MASS HAUL DIAGRAM

General Objective : To understand the basic concepts of the use of planning in the haulage of large volumes of earthwork for construction. Specific Objectives : At the end of the unit you should be able to :- > explain the basic concepts of the Mass Haul Diagram. > define the use of Mass Haul Diagram. > explain the methods used in the haulage of large volumes of earthwork calculation . > explain the method used in calculating the haulage of large volumes of earthwork for construction

 

Prepared by: SAIFULBAHARI B MOHD RASID

INTRODUCTION

Mass-haul diagrams (MHD) are used to compare the economics of the various methods of earthwork distribution on road or railway construction schemes. With the combined use of the MHD plotted directly below the longitudinal section of the survey centre-line, one can find :

The distances over which ‘cut and fill’ will balance.

Quantities of materials to be moved and the direction of movement.

Areas where earth may have to be borrowed or wasted and the amounts involved.

The best policy to adopt to obtain the most economic use of plan.

DEFINITION AND IMPORTANT PHRASES

Bulking An increase in volume of earthwork after excavation

Shrinkage A decrease in volume earthwork after deposition and compaction.

Haul distance (d)The distance from the working face of the excavation to the tipping point.

Average haul distance (D)The distance from the centre of gravity of the cutting to that of the filling.

Freehaul DistanceThe distance, given in the Bill of Quantities, included in the price of excavation per cubic metre.

Overhaul DistanceThe extra distance of transport of earthwork volumes beyond the freehaul distance.

HaulThe sum of the product of each load by its haul distance. This must equal the total volume of excavation multiplied by the average haul distance, i.e. ∑ vd = VD

OverhaulThe product of volumes by their respective overhaul distance. Excess payment will depend upon overhaul.

Station Metre A unit of overhaul, viz. 1 m3 x 100 m.

Borrow The volume of material brought into a section due to a deficiency.

Waste The volume of material taken from a section due to excess

Mass-haul diagram

Freehaul and Overhaul

CONSTRUCTION OF THE MASS-HAUL DIAGRAM 

Volumes of cut and fill along a length of proposed road are as follows :  

Volume

Chainage Cut Fill0

100 290200 760300 1680400 620480 120500 20600 110700 350800 600900 780

1000 6901100 4001200 120

Draw a mass-haul diagram and exclude the surplus excavated material along this length. Determine the overhaul if the freehaul distance is 300 m.

Volume Aggregate

Chainage Cut Fill volume0

100 290 + 290200 760 + 1050300 1680 + 2730400 620 + 3350480 120 + 3470500 20 + 3450600 110 + 3340700 350 + 2990800 600 + 2390900 780 + 1610

1000 690 + 9201100 400 + 5201200   120 + 400

3470 30703070

check 400

Draw mass haul diagram

Figure 3.3 : Mass-haul diagram (Source : Land Survey, Ramsay)

Graphical Method (figure 3.3)

1.As the surplus of 400 m³ is to be neglected, the balancing line is drawn from the end of the mass-haul curve, parallel to the base line, to form a new balancing line ab.2. As the freehaul distance is 300 m, this is drawn as a balancing line cd. 3. From c and d, draw ordinates cutting the new base line at c1d1.

4.To find the overhaul :i.Bisect cc1, to give c2 and draw a line through c2 parallel to the base line

cutting the curve at e and f, which now represent the centroids of the masses acc1 and dbd1.

ii.The average haul distance is the centroids of the masses acc1 and dbd1.

iii.The overhaul distance = the haul distance – the free haul distance

Planimetric method

Distance to centroid = haul / volume = (area x horizontal scale x

vertical scale) / volume ordinate from area acc1

area scaled from mass-haul curve = 0.9375 cm² horizontal scale = 1 cm = 200 cm vertical scale = 1 cm = 1600 cm³   Therefore haul = 0.9375 x 200 x 1600 = 300 000 volume (ordinate acc1 ) = 2750

distance to centroid = 300 000 / 2750 = 109.1 m chainage of centroid = 350 – 109.1 = 240.9 m   for area dbd1

area scaled = 1.9688 cm²   Therefore haul = 1.9688 x 320 000 = 630 016 volume (ordinate dd1 ) = 2750 distance to centroid = 229.1 m chainage of centroid = 650 + 229.1 = 879.1 m average haul distance = 879.1 -240.9 = 638.2 m overhaul distance = 638.2 – 300 = 338.2 m Therefore overhaul = 338.2 x 2750 = 9300 station

metres

Instead of the above calculations, the overhaul can be obtained direct, as the sum of the two mass-haul curve areas acc1 and dbd1 is:

Area acc1 = (302 950) / 100 station metre Area dbd1 = (634 950) / 100 station metre Total area = overhaul = (936 900) / 9369 station metre Proof : Take any cutoff by a balancing line, Figure 5.4. Let a small increment of area δ A =

(say) 1 m³ and length of haul be L. Then δA = 1 m³ x L / 100 station metre    

Therefore A = n x 1 m³ x ( ∑L / n ) = total volume x average haul distance Therefore, Area = total haul