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.
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.
(1)
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)
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.
Ajay Agnihotri said:
10 years ago
However, I believe that the density of cement is 1440 Kg/Cum.
Therefore, the volume of a 50 kg cement bag works out to be [ 50/1440 ]*1000 = 34.722 Liters.
Therefore, the volume of a 50 kg cement bag works out to be [ 50/1440 ]*1000 = 34.722 Liters.
Atul said:
8 years ago
30%of agg+ 10%of cement.
.3*9*50+.1*50= 140.
.3*9*50+.1*50= 140.
Kayathri said:
10 years ago
Why we are taking 5% of aggregate and 30% of cement?
Sengathir said:
8 years ago
What is that 1bag cement~36 liter (or) 0.036cum please explain?
Rohit said:
8 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:
8 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:
8 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.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers