UNESCO-NIGERIA TECHNICAL & VOCATIONAL EDUCATION

REVITALISATION PROJECT-PHASE II

YEAR 2- SE MESTER 2

THEORY/PRACTICAL

Version 1: December 2008

NATIONAL DIPLOMA IN

CIVIL ENGINEERING TECHNOLOGY

ENGINERING SURVEYING I

COURSE CODE: SUG 208

CIVIL ENGINEERING TECHNOLOGY

ENGINEERING SURVEYING I (SUG 208)

THEORETICAL COURSE INDEX.

WEEK 1. 1. GENERAL INTRODUCTION

1.1 INTRODUCTION………………..………..…1 1.2 TYPES OF LAND SURVEY……………… 1 WEEK 2.

2.1 SCALE……………………………………...…5 2.2 PREFERED SCALES……………………… 5 2.3 ACCURACY AND PRECISION……………7 2.4 ERRORS IN SURVEYING………………….8

WEEK 3. 3.1 GEOMETRIC DESIGN OF ROUTES…….10

3.2 CURVES……………………..……….….…..10 3.3 CIRCULAR CURVES………………………11 3.4 SETTING OUT HORIZONTAL CURVES..12

WEEK 4. 4.1 WORKED NUMERICAL EXAMPLE...….16 WEEK 5. 5.1 COMPOUND CURVE…..…………………22

5.2 LOCATION OF THE INTERSECTION AND TANGENT POINTS IN THE FIELD..26

WEEK 6. 6.1 LOCATION OF THE TANGENT POINTS WHEN THE INTERSECTION POINT IS INACCESSIBLE ……………..….28 6.2 OBSTRUCTION IN THE LINE OF SIGHT..29 WEEK 7. 7.1 SETTING OUT………………..………… …30

7.2 AIMS OF SETTING OUT …………………30 7.3 STAGES IN SETTING OUT……… ……..31

WEEK 8. 8.1 EQUIPMENT FOR SETTING OUT

BUILDI………………………………………33 WEEK 9. 9.1 SETTING OUT OF ABUILDING… ……..38 WEEK 10. 10.1 EARTH WORK QANTITIES..…………..41

10.2 AREAS ENCLOSED BY STRAIGHT LINES………………………………………42 10.3 COMPUTATION OF AREAS FROM PLOTTED PLAN……………… 45

WEEK 11. 11.1 AREAS ENCLOSED BY IRREGULAR LINE………………………………………...46 WEEK 12. 12.1 THE PLANIMETER……………………..50 12.2 MECHANICAL PLANIMETER…………50 WEEK 13. 13.1 VOLUME CALCULATIONS …………...52

13.2 ESTIMATION OF VOLUMES FROM CONTOUR LINES…….………….52 13.3 THE END – AREA FORMULA……….….53

WEEK 14. 14.1 PRISMODAL FORMULA……………….55

14.2 ESTIMATION OF VOLUMES FROM SPOT LEVELS…………………….……….55

WEEK 15. 15.1 WORKED EXAMPLES……………….…..58 15.2 ESTIMATION OF VOLUMES FROM CROSS - SECTION………………………....60

WEEK 1.

INTRODUCTION

Survey is the process of examining and recording data i.e. taking measurements of

lengths, height differences and angles on site for either the preparation of maps, large

scale plans or in order that engineering works can be located in their correct positions on

the ground. This process is termed engineering surveying and falls under the general title

of land surveying.

Basically, land surveying involves measuring land and its physical features accurately

and records of these presented in the form of a map or plan. Such information is used by

commerce and industry for planning new buildings and by local authorities in managing

facilities. Maps that are used to locate features and/or places are made possible by the

information gathered by land surveyors. Land surveying also undertakes the accurate

positioning and monitoring of Civil Engineering Construction Works on Sites.

TYPES OF LAND SURVEYING.

Land surveying can be broken down into the following sub – sections:

Geodetic survey:

This covers such large areas that the curved shape of the earth is taken into consideration

on area greater than 250km2.

Topographical survey: This establishes the positions and shapes of natural and artificial

features found on a given area. Usually, for the purpose of producing a map of that area

or for establishing a geographic information system.

Hydrographic survey:

This sub – section of land surveying is concerned with gathering information in the

marine environment and mapping coast lines and sea bed in order to produce navigation

charts. It is also found useful in offshore oil exploration and production. Hydrographic

surveys are also used in the design, construction and maintenance of harbours in land

water routes, river and sea defences, in control of pollution and in scientific studies of the

ocean.

Cadastral survey: This establishes and records the boundaries and ownership of land and

property. Cadastral surveys are based on the topographical detail appearing on ordnance

survey maps. This work is mainly limited to overseas countries where National Land

Registry Systems are under development.

Photogrammentry:

This land survey is basically topographic surveys taken by the aid of photographs using

special cameras mounted in an air craft viewed in pairs, the photographs produce three

dimensional images of ground features from which maps or numerical data can be

produced, usually with the aid of stereo plotting machines and computers. Close range

photogrammetry uses photographs taken with cameras on the ground and is used in many

applications.

Engineering surveying:

This deals with any survey wok carried out in connection with the construction of Civil

Engineering and building projects. Engineers and surveyors involved in surveying site are

responsible for all aspects of dimensional control on such schemes. The main purposes of

engineering surveying are for design reasons, construction and monitoring of structures

e.g. Dam, roads, airports, bridges, e.t.c.

Engineering surveys are usually based on horizontal and vertical control frameworks

which consist of fixed points called control stations. Horizontal control, as its title

suggests defines points on an arbitrary two – dimensional horizontal plane which covers

the area of interest. Vertical control, although usually treated separately from horizontal

control as far as field work and calculations are concerned, is the third dimension added

to the chosen horizontal Datum.

Horizontal and vertical controls are established by measuring angles, distances or a

combination of both in well established techniques such as traversing, triangulation,

intersection, resection and levelling.

On site, a wide variety of equipment is used for establishing control and for setting out.

These include theodolites for measuring angels, levels for measuring vertical distances

(heights) tapes and electronic instruments for measuring distances, computers for

applications such as network analysis, automated data processing for plan production and

computation of setting out data and quantities.

In order to ensure that reliable measurements are taken for engineering surveys,

equipment and techniques of sufficient precision should be used both before and during

construction. However, it is not always necessary to the highest possible precision; some

projects may only require angles and distances to be measured to 1’ and 0.1m, whereas

others may require precisions of 1’ and 0.001m. It is very important and necessary that

the engineer realises this and chooses equipment and techniques accordingly. This will

not only ensure qualify works but also meet set standards of engineering, design and

constructions.

City surveying: Conducted in order to locate streets water supply and sanitary systems

etc. of a city.

MAPS AND PLANS

Figure 1.1 :WORLD MAP.

The ordnance survey:

This is the principal surveying and mapping organisation in Great Britain. Its work

includes geodetic surveys and associated scientific studies. I t does Topographical

surveys and the production of maps of Great Britain at various scales.

Ordnance survey maps: The range of map production from the ordnance survey is

extremely wide and maps are available from the small – scale route planner map, which

is revised every year and contains the whole of Great Britain on one sheet at a scale of

1:625000 to super plan products, some of which are available at 1:200 scale.

As far as engineering surveying is concerned, the ordnance surveying maps of particular

interest are those at the basic scale of 1:1250, 1:2500 and 1:10000.

Figure 1.2 :A MAP

WEEK 2.

SCALE :

A scale refers to the ratio of ground length to plan length. Thus, a scale of 1:1000

indicates that 1cm on the plan (paper) represents 1000cm on the ground. Scale can be

expressed in the following ways;

a. By statement such as 2cm to 5cm

b. By a representative function such as 1:1/R

c. By a divided line which is usually open divided.

PREFERED SCALE

The preferred scales for use in metric system are as follows;

Small scale maps: 1:1000,000, 1:500,000

Large scale maps: 1:1250, 1:20,000

Site plans: 1:100 1:50, 1:200 etc.

Detail plan: 1:20, 1:10, 1:15, 1:5, 1:1 (full size)

The difference between maps and plans is that on maps, scale is too small to allow every

features to be properly represented to scale for this reason, conventional symbols are used

to represent features which would otherwise be too small to be recognised on the maps.

Basic scales can range from 1:100,000. The larger scales being used for detailed

drawings and plans while the smaller scales are used for mapping.

Basic works and their appropriate scales are summarised below;

Architectural works: components and assembly detail drawings etc: - 1:1, 1:5, 1:10, 1:20,

1:50, 1:100, 1:200

Civil engineering works: site plans, key plans survey and lay out: - 1:5000, 1:1000, 1:250

Town surveys highways and route surveys; 1:2500, 1:5000

Maps: 1:2500, 1:5000, 1:10000, 1:20000, 1:30000

All engineering plans and drawings are produced at a particular scale for example;

1:5000, 1:200, 1:100 etc.

The scale value indicates the ratio of horizontal and/or vertical plan distance to horizontal

and/or vertical ground distance that was used when the drawing was produced, for

example, a horizontal plan having a scale of 1:100 indicates that for any line

AB = horizontal plan length AB = 1

Horizontal ground length AB 100

And if a line AB as measured on the plan is 13:50mm, then horizontal ground length AB

= 13.50 x 100 = 1350mm.

The term “large scale’’ indicates a small ratio e.g. 1:10 and 1:20, where as the term

“small – scale” indicates large ratio e.g. 1:5000.

On engineering drawings, scales are usually chosen to be as large as possible to enable

features to be drawn as they actually appear on the ground. If too small a scale chosen,

then it may be impossible to draw the true representations of features and in such cases,

conventional symbols are used; this is a technique commonly adopted by the ordnance

survey.

It must be stressed that the scale value of any engineering drawing or plan must always

be indicated on the drawing or plan, without this, it is incomplete and it’s impossible to

scale dimensions from the plan with complete confidence.

ACCURACY AND PRECISION

Accuracy allows for a certain amount of tolerance of either plus or minus, in

measurements while precision demands exact measurement since there is no such things

as absolutely exact measurement.

Plane survey work is usually described as being to a certain standard of accuracy which

in turn is suited to the work at hand. Bearing in mind the basic aim of carrying out the

measurement, it is better to achieve a high level of accuracy than aim for precision

(exactness) which if it were to be altered would depend not only on the instrument used

but also on the care taken by the operator to ensure that his work (measurement) is free

from mistakes. Always remember however that the greater the effort and time needed

both in the field and in the office and of course the more expensive to survey will be for

the client. The standard accuracy attained in the field must be in keeping with the size of

the drawings.

ERRORS IN SURVEYING

In all surveying operations and indeed in any operation involving measurements, errors

are likely to occur and so far as is possible, such errors must be guarded against or their

effects corrected for.

There are three types of errors, which can occur in surveying measurements;

Mistakes (gross errors)

Systematic (cumulative errors)

Accidental (random errors)

GROSS ERRORS:

These are often called mistakes or blunders, and are usually much larger that the other

categories of errors.

On construction sites, inexperienced engineers and surveyors who are unfamiliar

with the equipment and/or methods that are being used or employed frequently make

mistakes. Gross errors are due then, to carelessness or incompetence and many examples

can be given of these common mistakes which include reading theodolite micrometer

scale or tape graduation incorrectly or wrong booking either by transposing numbers (e.g.

28.432mm instead of 28.342mm) or otherwise. Failure to detect a gross error in survey or

in setting out can lead to serious problems and for this reason, it is important that all

survey work has observational and computational procedures that can be checked so that

mistakes can be corrected and hence, eliminated.

SYSTEMATIC ERRORS.

These errors are those, which follow some mathematical law and will have the same

magnitude and sign in a series of measurements that are repeated under the same

conditions. If an appropriate mathematical model can be derived for a systematic error, it

can be eliminated from a measurement using corrections. For example, the effects of any

temperature and tension variations in steel taping can be eliminated from a measurement

by calculation using simple formulae. Another method of removing systemic errors is to

calibrate the observing equipment and to quantity the error allowing corrections to be

made to further observations.

RANDOM ERRORS.

When all gross and systemic errors have been removed, a series of repeated measurement

taken of the same quantity under the same conditions would still show some variation

beyond the control of the observer. These variations are inherent in all types of

measurements and are called random errors, the magnitude and sign of which are not

constant. Random errors cannot be removed from observations but methods can be

adopted to ensure that they are kept within acceptance limits. In this context, the use of

the word error does not always imply that something has gone wrong, it simply tells us

that a difference exist between the true value of a quantity and a measured value of that

quantity. It is imperative to realize that for surveying measurements, the true value of a

quantity is usually not known and therefore, the exact errors in a measurement or

observation can never be known.

In order to analyze random error or variables, statistical principles must be used and in

surveying, it is usual to assume that random variables are normally distributed.

WEEK 3.

GEOMETRIC DESIGN OF ROUTES.

Geometric design of route simply means the arrangement of the physical elements of the

route according to the design controls of traffic, topographic features, physical features,

capacity, safety and economy. The basic objective of the design is to provide

facility/facilities that will enable movement of vehicles at capacity level at the highest

possible speed within allowable limits and with maximum safety.

Geometric elements of routes include structures like bridges, drainages, cross drainages,

embankment, vertical and horizontal alignments, e.t.c.

CURVES .

In route design, as much as possible straight course are maintained, but where inevitable,

a change in direction may occur. Since it is not convenient for the intersection of to

straights to form part of our route, a smooth curve is used to connect any two intersecting

straights.

Generally, curves may be categorized into three classed, namely;

Circular curves: These include all simple curves, compound curves and reverse curves

Transition curves: These curves have continuous changing radii.

Vertical curves: These are of simple parabola used as either crest or sag.

These curves are carefully inserted at the appropriate places in the geometric design of

motor ways, rail ways, pipe lines etc.

In each case, the initial design is usually based on series of straight sections whose

positions are defined largely by the topography of the area.

In the case of vertical curves – design intersecting gradients are connected by vertical

curves in the vertical plane.

Circular curve.

Simple curves: Two straights D1 T1 and D1 T2 as shown in figure 1 below are connected

by a simple curve of radius R

FIGURE 3.1 A Simple.

From the above figure, are essential elements of simple curve can be summarized as

follows

I: intersection point of the two straights

Ø: Angle of intersection or deflection angle. This is equal to the angle subtended at

the centre of the curve

Ф: Apex angle

The curve commences from T1 and ends at T2. These points are called tangent

points.

T1 I and T2 I are called tangent lengths. Each is equal to R tan Ø/2

The curve length T1 A T2 is equal to RØ, where Ø is expressed in radius. It is called

arc length.

D1

T1

φ

θ/2

θ/2

θ

T2

D2

O

I

R

θ

B

A

The straight distance T1 T2 is called the main cord, denoted by C = 2R sin Ø/2

Through drainage is the horizontal distance from the start of a construction scheme to any

point of interest.

SETTING OUT OF THE HORIZONTAL CURVES.

The process of route survey comprises of three stages;

Reconnaissance survey

Preliminary survey

Final survey

The reconnaissance survey is the practical visit to the construction site in order to acquire

the physical knowledge of the nature of the area as a whole. Station points are also

selected and indivisibility ensured. The survey method is strictly determined by the

topography of area.

The preliminary survey is the initial survey carried our in order to determine the actual

topography and details of the area concerned. At the end of this exercise, the physical

undulation and existing features on the ground are provided on the working plan. This

information helps the designer to plan and determine the course of the route, taking into

consideration, the purpose, safety and economy.

The final survey consists of the actual setting out exercise on the ground to locate the

course of the route including the course of designed curves.

Example 1:

The centre lines of two straights are projected forward to meet at I and the deflection

angle is measured to be 300. If the straights are to be connected by a simple circular curve

of radius 200m, tabulate all the setting that data taking 20m chords on a through chainage

basis. The chainage of I is determined to be 2259.5m.

Solution:

Data: Chainage at I = 2259.59m

Radius curve = 200m

Deflection angle = 300

Standard chord length = 20m

O

Figure 3.2 A simple Curve.

Tangent length, T1,I = R tan Ø/2

= 200 tan 30/2

= 200 tan 150

= 53.59m

Since the chainage of I is known, we can now determine the drainage at T1

Chainage at T1 = chainage I – tangent length

= 2259.59 – 53.59

T1

θ = 300

T2

R = 200m

θ

= 2206.00m

Length of arc= R Ø (radius)

= 200 x 300(radians) = 200 x 5.2359878 x 10-1

= 104.72m

For a standard chord length of 20.00m, the first sub chord is 14.00m. The second, third,

fourth and fifth chords (standard) = 20m each, total = 80.00m

The final sub chord = 10.72m

Check: 14.00 + 80.00 + 10.72 = 104.72m (ok)

Deflection angle, δmm = chord length x 180 x 60

2pR

= 1718.9 x chord length/R

Hence for the first sub chord,

Deflection angle δ mm= 1718.9 x 14.00/200

= 120.30min = 20 00’ 19’’

For standard chord, δ mm = 1718.9 x 20.00/200

= 171.90min = 20 51’ 53’’

For final sub chord δ mm = 1718.9 x 10.72/200

= 92.00min= 10 32’ 08’’

Sum of deflection angles = 140 59’ 59’’

The setting out table is presented as follows;

Figure 3.1 setting out data.

Chords

No.

Chord

length(m)

Chainage

(m)

Deflection angle

0 ‘ ‘’

Setting out angle

0 ‘ ‘’

Remarks

1 14.00 2220.00 2 00 19 2 00 19 Peg 1

2 20.00 2240.00 2 51 53 4 52 12 Peg 2

3 20.00 2260.00 2 51 53 7 44 05 Peg 3

4 20.00 2280.00 2 51 53 10 51 58 Peg 4

5 20.00 2300.00 2 51 53 13 27 51 Peg 5

6 10.72 2310.72 1 32 08 14 59 59 Peg 6 (T2)

Assignment 1

Solve example No. 1 (above), taking a standard chord length of 10.00m instead of the

20.00m used.

WEEK 4.

Example 2.

Produce a setting out table for a horizontal curve having a radius of 350m and chainage

of intersection point set at 6527.42m. The deflection angle is set at 72.590. Take standard

chord length of 100.00m

Solution:

Data: chainage at I = 6527.42m

Radius of curve = 350.00m

Deflection angle = 72.590

Standard chord length = 100.00m

0

Figure 4.1 A SIMPLE CURVE.

T1

θ = 72.590

T2

R = 350m

θ

Tangent length = R tan Ø/2

= 350 tan 72.59/2

= 350 tan 36.290

= 257.04m

Chainage at T1 = drainage at I – T1 I

= 6527.42 – 257.04

= 6270.35m

Length of circular arc = R Ø (radians)

= 350.00 x 72.590 (radius)

= 350.00 1.266934504

= 443.43m

Chainage at T2 = drainage at T1 + length of arc

= 6270.38 + 443.43

= 6713.81m

For standard chord of 100.00m, the first sub chord is 29.62m Standard chords = 100m x 4

= 400.00m

The last sub chord = 13.81m

Check: 29.62 + 400.00 + 13.81 = 443.43m (O.K)

Deflection angle: chord length x 1800 x 60

2p R

The deflection angle for the first sub chord = 171.81 x 29.62/350 = 145.47’ = 20 25’ 28’’

The deflection for the standard chord = 171.81 x 100/350 = 491.11’= 50 11’ 07’’

The deflection angle for the last sub chord = 171.81 x 13.81/350 = 67.82’ = 10 07’ 49’’

The setting out table is presented as follows;

Chords

No.

Chord

length(m)

chainage

(m)

Deflection angle

0 ‘ ‘’

Setting out angle

0 ‘ ‘’

Remarks

1 29.62 6300.00 2 25 28 2 25 28 Peg 1

2 100.00 6400.00 8 11 07 10 36 35

Peg 2

3 100.00 6500.00 8 11 07 18 47 42

Peg 3

4 100.00 6600.00 8 11 07 26 58 49

Peg 4

5 100.00 6700.00 8 11 07 35 09 56

Peg 5

6 13.81 6713.81 1 07 49 36 17 45

Peg 6 (T2)

Example 3:

It is required to connect two straights whose deflection angle 130 16’ 00’’ by a simple

curve of radius 600m. Prepare a detailed setting out table if the through drainage of the

intersection point is 2745.72m.Adopt a chord length of 25m and sub chords at the

beginning and end of the curve.

Solution:

Data: Chainage at I = 2745.72m

Radius of curve = 600m

Deflection angle = 13 0 16 00’’

Standard chord length = 25m

Tangent length = R tan Ø/2

= 600 tan 130 16’ 00’’

2

= 600 x 1.162935523 x 10-1

= 69.76m

Chainage at T1 = 2745.72 – 69.78

= 2675.94m

Length of arc = R Ø (radians)

= 600 x 13 0 16’ 00’’

= 600 x 2.315470141 x 10-1

= 138.93m

Chainage at T2 = chainage at T1 + length of arc

T1

θ = 30016’ 00’’

T2

R = 600m

θ

O

= 2675.94 + 138.93

= 2814.87m

First sub chord = 2700.00 – 2675.94

= 24.06m

Standard chord = 25.00m (x 4)

Final sub chord = 14.87m

Check: 24.06 + (25.00 x 4) + 14.87 = 138.93m (O.K)

First deflection angle = 1718.9 x 24.06

600

= 68.93’ = 10 08’ 56’’

Standard deflection angle = 1718.9 x 25.00

600

= 71.62’ = 10 11’ 37’’

Final deflection angle = 1718.9 x 14.87

600

= 42.60’ = 000 42’ 36’’

Chords

No.

Chord

length(m)

chainage

(m)

Deflection angle 0 ‘ ‘’

Setting out 0 ‘ ‘’

Remarks

1 24.06 2700.00 01 08 56 01 08 56 Peg 1

2 25.00 2725.00 01 11 37 02 20 33 Peg 2

3 25.00 2750.00 01 11 37 03 32 10 Peg 3

4 25.00 2775.00 01 11 37 04 43 47 Peg 4

5 25.00 2800.00 01 11 37 05 55 24 Peg 5

6 14.87 2814.87 00 42 36 06 38 10 Peg 6 (T2)

WEEK 5.

COMPOUND CURVE.

Figure 5.1 : Compound Curve.

It is advisable to treat compound curve in each case as two simple curves with common

tangent point t. The total tangent lengths T1 I and T2 I are found as follows;

R1 tan Ø1/2 = T1t1 = t1 t

R2 tan Ø2/2 = T2t2 = t2 t

Then tangent t1 I t2 may be solved from length t1 I and t2 I if added to the known lengths

T1t1 andT2t2 respectively. The first curve R1 is set out in the usual back sighted to T1 with

the horizontal circle reading (180 – Ø1/2). Turn the instrument to read zero and it will then

be pointing to t2. Hence, curve R2 can now be set out. In case of reverse curve, both axes

can be set out from the common tangent point t.

A

T1

θ1

θ = θ1 + θ2

T2

B

O

I

R2

θ2

θ1 R1

t θ2 t2

t1

Example:

The centre line of a road way is to be set out along a valley. The first straight, A I bears

750 whilst the connecting straight, IB bears 1200. Due to the site condition, it has been

decided to join the straights with a compound curve. The first curve of 500.00m radius

commences at T1, situated at 300.00m from I on straight AI, and deflects through angle

250 before joining the second curve.

Required: Calculate the radius of the second curve and the distance of the tangent point

T2 from I on the straight IB.

Solution:

Figure 5.2 : Compound Curve.

Ø =120O -75O =450

The total angle in a triangle = 1800

= Ø2 = (135 + 25) out of 1800

= 200

A

T1

θ1 = 250

θ = 1200 – 750 = 450

T2

B

O2

I

R2 = ?

θ2 θ1 R1 = 500m

t

t2 t1

θ2 = 200

O1

1350

t

Tangent length T1t1 = R1 tan Ø1/2

= 500 tan 120 30I 00II

= 110.85m

Length It1 = T1I – T1t1

= 300 – 110.85 = 189.15m

By sine rule:

t1t2 = (300 – 110.85)

Sin 135O sin 200

» t1t2 = sin1350 x 189.15 = 319.06m

Sin 200

Similarly, It2 = 391.06

Sin 25O sin 135O

» It2 = 391.06 x sin 25O = 233.73m

Sin 1350

T1t2 = t1t2 – T1t1 or t1t

= 391.06 – 110.85

= 280.21m = (tangent length for second curve)

Tangent length for second curve is given by

R2 tan Ø2/2 = R2 tan 100 = 280.31m

= R2 = 280.21

tan 10

= 1589.15m

Distance IT2 = It2 + t2T2

= 233.73 + 280.21

= 513.94m

Questions.

(1) Prepare a setting out table for horizontal curve on a roadway having a radius of

350.00m to accommodate two straights meeting at an intersection point whose drainage

is 8526.80m with a deflection angle of 220 33’ 00’’ use a chord length of 50.00m

(2) Given:

Figure 5.3 : A Horizontal Curve.

Data: R = 115.00m

Ø = 270 57’20’’

O

T1 T2

θ

R

C

L

I

Chainage of I = 1330.00m

Required: Construct a setting out table showing all your calculations

LOCATION OF THE INTERSECTION AND TANGENT POINTS IN

THE FIELD

It is not sufficient to scale the positions of the tangent points from a plan; they must be

accurately set out on the site. The procedure is as follows; with consideration to the figure

below;

T

Figure 3.4: location of intersection and tangent points.

Locate the two tangents lines AC and BD and define them by means of a suitable target.

Set a theodolite up on one of the lines (say AC) and sight towards the intersection of the

two tangents at I

Drive in two pegs x and y on the line AC such that BD will intersect the line xy. Nails in

the top of the pegs should mark the exact position of the tangent line.

θ/2

θ

U

I C D

A B

x

y

Join pegs x and y by means of a string line

Set up the theodolite on BD pointing towards I and fix the position of I by driving a peg

where the line of sight from BD intersects the string line.

Set up the theodolite over I and measure angle AIB, hence angle Ø

Calculate tangent lengths IT and IU using R tan Ø/2

Measure back from I to T and U, drive in pegs and mark the exact points by nails in the

tops of the pegs.

Check the setting out by measuring angle ITU, which should be equal to Ø/2.

However, the use of two theodolites simplifies the procedure by eliminating steps (3) and

(4).

WEEK 6.

LOCATION OF THE TANGENT POINTS WHEN THE

INTERSECTION POINT IS INACCESSIBLE.

Due to site conditions e.g. marshy ground, river, hill etc. obstructing the accessibility of

the intersection point on site, the following steps should be adopted to determine Ø and

locate the tangent points T and U as shown in figure below

Figure 6.1 : Location of tangent points when intersection point is inaccessible.

PROCEDURE.

Choose points A and B somewhere on the tangents such that it is possible to sight A from

B and vice versa.

Measure distance AB.

Measure angle а and β, deduce & and hence Ø.

Use the sine rule to calculate IA and IB

θ

ϒ

α β

T U

I

A B

Calculate IT and IU from R tan Ø/2

AT = IA – IT and BU = IB – IU, hence set out T and U. If A and B are chosen to be on

the other side of T and U, AT and BU will have negative values

If possible, sight from T to U as a check, measure angle ITU which should be equal to Ø/2

Figure 6.2 :OBSTRUCTION IN THE LINE OF SIGHT

OBSTRUCTION IN THE LINE OF SIGHT.

If peg 4 for example can not be set out due to obstruction in the line of sight shown in

the figure above, after setting out pegs 1, 2, and 3 ,move the instrument to peg 3 and sight

back to T1 at a zero setting. Turn off an angle equal to 1800 + δ (4 – 1) i.e. δ must be

multiplied by the number of standard chord lengths between the two points being sighted

to and if there is a sub chord at the beginning of the curve, the deflection angle must be

added.

T1 T2

θ

1 2 3 4

5 δ1

δ2 δ3 δ4

δ5

WEEK 8.

EQUIPMENT FOR SETTING OUT OF BUILDING.

The nature and complexity of the building or any engineering work like; bridges, dams,

roads etc. determine the accuracy that need to be achieved, which in turns, defines or

determines which types of equipment will be selected for the task e.g. dumpy level or

theodolite, fibre glass tape or steel tape, plumb bob and line or optical plummet etc.

Figure 8.1 :Digital Tape.

Figure 8.2 : Steel Tape.

Figure 8.3: Measuring Tape.

1. PEGS:

Figure 8.4 : Wooden Pegs.

These are usually made of two materials;

i. Timber pegs 50mm section of variable length but having a pointed end to facilitated

driving into the ground by hammering. A timber peg may have a nail fixed to its top at

the centre to locate exactly the station point. All setting out pegs should be clearly

marked with a 50mm deep and red paint and should have a board of blue paint.

ii. Steel pegs they are usually formed from lengths of steel reinforcement rods, cut to a

suitable lengths and may have one edge sharpened to facilitate careful driving positions

have been checked, they are normally surrounded by concrete. Identification works may

be made into the surface of the concrete before it sets hard

2. LINES:

They are strings, wire, nylon etc. the weather condition plays a very vital role in selecting

which material to be used so that the line is safe from damage, stretch, sag in prevailing

working conditions of the weather. The lines provides straight out lines from a peg to

another. They define straight lines from points or stations.

3. PROFILE BOARDS:

Figure 8.5 : Profile board

These are used in conjunction with pegs so that extended lines positions may be marked

by using profile boards, the string or wire lines can be removed in the knowledge that

when they are required again, they can be positioned exactly as they were originally.

Normally, a profile boards is erected near each off set peg and used in exactly the same

way as a sight rail, a traveler are being used between profile boards to monitor

excavation.

4. SITE SQUARE:

This is an optical device used for setting out right angles whereby unskilled labour can

attain an accuracy of ± 5mm in 30m.

The instrument is basically of two telescopes mounted one above the other and with their

lines of sight set at 900 to each other. The site square is supported on a tripod stand,

which can be set over a fixed mark on the ground. The lower telescope is aimed along the

line from which the right angle is to be established being brought to bear on any site mark

in the line by moving the telescope:

In the vertical plane

Laterally by means of a fine – turning screw. Once the adjustment of the lower telescope

is complete, the upper telescope will trace out a line at right angles to the original line and

a further site mark can be positioned as required by moving this telescope in the vertical

plane only.

Fifure 8.6: peg.

Figure8.7 :TRAVELLERS USED IN SETTING OUT SLOPING GROUNDS

WEEK 9.

SETTING OUT OF BUILDING .

Setting out the base line: The base line adopted in setting out a building is usually the

building line, although on extensive factory layouts are centre lines of buildings are

sometimes runs of machinery. In either case, the location of such lines is reacted to the

physical features of the site. The building line is the line of the front face of the building

as indicated in figure below (line AB).

Figure 9.1 Plan of a simple building for setting out.

The position of the building line may be defined on the working plan by measurements

from any of the following;

The property boundary

The edge of the road kerb

The centre line of the road.

Proposed building (structure)

EXISTING BUILDING

EXISTING BUILDING

It is important to note that

Where there is no indication of the building line, its position must be agreed on site with

the local authority-building inspectors.

Where there is an obvious line of existing building frontages, this line is usually adopted

as the building line.

The building line is first ranged by eye and pegs are placed at the two front corners of the

outer face of the proposed building. Critical measurements are made from the boundary

to the building corners as shown in figure above or defined by local regulations and along

the face of the proposed building by nails hammed into the pegs.

B. Setting out the subsidiary lines: From the two front pegs, A and B, angles are set

out in accordance with the building plan to follow the outer face of the flank walls. This

could be done with a theodolite, setting up over each peg in turn and turning off the

required angle from the building line in each case. As the angle of the flank wall is most

often 900 this could be set out without a theodolite using the following;

a. A 3:4:5 taped triangle

b. A builder’s square, which is a 3:4:5-ratio triangle made out of timber

c. An optical square

d. A site square, which is a proprietary instrument consisting of two small telescopes

fixed rigidly at right angles on a small stand.

e. A level incorporating a horizontal circle like a theodolite, but reading by vernier

to about 5’only. When the two rear pegs; C and D are placed and nail – marked, they are

checked by measuring between them and by measuring the diagonals.

In a rectangle building, the two diagonals must be equal to prove the positioning

of the pegs.

After the main outline has been pegged ,any minor extensions or returns from the

main figure are pegged and checked, such as the pegs at e, f, g, h, j and k, when the

complete outline of the outer face of the building has been pegged and checked.

Setting out the reference marks: The pegs now placed will be destroyed as the

foundations are excavated and the reference system must be adopted. This can be

achieved by the use of profile boards, illustrated in the above figure.

Profile boards are constructed of 150 – or 200mm by 25mm, boards supported on

50mm square posts hammered firmly into the ground, well clear of the working area. On

well organized sites, the boards are placed at one level, usually finished floor level or

dam – proof course level. The advantages of these are as follows;

They help to keep the tape horizontal when making measurements.

They provide a level datum around the site so that less check leveling is needed

subsequently.

Disturbance of the boards can easily be noted visually.

Levels of work below ground can be controlled by travelers using the profile boards as

sight rails.

Approximate levels can be obtained by direct vertical measurements up or down from

lines strung between the profile boards.

Once all boards have been placed in position, all at one level, lines are strung

between them and positioned vertically above the nail markers defining the building

outline. If the line is some distance above the peg, the peg position must be plumbed

upward, using plumb line in reference to the less accurate bricklayer’s spirit level. When

the lines have been accurately strung across the profile boards, positions are marked with

a nail or saw – cut so that they may be replaced at any time. The intersections of the

various strung lines will then define the peg positions when they are removed for

excavations. Profile boards for minor buildings or projections are not always erected or

needed. Full foundation width is marked on the boards and two lines strung between

these points to define the width of the foundation trench to be dug. Once the trench has

been started, the lines are removed.

Alternatively, the lines may be temporarily defined along the ground by means of

strips of lime or sand to guide excavation of the trench.

WEEK 10.

EARTHWORK QUANTITIES.

In many engineering projects, large parcels of land are required for the site and huge

amounts of construction materials have to be moved ignored to form the necessary

embankments, cuttings, foundations, basements, lakes and so on, that have been specified

in the design. Suitable land and construction materials can be very expensive and if a

project is to be profitable to the contractor or the construction company, it must involve

accurate measurements as possible of any areas and volumes in order that appropriate

estimates for such earthwork quantities can be included in the tender documents.

Also, for certain projects, such as the construction of a new highway, where large amount

of material have to be excavated and moved around the site, careful planning of this

movement is essential since charges may be levied not only on the volumes involved but

also on the haulage distances.

UNITS OF MEASUREMENTS.

Although systeme International (SI) units are widely used, there are times where other

acceptable units are equally appropriate. Area calculation represents one such occasion

since the SI unit of area is square meter (m2), the figure involved sometimes may be very

big for large areas. To overcome this, the following units system is often adopted;

100 m2 =1 are

100ares =1hectrare (10000m2)

100hectares = 1 square kilometer (106m2)

For volumes, the SI unit is the cubic meter (m3) and this is used through out most civil

engineering works for all volumes of materials, no matter how large or small.

AREAS ENCLOSED BY STRAIHGT LINES.

This category of areas form all areas enclosed by transverse, triangulation, trilateration or

detail survey lines. The results obtained for such areas will be exact since correct

geometric equations and theorems can be applied directly.

i. AREAS ENCLOSED BY STRAIGHT LINES.

The straight-sided figre can be divided into well-conditioned triangles, the areas of which

can be calculated using one of the following formulae;

(a) Area = √[S(S-a) (S-b) (S-c)]

(b) Are = ½(Base of triangle x Height)

© Area = ½ ab sin C where C is the angle contained between sides lengths a and b.

AREAS FROM CO-ORDINATES.

In traverse and triangulation calculations, the coordinate of the junctions of the sides of

a straight – sided figure are calculated and it is possible to use them to calculated area

enclosed by the control network lines. This is achieved using the cross coordinate

method.

Consider the figure below;

The three-sided clockwise control network lines shown above; ABC has an area =

ABC.

The area of ABC has = Area of ABQP + Area of BCRQ – Area of ACRP

Note that ABQP, BCRQ and ACRP are all Trapeziums for which areas can be

calculated from the relation;

Area of Trapezium = (mean height) x (width) or

Area of Trapezium = ½ (sum of parallel sides) x height.

Therefore,

Area of ABQP = ½ (N1 +N2) (E2-E1)

Area of ACRQ = ½ (N2 + N3) (E3-E2)

Area of ACRP = ½ (N1 +N3) (E3-E2)

Hence, Area of ABC = ½�N1+N2� �E2-E1� + ½� N2+ N3� �E3-E2�

- ½�N1+N3� �1E - 3 E �

»Area of ABC =½[�N1+N2� �E2-E1� + � N2+ N3� �E3-E2�

- �N1+N3� �1E - 3 E � ]

2XArea of ABC = (N1+N2)(E2-E1)+(N2+N3)(E3-E2)-(N1+N3)(E3-E1)

Opening the brackets,

2xAreaofABC=N1E2-N1E1+N2E2-N2E1+N2E3-N2E2+N3E3-N3E2-

N1E3+N1E1-N3E3+N3E1.

B (E2N2) N

E

C (E3N3)

A (E1N1)

R Q P O

O

Re-arranging,

2xArea of ABC = N1E2=N2E3+N3E1-E1N2-E2N3-E3N1

= (N1E2+N2E3+N3E1) – (E1N2+E2N3+E3N1)

» 2XArea of ABC = (N1E2+N2E3+N3E1) – (E1N2+E2N3+E3N1)

The similarity between the two brackets should be taken note of.

The illustration above is for a three sided figure but the formula can be used

appropriately for any figure containing N-sides and the general formula for such a

case is given by;

2xArea = (N1E2+N2E3+N3E4+…...NN-1 EN+NNE1)

-

(E1N2+E2N3+E3N4+…...EN-1 NN+ENN1).

The area of figure shown below can be calculated, using the cross-co-ordinate method as

follows;

Area =1/2 [(N1E2+N2E3+N3E4+N4E1) – (E1N2+E2N3+E3N4+E4N1)]

A (N1E1) B (N2E2)

C (N3E3) D (N4E4)

QUESTION (Class work).

Use the cross-coordinate method to calculate the total area of a triangle whose co-

ordinates are given as follows;

Where;

EA = 806.71, NA = 366.84

EB = 203.18, NB = 203.18

EC = 314.24, NC = 251.62

COMPUTATION OF AREA FROM PLOTTED PLAN.

By dividing the area into squares: In this method, square of equal size are ruled out a

piece of tracing paper. Each square represents unit area, which could be 1 cm2 or 1 m2.

the tracing paper is placed over the plan and the number of full squares are counted. The

total area is then calculated by multiplying the number of squares by the unit area of each

square.

B

C

A

WEEK 11.

AREAS ENCLOSED BY IRREGULAR LINES.

For such cases, only approximate results are achieved. However, there are many methods

in use which give best approximations. Give and take lines, Graphical methods and

mathematical approach are some of the available methods which can be adopted.

Nevertheless, under the mathematical methods, Trapezoidal rule and Simpson’s rule are

common.

y irregular figure or shape, which can be divided into strips most especially when the area

is bounded by a traverse line and irregular boundary, the total area can be found using

the Trapezoidal rule.

(I) THE TRAPEZOIDAL RULE.

Consider the figure below, showing a figure bounded by a survey line and an irregular

boundary.

x x x x x x x x x

A1 A2 A3 A4 A5 A6 A7 A8

The traverse line is divided into a number of small strips of equal intercepts of lengths; x

and the offsets O1, O2, O3, e.t.c. and are measured directly on the ground or by scaling

O1

O2

O3

O4 O5

O6 O7

O8

O9

Traverse line

Offsets at regular intervals

from the plan. If x is so short enough for length of boundary between offsets to be

assumed straight, then, the area is divided into a series of trapezoids, as shown in the

figure above.

The area between each strip is calculated, thus;

A1 = (O1+O2) X

2

A2 = (O2+O3) X

2

A3 = (O3+O4) X , e.t.c.

2

Hence, for N offsets, the total area (A) is given by;

Area = (O1+O2) X + (O2+O3) X + …. (ON-1+ ON) X

2 2 2

This leads to the general trapezoidal rule as follows;

Total area = x/2 [O1+ON+2(O2+O3+O4+…. +ON-1)]

The trapezoidal rule applies any number of offsets.

Example: The following offsets, 8m apart were measured at right angles from a traverse

line to an irregular boundary.

0m,2.3m,5.5m,7.9m,8.6m,6.9m,7.3m,6.2m,3.1m,0m.

Required; calculate the area between the survey line and the irregular boundary using the

trapezoidal rule.

Solution;

Total Area = 8.0/2 [0+0 + 2(2.3+5.5+7.9+8.6+6.9+7.3+6.2+3.1)]

= 4 x 2(47.8) = 382.4m2

Question;

Calculate the total area of the plot between the survey line the boundary if the offsets,

scaled from the plan of the intervals of 10.00m are shown below.

Present your answer in both m2 and hectares.

(II) SIMPSON’S RULE:

This method assumes that instead of being made of a series of straight lines, the

boundary consists of a series of parabolic arcs. A more accurate result is obtained

since a better approximation of the true shape of the irregular boundary is achieved.

Figure below shows an illustration.

Offset O1 O2 O3 O4 O5 O6 O7 O8 O9 O10

Length(m) 16.76 19.81 20.42 16.76 17.69 17.68 17.37 18.53 16.76 17.68

A1

A2

A3 A4

A5 A6

A7

X X X X X X X

O1

O2

O3

O4 O5 O6 O7

O8

Offset at regular intervals.

Transverse line

Simpson’s rule considers offsets in sets of three and it can be shown that the area between

offset 1 and 3 is given by;

A1 + A2 = L/3 (O1 + 4 O2 + O3)

Similarly,

A3+ A4 = L/3 (O3 + 4 O4 + O5)

Hence, in general,

Total Area = L/3(O1 + ON + 4∑ Even Offsets + 2∑ Remaining Odd Offsets)

It is important to note that N must be an Odd number for Simpson’s rule to be applicable.

When faced with an Even number of N offsets, the final offset should be omitted (for

example O8 or O10), the rest of the area calculated and the last (omitted) area calculated

separately as a trapezium using the trapezoidal rule.

Example; Use the Simpson rule to solve the last worked example on page 63.

Solution; There are 10 (even) number of offsets. Thus, area between offsets 1-9 would be

calculated using the Simpson’s rule while the area between offsets 9-10 would be

calculated by trapezoidal rule.

Area (1-9) = 8/3[0 + 3.1 + 4(2.3+7.9+6.9+6.2) + 2(5.5+8.6+7.3)]

= 8/3[3.1 + 4(23.3) + (2(21.4)]

= 8(139.1)/3

= 370.9m2

Area (9-10) =8/2(3.1 + 0)

= 12.4m2

Hence, the Total Area = 370.9 + 12.4 = 383.3m2

Note the difference between this result and that obtained previously. This is so because

the Simpson’s rule will give more accurate results when the irregular boundary is

genuinely irregular and the trapezoidal rule will give the most accurate results if the

irregular boundary is almost a series of straight lines.

In general, for irregular – sided figure, Simpson’s rule should be used.

WEEK 12.

iii. The Planimeter.

A Planimeter is an instrument which is used to automatically measure the total area of

any irregular sided plane figure. Traditionally, mechanical devices were used but

although these are still being manufactured, they have been largely super ceded by digital

instruments like the Planimeter.

When using the Planimeter, a high degree of accuracy can be achieved no matter how

complex the shape of the irregular area in question may be.

Mechanical Planimeter;

This consists of two arms, the pole arm and the tracing arm, which are joint at a pivot.

At the other end of the pole is a heavy weight known as the pole block and at the other

end of the tracing arm is the tracing point which normally consists of magnifying

eyepiece containing an index mark.

The tracing arm also incorporates a measuring unit which contains an integrating disc.

The area is obtained from the integrating disc which revolves and alters the reading on

the measuring unit as the tracing point is moved round the perimeter of the figure. It is

possible to read 1/1000th of a revolution of the disc. The reading obtained on the

measuring unit is directly related to the length of the tracing arm. There are two types of

the mechanical planimeter; those with fixed tracing arms and those with movable tracing

arms.

On a fixed tracing arm type, the readings are obtained directly in m2 and then have to be

converted according to the plan scale to get the ground area.

While on the other hand, on the movable arm instrument, the tracing am length can be set

to particular values, depending on the plan scale such that the reading obtained give the

area directly.

Example:

A cross-sectional area was measured using a fixed arm mechanical planimeter which

gave readings in mm directly. The initial planimeter reading was 88 and the final reading

was 7450. If the horizontal scale of the cross-section was 1 in 200 and the vertical scale is

1 in 100, calculate the true area represented by the cross-section.

Solution:

The planimeter difference = Final reading – Initial reading

= 7450 – 88 = 7362mm2

However, 1mm2 actually represents an area (200mm x 100mm) since the horizontal and

vertical scales are 1 in 200 and 1 in 100 respectively.

Thus, 7362mm2 = (7362 x 200 x 100)

= 147240000mm2

=147.24m2

Note that once the areas of all the cross-sections have been obtained they are used to

calculate the volumes of material to be either excavated (cut) or imported (fill) between

consecutive cross-sections.

ASSIGNMENT.

Draw the different types of planimeter you know and explain fully, the working

principles of one (how to find area on the ground).

WEEK 13.

VOLUME CALCULATIONS.

The excavation, removal and dumping of earth is a frequent operation in

building/constructions sites in civil engineering practice.

In the construction of sewer, for example, a trench of sufficient width is excavated to a

given depths and gradients. The earth being stored in some convenient place (usually the

side of the trench) and then returned to the trench after the laying of the pipe. Any

material left over after the re-instatement must be carted away and disposed of. In

basement excavation, probably all the material removed will have to be required to be

carted away. But for embankment, the material required will have to be brought from

some other places.

In each case, however, payment will have to be made for labour, plant e.t.c; and this is

done on the basis of the calculated volume of the material handled. It is therefore

essential that the engineer or surveyor concerned is able to make good estimation of

volumes of earth works.

There are three general methods for calculating earth work volumes;

By Contours.

By Spot heights.

By Cross-sections.

ESTIMATION OF VOLUMES FROM CONTOUR LINES.

It is possible to calculate volumes using the horizontal areas contained in contour lines.

Due to the relatively high cost of accurately contouring large areas, the method is of

limited use. But where accurate contours are available, for instance, in reservoir sites,

they may be conveniently used.

The contour interval will determine the distance; D in the ‘End-area’ method or

‘Prismoidal formula’, however, for accuracy, this should be as small as possible,

preferably, 1 or 2 metres.

(A) THE END-AREA FORMULA.

The areas enclosed by individual contour lines are best taken off the plan by means of a

planimeter. In computing the volumes, the areas enclosed by two successive contour lines

are used in the ‘end area ‘formula, whence,

Volume = D. (Ax+Ay)/2

Where D = Vertical interval, A = Area enclosed by the contours.

Example:

The area within the underwater contour lines of a lake are as follows;

Contour(m) 190 188 186 184 182

Area (m2) 3150 2460 1630 840 210

Required: Calculate the Volume of water in the Lake between the 190 and 182 contour

lines.

Solution: Using the End-area method, Volume = D (Ax + Ay)/2 for two successive

contour; x and y.

Volume=2(3150+2460) + 2(2460+1630) + 2(1630+840) + 2(840+210)

2 2 2 2

= 5610 + 4090 + 2470 + 1050

= 13,220m3.

In general, Volume = D/2 [A1+2(A2+A3+A4+……. +AN-1) +AN]

Applying the above general formula to solve the last worked example, Volume =

2/2[3150+2(2460+1630+840) +210]

= 1[3150+9860+210]

= 13,220m2

Question (1):

Use the End-area method to calculate the volume of the following contour data for a

given site.

Contour(m) 0 15 30 45 60 75 90

Area(m2) 11 42 64 72 160 180 220

Question (2):

A trench, 80m long and with end faces vertical is excavated in a flat land. Subsequent to

excavation, cross-sections were taken at 20m intervals beginning at one end of the trench

and ending at the other end. The areas of the cross sections are progressively 63,71,76,72

and 69 m2. Determine the amount of excavation required in cubic meters, using the End –

area method.

WEEK 14.

PRISMOIDAL FORMULA.

If the volume of earth between two successive cross-sections be considered, a prismoid,

then a more precise formula (The Prismoidal formula) may be used. It is generally

considered that; all things being equal, the use of this formula gives the most accurate

estimate of volume which must be parallel plane figure not necessary of top and bottom

must be formed by straight contour lines running from one end face to the other.

The volume of a prismoid is given by;

Volume = D/6 (A1 + 4M + A2)

Where A1 and A2 are the areas of the two end faces, distance; D apart, M = area of the

section mid-way between the end faces.

Question:

A trench, 80m long and with end faces vertical is excavated in a flat land. Subsequent to

excavation, cross-sections were taken at 20m intervals beginning at one end of the trench

and ending at the other end. The areas of the cross sections are progressively 63,71,76,72

and 69 m2. Determine the amount of excavation required in cubic meters, using the

Prismoidal method.

ESTIMATION OF VOLUMES FROM SPOT LEVELS.

This is another method by means of which the earth involved in the construction of large

tanks, basements, borrow-pits e.t.c; and similar works with vertical sides may be

calculated. The computation is simplified if the formation level is to be a fixed level or to

fixed falls. But even basement with several levels present little difficulty.

Having located the outline of the structure on the ground, the engineer divides up the area

into squares or rectangles, marking the corner points. Levels are taken at each of these

corner points and by subtracting from these, the corresponding formation levels; a series

of heights is obtained from which the mean height of a series of vertical truncated prisms

of earth can be found. It is important to note that the prisms are called truncated because

unless the ground and formation levels are parallel, the end planes are not parallel to each

other.

The volume of each prism is given by he plan area (or area of the normal section)

multiplied by the mean height of the prism. The prisms may, ofcorse, be considered as

either rectangular or triangles as shown in the following examples.

Example; Given the layout of a construction site as shown in figure below, calculate

the volume of exaction required if the formation level is designed as 90.00m.

10M 10M

A B C

10M

D EEE F F

10M

G H I

91M 93M 94M

92M E 92M 93M 93M 91M 92M

SOLUTION:

POINT DIFFERENCE IN HEIGHT

ABOVE FORMATION

LEVEL(M)

NO. OF TIMES

TRIANGLE

OCCURS

PRODUCT

A 91-90 = 1 1 1

B 93-90 = 3 3 9

C 94-90 = 4 2 8

D 92-90 = 2 3 6

E 92-90 = 2 6 12

F 93-90 = 3 3 9

G 93-90 = 3 2 6

H 91-90 = 1 3 3

I 92-90 = 2 1 2

- ∑ TOTAL 24 56

Volume = ( ∑ product /∑ No. of times triangle occurs) x total area

Where; ∑ product = 56

∑ No. of times triangle occurs = 24

Total area = 20 x 20 = 400m2

Hence, Volume = (56/24) x 400

= 2.33 x 400

= 933.33m3

WEEK 15.

Example:

The figure below shows a rectangular plot of land which is to be excavated for

development. If the sides of the excavations are to be vertical, estimate the total volume

of earth to be removed. The depth of excavation at each point is indicated on the figure.

15M 15M

A B C

12.5M

D EEE F F E

12.5M

G H I

3.15M 3.70M 4.33M

3.94M 4.80M E 4.97M 5.17M 6.10M 4.67M

SOLUTION:

POINT DIFFERENCE IN HEIGHT

ABOVE FORMATION

LEVEL(M)

NO. OF TIMES

TRIANGLE

OCCURS

PRODUCT

A 3.15 1 3.15

B 3.70 3 11.10

C 4.33 2 8.66

D 3.94 3 11.82

E 4.80 6 28.80

F 4.97 3 14.91

G 5.17 2 10.34

H 6.10 3 18.30

I 4.67 1 4.67

- ∑ TOTAL 24 111.75

Volume = (∑ product /∑ No. of times triangle occurs) x total area

Where; ∑ product = 111.75

∑ No. of times triangle occurs = 24

Total area = 25 x 30 = 750m2

Hence, Volume = (111.75/24) x 750

= 3492.19m3

ESTIMATION OF VOLUMES FROM CROSS SECTIONS.

In this method, cross-sections are taken at right angles to some convenient line which

runs longitudinally through the earthworks and, although it is capable of general

application, it is probably most useful on long narrow works such as roads, ail ways,

canals, embankments, and pipe lines excavations e.t.c. The volumes of earthwork

between successive cross-sections are calculated from a consideration of the

cross-sectional areas, which in turn are measured by the general methods already

explained i.e. planimeter, division into triangles, counting squares e.t.c.

In long constructions, which have constant formation width ad side-slopes, it is possible

to simplify the computation of the cross-sectional areas by the use of formulae. These are

especially useful for railways, long embankments e.t.c. and formulae are available for the

following types of cross-sections:

(a) Sections level across.

(b) Section with a cross fall.

© Section part in cut part in fill.

(d Section of variable levels.