0.85*28))
0
0.0033333333
or Tension Reinforcement
ρbd Ab= πdb2/4
0.0033*350*389 Ab= π*22^2/4
453.83333333 mm2 Ab= 380.1327 mm2
for Spacing
b - 2cc - 2dbs - N(db)
N-1
530 mm
Sample Beam
Given
Mu= 0
Vu= 231.624
Concrete Strength f'c= 28 Mpa
Yield Strength of Steel fy= 420 Mpa
Reduction Factor ϕ= 0.9 (Flexure)
ϕ= 0.75 (Shear)
β1 = 0.85
Es= 200000 Mpa
Total Depth h= 450 mm
Beam Width b= 350 mm
for
Distance of extreme
doubly
fiber to centroid of d'= 50 mm reinforce
compresion rebar
d only
Clear Cover cc= 50 mm
Diameter of Bar db = 25 mm ∅ Rebar tension
db = 25 mm ∅ Rebar compresion
Using 10 mm ∅ stirrups
Assumed no of layer/s n= 1 number of layer/s of bars
of tension bars spacing of bars between layers a
Assumed no of layer/s n= 1 number of layer/s of bars
of compression bars
d= 387.5 mm
ρbal = 0.85*28*0.85*600
420(600+420)
ρbal = 0.0283333333
ρmax = 0.75ρbal
ρmax = 0.75*0.03
ρmax = 0.02125
use ρ= 0.02125
Mu = (ϕ)(fc')(ω)(b)(d2)(1 - 0.59ω)
Mu = (0.9)(28)(0.31875)(350)(387.5^2)(1 - 0.59*0.31875)
Mu = 342.75578411 kN-m
Mn = Mu/ϕ
Mn = 342.76/0.9
Mn = 380.83976012 kN-m
380.8398>0.0000
Therefore, design as singly reinforced
IF DOUBLY IF SINGLY
Check if Compression Steel Yields, when f's >fy Solve for Coefficient of Resista
As1 = ρmaxbd *rho max is used not rho M u=
As1 = 0.02125*350*387.5 Rn =
As1 = 2882.03125 mm2 Rn =
Mn1 = Mn Rn=
Mn1 = 380.83976012 kN-m
Solve for Steel Ratio
Mn2 = Mn - Mn1 ρ=
Mn2 = 0.0000-380.8398 ρ=
Mn2 = -380.8398 kN-m ρ=
ρmax =
Mn2 = As2*fy*(d-d')
use ρ=
As2 = Mn2 = -380.84
fy*(d-d') 420*(387.5-50) Solve for Tension Reinforcem
A s=
As2 = -2686.700248 mm2 A s=
A s=
0.85*f'c*a*b = As1*fy
a= As1*fy = 2882.03125*420 No. of Bars = As/Ab
0.85*f'c*b 0.85*28*350 As/Ab=
N=
a= 145.3125 mm
c= a/β1
c= 145.3125/0.85 Check for Spacing
c= 170.95588235 mm S=
f's= 600(c-d')/c
f's= 600(170.956-50)/170.956 S=
f's= 424.51612903
S=
424.5161>420
Therefore, compression reinforcement yields The proponents reco
reinforcement
Solve Area of Reinforcement
A's= As2
A's= -2686.700248
Vc= 0.17λ(√f'c)bwd λ= 1
Vc= 0.17*1*(√28)*350*387.5
Vc= 122.00220733 kN
ϕVc = 91.501655499 kN
Vu= 231.624 kN
Vs= Vn - Vc
Vs= 308.83 - 122.00 s=
Vs= 186.82979267 kN
use s=
0.66(√f'c)bwd= 0.66(√28)*350*387.5
= 473.656 kN
Vs<0.66(√f'c)bd; Okay!
3
of bars between layers assumed to be equal to size of bar
or Coefficient of Resistance
ϕRnbd2
Mu/ϕbd2
0.00/0.9*350*387.5*^2
0 Mpa
or Steel Ratio
0.85(f'c/fy)(1-√(1-2Rn/0.85f'c))
0.85(28/420)(1-√(1-2*(0.000)/0.85*28))
0
0.0033333333
or Tension Reinforcement
ρbd Ab= πdb2/4
0.0033*350*387.5 Ab= π*25^2/4
452.08333333 mm2 Ab= 490.8739 mm2
for Spacing
b - 2cc - 2dbs - N(db)
N-1
530 mm
Sample Beam
Given
Mu= 998.985
Vu= 231.624
Concrete Strength f'c= 28 Mpa
Yield Strength of Steel fy= 420 Mpa
Reduction Factor ϕ= 0.9 (Flexure)
ϕ= 0.75 (Shear)
β1 = 0.85
Es= 200000 Mpa
Total Depth h= 450 mm
Beam Width b= 350 mm
for
Distance of extreme
doubly
fiber to centroid of d'= 50 mm reinforce
compresion rebar
d only
Clear Cover cc= 50 mm
Diameter of Bar db = 28 mm ∅ Rebar tension
db = 25 mm ∅ Rebar compresion
Using 10 mm ∅ stirrups
Assumed no of layer/s n= 2 number of layer/s of bars
of tension bars spacing of bars between layers a
Assumed no of layer/s n= 1 number of layer/s of bars
of compression bars
d= 358 mm
ρbal = 0.85*28*0.85*600
420(600+420)
ρbal = 0.0283333333
ρmax = 0.75ρbal
ρmax = 0.75*0.03
ρmax = 0.02125
use ρ= 0.02125
Mu = (ϕ)(fc')(ω)(b)(d2)(1 - 0.59ω)
Mu = (0.9)(28)(0.31875)(350)(358^2)(1 - 0.59*0.31875)
Mu = 292.55493737 kN-m
Mn = Mu/ϕ
Mn = 292.56/0.9
Mn = 325.06104152 kN-m
325.0610<1,109.9833
Therefore, design as doubly reinforced
IF DOUBLY IF SINGLY
Check if Compression Steel Yields, when f's >fy Solve for Coefficient of Resista
As1 = ρmaxbd *rho max is used not rho M u=
As1 = 0.02125*350*358 Rn =
As1 = 2662.625 mm2 Rn =
Mn1 = Mn Rn=
Mn1 = 325.06104152 kN-m
Solve for Steel Ratio
Mn2 = Mn - Mn1 ρ=
Mn2 = 1,109.9833-325.0610 ρ=
Mn2 = 784.9223 kN-m ρ=
ρmax =
Mn2 = As2*fy*(d-d')
use ρ=
As2 = Mn2 = 784.9223
fy*(d-d') 420*(358-50) Solve for Tension Reinforcem
A s=
As2 = 6067.7357128 mm2 A s=
A s=
0.85*f'c*a*b = As1*fy
a= As1*fy = 2662.625*420 No. of Bars = As/Ab
0.85*f'c*b 0.85*28*350 As/Ab=
N=
a= 134.25 mm
c= a/β1
c= 134.25/0.85 Check for Spacing
c= 157.94117647 mm S=
f's= 600(c-d')/c
f's= 600(157.941-50)/157.941 S=
f's= 410.05586592
S=
410.0559<420
Therefore, compression reinforcement does not yields
Err:502
Solve Area of Reinforcement
A's= As2 (fy/f's)
A's= 6214.8824372
Vc= 0.17λ(√f'c)bwd λ= 1
Vc= 0.17*1*(√28)*350*358
Vc= 112.71429735 kN
ϕVc = 84.535723015 kN
Vu= 231.624 kN
Vs= Vn - Vc
Vs= 308.83 - 112.71 s=
Vs= 196.11770265 kN
use s=
0.66(√f'c)bwd= 0.66(√28)*350*358
= 437.597 kN
Vs<0.66(√f'c)bd; Okay!
3
of bars between layers assumed to be equal to size of bar
or Coefficient of Resistance
ϕRnbd2
Mu/ϕbd2
998.99/0.9*350*358*^2
24.744709531 Mpa
or Steel Ratio
0.85(f'c/fy)(1-√(1-2Rn/0.85f'c))
0.85(28/420)(1-√(1-2*(24.745)/0.85*28))
Err:502
Err:502
or Tension Reinforcement
ρbd Ab= πdb2/4
Err:502 Ab= π*28^2/4
Err:502 mm2 Ab= 615.7522 mm2
for Spacing
b - 2cc - 2dbs - N(db)
N-1
Err:502
Err:502
Err:502 Err:502
Err:502
or Spacing of Stirrups
2Ab = 2(π/4)*10^2
157.07963268
530 mm