calculating+bulk+volumes+using+simpson+rule+
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simpson ruleTRANSCRIPT
1
Calculating Bulk Volumes Using Simpson Rule
Simpson’s Rule
Previously when we drew our sections we assumed that the ground was straight between the spot
heights. In the real world natural land will undulate between the spot heights you have taken.
Simpson’s rule is a method that tries to estimate this undulation.
The formula is expressed as follows,
Area of the section = D ÷ 3 x (X1 + 4X2 + 2X3 + 4X4……..+ 2Xn-2 + 4Xn-1 + Xn) Where,
D= Spacing between spot heights (Note this method will only work if an even grid method is used)
Xx = Cut/Fill value calculated
subscript is the grid position along the section
n = The number of spot heights along the grid – The number of grids must be an odd number
Simpson’s Rules uses the sections developed from either the longitudinal or cross sections.
The formula may also be expressed as
D ÷ 3 x ( First Cut/Fill + 4 x Second Cut/Fill + 2 x Third Cut/Fill + ….4 x Even Numbered Cut/Fill + 2 x Odd
Numbered Cut/Fill …… + 4 x Second Last Cut/Fill + Last Cut/Fill)
From the previous grid layout
Step 1
Determine Formation Height = 100.500
Step2
2
Determine Cut/Fill for each spot height.
RL 100.000
+/- above Datum 0.9
50
0.8
00
0.7
30
0.7
35
0.7
24
0.6
80
0.8
30
Design Level 10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
Cut -0.4
50
-0.3
00
-0.2
30
-0.2
35
-0.2
24
Fill
-0.1
80
-0.3
30
10
0.5
00
Existing Ground Level 10
0.9
50
10
0.8
00
10
0.7
30
10
0.7
35
10
0.7
24
10
0.6
80
10
0.8
30
Chainage 0.0
00
5.0
00
10
.00
0
15
.00
0
20
.00
0
25
.00
0
30
.00
0
RL 100.500
Longitudinal Section – Grid 1
Vertical Scale 1:20
Horizontal Scale 1:100
= D ÷ 3 x (X0 + 4X1 + 2X2 + 4X3 + 2Xn4 + 4X5 + X6)
=5 ÷ 3 x (-0.450 + 4 x -0.300 + 2 x -0.230 + 4 x -0.235 + 2 x -0.224 + 4 x -0.180 + -0.330)
= -7.580m2
3
To remove any doubt if this formula works lets substitutes the following example,
RL 100.000
+/- above Datum 0.7
00
0.7
00
0.7
00
0.7
00
0.7
00
0.7
00
0.7
00
Design Level 10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
Cut -0.2
00
-0.2
00
-0.2
00
-0.2
00
-0.2
00
Fill
-0.2
00
-0.2
00
10
0.5
00
Existing Ground Level 10
0.7
00
10
0.7
00
10
0.7
00
10
0.7
00
10
0.7
00
10
0.7
00
10
0.7
00
Chainage 0.0
00
5.0
00
10
.00
0
15
.00
0
20
.00
0
25
.00
0
30
.00
0
RL 100.500
Longitudinal Section – Grid X
Vertical Scale 1:20
Horizontal Scale 1:100
The example is a longitudinal section which shows a perfectly flat section with an RL 100.700.
To reduce it to its design level there is a constant cut of 0.200m.
This may be easily calculated as 30m x 0.200m = 6m2
Using Simpson’s we can write the formula as follows
Area = D ÷ 3 x (X0 + 4X1 + 2X2 + 4X3 + 2Xn4 + 4X5 + X6)
=5 ÷ 3 x (-0.200 + 4 x -0.200 + 2 x -0.200 + 4 x -0.200 + 2 x -0.200 + 4 x -0.200 + -0.200)
=- 6m2
= -7.580m2
4
RL 100.000
+/- above Datum 0.9
00
0.8
30
0.7
20
0.7
00
0.6
50
0.6
50
0.6
00
Design Level 10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
Cut -0.4
00
-0.3
30
-0.2
20
-0.2
00
-0.1
50
Fill
-0.1
00
10
0.5
00
Existing Ground Level 10
0.9
00
10
0.8
30
10
0.7
20
10
0.7
00
10
0.6
50
10
0.6
50
10
0.6
00
Chainage 0.0
00
5.0
00
10
.00
0
15
.00
0
20
.00
0
25
.00
0
30
.00
0
RL 100.500
Longitudinal Section – Grid 2
Vertical Scale 1:10
Horizontal Scale 1:100
-0.1
50
= D ÷ 3 x (X0 + 4X1 + 2X2 + 4X3 + 2Xn4 + 4X5 + X6)
=5 ÷ 3 x (-0.400 + 4 x -0.330 + 2 x -0.220 + 4 x -0.200 + 2 x -0.150 + 4 x -0.150 + -0.100)
= -6.600m2
5
RL 100.000
+/- above Datum 0.7
50
0.7
30
0.6
90
0.5
50
0.5
30
0.4
50
0.4
00
Design Level 10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
Cut -0.2
50
-0.2
30
-0.1
90
-0.0
50
-0.0
30
Fill +0.0
50
+0.1
00
10
0.5
00
Existing Ground Level 10
0.7
50
10
0.7
30
10
0.6
90
10
0.5
50
10
0.5
30
10
0.4
50
10
0.4
00
Chainage 0.0
00
5.0
00
10
.00
0
15
.00
0
20
.00
0
25
.00
0
30
.00
0
RL 100.500
Longitudinal Section – Grid 3
Vertical Scale 1:20
Horizontal Scale 1:100
= D ÷ 3 x (X0 + 4X1 + 2X2 + 4X3 + 2Xn4 + 4X5 + X6)
=5 ÷ 3 x (-0.250 + 4 x -0.230 + 2 x -0.190 + 4 x -0.050 + 2 x -0.030 + 4 x 0.050 + 0.100)
= -2.517m2
6
RL 100.000
+/- above Datum 0.5
50
0.4
70
0.3
30
0.3
20
0.3
50
0.6
00
0.2
90
Design Level 10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
Cut -0.0
50
+0.0
30
+0.1
70
+0.1
80
+0.1
50
Fill
-0.1
00
+0.2
10
10
0.5
00
Existing Ground Level 10
0.5
50
10
0.4
70
10
0.3
30
10
0.3
20
10
0.3
50
10
0.6
00
10
0.2
90
Chainage 0.0
00
5.0
00
10
.00
0
15
.00
0
20
.00
0
25
.00
0
30
.00
0
RL 100.500
Longitudinal Section – Grid 4
Vertical Scale 1:20
Horizontal Scale 1:100
= D ÷ 3 x (X0 + 4X1 + 2X2 + 4X3 + 2Xn4 + 4X5 + X6)
=5 ÷ 3 x (-0.050 + 4 x 0.030 + 2 x 0.170 + 4 x 0.180 + 2 x 0.150 + 4 x -0.100 + 0.210)
= 2.067m2
7
RL 100.000
+/- above Datum 0.3
94
0.3
30
0.3
00
0.2
00
0.1
90
0.1
40
0.0
40
Design Level 10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
10
0.5
00
Cut
+0.1
06
+0.1
70
+0.2
00
+0.3
00
+03
10
Fill +0.3
60
+0.4
60
10
0.5
00
Existing Ground Level 10
0.3
94
10
0.3
30
10
0.3
00
10
0.2
00
10
0.1
90
10
0.1
40
10
0.0
40
Chainage 0.0
00
5.0
00
10
.00
0
15
.00
0
20
.00
0
25
.00
0
30
.00
0
RL 100.500
Longitudinal Section – Grid 5
Vertical Scale 1:20
Horizontal Scale 1:100
= D ÷ 3 x (X0 + 4X1 + 2X2 + 4X3 + 2Xn4 + 4X5 + X6)
=5 ÷ 3 x (0.106 + 4 x 0.170 + 2 x 0.200 + 4 x 0.300 + 2 x 0.310+ 4 x 0.360 + 0.460)
= 8.177m2
8
So far we have calculated the cross sectional area for each Longitudinal Sections 1 to 5.
This is represented by the following table.
A B C D E F G
1 -0.450 -0.300 -0.230 -0.235 -0.224 -0.180 -0.330
2 -0.400 -0.330 -0.220 -0.200 -0.150 -0.150 -0.100
3 -0.250 -0.230 -0.190 -0.050 -0.030 0.050 0.100
4 -0.050 0.030 0.170 0.180 0.150 -0.100 0.210
5 0.106 0.170 0.200 0.300 0.310 0.360 0.460
Cross Sectional Area
-7.580
-6.600
-2.517
2.067
8.177
We now need to calculate the volume.
We could represent our answers as a polygon as displayed on the following page
Longitudinal Section – Grid 1 Longitudinal Section – Grid 2
Longitudinal Section – Grid 3 Longitudinal Section – Grid 4
Longitudinal Section – Grid 5
9
As there are an odd number of cross sections we again can use Simpson’s Rule to determine the
volume.
= D ÷ 3 x (X0 + 4X1 + 2X2 + 4X3 + 2Xn4 + 4X5 + X6)
=5 ÷ 3 x (-7.580 + 4 x -6.600 + 2 x -2.517 + 4 x 2.067 + 8.177)
= -37.615m2
If there were an even amount of spot heights the next best options would be to use the Trapezoid
rule to finish of the calculation as shown on the next page.
10
= D x ((X1 x 0.5) + X2 + X3……..+ Xn-1 + (Xn x 0.5))
=5 x ((-7.780 x 0.5) + -6.600 + -2.517 + 2.067 + (8.177 x 0.5)
= -34.258