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 1 of 5.
Sudip kandel said:
3 years ago
We have the relation of:
0.3 * cement + 0.1 * sand + 0.01 * aggregate = w/c * cement.
Here,
0.3 * c + 0.1 * 50 +100 * 0.1 = 0.6 * c.
c = 20 kg.
w/c = 0.6.
So, water = 0.6 * 20 = 12 kg.
0.3 * cement + 0.1 * sand + 0.01 * aggregate = w/c * cement.
Here,
0.3 * c + 0.1 * 50 +100 * 0.1 = 0.6 * c.
c = 20 kg.
w/c = 0.6.
So, water = 0.6 * 20 = 12 kg.
(24)
B.Singh said:
3 years ago
W/C * P = 0.3P + 0.1F.A +.01C.A.
0.6 * P = 0.3P + 0.1(50) + 0.01(100),
0.6P - 0.3P = 5 + 1,
P = 6/0.3.
P = 20 => Where P= Cement.
Now;
W/C = 0.6,
W = C * 0.6,
W = 20 * 0.6,
W = 12Kg.
0.6 * P = 0.3P + 0.1(50) + 0.01(100),
0.6P - 0.3P = 5 + 1,
P = 6/0.3.
P = 20 => Where P= Cement.
Now;
W/C = 0.6,
W = C * 0.6,
W = 20 * 0.6,
W = 12Kg.
(15)
Aryan said:
2 years ago
@All.
W/c * cement = 0.3 * cement + 0.1 * fine aggregate + 0.01 * coarse aggregate.
0.6 * C = 0.3 * C + 0.1 * 50 + 0.01 * 100.
C = 20KG.
W/C * Cement = wt of water.
= 0.6 * 20 = 12 kg.
W/c * cement = 0.3 * cement + 0.1 * fine aggregate + 0.01 * coarse aggregate.
0.6 * C = 0.3 * C + 0.1 * 50 + 0.01 * 100.
C = 20KG.
W/C * Cement = wt of water.
= 0.6 * 20 = 12 kg.
(11)
Ejaz Ahmad said:
5 years ago
Wt. Of water = (5% of Coarse aggregate + 30% of fine aggregate) * water cement ratio.
Therefore,
Wt. Of water = {5 * 100/100 +30 * 50/100} * 0.6,
Wt. Of water= 12kg.
Therefore,
Wt. Of water = {5 * 100/100 +30 * 50/100} * 0.6,
Wt. Of water= 12kg.
(10)
Anil said:
3 years ago
I can't understand please help me. What is 30% & how it come?
(8)
Er vikash saini said:
4 years ago
Here 50 kg weight of cement:
M 25 ratio 1:1:2.
(5% weight of aggregate + 30 weight of cement)w/c ratio.
M 25 ratio 1:1:2.
(5% weight of aggregate + 30 weight of cement)w/c ratio.
(4)
Mir Imroz said:
4 years ago
In harsh mix weight of water =. 30% wt of fine aggregate + .05% wt of coarse aggregate) water-cement ratio.
(3)
Sameer Khan said:
4 years ago
Where does this formula comes from?
W/C *P = 0.3P + 0.1F.A +.01C.A
How do you justify it? Please explain me.
W/C *P = 0.3P + 0.1F.A +.01C.A
How do you justify it? Please explain me.
(3)
Muhtasim said:
4 years ago
What will be formula for rich mix?
(1)
ASHISH SHARMA said:
5 years ago
W/C *P = 0.3P + 0.1F.A +.01C.A.
0.6*P =0.3P + 0.1(50) + 0.01(100),
0.6P - 0.3P = 5 + 1,
P = 6/0.3,
P = 20 => Where P= Cement.
Now;
W/C = 0.6,
W = C * 0.6,
W = 20 * 0.6,
W = 12Kg.
0.6*P =0.3P + 0.1(50) + 0.01(100),
0.6P - 0.3P = 5 + 1,
P = 6/0.3,
P = 20 => Where P= Cement.
Now;
W/C = 0.6,
W = C * 0.6,
W = 20 * 0.6,
W = 12Kg.
(1)
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