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 3 of 5.
Hussain said:
8 years ago
W= 30%cement +4%of total aggregates.
W=. 3W/.6+.04*150 (Given- Water/Cement =0.6).
W=. 5W+6.
W=12kg.
W=. 3W/.6+.04*150 (Given- Water/Cement =0.6).
W=. 5W+6.
W=12kg.
PRIYA said:
8 years ago
W/C*P = 0.3P + 0.1Y + 0.01Z.
Ak jain said:
8 years ago
Please solve,
How to find the water required for concrete 1:2:4 if w/c =.6 and unit w.=2400kg/m3?
How to find the water required for concrete 1:2:4 if w/c =.6 and unit w.=2400kg/m3?
Qazi said:
8 years ago
Why water cement ratio always used in weight?
Pakkirappa said:
8 years ago
Thank you very much @Sanjeev Kr Anad.
Ram soni said:
8 years ago
Thanks @Apcivilian & @Prem. Vit.
SANJEEV KR ANAD said:
9 years ago
W/c *p = 0.3p +0.1F +0.01C.
0.6*p = 0.3p + 0.1*50 +0.01*100,
0.3p = 5+1,
P = 6 ÷ 0.3,
P = 20.
Then, water= w/c*P =0.6 * 20 = 12kg ans.
0.6*p = 0.3p + 0.1*50 +0.01*100,
0.3p = 5+1,
P = 6 ÷ 0.3,
P = 20.
Then, water= w/c*P =0.6 * 20 = 12kg ans.
Bivek Pradhanang said:
9 years ago
D. w= 5%aggregate + 30%cement and here w/c=0.6.
So, here w= .05 * [50+150]+.3 * [w/0.6], so w = 15kg.
So, here w= .05 * [50+150]+.3 * [w/0.6], so w = 15kg.
Prem. vit jaipur said:
9 years ago
0.3 * cement + 0.1FA + 0.01CA = w/c * cement.
0.3 * cement + 0.1 * 50 + .01 * 100 = 0.6*cement.
cement = 20.
W/c = 0.6 water = 0.6 * 20 = 12kg.
0.3 * cement + 0.1 * 50 + .01 * 100 = 0.6*cement.
cement = 20.
W/c = 0.6 water = 0.6 * 20 = 12kg.
ConcreteSukan said:
9 years ago
(5% Coarse aggregate + 30% of cement (50kg here))* w/c is correct.
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