EARTHWORKSEARTHWORKS

• EARTHWORKS• Areas and Volumes of earthworks• Distribution Analysis (HAQL and MASS

DIAGRAM)

• ROUTE SURVEYING• DEFINITION• Route Surveying is a survey which supplies data

necessary to determine the alignment, grades, and earthworks quantities necessary for the location and construction of engineering projects. This includes highways, drainage, canal, pipelines, railways, transmission lines, and other civil engineering projects that do not close upon the point of beginning

ROUTE LOCATIONSTAGES OF HIGHWAY SURVEYS• Development of the interstate highway system

and more general acceptance of the limited access principle for major highways have resulted in a more and more highway projects being to serve local traffic, surveys for highway projects where new location is being considered start with a general study of the entire area between termini, proceed to more specific studies of possible alternative routes, and finally conclude with a detailed survey of the selected route and staking of the final centerline on the ground.

• These procedures are generally carried in three stages:

• RECONAISSANCE

• PRELIMINARY SURVEY

• LOCATION SURVEY

RECONAISSANCE

• Includes a general study of the entire area the development of one or more alternative routes or corridors, and the study of each of these corridors in sufficient detail to enable the proper officials to decide which will provide the optimum location.

PRELIMINARY SURVEY

• Is a survey of selected corridors in sufficient detail to permit staking of the final centerline on the ground in some cases, the preliminary survey may be completed and staked in the field without variation in other instances, Minor adjustments may be required during the location survey.

LOCATION SURVEY

• Consists in staking the final centerline and obtaining all additional information necessary to enable the design engineer to prepare completed plans, specifications, and estimates of earthwork quantities and to prepare deeds and descriptions covering the rights of way to be acquired.

EARTHWORKSEARTHWORKS• EARTHWORKS – the construction of large open

cuttings or excavations involving both cutting and filling of material other than rock.

• EXCAVATION – is the process of loosening and removing earth or rock from its original position in a cut and transporting it to a fill or to a waste deposit.

• EMBANKMENT – the term embankment describes the fill added above the low points along the roadway to raise the level to the bottom of the pavement structure material for embankment commonly comes from roadway cuts or designated borrow areas.

• SETTING STAKES FOR EARTHWORK

• The first step in connection with earthwork is staking out or setting slope stakes as it is commonly called.

• Two important parts of the work of setting slope stakes:

• Setting the Stakes

• Keeping the Notes

The data for setting the stakes are:• The ground with center stakes set at every

station.• A record of benchmarks and of elevations

and rates of grades established.• The base and side slopes of the cross

section for each class of material.• In practice, notes of alignment, a full

profile, and various convenient data are commonly given in addition to the above mentioned data.

• Side Slopes most commonly employed for cuts and fills.

• MATERIAL EXCAVATION SIDE SLOPE• ORDINARY EARTH 1.50 : 1.00• COURSE GRAVEL 1.00 : 1.00• LOOSE ROCK 0.50 : 1.00• SOLID ROCK 0.25 : 1.00• SOFT CLAY OR SAND 2 or 3 : 1.00

• SETTING THE STAKES

• Setting the stakes work consists of:

• Making upon the back of the center stakes the cut or fill in feet or meters and tenths, as C 2 3 or F 4 7

• Setting side stakes or slope stakes at each side of centerline at the point where the side slope intersects the surface of the ground and marking upon the inner side of the stake, cut or fill at that point.

FIGUREFIGURE

FIGUREFIGURE

• Process of determining the height of cut or fill at the center stake or at any other points between the center space and slope stake.

FIGUREFIGURE

• Let HI = elevation of the line of sight or telescope refereed fro known or assumed datum.

• Grade ROD = difference in elevations between the line of sight (HI) and the grade elevation

• Ground ROD = HI – Grad Elevation

• CUT = Grade ROD – Ground ROD

FIGUREFIGURE

• When the instrument is set up above the grade or subgrade:

• Grade ROD A = (HI)A – Grade Elevation

• FILL = Ground ROD A – Grade ROD A– When the instrument is set up below the

grade or subgrade:

• Grade ROD B = Grade Elevation – (HI)B

• FILL = Grade ROD B + Ground ROD B

SETTING SIDE OF SLOPE

(FIELD PROCEDURES)

• The cross – sectioning is done after the grade lines have been determined in the office. The amounts of cut and fill at the center are computed, the distances and their heights above the base, or below it of the slope stakes are determined as follows:

An engineer’s level is set up and rod readings are taken at the center and at trial point. Assume that the third trial point is on the slope, compute the distance fro the center using the following formulas:DL = B / 2 + SHL S = Horizontal / VerticalDR = B / 2 + SHR

Where:S = Side Slope HL = Side Height LeftB = Base pr Width HR = Side Height RightDL = Distance out leftDR = Distance out right

• Measure the distance from the center to the trial point, if this distance is less than the calculated distance, the rod is to be moved out for another trial point; if greater, the rod is to be moved in, if equal, the point is correctly located. A stake is placed here indicating the right of the slope point above or below the base or sub grade.

• ILLUSTRATION:– If the measured distance is greater than the

calculated distance, then the trial point is too far out the center line of the roadway and the direction to the rodman is to move in.

• Figure

– If the measured distance is less than the calculated distance, the trial point is too near to the centerline of the roadway and the direction to the rodman is to move out.

• Figure

• C. If the measured distance is exactly equal to the calculated distance, the point is correctly located and the slope stake is at on the ground indicating the height of the slope point above or below the ground.

• Figure

ROAD CROSS SECTIONROAD CROSS SECTION

A. LEVEL SECTION• If the ground level in a direction

transverse to the centerline, the only rod reading necessary is that the centerstake, and the distance to the slope stake can be calculated once the center cut or fill has been determined, such a cross-section is called level section.

1.LEVEL SECTION IN CUT1.LEVEL SECTION IN CUT• Figure

Centerheight = 1.83m

Base for Cut = 8.00m

SS for Cut = 1:1

DR = DL = B / 2 + SC

= 4 + 1 (1.83)

= 5.83

2.LEVEL SECTION IN FILL2.LEVEL SECTION IN FILL• FigureCenterheight = 1.50m Base for Fill = 7.00m

SS for Fill = -1.50 : 1.00 DR = DL = B / 2 + SC

= 3.50 + 1.50 (1.50) = 5.75

B.THREE LEVEL SECTIONB.THREE LEVEL SECTION

• When Rod readings are taken at each slope stake in addition to readings taken at the center as will normally be done whEre the ground is sloping the cross-section is called Three Level Section.

• FIGURE:

• Base for Cut = 8.00m SS for Cut = 1.00:1.00

• DL = B / 2 + SHL = 4.00 + 1(0.63) = 4.63m

• DR = B / 2 + SHR = 4.00 + 1(4.96) = 8.96m

• FIGURE:

• Base for Fill = 7.00m SS for Fill = 1.50:1.00• DL = B / 2 + SHL = 3.50 + 1.50(3.12) = 8.18m• DR = B / 2 + SHR = 3.50 + 1.50(2.62) = 7.43

C.FIVE LEVEL SECTIONC.FIVE LEVEL SECTION

• When rod reading is taken at the centerside the slope stake and at points on each side of the center of the distance of half the width of the road bed, the cross section is called a FIVE LEVEL SECTION.

• FIGURE:

• Base for Fill = 7.00m SS for Fill = 1.50:1.00

• DL = B / 2 + SHL = 3.50 + 1.50(2.42) = 7.13m• DR = B / 2 + SHR = 3.50 + 1.50(3.28) = 9.23m

SAMPLE PROBLEM (Setting SAMPLE PROBLEM (Setting Slope Stakes)Slope Stakes)

In setting slope stakes, the height of cut at the center has been found to be 1.43m, the ground readings at center “M” and trial point A on the slope are 2.33m and 1.46m, respectively, and the measured distance from the center line of the roadway to the trial point is 8.24m. If the base of the roadway is 9m and the side slope is 1.50:1.00, should the trial point be moved in or out?

• FIGURE:

SOLUTIONSOLUTION• Grade Rod – Grade Rod @ M = 2.33+ 1.43 =

3.76• Measured Distance (DM) = 8.24• Calculated Distance (DC) = B / 2 + SHR • Where: B / 2 = 4.5m• HR = 3.76 – 1.46 = 2.30m• 1.50 / 1.00 = SHR / 2.30• SHR = 2.30 (1.50) / 1.00• SHR = 3.45• DC = 4.5 + 3.45 = 7.95• Since DC < DM --- Move In

• FIGURE:

Base for Cut = 8.00m SS for Cut = 1:1

DL = B / 2 + SHL = 4.00 + 1(2.75) = 6.75m

DR = B / 2 + SHR = 4.00 + 1(3.60) = 7.60m

• FIGURE:

• Base for Fill = 7.00m SS for Fill = 1.50:1.00

• DL = B / 2 + SHL = 3.5 + 1.50(2.84) = 7.76m• DR = B / 2 + SHR = 3.50 + 1.50(2.92) = 7.88m

D. IRREGULAR SECTION IN CUTD. IRREGULAR SECTION IN CUT

• A cross section for which observation is taken to points between center and slope stakes at irregular intervals is called irregular section.

• FIGURE:

• Base for Cut = 8.00m SS for Cut = 1:1

• DL = B / 2 + SHL = 4.00 + 1(2.60) = 6.60m

• DR = B / 2 + SHR = 4.00 + 1(3.47) = 7.47m

E. SIDE HILL SECTIONE. SIDE HILL SECTION• Where the cross-section passes through

from cut to fill, it is called a SIDE HILL SECTION and an additional observation is made to determine the distance from center to the grade point. That is the point where subgrade will intersect the natural ground surface. A peg is usually driven to grade at this point and its position is indicated by a guard stake marked “Grade”. In this case also cross-section is taken additional plus station.

• Base for Cut = 8.00m Base for Fill = 7.00m• DL = B / 2 + SHL = 3.50 + 1.50(3.60) = 8.99m• DR = B / 2 + SHR = 4.00 + 1(3.67) = 7.47m

• PROBLEMS:

• In two ways, find the areas of each of the following cross-section note, given the corresponding bases and side slope if not given they are to be computed

• A. BASE WIDTH 8.00m• SIDE SLOPE 1.50:1.00 ? / 2.84• ? / 12.84 + 2.84

• B. BASE WIDTH?• SIDE SLOPE ?• 5.79 / -1.86 – 1.27 6.03 / -2.02

• C. BASE WIDTH ?• SIDE SLOPE ?• 7.85 / 3.08 4.00 / 3.65 + 3.27

4.00 / 2.83 8.05 / 3.24

• D. BASE WIDTH 8.00m• SIDE SLOPE 1.50:1.00• ? / -3.56 6.28 / -2.28 -2.32• 1.00 / -1.11 7.50 / -3.82• ? / -2.74

• E. BASE WIDTH8.00m (Cut) 5.00 (Fill)• SIDE SLOPE 1.00:1.00• 6.97 / -3.47 -0.61 1.04 / 0.00• 3.66 / -5.94 3.44 / 2.44

a. LEVEL SECTION IN CUTa. LEVEL SECTION IN CUT• figure

b. THREE LEVEL SECTION IN b. THREE LEVEL SECTION IN FILLFILL

• figure

c. FIVE LEVEL SECTION IN CUTc. FIVE LEVEL SECTION IN CUT

• Figure:

d. IRREGULAR SECTION IN FILLd. IRREGULAR SECTION IN FILL

• Figure:

e. SIDE HILL SECTIONe. SIDE HILL SECTION• Figure:

METHODS OF DETERMINING METHODS OF DETERMINING VOLUMES OF EARTHWORKSVOLUMES OF EARTHWORKS

figurefigure

A. By Average End AreasA. By Average End Areas

• V = L / 2 (A1 + A2)• Where:• V = Volume of Section of

Earthworks between Sta 1 and 2, m³• A1 , A2 = Cross – sectional area of

end stations, m²• L = Perpendicular Distance between

the end station, m

• NOTES:• The above volume formula is exact only when

A1 = A2 but is approximate A1 <> A2. • Considering the facts that cross-sections are

usually a considerable distance apart and that minor inequalities in the surface of the earth between sections are not considered, the method of end areas is sufficiently precise for ordinary earthwork.

• By where heavy cuts or fills occur on sharp curves. The computed volume of earthwork ay be corrected for curvature out of ordinarily the corrected is not large enough to be considered.

B. By Prismoidal FormulaB. By Prismoidal Formula

• V = L / 6 (A1 + 4AM + A2)• Where:• V = Volume of section of earthwork

between Sta 1 and 2 of volume of prismoid, m³

• A1 , A2 = cross – sectional area of end sections, m²

• AM = Area of mid section parallel to the end sections and which will be computed as the averages of respective end dimensions, m³

• NOTES:• A Prismoidal is a solid having for its two ends

any dissimilar parallel plane figures of the same no. of sides, and all the sides of the solid plane figures. Also, any prismoid may be resolve into prisms, pyramids and wedges, having a common altitudes the perpendicular distance between the two parallel end plane cross – section.

• As far as volume of earthworks are concerned, the use of Prismoidal formula is justified only if cross-section are taken at short intervals, is a small surface deviations are observed, and if the areas of successive cross-section cliff or widely usually it yields smaller values than those computed from average end areas.

C. PRISMOIDAL CORRECTION C. PRISMOIDAL CORRECTION FORMULAFORMULA

• Figure:

• CD = L / 12 (b1 – b2)(h1 – h2)• Where:• CD = Prismoidal Correction, It is subtracted

algebraically from the volume as determined by the average and the areas method to give the more nearly correct volume as determined by the Prismoidal formula, m³

• L = Perpendicular distance between 2 parallel and sections, m

• b1 = Distance between slope stakes at end section ABC where the altitude is h1, m

• b2 = Distance between slope stake at end section DEF where the altitude is h2, m

• h1 = Altitude of end section ABC at Sta 1, m• h2 = Altitude of end section DEF at Sta 2, m

PRISMOIDAL CORRECTION FOR IRREGULAR SECTION

• In prismoid, there should be equal number of slope in both bases so that on equal number triangles can be found. The Prismoidal correction can then be found. The Prismoidal correction can then be found using either the fundamental formula of correction, CD = L / 12 (b1 – b2)(h1 – h2)

• or any of the formulas derive from it, where, however one base or any a five level section or three level section and other.

• A five level section (or irregular section) or both bases are irregular sections or, if one base is a five level section and the other irregular section, the formulas cannot be directly applied without making certain assumptions because there are more triangles formed in one section than in the other. The determination of the correction is at best only approximate. For the purpose of determining the Prismoidal correction, the following may be used:

– Neglect the intermediate heights thereby reducing the sections into three level or level sections this is the most convenient method.

– Plot the irregular or five level sections on cross sections paper. Draw on this section two equalizing lines starting from the same point or the center height such that the error added equal the areas subtracted approximately by estimating the center height as well as the distances in the right or in the left can then be scaled. This is more accurate than method A but involves more work.

– Reduce the five level or irregular section by calculation to equivalent level or three level sections as follows:

1. To LEVEL SECTIONS

• The area of a level section BC + SC (B is the base, C is the center point, and S is the side slope.)

• Equate this area forced per the irregular or five-level section

• Base SS being known, a quadratic formula in one unknown is formed from which C is determined.

• Solve for the corresponding value of C.

2.To THREE LEVEL SECTIONS2.To THREE LEVEL SECTIONS• Figure:

• Total Area of three level section in cut• A = A1 + A2 • Where:• A1 = B / 4 (HL + HR)• A2 = C / 4 (B + S) (HL + HR)• Then K = BC / 2 + (HL + HR) (B / 4 – CS /

2)• NOTE:• The unknowns are C, HR and HL. Two

these should be assumed and the third computed. It is simpler to covert to level section.

• PROBLEM:– Given the following cross-section notes of a

roadway with a base of 6m and SS of 1.25:1.00, between the volume of the prismoid between the two-end sections by the following methods:

• END AREA METHOD

• PRISMOIDAL FORMULA

• END AREA METHOD and PRISMOIDAL CORRECTION FORMULA

• Given:

STATION CROSS – SECTION NOTES

10 + 000 +6.55 + 2.84+2.84 +6.55 + 2.84

10 + 020 +7.55 + 3.64+1.85 +3.65 + 0.52

• SOLUTION:• Compute for the area at each station cross-

section and at mid-section

Figure:

• Check for Cut distances

• DR1 = DL1

• = B / 2 + SHR

• = 1 / 2 (6m) + 1.25(2.84)

• = 6.55m

• Area by method of triangle and rhombus

• A1 = BC + SC² = 27.12m²

• Figure:

• Check for the distances• DR2 = B / 2 + SHR2 • = 1 / 2 (6) + 1.25(0.52)• = 3.65m• DL2 = B / 2 + SHL2 • = 1 / 2 (6) + 1.25(3.64)• = 7.55m• Area by method of triangle• A2 = Aa + AL + Ac + Ad • = 1 / 2 (3)(3.64) + 1 / 2 (1.85)(7.55) + 1 / 2 (1.85)

(3.65) + 1 / 2 (0.52)(3) • • A2 = 16.60m²

• Compute for the dimensions of the mid sections

Figure

DRm = 1 / 2 (DR1 + DR2) = 1 / 2 (2.84 + 0.52)

DRm = 5.10mDLm = 1 / 2 (DL1 + DL2)

= 1 / 2 (2.84 + 3.64)DLm = 7.05mHCm = 1 / 2 (HC1 + HC2) = 1 / 2 (2.84 + 1.85)HCm = 2.345mHRm = 1 /2 (HR1 + HR2) HLm = 1 / 2 (HL1 + HL2)= 1 / 2 (6.55 + 3.65) = 1 / 2 (6.55 +

7.55) HRm = 1.68m HLm = 3.24m

Check for Cut distancesDRm = B / 2 SHRm = 1 / 2 (6) + 1.25(1.68)DRm = 5.10mDLm = B / 2 SHLm = 1 / 2 (6) + 1.25(3.22)DLm = 7.05mArea by method of triangleAm = Ae + Af + Ag + Ah = 1 / 2 (3)(3.24) + 1 / 2 (7.05)(2.345) + 1 / 2

(5.10)(2.345) + 1 / 2 (3)(1.68)Am = 21.68m

COMPUTE FOR THE VOLUME OF COMPUTE FOR THE VOLUME OF EARTHWORK VOLUME OF CUT INEARTHWORK VOLUME OF CUT IN

BETWEEN THE TWO STATIONSBETWEEN THE TWO STATIONS• Figure

1. By End Area Method• Ve = L / 2 (A1 + A2)• Where:• L = (10 + 020) – (10 + 000)• = 20m• A1 = 27.12m²• A2 = 16.60m²• Then,• Ve = 20 / 2 (27.12 + 16.60)• = 437.20m²

2. By Prismoidal Formula

• Vp = L / 6 (A1 + 4Am + A2)

• Where:

• L = 20m

• A1 = 27.12m²

• A2 = 16.60m²

• Am = 21.67m²

• Then,

• Vp = 20 / 6 (27.12 + 4*21.67 + 16.60)

• = 434.13m³

Prismoidal Formula for CorrectionCp = L / 12 (A1 + A2)(b1 – b2)Note:Resolve the given prismoid into a series of triangular prismoid into a series of triangular prismoid.Cp = Cpa + Cpb + Cpc + Ppd

Where:

Cpa = 20 / 12 (2.84 – 3.64)(3-3) = 0

Cpb = 20 / 12 (2.84 – 1.85)(6.55 – 7.55) = -1.65m³

Cpc = 20 / 12 (2.84 – 1.85)(6.55 – 3.65) = 4.785m³

Ppd = 20 / 12 (2.84 – 0.52)(3-3) = 0

Then,

Cp = -1.65 + 4.785

= 3.135m³

4. Corrected Volume

• Vc = Ve - Cp

• = 437.20 – 3.135

• Vc = 434.065m³

• Given the following cross section notes, determine the volume of the prismoid b end areas method and apply the Prismoidal formula. The roadway base is 6m with side slope of 1.25:1.00

• STATIONSCROSS-SECTION NOTES• 10 + 040 +4.05 +3.00 +2.85 +3.00 +7.05• +0.84 +3.50 +2.12 +3.24• 10 + 050 +7.80 +2.00 +3.25 +4.00 +5.65

+3.84 +2.24 +2.50 +2.12

• SOLUTION:• Compute for the end areas of the end sections• figure:

• Check for distances:• DR1 = B / 2 + SHR1 DL1 = B / 2 + SHL1• = 3 + 1.25(3.24) = 3 + 1.25(0.84)• DR1 = 7.05 DL1 = 4.05

• A1 = A1 + A2 + A3 + A4 • = 1 / 2 (3.5)(1.05) + 1 / 2 (2.85) + 3.5(3) + 1 / 2

(2.12 + 2.85)(3) • + 1 / 2 (2.12)(4.05)• A1 = 23.11m²

• Figure:

• Check for distances:• DR2 = B / 2 + SHR2 DL2 = B / 2 + SHL2• = 3 + 1.25(2.12) = 3 + 1.25(3.85)• DR2 = 5.65m DL2 = 7.80m

• A2 = Aa + Ab + Ac + Ad + Ae + Af • = 1 / 2 (2.12 + 3.84)(5.80) + 1 / 2 (2.42 +

3.25)(2) • + 1 / 2 (2.50 + 3.25)(4) + 1 / 2 (2.12 +2.50)

(1.65) • – 1 / 2 (4.00)(3.84) – 1 / 2 (2.65)(2.52)• A2 = 27.11m²

CONVERT THE END SECTIONS TO AN

EQUIVALENT LEVEL SECTIONS

• Figure

• A1 = five level section – A1 (equivalent level section)

• 1.11= 13 HCe1 + S (HCe1)²• 3.11 = 6 HCe1 + 1.25 (HCe1)²; Let C1 = HCe1 • 3.11 = 6 C1 + 1.25 C1²

• By quadratic formula• C1 = 2.52m• DRe1 = B / 2 + SHCe1; DRe1 = DLe2 • = 3 + 1.25(2.52)

• Figure:

• A2 = irregular section = A2 (equivalent level section)

• 27.11 = BHCe2 + S(HCe2)²

• 27.11 = 6HCe2 + 1.25(HCe2)²

• Let Ce = HCe2

• By quadratic formula

• C2 = HCe2

• Dre2 = B / 2 + SHRe2

• = 3 + 1.25

• DLe2 = DRe2