C2005 Ukur Kejuruteraan UNIT2

8/3/2019 C2005 Ukur Kejuruteraan UNIT2

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AREA AND VOLUME

2.1 INTRODUCTION

Estimation of area and volume is basic to most engineering schemes such asroute alignment, reservoirs, construction of tunnels, etc. The excavation andhauling of material on such schemes is the most significant and costly aspect ofthe work, on which profit or loss may depend. Area may be required inconnection with the purchase or sale of land, with the division of land or with thegrading of land. Earthwork volumes must be estimated :

to enable route alignment to be located at such lines and levels that cutand fill are balanced as far as practical.

to enable contract estimates of time and cost to be made for proposedwork.

to form the basis of payment for work carried out.

It is frequently necessary as part of engineering surveying projects to determinethe area enclosed by the boundaries of a site or the volume of earthworkrequired to be moved. Many of the figures involve accepted mensurationformulae (see 1.6 ) but it is more common to meet irregular shapes and theserequire special attention.

2.2 PLAN AREAS

The basic unit of area in SI units is the square metre (m) but for large areas thehectare is a derived unit.

1 hectare (ha) = 10 000 m = 2.471 05 acres

2.2.1 Conversion Of Plannimetric Area Into Actual Area

Let the scale of the plan be 1 in H (or as representative fraction1/H). Then 1 mm is equivalent to H mm and 1 mm is equivalent toH mm is equivalent to H mm, i.e.

H x 10-6 m

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2.3 AREA CALCULATION

Areas of ground may be obtained from the plotted plan but results are only asaccurate as it is possible to scale off the drawings. Accuracy is greatly increased

by using the measurements taken in the field. In most surveys the area isdivisible into two parts :a) The rectilinearareas enclosed by the survey lines

b) The irregular areas of the strips between these lines and theboundary

In order to calculate the area of the whole, each of these areas must beevaluated separately because each is defined by a different form of geometricalfigure.

2.3.1 Rectilinear Areas

The method of evaluating the rectilinear area enclosed by surveylines depends on the method of survey.

a) If chain surveying is used, the areas of the triangles forming the surveynetwork are calculated from the field dimensions from the formula :

Area = (s(s a) (s b) (s c))

Where a, b and c = the lengths of the triangles sides

and

s = (a + b + c) / 2

b) If traversing is used and the survey stations are coordinated, the computedcoordinated are used in the area calculation.

Whichever calculation method is used, checks must be applied to provethe area calculations. In a chain survey network the work must bearranged so that two different sets of the triangles forming the rectilinear

figure are used in evaluating the total area, which is thus twice calculated.These two results will not normally agree precisely because the networkwill not be geometrically perfect. Owing to observational errors, the tworesults are meaned to produce the final rectilinear area. When areas arecalculated from coordinates, the calculation must be repeated another wayto prove the result.

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2.3.2 Irregular Areas

Unless boundaries are straight and the corner points coordinated there

are usually irregular strips of ground between the survey lines and theproperty boundaries. The area of the irregular strips are either positive ornegative to the rectilinear area and since they are divided up by offsetsbetween which the boundary is supposed to run straight, they arecomputed as a series of trapezoids. The mean of each pair of offsets istaken and multiplied by the chainage between them. Where the offsets aretaken at regular intervals, the trapezoidal rule or Simpsons rule for areasis used, (see section 2.6).

NOTE

a. The field work should be arranged to overcome difficulties with

corners. This is usually achieved by extending the survey line to

the boundary, allowing for the triangular shape which may occur.

b. In order to check the irregular area the calculations should be

repeated by another person, or a check against gross error may be

made taking out a planimeter area of the plot.

2.4 CALCULATING AREA FROM A CHAIN SURVEY

The figure shows the rectilinear area ABCD, which is calculated first. Theirregular strips between thechain lines and theboundary must beseparately evaluated andeither added or subtracted asnecessary from the mainrectilinear area calculation

result. The following datawere obtained from thechain survey of the site :

AB - 63.0 mBC - 45.0 m

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CD - 60.0 mDA - 78.0 mBD - 93.3 mAC - 76.0 m

SOLUTION

The rectilinear area from A = ((s a) (s b) (s c))

4

Chainage AD Offset

A 0.0 0.016.0 6.0

33.0 7.040.0 0.049.0 7.061.0 7.068.0 0.0

B 78.0 11.089.0 5.093.0 9.0

Chainage CD

Offset

C 0.0 0.0

10.0 4.220.0 6.430.0 8.140.0 10.350.0 11.3

D 60.0 13.2

AB and BC are straight boundaries. Offsets to the irregular

boundaries are as follows :


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The area of triangle ACD = (107(31) (47) (29))

= 2126.3 m2

The area of triangle ABC = (92(29) (47) (16))

= 1416.4 m2

Area of ABCD = 2126.3 + 1416.4

= 3542.7 m2

Check :

The area of triangle ABD = ((117.15 (54.15) (39.15) (23.85))

= 2433.8 m2

The area of triangle ABD = (( 99.15 (39.15) (54.15) (5.85)))

= 1108.9 m2

Area of ABCD = 2433.8 + 1108.9

= 3542.7 m2

Area of triangle ABD: Plus Minus

(0+6) x 2 x 16 = 48.0

(6+7) x 2 x 17 = 110.5(7+0) x 2 x 7 = 24.5

(0+7) x 2 x 9 = 31.5

(7+7) x 2 x 12 = 84.0

(7+0) x 2 x 7 = 24.5

(0+11) x 2 x 10 = 55.0

(11+9) x 2 x 15 = 150.0

388.5 140.0

- 140.0248.5 m2 (total plus area on AD)

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2.5 CALCULATING AREAS FROM COORDINATES

A = Area

SPECIMEN QUESTION

Calculate the area of the figure ABCDEF of which the coordinates are listed

below.

SOLUTION

The calculation is tabulated as shown :

Station Easting Northing

E + EDoubleLongitude

N

A 150 100B 95.2 164.3 245.2 64.3 15 766.36C 127.9 210.7 223.1 46.4 10 351.84D 176.3 239.8 304.2 29.1 8 852.22E 219.4 222.4 395.7 -17.4 6 885.18

F 237.5 163.8 456.9 -58.626

774.34

A 150 100 387.5 -63.8 24722.50

34 970.4258

382.0234

970.42

2A =23

411.60

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Area = 11 705.8 m2

= 1.1706 ha

2.6 AREAS OF IRREGULAR FIGURES

There are several practical situations where it is necessary to estimate the areaof irregular figures. Examples include estimation of areas of plots of land bysurveyors, areas of indicator diagrams of steam engines by engineers and areasof water planes and transverse sections of a ship by naval architects. There aremany methods whereby the area of an irregular plane surface may be found andthese include:

(a) Use of a planimeter,

(b) Trapezoidal rule,(c) Mid-ordinate rule and(d) Simpsons rule.

2.6.1 The planimeter

A planimeter is an instrument for directly measuring areas bounded by anirregular curve. There are many different types of the instrument but allconsist basically of two rods AB and BC, hinged at B (see Fig. 2.1). Theend labelled A is fixed, preferably outside of the irregular area beingmeasured. Rod BC carries at B a wheel whose plane is at right angles tothe plane formed by ABC. Point C, called the tracer, is guided round theboundary of the figure to be measured. The wheel is geared to a dialwhich records the area directly. If the length BC is adjustable, the scalecan be altered and readings obtained in mm2, cm2, m2 and so on.

FIGURES 2.1 : Planimeter

(Source : Mathematics for

Technicians, S. Adam)

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2.6.2 Trapezoidal rule

To find the area ABCD in Fig. 2.2, the base AD is divided into a number ofequal intervals of width d. This can be any number; the greater the

number the more accurate the result. The ordinates y1, y2, y3, etc. areaccurately measured. The approximation used in this rule is to assumethat each strip is equal to the area of a trapezium.

FIGURE 2.2 : Trapezoidal rule (Source : Mathematics for Technicians, S.

Adam)

The area of a trapezium = (sum of the parallel sides) (perpendiculardistance between the parallel sides).

Hence for the first strip, shown in Fig. 2.2, the approximate area is (y1 +y2)d. For the second strip area is (y1 + y2)d and so on. Hence theapproximate area of

ABCD = (y1 + y2)d+ (y3 + y4)d + (y3 + y4)d + (y4 + y5)d+ (y5+ y6)d + (y6 + y7)d

= y1 d + y2d+ y2d + y3 d + y3 d + y4 d + y4 d+ y5d + y5d + y6d + y6d + y7d

= y1 d + y2d + y3 d + y4 d + y5d + y6d + y7d= d[ ( y1 + + y7 ) / 2 + y2+ y3 + y4 + y5+ y6 ]

Generally, the trapezoidal rule states that the area of an irregular figure isgiven by:Area = (width of internal) [ (first + last ordinate) + sum of remaining

ordinates]

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2.6.3 Mid-ordinate rule

FIGURE 2.3 : Mid-ordinate

rule method

(Source : Mathematics for

Technicians, S. Adam)

To find the area of ABCD in Figure 2.3 the base AD is divided into anynumber of equal strips of width d. (As with the trapezoidal rule, the greaterthe number of intervals used the more accurate the result.) If each strip isassumed to be a trapezium, then the average length of the two parallelsides will be given by the length of a mid-ordinate, i.e. an ordinate erectedin the middle of each trapezium. This is the approximation used in themid-ordinate rule.

The mid-ordinates are labelled y1, y2, y3, etc. as in Fig. 18.3 and each isthen accurately measured. Hence the approximate area of ABCD

= y1 d + y2 d + y3 d + y4 d + y5 d + y6 d= d(y1 + y2 + y3 + y4 + y5 + y6)

where d = ( length of AD / number of mid-ordinates )

Generally, the mid-ordinate rule states that the area of an irregular figureis given by:

2.6.4 Simpsons rule

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Area = (width of interval) (sum of mid-ordinates)


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FIGURE 2.4 : Simpsons rule (Source : Mathematics for Technicians, S.

Adam)

To find the circa A BCD in Figure 2.4 the base AD must be divided into an

even number of strips of equal width d. Thus producing an odd number ofordinates. The length of each ordinate, y1, y2, y3, etc., is accuratelymeasured. Simpson's rule states that (the area of the irregular area ABCDis given by;

Area of ABCD = d/ 3 [(y1 + y7 ) + 4(y2 + y4 + y6) + 2(y3 + y5)]

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More generally, the calculation of the area of:

Area = 1/3 (width of interval) [(first and last ordinates) + 4( sum of

even ordinates) + 2 (sum of remaining odd ordinates)]

When estimating areas of irregular figures, Simpson's rule is generallyregarded as the most accurate of the approximate methods available.

Activity 2a

2.1 The values of the y ordinates of a curve and their distance x from theorigin are given in the table below. Plot the graph and find the area underthe curve by :

x 0 1 2 3 4 5 6y 2 5 8 11 14 17 20

a) The trapezoidal rule

b) The mid-ordinate rulec) Simpsons rule

2.2 Sketch a semicircle of radius 10cm. Erect ordinates at intervals of 2 cmand determine the lengths of the ordinates andmid-ordinates. Determinethe area of the semicircle using the three approximate methods. Calculatethe true area of thesemicircle.

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Feedback 2a

2.1)

FIGURE 2.5 : Graph of y against x

a) Trapezoidal rule

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Using 7 ordinates with interval width of 1 the area under the curve is:

Area = 1 [ (2 + 20) + 5 + 8 + 11 + 14 + 17 ]= [ 11 + 5 + 8 + 11 + 14 + 17 ]= 66 square units

b) Mid-ordinate rule

Using 6 intervals of width 1 the mid-ordinates of the 6 strips are measured.

The area under the curve is:

Area = 1 (3.5 + 6.5 + 9.5 + 12.5 + 15.5 + 18.5)= 66 square unit

c) Simpsons rule

Using 7 ordinates, given an even number of strips, i.e. 6, each of width 1, thusthe area under the curve is:

Area = 1 / 3 [ (2 + 20) + 4(5 + 11 + 17) + 2 (8 + 14) ]= 1 / 3 [ 22 + 4(33) + 2(22)]= 1 / 3 [ 22 + 132 + 44 ]= 198 / 3= 66 square units

The area under the curve is a trapezium and may be calculated using the formula(a+b)h, where a and b are the lengths of the parallel sides and h theperpendicular distance between the parallel sides.

Hence area = (2 + 20)(6) = 66 square units. This problem demonstrates themethods for finding areas under curves. Obviously the three 'approximate'methods would notnormally be used for an area such as in this problem since it isnot 'irregular'.

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2.2). The semicircle is shown in Fig. 2.6 with the lengths of the ordinates andmid-ordinates marked, the dimensions being incentimetres.

FIGURE 2.6 : Sketch a semicircle

a) Trapezoidal rule

Area = 2 [ (0 + 0) + 6.0 + 8.0 + 9.15 + 9.80 + 10.0 + 9.80 + 9.15 + 8.0 +6.0 ]= 2 (75.90)= 151.8 square units

b) Mid-ordinate rule

Area = 2 [ 4.3 + 7.1 + 8.65 + 9.55 + 9.95 + 9.95 + 9.95 + 8.65 + 7.1 + 4.3 ]= 2 (79.10)= 158.2 square units

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c) Simpsons rule

Area = 2/3 [ (0 + 0) + 4(6.0 + 9.15 + 10.00 + 9.15 + 6.00) + 2(8.0 + 9.8 +8.0)]= 2/3 [0 + 4(40.3) + 2(35.6)]= 2/3 (161.2 + 71.2)= 2/3 (232.4)= 154.9 square units

The true area is given by r / 2, i.e (10) / 2 = 157.1 square units2.7 VOLUME CALCULATION

In construction works, the excavation, loading, hauling and dumping of earthfrequently forms a substantial part of the project. Payment must be made for the

labour and plan needed for earthworks and this is based on the quantity orvolume handled. These volumes must be calculated and depending on the shapeof the site, this may be done in three ways :

i) by cross-sections, generally used for long, narrow works such as roads,railways, pipelines, etc.

ii) by contours, generally used for larger areas such as reservoirs,landscapes, redevelopment sites, etc.

iii) by spot height, generally used for small areas such as underground tanks,

basements, building sites, etc.

2.8 CROSS SECTION VOLUME CALCULATION

Cross-sections are established at some convenient intervals along a centre lineof the works. Volumes are calculated by relating the cross-sectional areas to thedistances between them. In order to compute the volume it is first necessary toevaluate the cross-sectional areas, which may be obtained by the followingmethods:

i) by calculating from the formula or from first principles the standard cross-sections of constant formation widths and side slopes.

ii) by measuring graphicallyfrom plotted cross-sections drawn to scale, areasbeing obtained by plannimeter or division into triangles or square.

NOTE :

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The graphic measure of the cross-sectional area is most often used and provides

a sufficiently accurate estimate of volume, but for railways, long embankments,

breakwaters, etc., with fairly regular dimensions, the use of formulae may be

easier and perhaps more accurate.

2.8.1 Prismoidal Method

In order to calculate the volume of a substance, its geometrical shape andsize must be known. A mass of earth has no regular geometrical figure inmost approaches. The prismoid is a solid, consisting of two ends whichform plane, parallel figures, not necessarily of the same number of sidesand which can be measured as cross-sections. The faces between theparallel ends are plane surfaces between straight lines which join all thecorners of the two end faces. A prismoid can be considered to be made up

of a series of prisms, wedges and pyramids, all having a length equal tothe perpendicular distance between the parallel ends. The geometricalsolids forming the prismoid are described as follows :

i) Prism, in which the end polygons are equal and the side faces areparallelograms.

ii) Wedge, in which one end is a line, the other end a parallelogram,and the sides are triangles and parallelograms.

iii) Pyramids, in which one end is a point, the other end a polygon

and the side faces are triangles.

The Prismoidal Formula

Let D = the perpendicular distance between the parallel endplanes

A1 andA2 = the areas of these end planesM = the mid-area, the area of the plane parallel to the end

planes and midway between them,V = the volume of the prismoid and

a1, a2, m, v = the equivalent for any prism, wedge or pyramid formingthe prismoid

then in aprism a1, = a2, = mand in a wedge a2 = 0 and m = 1/2 a1

and in a pyramid a2 = 0 and m = 1/4 a1Prism volume v = D . a1 = D/6 (6 . a1 ) = D/6 (a1, + 4m + a2)

Wedge volume v = D . a1 = D/6 (3 . a1 ) = D/6 (a1, + 4m + a2)

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Pyramid volume v = 1/3 D . a1 = D/6 (2 . a1 ) = D/6 (a1, + 4m + a2)

As the volume of each part can be expressed in the same terms, thevolume of the whole can take the same form. Thus the prismoidal formulais expressed in the following way :

V= D/6 (A1, + 4M+A2)

Note :

A. M does not represent the mean of the end areas A 1 and A2except

where the prismoid is composed of prisms and wedges only.

B. The formula gives the volume of one prismoid of which the end and

mid-sectional areas are known.

The prismoidal formula may be used to calculate volume if a series ofcross-sectional areas, A1, A2, A3,. An, have been established adistance d apart. Each alternate cross-section may be considered to bethe mid-area M of a prismoid of length 2d.

Then the volume of the first prismoid of length 2d:= 2d/ 6 (A1, + 4A2+A3)

and of the second = 2d/ 6 (A3, + 4A4+A5)and of the nth = 2d/ 6 (An-2, + 4An-1+An)

summing up the volumes of each prismoidal :

V= d/ 3 (A1, + 4A2+ 2A3 + 4A4 + 2An-2+ 4An-1 +An)Which is Simpsons rule for volumes.

Specimen Question

Calculate, using the prismoidal formula, the cubic contents of anembankment of which the cross-sectional areas at 15m intervals are asfollows :

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

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Solution,

V = 15 / 3 (11 + 220 + 4 ( 42 + 72 + 180 ) + ( 64 + 160))V = 5 ( 231 + 1176 + 448 )V = 9275 m3

Note :

A. The 15m interval is divided by 3, as the length of the individual

prismoids used is 30m, which in the prismoidal formula is divided

by 6.

B. A mass of earth, length double the usual cross-sectional interval of

15m, 20m or 25m, is considerably different from a true prismoid, so

this method is not as accurate as it would be if the true mid-

sectional area had been measured. This results in the use of

prismoids of length equal to, instead of double, the interval

between cross-sections.

2.8.2 End Areas Method

It is no more accurate to use the prismoidal formula where the mid-sectional areas have not been directly measured than it is to use the endareas formula, particularly as the earth solid is not exactly represented bya prismoid. Using the same symbols the volume may be expressed as :

v = d[ ( A1 + A2) / 2 ]

although this is only correct where the mid-area is the mean of the endareas :

M = ( A1 + A2) / 2

However, in view of the inaccuracies that arise in assuming anygeometric shape between cross-sections and because of bulking andsettlement and the fact that the end areas calculation is simple to use, it isgenerally used for most estimating purposes.

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Note :

A. The summation of a series of cross-sectional areas by this method

provides a total volume :

V = d{[( A1 + A2) / 2 ] + A2+ A3 + An-1}

Specimen Question

Calculate, using the end areas method, the cubic contents of theembankment of which the cross-sectional areas at 15m intervals are as

follows :

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

Solution

V = 15{[ (11 + 220) / 2 ] + 42 + 64 + 72 + 160 + 180 }V = 9502.5 m3

2.9 VOLUME CALCULATION FOR CONTOUR LINES

Contour lines may be used for volume calculations and theoretically this is themost accurate method. However, as the small contour interval necessary foraccurate work is seldom provided due to cost, high accuracy is not oftenobtained. Unless the contour interval is less than 1m or 2m at the most, theassumption that there is an even slope between the contour is incorrect andvolume calculation from contours become unreliable.

The formula used for volume calculation is the end areas formula of Simpsonsrule for volumes, the distance d in the formula being contour interval. The areaenclosed by each contour line is measured, usually by plannimeter, and theseareas A1, A2, etc., are used in the formula as before ( see the end areasmethod). If the prismoidal method is used, each alternate contour line is

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assumed to enclose a mid-area or the outline of the mid-area can be interpolatedbetween the existing contour intervals.

This illustration shows an area contoured at 5m intervals and how the contours ofproposed works, in this case a dam wall with an access road through a cutting,shown as packed lines, define the plan outline of the works. This also allows thevolume of the earthworks to be calculated using the positions of the contourlines.

Picture 4.1 : Intersection Of Contoured Surfaces (Source : Land Survey,Ramsay)

The volume of the dam wall and the amount of cut may be obtained from thecontour lines by calculating the volume of ground within the working area down toa common level surface and then calculating the new volume from the formationcontour lines, the difference being the change in volume due to the works. Thisvolume calculation is more usually carried out by using the cross-sectional

method. The use of contours is a practical method of calculating volumes inseveral cases, one of which being the calculation of water at various levels in areservoir. For example, in picture 4.1 the volume of water which could becontained up to the level of the 60m, contours could be calculated as followsfrom these data :

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Contour above datum (m) 50 52.5 55 57.5 60Area (m2) 12 135 660 1500 1950

Using the end areas method :

V = 2.5 { [ (12 + 1950)/2] + 135 + 660 + 1500 }= 8190 m3

Using Simpsons rule from the prismoidal formula :

V = ( 2.5 / 3 ) [ 12 + 1950 + 4(135 + 1500) + 2 (660) ]= (2.5 / 3 ) (9822)= 8185 m3

Note :The small volume of water below 50m (not included in the above calculation)

would be estimated from the interpolated depth of 2m at the deepest point, using

the end areas formula, the lowest end area being 0, thus :

V = [ ( 12 + 0 ) / 2 ] x 2

V = 12m3

This would then be added to either of the results above.

2.10 VOLUME CALCULATION FROM SPOT HEIGHT

This is a method of volume calculation frequently used on excavations wherethere are vertical sides covering a fairly large area, although it can be used forexcavation with sloping sides. The site is divided into squares or rectangles, andif they are of equal size the calculations are simplified. The volumes arecalculated from the product of the mean length of the sides of each verticaltruncated prism ( a prism in which the base planes are not parallel ) and the

cross-sectional area. The sizes of the rectangles is dependent on the degree ofaccuracy required. The aim is to produce areas such that the ground surfacewithin each can be assumed to be plane.

Specimen Question

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Picture 4.2 shows the reduced levels of a rectangular plot which is to beexcavated to a uniform depth of 8m above datum. Calculate the mean level ofthe ground and the volume of earth to be excavated.

Note :

a) The mean or average level of the ground is that level of ground which would

be achieved by smoothing the ground off level, assuming that no bulking

would take place.

b) The mean level of the ground is the mean of the mean height of each prism.

It is not the mean of all the spot heights.

Picture 4.2 : Calculating volume from spot heighton a levelling grid. (Source : Land Survey, Ramsay)

Solution

(a) Calculation from rectangles :

Station R.L.

Number oftimes the Product

R.L. is used = n (R.L.) x nA 12.16 1 12.16B 12.48 2 24.96C 13.01 1 13.01D 12.56 2 25.12E 12.87 4 51.48F 13.53 2 27.06G 12.94 1 12.94H 13.27 2 26.54J 13.84 1 13.84

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n = 16 207.11

Mean level = 207.11 / 16= 12.944 m

Depth of excavation = 12.944 8.00= 4.944

Volume = Total area x Depth

= 30 x 20 x 4.944= 2966.4 m3

(b) Calculation from triangles

It is usually more accurate to calculate from triangles as the upper base of thetriangular prism is more likely to correspond with the ground plane than the larger

rectangle. The mean level of each prism is then the mean of the three heightenclosing the triangle instead of four as before.

Station R.L.Number of timesthe Product

R.L. is used = n (R.L.) x n

A 12.16 1 12.16B 12.48 3 37.44C 13.01 2 26.02D 12.56 3 37.44

E 12.87 7 90.09F 13.53 2 27.06G 12.94 2 25.88H 13.27 2 26.54J 13.84 2 27.68

n = 24 310.55

Mean level = 310.55 / 24= 12.940 m

Depth of excavation = 4.960

Volume = 30 x 20 x 4.944

= 2966.4 m3

Note :The diagonal forming the triangles would be noted in the field book on the grid

layout to conform most suitably with the ground planes.

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Activity 2b

2.3) An embankment is to be formed with its centre line on the surface (in theform of a plane) on full dip of 1 in 20. If the formation width is 12.00m andthe formation heights are 3.00m, 4.50m and 6.00m at intervals of 30.00m,with side slopes 1 in 2, calculate the volume between the end sections.

Calculatea). Volume by mean areasb). Volume by end areasc). Volume by prismoidal rule

2.3Given the previous example but with the centre line turned through 90,calculate volume

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a) By mean areasb) By end areasc) By prismoidal

Feedback 2b

2.3 Area (1) = h1 (w + mh1)= 3.00 [ 12.00 + (2 x 3.00) ] = 54.00 m2

Area (2) = 4.50 [ 12.00 + (2 x 4.50) ] = 94.50 m2

Area (3) = 6.00 [ 12.00 + (2 x 6.00) ] = 144.00 m2

Volume :

a). By mean areas

V = W(A/n) = 60.00 ( 54.00 + 94.50 + 144.00 ) / 3

= 5850.0 m3

b). By end areas

V = w ( A1 + 2A2 + A3) / 2= 30.00 (54.00 + 189.00 + 144.00) / 2

= 5805.0 m3

c). By Prismoidal Rule

V= w ( A1 + 4A2 + A3) / 3

= 30.00 (54.00 + 378.00 + 144.00)/3

= 5760.0 m3

2.4 A = m ( h0 k + w / 4 + wh0 m) + wh0

( k - m )

Cross-sectional areas

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A1 = 2 [ (3.00 x 20 ) + ( 0.25 x 12.00) + ( 12.00 x 3.00 x 2 ) + (12 x 3.00)

( 20 - 2 )

= [ (3600.00 + 36.00 + 72.00) / 198 ] + 36.00 = 54.73 m

A2 = [ ( 8100.00 + 36.00 + 108.00 ) / 198 ] + 54.00 = 95.64 mA3 = [ ( 14400.00 + 36.00 + 144.00 ) / 198 ] + 72.00 = 145.64 m

Volume

a). By mean areasV= 60.00 ( 54.00 + 95.64 + 145.64 ) / 3 = 5920.2 m

b). By end areasV= 30.00 ( 54.73 + 191.28 + 145.64 ) / 2 = 5874.8 m

c). By prismoidal ruleV= 30.00 ( 54.73 + 382.56 + 145.64 ) / 3 = 5829.3 m

Self Assessment

Calculate the volumes in Figure 1 and Figure 2.

Figure 1

Figure 2

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Feedback to Self Assessment

Figure 1

A1= [ ( 0.75 + 4.75 ) / 2 ] x 7 = 19.25 ftA2= [ ( 0.75 + 3.75 ) / 2 ] x 5 = 11.25 ftA3= [ ( 0.75 + 2.75 ) / 2 ] x 3 = 5.25 ft

Volume, V = L / 6 ( A1 + 4Am + A2 )= 17 / 6 (19.25 + 4 x 11.25 + 5.25)= 17 / 6 ( 19.25 + 45.00 + 5.25 )= 17 / 6 ( 69.50 )= 1181.50 / 6= 196.92 ft= 7.29 yrd

Figure 2

Volume, V = h / 3 ( area of base )= 27.4 / 3 ( 13.5 x 13.5 )= 1664.6 m

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C2005 Ukur Kejuruteraan UNIT2

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