SURVEY NOTES N5

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BUILDING & STRUCTURAL

SURVEYING

N5

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CHAPTER 1

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1. BASIC PRINCILPES

1.1 Terminology

1.1.1 Surveying

Surveying is the art of taking measurements upon the surface of the earth either

in the horizontal or vertical plane, the results are shown in the formof a map or plan or as calculated figure. The plan is later set out on the ground

1.1.2 Plane surveying

Plane surveying is the surveying whereby the are to be surveyed is small that

the curvature (shape) of the earth is not taken into account 1.1.3 Gravity

Gravity is the force the keeps the earth in equilibrium, keep us on earth and

gives us the sense of balance. It acts towards the centre of the earth

1.1.4 Topography

Topography is a survey done to locate the main natural and artificial features

earth surface, e.g hills, rivers, roads, buildings etc

1.1.4.1 Planimetry

It is the representation, in plan, of the natural and man made features

1.1.4.2 Relief

It is the indication, in plan, of variations in elevation of the

surface of the land. Relief maybe shown in the following methods a) Colour layering b) Shading c) Contours d) Form lines 1.1.5 Contour line

It is an imaginary line which links up a series of points of the same level on the

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1.1.6 Reduced level

It is the calculated height of a point above or below a datum as deduced from

survey observation 1.1.7 Change point

Change point is the staff position at which a foresight reading was taken and

later a backsight was taken 1.1.8 Invert level

It is the level of the inside bottom surface of the pipe

1.1.9 Field book

It is the book in which all the details of the survey are recorded by the surveyor

1.1.10 Chainage

It is the distance to a point along a survey line, even though one talk of taping a

line than say chaining a line

1.1.11 Booking

It is the entering of all survey information and measurements in the field book

1.1.12 Setting out

Is the transfer of details from a drawing to a piece of ground 1.1.13 Zenith angle (Zenith distance)

It is the vertical angle measured from the vertical line (zenith line) downward

1.1.14 Offset

It is a distance measured at right angle to a chain line to some feature of the site

such as a tree, building etc 1.1.15 Height

It is the vertical distance of a feature above or below the datum or reference

surface

1.1.16 Optical square

Is a hand held instrument used to set out right angles in a building site

1.1.17 Dumpy

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1.1.18 Instrument

It is a common site name given to a surveying instrument on a tripot

1.1.19 Distometer

It is an used together with a theodolite, it measures distance 1.1.20 Tri-beacon

It is the highest point of known height above sea level and of known co-ordinates

1.1.21 Grid

It is the representation on a map, of a system of equally spaced straight parallel lines

to Y and X of the co-ordinate system

CHAPTER 2

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CHAIN SURVEYING

2. CHAIN SURVEYING

2.1 METHODS OF FIXING A POINT

2.1.1 Offset method (rectangular)

Measure the perpendicular distance DC and the length of AD and BD

C

A B

D

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Measure angle A and the length of AC

C

A B

2.1.3 Intersecting arcs

Measure the length of AC and BC C

A B

2.1.4 Triangulation (forward Intersection)

Measure angle A and angle B

C

A B

2.1.5 Trilateration (Resection)

Measure the angles formed at P by the rays to three known points A , B and C

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a

P b B c

C

2.2 Field problems

2.2.1 Ranging and measuring over a hill

(obstacle to line of sight but not to measurement)

C D

A B

Step 1 Rod holders C and D position themselves so that D can see A, and C can see B

Step 2 C directs the rod holder of D into line between C and B

Step 3 D then directs the rod holder of C into line between A and D

Step 4 This will continue until no further movement is possible and they all are in line

2.2.2 Measuring around a pond / Building

(obstacles to measurement but not to sight) C D

A B E F a) Bring the survey line to from A to B

b) Set off line BC at right angle to AB c) Set off line CD at right angle to BC d) Set off line DE at right angle to CD e) At E set off line EF at right angle to DE

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2.2.3 Measuring distance across a river

(obstacles to measurement but not to sight) B

A D C

F E To find distance AB

a) Set off line AC at right angle to AB b) Find D midpoint of AC

c) Set off line CE at right angle to AC

d) Move along line CE until point B and D are in line e) This will be point F

Distance AB = CF

2.3 Methods of taping

2.3.1 Surface taping

Measuring with the tape laid on the ground and fully supported by the ground

2.3.2 Catenary taping

Measuring with the tape suspended clear off the ground

Used when surface is very bad and involves the removal of grass, shrubs etc along

The line of suvey

2.4 How step chaining is conducted on site

1. When step chaining is conducted the tape should be stretched out horizontally from point A

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2. The distance is limited to the tape being held comfortably at waist height

3. To obtain a true horizontal a plumb bob is held on the tape reading and at the same point a

peg is put on the ground.

4. The zero point of the tape is place on the second peg and the horizontal distance measured

using the waist height as our level.

5. This is repeated up until point B is reached

2.5 How the 3 – 4 – 5 method of setting out a right angle is

applied

1. Three people are needed

2. from the point at which the perpendicular line is required the tape is held at zero and stretch

along the existing straight line for 4m and held by the second assistant at that point

3. The third assistant pulls the tape at an acute angle towards the first assistant holding the tape

at 7m

4. The remainder of the tape is held by the first assistant at 12m and the whole system is

Stretched to form a right angle triangle

2.6 Five requirements to obtain sufficient accuracy when

taping

1. Tape must be held horizontal

2. Tape must be held on its correct zero mark 3. The correct tension must be applied to the tape 4. Remove all kinks

5. Tape must be held on correct pegs 6. View tape vertically over the peg 7. Measure the centre of ranging rod

2.7 Care of steel tape

1. Pull tape straight in the direction in which it is curled 2. Ensure that no kinks are present

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4. Wipe tape with oily cloth after use to prevent rust 5. Do not allow vehicle to run over the tape

2.8 CHAINAGE CORRECTIONS

2.8.1 Constants ( length of tape too short or too long) GTL ATL MD CD = × CD = correct distance

MD = measured distance (length) ATL = Actual tape length

GTL = Graduated tape length

2.8.2 Temperature C=L×e(Tm −Ts)

C = Temperature Correction L = Measured distance (length) e = Coefficient of linear expansion Tm = Measured temperature Ts = Standard temperature 2.8.3 Sag 2 ₂3 24T L w C = for bay 3 24 3 2 3 2 ×       × = T L w C

For Three bays C = Sag correction

w

= Mass of the tape in kg/m L = Measured length (distance) in m T = Measured tension in kgF

2.8.4 Slope

To calculate the horizontal distance To calculate the slope distance

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C=L(1−cosθ) ) 1 (sec − =H A C

Sc = Slope correction SC = Correction

L = Measured length of the slope H = Measured horizontal distance

θ = Slope angle θ = Slope angle

2.8.5 Height at sea level R H L C = × CH = Correction

L = Measured length (distance) H = Height above or below sea level

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OBSERVATION ANGLE

Z = Zenith distance

θ

= Angle of elevation

Vertical Zenith line of sight Zenith distance Z Angle of elevation θ Horizontal

Z = 90º – θ

θ = 90º – Z

Zenith distance Z θ Angle of depression line of sight

Z = 90º + θ

θ = Z – 90º

NB: ALWAYS use angle θ which represents the angle of

Elevation or Depression

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EXAMPLE 1

QUESTION

Calculate the reduced horizontal distance, if the measured distance is 87,281 m at a slope if 86º 22’ 44”

the distance is measured is three bays at a temperature of 16 ºC Standard temperature is 20ºC

Co-efficient of expansion is 1,13 x ₁₀−5_/ºC

Tension 65N

Mass of the tape is 0,015 kg/m Earth radius is 6373 km

Height above sea level 2 280,544 m

SOLUTION

Given : ts = 20ºC e = 1,13 x 10−5/ºC T = 65N w = 0,015 kg/m R = 6373 km H = 2280,544 m L = 87,281 m θ = 90 – 86º 22’ 44”= 3º 37’ 16” tm = 16ºC ) (Tm Ts e L C = × − ) 20 16 ( 10 13 , 1 281 , 87 _× _× 5 ₋ = − C 0039 , 0 − = C 3 24 3 2 3 2 ×       × = T L w C 3 ) 5 , 6 ( 24 3 281 , 87 015 , 0 2 3 2 ×       × = C ( 65 N = 10 65 = 6,5 kgF ) 0164 , 0 = C ) cos 1 ( − θ =L C

)

"

1 6

'

3 7

3 c o s

1 (

2 8 1

,

8 7

−

°

=

C

1743 , 0 = C _m

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R H L C = × 6373000 544 , 2280 281 , 87 × = C 0312 , 0 = C _m CD = 87,281 – 0,0039 – 0,0164 – 0,1743 – 0,0312 = 87,055 m

NB: CD = reduced distance or correct horizontal distance

EXAMPLE 2

QUESTION

Calculate the slope distance if the measured correct horizontal distance to be measured is 98,285 and the slope is 2º 22’ 15”

SOLUTION

) 1 (sec − =H A C

)

1 "

1 5

'

22

2 (sec

2 85

,

9 8

°

−

=

C

A

_{c o s}1

s e c

=

0842 , 0 = C _m Slope distance = 98,285 + 0,0842 = 98,369 m OR Slope =

"

1 5

'

2 2

2 co s

2 8 5

,

9 8

°

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= 98,369 m

EXERCISE 1

Question 1.1

The standard temperature of a tape is 14ºC and the coefficient of expansion is 0,000012/ºC

What is the reduced horizontal distance, if the measured distance is 32,40 m on a slope of

5º 40’ and a temperature of 30ºC

Answer 32,248 m

Question 1.2

Given: Length of tape = 100m Standard temperature = 20ºC

Coefficient of expansion = 1,13

×

10–5_/ºC

Tension = 7 kg.f

Mass of tape = 0,015 kg/m Radius of the earth = 6,373 km

1.2.1 Calculate the reduced horizontal distance, if the measured distance is 94,151 m

at a slope of 95º 15’. The distance is measured in three bays at a temperature of

14ºC Answer: 93,732 m

1.2.2 Calculate distance at sea level if the height was 2 017,443 above sea level,

use the distance in Question 2.1. Answer: 93,702 m

Question 1.3

A line A-D was measured in three sections A-B 90,288 @ slope 3º 44’ 20”

B-C 72,408 @ slope 4º 32’ 59” C-D 47,652 @ Slope 2º 09’ 07”

Calculate the horizontal distanceA-D Answer: 209,459 m

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1.4.1 M to R is a distance to be pegged (as viewed on plan)

M to R lies on a slope of 0º 34’ 23” Calculate the distance on the slope M to R to be measured to obtain the correct horizontal distance of 1000 m

Answer: 1000,050 m

1.4.2 Determine the horizontal distance of a measurement of 136,57 m that was done on a

Slope of 3,64%

Answer: 136,479 m

Question 1.5

A measured distance of 164,81m was measured from T1 to T2 the reduced distance is 160,49 m

Calculate the angle of the slope and the zenith distance

Answer: 13º 8′ 50,8″ ; 76º 51′ 7,24″

Question 1.6

A steel tape is used to measure a baseline A-B. Each measurement is done in THREE bays

Calculate the correct baseline distance. Give the formula for each of your calculations A-B 94,01 m /96º 10’ @ 35ºC 83,14 m / 87º 30’ @ 11ºC Length of tape = 100m Standard temperature = 20ºC Coefficient of expansion = 0,0000112/ºC Tension = 70 N Mass of tape = 0,015 kg/m Answer: 176,504 m Question 1.7

A 100 m steel tape was compared with a 100 m standard base. The length of the base as

read on the tape was 99,997 m at a temperature of 22,4ºC. The coefficient of linear

expansion is 0,000014/ºC. Calculate the standard temperature of the tape.

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CHAPTER 3 LEVELLING

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3. HEIGHT MEASUREMENT ( REDUCED LEVEL)

3.1 RISE AND FALL METHOD

Poin

t BS IS FS Rise Fall Reduced level

A 2,50 50,00 B 2,00 0,50 50,50 C 2,25 0,25 50,25 D 1,75 1,80 0,45 50,70 E 1,60 0,15 50,85 4,25 –3,40 0,85 3,40 1,10 –0,25 0,85 0,25 50,85 –50,00 0,85 Steps to follow:

a) Enter the backsight, intermediate sight and the foresight in the appropriate column

in different rows. Except that at change point the FS and the BS are entered in the

same row ( as in row D)

b) The first reduced level is the height of the O.B.M or other datum which has been

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c) If the IS or FS in smaller than the immediately preceding staff reading,

The difference between the two readings is placed in the rise column

Example: immediately preceding reading of 2,00 in row B is 2,50 in row A

Therefore 2,50 – 2,00 = 0,50

d) If the IS or FS in larger than the immediately preceding staff reading,

The difference between the two readings is placed in the fall column

Example: immediately preceding reading of 2,25 in row C is 2,00 in row B

Therefore 2,00 – 2,25 = – 0,25

e) A rise is added to the immediately preceding reduced level entry to obtain the

reduced level of a station.

Example: immediately preceding reduced level of row B is row A = 50,00

Reduced level of B= 50,00 + 50 = 50,50

f) A fall is subtracted from the immediately preceding reduced level entry

Example: immediately preceding reduced level of row C is row B = 50,50

Reduced level of C = 50,50 – 0,25 = 50,25

Checking on booking

1) Sum up the BS readings and the FS readings Subtract FS from the BS

Total BS = 2,50 + 1,75 = 4,25 Total FS = 1,80 + 1,60 = 3,40 Difference = 0,85

2) Sum up Rise and Fall Subtract Fall from the Rise

Total Rise = 0,50 + 0,45 + 0,15 = 1,10 Total Fall = 0,25 Difference = 0,85

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3) Subtract the first reduced level from the last reduced level

Last reduced level = 50,85 First reduced level = 50,00

0,85

NOTE:

If all their differences are the same then the

calculation are correct

NOTE:

Any entry that is underlined 3,456 or has a line

ontop 3,456

means the entry is negative

Example :

Rise and Fall method

NOTE: The numbers in brackets indicate the steps from the calculations below

Poin

t BS IS FS Rise Fall Reduced level

A 2,10

9 (A1) 130,50

B 3,20

4 (B1) 1,095 129,405(B2)

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9 131,235 D 1,28 1 (D1) 4,400 135,635(D2) E 2,05 4 0,773(E1) 136,408(E2) F 1,90 6 2,200 (F1) 4,254 132,154(F2) G 1,65 4 1,812 (G1) 3,718 135,872(G2) H 2,850 (H1) 4,504 (H2) 131,368 5,48 0 – 4,61 2 0,86 8 4,612 10,721 –9,853 0,868 9,853 131,368 –130,50 0,868

Use the table above with the steps below to understand

A1. The BM is the reduced level in A which is 130,50

B1. To find the Rise or Fall in row B

BS in row A – IS in row B = Rise/Fall

2,109 – 3,204 = – 1,095 ( since its negative it is a Fall)

B2. To find the reduced level at B

Reduced level at A – Fall in B = Reduced level at B

2,109 – 3,204 = 129,405

C1. To find the Rise or Fall in row C

IS in row B – FS in row C = Rise/Fall

3,204 – 1,374 = 1,830 ( since its positive it is a Rise)

C2. To find the reduced level at C

Reduced level at B + Rise in C = Reduced level at C 129,405 + 1,830 = 131,235

D1. To find the Rise or Fall in row D

BS in row C – IS in row D = Rise/Fall

3,119 – ( –1,281) = 4,400 ( since its positive it is a Rise)

D2. To find the reduced level at D

Reduced level at C + Rise in D = Reduced level at D 131,235 + 4,400 = 135,635

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E1. To find the Rise or Fall in row E

IS in row D – IS in row E = Rise/Fall

(–1,281) – (–2,054) = 0,773 ( since its positive it is a Rise)

E2. To find the reduced level at E

Reduced level at D + Rise in E = Reduced level at E 135,635 + 0,773 = 136,408

F1. To find the Rise or Fall in row F

IS in row E – FS in row F = Rise/Fall

–2,054 – 2,200 = – 4,254 ( since its negative it is a Fall)

F2. To find the reduced level at F

Reduced level at E – Fall in F = Reduced level at F 136,408 – 4,4254 = 132,154

G1. To find the Rise or Fall in row G

BS in row F – FS in row G = Rise/Fall

1,906 – (–1,812) = 3,718 ( since its positive it is a Rise)

G2. To find the reduced level at G

Reduced level at F + Rise in G = Reduced level at C 132,154 + 3,718 = 135,872

H1. To find the Rise or Fall in row H

BS in row G – FS in row H = Rise/Fall

–1,654 – 2,850 = –4,504 ( since its negative it is a Fall)

H2. To find the reduced level at H

Reduced level at G – Fall in H = Reduced level at H

135,872 – 4,504 = 131,368

CHECK ON BOOKING

Total BS – Total FS = 5,480 – 4,612 = 0,868 Total Rise – Total Fall = 10,721 – 9,853 = 0,868

Last R/level – First R/Level = 131,368 – 130,50 = 0,868

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EXERCISE 2

Question 2.1

Reduce the field notes using the rise and fall method. Do the necessary checks

(If your calculation are correct then the check must give an answer of –1,6)

Poi

nt BS IS FS Rise Fall Reduced level Remarks

A

2,40

TBM 150,00

B

2,00

C

1,90

D

2,80

1,40

E

2,00

F

1,30

2,60

G

0,60

3,00

H

1,70

7,10

–

8,70

–1,6

8,70

Question 2.2

(If your calculation are correct then the check must give an answer of 4,64)

Poi

nt BS IS FS Rise Fall Reduced level Remarks

A

0,49

0,27

3,29

0,39

3,77

3,72

3,59

1,1

1 3,56

0,82

BM

3,89

1,36

BM 1

025,00

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3,72

0,99

3,69

1,02

3,86

1,31

3,90

1,56

B

2,40

23,93

–

19,29

4,64

19,2

9

Question 2.3

(If your calculation are correct then the check must give an answer of 4,09)

Poin

t BS IS FS Rise Fall Reduced level

BM

1 3,14

1 401,261

A

0,08

6 B

2,11

1 4,28

3 C

1,40

6 D

2,03

2 E

3,10

8 3,63

8 F

1,82

3 2,11

0 G

3,11

1 H

4,12

3 2,10

9 I

3,28

1 BM

2 3,08

3

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3.2 COLLIMATION HEIGHT METHOD

NOTE:

1. BS, IS and FS reading are entered as in the rise and fall method 2. The first reduced level entry is the height of the OBM

3. The first BS reading which is taken with the staff held on the OBM is added to

the first reduced level to give the height of collimation and the entries are place

in the same row

4. The height of collimation is change only when the levelling instrument is moved

to a new position. The new height of collimation is obtained by adding the new BS to the reduced level at the change point

EXAMPLE

Point BS IS FS Collimation height Reduced level

A 3,16 (A2) 141,66 (A1) 138,50

B 1,95 141,66 (B1) 139,71

C 4,16 3,97 (C2) 141,85 (C1) 137,69

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NOTE: Collimation height at A is used to find all the reduced level of B and C

But since C is a change point, the collimation height change

Step to follow

A1. The reduced level entered is the OBM or TBM etc given A2 How find the collimation height at A

Reduced level at A + Backsight at A = Collimation height at A 138,50 + 3,16 = 141,66

B1. How to find reduced level at B

Collimation height at A – Inter-sight at B = Reduced level at B 141,66 – 1,95 = 139,71

C1 How to find the reduced level at C

Collimation height at A – Foresight at C = Reduced level at C 141,66 – 3,97 = 137,69

C2 How find the collimation height at C (Change point has both BS and FS)

Reduced level at C + Backsight at C = Collimation height at C 137,69 + 4,16 = 141,85

D1 How to find the reduced level at D

Collimation height at C – Foresight at D = Reduced level at D 141,85 – 1,62 = 140,23

EXERCISE 3

Question 3.1

Reduce the following notes by using the collimation height method Do the necessary checks

Poin

t BS IS FS Collimation height Reduced level

BM

2,4

0 150,00

A

2,0

0 B

1,9

0 C

2,8

1,40

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0 D

2,0

0 E

1,3

0 2,60

F

0,6

0 3,00

G

1,70

Question 3.2

Poin

BM

1 4,61

884,86

A

3,54

B

1,69

C

1,80

3,21

D

2,40

E

2,89

3,62

F

1,20

G

2,20

H

1,46

BM

2 1,20

5,70

–1,61

4,09

1,61

Question 3.3

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Poin

A

4,42

0 B

5,50

0 C

3,88

0 D

3,16

0 1,47

0 E

1,95

0 F

4,55

0 G

4,65

0 3,97

0 145,990

H

3,90

0 I

6,32

0 J

2,34

0 K

4,16

0 1,62

0 L

1,62

0 M

5,20

0

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3.3 RECIPROCAL LEVELLING

y1 y x x1 station 1 C station 2 3m D 3m

Let us assume the following staff reading

Reading taken 3m from C Reading taken 3m from D

C (y = 1,480m) D (x =

1,738m)

D (y1 = 1,852m ) C (x1 = 1,380)

One can see that all reading taken when staff was at C are larger than that of staff at D

We can therefore say a point with larger reading is lower and the one with smaller

readings is high. In this case we can say point D is lower and C is higher Therefore to find the reduced level of point use the following

1. Calculate the difference from readings taken at station C (e.g between

x

and x1)

2 Calculate the difference from readings taken at station D (e.g between y and y2

3 Calculate the mean difference between the stations 4 Calculate the Reduced level required

4.1 R/L of a lower point = R/L of a higher point – mean difference

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4.1 R/L of a higher point = R/L of a lower point + mean difference

EXAMPLE

Using the data given let, let us say that the reduced level of D is 352,710m

And we need to calculate the reduced level of C Solution

Difference between readings taken 3m from C = 1,852 – 1,480 = 0,372 m

Difference between readings taken 3m from D = 1,738 – 1,380 = 0,358 m Mean difference = 2 358 , 0 372 , 0 + = 0,365 m Reduced level of C = 352,710 – 0,365 = 352,345

EXERCISE 4

Question 4.1

4.1.1 The following readings were taken by ‘reciprocal’ levelling across a

swamp.

What is the correct elevation of point B if the elevation of A is 100,00 m Reading taken 3m from A Reading taken 3m from D

A 2,91 B 1,71

B 2,49 A 1,39

4.1.2 Make a neat sketch showing the above mentioned set ups and readings

Question 4.2 3,35 1,95 2,85 1,55 B 3m A 3m

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The above sketch shows the reading taken by reciprocal levelling across a river. If the elevation

Of B is 139,50 Calculate the Reduced level of A

Question 4.3

4.3.1 The following readings were taken by ‘reciprocal’ levelling across a

swamp.

What is the correct elevation of point B if the elevation of A is 123,50 m Reading taken 3m from A Reading taken 3m from D

A 1,50 B 3,30

B 1,00 A 3,00

4.3.2 Make a neat sketch showing the above mentioned set ups and readings

CHAPTER 4

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LEVELLING

4. TACHEOMETRY

Use of Vertical angles

MH VD B θ Δh Height of B IH Height of A Datum line

When the instrument is at a lower point

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Higher point = Lower point + IH + VD – MH

This means :

Reduced level B = height of A + Instrument height + vertical distance

– mean height θ IH VD A Δh Height of A MH B Height of B Datum line

When the instrument is at a higher point

( that means the angle given is more than 90º measured from the vertical line)

Lower point = Higher point + IH – VD – MH

This means :

Reduced level B = height of A

+

Instrument height

–

vertical distance

–

mean height

FORMULAE

1. I = TH – BH TH =Top hair BH = Bottom hair

2. VD = KI cosθ Sinθ or _{VD = 2}1 KI sin 2θ (cosθ Sinθ = 21 sin 2θ)

VD = Vertical distance

K = 100

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θ = Slope angle 3. 2 BH TH MH = + MH = Mean height 4. HD = KI cos2_θ HD = Horizontal distance 5. HD VD m= or _m₌_tan−1θ

m

= gradient

Example

Given The height of point A is 24,135m, height of instrument at A is 1,60m

TH = 3,01 BH = 2,41

vertical angle is 86º 30’

Required to calculate 1.1 The reduced level of B 1.2 The horizontal distance A-B

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SOLUTION 1.1 θ = 90 – 86º 30’ ( A is lower than B) = 3º 30’ 2 BH TH MH = + 2 41 , 2 01 , 3 + = MH MH =2,71 VD = KI cosθ Sinθ VD = 100(3,01 – 2,41) cos3º 30’ Sin 3º 30’ = 3,656m Reduced level at B = 24,135 + 1,60 + 3,656 – 2,71 = 26,681m 1.2 HD = KI cos2_θ HD = 100(3,01 – 2,41) (cos 3º 30’)2 = 59,776m

EXERCISE 5

Question 5.1

If the height of point A is 900,00m and the height of the instrument at A is 1,162m

Observation to point B TH = 3,492

BH =1,250

Vertical angle = 97º 24’ 00”

5.1.1 Calculate the Reduced level of point B 5.1.2 Calculate the horizontal distance A-B

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Answers 870,156m 220,480m

Question 5.2

A surveyor obtained the following information between two points, A and B: Height of A = 430,17m

Instrument height at A = 1,56m Stadia reading 0,97 1,43 1,89 Vertical angle = 82º 10’ 40”

5.2.1 Calculate the horizontal distance A-B 5.2.2 Calculate the Reduced level of B

Answer : 90,296m 442,705m

Question 5.3

A theodolite is set up at point P and the height of the instrument is 1,61m. The elevation of

Point Q is 1 718,440m TH = 2,14

BH = 0,93

Vertical angle = 96º 10’

Calculate the height of point P

Answer : 1 731,288m

Question 5.4

A theodolite is set up at point A and the height of the instrument is 1,20m. When sighted to a staff held at B, the upper and lower stadia lines read 2,00 and 0,80 respectively. The angle of elevation of the instrument is 6º 20’ Calculate the Reduced level of A if the reduced level of B is 104,68m

Answer : 917,23m

Question 5.5

A theodolite is set up at point C and the instrument height is 1,58m. The elevation of C is

2 015,67m. Sighting staff held at D, the following observation are recorded

STADIA CIRCLE READINGS

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2,55 1,6 0,65 98º 16’ 00” 210º 50’ 40” Calculate: 5.5.1 Vertical angle 5.5.2 Vertical distance 5.5.3 Horizontal distance 5.5.4 Height of D

5.5.6 The slope distance

5.5.7 The gradient of the slope

Answer: ---- 27,034 186,072 1988,616 188,026 0,145

Question 5.6

6.1 If the angle of elevation is of the instrument 5º 22’ 30”. Calculate the zenith distance

6.2 If the angle of depression is of the instrument 6º 25’ 32”. Calculate the zenith distance

6.3 If the vertical circle reading 97º 30’ 35”. Calculate the vertical angle and state whether

it is an angle of elevation or depression

6.4 If the vertical circle reading 82º 18’ 45”. Calculate the vertical angle and state whether

it is an angle of elevation or depression

Answer not given Question 5.7

The following information was recorded during field survey The slope distance from point C to D is 500m

The reduced level at point C is 1,284m The reduced level at point D is 12,484m Calculate the following:

5.7.1 The horizontal distance A to B 5.7.2 The slope angle at A

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CHAPTER 5 TRAVERSING

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CIRCLE LEFT

NORTH

When this circle on the left of the telescope is turned to face left The readings will be between º and 180º

Usually in the vicinity of 90º Direction of sight 360º start position 0º B 25º 40’ 05” G 289º 34’ 00” C 80º 22’ 40” 270º A 90º F 229º 20’ 10” E 186º 31’ 50” D 135º 14’ 20” 180º

CIRCLE RIGHT

NORTH

With the circle on the right of the telescope The readings will be between 180º and 360º Usually in the vicinity of 270º

180º Direction of sight Start position B 205º 40’ 25” G 109º 34’ 14” C 260º 22’ 40” 90º 270º

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F 49º 21’ 10”

E 06º 31’ 20” D 315º 14’ 30”

0º

360º

ENTRY OF THE READINGS

Poin

t Circle Left Circle Right Mean angle Correction Corrected Angle

B 25º 40’ 05” 205º 40’ 20” C 80º 22’ 40” 260º 22’ 40” D 135º 14’ 20” 315º 14’ 30” E 186º 31’ 50” 06º 31’ 20” F 229º 20’ 10” 49º 21’ 10” G 289º 34’ 00” 109º 34’ 14” B 25º 40’ 05” 205º 40’ 25”

5.1.1 How to calculate the mean observation

angle

5.1.1.1. Given the reading of

one point ONLY

When the Circle right is Smaller than Circle left add 180º to Circle right

When the Circle right is larger than Circle left subtract 180º from Circle right

Example

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128º 45’ 12” 308º 45’ 20”

Solution (Circle right is larger 308º – 180º = 128º)

This is enter as in the following table

Point Circle Left Circle Right Mean angle

128º 45’ 12” 308º 45’ 20”

128º 45’ 12” 128º 45’

20” 128º 45’ 16”

NOTE: Mean angle =

2 " 20 ' 45 128 " 12 ' 45 128° + ° = 128º 45’ 16”

5.1.1.2

Given reading from

more than one point

Example Given : Circle left Circle right A 87º 40’ 20” 267º 40’ 35”

B 110º 32’ 30” 290º 32’ 45” C 184º 14’ 20” 04º 13’ 33” D 286º 25’ 30” 106º 25’ 50” A 87º 40’ 40” 267º 40’ 55” Solution: (Use the table with the calculations below)

Circle left Circle right Mean angle Correct

ion Corrected angle

A 87º 40’ 20” 267º 40’ 35” B 110º 32’ 30” 290º 32’ 45” (1) 22º 52’ 10” (2) 22º 52’ 10” (3) 22º 52’ 10”

5 "

22º 52’ 05” B 110º 32’ 30” 290º 32’ 45” C 184º 14’ 20” 04º 13’ 33” (4) 73º 41’ 50” (5) 73º 41’ 48” (6) 73º 41’ 49”

5 "

73º 41’ 44” C 184º 14’ 20” 04º 13’ 33” D 286º 25’ 30” 106º 25’ 50”

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(7) 102º 11’ 10” (8) 102º 11’ 17” (9) 102º 11’ 13,5”

5 "

102º 11’ 8,5” D 286º 25’ 30” 106º 25’ 50” A 87º 40’ 40” 267º 40’ 55” (10) 161º 15’ 10” (11) 161º 15’ 05” (12) 161º 15’ 7,5”

5 "

161º 15’ 2,5” Total 360º 00’ 20” 360º 00’ 00” Correction =

4 "

2

0

=

5 "

STEPS TO FOLLOW

NOTE: When the number subtracting is smaller add 360º to the number B minus A 1. 110º 32’ 30” – 87º 40’ 20” = 22º 52’ 10” 2. 290º 32’ 45” – 267º 40’ 35” = 22º 52’ 10” (1) + (2) = 45º 44’ 20” 3. Mean angle =

2 "

20 '

44

44 °

= 22º 52’ 10” C minus B 4. 184º 14’ 20” – 110º 32’ 30” = 73º 41’ 50” 5. 364º 13’ 33” – 290º 32’ 45” = 73º 41’ 48” ( add 360º to the 4º) (4) + (5) = 147º 23’ 38” 6. Mean angle =

2 "

48 '

41

147 °

= 73º 41’ 49” D minus C 7. 286º 25’ 30” – 184º 14’ 20” = 102º 11’ 10” 8. 106º 25’ 50” – 04º 13’ 33” = 102º 11’ 17” (7) + (8) = 204º 22’ 27”

(45)

9. Mean angle =

2 "

27 '

22

204 °

= 102º 11’ 13,5” A minus D 10 447º 40’ 40” – 286º 25’ 30” = 161º 15’ 10” ( add 360º to the 87º) 11 267º 40’ 55” – 106º 25’ 50” = 161º 15’ 05” (10) + (11) = 322º 30’ 15” 12. Mean angle =

2 "

15 '

30

322 °

= 161º 15’ 7,5”

13. Add ALL the mean angles 14. If the answer is less than 360

Divide the difference by the number of mean angle Add the answer to each mean angle

15. If the answer is more than 360

Divide the difference by the number of mean angle Subtract the answer from each mean angle

From the example above :

In this case the total mean angles is 360º 00’ 20” which is 20” more Therefore divide 20” by 4 and the answer is 5”

We then subtract 5” from each mean angle

OTHER FORMULAE Collimation error (E)

E =

[

(

CR−180°

)

−CL

]

2 1 CL = Circle left CR = Circle right Index error E = ( CL + CR) – 360º

EXERCISE 6

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QUESTION 6.1

The following reading were obtained from a survey station “P”. Calculate the mean observed angles QPR , RPS, SPT, TPQ Check the calculations

Target

station Circle left Circle right

Q 168 : 11 : 43 348 : 11 : 42 R 17 : 11 : 50 197 : 11 : 56 S 104 : 03 : 42 284 : 03 : 43 T 105 : 04 : 43 285 : 04 : 43 Q 168 : 11 : 45 348 : 11 : 42 QUESTION 6.2

The following reading were obtained from a survey station “A”.

Calculate the mean observed angles BAC , CAD, DAE, EAF, FAG, GAB Check the calculations

Target

station Circle left Circle right

B 07 : 00 : 40 187 : 01 : 00 C 96 : 43 : 20 276 : 43 : 00 D 187 : 22 : 10 07 : 21 : 50 E 204 : 15 : 15 24 : 15 : 40 F 276 : 35 : 10 96 : 35 : 50 G 342 : 20 : 05 162 : 20 : 00 B 07 : 01 : 00 187 : 00 : 40 Question 6.3

6.3.1 A direction on a circle left observation of a theodolite was 36º 10’ 21”. The same direction on a

circle right observation was 216º 11’ 08”. Calculate the collimation error of the theodolite

6.3.2 Vertical angle on the circle left and circle right to a distant point are CL 93º 56’ 10”

CR 266º 04’ 40”

(47)

6.3.3 The following directions were observed from A to R. Reduce the mean direction

Cirlce left Circle right

96º 43’ 20” 276º 43’ 00”

6.3.4 The following directions were observed from P toQ. Reduce the mean direction

Cirlce left Circle right 7º 01’ 00” 187º 00’ 40”

5.2 CO – ORDINATES (JOIN CALCULATIONS)

SA co-ordinate system –x (180º) ((θ + 90) (θ + 180) +y (90º) – y (270º) θ (θ + 270) +x (0º and 360º) FORMULAE ∆yAB =yB −yA ∆xAB =xB −xA

Note: If given AB ( meaning from A to B) then its B minus A If given BA ( meaning from B to A) then its A minus B

Distance or length

AB = ₍_∆_y₎2 ₊₍_∆_x₎2

Direction or angle

θ =       ∆ ∆ − x y 1 tan NOTE:

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2.1 If is positive and ∆ is positive, we are in the 1 quadrant D =       ∆ ∆ − x y 1 tan

2.2 If ∆yis positive and ∆xis negative, we are in the 2nd_quadrant

D =       ∆ ∆ − x y 1 tan _{+ 90}

2.3 If ∆yis negative and ∆xis negative, we are in the 3rd_quadrant

D =       ∆ ∆ − x y 1

tan _{+ 180} _2.4 _If∆yis negative and ∆xis positive, we are in the 4th_{quadrant D =} _      ∆ ∆ − x y 1 tan _{+ 270} EXAMPLE Question

The co-ordinates of point A and B are

Y X

A – 248,17 – 58,47

B – 150,27 – 260,80

Calculate the orientated direction and distance between A and B (

Answer ( remember it’s A to B )

A B AB y y y = − ∆ ) 17 , 248 ( 150 − − − = ∆yAB = 97,9 A B AB x x x = − ∆ ) 47 , 58 ( 80 , 260 − − − = ∆xAB = −202 ,33 Distance from A to B 2 2 ₍ ₎ ) ( y x AB = ∆ + ∆ 2 2 ₍ ₂₀₂_,₃₃₎ ) 9 , 97 ( + − = AB 771 , 224 = AB m θ =       ∆ ∆ − x y 1 tan θ =       − 33 , 202 9 , 97 tan 1 θ = 25º 49′ 14,24″

Since ∆yis negative and ∆xis positive, we are in the 4th_{quadrant D =}

      ∆ ∆ − x y 1 tan _{+ 270}

(49)

∴

Direction = 25º 49′ 14,24″ + 270 = 295º 49′ 14,24″

Question

The co-ordinates of point A and B are

Y X

A – 248,17 – 58,47

B – 150,27 – 260,80

Calculate the orientated direction and distance between B and A (

Answer ( remember it’s B to A )

B A BA y y y = − ∆ ) 150 ( 17 , 248 − − − = ∆yBA = −97,9 B A BA x x x = − ∆ ) 80 , 260 ( 47 , 58 − − − = ∆xBA = 202 ,33 _m Distance from A to B 2 2 ₍ ₎ ) ( y x AB = ∆ + ∆ 2 2 ₍₂₀₂_,₃₃₎ ) 9 , 97 (− + = AB 771 , 224 = AB m θ =       ∆ ∆ − x y 1 tan θ =       − 33 , 202 9 , 97 tan 1 θ = 25º 49′ 14,24″

Since ∆yis positive and ∆xis negative, we are in the 2nd_{quadrant D =}

      ∆ ∆ − x y 1 tan _{+ 90}

∴

Direction = 25º 49′ 14,24″ + 90

(50)

= 115º 49′ 14,24″

EXERCISE 7

7.1 The co-ordinates of point A and B are

Y X

A +310 248,17 – 1 058,47

B +309 295,17 – 1 688,04

Calculate the orientated direction and distance between A and B

7.2 The co-ordinates of point A and B are

Y X

A –467,89 +120,45

B –120,45 +467,89

Calculate the orientated direction and distance between B and A

7.3 Study the information below showing point that formulate boundaries of a property. Calculate the lengths of all the side of the property

SIDES (Metres) ANGLES OF DIRECTION CO-ORDINATES Y AB BC CD DE EF FA 335 : 24 : 20 66 : 04 : 20 155 : 24 : 20 246 : 04 : 20 155 : 24 : 20 246 : 04 : 20 A B C D E F Constant: 0,00 – 53 937,40 – 53 963,29 – 53 930,38 – 53 918,31 – 53 921,96 – 53 908,23 +3 700 000,00 + 60 210,88 + 60 267,26 + 60 281,86 + 60 255,49 + 60 253,86 + 60 223,86 * 377 P5 – 54 006,63 + 60 267,44 * 378 P5 – 53 879,14 + 60 320,80

All beacons are 12 mm iron pegs

A

723 F 726 E

(51)

B 728

C Scale 1/1000

The figure A B C D E F represents 2100 square metres of land being

ERF PORTION OF ERF WALMER

Situated in the municipality and Administrative District of Port Elizabeth surveyed in June 1981

This diagram is annexed to

No Dated i.f.o

Registrar of Deeds

The original diagram is

No 7494/1981 annexed to Transfer/Grant No File No s/7902/94 S.R No Comp. BO-8CC/x43

CHAPTER 6 SETTING OUT

(52)

6.1

How a traveller is used with profiles to control excavation and foundation levels

a) Two profiles ate place at each end of the excavation.

b) A small cross bar is fixed at each profile at a level equal to the invert level of the

excavation plus the traveller (follower)

c) The depth if the excavation is therefore controlled by dipping the traveller such

that the line of sight between the two bars of the profiles is in line with that of the

cross bar of the traveller

6.2

How to set up a rectangular site along a road and use the road as a reference on the site plan

a) Offsets must be taken from the road reserve to the boundary of the site that is required

b) A baseline parallel to the road drawn.

c) The corresponding chainage of the site boundary are then identified

d) the required site shape is then set out from the baseline

6.3

On site, how to set out two points A and B stationed on a third point known co-ordinates

a) Set up the instrument and calculate the direction between the point on which the instrument

is set up and point A and the distance between the two points b) Also calculate the direction between the point on which the instrument is set up and point B

(53)

c) Orientate direction from instrument to point A and measure the distance and put a peg

d) Swing the instrument to point B using the calculated direction of point B and measure the

distance from the instrument to point B and put a peg

6.4

How to set out a rectangular proposed building using simple surveying instrument for

earthworks

Equipment: 100m tape, Four steel pegs or droppers ±2m long, Lime, Fish line,

Levelling instrument. Traveller ±2m high

Measure the distance of the proposed structure from all the four corners and make it ±1m less

From each side of the building.

Put the steel pegs or droppers on these new found points and mark them for a 2m traveller

taking the 150m depth of the top soil into consideration

The area to be removed top soil is the one that is indicated by ABCD

A B

C D

6.5 CUT AND FILL

Ground level Depth of excavation at A A Depth of excavation at B Invert level at A Δh B Invert level at B Steps to follow:

(54)

Δh = gradient x distance from first to the last point 2. Calculating the invert level of the first point

Invert level at A = ground level – excavation

3. Calculating the invert level of the last point

Invert level at B = invert level at A – Δh 4 Calculating invert level at other points

e.g. chainage is 00 20 40 55 60

invert level at 00 is the invert level of the first point invert level at 60 is the invert level of the last point Invert level at 20 = invert level at A – gradient (20) Invert level at 40 = invert level at A – gradient (40) Invert level at 55 = invert level at 40 – gradient (55) 5. How to determine Cut or Fill

Ground level of a point – invert level of that point = Cut or Fill If the answer is positive then it is a Cut

If the answer is negative then it is a Fill

EXAMPLE

Given : Chainage Ground level A 0 50,76 20 50,00 40 45,75 45 47,28 B 60 49 20

The information above refers to a drain between point A and B. The depth of excavation at A is 1,96m

and the fall from A to B is 1:40. Calculate the Cut and Fill in metres 1. Δh = gradient x distance

(55)

Δh = 1,5m

2. Invert level at A = 50,76 – 1,96

= 48,8 m (write this figure in the invert level column as shown in table 1)

3. Invert level at B = 48,8 – 1,5

4. Invert level at 20 = 48,8 – 0,025(20)

5. Invert level at 40 = 48,8 – 0,025(40)

6. Invert level at 45 = 48,8 – 0,025(45)

calculating the cut and fill

At 00 chainage 50,76 – 48,8 = 1,96 Cut At 20 chainage 50,00 – 48,3 = 1,70 Cut At 40 chainage 45,75 – 47,8 = –2,05 Fill At 45 chainage 47,28 – 47,675 = –0,395 Fill At 60 chainage 50,76 – 48,8 = 1,96 Cut

Table 1

Chainage Ground

level Invert levels Cut Fill

A 0 50,76 48,8 1,96 20 50,00 48,3 1,70 40 45,75 47,8 2,05 45 47,28 47,675 0,395 B 60 49,20 47,3 1,90

EXERCISE 8

Quesrtion 8.1

A drain is laid between A and B. The excavation at A is 1,58m and the fall from A to B is 1:75

Calculate the cut and fill in metres. Use the data given below Chainage Ground

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A 0 250,15 20 251,26 40 247,99 60 252,56 65 253,01 80 247,323 100 250,08 B 120 249,67 Question 8.2

A drain is laid between a House (H) and the manhole (M). the depth of the excavation at House

1,368m and the fall from the house to the manhole is 0,76%. Use the information given below to

calculate the cut and fill in metres

Chainange Ground level

House 0 322,80 25 326,42 40 327,42 60 320,42 75 319,46 80 324,22 100 325,44 105 326,47 Manhole 120 322,10 Question 8.3

The information below refers to the details of a drain which is to be laid between P and Q. the depth

of the excavation at Q is 2,4m the rise from Q to P is 0,82%. Calculate the invert level at the different

points and also the Cut and Fill

Chainange Ground level

P 0 420,90 15 424,40 25 425,43 40 420,42 60 419,46 75 424,22 80 425,44 100 426,47 Q 120 420,20 Question 8.4

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A drain is laid between A and B. The Reduced level at B is 47,3m and the fall from A to B is 1:80

Calculate the cut and fill in metres. Use the data given below

6.6 LENGTH AND THE SLOPE OF DRAINAGE PIPE

]

1. Slope = _∆_chainage∆h Chainage Ground

level Invert level A 0 50,76 20 50,00 40 47,16 60 49,57 80 49,00 100 48,80 B 120 48,50 47,30

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Δh = difference between invert levels

Δchainage (distance) = difference between chainage

2. Length of the pipe = ₍_∆_h₎2 ₊₍_∆_chainage ₎2

NOTE: Length of pipe is the same as the length of the slope

3. Slope as a percentage of the pipe = ×100% ∆ ∆ chainage h

EXAMPLE

QUESTION

A B

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C DATUM MSL 300 GROUND LEVEL INVERT LEVEL DISTANCE ( CHAINAGE) 4 6 0 0 ,0 0 3 0 9 ,2 7 8 3 1 0 ,8 0 4 6 6 0 ,0 0 3 1 1 ,1 0 0 3 1 2 ,4 0 4 7 3 6 ,6 0 3 0 7 ,9 0 0 3 0 9 ,2 2

The diagram below shows a longitudinal section of a pipeline 1. Calculate the total length of the piping required from A to C 2. Calculate the slope as a percentage of the pipeline

(60)

SOLUTION

1.1 _AB ₌ ₍_∆_h₎2₊₍_∆_chainage ₎2 ( )2 ( )2 4600 4660 278 , 309 10 , 311 − + − = AB ( ) ( )2 2 60 822 , 1 + = AB 028 , 60 = AB m 2 2 ₍ ₎ ) ( h chainage BC = ∆ + ∆ ( )2 ( )2 4660 6 , 4736 9 , 307 10 , 311 − + − = BC ( ) (2 )2 6 , 76 2 , 3 + = BC 667 , 76 = BC _m Total length AC = 60,028 + 76,667 = 136,695 m 2. slope A to B 100 4600 4660 278 , 309 10 , 311 _× − − = = 3,037 % slope B to C 100 4660 6 , 4736 9 , 307 10 , 311 × − − = = 4,178 % 3 slope A to B 4600 4660 278 , 309 10 , 311 − − = = 60 822 , 1 =₃₂_,1₉₃₁ Ratio = 1 : 32,931 slope B to C =311₄₇₃₆,10_,₆−₋307₄₆₆₀,9 =₇₆3,2_,₆ =₂₃_,1₉₃₈ Ratio = 1 : 23,938

(61)

EXERCISE 9

Question 9.1

The diagram below is a longitudinal section of a pipeline Carefully study the drawing and calculate the following:

9.1.1 the length of the required piping between point A and B and B and C and C and D

9.1.2 The slope as a percentage of the pipeline between A and B and B and C and C and D

9.1.3 The slope as a ratio of the pipeline between A and B and B and C

B C A D E

(62)

200 DATUM MSL

(63)

GROUND LEVEL INVERT LEVEL DISTANCE 3 8 0 0 ,0 2 0 9 ,4 8 9 2 1 0 ,9 0 3 8 6 0 ,0 2 1 1 ,2 0 0 2 1 2 ,5 0 3 9 2 3 ,0 2 1 1 ,0 2 2 1 2 ,2 0 3 9 3 6 ,6 2 0 8 ,0 0 2 0 9 ,3 2 9 6 7 ,5 2 0 7 ,9 0 2 0 9 ,4 8 63

(64)

Question 9.2

A sewer internal reticulation longitudinal section MHO20 – MHO26 is shown below

9.2.1 Calculate the length on the pipeline between MHO23 and MHO25 9.2.2 Calculate the slope of the terrain between MHO24 and MHO25

PIPE NGL MHO20 MHO21 MHO22 MHO23 MHO24 MHO25 MHO26 Datum 55m

(65)

INVERT LEVEL GROUND LEVEL CHAINAGE (m) 0 0 ,0 0 0 6 7 ,3 0 9 6 5 ,9 0 0 1 8 ,1 9 4 6 5 ,5 9 7 4 5 ,9 5 6 6 6 ,5 8 1 6 5 ,1 3 4 6 3 ,9 1 4 5 4 ,8 9 0 9 1 ,9 1 2 6 5 ,8 9 4 6 4 ,5 1 0 1 5 0 ,9 1 2 6 3 5 9 2 6 2 ,3 3 7

(66)

Question 9.3

From ANNEXURE C on REESTON INTERNAL SERVICES AREA C’-SEWER LAYOUT details junction C1 – C7

(NB: There are five sections of the pipeline to be considered) Calculate the total length of pipework C1 to C7

ANNEXURE C

REESTON INTERNAL SERVICES AREA C – AS-BUILT SEWER LEVELS

(67)

Question 9.4

The information below is for a layout plan of a pipeline from MHO10 – MHO134 Calculate the pipe length required from

9.4.1 MHO10 to MHO16 9.4.2 MHO16 to MHO15 9.4.3 MHO15 to MHO14 9.4.4 MHO14 to MHO13

(68)

6.7 Calculating the staff reading and height of sight

rails (SR)

Collimation line

Staff staff reading at A

Staff reading at B datum A Traveller sight rail Invert level A B Δh Invert level B

NOTE: This is base on the assumption that we are given the Depth of excavation

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Step 1 Calculate the collimation height

Collimation height = BS + BM

Step 2 Calculate the fall (difference in height)

Δh = gradient x distance

Step 3 Calculate the Invert level at A (if not given invert level

of A)

Invert level at A = Ground level – Excavation Step 4 Calculate the Invert level at B

Invert level at B = Invert level at A – Δh Step 5 Calculate the staff reading at A

Staff reading at A = Collimation height – (Invert level at A + height of the traveller)

Step 6 Calculate the staff reading at B

Staff reading at B = Collimation height – (Invert level at B + height of the traveller)

EXAMPLE

Question

A drain PQ is to be set out using a tilting level, given the following information Length of the drain 225m

Gradient PQ falling from P to Q at 0,80% Ground level at P is 229,38m

Depth of excavation at P 0,90m

Length of the traveller (Boning rod) is 1,90m

A backsight of 1,50m is taken on a benchmark with an elevation of 230,31m Calculate the staff reading necessary to locate sight rails over P and Q

SOLUTION

Collimation height = 1,5 + 230,31 = 231,81

(70)

Falling = 200 100 × = 1,6 Invert level at P = 229,38 – 0,9 = 228,48 Invert level at Q = 228,48 – 1,6 = 226, 88 Staff reading at P = 231,81−

(

228,48+1,9

)

= 1,43 Staff reading at Q = 231,81−

(

226,88 +1,9

)

= 3,03

EXERCISE 10

Question 10.1

A drain AB is to be set out using a tilting level, given the following information Length of the drain 200m

Gradient AB falling from A to B at 1 : 125 Ground level at A is 207,49m

Depth of excavation at A 1,00m

A backsight of 1,50m is taken on a benchmark with an elevation of 208,21m Calculate the staff reading necessary to locate sight rails over A and B

Question 10.2

(71)

Length of the drain 225m

Gradient PQ falling from P to Q at 0,80% Ground level at P is 229,38m

Depth of excavation at P 0,90m

A backsight of 1,60m is taken on a benchmark with an elevation of 230,31m Calculate the staff reading necessary to locate sight rails over P and Q

Question 10.3

A drain AB is to be set out using an automatic levelling instrument from the following information

Length of the drain pipe 155m

Gradient AB falling from A to B at 1 : 180 Invert level of pipe at A 100,400m

Thickness of the pipe 30mm Bedding depth 50mm

Question 10.4

Calculate the staff readings necessary to locate sight rails over N and M. Given the following information

Gradient NM rising from N to M at 1 : 100 Invert level of pipe at N is 280,400m

A backsight of 2,60m is taken on a benchmark with an elevation of 282,00m Calculate the staff reading necessary to locate sight rails over N and M

(72)

Calculate the staff readings necessary to locate sight rails over A and B Given the following information

Gradient AB rising from A to B at 0,75% Ground level at P is 312,50m

Depth of excavation at P is 1,00m

(73)

CHAPTER 7 AREAS AND VOLUMES

7.1 AREA

Difference between the ordinate and mid-ordinate Given the following Sketch

5m 6m 7m

3m 5m and 7m are ordinate

6 2

7 5

=

(74)

mid-ordinate = sum of two ordinate divide by two FORMULAE

7.1.1 Mid ordinate rule

a) If given mid-ordinates

Area = width x sum of the mid ordinates

Example

Given: mid-ordinate 6, 8, 7, 4 all in m and the width between the ordinate is 10m

Area = 10( 6 + 8 + 7 + 4) Area = 250 m2

b) If given ordinates

Area = width st last ord sum..of ..other ..ordinates

2 .. 1 + + Example

Given: ordinate 8, 7, 9, 6, 5, 3, 0 all in m and the width between the ordinate is 15m Area = 15       + ₊₇₊₉₊₆₊₅₊₃ 2 0 8 Area = 15 x 34 Area = 510 m2

7.1.2 Trapezoidal rule

(ONLY when given ordinates)

Area =

[

1 2( .. .. .. )

]

2 last sum of other ordinates

width st ₊ ₊

Example

Given: ordinate 8, 7, 9, 6, 5, 3, 0 all in m and the width between the ordinate is 10m Area =

[

8 0 2(7 9 6 5 3)

]

2 15 ₊ ₊ ₊ ₊ ₊ ₊ Area = 7,5 x 68 Area = 510 m2

(75)

7.1.3 Simpson’s rule

(ONLY when given ordinates)

Area =

[

1 4( .. .. .. ) 2( .. .. .. )

]

3 last sum of even ord sum of odd ord

width st ₊ ₊ ₊

Example

Given: ordinate 8, 7, 9, 6, 5, 3, 0 all in m and the width between the ordinate is 15m Area =

[

8 0 4(7 6 3) 2(9 5)

]

3 15 ₊ ₊ ₊ ₊ ₊ ₊ Area = 5 x 100 Area = 500 m2

Exercise 11

Question 11.1

From a straight line AB Offsets to the bank of a river are

A B

Distance (m) 00 50 100 150 200

0ffsets (m) 21 26 22 23 21

Calculate the area between AB and the river using the following methods 11.1.1 Trapezoidal rule

11.1.2 Mid-ordinate rule 11.1.3 Simpson’s rule

SURVEY NOTES N5

BUILDING & STRUCTURAL

SURVEYING

N5

CONTENTS

CHAPTER 1

1.1 Terminology

CHAPTER 2

CHAIN SURVEYING

2.

CHAIN SURVEYING

2.1 METHODS OF FIXING A POINT

2.2 Field problems

2.3 Methods of taping

2.4 How step chaining is conducted on site

2.5 How the 3 – 4 – 5 method of setting out a right angle is

applied

2.6 Five requirements to obtain sufficient accuracy when

taping

2.7 Care of steel tape

2.8 CHAINAGE CORRECTIONS

w

OBSERVATION ANGLE

θ

Z = 90º – θ

θ = 90º – Z

Z = 90º + θ

θ = Z – 90º

NB: ALWAYS use angle θ which represents the angle of

Elevation or Depression

EXAMPLE 1

QUESTION

SOLUTION

)

"

1 6

'

3 7

3

c o s

1

(

2 8 1

,

8 7

−

°

=

C

EXAMPLE 2

QUESTION

SOLUTION

)

1

"

1 5

'

22

2

(sec

2 85

,

9 8

°

−

=

C

A

s e c

=

"

1 5

'

2 2

2

co s

2 8 5

,

9 8

°