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.
Lalit Pathade said:
6 years ago
Right answer:
Simple hai - assume 10m^3 , Dry mortar volume =30x...*10=3m^3.
Cement = (3) /(1+3+6) = 0.3m^3.
No.of bags - 0.3/(0.034) =8.8235 bags.
Quantity of water per bag = 8.8235/(0.6) = 14.70 = 14kg.
Simple hai - assume 10m^3 , Dry mortar volume =30x...*10=3m^3.
Cement = (3) /(1+3+6) = 0.3m^3.
No.of bags - 0.3/(0.034) =8.8235 bags.
Quantity of water per bag = 8.8235/(0.6) = 14.70 = 14kg.
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.
Manoj Balakrishnan said:
7 years ago
Here the question is "the quantity of water required per bag and W/C ratio is 0.6" then the water required for 1 bag of cement is 30 Kg.
Wani saqib said:
7 years ago
You are right @Nitu Bharti. Thanks.
Gauri shankar mandal said:
7 years ago
Of course, 30kg is correct. I also agree.
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.
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.
Sengathir said:
8 years ago
What is that 1bag cement~36 liter (or) 0.036cum please explain?
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.
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