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 1 of 6.
Abdur Rehman said:
4 years ago
Water required = (5% of aggregate + 30% of cement) * w/c.
Cement per bag = 50 kg.
Aggregate = 50*3 = 150 kg.
5% of aggregate = 7.5 kg.
30% of cement= 15 kg.
Water required = (15+7.5)*0.6 = 13.5 kg which may be approximately 14 kg.
Cement per bag = 50 kg.
Aggregate = 50*3 = 150 kg.
5% of aggregate = 7.5 kg.
30% of cement= 15 kg.
Water required = (15+7.5)*0.6 = 13.5 kg which may be approximately 14 kg.
(30)
Tanmoy Ghosh said:
3 years ago
Here, 0.6 * 50 = 30.
(9)
Shubham Angre said:
4 years ago
28% of water is required for the concrete mix and 1 bag of cement = 50 kg.
So, 50*28% = 14 kg.
So, 50*28% = 14 kg.
(9)
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)
Ramchandra chaulagain said:
4 years ago
We know, W/C=weight of water in kg/weight of cement in kg.
so, the weight of water = weight of cement * w/c.
= 50 * 0.6.
= 30 kg.
That's why Ans is 30kg.
so, the weight of water = weight of cement * w/c.
= 50 * 0.6.
= 30 kg.
That's why Ans is 30kg.
(4)
Anne said:
5 years ago
Water required = (5% of aggregate + 30% of cement ) * w/c.
Ratio given = 1:3:6 ( c:s:a).
1 bag cement = 50kg.
Sand = 3*50 = 150kg (3 parts of sand).
Therefore,
Water required = (5/100*150) + (30/100*50) * 0.6.
= 13.5 = 14 kg.
Ratio given = 1:3:6 ( c:s:a).
1 bag cement = 50kg.
Sand = 3*50 = 150kg (3 parts of sand).
Therefore,
Water required = (5/100*150) + (30/100*50) * 0.6.
= 13.5 = 14 kg.
(4)
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)
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)
Prahumac said:
1 decade ago
Can anybody please give a proper explanation for this one?
(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)
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