Question

Fixed load bending moment MD = 130 kN·m, live load bending moment ML = 140 kNNm
Rectangular short beams with width b = 400 mm and depth h = 600 mm are designed as follows:
Determine design suitability. Assume the coating thickness is 40 mm.(f ck = 27 MPa, f y = 300 MPa , tensile reinforcement 6-D22, shear reinforcement D10)

Answer 1

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

dead load bending moment(MD)=130 kN-m

live load bending moment(ML)=140 kN-m

width of beam(b)=400 mm

depth of beam(h)=600 mm

clear cover(coating) thickness=40 mm

characteristic strength of concrete(fck)=27 MPa

yield strength of steel(fy)=300 MPa

tensile reinforcement(Ast)=6D22

diameter of design shear bar=10 mm

effective depth(d)=h-effective cover

effective depth(d)=600-40-22/2

effective depth(d)=549 mm

Check suitability of design:-

total bending moment on beam(M)=dead load bending moment(MD)+live load bending moment(ML)

total bending moment on beam(M)=130+140

total bending moment on beam(M)=170 kN-m

design of beam is safe when the moment of resistance(Mr) is greater than the total bending moment on beam.

According to ACI-

moment of resistance of beam(Mr)=0.85*fck*b*a*(d-a/2)

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

area of steel(Ast)=6*(3.14/4)*22^2

area of steel(Ast)=2280.796 mm^2

a=2280.796*300/(0.85*27*400)

a=74.535 mm

b=400 mm

d=549 mm

values put in above equation-(1) and calculate the moment of resistance

moment of resistance of beam(Mr)=0.85*27*400*74.535*(549-74.535/2)

moment of resistance of beam(Mr)=350143393.7 N-mm

moment of resistance of beam(Mr)=350.1433 kN-m

moment of resistance of beam(Mr) total bending moment on beam(M) , [so design is safe and good.]

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