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Example 2: Design of rectangular beam(SI units)

Design a rectangular beam for a 10-m simple span to support a dead load of 20 kN/m (not including beam weight) and a live load of 30 kN/m. Use ρ = 0.01415, fc'= 28MPa, and fy= 420 MPa, and concrete weight is 23.5 kN/m3. Do not use the ACI thickness limitation.






assuming beam weight=10/kn/m
dead load for beam=10+20=30kn
Wu=1.2*dl+1.6*LL=1.2*30+1.6*30=84kn/m
Mu=(Wu*L^2)/8
Mu=(84*10^2)/8=1,050kn-m=10,050,000,000n-mm


Mu/(b*d^2)=Ф*ρ*fy(1-ρ*fy/(1.7*fc'))
b*d^2= Mu/(Ф*ρ*fy(1-ρ*fy/(1.7*fc')))/Mu
assuming Ф=0.9



b*d^2=10,050,000,000/0.9*0.01415*420*(1-.01415*420/(1.7*28))
b*d^2=2.243x10^8=   400 mm × 749 mm
                                                   450 mm × 706 mm                                                    500 mm × 670 mm

selecting 500 x 800 (d=680mm)

Beam weight=0.5*0.8*23.5=9.4 kn/m this less than assumed so ok

As= ρ *b*d=0.01415*500*670=4811mm2

use 6#32 in two rows, As=4914mm2

now checking if Ф=0.9 is valid

a=As*fy/(0.85*fc'*b)
a=4914*420/(0.85*28*500)=173.43 mm
β1=0.85 for fc'=28 Mpa
C=a/β1=204 mm

εt=(0.003*d/c)-0.003=0.007>0.005    ok


checking Ф*Mn>Mu
Ф*Mn=Ф*As*fy*(d-a/2)
Ф*Mn=0.90*4914*420*(680-174.43/2)=1,101,093,395.22>Mu   ok




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