CIVE 2004: GIS, Surveying and Graphics Fall 2022 PDF

CIVE 2004: GIS, Surveying and Graphics Fall 2022  Chapter 24, 25, 26 Horizontal highway curve Vertical highway curve Volume  Lecture Outline 3 Introduction - Why we need to construct highway curves? Overview of horizont...

CIVE 2004: GIS, Surveying and Graphics Fall 2022  Chapter 24, 25, 26 Horizontal highway curve Vertical highway curve Volume  Lecture Outline 3 Introduction - Why we need to construct highway curves? Overview of horizontal and vertical curves alignment design Degree of curvature Elements and terminologies of horizontal curves Laying out and stationing on horizontal curve Methods for laying out horizontal circular curve: Deflection angles method Vertical curves; introduction and summary of factors considered in designing a vertical curve Implication of proper vertical curve Vertical curve geometry and elements and terminologies of vertical curves Volume measurements  Articles 4 Article: 24.1 to 24.8 Article: 25.1 to 25.7 Article 26.1 to 26.8 5 Why can we not always make our roads straight?  Notes on highway curves 6 Highway curves are provided a) to avoid route obstacles such as waterways/mountains b) to provide smooth transitions between straight segments c) to avoid excess cuts or fills. There are many types of curves (horizonal curve, vertical curve). The least costly highway alignment is one that takes the form of the natural topography. It is not always possible to select the lowest cost alternative design because the designer must adhere to certain standards that may not exist on the natural topography. The design speed primarily affect the design of the highway alignments Curves are constructed to provide for a smooth flow of traffic while ensuring safety features and design speed  Highway curves 7 Horizontal curve Vertical curve  Design of highway alignments: Notes 8 Highway alignments Vertical Horizontal (selection of grades for (selection of straight/tangent sections straight/tangent sections and the length of and the length and radius parabolic curve) of circular curve/arc)  Horizontal curves 9 Horizontal curves refer to the curves that are laid on horizontal planes to connect two straight tangent sections In design of horizontal curve; length of radius, length of arce/curve and horizontal offsets from tangents are determined based on design speed and roadway characteristics Curves need to be long enough to avoid unsafe condition (minimum radius for highway standards Horizontal curve ~350 m or 5o degree of curvature) Additional features can help reduce driving effort o Superelevation  Concept of Superelevation of horizontal curve 10 𝑊𝑢2 cos 𝛼 = W sin 𝛼 + 𝑊𝑓𝑠 cos 𝛼 𝑔𝑅 Dividing both sides by W and 𝑐𝑜𝑠 𝛼 𝑢2 = tan 𝛼 + 𝑓𝑠 𝑔𝑅 𝑡𝑎𝑛 𝛼 = 𝑒 (𝑐𝑎𝑙𝑙𝑒𝑑 𝑠𝑢𝑝𝑒𝑟𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛) 𝑢2 𝑅= 𝑔(𝑒 + 𝑓𝑠 ) 𝑢2 𝑒= − 𝑓𝑠 𝑔𝑅 No need to review this slide for exam. It is discussed in Highway Engineering course  Horizontal curves: Type 11 Simple Curve Compound Curve R R R r spiral R Reverse Curve R R R Easement or spiral Transitional Curve  Notes on spiral curves 12 Transition (or spiral) curves, which slowly transition from an infinite radius (a tangent) to the radius of the circular curve. Spiral/transition curves  Terminologies: Degree of curvature 13 Degree of curvature defines the severity of a curve and measured by the angle subtended by an arc/chord on the curve of a given length. Minimizing the “degree of curvature” and maximizing the “radius of a curve” both produce long gradual curves. Radius of curvature is also used to show the severity of curve. They are mathematically related. Metric Design of roadways tend to specify the curve radius while English unit design tends to specify the degree of curvature or the angle at the center of a circular arc subtending a i) chord length of 100ft or ii) arc length of 100ft. ARC DEFINITION CORD DEFINITION “Degree of Curve” “Degree of Curve” Central angle subtended by Central angle subtended by a circular ARC of 100 ft A CHORD of 100 ft (highways) (railways) Derivation is available in the chapter 15 in Highway Book by Garber and Hoel. Not needed for exam.  Circular curve – degree of curve and radius (in feet) 14  Example of degree of curvature 15 For a curve radius of 350m, determine the degree of curvature. 5729.58 ft ´ 0.3048m / ft = 4 59¢21¢¢ 350m  Elements and terminologies of a horizontal curve 16 PI = Point of Intersection of tangent lines PC = Point of Curvature also called the beginning of curve (BC) PT = Point of Tangency also called the end of curve (EC) L L = Length of Curve LC = Length of Long Chord M = Middle Ordinate, is the distance, from mid- point of LC to mid-point of curve E = External Distance from PI to mid-point of curve T = Tangent Distance from PI to PC (or PT) I = Intersection Angle R=Radius D=Subtended angle for curve length “s” POC = Point on Curve POT = Point on Tangent  Elements and terminologies of a horizontal curve 17 æIö T = R * tanç ÷ è 2ø L = RI (I in rads) p L L = RI (I in degrees) 180 æIö LC = 2R * sinç ÷ è 2ø  Elements and terminologies of a horizontal curve 18 æIö M = R - R * cosç ÷ è2ø L æIö R cosç ÷ = or è 2ø E + R æ ö ç 1 ÷ E = Rç -1÷ æIö Other formula from ç cosæç I ö÷ ÷ E = T * tanç ÷ Trigonometry: è 4æ øI ö ç ÷ E = T * tanç ÷ è è 2 ø ø è 4ø æIö M = E * cosç ÷  Laying out and stationing on a horizontal curve (Art 24.4) 19 (Example, every 20 m) 0+160.00 T PI L PC sta = PI sta – T PT sta = PC sta + L Note: PT may have two stations, one from curve calculation above and another if the route stationing was first completed using tangent stations (PT sta = PI stat + T).  Problem on horizontal curve (Example 24.1) 20 D 100 ft Assume that I= 8°24’, the station of the PI 360 = 2pR 64+27.46, and terrain conditions require L = RI (I in rads) the minimum radius permitted by the æIö T = R * tanç ÷ specifications of, say, 2864.79 ft (arc è 2ø æIö definition). Calculate the PC and PT LC = 2R * sinç ÷ è2ø stationing and the external and middle é ù æIö 1 ordinate distances for this curve. E = T * tanç ÷ = Rê è4ø ( I ) -1ú ë cos /2 û æIö é æ I öù M = E * cosç ÷ = Rê1- cosç ÷ú è 2ø ë è 2 øû PC sta = PI sta - T PT sta = PC sta + L D æ I ö d = = sç ÷ = sk 2 è 2L ø 𝐼 𝐾= ቇ c = 2R sin(d ) 2L  Problem 24.4 21 Compute L, T, E, M, LC, R, and stations of the PC and PT for the highway circular curve with following curve information (D, I and PI). D = 4d00m, I = 24d00m, and PI station = 36+45.00ft Ans: L 600ft, R 1433.12ft T 304.53ft E 32.01ft M 31.31ft LC 595.74ft PC 33+40.47ft PTBack 42+45.00 PTForward 39+49.53  Methods for laying a horizontal circular curve 22 Methods Deflection angles (This is the most common method. It Tangent Chord Middle Ordinates can be done by Coordinates offsets offsets ordinates from long chord incremental chord method or total chord method)  Methods for laying a horizontal circular curve: Deflection angles method 23  Methods for laying a horizontal circular curve: Deflection angles method 24 𝑑𝑎 𝐷 = 𝑠𝑎 100 𝑠𝑎 𝐷 𝑑𝑎 = (𝑑𝑒𝑔𝑟𝑒𝑒𝑠ቇ 100 𝑠𝑎 𝐷 The angle at a point 𝛿𝑎 = (𝑑𝑒𝑔𝑟𝑒𝑒𝑠ቇ between a tangent and 100 x2 any chord is equal to half Ca the central angle sinδa = subtended by the chord 2R 𝑐𝑎 = 2𝑅 𝑠𝑖𝑛 𝛿𝑎 Figure 24.6 Subchords and subdeflections.  Problem on deflection angles (Example 24.2) 25 Compute subdeflection angles and subchords 𝛿𝑎 , 𝑐𝑎, 𝛿𝑏 and 𝑐𝑏 and calculate chord 𝑐 of Example 24.1 Hints: Use these formulas 𝑠𝑎 𝐷 𝛿𝑎 = (𝑑𝑒𝑔𝑟𝑒𝑒𝑠ቇ 100 x2 𝑐𝑎 = 2𝑅 𝑠𝑖𝑛 𝛿𝑎  Example 24.3 26 Assume that a metric curve will be used at a PI where I is 8°24’. Assume also that the station of the PI is 6+427.464 , and that terrain conditions require a minimum radius of 900 m. Calculate the PC and PT stationing, and other defining elements of the curve. Also compute notes for staking the curve using 20-m increments. Review it in home. See me in office hour if you face any problem to understand.  Problem# 24.7 27 Tabulate R or D, T, L, E, M, PC, PT, deflection angles, and incremental chords to lay out the circular curve at full stations (100 ft or 30 m) with the following given information. Highway curve with Da = 2°30′, I = 10°30′, and PI station = 36 +44.50 ft. Ans: Intersection Angle = 10°30'00" Degree of Curvature = 2°30'00" Radius = 2,291.83 Circular Curve Length = 420.00 Tangent Distance = 210.59 Circular Curve Long Chord = 419.41 Middle Ordinate = 9.61 External = 9.65 PI Stationing = 36+44.50 38+53.91 Back = 38+55.09 Ahead And field note like one presented in Example 24.3 28 Ch 25 Vertical highway curve 29 Vertical curves In addition to horizontal curves that go to the right or left, roads also have vertical curves that go up or down Vertical curves are used to join tangents (eg: tangent 1, 2 and 3 below) in order to provide a gradual change in grade from the initial (back) tangent to the grade of the second (forward) tangent Vertical curves at the top of a hill are called crest curves and at the bottom of a hill or dip are called sag curves.  Summary of factors considered in 30 designing a vertical curve Providing a good fit with the existing ground profile, thereby minimizing depths of cuts and fills Balancing the volume of cut materials against fill Maintaining adequate drainage Not exceeding maximum specified grades (G/g) and meeting fixed elevations such as intersections with other roads The curves must have lengths sufficient to meet specifications covering a maximum rate of change of grade (which affects the comfort of vehicle occupants) The curves should provide sufficient sight distance for safe vehicle operation.  Vertical curves: Important Notes 31 Design objective of vertical curve is to provide a gradual change from one grade to another so that vehicles may run smoothly as they traverse the curve. Vertical curve is usually provided in parabolic shape G1 (g1) and G2 (g2) are selected based on terrain type and design speed. Lmin is calculated using parabolic formula and minimum stopping sight distance/SSD (indicated with “S” as well) Crest vertical curve Sag vertical curve  Implication of proper vertical curve 32 https://www.youtube.com/watch?v=7LSm79HWsK4  Vertical curve geometry 33 Parabolas provide a constant rate of change of grade, they are ideal and almost always applied for vertical alignments used by vehicular traffic. The general mathematical expression of a parabola: YP = the ordinate at any point p of the parabola at a distance Xp from the origin of the curve a = the ordinate at the beginning of the curve (X = 0) b = the slope of the tangent to the curve (X = 0) bXp= the change in ordinate along the tangent over distance Xp c𝑋𝑝2 = the parabola’s departure from the tangent (tangent offset) in distance Xp  Elements and terminologies of a vertical curve 34 These terms used by engineers/surveyors BVC = beginning of vertical curve OR VPC = vertical point of curvature V = the vertex, often called VPI VPI = vertical point of intersections EVC = end of vertical curve OR VPT = vertical point of tangency g1 = grade of the back tangent (%) g2 = grade of the forward tangent (%) L = horizontal distance (BVC to EVC)  Property of Equal Tangent Vertical Parabolic Curve (Both for crest and sag curves) 35 Equal Tangent Vertical Parabolic curve: An equal tangent vertical parabolic curve means the vertex occurs at a distance X = L/2 from the BVC (See proof at section 25.3)  Elements and terminologies of a vertical curve 36 Using surveying terminology the equation becomes: We need to express c values in terms of known values. To do so, we can write: So, the final equation is Explanation is available on Page 761(Textbook) Elements and terminologies of a vertical curve 37 The rate of change of grade, r, for an equal tangent parabolic curve equals the total grade change from BVC to EVC divided by length L (on stations for the English system, or 1/10th stations for metric units), over which the change occur. This is the same as the second derivative of the vertical curve equation. The slope of the curve at any point: The rate of change grade (r) is given by the second derivative: The value of r (which is negative for a crest curve and positive for a sag type) is an important design parameter because it controls the rate of curvature, and hence rider comfort  Example 25.1: Vertical curve stakeout 38 A grade g1 of +3.00% intersects grade g2 of -2.4% at a vertex whose station and elevation are 46+70 and 853.48 ft, respectively. An equal-tangent parabolic curve 600 ft long has been selected to join the two tangents. Compute and tabulate the curve for stakeout at full stations. (Figure 25.4 shows the curve.)  Example 25.1: Vertical curve stakeout 39 40 41 Ch 26 Volume  Methods for volume measurement 42 Methods Unit area Contour Cross-section (Borrow- area method pit) method method  Cross-section method 43 The cross-section method for computing volumes used in linear construction projects such as highways, railroads, and canals. In this procedure, after the centerline has been staked, ground profiles called cross sections are taken (at right angles to the centerline), usually at intervals of full or half stations Cross-sectioning consists of observing ground elevations and their corresponding distances left and right perpendicular to the centerline. Readings must be taken at the centerline, at high and low points, and at locations where slope changes occur to determine the ground profile accurately. This is done in the field using a level, level rod, and tape  Cross-section method 44 Cross section leveling Appendix B.5  Cross-section method 45 Figure 26.1 Section of roadway illustrating excavation (cut) and embankment (fill).  Types of cross-sections 46 Level section Three-level section Five-level section Irregular section Transition section Side-hill section  Types of cross-sections 47 Figure 26.2: Earthwork sections.  Volume by average-end-area-method 48 Example 26.1: Compute the volume of excavation between station 24+00 (ft) with an end area of 711sft and station 25+00 (ft) with an end area 515sft. Ans: 2270 yd3  Determining End Areas by Simple Figures 49 Elevation (ft) Rod reading Distances from the centerline (ft) Assume: A level roadbed of 30-ft width Cut slopes of 1-1/2:1 Subgrade elevation at station of 858.9 ft. Figure 26.4: End-area computation  Computation of slope intercept 50 Figure 26.5: Computation of slope intercept R of Figure 26.4  Example 26.2 on computation of slope intercept 51 Determine the coordinates of point R in Figure 26.5 using Line Equation Point X Y E 12 869.0 F 50 872.8 G 15 858.9 Figure 26.5: Computation of slope intercept R of Figure 26.4  Prismodial method for volume estimate 52 Acknowledgements: Dr. Snelgrove Dr. Charles Ghilani’s book Pearson Canada Sitotaw Yirdaw  Other Photo Sources/Credits 54 https://www.mathsisfun.com/equation_of_line.html https://en.wikipedia.org/wiki/Conformal_map#/media/File:Conformal_map.svg I will add later