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 2 of 6.
Asif wazir said:
7 years ago
30 kg is correct because for 14 kg the WC ratio becomes 27 which is less than even the hydration water required for a single bag of cement.
Garry said:
8 years ago
36 liter is the correct answer as per empirical formulas.
NIKUNJ said:
9 years ago
Water required per bag means:
Cement bag = 50 kg.
W/c ratio = 0.6.
Water = 50 x 0.6.
Hence water = 30 kg.
If we assume water absorption in calculation then water require more than 30 kg.
Cement bag = 50 kg.
W/c ratio = 0.6.
Water = 50 x 0.6.
Hence water = 30 kg.
If we assume water absorption in calculation then water require more than 30 kg.
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.
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.
Gauri shankar mandal said:
7 years ago
Of course, 30kg is correct. I also agree.
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