Area & volume 2

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TOPIC 2TOPIC 2

AREA AREA & &

VOLUMEVOLUMEEr. Mital Damani

LOGO OBJECTIVESOBJECTIVES

Explain the basic concept of Area andExplain the basic concept of Area and Volume Method.Volume Method.11 Define the usage of Area And Volume Define the usage of Area And Volume Calculation.Calculation.22 Describe the methods that have been usedDescribe the methods that have been used in Area and Volume Calculation .in Area and Volume Calculation .33

INTRODUCTIONINTRODUCTION

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 To enable route alignment to be located at such lines and levels that cut and fill are balanced as far as and levels that cut and fill are balanced as far as practical. practical. •To enable contract estimates of time and cost to be To enable contract estimates of time and cost to be made for proposed work.made for proposed work.•To form the basis of payment for work carried out.To form the basis of payment for work carried out.

AREA CALCULATIONAREA CALCULATION

1

The rectilinear areas enclosed by the survey lines

2

The irregular areas of the strips between these lines and the boundary

The Rectilinear areasThe Rectilinear areas

•Mathematical equation

•Coordinates station traverse

Method

Mechanical - use of a planimeter

MECHANICAL - PLANIMETERMECHANICAL - PLANIMETERCont..

A

B

C

ac

b

A

B

Cb

h

i) Area = [S(S-a)(S-b)(S-c)] where; S = ½ (a+b+c)

ii) Area = ½ (height x width) = ½ (b x h)

a

c0

b

iii) Area = ½ a b sin c0

Triangular equation Rectangular equation

b

a

i) Area = a x b

b

a

h

Trapezium equationi) Area = ½ (a + b) x h

MATHEMATICAL EQUATIONMATHEMATICAL EQUATIONCont..

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 ordinateE = easting of that ordinate

BY COORDINATESBY COORDINATESCont..

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Text

Trapezoidal rule

Mid-ordinate ruleSimpson’s rules

Irregular plane surface

The irregular areasThe irregular areas

Trapezoidal RuleThis 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 ordinatesF = first ordinateL = last ordinateO1 = first offsetOn = last offset

o1 o2 o3 o4 o5 o6 o7

Trapezoidal RuleCont..Example

01 02 03 04 05 06

8 m 8 m 8 m8 m 8 mDistance 0 8 16 24 32 40Offset 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

The total area = d /2 x [(F + L) + 2 (other ordinates)]

Mid-ordinate rule

o1 o2 o3 o4 o5 o6 o7

O2+ O3

2O3+ O4

2O4+ O5

2O5+ O6

2O6+ O7

2O1+ O2

2

The total area = d x [sum of mid-ordinates]

Mid-ordinate ruleCont..Example

01 02 03 04 05 06

8 m 8 m 8 m8 m 8 mDistance 0 8 16 24 32 40Offset 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

The total area = d x [sum of mid-ordinates]

0.75 1.85 2.10 2.05 1.60

Simpson RuleThe total area = 1 / 3 d [F + L + 4 (Es) + 2 (Os)]The total area = 1 / 3 d [F + L + 4 (Es) + 2 (Os)]

o1 o2 o3 o4 o5 o6 o7

whereD = equal distance between ordinatesD = equal distance between ordinatesF = first ordinateF = first ordinateL = last ordinateL = last ordinateE = even numbered ordinatesE = even numbered ordinatesO = odd numbered ordinatesO = odd numbered ordinates

Example formulaExample formula

The total area = The total area = dd / 3 [ / 3 [OO11 + + OO77 + 4 ( + 4 (OO22 + + OO44 + + OO66) + 2 () + 2 (OO33 + O + O55)])]

Simpson RuleCont..Example

01 02 03 04 05 06

8 m 8 m 8 m8 m 8 mDistance 0 8 16 24 32 40Offset 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

The total area = 1 / 3 d [F + L + 4 (Es) + 2 (Os)]The total area = 1 / 3 d [F + L + 4 (Es) + 2 (Os)]

Calculation of cross sectional areaCalculation of cross sectional area

1) Sections with level across

2) Sections with cross-fall

3) Sections part in cut and part in fill

“Cut” means an excavation of the earth“fill” means the filling or raising of the original ground surface.

4) Cross sections of variable level or three

level sections

Calculation of cross sectional areaCalculation of cross sectional area1)Sections with

level across

Cont..

Depth of centre line or height of embankment = hFormation width = bSide width = wArea = h(b + mh)

2)Sections with cross-fall

Area = 1/2m [(b/2 + mh)(w1 + w2) – b2/2]

Calculation of cross sectional areaCalculation of cross sectional areaCont..

Area of fill = ½ [(b/2 + kh)2/(k-m)]

Area of cut = ½ [(b/2 - kh)2/(k-n)]Area = 1/2m[(w1 + w2)(mh + b/2) – b2/2]

3) Sections part in cut and part in fill

4) Cross sections of variable level or three

level sections

Volume calculationVolume calculation

volumevolumeby spot height

generally used for small areas such as underground tanks, basements,

building sites, etc.

by cross-sectionsgenerally used for long, narrow works such as roads, railways, pipelines, etc.

by contoursgenerally used for larger areas such

as reservoirs, landscapes, redevelopment sites, etc.

These volumes must be calculated and depending on the shape of the site, this may be done in three ways :

Computational of volumes based on Computational of volumes based on area of area of CROSS SECTIONSCROSS SECTIONS

Mean Mean areasareas

End End areasareas

Prismoidal Prismoidal formulaformula

Vol. = {[A1 + A2 + A3 + ……… A n+1 + An] / n} . L Vol. = D/2 {(A1 + An) + 2(A2 + A3 + …… A n-1)}

Vol = D/3 (A1 + An + 4Even Areas + 2odd Areas)

Computational of volumes based on Computational of volumes based on area of area of CROSS SECTIONSCROSS SECTIONS

Solution - Mean areas methodSolution - Mean areas method

Vol. = {[A1 + A2 + A3 + ……… A n+1 + An] / n} . L

V = {(11 + 42 + 64 + 72 + 160 + 180 + 220)/ 7 } . 90V = 9630 m3

Solution - Solution - End areas methodEnd 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

Example calculationExample calculation

Distance (m) 0 15 30 45 60 75 90Area (m2) 11 42 64 72 160 180 220

Calculate, using the prismoidal formula, the cubic contents of an embankment of which the cross-sectional areas at 15m intervals are as follows :

Solution – Prismoidal methodSolution – Prismoidal method

V = D/3 (A1 + A7 + 4( A2 + A4 + A6) + 2 ( A3 + A5) =15 / 3 (11 + 220 + 4 ( 42 + 72 + 180 ) + 2( 64 + 160))V = 5 ( 231 + 1176 + 448 )V = 9275 m3

A1 A2 A3 A4 A5 A6 A7

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.

Contour (m) 184 182 180 178 176 174 172

Areas (m2) 3125 2454 1630 890 223 110 69

Example: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.

Answer:-Answer:-

End area method;Volume = 2/2 [3125+69 + 2(110 + 223 + 890 + 1630 + 2454)] = 13808 m3

Volume from SPOT LEVELSSPOT LEVELSThis 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 heightl and b are the side dimensions

Volume from SPOT LEVELS SPOT LEVELS – Square method– Square method

A(16.54m) B(17.25m) C(15.40m)

D(16.32m) E(12.95m) F(15.55m)

G(16.17m) H(15.84m) I(13.38m)

25.5 m

30.0 m

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 )

Station Reduced Level

Excavated Level

Depth Of excavated

(hn)

No. Of Rectangles

(n)

Product( hn x n )

AABBCCDDEEFFGGHHII

16.5416.5417.2517.2515.4015.4016.3216.3212.9512.9515.5515.5516.1716.1715.8415.8413.3813.38

10.0010.0010.0010.0010.0010.0010.0010.0010.0010.0010.0010.0010.0010.0010.0010.0010.0010.00

6.546.547.257.255.405.406.326.322.952.955.555.556.176.175.845.843.383.38

112211224422112211

6.546.5414.5014.505.405.4012.6412.6411.8011.8011.1011.106.176.1711.6811.683.383.38

Total 16 83.21

Average excavated depth = h x n n

= 83.21 = 5.2 m 16

Base area = 25.5 x 30.0 = 765 m2

Volume to excavated = 5.2 x 765 = 39783978 m3

Solution:

Volume from SPOT LEVELS SPOT LEVELS – Triangle method– Triangle method

A(16.54m) B(17.25m) C(15.40m)

D(16.32m) E(12.95m) F(15.55m)

G(16.17m) H(15.84m) I(13.38m)

25.5 m

30.0 m

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 )

Station Reduced Level

Excavated Level

Depth Of excavated

(hn)

No. Of Rectangles

(n)

Product( hn x n )

AABBCCDDEEFFGGHHII

16.5416.5417.2517.2515.4015.4016.3216.3212.9512.9515.5515.5516.1716.1715.8415.8413.3813.38

10.0010.0010.0010.0010.0010.0010.0010.0010.0010.0010.0010.0010.0010.0010.0010.0010.0010.00

6.546.547.257.255.405.406.326.322.952.955.555.556.176.175.845.843.383.38

223311336633113322

13.0813.0821.7521.755.405.4018.9618.9617.7017.7016.6516.656.176.1717.5217.526.766.76

Total 24 123.99

Average excavated depth = h x n n

= 123.99 = 5.17 m 24

Base area = 25.5 x 30.0 = 765 m2

Volume to excavated = 5.17 x 765 = 39553955 m3

Solution:

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