Civil Engineering - Estimating and Costing - Discussion
Discussion Forum : Estimating and Costing - Section 2 (Q.No. 32)
32.
The ground surface slopes 1 in 50 along a proposed railway embankment 150 m in length. The height of the embankment at zero chainage is 0.5 m, the width is 11 m and side slopes 2:1. If the falling gradient of the embankment is 1 in 150, the quantity of the earthwork calculated by prismoidal formula, is
Discussion:
17 comments Page 1 of 2.
Mannu verma said:
3 years ago
A1 = (11+13) * 0.5/2,
Am = (11+17) * 1.5/2,
A2 = (11+22) * 2.5/2.
Put these values in prismatic formula.
Then, I find;
3250m3 is the right answer
We can find by given details
We can create diagrams in the mind.
The ground on slop 1/50 we will take bottom elevation,
Road slop will be given 1/150 top elevation.
Top and bottom elevation of A1 = 0.5, 0.
Top and bottom of elevation of Am = 0, - 1.5.
Top and bottom elevation of A3= -0.5, -3.
Am = (11+17) * 1.5/2,
A2 = (11+22) * 2.5/2.
Put these values in prismatic formula.
Then, I find;
3250m3 is the right answer
We can find by given details
We can create diagrams in the mind.
The ground on slop 1/50 we will take bottom elevation,
Road slop will be given 1/150 top elevation.
Top and bottom elevation of A1 = 0.5, 0.
Top and bottom of elevation of Am = 0, - 1.5.
Top and bottom elevation of A3= -0.5, -3.
(2)
Eng.RAJ said:
1 decade ago
Prismoidal formula = L/6 *( A1+A2+4AM).
Where,
A1= Cross sectional area at 1 end = Bd1+Sd1*d1 - Sqm.
A2= Cross sectional area at another end = Bd2+Sd2*d2 - Sqm.
AM= Cross sectional area at middle = Bdm+Sdm*dm - Sqm.
dm= (d1+d2)/2 - m.
Where d1= depth at starting - m.
d2= depth at ending - m.
L= length of the section - m.
Where,
A1= Cross sectional area at 1 end = Bd1+Sd1*d1 - Sqm.
A2= Cross sectional area at another end = Bd2+Sd2*d2 - Sqm.
AM= Cross sectional area at middle = Bdm+Sdm*dm - Sqm.
dm= (d1+d2)/2 - m.
Where d1= depth at starting - m.
d2= depth at ending - m.
L= length of the section - m.
Tarun said:
4 years ago
B is right ans. The first trapezoidal area will be 5.75 with height .5m and second one's area will be 15.3125 with height 1.25m at 37.5m away.
A3 will be 26. Height 2m
A4 will be 28 Height 3m
A5 will be 60 Height 4m
Slope 2/75.
Area is calculated as trapezoid base width will be;
2 Height+ width.
A3 will be 26. Height 2m
A4 will be 28 Height 3m
A5 will be 60 Height 4m
Slope 2/75.
Area is calculated as trapezoid base width will be;
2 Height+ width.
(1)
Adnan said:
4 years ago
There is not clear idea in them.
Trapezoidal base area is a+b/2 into h.
Then it is getting the wrong value.
is there someone to do step by step it? Anyone explain it, please.
Trapezoidal base area is a+b/2 into h.
Then it is getting the wrong value.
is there someone to do step by step it? Anyone explain it, please.
Biswajit Mohanty said:
5 years ago
I solved this problem using prismoidal formula answer will be 3250 cum i.e. option A.
(1)
Soumya Ranjan Sahoo said:
2 years ago
{(11+13)/2×0.5 + (11+21)/2×2.5}/2 × 150 = 3450cum is the right answer.
(5)
Sb221 said:
4 years ago
I'm not getting this, Please explain the answer.
Egide Ndatimana said:
2 years ago
Please explain to us how A be the right answer?
Irfan said:
6 years ago
Not getting it, please solve it clearly.
Santosh said:
5 years ago
Please explain the solution clearly.
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