Tiêu đề MSTE 36 Solutions.pdf ✅

36Surveying: Vertical Curves Solutions SITUATION 1. A descending grade of 4.2% intersects an ascending of 3% grade at station 12+125 at elevation 14.2 m. These two grades are to be connected by a 260 m vertical parabolic curve. A reinforced concrete culvert pipe with overall diameter of 105 cm is to be constructed with its top 30 cm below the subgrade. ▣ 1. What is the stationing of the lowest point of the curve? [SOLUTION] x 0.042 = 260 − x 0.03 → x = 151.6667 m Station of lowest point Sta LP = Sta PI − L 2 + x Sta LP = (12 + 125) − 130 + 151.6667 Sta LP = 112 + 146.67 ▣ 2. What is the elevation of PC? [SOLTUION]
Elev PC = Elec PI + g1L1 Elev PC = 14.2 + (0.042)(130 m) Elev PC = 19.66 m ▣ 3. What will be the invert elevation? [SOLUTION] Elev LP = 19.66 − 1 2 (0.042)(151.6667) = 16.475 m Elev Invert = 16.475 m − 0.3 m − 1.05 m = 15.125 m SITUATION 2. An unsymmetrical parabolic curve connects a +4.2% grade and a -3.4% grade. The length of curve on the left side of the vertex is 80 m and 110 m on the other side. If the stationing of the point of intersection is 4+460 and its elevation is 145.2 m. Determine the following. ▣ 4. Location of the summit form PT. [SOLUTION] g3 = g1L1 + g2L2 L1 + L2 g3 = (0.042)(80) + (−0.034)(110) 80 + 110 = −0.002