Answer / ranga

take the ratio 1:1.5:3
cement =1
sand =1.5
metal =3
So what is the ratio we want to mixing 1+1.5+3=5.5
Volum of wet cement concret is 1.54 to 1.57
unit weight of cement bag is 1440 kg/cum
one bag of cement = 50/1440 = 0.034722 cum

SOLUTION : CEMENT = 1.54/5.5 =0.28
=0.28/0.0347 = 8.069 ( 8 bags)
one bag of cement =50 kg X 8.069 = 403.45 kgs.

SAND solution =0.28 X 1.5 = 0.42 Cum
1.5 is the mixing proportion
METAL SOLUTION = 0.28 x 3 = 0.84 cUM.
3 is the mixing proportion
1 Cum cement concrete is 1:1.5:3 ratio

cement = 403.45 kgs.
sand = 0.42 Cum.
metal = 0.84 Cum.

Answer / sachdeva d.p

M30- cement=1/4*1.54=0.385*28.8=11bags
sand-0.385*1=0.385cum
agregate 0.385*2=0.77cum

Notes;why we multiply by 1.54
dry dencity=1.54*wet denciti in conct.work

Answer / rajarshi basu

The approach by Sjtbehera is totally correct but has some
little silly mistakes...I have tried to give a rectified
solution.

In the case M30 design mix, the quantities of cement,sand
and aggregates cannot be calculated as it is a variable
quantity.It depends upon the Mix designer how much quantity
of cement,sand and aggregates he will employ.His only
intention would be to design a mix whose specified
characteristic compressive strength at 28 days=30 N/sq.mm
But in case of nominal mix we can use 1:1:2
Let us consider a volume of 10 cu. m (Wet concrete)

It is difficult to access exactly the amount of each
material required to produce 10 cu m of wet concrete when
deposited in place.
Hence to convert the wet volume into dry volume,
Increase by 54 % to account for shrinkage and wastage

Thus it becomes=15.4 cu. m (Variable upto 15.70)

Calculating cement, sand and aggregates in the mix:
Total summation of proportion=1+1+2=4

Cement = 15.4/4= 3.85 cu. m =3.85/0.0347=110 bags
Sand=15.4/4=3.85 cu. m
Aggregates(20mm to 6mm)=15.4 x (2/4)=7.70 cum

Hope u have understood and the mistake is clarified!!!

Answer / ratul

1:1:2

Answer / deepa

how do we calculate quantity of cement , sand and
aggregates in 1 m3 of M7.5 grade concrete ?

Answer / samsulhoque@jadavpur universit

for m30 grade .we have to calculate the quantity of all
ingredients of concrete by design mix as per IS
10262-2009.it depends upon the whether admixture is used or not.

Answer / rahul mehta j.e

Take a Ratio 1:1.5:3
Now 1+1.5+3= 5.5
Now Divide This By 1.52 Which Is wet Volume
Now 1.52/5.5 =0.276
Now Multiply This by 30 Coz 1 Cum Of Cement Contains 30
bags dry.
==> 0.276 X 30 = 8.29 Bags Or 414 Kg cement.

Answer / enayatullah

Take the retio 1:1.5:3
Cement = 1
Sand = 1.5
Aggregates = 3
The volume of none mixed material is 1.52 m3 for one cu.m
mexid concret becouse the the aggregate sand and cement
have porosity that will fill with some of sand and
cement,so 1.52/(1+1.5+3)=1.52/5.5=0.276
Cement=0.276*1=0.276 cu.m = 0.276/0.036*50 bag = 383kg
0.036 is one bag volume
Sand = 0.276*1.5 = 0.414 cu.m
Aggregate = 0.276*3 = 0.828 cu.m
0.276+0.414+0.828=1.518 near to 1.52

Answer / subilash .s(civil engineer)

for designing 1m3 concrete of M25 mix the ratio is 1:1:2
for calculating cement,sand and course aggregate ,
cement= 1/(1+1+2)=.25
density of cement =1440kg/m3
cement required for this mix=1440*.25=360kg=6bags+10kg
sand=1/4=.25m3
course aggregate=1-.5=.5m3
this methode of designing is known as volume batching ,this
methode will be ues to design mixes upto m25.

Answer / fsdfsdfsdfs

For m30 8 bags cement

hi every body..i have applied for RRB section engineers CIVIL.i wanna prepare for that,so can any one suggest me some books and where can i get the syllabus for this..i would be vry greatful if any one can help in this regard

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