1 / 49

AN EDUSAT LECTURE ON

AN EDUSAT LECTURE ON. AN EDUSAT LECTURE ON. CURVES-I. BY:. April 16,2013 . Er.Mohinder Kumar Senior Lecturer Civil Engg . GOVT. POLYTECHNIC COLLEGE BATALA.( GURDASPUR). B. CURVES.

dalit
Download Presentation

AN EDUSAT LECTURE ON

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. AN EDUSAT LECTURE ON AN EDUSAT LECTURE ON CURVES-I BY: April 16,2013. Er.Mohinder Kumar Senior Lecturer Civil Engg. GOVT. POLYTECHNIC COLLEGE BATALA.( GURDASPUR)

  2. B CURVES Curves are regular bends provided in the lines of communication like roads, railways and canals etc. to bring about gradual change of direction. T2 T1 A C O Fig. 1 . A CURVE CURVES

  3. CURVES B They enable the vehicle to pass from one path on to another when the two paths meet at an angle. They are also used in the vertical plane at all changes of grade to avoid the abrupt change of grade at the apex. T2 T1 A C O Fig. 2. A CURVE CURVES

  4. Curves provided in the horizontal plane to have the gradual change in direction are known as horizontal curves. VERTICAL CURVES Curves provided in the vertical plane to obtain the gradual change in grade are called as vertical curves. Curves may be circular or parabolic. Curves are generally arcs of parabolas. Curves are laid out on the ground along the centre line of the work. HORIZONTAL CURVES CURVES

  5. NEED OF PROVIDING CURVES Curves are needed on Highways, railways and canals for bringing about gradual change of direction of motion. They are provided for following reasons:- i) To bring about gradual change in direction of motion. ii) To bring about gradual change in grade and for good visibility. CURVES

  6. NEED OF PROVIDING CURVES Contd… iii) To alert the driver so that he may not fall asleep. iv) To layout Canal alignment. v) To control erosion of canal banks by the ……thrust of flowing water in a canal. CURVES

  7. CLASSIFICATION OF CIRCULAR CURVES • Circular curves are classified as : • Simple Curves. • Compound Curves. • Serpentine Curves. • Deviation Curves. CURVES

  8. B i) Simple Curve: A simple curve Consists of a single arc of circle connecting two straights. It has radius of the same magnitude throughout. T2 T1 A C R R O Fig. 3. A SIMPLE CURVE CURVES

  9. ii) COMPOUND CURVE M P N T1 R1 O1 R2 Fig.4 Compound Curve A C O2 A compound Curve consists of two or more simple curves having different radii bending in the same direction and lying on the same side of the common tangent. Their centres lie on the same side of the curve. CURVES

  10. iii) REVERSE OR SERPENTINE CURVE A reverse or serpentine curve is made up of two arcs having equal or different radii bending in opposite direction with a common tangent at their junction . B O2 R2 T2 R2 Their centres lie on opposite sides of the curve. Reverse curves are used when the straights are parallel or intersect at a very small angle. M p N T1 R1 R1 A O1 Fig. 5. A Reverse or Serpentine Curve. CURVES

  11. REVERSE OR SERPENTINE CURVE They are commonly used in railway sidings and sometimes on railway tracks and roads meant for low speeds. They should be avoided as far as possible on main lines and highways where speeds are necessarily high. B O2 T2 P M N T1 A O1 Fig.6 A Reverse or Serpentine Curve. CURVES

  12. O3 iv) DEVIATION CURVE O1 A deviation curve is simply a combination of two reverse curves. it is used when it becomes necessary to deviate from a given straight path in order to avoid intervening obstructions such as bend of river, a building , etc. Building T1 T2 O2 Fig. 7 A Deviation Curve CURVES

  13. B’ B φ I F T1 E T2 A C φ/2 R φ O Fig. 8 SIMPLE CIRCULAR CURVE CURVES

  14. NAMES OF VARIOUS PARTS OF CURVE • The two straight lines AB and BC which are connected by the curve are called the tangents or straights to the curve. • The point of intersection of the two straights (B) is called the intersection point or the vertex. • When the curve deflects to the right side of the progress of survey ,it is termed as righthanded curve and when to the left , it is termed as left handed curve. CURVES

  15. NAMES OF VARIOUS PARTS OF CURVE (iv) The lines AB and BC are tangents to the curve. AB is called the first tangent or the rear tangent . BC is called the second tangent or the forward tangent. (v) The points ( T1and T2 ) at which the curve touches the tangents are called thetangent points. The beginning of the curve ( T1) is called the tangent curve point and the end of the curve (T2) is called the curve tangent point. CURVES

  16. NAMES OF VARIOUS PARTS OF CURVE (vi) The angle between the lines AB and BC (└ABC) is called the angle of intersection (I). (vii) The angle by which the forward tangent deflects from the rear tangent (└B’BC) is called the deflection angle (φ) of the curve. (viii) The distance from the point of intersection to the tangent point is called tangent length ( BT1 and BT2). (ix) The line joining the two tangent points (T1 and T2) is known as the long chord. CURVES

  17. The arc T1FT2 is called the length of curve. • The mid point(F) of the arc (T1FT2) is called the summit or apex of the curve. • (xii) The distance from the point of intersection to the apex of the curve BF is called the apex distance. • (xiii) The distance between the apex of the curve and the mid point of the long chord (EF) is called versed sine of the curve. • (xiv) The angle subtended at the centre of the curve by the arc T1FT2 is known as central angle and is equal to the deflection angle (φ) . CURVES

  18. ELEMENTS of a Simple Circular Curve • Angle of intersection +Deflection angle = 1800. • or I +φ =1800 • (ii) └ T1OT2 = 1800 - I = φ • i.ethe central angle = deflection angle. • Tangent length = BT1 =BT2= OT1 tan φ/2 • = R tan φ/2 CURVES

  19. ELEMENTS of a Simple Circular Curve • (iv) Length of long chord =2T1E • =2R sin φ/2 • Length of curve = Length of arc T1FT2 • = R X φ (in radians) • = πR φ/1800 • (vi) Apex distance = BF = BO – OF • = R sec. φ/2 - R • = R (1 – cos φ/2 )=R versine φ/2 CURVES

  20. DESIGNATION OF CURVE A curve may be designated either by the radius or by the angle subtended at the centre by a chord of particular length. In India, a curve is designated by the angle (in degrees) subtended at the centre by a chord of 30 metres (100 ft.) length. This angle is called the degree of curve (D). The degree of the curve indicates the sharpness of the curve. CURVES

  21. DESIGNATION OF CURVES. In English practice , a curve is defined by the radius of the curve in terms of chains, such as a six chain curve means a curve having radius equal to six full chains, chain being 30 metres unless otherwise specified. In America,Canada,India and some other countries a curve is designated by the degree of the curve. For example a 40 curve means a curve having angle of 90 degrees at the centre subtended by a chord of 30m length unless otherwise specified. CURVES

  22. RELATION between the Radius of curve and Degree of Curve. The relation between the radius and the degree of the curve may be determined as follows:- M P N D R R Let R = the radius of the curve in metres. D = the degree of the curve. MN = the chord, 30m long. P = the mid-point of the chord. In OMP,OM=R, MP= ½ MN =15m └MOP=D/2 Then, sin D/2=MP/OM= 15/R D/2 O Fig.9 Degree of Curve PTO CURVES

  23. RELATION between the Radius of curve and Degree of Curve. Then,sin D/2=MP/OM= 15/R Or R = 15 sin D/2 But when D is small, sin D/2 may be assumed approximately equal to D/2 in radians. Therefore: R = 15 X 360 πD = 1718.87 D Or say , R = 1719 D M P N (Exact) D R R D/2 O Fig. 10 Degree of Curve This relation holds good up to 50 curves.For higher degree curves the exact relation should be used. (Approximate) CURVES

  24. METHODS OF CURVE RANGING A curve may be set out (1) By linear Methods, where chain and tape are used or (2) By Angular or instrumental methods, where a theodolite with or without a chain is used. Before starting setting out a curve by any method, the exact positions of the tangents points between which the curve lies ,must be determined. Following procedure is adopted:- CURVES

  25. METHODS OF SETTING OUT A CURVE • Procedure :- • After fixing the directions of the straights, produce them to meet in point (B) • Set up the Theodolite at the intersection point (B) and measure the angle of intersection (I) .Then find the deflection angle ( ) by subtracting (I) from 1800i.eφ=1800 – I. • iii) Calculate the tangent length from the following equation • Tangent length = R tanφ/2 φ CURVES

  26. METHODS OF SETTING OUT A CURVE Procedure :- iv) Measure the tangent length (BT1) backward along the rear tangent BA from the intersection point B, thus locating the position of T1. vi) Similarly, locate the position of T2 by measuring the same distance forward along the forward tangent BC from B. CURVES

  27. METHODS OF SETTING OUT A CURVE Procedure (contd…) :- After locating the positions of the tangent points T1 and T2 ,their chainages may be determined. The chainage of T1 is obtained by subtracting the tangent length from the known chainage of the intersection point B. And the chainage of T2 is found by adding the length of curve to the chainage of T1. Then the pegs are fixed at equal intervals on the curve.The interval between pegs is usually 30m or one chain length. ……............... CURVES

  28. METHODS OF SETTING OUT A CURVE Procedure (contd…) :- This distance should actually be measured along the arc ,but in practice it is measured along the chord ,as the difference between the chord and the corresponding arc is small and hence negligible. In order that this difference is always small and negligible ,the length of the chord should not be more than 1/20th of the radius of the curve. The curve is then obtained by joining all these pegs. ……............... CURVES

  29. METHODS OF SETTING OUT A CURVE Procedure (contd…) :- 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. 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 …. CURVES

  30. METHODS OF SETTING OUT A CURVE Procedure (contd…) :- … at such a distance from the first tangent point (T1) 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. CURVES

  31. Example : A simple circular curve is to have a radius of 573 m .the tangents intersect at chainage 1060 m and the angle of intersection is 1200. Find, • Tangent Distance. • Chainage at beginning and end of the curve. • Length of the long chord. • Degree of the curve. • Number of full and sub chords. • Solution: Please see fig.11 • Given, • The deflection angle, φ= 1800 – 1200 =600 • Radius of curve = R = 573 m PTO CURVES

  32. 1060 m = φ 600 1200 330.85 L=600m 729.15 1329.15 T2 T1 R=573m O Fig.11 CURVES

  33. We know ,tangent length = R tan φ /2 • = 573 x tan 300 • = 573x 0.5774 • = 330.85 m (Ans.) • (ii) Length of curve is given by:π R φ • 1800 • = π x 573x600 • 1800 • = 600 m (Ans.) • Chainage of first tangent point (T1) • = Chainage of intersection point – tangent length. • = 1060 – 330.85 • = 729.15 m (Ans.) PTO CURVES

  34. (iii) The length of long chord is given by: • L = 2R sin φ /2 • = 2 x 573 x sin 300 • = 573 m ( Ans.) • Degree of Curve • We know the relation , R= 1719 • D • or D = 1719 • R • =30 • Therefore , degree of curve is =30 (Ans.) PTO CURVES

  35. (v) Number of Full and sub chords: Assuming peg interval =30m Chainage of T1 = 729.15 m = 729.15 30 = 24 full chain lengths + 9.15 m Chainage of Ist peg on the curve should be 25 full chain lengths. The length of Ist sub chord= (25+00) – (24 + 9.15) = 20.85 m Chainage of T2 = 1329.15 Chain lengths. 30 = 44 full chain lengths + 9.15 m. Chainage of last peg on the curve =44 full chains. Therefore length of last sub chord = (44+9.15) – (44+00) = 9.15m Chain lengths. PTO CURVES

  36. No. Of full chords = chainage of last peg – chainage of Ist peg = 44 – 25 = 19 So, there will be 19 full chords and two sub chords. Check: Length of full chords = 19x30 =570.00m ” ” Ist sub chord = 20.85m ” ” last sub chord = 9.15m Total length of all chords = 600.00m (Same as length of curve) PTO CURVES

  37. LINEAR METHODS of setting out Curves • The following are the methods of setting out simple circular curves by the use of chain and tape :- • By offsets from the tangents. • By successive bisection of arcs. • By offsets from chords produced. CURVES

  38. LINEAR METHODS of setting out Curves 1. By offsets from the tangents. When the deflection angle and the radius of the curve both are small, the curves are set out by offsets from the tangents. Offsets are set out either (i) radially or (ii) perpendicular to the tangents according as the centre of the curve is accessible or inaccessible CURVES

  39. LINEAR METHODS of setting out Curves B’ B φ x P Ox P1 T2 900 T1 A C R O Fig. 12 By Radial Offsets CURVES

  40. LINEAR METHODS of setting out Curves B’ Offsets is given by : Ox = R2 +x2 – R …….. (Exact relation.) When the radius is large ,the offsets may be calculated by the approximate formula which is as under Ox = x2 ……… (Approximate ) 2R By Radial Offsets CURVES

  41. LINEAR METHODS of setting out Curves B’ B P Ox x P1 T2 T1 A P2 B Fig. 13. O (ii) By offsets perpendicular to the Tangents CURVES

  42. LINEAR METHODS of setting out Curves 1. (ii) By offsets perpendicular to the Tangents Ox= R – R2 – x2 ……………(Exact) Ox = x2 ……… (Approximate ) 2R CURVES

  43. LINEAR METHODS of setting out Curves • By offsets from the tangents: Procedure • Locate the tangent points T1 and T2. • Measure equal distances , say 15 or 30 m along the tangent fro T1. • (iii) Set out the offsets calculated by any of the above methods at each distance ,thus obtaining the required points on the curve. CURVES

  44. LINEAR METHODS of setting out Curves • By offsets from the tangents: Procedure…. • Continue the process until the apex of the curve is reached. • (v) Set out the other half of the curve from second tangent. • (vi) This method is suitable for setting out sharp curves where the ground outside the curve is favourable for chaining. CURVES

  45. Example. Calculate the offsets at 20m intervals along the tangents to locate a curve having a radius of 400m ,the deflection angle being 600 . Solution . Given: Radius of the curve ,R = 400m Deflection angle, φ = 600 Therefore tangent length = R. tanφ/2 = 400 x tan 600 = 230.96 m Radial offsets. (Exact method) Ox= R2 + x2 - R……………(Exact) CURVES

  46. Radial offsets. (Exact method) Ox= R2 + x2 - R……………(Exact) O20 = 4002+202 - 400 = 400.50 - 400 = 0.50 m O40 = 4002+402 - 400 = 402.00 - 400 = 2.00 m O60 = 4002+602 - 400 = 404.47 - 400 = 4.47 m O80 = 4002+802 - 400 = 407.92 - 400 = 7.92 m O100 = 4002+1002- 400 = 412.31 - 400 = 12.31 m And so on…. CURVES

  47. B) Perpendicular offsets (Exact method) Ox = R – R2 – x2 ……………(Exact) O20 = 400 - 4002 - 202 = 400 -399.50 = 0.50 m O40 = 400 - 4002 - 402 = 400 -398.00 = 2.00 m O60 = 400 - 4002 - 602 = 400 -395.47 = 4.53 m O80 = 400 - 4002 - 802 = 400 -391.92 =8.08 m O100 = 400 - 4002 -1002 = 400 -387.30 =12.70 m And so on….. CURVES

  48. By the approximate Formula • (Both radial and perpendicular offsets) Ox = 2R Therefore O20 = 202 = 0.50 m 2x400 x2 O40 = 402 = 2.00 m 2x400 O60 = 602 = 4.50 m 2x400 O80 =802 = 8.00 m 2x 400 O100 = 1002 = 12.50 m 2 x 400 and so on…. CURVES

  49. THANKS

More Related