Curves.pptx

Curves

1. Advance Surveying 1CVPC202 1 Unit no 6 : CURVES Department of C ivil E ngineering Prepared By Mr. Kiran K. Shinde

2. Curves 2 • C u r v es ar e u s u all y em ploye d i n l i nes of communication in order that the change in direction at the intersection of the straight lines shall be gradual. The lines connected by the curves are tangent to it and are called Tangents or Straights. The curves are generally circular arcs but parabolic arcs are often used in some countries for this purpose. Most types of transportation routes, such as highways, railroads, and pipelines, are connected by curves in both horizontal and vertical planes.

3. Curves The purpose of the curves is to deflect a vehicle travelling along one of the straights safely and comfortably through a deflection angle θ to enable it to continue its journey along the other straight. θ 3

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8. Horizontal Alignment Plan View Vertical Alignment Profile View Curves

9. Curves Horizontal Curves Vertical Curves Simple Curves Compound Curves Reverse Curves Transition Curves Circular Curves Spiral Curves (non-circular curves) Classification of Curves Curves 9

10. Curves Classification of Curves Simple Curves: Consist of single Arc Connecting two straights. Compound curves: Consist of 2 arcs of different radii, bending in the same direction and lying on the same sides of their common tangents, their centers being on the same side of the curve. Reverse curves: Consist of 2 arcs of equal or unequal radii, bending in opposite direction with common tangent at their junction (meeting Point), their center lying on the opposite sides of the curve. 10

11. Curves Simple Curve Compound Curve Reverse Curve 11

12. Curves A C R R O E 90 o 90 o T1 T2 ∅ 𝟐 90 o - ∅ /2 Nomenclature of Simple Curves B’ B ∅ I F ∅ 𝟐 ∅ ∅ 𝟐 12

13. Curves Nomenclature of Simple Curves Tangents or Straights: The straight lines AB and BC which are connected by the curves are called the tangents or straights to curves. Point of Intersection: (PI.) The Point B at which the 2 tangents AB and BC intersect or Vertex (V) . Back Tangent: The tangent line AB is called 1st tangent or Rear tangents or Back tangent. Forward Tangent: The tangents line BC is called 2nd tangent or Forward tangent. 13

14. Curves Nomenclature of Simple Curves Tangents Points: The points T 1 and T 2 at which the curves touches the straights. Point of Curve (P.C): The beginning of the curve T 1 is called the point of curve or tangent curve (T.C). b) Point of tangency (C.T): The end of curve T 2 is called point of tangency or curve tangent (C.T). 6) Angle of Intersection: ( I ) The angle ABC between the tangent lines AB and BC. Denoted by I . 14

15. Curves Nomenclature of Simple Curves Angle of Deflection ( ∅ ): Then angle B`BC by which the forward (head tangent deflect from the Rear tangent. Tangent Length: ( BT 1 and BT 2 ) The distance from point of intersection B to the tangent points T 1 and T 2 . These depend upon the radii of curves. Long Cord: The line T 1 T 2 joining the two tangents point T 1 and T 2 is called long chord. Denoted by l . 15

16. Curves Nomenclature of Simple Curves Length of Curve: the arc T 1 FT 2 is called length of curve. Denoted by L . Apex or Summit of Curve: The mid point F of the arc T 1 FT 2 is called Apex of curve and lies on the bisection of angle of intersection. It is the junction of lines radii. External Distance (BF): The distance BF from the point of intersection to the apex of the curve is called Apex distance or External distance. 16

17. 1 2 2 1 ∟ BT T = ∟ BT T = If the curve deflect to the right of the direction of the progress of survey it is called Right-hand curve and id to the left , it is called Left-hand curve. The ∆ BT 1 T 2 is an isosceles triangle and therefore the angle ∅ 2 Curves Nomenclature of Simple Curves Central Angle: The angle T 1 OT 2 subtended at the center of the curve by the arc T 1 FT 2 is called central angle and is equal to the deflection angle. Mid ordinate (EF): It is a ordinate from the mid point of the long chord to the mid point of the curve i.e distance EF . Also called Versed sine of the curve. 17

18. 2 1 1 In ∆ T OB, tan ( ∅ ) = BT / OT 1 1= 1 2 BT OT tan ( ∅ ) 1 = 2= 2 BT BT R tan ( ∅ ) c) Length of Chord( l ): 2 1 1 In ∆ T OE, sin ( ∅ ) = T E / OT 1 1 1 2 T E = OT sin ( ∅ ) 1 2 T E = R sin ( ∅ ) Curves Elements of Simple Curves a) ∟T 1 BT 2 + ∟B`BT 2 = 180 o I + ∅ = 180 o ∟T 1 OT 2 = ∅ = 180 o – I b) Tangent lengths: ( BT 1 , BT 2 ) 18

19. 1 2 l = 2 T E = 2 R sin ( ∅ ) d) Length of Curve (L): L = length of arc T 1 FT 2 L = R ∅ (rad) = 𝜋 R ∅ 180 Or L /2 𝜋 R = ∅ / 360 L =2 𝜋 R ∅ / 360 = 𝜋 R ∅ 180 e) Apex distance or External distance: BF = BO – OF 2 1 1 In ∆ OT B, cos ( ∅ ) = OT / BO Curves Elements of Simple Curves 19 r S = r x 𝜃 𝜃 ∅ L 360 o 2 𝜋 R R R

20. 1 2 2 BO = OT / cos ( ∅ )= R / cos ( ∅ ) 2 BO = R sec ( ∅ ) BF = R sec (∅/ 2) – R 2 BF = R ( sec ( ∅ ) – 1) BF = R ( 1 cos ( ∅ ) – 1) 2 f) Mid ordinate or Versed sine of curve : EF= OF – OE In ∆ T 1 OE, cos (∅/ 2) = OE / OT 1 OE = OT 1 cos (∅/ 2) = R cos (∅/ 2) EF = R – R cos (∅/ 2) EF = R (1 – cos ( ∅ )) Curves Elements of Simple Curves 20 2

21. Curves Designation Of Curves In U.K a curve is defined by Radius which it expressed in terms of feet or chains(Gunter chain) e.g 12 chain curve, 24 chain curve. When expressed in feet the radius is taken as multiple of 100 e.g 200, 300, 400.. . In USA, Canada, India and Pakistan a curve is designated by a degree e.g 2 degree curve , 6 degree curve. Degree Of Curves Degree of curve is defined in 2 ways Arc Definition Chord Definition 21

22. 1) Arc Definition: “ The degree of a curve is the central angle subtended by 100 feet of arc”. Let R = Radius of Curve D = Degree of Curve Then = 𝐷 100 360 2 𝜋 R R = 5729. 5 8 𝐷 ( f ee t ) It is used in highways. Curves Degree Of Curves R R O 100 feet D D 100 feet 360 o 2 𝜋 R R R 22

23. From ∆ OPM sin( 2 ) = 𝑂𝑀 = 𝐷 𝑀 𝑃 50 𝑅 50 sin( 𝐷 ) R = ( f ee t ) 2 It is used in Railway. Example : D = 1 o 1) Arc Def: 2) Chord Def: R = 5729.58 / D = 5729.58 feet R = 50 / sin (D/2) Curves R R O Degree Of Curves 2) Chord Definition: “ The degree of curve is the central angle subtended by 100 feet of chord”. 100 feet D M N 50 ‘ P 50 ‘ D/2 23 = 5729.65 feet

24. Simple Curves Method of Curve Ranging There are a number of different methods by which a centerline can be set out, all of which can be summarized in two categories: Traditional methods: which involve working along the centerline itself using the straights, intersection points and tangent points for reference. The equipment used for these methods include, tapes and theodolites or total stations. Coordinate methods: which use control networks as reference. These networks take the form of control points located on site some distance away from the centerline. For this method, theodolites, totals stations or GPS receivers can be used. 24

25. Along the arc it is practically not possible that is why measured along the chord. Simple Curves Method of Curve Ranging The methods for setting out curves may be divided into 2 classes according to the instrument employed . Linear or Chain & Tape Method Angular or Instrumental Method Peg Interval: Usual Practice--- Fix pegs at equal interval on the curve 20 m to 30 m ( 100 feet or one chain) 66 feet ( Gunter’s Chain) Strictly speaking this interval must be measured as the Arc intercept b/w them, however it is necessarily measure along the chord. The curve consist of a series of chords rather than arcs. 25

26. Peg Interval: For difference in arc and chord to be negligible 𝑅 Length of chord > 20 of curve R = Radius of curve Length of unit chord = 30 m for flate curve ( 100 ft) (peg interval) 20 m for sharp curve (50 ft) 10 m for very sharp curves (25 ft or less) Simple Curves Method of Curve Ranging 26

27. 1 2 2 BT = BT = R tan ( ∅ ) 4) Locate T 1 and T 2 points by measuring the tangent lengths backward and forward along tangent lines AB and BC . Method of Curve Ranging Location of Tangent points: To locate T 1 and T 2 Fixed direction of tangents, produce them so as to meet at point B . Set up theodolite at point B and measure T 1 BT 2 (I) . Then deflection angle ∅ = 180 o – I Calculate tangents lengths by Simple Curves 27

28. Procedure: After locating the positions of the tangent points T 1 and T 2 ,their chainages may be determined. The chainage of T 1 is obtained by subtracting the tangent length from the known chainage of the intersection point B . And the chainage of T 2 is found by adding the length of curve to the chainage of T 1 . Then the pegs are fixed at equal intervals on the curve. The interval between pegs is usually 30m or one chain length. The distances along the centre line of the curve are continuously measured from the point of beginning of the line up to the end .i.e the pegs along the centre line of the work should be at equal interval from the beginning of the line up to the end. Simple Curves Method of Curve Ranging 28

29. Procedure: There should be no break in the regularity of their spacing in passing from a tangent to a curve or from a curve to the tangent. For this reason ,the first peg on the curve is fixed at such a distance from the first tangent point (T 1 ) that its chainage becomes the whole number of chains i.e the whole number of peg interval. The length of the first sub chord is thus less than the peg interval and it is called a sub-chord. Similarly there will be a sub-chord at the end of the curve. Thus a curve usually consists of two sub-chords and a no. of full chords. Simple Curves Method of Curve Ranging 29

30. Method of Curve Ranging Simple Curves 30 1 2 2 1 ∟ B T T = ∟ B T T = Important relationships for Circular Curves for Setting Out The ∆ BT 1 T 2 is an isosceles triangle and therefore the angle ∅ 2 The following definition can be given: The tangential angle 𝛼 at T 1 to any point X on the curve T 1 T 2 is equal to half the angle subtended at the centre of curvature O by the chord from T1 to that point. The tangential angle to any point on the curve is equal to the sum of the tangential angles from each chord up to that point. I.e. T 1 OT 2 = 2(α + β + 𝛾 ) and it follows that BT 1 T 2 = (α + β + 𝛾 ).

31. Problem 01: Two tangents intersect at chainage of 6 +26.57. it is proposed to insert a circular curve of radius 1000ft. The deflection angle being 16 o 38’. Calculate chainage of tangents points Lengths of long chord , Mid ordinate and External distance. Solution: 1 2 2 Tangent length = BT = BT = R tan ( ∅ ) BT 1 = BT 2 = 1000 x tan(16 o 38`/2) = 146.18 ft Length of curve = L = 𝜋 R ∅ 180 o L = 𝜋 x 1000 x 16 o 38` = 290.31ft 180 o Chainage of point of intersection minus tangent length chainage of T 1 plus L Chainage of T 2 = 6 + 26.56 = -1 + 46.18 = 4 + 80.39 = + 2 + 90.31 = 7 + 70.70 Simple Curves 31

32. Problem 01: Solution: Length of chord = l = 2 R sin ( ∅ ) 2 l = 2 x 1000 x sin(36 o 38`/2) = 289.29 ft Mid ordinate = EF = R (1 – cos ( ∅ ) 2 EF= 1000 x ( 1 – cos(36 o 38`/2)) = 10.52 ft 2 Ex. distance = BF = R ( sec ∅ - 1) 2 BF = 1000 x ((1/cos ( ∅ ) – 1) = 10.63 ft Simple Curves 32

33. Problem 02: Two tangents intersect at chainage of 14 +87.33, with a deflection angle of 11 o 21`35``. Degree of curve is 6 o . Calculate chainage of beginning and end of the curve. Solution: D = 6 o R = 5729.58 / D ft = 954.93 ft 1 2 2 Tangent length = BT = BT = R tan ( ∅ ) BT 1 = BT 2 = 954.93 x tan(11 o 21`35``/2) BT 1 = BT 2 = 94.98 ft Length of curve = L = 𝜋 R ∅ 180 o L = 𝜋 x 954.93 x 11 o 21`35`` = 189.33 ft 180 o Chainage of intersection point B minus tangent length BT 1 Chainage of T 1 plus L Chainage T 2 = 14 + 87.33 = - 0 + 94.96 = 13 + 92.35 = + 1 + 89.33 = 15 + 81.68 Simple Curves 33

34. Method of Curve Ranging 1) Linear or Chain & Tape Method These methods use the chain surveying tools only. These methods are used for the short curves which doesn’t require high degree of accuracy. These methods are used for the clear situations on the road intersections. By offset or ordinate from Long chord By successive bisections of Arcs By offset from the Tangents By offset from the Chords produced Simple Curves 34

35. 1 1 OT = R , T E = l 2 OE = OD – DE = R – O o R 2 = ( l /2) 2 + (R – O o ) 2 Method of Curve Ranging 1) Linear or Chain & Tape Method a) By offset or Ordinate from long chord ED = O o = offset at mid point of T 1 T 2 PQ = O x = offset at distance x from E , so that EP = x OT 1 = OT 2 = OD = R = Radius of the curve Exact formula for offset at any point on the chord line may be derived as: By Pathagoras theorem ∆ OT 1 E, OT 1 2 = T 1 E 2 + OE 2 Simple Curves 35

36. In ∆ OQQ 1 , 2 2 OQ 2 = QQ 1 + OQ 1 OQ 1 = OE + EQ 1 = OE + O x OQ 1 = ( R – O o ) + O x R 2 = x 2 + { (R – O o ) + O x ) } 2 O x = √ (R 2 –x 2 ) – (R - O o ) O x = √ (R 2 –x 2 ) – (R – (R- √ (R 2 – (l/2) 2 )) x l 2 2 2 2 2 O = √ (R –x ) – ( √ (R – ( ) )) -- 1 exact formula Method of Curve Ranging 1) Linear or Chain & Tape Method a) By offset or Ordinate from long chord DE = O o = R – √ (R 2 – ( l ) 2 ) ------- A 2 In eqn A two quantities are usually or must known. Simple Curves 36

37. formula or x 2 𝑅 O = 𝑥 (𝐿 −𝑥) -------- 2 (Approximate formula) In eqn 1 the distance x is measured from the mid point of the long chord where as eqn 2 it is measured from the 1 st tangent point T 1 . This method is used for setting out short curves e.g curves for street kerbs. . Method of Curve Ranging 1) Linear or Chain & Tape Method a) By offset or Ordinate from long chord When the radius of the arc is larger as compare to the length of the chord, the offset may be calculated approximately by Simple Curves 37

38. Method of Curve Ranging 1) Linear or Chain & Tape Method a) By offset or Ordinate from long chord Working Method: To set out the curve Divided the long chord into even number of equal parts. Set out offsets as calculated from the equation at each of the points of division. Thus obtaining the required points on the curve. Since the curve s symmetrical along ED, the offset for the right half of the curve will be same as those for the left half. Simple Curves 38

39. Problem 03: calculate the ordinate at 7.5 m interval for a circular curve given that l = 60 m and R = 180 m, by offset or ordinate from long chord. Solution: Ordinate at middle of the long chord = verse sine = O o o l 60 2 2 O = R – √ (R – ( ) ) = 1 8 – √ ( 1 80 – ( ) 2 2 2 2 O o = 2.52 m Various coo2.52 rdinates may be calculated by formula x 2 2 O = √ (R –x ) – ( √ l 2 2 2 (R – ( ) )) x = distance measured from mid point of long chord. 7.5 2 2 O = √ (180 –7.5 ) – ( √ 60 2 2 2 (180 – ( ) )) = 2.34 m 15 2 O = √ (180 2 –15 2 ) – ( √ (180 2 – ( 60 ) 2 )) = 1.89 m 22.5 2 2 O = √ (180 –22.5 ) – ( √ 60 2 2 2 (180 – ( ) )) = 1.14 m 30 2 O = √ (180 2 –30 2 ) – ( √ (180 2 – ( 60 ) 2 )) = 0 m X (m) Ox (m) 2.52 7.5 2.34 15 1.89 22.5 1.14 30 Simple Curves 39

40. ED = y’ = R ( 1 – cos ( ∅ ) 𝟐 Join T 1 D and DT 2 and bisect them at F and G respectively. Set out offset HF(y’’) and GK(y’’) each eqn be 𝟒 FH = GK = y’’= R ( 1 – cos ( ∅ ) Obtain point H and K on a curve. By repeating the same process, obtain as many pints as required on the curve. Method of Curve Ranging 1) Linear or Chain & Tape Method b) By Successive Bisection of Arcs Let T 1 and T 2 be the tangents points. Join T 1 and T 2 and bisect it at E . Setout offsets ED(y’) , determined point D on the curve equal to Simple Curves 40

41. Method of Curve Ranging 1) Linear or Chain & Tape Method By Offsets from the Tangents In this method the offsets are setout either radially or perpendicular to the tangents BA ad BC according to as the center O of the curve is accessible or inaccessible. By Radial Offsets: (O is Accessible) Working Method: Measure a distance x from T 1 on back tangent or from T 2 on the forward tangent. Measure a distance O x along radial line A 1 O. The resulting point E 1 lies on the curve. Simple Curves 41

42. a) By Radial Offsets: EE 1 = O x , offsets at distance x from T 1 along tangent AB. In ∆ OT 1 E 2 2 2 OT 1 + T 1 E = OE OT 1 = R, T 1 E = x, OE = R + O x R 2 + x 2 = (R + O x ) 2 R + Ox = √ (R 2 + O x 2 ) O x = √ (R 2 + X 2 ) – R ---------A (Exact Formula) O R E1 A B R Ox T 1 E x Method of Curve Ranging 1) Linear or Chain & Tape Method 3) By Offsets from the Tangents In this method the offsets are setout either radially or perpendicular to the tangents BA ad BC according to as the center O of the curve is accessible or inaccessible. Simple Curves 42

43. x Expanding √ (R 2 + x ) , O = R ( 1 + x 2 + x 4 2R 2 8R 4 + …..) - R Taking any two terms x 2 R x 2 eqn (A) O x = R ( 1 + 2R 2 ) – R O x = 2 R 2 x 2 R O = x 2 -------B ( Approximate formula) Used for short curve O R E1 A B T 1 R Ox E x Method of Curve Ranging 1) Linear or Chain & Tape Method By Offsets from the Tangents By Radial Offsets: Simple Curves 43

44. Method of Curve Ranging 1) Linear or Chain & Tape Method By Offsets from the Tangents By Radial Offsets:: Example Simple Curves 44

45. Method of Curve Ranging 1) Linear or Chain & Tape Method 3) By Offsets from the Tangents In this method the offsets are setout either radially or perpendicular to the tangents BA ad BC according to as the center O of the curve is accessible or inaccessible. b) By Offsets Perpendicular to Tangents (O is Inaccessible) Working Method: Measure a distance x from T 1 on back tangent or from T 2 on the forward tangent. Erect a perpendicular of length O x . The resulting point E 1 lies on the curve. Simple Curves 45

46. tangent AB ∆ OE 1 E 2 , 2 2 2 OE 1 = OE 2 + E 1 E 2 OE = R, E E = x, OE = R – O 1 1 2 2 x R 2 = (R – O x ) 2 + x 2 (R – O x ) 2 = R 2 – x 2 O x = R – √ (R 2 – x 2 ) ------- A (Exact Formula) Expanding √ (R 2 – x 2 ) and neglecting higher power x O = x 2 2 R ------- Approximate Formula O R E1 A B T 1 R Ox E x E2 Method of Curve Ranging 1) Linear or Chain & Tape Method 3) By Offsets from the Tangents b) By Offsets Perpendicular to Tangents EE 1 = O x = T offset at a distance of x measured along Simple Curves 46

47. Method of Curve Ranging 1) Linear or Chain & Tape Method 3) By Offsets from the Tangents b) By Offsets Perpendicular to Tangents : Example Simple Curves 47

48. 1 2R T E = R 2 𝛼, 𝛼 = T 1 E --------- 1 Method of Curve Ranging 1) Linear or Chain & Tape Method 4) By Offsets from Chord Produced T 1 E = T 1 E 1 = b o --- 1st chord of length “b 1 ” EF, FG, etc = successive chords of length b 2 and b 3 , each equal to length of unit chord. BT 1 E = 𝛼 = angle b/w tangents T 1 B and 1 st chord T 1 E E 1 E = O 1 = offset from tangent BT 1 E 2 F = O 2 = offset from preceding chord T 1 E produced. Arc T 1 E = chord T 1 E T 1 E = OT 1 2 𝛼 Simple Curves 48

49. 1 1 2 O = T E /2R = b 1 2 2 𝑅 ----- 2 ∟E 2 EF 1 = ∟DET 1 (vertically opposite) ∟DET 1 = ∟DT 1 E since DT 1 = DE ::: ∟E 2 EF 1 = ∟DT 1 E = ∟E 1 T 1 E The ∆ s E 1 T 1 E and E 2 EF 2 being nearly isosceles may be considered similar Method of Curve Ranging 1) Linear or Chain & Tape Method 4) By Offsets from Chord Produced Similarly chord EE 1 = arc E 1 E 1 st offset O 1 = E 1 E = T 1 E x 𝛼 O 1 = T 1 E T 1 E / 2R Simple Curves 49

50. 𝐸𝐸 2 𝑇 1 𝐸 i.e 𝐸 2 𝐹 1 = 𝐸 1 𝐸 𝐸 2 𝐹 1 = 𝑂 1 2 1 𝐸 𝐹 = 𝑏 𝑂 2 1 2 𝑏 𝑏 1 𝑏 2 𝑏 1 2 = x = 𝑏 𝑏 2 1 𝑏 1 𝑏 1 2𝑅 2𝑅 F 1 F being the offset from the tangent at E is equal to : F 1 F = E F 2 2 𝑅 = 2 b 2 2 𝑅 Method of Curve Ranging 1) Linear or Chain & Tape Method 4) By Offsets from Chord Produced The ∆ s E 1 T 1 E and E 2 EF 2 being nearly isosceles may be considered similar Simple Curves 50

51. O 2 = E 2 F = 2 1 + 𝑏 𝑏 𝑏 2 2 2 𝑅 2 𝑅 O 2 = E 2 F = 2 1 2 𝑏 (𝑏 +𝑏 ) 2 𝑅 Similarly 3 rd offset 3 O = 3 2 3 b ( b + b ) 2 𝑅 2 3 4 , since b =b =b … O 3 = 2 b 2 𝑅 , so O 3 =O 4 =O 5 except for last offset. n O = n n b ( bn −1 ……… ...+ b ) 2 𝑅 Method of Curve Ranging 1) Linear or Chain & Tape Method 4) By Offsets from Chord Produced Now 2 nd offset O 2 O 2 = E 2 F = E 2 F 1 + F 1 F Simple Curves 51

52. Method of Curve Ranging 2) Angular or Instrumental Methods Rankine’s Method of Tangential Angles Two Theodolite Method 1) Rankine’s Method of Tangential Angles In this method the curve is set out tangential angle often called deflection angles with a theodolite, chain or tape. Simple Curves 52

53. Method of Curve Ranging 2) Angular or Instrumental Methods Rankine’s Method of Tangential Angles Working method: Fix the theodolite device to be at point T 1 and directed at point B. Measure the deflection angles 𝛿 1 and the chords C 1 . Connect the ends of the chords to draw the curve. Deflection Angles: The angles between the tangent and the ends of the chords from point T 1 . Simple Curves 53

54. Method of Curve Ranging 2) Angular or Instrumental Methods 1) Rankine’s Method of Tangential Angles AB = Rear tangent to curve D,E,F = Successive point on the curve δ 1 , δ 2 , δ 3 ….. The tangential angles which each of successive chord …T 1 D,DE,EF…… makes with the respective tangents at T 1 , D, E. ∆ 1 , ∆ 2 , ∆ 3 …… Total deflection angles C 1 , C 2 , C 3 …….. Length of the chord. T 1 D, DE, EF. Simple Curves 54

55. C 1 = T 1 D = 1 2 π R δ 180 A n d δ 1 = C 1 ∗ 180 2 π R degree δ 1 2 π R = C 1 ∗ 180 ∗ 60 min 1 δ = 1718.9 ∗ C 1 𝑅 2 δ = 1718.9 ∗ C 2 𝑅 n δ = 1718.9 ∗ Cn 𝑅 Method of Curve Ranging 2) Angular or Instrumental Methods 1) Rankine’s Method of Tangential Angles Arc T 1 D = chord T 1 D = C 1 So, T 1 D = R x 2 δ 1 Simple Curves 55

56. But B T 1 E = B T 1 D + D T 1 E ∆ 2 = δ 1 + δ 2 = ∆ 1 + δ 2 ∆ 3 = δ 1 + δ 2 + δ 3 = ∆ 2 + δ 3 ∆ 4 = δ 1 + δ 2 + δ 3 + δ 4 = ∆ 3 + δ 4 ∆ n = δ 1 + δ 2 +…………. + δ n = ∆ n-1 + δ n Check: 1 2 2 Total deflection angle BT T = φ , φ = Deflection angle of the curve This Method give more accurate result and is used in railway & other important curve. Method of Curve Ranging 2) Angular or Instrumental Methods 1) Rankine’s Method of Tangential Angles Total deflection angle for the 1 st chord --T 1 D = BT 1 D :. ∆ 1 = δ 1 2 nd Chord --DE = B T 1 E Simple Curves 56

57. Problem 04: Two tangents intersect at chainage 2140 m . ∅ = 18 o 24`. Calculate all the data necessary for setting out the curve, with R = 600 m and Peg interval being 20 m by: By deflection angle ∅ offsets from chords. Solution: BT 1 = BT 2 = R tan ∅ 2 = 600 tan o 18 24` 2 BT 1 = BT 2 = 97.18 m 180 𝑜 Length of curve = L = 𝜋 R ∅ = 𝜋 600 18 𝑜 24` 180 𝑜 L = 192.68 m Chainage of point of intersection Minus Tangent length Chainage of T 1 Plus L Chainage of T 2 = 2140 m = - 97.18 m = 2042.82 m = + 192.68 m = 2235.50 m Simple Curves 57

58. Problem 04: Solution: Length of 1 st chord = C 1 = 2060 – 2042.82 = 17.18 m C 2 = C 3 = C 4 = C 5 = C 6 = C 7 = C 8 = C 9 = 20 m C 10 = 2235.50 – 2220 = 15.15 m 1) By deflection angle 1 δ = 1 1718.9 C 1718.9 C 1 𝑅 60 𝑅 ( m in) = ( d eg r ee) 1 δ = 1718.9 x 17.18 60 𝑥 600 = o 49`13.07`` 2 δ = 1718.9 x 20 60 𝑥 600 = o 57`17.8 δ 2 =δ 3 =δ 4 =δ 5 =δ 6 =δ 7 =δ 8 =δ 9 δ 10 = 1718.9 x 15.15 60 𝑥 600 = o 44`24.3`` 𝑙𝑒𝑛𝑔ℎ𝑡 𝑜𝑓 𝑐𝑢𝑟𝑣𝑒 −𝐶 1 Note: No of chords = 𝐼 𝑛 𝑡 𝑒𝑟𝑣𝑎𝑙 = (192.68 – 17.18)/ 20 = 8.77=8 Simple Curves 58

59. Problem 04: Solution: 1) By deflection angle Total deflection (tangential) angle for the chords are: ∆ 1 = δ 1 = o 49`13.07`` ∆ 2 = δ 1 + δ 2 = ∆ 1 + δ 2 = 1 o 46`30.87`` = ∆ 2 + δ 3 = 2 o 43`48.67`` ∆ 3 = δ 1 + δ 2 + δ 3 ∆ 4 = 3 o 41`6.4 ∆ 5 = 4 o 38`24.27`` ∆ 6 = 5 o 35`42.07`` ∆ 7 = 6 o 32``54.87`` ∆ 8 = 7 o 30`17.67`` ∆ 9 = 8 o 27`35.47`` ∆ 10 = ∆ 9 + δ 10 = 9 o 11`54.77`` Check: ∆ 10 = ∅ = 18 o 34` 2 2 o = 9 12`0`` Simple Curves 59

60. Problem 04: Solution: 2) By Offsets from Chords n O = n b ( bn −1 ……… ...+ bn) 2 𝑅 1 O = n 2 = b 17.18 2 2𝑅 2 𝑥 600 = 0.25 m 2 O = 2 1 2 b (b + b ) 2 𝑅 = 20 17.18+20 2 𝑥 600 = 0.62 m 3 O = 3 2 3 b (b + b ) b 2 2 2 𝑅 𝑅 = = 0.67 m O 3 = O 4 = O 5 = O 6 = O 7 = O 8 = O 9 O 10 = b 10 (b 9 + b 10 ) = 15.50 20 +15.50 = 0.46 m 2𝑅 2 𝑥 600 Simple Curves 60

61. ∆1 is b/w T 1 E = ∆ 2 tangent T1B & T 1 D => BT1D = T 1 T 2 D = ∆ 1 = T 1 T 2 E The total tangential angle or deflection angle ∆ 1 , ∆ 2 ∆ 3 … ,As calculate in the 1st method. Method of Curve Ranging 2) Angular or Instrumental Methods 2) Two Theodolite Method This method is used when ground is not favorable for accurate chaining i.e rough ground , very steep slope or if the curve one water It is based on the fact that angle between tangent & chord is equal to the angle which that chord subtends in the opposite segments. Simple Curves 61

62. Obstacles in Setting Out Simple Curve The following obstacles occurring in common practice will be considered. When the point of intersection of Tangent lines is inaccessible. When the whole curve cannot be set out from the Tangent point, Vision being obstructed. When the obstacle to chaining occurs. Simple Curves 62

63. Obstacles in Setting Out Simple Curve 1) When the point of intersection of Tangent lines is inaccessible When intersection point falls in lake, river , wood or any other construction work To determine the value of ∅ To locate the points T 1 & T 2 Calculate θ 1 & θ 2 by instrument (theodolite). ∟ BMN = α = 180 – θ 1 ∟ BNM = β = 180 – θ 2 Deflection angle = ∅ = α + β ∅ = 360 o – (sum of measured angles) ∅ = 360 o - ( θ 1 + θ 2 ) Simple Curves 63

64. = 𝐵 𝑀 𝑀 𝑁 sin β sin (180 o − ( α+β)) BM = 𝑀𝑁 sin β sin (180 o − (α+β)) BN = 𝑀𝑁 sin α sin (180 o − (α+β)) 1 2 2 BT & BT = R * tan ( ∅ ) MT 1 =BT 1 - BM NT 2 = BT 2 - BN Obstacles in Setting Out Simple Curve 1) When the point of intersection of Tangent lines is inaccessible Calculate the BM & BN from ∆BMN By sine rule Simple Curves 64

65. Obstacles in Setting Out Simple Curve 2) When the whole curve cannot be set out from the Tangent point, Vision being obstructed As a rule the whole curve is to be set out from T 1 however obstructions intervening the line of sight i.e Building, cluster of tree, Plantation etc. In such a case the instrument required to be set up at one or more point along the curve. Simple Curves 65

66. Obstacles in Setting Out Simple Curve 3) When the obstacle to chaining occurs Simple Curves 66

67. Simple Curves 67 Assignment Obstacles in Setting Out Simple Curve (Detail procedure) Page 130 Part II Example 1 (approximate method) Example 2 Example 3 Page 135 Part II

68. Curves Compound Curves A compound curve consist of 2 arcs of different radii bending in the same direction and lying on the same side of their common tangent. Then the center being on the same side of the curve. R S = Smaller radius R L = Larger radius T S = smaller tangent length = BT 1 T L = larger tangent length = BT 2 𝛼 = deflection angle b/w common tangent and rear tangent 𝛽 = angle of deflection b/w common tangent and forward tangent N = point of compound curvature KM = common tangent 68

69. Curves Compound Curves Elements of Compound Curve ∅ = 𝛼 + 𝛽 1 S 2 KT = KN =R tan( 𝛼 ) 2 L 2 MN = MT = R tan ( 𝛽 ) From ∆ BKN, by sine rule = 𝐵 𝐾 𝑀 𝐾 sin 𝛽 sin 𝐼 = 𝐵 𝐾 𝑀 𝐾 sin 𝛽 sin(180 o – ( 𝛼 + 𝛽 )) BK = 𝑀𝐾 sin 𝛽 sin(180 o – ( 𝛼 + 𝛽 )) BM = 𝑀𝐾 sin 𝜶 sin(180 o – ( 𝛼 + 𝛽 )) 69 180 o - ( 𝛼 + 𝛽 ) = I 180 o – ∅ = I

70. Curves Compound Curves Elements of Compound Curve T S = BT 1 = BK + KT 1 T S = BT 1 = 𝑀𝐾 sin 𝛽 sin(180 o – ( 𝛼 + 𝛽 )) S 2 + R tan ( 𝛼 ) T L = BT 2 = 𝑀𝐾 sin 𝜶 sin(180 o – ( 𝛼 + 𝛽 )) L 2 + R tan ( 𝛽 ) 70 Of the seven quantities R S , R L , TS, T L , ∅ , 𝛼 , 𝛽 four must be known.

71. Curves 1 L 𝛼 2 o KT = KN = R tan ( ) = 600 tan(40 /2) KT 1 = KN = 218.38 m 2 S 𝛽 2 o MN = MT = R tan( ) = 400 tan(35 /2) MN = MT 2 = 126.12 m KM = MT 2 + MN = 218.38 + 126.12 KM = 344.50 m Compound Curves Problem: Two tangents AB & BC are intersected by a line KM. the angles AKM and KMC are 140 o & 145 o respectively. The radius of 1 st arc is 600m and of 2 nd arc is 400m. Find the chainage of tangent points and the point of compound curvature given that the chainage of intersection point is 3415 m. Solution: 𝛼 = 180 o – 140 o = 40 o 𝛽 = 180 o – 145 o = 35 o ∅ = 𝛼 + 𝛽 = 75 o I = 180 o – 75 o = 105 o 71

72. Curves Solution: Fin ∆ BKM, by sin rule = 𝐵 𝐾 𝑀 𝐾 sin 𝛽 sin( 𝐼) sin ( 𝐼 ) sin 105 𝑜 BK = 𝑀𝐾 sin 𝛽 = 344.50 𝑥 𝑠𝑖𝑛35 𝑜 = 204.57 m sin ( 𝐼 ) sin 105 𝑜 BM = 𝑀𝐾 sin 𝛼 = 344.50 𝑥 𝑠𝑖𝑛40 𝑜 = 229.25 m L 180 𝑜 180 𝑜 T L = KT 1 + BK = 218.38 + 204.57 = 422.95 m T S = MT 2 + BM = 126.12 + 229.25 = 355.37 m L = 𝜋 𝑅𝐿 𝛼 = 𝜋 𝑥 600 𝑥4 = 418.88 m S L = 𝑆 180 𝑜 𝜋 𝑅 𝛽 𝜋 𝑥 400 𝑥 30 180 𝑜 = = 244.35 m Compound Curves 72

73. Curves S o lution: Chainage of intersection point = 3415 m Minus T L = - 422.95 m Chainage of T 2 = 2992.05 m Plus L = + 418.88 m Chainage of compound curvature (N) = 3410.93 m Plus L S = + 244.35 m Chainage of T 2 = 3655.25 m Compound Curves 73