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THE INSTITUTION OF ENGINEERS SRI LANKAPART 11 EXAMINATION -June 2012

203- SURVEYING

Instructions:

Answer any five questionsTime allowed: 3 hours June 2012

INSTRUCTIONS TO CANDIDATES:

This paper contains 7 questions on 8 pages (including this page).

Answer any five questions.

This examination accounts for 100% of the subject assessment. The marks assigned for eachquestion and sections there of are indicated in square brackets.

This is a closed book examination.

It is an offence to be in possession of unauthorized material during the examination

Only approved calculators are permitted .programable calculators are not allowed.

Assume reasonable values for any data not given in or with the examination paper. Clearly statesuch assumptions made on the script.

Question 1.

1. (a). A base line length AB in a chain survey was measured by using a steel tape of nominallength 30.0 m, and the length was recorded as 321.45m.(i). If the length of the tape was actually 29.75 m, what is the corrected length of AB?(ii).If, in addition to the error in (i), the length was measured in a slope of 1 in 35, what is

the corrected length of AB?(iii). If, in addition to the errors in (i) and (ii), the length AB was measured in catenary

(suspended tape), what is the corrected length of AB?Weight of tape = 0.1 kg/m; pull exerted on tape = 10 kg.Formula for catenary correcton, in the usual notation, is = I(wlll (24/)

[9 marks]

(b). Briefly explain the following (you may use sketches):(i). Triangulation.(ii). Trilateration.

[6 marks]

(c). On a cross-sections drawing, a horizontal scale of 1 em = 10m and a vertical scale of 1m = 2.5 ern has been used. A fill cross sectional area was measured by a planimeter as38.75 cm2

.

(i). What is the actual cross-sectional area of the fill, in square metres?(ii). It was later discovered that the above cross-sectional drawing had been enlarged by

a photocopier by 10%. What is the corrected cross sectional area of the fill in thefield (in square metres)?

[5 marks]Question 2.

(a). It was required to ascertain the elevations of two points, A and B, and a line of levels wasrun from A to B. A is a point on the ground, and B is an overhead location, on the soffit(underside) of an overhead bridge. The leveling was continued from B to a bench mark ofelevation 140.65 m MSL, and the readings obtained are shown below. Obtain the reducedlevels of A and B.

Back Sight Intermediate Fore Sight Reduced Level RemarksSight

4.35 A1.40 7.767.28 3.35

2.32 B (Invert level)4.82 1.854.86 2.35

0.965.32 140.65 Bench Mark

[13 marks]

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[02 marks]

(b). An observer of height 2m stands on the sea-shore, gazing (looking) at the horizon througha telescope. A ship which has a mast 30m in height sails on the sea away from him. Whatis the distance between the observer and the ship at which the mast of the ship disappearsbelow the horizon for the observer?

Take the radius of the Earth as 6371 km; clearly state any assumptions that you make.[07 marks

Question 3

(a). (i). List the main steps of theodolite traversing field work.

(ii). List the temporary adjustments that need to be made to the theodolite at each set up ofthe instrument, and briefly explain the procedure of one temporary adjustment.

[03 marks]

(iii), List the errors that need to be adjusted in a theodolite traverse, and briefly explain howto adjust (or correct) one of those errors.

[04 marks]

(b) The Table Q .02 indicates the lines, mean included angles and lengths of the legs of a closedtraverse. The whole circle bearing of line AB is 240 ° 02' 25".

Table Q.02

Line Mean Included Angle Length (m)AB 8A 108 ° 22 12" 124.519BC 8B 94° 32 18" 103.750CD 8c 110°32'10" 96.041DE 80 120° 40' 40" 70.302EA 8E lOT 52 50" 79.233

I.Use the appropriate method of checking and show whether there is any error in angularmeasurements. [2 Marks]

II.Compute the closing error in distance measurements. [2 Marks]III.Calculate the final coordinate values of all traverse points. Co-ordinate values of point A

are 1000.0 Nand 1000.0 E (in meters). [7 Marks]

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Question 4

(a). List advantages of tach eometry (optical distance measurement).[03 marks]

(b). Prove the following formulae for inclined sights.D=CsCos8+K;

H=Cs Cos2 8+ KCos8;

V=1I2 Cs Sin28+KSin 8;

where,D= Inclined Distance.V= Vertical Distance.8= Vertical Angle.C= Multiplying Constant.s= Staff interval intercepted by the stadia lines.K= additive constant

[08 marks]

(c). In a tacheometric traverse, following readings were observed;

Station Point Vertical Circle Stadia readings Height of

Instrument

P Q +5' 24" 2.68;2.44; 2.20 ?

Q P -4' 12" 2.68;2.34; 2.00 3.23

Q R -4' 12" 1.43; 1.07; 0.71 3.23

In above observations, the instrument height at station P had not been measured. Find theinstrument height at P and the reduced level of the ground at R, if the reduced level at P is 100.23m. Assume that the multiplicative constant and the additive constant are 100 and 0.005,respectively.

[09 mark]

Question 5

(a). The following offsets were taken from a chain line to an irregular boundary:

Distance in meters 0Offset in meters 23

3028

5016

7024

9028

11021

14025

17027

Distance in meters 200Offset in meters 24

23008

27022

31018

5

Calculate the area enclosed by the chain line, the irregular boundary and the end offsets by:

(i). Simpson's Rule(ii). Trapezoidal Rule

[10 marks]

(b). Determine the area in hectares enclosed by the lines of a closed traverse survey ABCDEAfrom the following data:

StationABCDE

E (Easting) (m)500.00713.96820.68889.74675.92

6

N (Northing) (m)600.00791.52577.12490.42365.10

[10 marks]

[10 marks]

Question 6

(a). Following are the reduced levels taken at intervals of 30 m along the centre line of part of aproposed railway embankment.

Chainage (m) 100 130 160 190 220 250Reduced Level ofnatural ground, 42.5 43.2 43.8 42.7 43.2 44.1mMSL

The formation level of the (top of) embankment at chainage 100 m is 45.0m above MSL. Therailway embankment is to rise at a gradient of 1 in 60 from that point onwards along thecentre line of the embankment, and is to have side slopes of 1 vertical to 2 horizontal. Theground is level in a direction transverse to the centre line of the embankment, thus giving alevel cross section for the embankment at all points along the centre-line.

Calculate the earth volume required for the embankment in this section of the railway, byusing the Prismoidal Formula as far as possible.Note: Area of a level section is given by

A = (b + sc) cwhere, b= formation width

s = defines the slops, i.e., side slope is given by l:sc = central height

(b). The spot levels (RLs with respect to MSL) taken on grid points of a 20m x 20m grid overan area ABCD are as shown in Figure Q-6. Calculate the volume of earth contained betweenthe surface and a level plane of 79.0m MSL within the area ABCD.

[09 marks]

(c). Brieflv explain possible sources of error in the volume calculation you did in (b) above.[01 mark]

76.4 78.5 67.9

79.

5·0 S·DA4 ~< "8d>

50

h~

It ),(

C D

6

77.7

7889.7 78.8

Figure-Q-6

Question 7

(a). Briefly explain the following (you may use sketches):(i). True Meridian (North).(ii). Chainage and Offset of a point.(iii). Systematic errors in measurements.

[06 marks](b). Briefly explain, by using a neat sketch, how the position on the earth's surface can befixed by a coordinate system comprising parallels of latitude and meridians. Show the originof the coordinate system on the sketch.

[04 marks]

(C) Tabulate data needed to set out using theodolite and tape a circular curve of radius 500 mto connect two straights having a deflection angle of 30° 18'. The chainage of theintersection point is 2160.0 m

[10marks]

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