Download pdf - Mass and Haul

8/2/2019 Mass and Haul

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Problem:

Determine the total volume of earth to be excavated up to elevation 0.

10m 10m

10m

10m

6 8 5

0-315

11 10 6

I II

III IV


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Mass and Haul DiagramMass and Haul Diagram

HAUL

: product of earth excavated and the distance it was transported to form an

embankment or to be disposed as waste.

unit: meter station (metric) = 1 m3 to be transported

FREE HAUL DISTANCE (FHD)

: fixed distance within which the hauling of material is not paid for but it isassumed to be included in the cost of excavation.

OVERHAUL (OH)

: product of volume in excess of the free haul mass and the length of overhaul

in which payment is already required.

LENGTH OF OVERHAUL (LOH)

: distance between the center of gravity of mass excavation beyond the free

haul mass and the center of gravity of the resulting embankment.


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LENGTH OF ECONOMICAL HAUL

: distance within which it is economical to haul materials than to throw them

as waste.

WASTE

: excavated earth to be thrown away and not to be used as filling materials.

BORROW

: the excavated materials obtained from borrow pits located beyond the limits

of the road when the excavation for the roadway is not sufficient to formembankments.

Mass and Haul DiagramMass and Haul Diagram


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Vcut=VfillVcut=Vfill

FHD FHD

xLxR xL xR

FGL

cg cg

cgcg

NGL

wasteOHM=Vcut

FHM

borrow

FHM OHM waste

LEH LEH

Profile DiagramProfile Diagram


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Problem:

Here under shows a table of quantities of earthworks of a

portion of a proposed national road to be constructed toconnect some provinces in the Philippines. The length of thefree haul distance is specified to be 50m long. It is requiredto determine the stationing limits of free haul distance so asnot to include in the computation of overhaul.

Station Cut (m2) Fill (m2)2+000 60

+020 35+040 20+060 10

+080 0 0

2+100 15+120 60

+140 100


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2+140

2+000

2+080

Station Cut (m2) Fill (m2)2+000 60

+020 35

+040 20+060 10+080 0 02+100 15+120 60

+140 100

60

3520

10

15y2

60 100

y1

FHD = 50m


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1

2

3

4

1


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sta2+060

sta1+760s

ta1+460

Problem:

The cross-sectional area of sta 1+460 is 40m2 in fill, at sta 2+060 is

60m2 in cut. The free haul distance is 50m. The cost of borrow isPhP4.00/m3 while the cost of excavation is PhP3.50/m3. Cost of haulis PhP0.20 per meter-station. Balancing point is at sta 1+760.Assume the ground surface to be sloping upward uniformly up tostation 1+760 and then with a slightly steeper slope to sta 2+060.

60

40

y1y3

y2

y4

300 300

xRxL

x 50-x

FHD=50mz

400-z

LEH=450m


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Find the following:

a.Sta limits of FHDb.Sta limits of LEHc.Length ofOverhauld.Overhaule.Cost of Haul

f.Cost of Wasteg.Cost of Borrow

1

2

3


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Properties of a Mass DiagramProperties of a Mass Diagram

1. Grade point on the profile corresponds to a max ormin point on the mass diagram.

2. Ascending line denotes cut or excavation anddescending line denotes fill or embankment.

3. The difference in length between any two verticalordinates of the mass diagram is the volumebetween the stations at which the ordinates wereerected.

4. Between any points, where the curve is intersectedby a horizontal line, excavation equals embankment.

5. In the mass diagram, a loop that forms a peakindicates a haul forward and a loop that forms a sagindicates a haul backward.


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Problem:

The following data represents

a single summit massdiagram of a proposedexpansion of the TolosaExpressway.

Free Haul Distance = 50m

Limit of Economical Haul =400mMass Ordinate of Initial Limitof the Free Haul = +910m3

Mass Ordinate of the initialLimit of Economical Haul =+350

1.Compute the volume ofwaste in m3.2.Compute the volume ofoverhaul in m3.3.Compute the volume of

borrow in m3.

Station Volumes

Cut Fill

10+000 +200

10+040 +100

10+080 +150

10+120 +140

10+160 +110

10+200 +190

10+240 +5010+280 -40

10+320 -120

10+360 -90

10+400 -80

10+440 -200

10+480 -22010+520 -110

10+560 -320

10+600 -280


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Station Volumes Mass

Ordinate

Cut Fill

10+000 +200 +20010+040 +100 +300

10+080 +150 +450

10+120 +140 +590

10+160 +110 +700

10+200 +190 +890

10+240 +50 +940

10+280 -40 +900

10+320 -120 +780

10+360 -90 +690

10+400 -80 +610

10+440 -200 +41010+480 -220 +190

10+520 -110 +80

10+560 -320 -240

10+600 -280 -520


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1. Volume of wasteVolume of Waste = 350 200Volume of Waste = 150 m3

2. Overhaul VolumeOverhaul Volume = 910 350Overhaul Volume = 560 m3

3. Volume of Borrow

Volume of Borrow = 350 + 520Volume of Borrow = 870 m3