ch6&7.pdf

Addis Ababa University Ethiopian institute of Architecture, Building
construction and City Development
Compiled by Ebisa Tesfaye (Msc.) Page 1
UNIT SIX
AREA COMPUTATION
6.1 INTRODUCTION
One of the primary objects of cadastral surveys is to determine the area of land in plan. The land
under consideration can be the property of the person, or a site on which building is to be erected
or to be used as a parking lot.
It’s often necessary to compute the area of a tract of land which may be regular or irregular in
shape. Land is frequently bought and sold on the basis of cost per unit area.
The most common unit of area for lots is the m². Large tracts such as huge tracts of forest, farm
land etc are measured in hectares (ha) or” gashas’’.
1 hectare =10,000m²
1 gasha = 40h
= 400,000 m²
Areas of land may be obtained from field measurements and plotted maps or plans.
6.2 METHODS OF MEASURING AREA
Field measurement method of areas includes, but not limited to, the following
1. Division of the tract or plot in to simple figures (i.e. triangles, rectangles and trapeziums)
2. Measuring offsets from a chain line which is also called a base line
3. Coordinate method
When the plan or map of an area is available, however irregular it may be, planimeter can be run
over the enclosing lines to compute the area of the plot.
6.2.1 THE SIMPLE TRIANGLE
Where an area is triangular in shape or is made up of a series of triangles, the following formulae
are used in computing the area.
a) If the lengths of the three sides have been field measured or given, the following Hero’s
formula can be used to compute the area: Area=√s(s-a)(s-b)(s-c)
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Where a, b, c are sides of the triangle and S=1/2 (a+ b+ c)
fig 1
b) Two sides and the included angle is field measured or given: let us suppose that sides a, b
and <C have been measured.
Area= ½ base x height
=1/2 b*h but h=a* sinC
Therefore, area=1/2 b*a SinC
C) Any side and the three angles are field measured or given
Let us suppose that side ‘a’ and the angles A, B, C have been measured.
Area of the Δ=1/2 b a sinC
Applying the law of sine’s to fig 1
b/ sin B=a/sin A then b=a sin B/sin A
By substituting the value of b, we have
Area of the Δ=1/2 (a sin B/ sin A) a sinC
= (a² sin B sinC)/2sinA

Ex1 A chain survey provides the length of the sides of the closed polygon shown in fig 2.
Compute the total area of the plot of land.
fig 2
Soln
Let A1, A2, A3 denote the areas of the triangle as shown in fig 2, and S1, S2, S3 denote the half
perimeters of the triangles A1,A2 and A3 respectively.
Step1: computation of half perimeters
S1=1/2 (25+18+26) =34.5m
S2=1/2(26+25+32) =41.5m
S3=1/2 (32+20+14) =33m
Step2 : computation of areas
A=√s(s-a) (s-b) (s-c)
A1=√34.5 (34.5-25)(34.5-18)(34.5-26)=214.40m²
in a similar manner , A2=317.54m², A3=90.28m²
Total Area= A1+A2+A3
=622.22m² =0.06ha

6.2.2 BY TAKING OFFSETS FROM A BASE LINE
If the boundaries of a tract are irregular, it is not possible to run the traverse along the
boundaries.
The traverse is usually run at convenient distance from the actual boundaries. The offsets
from the traverse to the irregular boundary are taken at regular intervals or if necessary at
irregular intervals. The area between the traverse line and the irregular boundary is determined
by:-
1) Average ordinate rule
2) Trapezoidal rule and
3) Simpsons rule
1) Average ordinate rule
For approximate results the area between a curve and a straight line i.e., base line, may be found
as follows (fig. 3).
Rule- Divide the base line in to any number of equal parts. At the centre of these parts; draw the
ordinates. Take the average length of these ordinates and multiply by the length of the base line.
The result is the area enclosed (approximately).
If h1, h2……….h11are the ordinates to the boundary from the baseline
Average ordinate = (h1+h2+……+h11)/11
In this case Total length = 300m
Area= total length * average ordinate
=L * (h1+h2+……h11)/11

2. Trapezoidal rule
This is obtained by considering each part as a trapezium and then adding the part mass together.
Area is equal to product of the common interval d and sum of intermediate ordinates plus
average of the first and last ordinates. If the intervals are not equal the areas of the trapeziums
have to be computed separately and added together.
Area= d/2(he+ h’e +2∑ ℎ )
Where d= distance between the offsets (m), he and h’e= offsets at two ends (m), and
∑ ℎ = sum of the intermediate offsets (m)
3. Simpsons rule
In the rules stated above the irregular boundary consists of a number of straight lines. If the
boundary is curved, it can be approximated as a serious of straight lines. Alternatively, Simpsons
rule is applied which assumes that the short lengths of boundaries between the ordinates are
parabolic arcs.
Since the trapezoidal rule presupposes very short intercepts along the base line. The area
computed by this method is less accurate than that computed by Simpsons rule for curved
boundary.
In order to apply Simpsons rule for the computation of areas of tracts, two conditions must be
fulfilled:-
i) There must be an even number of equal intercepts; say 8, along the base line i.e. the number
of offsets must be odd.
ii) The offsets must be at regular intervals
To get the area by Simpson’s rule, add first and last ordinates to four times the even ordinates
and two times the odd ordinates and multiply the sum by one third the common interval.
Area=d/3(he+h’e+2∑hodd+4∑ h even)
Where he, h’e=offset at the two ends
∑h odd=the sum of the odd offsets except the first and the last (the 3rd
, 5th
, 7th
, etc.)
∑h even=the sum of the even offsets (the 2nd
, 4th
, 6th
, etc.)
Ex. The following offsets were taken from a chain line AB to a hedge
Calculate the area enclosed by the chain line, the hedge and the end offsets by
i) Simpsons rule, ii) trapezoidal rule

Soln
Let A be the required total area, A1 be the area of section C-G, A2 be the area of section G-K,
and A3 be the area of section K-M
1) By Simpsons rule
A1=20/3 ((24+8) +2(16) +4(20+12)) =1280m²
A2=40/3((8+20) +2(14) +4(10+16)) =2133.33m²
A3=30/3((20+26) +2(0) +4(22)) =1340m²
A=A1+A2+A3=4753.33m²
ii) By Trapezoid rule
A1=20/2((24+8) +2(20+16+12)) =1280m²
A2=40/2((8+20) +2(10+14+16)) =2160m²
A3=30/20((20+26) +2(22)) =1350m²
A=A1+A2+A3=4790m²

6.2.3 Coordinate Method
In this method independent coordinates of the points are used in the computation of areas.
Total area of the traverse
A=1/2(XC+XB) (YC-YB) +1/2 (XD+XC) (YD-YC) -1/2 (XA+XB) (YA-YB)-1/2(XD+XA)
(YD-YA)
Or
2(A) = XAYB+XBYC+XCYD+XDYA-XBYA-XCYB-XDYC-XAYD
= (XAYB-XBYA) + (XBYC -XCYB)+(XCYD-XDYC)+(XDYA-XAYD)
The above relation can be expressed as follows for easy remembrance.
XA YA
XB YB
XC YC
XD YD
XA YA
Two sums of the product should be taken.
i) Product of all adjacent terms taken down to the right, i.e.
XAYB, XBYC, XCYD, XDYA

ii) Product of all adjacent terms taken down to the left, i.e.
YAXB, YBXC, YCXD, YDXA
The traverse area is equal to half the absolute value of the difference between these two sums. In
applying this procedure, it’s to be observed that first coordinate listed must be repeated at the end
of the list.
Ex3: using the following data, calculate the area of the closed traverse by the coordinate method.
Solution
i) Let S1 be the sum of the product of adjacent diagonal terms taken down to the right.
S1=300x1200+800x1000+1100x600+1300x200+900x500+300x1000=2,830,000
ii) Let S2 be the sum of the product of adjacent diagonal terms taken down to the left
S2=800x1000+1100x1200+1300x1000+900x600+300x200+300x500=4,170,000
The algebraic summation S of these two terms is
S=2,830,000-4,170,000
=-1,340,000
Therefore Area=1/2(1,340,000) = 670,000m² = 67ha

CHAPTER 7
VOLUMES OF EARTH WORK
7.1 INTRODUCTION
On almost every construction site , some form of cutting or embankment is necessary to
accommodate roads, railways, buildings, airports, dams, etc in general earthworks fall in to one
of the two categories. They are
a) Long narrow earthworks of varying depths. E.g. road way cutting and embankment
b) Wide flat earthworks, e.g. sports ground, airports, car parks, etc
7.2 Computation of volumes from cross sections
Having determined cross sectional areas, the volumes of earth work involved in construction
must be computed. The solid between the cross sections approximates a prismoid and is
considered as such in computing earth work quantities.
Embankment or road in cut considered as a prismoid
From solid geometry we know that a prism is a solid whose ends (or bases) are parallel and equal
polygons and whose sides or faces are parallelograms.
A prismoid is a solid resembling a prism in form. A prismoid has two parallel polygons at its
two ends and its sides or faces are quadrilaterals.
In fig below, we note that the solid of earth enclosed by A1 and A2 and the four quadrilaterals
resembles a prism, i.e. it is prismoid. By inference any solid of earth enclosed between any other
two adjacent cross sections and four quadrilaterals is a prismoid.
Among many formulae that exist for computing earth work quantities, we will consider
only the following ;a) average End-area volume rule, b) trapezoidal volume rule


A1
A2
A3

a. Average End-area Volume rule
The average end-area formula is most commonly used for computing the volume between two
cross sections or end areas. The formula gives good results when the two end sections are
approximately the same shape and size.
V=1/2(A1+A2) D; where v=Volume (m3
) A1 and A2= end areas (m2
), and
b. The trapezoidal volume rule
For a series of “n” consecutive cross sections at a constant distance D between any two adjacent
cross sections, the volume of earthwork can be computed by the trapezoidal volume rule.
Let A1, A2, A3…….. An be the cross sections, the total volume of earth work is given by:
V=d/2(A1+An+ 2 ∑ )
7.3 Computation of Volumes from spot levels
When the excavated area is square, rectangular or consists of a number of vertical sides,
as in the case of foundation of a water tank, underground reservoir, etc. the volume can be
computed by taking levels of number of points along a grid. The difference between the
formation level and the existing level of the ground will give the height of fill or cut at the
corresponding points.
The volumes of any square say a, b, c, d
Volume from spot levels
Volume of abcd = average height *area of the square abcd
= 1/4 (ha+hb+hc+hd) * area of the square abcd

Similarly volume of bcef
=1/4 (hb+hc+he+hf) * area of the square bcef
If the areas of the squares are all equal and is given by A, summation of volumes can be
expressed as
V=A/4 (h1+2hs+3h3+4h4)
Where h1=sum of depth used once
h2=sum of depth used twice
h3=sum of depth used thrice
h4=sum of depth used four times
If the plan area divided into a number of triangles, the volume of triangular prism V will be,
V=1/3(ha+hb+hc)*A
If all triangular areas are equal
V=A/3({h1+2h2+3h3+…+8h8)
Where h1=sum of depth used once
h2=sum of depth used twice and so on

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