Answer / himanth kumar

Consider for 1CUM of concrete is to be prepare,For that

M15 is a nominal mix, its proportion is 1:2:4

for 1cum of final mix(Dence) is required, 1.50 to 1.55 Cum
of loose ingredients with the required proportion we want
to add.

Based on that take avg of 1.50 & 1.55 is 1.52

Total proportion = 1+2+4 = 7

For Cement Qty = 1.52/7=0.217 Cum of cement required to
prepare 1Cum of M15 concrete.But cement is always taken in
Weight.For that the unit weight of cement is 1440 KG/Cum
Total cement required = 0.217x1440= 312.68 Kgs.

Sand = 0.217 x 2 =0.434 Cum

Metal or Aggregate = 0.217 x 4 =0.868 Cum.

Answer / adishankar (quantity surve

DETAILS:

cement = 1,
sand = 2
metal = 4
so wat is the ratio we want mixing (1:2:4)=1+2+4=7
7 is mixing propotion
volume of wet cement conceret is 1.54 to 1.57

unit weigth of cement is 1440kg/cum

one bag of cement =50/1440=0.034722 cum

solution : cement = 1.54/7=0.22
0.22/0.03472=6.33 bags
one bag of cement = 50kgs
=6.33*50kgs=316.82 kgs

solution of sand = 0.22*2=0.44cum

2 is the mixing of propotion

soolution of metal = 0.22*4=0.88cum

4 is the mixing of propotion

say i cum of cement conceret 1:2:4 ratio

cement = 6.33 bags

sand = 0.46 cums

metal = 0.88 cums


this is one type of solution

unit weight of cement is 1440kg/cum
sand 1660 to 1800kg/cum
Unit weight of metal is 1630kg to 2000kg/cum

so cenment = 1.54/7=0.22
0.22*1440=316.8 kgs/cum
sand = 0.22*1750=385kgs/cum

metal = 0.22*2000=440kgs/cum

Answer / pinkal

Here volume is (1 cement bag=1.25 cu.ft)
cement 1 1.25 cu.ft
sand 2 2.50 cu.ft
metal 4 5.00 cu.ft
-----------
8.75 cu.ft total volume

Answer / raja babu.k

Sample = 1 M3 of M15 i.e 1:2:4

Aggregate = 20mm

Final Mixture and Loose Aggregate Ratio is 1.57(in practies)

SO CALCULATION FOR 1:2:4

1+2+4=7

CEMENT = (1.57/7)X1=0.2242 M3X1440=322.8Kg= 6.45 Bags
RIVER SAND= (1.57/7)X2=0.4485 M3X35.314 = 15.83 Cft
AGGREGATE = (1.57/7)X4=0.8971 M3X35.314 = 31.68 Cft

Answer / sjtbehera

The ratio is 1:2:4, so total ratio is 1+2+4=7
considering 1cum.
cement = (1 x 15.4)/7= 2.2 cum
= 2.2/.347=6.3bags
sand=2.2x2=4.4cum
metal=2.2x4=8.8cum

Answer / arun

DETAILS:

cement = 1,
sand = 2
metal = 4
so wat is the ratio we want mixing (1:2:4)=1+2+4=7
7 is mixing propotion
volume of wet cement conceret is 1.54 to 1.57

unit weigth of cement is 1440kg/cum

one bag of cement =50/1440=0.034722 cum

solution : cement = 1.54/7=0.22
0.22/0.03472=6.33 bags
one bag of cement = 50kgs
=6.33*50kgs=316.82 kgs

solution of sand = 0.22*2=0.44cum

2 is the mixing of propotion

soolution of metal = 0.22*4=0.88cum

4 is the mixing of propotion

say i cum of cement conceret 1:2:4 ratio

cement = 6.33 bags

sand = 0.46 cums

metal = 0.88 cums


this is one type of solution

unit weight of cement is 1440kg/cum
sand 1660 to 1800kg/cum
Unit weight of metal is 1630kg to 2000kg/cum

so cenment = 1.54/7=0.22
0.22*1440=316.82 kgs/cum
sand = 0.44*1750=770kgs/cum

metal = 0.88*2000=1760kgs/cum

Answer / sandeepthakur1981

for 01 cu.m.

cement=1.52/7 =.217
= .217x30= 6.51 bags.

sand= .217x2= 0.434cu.m

aggregate= .217x4= 0.868cu.m

Answer / the engineer

In order to estimate the quantity of loose materials to have
100 cubic feet wet concrete we first multiply the wet
concrete by 1.54.
e.g 100x1.54=154 cubic feet of dry materials we need.

now come to the ratio of concrete which is 1:2:4 so sum of
the ratio will be 1+2+4=7

calculating separately each material:

Cement: 1/7x154 = 22 cubic feet
where 1 bag of cement measured 1.25 cubic feet approx
therefore 22/1.25 = 17.6 bags of cement. 1 bag contains 50kg
cement so total cement by weight we need for 100 cubic feet
concrete = 17.6x50 = 880kgs

sand: 2/7x154 = 44 cubic feet

coarse aggregate: 4/7x154 = 88 cubic feet

Answer / m

ratio is 1:2:4
1+2+4=7

100cft concrete
100*1.54=
then divide by 7
then divide by 1.25 you wil have cement bag

Answer / chetan khairnar

Note: For 1m3 wet concrete = 1.52m3 dry concrete approximately
SP.Wt of concrete= 1440 kg/m3 (or) 1.44 t/m3
1 bag of cement = 50 Kg
Example:- for 20 m3 volume of concrete.

a) Quantity of cement required = (1:2:4)
=1/(1+2+4) x 1.52 × 20 = 4.14m3 x 1440/50 =119.26 bags

b)Quantity of Sand required = (1:2:4)
=2/(1+2+4)×1.52x20=8.28m3 OR =4.14 x 2=8.28 m3.

c)Quantity of course aggregate = (1:2:4)
=4/(1:2;4) x 1.52 x20 = 16.56m3 OR =4.14 x 4= 16.56m3.

If qty. calculate in Kg just multiply by its Sp. wt of sand
or aggregate.

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