what"s the formula to find the quantity (consumption) of
cement,sand,aggregate for 1m3(1:2:4)?
Answers were Sorted based on User's Feedback
Answer / prakash kale
for 1m3 conctre --
Take a 52% added dry surface of concrete
1+.52 = 1.52
i.e.
For 1.52 cu.m. concrete
Then
1.52/1+2+4=1.52/7
=.21714
=.22
so.22/.0354 =6.2 bags(1 bag = .0354 cu.m cement)
6.2 bags/cu.m in 1:2:4
Sand--
.22 x 2 =.44 cu.m.
Is This Answer Correct ? | 118 Yes | 13 No |
Answer / mrinal kumar brahma
Mix ratio 1(cement):2(sand):4(aggregate)
For 1m3 concrete:
1.51(constant)/(Sum of ratio, i.e 1+2+4=7)=0.216
For cement= 0.216x28.57(constant)=6.2 bag
For sand=0.216x2(sand ratio)=0.43 m3
For aggregate=0.216x4(aggregate ratio)=0.86 m3
Is This Answer Correct ? | 60 Yes | 10 No |
Answer / vishal rana
Ratio=1:2:4(cement:fine
agg.:coarse agg.)
Sum of the ratio=1+2+4=7.
Total dry mortar for 1cum
cement conc.=1.54(it is fixed)
There for=1.54/7=.22cum
Then quantiy of,cement=.22 into
28.8=6.24 bags.
Sand=.22 into 2(ratio of
sand)=.44cu m.
Aggregate(shingle or bajri)=.22
into 4=.88 cu m.
Materialr for 1cum cement conc.
1:2:4.
Cement=6.24bags
Sand= 0.44cum
Agg.=0.88cum
Is This Answer Correct ? | 47 Yes | 3 No |
Answer / ajay kumar pal.(j.e)
for finding the total value of
take 1.52/(1+2+4)
=.22
hence total qty of cement .22*1= .22cum
cement= .22/.0354 = 6.2 bags of cement
sand = .22*2 = .44cum
Aggr =.22*4 =.88cum
ANS.
Is This Answer Correct ? | 17 Yes | 6 No |
Answer / ranjithkumar.t
M15= 1:2:4 (for 1 m^3)
1m^3 = 1440kg, 1 bag = 50 kg
so, 1m^3 bags=1440/50
= 28.8 bags
Cement = 1/7*1.52*28.8
= 6.25 bags
Sand = 2/7*1.52
= 0.434 cubic meter
Aggregate = 4/7*1.52
= 0.868 cubic meter
Is This Answer Correct ? | 6 Yes | 0 No |
Answer / bhupendra potpose
Consumption & Rate analysis RCC(Cement Concrete) 1:2:4
Providing and placing cement concrete excluding formwork &
reinforcement
1.475 Ref: Always use 1..475 standard
Proportion of materials
(1440 Kg / Cum ) Cement 1
(Dry 1600 Kg / Cum & Wet 1850 Kg /Cum ) Sand
2
(1650 Kg / Cum ) Metal 4
Total 7
Std constant(1.5) / Total of proporttion(7)
0.211
1) Required volume of cement for 1 cum
1 Proportion of cement 0.211
Add wastage 2.50% 0.005
Vol of cement for 1 cum 0.216
0.035 is cum qty of per 50kg of bags, no of bags
0.035 6.171
Cost of cement per bags 325
Total of -1, cost of bags 2005.5
2) Required volume of Sand for 1 cum
2 Proportionof Sand 0.421
Add wastage 2.50% 0.011
Vol of Sand for 1 cum 0.432
Cost of sand per Cum 2125
Total of -2, cost of sand 917.9
3) Required volume of Aggregate for 1 cum
3 Proportionof Metal 0.843
Add wastage 2.50% 0.021
Vol of Metal for 1 cum 0.864
Cost of Metal per cum 975
Total of -3, cost of metal 842.3
A- Total of (1+2+3), cost of Material (Concrete)
3765.8
4)Cost for Manpower ,Mixing ,Shifting & Machinary ,
0.35%
B-Total of -4, Cost of Labour per cum (35%)
13.2
C-Total of (A + B) cost of Material & Labour
3779.0
5) Add Over head if any 5% 188.95
6) Profit 15% 595.19
D- Total of (A+B+C)cost P/A 1:2:4 Concreting
4563.12
Is This Answer Correct ? | 8 Yes | 4 No |
Answer / gilbertrufino
calculate first the total volume of concrete
1:2:4
for cement = vol. X 9 = ____ bags of 40kg portland cement
for sannd = vol. X 0.5 = ____ m^3
for gravel = vol. X 1.0 = ____ m^3
Is This Answer Correct ? | 11 Yes | 10 No |
Answer / vigneshwaran s
Generally its easy to find it.
Find Total Parts = (1+2+4)=7 Parts.
For 1 cu.m of concrete the total parts = 1.57
Unit weight of cement = 1440 kgs
Cement = (1440/7)*1.57 = 322.97/50kgs = 6.46bags.
1 bag cement= 1.25cft
Sand = 6.46*1.25 = 8.075 cft*2 =16.15 cft
Aggregate = 8.075 cft*4 =32.3 cft.
Thats all...
Result:
Cement = 6.46 bags
Sand = 16.15 cft
Aggregate = 32.3 cft.
Regards,
Er.Vigneshwaran S
Is This Answer Correct ? | 5 Yes | 9 No |
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