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1(a) Explain the "Uses of Surveying". (10 marks)
(b) The distance between two stations was 1200m when measured
with a 20m chain.
The same distance when measured with 30m chain was found to be
1195m. If the 20m
chain was 0.05m too long, what was the error in the 30m chain?
(10 marks)
(a) Uses of Surveying
1. To prepare a topographical map which shows the hills,
valleys, rivers, villages, towns,
forests, etc. of a country.
2. To prepare a cadastral map showing the boundaries of the
fields, houses and other
properties.
3. To prepare an engineering map which shows the details of
engineering works such
as roads, railways, reservoirs, irrigation canals, etc.
4. To prepare a military map showing the road and railway
communications with
different parts of a country.
5. To prepare a contour map to determine the capacity of a
reservoir and to find the best
possible route for roads, railways, etc.
6. To prepare a geological map showing areas including
underground resources.
7. To prepare an archeological map including places ancient
relics exist.
(b) For 20m chain:
L = 20m, L' = 20 + 0.05 = 20.05 m
measured length (M.L) = 1200 m
True length of line =L'
L M.L
=
20.05
20
1200
= 1203 m
For 30 m chain:
L = 30m , L' = ?
True length (T.L) = 1203 m
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Ministry of Science and Technology
Department of Technical and Vocational Education
Civil Engineering
A .G .T .I - Year I
Semester - I
CE. 1011 Surveying (Civil, Architecture)
Sample Questions and Answers
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The amount 100 (sec - 1) is said to be the hypotenusal
allowance.
(ii) Compensating errors
Errors which may occur in both directions (i.e both positive and
negative) and
which finally tend to compensate are known as compensating
errors. Such errors may be
caused by:
(a) Incorrect holding of the chain
(b) Horizontality and verticality of steps not being property
maintained during the
stepping operation.
(c) Fractional parts of the chain or tape not being uniform
throughout its length.
(d) Inaccurate measurement of right angles with chain and
tapes.
(b)
AB =
2 217.5 - 2.35 17.34 m=
B1C = 2 219.3 - 4.2 18.84 m=
C1D = 2 217.8 - 2.95 17.56 m=
D1E = 2 213.6 - 1.65 13.49 m=
E1F = 2 212.9 - 3.25 12.48 m=
Total horizontal distance = AB + B1C + C
1D + D
1E + E
1F
= 79.71 m
20 m steel tape was 2.5 cm too short.
L = 20 m, M.L = 79.71 m
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L' = 20 - 0.025 = 19.975 m
True Length =L'
L M.L
=
19.975
20
79.71
= 79.61 m
3 (a) Explain the following terms:
(i) By Pacing or stepping (3 marks)
(ii) Engineers' chain (3 marks)
(iii) Gunter's chain (3 marks)
(iv) Steel Band (3 marks)
(v) Arrows (3 marks)
(b) The length of a line measured on a slope of 15 was recorded
as 550 m. But it was
found that the 20 m chain was 0.05 m too long. Calculate the
true horizontal distance of
the line. (5 marks)
Solution:
(a) (i) By pacing or stepping
For rough and speedy work, distances are measured by pacing, i.e
by counting the
number of walking steps of a man. The walking step of a man is
considered as 2.5 ft
or 80 cm. This method is generally employed in the
reconnaissance survey of any
object.
(ii) Engineers' Chain
The engineer's chain is 100 ft long and is divided into 100
links. So, each link is of
1 ft. Tallies are provided at very 10 links (10 ft), the central
tally being round. Such
chains were previously used of all engineering works.
(iii) Gunters' Chain
It is 66 ft long and divided into 100 links. So, each link is of
0.66 ft. It was previously
used for measuring distances in miles and furlongs.
(iv) Steel Band
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It consists of a ribbon of steel of width 16 mm and of length 20
or 30 m. It has a
brass handle at each end. It is graduated in meters, decimeters
and centimeters on
one side and has 0.2 m links on the other. The steel band is
used in projects where
more accuracy is required.
(v) Arrows
Arrows are made of tempered steel wire of diameter 4mm. One end
of the arrow is
bent into a ring, diameter 50 mm and the other end is pointed.
Its overall length is
400mm. Arrows are used for counting the number of chains while
measruing a
chain line.
(b)
Horizontal distance AB = AB1
cos 15
= 550 cos 15
= 531.25m
20m chain was 0.05 m too long.
L = 20 m
L' = (20 + 0.05) m = 20.05m
M.L = 531.25 m
True Length =
L'
L
M.L
=
20.05
20
531.25
= 532.6 m
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4 (a) Explain the "Adjustment of chain". (8 marks)
(b) A line was measured by a 20m chain which was accurated
before starting the day's
work. After chaining 900m, the chain was found to be 6cm too
long. After chaining a
total distance of 1575m, the chain was found to be 14cm too
long. Find the true distance
of the line. (12 marks)
Solution:
(a) Adjustment of chain
(i) When the chain is too long, it is adjustment by:
(a) Closing up the joints of the rings.
(b) Hammering the elongated rings.
(c) Replacing some old rings by new rings
(d) Removing some of the rings.
(ii) When the chain is too short, it is adjustment by:
(a) Straightening the bent links.
(b) Opening the joints of the rings.
(c) Replacing the old rings by some larger rings,
(d) Inserting new rings where necessary.
(b) First part:
L = 20 m
L' = 20 +0 0.06
2
+
= 20.03 m
M.L = 900 m
TL = ?
TL =
L'
L
M.L
=
20.03
20
900
= 901.35 m
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Second part:
L = 20 m
L' = 20 +
0.06 + 0.14
2
= 20.1 m
ML = 1575 - 900 = 675 m
TL =
L'
L
M.L
=
20.1
20
675
= 678.375 m
Total true distance = 901.350 + 678.375
= 1579.725 m
5(a) Describe the advantages and disadvantages of chains. (6
marks)
(b) A line was measured by a Engineers' chain which was found to
be 3 in too short,
before starting the day's work. After chaining 3000 ft, the
chain was accurate. After
chaining a total distance of 7500 ft, the chain was found to be
6 in too long. Find the true
distance of the line. (14 marks)
Solution:
(a) Advantages of chains
(i) They can be read easily and quickly.
(ii) They can withstand wear and tear.
(iii) The can be easily repaired or rectified in the field.
Disadvantages of chains
(i) They are heavy and take too much time to open or fold.
(ii) They become longer or shorter due to continuous use.
(iii) When the measurement is taken in suspension, the chain
sags excessively.
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(b) In first part:
L = 100ft (
Q
engineer's chain)
L' = 100 -
3/12 + 0
2
= 99.875 ft
M.L = 3000 ft
T.L =
L'
L
M.L
=
99.875 3000
100
= 2996.25 ft
In second part:
L = 100 ft
L' = 100 +
0 + 6/12
2
= 100.25 ft
M.L = 7500 - 3000 = 4500 ft
T.L =
L'
L
M.L
=
100.25
100
4500 = 4511.25 ft
The total true distance = 2996.25 + 4511.25
= 7507.5 ft
6(a) Describe the advantages and disadvantages of steel band. (6
marks)
(b) An old map was plotted to a scale of 40 m to 1 cm. Over the
years, this map has
been shrinking and a line originally 20 cm long is only 19.5 cm
long at present. Again
the 20 m chian was 5 cm too long. If the present area of the map
measured by planimeter
is 125.50 cm2, find the true area of the land surveyed. (14
marks)
Solution:
Advantages of Steel Band
(i) They are very light and easy to open or fold.
(ii) They maintain their standard length even after continuous
use.
(iii) When the measured is taken in suspension, they sag
slightly.
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Solution:
(a) Testing a Chain
For testing the chain, a test gauge is established on a level
paltform with the help
of a standard steel tape. The steel tape is standardized at 20C
and under a tension of
8kg. The test gauge consists of two pegs having nails at the top
and fixed on a level
platform a required distance apart (say 20 or 30m). The
incorrect chain is fully stretched
by pulling it under normal tension (say about 8 kg) along the
test gauge. If the length of
the chain does not tally with standard length, then an attempt
should be made to rectify
the error. Finally, the amount of elongation or shortening
should be noted.
The allowable error is about 2mm per 1m length of the chain. The
overall length
of the chain should be within the following permissible
limits.
20 m chain:
5mm, 30m chain
8mm
(b) Measured length ML = 327 m
20 m chain was 3cm too long.
L = 20m, L' = 20 +
3
100
= 20.03m
True length =
L'
L
ML
=
20.03
20
327
= 327.49m
(c)Correct scale = 50m to 1cm =
1cm
50m
Wrong scale = 100m to 1cm =
1cm
100m
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Measured length "ML" = 3500 m
True length =wrong scale
correct scale ML
=
1cm/100m
1cm/50m
3500
=
50
100
3500 = 1750m
8 A steel tape was exactly 30m long at 20 C when supported
throughout its length under
a pull of 10 kg. A line was measured with this tape under a pull
of 15kg and at a mean
temperature of 32C and found to be 780m long. The
cross-sectional are of the tape =0.03cm2 and its total weight =
0.693 kg,
for steel = 11 10-6 per C and E for steel =
2.1 106 kg/cm2. Compute the true length of the line if the tape
was supported during
measurement (a) at every 30m (b) at every 15m. (20 marks)
Solution:
(a) When supported at every 30m
span, n = 1
(i) Temperature correction
Ct
=
(Tm
- To) 30
= 11 10-6 (32 - 20) 30
= 0.00396m ( + ve )
(ii) Pull correction
Cp
=
m o(P - P ) L
AE
=
6
(15 10) 30
0.03 2.1 10
= 0.00238m ( + ve )
(ii) Sag correction
Cs
=
2
22m
LW
24n P
=
3
2 2
30 (0.693)
24 1 15
= 0.00267m (-ve)
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Total correction, e = 0.00396 + 0.00238 - 0.00267
= + 0.00367 m
L' = L + e = 30 + 0.00367 = 30.00367m
True Length =
L'
L
ML
=
30.00367
30
780
= 780.095 m
(b) When supported at every 15m
Span, n = 2
(i) Temperature correction Ct= 0.00396m (+ ve)
(ii) Pull correction Cp
= 0.00238m (+ ve)
(iii) Sag correction Cs=
2
22m
LW
24 n P=
2
2 2
30 (0.693)
24 2 15
= 0.00067 m (- ve)
Total correction, e = 0.00396 + 0.00238 - 0.00067
= 0.00567 m
L' = L + e = 30.00567m
True Length =
L'
L
ML
=
30.00567
30
780
= 780.147m
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9(a) Define the following terms: (i) True Meridian (ii) Azimuth
(iii) Magnetic Meridian
(iv) Magnetic bearing. (8 marks)
(b) A closed traverse is conducted with five stations A, B, C, D
and E taken in
anticlockwise order, in the form of a regular pentagon. If the
FB of AB is 30 0', find the
FBs of the other sides. (12 marks)
Solution:
(a) (i) True Meridian
A line or plane passing through the geographical north pole,
geographical south
pole and any point on the surface of the earth is known as the
"true meridian".
(ii) Azimuth
The angle between the true meridian and a line is known as "
true bearing " of the
line. It is also known as the "azimuth".
(iii) Magnetic Meridian
When a magnetic needle is suspended freely and balanced
property, unaffected
by magnetic substances, it indicates a direction. This direction
is known as the "magnetic
meridian".
(iv) Magnetic bearing
The angle between the magnetic meridian and a line is known as
the "magnetic
bearing" of the line.
(b)
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Each interior angle of pentagon =(2N - 4) 90
5
=
(2 5 - 4) 90
5
=
540
5
= 108
FB of AB = 30 0'
Add
B = +108 0'
= 138 0'
Add 180 = +180 0'
FB of BC = 318 0'
Add
C = +108 0'
= 426 0'
Deduct 180 = -180 0'
FB of CD = 246 0'
Add D = 108 0'
= 354 0'
Deduct 180 = -180 0'
FB of DE = 174 0'
Add
E = +180 0'
= 282 0'
Deduct 180 = -180 0'
FB of EA = 102 0'
Add
A = +108 0'
= 210 0'
Deduct 180 = -180 0'
FB of AB = 30 0'
(
ok!)
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10(a) Define the following terms: (i) Whole circle Bearing
(WCB)
(ii) Quadrantal Bearing (QB)
(iii) Reduced Bearing (RB) (10 marks)
(b) The following are the berings of a closed traverse:
Side FB BB
AB N 45 30' E S 45 30' W
BC S 60 0' E N 60 0' W
CD S 10 30' W N 10 30' E
DA N 75 45' W S 75 45' E
Calculate the interior angles of the traverse. (10 marks)
Solution: (a)
(i) Whole Circle Bearing (WCB)
The magnetic bearing of a line measured clockwise from the north
pole towards
the lines is known as the whole circle bearing of that line.
Such a bearing may have any
value between 0 and 360.
(ii) Quadrantal Bearing (QB)
The magnetic bearing of a line measured clockwise or
counterclockwise from the
Nroth Pole or South Pole (whichever is nearer the line) towards
the East or West is
known as the 'quadrantal bearing' of the line. This system
consists of four quardants -
NE, SE, SW and NW. The value of a quadrantal bearing lies
between 0 and 90.
(iii) Reduced Bearing (RB)
When the whole circle bearing of a line is converted to
quadrantal bearing it is
termed the 'reduce bearing'. Thus, the reduced bearing is
similar to the quadrantal bearing.
Its value lies between 0 and 90.
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(b)
Interior
A = 180 - (FB of AB + BB of DA)
= 180 - (45 30' + 75 45')
= 58 45'
Interior
B = BB of AB + FB of BC
= 45 30' + 60 0'
= 105 30'
Interior
C = 180 - (BB of BC + FB of CD)
= 180 - (60 0' + 10 30')
= 109 30'
Interior
D = BB of CD + FB of DA
= 10 30' + 75 45'
= 86 15'
Check: (2N - 4) 90 = 360
i nterior angles = A + B + C + D
= 360 (
ok!)
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(11) (a) Define the following terms (i) Fore and Back Bearng
(ii) Magnetic Declination (10 marks)
(b) The following are the fore and back bearings of the sides of
a closed traverse:
Side FB BB
AB 150 15' 330 15'
BC 20 30' 200 30'
CD 295 45' 115 45'
DE 218 0' 38 0'
EA 120 30' 300 30'
Calculate the interior angles of the traverse. (10 marks)
Solution: (a)
(i) Fore and Back Bearing
The bearing of a line measured in the direction of the progress
of survey is called
the "fore bearing" (FB) of the line.
The bearing of a line measured in the direction opposite to the
survey is called the
'back bearing' (BB) of the line.
(ii) Magnetic declination
The horizontal angle between the magnetic meridian and true
meridian is known
as "magnetic declination".
When the north end of the magnetic needle is pointed towards the
west side of the
true meridian, the position is termed "Declination West" (
W).
When the north end of the magnetic needle is pointed towards
east side of the
true meridian, the position is termed "Declination East (
E)".
(b)
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Interior A = 360 - (BB of EA - FB of AB)
= 360 - (300 30' - 150 15')
= 209 45'
Interior
B = 360 - (BB of AB - FB of BC)
= 360 - (330 15' - 20 30')
= 50 15'
Interior
C = FB of CD - BB of BC
= 295 45' - 200 30'
= 95 15'
Interior
D = FB of DE - BB of CD
= 218 0' - 115 45'
= 102 15'
Interior E = FB of EA - BB of DE
= 120 30' - 38 0'
= 82 30'
Check:
interior angles (2N - 4) 90 = 540=
A +
B +
C +
D +
E = 540 (
ok!)
12(a) The follow are the interior angels of a closed traverse
ABCD.
A = 87 50' 20",
B = 114 55' 40",
C = 94 38' 50",
D = 129 40' 40",
E = 112 54' 30".
If the observed bearing of AB is 221 18' 40", Calculate the
bearings of the remaining
side of the traverse. (12 marks)
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(b) The following bearings were observed in a compase
traverse:
Line Bearing
AB N 52 45' E
BC N 35 30' E
CD S 85 15' E
DE N 46 45' E
EF S 82 0' E
Calculate the deflection angles. (8 marks)
Solution:
(a) Bearng of AB = 221 18' 40"
Add
B = +114 55' 40"
= 336 14' 20"
Deduct 180 = -180 0' 0"
Bearing of BC = 156 14' 20"
Add
C = +94 38' 50"
= 250 53' 10"
Deduct 180 = -180 0' 0"
Bearing of CD = 70 53' 10"
Add
D = +129 40' 40"
= 200 33' 50"
Deduct 180 = -180 0' 0"
Bearing of DE = 20 33' 50"
Add
E = +112 54' 30"= 133 28' 20"
Add 180 = + 180 0' 0"
Bearing of EA = 313 28' 20"
Add
F = +87 50' 20"
= 401 18' 40"
Deduct 180 = -180 0' 0"
Bearing of AB = 221 18' 40" (
ok)
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(b)
Deflection angle at B = Bearing of AB - Bearing of BC
= 52 45' - 34 30'
= 18 15' L
Deflection angle at C = 180 - (bearing of BC + bearing of
CD)
= 180 - (34 30' + 85 15')
= 60 15' R
Deflection angle at D = 180 - (bearing of CD + bearing of
DE)
= 180 - (85 15' + 46 45')
= 48 0' L
Deflction angle at E = 180 - (bearing of DE + bearing of EF)
= 180 - (46 45' + 82 0')
= 51 15' R
13(a) The following angles were measured in running a closed
traverse ABCDE in a
clockwise direction:
Station Included angle (exterior)
A 291 33'
B 225 13'
C 211 36'
D 300 26'
E 231 12'
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Compute the bearing of the remaining sides of the traverse given
that the observed
bearing of AB was 10 12. (12 marks)
(b) A traverse is run form A to G and the deflection angles are
as follows:
At station B, 32 16' L; C, 18 34' R; D, 22 12' L; E, 42 24' R;
F, 52 42' R. Compute
the bearings of the remaining lines of the traverse, given that
the forward bearing of the
AB is 110 6'. (8marks)
Solution:
(a) Bearing of AB = 10 12'
Add B = +225 13'
= 235 25'
Deduct 180 = -180 0'
Bearing of BC = 55 25'
Add
C = +211 36'
= 267 1'
Deduct 180 = -180 0'
Bearing of CD = 87 1'
Add
D = +300 26'
= 387 27'
Deduct 180 = -180 0'
Bearing of DE = 207 27'
Add
E = 231 12'
= 438 39'
Deduct 180 = -180 0'Bearing of EA = 258 39'
Add
A = 291 33'
= 550 12'
Deduct 180 = -180 0'
= 370 12'
Deduct 360 = -360 0'
Bearing of AB = 10 12' (ok)
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(b) Bearing of AB = 110 6'
Deduct
B = -32 16'
Bearing of BC = 77 50'
Add
C = +18 34'
Bearing of CD = 96 24'
Deduct
D = -22 12'
Bearing of DE = 74 12'
Add
E = +42 24'
Bearing of EF = 116 36'
Add
F = +52 42'
Bearing of FG = 169 18'
Check
Bearing of last line = Bearing of first line +
Right deflection angle -
left deflection angle
= 110 6' + ( 18 34' 0 42 24' + 52 42') - (32 16' + 22 12')
= 169 18' (ok)
14(a) The following deflection angles were measured in running a
traverse from A toG.
Station Deflection angle
B 23 47' R
C 18 19' L
D 37 20' R E 15 38' R
F 10 12' L
If the true bearing of AB is N 62 18' E, Calculate the true
bearing of the remaining
sides. (12 marks)
(b) Below are given the back angles:
Station B: 164 36'; station C: 196 12'; station D: 170 24'. Find
the azimuths of the
remaining lines, given that the fore azimuth of AB is 36 18'. (8
marks)
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Solution:
(a) True Bearing of AB = N 62 18' E
= 62 18'
Add
B = +23 47'
True Bearing of BC = 86 5' (N 86 5' E)
Deduct
C = -18 19'
True Bearing of CD = 67 46' (N 67 46' E)
Add
D = + 37 20'
True Bearing of DE = 105 6' (S 74 54' E)
Add
E = +15 38'
True Bearing of EF = 120 44' (S 59 16'E)
Deduct
F = -10 12'
True Bearing of FG = 110 32' (S 69 28'E)
Check
Bearing of Last line = Bearing of first line +
right deflection angle -
left deflection angle
= 62 18' + (23 47' + 37 20' + 15 38') - (18 19' +
10 12')
= 110 32' (S 69 28' E) (ok)
(b) Azimuth of AB = 36 18'
Add
B = +164 36'
= 200 54'
Deduct 180 = -180
Azimuth of BC = 20 54'
Add
C = +196 12'
= 217 6'
Deduct 180 = -180
Azimuth of CD = 37 6'
Add LD = 170 24'
= 207 30'
Deduct 180 = -180
Azimuth of DE = 27 30'
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15 (a) Define the "Local Attraction". (4 marks)
(b)The following are the observed bearings of the lines of a
tranverse ABCDEA with
a compass in a place where local attraction wass suspected.
Line FB BB
AB 191 45' 13 0'
BC 39 30' 222 30'
CD 22 15' 200 30'
DE 242 45' 62 45'
EA 330 15' 147 45'
Find the correct bearing of the lines. (16 marks)
Solution:
(a) Local Attraction
A magnetic needle indicates the north direction when dreely
suspended or pivoted.
But if the needle comes near some magnetic substances, such as
iron ore, steel structures,
electric cables conveying current, etc. It is found to be
deflected from its true direction,
and does not show the actual north. This distrubing influence of
magnetic substances is
known as "Local attractin".
(b)
AB 191 45' 13 0' +2 30' at A 194 15' 14 15'
BC 39 30' 222 30' +1 15' at B 40 45' 220 45'
CD 22 15' 200 30' -1 45' at C 20 30' 200 30'DE 242 45' 62 45' 0
at D 242 45' 62 45'
EA 330 15' 147 45' 0 at E 330 15' 150 15'
Lin
eObserved
Correct
ion RemarksFB BB FB BB
Stations D
and E are
free fromLocal
Attraction
Correct
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16 The following are the bearings observed in traversing, with a
compass, an area where
local attraction was suspected. Calculate the interior angles of
the traverse and correct
them if necessary.
Line FB BB
AB 150 0' 330 0'
BC 230 30' 47 0'
CD 306 15' 127 45'
DE 298 0' 120 0'
EA 49 30' 229 30' (20 marks)
Solution:
Interior
A = BB of EA - FB of AB
= 229 30' - 150 0' = 79 30'
Interior
B = BB of AB - FB of BC
= 330 0' - 230 30' = 99 30'
Exterior
C = FB of CD - BB of BC
= 306 15' - 47 0' = 259 15'
Interior
C = 360 - 259 15' = 100 45'
Exterior
D = FB of DE - BB of CD
= 298 0' - 127 45' = 170 15'
Interior
D = BB of DE - FB of EA
= 120 0' - 49 30' = 70 30'
interior angles =
A +
B +
C +
D +
E = 540
(2N - 4) 90 = (2 5 - 4) 90 = 540 (ok)
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Lines AB and EA are free from local attraction.
FB of AB = 150 0'
FB of BC = 230 30'
FB of EA = 49 30'
Then,
FB of CD = (BB of BC) + Exterior
C
= (230 30' - 180) + 259 15'
= 309 45'
FB of DE = (BB of CD) + Exterior
D
= (309 45' - 180) + 170 15'
= 300 0'
17 The following bearings were observed in traversing with a
compass, an area where
local attraction was suspected. Find the amounts of local
attraction at different stations,
the correct bearings at lines and the included angles.
Line FB BB
AB 68 15' 248 15'
BC 148 45' 326 15'
CD 224 30' 46 0'
DE 217 15' 38 15'
EA 327 45' 147 45' (20 marks)
LineCorrected
FB BB
AB 150 0' 330 0'
BC 230 30' 50 30'
CD 309 45' 129 45'
DE 300 0' 120 0'
EA 49 30' 229 30'
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Solution:
(i) Interior
A = BB of EA - FB of AB
= 147 45' - 68 15' = 79 30'
(ii) Interior
B = BB of AB - FB of BC
= 248 15' - 148 45' = 99 30'
(iii) Interior
C = BB of BC - FB of CD
= 328 45' - 227 0' = 101 45'
(iv) Interior
D = 360 - (FB of DE - BB of CD)
= 360 - (218 15' - 47 0') = 188 45'
(v) Interior
E = 360 - (FB of EA - BB of DE)
= 360 - (327 45' - 38 15') = 70 30'
Check
interior angles =
A +
B +
C +
D +
E = 540
(2N - 4) 90 = 540 (ok)
LineObserved
FBCorrection Remark
AB 68 15' 248 15' 0 at A 68 15' 248 15'
BC 148 45' 326 15' 0 at B 148 45' 328 45'
CD 224 30' 46 0' + 2 30' at C 227 00' 47 0'
DE 217 15' 38 15' + 1 0' at D 218 15' 38 15'
EA 327 45' 147 45' 0 at E 327 45' 147 45'
Stations A, B
and E are free
from local
attraction
BB
Correct
FB BB
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18(a) What are the accessories of plane tabling?
Describe the plane table. (8marks)
(b) Describe the method of orientation by backsighting. (12
marks)
Solution:
(a) The accessories of plane tubling are
(1) Plane Table
(2) Alidade
(3) Spirit level
(4) Compass
(5) U-fork of Plumbing Fork with Plumb Bob
The Plane Table
The plane table is a drawing borad of size 750mm 600mm made of
well-seasoned
wood like teak.pine.etc. The top surface of the table is well
leveled. The bottom surface
consists of a threaded circular plate for fixing the table on
the tripod stand by a wing nut.
The plane table is meant for fixing a drawing sheet over it. The
positions of the
objects are located on this sheet by drawing rays and plotting
to any suitable scale.
(b) Orientation by backsighting
This method is accurate and is always preferred.
Procedure
(a) Suppose A and B are two stations. The plane table is set up
over A. The table is
levelled by spirit level and centered by U-fork so that point a
is just over station A.
The north line is marked on the right-hand top corner of the
sheet by trough compass.
(b) With the alidade touching a the ranging rod at B is bisected
and a ray is drawn. The
distance AB is measured and plotted to any suitable scale. So,
the point b represents
station B.
(c) The table is shifted and set up over B. It is levelled and
centered so that b is just over
B. Now the alidade is placed along the ba and the ranging rod at
A is bisected by
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rays are drawn through the fiducial edge of the alidade.
(d) The distance AB is measured and plotted to any suitable
scale to obtain the point b.
(e) The table is shifted and centred over B and levelled
properly. Now the alidade is
placed along the line ba and orientation is done by
backsighting. At this time it
should be remembered that the centring, levelling and
orientation must be perfect
simulataneously.
(f) With the alidade touching b, the object p is bisected and a
ray is drawn. Suppose this
ray intersects the previous ray at a pointp. This pointp is the
required plotted position
of P.
20 Describe, with a neat sketch, the "Traversing method". (20
marks)
Solution:
The Traversing Method
This method is suitable for connecting the traverse stations.
This is similar to
compass traversing or theodolite traversing. But here fielding
and plotting are done
simulataneously with the help of the radiation and intersection
methods.
Procedure
(a) Suppose A.B.C and D are the traverse stations.
(b) The table is set up the station A. A suitable point a is
selected on the sheet in such a
way that the whole area may be plotted in the sheet. The table
is centred, levelled
and clamped. The north line is marked on the right-hand top
corner of the sheet.
(c) With the alidade touching point a the ranging rod at B is
bisected and a ray is drawn.
The distance AB is measured and plotted to any suitable
scale.
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(d) The table is shifted and centred over B. It is then
levelled,oriented by back-sighting
and clamped.
(e) With the alidade touching point b, the ranging rod at C is
bisected and a ray is
drawn. The distance BC is measured and plotted to the same
scale.
(f) The table is shifted and set up at C and the same procedure
is repeated.
(g) In this manner, all station of the traverse are
connected.
(h) At the end, the finishing point may not coincide with the
starting point and there
may be some closing error. This error is adjusted graphically by
Bowditchs rule.
(i) After making the corrections for closing error the table is
again set up at A. After
centering levelling and orientation the sorrounding details are
located by radiation.
(j) The table is then shifted and set up at all the stations of
the traverse and after proper
adjustments the details are located by the radiation and
intersection method.
21 The following consecutive readings were taken with a dumpy
level along a chain line
at a common interval of 15 m. The first reading was at a
chainage of 165m where the RL
is 98.085. The instrument was shifted after the fourth and ninth
readings. 3.150, 2.245,
1.125, 00860, 3.125, 2.760, 1.835, 1.470, 1.965, 1.225, 2.390
and 3.035m.
Mark rules on a page of your notebook in the form of a level
book page and enter
on it the above readings and find the RL of all the points
by:
1. The collimation system and
2. The rise-and-fall system. Apply the usual checks. (20
marks)
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1. By the collimation system.
Arithmetical check:
BS - FS = 7.500+5.860 = 1.640Last RL - 1st RL = 99.725 - 99.085
= 1.640 (ok)
2. By the rise-and-fall system
Station
pointChainage BS IS FS Rise RL RemarkFall
1 165 3.150 98.085
2 180 2.245 0.905 98.990
3 195 1.125 1.120 100.110
4 210 3.125 0.860 0.265 100.375 Change point
5 225 2.760 0.365 100.740
6 240 1.835 0.925 101.665
7 255 1.470 0.365 102.030
8 270 1.225 1.965 0.495 101.535 Change point
9 285 2.390 1.165 100.370
10 300 3.035 0.645 99.725Total= 7.500 5.860 3.945 2.305
Station
pointChainage BS IS FS
RL of
collimation
line (HI)
RL Remark
1 165 3.150 101.235 98.085
2 180 2.245 98.990
3 195 1.125 100.110
4 210 3.125 0.860 103.500 100.375 changed point
5 225 2.760 100.740
6 240 1.835 101.665
7 255 1.470 102.030
8 270 1.225 1.965 102.760 101.535 Change point
9 285 2.390 100.370
10 300 3.035 99.725
Total 7.500 5.860
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Arithmetical checl:
BS -
FS = 7.500-5.860 = + 1.640
Rise -
fall = 3.945-2.305 = + 1.640
Last RL - 1st RL = 99.725 - 98.085 = + 1.640
22 (a) Define the following terms
(i) Levelling (2 marks)
(ii) Level surface (2 marks)
(b)The following consecutive readings were taken with a level
and a 4-meter levelling
staff on a continuously sloping ground at common intervals of
30m:
0.855 (on A) 1.545m, 2.335, 3.115, 3.825, 0.455, 1.380, 2.055,
2.855, 3.455,
0.585, 1.015, 1.850, 2.755, 3.845 (on B).
The RL of A was 380.500. Mark entires in a level book and apply
the usual
checks. Determine the fradient of AB. (16 marks)
(i) Levelling
The art of determining the relative heights of different point
on or below the
surface of the earth is known as levelling. Thus, levelling
deals with measurement in the
vertical plane.
(ii) Level surface
Any surface parallel to the mean spheroidal surface of the earth
in side to be a
level surface. Such a surface is obviously curved. The water
surface of a still lake is also
considered to be level surface.
(b)
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Check: BS - FS = 1.895 - 11.125 = - 9.230Rise -
fall = 0 - 9.230 = - 9.230
Last RL - 1st RL = 371.270 - 380.500 = - 9.230
Falling gradient of AB =
different of level
horizontal distance
=
9.230
360
=
1
39
(i.e 1 in 39)
23 (a) Define the following terms.
(i) Datum surface (or) line (2marks)
(ii) Reduced level (2 marks)
(b)The following consecutive readings were taken with a level
and a 4 metre levelling
staff on continuously sloping ground at a common interval of
30m.
0.585 on A, 0.936, 1.953, 2.846, 3.644, 3.938, 0.962, 1.035,
1.689, 2.534, 3.844, 0.956,
1.579, 3.016, on B.
Station
pointChainage BS IS FS Rise(+) RL RemarkFall (-)
A 0 0.855 380.500
30 1.545 0.690 379.810
60 2.335 0.790 379.020
90 3.115 0.780 378.240
120 0.455 3.825 0.710 377.530 Change points
150 1.380 0.925 736.605
180 2.055 0.675 375.930
210 2.855 0.800 375.130
240 0.585 3.455 0.600 374.530 Change points
270 1.015 0.430 374.100
300 1.850 0.835 373.265
330 2.755 0.905 372.360
B 360 3.845 1.090 371.270
Total 1.895 11.125 0 9.230
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The elevation of A was 520, 450. Make up a level book and apply
the usual checks.
Determine the gradient of the line AB. (16 marks)
Solution:
(i) Datum surface or line
This is an imaginary level surface or level line from which the
vertical distances
of different points (above or below this line) are measured.
(ii) Reduced level (RL)
The vertical distance of a point above or below the datum line
is known as the
reduced level (RL) of that point. The RL of a point may be
positive or negative according
as the point is above or below the datum.
(b)
Station
A 0 0.585 521.035 520.450
30 0.936 520.099
60 1.953 519.082
90 2.846 518.189
120 3.644 517.391
150 0.962 3.938 518.059 517.097 Change point
180 1.035 517.024
210 1.689 516.370
240 2.534 515.525
270 0.956 3.844 515.171 514.215 Change point
300 1.579 513.592
B 330 3.016 512.155
Readings
Backsight
Intersight
Foresight
CollimationReduced
LevelRemarks
Distance
Arithmetical
check
2.503 10.798- 8.295
- 8.295
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R.L of A = 520.450
R.L of B = 521.155
Difference = - 8.295
There is a fall of 8.925 metres from A to B. The distance AB =
330 metres.
Gradient of the line AB =
8.295 1 1= =
330 39.77 39.8
i.e. 1 in 39.8 (falling)
24 (a) Define the following terms.
(i) Foresigh reading (2 marks)
(ii) Intermediates sights reading. (2 marks)
(b) Following consecutive readings were taken with a dumpy
level:
0.894, 1.643, 2.896, 3.016, 0.954, 0.692, 0.582, 0.251, 1.532,
0.996, 2.135.
The instrument was shifted after the fourth and the eighth
readings. The first reading
was taken on the staff held on the bench mark of R.L
820.765.
Rule out a page of a level field book and enter the above
readings. Calculate the reduced
levels of the points and show the usual checks.
What is the difference of level between the first and last
points? Using the collimation
system. (16 marks)
Solution:
(i) Foresigh reading (FS)
It is the last staff reading in any set up of the instrument and
indicates the shifting
of the latter .
(ii) Intermediates sights reading (IS)
It is any other staff reading between the BS and FS in the same
set up of the
instrument.
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The difference of level between the first and last points =
820.765 - 818.143 = 2.022,
which indicates that there is a fall form the 1st point to the
last point.
25 (a) Define the following terms.
(i) Bench - marks (2 marks)
(ii) Backsight reading (2 marks)
(b) The following consecutive readings were taken with a
levelling instrument of
intervals of 20m.
2.375, 1.730, 0.615, 3.450, 2.835, 2.070, 1.835, 0.985, 0.435,
1.630, 2.255 and 3.630m.
The instrument was shifted after the fourth and eight readings.
The last reading was
taken on a B.M of R.L 110.200m. Find the R.Ls of all the points.
Using Rise and Fall
system. (16 marks)
Solution:
(a) (i) Bench-marks (BM)
Station
1 0.894 820.765 B.M
2 1.643 0.749 820.016
3 2.896 1.253 818.763
4 0.954 3.016 0.120 818.643 C.P
5 0.692 0.262 818.905
6 0.582 0.110 819.015
7 1.532 0.251 0.331 819.346 C.P
8 0.996 0.536 819.882
9 2.135 1.139 818.743
Station
Back
sight
Inter
sight
Fore
sightRise Fall
Reduced
LevelRemarks
3.380 5.402 1.239 3.261
- 2.022 - 2.022
Arithmetic
Check- 2.022
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These are fixed points or marks of known RL determined with
reference to the
datum line. These are very important marks. They serve as
reference points for finding
the RL of new points or for conducting levelling opreations in
projects involving roads
railways etc.
(ii) Backsight reading (BS)
This is the first staff reading taken in any set up of the
instrument after the levelling
has been perfectly done. This reading is always taken on a point
of known RL i.e on
bench-mark or change points.
(b) Rise and Full system
Check
BS -
FS = 5.645 - 8.065 = -2.42
Rise -
Fall = 3.61 - 6.03 = -2.42
Last RL - 1st RL = 110.2 - 112.62 = -2.42 (ok)
Chainage(m)
Station BS FS Rise Fall RL Remarks
0 2.375 112.62
20 1.73 0.645 113.265
40 0.615 1.115 114.38
60 2.835 3.45 2.835 111.545 C.P
80 2.07 0.765 112.31
100 1.835 0.235 112.545
120 0.435 0.985 0.85 113.395 C.P
140 1.63 1.195 112.2
160 2.255 0.625 111.575
180 3.63 1.375 110.2 B.M
Total 5.645 8.065 3.61
IS
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26 (a) Define the following terms.
(i) Change point (2 marks)
(ii)Height of instrument (2 marks)(b) The following successive
readings were taken with a dumpy level along a chain
line at common intervals of 20m. The first reading was taken on
a chainage 140m. The
R.L of the second change point was 107.215m. The instrument was
shifted after the
third and seventh readings. Calculated the RLs, of all the
points;
3.150, 2.245, 3.860, 2.125, 0.760, 2.235, 0.470, 1.953, 3.225,
and 3.890m. Using Rise
and Fall method. (16 marks)
Solution:
(a) (i) Change point (CP)
This point indicates the shifting of the instrument. At this
point, an FS is taken
from one setting and a BS from the next setting.
(ii) Height of instrument (HI)
When the levelling instrument is property levelled the RL of the
line of collimation
is known as the height of the instrument. This is obtained by
adding the BS reading tothe RL of the BM or CP on which the staff
reading was taken.
(b) Rise and Fall method
Chainage(m)
Station BS FS Rise Fall RL Remarks
140 3.15 103.565160 2.245 0.905 104.47
180 3.86 1.125 1.12 105.59 CP
200 20125 1.735 107.325
220 0.76 1.365 108.69
240 0.47 2.235 1.475 107.215 CP
260 1.935 1.465 105.75
280 3.225 1.29 104.46
300 3.89 0.665 103.795
Total 7.48 7.25 5.125 4.895
IS
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Check
BS -
FS = 7.48 - 7.25 = 0.23
Rise -
Fall = 5.125 - 4.895 = 0.23
Last RL - 1st RL = 103.795 - 103.565 = 0.23 (ok)
27 (a) Define the following terms.
(i) Horizontal plane (2 marks)
(ii) Line of the collimation (2 marks)
(b)In fly levelling from a BM of R.L 140.605, the following
readings were observed;
Backsight - 1.545, 2.695, 1.415, 2.925
Foresight - 0.575, 1.235, 0.595
From the last position of the instrument, six pegs at 20m
intervals are to be set out on a
uniformly rising gradient of 1 in 50, the first peg is to have
an RL of 144.000. Find the
staff readings and RLs of the pegs. (16 marks)
Solution
(a) (i) Horizontal plane
Any plane tangential to the level surface at any point is known
as the horizontal
plane. It is perpendicular to the plumb line which indicates the
direction of gravity.
(ii) Line of the collimation
It is an imaginary line passing through the intersection of the
cross-hairs at the
diaphragm and the optical centre of the object glass and its
continuation. It is also known
as the line of sight.
(b) For pegs, rising gradiend = 1 in 50 =
1
50
H = 50
V = 1
H = 20
V = ? (
1
50
20 = 0.4m)
RL of peg (2) = 144 + 0.4 = 144.4
RL of peg (3) = 144.4 + 0.4 = 144.8 and so on.
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Check
BS -
FS = 8.58 - 3.185 = 5.395
Last RL - 1st RL = 146 - 140.605 = 5.395 (ok)
28 Define the terms "contour line", "contour interval" and
"horizontal equiralent".
(20 marks)
Solution:
1. Contour line
The line of intersection of a level surface with the ground
surface is known as the
contour line or simply the contour. It can also be difined as a
line passing through points
of equal reduced levels.
For example, a contour of 100m indicates that all the points on
this line have an
RL of 100m. Similarly, in a contour of 99m all points have an RL
of 99m, and so on. Amap showing only the contour lines of an area
is called a contour map.
Station Distance BS IS FS HI RL Remark
1.545 142.15 140.605 BM
2.695 0.575 144.27 141.575 CP
1.415 1.235 144.45 143.035 CP
2.925 0.595 146.78 143.855 CP
0 2.78 146.78 144 peg - 1
20 2.38 144.4 peg - 2
40 1.98 144.8 peg - 3
60 1.58 145.2 peg - 4
80 1.18 145.6 peg - 5
100 0.78 146 peg - 6
8.58 3.185
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2. Contour interval
The vertical distance between any two consecutive contours is
known as a contour
interval. Suppose a map includes contour lines of 100m, 98m, 96m
and so on. Thecontour interval here is 2m. This interval depends
upon: (i) the nature of the ground (i.e
whether flat or sleep). (ii) the scale of the map and (iii) the
purpose of survey.
Contour intervals for flat country are generally small, e.g.
0.25m, 0.50m, 0.75m,
etc. The contour interval for a sleep slope in a hilly area is
generally greater, e.g. 5m,
10m, 15m, etc.
Again for a small-scale map,the interval may be of 1m, 2m, 3m,
etc and for large
scale map, it may be of 0.25m, 0.05m, 0.75m, etc.
It should be remembered that the contour interval for a
particular map is constant.
3. Horizontal equivalent
The horizontal distance between any two consecutive contours is
known as
horizontal equivalent. It is not constant. It varies accordin to
the steepness of the ground.
For steep slopes, the contour lines run close together, and for
flatter slopes they are
widely spaced.
29 Describe fully, with sketches, the characteristics of
contours. (20 marks)
Solution:
Characteristics of Contours
1. In Fig the contour lines are closer near the top of a hill or
high ground and wide apart
near the foot. This indicates a very steep slope towards the
peak and a flatter slope
towards the foot.
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2. In Fig the contour lines are closer near the bank of a pond
or depression and wide
apart towards the centre. This indicates a steep slope near the
bank and a flatter slope at
the centre.
3. Uniformly spaced, contour lines indicate a uniform slope.
4. Contour lines always form a closed circuit. But these lines
may be within or outside
the limits of the map.
5. Counter lines cannot cross one another, except in the case of
an overhanging cliff. But
the overlapping portion must be shown by a dotted line.
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6. When the higher values are inside the loop, it indicates
ridge line. Contour lines cross
ridge at right angles.
7. When the lower values are inside the loop, it indicated a
valley line. Contour lines
cross the valley line at right angles.
8.A series of closed contours always indicates a depression or
summit. The lower values
being inside the loop indicates a depression and the higher
values being inside the loop
indicates a summit.
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9. Depression between summits are called saddles.
10. Contour lines meeting at a point indicate a vertical
cliff.
30 Describe the direct method of contouring with sketch. (20
marks)
Solution:
Direct Method
There may be two cases, as outlined below.
Case I: When the area is oblong and cannot be controlled from a
single station:
Procedure
1. Suppose a contour map is to be prepared for an oblong area. A
temporary beneh-
mark is set up near the site by taking fly level readings from a
permanent bench-
mark.
2. The level is then set up at suitable station I from where
maximum area can be covered.
3. The plane table is set up at a suitable station P from where
the above area can be
plotted.
4. A back sight reading is taken on the TBM. Suppose the RI of
the TBM is 249.500m
and that the BS reading is 2.250m. Then theRI of III is
251.750m. If a contour of
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250.000m is required the staff reading should be 1.750m. If a
contour of 249.000m is
required the staff reading should be 2.750m and so on.
5. The staffman holds the staff at different points of the area
by moving up and down or
left and right, until the staff reading is exactly 1.750. Then
the points are marked by
pegs. Suppose, these points are A, B, C, D.
6. A suitable point p is selected on the sheet to represent the
station P. Then, with the
alidade touching p rays are drawn to A.B.C and D. The distances
PA, PB, PC and PD
are measured and plotted to a suitable scale. In this manner,
the points a, b, c and d of
the contour line of RI 250.000m are obtained. These points are
jointed to obtain the
contour of 250.000m.
7. Similarly, the points of the other contours are located.
8. When reqired, the levelling instrument and the plane table
are shifted and set up a
new position in order to continue the operation along the oblong
area.
Case II
Procedure
1. The plane table is set up at a suitable station P from where
the whole are can be
commanded.
2. A point p is suitably selected on the sheet to represent the
station P. Radial lines are
then drawn in different directions.
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3. A temporary bench-mark is established near the site. The
level is set up at a suitable
position I and a BS reading is taken on the TBM. Let the HI in
this setting be 153.250m
so to find the contour of RI 152.000m a staff reading of 1.250m
is required at a
particular point, so that the RI of contour of that point comes
to 152.000m.
RI = HI - staff reading
= 153.250 - 1.250 = 152.000m
4. The staffman holds the staff along the rays drawn from the
plane table station in such
a way that the staff reading on that point is exactly 1.250. In
this manner points A, B,
C, D and E are located on the ground, where the staff readings
are exactly 1.250.
5. The distances PA, PB, PC, PD and PE are measured and plotted
to any suitable scale.
Thus the points a, b, c are obtained which are jointed in order
to obtain a contour of
152.000.
6. The other contours may be located in similar fashion.
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BY
TU (Hmawbi)
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09-5030281,01-620072/620454
