BUILDING SITE SURVEY AND SET OUT

 Two Peg test Grid levels Volume calculation Rectangular base method Triangular base method MS-Excel Volume calculation Contour lines Assignment

Two Peg Test

All instruments are subject to errors. The checking of the instrument (level) is therefore important. The main error is where the line of sight is not parallel to the horizontal line of collimation. In this case your levels will not be correct. A test for checking the level is known as the two peg test. This test determines the amount of error and if an error occurs notify the technician (the level must be serviced).

Figure 1

• Establish 2 points approximately 50 metres apart on level ground as shown below. Set the level half way between the 2 points.
• Take the 2 staff readings. In our example an error will exists (line of sight does not coincide with line of collimation).

Figure 2
• Move the level as close as possible to one of the peg. (In the case above 'Peg A').
Take the 2 staff readings again.
• If the difference in height is the same the level is okay. If not, as shown in the example above, the instrument needs to be serviced.

Grid levels

Grid leveling is used for site investigation, for drawing contour lines and for the easy calculation of volumes.

The opposite figure shows a typical survey of a site using grid point levels. The area of the site is divided into a number of squares 5 × 5 metres (triangles or rectangles can also be used) and levels are taken at corner points. The grid levels enable us to calculate the volume of material above or below a certain reduced level (RL) and to draw contour lines. In the example one 12.9 m contour line is shown.

Figure 3

The lowest level of the site is RL = 10.000 m and the highest is RL = 15.831 m.
Volume calculation from spot heights - Grid leveling

Isometric view of a building site 20 × 20 m

To determine volumes a grid level system is superimposed on a Lot. Figure 4 shows a grid level system. In the figure the grid interval is equal in both directions. However, whether the interval is square or rectangular depends on the contour and the lot. The area of a grid element in Figure 4 is 5 m by 5 m (25 m²).

A reference RL must be defined to calculate the volume of the truncated prism A, B, C, etc. The volume of a grid element (A, B, C etc). is the area of the grid element multiplied by the average oft the four corner heights. To calculate the volume correctly each prism is to be considered on its own. Figure 4 below shows the site in isometric view. As can be seen the horizontal plane (green rectangle) at RL = 12.900 lies between the RL = 10.000 and RL = 15.831. If a house is placed on this level then cut (blue area) equals the fill (green area). However, if the FGL (finished ground level) is at RL  10.00 then all soil of the site need to be removed and if RL  15.831 soil must be transported to the site.

Figure 4

The two grid elements F and R are truncated prism of which the area is 5 × 5 metres and the height of the prisms is to be the mean of the four corner heights. Therefore for the volume of the prism F is:

The volume in the opposite equation is calculated for a level above RL of 10 metre. The average RL numbers are therefore reduce by 10 m. The volume of prism F is therefore 24.527 m³

Rectangular Base Method

However, the calculation method of the privious example can be simplified. To calculate the volume of each separate prism would take considerable time. Instead of calculating each prism we can simplify the calculation by taking the average of all levels.
As can be seen prism F is surrounded by other prisms and each corner must be counted four times. The corners of the prism R on the other hand must be counted one (corner), two (at the boundaries) and four (internal corner) as shown in Figure  4.

If you consider the whole lot (20 m x 20 m) and take an average of all the grid points (25 RL's) the volume quantity would be incorrect. Instead of calculate 16 single prism we could record a total of 64 points (RL's) with the number occurring at each intersecting grid point as shown in Figure 4. Each corner RL occurs 1 time, all other RL's on the boundary 2 times and all internal RL's 4 times.

Then the RL's need to be multiplied by the times occurring on the grid points. Both columns added up and the sum of the levels is then divided by the sum of the points as shown in Table 2.
A MS-Excel program can quickly do the work for you. See below for additional information and a complete calculation example is shown in Table 2.

 Figure 5 Table 1
Table 1 has 4 columns which are required to calculate the mean height . The procedure for volume calculation is shown below in Step 1 to Step 3. If you want you can select additional columns as shown in Table 2 to suit your own design.

Triangular Base Method

This is even a more accurate calculation method. Triangular elements can better define the surface because three levels define a plane. (A desk with three legs cannot wobble which happens with four leg tables.) In this method the volume is calculated of prisms with triangular bases. The calculation process is similar to the above method but with a higher number occurring at each intersecting grid point. Each corner RL occurs 1 or 2 times, all other RL's on the boundary 3 times and all internal RL's 6 times with a total of 96 occurrences.

Figure 6

MS - Excel Volume calculation

At each grid point a level reading is taken and recorded in the rise & fall sheet. The RL's are then recorded in the column shown in Table 1

The work of getting the average of the spot levels can be simplified by the use of a suitable table as illustrated below:

Step 1
Sum up the column 5 & 6 and find the mean height (MH) by dividing the sum of the levels by the sum of the number of common points.
```	MH = 824.999/64
= 12.891```
Step 2
Find the difference between the mean height (MH) and the proposed finish ground level (FGL).
a) the difference can be positive (+ve) or negative (-ve).
b) if the mean height is greater (+ve) than the finish ground level (FGL) then it is a cut.
c) if the mean height is smaller (-ve) than the finish ground level (FGL) then it is a fill .

MH < FGL = FILL
MH > FGL = CUT
MH = FGL then CUT = FILL

Step 3

Multiply the difference (from Step 2) with the area of the lot to find the required volume.

In the isometric view the finish ground level is: FGL = 12.900
Therefore the mean height from Step 2 minus the given RL MH 12.891-12.900=—0.009

The MH (12.891) < FGL (12.900) therefore some fill is required. The fill that is required is minimal because the FGL ~ MH. The amount of fill required is —0.009 × 400 = 3.6 mē
Contour lines

Contour lines are an essential part of the assessment and this subject matter needs to be clearly understood. Contour lines are lines drawn on a plan connecting points of equal elevation. If you walk along a contour line you neither gain or lose elevation. The contour interval is constant for each plan. The interval will be indicated for each exercise or assessment. Depending on the slope of the lot, intervals can be 100 mm, 200 mm or 500 mm.

Contour lines show the vertical dimension (the third dimension) of the ground on site plans. The vertical distance separating contour lines gives an indication of the steepness of the slopes. A few simple rules for contour lines will be helpful in interpreting the vertical dimension of a building site.

In general gentle slopes are represented by widely spaced contour lines.
Steep slopes are represented by closely spaced contour lines.

The contour lines shown in Figure 7 spacing at 500 mm interval.

Figure 7

There are two methods to plot contour lines:      a) by estimation      b) by calculation

a) is the quickest method of plotting the contours. Estimate by visual inspection the position of a contour between two adjacent spot levels.

b) is more accurate if the ground between the RL's is nearly straight. Similar triangle rule is then used to calculate the contour line.

Method a)
Contours lines are at 500 mm interval, that means there are two contour lines between grid point 21 (RL 10.541) and grid point 16 (RL 11.687) the 11.000 and 11.500 respectively. The distance between the two RL's is 5 metres. By estimation the 11.000 contour line would be approx. 0.4 × 5 m 2.000 m from RL 10.541 and the 11.500 contour line would be close to RL 11.687

Method b)
We use similar triangles to to find the intersection of the contour lines on the level grid. For example the difference between RL 10.541 (grid point 21) and RL 11.687 (grid point 16) is 1.146 metres. The difference between contour line 11.000 and RL 10.541 is 0.459. This figure will be used to calculate the intersection of the 11.000 m contour line. As the increment for contour lines is 0.500 m we add 0.500 + 0.459 = 0.959 to find the 11.500 contour line intersection. 0.459 and 0.959 is now used to find the distance d1 for the 11.000 and d2 for the 11.500 contour lines.

This principle of calculation is show below and should be used for all your exercises and assessments that involves drawing contour lines.

Using similar triangles:

Using the above equation the actual calculation for the contour lines 11.000 m and 11.500 m is:

 Sometimes you know the distance for a specific point on a grid system and want to know the reduced levels (RL's) that belongs to it. In this case use the formula to the right to find the value that need to be added to the reduced level (RL). RL-Calculation

Example:

Determine the RL at the distance of 2.003 m from point 21 (RL 10.541). The distance between grid point 21 and 16 ( RL 10.541 and 11.687) is 5 metre. Using the above equation we can calculate the number that must be added to the RL of 10.541.

 adding 0.459 to RL 10.541 equals to RL 11.000 m, which is also the 11.000 m    contour line height.

Assignment (Volumes and contour lines)

The opposite figure shows a building site.

The lot is 50 × 40 metres and has spot levels at 10×10 metre grid points as indicated.

1) For this site draw all contour lines at 1 metre interval. The distance on the grid lines must be calculated using the formula for similar triangles.

2) Find the reduced level to establish a finish ground level (FGL) by utilising the cut and fill method (cut equals fill)

3) How much soil is required if the RL of the building pad (FGL) is to be 16.300. Calculate the volume.

If submitted to the lecturer the correct answers will be emailed to the student

Figure 8