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Volume and Area Calculation

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A brief presentation about earthwork
Lecturer : Khan Mohammad Shinwaray

Published in: Engineering

Volume and Area Calculation

  1. 1. ~_ 0*, BAKI-ITAR UNIVERSITY Area and Volume
  2. 2. Areas 1.Regu| ar Figures - Mathematical Formulae - Method of coordinates 2.Irregular Figures - Graphical Method - Trapezoidal Rule - Simpson’s One-Third Rule Kl A, “ S«. u‘~. ‘v3v [r'ag; r.«3.3v>| ag
  3. 3. Planimeters Digital Planimeter Optical Polar Planimeter -L3,; ,-, »_vg1; g, ;,_. 3,” Sur‘~. 'e, ' Engin-39: mg
  4. 4. Planimeters 13,; -,; ,_m1.1 sm. ,.,3.a, . Survey Engmeenng
  5. 5. Mathematical Formulae 3 a h ‘ - A ‘ b ‘ C (a) Triangle Square (c) Rectangle 2 lbh or la-b-sin(C) 3 3'19 2 2 8 J ( d) Trapezoid (2) Regular polygon (I) Circle 1 , ° —h-(a+b) Ln-a‘-cot m £7[. d2 2 4 n 4 Sl'w1r. '/avay Survey Engmeeung
  6. 6. Mathematical Formulae V o (g) Circular ring (h) Circular sector (i) Circular segment 1 1 1tA , 1r(r22—r12) —1t-A-r2 —r2 ——s1nA 360 2 180 . £ - {L} (J) Ellipse (k) Parabola 2 I§(. a’2<bA) — Sl'w1r. uamy Survey Engineer mg 3
  7. 7. Areas by Method of Coordinates Survey Engineering
  8. 8. Areas by Method of Coordi Positive Negative Point y x product product (Solid product) (Dashed product) Sur‘~. rey [rrwre-zrrrrg Area= —12—(lSum1-Sum2|) 4
  9. 9. Example Areas by Method of Coordinates Find the area of the following closed loop traverse (ABCDEA):
  10. 10. Example Areas by Method of Coordinates Positive Negative product product (Solid product) (Dashed product) -57.41 1,116.05 -40,021.58 = o.5(| 136840.01 — (-84890.94) | ) = 110865.48 fiz 136,840.01 -84,890.94
  11. 11. Graphical Method IIIIII_I! III IIZIIIIIEHI For example: If the number of squares ~ 42.5 (Figure 8.8), area of each square and the scale = L = > Area= 42.5 x 1 x (1ooo)’ = 4.25 x 10’ cm’ l"rlri‘«RlFi‘ In: ey Engiiieeiriig
  12. 12. Area Calculation A Trapezoidal Rule Area= %[h, + hn + 2(hz + 113 + + h, ,_, )] Simspon’s One-Third Rule Area= E(X + 20 + 4B) 3
  13. 13. Trapezoidal Rule Survey line
  14. 14. Trapezoidal Rule Area oftrapezoid (1) = h’ E h’ -1) Area of trapezoid (2) = h’ E h’ -b Area of trapezoid (6) = h‘ 3 h’ -b ; g.,7i_2ri: _i "lH"‘VA(, ?'i Survey Erigiiieerriig
  15. 15. Trapezoidal Rule Summing up, we get for n offsets Area= Elia, + 11,, + 2(h, + 113 + + h, ,_, )] J: : r, ,rr, r14 5l'iir. ,',31g. y Survey Engineer mg
  16. 16. Example — ' Calculate the area between offsets 1 and 10 in Figure 8.10 where these offsets, scaled from the plan at intervals of 10 m, are: EIMIEIEIEIEIEIE Efilfiliflllmlllifil 17.37 16.76 17 68 SOLUTION: Using Equation (8.26), Area= -I-29[l6.76 + 17.68 + 2(l9.8l + 20.42 + 18.59 + 16.76 + 17.68 + 17.68 + 17.37 + 16.76)] = 1622.9 m’ = 0.1623 ha irvey [r'r; J'iis. =-zrriiiz
  17. 17. Simspon’s One-Third Rule Intercept E Oi~! LY used with Odd number of offsets (i. e. Even number of Intercepts) Survey
  18. 18. Simspon’s One-Third Rule A Area= E-(h, + 11-, + 4(h, + 114 + 116) + 2(ii, + h, » In the general case, Area= 1’-(X + 20 + 45) . . . . . . . . . . . . . . . .. 3 where X = sum of first and last offsets 0 = sum of the remaining odd offsets E = sum of the even offsets
  19. 19. Example In a chain survey, the following offsets were taken to a fence from a chain line: IIEKWEIIWWEIEIIHHEEII HIIZHIISITWIEIUH ; g.,7i_iri: _i i—r, ,._. ,s, ;(. Survey Erigiiieerriig
  20. 20. Here, we have 10 offsets and 9 intercepts of length = .20 m, and since Simpson's one- third rule applies to an odd number of offsets (even number of intercepts), it will be used here to calculate the area contained between the lst and 9th olfsets. The residual triangular area between the 9th and 10th offsets is calculated separately. E 1 2 3 4 5 6 7 3 9 z=1s5.31 Area (11, - 11,) = ? x1ss.31 = 1235.40 m’ Area (11, - 11.0) = 329 x 1.83 = 18.30 m’ Ejlgifjlllfi | "1mua1av l'99ll"g Total area = 1253.70 m’ = 0.1254 ha
  21. 21. BAKI. -{TAR UNIVERSITY
  22. 22. Content - Volumes by: 1 . Average-End-Area Method 2.Prismoida| Method 3.Contour Maps ~ Volume from Spot Levels t. ,__, _,q, ;. 5Lll‘~JE‘V [r'19_1r. v3.; »H11g
  23. 23. Cut and Fill (a) 1-111' section (c) Cut_and fill section , . 4 -
  24. 24. Average-End-Area Method where V = volume A, and A, are the areas of the end sections L = distance between sections For n sections, distance L apart, V = £(A, + An + 2[A, + A3 + + A, ,_, )) 2 1} 2/1 1.‘ V. — 1 11 , l1>1r. ,-may 51111./ evE11g1r1ee1r11g
  25. 25. ""“’F"v' ‘*‘-*. - 3"‘ "‘ . "'3': -: ul ‘:1 ; ~ 5 ’ ‘ 5 fi 4' 3‘! -vi‘: fir‘; ‘- , ~. --5:: _-r’ I -T The planimetered areas in m’ of two side-hill cross-sections of a proposed highway ’ are as follows: Chainage 4200.0 m C82 F112 Chainage 4250.0 m C214 F78 C denotes cut, and F denotes fill. Calculate the quantities of earthwork SOLUTION: Distance L = 4250.0 - 4200 = 5o. o m Cut Volume = X 50 = 7400 ms 2 Fill Volume = x 50 = 4750 ms 2 . .
  26. 26. Prismoidal Method For n cross-sectional areas separated by equal strips of length L, the prismoidal formula is: 1 = %lA, + A, + 2(A, + A, + + A-, _,) + 4(A, + A. + + A, __. )] . . . . . . . . . . . . . . . . . . . . . . . . . .. (8.30) where n = odd number J‘: RN11 5‘. .”__, VaW S11l'= .revE11g1neev111g
  27. 27. Calculating Volumes from Contour M S«.1r= ./ev [r'1;J_111s3s3llI1g
  28. 28. Calculating Volumes from ( A I — —l ’EEIIEIIEEEI 1% Knowing that the bottom of the reservoir is flat with a level of 800 m AMSL, calculate the volume of water in the reservoir when it reaches the maximum allowed level which is 840 m AMSL, Use both A-E—A and prismoidal methods. Da ‘ 7sK, r7‘k}ll'1" _ “ l“-'4'“ 51.1%. /ey [1’1;J: r1s; .gm, g
  29. 29. Calculating Volumes from Contour M (1) A-E-A Method: Contour interval = L = 10 m v = £2'—[Am + Am + 2(A, ,., + Am + A, ,,, )] = > V = 1—2g[20365 + 211210 + 2(4l375 + 117120 + 160340)] = 4346225 m3 (2) By Prismoidal Formula 7 = -I3—‘[Am + Am + 2(Am) + 4(Am + A, ,., )] = %[2036S + 211210 + 2x 117120 + 4(41375 + 160340)] = 4242250 rn’ “"~= r-Wax‘ S111‘. /ev [1'1;3,1r1e+. =1l11g
  30. 30. Volume from Spot Le ; ,,, «_; ,r14 g.4_, _., q,; , S111‘-. reyE11g111eevr1'1g
  31. 31. I31‘; Willi. ‘ Volume from Spot Levels The volume removed fi'om any rectangle is computed as the average of the heights of the four corners times the area. Some comers are common to more than one rectangle; thus A1,’ in the table, is common to one, A2 is common to two, 133 is common to three, and B2 is common to four rectangles. By summing up the volumes within all the rectangles, the following equation results: v= ¥A—(Xh, ’ + 2211, + 231., + 2411.) 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . (8.32) where V is the volume A is the area of one rectangle or square h, , h, , h, and h, are the comer heights common to one, two, three and four rectangles respectively. l"11,. .Al: .1 S111‘. /ev [1'1g1ne<. =u11g
  32. 32. Volume from Spot Levels Figure 8.15 shows a borrow pit in the shape of a grid with all the dimensions and reduced levels. If this area is to be leveled to a constant level of 200.0 m for all the grid comers, calculate the volume of cut. FIGURE 8.15 . .,, _V, ,;. 5111'». /ev [r'1;3,1r1e+. =1l11g
  33. 33. ‘.112/“1lll—‘ A 1 A2 A3 A4 Old New Elev. Elev. m m 75.8 2.1 1.9 Volume ofcut= 8 x 12 x — = 41814902 m’ 4 Cut c 3.5 5 6 No. of Rectangles 7?) . ..a| )y. ———I! .n). hl)v-df. )KJ>-I Sum = 51.111./ ev 11112111991 mg
  34. 34. Survey

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