Civil Engineering - Estimating and Costing - Discussion

Actually, none of the answers is correct.

From calculations, we get that the formation line intersects the ground surface at a distance of 12m from chainage point 6. (let us call this point 'X').

Also, when chainage is represented as (A+B) chains, it refers to "A chains" and "B links".

Since the distance between consecutive chainage points is 30m, we can say that a 30m metric chain is used for chaining and a 30m chain has 150 links, each 20cm (or 0.2m) long.

So, number of links between the points '6' and 'X' = (distance between points '6' and 'X') / (link length).
= (12)/(0.2)
= 60.

Hence, the correct way of representing the point 'X' is (6+60) chains.

Gradient given is 1:120
So for 30m rise of = (1/120*30),
= 0.25 m.

RL of formation level @6 chain = 100.0 + 0.25.
= 100.25 m.
RL of formation level @7 chain = 100.25+ 0.25
=100.5 m.

So difference = 102-100.5 = 1.5.

By similar triangles -
(1/x) = 1.5/(30-x)
X = 12.

So the answer is (6+12) chains.

@6 difference b/w level is (100.25-99.25)=1.
@7 RL=100.25+0.25=100.5.

So difference= 102-100.5=1.5,
let x be distance at no filling or cutting,
distance b/w 7 and 6 is 30,
By similar triangle,
1 -> x,
1.5 ->30-x.

1 in 120.

So for 30 m there is rise of 0.25 m.

R.L of formation level at 6 chain = 100+0.25 = 100.25.

Now apply the similar triangles as 1/x = 1.5/(30-X).

Where, 1 = 100.25-99.25.

x = 12.

So the answer B.

99.25+(2.75/30)*x = 100+(x+30)/120.

By solving it x=12m from chainage 6 towards ch. 7, so answer will be (6+12).

The number of neutrons will be 6 so the total number becomes 6 + 6 is equal to 12.

I think the correct answer is 6+14 chains.

Thanks for your explanation @Gance.

I think the answer is (6+8) chains.

@Gance.

Where is 1.5 from?