Introduction

Route surveys involve measuring and computing horizontal and vertical angles, elevations, and horizontal distances. The results of these surveys are used to prepare detailed plan and profile base maps of proposed roadways. In addition, the elevations determined in the survey serve as the basis for calculation of construction cut and fill quantities, and in determining roadway banking. This section presents a review of basic terminology, concepts, and standard procedures used in highway surveys. The review begins with some basic definitions.

Highway curves can be either circular arcs or spirals. A simple curve is a circular are connecting two straight lines (tangents). A compound curve consists of two or more circular arcs of different radii tangent to each other with their centers on the same side of the common tangent. Compound curves where two circular ares having centers on the same side are connected by a short tangent are called broken-back curves. A reverse curve is two circular arcs tangent to each other but with their centers on opposite sides of the common tangent. A curve whose radius decreases uniformly from infinity to that of the curve it meets is called a spiral curve. Spiral curves with the proper superelevation (banking) provide safe and smooth riding qualities. Circular and spiral curves are used for curves in the horizontal plane. Tangents in the vertical plane are joined by parabolic curves (also referred to simply as vertical curves)

A route surveying system usually contains four separate but interrelatedprocesses:

• Reconnaissance and planning

• Works design

• Right of way acquisition

• Construction of works

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DEFINITION OF TERMS

SIMPLE CURVE Most commonly used for highways and railroads construction. Circular arc, extending from one tangent to the next

PC Point of the curvature The point where the curve leaves the first tangent

PT Point of the tangency The point where the curve leaves the second tangent

PC and PT Tangent points

VERTEX Point of the intersection of the two tangents

TANGENT DISTANCE (T) Distance from the vertex to the PC and PT

EXTERNAL DISTANCE (E) Distance from the vertex to the curve

MIDDLE ORDINATE (M) The line joining the middle of the curve and the mid-point of the chord

joining the PT and PC

DEGREE OF CURVE (D) Generally used for highway practice (when the radius of the curve is

usually small) It is the angle of the center subtended by an arc of 20m (SI) or

100’(English)

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A. ARC BASIS

1. SI 2. ENGLISH

B. CHORD BASIS

The degree of the curve is the angle subtended by a chord of 20m (SI) or 100’ (English)

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ELEMENTS OF A SIMPLE CURVE

Tangent distance (T) External distance (E)

Tan I/2 = T/R cos I/2 = R/R+E T=R tan I/2 (R+E) cosI/2 = R R+E = Rsec I/2 E=RsecI/2-R E= R(secI/2-1)

Middle ordinate (M) Length of chord (LC)

cosI/2 = R-M/R sin I/2 = LC/2 /R Rcos I/2 = R-M LC = 2Rsin I/2 M = R-Rcos I/2 M = R(1-cos I/2)

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Length of curve (LCu) LCu/I = 20/D LC = 20I/D EXAMPLES:

1. The tangent distance of a 3˚ simple curve is ½ of its radius.Determine:

Angle of intersection (I) LC Area of the fillet of the curve

Solution:

D=3˚, T=1/2R LC=20I/DT=RtanI/2 LC=20(53.13)/3˚1/2R=RtanI/2 LC=354.2mtanI/2=0.5I/2=tan 0.5I=53.13˚

A=T(R)- R² o/360˚ A=(190.99)(381.97)-R=1145.916/D (381.97) (53.13)/360˚R=381.97 sqm A=5305.89sqm

2. The point of intersection of tangents on a simple curve is inaccessible falling within a river.

Points B and C on the tangents are connected by measurements on the ground.

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Figure:

REQUIRED: Distance CD and the length of curve Area of the cross-hatched section

Solution:

use sine law

I=180˚-45.48˚ o=180˚-47.5˚-87.02˚I=134.52˚ o=45.48 128.015/sino=T-53.58/sin87.02˚ T=232.88m

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Sine law: 232.86-CD/sin47.5˚=128.015/sin45.48˚ CD=100.51

LCu=RI, 20I/D LC=97.61(134.52˚)(‼/180) T=RtanI/2 LC=229.7m232.88=Rtan134.52˚/2R=97.61m

Area Asec-Atriangle LC=2(97.61)sin134.52˚/2 R²o/360˚-1/2(180.05)(37.73) LC=180.05m A=384.79sqm cosI/2=x/R x= 37.73m

If station PI=sta1+054 Req’d: staPC=1+054-232.88 staPC=0+821.21 staPT=staPC+LC staPT=0+821.12+229.17 staPT=1+050.29

3. Two tangents AB and BC intersecting in at an angle of 240˚. A point P is located 21.03 from point B and has a parallel distance of 2.79m from line AB.

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Figure:

sine law: sino=2.79/21.03 R/sin 70˚23’=R+E/sinβ o=7˚57’ ℓ=90˚-12˚-7˚37’ ℓ=70˚23’

E=R(secI/2) R/sin70˚23’=R+E/sinβE=R(sec 24˚/2-1) R/sin70˚23’=R+0.022312/sinβE=R(12˚-1) β=74˚21’ E=0.022312 180˚-74˚21’ =105˚39’

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o=180˚-β-ℓ o=180˚-105˚39’-70˚23’o=3˚58’

using sine law: R/sin 70˚23’=21.03/sin 3˚58’ R=286.36m

Length of chord connecting A and P sin8˚2’=LC/2/R LC=40.02m

Area of the fillet T=RtanI/2 A=TR- R²I/360˚

GIVEN:

AB=S65˚30’E required:BC=S25˚30’E; 170.75m R=?CD=S54˚20’W I=? Station PT if V is at sta 20+140

I=119˚50’

170.75m=T1+T2 170.75m=RtanI1/2+RtanI2/2

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170.75m=Rtan40˚/2+Rtan79˚50’/2170.75m=R(tan20˚+tan37˚25’)R=142.22m

T=RtanI/2T=142.22tan(119˚50’/2)T=245.57m

sta PC=sta V-Ista PC= 20+140-295.57msta Pc=19+894.44

sta PT=sta PC+LCLC=RI /180˚LC=142.22(119˚50’) /180˚sta PT=19+894.44+297.57sta PT=20+192.06

GIVEN/FIGURE:

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tan20˚=600m-OI/587.96m OI=386m

IP= (587.96)²+ (214)² IP= 625.69m

cosine law: OP²=625.69²+386² OP=2(625.69)(386)cos110˚ OP=840.05m

o=? sine law: 386/sino=840.05/sin110˚ o=25.58˚ β=180˚-110˚-25.58˚ β=44.42˚

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cosine law: 600²=840.05²+(Px)-2(840.05) (Px)cos25.58˚ Px=279.75m

LC=(600)(32.81˚)(‼/180˚)LC=343.59m

sta x=staA+LCsta x=50+000+343.59msta x=50+343.59

COMPOUND CURVES

Composed of two or more consecutive simple curve having different radii but whose center lie on the same side of the curve.

Any two consecutive curves must have a common tangent on their meeting PT.

PCC Point of compound curvature the PT on the common tangent the through which the two curves join.

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EXAMPLES:

1. The long chord from the PC to the PT of a compound curve is 300m long and the angle that it makes the longer and shorter tangents are 12˚ and 15˚ respectively. If the common tangent is parallel to the long chord.

Required: R1

R2

Station PT if PC is at sta 10+204.30

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sine law: 300m/sin166˚30’=LC1/sin7˚30’=LC2/sin6˚ LC1=167.74m LC2=134.33m

LC=2RsinI/2LC1=2R1sinI1/2 LC2=2R2sinI2/2167.74m=2(R1)sin6˚/2 134.33m=2(R2)sin7˚30’/2R1=802.36m R2=514.57m

LCu1=R1I1( ‼/180˚) LCu2=R2I2( ‼/180˚)LCu1=802.36n(6˚)( ‼/180˚) LCu2=514.57m(7˚30’)( ‼/180˚)LCu1=168.05m LCu2=134.71m

sta PT=staPC+LCu1+LCu2

sta PT=10+204.30+168.05+134.71sta PT=10+507.06

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I1=268˚30’-247˚50’I1=20˚40’I2=282˚50’-268˚30’I2=14˚20’ I=35˚

R1=1145.916/D1 T1=R1tanI1/2R1=1145.916/4˚ T1=286.479tan(20˚40’/2)R1=286.479m T1=52.23m

T2=76.42m-T1 T2=R2tanI2/2T2=76.42m-52.43m 24.19=R2tan(14˚20’/2)T2=24.19m R2=192.38m

staPC=staA-T1+LCu1 sta PC=10+010.46-52.23+103.33sta PC=10+061.56

LCu1=( R1I1)( ‼/180˚) LCu1=(286.479)(20˚40’)( ‼/180˚)LCu1=103.33msta PT=staPCC+LCu2

staPT=10+061.56+48.13staPT=10+109.69

LCu2=(R2I2)( ‼/180˚)

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LCu2=(192.38)(14˚20’)( ‼/180˚)LCu2=48.13m

REVERSE CURVES Composed of two consecutive circular simple curves having a common

tangent but lie on the opposite side.

PRC Point of the reverse curvature. The point along the common tangent to which the curve reversed in its

direction.

FOUR TYPES OF REVERSE CURVES:

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17

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EXAMPLE: The parallel tangent of a reversal curve are 10m apart the long chord from the PC to the PT is equal to 120m determine the following:

Radius of the curve Length of the common tangent

Solution: sinI/2=10/120 I=9˚33’

sinI=10/2T T=RtanI/2 2T=60.27m 30.14=Rtan(9˚33’/2) T=30.14m R=360.82m

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EXAMPLE:

Two converging tangent have azimuth of 300˚ and 90˚ respectively common tangent AB has an azimuth of 320˚. The distance from the point of intersection of two converging tangent and that of the vertex of the second curve has a distance of 100m. if the radius of the first curve is 285.4m between.Determine:

R2

sta PRC and sta PT if station of V1 is 10+040

Isolate triangle ABC

By sine law: 100/sin20˚=AB/sin30˚ AB=146.19m

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AB=T1+T2

146.19=R1tanI1/2+R2tanI2

R2=205.59m

station PRC= staV1-T1+LC1

station PT= staPRC+LC2

sta PRC=10+040-50.32+285.40(20)( ‼/180˚)sta PRC= 10+089.30

sta Pt=10+089.30+(205.59)(50)( ‼/180˚)sta PT= 10+268.71

EXAMPLE:

Two tangents 20m apart are to be connected by a reversed curve. The radius of the curve passing thru PC is 800m. if the total length of chord from PC to PT is 300m and stationing of PC is 10+620.Determine:

I R2

Station of PT

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sin I/2=20/300I=7˚38’

1st way to get the R2: AB=150-56=T1+T2

150-56=R1tanI1/2+R2tanI2/2 R2=1456.89

2nd way:

300=2R1sinI/2+2R2sinI/2 R2=1453.47

3rd way:

cosI=800-b/800 b=7.09 a=12.91 cosI=R2-12.91/R2

R2=1456.85

station PT=staPC+LC1+LC2 LC1=R1I( ‼/180˚) LC2=R2I( ‼/180˚) sta PT=10+920.67

Example:

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Solution:

X²=R²+100²…….eqn1 X²=(R-100)²+400²…….eqn2 R²+100²=R²-200R+100²+400²

200R=400² R=800m

tano=100/800 I=β-oo=7.13˚ I=29.74˚-7.13˚tanβ=400/200 I=22.61˚β=29.74˚

VERTICAL PARABOLIC CURVES

A curve used to connect two intersecting gradelines A curve tangent to two intersecting gradelines

TYPES OF VERTICAL PARABOLIC CURVES

1. SYMMETRICAL PARABOLIC CURVES

A parabolic curve wherein the horizontal length of the curve from the PC to the vertex is equal to the horizontal length from the vertex to the PT.

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ELEMENTS OF A SYMMETRICAL PARABOLIC CURVE

1. VERTEX (PI)2. PC3. PT4. BACKWARD TANGENT5. FORWARD TANGENT6. g1 and g2 (GRADES)

GUIDING PRICIPLES FOR SYMMETRICAL PARABOLIC CURVES

1. A given grade or slope ( in %) is numerically the rate at which an elevation changes in a horizontal distance. eg 5% = g

2. The vertical offset fro the tangent to the curve is proportional to the squares of the distances from the point of tangency. (Squared Property of a Parabola)

y1 / x1 = H / (L/2)2 = y2 / (x2)²2. The curve bisects the distance between the vertex and the midpoint of the

long chord.

BF / (L/2)² = CD / (L)²

3. If g1 - g2 (+) = “summit” g1 - g2 (+) = “sag”

4. No of stations to the left equal to the no of stations to the right.

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5. The slope of the parabola varies uniformly along the curve.

r = g2 - g1 / n ; n = 20m stationing

LOCATION OF THE HIGHEST OR LOWEST POINT OF THE CURVE

1. FROM PC

S1 = g1L / g1 - g2

2. FROM PT

S2 = g2L / g2 - g1

UNSYMMETRICAL PARABOLIC CURVES

Consist of a symmetrical parabolic curve from PC to PT. A,B another symmetrical parabolic curve tangent to that point A and PT

Used in provide a smooth and continues curve transition from PC to PT Point A is the common tangent point

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EXAMPLE:

Given: g2=-8% g1=5% L1=40m L2=60m

Required: Height of fill needed to cover the outcrop Elevation at station 6+820 Elevation of the HP

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Required: Elevation of the curve of the underpass If elevation curve is 22.6835m Stationing of the HP of the curve for question#2

2H/L1=(g1-g2)L2/L1+L2

L1=2HL2/(g1-g2)L2-2H160=2H(120)/(0.11)L2-2HH=3.77m

y/(60)²=H/(120)²y=0.94

elevation of the curve= elevation V-(60)(0.04)-y-helev curve=30-2.4-0.04-4.42elev curve=22.24m

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elev curve=22.6835mL2=?H=4.42m(remains the same)

L1=2HL2/(g1-g2)L2-2H160m=2HL2/(0.11)L2-2H……..eq’n1

Elev 22.6835=elevV-(60)(0.04)-y-4.4222.6835=30-2.4-y-4.42y=0.4985

0.4965/(L2-60)²=H/(L2)²H=0.4965(L2)²/(L2-60)²H=3.10m

160=2(3.10)L2/(0.11)L2-2(3.10)L2=100m

g1L1/2 ? H(0.07)(60)/2 ? 3.105.6 > 3.10

S2=g2L2/2H (from point PT)S2=(0.04)(100)/2(3.10)S2=64.52m

station HP=staV+35.48mstation HP=12+200+35.48mstation HP=12+235.48

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SPIRAL BASEMENT CURVE (TRANSITION SPIRAL CURVE)

A curve of ranging radius introduced at the outer edges of the roadway or track in order to allow the vehicle or train to pass gradually from the tangent to the circular curve.

A curve provided to smooth the elevation from the super elevation of the tangent to the maximum super elevation at the circular curve.

PRINCIPLES OF A SPIRAL CURVE

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The super elevation varies directly with the length of the space.

e/ec=L/Lc where: e super elevation of the spiral curve at any point ec super elevation at SC L length of the spiral from TS to any point Lc length of the spiral curve

The degree of curve varies directly with the length of the spiral

D/Dc=L/Lc where: D degree of the curve of the spiral at any point Dc degree of the spiral at SC

The spiral angle at any point on the spiral curve S=L²/2RcLc

The deflection angle varies directly at the square at the lengths from TS

i/ic=(L)²/Lc²where: i deflection angle at any point ic deflection angle at SC

The deflection angle is 1/3 of the spiral angle

i=S/3

FORMULAS

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Spiral angle, S where: s spiral angle of any point along the spiral Sc spiral angle at SC

D=1145.916/R Dc=1145.916/Rc

D=K/R let K 1145.916

D/Dc=L/Lc (K/R)/(Kc/Rc)=L/Lc

Solving R: R=RcLc/L …….eq1 L=ro dL=Rds……..eq2 dL=RcLc/L ds ∫ ds=∫ LdL/RcLc S=1/(RcLc)∫ LdL S=L²/2RcLc (in radius)

At SC: S=Sc : L=Lc Sc=Lc/2Rc (in radius)

Recall full arc length is equal to 20m

20m=DcRc Rc=20m/Dc

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Substitution to Sc=Lc/2Rc Sc=LcDc/40 (in degree)

Offset from tangent to spiral curve (x,Xc) where: x offset from tangent to any point along the spiral curve

Xc offset at SC or CS

Note: for small angle S sinS almost= S sinS=dx/dL dx=sins dL dx=SdL ∫dx=∫ L²/2RcLc dL x=1/2Rdc∫ L²dL x=L³/6RcLc

at SC or CS : X=Xc, L=Lc Xc=Lc²/6Rc

Distance along the tangent

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Slope correction formula:

c²=a²+h² (2c)(c-a)=h² c²-a²=h² c almost equal to a(c+a)(c-a)=h²

For spiral curves:

a=dy ∫ dy= ∫ dL-∫L dL/8Rc²Lc² b=dx y=∫ dL-(1/8Rc²Lc²)c=dL ∫ LydL y=L-(L /40Rc²Lc²)dy=dL-(dx²/2dL) sinS=dx/dLS=dx/dLdy=dL-S²(dL)²/2dLdy=dL-(L²/2RcLc)²dL

at SC:y=yc

L=Lc yc=Lc-(Lc³/40Rc²)

Deflection angle, i

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where: i deflection angle at any point along the spiral curve

sini=X/L X=L³/6RcLci=X/L i=L²/6RcLc

S=L²/2RcLc i=S/3

Length of throw, P

P=?Ts=?Es=?

180˚=I+2β Ts=?Β=90˚-Ic/2-Sc sinI/2=Ts-9/Rc+Es180˚=I+(90˚-Ic/2-Sc) Ts=Rc+Es(sinI/2+9)

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0=I-Ic-2ScIc=I-2Sc

P=?cosSc=Rc-(Xc-P)/RcP=Xc-Rc(1-cosSc)or P=Xc/4; P=Lc²/24Rc

Es=?cosI/2=Rc+P/Rc+EsEs=(Rc+P)secI/2-Rc

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 LOCATION

STAGES 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.

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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.

EARTHWORKS

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

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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:

I. Setting the StakesII. Keeping the Notes

The data for setting the stakes are:

A. The ground with center stakes set at every station.B. A record of benchmarks and of elevations and rates of grades established.C. The base and side slopes of the cross section for each class of material.D. 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 SLOPEORDINARY EARTH 1.50 : 1.00COURSE GRAVEL 1.00 : 1.00LOOSE ROCK 0.50 : 1.00SOLID ROCK 0.25 : 1.00SOFT CLAY OR SAND 2 or 3 : 1.00

SETTING THE STAKES

Setting the stakes work consists of:

a) 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

b) 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.

Figure

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Figure

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.

Figure

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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 elevationGround ROD = HI – Grad ElevationCUT = Grade ROD – Ground ROD

Figure

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

Grade ROD A = (HI)A – Grade ElevationFILL = Ground ROD A – Grade ROD A

1. 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:

1. 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 / Vertical

DR = B / 2 + SHR

Where:S = Side Slope HL = Side Height LeftB = Base pr Width DR = Distance out rightHR = Side Height Right DL = Distance out left

2. 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:

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A. 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

B. 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

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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 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.

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1. LEVEL SECTION IN CUTFigure

Centerheight = 1.83mBase for Cut = 8.00mSS for Cut = 1:1DR = DL = B / 2 + SC

= 4 + 1 (1.83) = 5.83

2. LEVEL SECTION IN FILLFigure

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Centerheight = 1.50mBase for Fill = 7.00mSS for Fill = -1.50 : 1.00DR = DL = B / 2 + SC

= 3.50 + 1.50 (1.50) = 5.75

B. 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 whre the ground is sloping the cross-section is called Three Level Section.

Figure

Base for Cut = 8.00mSS for Cut = 1.00:1.00DL = B / 2 + SHL

= 4.00 + 1(0.63) = 4.63m

DR = B / 2 + SHR

= 4.00 + 1(4.96) = 8.96m

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Figure

Base for Fill = 7.00mSS for Fill = 1.50:1.00DL = 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 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

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Base for Fill = 7.00mSS for Fill = 1.50:1.00DL = 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 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

Solution:

Grade Rod – Grade Rod @ M = 2.33+ 1.43 = 3.76

46

Measured Distance (DM) = 8.24Calculated Distance (DC) = B / 2 + SHR

Where: B / 2 = 4.5m HR = 3.76 – 1.46 = 2.30m

1.50 / 1.00 = SHR / 2.30SHR = 2.30 (1.50) / 1.00SHR = 3.45DC = 4.5 + 3.45 = 7.95Since DC < DM

--- Move In

Figure

Base for Cut = 8.00mSS for Cut = 1:1DL = B / 2 + SHL

= 4.00 + 1(2.75) = 6.75m

DR = B / 2 + SHR

= 4.00 + 1(3.60) = 7.60m

47

Figure

Base for Fill = 7.00mSS for Fill = 1.50:1.00DL = 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 CUT

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

Figure

48

Base for Cut = 8.00mSS for Cut = 1:1DL = B / 2 + SHL

= 4.00 + 1(2.60) = 6.60m

DR = B / 2 + SHR

= 4.00 + 1(3.47) = 7.47m

E. 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 = 8.00mBase for Fill = 7.00mDL = B / 2 + SHL

= 3.50 + 1.50(3.60) = 8.99m

DR = B / 2 + SHR

= 4.00 + 1(3.67) = 7.47m

49

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 8.00mSIDE 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 WIDTH 8.00m (Cut) 5.00 (Fill)SIDE SLOPE 1.00:1.006.97 / -3.47 -0.61 1.04 / 0.00 3.66 / -5.94 3.44 / 2.44

50

A. LEVEL SECTION IN CUT

FIGURE

B. THREE LEVEL SECTION IN FILL

FIGURE

51

C. FIVE LEVEL SECTION IN CUT

FIGURE

D. IRREGULAR SECTION IN FILL

FIGURE

52

E. SIDE HILL SECTION

FIGURE

METHODS OF DETERMINING VOLUMES O EARTHEWORKS

FIGURE

A. By Average End Areas

53

V = L / 2 (A1 + A2)Where:V = Volume of Section of Earthworks between Sta 1 and 2, m³A1 , A2 = Cross – sectional area of end stations, m²L = Perpendicular Distance between the end station, m

NOTES:

1. The above volume formula is exact only when A1 = A2 but is approximate A1 <> A2.

2. 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.

3. 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 Formula

V = L / 6 (A1 + AM + A2)Where:V = Volume of section of earthwork between Sta 1 and 2 of volume of prismoid, m³A1 , A2 = cross – sectional area of end sections, m²AM = Area of mid section parallel to the end sections and which will be computed as the averages of respective end dimensions, m³

NOTES:

1. 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.

2. 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.

PRISMOIDAL CORRECTION FORMULA

54

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, m³L = Perpendicular distance between 2 parallel and sections, mb1 = Distance between slope stakes at end section ABC where the altitude is h1, mb2 = Distance between slope stake at end section DEF where the altitude is h2, mh1 = Altitude of end section ABC at Sta 1, mh2 = 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:

55

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

B. 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.

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

1. To LEVEL SECTIONSa. The area of a level section BC + SC (B is the base, C is the center point,

and S is the side slope.)b. Equate this area forced per the irregular or five-level sectionc. Base SS being known, a quadratic formula in one unknown is formed from

which C is determined.d. Solve for the corresponding value of C.

2. To THREE LEVEL SECTIONS

Figure

Total Area of three level section in cutA = A1 + A2

56

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:

1. 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:

A. END AREA METHODB. PRISMOIDAL FORMULAC. END AREA METHOD and PRISMOIDAL CORRECTION FORMULA

SOLUTION:

Compute for the area at each station cross-section and at mid-sectionFigure

57

STATION CROSS – SECTION NOTES10 + 000 +6.55 + 2.84 +2.84 +6.55 + 2.8410 + 020 +7.55 + 3.64 +1.85 +3.65 + 0.52

Check for Cut distances

DR1 = DL1

= B / 2 + SHR

= 1 / 2 (6m) + 1.25(2.84) = 6.55mArea by method of triangle and rhombusA1 = BC + SC² = 27.12m²

Figure

58

Check for the distancesDR2 = 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

59

DRm = 1 / 2 (DR1 + DR2) HRm = 1 /2 (HR1 + HR2) = 1 / 2 (6.55 + 3.65) = 1 / 2 (2.84 + 0.52)DRm = 5.10m HRm = 1.68m

DLm = 1 / 2 (DL1 + DL2) HLm = 1 / 2 (HL1 + HL2) = 1 / 2 (6.55 + 7.55) = 1 / 2 (2.84 + 3.64)DLm = 7.05m = 3.24m

HCm = 1 / 2 (HC1 + HC2) = 1 / 2 (2.84 + 1.85) HCm = 2.345m

Check for Cut distancesDRm = B / 2 SHRm

= 1 / 2 (6) + 1.25(1.68)DRm = 5.10m

DLm = B / 2 SHLm

= 1 / 2 (6) + 1.25(3.22)DLm = 7.05m

Area by method of triangle

Am = 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 EARTHWORK VOLUME OF CUT IN BETWEEN THE TWO STATIONS

Figure

60

1. By End Area MethodVe = L / 2 (A1 + A2)

Where:L = (10 + 020) – (10 + 000) = 20mA1 = 27.12m²A2 = 16.60m²

Then,Ve = 20 / 2 (27.12 + 16.60) = 437.20m²

2. By Prismoidal FormulaVp = L / 6 (A1 + 4Am + A2)

Where:L = 20mA1 = 27.12m²A2 = 16.60m²Am = 21.67m²

Then,Vp = 20 / 6 (27.12 + 4*21.67 + 16.60) = 434.13m³

3. Prismoidal Formula for CorrectionCp = L / 2 (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) = 0Cpb = 20 / 12 (2.84 – 1.85)(6.55 – 7.55) = -1.65m³Cpc = 20 / 12 (2.84 – 1.85)(6.55 – 3.65) = 4.785m³

61

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 – 3.135Vc = 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

STATIONS CROSS-SECTION NOTES10 + 040 +4.05

+0.84+3.00 +

3.50+2.85 +3.00

+2.12+7.05 +3.24

10 + 050 +7.80 +3.84

+2.00 +2.24

+3.25 +4.00 +2.50

+5.65 +2.12

SOLUTION:

1. Compute for the end areas of the end sectionsFigure

62

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

63

CONVERT THE END SECTIONS TO AN EQUIVALENT LEVEL SECTIONS

Figure

A1 = five level section – A1 (equivalent level section)1.11 = 13 HCe1 + S (HCe1)²3.11 = 6 HCe1 + 1.25 (HCe1)²; Let C1 = HCe1

3.11 = 6 C1 + 1.25 C1²

By quadratic formulaC1 = 2.52mDRe1 = B / 2 + SHCe1; DRe1 = DLe2

= 3 + 1.25(2.52)

Figure

64

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 formulaC2 = HCe2

Dre2 = B / 2 + SHRe2

= 3 + 1.25DLe2 = DRe2

Ex. Solve for the following:a.) Stationing of the limits of free haulb.) Stationing of the limits of economical haulc.) Vertical volumed.) Length of overhaule.) Cost of haulf.) Cost of wasteg.) Cost of borrow

Given:FHD = 50mCost of borrow = P 4.00/m3

Cost of excavation = P 3.50/ m3

Cost of haul = P 0.20/m station

StationEND AREA M2

CUT FILL

1+460 40

1+760 0 Balance point

2+060 60

65

LEH = Cb

Ch(C) + FHD =

4 (20)0.2

+ 50

LEH = 450 max

= 60

300 ; a =

60 x300

- eq. (1)

b50−x

= 40

300 ; b =

2000−40 x300

0.5ax = 0.5b(50 – x) - eq. (2)Equate 1 and 3; 2 and 3:x = 22.47 m ; a = 4.49 m2

50 – x = 27.53 m ; b = 3.67 m2

a.) Left limit = 1+760 – 27.53 = 1+732.47Right limit = 1+760 + 22.47 = 1+782.47

S = 50 m = FHD

a 'y

= 60

300 ; a’ =

60 y300

- eq. (1)

b'400− y

= 40

300 ; b’ =

16,000−40 y300

- eq. (2)

0.5(b’ + b + b)(400 - y) = 0.5(a’ + a + a)(y)(b’ + 7.34)(400 - y) = (a’ + 8.98)(y) - eq. (3)Equating (1) and (3):

y = 177.81mequating (2) and (3):

400 – y = 220.19 ma’ = 35.96 m2

b’ = 29.36 m2

b.) Limits of LEH:Left limit = 1+760 – 27.53 – 220.19 = 1+512.28Right limit = 1+760 + 22.47 + 179.81 = 1+962.28

LEH = 450

c.) OH Vol. = 0.5(a’ + a + a)(y) = 0.5(35.96 + 8.98)(179.81)= 4040.33 m3

OH Vol. = 0.5(b’ + b + b)(400 - y) = 0.5(29.36 + 7.39)(220.19)= 4040.49 m3

OH Vol. = 0.5(4040.33 + 4040.49)= 4040.41 m 3

66

d.) ATXL = ∑ AiXi

4040.49XL = 0.5(b’)(400 - y)(2/3)(400 - y) + b(400 - y)(0.5)( 400 - y) = 0.5(29.36)(220.19)( 2/3)( 220.19) + (3.67)( 220.19)(0.5)( 220.19)XL = 139.45 m

ATXR = ∑ AiXi4040.33XR = 0.5(a’)(y)(2/3)(y) + (a)(y)0.5(y)

= 0.5(29.36)(220.19)(2/3)(220.19) + (3.67)(220.19)(0.5) (220.19)XR = 113.89 m

Length of OH = XL + XR = 139.45 + 113.89Length of OH = 253.34 m

e.) Cost of haul = (OH Vol . ) (cost of haul )(lengthof OH )

C

= (253.34 ) (0.20 )(4040.41)

20

Cost of haul = P 10,235.97

f.) cost of waste = cost of excavation x vol. of waste

vol. of waste = 0.5(60 + a’ c)(97,72) = 0.5(60 + 35.96 + 4.49)(97.72) = 4908 m3

Cost of waste = P 3.50(4908)Cost of waste = P 17,178.00

g.) cost of borrow = vol. of borrow x cost of borrow

vol. of borrow = 0.5(40 + b’ + b)(52.28) = 0.5(40 + 29.36 + 3.67)(52.28) = 1909.0 m3

67