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Surveying 2 Lecture Complete

Bachelor of Science in Civil Engineering (Technological University of the Philippines)

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INTRODUCTION TO ROUTE SURVEYING CONCEPTS

Route surveying is comprised of all survey operations required for the design and CONCEPTS AND TERMINOLOGY
construction of lines of transportation or communication such as highways, Right of way the area occuat vlities such has roads, pipelines and
railroads, pipelines, and power lines transmission lines
Centerline a line running along the center of a road or highway that
The work may consist of the following: divides it into two sections
○ Locating the center line, usually marked by stakes at constant intervals Horizontal curve provides a transition between two tangent strips of roadway,
called stations allowing a vehicle to negotiate a turn at a gradual rate rather
○ Determining elevations along and across the center line for plotting profile than a sharp curve
and cross sections Degree of curve defines the sharpness of the curve
○ Computing the volumes of earthwork and preparing a mass diagram Stationing used to define distances relative to the alignment of the
○ Staking out the extremities for cuts and fills roadway.
○ Determining drainage areas to be used in the design of ditches and culverts ○ English - incremented every 100 ft. A station of
○ Laying out structures, such as bridges and culverts 3+56.00 represents a point 356.00 feet from the
○ Locating right-of-way boundaries, as well as staking out fence lines, if beginning point.
necessary ○ Metric – incremented every 1 km. A station of
2+045.011 represents a point 2,045.011 meters from the
✘ Locating the center line beginning point.
✘ Plotting profile and cross sections ○ Stationing shall be sequential from south to north and
✘ Computing the volumes of earthwork and preparing a mass diagram.
from west to east.
✘ Staking out the extremities for cuts and fills
Vertical curves used in a vertical plane to provide a smooth transition
between the grade lines of highways and railroads
✘ Determining drainage areas for ditches and culverts Grade measured in terms of percent; that is, the number of feet of
✘ Locating right of way boundaries rise or fall in a 100-foot horizontal stretch of the road
Sag curve curves that connect descending grades, forming a bowl or a
sag
Summit Curve curves that connect inclined sections of roadway, forming a
crest.
Slope stakes indicate the outer limits of grading on a construction project,
the intersection of the cut-or-fill slope with the existing
natural ground line and limit of earthwork on each side of the
center line. The information normally found on a slope stake
is any cutor-fill requirements, the distance from the center
line, and the slope ratio.

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IMPORTANCE TO CONSTRUCTION, RELATION TO TRANSPORTATION AND COMMUNICATION

IMPORTANCE TO CONSTRUCTION, RELATION TO RELATION TO TRANSPORTATION AND COMMUNICATION
TRANSPORTATION AND COMMUNICATION
✘ Each project site has a unique mix of natural and man-made features. Existing ✘ Surveyors have a crucial role in acquiring accurate data that are important in
roads, trees, hills, rivers, landmarks, buildings and other structures could make planning transportation infrastructure.
construction of linear facilities much more complex. The identities and locations ✘ Surveyors help decide the best route through certain types of landscapes. 9
of these structures should be accurately determined and visualized. ✘ A deep valley may be cheaply traversed with a bridge, while a large mountain
✘ A route survey provides information and data pertaining to general location might be too big to tunnel through, necessitating a diversion of the route around
possibilities, feasibility, and probable cost of right-of-way, construction, use, and it.
maintenance for any linear facility. ✘ Grading can be an important topic for trains, as very steep grades can be
extremely hard on engines
✘ Surveying can play a crucial role in setting up transmission infrastructure in a
IMPORTANCE TO CONSTRUCTION
✘ Route surveyors play an integral role in land development, from the planning and timely and cost-effective manner.
✘ It enables the selection of the shortest possible route for laying a transmission
design of land subdivisions through to the final construction of roads, utilities
and landscaping.
✘ One purpose of route surveying is to clearly set boundaries for which linear
line.
✘ The objective is to optimize the cost of transmission based on several factors such
facilities may be constructed.
✘ Surveyors are the first people on any construction site, measuring and mapping
as number of river crossing towers and type of terrain,
the land.
✘ These primary measurements are then used by architects to understand and make
the most of the unique landscape when designing and engineers to plan structures
accurately and safely, ensuring buildings not only fit with the landscape but are
able to be constructed.

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SIMPLE CURVES
• The simple curve is an arc of a circle. SIMPLE CURVE ELEMENTS OF A SIMPLE CURVE
• It is the most commonly used.
E External Distance Degree of Curve
• The radius of the circle determines the
“sharpness” or “flatness” of the curve. T Tangent Distance Arc Basis
20 = 𝑅𝐷
• The larger the radius, the “flatter” the curve.
m Middle Ordinate

L Long Chord
ELEMENTS OF A SIMPLE CURVE Lc Length of Curve

PC Point of curvature. It is the beginning of curve. PC Point of curvature

PT Point of tangency. It is the end of curve. PT Point of tangency
PI Point of intersection of the tangents. Also PI Point of intersection
called vertex
T Length of tangent from PC to PI and from PI to T Length of tangent
PT. It is known as subtangent. R Radius Chords Basis
𝐷 10
R Radius of simple curve, or simply radius L Length of chord sin =
2 𝑅
L Length of chord from PC to PT. Point Q as
Lc Length of curve
shown below is the midpoint of L. Also known E External distance
as Long Chord m Middle ordinate
Lc Length of curve from PC to PT. Point M in the I Intersection angle
figure is the midpoint of Lc.
E External distance, the nearest distance from PI x offset distance
to the curve. θ deflection angle
m Middle ordinate, the distance from midpoint of C1 Sub chord.
curve to midpoint of chord.
I Intersection angle (also called deflection angle STATIONS
and central angle). It is the angle of intersection 𝑺𝒕𝒂. 𝑷𝑪 = 𝑆𝑡𝑎. 𝑃𝐼 − 𝑇
of the tangents. The angle subtended by PC and
𝑺𝒕𝒂. 𝑷𝑻 = 𝑆𝑡𝑎. 𝑃𝐶 + 𝐿𝑐
PT at O is also equal to I, where O is the center
of the circular curve
x offset distance from tangent to the curve. Note:
x is perpendicular to T.
θ deflection angle or offset angle from the
tangent to any point on the curve
D Degree of curve. It is the central angle
subtended by an arc (arc basis) or chord (chord
basis) of one station.
C1 Sub chord. It is the chord distance between two
adjacent stations

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SIMPLE CURVES
A 3° curve has an angle of intersection of 24°, what A simple curve with tangents AV and VE have A simple curve the azarailgents that intersect at an
is the length of the long chord and the length of the azimuth of 260°48’ and 285°40’ respectively. Point B angle of 30°. A point P on the curve has a chord
curve? is taken along AV and C along VE. The azimuth and length of 42.8 m from the PT and the deflection angle
distance of BC are 272°16’ and 61.22 m respectively. from the forward tangent to the same point is 4°20’.
The degree of curve is 5°. If stationing of point B is 8 Determine the middle ordinate.
+ 126.3, determine the stationing of PT.

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SIMPLE CURVE
The deflection angles of two intermediate points A Two tangents having an azimuth of 265° and 289°
and B of a highway curve are 5°15’ and 10°15’ respectively intersect at station 10 + 195.35 and are
respectively. The chord distance between points A to be connected with a 5° simple curve. Without
and B is 20 m, while the long chord is 150 m. changing the direction of the tangent, it is required to
Stationing of PI is 8 + 060. Find the stationing of PC find the radius and the station of the PC of the new
and PT curve such that the new PT is 5 m directly opposite
but outside the old PT.

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SIMPLE CURVE
A 7° circular turnout is to connect a railway track,
loading due east to the mouth of the tunnel which is
70 m from station 7 + 812 as shown on the figure.
Use chord basis

a. Determine the stationing of the point of deviation

b. Determine stationing of the mouth of the tunnel

c. What is the direction of the tunnel if it is used for

hauling?

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SIMPLE CURVE
Piers of the proposed overpass in Nagtahan are to be placed with clearance of
2.5 m of the existing Ramon Magsaysaty Avenue as shown

a. Determine the minimum distance between the piers when the radius of the
curve is 100m width of the roadway is 20m.

b. Find the angle θ

c. Find the area of the road between A and B

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COMPOUND CURVE
A compound curve consists of two (or more) circular curves between two main ELEMENTS OF A COMPOUND CURVE
tangents joined at point of compound curve (PCC). Curve at PC is designated as 1
(R1, L1, T1, etc) and curve at PT is designated as 2 (R2, L2 , T2 , etc). PC point of curvature I angle of intersection = I1 +
I2
PT point of curvature I1 central angle of the first
curve
PI point of intersection I2 central angle of the second
curve
PCC point of compound curve L length of long chord from
PC to PT
LC1 length of first curve
T1 length of tangent of the first LC2 length of second curve
curve
T2 length of tangent of the L1 length of first chord
second curve
T1 + T2 = length got7of common L2 length of second chord
tangent measured from
V1 to V2

V1 vertex of the first curve θ = 180° - I

V2 vertex of the second curve x and y can be found from triangle
V1-V2-PI
L can be found from triangle
PC-PCC-PT

Finding the stationing of PT

Given the stationing of PC
𝑆𝑡𝑎 𝑃𝑇 = 𝑆𝑡𝑎 𝑃𝐶 + 𝐿𝐶1 + 𝐿𝐶2
Given the stationing of PI
𝑆𝑡𝑎 𝑃𝑇 = 𝑆𝑡𝑎 𝑃𝐼 − 𝑥 − 𝑇1 + 𝐿𝐶1 + 𝐿𝐶2

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COMPOUND CURVE
1. A long chord from PC to PT of a compound curve 2. A village in Mindoro is to be protected by a 3. A compound curve laid on their tangents has the
is 180 m long and the angle it makes with the longer floodwall and should start at station 20 + 304 forming following data shown below. Without changing the
and shorter tangents are 12° and 18° respectively. two consecutive simple curves ending at Sta 20 + direction of the two tangents, it is required to change
Find the difference in radius of the compound curve this compound with a simple curve that shall end at
544. At point V on the opposite bank of the river, the
if the common tangent will be parallel to the long the same point. Find the radius of the new curve and
chord azimuth of lines tangent to the PC and PT were taken stationing of PC if PI1 is at 6 + 425.82. I1 = 22°, D1
to be 60° and 295° respectively. The degree of the = 3°, I2 = 36°, D2 = 5°
first curve is 4° and that of the second curve is 5°.
Compute the stationing of PCC.

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REVERSE CURVE
A reverse curve consists of two Reversed Curve for Nonparallel Tangents
simple curves joined together but The following figure is an example reversed curves of unequal radii connecting non-
curving in opposite directions. parallel tangents.
The instant change in direction at
the point of reversed curvature
(PRC) brought some safety
problems. Despite this fact,
reversed curves are being used
with great success on park roads,
formal paths, waterway channels,
and the like.

ELEMENTS OF A REVERSE CURVE

PC point of curvature I1 central angle of the first

curve
PT point of curvature I2 central angle of the second Reversed Curve for Parallel Tangents
curve The figure below is an example of reversed curves of unequal radii connecting two
PI point of reverse parallel roads.
curvature
LC1 length of first curve
T1 length of tangent of the LC2 length of second curve
first curve
T2 length of tangent of the L1 length of first chord
second curve
T1 + T2 = length of common L2 length of second chord
tangent measured from
V1 to V2
V1 vertex of the first curve
V2 vertex of the second curve
Finding the stationing of PT

Given the stationing of PC

𝑆𝑡𝑎 𝑃𝑇 = 𝑆𝑡𝑎 𝑃𝐶 + 𝐿𝐶1 + 𝐿𝐶2

Given the stationing of V1

𝑆𝑡𝑎 𝑃𝑇 = 𝑆𝑡𝑎𝑉1 − 𝑇1 + 𝐿𝐶1 + 𝐿𝐶2

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REVERSE CURVES
The perpendicular distance between parallel tangents The common tangent BC of a reverse curve is 280.5 A reverse curve connects two converging tangents
of the reverse curve is 35 m. The back tangent has an m and has a bearing of S 47°31’ E. AB is the tangent that intersect at an angle of 30°. The distance of this
azimuth of 270° while the common tangent is 300°. If of the first curve whose bearing is N 72° 45’ E. CD is intersection from the PI of the second curve is 150 m.
the radius of the first curve is 160 m, determine the the tangent of the second curve whose aziming is N The deflection angle of the common tangent from the
radius of the 2nd curve. 38° 13’ E. A is at the PC while D is at the PT. The back tangent is 20° and the degree of curve of the 2nd
radius of the first curve is 180 m and its PI is at curve is 6°. Find the degree of the first curve
station 12 + 523.37. Find the stationing of PT.

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SYMMETRICAL PARABOLIC CURVES

✘ Symmetrical Parabolic Curve - Symmetrical parabolic curve does not necessarily ELEMENTS OF VERTICAL CURVE
mean the curve is symmetrical at L/2, it simply means that the curve is made up
PC point of curvature, also known as BVC (beginning of vertical curve)
of single vertical parabolic curve.
PT point of tangency, also known as EVC (end of vertical curve)
✘ Using two or more parabolic curves placed adjacent to each other is called PI point of intersection of the tangents, also called PVI (point of vertical
unsymmetrical parabolic curve. intersection)
L length of parabolic curve, it is the projection of the curve onto a horizontal
ELEMENTS OF VERTICAL CURVE surface which corresponds to the plan distance.
S1 horizontal distance from PC to the highest (lowest) point of the summit
(sag) curve
S2 horizontal distance from PT to the highest (lowest) point of the summit
(sag) curve
h1 vertical distance between PC and the highest (lowest) point of the summit
(sag) curve
h2 vertical distance between PT and the highest (lowest) point of the summit
(sag) curve
g1 grade (in percent) of back tangent (tangent through PC)
g2 grade (in percent) of forward tangent (tangent through PT)
A change in grade from PC to PT
a vertical distance between PC and PI
b vertical distance between PT and PI
H vertical distance between PI and the curve. Sometimes known as the
maximum offset.

PROPERTIES OF VERTICAL PARABOLIC CURVES

• The stationing if vertical parabolic curves is measured along a horizontal line
• The slope of a parabola varies uniformly along the curve
• PI is midway between PC and PT.
• The vertical offsets from the tangent to the curve are proportional to the squares of
the distances from PC or PT
PROPERTIES OF VERTICAL PARABOLIC CURVES • The curve is midway between PI and the midpoint of the chord from PC to PT.
𝒈𝟐 −𝒈𝟏 𝟏 𝒚 𝑯 • The vertical distance between any two points on the curve is equal to area under the
𝒓= 𝑯 = 𝑳(𝒈𝟏 − 𝒈𝟐 ) = 𝑳𝟐
𝑳 𝟖 𝒙𝟐 grade diagram. The vertical distance c = Area.
𝟐

𝟏 𝟏 𝒈𝟏 𝑳
𝒂 = 𝒈𝟏 𝑳 𝒉𝟏 = 𝒈𝟏 𝑺𝟏 𝑺𝟏 =
𝟐 𝟐 𝒈𝟏 −𝒈𝟐

𝟏 𝟏 𝒈𝟐 𝑳
𝒃 = 𝒈𝟐 𝑳 𝒉𝟐 = 𝒈𝟐 𝑺𝟐 𝑺𝟐 =
𝟐 𝟐 𝒈𝟐 −𝒈𝟏
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SYMMETRICAL PARABOLIC CURVES

A descending grade of 6% and an ascending grade of A grade line AB having a slope of +5% intersect
2% intersect at Sta 12 + 200 km whose elevation is at another grade line BC having a slope of –3% at B. The
14.375 m. The two grades are to be connected by a elevations of points A, B and C are 95 m, 100 m and 97
parabolic curve, 160 m long. Find the elevation of the m respectively. Determine the elevation of the summit
first quarter point on the curve. of the 100 m parabolic vertical curve to connect the
grade lines.

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SYMMETRICAL PARABOLIC CURVES

A highway engineer must stake a symmetrical vertical A grade of -4.2% grade intersects a grade of +3.0% at A grade descending at a rate of -4% intersecting
curve where an entering grade of +0.80% meets an Station 11 + 488.00 of elevations 20.80 meters. These another grade ascending at the rate of +8% at station
existing grade of -0.40% at station 10 + 100 which has two center grade lines are to be connected by a 260 2+000, elevation 100 m. A vertical curve is to connect
meter vertical parabolic curve. the two such that the curve will clear a boulder located
an elevation of 140.36 m. If the maximum allowable
at station 1+980, elevation 101.34
change in grade per 20 m station is -0.20%, what is the a) At what station is the cross-drainage pipes be a) Determine the necessary length of the curve
length of the vertical curve? situated? b) Determine the station of the location of a sewer to
b) If the overall outside dimensions of the reinforced be laid out
concrete pipe to be installed is 95 cm, and the top c) Compute the elevation of station where the sewer
of the culvert is 30 cm below the subgrade, what is to be placed.
will be the invert elevation at the center?

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SYMMETRICAL PARABOLIC CURVES

Forward and6. A vertical symmetrical sag curve has a An underpass crossing a reinforced concrete bridge
descending grade of -4.2% and an ascending grade of along the Shaw Blvd. has a downward grade of -4%
+3% intersecting at station 10 +020, whose elevation is meeting an upward grade of +8% at the vertex V at
100 m. The two grade lines are connected by a 260 m elevation 70 m, and stationing of 7 + 700, exactly
vertical parabolic sag curve. underneath the center line of the bridge having a width
a) At what distance from PC is the lowest point of the of 10 m. if the required minimum clearance under the
curve located? bridge is 5 m, and the elevation of the bottom of the
b) What is the vertical offset of the parabolic curve to bridge is 78.10 m.
the point of intersection of the tangent grades?
c) If a 1 m diameter culvert is placed at the lowest a) Determine the length of the vertical parabolic curve
point of the curve with the top of the culvert buried that shall connect the two tangents
0.60 m below the subgrade, what will be the b) Determine the stationing of the point where a catch
elevation of the invert of the culvert? basin will be placed
c) Determine the elevation of the point where a catch
basin will be placed

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UNSYMMETRICAL PARABOLIC CURVES

A vertical highway curve is at times designed to include a particular elevation at a
certain station where the grades of the forward and backward tangents have already
been established. It is therefore necessary to use an “unsymmetrical” or
asymmetrical parabolic curve where one parabola extends from the P.C to a point
directly below the vertex and a second parabola which extends from this point to the
PI.

ELEMENTS OF UNSYMMETRICAL PARABOLIC CURVE

𝑳𝟏 𝑳𝟐 |(𝒈𝟏 −𝒈𝟐 )| 𝒈𝟏 𝑳𝟐
𝟏 𝒈𝟐 𝑳𝟐
𝟐
𝑯= 𝒔𝟏 = 𝒔𝟐 =
𝟐(𝑳𝟏 +𝑳𝟐 ) 𝟐𝑯 𝟐𝑯

𝒚 𝑯
=
𝒙 𝟐 𝑳𝟐
𝟐

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UNSYMMETRICAL PARABOLIC CURVES

In a certain road construction undertaken by the An unsymmetrical parabolic curve has a forward A -3% grade meets a +5% grade near an underpass. In
Bureau of Public Highway it was decided to connect a tangent of -8% and a back tangent of +5%. The length order to maintain the minimum clearance allowed
forward tangent of 3% and a back tangent of – 5% by of the curve on the left side of the curve is 40 m long under the bridge and at the same time induce a vertical
a 200 m symmetrical parabolic curve. It was while that of the right side is 60 m long. The PC is at transition curve in the grade line, it is necessary to use
discovered that the grade intersection at station 10 + station 6 + 780 and has an elevation of 110 m. An a curve that lies 200 m on one side of the vertex of the
100 whose elevation is 100 m falls on a rocky section outcrop is found at station 6 + 800 and has an straight grade and 100 m on the other. The station of
with an outcrop of 2.67 m directly above the grade elevation of 108.40 m the beginning of the curve is 10 + 000 and its elevation
intersection. To avoid rock excavation, the project a) Compute the height of fill needed to cover the is 228 m.
engineer decided to adjust the vertical parabolic curve outcrop a) Determine the elevation at station 10 + 040.
in such a way that the curve will just clear the rock b) Compute the elevation of curve at station 6 + b) If the uphill edge of the under side of the bridge is
without altering the position of PC and the grade of 820 at station 10 + 220 and at elevation 229.206 m,
the tangents. c) Compute the elevation of the highest point on what is the vertical clearance under the bridge at
a) Determine the total length of the new the curve. this point?
parabolic curve. c) Determine the stationing of lowest point of the
b) Determine the stationing and elevation of the curve.
new PT

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UNSYMMETRICAL PARABOLIC CURVES

A forward tangent having a slope of -4% intersects the V1 is at station 12 + 200 and has an elevation of 30 m. A forward tangent of +6% was designed to intersect a
back tangent having a slope of +7% at point V at It is required to connect the two tangents with an back tangent of -3% at a proposed underpass along
station 6 + 300 having an elevation of 230 m. It is unsymmetrical vertical parabolic curve 160 m on the Epifanio de los Santos Avenue so as to maintain a
required to connect the two tangents with an side of the back tangent and 120 m on the side of the minimum clearance allowed under a bridge which
unsymmetrical parabolic curve that shall pass through forward tangent. The curve must provide a vertical crosses perpendicular to the underpass. A 200 m curve
point A on the curve having an elevation of 227.57 m clearance of at least 4.42 m above the right curb of an lies on the side of the back tangent while a 100 m
at station 6 + 270. The length of the curve is 60 m on underpass, the stationing of which is 12 + 260. curve lies on the side of the forward tangent. The
the side of the back tangent. a) Determine the elevation of the curb. stationing and elevation of the grade intersection is
a) Determine the length of the curve on the side b) If the curb elevation is 22.6385 m, compute 123 + 530.20 and 100 m respectively. The centerline
of the forward tangent. the length of the curve of the side of the of the bridge fails at station12 + 575.20. The elevation
b) Determine the stationing of the highest point forward tangent in order to fulfill the of the underside of the bridge is 117.48.
on the curve. requirement. a) Determine the clearance of the bridge at the
c) Determine the elevation of the highest point c) Compute the stationing of the highest point of left side if it has a width of 10 m.
on the curve the curve for the second condition b) Determine the clearance of the bridge at the
right side
c) Determine the clearance of the bridge at the
center of the bridge.

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EARTHWORKS
Earthworks surveying is important especially in large-scale projects such as EXCAVATION
residential development, roads, railways and other government infrastructure.
Excavation is the loosening of the soil from its natural, in situ, stage.
Earthworks usually involve either the removal or placement of soil or land mass on a
particular area. • Levelling This word is used to describe a situation where only a
levelling of bumps and depressions is necessary to provide a sufficient
• For this purpose, surveyors calculate the volumes of cuts and embankment width for the road at a uniform level.
fills • Cut to crossfill means that the soil is excavated (cut) from one side of the
road and used as fill material at the other.
The principal activities in earthworks are: • The U-cut is a cut which is roughly shaped as a U.
• Suitable soil has to be "borrowed" from outside when the excavated soil
• measuring and calculation of volumes
within the road width is not sufficient for a fill or embankment
• excavation which includes levelling, cut to crossfill, U-cut and borrow
excavation
• loading, hauling and unloading
• filling, including spreading and compaction

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EARTHWORKS
MEASURING AREAS AND VOLUMES

HAULING AND UNLOADING

Hauling is the transport of soil, the modes of haulage vary with the distance.

• Haulage can be done by stretchers, wheelbarrows, animal carts, tractor-

trailers and trucks.
• Wheelbarrows and tractor-trailers are most commonly used for labour-based
road construction.
• In flat areas where the road needs to be elevated the soil can often be taken
from pits alongside the road (borrowpits).

Unloading is usually done by tipping. Tipping vehicles can also spread the
soil/gravel.

Measuring Areas of Irregular Cross Section

𝒘 𝑪
𝑨= (𝒉𝑳 + 𝒉𝑹 ) + (𝒅𝑳 + 𝒅𝑹 )
𝟒 𝟐

𝑳
Volume by End Area 𝑽𝒆 = (𝑨𝟏 + 𝑨𝟐 )
𝟐

𝑳
Volume by Prismoidal 𝑽𝑷 = (𝑨𝟏 + 𝟒𝑨𝒎 + 𝑨𝟐 )
𝟔

Volume by Prismoidal Correction 𝑽𝒄𝒑 = 𝑽𝒆 − 𝑪𝒑

𝑳
𝑪𝒑 = (𝑽 − 𝑽𝟐 )(𝑯𝟏 − 𝑯𝟐 )
𝟏𝟐 𝟏

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 lOMoARcPSD|54785873

EARTHWORKS
The cross section notes shown below are for ground A 60 x 60 square lot is to be divided into 9 square
excavation for a 10 m road. sections. Determine the volume of earthworks
excavated if the ground surface is to be leveled to
Station 25 +100 Station elevation 10 m. The following data are the elevations
25 + 150 of the ground surface

7.85 0 8.45 AI = 13.3 BI = 12.6 CI = 12.2 DI = 12.2

9.35 0 10.7
AII = 12.9 BII = 14.2 CII = 15.2 DII = 11.2
1.90 3.2 2.30
2.90 2.6 3.80 AIII = 15.8 BII = 15.6 CIII = 14.4 DIII = 14.6

What is the volume of excavation between the two AIV = 11.8 BIV = 13.8 CIV = 11.9 DIV = 13.9

stations using
a) Volume by end area method
b) Volume by prismoidal formula
c) Volume by prismodial correction (assume 2:1)

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