OCTOBER -2019 Year/Term: II/III Department: Civil Engineering Question

Code: 285

Subject : Surveying –I Maximum marks :75 [N.B: (1) Q. No. 8 in PART-A and Q. No. 16 in PART-B are compulsory. Answer any FOUR

questions form the remaining in each PART-A and PART-B

(2) Answer division (a) or division (b) of each question in PART-C (3) Each question carries 2 marks in PART-A, 3 marks in PART-B and 10 marks in PART-C ]

Part-A ----------5 X 2 =10

1. What is geodetic surveying? When the extent of the area surveyed is large more than about 260 km2, Or degree of accuracy required is more, the effect of curvature of earth is taken in to consideration and the survey is called Geodetic surveying this is also called Trigonometrically surveying since spherical angles and spherical trigonometry are used.

2. What are the correction in tape?------------------Any two correction each one mark The following five corrections may be found for the measured lengths of tape: (i) Corrections for absolute length (ii) Corrections for pull (iii) Corrections for temperature (iv) Corrections for slope and (v) Corrections for sag.

3. Why is meant by whole circle bearing? In this system bearing are measured clockwise from the direction of the magnetic North.

Whole circle Bearing are expressed from 0° to 360°. Example:

4. What is back sight in levelling? It is the sight taken on a staff held at a point of known elevation. It is the first staff reading taken after the level is set up and leveled.

5. What is check levelling? It is the operation of running levels for the purpose of checking the series of levels, which have been previously fixed. At the end of each day’s work, a line of level is run, returning to the starting point of that day with a view to check the work done on that day.

6. What is contour interval? The constant vertical distance between two successive contours is called the Contour Interval

Contour lien showing contour interval of

1m

71-70=1m Or 72-71=1m etc.,

7. Name any two purpose of GPS--------------------Any two purpose each one mark

• Establishment of high precision zero order Geodetic National Survey Control Network of GPS stations.

• Geophysical positioning, mineral exploration and mining.

• Survey control for topographical and cadastral surveys.

• Ground control for photogrammetric control surveys and mapping.

• Offshore positioning: Shipping, offshore platforms, fishing boats etc.

• Space craft tracking

• Military ; Improved weapon delivery accuracies i.e. for missiles etc.,

8. What is local attraction in compass surveying?(Compulsory) The deviation of the compass needle arising from local source is called Local attraction. Steel tape. Chaining pins, axe, bunch of keys, knife, steel frames of spectacles are some common source of local attraction.

PART-B----------------------5 X 3= 15

9. What is offset? State the types of offset. The lateral measurement taken to the left or right of a survey line to locate the details such as buildings, boundaries, fences, roads etc are called “offsets” ---------2marks The offsets are of two types------------------------1 mark

• Oblique offsets

• Perpendicular offsets. An offset taken at an angle other than 900 to the survey line is called “oblique offset” Perpendicular offsets: The offset taken at right angle to the survey line is called

10. What is dip and declination in compass?

Magnetic Dip: ------------1mark The angle of inclination of magnetic needle with horizontal is known as dip of the needle.

Magnetic Declination-----------2 mark The magnetic declination at a station is the horizontal angle between the true meridian and the magnetic meridian through the station.

Sketch -1 mark

11. What is GTS benchmark and permanent benchmark?

G.T.S. Bench Marks: These bench marks have been established by the Great Trigonometrical Survey of India department. These bench marks have been established all over India with reference to the mean sea level of Mumbai as a datum. Their position and reduced levels are shown on G.T.S. maps. A concrete pedestal with bronze plate at its top is provided at a point where GTS benchmark is located. The bronze plate consists the information or value of the benchmark on its top. These pedestals are generally protected by masonry structure build around it. ----------------Three points each one mark

12. What is curvature and refraction correction in levelling? For long sight and accurate leveling work. The effect of curvature of the earth and refraction of the line of sight shall have to be taken into consideration. Earth has a curved face which is assumed to be a level surface but the line of sight as furnished by the leveling instrument is horizontal and not the level line-----1 mark Through level Rays of light from the staff to the instrument passing through layers of different densities do not remain straight but are refracted or bent down the denser medium. Due to curvature, the points appear to be lower than they actually are, while due to refraction, they appear to be higher than the

Fig. Effect of curvature and refraction in levelling

----------------------1 mark y actually are.-----------1 mark

13. Mention the different types of map.----------------Any three maps each one mark

• Political Map

• Physical Map

• Topographic Map

• Climatic Map

• Economic or Resource Map

• Road Map

14. Define the desired relationship between the fundamental lines. ------------------Three relation each one mark

1. Axis of the level tube is perpendicular to the Vertical axis

2. Horizontal cross hair should lie in a plane perpendicular to the Vertical axis, so that it will lie a Horizontal plane when the instrument is properly leveled.

3. The Line of sight is parallel to the axis of the level tube.

15. State the application of GPS?----------------Any three application each one mark Surveying and Mapping : Many Broad Companies in India and abroad uses GPS

• to locate different points,

• preparing Contour maps,

• giving Alignments of Roads, Bridges

• Telecom tower placement

• Pipeline installation

• Dam construction and deformation monitoring of building foundations, structural retaining walls, and embankments etc.,

• mining operations such as drilling, shoveling, vehicle tracking

• GPS-equipped fleet vehicles, public transportation systems, delivery trucks, and courier services use receivers to monitor their locations at all times for both efficiency and driver safety

• GPS-equipped balloons monitor holes in the ozone layer over the polar regions as well as air quality across the nation as part of environment studies

• Commercial fishing – Enforcement of fishery boundaries

• Forestry – Monitoring of illegal deforestation

• Helps relief workers navigate disaster areas devoid of landmarks

• GPS-equipped buoys for tsunami warnings

16. List out the errors in chaining.(Compulsory)------------Any three errors each one marks Errors in chaining broadly classified in to a. Instrumental errors. b. Natural errors. c. Personal errors.

Various sources of these errors are

Error in length of chain or tape.

• The chain getting out of the survey line due to wrong ranging Chain or tape not

being stretched straight.

• Sag in chain or tape.

• Variation in temperature.

• Variable tension in the chain.

PART-C ---------------------------------5 X 10 =50

17.(a) Describe briefly the main classification of surveying. The general classification of surveying is based on

(a) Nature of field of surveying-----------------4 marks

(b) Object of surveying --------------------------2 marks

(c) Method of surveying--------------------------2 marks

(d) Instrument used in surveying----------------2 marks (a) Based on nature of field of surveying 1. Land surveying: the surveys conducted on land are

• Topographical survey to locate the features like roads, railway, towns, Hills, valleys. Rivers etc., on land

• Cadastral survey to locate property boundary

• City survey to locate Buildings Street. Layouts water supply and drainage lines etc.,

• Engineering surveys to collect data for the design and construction of engineering works such as buildings, bridges, dams. Roads. Railway etc., it consists of reconnaissance survey preliminary survey and final location survey.

2. Marine surveying of Navigation surveying: Survey conducted on or near the body of water such as river, lake, bay harbor etc., to estimate water flow and to determine shape of area below the water surface. 3. Astronomical surveying:survey conducted to determine the latitude, longitude, Azimuth, local time etc., on the earth by observing sun or star (b)Based on the object of surveying: The surveys are

• Archaeological survey to bring out the relics of antiquity

• Geological survey to determine various strata of earth

• Military survey to identify the offensive and defensive strategic point and

• Mine survey to fine position of mineral wealth such as gold, iron. Coal etc., (c) Based on the method of surveying

• Triangulation survey: The area covered with a network of triangles and details are collected.

• Traverse survey: The area covered by series of survey lines forming traverse and details are collected.

(d) Based on the instrument used in surveying

▪ Chain survey

▪ Compass survey

▪ Plane table survey

▪ Levelling

▪ Theodolite survey

▪ Tachometric survey

▪ Total station

▪ GPS:Global Positioning System

▪ Remote sensing

▪ Photogrammetry

▪ Hydrographic survey

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[Or]

b) (i) State and explain the types of obstacles. Obstacles in chain surveying There are 3 types of obstacles (a) Obstacle to ranging-------------2 marks (b) Obstacle to chaining and----- 2 marks (c) Obstacle to both ranging & chaining. –1 mark a) Obstacle to Ranging: The type of obstacle in which the ends are not inter visible is quite common except in flat country. These may be two cases. i) Both end(A and B) of the line may be visible form intermediate points on the line ii) Both ends of the line may not be visible from intermediate points on the line

Any one of the following methods may be employed:

b) Obstacle to chaining but not ranging There may be two cases of this obstacle

i) When it is possible to chain round the obstacle ex: a pond ii)

When it is not possible to chain round the obstacle ex: a river

Any one of the following methods may be employed:

c) Obstacles to both chaining & ranging A building is the typical example of this type of obstacle the problem lies in prolonging the line beyond the obstacle & determining the distance across it.

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Any one of the following methods may be employed:

(ii) The distance between two stations is 1200m when measured with a 20m chain. The distance when measured with 30m chain is 30cm too long. What is the error in 30m chain? Solution: Solution:1 Using given data without

assuming a) Consider 30 m chain: Measured

length (ML)= 1200 m

Error in 30 m chain= 30 cm too long= 0.30 m L= 30 m L’= 30 + 0.30= 30.30m True length (TL)= L’ x Measured Length (ML)= (30.30/30 )x 1200= 1212 m----2 marks

L Result:

True Length (TL)= 1212 m Error in 30 m chain= 30 cm too long-----------------------------3 marks [Or] Solution:2 By assuming 1200 m distance between two station using both 20m and 30m chain

i) While using 30m chain Designed length (L) =30m Error in 30 m chain = 30 cm too long = +0.30m Incorrect length of chain(L1)= 30+0.3 = 30.3m Measured length (D) has not given and taken as same 1200m when measured with 20 m chain True length of the line (l1) = (L1/L)X D = (30.30/30)X1200 = 1212m--------------------------2 marks ii)

While using 20 m chain

Designed length (L) =20m Error in 20 m chain = x Measured distance between two station = 1200m Take the true distance between two station as while using 30 chain =1212m True length of the line (l1) = (L1/L)X D 1212 = (L1/20)X1200 L1 =(1212/1200) X 20 L1= 20.20 m

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Hence errors is 20 m chain = 20 cm too long ---------------------3 marks Result:: Error in 20m chain = 20cm too long or 0.20 m too long [Or] Solution:3 Corrected Question: The distance between two stations is 1200 m when measures with a 20 m chain. The distance measured with 20 m chain is 3 cm too long. What is the error in 30 m chain? (5 Marks) Solution:

a) Consider 20 m chain: Measured Length (ML)= 1200 m Error in 20 m chain= 3 cm too long. = 0.03 m L= 20m L’= 20 +0.03= 20.03 m True Length (TL)= (L’/L) x Measured Length=(20.03/20)x1200= 1201.80m------2marks

b) Consider 30 m chain: Measured Length (ML)= 1200 m

L= 30m L’= ? True Length (TL)= (L’/L) x Measured Length(ML) 1201.80=(L’/30) x 1200 L’= 30.045 m Result: Error in 30 m chain= 0.045 m (OR) 4.5 cm too long---------------------------3 marks

18.(a) Explain the temporary adjustment of prismatic compass. Temporary adjustment of Prismatic compass: To measure bearing of a station B from a station A. The following steps are adopted. Temporary Adjustment of the compass: This process consists of centering the instrument over the Center of he peg at station A, and leveling the instrument The step by step method for making Temporary adjustments in Prismatic Compass necessary for carrying out Compass Surveys.

1. Fixing the compass on the tripod-----2 marks

2. Centering the compass------------------2 marks

3. Levelling the compass-------------------2 marks

4. Sighting the object-----------------------2 marks

5. Observation of bearings---------------- 2 marks 1. Fixing the compass to the tripod The box of prismatic compass is fixed to a spindle of ball and socket joint. By the ball and socket arrangement, this can be quickly levelled and rotated in any direction.

2. Centering the compass The prismatic compass is centered over a survey station correctly by means of a plumb bob or by dropping a pebble from the centre of the instrument.

3. Levelling the compass The compass is quickly levelled by ball and socket arrangement by eye judgment.

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It should be levelled in such a way that dial moves freely and does not touch the rim of the bob.

4. Sighting the object The object is sighted with the help of eye vane and object vane in the compass. The surveyor views through the eye vane and rotate the box until the ranging rod at a station is bisected.

5. Observation of Bearing After citing the object correctly, the bearing of the survey lines is noted through prism at which the line of sight and object cuts the image of the graduation on the dial. Take down the reading. The reading indicates the bearing of the line AB.

[Or]

(b) The following are bearing observed in running compass traverse. Find the included angle and apply usual check.

Line F.B B.B

AB S 85° 30’ W N 85° 30’ W

BC N 34° 30’ E S 34° 30’ W

CD N 70° 00’ W S 70° 00’ E

DE S 65° 45’ W N 65° 45’ W

EA S 47° 30’ E S 47° 30’ E

Solution:1

a) Calculation of included angle after converting QB in to WCB Solution: (Student can solve without changing the given data )

Line FB BB

AB 265° 30’ 274°30’

BC 34° 30’ 214°30’

CD 290° 00’ 110° 00’

DE 245° 45’ 294° 15’

EA 132° 30’ 132°30’

Anticlockwise traverse,

Included angle=FB of forward line –BB of previous line

Angle A= 265°30’-132°30’= 133° 0’---------2 marks

Angle B= 34°30’-274°30’+360°= 120° 0’-----------2 marks

Angle C= 290°0’-214° 30’= 75°30’-----------2 marks

Angle D= 245° 45’- 110°0’= 135°45’--------2 marks Angle

E= (132°30’- 294°15’)+360°= 198°15’-----2 marks

Result:

Total Included angle= 662° 30’

Check: Total Included angle= (2N-4) x 90= (2x5-4)x 90= 540° 0’

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19.(a) The following reading is taken from a page of a levelling field is. Fill the missing readings. Apply usual check.

Point BS m

IS m

FS m

Rise m Fall m

R.L m

B.M 1.450 420.690

1 X 1.670 X

2 4.320 417.820

3 X 417.440

4 X 1.320

5 4.320 X 2.520

6 2.380

7 X

8 X 1.040 422.090

9 X 420.820

Solution :

Point BS m

IS m

FS m Rise m Fall m

R.L m

B.M 1.450 420.690

1 3.120 1.670 419.020

2 4.320 1.200 417.820

3 4.700 0.380 417.440

4 3.380 1.320 418.760

5 4.320 0.860 2.520 421.280

6 2.380 1.940 423.220

7 2.470 0.090 423.130

8 3.510 1.040 422.090

9 4.780 1.270 420.820

Arithmetic Check: ΣB.S- ΣB.S= ΣRise- ΣFall = Last R.L –First R.L 5.770-5.640=5.780-5.650=420.820-420.690 0.130 =0.130=0.130 (o.k) Detailed calculation: At Point 1 I.S of (1) – B.S of B.M = Fall(1) I. S(1)= B.S of B.M+ Fall(1) = 1.450+1.670= 3.120 R.L of (1) = R.L of B.M –Fall= 420.690-1.670=419.020 At Point 2 Fall(2) = I.S (2)-I.S(1)= 4.320-3.120= 1.200 At Point 3 Fall (3) = R.L of (2)-R.L of (3) = 417.820-417.440=0.380 Fall (3) = I.S of (3) - I.S of (2)

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I.S of (3)= Fall of (3)+ I.S of (2)= 0.380+4.320=4.700 At Point 4 Rise of (4) = I.S of (3)- I.S of (4) Hence I.S(4)= I.S of (3)- Rise of (4) = 4.700-1.320=3.380 R.L of (4) = R.L of (3) + Rise of (4)

= 417.44-+1.320= 418.760 At Point 5 Rise of (5) = I.S of (4)- F.S(5) Hence F.S(5)= I.S of (4)- Rise of (5)= 3.380-2.520=0.860 R.L of (5) = R.L of (4) + Rise of (5) = 418.760+2.520=421.880 At Point 6 Rise of (6)= B.S of (5)- I.S of (6)= 4.320-2.380=1.940 R.L of (6) = R.L of (5) +Rise of (6) = 421.280+1.940=423.220 At Point 7 R.L of (7) = R.L of (8)+Fall of (8) = 422.090+1.040 = 423.130 Fall of (7) = R.L of (6)- R.L of (7) = 423.220-423.130= 0.090 Fall of (7) = I.S of (7)- I.S of (6) Hence I.S (7) =Fall of (7)+ I.S of (6) = 0.09+2.380 = 2.470

At Point 8

Fall of (8) = I.S of (8) – I.S of (7) 1.040 = I.S of (8) – 2.470 Hence I.S of (8) = 1.040+2.470 = 3.510 At point 9 Fall of (9) = R.L of (8)- R.L of (7) = 422.090 – 420.820 = 1.270 Fall of (9) = F.S (9)- I.S of (8) 1.270 = F.S of (9) -3.510 Hence F.S of (9) = 1.270+3.510 = 4.780

Mark allocation: For eight omitted value each one marks and arithmetic check 2 marks.

[Or]

(b) Draw the dumpy level with neat sketch and explain the components parts. 10 A schematic diagram of an engineer's level is

shown Figure. An engineer's level primarily consists of a telescope mounted upon a level bar which is rigidly fastened to the spindle. Inside the tube of the telescope, there are objective and eye piece lens at the either end of the tube. A diaphragm fitted with cross hairs is present near the eye piece end. A focusing screw is attached with the telescope. A level tube housing a sensitive plate bubble is attached to the telescope (or to the level bar) and parallel to it. The spindle fits into a cone-shaped bearing of the leveling head.

The leveling head consists of tribrach and trivet with three foot screws known as leveling screws in between. The trivet is attached to a tripod stand.

Mark allocation: For description of parts of level 6

marks and sketch 4 marks.

20. ( a) Explain the field procedure for carrying out cross - sectional levelling (C.S).

Field procedure of cross section levelling

F igure: Field procedure for a typical Cross section obs ervation at 0 m Chainage

Field book For Cross - section levelling

1. Two perpendicular lines are erect with the help of cross - staff at each L.S point on both sides with respect to the alignment.

2. An additional cross – section may require at each turning point perpendicular to the proceeding line.

3. Mark the points along each perpendicular line say L1, L2, L3 etc. at left R1, R2, R3 at right hand side at specified intervals.

4. Setup up the instrument by the side of alignment on firm ground at some suitable places so as to cover a maximum number of points.

5. Take a B.S on the given A.B.M point to determine the height of instrument.

6. Hold the leveling staff of all those marked points and observe the staff reading and recorded it in field table.

7. Complete staff reading obse rvation with the change points if required

8. Reduce the R.L of each points in the order of occurrence.

Table above for taking the cross - section at 0 m and 5m Chainage. Reduction of levels,

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] OR [ s

( b ) In Levelling between two points A and B on o pposite banks of a river, a level is set up near A and the staff readings on A and B are 1.710m and 2.950m respectively. The level is then moved and set up near B, and the respective readings on A and B are 0.905m and 2.225 m. Find the true difference of le vel between A and B, if the R.L of A is +150.000m. Find the R.L of B.

Data Given :

R.L of A is +150.000m

Staff reading w hen level is set up near point A a1=1.710m b1 =2.950 m ------------------- 2 marks

Staff reading When level is set up near point B a2 =0.905 m b2=2.225 m ------------------ 2 mark

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Mark allocation: For procedure 3 marks and sketch and table one mark each

Staff reading when level is set up near point A Staff reading When level is set up near point B

Formula for true difference between given points A and B h= [(b1 – a1) + (b2 – a2)] / 2 since B is fall--------------2 mark h= [2.950 – 1.710) + (2.225 – 0.905)] / 2 h= [1.240 + 1.320] / 2 h= [2.560] / 2 h= 1.280m -------------------------------------2 mark R.L of B= R.L of A- True difference R.L of B= 150.000 – 1.280= 148.720m ------2 mark Result:

a) True difference between point A and B = 1.280 m

b) R.L of point B =148.720 m

21.(a) Describe various characteristics of contour with neat sketches

1. The variation of vertical distance between any two contour lines is assumed to be uniform.

2. The horizontal distance between any two contour lines indicates the amount of slope and varies inversely on the amount of slope.

3. Thus, contours are spaced equally for uniform slope (Figure:1 ); closely for steep slope contours (Figure 2) and widely for moderate slope (Figure 3)

Mark allocation:

Any five character description each one mark and any five character sketch each one mark

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Figure:1 Contour of Terrain having uniform slope Figure:2 Contour of Terrain having Steep slope

Figure:3 Contour of Terrain having Gentle slope

The steepest slope of terrain at any point on a contour is represented along the normal of the contour at that point (Figure: 4) they are perpendicular to ridge and valley lines where they cross such lines.

Figure:4 Steepest slope of a terrain indicated at contour lines

4. Contours do not pass through permanent

structures such as buildings. (Figure: 5.)

5. Contours of different elevations cannot cross each other (caves and overhanging cliffs are the exceptions). (Figure 6)

6. Contours of different elevations cannot unite to form one contour (vertical cliff exception). (Figure 7)

Figure: :7

7. Contour lines cannot begin or end on the plan.

8. A contour line must close itself but need not be

necessarily within the limits of the map.

9. A closed contour line on a map represents either d of ring contours with higher values inside, depicts a depression Figure (9) (without an outlet)

Figure: :6 Over hanging cliff and the contour

Figure : 8 Hill and its contour

Figure: 5 contour lines at buildings

is an

Vertical cliff and the contour

epression or hill (Figure 8). A set depicts a hill and the lower value inside,

Ponds and its contour

ridge lines. Contour lines in U-

shape cross a valley at right angles. The concavity in

Figure:9

10. Contours deflect uphill at valley lines and downhill at shape cross a ridge and in V-

contour lines is towards higher ground in the case of ridge and towards lower ground in the case of valley (Figure 10).

Figure: 10. Valley and Ridge lines

11. Contours do not have sharp turning

[Or]

(b) The area of enclosed by contours at the site of a reservoir and the proposed dam as computed by a planimeter are as shown in below.

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Contour

s (m)

100 105 110 115 120 125 130

Area m2 400 1500 3000 8000 10800 2050 4000

Take 1000m as bottom level of the reservoir and 1030m as highest achievable. Compute the capacity of reservoir by trapezoidal rule and prismoidal formula. Solution: Contour interval d= 100m ~ 105m= 5m Let

Contour in m

Given area in m2

Area ID

100 400 A1

105 1500 A2

110 3000 A3

115 8000 A4

120 10800 A5

125 2050 A6

130 4000 A7

a) Capacity of the reservoir by Trapezoidal formula -----For trapezoidal rule total 5marks =d [ {(A1+ A7)/2} +( A2 + A3+ A4+ A5+ A6)] ----------------------For formula (1 mark) = 5[ {(400+ 4000)/2} +( 1500 + 3000+ 8000+ 10800+2050)]----(1 marks) = 5[(4400/2) + (25350)] ----(1 mark) = 5 [2200+25350] = 5 [27550] ----(1 mark) =137750m3-----------------1 marks OR

=d/2 [ (A1+ A7) +2( A2 + A3+ A4+ A5+ A6)] =5/2 [ (400+ 4000) +2( 1500 + 3000+ 8000+ 10800+ 2050)] = 2.5 [ (4400) +2( 25350)] = 2.5 [ (4400) +50700)] = 2.5 [ 55100] =137750m3