Civil Engineering - Concrete Technology - Discussion
Discussion Forum : Concrete Technology - Section 5 (Q.No. 29)
29.
For a concrete mix 1:3:6 and water cement ratio 0.6 both by weight, the quantity of water required per bag, is
Discussion:
57 comments Page 1 of 6.
Waleed said:
5 years ago
The volume of 1 cubic meter of wet concrete = 1.54 x Volume of dry ingredients of concrete.
Volume of cement req. for 1 cubic meter of wet concrete = 1/(1+3+6) x 1.54 = 0.154 cubic meter.
Volume of 1 bag (50 kg) of cement = 0.0347 cubic meter.
No. of bags of cement = 0.154/0.0347=4.438 bags (220 kg)
Weight of water required = w/c ratio x weight of cement = 0.6 x 220 = 133 kg
Weight of water required per bag of cement= 133/4.438 = 30 kg.
Volume of cement req. for 1 cubic meter of wet concrete = 1/(1+3+6) x 1.54 = 0.154 cubic meter.
Volume of 1 bag (50 kg) of cement = 0.0347 cubic meter.
No. of bags of cement = 0.154/0.0347=4.438 bags (220 kg)
Weight of water required = w/c ratio x weight of cement = 0.6 x 220 = 133 kg
Weight of water required per bag of cement= 133/4.438 = 30 kg.
(2)
Abhinav Kumar said:
8 years ago
Thumb rule for the calculation of quantity of water in Concrete is ;
Amount of water = (5% of Fine aggregate + 30% of Cement)* Water Cement ratio.
So for a concrete mix of 1:3:6, Cement per bag = 50 Kg, Fine aggregate = 3*50 = 150 Kg.
So amount of water required will be;
W = (5% of 150 Kg + 30% of 50 Kg)* 0.6,
= (0.05*150 + 0.3*50)*0.6,
= (7.5 Kg + 15 Kg)*0.6 = 13.5 Kg approximately equal to 14 Kg.
Amount of water = (5% of Fine aggregate + 30% of Cement)* Water Cement ratio.
So for a concrete mix of 1:3:6, Cement per bag = 50 Kg, Fine aggregate = 3*50 = 150 Kg.
So amount of water required will be;
W = (5% of 150 Kg + 30% of 50 Kg)* 0.6,
= (0.05*150 + 0.3*50)*0.6,
= (7.5 Kg + 15 Kg)*0.6 = 13.5 Kg approximately equal to 14 Kg.
Saideepak Nagajjanavar Hangal said:
8 years ago
Let's discuss with the calculation of material for 1cum concrete (1:2:4).
Dry volume m = 1X1.45 (45 % to 55 % Increase in volume = 1.45 cum.
Cement = 1.45/(1+2+4) = 0.207 cumX1470 = 304.29 kg. (Density 1470)
Sand = 1.45/(1+2+4) X2 = 0.414 cum/X1350(Density) = 558.9 Kg.
CA = 0.207X4 = 0.828cumX1300 = 1076.4 Kg.
Therefore 1 : 2 : 4.
304.29 Kg : 558.9 Kg : 1076.4 Kg for 1 CUM.
Dry volume m = 1X1.45 (45 % to 55 % Increase in volume = 1.45 cum.
Cement = 1.45/(1+2+4) = 0.207 cumX1470 = 304.29 kg. (Density 1470)
Sand = 1.45/(1+2+4) X2 = 0.414 cum/X1350(Density) = 558.9 Kg.
CA = 0.207X4 = 0.828cumX1300 = 1076.4 Kg.
Therefore 1 : 2 : 4.
304.29 Kg : 558.9 Kg : 1076.4 Kg for 1 CUM.
AMIT ANAND said:
8 years ago
Here we have to know only about the requirement of water for 1-bag of cement.
We know 1bag contain 50 kg cement.
As per w/c ratio its value 0.6 i.e 60 litre pe 100kg cement,
So the density of cement nearly 1440kg/cum,
Density = mass/volume.
So, volume=[mass]/density * 1000,
i.e 50/1440 * 1000,
Ans is 34.7222 which is nearly 35.
Hence 35 kg answer is right.
We know 1bag contain 50 kg cement.
As per w/c ratio its value 0.6 i.e 60 litre pe 100kg cement,
So the density of cement nearly 1440kg/cum,
Density = mass/volume.
So, volume=[mass]/density * 1000,
i.e 50/1440 * 1000,
Ans is 34.7222 which is nearly 35.
Hence 35 kg answer is right.
Wasim shaikh said:
8 years ago
The density of cement is 1440kg per 1000L(volume).
we know 1 bag of cement = 50 kg.
so how much volume(L) it will require (50x1000) ÷ (1440) =34.722.
which is approximately 35 kg.
But if we use w/c ratio= 0.60 =(wt of water) ÷ (wt of cement)
Then wt of water=.6 x 50 kg=30 kg.
Can anyone clear this?
we know 1 bag of cement = 50 kg.
so how much volume(L) it will require (50x1000) ÷ (1440) =34.722.
which is approximately 35 kg.
But if we use w/c ratio= 0.60 =(wt of water) ÷ (wt of cement)
Then wt of water=.6 x 50 kg=30 kg.
Can anyone clear this?
Vinay said:
8 years ago
For 1:3:6 mix for 1 cu.m.
Total dry volume = 1.5.
Now total cement required is,
--- 1/(1+3+6)*1.5 = 0.15 volume of cement.
= 0.15/0.0347 = 4.33 bags.
1bag- 50kgs,
50*4.333=216 kg,
Approximate 4 bags,
W/c=0.6.
W = 0.6 *200= 120kg.
Water required is 120kg for 4 bag of cement.
For one bag = 120/4 = 30 KGS.
Total dry volume = 1.5.
Now total cement required is,
--- 1/(1+3+6)*1.5 = 0.15 volume of cement.
= 0.15/0.0347 = 4.33 bags.
1bag- 50kgs,
50*4.333=216 kg,
Approximate 4 bags,
W/c=0.6.
W = 0.6 *200= 120kg.
Water required is 120kg for 4 bag of cement.
For one bag = 120/4 = 30 KGS.
Himanshu gupta said:
7 years ago
1m^3 wet concrete is equal to 1.52 m^3 dry material {include wastage }.Then quantity
Cement=1.52/(1+3+6)=.152m^3.
Then,1m^3 of cement contains in 28.8 bags then no of bags is 4.37.
Mass of cement =4.36*50=218kg,
Water/cement=.6,
Then water =218*6/9.81=13.86kg~14kg,
Hence, the correct answer is 14kg.
Cement=1.52/(1+3+6)=.152m^3.
Then,1m^3 of cement contains in 28.8 bags then no of bags is 4.37.
Mass of cement =4.36*50=218kg,
Water/cement=.6,
Then water =218*6/9.81=13.86kg~14kg,
Hence, the correct answer is 14kg.
Himanshu gupta said:
7 years ago
1m^3 wet concrete is equal to 1.52 m^3 dry material {include wastage }.Then quantity
Cement=1.52/(1+3+6)=.152m^3.
Then,1m^3 of cement contains in 28.8 bags then no of bags is 4.37.
Mass of cement =4.36*50=218kg,
Water/cement=.6,
Then water =218*6/9.81=13.86kg~14kg,
Hence, the correct answer is 14kg.
Cement=1.52/(1+3+6)=.152m^3.
Then,1m^3 of cement contains in 28.8 bags then no of bags is 4.37.
Mass of cement =4.36*50=218kg,
Water/cement=.6,
Then water =218*6/9.81=13.86kg~14kg,
Hence, the correct answer is 14kg.
Sudhansu said:
5 years ago
1:3:6 = 7.
1.52/7= 0.217 cm Cement.
1 cm Cement = 1440 kg cement.
0.217 cm Cement = 1440 x 0.217 = 312.48 kg.
For 312.48 kg Cement water required = 312.48 x 0.6 = 187.2 kg.
1 bag of Cement =50 kg So, 312.48 / 50 = 6.25 bags.
For1 bag, cement required = 187.2/6.25 = 30 kg (approx).
1.52/7= 0.217 cm Cement.
1 cm Cement = 1440 kg cement.
0.217 cm Cement = 1440 x 0.217 = 312.48 kg.
For 312.48 kg Cement water required = 312.48 x 0.6 = 187.2 kg.
1 bag of Cement =50 kg So, 312.48 / 50 = 6.25 bags.
For1 bag, cement required = 187.2/6.25 = 30 kg (approx).
Rakesh Nhavkar said:
9 years ago
It's Thumb Rules for deciding the quantity of water in concrete:
(i) Weight of water = 28% of the weight of cement + 4% of the weight of total aggregate.
(ii) Weight of water = 30% of the weight of cement + 5% of the weight of total aggregate.
(i) Weight of water = 28% of the weight of cement + 4% of the weight of total aggregate.
(ii) Weight of water = 30% of the weight of cement + 5% of the weight of total aggregate.
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