Pearl Angelie Villanueva Mahusay

Student’s Name Professor’s Name Subject DD MM YYYY Given: Design speed = 80 km/h Terrain is typically clay with some localised sandstone Horizontal curve radii: Curve 1 (chainage 60 to 200) 260 metres; Curve 2 (chainage 420 to 600) 240 metres; Curve 3 (chainage 870 to- metres Drainage structures: Chainage 90 - 1 x 900 mm diameter pipe; Chainage 562 - 2 x 1200 mm diameter pipes; Chainage 1000 - 1 x 900 mm diameter pipe Maximum depth of fill = 6.5 metres Maximum depth of cut = 5 metres Minimum cover over drainage pipes = 1.25 metres Pavement width = 7.0 m (2 x 3.5 m lanes) with shoulder width = 1.5 m. Verge has 1m in fill (allowing for safety barrier) and 2.0 m in cut (includes table drain) Table drain 2.0 m wide, 1:4 slope Batter slopes 1:1 in cut Embankment = 2 (h): 1 (v) Maximum grade = 7% and Minimum grade = 1% Maximum superelevation = 7% and Minimum superelevation = 3% Crossfall on straights 3% Problem: 1. Horizontal Design a. Design the superelevation required for the three horizontal curves. For Curve 1 with R = 260 m; Curve 2 with R = 240 m; and Curve 3 with R = 230 m. For the design speed of 80 km/h, side friction (f) = 14 (WSDOT Design Manual, 3). To solve for the superelavation: R = 6.68V2 / (e + f); where R is the minimum allowable radius; V is the design speed; e is the superelevation rate; and f is the side friction factor. Substituting values: For Curve 1: e = (V2/127R) – f = (802/127(260)) – 0.26; e =- For Curve 2: e = (V2/127R) – f = (802/127(240)) – 0.26; e =- For Curve 3: e = (V2/127R) – f = (802/127(230)) – 0.26; e =- b. Design the superelevation development (using rate of rotation criterion), the plan transition and the shift (if needed) for the 230 metre curve -). For the design speed of 80 km/h the minimum radius for normal crown section is 2.5 km (WSDOT Design Manual, 3) and side friction (f) = 14. To solve for the superelavation: R = 6.68V2 / (e + f); where R is the minimum allowable radius; V is the design speed; e is the superelevation rate; and f is the side friction factor. Substituting values: R = 6.68V2 / (e + f) => 2.5 km = 6.68 (80)2 / (e + 14) => 2.5 km = 42,752 / (e + 14) e = 11.99% c. Draw the cross sections at chainages 900 and 940 using the attached cross section sheet. The cross sections show the natural surface levels. At chainage 900, the assumed design level at the centreline is 289.48. At Chainage 940, the assumed design level is 291.10. On the cross sections show the dimensions of the lanes, the shoulders, verge and shift. At Chainage 900, with assumed design level at the centreline is 289.48: At Chainage 940, with assumed design level at the centreline is 291.10: 2. Vertical Design a. The attached PDF shows the plan of the road and the long section of the natural surface. What are the control points, what are the target levels at the control points and why? Control points are large marked targets on the ground, spaced strategically throughout the area of interest. If the control points are used in the plan, there is a need to determine the coordinates at the center of each. For this problem, the control points shown in the plan are:-N, 281500E, 281750E, 282000E, 282250E, 282500E, and-N. These control points with their coordinates are then used to accurately position the plan in relation to the real map. b. What are the k values for the crests and the sags. K values are calculated by the equation: K = S2 / 2158; where K is the rate of vertical curvature, and S is sight distance. Since the design speed is 80 km/h and the gradients are 7% maximum and 1% minimum, the calculated values of k are 25.7 and 26. Table 1. K Values Design Speed (kph) Based on Stopping Sight Distance for a Car Absolute Minimum Values K Value Desirable Minimum Values K Value R­­T=1.5 s R­­T=2.0 s R­­T=2.5 s R­­T=1.5 s R­­T=2.0 s R­­T = 2.5 s - - - 26 c. Select trial grades between the control points and select vertical curves which meet the k value criteria. Show how the grades and the length of vertical curves meet the requirements for minimum k values. Note that the start and finish the design levels at the existing levels at chainages 00 and 1180. The trial gradient is assumed either by a distance-elevation computation or by use of an established maximum limit. Since the contours have 400-m interval, a pair of dividers can be set equal to the horizontal distance in which the given grade rises 400-m in 1000-m. Assuming the grade is 0.90%, the distance will be 400/0.009, which meet the requirements for minimum k values. Works Cited Austroads Guide to Road Design Part 3. Geometric Design 2016 and the relevant Roads and Maritime Services NSW Supplements WSDOT Design Manual. “Chapter 1250 Cross Slope and Superelevation.” Design Manual M 22-01, pp. 1-16, July 2017, https://www.wsdot.wa.gov/publications/manuals/fulltext/M22-01/1250.pdf