In the case M30 concrete we calculating by design mix.but
incase of norminal mix we can use 1:1:2.
calculating cement, sand and concrete.
toal ratio=1+1+2=4
cement = 15.4/4=3.85cum=11bags
sand=3.85*1=3.85 cum
chips=3.85*2=7.7 cum
Notes: why we divided 15.4
but it is difficult to acess exactly the amount of each
material recquired to produce 1cum.of wet concrete when
deposite in place.
to find out volumes of cement sand & coarse aggregate
divided a numerical no. 15.4 variable up to 15.7 according
to proportioning and water cement ratio by the summation of
the proporation of the ingrediants used and then multiply
the result thus obtained their respective srength of
proportion.
If you Consider M25 Grade Concrete Nominal mix Design is
1:1:2(1 Cement:1 Sand:2 Aggregates)
1+1+2=5
Volume of Dry Mix=1.5% X Volume of wet mix
If you Consider 1 Cum Concrete
Cement=1/5X1.5(1)=0.3 Cum(1 Bag of Cement=50 Kgs=1.235
Cft=0.035 Cum)
Then 0.3 /0.035=8.57 bags=428 kgs
if we consider concrete grade M20(1:2:3)
1+2+3=6
wet volume 1.58cum
cement = 1/6x1.58=0.26cum
volume of one bag of cement = 0.033cum
total quantity of cement bag one CUM concrete =0.26/.033=7.87bags
7.87x50 kg=393kg
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!!!
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.
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.
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.
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
Structural steel: For a thumb rule check in field, the
allowable weight which can be hung down for 10mm dia mild
steel rod, of yield value of 0, 25 KN/mm2 and applying a
factor of safety 2.00, will be how many kg?
along with your userid / name. ThanQ
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