Download - 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

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])

Computation of Areas (Irregular bounded fields)

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

methods:

• Mid-ordinate method

• Average ordinate method

• Trapezoidal rule

• Simpson’s rule


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.

𝐴𝑟𝑒𝑎 ∆ = 𝑆𝑢𝑚 𝑜𝑓 𝑚𝑖𝑑 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 × 𝑑

𝑤𝑕𝑒𝑟𝑒 𝑑 = 𝑡𝑕𝑒 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑒𝑛 𝑡𝑕𝑒 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠

𝑀𝑖𝑑 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 =𝑂1 + 𝑂2

2

𝑤𝑕𝑒𝑟𝑒 𝑂1, 𝑂2, 𝑂3, … . . , 𝑂𝑛 𝑎𝑟𝑒 𝑡𝑕𝑒 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑎𝑡 𝑒𝑎𝑐𝑕 𝑜𝑓 𝑡𝑕𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛



The mid ordinate rule can also be in other way.

𝐴𝑟𝑒𝑎 ∆ =ℎ1+ℎ2+ℎ3+⋯ℎ𝑛

𝑛×L

𝑤𝑕𝑒𝑟𝑒 𝑕1, 𝑕2, 𝑕3, … . . , 𝑕𝑛 𝑎𝑟𝑒 𝑡𝑕𝑒 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑎𝑡 𝑡𝑕𝑒 𝑚𝑖𝑑 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑒𝑎𝑐𝑕 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛

𝐿 = 𝑇𝑕𝑒 𝑙𝑒𝑛𝑔𝑡𝑕 𝑜𝑓 𝑡𝑕𝑒 𝑏𝑎𝑠𝑒 𝑙𝑖𝑛𝑒

𝑛 = 𝑡𝑕𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑞𝑢𝑎𝑙 𝑝𝑎𝑟𝑡𝑠 𝑖𝑛𝑡𝑜 𝑤𝑕𝑖𝑐𝑕 𝑡𝑕𝑒 𝑏𝑎𝑠𝑒 𝑙𝑖𝑛𝑒 𝑖𝑠 𝑑𝑖𝑣𝑖𝑑𝑒𝑑


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


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 𝑚

𝑛 = 6

𝐴𝑟𝑒𝑎 ∆ =ℎ1+ℎ2+ℎ3+⋯ℎ𝑛

𝑛×L

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

6× 60 = 188 𝑚2


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.

𝐴𝑟𝑒𝑎 ∆ =𝑆𝑢𝑚 𝑜𝑓 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠

𝑛 + 1× 𝑙

𝐴𝑟𝑒𝑎 ∆ =𝑂1 + 𝑂2 + 𝑂3 + ⋯ , +𝑂𝑛

𝑛 + 1× 𝑙

𝑙 = 𝑡𝑕𝑒 𝑙𝑒𝑛𝑔𝑡𝑕 𝑜𝑓 𝑡𝑕𝑒 𝑏𝑎𝑠𝑒 𝑙𝑖𝑛𝑒

𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑞𝑢𝑎𝑙 𝑜𝑟𝑑𝑖𝑎𝑛𝑡𝑒𝑠 𝑖𝑛𝑡𝑜 𝑤𝑕𝑖𝑐𝑕 𝑡𝑕𝑒 𝑏𝑎𝑠𝑒 𝑙𝑖𝑛𝑒 𝑖𝑠 𝑑𝑖𝑣𝑖𝑑𝑒𝑑

𝑤𝑕𝑒𝑟𝑒 𝑑 = 𝑡𝑕𝑒 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑒𝑛 𝑡𝑕𝑒 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠

𝑤𝑕𝑒𝑟𝑒 𝑂1, 𝑂2, 𝑂3, … . . , 𝑂𝑛 𝑎𝑟𝑒 𝑡𝑕𝑒 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑎𝑡 𝑒𝑎𝑐𝑕 𝑜𝑓 𝑡𝑕𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛


The Average ordinate method


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 = 𝑡𝑕𝑒 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑒𝑛 𝑡𝑕𝑒 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒,

𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑞𝑢𝑎𝑙 𝑜𝑟𝑑𝑖𝑎𝑛𝑡𝑒𝑠 𝑖𝑛𝑡𝑜 𝑤𝑕𝑖𝑐𝑕 𝑡𝑕𝑒 𝑏𝑎𝑠𝑒 𝑙𝑖𝑛𝑒 𝑖𝑠 𝑑𝑖𝑣𝑖𝑑𝑒𝑑

𝑎𝑛𝑑 𝑂1, 𝑂2, 𝑂3, … . . , 𝑂𝑛 𝑎𝑟𝑒 𝑡𝑕𝑒 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑎𝑡 𝑒𝑎𝑐𝑕 𝑜𝑓 𝑡𝑕𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛

𝐴𝑟𝑒𝑎 ∆ =𝑂1 + 2𝑂2 + 2𝑂3 + ⋯ , +2𝑂𝑛−1 + 𝑂𝑛

2×

𝑙

𝑛

OR

𝐴𝑟𝑒𝑎 ∆ =𝑑

2× 𝑂1 + 2𝑂2 + 2𝑂3 + ⋯ , +2𝑂𝑛−1 + 𝑂𝑛

OR

𝐴𝑟𝑒𝑎 ∆ = 𝑑 ×𝑂1 + 𝑂𝑛

2+ 𝑂2 + 𝑂3 + ⋯ , +𝑂𝑛−1


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:


Continue..

𝑂1, 𝑂2, 𝑂3 are the consecutive coordinates

𝑑 = 𝑡𝑕𝑒 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑕𝑒 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠

A𝑟𝑒𝑎 𝐴𝐹𝑒𝐷𝐶 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚 𝐴𝐹𝐷𝐶 + 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 𝐹𝑒𝐷𝐸𝐹

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑇𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚 =𝑂1 + 𝑂3

2× 𝐴𝐶

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑇𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚 =𝑂1 + 𝑂3

2× 2𝑑

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 =2

3× 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑝𝑒𝑟𝑎𝑙𝑙𝑎𝑙𝑜𝑔𝑟𝑎𝑚𝑒 𝐹𝑓𝑑𝐷

=2

3× 𝑒𝐸 × 𝐹𝐷

=2

3× 𝑒𝐸 × 2𝑑

𝑤𝑕𝑒𝑟𝑒 𝑒𝐸 = 𝑂2 − (𝑂1+𝑂3

2) So

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 =2

3× 𝑂2 − (

𝑂1 + 𝑂3

2) × 2𝑑


Continue..

So the area between the first Two divisions,

𝐴𝑟𝑒𝑎 ∆ =𝑂1 + 𝑂3

2× 2𝑑 +

2

3× 𝑂2 − (

𝑂1 + 𝑂3

2) × 2𝑑

= 𝑂1 + 𝑂3 𝑑 +4𝑑

3

2𝑂2 − 𝑂1 − 𝑂3

2

= 𝑂1𝑑 + 𝑂3𝑑 +4𝑑

62𝑂2 − 𝑂1 − 𝑂3

= 𝑂1𝑑 + 𝑂3𝑑 +2𝑑

32𝑂2 − 𝑂1 − 𝑂3

=3𝑂1𝑑 + 3𝑂3𝑑 + 4𝑂2𝑑 − 2𝑂1𝑑 − 2𝑂3𝑑

3

=3𝑂1𝑑 − 2𝑂1𝑑 + 4𝑂2𝑑 + 3𝑂3𝑑 − 2𝑂3𝑑

3

=𝑂1𝑑 + 4𝑂2𝑑 + 𝑂3𝑑

3

=𝒅

𝟑× 𝑶𝟏 + 𝟒𝑶𝟐 + 𝑶𝟑


Continue..

Similarly, the area of next two divisions

=𝒅

𝟑× 𝑶𝟑 + 𝟒𝑶𝟒 + 𝑶𝟓

𝑇𝑜𝑡𝑎𝑙 𝐴𝑟𝑒𝑎 =𝑑

3× 𝑂1 + 4𝑂2 + 2𝑂3 + 4𝑂4 + ⋯ … + 2𝑂𝑛−2 + 4𝑂𝑛−1 + 𝑂𝑛

OR


3× 𝑂1 + 𝑂𝑛) + 4(𝑂2 + 𝑂4 + ⋯ ) + 2(𝑂3 + 𝑂5 + ⋯


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


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


Continue..

By Using Trapezoidal Rule

𝐴𝑟𝑒𝑎 ∆ = 𝑑𝑂1 + 𝑂𝑛

2+ 𝑂2 + 𝑂3 + ⋯ , +𝑂𝑛−1

𝐴𝑟𝑒𝑎 ∆ = 𝑑𝑂1 + 𝑂10

2+ 𝑂2 + 𝑂3 + ⋯ , +𝑂9

𝐴𝑟𝑒𝑎 ∆ = 15 ×3.50 + 3.25

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

𝑨𝒓𝒆𝒂 ∆ = 𝟖𝟐𝟎. 𝟏𝟐𝟓𝒎𝟐


Continue..

By Using Simpson’s Rule


3× 𝑂1 + 𝑂𝑛) + 4(𝑂2 + 𝑂4 + ⋯ ) + 2(𝑂3 + 𝑂5 + ⋯

𝐴𝑟𝑒𝑎 (∆1) =15

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

= 756.00𝑚2

𝐴𝑟𝑒𝑎 ∆2 =𝑂9 + 𝑂10

2× 𝑑 =

4.40 + 3.25

2× 15 = 57.38𝑚2

𝑻𝒐𝒕𝒂𝒍 𝑨𝒓𝒆𝒂 = ∆𝟏 + ∆𝟐

𝑻𝒐𝒕𝒂𝒍 𝑨𝒓𝒆𝒂 = 𝟕𝟓𝟔. 𝟎𝟎 + 𝟓𝟕. 𝟑𝟖 = 𝟖𝟏𝟑. 𝟑𝟖𝒎𝟐


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