Answer / benakappa

sir pls tel me wts thge value of 1.52 while calculating the ingredients of 1 m 3 concrete pls explain me...

Answer / jj r

For M-15- 1:2:4 you just add the all ratio
1+2+4=7 then for 1 m3(cement=1/7,F.A=2/7,C.A=4/7) in CUM...
tats all

Answer / mufaddal hamdard

1.5m3 of dry material is required to get 1m3 of concrete.
and 2% or 5% is the amount of waste taken in account

1.5 + 0.02 is 1.52
or
1.5 + 0.05 is 1.55 thats it.

and now how 1.5 is derived is the question
As it is considered that when materials are mix in mixer the voids are finished after mixing and the output is 67% of the input in mixer
so
1/.67 is 1.5

Answer / karthiknitw2004

I heard DLBD method is more accurate that using the factor of 1.54 to convert wet volume to dry volume. In DLBD method, To calculate dry ingredients volume in 1:2:4 concrete

Step-1: Calculate Volume of materials required
Density of Cement = 1440 kg/cum (Approx)
Volume of 1 Kg of Cement = 1/1440 = 0.00694 cum
Volume of 01 bag (50 kg) of cement = 50 X 0.00694 = 0.035 cubic meter (cum)
Since we know the ratio of cement to sand (1:2) and cement to aggregate (1:4)
Volume of Sand required would be = 0.035*2 = 0.07 cubic meter (cum)
Volume of Aggregate required would be = 0.035*4 = 0.14 cubic meter (cum)


Step-2: Convert Volume requirement to weights
To convert Sand volume into weight we assume, we need the dry loose bulk density (DLBD). This density for practical purposes has to be determined at site for arriving at the exact quantities. We can also assume the following dry loose bulk densities for calculation.
DLBD of Sand = 1600 kgs/cum
DLBD of Aggregate = 1450 Kgs/Cum
So, Sand required = 0.072*1600 = 115 kgs
and Aggregate required = 0.144*1450 = 209 kgs
Considering water/cement (W/C) ratio of 0.55
We can also arrive at the Water required = 50*0.55 = 27.5 kg
So, One bag of cement (50 Kgs) has to be mixed with 115 kgs of Sand, 209 Kgs of aggregate and 27.5 kgs of water to produce M20 grade concrete.

Cement Sand Aggregate Water
1 bag (50kg) 115 Kgs 209 Kgs 27.5 Kgs


Step-3: Calculate Material requirement for producing 1 cum Concrete
From the above calculation, we have already got the weights of individual ingredients in concrete.
So, the weight of concrete produced with 1 Bag of cement (50 Kgs) =50 kg + 115 kg + 209 kg + 27.5 kg = 401.5 kg ~ 400 kgs
Considering concrete density = 2400 kg/cum,
One bag of cement and other ingredients can produce = 400/2400 = 0.167 Cum of concrete (1:2:4)
01 bag cement yield = 0.167 cum concrete with a proportion of 1:2:4
01 cum of concrete will require
Cement required = 1/0.167 = 5.98 Bags ~ 6 Bags
Sand required = 115/0.167 = 688 Kgs or 14.98 cft
Aggregate required = 209/0.167 = 1251 kgs or 29.96 cft

Procedure Posted on http://www.happho.com/

Answer / munna kumar das

suppose concrete mix ratio=1:1.5:3
and dry material required for 1cum of concrete =1.54
Let 1x+1.5x+3x=1.54
x=0.28
cement=0.28*1=0.28cum=0.28*1440/50=8.06bags

sand=0.28*1.5=0.42cum=0.42*1500=630Kg
coarse agg=0.28*3=0.84cum=0.84*1550=1302kg
add 6 to 8% extra material.
This process very easy to understand.

Answer / mohan venkatesh

PLAIN CEMENT CONCRETE IN EASILY FIND INGREDIENT QUANTITY OF 1M3 CONCRETE

Answer / abdul marig

plzzz tell me what is the reason behind using this constant factor 1.54 or 1.52 and where it comes from???

Answer / arbab_91

simple formula 1:2:4 ration is

square foot / 4.46 x 1 = cement
square foot / 4.46 x 2 = sand
square foot / 4.46 x 4 = gravel, aggregat

Answer / arbab_91

M 20 Ratio 1:1.5:3

square feet / 3.62 x 1 = cement
square feet / 3.62 x 1.5 = sand
square feet / 3.62 x 3 = aggregate

Answer / ranjith679@gmail.com

Ref: M20=1:1.5:3
Cemet = 1/5.5*1.52*28.8= 7.95 say 8 Bags
Sand = 1.5/5.5*1.52= 0.414cum
Aggregate= 3/5.5*1.52=0.829cum

28.8 means 1 cum unit bags of concrete

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