For a beam of span 5m and cross section 230mm*500mm , is reinforced with 2-12Φ at top and 4-16 Φ at bottom out of which two bars are bent up and stirrups 8 Φ @ 150c/c. Calculate the quantity and cost of Materials.(Steel, cement, sand , aggregates).

Assume clear cover of 40mm

1) Length of top straight bars

L= Length of beam – 2×cover + (2×9×d)

=5000 - (2×40) + (2×9×12)

=5135m

2) Length of bottom straight bars

L= Length of beam – 2×cover + (2×9×d)

=5000 - (2×40) + (2×9×16)

=55208m

3) Length of bent up bars

L= Length of beam – 2×cover + (2×9×d) + [No.of bent up×0.42D]

L= 5000 – 2×40 + (2×9×16) +[2×0.42×420]

=5555.76m

4) Length of Stirrups

L= {[2×(depth – 2×cover)]+ [2×(depth – 2×cover)]}×24d

L= {[2×(500 – 2×40)]+ [2×(230 – 2×40)]}×24×8

L=1332m

No. of Stirrups $=\frac{L-2*cover}{spacing}+1 \\ =\frac{1332-2*40}{150}+1 \\ =9.34m =10 no.s (say)$

Total weight = 48.88kgs = 49kgs (Say)

To find out quantities

Assume M20 grade (1:1.5:3)

1) Concrete Quantity = 50.230.5=0.575m³

2) Consumptive steel $=\frac{\text{Total weight}}{\text{concrete quantity}}=\frac{48.88}{0.575} \\ =85.02 kg/m³$