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Area and volume calculation and measurement

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Surveying Engineering Area and volume calculation and measurement 2022 -2021 Area of cross -section in cut and fill case 1:center line at fill section Case 2: center line at cut section Example of the area calculation by midpoint ordinate rule The following perpendicular offsets were taken at 10m interval from a survey line to an irregular boundary line. The ordinates are measured at midpoint of the division are 10, 13, 17, 16, 19, 21, 20 and 18m. Calculate the are enclosed by the midpoint ordinate rule. Example: The following offsets are taken from a chain line to an irregular boundary towards right side of the chain line. O1 O2 O3 O4 O5 O6 O7 Chainage(m) 0 25 50 75 100 125 150 Offset m 3.6 5.0 6.5 0.5 7.3 6.0 4.0 Common distance, d = 25m -Volume Of Spot Levels Example on Volume A road at constant reduced level of 180.00 m runs NorthSouth direction. The ground in direction East West is Level. The surface elevation along the center line of the road is as follows. Chainage (m) 0 50 100 150 200 250 300 Elevation (m) 190 188 187 185 183 186 182 Calculate the volume of the cutting given that the width of the formation level is (24m) and side slope is 1.5:1. Home Work: Calculate the volume of cutting in m3 for a road from the following data: Distance Depth of cutting Natural Ground Level ------------------------------------------------------------------- 0 8 1:12 100 10 1:8 200 12 1:10 Formation width =30m side slope= 1.5:1. Prepared by: Asst. Prof. Salar K.Hussein Mr. Kamal Y.Abdullah Asst.Lecturer. Dilveen H. Omar Erbil Polytechnic University Technical Engineering College Civil Engineering Department

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Area and volume calculation and measurement 2022 -2021 Erbil Polytechnic University Technical Engineering College Civil Engineering Department Surveying Engineering 1 Prepared by Asst. Prof. Salar Khudhur Hussein Asst. Lecturer Mr. Kamal Yaseen Asst. Lecturer Ms. Dilveen H. Omar
2 Area of cross-section in cut and fill case 1:center line at fill section Area of fill=1/2 h*w w= b/2 - c*n d=b/2+h*s= h*n + c*n h=(b/2-c*n)/(n-s) d=(b/2-c*s)*(n/(n-s)) Area of fill = 遜(b/2-c*n)2 /(n-s) Area of cut=1/2(b/2+c*n)2/(n-s) Case 2: center line at cut section w= b/2 + c*n d=b/2+h*s= h*n - c*n h=(b/+-c*n)/(n-s) d=(b/2+c*s)*(n/(n-s)) Area of fill = 遜(b/2+c*n)2 /(n-s) Area of cut=1/2(b/2-c*n)2/(n-s)
3 Example of the area calculation by midpoint ordinate rule The following perpendicular offsets were taken at 10m interval from a survey line to an irregular boundary line. The ordinates are measured at midpoint of the division are 10, 13, 17, 16, 19, 21, 20 and 18m. Calculate the are enclosed by the midpoint ordinate rule. Given: Ordinates O1 = 10, O2 = 13, O3 = 17, O4 = 16, O5 = 19, O6 = 21, O7 = 20, O8 = 18 Common distance, d =10m Number of equal parts of the baseline, n = 8 Solution: Length of baseline, L = n *d = 8*10 = 80m Area = [(10+13+17+16+19+21+20+18)*80]/8 = 1340sqm
4 Example: The following offsets are taken from a chain line to an irregular boundary towards right side of the chain line. O1 O2 O3 O4 O5 O6 O7 Chainage(m) 0 25 50 75 100 125 150 Offset m 3.6 5.0 6.5 0.5 7.3 6.0 4.0 Common distance, d = 25m Area = d/3[(O1+O7) + 2 (O3+O5)+4(O2+O4+O6)] = 25/3[(3.6+4)+2(6.5+7.3)+4(5+0.5+6)] Area = 843.33sqm
5 Volume Of Spot Levels: L2 V = ------[ O1+O2+O3+O4] 4 L2 V = ------[ O1+ 2O2+3 O3+ 4O4 ] 4
6 Example on Volume A road at constant reduced level of 180.00 m runs North- South direction. The ground in direction East West is Level. The surface elevation along the center line of the road is as follows. Chainage (m) 0 50 100 150 200 250 300 Elevation (m) 190 188 187 185 183 186 182 Calculate the volume of the cutting given that the width of the formation level is (24m) and side slope is 1.5:1
7 A = (b+cs)*c A0= (24+10*1.5)*10= 390 C=190-180=10m A1= (24+ 8*1.5)*8 = 288 C=188-180=8m A2=241.5 C=187-180=7 A3=157.5 C=185-180=5 A4=85.5 C=183-180=3 A5=198 C=186-180=6 A6=54 C=182-180=2 Volume Prismoid method = L/3 {(A0+A6)+4(A1+A3+A5)+2(A2+A4)} V= 50/3 {(390+54)+4(288+157.5+198)+2(241.5+85.5)} V= 61200 m3
8 Home Work: Calculate the volume of cutting in m3 for a road from the following data: Distance Depth of cutting Natural Ground Level ------------------------------------------------------------------- 0 8 1:12 100 10 1:8 200 12 1:10 Formation width =30m side slope= 1.5:1

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Area and volume calculation and measurement

  • 1. Area and volume calculation and measurement 2022 -2021 Erbil Polytechnic University Technical Engineering College Civil Engineering Department Surveying Engineering 1 Prepared by Asst. Prof. Salar Khudhur Hussein Asst. Lecturer Mr. Kamal Yaseen Asst. Lecturer Ms. Dilveen H. Omar
  • 2. 2 Area of cross-section in cut and fill case 1:center line at fill section Area of fill=1/2 h*w w= b/2 - c*n d=b/2+h*s= h*n + c*n h=(b/2-c*n)/(n-s) d=(b/2-c*s)*(n/(n-s)) Area of fill = 遜(b/2-c*n)2 /(n-s) Area of cut=1/2(b/2+c*n)2/(n-s) Case 2: center line at cut section w= b/2 + c*n d=b/2+h*s= h*n - c*n h=(b/+-c*n)/(n-s) d=(b/2+c*s)*(n/(n-s)) Area of fill = 遜(b/2+c*n)2 /(n-s) Area of cut=1/2(b/2-c*n)2/(n-s)
  • 3. 3 Example of the area calculation by midpoint ordinate rule The following perpendicular offsets were taken at 10m interval from a survey line to an irregular boundary line. The ordinates are measured at midpoint of the division are 10, 13, 17, 16, 19, 21, 20 and 18m. Calculate the are enclosed by the midpoint ordinate rule. Given: Ordinates O1 = 10, O2 = 13, O3 = 17, O4 = 16, O5 = 19, O6 = 21, O7 = 20, O8 = 18 Common distance, d =10m Number of equal parts of the baseline, n = 8 Solution: Length of baseline, L = n *d = 8*10 = 80m Area = [(10+13+17+16+19+21+20+18)*80]/8 = 1340sqm
  • 4. 4 Example: The following offsets are taken from a chain line to an irregular boundary towards right side of the chain line. O1 O2 O3 O4 O5 O6 O7 Chainage(m) 0 25 50 75 100 125 150 Offset m 3.6 5.0 6.5 0.5 7.3 6.0 4.0 Common distance, d = 25m Area = d/3[(O1+O7) + 2 (O3+O5)+4(O2+O4+O6)] = 25/3[(3.6+4)+2(6.5+7.3)+4(5+0.5+6)] Area = 843.33sqm
  • 5. 5 Volume Of Spot Levels: L2 V = ------[ O1+O2+O3+O4] 4 L2 V = ------[ O1+ 2O2+3 O3+ 4O4 ] 4
  • 6. 6 Example on Volume A road at constant reduced level of 180.00 m runs North- South direction. The ground in direction East West is Level. The surface elevation along the center line of the road is as follows. Chainage (m) 0 50 100 150 200 250 300 Elevation (m) 190 188 187 185 183 186 182 Calculate the volume of the cutting given that the width of the formation level is (24m) and side slope is 1.5:1
  • 7. 7 A = (b+cs)*c A0= (24+10*1.5)*10= 390 C=190-180=10m A1= (24+ 8*1.5)*8 = 288 C=188-180=8m A2=241.5 C=187-180=7 A3=157.5 C=185-180=5 A4=85.5 C=183-180=3 A5=198 C=186-180=6 A6=54 C=182-180=2 Volume Prismoid method = L/3 {(A0+A6)+4(A1+A3+A5)+2(A2+A4)} V= 50/3 {(390+54)+4(288+157.5+198)+2(241.5+85.5)} V= 61200 m3
  • 8. 8 Home Work: Calculate the volume of cutting in m3 for a road from the following data: Distance Depth of cutting Natural Ground Level ------------------------------------------------------------------- 0 8 1:12 100 10 1:8 200 12 1:10 Formation width =30m side slope= 1.5:1