Civil Engineering - Estimating and Costing - Discussion
Discussion Forum : Estimating and Costing - Section 1 (Q.No. 2)
2.
The expected out turn of 2.5 cm cement concrete floor per manson per day
Discussion:
45 comments Page 2 of 5.
Jithin said:
6 years ago
Can anyone explain on brief with a reference?
Akshy said:
6 years ago
1 mason 1 day work = 0.1875m^3.
For 2.5cm, it will be(0.1875/0.025) = 7.5m^2.
For 2.5cm, it will be(0.1875/0.025) = 7.5m^2.
(1)
S Raju said:
6 years ago
The value of 1.875 coming from where?
Ravi said:
6 years ago
Can anyone please explain me to get it?
Ravi said:
6 years ago
Can anyone please explain me to get it?
Pritam rajak said:
7 years ago
It means the avg floor rough casting of thickness 2.5cm, per Masson/day.
Milton shaikh said:
7 years ago
It should not understand how can do one masonry per day 1.876?
Naresh said:
7 years ago
I am not getting it. Please explain.
Abcd said:
8 years ago
1 : 2 : 4 concrete, 1m3 need Labour constant.
Mixing concrete --------> 3.00 hour/m3.
Lifting and carrying concrete --------> 1.20 hour/m3.
c.compacting concrete --------> 0.80 hour/m3.
d.leveling surface of concrete --------> 0.10 hour/m3.
Total=(3+1.2+0.8+0.1) = 5.
Mixing concrete --------> 3.00 hour/m3.
Lifting and carrying concrete --------> 1.20 hour/m3.
c.compacting concrete --------> 0.80 hour/m3.
d.leveling surface of concrete --------> 0.10 hour/m3.
Total=(3+1.2+0.8+0.1) = 5.
Satya.M said:
8 years ago
.1875/.025 = 7.5.
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