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# Horizontal curves pdf

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### Horizontal curves pdf

1. 1. Horizontal Curves Kaila Marie Joy D.R. Turla CE 41 FA1
2. 2. introductionHORIZONTAL CURVES As a highway changes horizontal direction, turning tochange the vehicle direction at the point of intersectionbetween the two straight lines is not feasible. The change indirection would be too abrupt and too risky for the safety ofmodern, high-speed vehicles, the driver and its passengers. Itis therefore necessary to interpose a curve between thestraight lines. Horizontal curves occur at locations wheretwo roadways intersect, providing a gradual transitionbetween the two. The straight lines of a road are calledtangents because the lines are tangent to the curves used tochange direction. kaila marie joy d.r. turla
3. 3. kaila marie joy d.r. turla
4. 4. Simple CurvesThe simple curve is an arc of a circle. The radiusof the circle determines the sharpness or flatnessof the curve. PI I T E M PC PT C R I kaila marie joy d.r. turla
5. 5. Elements of a Simple Curve CIRCULAR CURVE ELEMENTS R = radius of the curve T = tangent distance I = intersection of deflection angle E = external distance M = middle ordinate LC = long chord or line connecting PC and PT PC = point of curvature or beginning point PT = point of tangency or end of the curve D = degree of curve L = length of curve kaila marie joy d.r. turla
6. 6. kaila marie joy d.r. turla
7. 7. kaila marie joy d.r. turla
8. 8. Formula for Simple Curve kaila marie joy d.r. turla
9. 9. Sample Problem:The angle of intersection of acircular curve is 36030. Compute R,T, sta PC and PT, and the Degreeof Curve if the external distance is PI12.02m, point of intersection is atsta 75+040. Use 10 m per station. T E M PC PTGiven:I= 36030 RE= 12.02mSta PI= 75+040 I kaila marie joy d.r. turla
10. 10. Solution:R = E/(sec I/2 - 1) = 12.02 . = 226.942m sec (36 30/2) -1T = R (tan I/2) = 226.942 tan (36 30/2) = 74.83mSta PC= Sta V - T = (75+040) - 74.83 = 74 + 965.17D = 10(360) . = 10(360) . = 2031 2(3.1416)R 2(3.1416)(226.94)LC = 10I = 10 (36031) = 145.03 m D 2031Sta PT= Sta PC + LC = (74+965.17) + 145.03 = 75 + 110.20 kaila marie joy d.r. turla
11. 11. Laying Out ofSimple Curves
12. 12. Laying Out of Simple CurvesINSTRUMENTS AND ACCESSORIES: as reference for the different abbreviations and terminologies used in this exercise.Theodolite or TransitSteel Tape 3. All values needed to lay out the curve should beChaining Pins tabulated accordStakes or Hubs 4. Set up and level the instrument at the designated vertex or point or intersection (PI).PROCEDURE:1. Before proceeding to the designated survey site, the 5. Establish on the ground the PC by laying out with alab instructor should be consulted with respect to the steel tape the computed tangent distance (T) from the PI.following curve elements which will be needed to define the The intersection angle (I) at the PI and the distancecircular curve to be laid out: carried through the forward tangent will also be needed a. Radius of the curve (R) to set a stake at the PT. b. Intersection or deflection angle (I) c. Stationing of the point of intersection (PI) 6. Transfer and set up the instrument at the PC. At the PC lay off the total deflection angle from PI to PT and2. Similarly, the different elements of the circular check if the stake previously set up at the PT is along thecurve such as: T, L, LC, E, M, and the stationings of the PC line of sight. If it doesnt check, an error exists in eitherand PT should be determined by calculations. The measurement or computation. As an added check, stake outcomputations should also include the deflection angles and the midpoint of the curve before beginning to setchord lengths which will be needed when staking out the intermediate stations. By intersecting the angle (180-I) atcurve by half section intervals. The accompanying sketch is the PI and laying off the external distance (E), the midpointgiven to serve can be established. A check of the deflection angle from the PC to the midpoint should equal to I/4. kaila marie joy d.r. turla
13. 13. 7.. To establish the first curve station, first set thehorizontal circle reading of the instrument to zero andsight along the back tangent. Then turn the instrumentabout its vertical axis and lay off the required sub-deflection angle and the corresponding chord distance forthe first station. Set a hub to mark the located station.8. With the first station already established, now layout the next chord length from it, and locate the secondstation on the intersection of the line of sight (defined bythe next deflection angle) and the end of the chord. Alsoset a hub to mark this located station. CIRCULAR CURVE ELEMENTS9. Repeat the process of locating stations on the curve R = radius of the curveby laying out the computed deflection angles and the chorddistances from the previously established station. Do this T = tangent distanceuntil all the required stations of the curve are laid out I = intersection of deflection angleand properly marked on the ground. E = external distance10. When the final station is established, the closing PT M = middle ordinateshould be staked out using the final deflection angle and LC = long chord or line connecting PC and PTsubchord, to determine the misclosure in laying out the PC = point of curvature or beginning pointcurve. PT = point of tangency or end of the curve D = degree of curve L = length of curve kaila marie joy d.r. turla
14. 14. Sample Problem:It is required to lay-out asimple curve by deflectionangles. The curve is toconnect two tangents withan angle of 300 and radiusof 130m. Compute thetangent distance, length ofcurve, deflection angles toeach station of the curve.Sta PC is at 5 + 767.20. } Radii Tangent LC Deflection Angles Chord Distance kaila marie joy d.r. turla
15. 15. Solution:T = R (tan I/2) = 130 tan (30/2) = 34.83mD = 1145.916 . = 1145.916 . = 8049 R 130LC = 20I = 20 (300) . = 68.07 m D 80 49Sta PT= Sta PC + LC = (5 + 767.20) + 68.07 = 5 + 835.27 20 = 68.07 = 12.8 = 15.27 B1= 8o49 B1 300 B2 B3 B2= 5o28 B3= 6o49Cn = 2R(sin I/2 )C1 = 2(130)(sin 20 49) = 12.78C2 = 2R(sin 70 13) = 32.66C3 = 2R(sin 110 38) = 52.43C4 = 2R(sin 150 ) = 67.29 kaila marie joy d.r. turla
16. 16. Tabulation: Stations Central Angles Deflection Chord Distance Remarks Angles 5 + 780 5038 2049 12.78 5 + 800 14027 7013 32.66 5 + 820 23016 11038 52.73 5 + 835.67 300 150 67.29 Sta PT kaila marie joy d.r. turla
17. 17. Compound curves kaila marie joy d.r. turla
18. 18. Compound CurvesFrequently, the terrain will require the use of the compound curve. This curvenormally consists of two simple curves joined together and curving in thesame direction. I1 + I 2 T I1 I2 T1 + T2 T1 T2 R1 R1 R2 kaila marie joy d.r. turla
19. 19. Elements of Compound Curve R = radius of the curve T = tangent distance I = intersection of deflection angle E = external distance M = middle ordinate LC = long chord or line connecting PC and PT PC = point of curvature or beginning point PT = point of tangency or end of the curve D = degree of curve L = length of curve PCC= Point of Compound Curvature kaila marie joy d.r. turla
20. 20. Formula for Compound Curve kaila marie joy d.r. turla
21. 21. Sample Problem:A long chord from PC to PT of a compoundcurve is 180m long and the angle it makeswith the longer tangents are 12o and 18orespectively. Find the radius of thecompound curve if the common tangent isparallel to the long chord. kaila marie joy d.r. turla
22. 22. Solution: 30 o 12 o 18 o c1 165 o c2 165 o 6o 12 o 6o 180 m 9o 180 m 18 o 9o 18 o R1 R2 12 o kaila marie joy d.r. turla
23. 23. Solution: 180m . = ___C1____ = ____C2____ sin 1650 sin 90 sin 60 Cn = 2R(sin I/2 ) C1 = 2R1(sin I/2 ) 108. 79 = 2 R1(sin I2/2 ) R1= 520.38 m C2 = 2R2(sin I/2 ) 72.70= 2R2(sin 18/2) R2= 232.37 m kaila marie joy d.r. turla
24. 24. Restationing kaila marie joy d.r. turla
25. 25. Restationing kaila marie joy d.r. turla
26. 26. Sample Problem: A 2.5 km roadway centerline isestablished during a preliminary survey. Thethree tangent section are to be connected bytwo simple curves with radius of 213.5 m andthe second with a radius of 182.9 m.Determine the station of the PCs and PTs,the total length of the curve and the laststation of the final route. kaila marie joy d.r. turla
27. 27. 0 + 400 V2 R1 =213.5m S2 PC1 PT1 S1 V1 R1 =182.9 m 35 o 0 + 500 Last Sta. 2 + 500 m 700 kaila marie joy d.r. turla
28. 28. T1 = R1 (tan I/2) Sta PC2= Sta PT1 + S1 = 213.5 tan (35/2) = ( 0 + 563.11 ) + 585.38T1 = 67.32 m Sta PC2 = 1 + 148.49Sta PC1= Sta V1 - T1 D2 = 1145.916 = 1145.916 . = 6015 = ( 0 + 500 ) - 67.32 R2 182. 40 Sta PC1 = 0 + 432.68D = 1145.916 = 1145.916 . = 5022 LC2 = 20I2 = 20 (290) = 92.8m R 213.5 D2 6015LC1 = 20I1 = 20 (350) = 130.43 Sta PT2= Sta PC2 + LC2m = ( 1 + 148.49) + 92.8 0 D1 5 22 Sta PT2 = 1 + 241.29Sta PT1= Sta PC1 + LC1 S2 = 1300 - T2 = ( 0 + 432.68 ) + 130.43 = 1300 - 47.30 Sta PT1 = 0 + 563.11 = 1252.70 mS1 = 700 - (T1+T2) Last Station = Sta. PT2 + S2T2 = R2 (tan I/2) = ( 1 + 241.29 ) + 1252.70 = 182.90 tan (29/2) Last Station = 2 + 493.99 therefore,T2 = 47.30 m Total Length = 2493.99mS1 = 700 - (67.32 + 47.30 )S1 =585. 38m kaila marie joy d.r. turla
29. 29. Reversed Curve kaila marie joy d.r. turla
30. 30. Reversed Curve A reverse curve consists of two simple curvesjoined together, but curving in opposite directions.The curves are connected at the Point of ReversedCurve, PRC, which is the PC of the first curve andalso the PT of the succeeding curve. For safety reasons, this curve is seldom used inhighway construction as it would tend to send anautomobile off the road. kaila marie joy d.r. turla
31. 31. rsed CurvesElements of Reve kaila marie joy d.r. turla
32. 32. ELEMENTSOFREVERSEDCURVES kaila marie joy d.r. turla
33. 33. F O D S E E S P R S Y E ET V V E RR U C kaila marie joy d.r. turla
34. 34. Since the tangents are parallel, the deflection angles are also equal. kaila marie joy d.r. turla
35. 35. From a common point, the backward and forward tangents lead indifferent directions. kaila marie joy d.r. turla
36. 36. The backward and forward tangents meet or join at a particular pointcalled point of convergence. kaila marie joy d.r. turla
37. 37. An intermediate tangent lies at the common in between the PT of thepreceding curve and the PC of the next curve. kaila marie joy d.r. turla
38. 38. Sample Problem: The perpendicular distance between twoparallel tangents of a reversed curve is 35m.The azimuth of the common tangent is 300o.If the radius of the first curve is 150m,determine the radius of the second curve. kaila marie joy d.r. turla
39. 39. Solution: a y R2 35mR1 b 30o x kaila marie joy d.r. turla
40. 40. Solution:R1 = a + x ; a = R 1- xR2 = b + y ; b = R2 - y35 = a + b35 = (R1- x) + (R2 - y)cos 30o = X/150 ; x= 150 cos 30ocos 30o = y/R2 ; y= R2 cos 30o35 = (150 - 150 cos 30o) + (R2 - R2 cos 30o)solving for R2 R2 = 111.24m kaila marie joy d.r. turla
41. 41. The end.... kaila marie joy d.r. turla