civil iv surveying ii [10cv44] solution

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Surveying II 10CV44 Department of Civil Engineering, S J B Institute of Technology Page 1 Question paper solution Unit 1: 1&5 Definition (Dec 2010 ,June 2011, June 2013) Transiting/Plunging: It’s a process of rotating the telescope about horizontal axis along the vertical plane through 180 0 . Swing: It is a direction of rotation of the instrument about vertical axis in the horizontal plane. When the instrument is rotated in the clockwise direction it is called right swing. When the instrument is rotated in the anticlockwise direction it is called left swing. Double sighting: Prolonging a line is an important supplementary work carried out by theodolite. This method is adopted when the instrument is not in adjustment. Fixing & Centering: It is an adjustment in which the instrument is attached to the tripod stand and then placed exactly over the identified ground point. Levelling: It is a stage of adjustment in which the instrument in made level w.r.t the mean ground at the station. Trunion axis: It is an axis parsing though the centre of vertical circle and A. frame. Telescope is supported and rotated about this axis in the vertical place. Vertical axis: It’s an axis passing through the centre of the level plates. Instrument is rotated about this axis in the horizontal place. Axis of collimation: It is an axis passing through the centre of cross hair of the eyepiece of the and the objective. This should run along the centre of the telescope tube. 2&6) Temporary adjustments (Dec 2010 ,June 2011, june 2013) Before the instrument is put to field usage certain adjustments one to be carried out so that the instrument is ready. Fixing & Centering: It is an adjustment in which the instrument is attached to the tripod stand and then placed exactly over the identified ground point. Levelling: It is a stage of adjustment in which the instrument in made level w.r.t the mean ground at the station.

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Page 1: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 1

Question paper solution

Unit 1:

1&5 Definition (Dec 2010 ,June 2011, June 2013)

Transiting/Plunging: It’s a process of rotating the telescope about horizontal axis

along the vertical plane through 1800.

Swing: It is a direction of rotation of the instrument about vertical axis in the

horizontal plane. When the instrument is rotated in the clockwise direction it is

called right swing. When the instrument is rotated in the anticlockwise direction it

is called left swing.

Double sighting: Prolonging a line is an important supplementary work carried

out by theodolite. This method is adopted when the instrument is not in

adjustment.

Fixing & Centering: It is an adjustment in which the instrument is attached to the

tripod stand and then placed exactly over the identified ground point.

Levelling: It is a stage of adjustment in which the instrument in made level w.r.t

the mean ground at the station.

Trunion axis: It is an axis parsing though the centre of vertical circle and A.

frame. Telescope is supported and rotated about this axis in the vertical place.

Vertical axis: It’s an axis passing through the centre of the level plates.

Instrument is rotated about this axis in the horizontal place.

Axis of collimation: It is an axis passing through the centre of cross hair of the

eyepiece of the and the objective. This should run along the centre of the

telescope tube.

2&6) Temporary adjustments (Dec 2010 ,June 2011, june 2013)

Before the instrument is put to field usage certain adjustments one to be

carried out so that the instrument is ready.

Fixing & Centering: It is an adjustment in which the instrument is attached to the

tripod stand and then placed exactly over the identified ground point.

Levelling: It is a stage of adjustment in which the instrument in made level w.r.t

the mean ground at the station.

Page 2: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 2

1. The 3 foot screens are brought to the certain of their run

2. By adjusting the legs the head plate is made horizontal by eye judgment

3. The plate level is brought parallel to any of the two foot screens and the

corresponding foot screens are turned inwards or outwards simultaneously till

the bubble is in the center of the run

4. Plate level is turned perpendicular to its earlier position and the 3rd

foot screen

is turned inwards or outwards till the bubble comes to the centre.

5. The steps 3 & 4 are repeated till the bubble is at the centre for any direction

Elimination of Parallax: It is an adjustment in which the image of the bisected

object is made to fall on the plane of cross hair.

It is done through the following steps:

1. Looking through the eyepiece lye piece is turned clockwise or anticlockwise

till the cross hairs are seen dark.

2. Telescope is turned to a far off object and looking through the eye piece the

focusing screen is turned till the clean image of the object is seen.

3&7) Measurement of Horizontal angle by repitition method

(June 2011, June 2013)

Horizontal angles are measured in 2 method

A) Method of repetition:

It is a method in which the angle between 2 points on objects in measured repeatedly for n no. of times, in different formats the actual angle in each format will be.

The method is adopted i) When there are few objects between which angle is required

ii) Very accurate value of the angle is required.

Tabulation

Page 3: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 3

Object Bisected Face:-

Scale A

0 , ,,

Scale B

, ,,

Mean

0 , ,,

No. of repetition

P

Q

0 0 0

42 25 20

0 0

26 40

0 0 0

42 26 00

1

P

Q

42 25 20

84 48 40

26 40

48 20

84 48 30 2

P

Q

84 48 40

126 22 20

48 20

21 0

126 21 40 3

Procedure:

1. Instrument is fixed and centered at R ( the reference point)

2. Releasing UCS and LCS the horizontal plate reading is made 00 0

’ 0

’’ the clamp

screws are tightened.

3. Releasing UCS the telescope is turned to Q and UCS is clamped.

4. With Upper tangential screw bisection of point is made.

5. The reading of scale A and scale B are interred in the corresponding column of the

tabular column.

This completes 1st repetition.

6. Releasing LCS the telescope in turned back to of ‘P’ LCS is clamped after

bisection with this the same reading which was at Q now will be at P.

7. The above procedure listed in step 3 and 4 is repeated to the required No. of

repetitions

P R

Q

Page 4: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 4

The accurate angle PRQ will be equal to the final reading/No.of repetitions

4) (Dec 2010)

Face: It is a condition that tells informs the side or position of the vertical circle to

the observer

Face left: If the vertical circle is to the left of the observer it is called face left

observation

Face right: It’s a condition when the vertical circle is to the light of the observer

8) Measuring direct angles, deflection angles, prolonging a straight line, running of a

straight lines between two points, locating point of intersection of two straight lines,

laying of horizontal angle, laying of angle by repetition. (Jan 2014)

9) Prolonging a line is an important supplementary work carried out by theodolite.

a) When theodolite is in adjustment: (Jan 2014)

This procedure helps in continuing a line during base line measurement in

laying out pipe line or roadway, setting out curves e.t.c the work is accurate and is

faster.

Procedure:

1. Let AB be the line to be prolonged

2. Instrument is placed at pt B with all temporary adjustments

3. Releasing the UCS and LCS telescope is turned about the vertical axis till the

peg or arrow at pt A is bisected then the plate ensues i.e (UCS & LCS) are

clamped.

4. Telescope is plunged there by line of sight shifts after B in line with AB.

5. Telescope is now rotated to bisect the tip of the ranging rod at a convenient

distance. This gives point C.

6. Now again the instrument is shifted to point C and the procedure in step 3 & 4

us repeated to get further pints D,E,F etc and hence the line is prolonged.

b) When the theodolite is not in adjustment (Poor adjustment)

Page 5: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 5

This method is adopted when the instrument is not in adjustment.

Procedure:

1. Let AB be the line which is to be continued on prolonged.

2. Instrument is centered about point B with all temporary adjustments.

3. The arrow kept at A is bisected and the horizontal plate is clamped

4. Telescope is transited and a pt C on other side of B is bisected. Releasing the

horizontal plate, pt ‘A’ is again bisected by swinging the telescope and the

horizontal plate is clamped

5. Transiting the telescope to bisect the earlier pt C, when we don’t get C we

bisect another point C’’ in the same line (Now instrument is said to be not in

adjustment.

6. Distance C’ C’’ is measured pt. C is obtained by measuring distance C’C or C’’

C = C’C’’/2

Instrument is now shifted to point ‘C’ which is in line with AB and the above procedure

from step 2 to 5 is repeated

Unit 2:

1) List of permanent adjustments of a transit, spire test (June 2011 & Dec 2010)

Adjusting the horizontal axis (Spire Test):

Aim: Horizontal axis perpendicular to V.A

Necessity: If the horizontal axis is not prospered to V.A when the telescope is

transited we do not get the vertical plane but it will be scant thus during

prolonging a line or while transferring foundation points we do not get the

designed straightness.

Testing:

Page 6: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 6

1. Instrument is established nears to high rise object like transmission town or

multistoried building with all temporary adjustments.

2. Top of the object is bisected and horizontal plate is clamped. Telescope is rotated

down to get a point on the ground mares to the station pt.(B)

3. Telescope is made to transit & the instrument swing back to bisect B again

4. Clamping the horizontal plate telescope is lifted up to bisect earlier point A if

bisected instrument is said to be another pt.C is noted and in adjustment if not

corrections has to be applied.

Rectification or Adjustments:

1) Looking through the telescope in the last stage trunion screw of the vertical frame

is released and the telescope is physically adjusted till the midpoint of AC is

bisected.

2) The tasting procedure is repeated again with the corrections applied at the end of

tartly till point A is bisected. When telescope is transited in the different face.

2&3) List the conditions to be satisfied by a vernier theodolite to be in permanent

adjustments? (Dec 2010 & June 2011)

Following are the desired relationship b/w the various axes for the proper

functioning of theodolite.

1. Plate level axis must be perpendicular to vertical axis: If this relation does not

exist vertical axis will not be truly vertical and hence we get error in centering

of the instrument which gives a very wrong observed value.

2. Horizontal axis must be perpendicular to vertical axis: This is essential to get

accurate horizontal angle measurements in any face.

3. Horizontal axis shall be perpendicular to axis of collimation (AOC): This is

important while prolonging a line or to eliminate the index error in the vertical

circle.

4. Altitude bubble shall be parallel to axis of collimation (AOC): This

relationship will reduce the index error in the vertical forms as well as in

getting concurrent values in different face readings. C

Page 7: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 7

4) Desired relation (June 2011)

i. Collimation axis perpendicular to axis of bubble tube

ii. Axis of bubble tube perpendicular to vertical axis

iii Vertical axis perpendicular to horizontal cross

5) Let us assume line of sight inclined upwards. (June 2011)

Let 'e' be the error in 25 m

Instrument at L1 ,

Correct difference (h) = 1.40 – 2.40 = -1.0 m -------(1)

h is negative, B is lower than A

Instrument at L2, La = 10 m

Correct Staff reading at A = 1.5 – 0.4e

Correct staff reading at B = 2.6 – 2.4e

Difference of levels = (1.5 – 0.4e) – (2.6 – 2.4e)

h = -1.10 +2e --------(2)

Equating (1) & (2)

e = +0.05 0.4e = 0.02 m

2.4e = 0.12 m

Correct S.R at A = 1.48 m

Correct S.R at B = 2.48 m

R.L of B = 200 – 1.00 = 199.0 m

relationship will reduce the index error in the vertical forms as well as in getting

concurrent values in different face readings.

6. Testing: Two peg tests adopted to rectify the collimation axis (Jan 2014)

Page 8: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 8

Two level staffs are kept at 100m apart on a level ground say at A & B

Dumpy level to be tested in placed closed to staff A and reading on staff B is taken as ha

& hb. The difference in elevation between A and B is having which is partially correct \

Instrument is shifted and placed very close to staff B, such that telescope is in contact

with level staff

The staff readings on B and A are taken as ha’ and hb’, the true difference between A and

B is obtained as ha’ ~ hb’ which is partially correct.

The actual difference in elevation between A and B is calculated as

(ha ~ hb) + (ha’ ~ hb’)/2

If the difference in elevation obtained when instrument is at P is equal to difference in

elevation when the instrument in at Q instrument is said to be in adjustment. If not

correction has to be applied.

7) True difference in elevation between A & B = H= 2.250-2.025=0.225 (Jan 2014)

Apparent difference in elevation between A & B=H’=1.875-1.670=0.205

Calculated reading on B at the same level of the staff = 1.875-0.225=1.650

Collimation error for 100m=1.670-1.650=0.020m

Correction on A = 20/100*0.020=0.004m

Correction on B=120/100*0.020=0.024m

Page 9: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 9

Correct reading on staff A=1.875-0.004=1.871m

Correct reading on staff B=1.670-0.024=1.646m

True difference = 1.871- 1.646=0.225m

8). AC=40.76m (June 2012)

BC=46.08m

H1=23.23m

H2= 25.04m

RL of C = 20+2.5+23.53=46.03m

Unit 3:

1) Derivation of expression (4 ) (Dec 2010 ,June 2011, June 2013)

When the base of the object is “Inaccessible” (single plane method)

i. When the perpendicular inst. Is at a lower level than 2nd

inst. St

Data collected:

Instrument at A

O1 – Vertical angle to top of object/tower

S1 – Staff reading on BM

Inst. At B

O2 – Vertical angle to top of object/tower

S2 – staff reading on BM

b – Horizontal distance between st A and st B

70

60 a

50

A B

Page 10: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 10

Calculation:

RL of tower or height of tower = HIA + h1

h1 = D tan θ1 ----------- (1)

h2 = (b + D) tan θ2 ---------- (2)

(2) – (1)

D = [S + b tan θ2] / [tan θ1 - tan θ2]

This is a condition when the object whose height is required will be in a thick forest on

mid of a pond and we cannot reach the base of the object. In such case, we take the help

of the data collected from 2 inst. Stations and calculate the height of the object.

Procedure

1. Let Q be the top of the lower whose height of is required

Page 11: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 11

2. Instrument is set at a convenient point with all temporary adjustments.

3. Bisecting the tip, the vertical angle to the object and staff reading on the BM is

observed as Q1 and S1 respectively.

4. When the top of the object is bisected horizontal plates are clamped

5. Telescope is transited and a ground point B for the second instrument station is

bisected.

6. Instrument is shifted and centered over B.

7. With all temporary adjustments vertical angle Q2 to the top of the object and staff

reading with telescope horizontal on BM as S2 are observed.

8. These data’s are utilized to find the height of the object as shown in the

calculations.

Data Collected:

Inst. At ‘A’ : Q1 – vertical angle to top of the lower h1-ht of top of tower above LOS.

Inst At ‘B’ : Q2 – vertical angle to top of lower h2 – ht of top of lower above LOS

b- Distance between 1st and 2

nd inst

Calculation:

h1 = D tan θ1

h2 = (D + b) tan θ2

But h1- h2 S1- S2 = S

(D + b) tan θ2 – D tan θ1 = h1- h2 =S

D (tan θ2- tan θ1) + b tan θ2 = S

D = (S – b tan θ2) / (tan θ2 – tan θ1)

Substituting D is h1 we get

Page 12: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 12

H1 = (S – b tan θ2) tan θ1 / (tan θ2 – tan θ1)

R.L of top of tower = S1 + BM + h1

2&3,5) List of advantages of total station (June 2011 & Dec 2010)

Advantages of totalstation.

1.reduce error

2.time saving

3.accurate outcome

4.precise data

6) α1 = 280 42' α2 = 18

0 6' b = 100 m (June 2011)

S1 = 1.67 S2 = 2.550

S = 2.550 – 1.670 = 0.88 m

D = (b + S cot α2) tan α2 / (tan α1 – tan α2)

D = 152.13 m

h1 = Δ tan α1 = 152.32× tan 280 42'

h1 = 83.29 m

h2 = (b + Δ) tan α2 => h2 = 82.40 m

R.L of the top of hill

= R.L of B.M + S1 + h1

= 345.580 + 83.29 + 1.67 – 5 = 425.539 m

or 345.580 + 82.40 + 2.55 – 5 = 425.539 m

7) Height of vane about instrument axis is equal to Dtanθ=2000tan9030’=334.68m

CC=0.06728D2=0.06728*4=0.27 (Jan 2014)

Height of vane above instrument axis = 334.68+0.27=334.95m

RL of vane= RL of BM+BS+Height of vane=50.217+0.880+334.95=386.047m

RL of Q=386.047-4.00=382.047m.

8) Distance D=bsinθ2/sin(θ1+ θ2)=60*sin68018’/sin(60

0 30’ + 68

0 18’)=71.532m

Height h1=dtanα1 = 71.532tan100 12’=12.87m

Page 13: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 13

RL of Q= 264.910m

Height h2= bsinθ1 / sin(θ1+ θ2)tanα2=60sin60030’*tan10

018’/sin(60

030’+68

018’)

= 12.78m

RL of Q = 264.910m

Unit 4: (Dec 2010 ,June 2011, June 2013)

1) Definition + Systems of tacheometry

Also called as indirect leveling to find the height of object / RL of object only

with one instrument station RL is found & θ is calculated. Compared to previous

measuring methods this is very fast. Tachometric surveying is a type of surveying in

which we determine the height or elevation of the objects similar to leveling work.

In this method, the horizontal distance to the object base is not measured but

calculated from the observed data. Hence the method is fast easy and convenient. Thus

the method is suitable to find out the elevation sin hilly areas, river valley, rough terrain

where the distance to the object from the instrument cannot be measured.

As all the calculations depends only on observed data. The method may not be

accurate. The staff man should be able to reach the point whose elevation is required. In

tachometric surveying and instrument called tachometer is adopted this is nothing but

normal venire theodolite fitted with stadia haired.

Stadia hairs are the additional cross hairs placed one above & one below the

regular horizontal cross hair.

`

2) pitch p = 1/100 = 0.01 cm, K = f/p = 2250 (June 2011)

m = 3.425 + 3.930 = 7.355

c = f + d = 0.425 m

Δ = KS/m + C S = 3 m

Δ = 2250 x 3/7.355 + 0.425

Δ = 918.17 m

Page 14: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 14

3) K = 100, C = 0, Δ = KS cos2 θ (June 2011)

CA = 100 x cos2 4

0 20' V = (KS sin2θ) /2

CA = 61.64 m --> V1 = 4.671 m

CB = 61.99 m --> V2 = 0.192 m

AB = √( CA2 + CB

2 – 2CA.CB cos 35

0 20')

AB = 37.25 m

Assuming RL of instrument axis as 100.0

R.L of A = RL of instrument axis + V1 – R1 = 103.061

R.L of B = 98.782 m

Difference in elevation between A & B = 103.061 – 98.782

= 4.279 m

Gradient AB = 1 in 8.77

4) determining the constants of a tacheometer (Dec 2010)

Tachometric constants K and C are determined in the field with the following

procedure.

1. Instrument is set at a convenient point on a level ground withal temporary

adjustments.

2. Horizontal plate of the instrument is clamped and along the plane ground points

are marked at regular intervals from the instrument ground point.

3. Staff is held at all the marked points say A,B,C …. And the corresponding staff

intercept (Upper hair heading lower hair reading) say SA, SB, SC,…….are

determined.

4. Knowing the distance to each staff point form the instrument the constant K & C

are determined as shown in the calculations

We have,

D1= K SA+C

D2= K SB+C

D3= K Sc+C

A

B

C 350 20'

Page 15: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 15

Solving above equations

Equation2 – equation 1 gives

D2- D1 = (KSB+C) - (KSA+C)

= KSB - KSA

D2- D1 = K(SB – SA)

K1 = D2 – D1 / SB – SA

K2 =D2 – D1 / SC - SA

K3 = D3 – D2 / SC - SB

True value of K = (K1 + K2 + K3) / 3

In the same way values are substituted in corresponding equations to get value of C1,

C2, C3

True value of C is obtained by taking the average

C = (C1 + C2 + C3) / 3

5. anglAPB=76˙30’ (June 2011)

PA=216.51m

V1= 16.66m

PB=197.16m

V2=17.94m

AB=256.55m, Gradiant=1/216.8m

(Jan 2014)

7) Also called as indirect leveling to find the height of object / RL of object only

with one instrument station RL is found & θ is calculated. Compared to previous

measuring methods this is very fast. Tachometric surveying is a type of surveying in

which we determine the height or elevation of the objects similar to leveling work.

In this method, the horizontal distance to the object base is not measured but

calculated from the observed data. Hence the method is fast easy and convenient. Thus

the method is suitable to find out the elevation sin hilly areas, river valley, rough terrain

A B

P

Page 16: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 16

where the distance to the object from the instrument cannot be measured.

As all the calculations depends only on observed data. The method may not be

accurate. The staff man should be able to reach the point whose elevation is required. In

tachometric surveying and instrument called tachometer is adopted this is nothing but

normal venire theodolite fitted with stadia haired.

Stadia hairs are the additional cross hairs placed one above & one below the

regular horizontal cross hair.

Following are the different arrangements of stadia hairs.

Page 17: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 17

8) Principle of tacheometric surveying:- Tachometric equation with usual notation.

(Jan 2014)

Let us take and internal focusing telescope to obtain tachometric equation.

As in the fig.F is a focal point M is a midpoint and O is a vertical axis pt for the

telescope.

Proof: Let AB be the two pts bisected on a level staff.

The distance between AB is S its inverted image is seen on the diaphragm as ab

the distance between ab will be i(stadia internal) i.e., (Distance between top and bottom

stadia hair) from the ∆le AMB & AMI

AB / ab = u / v -------------- (1)

Here u & v are the conjugate distances and f is the focal length distance, d is the distance

between vertical axis & midpoint.

The lens formula can be used i.e

1/f = 1/u + 1/v

Multiplying uf on both sides

u = f + uf / v

u = (AB / ab) v

Substituting values

u = f + (s/i)f

Where AB = s, ab = i

But D = Horizontal distance to the object is from the vertical axis of the inst.

D = u+d

Substring the value of U from (2) we get.

D = (f/i) s + (f + d)

Here f, i, d are constants for a given instrument.

Page 18: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 18

9) First observation:

VA = f/q*sA*sin2θA/2 = 100*(2.355-1.155)*sin90/2 = 9.386m

DA= f/q*sA*cos2θA= 100*(2.355-1.155)*cos

24

030’ = 119.261m

Second observation:

VB = 26.265m

DB = 142.250m

RL of instrument axis = 150.00+1.550=151.550m

RL of A = 151.550.9.386-1.755= 159.181m

RL of B = 151.550+26.265-2.0=175.81m

Distance between A B =104.050m

Difference of level between A and B = -16.635

Gradient from A to B= 1 in 6.25

Part – B

Unit 5:

1)Definition (Dec 2010 & June 2011)

1. Mid ordinate – Mp (O0) This is the line joining centre of curve to mid point to the

long chord.

O0 = R (1 - cos Δ/2)

2. Vertex distance/external distance : PV= (V)

This is the distance between centre of the curve to intersection angle

VP = R (sec Δ/2 – 1)

3. Point of tangency: T2 (P.T):

This is the point where the curve ends and changes to straight stretch

2. By successive bisection of chord:

Page 19: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 19

In this method the major chord is bisected successively to get the midpoints of small

curves. These midpoints when jointed will give the curvature of simple circular curve.

Procedure:

1. After getting T1, T2 they are connected to get the major chord.

2. The midpoint is identified and midordinate is erected to get P.

3. T1 P is taken as an independent curve for which T1 P is the major chord and the

midpoint M1 is identified.

4. At M1 the midordinate O1 further curve is constructed

5. With this T1 P1 and P1 P will now be 2 independent curves for which we set out

midordinates to get curve points P2 P3 .

6. This process can be continued to get number of midodirate points joining which,

we get smooth curve.

Calculations

Midordinate MP = R (1 - cos Δ/2) = R - √(R

2 – (L/2)

2)

Ordinate M1P1 = R(1 – cos Δ/4) = R - √(R2 – (L/4)

2)

Page 20: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 20

3) O= R - √(R2 – (L)

2)

O1 = 400 - √(4002 – 0

2) = 0

O2 = 0.125m

O3 = 0.50m

O4 = 1.12m

O5 = 2.00m

O6 = 3.13m

4) Various methods of setting out simple curves (Dec 2010)

Simple circular curves can be set in two ways.

a. Linear method: Here, only chain and tape is used as a major device for

setting out curves. The method is adopted for small curves and in roadway

construction.

b. Angular / Instrumental method: Here theodolite is used as a major device

along with tape and chain the method is adopted in setting out large curves,

in railway line construction and where accuracy is preferred.

5) O0 = R- √(R2-(L/2)

2); L = 100m, O0 = 5m, x = 10m (June 2011)

R = 252.5 m

Ox = √(R2-x

2) – (R-O0)

x 10 20 30 40 50

Ox 4.8 4.21 3.21 1.81 0

6&7) Tangent length = Rtan(Δ/2) = 97.48m (June 2011)

curve length = (Π/180) R x Δ = 188.50 m

Chainage of T1 = 1190 – 97.48 = 1092.52

02

Chainage of T2 = 1281.02m

δn = 1718.89 Cn/R Δn = Δn-1 + δn

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Department of Civil Engineering, S J B Institute of Technology Page 21

Peg No Chainage (m) Chord Length(m) Tangential Angle Deflection Angle Actual Thread

Reading

T1 1092.52

1 1110 17.48 10 40' 09'' 1

0 40' 09'' 1

0 40' 0''

2 1140 30 20 51' 53'' 4

0 32' 02'' 4

0 32' 0''

3 1170 30 20 51' 53'' 7

0 23' 55'' 7

0 24' 0''

4 1200 30 20 51' 53'' 10

0 15' 48'' 10

0 15' 40''

5 1230 30 20 51' 53'' 13

0 07' 41'' 13

0 07' 40''

6 1260 30 20 51' 53'' 15

0 59' 34'' 15

0 59' 40''

7 1281.02 21.02 20 0' 26'' 18

0 0' 0'' 18

0 0' 0''

Check = Δ7 = 36/2 = 180

8. Offsets from chord produced (Deflection distance method) (June 2013, Jan 2014)

This method is an important one which is used to set out large curves and cures on high ways or

road constructions.

Page 22: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 22

Procedure:

1. The total curve length is divided into no. of equal parts such that joining these parts we get sub

chords and the sub chord length will be equal to the corresponding curve length ie., T1A = T1A

2. The length of 1st chord C1 is measured along tangent to get A1 At A1 an arc of length o1 is

drawn to cut an arc of radius C1 drawn from T1 this will give point A on the curve

3. The chord T1A is produced to B1 so that A B1 = C2 At B1 and ordinate of length O2 is cut

through arc to cut an arc of radius C2 from pt A this gives pt B on the curve.

4. Step 3 is repeated till we reach T2 by getting points on the curve

Data and Calculation:

T1v = Rare tangent/ Backward tangent

T1A = T1A = C1 = First sub chord

A B1= AB = C2 = Normal chord

O1 = A1a = First ordinate = T1A* 8

O2 = B1b = Second ordinate

= On-1

VT1A = δ = AA1 T1= B1A B2

T1OA = 2 δ

Let the first sub chord T1A Make an angle with the back tangent (race tangent)

Angle V T1A== T1A A1

Now the arc length A1A = 01 = T1A x δ

Here

T1A = T1O x 2 δ

T1OA=2A T1 A1

δ = C1 / 2R

Page 23: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 23

O1 = C12 / 2R

δ' = C2 (C1 + C2) / 2R

On = Cn (Cn + Cn-1) / 2R

9) Degree of curve is expressed in terms of either

a. Radius : (Adopted is Britain)

b. Degree of curvature : (Adopted in India, USA, Germany)

To express the degree of curvature there are 2 definitions

b. Arc definition: According to this degree of curve is the central angle subtended at the centre by

an arc of 30m length.

D0 = 1718.87 / R (degree)

c. Chord definition: According to this degree of curve is the angle subtended at the centre by a

chord of 30M or 10M length.

10) R = 1719 / D

Point Chainage Chord

length

Deflection

angle

Total

deflection

angle

T1 1183 - - -

P1 1190 7 0048’8” 0

048’8”

P2 1210 20 2017’31” 3

05’39”

P3 1230 20 2017’31” 5

023’10”

P4 1250 20 2017’31” 7

040’41”

P5 1270 20 2017’31” 9

058’12”

P6 1290 20 2017’31” 12

015’43”

P7 1310 20 2017’31” 14

033’14”

T2 1313.89 3.89 0026’45” 14

059’59”

Page 24: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 24

Unit 6:

1) (June 2011)

R1 = 350m R2 = 500m

Length of common tangent

R1tan 55/2 + R2 tan 25/2 = PQ

PQ = common tangent = 293.04 m

length of First tangent = 125.75 + 350 tan 55/2

T1I = 307.95m

Length of second tangent = 243.74+500 tan 25/2

T2I = 354.59m

I Arc Length = (Π/180) 350 x 550 = 335.93

II Arc length = 218.14m

2) Sin (Δ/2) = V/L = 25/220 (June 2011)

Δ = 130 03'

V = 2R(1- cos Δ) R = 484m L = 110m

Setting of I arc offsets from long chord

X m 0 10 20 30 40 50 55

Ox 3.14 3.04 2.73 2.21 1.48 0.55 0

O0 = R- √(R2-(L/2)

2) O0 = 3.14m

Ox = √(R2-x

2) – (R-O0)

3) Compound curves: A kind of circular curves which has more than one are of a

circle having different radio with centers of the arc on one side of common

tangent.

Length of the common tangent = R=(R1tanΔ1/2+ R2 tanΔ2/2)

Length of the arc = πR1 Δ1/180

I

P Q

T1 T2

250

550

Page 25: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 25

4) Reverse curve: It’s a kind of circular curve which has more than one arc of a

circle with different on equal radii having the centers of are on either side of the

common tangent.

They are used when either the two straights are parallel or their angle of

intersection is too small. They are used in hilly terrains, in railway sidings, on

highways when low speed is required. When r1 and r2 are the radius of two

circular arms total angle of deflection is Δ the central angle is Δ1 and Δ2. δ1 and δ2

are the angle between the straights.

R=d/(tanΔ1/2 + tanΔ2/2)

5) θ1= 1800- 150

0 = 30

0

θ2 = 1800- 140

0 = 40

0

θ = 300+40

0=70

0

Length of the tangent for first arc = 200tan150=53.58m

Length of the tangent for second arc = 300tan200=109.19m

Common tangent length = 162.77m

Chainage of T1= 950-164.92=785.08m

Curve length = 104.72m

Chainage of T2= 785.08+104.72= 889.80m

Long curve length = 209.44m

Chainage of T3=1099.24m

Deflection angle of full chord = 1718.9*20/200=2051’53”

Deflection angle of sub-chord= 1718.9*4.12/200 = 0040’34”

Deflection angle of full chord for 30m= 1718.9*30/300=2051’53”

Deflection angle of final sub-chord = 1718.9/300*29.44=2048’41”

Page 26: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 26

Unit 7:

1) Transition curve (Dec 2010, June 2011, June 2013)

Characteristics of Transition Curve

1. In order to fit in the transition curve at the ends a circular imaginary curve of

slightly greater radius has to be shifted towards the center. The distance

through which the curve is shifted is known as shift (S) of the curve equal to

L2/24R, where L is the length of each transition curve and R is the desired

circular curve

2. Length of the combined curve is equal to

= (R + S) tan Δ/2 + L/2

3. Spiral angle φ1 = L /2R radians

4. The central angle for the curve = Δ - 2φ1

5. Length of circular curve is equal to πR(Δ - 2φ1) / 1800

6. Length of combined curve = πR(Δ - 2φ1) / 1800 + 2L

7. Chainage of beginning of combined curve = chainage of intersection point –

total length for combined curve

8. Chainage of junction point of transition curve and circular curve = chainage of

tangent point + length of transition curve

9. Chainage of other junction point of the circular curve and the other transition

curve is equal to chainage of E + length of circular curve.

10. Chainage of the end point of the combined curve = chainage of T + length of

combined curve

11. The deflection angle for any point on the transition curve distant l from the

beginning of combined curve = L/6R radians

Page 27: Civil IV Surveying II [10cv44] Solution

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Department of Civil Engineering, S J B Institute of Technology Page 27

Length of Transition Curve

1. Length may be assumed on the basis of experience and judgment.

2. Length may be such that super elevation is applied at a uniform rate 1 in

300 to 1 in 1200.

3. The length of the transition curve may be such that super elevation is

applied at an arbitrary time rate of “a” cm/sec. The super elevation attained

= (L/v) x a = h.

4. The radial acceleration on the circular curve is L = v3/ CR, where v is the

speed in m/sec , C is the rate of change of radial acceleration in m/sec2 , R

is the radius of curve in meters.

2&4) Length of vertical curve (June 2011)

Vertical curves are introduced at changes of gradient to avoid impact and to

maintain good visibility. They are set out in a vertical plane to round off the angle and to

obtain gradual change of gradient. They are also called as summit curves if they have

convexity upwards and valley curves if they have concavity upwards.

Page 28: Civil IV Surveying II [10cv44] Solution

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Department of Civil Engineering, S J B Institute of Technology Page 28

3) V = 50kmph = 13.88m/s α = 0.5m/s R = 300m (June 2011)

1. length of transition curve = V3/(αR) = 17.867m

2. Super elevation BV2/(9R) = (8 x 13.88

2)/(9.81 x 300) = 0.5243

3. Shift δ = l2/(24R) = 17.869

2/(24 x 300) = 0.044m

5). CR= V2/8R R=315m (June 2013)

L= V3/αR = 227m

Δs = L/2K = 20˙38’

Δc = Δ-2 Δs=38˙42’24”

lc = 212.8m S= L2/24R=6.82m

T= (R+S)tanΔ/2+L/2=383.5m

Ch 0f T1=42478.5m

Ch of D= 42705.5m

Ch of D’= 42918.3m

Ch of T2 = 43145.3m

6)V= 60 kmph= 16.66m/s (Jan 2014)

K=0.3m/s3

Page 29: Civil IV Surveying II [10cv44] Solution

Surveying II 10CV44

Department of Civil Engineering, S J B Institute of Technology Page 29

R=200m

Length of transition curve = v3/KR = 77m

7) Vertical curves (Jan 2014)

Vertical curves are introduced at changes of gradient to avoid impact and to

maintain good visibility. They are set out in a vertical plane to round off the angle

and to obtain gradual change of gradient. They are also called as summit curves if

they have convexity upwards and valley curves if they have concavity upwards.

Unit 8:

1) (June 2011)

i. C = 0 A = M(FR – IR + 10N)

N = +1

A = 100 (4.325 – 80436 + 10)

A = 588.9 cm2

ii. N = -2

A = M(FR – IR + 10N + C)

588.9 = 100[5.434 – 2.844 – 20 +C]

C = 23.299

A0 = MC = 2329.9 cm2

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Department of Civil Engineering, S J B Institute of Technology Page 30

2) (June 2011)

Chainx 0 100 200 300 400 500

F.L 110.5 111.7 112.9 114.1 115.3 116.5

G.L 107.8 106.3 110.5 111 110.7 112.2

Height h 2.7 5.4 2.4 3.4 4.6 4.3

Area of c/s m2

38.8 106.92 33.12 47.12 83.72 75.68

A = ((9+9+4h)/ 2) h = (9 + 2h) h

V= 100[(39.88+75.68)/2 + 106.92 + 33.12 + 47.12 + 83.72]

3) It is a mechanical integrator used for the measurement of areas of figures,

plotted to a scale. There are tw types of planimeters: the Amsler polar planimeter

and the rolling planimeter. If areas that do not have straight line boundaries are

drawn to some scale on a map or plan, their values can be very easily found by the

planimeter. Some of the other uses of planimeter are measuring areas of cross-

sections for highways and railways, and checking computed areas in property

surveys.

The most common type of a planimeter is the polar planimeter. The area is

computed by utilizing the relationship between the tracing arm point, moved over

the outline of the figure, and the connected recording wheel which records the

displacement.

A planimeter essentially consists of two bars hinged together. At the extreme end

on one of the bars, a weight is suspended over a 'needle point' or 'anchor point'

which is used to anchor the bar outside the area to be measured. The other bar

known as the tracing arm, has a 'tracing point' at its extreme end. The tracing point

is moved as desired, about the needle point. At the other end of the tracing arm

there is a roller which rolls on the surface of the plan as the pointer is moved.

Thus, when in use the planimeter has three contact points on the surface of the

plan, the anchor point, the tracing point, and the roller circumference. A fixed

vernier is attached to the roller drum. A disk is also connected to the roller, by a

work drive such that one revolution of the roller turns the disk trough one part.

2

1

2

1

9+4h

9m

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Department of Civil Engineering, S J B Institute of Technology Page 31

In large planimeters the tracing arm is made adjustable. This arrangement has two

distinct advantages. Firstly, the arm can be adjusted to the unit area of the plan; and

secondly, the planimeter may be tested for any unit area, and if found incorrect , the

error can be eliminated by adjusting the arm. Consider a moving line AB, of length L as

shown in the fig. the ends of which move in a given loci. The ends A and B, of the line

AB, are displaced to A1 and B2 respectively. Let the normal displacement of end A be dz

and the rotation of B1 as dӨ.

The area swept ABB2A1,

dA' = Ldz + ½ L× L dӨ

or dA' = Ldz + ½ L2 dӨ

Now suppose a wheel F fixed upon AB, its plane being perpendicular to that line, so that

in the displacement of AB, the wheel rolls when the point F moves perpendicularly to

AB. Let dα be the angle through which the wheel turns upon its axis in passing from F to

F'. If r is the radius of the wheel, rdα is the length of the arc applied to the paper. This

length is equal to dz + the arc L'dӨ

rdα = dz + L'dӨ

Page 32: Civil IV Surveying II [10cv44] Solution

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Department of Civil Engineering, S J B Institute of Technology Page 32

Eliminating dz from Eq.(1)

dA' = rLdα + (L2/2 – LL')dӨ

∫dA' = ∫rLdα + ∫(L2/2 – LL')dӨ

When the directing curve AA1 is exterior to the area X,

∫rLdα = rLα = Lz (where z = rα)

∫ dӨ = 0, since AB returns to its original position without having made a circuit about O.

Therefore integrating Eq.(4)

A' = A = Lz

4.

Trapezoidal rule A=d/2(O0+On+2(O1+O3+…….On-1) (June 2013)

A=188sqm

Simpsons rule A=d/3(O0+On+4(O1+O3+…….)+2(O2+O4+…………)

A=199.66 sqm

5. Cross Sectional area A=(b+nh)h (June 2013)

A0=10.2sqm,

A1=14.84sqm,

A2=28.43sqm,

A3=34.38sqm,

A4=23.63sqm,

A5=16.23sqm,

A6=9.58sqm.

Trapezoidal rule =5096.4 cub.m

Prismoidal rule = 5142.9 cub.m

6) (Jan 2014)

Sl.no Component Chainage Base Offset Mean Offset Area

1 Triangle 0 & 20 20 0 & 42 21 420

2 Trapezoidal 20& 65 45 42 & 58 50 2250

3 Triangle 65 & 110 45 58 & 0 29 1305

4 Triangle 90 & 110 20 0 & 60 30 600

5 Trapezoid 40 & 90 50 60& 20 40 2000

6 Triangle 0 & 40 40 20 & 0 10 400

Total area= 6975m2

Page 33: Civil IV Surveying II [10cv44] Solution

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Department of Civil Engineering, S J B Institute of Technology Page 33

7) Cross Sectional area A=(b+nh)h (Jan 2014)

A0=10.2sqm,

A1=14.84sqm,

A2=28.43sqm,

A3=34.38sqm,

A4=23.63sqm,

A5=16.23sqm,

A6=9.58sqm.

Trapezoidal rule A=d/2(O0+On+2(O1+O3+…….On-1)

V=5096.4 m3

Simpsons rule A=d/3(O0+On+4(O1+O3+…….)+2(O2+O4+…………)

V=5142.9 m3