This document discusses different methods for computing the area and volume of land surfaces in civil engineering. It provides formulas for calculating areas using the mid-ordinate rule, average ordinate rule, trapezoidal rule, and Simpson's rule based on taking offset measurements. It also describes calculating cross-sectional areas of surfaces using formulas for level sections, two-level sections, three-level sections, side hill two-level sections, and multi-level sections. The document states that volume is calculated using the trapezoidal rule or prismoidal rule, and the prismoidal rule is more accurate by applying a correction.

1. COMPUTATION OF AREAAND VOLUME

2. Content :
• Introduction
• Methods of computing area
• Average ordinate rule
• Mid ordinate rule
• simpson’s rule

3. INTRODUCTION
In civil engineering work such as design of bridges , dam
,reservoirs etc. The area of catchment of a river is required. For road
and railways land is to be acquired on the basis of area. Thus , finding
areas is essential part of surveying. It may be noted that the area to be
found is the projected area upon the horizontal plane.

4.  units used for finding the area are square , meters , hectare , acres
etc.
100 Sq. m=1 are
100 are=1 hectare =10000 Sq. m
1 acre = 4047 Sq. m = 2.5 vigha
1 vigha =16 guntha
1 acre = 40 guntha
1 Hectare =2.471 acres
1 Sq. m=10000000 Sq. m

5. COMPUTATION OF AREA FROM PLOTTED PLAN
• Boundary area can be
calculated as one of the
following rule:
– The mid-ordinate rule
– The average ordinate
rule
– The trapezoidal rule
– Simpson’s rule

6. METHODS OF COMPUTING AREA
 Computation of area by taking offsets
1. Mid-ordinate rule
2. Average ordinate rule
3. Trapezoidal rule
4. Simpson’s rule

7. MID-ORDINATE RULE
In this method the base line is divided into a
number of divisions and the ordinates are measured at the points
of each divisions . Boundaries between the offsets are considered
straight lines.

9. Where h1,h2,h3,…………=mid ordinates
d=distance of each division
L=length of base line= nd
n=number of division
 
 
 
1 2 3
1 2 3
1 2 3
............
............
............
n
n
n
h h h h
Area L
n
h h h h
Area nd
n
Area h h h h d
   
 
   
 
     

10. AVERAGE ORDINATE RULE
This rule also assumes that the boundaries between the
extremities of the ordinates are straight lines.

12. Where h0,h1,h2,……=ordinates of offsets
d=distance of each division
n=number of division
n+1=number of offsets
L=length of base line=nd
 
 
1 2 3
1 2 3
............
1
............
1
n
n
h h h h
Area L
n
h h h h
Area nd
n
   
 

   
 


13. TRAPEZOIDAL RULE
In this method , entire area is divided in to trapezoids .
The rule is more accurate than the previous two rules.

15. which is known as trapezoidal rule.
   
   
0 1 2 3 1
2
15
2 ..........
2
15
0 4.85 2 1.65 3.50 2.70 4.65 3.60 3.95
2
15 22.475
337.125
n n
d m
d
Area h h h h h h
Area
Area
Area m


       
 
 
        
 
 
 


16.  Example: series of offsets were taken from a chain line to an
boundary , interval of 15 m , in the following order.
0,1.65,3.50,2.70,4.65,3.60,3.95,4.85m
Compute the area by trapezoidal rule.
Solution:
   
   
0 1 2 3 1
2
15
2 ..........
2
15
0 4.85 2 1.65 3.50 2.70 4.65 3.60 3.95
2
15 22.475
337.125
n n
d m
d
Area h h h h h h
Area
Area
Area m


       
 
 
        
 
 
 


17. SIMPSON’S RULE
This rule assumes that the short lengths of boundary
between the ordinates are parabolic arcs.

19. For simpson’s rule , the number of ordinate must be odd.
simpson’s rule is:
   
 
0 1 3 1 2 4 2
4 ........ 2 ..........
2
n n n
d
Area h h h h h h h h
 
         

20. APPLICATION:
• Simpson’s rule used for find the earthwork volume using
contour maps.it gives more accurate area.
• Trapezoidal rule can be applied for any number of ordinates. It
gives an approximate area
• A planimeter is used to measure the area of any shape with
more accuracy.
• Zero circle is used when the tracing point is moved , no
rotation of wheel will take place .

21. Example : Following perpendicular offsets were taken from a
chain line a curved boundary line at an interval of 10 m.
0,7.26,5.83,6.45,7.20,8.18,8.0,0
compute the area by simpsons rule
Solution:
To find area by simpson’s rule , number of offsets must be
odd. Here we have 8 offsets. Therefore , for offsets h0 to h6 apply
simpson’s rule and for offsets h6 and h7 apply trapezoidal rule.

22. (continue)
   
   
6 7
0 6 1 3 5 2 4
2
4 2
3 2
10 8 0
0 8 4 7.26 6.45 8.18 2 5.83 7.20 10
3 2
405.4 40
445.4
h h
d
Area A h h h h h h h d
A
A
A m

 
        
 
   
 

 
       
 
   
 
 


23. COMPUTATION OF VOLUME
 Content :
• Formulae for Calculation of Cross-Sectional Area
(a) Level Section
(b) Two level section
(c) Three Level Section
(d) Side Hill Two-Level Section
(e) Multi-Level Section

24. • Introduction:
• For computation of volume of earth work, the sectional area of the
cross section which are taken transverse to the longitudinal section
during profile leveling are first calculated. Again cross section may
be different types namely…
(a) Level Section
(b) Two level section
(c) Three Level Section
(d) Side Hill Two-Level Section
(e) Multi-Level Section

25. • The method of calculating areas of such sections are calculated
• After calculation of cross-sectional areas, the volume of earth work
calculated by…
• (a) the trapezoidal (or average end area)
• (b) the prismoidal rule
• Note : 1. The Prismoidal rule gives the correct volume directly
2. the trapezoidal does not give the correct volume. Prismoidal
correction should be applied for this purpose . This
correction is always subtractive.
3. cutting is denoted by a positive sign and filling by a Negative
sign

26. Level Section :
When the ground is level along the transverse section

27. Example: Calculate the sectional area of an embankment 10 m wide
with a side slope of 2:1. The ground is level in a transverse direction
to the centre line. The central height of the embankment is 2.5m
Here b = 10 m
s = 2
h = 2.5
Cross sectional area = (b + s X h)h
= (10+2 X 2.5)X2.5
= 37.5 m2

28. THANK YOU
Two-Level Section :
When the ground surface has transverse slope

30. Example : The width of the formation level of a certain cutting
is 10m and the side slope is 1 :1. The surface of the ground has
a uniform slope of 1 in 6 in transverse direction. Let us find the
cross sectional area when the depth of the cutting at the center
is 3m