surveying and levelling 2
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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
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 (shams@cecos.edu.pk)
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
3 Engr.Shams Ul Islam (shams@cecos.edu.pk)
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
𝑤𝑒𝑟𝑒 𝑑 = 𝑡𝑒 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑒𝑛 𝑡𝑒 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠
𝑀𝑖𝑑 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 =𝑂1 + 𝑂2
2
𝑤𝑒𝑟𝑒 𝑂1, 𝑂2, 𝑂3, … . . , 𝑂𝑛 𝑎𝑟𝑒 𝑡𝑒 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑎𝑡 𝑒𝑎𝑐 𝑜𝑓 𝑡𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛
4 Engr.Shams Ul Islam (shams@cecos.edu.pk)
The mid-ordinate rule
5 Engr.Shams Ul Islam (shams@cecos.edu.pk)
The mid-ordinate rule
The mid ordinate rule can also be in other way.
𝐴𝑟𝑒𝑎 ∆ =ℎ1+ℎ2+ℎ3+⋯ℎ𝑛
𝑛×L
𝑤𝑒𝑟𝑒 1, 2, 3, … . . , 𝑛 𝑎𝑟𝑒 𝑡𝑒 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑎𝑡 𝑡𝑒 𝑚𝑖𝑑 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑒𝑎𝑐 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛
𝐿 = 𝑇𝑒 𝑙𝑒𝑛𝑔𝑡 𝑜𝑓 𝑡𝑒 𝑏𝑎𝑠𝑒 𝑙𝑖𝑛𝑒
𝑛 = 𝑡𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑞𝑢𝑎𝑙 𝑝𝑎𝑟𝑡𝑠 𝑖𝑛𝑡𝑜 𝑤𝑖𝑐 𝑡𝑒 𝑏𝑎𝑠𝑒 𝑙𝑖𝑛𝑒 𝑖𝑠 𝑑𝑖𝑣𝑖𝑑𝑒𝑑
6 Engr.Shams Ul Islam (shams@cecos.edu.pk)
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 (shams@cecos.edu.pk)
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
8 Engr.Shams Ul Islam (shams@cecos.edu.pk)
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 = 𝑡𝑒 𝑙𝑒𝑛𝑔𝑡 𝑜𝑓 𝑡𝑒 𝑏𝑎𝑠𝑒 𝑙𝑖𝑛𝑒
𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑞𝑢𝑎𝑙 𝑜𝑟𝑑𝑖𝑎𝑛𝑡𝑒𝑠 𝑖𝑛𝑡𝑜 𝑤𝑖𝑐 𝑡𝑒 𝑏𝑎𝑠𝑒 𝑙𝑖𝑛𝑒 𝑖𝑠 𝑑𝑖𝑣𝑖𝑑𝑒𝑑
𝑤𝑒𝑟𝑒 𝑑 = 𝑡𝑒 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑒𝑛 𝑡𝑒 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠
𝑤𝑒𝑟𝑒 𝑂1, 𝑂2, 𝑂3, … . . , 𝑂𝑛 𝑎𝑟𝑒 𝑡𝑒 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑎𝑡 𝑒𝑎𝑐 𝑜𝑓 𝑡𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛
9 Engr.Shams Ul Islam (shams@cecos.edu.pk)
The Average ordinate method
10 Engr.Shams Ul Islam (shams@cecos.edu.pk)
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 = 𝑡𝑒 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑒𝑛 𝑡𝑒 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒,
𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑞𝑢𝑎𝑙 𝑜𝑟𝑑𝑖𝑎𝑛𝑡𝑒𝑠 𝑖𝑛𝑡𝑜 𝑤𝑖𝑐 𝑡𝑒 𝑏𝑎𝑠𝑒 𝑙𝑖𝑛𝑒 𝑖𝑠 𝑑𝑖𝑣𝑖𝑑𝑒𝑑
𝑎𝑛𝑑 𝑂1, 𝑂2, 𝑂3, … . . , 𝑂𝑛 𝑎𝑟𝑒 𝑡𝑒 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑎𝑡 𝑒𝑎𝑐 𝑜𝑓 𝑡𝑒 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
11 Engr.Shams Ul Islam (shams@cecos.edu.pk)
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 (shams@cecos.edu.pk)
Continue..
𝑂1, 𝑂2, 𝑂3 are the consecutive coordinates
𝑑 = 𝑡𝑒 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑒 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
𝑤𝑒𝑟𝑒 𝑒𝐸 = 𝑂2 − (𝑂1+𝑂3
2) So
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 =2
3× 𝑂2 − (
𝑂1 + 𝑂3
2) × 2𝑑
13 Engr.Shams Ul Islam (shams@cecos.edu.pk)
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
14 Engr.Shams Ul Islam (shams@cecos.edu.pk)
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
𝑇𝑜𝑡𝑎𝑙 𝐴𝑟𝑒𝑎 =𝑑
3× 𝑂1 + 𝑂𝑛) + 4(𝑂2 + 𝑂4 + ⋯ ) + 2(𝑂3 + 𝑂5 + ⋯
15 Engr.Shams Ul Islam (shams@cecos.edu.pk)
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 (shams@cecos.edu.pk)
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 (shams@cecos.edu.pk)
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
18 Engr.Shams Ul Islam (shams@cecos.edu.pk)
Continue..
By Using Simpson’s Rule
𝑇𝑜𝑡𝑎𝑙 𝐴𝑟𝑒𝑎 =𝑑
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
19 Engr.Shams Ul Islam (shams@cecos.edu.pk)
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