9*8m slab 1:2:4 ratio use for M20 mix how much cent is
require it will become in cum but i want bags.
Answers were Sorted based on User's Feedback
Answer / k. bu.babu
basically slab size given;9*8
what is thickness of slab?
O.k Assume thick:0.100 m
volume of concrete=9*8*.1=7.2 Cu.m
dry materiels coeff. =1.52
M20 mix 1:2:4
cement=1.52*(1/(1+2+4))*1440=312.68 kgs.=6.25 bags
total bags required to volume=7.20cum*6.25babgs=45bags
| Is This Answer Correct ? | 44 Yes | 0 No |
Answer / harathi nitin
9m x 8m x 0.1m = 7.2cum
1.52/(1+2+4) = 0.217cum
sand = 0.217 x 2 = 0.434cum
agg = 0.217 x 4 = 0.868 cum
cement = 0.217 x 30 = 6.51bags
for 1cum = 6.51 bags
for 7.2cum = 6.51 x 7.2 = 47 bags
| Is This Answer Correct ? | 4 Yes | 0 No |
Answer / sandeep bhardwaj
The Nominal mix for M20 is 1:1.5:3 not 1:2:4,all above answers
are correct,but you shoudd have to calculate by this ratio.
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / sahilshah.civil@gmail.com
Question was wrong M 20 have nominal mix proportion (1:1.5:3) not mix proportion (1:2:4)
For Mix M 20
Qty of Slab 7.2 Cu m after consider slab thickness 0.1 m
Cement : 57 Bag
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / sundarapandiyan
Bulk material qty for one cum conc=1.52 cum(using 20mm and lesser size aggrigate)
Concrete ratio 1:2:4 = 7 part
cement =1 part
for 9x8m = 72 sqm
assume thickness as =100mm
so cc qty=9X8x0.1=7.2 cum
cement= (7.2x1x1.52)/(1+2+4)=1.563 cum
bulk cement density= 1440 kg ( for clarification cement density mean specific gravity =3.15 mean 3150kg /cum)
so cement in kg=1440X1.563 =2250.72kg
cement in bag= 2250.72/50=45.014 bag can assume as 45 bag.
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / johnmiya shaik
M.20 mix(1:2:4)
1 2 4=6
1.52/6=0.253333
(Cement wet density to dry density ratio)
(One cement bag=0.035cum/1.225 cft)
0.253333/0.035=7.238095
Slab area=72sqm
Assume thickness=100mm
So concrete qty=9x8x.1=7.2cum
Cement bags=7.2 cumx7.238095=52.11 bags/2605.714kg
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / binubal123
Volume of Conc needed=9*8*0.01(assume)
=7.2Cum
M20 mix (1:2:4)
material
cement by vol=7.2*1/(1+2+4)
=1.028
Sand By Vol=7.2*2*1/(1+2+4){2*1.028}
=2.05
Agg By vol~4*1.028
=4.114
Cement in bag req=1.028*308.53(Cement Constant frm Practical Info for Quantity Surveyor P.T.Joglekar) =317.34Kg=317.34/50=6.36bags
| Is This Answer Correct ? | 1 Yes | 5 No |
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