Surveying for 1-civil engineering

1. Surveying

2. Linear measurement

3. Surveying: it is the art of determining the relative positions of different objects
on surface by measuring the horizontal distance between them
Using of surveying:-
1-to prepare topographical maps.
2-to prepare engineering maps.
3-to prepare military maps.
4-to prepare contour maps.
5-to prepare geological maps .

4. Classification of surveying :-
Primary class. Secondary class.
Plane surveying Based on instrument
Based on method
Geodetic surveying Based on object
Based on nature of field

5. Linear measurement:-
The measurement of this linear
horizontal distance between two
points on the earth surface is known
as linear measurement.

6. There are three main methods of
determined the distance on earth:-
-Direct measurement
-Computative
-E.D.M (electromagnetic-distance-measuring method)

7. Direct measurement instrument for
measuring:-
- Tapes
-Chains
-Arrows
-Offset rods
-Ranging rods poles
-Pegs

8. Common instrumental errors in linear
measurement:
–Length other than standard:-
Tape manufacturers do not guarantee 100 ft steel tapes to be exactly 100.00 ft.
–Temperature other than standard:-
A temperature higher or lower will change the length of the tape. (Steel tapes are standardized at
20°C )
–Tension:-
If tension is greater than standard the tape will stretch.
–Sag:-
A tape not supported along its entire length will sag. By applying correct tension the sag can be
reduced.

9. Poor alignment:-
This error occurs when one end of the tape is off-line or there is an obstruction in-
line.
Faulty marking:-
This error is random as the result of incorrect placement of chaining pins.

10. Correction for standard length:
CL=((l-l’)/l’)*L
CL : Correction applied to recorded measurement.
l:the actual tape length.
l‘ : the nominal tape length.
L:the recorded measurement.

11. Correction for slope:-
–When you take a measurement with a tape along an inclined plane (along the
natural slope of the ground),obviously, the taped distance is greater than the
horizontal distance. This taped distance is represented by (s) in the figure
.The difference between the slope distance and the horizontal distance (s – d)
is called the slope correction. This correction is always subtracted from the slope
distance. To compute for the slope correction, you should know either the
vertical

12. Ch=s(1-cosØ).
d²=s²-h²
so, h²=s²-d²
h²=(s+d)(s-d)
(s-d)=h²/(s+d)
at Ø is very small
s+d=2s
h²=2s(s-d)
since. Ch=s-d from the figure.

13. where:
Ch : the slope correction
s : slope distance
d: horizontal distance
h: difference in elevation
Ø: angle of slope

14. The Scale

15. The scale : Scale is a fixed ratio that every distance on
the plan bears with corresponding distance on the ground.
Scale can be represented by the following methods:-
1)Engineering Scale:
One centimeter on the plan represents some whole
number of the meters on the ground.
EX: 1cm : 10m

16. 2)Representative Scale :
one unit in the plan represents some number of same units of
length on the ground
the scale is represented as R.f or representative fraction.
This ratio of the map distance to the corresponding ground distance is independent of the
measurement units.
Ex: if the scale is 1cm = 50m;
So , R.F = 1 / (50 x 100 ) = 1/5000

17. 3)Graphical scale
Which can be divided into :
-- linear scale : used when the number of divisions less than or equals to 5
--diagonal scale :used when the number of divisions more than 5
No. of div. = Mean unit / less count

18. Example for the diagonal scale:

19. Bearings
Define
Fore bearing (F.B)
Back bearing (B.B)

20. Compass
this article is about the direction finding
instrument used in navigation. shows the
directions North corresponds to 0°, and
the angles increase clockwise, so east is
90° degrees, south is 180°, and west is
270°.
These numbers allow the compass to
show the bearings

21. Bearing
Whole circle bearing Quadrant ( reduced )
bearing

23. Methods of solving bearing
Example of mean method
Line FB BB Diff Correction
(FB)
Correction
(BB)
Corrected
(FB)
Corrected
(BB)
AB 180 20 001 05 179 15 000 22 30 -000 22 30 180 42 30 00 42 30
BC 111 45 291 45 180 000 000 000 000 111 45 291 45
CD 340 50 160 180 50 -00 25 00 25 340 25 160 25
DE 330 35 150 05 180 30 -00 15 00 15 330 20 150 20
EA 287 106 181 -00 30 00 30 286 30 106 30

24. Example of local attraction
method
line FB
(angle)
BB
(angle)
Diff Correction
(FB)
Correction
(BB)
Corrected
(FB)
Corrected
(BB)
AB 29 205 30 176 30 -01 30 02 27 30 207 30
BC 65 12 245 42 180 30 02 01 30 67 12 247 12
CD 117 06 298 36 181 30 01 30 00 00 118 36 298 36
DE 135 18 315 18 180 00 00 00 00 135 18 315 18
EF 150 12 329 12 179 00 00 01 150 12 330 12
FG 204 36 25 12 179 24 01 00 24 205 36 25 36
GH 277 24 97 180 24 00 24 00 48 277 48 97 98
HA 337 42 160 177 42 00 48 -01 30 338 30 158 30

25. •True and magnetic north

26. Traverse

27. Traverse
Types of traverse :
1- closed traverse 2- connected traverse 3- open traverse

28. Angles :
Summation of interior angles = (2n-4)*90
Summation of exterior angles = (2n+4)*90
Correction angle = - Error / n where n is No. of angles
Corrected angle = correction angle + angle before correction

29. Open Traverse
Departure
•EB= EA+L(AB)sinα(AB)
Latitude
•NB= NA+ L(AB)cosα(AB)
E
L(AB)cosα(AB)
L (AB) sin α (AB)
N
B
A

30. Correction of departure = - ( total error in dep. / perimeter of traverse ) * Length
of that line
Correction of latitude = - ( total error in Lat. / perimeter of traverse ) * Length of
that line
Linear closing error = √(dep)^2+( lat. )^2
The accuracy ratio = the linear closing error / perimeter of traverse

31. Example :

32. Answer :
•Summation of angles = (10-4)* 90 = 540
A+B+C+D+E = 101 03 19 + 101 41 49 + 102 22 03 + 115 57 20 + 118 55 29 =540
FB of (AB) = 75 05 30  BB of (AB) = 225 5 30
FB of (BC) = 225 5 30 – 101 41 49 = 153 23 41  BB of ( BC ) = 333 23 41
FB of (CD) =333 23 41 – 102 22 03 = 231 1 38  BB of ( CD ) = 51 1 38
FB of (DE) = 51 1 38 – 115 57 20 + 360 = 295 4 18  BB of ( DE ) = 115 4 18
FB of (EA) = 115 4 18 – 118 55 29+360 = 356 8 49  BB of ( EA ) = 176 8 49
FB of (AB) = 176 8 49 – 101 03 19 = 75 05 30

33. NELat.Dep.Lat.Dep.Lat.DepBearingLengthLineST
486.686749.981
+13.314+50.019-3.952*10^-3-4.256*10^-3+13.318+50.02375 05 3051.766AB
A
500800
-24.096+12.066-2.057*10^-32.216*10^-3-24.094+12.068153 23 4126.947BC
B
475.904812.066
-23.318-28.823-2.830*10^-3-3.048*10^-3-23.315-28.82231 1 3837.070CD
C
452.586783.243
+14.952-31.97-2.695*10^-3-2.902*10^-3+14.955-31.967295 4 1835.292DE
D
467.538751.273
+19.148-1.292-1.465*10^-3-1.578*10^-3+19.149-1.29356 8 4919.192EA
E
0 00 00.0130.014170.267Sum
CoordinateCorrectedCorrection

34. area

36. A= n ∑O / n+1
Average rule :

39. Levelling
In SURVAY

40. Definition
–An art of determining the relative height of
different points above or below the surface

41. Object of levelling:
1)is to find the elevation of given point with respect to
some assumed reference line called datum
2)to establish point at required elevation with respect
to datum

42. Important Terms
Datum: an arbitrarily assumed level surface or line with reference to which level of
other line or surface are calculated.
Reduced level: height or depth of a point above or below the assumed datum.
Bench mark (B.M.) : a fixed reference point of known elevation.
Mean sea level (M.S.L.): obtained by making hourly observations of the tides et any
place over a period of 19 years.

43. Types of Levelling :
(I ) Direct Levelling
* Simple levelling
* Differential levelling
*Fly levelling
* Precise levelling
*Profile levelling
*Reciprocal levelling
(II) Indirect or Trigonometric Levelling: By measuring vertical angles and horizontal distance; Less precise.
(III) Stadia Levelling: Using Tachometer principles.
(IV) Barometric levelling: Based on atmospheric pressure difference

44. Simple levelling:
When the difference between two nearby points is
required Then simple levelling is performed.
Assume that the datum at point A of elevation 0.00
Datum 0.00 Elev.
A
B

45. Differential levelling
Differential levelling is adopted when :
(i) the points are at a great difference apart.
(ii) the difference of elevation between the points is
large.
(iii) there are obstacles between the points.

46. Instruments for levelling
Level
Level staffs: scales on which these distances are measured
Two types of staffs 1)self reading staff
2)target staff

47. Basic components of level
1)telescope:to provide a line of sight.
2)level or bubble tube: to mark line of sight horizontal.
3)levelling head:to bring the bubble of tube level at the center of its run.
4)tripod:to support the above three parts of the level.

48. Basic components of level

49. Adjustments of a level
– Temporary adjustment: are performed at every
setup of instrument
– setting up of level
– levelling of telescope
– focusing of the eye piece
– Focusing of object glass
– Ex: setting up the level 1)fixing the instrument on
tripod
– 2) levelling the instrument approximately by tripod

50. Permanent Adjustment
– The following relationship between the lines are
desirable
– The line of collimation should be parallel to the axis of
the bubble.
– The line of collimation should coincide with the axis of
the telescope.
– The axis of the bubble should be perpendicular to the
vertical axis. That is, the bubble should remain in the
central position for all the directions of the telescope.

53. –There are two methods for obtaining the
elevations at different points : -
–1)Height of instrument (or plane of collimation)
method.
–2)Rise and fall method.

54. Height of instrument method:
The basic equations are
Height of instrument for the first setting=RL of BM + BS (at
BM)
Subtract the BS and FS from HI to get RL of intermediate
stations and change points
Checking: ΣBS - ΣFS = Last RL – First RL. This is -ve for fall and
+ve for RISE

56. Rise and Fall method:
Compute all rises and falls
Start at a BM with known RL
To get RL of next stations: add rise to previous RL, or
subtract fall from previous RL
Repeat for all subsequent stations.

58. Two – pegs test :
– Place two pegs about L=30 m (or 40m) apart.
– Set up level midway between the two pegs.
– Read staff on each peg, and calculate true height difference(ΔhT).
– Move level about L/10=3m (or 4m) beyond one of the pegs.
– Read staff on each peg again, and calculate height difference (ΔhA).
– If ΔhA = ΔhT then the instrument is ok.
– If not, then the error is Σ=(s1-S2)-(S3-S4)/L Mm/m

62. Cut
and
Fill

63. Cut : the removal of soil or rock from its natural
location
Fill : the placement and compaction of layers of
earth or rock to form a roadbed of the
planned shape, density, and profile grade

64. Cut and Fillcalculation
Such
Railwayroad buildings

66. A=[(w+2s+w)/2 ] * H
A=(w + s) * H= (4+3.71)*3.71= 28.60 m
To calculate area
at chainage 0
H=3.71 m “height of Cut”

67. To calculate area
at chainage 8
H=3.44 m “height of Fill”
S=H as slope = 1:1
A=[(w+2s+w)/2 ] * H
A=(w + s) * H = (4+3.44)*3.44= 25.59m
To calculate area
at chainage 16
H=3.40 m
S=H as slope = 1:1
A=[(w+2s+w)/2 ] * H
A=(w + s) * H= (4+3.40)*3.40= 25.16 m

68. To calculate area
at chainage 32
H=0.5 m
S=H as slope = 1:1
A=[(w+2s+w)/2 ] * H
A=(w + s) * H = (4+0.5)*0.5= 2.25 m
To calculate area
at chainage 32
H=0.35 m
S=H as slope = 1:1
A=[(w+2s+w)/2 ] * H
A=(w + s) * H = (4+0.35)*0.35= 1.52 m
To calculate area
at chainage 32
H=1 m
S=H as slope = 1:1
A=[(w+2s+w)/2 ] * H
A=(w + s) * H= (4+1)*1= 5 m

69. Volume
Volume= (( A1 +A2)/2 )*L
L

71. 0
5
10
15
20
25
0 50 100 150 200 250 300 350 400 450
Y-VALUES
400350300250200150100500
14.414.31519.815.21615.715.916.5GL
1615.87315.75015.62515.515.37515.2515.12515DL
0.6250.950.7751.9H.cut
1.61.5750.3500.8250.3H.Fill
7.034.98.9519.5A cut
21.128.0718.6259.613.188.0718.6259.613.18A fill
Total volume = -1125.95 m^3 (-ve : means we need this amount of soil)

72. Contouring

73. Contouring
Contouring is the science of representing the
vertical dimension of the terrain on a two
dimensional map. We can understand contouring
by considering a simple example.

74. Example:
–Let us assume that a right
circular cone of base 5m
diameter and vertical height 5m
is standing upright on its base.
Let the base be resting on a
horizontal plane at zero level as
shown in the Figure 1.

76. Contour Line :-
A Contour line is an imaginary outline of the
terrain obtained by joining its points of equal
elevation. In our example of the cone, each
circle is a contour line joining points of same
level.

77. Contour Interval (CI) :-
Contour interval is the vertical difference between the
levels of consecutive contour lines on a map. The contour
interval is a constant in a given map. In our example, the
contour interval is 100m.

78. Horizontal Equivalent (HE) :-
Horizontal equivalent is the horizontal distance between two
consecutive contour lines measured to the scale of the map.

79. What are The uses of contour?
i) The nature of the ground and its slope can be estimated
ii) Earth work can be estimated for civil engineering projects like road works,
railway, canals, dams etc.
iii) It is possible to identify suitable site for any project from the contour map of the
region.
iv) Inter-visibility of points can be ascertained using contour maps. This is most
useful for locating communication towers.
v) Military uses contour maps for strategic planning.

81. Methods of Contour Surveying
There are two methods of contour surveying:
************************************************
Direct method
Indirect method

82. Direct Method
* Marking the points of the same elevation and plotting
these points on plan
*Accurate but slow
Small area*

83. Procedure :
* a temporary B.M is established
* Setting up the level (in such a position so that the maximum number of points can be
commanded from the instrument station.)
* The height of instrument is determined
* The staff reading required to fix points on the various contours is determined by
subtracting the R.L. of each of the contours from the height of instrument.

84. Example
–If the height of instrument is 82.48m., then the staff readings required to locate
82, 81 and 80m contours are 0.48, 1.48 and 2.48m respectively. The staff is held
on an approximate position of point and then moved up and down the slope
until the desired reading is obtained. The point is marked with a peg

85. Indirect Method
1-

86. –2- By Cross- Sections:
–This method is most suitable for the survey of long narrow strips such as a
road, railway or canal etc.