Answer / ronak shah

first of all i m not understand 9*9.
so i consider 1m3
52% more shrinkage and wastage.
1.52m3
cement= 1.52/(1+4+8)=0.1169/(cement bag volume in m3)=
0.1169/0.0348= 3.35 bags so 168 kg/m3.

sand=1.52/(1+4+8)=0.1169*4=0.47m3

aggregate=1.52/(1+4+8)=0.1169*8=0.94m3

Answer / sravan875

Cement =182.3 Kgs

Fine aggregate =182.3*4=729.2Kgs

COarse aggregate=182.3*8=1458.4Kgs

Answer / sanjeev kumar singh

Dear sir
1cum cement concrete me material quantity
.9 cum.aggregate
.45 cum. Sand
.12 cum. Cement
Right answer kya hai

Answer / sravan875

As per ratio 1:4:8
1=cement
4=fine aggregate
8=Coarse

Cement=182.3 Kgs

Fine aggregate =0.3076924*1750=538.46 Kgs

Coarse Aggregate =0.6153848*1950=1200 Kgs

Answer / sravan875

1:4:8 Ratio means 1%cement,4%aggregate,8%sand.
cement percentage--1/13=0.0769231cum.
aggregate %---4*0.0769...=0.3076924cum.
Sand % ----8*0.0169...=0.6153848cum..

Cement=0.0769*2370kg/cum (unit weight of concrete)=182.30775kg

40mm Aggregate=(density 1950kg/cum).
aggregate needed=0.3076924*1950=600.00kg of aggregate

Sand density(dry,packed)---1600-1900kg/cum
sand(dry,packed)-0.6153848*1750=1076.9234 Kg of sand is required.

Cement=182.3 Kgs
Aggregate=600Kgs
Sand=1077 Kgs

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