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 3 of 6.
Atul said:
8 years ago
30%of agg+ 10%of cement.
.3*9*50+.1*50= 140.
.3*9*50+.1*50= 140.
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)
Garry said:
8 years ago
36 liter is the correct answer as per empirical formulas.
Kishanjee Gujarath said:
8 years ago
Very good information for civil Engineers.
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.
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?
Wasim Shaikh said:
8 years ago
Well, the correct answer is 35kg.
Chijioke said:
8 years ago
Determine the quantities of water needed to mix 150kg of cement used for concrete mix.
Inayat said:
8 years ago
Thanks to all. It was very helpful.
Vinay said:
8 years ago
For 1:3:6 mix for 1 cu.m.
Total dry volume = 1.5.
Now total cement required is,
--- 1/(1+3+6)*1.5 = 0.15 volume of cement.
= 0.15/0.0347 = 4.33 bags.
1bag- 50kgs,
50*4.333=216 kg,
Approximate 4 bags,
W/c=0.6.
W = 0.6 *200= 120kg.
Water required is 120kg for 4 bag of cement.
For one bag = 120/4 = 30 KGS.
Total dry volume = 1.5.
Now total cement required is,
--- 1/(1+3+6)*1.5 = 0.15 volume of cement.
= 0.15/0.0347 = 4.33 bags.
1bag- 50kgs,
50*4.333=216 kg,
Approximate 4 bags,
W/c=0.6.
W = 0.6 *200= 120kg.
Water required is 120kg for 4 bag of cement.
For one bag = 120/4 = 30 KGS.
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