LESSON Note on Computation
of area and volume
The main objective of the surveying is to
compute the areas and volumes.
compute the areas and volumes.
Generally, the lands will be of irregular shaped
polygons.
polygons.
There are formulae readily available for
regular polygons like, triangle, rectangle, square and other polygons.
regular polygons like, triangle, rectangle, square and other polygons.
But for determining the areas of irregular
polygons, different methods are used.
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.
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.
Calculation of area is carried out by any
one of the following methods:
one of the following methods:
a) Mid-ordinate method
b) Average ordinate method
c) Trapezoidal rule
d) Simpson’s rule
The mid-ordinate rule
Consider figure.
Let O1, O2, O3,
O4……….On= ordinates at equal intervals
O4……….On= ordinates at equal intervals
l=length of base line
d= common distance between ordinates
h1,h2,……..hn=mid-ordinates
Area = common distance* sum of
mid-ordinates
mid-ordinates
Average ordinate method
Let O1, O2, …..On=ordinates
or offsets at regular intervals
or offsets at regular intervals
l= length of base line
n= number of divisions
n+1= number of ordinates
THE TRAPEZOIDAL RULE
While applying the trapezoidal 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.
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 O1, O2, …..On=ordinate at equal
intervals, and d= common distance between two ordinates
intervals, and d= common distance between two ordinates
Total area=d/2{ O1+2O2+2O3+…….+2On-1+On}
Thus the trapezoidal rule may be stated as
follows:
follows:
To the sum of the first and last ordinate,
twice the sum of intermediate ordinates is added. This total sum is multiplied by
the common distance. Half of this product is the required area.
twice the sum of intermediate ordinates is added. This total sum is multiplied by
the common distance. Half of this product is the required area.
Limitation: There is no limitation for this rule. This
rule can be applied for any number of ordinates
rule can be applied for any number of ordinates
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:
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:
Let
O1, O2, O3=
three consecutive ordinates
three consecutive ordinates
d= common distance between the ordinates
area AFeDC= area of trapezium AFDC+ area
of segment FeDEF
of segment FeDEF
Here,
Area of segment= 2/3* area of
parallelogram FfdD
parallelogram FfdD
= 2/3* eE*2d
= 2/3 *{ O2– O1+O3 /2 }*2d
So, the area between the first two
divisions,
divisions,
= d/3(O1+4O2+O3)
Similarly, the area of next two divisions
∆2 = d/3(O1+4O2+O3)
and so on
Total area = d/3[O1+On+4(O2+O4+……)
+ 2(O3+O5)]
+ 2(O3+O5)]
Thus the rule may be stated as the follows
To the sum of the first and the last
ordinate, four times the sum of even ordinates and twice the sum of the
remaining odd ordinates are added. This total sum is multiplied by the common
distance. One third of this product is the required area.
ordinate, four times the sum of even ordinates and twice the sum of the
remaining odd ordinates are added. This total sum is multiplied by the common
distance. One third of this product is the required area.
|
Trapezoidal rule
|
Simpson’s rule
|
|
1.
The boundary between the ordinates is considered to be straight
1.
There is no limitation. It can be applied for any number of ordinates
1. It gives an approximate result
|
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. |
Limitation: This rule is applicable only when the
number divisions is even i.e. the number of ordinates is odd.
number divisions is even i.e. the number of ordinates is odd.
The trapezoidal rule may be compared in
the following manner:
the following manner:
Note: sometimes one or both the end of the
ordinates may be zero. However they must be taken into account while applying
these rules.
ordinates may be zero. However they must be taken into account while applying
these rules.
Worked- out problems
Problem 1: The following offsets were taken from a
chain line to an irregular boundary line at an interval of 10 m:
chain line to an irregular boundary line at an interval of 10 m:
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:
the irregular boundary line and the end of offsets by:
a) mid ordinate rule
b) the average –ordinate rule
c) the trapezoidal rule
d) Simpson’s rule
Solution: (Refer fig)
Mid-ordinate rule:
Required area=
10(1.25+3.00+4.25+3.90+1.60)
10(1.25+3.00+4.25+3.90+1.60)
= 10*18.80=188 m2
By average-ordinate rule:
Here d=10 m and n=6(no of devices)
Base length= 10*6=60 m
Number of ordinates= 7
Required
area=10((1.25+3.00+5.00+4.60+3.20+0)/7)
area=10((1.25+3.00+5.00+4.60+3.20+0)/7)
By trapezoidal rule:
Here d=10m
Required
area=10/2{0+0+2(2.50+3.50+5.00+4.60+3.20+)}
area=10/2{0+0+2(2.50+3.50+5.00+4.60+3.20+)}
= 5*37.60=188 m2
By Simpson’s rule:
d=10m
required
area=10/3{0+0+4(2.50+5.00+3.20)+2(3.50+4.60)}
area=10/3{0+0+4(2.50+5.00+3.20)+2(3.50+4.60)}
= 10/3{ 42.80+16.20}=10/3*59.00
10/3*59= 196.66m2
Problem 2: The following offsets were taken at 15 m
intervals from a survey line to an irregular boundary line
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
6.40, 4.40, 3.25 m
Calculate the area enclosed between the
survey line, the irregular boundary line, and the offsets, by:
survey line, the irregular boundary line, and the offsets, by:
a) the trapezoidal rule
b) simpson’s rule
solution:
a) the trapezoidal rule
required
area=15/2{3.50+3.25+2(4.30+6.75+5.25+7.50+8.80+7.90+6.40+4.40)}
area=15/2{3.50+3.25+2(4.30+6.75+5.25+7.50+8.80+7.90+6.40+4.40)}
= 15/2{6.75+102.60} = 820.125 m2
c) simpson’s rule
if this rule is to be applied, the number
of ordinates must be odd. But here the number of ordinates must be odd. But
here the number of ordinate is even(ten).
of ordinates must be odd. But here the number of ordinates must be odd. But
here the number of ordinate is even(ten).
So, simpson’s rule is applied from O1 to O9 and the area
between O9 and O10
is found out by the trapezoidal rule.
between O9 and O10
is found out by the trapezoidal rule.
A1= 15/3{ 3.50+4.40+4(
4.30+5.25+8.80+6.40)}+2(6.75+7.50+7.90)
4.30+5.25+8.80+6.40)}+2(6.75+7.50+7.90)
= 15/3(
7.90+99.00+44.30)= 756.00 m2
7.90+99.00+44.30)= 756.00 m2
A2= 15/2(4.40+3.25)=
57.38 m2
57.38 m2
Total area= A1+ A2 =756.00+57.38 = 813.38 m2
Problem 3: the following offsets are taken from a survey line to a
curves boundary line, and the first and the last offsets by:
curves boundary line, and the first and the last offsets by:
a) the trapezoidal rule
b) simpson’s rule
solution:
here the intervals between the offsets are
not reglar through out the length.
not reglar through out the length.
So, the section is divided into three
compartments
compartments
Let
∆I= area of the first
section
section
∆II= area of 2nd section
∆III= area of 3rd section
Here
d1= 5 m
d2=10 m
d3=20 m
a) by trapezoidal rule
∆I= 5/2{2.50+6.10+2(3.80+4.60+5.20)}
= 89.50 m2
= 89.50 m2
∆II= 10/2{6.10+5.80+2(4.70)}
=106.50 m2
=106.50 m2
∆III= 20/2{5.80+2.20+2(3.90)} =
158.00 m2
158.00 m2
Total area = 89.50+106.50+158.00 = 354.00
m2
m2
b) by simpson’s rule
∆I= 5/3{2.50+6.10+4(3.8+5.20) +
2(4.60)} = 89.66 m2
2(4.60)} = 89.66 m2
∆II= 10/3{6.10+5.80+4(4.70)}
=102.33 m2
=102.33 m2
∆III= 20/3{5.80+2.20+4(3.90)} =
157.33 m2
157.33 m2
Total area= 89.66+102.33+157.33 = 349.32 m2
FORMULA FOR CALCULATION OF VOLUME:
D= common distance between the sections
A. trapezoidal rule
volume (cutting or filling),
V=D/2(A1+An+2(A2+A3+….+An-1))
V=D/2(A1+An+2(A2+A3+….+An-1))
1.
Prismoidal
formula
Prismoidal
formula
Volume( cutting or filling), V= D/3{A1+
An +4(A2+ A4+
An-1)+ 2(A3+ A5+….+ Ann-1)}
An +4(A2+ A4+
An-1)+ 2(A3+ A5+….+ Ann-1)}
i.e. V=common distance {area of 1st section+ area of last section+ 4(sum
of areas of even sections)
of areas of even sections)
3
+2(sum of areas of odd
sections)
Note: the prismoidal formula is applicable
whrn there is an odd number of sections. If the number of sections is even, the
end strip is treated separately and the area is calculated according to the
trapezoidal rule. The volume of the remaining strips is calculated in the usual
manner by the prismoidal formula. Then both the results are added to obtain the
total volume.
whrn there is an odd number of sections. If the number of sections is even, the
end strip is treated separately and the area is calculated according to the
trapezoidal rule. The volume of the remaining strips is calculated in the usual
manner by the prismoidal formula. Then both the results are added to obtain the
total volume.
Works out problems
Problem 1: an embankment of width 10 m and side
slopes 1 ½:1 is required to be made on a ground which is level in a direction
transverseto the centre line. The central heights at 40 m intervals are as
follows:
slopes 1 ½:1 is required to be made on a ground which is level in a direction
transverseto the centre line. The central heights at 40 m intervals are as
follows:
0.90,1.25,2.15,2.50,1.85,1.35, and 0.85
Calculate the volume of earth work
according to
according to
i) Trapezoidal formula
ii) Prismoidal formula
Solution: the c/s areas are
calculated by
calculated by
∆= (b+sh)*h
∆1= (10+1.5*0.90)*0.90 = 10.22
m2
m2
∆2= (10+1.5*1.25)*0.90 = 14.84
m2
m2
∆3= (10+1.5*1.25)*2.15 = 28.43
m2
m2
∆4= (10+1.5*2.50)*2.50 = 34.38
m2
m2
∆5= (10+1.5*1.85)*1.85 = 23.63
m2
m2
∆6=(10+1.5*1.35)*1.35 = 16.23 m2
∆7=(10+1.5*0.85)*0.85= 9.58 m2
(a) Volume according to trapezoidal
formula
formula
V= 40/2{10.22+
9.58+2(14.84+28.43+34.38+23.63+16.23)}
9.58+2(14.84+28.43+34.38+23.63+16.23)}
= 20{19.80+235.02} = 5096.4 m2
(b) Volume calculated in prismoidal
formula:
formula:
V = 40/3
{10.22+9.58+4(14.84+34.38+16.23)+2(28.43+23.63)}
{10.22+9.58+4(14.84+34.38+16.23)+2(28.43+23.63)}
= 40/3 (19.80+
261.80+104.12) = 5142.9 m2
261.80+104.12) = 5142.9 m2
Problem the areas enclosed by the contours
in the lake are as follows:
in the lake are as follows:
|
Contour (m)
|
270
|
275
|
280
|
285
|
290
|
|
Area (m2)
|
2050
|
8400
|
16300
|
24600
|
31500
|
Calculate the volume of water between the
contours 270 m and 290 m by:
contours 270 m and 290 m by:
i) Trapezoidal formula
ii) Prismoidal formula
Volume according to trapezoidal formula:
=5/2{2050+31500+2(8400+16300+24600)}
=330,250 m3
google-site-verification=2j8Us0psP3GGjD12h0fH6_Z08_sRyWQjNbWcaMV4E9A