Civil Engineering - Highway Engineering - Discussion
Discussion Forum : Highway Engineering - Section 5 (Q.No. 34)
34.
Minimum number of 50 kg cement bags per cubic metre of concrete for a mix corresponding to crushing strength 280 kg/cm2 at 28 days, are
Discussion:
11 comments Page 1 of 2.
Ajay Sharma said:
5 years ago
8 bags.
M20(1:1.5:3)=(1/5.5)*1.54=0.28m3.
Now, 1bag = 0.035m3.
0.28/0.035=8bags.
M20(1:1.5:3)=(1/5.5)*1.54=0.28m3.
Now, 1bag = 0.035m3.
0.28/0.035=8bags.
(2)
Vishesh said:
7 years ago
W/c ratio = wt Of water/wt Of cement.
For min Value, wt of water/cub.m of conc = 180kg.
In mild condition w/c ratio = 0.55.
Wt Of cememt= 180/0.55 = 327.27kg.
No of bags = 327.27/50=6.54.
For min Value, wt of water/cub.m of conc = 180kg.
In mild condition w/c ratio = 0.55.
Wt Of cememt= 180/0.55 = 327.27kg.
No of bags = 327.27/50=6.54.
(1)
Zara said:
6 years ago
If it is M20 then the answer is 8.
For 1:1.5:3, quantity of cement = 1.54x(1/5.5)= .28m3,
to convert to kg, .28x1440kg/m3 = 403.2kg total, per bag 403.2/50 = 8bags.
For 1:1.5:3, quantity of cement = 1.54x(1/5.5)= .28m3,
to convert to kg, .28x1440kg/m3 = 403.2kg total, per bag 403.2/50 = 8bags.
(1)
Sunny Thakur said:
8 years ago
Can anyone explain?
Diesel said:
8 years ago
1,470 ÷ 280 = 5.25.
1470 is the density of cement.
1470 is the density of cement.
Hit said:
8 years ago
I think it should be 7.5 bag.
Gubendhiran said:
8 years ago
280kg/cm^2=27.5 N/mm^2.
For M20 grade concrete mix design the Target mean strength is 27.6 N/mm^2,
So this can be considered as M20 grade mix design,
As per IS: 10262 water-cement ratio is 0.5,
For 20mm nominal max size aggregate and sand of zone 2,
Water required for 1m^3 of concreting is;
186 kg.
W÷C=0.5.; C=W÷0.5=2W.
Cement required=2*186=372 kg,
No.of bags=372/50=7.44 bags.
For M20 grade concrete mix design the Target mean strength is 27.6 N/mm^2,
So this can be considered as M20 grade mix design,
As per IS: 10262 water-cement ratio is 0.5,
For 20mm nominal max size aggregate and sand of zone 2,
Water required for 1m^3 of concreting is;
186 kg.
W÷C=0.5.; C=W÷0.5=2W.
Cement required=2*186=372 kg,
No.of bags=372/50=7.44 bags.
Sumit said:
7 years ago
Explain the answer.
Kashif khan said:
6 years ago
Thanks @VISHESH.
N. K. Varsani said:
4 years ago
W/c = 0.85W.
Number of bags = 280 * 0.85/50 = 6.5.
Number of bags = 280 * 0.85/50 = 6.5.
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