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Week No.2 Advance Engineering Surveying Lecture No.2 B-Tech By: Engr. Shams Ul Islam Lecturer, Civil Engg. Department CECOS University Peshawar

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Page 1: surveying and levelling 2

Week No.2

Advance Engineering Surveying

Lecture No.2

B-Tech

By:

Engr. Shams Ul Islam

Lecturer, Civil Engg. Department

CECOS University Peshawar

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Computation of Areas (Irregular bounded fields)

The main objective of the surveying is to compute the areas and volumes.

Generally, the lands will be of irregular shaped polygons. There are formulae

readily available for regular polygons like, triangle, rectangle, square and other

polygons.

But for determining the areas of irregular polygons, different methods are used.

Earthwork computation is involved in the excavation of channels, digging of

trenches for laying underground pipelines, formation of bunds, earthen

embankments, digging farm ponds, land levelling and smoothening. In most of

the computation the cross sectional areas at different interval along the length of

the channels and embankments are first calculated and the volume of the

prismoids are obtained between successive cross section either by trapezoidal or

prismoidal formula.

2 Engr.Shams Ul Islam ([email protected])

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Computation of Areas (Irregular bounded fields)

Computation of areas is carried out by any one of the following

methods:

β€’ Mid-ordinate method

β€’ Average ordinate method

β€’ Trapezoidal rule

β€’ Simpson’s rule

3 Engr.Shams Ul Islam ([email protected])

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The mid-ordinate rule

In this method, the ordinates are measured at the mid-points of

each division and the area is calculated by the formula.

π΄π‘Ÿπ‘’π‘Ž βˆ† = π‘†π‘’π‘š π‘œπ‘“ π‘šπ‘–π‘‘ π‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘‘π‘’π‘  Γ— 𝑑

π‘€π‘•π‘’π‘Ÿπ‘’ 𝑑 = 𝑑𝑕𝑒 π‘π‘œπ‘šπ‘šπ‘œπ‘› π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ 𝑏𝑒𝑑𝑀𝑒𝑒𝑒𝑛 𝑑𝑕𝑒 π‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘‘π‘’π‘ 

𝑀𝑖𝑑 π‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘‘π‘’ =𝑂1 + 𝑂2

2

π‘€π‘•π‘’π‘Ÿπ‘’ 𝑂1, 𝑂2, 𝑂3, … . . , 𝑂𝑛 π‘Žπ‘Ÿπ‘’ 𝑑𝑕𝑒 π‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘‘π‘’ π‘Žπ‘‘ π‘’π‘Žπ‘π‘• π‘œπ‘“ 𝑑𝑕𝑒 π‘‘π‘–π‘£π‘–π‘ π‘–π‘œπ‘›

4 Engr.Shams Ul Islam ([email protected])

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The mid-ordinate rule

5 Engr.Shams Ul Islam ([email protected])

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The mid-ordinate rule

The mid ordinate rule can also be in other way.

π΄π‘Ÿπ‘’π‘Ž βˆ† =β„Ž1+β„Ž2+β„Ž3+β‹―β„Žπ‘›

𝑛×L

π‘€π‘•π‘’π‘Ÿπ‘’ 𝑕1, 𝑕2, 𝑕3, … . . , 𝑕𝑛 π‘Žπ‘Ÿπ‘’ 𝑑𝑕𝑒 π‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘‘π‘’ π‘Žπ‘‘ 𝑑𝑕𝑒 π‘šπ‘–π‘‘ π‘π‘œπ‘–π‘›π‘‘ π‘œπ‘“ π‘’π‘Žπ‘π‘• π‘‘π‘–π‘£π‘–π‘ π‘–π‘œπ‘›

𝐿 = 𝑇𝑕𝑒 𝑙𝑒𝑛𝑔𝑑𝑕 π‘œπ‘“ 𝑑𝑕𝑒 π‘π‘Žπ‘ π‘’ 𝑙𝑖𝑛𝑒

𝑛 = 𝑑𝑕𝑒 π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘’π‘žπ‘’π‘Žπ‘™ π‘π‘Žπ‘Ÿπ‘‘π‘  π‘–π‘›π‘‘π‘œ 𝑀𝑕𝑖𝑐𝑕 𝑑𝑕𝑒 π‘π‘Žπ‘ π‘’ 𝑙𝑖𝑛𝑒 𝑖𝑠 𝑑𝑖𝑣𝑖𝑑𝑒𝑑

6 Engr.Shams Ul Islam ([email protected])

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Example

The following offsets were taken from a chain line of 60 m to an irregular

boundary line at an interval of 10 m, the offsets are:

0, 2.50, 3.50, 5.00, 4.60, 3.20, 0 m

Compute the area between the chain line, the irregular boundary line and the

end of offsets by mid ordinate rule

7 Engr.Shams Ul Islam ([email protected])

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Solution

𝑕1 =0 + 2.5

2= 1.25

𝑕2 =2.5 + 2.5

2= 3.0

𝑕3 =3.5 + 5

2= 4.25

𝑕4 =5 + 4.60

2= 4.80

𝑕5 =4.60 + 3.20

2= 3.90

𝑕6 =3.20 + 0

2= 1.60

𝐿 = 60 π‘š

𝑛 = 6

π΄π‘Ÿπ‘’π‘Ž βˆ† =β„Ž1+β„Ž2+β„Ž3+β‹―β„Žπ‘›

𝑛×L

𝐴 =1.25 + 3.0 + 4.25 + 4.80 + 3.90 + 1.60

6Γ— 60 = 188 π‘š2

8 Engr.Shams Ul Islam ([email protected])

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Average Ordinate Method

In this method, the ordinates are drawn and scaled at each of the points

of division of the base line and the area is calculated by the formula.

π΄π‘Ÿπ‘’π‘Ž βˆ† =π‘†π‘’π‘š π‘œπ‘“ π‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘‘π‘’π‘ 

𝑛 + 1Γ— 𝑙

π΄π‘Ÿπ‘’π‘Ž βˆ† =𝑂1 + 𝑂2 + 𝑂3 + β‹― , +𝑂𝑛

𝑛 + 1Γ— 𝑙

𝑙 = 𝑑𝑕𝑒 𝑙𝑒𝑛𝑔𝑑𝑕 π‘œπ‘“ 𝑑𝑕𝑒 π‘π‘Žπ‘ π‘’ 𝑙𝑖𝑛𝑒

𝑛 = π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘’π‘žπ‘’π‘Žπ‘™ π‘œπ‘Ÿπ‘‘π‘–π‘Žπ‘›π‘‘π‘’π‘  π‘–π‘›π‘‘π‘œ 𝑀𝑕𝑖𝑐𝑕 𝑑𝑕𝑒 π‘π‘Žπ‘ π‘’ 𝑙𝑖𝑛𝑒 𝑖𝑠 𝑑𝑖𝑣𝑖𝑑𝑒𝑑

π‘€π‘•π‘’π‘Ÿπ‘’ 𝑑 = 𝑑𝑕𝑒 π‘π‘œπ‘šπ‘šπ‘œπ‘› π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ 𝑏𝑒𝑑𝑀𝑒𝑒𝑒𝑛 𝑑𝑕𝑒 π‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘‘π‘’π‘ 

π‘€π‘•π‘’π‘Ÿπ‘’ 𝑂1, 𝑂2, 𝑂3, … . . , 𝑂𝑛 π‘Žπ‘Ÿπ‘’ 𝑑𝑕𝑒 π‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘‘π‘’ π‘Žπ‘‘ π‘’π‘Žπ‘π‘• π‘œπ‘“ 𝑑𝑕𝑒 π‘‘π‘–π‘£π‘–π‘ π‘–π‘œπ‘›

9 Engr.Shams Ul Islam ([email protected])

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The Average ordinate method

10 Engr.Shams Ul Islam ([email protected])

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Trapezoidal Rule

This rule is more accurate than the first two ones. In this rule, boundaries between

the ends of ordinates are assumed to be straight. Thus the areas enclosed between

the base line and the irregular boundary line are considered as trapezoids.

Let

𝑑 = 𝑑𝑕𝑒 π‘π‘œπ‘šπ‘šπ‘œπ‘› π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ 𝑏𝑒𝑑𝑀𝑒𝑒𝑒𝑛 𝑑𝑕𝑒 π‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘‘π‘’,

𝑛 = π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘’π‘žπ‘’π‘Žπ‘™ π‘œπ‘Ÿπ‘‘π‘–π‘Žπ‘›π‘‘π‘’π‘  π‘–π‘›π‘‘π‘œ 𝑀𝑕𝑖𝑐𝑕 𝑑𝑕𝑒 π‘π‘Žπ‘ π‘’ 𝑙𝑖𝑛𝑒 𝑖𝑠 𝑑𝑖𝑣𝑖𝑑𝑒𝑑

π‘Žπ‘›π‘‘ 𝑂1, 𝑂2, 𝑂3, … . . , 𝑂𝑛 π‘Žπ‘Ÿπ‘’ 𝑑𝑕𝑒 π‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘‘π‘’ π‘Žπ‘‘ π‘’π‘Žπ‘π‘• π‘œπ‘“ 𝑑𝑕𝑒 π‘‘π‘–π‘£π‘–π‘ π‘–π‘œπ‘›

π΄π‘Ÿπ‘’π‘Ž βˆ† =𝑂1 + 2𝑂2 + 2𝑂3 + β‹― , +2π‘‚π‘›βˆ’1 + 𝑂𝑛

2Γ—

𝑙

𝑛

OR

π΄π‘Ÿπ‘’π‘Ž βˆ† =𝑑

2Γ— 𝑂1 + 2𝑂2 + 2𝑂3 + β‹― , +2π‘‚π‘›βˆ’1 + 𝑂𝑛

OR

π΄π‘Ÿπ‘’π‘Ž βˆ† = 𝑑 ×𝑂1 + 𝑂𝑛

2+ 𝑂2 + 𝑂3 + β‹― , +π‘‚π‘›βˆ’1

11 Engr.Shams Ul Islam ([email protected])

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Simpson’s Rule

In this rule, the boundaries between the ends of ordinates are assumed to

form an arc of parabola. Hence Simpson’s rule is some times called as

parabolic rule. Refer to figure:

12 Engr.Shams Ul Islam ([email protected])

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Continue..

𝑂1, 𝑂2, 𝑂3 are the consecutive coordinates

𝑑 = 𝑑𝑕𝑒 π‘π‘œπ‘šπ‘šπ‘œπ‘› π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ 𝑏𝑒𝑑𝑀𝑒𝑒𝑛 𝑑𝑕𝑒 π‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘‘π‘’π‘ 

Aπ‘Ÿπ‘’π‘Ž 𝐴𝐹𝑒𝐷𝐢 = π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘π‘Ÿπ‘Žπ‘π‘’π‘§π‘–π‘’π‘š 𝐴𝐹𝐷𝐢 + π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ π‘’π‘”π‘šπ‘’π‘›π‘‘ 𝐹𝑒𝐷𝐸𝐹

π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‡π‘Ÿπ‘Žπ‘π‘’π‘§π‘–π‘’π‘š =𝑂1 + 𝑂3

2Γ— 𝐴𝐢

π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‡π‘Ÿπ‘Žπ‘π‘’π‘§π‘–π‘’π‘š =𝑂1 + 𝑂3

2Γ— 2𝑑

π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ π‘’π‘”π‘šπ‘’π‘›π‘‘ =2

3Γ— π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘’π‘Ÿπ‘Žπ‘™π‘™π‘Žπ‘™π‘œπ‘”π‘Ÿπ‘Žπ‘šπ‘’ 𝐹𝑓𝑑𝐷

=2

3Γ— 𝑒𝐸 Γ— 𝐹𝐷

=2

3Γ— 𝑒𝐸 Γ— 2𝑑

π‘€π‘•π‘’π‘Ÿπ‘’ 𝑒𝐸 = 𝑂2 βˆ’ (𝑂1+𝑂3

2) So

π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ π‘’π‘”π‘šπ‘’π‘›π‘‘ =2

3Γ— 𝑂2 βˆ’ (

𝑂1 + 𝑂3

2) Γ— 2𝑑

13 Engr.Shams Ul Islam ([email protected])

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Continue..

So the area between the first Two divisions,

π΄π‘Ÿπ‘’π‘Ž βˆ† =𝑂1 + 𝑂3

2Γ— 2𝑑 +

2

3Γ— 𝑂2 βˆ’ (

𝑂1 + 𝑂3

2) Γ— 2𝑑

= 𝑂1 + 𝑂3 𝑑 +4𝑑

3

2𝑂2 βˆ’ 𝑂1 βˆ’ 𝑂3

2

= 𝑂1𝑑 + 𝑂3𝑑 +4𝑑

62𝑂2 βˆ’ 𝑂1 βˆ’ 𝑂3

= 𝑂1𝑑 + 𝑂3𝑑 +2𝑑

32𝑂2 βˆ’ 𝑂1 βˆ’ 𝑂3

=3𝑂1𝑑 + 3𝑂3𝑑 + 4𝑂2𝑑 βˆ’ 2𝑂1𝑑 βˆ’ 2𝑂3𝑑

3

=3𝑂1𝑑 βˆ’ 2𝑂1𝑑 + 4𝑂2𝑑 + 3𝑂3𝑑 βˆ’ 2𝑂3𝑑

3

=𝑂1𝑑 + 4𝑂2𝑑 + 𝑂3𝑑

3

=𝒅

πŸ‘Γ— π‘ΆπŸ + πŸ’π‘ΆπŸ + π‘ΆπŸ‘

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Continue..

Similarly, the area of next two divisions

=𝒅

πŸ‘Γ— π‘ΆπŸ‘ + πŸ’π‘ΆπŸ’ + π‘ΆπŸ“

π‘‡π‘œπ‘‘π‘Žπ‘™ π΄π‘Ÿπ‘’π‘Ž =𝑑

3Γ— 𝑂1 + 4𝑂2 + 2𝑂3 + 4𝑂4 + β‹― … + 2π‘‚π‘›βˆ’2 + 4π‘‚π‘›βˆ’1 + 𝑂𝑛

OR

π‘‡π‘œπ‘‘π‘Žπ‘™ π΄π‘Ÿπ‘’π‘Ž =𝑑

3Γ— 𝑂1 + 𝑂𝑛) + 4(𝑂2 + 𝑂4 + β‹― ) + 2(𝑂3 + 𝑂5 + β‹―

15 Engr.Shams Ul Islam ([email protected])

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Comparison of Trapezoidal rule with Simpson’s Rule

Trapezoidal rule

The boundary between the ordinates is considered to be straight

There is no limitation. It can be applied for any number of ordinates

It gives an approximate result

Simpson’s Rule

The boundary between the ordinates is considered to be an arc of a parabola

To apply this rule, the number of ordinates must be odd

It gives a more accurate result

16 Engr.Shams Ul Islam ([email protected])

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Problem

The following offsets were taken at 15 m intervals from a survey line to an

irregular boundary line.

3.50,4.30, 6.75, 5.25, 7.50, 8.80, 7.90, 6.40, 4.40, 3.25 m

Calculate the area enclosed between the survey line, the irregular boundary line,

and the offsets, by:

(1) The trapezoidal rule (2) Simpson’s rule

17 Engr.Shams Ul Islam ([email protected])

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Continue..

By Using Trapezoidal Rule

π΄π‘Ÿπ‘’π‘Ž βˆ† = 𝑑𝑂1 + 𝑂𝑛

2+ 𝑂2 + 𝑂3 + β‹― , +π‘‚π‘›βˆ’1

π΄π‘Ÿπ‘’π‘Ž βˆ† = 𝑑𝑂1 + 𝑂10

2+ 𝑂2 + 𝑂3 + β‹― , +𝑂9

π΄π‘Ÿπ‘’π‘Ž βˆ† = 15 Γ—3.50 + 3.25

2+ 4.30 + 6.75 + 5.25 + 7.50 + 8.80 + 7.90 + 6.40 + 4.40

𝑨𝒓𝒆𝒂 βˆ† = πŸ–πŸπŸŽ. πŸπŸπŸ“π’ŽπŸ

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Continue..

By Using Simpson’s Rule

π‘‡π‘œπ‘‘π‘Žπ‘™ π΄π‘Ÿπ‘’π‘Ž =𝑑

3Γ— 𝑂1 + 𝑂𝑛) + 4(𝑂2 + 𝑂4 + β‹― ) + 2(𝑂3 + 𝑂5 + β‹―

π΄π‘Ÿπ‘’π‘Ž (βˆ†1) =15

3Γ— 3.50 + 4.40) + 4 4.30 + 5.25 + 8.80 + 6.40 + 2(6.75 + 7.50 + 7.9

= 756.00π‘š2

π΄π‘Ÿπ‘’π‘Ž βˆ†2 =𝑂9 + 𝑂10

2Γ— 𝑑 =

4.40 + 3.25

2Γ— 15 = 57.38π‘š2

𝑻𝒐𝒕𝒂𝒍 𝑨𝒓𝒆𝒂 = βˆ†πŸ + βˆ†πŸ

𝑻𝒐𝒕𝒂𝒍 𝑨𝒓𝒆𝒂 = πŸ•πŸ“πŸ”. 𝟎𝟎 + πŸ“πŸ•. πŸ‘πŸ– = πŸ–πŸπŸ‘. πŸ‘πŸ–π’ŽπŸ

19 Engr.Shams Ul Islam ([email protected])