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.
Wasim Shaikh said:
8 years ago
Well, the correct answer is 35kg.
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?
Deepak sonu said:
8 years ago
Water required = (5% of fine aggregate+30%of cement)*water cement ratio.
Solve: 1:3:6(c:s:a).
Since one bag cement = 50 kg.
Fine aggregate = 50*3 = 150.
So(7.5+15)*0.6=13.5~14 kg.
Solve: 1:3:6(c:s:a).
Since one bag cement = 50 kg.
Fine aggregate = 50*3 = 150.
So(7.5+15)*0.6=13.5~14 kg.
Kishanjee Gujarath said:
8 years ago
Very good information for civil Engineers.
Garry said:
8 years ago
36 liter is the correct answer as per empirical formulas.
Jagmohan deol said:
8 years ago
@ALL,
Aggregate = 50*3 = 150 kg is wrong.
It should be 50*6 as the aggregate ratio is 6 to that of 1 of cement.
Aggregate = 50*3 = 150 kg is wrong.
It should be 50*6 as the aggregate ratio is 6 to that of 1 of cement.
(1)
Atul said:
8 years ago
30%of agg+ 10%of cement.
.3*9*50+.1*50= 140.
.3*9*50+.1*50= 140.
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.
Sengathir said:
8 years ago
What is that 1bag cement~36 liter (or) 0.036cum please explain?
Rohit said:
7 years ago
Water=water cement ratio* cement weight.
So, W=0.6*221=133/4.45=30.
So, W=0.6*221=133/4.45=30.
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