Answer / rocky

i wanna consruct a wall having length 1450 RFT so i wanna calculate how much cement and sand is required???

Answer / er. r k rajwar

SImply go to irc or morth site

Answer / rajasekar

Grade M30 Ratio is 1:1.5:3
I am telling simple concept
Add the ratio 1+1.5+3=5.5
assume approximate volume for 20mm aggregate =1.57
40mm aggregate =1.52
Qty of cement 1/5.5*1.57=0.28m3
Density of cement =1440kg/m3
Weight of cement per m3=1440*0.28=403.2kg/m3
per bag=403.2/50=8.02bag
Qty of fine aggregate=1.5/5.5*1.57=0.428kg/m3
Qty of coarse aggregate=3/5.5*1.57=0.856kg/m3

Answer / manish singh

M-30 Grade Concrete=
Ratio-1:1.5:3
1mq concrete
Cement => 1*1/5.5=.182 (5.5=Mixing)
=> .182*1.5=.273 (1.5 *dry moisture)
=> .273*30 = 8.19 (8 bags /409.5kg)
Send => 1*1.5/5.5 =.273
=> .273*1.5 =.410 mq
Aggregate => 1*3/5.5 =.545
=> .545*1.5 =.818 mq

Answer / sunil kumar yadav

Taken the ratio-1:1.5:3
cement-1
sand-1.5
metal-3
so ratio will be for mixing-1+1.5+3=5.5
Totals ingredients forusing to=1.54/5.5=o.28
cement solution=0.28 cum
send=0.28×1.5=0.42 cum
(1.5 is a maximum proportion)
metal=0.28×3=0.84 cum
(3 is a maximum proporties)
1 cum cement concrete ratio is 1:1.5:3
cement=0.28 cum
Sand=0.42 cum
metal=0.84 cum
so cemenbbage will be =50/1440=0.0347
(1440=unit wait of cement & 50=wait of 1bage cement)
so cement bags will be=0.28/0.034=8.24 bage.

Answer / k.nabadwip sharma

Calculation for Cement:-

100 cft of conc in loose 144 to 154 cft required,average of
144 and 154 is 149 cft say 150 cft.

M25 = 1:1:2=4

150cft/4=37.5cft
37.5cft/1.2=31.25bags [1 bags of cement is 1.2cft]
=31.25/2.83=11bags [1 cft = 0.0283cum x 100 cft=2.83cum]
per cum cement required=11bags

Calculation for Sand:-
11bags x 1.2 cft = 13.2 cft = 13.2 x 0.0283=0.374 cum
Per cum sand required = 0.374cum

Calculation for Aggregate:
=0.374 x 2 = 0.747 cum
per cum Aggregate required = 0.747 cum

Answer / kavinder

For M30 nominal mix, considering the ratio (1:1:2),

Cement = 1
Fine Agg = 1
Coarse Agg = 2

Total = 1+1+2=4

Considering the above ratio in grams i.e, (cement = 1 gm, Fine agg = 1 gm, Coarse agg = 2 gm) then

For 1 cum, 1/4 = 0.25 % will be filled.

Now, Volume of 1 bag of cement is 35 litres or 0.035 mL

Since, to maintain W/C ratio, the amount of cement & water should be equal.

Therefore, 1 gm of cement will require 0.035 mL of water.

0.25/0.035 = 7.15 approx. Bags or 357.5 kg of Cement.

or consider like this

1/((1+1+2)*0.035) = 7.15 approx. Bags or 357.5 kg of Cement (7.15 * 50 = 357.5), here ratio is 1 therefore 1*50=50

Fine Agg = 357.5 kg (7.15 * 50 = 357.5), here ratio is also 1 therefore 1*50=50

Coarse Agg = 715 kg (7.15 * 100 = 715), here ratio is 2 therefore 2*50=100

Adding all, 357.5+357.5+715 = 1430 kg

Answer / a p s chandel

Dry Density of the concrete mix=15.4 to15.7
If we consider M30 mix is 1:1:2 than sum of the proption =4
Now calculate the quantity of cement= 15.4/4=3.85cu m
1.00cum have 30bags cement
So 1 bag cement=0.034 cum
3.85/0.34=11.32 bags
11.32*50=339.6 KG`
Sand=3.85cum
coarse aggregate=3.85*2=7.7 cum

Answer / randeepchaudhary

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

Answer / sjtbehera

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.

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