Civil Engineering - Concrete Technology - Discussion

DEAR all, First of all, pick the question, In the question, we are asked for two quantities not for one. We have to fine cement as well as water.

Therefore We know that,

Weight of water / Weight of cement = Water to cement ratio.
And the water to cement ratio is 0.6.

Rearrange the above formula,

Weight of water = Weight of cement x Water to cement ratio.

So,
Weight of water = Weight of cement x 0.6.....................(i)
Here, We have in weight of water =
0.3 x weight of cement + 0.1 x fine aggregate + 0.1 x coarse aggregate=
0.3 x weight of cement + 0.1 x (50)+ 0.1 x (100) = 0.3 x weight of cement + 5 +1
0.3 x weight of cement + 0.1 x (50)+ 0.1 x (100) = 0.3 x weight of cement +6.

So,
Weight of water = 0.3 x weight of cement +6...........................(ii)

From equation 1,
Weight of cement x 0.6 = 0.3 x weight of cement + 6.
Let weight of cement be denoted by L
L x 0.6 =0.3 x L + 6
0.6 L-0.3 L = 6
0.3 L = 6
L = 6/ 0.3= 20 kg.
Where L is the weight of cement.

By using formula,
Weight of water = weight of cement x water to cement ratio.
Weight of water = 20 x 0.6.
= 12 kg.

@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.

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.

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.

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.

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.

0.3 * 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.

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.

0.3 * cement + 0.1fine + 0.01coarse = 0.6 * cement.
0.3 * cent + 0.1 * 50 + .0.01 * 100 = 0.6 * cement.
0.3 * cement = 6.

Cement = 20,
W/c = 0.6,
W = 0.6 * 20 = 12kg.

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.