surveying and levelling 2
TRANSCRIPT
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.
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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’ Simpsonβs rule
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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΄πππ β = ππ’π ππ πππ πππππππ‘ππ Γ π
π€ππππ π = π‘ππ ππππππ πππ π‘ππππ πππ‘π€ππππ π‘ππ πππππππ‘ππ
πππ πππππππ‘π =π1 + π2
2
π€ππππ π1, π2, π3, β¦ . . , ππ πππ π‘ππ πππππππ‘π ππ‘ ππππ ππ π‘ππ πππ£ππ πππ
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The mid-ordinate rule
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The mid-ordinate rule
The mid ordinate rule can also be in other way.
π΄πππ β =β1+β2+β3+β―βπ
πΓL
π€ππππ π1, π2, π3, β¦ . . , ππ πππ π‘ππ πππππππ‘π ππ‘ π‘ππ πππ πππππ‘ ππ ππππ πππ£ππ πππ
πΏ = πππ πππππ‘π ππ π‘ππ πππ π ππππ
π = π‘ππ ππ’ππππ ππ πππ’ππ ππππ‘π πππ‘π π€ππππ π‘ππ πππ π ππππ ππ πππ£ππππ
6 Engr.Shams Ul Islam ([email protected])
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
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Solution
π1 =0 + 2.5
2= 1.25
π2 =2.5 + 2.5
2= 3.0
π3 =3.5 + 5
2= 4.25
π4 =5 + 4.60
2= 4.80
π5 =4.60 + 3.20
2= 3.90
π6 =3.20 + 0
2= 1.60
πΏ = 60 π
π = 6
π΄πππ β =β1+β2+β3+β―βπ
πΓL
π΄ =1.25 + 3.0 + 4.25 + 4.80 + 3.90 + 1.60
6Γ 60 = 188 π2
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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΄πππ β =ππ’π ππ πππππππ‘ππ
π + 1Γ π
π΄πππ β =π1 + π2 + π3 + β― , +ππ
π + 1Γ π
π = π‘ππ πππππ‘π ππ π‘ππ πππ π ππππ
π = ππ’ππππ ππ πππ’ππ πππππππ‘ππ πππ‘π π€ππππ π‘ππ πππ π ππππ ππ πππ£ππππ
π€ππππ π = π‘ππ ππππππ πππ π‘ππππ πππ‘π€ππππ π‘ππ πππππππ‘ππ
π€ππππ π1, π2, π3, β¦ . . , ππ πππ π‘ππ πππππππ‘π ππ‘ ππππ ππ π‘ππ πππ£ππ πππ
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The Average ordinate method
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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
π = π‘ππ ππππππ πππ π‘ππππ πππ‘π€ππππ π‘ππ πππππππ‘π,
π = ππ’ππππ ππ πππ’ππ πππππππ‘ππ πππ‘π π€ππππ π‘ππ πππ π ππππ ππ πππ£ππππ
πππ π1, π2, π3, β¦ . . , ππ πππ π‘ππ πππππππ‘π ππ‘ ππππ ππ π‘ππ πππ£ππ πππ
π΄πππ β =π1 + 2π2 + 2π3 + β― , +2ππβ1 + ππ
2Γ
π
π
OR
π΄πππ β =π
2Γ π1 + 2π2 + 2π3 + β― , +2ππβ1 + ππ
OR
π΄πππ β = π Γπ1 + ππ
2+ π2 + π3 + β― , +ππβ1
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Simpsonβs Rule
In this rule, the boundaries between the ends of ordinates are assumed to
form an arc of parabola. Hence Simpsonβs rule is some times called as
parabolic rule. Refer to figure:
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Continue..
π1, π2, π3 are the consecutive coordinates
π = π‘ππ ππππππ πππ π‘ππππ πππ‘π€πππ π‘ππ πππππππ‘ππ
Aπππ π΄πΉππ·πΆ = ππππ ππ π‘πππππ§ππ’π π΄πΉπ·πΆ + ππππ ππ π ππππππ‘ πΉππ·πΈπΉ
π΄πππ ππ ππππππ§ππ’π =π1 + π3
2Γ π΄πΆ
π΄πππ ππ ππππππ§ππ’π =π1 + π3
2Γ 2π
π΄πππ ππ π ππππππ‘ =2
3Γ π΄πππ ππ ππππππππππππππ πΉπππ·
=2
3Γ ππΈ Γ πΉπ·
=2
3Γ ππΈ Γ 2π
π€ππππ ππΈ = π2 β (π1+π3
2) So
π΄πππ ππ π ππππππ‘ =2
3Γ π2 β (
π1 + π3
2) Γ 2π
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Continue..
So the area between the first Two divisions,
π΄πππ β =π1 + π3
2Γ 2π +
2
3Γ π2 β (
π1 + π3
2) Γ 2π
= π1 + π3 π +4π
3
2π2 β π1 β π3
2
= π1π + π3π +4π
62π2 β π1 β π3
= π1π + π3π +2π
32π2 β π1 β π3
=3π1π + 3π3π + 4π2π β 2π1π β 2π3π
3
=3π1π β 2π1π + 4π2π + 3π3π β 2π3π
3
=π1π + 4π2π + π3π
3
=π
πΓ πΆπ + ππΆπ + πΆπ
14 Engr.Shams Ul Islam ([email protected])
Continue..
Similarly, the area of next two divisions
=π
πΓ πΆπ + ππΆπ + πΆπ
πππ‘ππ π΄πππ =π
3Γ π1 + 4π2 + 2π3 + 4π4 + β― β¦ + 2ππβ2 + 4ππβ1 + ππ
OR
πππ‘ππ π΄πππ =π
3Γ π1 + ππ) + 4(π2 + π4 + β― ) + 2(π3 + π5 + β―
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Comparison of Trapezoidal rule with Simpsonβs 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
Simpsonβs 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
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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) Simpsonβs rule
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Continue..
By Using Trapezoidal Rule
π΄πππ β = ππ1 + ππ
2+ π2 + π3 + β― , +ππβ1
π΄πππ β = ππ1 + π10
2+ π2 + π3 + β― , +π9
π΄πππ β = 15 Γ3.50 + 3.25
2+ 4.30 + 6.75 + 5.25 + 7.50 + 8.80 + 7.90 + 6.40 + 4.40
π¨πππ β = πππ. πππππ
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Continue..
By Using Simpsonβs Rule
πππ‘ππ π΄πππ =π
3Γ π1 + ππ) + 4(π2 + π4 + β― ) + 2(π3 + π5 + β―
π΄πππ (β1) =15
3Γ 3.50 + 4.40) + 4 4.30 + 5.25 + 8.80 + 6.40 + 2(6.75 + 7.50 + 7.9
= 756.00π2
π΄πππ β2 =π9 + π10
2Γ π =
4.40 + 3.25
2Γ 15 = 57.38π2
π»ππππ π¨πππ = βπ + βπ
π»ππππ π¨πππ = πππ. ππ + ππ. ππ = πππ. ππππ
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