Lecture3( Earth Work and Mass Hual Diagram)2

University of Kerbala

College of Eng. Highway Engineering

Department of Civil Eng.

1 | 1 L 3 - E a r t h w o r k a n d m a s s h a u l d i a g r a m - - - - - - - - 2 0 1 3 / 2 0 1 4

EARTHWORKS AND MASS-HAUL DIAGRAM Estimation of Earthwork Quantities:-

1. ESTABLISHING OF THE VERTICL ALIGNMENT:

Determination of the grade for each section depends on: 1. Type of highway. 2. Location.

The grade may be chosen so as to balance the quantities of excavation (CUT) & embankment (FILL).

Then the vertical curves are: - Calculated. - Located on the profile. - Checked to ensure adequate sight distance.

2. CROSS-SECTION AREAS: The (-ve) sign indicates an embankment.

The (+ve) sign indicates an excavation.

Fill

Stripping

Stripping

Cut

Cut and Fill section

C.L.

Cut

Fill





Stripping: - the upper layer of organic material that must be removed before beginning an excavation or embankment. There are several different procedures for calculating areas of earthwork cross-sections. Recently, most earthwork calculations have been done by computer. The COORDINATE METHOD is used. In this method, points with known elevations are numbered consecutively around the cross-section (starting in either direction) as illustrated by the figure below.

112

1iii XXyA

Where n= no. of points for first point Xi-1= Xn for last point Xi+1= X1

Cross-sectional areas may be calculated by hand by dividing the cross-section into triangles and trapezoids.

Example: determine the cross section area of the figure below? Solution: A1=0.5(4(0-5)+0(11-0)+0(5-0)+6(0-11))=-43 m2 A2=0.5(7(0-19)+7(11-5)+0(19-0)+0(5-0)+6(0-11))=-157 m2 A1+A2=157+43=200 m2

1 2

3

4 5

6

7 Y- axis

X- axis

X1

y1

0,0

-11,-7 0,-6 11,-4

5,0 -5,0

-19,-7





3. DETERMINING THE EARTHWORK VOLUMES.

Earthwork volumes are normally calculated by the average end area method.

Average end area method.

)(2

21 AAL

V

where A1 and A2 are the end areas and L is the distance between them. This method is entirely accurate only if the two end areas are equal. If one end area is zero, the earthwork volume is a pyramid.

3

ALV (Pyramid formula)

Thus, as the area of one end of the earthwork volume approaches zero, the error in the volume calculated by the average end area method approaches 50 percent. A more accurate formula, known as the Prismoidal formula, is available. This is:-

6

)4( 21 AAALV m

p

(Prismoidal formula)

where Am is the area of a plane surface midway between cross-sections with areas A1 and A2. But use of the average end area method is more common because only approximate volumes are required for purposes of estimation.





4. DISTRIBUTION OF THE ERATHWORK QUANTITIES.

The Mass-Haul Diagram: The mass-Haul diagram is a graphical representation of the amount of earthworks involved in a highway schemes & the manner in which they may be most economically handled. It shows accumulated volume at any point along the center line. (+) cut (excavation). ( - ) fill (embankment).

Definitions: 1. Haul: Distance over which material is moved (m or station).

Volume-distance (m3.m or m3.sta.).

2. Free-Haul Distance (FHD): - The distance within which a contractor is paid a fixed amount per cubic meter of material irrespective of actual distance price, (free-haul charge=I.D./m3).

3. Over-Haul Distance (OHD):- Distance beyond free-Haul for which extra charges are required

for each (m3. sta.), (over-haul charge=I.D./m3.sta.).

4. Limit of Economical Haul Distance (LEHD): - The maximum Over-Haul distance plus Free-Haul distance beyond which it is more economical to waste & borrow than to pay for over hauling.

O.H.

F.H.

Waste

Borrow

Grade

Line

N.G.L.

OHD+FH

D





Economic Haul Limit (LEHD) = FHD + L

OH

B

C

CL sta.

where:- COH = cost of Over-Haul / m3. sta. CB = cost of borrow material / m3. FHD = Free-Haul distance. L = Maximum Over-Haul distance.

5. Shrinkage: A term used to explain the condition whereby a unit volume of excavation will

occupy less space when placed in compacted embankments.

e

f

V

VFS ..

Where: S.F = shrinkage factor. Vf = volume of fill. Ve = volume of excavation.

Correction for earthwork volumes





1. If the Subgrade soil is sand, silt, or clay then shrinkage (5-15) 10%. 2. If the Subgrade soil is rock, sand stone or lime stone then swell (bulking) (25-35) 30%.

Example:- Given the end areas below, calculate the volumes of cut and fill between stations 351+00 and 352+50. If the material shrinks 12 percent, how much excess cut or fill is there?

Station End areas, m2

Cut Fill

351+00 57.93

351+50 52.28

351+75 0 23.58

352+00 8.4 3.73

352+14 13.8 0

352+50 33.34

Solution:- Calculate earthwork volumes

351+00 to 351+50

3.2755)50(2

28.5293.57

Fill m3 (Average end area)

351+50 to 351+75

3.948)25(2

58.2328.52


351+75 to 352+00

4.341)25(2

73.358.23


703

)25(4.8Cut m3 (Pyramid)

352+00 to 352+14

4.17

3

)14(73.3Fill m3 (Pyramid)

4.155)14(2

8.134.8

Cut m3 (Average end area)

352+14 to 352+50

5.848)36(2

34.338.13

Cut m3 (Average end area)

Summary Station Cut, m3 Fill, m3





351+00

2755.3

351+50

948.3

351+75

70 341.4

352+00

155.4 17.4

352+14

848.5

352+50

Total 1073.9 4062.4

Excess fill = 4062.4-1073.9(1-0.12) = 3117.4 m3

Some characteristics of the mass-haul diagram shown at the next Figure are:- 1. All volumes are expressed in terms of excavated volumes and embankment data are adjusted to take this into account. 2. The ordinate at any station represents the earthworks accumulation to that point. The maximum ordinate (+) indicates a change from cut to fill, whilst the minimum ordinate (−) represents a change from fill to cut, proceeding along the centerline from the arbitrarily-assumed origin. 3. A rising curve at any point indicates an excess of excavation over embankment material, whilst a falling curve indicates the reverse. Steeply rising (or falling) curves indicate major cuts (or fills), whereas flat curves show that the earthworks quantities are small.





4. The shapes of the mass-haul loops indicate the directions of haul. Thus, a convex loop shows that the haul from cut to fill is from left to right, whilst a concave loop indicates that the haul is from right to left.

5. The curve starts with zero-accumulated earthworks and the baseline is the zero balance line, i.e. when the curve intersects this line again the total cut and fill will balance. A line that is drawn parallel to the baseline so as to cut a loop is called a 'balancing line’, and the two intersection points on the curve are called ‘balancing points’ as the volumes of cut and fill are balanced between them.

AB = FHD CD = FHV (volume) EF = LEHD GH = OHV (volume) EK = Waste (volume) FP = Borrow (volume)

6. The area between a balance line and the mass-haul curve is a measure of the haul (in station.m3) between the balance points. If this area is divided by the maximum ordinate between the balance line and the curve, the value obtained is the average distance that the cut material needs to be hauled in order to make the fill. (This distance can be estimated by drawing a horizontal line through the mid-point of the maximum ordinate until it intersects the loop at two points; if the loop is ‘smooth’ the length of this line will be close to the average haul).

7. Balance lines need not be continuous, i.e. a vertical break between two balance lines merely indicates unbalanced earthworks between the adjacent termination points of the two lines. Adjacent balance lines should never overlap as it means using the same part of the mass-haul diagram twice. In the Figure before the last Figure the economic-haul limits are drawn as the balance lines bd, fh, and km; this indicates that, generally, the earthworks volumes are not only balanced in volume but economically as well. The direction of haulage is shown by the arrows in the Figure; note that the haulage is downhill to the embankments so that the emptied vehicles can travel uphill to the excavation sites. The limits of free-haul are indicated by the balance lines 1–2, 3–4, and 5–6. The free-haul station-meterage is indicated by the dotted areas 1c2, 3g4, and 5l6. In this case, by chance, the balance line d–f is equal to the free-haul distance and, hence, the area def is also free-haul station-meterage. The overhaul volume for section BCD is given by the difference between the ordinates from c to b–d and from c to 1–2. The average length of overhaul is estimated by drawing the balance line 7–8 through the median of the overhaul ordinate; as the curve is smooth, the points 7 and 8 lie directly below the centers of mass of the overhaul volumes, and the average distance

Distance (sta.)

Cumulative

Volume (m3) A B

C

D

G

H E

K P

F

Convex

Concave





that this excavated material is moved is given by 7–8. Since the free-haul is given by 1–2, the average overhaul is the distance 7–8 minus the distance 1–2. In practice, selection of the optimum horizontal and vertical alignment rarely results in the earthworks being exactly balanced from beginning to end of the project. The earthworks are not fully balanced, and it can be seen that borrow material will have to be imported for the embankments between A and B and between M and P, and that the required quantities are given by the ordinates at b and m, respectively. Between H and K the excavated material will have to be wasted to a spoil tip as it is uneconomical to overhaul it for use in the embankment.

The LEHD is the longest distance material would ever be hauled and, hence, the longest balance line that would ever be used. There are cases, however, in which shorter balance lines are used. Most commonly these occur in multiple loops, for which use of the LEHD as the balance line would lead to ambiguous results. For the double loop shown in the figure below, an attempt to use the LEHD as the balance line for both loops leads to a situation in which material for a portion of the fill could be supplied from either of two cuts. In this case, it is necessary to determine a single balance line which will give the most economical point to reverse the haul.

To determine what this point is, consider the figure below, in which AB and BC are the lengths of the balance lines spanning the two loops. The change in the total cost CT when the balance line is moved up by a distance dy is given by:-

dyFHDBCFHDABCdC OHT

BCABCdy

dCOH

T

The minimum total cost may be found by setting this derivative equal to zero, so the condition for the minimum cost balance line is that:-

0BCABCOH

or AB = BC

LEHD

LEHD

A dy

B C





This may be generalized to any even number of loops. Similar logic can be used to determine the most economical balance line for a triple loop. Moving the balance line up by a distance of dy leads to a change in total cost of:-

dyCdyFHDBCFHDCDFHDABCdC BOHT

0 BOHT CFHDBCCDABC

dy

dC

FHDC

CBCCDAB

OH

B

LEHDBCCDAB Again, this can be generalized to any odd number of loops, with the condition for optimality being that the sum of the distances across the more numerous type of loop be equal to the sum of the distances across the less numerous type of loop plus the limit of economical haul. To calculate overhaul volume, the median haul distance for material involved in overhaul is measured by drawing a horizontal line halfway between the free haul line and the balance line, as shown in the figure below.

)( FHDMVOHV

where V = the total volume involved in the overhaul. M = median haul distance. FHD = free-haul distance.

EX: Given the following end area for cut & fill. Complete the earthworks using shrinkage of 90% then prepare the M.H.D. & find the following:

a) Limit of economical haul. b) Free-Haul volume. c) Over-Haul volume. d) Waste volume. e) Borrowing volume. f) Direction of hauling. g) Total cost of the earthwork.

Giving that cost of Over-Haul = 30 I.D./m3.sta.

A dy

B C D

M

Balance line

FHD

1/2 V V

1/2 V





Cost of Free-Haul = 70 I.D./m3 Cost of borrow = 120 I.D./m3 Free-Haul distance = 200 m.

Solution:-

Station Area (m2)

Volume (m3) Corrected Fill

Cumulated volume

Cut Fill Cut(+) Fill(-)

0 10 0

1100)100(2

1210

1100

1 12

1300 2400

2 14

1500 3900

3 16

1500 5400

4 14

350)50(2

014

250)50(2

010

275 5475

5 10

1300 1430 4045

6 16

1500 1650 2395

7 14

1300 1430 965

8 12

1000 1100 -135

9 8

300 200 220 -55

10 12

1400 1345

11 16

a) 30

120

b

CL = 4 Stations = 400 m.

Economic haul limit = L+FHD= 400+200 = 600 m = 6 Stations. b) Free-Haul volumes = y1+y2 = 700+900 = 1600 m3. c) Over-Haul volumes = y3+y4 = 3000+800 = 3800 m3. d) Waste volume = y5 = 1950 m3. e) Borrow volume = y6 = 600 m3. f) Total cost of the earthwork = y1*70+y3*70+y3*30*(XY-200)+y5*70+ y2*70+y4*70+y4*30*(mn-200)+y6*120 Cost of over-haul = y*cost of free-haul + y*cost of over-haul*distance





1000

2000

3000

4000

5000

0

- 1000

Station

Cumulated volume

(m3)

b

e f

FHD

OHD

6000

y2 c d

End of project

m n

2 sta.

y6

X Y

OH

FH

OH

FH

Bor

row

OH

Road Profile

XY=360 mn=280

Grade line

1 2 3 4 5 6 7 8 9 10 11

6 sta.

a

y3

y1

OHD

LEHD

y5 2 sta.

y4

<6 sta.

Waste





Home work:- For the given data

1. Draw the M.H.D and profile. 2. Indicate Free-Haul section. 3. Indicate all possible of over haul section and their

direction of haul. Choose the most economical. 4. Determine waste and borrow volumes between

stations 27-35. Cost of Over-Haul = 150 Fl. /m3 .sta. Cost of borrow = 1.5 I.D. /m3. Cost of Free-Haul = 0.8 I.D. /m3. Free-Haul distance = 3 sta. Solution:-

Sta. Vol. (m3)

20 0

21 500

22 1500

23 1800

24 2000

25 1620

26 1110

27 620

29 710

31 980

33 1140

35 1700





EX#1: For the data given below:

1. Find the cost of over-haul. 2. Plot the M.H.D. & profile. 3. Indicate free-haul & over-haul sections and their directions of haul. 4. Determine free-haul, over-haul, borrow & waste volumes. 5. Determine the total cost of the project.

Given that: Cost of free-haul = 130 I.D. /m3 Cost of borrow = 150 I.D. /m3 Free-haul distance = 2 stations Economic haul-distance = 700m

Sta. 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

Vol. 0 -700 -1500 -1612 -1580 -1380 -555 295 845 1245 1545 1575

1475 1308

908 -600

SOL: Economic haul distance = L + F 700 = L + 200 → L = 500

.//..30150

5

..cos

..cos

3 stamDIxx

overhouloft

borrwooftL

Free-haul volumes = y1 + y2 = 80 + 120 = 200m3

Over-haul volumes = y3 + y4 = 1520 + 880 = 2400m3 Waste volumes =y5 = 620m3 Borrow volumes =y6 = 1220m3

EX#2: For the data given below:

1. Plot the M.H.D. & profile 2. Indicate free-haul & over-haul sections & their direction of houl. 3. Determine free-haul, over-haul, borrow & waste volume 4. Determine the total cost of the project

Given that: Shrinkage factor = 0.9 Cost of free-haul = 0.3 I.D. /m3 Cost of over-haul = 0.2 I.D. / m3/sation





Cost of borrow = 0.8 I.D. /m3 Free-haul distance = 200m

Station 0 1 2 3 4 5 6 7 8 9

Cut (m3) 50 75 200 635 220 110 250 500

Fill (m3) 630 720 180 54

SOL:

station Cut (m3) Fill (m3) Corrected fill (m3) Exact volume(m3)

Cumulative volume(m3)

0 630 700 -700 0

1 -700

50 720 800 -750

2 -1450

75 180 200 -125

3 -1575

200 54 60 +140

4 -1435

635 +635

5 -800

220 +220

6 -580

110 +110

7 -470

250 +250

8 -220

500 +500

9 +280

.sta42.0

8.0

c

bL

Economic haul limit = L + F = 4+2 = 6sta. Free-haul volume = y1 = 80m3 Over-haul volume = y2 = 940m3 Borrow volume = y3 = 480m3 Waste volume = y4 = 760m3 Total cost = y1*0.3 + y2*0.3 + y2*0.2*(3.5-2) + y3*0.8 + y4*0.3 = 1200 I.D.





EX#3: for the data given below: 1. Plot the M.H.D. & profile 2. Indicate free-haul & over-haul section their direction of houl.2 3. Determine free-haul volume, over haul volume, borrow volume & waste

volume. 4. Determine the total cost of the project.

Given that: Cost of over-haul = 80 I.D. /m3.sta Cost of free-haul = 150 I.D. /m3

Cost of borrow = 400 I.D. /m3

Economic haul distance = 8 sta. = L + F

Sta. 1 2 3 4 5 6 7 8 9 10 11

Vol. -1100 -3200 -4500 -4150 -3500 -

1300 1500 3200 2600 -400 -2000

mstab

cL 500.5

80

400

Free-haul distance = 8 – 5 = 3sta. = 300m

Lecture3( Earth Work and Mass Hual Diagram)2

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