Civil Engineering - Concrete Technology - Discussion
Discussion Forum : Concrete Technology - Section 4 (Q.No. 34)
34.
If 50 kg of fine aggregates and 100 kg of coarse agregates are mixed in a concrete whose water cement ratio is 0.6, the weight of water required for harsh mix, is
Discussion:
41 comments Page 4 of 5.
Apcivilian said:
9 years ago
.3p+.1F+.01C= w/c xp
By using this equation, we can get the answer.
By using this equation, we can get the answer.
BUET-K said:
9 years ago
C = 1/7 * 150 = 21 (1;2;4) = 7,
W/C = 0.6,
W = 0.6 * 21 = 12.
W/C = 0.6,
W = 0.6 * 21 = 12.
Baloch said:
9 years ago
Please read question, it's giving 50kg of FINE aggregates not cement.
Bahram said:
9 years ago
@Sumit, 50 kg is the weight of fine aggregates, not cement.
Ruhi varshney said:
10 years ago
Water cement ratio are different for different grade of concrete.
Nabaratna said:
1 decade ago
WC ratio is always 0.6.
Nungsang said:
1 decade ago
If the mix is 1:3:6, Can we use same method keeping same wc ratio?
Vishal singh said:
1 decade ago
Its 30% of coarse aggregate.
Sumit said:
1 decade ago
100 (wt.of Coarse sand) *5/100 = 5 kg.
50 (wt.of cement)* 30/100 = 15 kg.
(5+15)*0.6 = 12 kg.
50 (wt.of cement)* 30/100 = 15 kg.
(5+15)*0.6 = 12 kg.
M.Pagavandoss said:
1 decade ago
Wt of water = (5% wt of coarse aggregate+30% wt of cement)*w/c ratio.
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