Question

12) of 7.5.km. What for overhaut, if the rate for overhaut in lor an average 2.2 Refer to Table 2-1 Plot the mass haul d 2 km

Please help with 2.2 (a) and (b)

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Answer #1

Here's the excel sheet containing the mass haul diagram in terms of cumulative cross sectional area (m2) with chainage:

CHAINAGE CUT AREA (m2) FILL AREA (mA2) CUMULATIVE AREA 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600
And here's the Mass haul diagram with directions of hauling (marked with arrows) and cut and fill sections:

Mass diagram 60 50 40 30 4 45 38 32 24 24 20 15 - Mass diagram 6 -2 2000 1000 1500 2500 -8 -10 -20 -30 -13 -1 -22
We can see that at the end of the Mass haul diagram, we are left with -2 m2 of balance which means we need to borrow that much form a borrow pit nearby.

Now, due to the shrinkage factor, 0.92, the fill sections will get compacted and the earth required to fill those sections will be increased by a factor of 1/0.92 = 1.087. Therefore, the new Mass haul diagram will look like:

8111 037 834 121 2527800 4443221 73008 - 25 14777653 D07 74 89500000000025 8897 6 89500000000025 8897 6 00009 12 7 9 9 8 6 63
Mass Haul Diagram (Corrected fill) 50.00 4.2.91 36.09 40.00 37.48 30.00 22.09 20.00 13.09 10.00 -- Cumulative Area 4.09 2.70
We can clearly see that now, the requirement of earth is 8.17 m2 due to shrinkage. The borrow pits aren't necessarily nearby to the project sites. Also, sometimes, the terrain makes it difficult to haul earth from a borrow pit to the work site. Therefore, spoiling is the better option.

Thank you. Upvotes are appreciated.

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