P.E. Civil Exam Review:

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1 P.E. Civil Exam Review: Geometric Design J.P. Mohsen

2 Horizontal Curves Slide 2

3 Back tangent Forward tangent T T Slide 3

4 Back tangent Forward tangent T T P.C. or T.C. P.T. Slide 4

5 Back tangent P.I. E I = intersection angle Forward tangent E T L = curve length T M M P.C. or T.C. P.T. LC = long chord R R Slide 5

6 Horizontal Curves: EQUATIONS Slide 6

7 Back tangent P.I. E I = intersection angle Forward tangent E T L = curve length T M M P.C. or T.C. P.T. LC = long chord R R Degree of curvature (D), chord basis: R 50 1 sin D 2 Slide 7

8 P.I. I = intersection angle Back tangent Forward tangent E E T L = curve length T M M P.C. or T.C. P.T. LC = long chord R R Curve Equations: arc basis R D Slide 8

9 P.I. I = intersection angle Back tangent Forward tangent E E T L = curve length T M M P.C. or T.C. P.T. LC = long chord R R Degree of curvature (D), arc basis: R 100 R D D 2 R D Slide 9

10 P.I. I = intersection angle Back tangent Forward tangent E E T L = curve length T M M P.C. or T.C. P.T. LC = long chord R R Curve Equations: T 1 R tan 2 I Slide 10

11 Back tangent P.I. E I = intersection angle Forward tangent E T L = curve length T M M P.C. or T.C. LC = long chord P.T. R R Curve Equations: LC 1 2 Rsin 2 I Slide 11

12 P.I. I = intersection angle Back tangent Forward tangent E E T L = curve length T M M P.C. or T.C. P.T. LC = long chord R R Curve Equations: E T tan I 4 E Rsec I E R 1 cos I 2 Slide 12

13 Back tangent P.I. E I = intersection angle Forward tangent E T L = curve length T M M P.C. or T.C. P.T. LC = long chord R R Curve Equations: M I R 1 cos 2 Slide 13

14 Back tangent P.I. E I = intersection angle Forward tangent E T L = curve length T M M P.C. or T.C. P.T. LC = long chord R R Curve Equations: L 100I D Slide 14

15 Horizontal Curves Equations D R D L 100I R 1 50 I R LC 1 sin 2 D D 2 1 sin I R LC 2 sin 2 I T E 4 tan I T E 2 cos 1 I R M I R T 2 1 tan 1 cos 1 I R E Slide cos

16 Degree of curvature, chord basis. The degree of curvature is defined as the central angle subtended by a chord of 100 ft. 100 ft 50 ft 50 ft R D R D/2 D/2 R = 50 / sin(1/2 D) Slide 16

17 Degree of curvature, arc basis. The degree of curvature is defined as the central angle of a circle which will subtend an arc of 100 ft. 100 ft R D R R = (360/D)(100/2π) ) = / D Slide 17

18 Example of a sharp curve vs. a flat curve. 100 ft 100 ft D R R D R R A sharp curve has a small D A flat curve has a large D and and a large R. a small R. Slide 18

19 EXAMPLE PROBLEM 1: The bearings of two tangents connected by a horizontal circular curve are N50 E and S35 E, respectively. The tangents intersect at station The curve radius is 800 ft. Slide 19

20 EXAMPLE PROBLEM: The bearings of two tangents t connected by a horizontal circular curve are N50 E and S35 E, respectively. The tangents intersect at station The curve radius is 800 ft. N 50 W E S Slide 20

21 EXAMPLE PROBLEM: The bearings of two tangents t connected by a horizontal circular curve are N50 E and S35 E, respectively. The tangents intersect at station The curve radius is 800 ft. N 50 W E 35 S N 50 W E S 35 Slide 21

22 EXAMPLE PROBLEM: The bearings of two tangents t connected by a horizontal circular curve are N50 E and S35 E, respectively. The tangents intersect at station The curve radius is 800 ft P.I. I = Slide 22

23 EXAMPLE PROBLEM 1: The bearings of two tangents t connected by a horizontal circular curve are N50 E and S35 E, respectively. The tangents intersect at station The curve radius is 800 ft. (a) What is the length of the curve? (b) What is the station of the PC? (c) What is the station of the PT? (d) What is the interior angle at the PI? (e) What is the tangent distance from the PI to the PC? (f) What is the long chord distance? (g) What is the external distance? (h) What is the degree of the curve (arc basis)? (i) What is the degree of the curve (chord basis)? (j) What is the chord length of a 100-ft arc (arc basis)? Slide 23

24 EXAMPLE PROBLEM 1: The bearings of two tangents t connected by a horizontal circular curve are N50 E and S35 E, respectively. The tangents intersect at station The curve radius is 800 ft I = 95 R D P.I. D D Slide 24

25 EXAMPLE PROBLEM 1: The bearings of two tangents t connected by a horizontal circular curve are N50 E and S35 E, respectively. The tangents intersect at station The curve radius is 800 ft. (a) What is the length of the curve? L 100I D L L ft (b) What is the station of the PC? R tan I 2 T T 800 tan47. 5 T ft PC PC station Slide 25

26 EXAMPLE PROBLEM 1: The bearings of two tangents t connected by a horizontal circular curve are N50 E and S35 E, respectively. The tangents intersect at station The curve radius is 800 ft. (c) What is the station of the PT? PC station L ft PT PT station Slide 26

27 EXAMPLE PROBLEM: The bearings of two tangents t connected by a horizontal circular curve are N50 E and S35 E, respectively. The tangents intersect at station The curve radius is 800 ft. (d) What is the interior angle at the PI? P.I. I = Slide 27

28 EXAMPLE PROBLEM 1: The bearings of two tangents t connected by a horizontal circular curve are N50 E and S35 E, respectively. The tangents intersect at station The curve radius is 800 ft. (e) What is the tangent distance from the PI to the PC? T R tan I T ft 2 (f) What is the long chord distance? LC I 2Rsin (800) sin 2 LC 1179 ft Slide 28

29 EXAMPLE PROBLEM 1: The bearings of two tangents t connected by a horizontal circular curve are N50 E and S35 E, respectively. The tangents intersect at station The curve radius is 800 ft. (g) What is the external distance? LC 95 Rsec I 1 800sec E 384 ft I E T tan ft 1 E R I cos 2 ft Slide 29

30 EXAMPLE PROBLEM 1: The bearings of two tangents t connected by a horizontal circular curve are N50 E and S35 E, respectively. The tangents intersect at station The curve radius is 800 ft. (h) What is the degree of the curve (arc basis)? D D R 800 (i) What is the degree of the curve (chord basis)? R 50 1 sin 2 D sin D D 2sin D 7.16 Slide 30

31 EXAMPLE PROBLEM 1: The bearings of two tangents t connected by a horizontal circular curve are N50 E and S35 E, respectively. The tangents intersect at station The curve radius is 800 ft. (j) What is the chord length of a 100-ft arc (arc basis)? D 7.16 chord length 2Rsin 2800sin chord length 100 ft 2 2 Slide 31

32 Example Problem 2 Convert the following angle to degrees 39 deg 41 min 54 sec Solving: 54/60 = 0.9 min 41.9/60= deg Answer = degrees Slide 32

33 Example Problem 3 Express the following angle in terms of degrees, minutes, and seconds degrees Solving; 0.74 (60) = 44.4 minutes 0.4 (60) = 24.0 seconds Answer: 91 deg 44 min 24 sec Slide 33

34 EXAMPLE PROBLEM 4: The two tangents shown intersect 2000 ft beyond Station The back tangent has a bearing of N W and the forward tangent has a bearing of N E. The decision has been made to design a 3000 ft radius horizontal curve between the two tangents. (a)what is the central angle of the curve? (b)compute the tangent distance and length of curve. (c)what is the station of the PC? (d)what is the station of the PT? (e)what is the degree of curvature of this curve using the arc definition? Slide 34

35 P.T. I I PI P.I. R=3000 P.C. Slide 35

36 EXAMPLE PROBLEM 4: 1. What is the central angle for this curve? (a) 32 degrees (b) 65 degrees (c) 45 degrees (d) 60 degrees (e) 15 degrees 2. What is the tangent distance for this curve? a) ft b) ft c) ft d) ft e) ft Slide 36

37 EXAMPLE PROBLEM 2: 3) What is the length of this curve? (a) ft (b) ft (c) ft (d) ft (e) ft 4) What is the station of the PC? a) b) c) d) e) Slide 37

38 EXAMPLE PROBLEM 4: 5) What is the station of the PT? (a) (b) (c) (d) (e) ) What is the degree of curvature of this curve using the arc definition? a) 60 degrees b) 3.5 degrees c) 2.9 degrees d) 1.9 degrees e) 3.9 degrees Slide 38

39 EXAMPLE PROBLEM 5: The long chord of a circular curve is 600 feet, the intersection angle is 110. Find the radius. The forward tangent of this curve needs to be moved in 5 feet due to a right of way dispute. What radius curve would you specify to hold the B.C. at the same location as in the original curve? Slide 39

40 Stations P.C. T T I = 110 C R =? P.T. 5 Slide 40

41 EXAMPLE PROBLEM 5: The long chord of a circular curve is 600 feet. The intersection angle is 110 degrees. What is the radius of this curve? The forward tangent of this curve needs to be moved in 5 feet due to a right of way dispute. What radius curve would you specify to hold the PC at the same location as in original first curve. I We know: L.C. = 600 ft LC 2R sin 2 Then 600 R sin 55 ft ft I T R tan tan 03 2 Slide 41

42 EXAMPLE PROBLEM 5: x X X 5 0 sin T new R new curve Please note T tan that new 55 the o answer is NOT ft ft Slide 42

43 EXAMPLE PROBLEM 5: The alignment of a proposed highway was adjusted in order to avoid a critical obstacle. This was accomplished by moving the forward tangent 120 ft forward. You are asked to find the radius of the new curve if it begins at the same location as the PC of the initial curve. You need to also determine the stations for PC and PT for the new curve alignment. Slide 43

44 P.I. Sta P.C. 6 Curve Initial Curve 120 I = 65 Tangent moved forward Original Tangent New P.T. 120 Slide 44

45 P.I. Sta P.C. 6 Curve Initial Curve 120 I = 65 Proposed new tangent Original Tangent New P.T. 120 Slide 45

46 65 x =? Slide 46

47 EXAMPLE PROBLEM 5: For the original curve, find R and T : R T ft D 6 I R tan ft PC station for the original curve is: ( ) ( ) = Slide 47

48 65 x =? X sin 65 ft Slide 48

49 EXAMPLE PROBLEM 5: T new = = ft R T new newcurve tan 65 / ft D new L 100 I 100 D Dnew ft new curve 10 0 PT Station = Station PC + L New PT Station = ( ) + ( ) = Slide 49

50 Vertical Curves Slide 50

51 P.V.I. Back tangent Forward tangent Slide 51

52 P.V.I. Back tangent Forward tangent P.V.C. LC = long chord P.V.T. Slide 52

53 Tangent offsets P.V.I. Back tangent B Forward tangent P.V.C. C LC = long chord P.V.T. Slide 53

54 PROPERTIES OF A PARABOLA: Applied to Vertical Curve Analysis Slide 54

55 PROPERTIES OF A PARABOLA: 1. The curve elevation at its midpoint is halfway from the elevation at the P.V.I. to the elevation at the midpoint of the long chord. E 1 PVI 2 PVC elevation elevation PVT 2 elevation The curve lies midway between the point of intersection of the grade lines and the middle point of the chord joining the BVC (beginning of vertical curve) and the EVC (end of vertical curve). 2. The tangent offsets vary as the square of the distance from the point of tangency Y x 2 E T 2 Slide 55

56 PROPERTIES OF A PARABOLA: E g g g 2 2 r 8 L 1 L g 1 rate of grade change T in sta. X in sta. y E y E 2 x T 2 g 2 g 1 P.V.C. Slide 56

57 The Algebraic Signs of r Slide 57

58 Example Problem 7 GIVEN: Station at PVI Elevation at PVI P.V.I. g 1 =+4% g 2 =-3% B C Curve Length 800 ft Slide 58

59 Stations P.V.I. g 1 =+4% g 2 =-3% B C Slide 59

60 Stations Elevations on grade lines P.V.I. g 1 =+4% 7.00 B g 2 =-3% C Slide 60

61 Stations Elevations on grade lines P.V.I. g 1 =+4% g 2 =-3% B C Slide 61

62 Calculations for example 7: Elevation of PVI = ft Elevation of C = ½( ) ) = ft Elevation of B = ½ ( ) = ft E g1 g2l r g2 g L 8 8 rate of grade change Slide 62

63 Tangent offsets P.V.I. Back tangent B Forward tangent P.V.C. C LC = long chord P.V.T. Slide 63

64 The tangent offsets vary as the square of the distance from the point of tangency. y 2 x E T 2 T in sta. X in sta. y E g 2 g 1 P.V.C. Slide 64

65 Stations Elevations on grade lines y E 2 x T y P.V.I B g 2 =-3% y C y = 3.94 tangent offset at station Slide 65

66 Stations Elevations on grade lines P.V.I. 94 y g 2 =-3% B C Slide 66

67 Stations Elevations on grade lines P.V.I. 94 Tangent offset at station y B C Slide 67

68 Stations Elevations on grade lines P.V.I g 1 =+4% g 2 =-3% B C Slide 68

69 Stations Elevations on grade lines P.V.I g 1 =+4% g 2 =-3% B C Slide 69

70 Stations Elevations on grade lines P.V.I g 1 =+4% g 2 =-3% B C Elevations on curve Slide 70

71 Highest and Lowest Points on Vertical Curves: X is the distance between PVC and station of high or Low point r g r X 1 For the vertical curve in example problem 5: g L g g1 4 X r Stations from PVC Elevation at highest point Station is equal to: ft Slide 71

72 Calculations for the Elevation at highest point on the vertical curve which is at station 65+57: Elevation of tangent line at station is, (.03)(57)= = ft Offset at station is calculated by 7 / [(4)(4)] = y / [(3.43)(3.43)] Offset at station is y=5.147 ft Elevation at station on the vertical curve is which is equal to ft Slide 72

73 Vertical Curve Problem #8 Elevation BVC A a B b C Elevation c M Elevation O d D V E e F f EVC Elevation Slide 73

74 Vertical Curve Problem #8 Elevation BVC A a B b C Elevation c M Elevation O d D V E e f F EVC Elevation Slide 74

75 Vertical Curve Problem #8 Solutions Station of BVC = 50 4=station 46 Elevation of BVC = (4 x 6) = ft Station of EVC = = station 54 Elevation of EVC = (4 x 2) = ft Slide 75

76 Vertical Curve Problem #8 Solutions Elevation of middle point of chord = = ft Offset to curve at intersection = = 8.00 ft Slide 76

77 Vertical Curve Problem #8 Solutions Offset at A and F = x 8.00 = ft Offset at B and E = x 8.00 = 200ft 2.00 Offset at C and D = x 8.00 = 4.50 ft Slide 77

78 Vertical Curve Problem #8 Solutions Elevation of A = = ft Elevation of B = = ft Elevation of C = = ft Elevation of V = = ft Elevation of F = = ft Elevation of E = = ft Elevation of D = = ft Elevation of V = = ft* *checks! Slide 78

79 Vertical Curve Problem #8 Solutions Elevation of a = station 47= = Elevation of b = station 48 = = Elevation of c = station 49 = = Elevation of o = station 50 = = Elevation of d = station 51 = = Elevation of e = station ti 52 = = Elevation of f = station 53 = = Slide 79

80 Vertical Curve Problem #8 Solutions OFFSET TANGENT ELEVATION STATION FROM ELEVATION CURVE TANGENT BVC = EVC = Slide 80

81 SIGHT DISTANCE: Minimum sight distance (s), where sight distance is less than the curve length (L) in stations and the height of driver s eye (h) is 3.5 ft to 3.75 ft above pavement. s 2 8Lh g 1 g 2 If the calculated sight distance (s) is greater than the curve length (L) then use the following equation: s L 4h 2 g g 1 2 Slide 81

82 EXAMPLE PROBLEM 9: Sight distance calculations l for the vertical curve in problem 5: L 800 ft g 1 4% g 2 3% h ft s stations s ft If the calculated sight distance (s) is greater than the curve length (L) then use the following equation: s L 4h 2 g g 1 2 Slide 82

83 EXAMPLE PROBLEM 10: A proposed 2-lane highway has a vertical alignment that is +3% grade intersecting a 2% grade at station at an elevation of The proposed alignment must bridge over an existing railroad track which crosses the proposed alignment at station The elevation of the railroad track at the point of intersection is The proposed highway alignment must have a vertical elevation difference of 26.0 ft at the point of intersection of the two alignment in order to satisfy vertical clearance requirements. You are to determine the longest length of vertical curve, rounded to the nearest 100 ft, that will fulfill these criteria. Determine the station of the high point on the vertical curve. Slide 83

84 26+00 El El. 223 g 1 =+3% El. 221 g 2 =-2% clearance requirement = 26.0 ft. El Slide 84

85 EXAMPLE PROBLEM 10: L For vertical curve to go through a fixed point, use the following formula : offset Aoffset 2 offset 2 A 4 g1 g2 g1 g2 g1 g2 2 offset = 2 A = 2.5 g1-g2 = 5 L = stations therefore use L= 1000 ft to maintain 26 ft minimum clearance Station of high point g1 3 x 6 r Stations from PVC Slide 85

86 EXAMPLE PRBLEM 10: Using design curve length L=1000 ft Location of high point g1 3 x 6 r Stations from PVC Station at PVC = (26+00 ) (5+00) = Station of high point = (21+00) + (6+00) = Slide 86

87 EXAMPLE PROBLEM 11: The proposed highway is to cross another highway at right angles. The elevation of the proposed crossing has been established and a minimum vertical clearance of 25 ft. will be required between the proposed highway and the existing highway. A. Determine the location and elevation of the low point on the existing vertical curve. B. Determine the minimum station at which the crossing may be located. C. Determine the maximum station at which the crossing may be located. Slide 87

88 Existing Vert. Curve of other highway Proposed Highway El g 1 =-4% g 2 =+3% El. 736 V.P.I Slide 88

89 EXAMPLE PROBLEM 11 : Calculate Low point on existing vertical curve g1 4 x r 3 ( 4) 12 Stations from PVC Station at low point = (82+00) + ( ) = Elevation of low point is calculated to be ft Proposed highway has an elevation of 777 ft and required clearance is 25 ft, therefore, there exists two locations on the curve with elevation of = 752 ft By trial and error, we need to find minimum and maximum station where curve elevation is 752 ft Slide 89

90 Existing Vert. Curve Proposed Highway El P.V.C g 1 =-4% Low point g 2 =+3% E.V..C El. 736 V.P.I. V.P P.I Slide 90

91 Trial and error for maximum station: Station Tangent elevation offset Curve elevation Maximum Station turns out to be Slide 91

92 Trial and error for minimum station: Station Tangent elevation offset Curve elevation Minimum Station turns out to be Slide 92

93 P.V.C El. 760 Min. Sta.? Sag Max. Sta.? 752 E.V..C El. 754 I Slide

94 References: M C Cormack, Jack, Surveying Fundamentals, Prentice Hall, 1983 Brinker, Russell, Wolf, Paul, Elementary Surveying, Sixth Edition, Harper and Row, 1977 Slide 94

95 Thank You. Any Questions? Good Luck! Slide 95

Symmetrical Parabolic Curve In highway practice, abrupt change in the vertical direction of moving vehicles should be avoided. In order to provide

Symmetrical Parabolic Curve In highway practice, abrupt change in the vertical direction of moving vehicles should be avoided. In order to provide gradual change in its vertical direction, a parabolic