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 5 of 6.
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.
Wani saqib said:
7 years ago
You are right @Nitu Bharti. Thanks.
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.
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.
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.
Kumar said:
6 years ago
Weight if water is given as: 30% of cement + 10% of fine aggregate + 1% of coarse aggregate. So, it should be 15+15+3 = 33 kg.
(3)
Maheshsinh said:
5 years ago
Dry volume of mix 1.54 cmt.
There fore qty of cement in mix of 1:3:6= 1.54*50/10*0.035.
= 220 kg.
Here w/c =0.6.
Qty of water= 0.6*220.
=13.3.
Rounding 14 kg.
There fore qty of cement in mix of 1:3:6= 1.54*50/10*0.035.
= 220 kg.
Here w/c =0.6.
Qty of water= 0.6*220.
=13.3.
Rounding 14 kg.
(9)
Waleed said:
5 years ago
The volume of 1 cubic meter of wet concrete = 1.54 x Volume of dry ingredients of concrete.
Volume of cement req. for 1 cubic meter of wet concrete = 1/(1+3+6) x 1.54 = 0.154 cubic meter.
Volume of 1 bag (50 kg) of cement = 0.0347 cubic meter.
No. of bags of cement = 0.154/0.0347=4.438 bags (220 kg)
Weight of water required = w/c ratio x weight of cement = 0.6 x 220 = 133 kg
Weight of water required per bag of cement= 133/4.438 = 30 kg.
Volume of cement req. for 1 cubic meter of wet concrete = 1/(1+3+6) x 1.54 = 0.154 cubic meter.
Volume of 1 bag (50 kg) of cement = 0.0347 cubic meter.
No. of bags of cement = 0.154/0.0347=4.438 bags (220 kg)
Weight of water required = w/c ratio x weight of cement = 0.6 x 220 = 133 kg
Weight of water required per bag of cement= 133/4.438 = 30 kg.
(2)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers