Answer / prabu

1 feet = 0.3048 meter
30'*40'*4' = 9.144 *12.192 * 1.2192 meter
total volume = 135.92 meter cubic

1 bag cement = 50 kgs
unit wt of cement = 1440 kg/m3

density = weight/volume = 50/1440 = 0.0347m3

if concrete grade is M20 and ratio is 1:1:2
1+1+2 = 4

cement = 135.92/4 = 33.98
33.98/0.0347 = 979.25 bags
1 cement bag = 50 kgs, so
979.25 bag = 979.25 * 50 = 48962.53 kgs

sand = 1* 33.98 = 33.98 m3
aggregate = 2*33.98 = 67.96 m3

for slab = 0.7 -1% of concrete volume (reinforcement provided )
for beam = 1 - 3%
for column = 1 - 5 %
for footing = 0.8 - 8%

in this problem = 135.92 m3 concrete
for slab = 135.92 * 0.007 = 0.95m3 volume reinforcement provided.

ans =
cement = 48962.53 kgs or 979.25 bags
sand = 33.98 m3
aggregate = 67.96 m3
for slab = 0.95m3 volume reinforcement provide

In IS:456-2000 the workability of concrete is given in mm (slump)in cl.7.1. In IS:10262-1982, cl.3.3.4 & table-6 says +/- 3 percent adjustment is required in the quantity of water (calculated as per table-4 or 5), for each increase or decrease in the value of compacting factor (i.e. workability of concrete) by 0.1. How it is possible to calculate the exact quantity of water required for a particular grade of concrete? What is the relation between Slump & Compacting factor of concrete?

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