Question

1.A reinforced concrete beam having a width of 500 mm and an effective depth of 750 mm is reinforced with 5 – 25mm φ. The beam has simple span of 10 m. It carries an ultimate uniform load of 50 KN/m. Use f’c = 28 MPa, and fy = 413 MPa. Determine the value of fs in MPa. Express your answer in two decimal places.

2.A reinforced concrete beam having a width of 500 mm and an effective depth of 750 mm is reinforced with 5 – 25mm φ. The beam has simple span of 10 m. It carries an ultimate uniform load of 50 KN/m. Use f’c = 28 MPa, and fy = 413 MPa. Determine the value of fs in MPa. Express your answer in two decimal places.

Answer 1

Solution:- the values given in the question are as follows:

width of beam(b)=500 mm

effective depth of beam(d)=750 mm

reinforcement=5-25 mm

span of simply suuported beam(L)=10 m

ultimate uniform load(Wu)=50 kN/m

characteristic strength of concrete(fc)=28 MPa

yield strength of steel(fy)=413 MPa

actual stress in steel(fc)=?

area of reinforcement(Ast)=number of bar*area of one bar

area of reinforcement(Ast)=5*[(Pi/4)*25^2]

area of reinforcement(Ast)=5*[(3.14/4)*25^2]

area of reinforcement(Ast)=2454.3692 mm^2

let bending moment due to ultimate losd is M

bending moment(M)=(Wu*L^2)/8

bending moment(M)=(50*10^2)/8

bending moment(M)=625 kN-m

bending moment(M)=625*10^6 N-mm

According to ACI-

Mu=fy*Ast*(d-a/2)

where, Mu=ultimate moment of resistance

[note:- in this case stress developed in reinforcement is equal to yield strength of steel]

a=(Ast*fy)/(0.85*fc*b)

a=(2454.3692*413)/(0.85*28*500)

a=85.18 mm

for bending moment(M)-

M=Ast*fs*(d-a/2) , [Eq-1]

where, fs=actual stress in beam due to udl on beam

M=bending moment due to udl on beam

[note:- in this case stress developed in reinforcement is less than the yield strength of steel]

values put in above equation-(1)

625*10^6=2454.3692*fs(750-85.18/2)

625*10^6=1736245.316*fs

fs=(625*10^6)/1736245.316

fs=359.97 MPa

actual stress in steel due to uniform load(fs)=359.97 MPa

[Ans]