For the curve above, (an equal-tangent vertical curve connectsa +3.2 % and a -1.1 % grade.The PVI is at station 100+20and elevation 950 ft and thePVT is at station 105+20), whatis the elevation of the highestcurve point?O 945.9 ftO 942.3 ft934.0 ft927.6 ftO 921.2 ft An equal-tangent vertical curveconnects a +3.2 % and a -1.1 %grade. The PVI is at station100+20 and elevation 950 ftand the PVT is at station105+20. Answer the following5 questions Question 3 toQuestion 7.The station of the PVC is at:sta 93+20sta 95+20sta 97+20sta 99+20sta 105+20

Answered: Image /qna-images/answer/b20f6aca-eee3-48e0-83f2-7a7a370b6b8b.jpg

For the curve above, (an equal- tangent vertical curve connects a +3.2 % and a -1.1 % grade. The PVI is at station 100+20 and elevation 950 ft and the PVT is at station 105+20), what is the elevation of the highest curve point? O 945.9 ft O 942.3 ft O 934.0 ft O 927.6 ft O 921.2 ft

For the curve above, (an equal- tangent vertical curve connects a +3.2 % and a -1.1 % grade. The PVI is at station 100+20 and elevation 950 ft and the PVT is at station 105+20), what is the elevation of the highest curve point? O 945.9 ft O 942.3 ft O 934.0 ft O 927.6 ft O 921.2 ft

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Assignment 3: Vertical Curve Design Question 2 The reduced level at the intersection of a rising gradient of P % and a falling gradient of Q % on a proposed road is XX.XX m. Given that K value is KK. The throught chainage of the intersection point is YYYY.YY m and proposed vertical curve is to have equal tangent. As a civil engineering you are required to prepare the following vertical curve design calculations: a) The through chainage of the points of the vertical curve if the minimum required length is to be used (Lv) and the vertical curve setting out data at 20 metres interval. b) The chainage and reduced level of the highest point on the curve. c) Through chainage at highest point was located exactly under a bridge, if the RL of soffit is SS.SS m, calculate the headroom based on proposed vertical curve design.
Compute and tabulate full-station elevations for an unequal-tangent vertical curve to fit the requirements shown below. A -3.40% grade meets a +2.75% grade at station 95+00 and elevation 1320.64 ft. Length of first curve 400 ft., second curve 600 ft. Match the correct elevation to the appropriate Station Value. Group of answer choices Sta. 91+00 (BVC) [ Choose ] Sta. 92+00 [ Choose ] Sta. 93+00 [ Choose ] Sta. 94+00 [ Choose
The curve elevation at station 20+00 is VC 1, PVI=21+72.41, PVI Elevation= 430.00 ft, L= 800 ft -3.20% -4.00% +1.50% VC 2, PVI 35+00 O 432.96 ft O 434.00 ft O 435.52 ft O 437.04 ft
A +2.5% grade meets a -1.75% grade at the station 44+25 and elevation 386.96 ft, the length of curve is 800 feet (8 stations). (1) Provide the stakeout for full stations using parabolic method STATION X g1 X (g2-g1) X2 2L Y (Elevation) PVT = PVC = (2) Determine the station and elevation at the high point of the curve
A 500-ft curve, grades of g₁ = +2.00% and g2 = -2.50%, VPI at station 96+80, and elevation 845.260 ft, stakeout at full stations. Part A Determine the elevation at the high point of the curve. Express your answer in feet to six significant figures. Elev = Submit Part B AΣo↓ vec Request Answer Determine the station at the high point of the curve. O 97+12.77 O 96+13.33 O 96+52.22 O 97+96.66 ? ft
A vertical summit curve is having tangents which intersects at point Pl at Station 12+618.37 and 133.75 meters elevation. The grade for the forward tangent is -513%. Station PC is at station 12+515.99 and having an elevation of 116 26 meters. What is the elevation of point A along the curve with Station 12+548.27? cs Seanned with CamScanner
Subject : Transportation engineering Please give me right solution according to the question clearly.. Urgent please A 6% downgrade meets a 3% upgrade at elevation 100m at station 10+000. If the length of the parabolic curve is 160m long, find the stationing of the lowest point of the curve.
18. An incoming grade of +3.00% meets an existing grade of -1.50% at a PVI having station 18+00 & elevation 120.00'. A perpendicular street crosses at station 19+50.00 & elevation 117.00'. What is the length of vertical curve required to match elevations at the intersection? A. 155.85' B. 600.00' C. 577.49' D. 639.51'
A Highway curve will pass through a railway at grade. The crossing must be at Sta. 4+310 and at elevation 220.82 m. the initial grade is +2% and meets a -3% grade at Sta. 4+235 at elevation of 223.38 m. The rate of change must not exceed at 2%. Compute the length of curve, station and elevation of highest point also check if the condition of rate of change if meet. 3 decimal places. ASAAP 30 MINS THANK YOU
A +3.0 percent grade intersects a - 2.0 percent grade at station 3 + 20 and at an elevation of 320.40 ft. Given that a 180-ft length of curve. The station of the PVC is A. 2+30 B. 3+30 C . 4+30 D. 5+30
A curve with a -6% downlade intersects an upward slope of +2%. The elevation at the intersection of these grades at station 50+00 is 550.97 feet. The length of the curve is 800 feet. What is the station of the EVC station? 55+00 58+00 54+00 54+00 56+00 57+00 50+00
B- Unequal-Tangent Parabolic (Vertical) Curve: Example: A grade gl of -2% intersects g2 of +1.6% at a vertex whose station and elevation are 87+00 and 743.24, respectively. A 400 vertical curve is to be extended back from the vertex, and a 600' vertical curve forward to closely fit ground conditions. Compute and tabulate the curve for stakeout at full stations. Stations 00 00 00 00 EVC (752.84) BVC (751.24) A CVC (748.04) (747.24) (747.56) -2.00% 400 +1.60% V(743.24) >とじ。 Solution: The CVC is defined as a point of Compound Vertical Curvature. We can determine the station and elevation of points A and B by reducing this unequal tangent problem to two equal tangent problems. Point A is located 200' from the BVC and Point B is located 300' from the EVC. Knowing this we can compute the elevation of points A and B. Once A and B are known we can compute the grade from A to B thus allowing us to solve this problem as two equal tangent curves. Point A STA 85+00: Elev. = 743.24+2 (2) = 747.24'…