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.

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

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

Answer / viplav

volume.

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.

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

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

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

Answer / peer mohd

90% aggregate
45% sand
25% cement

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

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