The Vertical Alignment of the Road - Assignment Example

Question 1 Design the vertical alignment of the road for a design speed of 100km/h. Draw an appropriate longitudinal section showing lengths of all vertical curves and tabulate levels at 20m chainages from CH100.00 to CH500.00. Calculation of Intersection Point (IP)’s Chainage (CH) and Reduced Level (RL) Derive three equations based on the given information: Solve for the three variables using the three equations: Therefore Chainage at intersection point (IP) = and Reduced Level (RL) at IP = Calculation of the circular curve length (L) Calculate the Stopping Sight Distance (SSD) based on the given design highway speed of 100km/h: Calculate the vertical curve length (L) using the formula L=KA and the K value using assumed parameters: Therefore the length of the circular curve is Calculation of the Chainage of TP1 (left hand side) and TP2 (right hand side) and their respective Reduced Level (RL) : Calculation of the e value at the Intersection Point (IP) and the Yt values along the 20m chainage intervals Parameters Values Chainage Range (m) 300 700 Chainage RL (m) 37 38 RL Difference 1 LHS Gradient -2.8%   RHS Gradient +3.0%   A 5.8   Design Speed (V) (km/h) 70   Sag Curve K Value 44   Length of curve (L) (m) 260   L/2 (m) 130   IP CH 490   IP RL 31.62 TP1 CH (LHS) 360   TP1 RL 35.26 TP2 CH (RHS) 620   TP2 RL 35.52 Low Point 490   Coefficient of side friction (f) 0.11   Radius of curve m 310   Tabulation of the Natural Surface RL can be performed by estimating the chainage RL that correspond with the contour map. The tabulation the RL of the circular curve along the 20m chainage intervals is completed by calculating the and values based on the RL and chainage of TP1 and TP2. The RL of the tangent make up the straight lines that lies before TP1 (CH360) and after TP2 (CH620). The road curve RL make up the circular curve that lies between TP1 and TP2.   Chainage (CH) x Tangent RL Yt Road Curve RL Estimated Contour RL (Natural Surface) Point A 300   37     35.3   320   36.44     35.2   340   35.88     35.1 TP1 360 0 35.32 0 35.320 35.15   380 20 34.76 0.0446 34.805 35   400 40 34.2 0.1785 34.378 34.8   420 60 33.64 0.4015 34.042 34.6   440 80 33.08 0.7138 33.794 34.4   460 100 32.52 1.1154 33.635 34.1   480 120 31.96 1.6062 33.566 33 IP 490 130 31.68 1.8850 33.565 32.8 TS 500 120 31.98 1.6062 33.586 33.2   520 100 32.58 1.1154 33.695 34.2   540 80 33.18 0.7138 33.894 34.8   560 60 33.78 0.4015 34.182 35.5   580 40 34.38 0.1785 34.558 36.1   600 20 34.98 0.0446 35.025 36.8 TP2 620 0 35.58 0 35.580 37.5   640   36.18     38.1   660   36.78     38.7   680   37.38     39.1 Point C 700   37.98     39.3 Question 2 Design the superelevation development. Superelevation diagram of a transition curve Calculation of the superelevation (e) Calculation of the length of superelevation development There are two methods calculating the superelevation development length: 1. Rate of rotation of pavement surface method The super elevation is applied to the right carriageway since the road pavement is turning left. Maximum Rate of rotation of pavement = 0.025m/m/sec = 2.5% per sec Total cross fall change for right carriageway = 3% + 3% = 6% Total cross fall change for right shoulder = 3% + 3% = 6% 2. Relative grade of the edge of pavement method According to Austroads GTRD3, the relative grade can be taken as with the operating speed of and applying to a two lane two way road with the axis of rotation on the centreline. Calculating the level at outer edge of road Where W = width of lane = 3.5+ 2 = 5.5 m e = design superelevation = 1.5% = 0.015 RL centre = 31.62 at IP CH490 Calculate the total rise of right edge of pavement relative to centreline (0.0825 below centreline and ended 0.0825 above centreline) (0.55% maximum relative grade allowed for design speed of 70km/h) Therefore the superelevation length can be assumed to be approximately 30m Calculation of the length of the transition curve: A = 0.45 V = 70km/h R = 310m L = Transition Curve in m L = 49.18m = 50m approximately Determine the chainage starting and ending points for superelevation: Looking at the map, the highway is curving to the left so this is a left hand curve scenario. Therefore, the cross falls at point 1, 2 and 3 should remain constant through the curve. Superelevation development of 1.5% will only affect the cross falls of point 4 and 5 in this case. Transition curve will be used in the scenario. When using transition curve, superelevation development is usually commences at some distances before the TS point (CH340) and become fully developed at the SC point (CH380). Starting before the TS point ensures that the superelevated part of the road cross section have zero cross fall after reaching the TS point (CH340). Cross fall change rotation is taken as 0.75% per 20m chainage distance. Therefore, superelevation development should begin at CH310 which lies 30m from the CH340 at the TS point. This ensure the right carriageway (-3%) and the right shoulder (-3%) reached zero cross fall after it has reached CH340 after passing the TS point at CH320. Superelevation should be fully developed when approaching the SC point at CH380. Question 3 Determine and tabulate the cross section levels of points 1 to 5 (shown below) at 20m intervals from CH100.00 to CH500.00. Right carriageway (3.5 m) Right shoulder (2m) RL of the cross section of the highway road g2 can alternate from negative to positive values during the development of the superelevation. Question 4 Determine the earthwork quantities at 20m intervals over the length of the road. (Pavement depth can be neglected. Fill batters 3:1. Cut batters 1:1 plus appropriate drain). Tabulate your calculations and results. Calculate the area of the Cut and Fill cross section along the 20m chainage intervals from CH100 to CH500 a = shorter length of trapezium b = longer length of trapezium h = height of trapezium which is the height difference between highway road RL and natural surface RL. Positive (+) indicate cut while negative (-) indicate fill. Fill cross section area (Batters 3:1) Cut cross section area (Batters 1:1) Calculate the volume of the Cut and Fill along the 20m chainage intervals from CH300 to CH700 Volume = average cross section areas of the two chainages multiply by the distance between the two chainages (20m). Fill volumes along the chainages are multiplied by the factor 1.15 to add an extra 15% volume to account for fill shrinkage effects. Cumulative cut and fill volumes along the chainages are tabulated and used to plot the Mass Haul Diagram which indicate the volume of cut and fill required. Question 5 The earthworks must be completed in 2 weeks and the equipment available are: Excavators with bucket capacity 1.2 m3 Scrapers with carry capacity 10 m3 Dump trucks with capacity 12 m3 Determine the number of each type of equipment required to complete the earthworks in 2 weeks. Volume of earthwork Total Cut volume = 4101.95 m3 = approximately 4100 m3 Total Fill volume = 1526.16 m3 = approximately 1530 m3 The amount of excess earth from cut =4100 – 1530 = 2570 m3 General assumptions for cut and fill operation Duration: assumed 10 days duration within a 2 weeks period for a five standard working days in one week and 8 standard hours of work per day and hence 40 hours week. Total hours per week: assumed 40 hours per week (80 hours for two weeks) but assuming a working period of 50 min/ hour. This means that 50 minutes out of the 60 minutes within a one hour period are used to perform earthwork. Fill operation: excavators are used to load volumes earth onto dump trucks. Then dump trucks are used to transport earth along the highway road. Upon reaching the destination, the trucks dump the earth over the areas that need to be filled. Cut operation: scrapers are used to remove the earth around the cut areas. Some of the earth volume cut can be used to cover some of the fill areas. Planning of excavators Bucket capacity = 1.2 m3 However, assuming 80% bucket efficiency rating (one scoop fill up 80% of bucket) Reduced bucket capacity = 1.2 x 0.8 = 0.96 m3 Number of load per truck Loading time = 0.8 minutes (assumed) Loading time per truck Truck positioning time = 2 minutes (assumed) Therefore, total loading time per truck = 10 + 2 = 12 minutes Assuming one working day consists of 8 hours. One hour is taken away for lunch break and 10 minutes from each remaining hour due to miscellaneous time expenditures. Therefore, one working week consists of approximately 29 hours of actual productivity (58 hours for 2 working weeks). Therefore, output in one week = Fill volume = 1530 m3 Total time of job for one excavator Therefore, number of excavator required However, use 1 excavators for this project to prepare for contingencies such as equipment breakdown/failures and inefficient operations. Planning for dump trucks Capacity of dump truck = 12 m3 Number of loads per truck = 13 loads Loading time = 0.8 minutes Loading time per truck Truck positioning time = 2 minutes Total loading time per truck = 10 + 2 = 12 minutes Assuming truck will have to travel an average distance of 1.5km with an average speed of 40km/h. Therefore, dump truck travel time Dumps trucks have to travel back and forth so total truck travel time per cycle is approximately 5 minutes. Dumping time = assuming 3 minutes Total time per truck cycle Number of trucks per excavator Therefore, a total of 2 dump trucks are needed to accommodate the 1 excavator. Planning for scrapers Cut volume = 4100 m3 Scraper capacity = 10 m3 Scraper loading time = assuming 0.8 minutes Scraper dumping time = assuming 0.8 minutes Travel time per cycle (assuming average distance is 100m) = 1.2 minutes Positioning time = 2 minutes Therefore, total time per cycle = 0.8+0.8+1.2+2 = 4.8 minutes = 5 minutes Therefore, output in one week = Total time of job for one scraper Hence, only 1 scraper is needed for this project Total earth moving equipment for the project = 1 excavator, 2 dump trucks and 1 scraper Read More