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 4 of 6.
Nitu bharti said:
1 decade ago
Water required = (5% of aggregate + 30% of cement)*w/c.
Cement per bag = 50 kg.
Aggregate = 50*3 = 150 kg.
5% of aggregate = 7.5 kg.
30% of cement= 15 kg.
Water required = (15+7.5)*0.6 = 13.5 kg which may be approximately 14 kg.
Cement per bag = 50 kg.
Aggregate = 50*3 = 150 kg.
5% of aggregate = 7.5 kg.
30% of cement= 15 kg.
Water required = (15+7.5)*0.6 = 13.5 kg which may be approximately 14 kg.
Robert said:
1 decade ago
How did you rear get them I don't get it?
P.s said:
1 decade ago
See 30%of cement means 0.3*50 kg = 15 kg.
5% of fine aggregate 0.05*150 = 7.5 kg.
0.6*(15+7.5) = 13.5 kg+14 kg.
5% of fine aggregate 0.05*150 = 7.5 kg.
0.6*(15+7.5) = 13.5 kg+14 kg.
Suman said:
1 decade ago
0.6 means 60 liter water per 100 kg cement as 1 bag of cement contains 50 kg of cement.
So 60/2 = 30 liter water required per bag of cement.
So 60/2 = 30 liter water required per bag of cement.
Kavita said:
1 decade ago
Please solve this problem. How this answer comes?
Saideepak Nagajjanavar Hangal said:
9 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.
Shivkumar said:
9 years ago
Yeah, I agree with the answer 30.
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.
Ar. SALMAN AFAQUE said:
9 years ago
1:3:6, so we multiply by 1*50 Kg cement and 3 fine aggregate * 50 kg, so.
We get 150 kg fine aggregate.
We get 150 kg fine aggregate.
Mehebub said:
9 years ago
@Salman.
But here told ratio 1:3:6,
Where, 1=Cement, 3=Sand, and 6=Stone.
Therefore Sand + Stone = Aggregate.
So, sum total 3 + 6 = 9. [50x9 = 450Kg].
But here told ratio 1:3:6,
Where, 1=Cement, 3=Sand, and 6=Stone.
Therefore Sand + Stone = Aggregate.
So, sum total 3 + 6 = 9. [50x9 = 450Kg].
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