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 interrelated processes:
• Reconnaissance and planning • Works design
• Right of way acquisition • Construction of works
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)
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)
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
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/D T=RtanI/2 LC=20(53.13)/3˚ 1/2R=RtanI/2 LC=354.2m tanI/2=0.5 I/2=tan 0.5 I=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.
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
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=229.7m 232.88=Rtan134.52˚/2 R=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
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/sinβ E=R(sec 24˚/2-1) R/sin70˚23’=R+0.022312/sinβ E=R(12˚-1) β=74˚21’ E=0.022312 180˚-74˚21’ =105˚39’
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’
R=142.22m
T=RtanI/2
T=142.22tan(119˚50’/2) T=245.57m
sta PC=sta V-I
sta PC= 20+140-295.57m sta Pc=19+894.44 sta PT=sta PC+LC LC=RI /180˚ LC=142.22(119˚50’) /180˚ sta PT=19+894.44+297.57 sta PT=20+192.06 GIVEN/FIGURE:
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˚
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+LC sta x=50+000+343.59m sta 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.
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
sine law: 300m/sin166˚30’=LC1/sin7˚30’=LC2/sin6˚ LC1=167.74m LC2=134.33m LC=2RsinI/2 LC1=2R1sinI1/2 LC2=2R2sinI2/2 167.74m=2(R1)sin6˚/2 134.33m=2(R2)sin7˚30’/2 R1=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+134.71 sta PT=10+507.06
I1=268˚30’-247˚50’ I1=20˚40’ I2=282˚50’-268˚30’ I2=14˚20’ I=35˚ R1=1145.916/D1 T1=R1tanI1/2 R1=1145.916/4˚ T1=286.479tan(20˚40’/2) R1=286.479m T1=52.23m T2=76.42m-T1 T2=R2tanI2/2 T2=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+103.33 sta PC=10+061.56 LCu1=( R1I1)( ‼/180˚) LCu1=(286.479)(20˚40’)( ‼/180˚) LCu1=103.33msta PT=staPCC+LCu2
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.
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
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
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
sin I/2=20/300 I=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:
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-o o=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.
ELEMENTS OF A SYMMETRICAL PARABOLIC CURVE 1. VERTEX (PI) 2. PC 3. PT 4. BACKWARD TANGENT 5. FORWARD TANGENT 6. 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)²
1. The curve bisects the distance between the vertex and the midpoint of the long chord.
BF / (L/2)² = CD / (L)² 2. If g1 - g2 (+) = “summit”
g1 - g2 (+) = “sag”
4. 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
EXAMPLE: Given: g2=-8% g1=5% L1=40m L2=60m Required:
Height of fill needed to cover the outcrop Elevation at station 6+820
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)L2-2H 160=2H(120)/(0.11)L2-2H H=3.77m y/(60)²=H/(120)² y=0.94
elevation of the curve= elevation V-(60)(0.04)-y-h elev curve=30-2.4-0.04-4.42
elev curve=22.6835m L2=?
H=4.42m(remains the same) L1=2HL2/(g1-g2)L2-2H 160m=2HL2/(0.11)L2-2H……..eq’n1 Elev 22.6835=elevV-(60)(0.04)-y-4.42 22.6835=30-2.4-y-4.42 y=0.4985 0.4965/(L2-60)²=H/(L2)² H=0.4965(L2)²/(L2-60)² H=3.10m 160=2(3.10)L2/(0.11)L2-2(3.10) L2=100m g1L1/2 ? H (0.07)(60)/2 ? 3.10 5.6 > 3.10 S2=g2L2/2H (from point PT) S2=(0.04)(100)/2(3.10) S2=64.52m station HP=staV+35.48m station HP=12+200+35.48m station HP=12+235.48
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
• 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/Lc²
where: i deflection angle at any point ic deflection angle at SC
• The deflection angle is 1/3 of the spiral angle i=S/3
• 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
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
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/dL S=dx/dL dy=dL-S²(dL)²/2dL dy=dL-(L²/2RcLc)²dL at SC: y=yc L=Lc yc=Lc-(Lc³/40Rc²)
• Deflection angle, i
where: i deflection angle at any point along the spiral curve
sini=X/L X=L³/6RcLc i=X/L i=L²/6RcLc S=L²/2RcLc i=S/3 • Length of throw, P P=? Ts=? Es=?
180˚=I+(90˚-Ic/2-Sc) Ts=Rc+Es(sinI/2+9) 0=I-Ic-2Sc Ic=I-2Sc P=? cosSc=Rc-(Xc-P)/Rc P=Xc-Rc(1-cosSc) or P=Xc/4; P=Lc²/24Rc Es=? cosI/2=Rc+P/Rc+Es Es=(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
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
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 Stakes
II. 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 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:
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
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
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
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
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:
Where:
S = Side Slope HL = Side Height Left B = Base pr Width DR = Distance out right HR = 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:
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
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.
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 CUT
Figure
Centerheight = 1.83m Base for Cut = 8.00m SS for Cut = 1:1 DR = DL = B / 2 + SC
Centerheight = 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
A. 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.
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
= 3.50 + 1.50(3.12) = 8.18m
DR = B / 2 + SHR = 3.50 + 1.50(2.62) = 7.43
B. 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 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
Figure
Solution:
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
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
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
C. 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)
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 ? 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 WIDTH 8.00m (Cut) 5.00 (Fill) SIDE SLOPE 1.00:1.00
A. LEVEL SECTION IN CUT
FIGURE
B. THREE LEVEL SECTION IN FILL
C. FIVE LEVEL SECTION IN CUT
FIGURE
D. IRREGULAR SECTION IN FILL
E. SIDE HILL SECTION
FIGURE
METHODS OF DETERMINING VOLUMES O EARTHEWORKS
A. 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:
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.
A. 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, 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:
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.
PRISMOIDAL CORRECTION FORMULA
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
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:
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 SECTIONS
a. 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 section
c. 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.
1. To THREE LEVEL SECTIONS
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:
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 METHOD B. PRISMOIDAL FORMULA
C. END AREA METHOD and PRISMOIDAL CORRECTION FORMULA
SOLUTION:
Compute for the area at each station cross-section and at mid-section Figure
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
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
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) 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 distances DRm = 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
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
1. Prismoidal Formula for Correction
Cp = 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:
Then, Cp = -1.65 + 4.785 = 3.135m³ 2. 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
STATIONS CROSS-SECTION NOTES
10 + 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 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
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)
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
Ex. Solve for the following:
a) Stationing of the limits of free haul b) Stationing of the limits of economical haul c) Vertical volume d) Length of overhaul e) Cost of haul f) Cost of waste g) Cost of borrow Given: FHD = 50m Cost of borrow = P 4.00/m3 Cost of excavation = P 3.50/ m3 Cost of haul = P 0.20/m station
Station END AREA M2 CUT FILL 1+460 40 1+760 0 Balance point 2+060 60 LEH = CbCh(C) + FHD = 4(20)0.2 + 50
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+732.47 Right limit = 1+760 + 22.47 = 1+782.47
S = 50 m = FHD
a'y = 60300 ; a’ = 60y300 - eq. (1)
b'400-y = 40300 ; b’ = 16,000-40y300 - 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 = 177.81m equating (2) and (3): 400 – y = 220.19 m a’ = 35.96 m2 b’ = 29.36 m2 b) Limits of LEH: Left limit = 1+760 – 27.53 – 220.19 = 1+512.28 Right 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 m3 d) ATXL = AiXi 4040.49X = 0.5(b’)(400 - y)(2/3)(400 - y) + b(400 - y)(0.5)( 400 - y)
Length of OH = XL + XR = 139.45 + 113.89
Length of OH = 253.34 m
e) Cost of haul = OH Vol.cost of haul(length of OH)C = 253.340.20(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