Answer / vikas p

Consider M20 (1:1.5:3) Grade Concrete

A) Suppose Total Volume of concrete= 1 Cum.
B) Add 55% volume for dry mortar=0.55 Cum.
C) Total wet volume of concrete A+B = 1.55 Cum.
D) Cement = 1.55/ (1+1.5+3) = 0.18 Cum. = (0.18/0.035=5.14 Bags)
E) Therefore Fine aggregate = 0.18 cum X 1.5 = 0.27 Cum.
F) Therefore Coarse aggregate = 0.18 cum X 3 = 0.54 Cum.

Answer / amit

For cement=0.158* required volume in cum/ratio factor (eg m20=1+1.5+3)
= answer/.035 which gives you no of bag
For sand=1st cement ratio answer* 1.5 which gives you cum of sand
Same for aggregate

Answer / nilesh kulkarni

If built up area of the building is known to you ,
You can calculate the qty as per constants for any ratio

Concrete Qty:- 0.03 Cum/Sqft

Cement For :- RCC Work = 0.19 Bag/Sqft
Masonary Work = 0.08 Bag/Sqft
Plastering Work = 0.15 Bag/Sqft
Other Work = 0.03 Bag/Sqft
---Total Cement--------- 0.45 Bag/Sqft----------------------------
-------------------------------------------
Sand For :- RCC Work = 0.75 Cft/Sqft
Masonary Work = 0.36 Cft/Sqft
Int. Plastering Work = 1.00 Cft/Sqft
Ext. Plastering Work = 0.65 Cft/Sqft
Tiling Work = 0.13 Cft/Sqft
W/p Work = 0.08 Cft/Sqft
Other Work = 0.02 Cft/Sqft
Total Sand---------------- 3.00 Cft/Sqft----------------------------
-------------------------------------------

Answer / nadim shaikh

for any quantity firstly you will need drawing of RCC members and architectural drawing
secondary you should know nominal mix ratio

WAP to print 1) 1 1 1 1 1 1 1 1 1 1 2) 1 2 3 4 5 6 7 8 9 10

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