Answer / dipak rakibe

TO DERIVE THE QUANTITY OF INGREADIENT MATERIAL FOR 1 CUBIC METER OF CONCRETE FOR M-15 GRADE

Proportion for M-15 Concrete is 1:2:4

For Easy Calculations Take/Assume 10 Cu. m. of concrete.

Proportion
10 X 1.52 = 15.2

(Here 1.52 is the dry volume factor)

A) Quantity of Cement =
For 10 cum of concrete
(1/ 1+2+4)= 0.143 X 15.2 = 2.17 Cum

1 Bag of Cement = 35 Liters = 0.035

Quantity of Cement Bags = 2.17/.035 = 62.10 Bags

for 1 cum of concrete = 6.21 Bags

B) Quantity of Fine Aggregates/Sand
For 10 cum of concrete
(2/ 1+2+4)=0.285 X 15.2=4.33 Cum

Quantity of Sand = 4.33 Cum

for 1 cum of concrete = 0.43 Cum

C) Quantity of Coarse Aggregates/ Gitti
For 10 cum of concrete
(4/ 1+2+4)= 0.571 X 15.2 = 8.68 Cum

Quantity of Gitti = 8.68 Cum

for 1 cum of concrete = 0.87 Cum

Summary =
A) Cement Bags= 6.210 Bags
B) Fine Agg./Sand=0.43 Cum
C) Coarse Agg./Gitti = 0.87 Cum
BY Dipak Rakibe

Answer / khan9219362784

1 cum of 1:1:4 concrete 
Dry volume equal is 1.52 
Cement equal is 1/6*28.8*1.52 =7.2 bags cements
        = 364.8 kg cements 

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