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# Topic 2 area & volume

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### Topic 2 area & volume

1. 1. TOPIC 2 AREA & VOLUME OGO L www.themegallery.com
2. 2. OBJECTIVES LOGO 1 Explain the basic concept of Area and Explain the basic concept of Area and Volume Method. Volume Method. 2 Define the usage of Area And Volume Define the usage of Area And Volume Calculation. Calculation. 3 Describe the methods that have been used Describe the methods that have been used in Area and Volume Calculation .. in Area and Volume Calculation
3. 3. INTRODUCTION 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 and levels that cut and fill are balanced as far as practical. •To enable contract estimates of time and cost to be made for proposed work. •To form the basis of payment for work carried out.
4. 4. AREA CALCULATION 1 The rectilinear areas enclosed by the survey lines 2 The irregular areas of the strips between these lines and the boundary
5. 5. The Rectilinear areas Method • Mathematical equation • Coordinates station traverse Mechanical - use of a planimeter
6. 6. MECHANICAL - PLANIMETER Cont..
7. 7. MATHEMATICAL EQUATION Cont.. Triangular equation i) Area = √[S(S-a)(S-b)(S-c)] where; S = ½ (a+b+c) Rectangular equation i) Area = a x b Trapezium equation i) Area = ½ (a + b) x h b B b a c A C b ii) Area = ½ (height x width) = ½ (b x h) B h A b C iii) Area = ½ a b sin c0 a c0 b h a a
8. 8. BY COORDINATES Cont.. 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 ordinate E = easting of that ordinate
9. 9. The irregular areas Trapezoidal rule Text Irregular plane surface Simpson’s rules Mid-ordinate rule www.themegallery.com
10. 10. Trapezoidal Rule This 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 ordinates F = first ordinate L = last ordinate O1 = first offset On = last offset o1 o2 o3 o4 o5 o6 o7
11. 11. Trapezoidal Rule Cont.. Example The total area = d /2 x [(F + L) + 2 (other ordinates)] 01 02 03 8m 8m 04 8m 05 8m 06 8m Distance 0 8 16 24 32 40 Offset 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
12. 12. Mid-ordinate rule o1 o2 o3 O1+ O2 2 O2+ O3 2 O3+ O4 2 The total area o4 O4+ O5 2 o5 o6 O5+ O6 2 o7 O6+ O7 2 = d x [sum of mid-ordinates]
13. 13. Mid-ordinate rule Cont.. Example The total area 0.75 01 = d x [sum of mid-ordinates] 1.85 02 2.10 03 8m 8m 2.05 04 8m 1.60 05 8m 06 8m Distance 0 8 16 24 32 40 Offset 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
14. 14. Simpson Rule where The total area = 1 / 3 d [F + L + 4 (Es) + 2 (Os)] D = equal distance between ordinates F = first ordinate L = last ordinate E = even numbered ordinates O = odd numbered ordinates o1 o2 o3 o4 o5 o6 o7 Example formula The total area = d / 3 [O1 + O7 + 4 (O2 + O4 + O6) + 2 (O3 + O5)]
15. 15. Simpson Rule Cont.. Example The total area = 1 / 3 d [F + L + 4 (Es) + 2 (Os)] 01 02 8m 03 8m 04 8m 06 05 8m 8m Distance 0 8 16 24 32 40 Offset 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
16. 16. Calculation of cross sectional area “Cut” means an excavation of the earth “fill” means the filling or raising of the original ground surface. 1) Sections with level across 2) Sections with cross-fall 4) Cross sections of variable level or three level sections 3) Sections part in cut and part in fill
17. 17. Calculation of cross sectional area Cont.. 1)Sections with level across Depth of centre line or height of embankment = h Formation width = b Side width = w Area = h(b + mh) 2)Sections with cross-fall Area = 1/2m [(b/2 + mh)(w1 + w2) – b2/2]
18. 18. Calculation of cross sectional area Cont.. 3) Sections part in cut and part in fill Area of fill = ½ [(b/2 + kh)2/(k-m)] Area of cut = ½ [(b/2 - kh)2/(k-n)] 4) Cross sections of variable level or three level sections Area = 1/2m[(w1 + w2)(mh + b/2) – b2/2]
19. 19. Volume calculation These volumes must be calculated and depending on the shape of the site, this may be done in three ways : by cross-sections generally used for long, narrow works such as roads, railways, pipelines, etc. by spot height generally used for small areas such as underground tanks, basements, building sites, etc. volume by contours generally used for larger areas such as reservoirs, landscapes, redevelopment sites, etc.
20. 20. Computational of volumes based on area of CROSS SECTIONS End areas Mean areas Vol. = {[A1 + A2 + A3 + ……… A n+1 + An] / n} . L Vol. = D/2 {(A1 + An) + 2(A2 + A3 + …… A n-1)} Prismoidal formula Vol = D/3 (A1 + An + 4ΣEven Areas + 2Σodd Areas)
21. 21. Computational of volumes based on area of CROSS SECTIONS Example calculation Calculate, using the prismoidal formula, the cubic contents of an embankment of which the cross-sectional areas at 15m intervals are as follows : Distance (m) Area (m2) 0 11 15 42 30 64 45 72 60 160 75 180 90 220 A1 A2 A3 A4 A5 A6 A7 Solution - Mean areas method Solution – Prismoidal method Vol. = {[A1 + A2 + A3 + ……… A n+1 + An] / n} . L V = D/3 (A1 + A7 + 4Σ( A2 + A4 + A6) + 2Σ ( A3 + A5) =15 / 3 (11 + 220 + 4 ( 42 + 72 + 180 ) + 2( 64 + 160)) V = 5 ( 231 + 1176 + 448 ) V = 9275 m3 V = {(11 + 42 + 64 + 72 + 160 + 180 + 220)/ 7 } . 90 V = 9630 m3 Solution - End 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
22. 22. 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. 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. Contour (m) 184 182 180 178 176 174 172 Areas (m2) 3125 2454 1630 890 223 110 69 Answer:End area method; Volume = 2/2 [3125+69 + 2(110 + 223 + 890 + 1630 + 2454)] = 13808 m3
23. 23. Volume from SPOT LEVELS This 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 height l and b are the side dimensions
24. 24. Volume from SPOT LEVELS – Square method 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 ) A(16.54m) 25.5 m B(17.25m) D(16.32m) E(12.95m) C(15.40m) F(15.55m) Solution: Station Reduced Level Excavated Level A B C D E F G H I 16.54 17.25 15.40 16.32 12.95 15.55 16.17 15.84 13.38 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 Depth Of excavated (hn) 6.54 7.25 5.40 6.32 2.95 5.55 6.17 5.84 3.38 Total G(16.17m) H(15.84m) I(13.38m) Average excavated depth = Σ h x n Σn = 83.21 16 No. Of Rectangles (n) 1 2 1 2 4 2 1 2 1 16 = 5.2 m Base area = 25.5 x 30.0 = 765 m2 30.0 m Volume to excavated = 5.2 x 765 = 3978 m3 Product ( hn x n ) 6.54 14.50 5.40 12.64 11.80 11.10 6.17 11.68 3.38 83.21
25. 25. Volume from SPOT LEVELS – Triangle method 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 ) A(16.54m) 25.5 m B(17.25m) D(16.32m) E(12.95m) C(15.40m) F(15.55m) Solution: Station Reduced Level Excavated Level Depth Of excavated (hn) No. Of Rectangles (n) A B C D E F G H I 16.54 17.25 15.40 16.32 12.95 15.55 16.17 15.84 13.38 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 10.00 6.54 7.25 5.40 6.32 2.95 5.55 6.17 5.84 3.38 2 3 1 3 6 3 1 3 2 13.08 21.75 5.40 18.96 17.70 16.65 6.17 17.52 6.76 24 123.99 Total G(16.17m) H(15.84m) I(13.38m) Average excavated depth = Σ h x n Σn = 123.99 24 = 5.17 m Base area = 25.5 x 30.0 = 765 m2 30.0 m Volume to excavated = 5.17 x 765 = 3955 m3 Product ( hn x n )
26. 26. LOGO End of topic Exercise