1. TOPIC 2
AREA
&
VOLUME
OGO
L
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2. OBJECTIVES
LOGO
1
Explain the basic concept of Area and
Explain the basic concept of Area and
Volume Method.
Volume Method.
2
Define the usage of Area And Volume
Define the usage of Area And Volume
Calculation.
Calculation.
3
Describe the methods that have been used
Describe the methods that have been used
in Area and Volume Calculation ..
in Area and Volume Calculation

3. INTRODUCTION
Estimation of area and volume is basic to
most engineering schemes
Earthwork volumes must be estimated :
•To enable route alignment to be located at such lines
and levels that cut and fill are balanced as far as
practical.
•To enable contract estimates of time and cost to be
made for proposed work.
•To form the basis of payment for work carried out.

4. AREA CALCULATION
1
The
rectilinear
areas
enclosed by the
survey lines
2
The irregular
areas
of the strips
between these
lines and the
boundary

5. The Rectilinear areas
Method
• Mathematical
equation
• Coordinates
station
traverse
Mechanical
- use of a
planimeter

6. MECHANICAL - PLANIMETER
Cont..

7. MATHEMATICAL EQUATION
Cont..
Triangular equation
i) Area = √[S(S-a)(S-b)(S-c)]
where; S = ½ (a+b+c)
Rectangular equation
i)
Area = a x b
Trapezium equation
i)
Area = ½ (a + b) x h
b
B
b
a
c
A
C
b
ii) Area = ½ (height x width)
= ½ (b x h)
B
h
A
b
C
iii) Area = ½ a b sin c0
a
c0
b
h
a
a

8. BY COORDINATES
Cont..
The position or location of a point / station in a plan can be described in terms of “Easting”
and “Northing” similar to x, y co-ordinates system.
The location of point P can be given by Np, Ep.
Area enclosed by co-ordinates ABCDE is given by:
= ½ [Ni (Ei+1 – Ei-1)]
or
= ½ [Ei (Ni+1 – Ni-1)]
where
N = northing of that ordinate
E = easting of that ordinate

9. The irregular areas
Trapezoidal rule
Text
Irregular plane surface
Simpson’s rules
Mid-ordinate rule
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10. Trapezoidal Rule
This rule assumes that the short lengths of boundary between the ordinates are
straight lines so that the area is divided into a series of trapezoids.
The total area = d x [(F + L) / 2 + other ordinates] where
or
= d/2 x [(F + L) + 2(other ordinates)]
or
= d/2 x [(O1 + On + 2(O3 + O4 +………+ On-1)]
D = equal distance between ordinates
F = first ordinate
L = last ordinate
O1 = first offset
On = last offset
o1
o2
o3
o4
o5
o6
o7

11. Trapezoidal Rule
Cont..
Example
The total area = d /2 x [(F + L) + 2 (other ordinates)]
01
02
03
8m
8m
04
8m
05
8m
06
8m
Distance
0
8
16
24
32
40
Offset
0
1.5
2.2
2.0
2.1
1.1
A = d/2 x [(O1 + O6 + 2 (O2 + O3 + O4 + O5)]
A = 8/2 x [(0 + 1.1 + 2 (1.5 + 2.2 + 2.0 + 2.1)]
A = 66.8 m2

12. Mid-ordinate rule
o1
o2
o3
O1+ O2
2
O2+ O3
2
O3+ O4
2
The total area
o4
O4+ O5
2
o5
o6
O5+ O6
2
o7
O6+ O7
2
= d x [sum of mid-ordinates]

13. Mid-ordinate rule
Cont..
Example
The total area
0.75
01
= d x [sum of mid-ordinates]
1.85
02
2.10
03
8m
8m
2.05
04
8m
1.60
05
8m
06
8m
Distance
0
8
16
24
32
40
Offset
0
1.5
2.2
2.0
2.1
1.1
A = 8 x [((0+1.5)/2)+[((1.5+2.2)/2) [((2.2+2.0)/2) [((2.0+2.1)/2) [((2.1+1.1)/2)]
A = 8 x [0.75 + 1.85 + 2.10 + 2.05 + 1.60]
A = 8 x 8.35 = 66.8m2

14. Simpson Rule
where
The total area = 1 / 3 d [F + L + 4 (Es) + 2 (Os)] D = equal distance between ordinates
F = first ordinate
L = last ordinate
E = even numbered ordinates
O = odd numbered ordinates
o1
o2
o3
o4
o5
o6
o7
Example formula
The total area = d / 3 [O1 + O7 + 4 (O2 + O4 + O6) + 2 (O3 + O5)]

15. Simpson Rule
Cont..
Example
The total area = 1 / 3 d [F + L + 4 (Es) + 2 (Os)]
01
02
8m
03
8m
04
8m
06
05
8m
8m
Distance
0
8
16
24
32
40
Offset
0
1.5
2.2
2.0
2.1
1.1
A = d/3 x [(O1 + O6 + 4 (O2 + O4) + 2 (O3 + O5)]
A = 8/3 x [(0 + 1.1 + 4 (1.5 + 2.0) + 2 (2.2 + 2.1)]
A = 8/3 x 23.7 = 63.2 m2

16. Calculation of cross sectional area
“Cut” means an excavation of the earth
“fill” means the filling or raising of the original ground surface.
1) Sections with
level across
2) Sections with
cross-fall
4) Cross sections of
variable level or three
level sections
3) Sections part in
cut and part in fill

17. Calculation of cross sectional area
Cont..
1)Sections with
level across
Depth of centre line or height of embankment = h
Formation width = b
Side width = w
Area = h(b + mh)
2)Sections with
cross-fall
Area = 1/2m [(b/2 + mh)(w1 + w2) – b2/2]

18. Calculation of cross sectional area
Cont..
3) Sections part in
cut and part in fill
Area of fill = ½ [(b/2 + kh)2/(k-m)]
Area of cut = ½ [(b/2 - kh)2/(k-n)]
4) Cross sections of
variable level or three
level sections
Area = 1/2m[(w1 + w2)(mh + b/2) – b2/2]

19. Volume calculation
These volumes must be calculated and depending on the
shape of the site, this may be done in three ways :
by cross-sections
generally used for long, narrow works
such as roads, railways, pipelines, etc.
by spot height
generally used for small areas such as
underground tanks, basements,
building sites, etc.
volume
by contours
generally used for larger areas such
as reservoirs, landscapes,
redevelopment sites, etc.

20. Computational of volumes based on
area of CROSS SECTIONS
End
areas
Mean
areas
Vol. = {[A1 + A2 + A3 + ……… A n+1 + An] / n} . L
Vol. = D/2 {(A1 + An) + 2(A2 + A3 + …… A n-1)}
Prismoidal
formula
Vol = D/3 (A1 + An + 4ΣEven Areas + 2Σodd Areas)

21. Computational of volumes based on
area of CROSS SECTIONS
Example calculation
Calculate, using the prismoidal formula, the cubic contents of an embankment of which the
cross-sectional areas at 15m intervals are as follows :
Distance (m)
Area (m2)
0
11
15
42
30
64
45
72
60
160
75
180
90
220
A1
A2
A3
A4
A5
A6
A7
Solution - Mean areas method
Solution – Prismoidal method
Vol. = {[A1 + A2 + A3 + ……… A n+1 + An] / n} . L
V = D/3 (A1 + A7 + 4Σ( A2 + A4 + A6) + 2Σ ( A3 + A5)
=15 / 3 (11 + 220 + 4 ( 42 + 72 + 180 ) + 2( 64 + 160))
V = 5 ( 231 + 1176 + 448 )
V = 9275 m3
V = {(11 + 42 + 64 + 72 + 160 + 180 + 220)/ 7 } . 90
V = 9630 m3
Solution - End areas method
Vol. = D/2 {(A1 + An) + 2(A2 + A3 + A4 + A5 + A6 )}
V = 15/2 {(11 + 220)+ 2 (42 + 64 + 72 + 160 + 180) }
V = 9502.5 m3

22. Volume calculation based on
CONTOUR LINES
The volume can be estimated by either end area method or prismoidal method. The distance
D is the contour interval, and for accuracy this should be as small as possible. If required,
the prismoidal formula can be used by treating alternate areas as mid area.
Example:
The areas within the underwater contour lines of a reservoir
are as follows:
Calculate the volume of water in the reservoir between 172 m
and 184 m contours.
Contour (m)
184
182
180
178
176
174
172
Areas (m2)
3125 2454 1630
890
223
110
69
Answer:End area method;
Volume =
2/2 [3125+69 + 2(110 + 223 + 890 + 1630 + 2454)]
= 13808 m3

23. Volume from SPOT LEVELS
This method is useful in the determination of volumes of large open
excavations for tanks, basements, borrow pits, and for ground levelling
operations such as playing fields and building sites. Having located the outline of
the sites, divide the area into squares or rectangles or triangles. Marking the
corner points and then determine the reduced level. By substracting from the
observed levels the corresponding formation levels, a series of heights can be
found.
The volume per square = {[ha + hb + hc + hd] / 4} 1 x b
where;
ha, hb, hc and hd are the side spot height
l and b are the side dimensions

24. Volume from SPOT LEVELS – Square method
Figure 1 shows a rectangular plot, which is to be
excavated to the given reduced level. Assuming
area is subdivided into square method, calculate
the volume of earth to be excavated ( Excavated
level = 10.00m )
A(16.54m)
25.5 m
B(17.25m)
D(16.32m)
E(12.95m)
C(15.40m)
F(15.55m)
Solution:
Station
Reduced
Level
Excavated
Level
A
B
C
D
E
F
G
H
I
16.54
17.25
15.40
16.32
12.95
15.55
16.17
15.84
13.38
10.00
10.00
10.00
10.00
10.00
10.00
10.00
10.00
10.00
Depth Of
excavated
(hn)
6.54
7.25
5.40
6.32
2.95
5.55
6.17
5.84
3.38
Total
G(16.17m)
H(15.84m)
I(13.38m)
Average excavated depth = Σ h x n
Σn
= 83.21
16
No. Of
Rectangles
(n)
1
2
1
2
4
2
1
2
1
16
= 5.2 m
Base area = 25.5 x 30.0 = 765 m2
30.0 m
Volume to excavated = 5.2 x 765
= 3978 m3
Product
( hn x n )
6.54
14.50
5.40
12.64
11.80
11.10
6.17
11.68
3.38
83.21

25. Volume from SPOT LEVELS – Triangle method
Figure 1 shows a rectangular plot, which is to be
excavated to the given reduced level. Assuming
area is subdivided into triangle method, calculate
the volume of earth to be excavated ( Excavated
level = 10.00m )
A(16.54m)
25.5 m
B(17.25m)
D(16.32m)
E(12.95m)
C(15.40m)
F(15.55m)
Solution:
Station
Reduced
Level
Excavated
Level
Depth Of
excavated
(hn)
No. Of
Rectangles
(n)
A
B
C
D
E
F
G
H
I
16.54
17.25
15.40
16.32
12.95
15.55
16.17
15.84
13.38
10.00
10.00
10.00
10.00
10.00
10.00
10.00
10.00
10.00
6.54
7.25
5.40
6.32
2.95
5.55
6.17
5.84
3.38
2
3
1
3
6
3
1
3
2
13.08
21.75
5.40
18.96
17.70
16.65
6.17
17.52
6.76
24
123.99
Total
G(16.17m)
H(15.84m)
I(13.38m)
Average excavated depth = Σ h x n
Σn
= 123.99
24
= 5.17 m
Base area = 25.5 x 30.0 = 765 m2
30.0 m
Volume to excavated = 5.17 x 765
= 3955 m3
Product
( hn x n )

26. LOGO
End of topic
Exercise