Answer / naresh thakur

soppose, concrete mix ratio = 1:1.5:3
then, sum = 1+1.5+3 = 5.5

as we know,dry material required for 1 cum. of concrete =
1.54 cum.
then we can find -:

(1) cement = 1*1.54/5.5 = .28 cum. or .28cum.* 28.8= 8.06
bags.

(2) sand = 1.5* 1.54/5.5 = .42 cum.
or .42cum.* 35.3 = 14.83 cubic feet.

(3) coase aggregate = 3* 1.54/5.5 = .84 cum.
or .84 cum.* 35.3 = 29.65 cubic feet.

so, dry ingredient quantity required for 1 m3 of concrete-:

cement = 8.06 bags.
sand = 14.83 cft.
coarse aggregate = 29.65 cft.

do not forget to add 5 to 8 % addition matreial.

Answer / b.santosh kumar

you have to take 1.52 as constant for calculating concrete
mix ,let me explain it this 1.52 is the quqntity of total
dry aggregate which is required.so when ever u require to
know the quantity for any given mix .u need to add up the
ratio of mix and divide 1.54 by the sum of mix ratio u ll
get the quantity of cement for one cum
eg: suppose m20 mix u need to calculate the concrete then
simply add the ratio of m20 i.e 1:1.5:3 so the total is 5.5
and divide 1.52 by 5.5 u ll get 0.27 cum of cement which ll
b required for 1 cum of concrete and if u multiply it(0.27)
by 30 u ll get the no. of cement bags .follow the same rule
for diff type of mix

Answer / b.subhash

Estimating Quantities of material needed.

1. Calculate the volume of concrete needed.
2. Estimate the total volume of dry material by
multiplying the required volume of concrete by 1.65 to get
the total volume of dry loose material needed (this includes
10% extra to compensate for losses).
3. Add the numbers in the volumetric proportion that you
will use to get a relative total. This will allow you later
to compute fractions of the total needed for each
ingredient. (i.e. 1:2:4 = 7).
4. Determine the required volume of cement, sand and
gravel by multiplying the total volume of dry material (Step
2) by each components fraction of the total mix volume (Step
3) i.e. the total amount cement needed = volume of dry
materials * 1/7.
5. Calculate the number of bags of concrete by dividing
the required volume of cement by the unit volume per bag of
cement (0.0332 m3 per 50 kg bag of cement or 1 ft3 per 94 lb
bag).

For example, for a 2 m x 2 m x 10 cm thick pump pad:

1. Required volume of concrete = 0.40 m3
2. Estimated volume of dry material = 0.4 x 1.65 = 0.66 m3
3. Mix totals = 1+2+4 = 7 (1:2:4 cement:sand:gravel)
4. Ingredient Volumes: 0.66 x 1/7 = .094 m3 cement
0.66 x 2/7 = .188 m3 sand
0.66 x 4/7 = .378 m3 gravel
5. # Bags of cement: 0.094 m3 cement / .0332 m3 per 50 Kg bag
= 2.83 bags of cement (use three bags)

Answer / khuram shahzad

mix proportions are

M15 = 1: 2 : 4
M20 = 1: 1.5 : 2
M25 = 1: 1 : 2

Ok lets go to the calculaton part
Now we are going to calculate the quantity of materials required for 1m3 Concrete (M15)

Quantity of Cement required for 1m3

[1/(1+2+4)] * 1.52 = 0.217m3
0.217 * 1440 = 312.48 kg/m3
To calculate in bags
312.48/50 = 6.24 bags/m3
Here 1.52 is the dry co-efficient and 1440 is the unit weignt of cement.

Quantity of Fine aggregate( sand) for 1m3

[2/(1+2+4)] * 1.52 = 0.434m3

Quantity of course aggregate for 1m3

[4/(1+2+4)] * 1.52 = 0.868m3

So this is the Quantity for 1m3. For example, If you want 6m3 concrete of M15 grade

for 6m3
cement = 312.48 * 6 = 1874.88 kg for 6m3 (37.5 bags )
fine aggregate = 0.434 * 6 = 2.604 m3 ( 2.604 m3 sand required for 6m3 concrete )
course aggregate = 0.868 * 6 = 5.208 m3 ( 5.208 m3 aggregate required for 6m3 concrete)

The same procedures should be followed for M20, M25 concrete
For example M20
[1/(1+1.5+2)]* 1.52
For M25
[1/(1+1+2)] * 1.52

Answer / oretan omowunmi olatunde

volume of concrete considering 1m3

ration 1:2:4

1+2+4= 7
cement
1/7+40%(compaction)+5%(waste)
0.142+0.0572+0.00715=0.20735m3
density of cement = 1440kg/cm3
1m3 of cement = 1440kg/cm3
xm3 contains 50kg = 50/1440 = 0.035m3
no of bags of cement =0.20735/0.035= 5.92bags

Sand
2/7 + 40% + 5%
0286 + 0.1144 + 0.143=0.4147m3
3.33m3 of lorry = 5cubic yard
no of load = 0.414/3.33 =0.125load/m3
Granite
4/7 + 40 + 5
0.572 + 0.229 + 0.0286 = 0.8296m3
3.33m3 of lorry = 5cubic yard
no of load = 0.8296/3.33 =0.25load/m3

cements = 6bags

Sand = 0.125load/m3

Granite = 0.250load/m3

Answer / waqar

you r confusing mr. some where u r using 1.52 constant and some where u r using 1.54. why you r confusing when 1.54 is correct?

Answer / sanjay rana

Divide the sum of ratio of any mix to dry cofficient eg.
1:1.5: 3= 1+1.5+3= 5.5, 1.54/5.5*3 you can find c.a.

Answer / wali

wath is the air content for concrete?

Answer / shilpa

2%

Answer / vikrant

(1.51/addition of mix of grade)

Target strength = fck +(1.65*4) What is this 4

4 Answers   Gammon,

how to check aggregate quantity in stock yard

1 Answers  

how much quantity of paint is required for 1 square meter wall for single and double coat..?

2 Answers   RK,

what is super elevation?

10 Answers   CCCL, KMC,

What does a construction cost covers and what does not?

0 Answers  

What are the methods in tiles flooring ? What are the methods in brick work? What are the methods in plastering ? What are the methods in wall tiling?

0 Answers   CECL, HTC,

how to calculate the 45 degree bent up length. if we given that whether the total length of the slab reinforcement will increase or not. mantion it clearly please. this is for bar bending schedule purpose?

4 Answers   Indian Army, L&T,

How do we say The LAP LENGHT is 50D to 60D ?????

3 Answers  

what do you mean by yeild stress of a material? What will happen when hit is given to cement?

0 Answers   Gammon,

why we use minimum cement as given in IS mix design code

5 Answers   BARC,

i have a interlock pavement work.i need to compact about 2000 sq metres of sub base grade c of thickness 15 cm.what is the amount of water required for compaction ??

0 Answers  

How to fix the reinforcement in footing shorter bar at bottom and longer fix at top.

0 Answers   JKC, L&T,