Answer / deepak

M20 = 1:1.5:3
Volume = 1+1.5+3=5.5
Total volume ingredients for using =1.57
Volume of broken stone Require = (3/5.5) x 1.57 = 0.856 m3
Volume of sand Require = (1.5/5.5) x 1.57 = 0.471 m3
Volume of cement = (1/5.5) x 1.57 = 0.285 m3
= 0.285 x1440 = 411 kg

For 1m3 of M20 (1:1.5:3)
Broken stone = 0.856 m3
Sand = 0.472 m3
Cement = 8.22 bag

Answer / sunil notiyal

M20 = 1:1.5:3
Volume = 1+1.5+3=5.5
Total volume ingredients for using =1.57
Volume of broken stone Require = (3/5.5) x 1.57 = 0.856 m3
Volume of sand Require = (1.5/5.5) x 1.57 = 0.471 m3
Volume of cement = (1/5.5) x 1.57 = 0.285 m3
= 0.285 x1440 = 411 kg

For 1m3 of M20 (1:1.5:3)
Broken stone = 0.856 m3
Sand = 0.472 m3
Cement = 8.22 bag

Answer / devender singh

For M20 Grade concrete the nominal mix is 1:1.5:3.
taking bulkage factor as 1.54,
volume of cement=(1.54*1/5.5)=.28cubic metre=.28*1440=403.2kg=8.06 bags
volume of sand=(1.54*1.5/5.5)=.42 cubic metre=14.83 cft
volume of coarse aggregate=1.54*3/5.5=.84 cubic metre=29.66 cft.

Answer / guest

cement used 1/5.5 sand
1.5/5.5 metal 3/5.5

Answer / mallikarjun

1.52/(1+1.5+3)=.276/.034=8.11bags of cement
volume of sand =(1.52x1.5/5.5)=0.414cum
volume of course aggregate=1.52x3/5.5=0.829 cum

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