Pls do number 3 The intersection of symmetrical summit parabolic curve with g₁= 7% and g -9% is at Sta 1+250 with elevationof 220 m. if the vertical increment at L/2 is H= a.bc m,9₁ = 7%$₂L/2Sta Pl1+250 eley 220 m92-9%L/2PTDetermine the followinga. The length of curve is Lb.The distance of highest point from PC, S.c. The Elevation of the Highest point is =

The intersection of symmetrical summit parabolic curve Up gradient, N1 = 7% and Down gradient, N2=…

me intersection of symmetrical summit parabolic curve with gi= 7% and g -9% is at Sta 1+250 with elevation of 220 m. if the vertical increment at L/2 is H= a.bc m, B₁ = 7% L/2 Sta Pl 1+250 eley 220 m Не m₂=-9% L/2 PT Determine the following a. The length of curve is L= b.The distance of highest point from PC, S c. The Elevation of the Highest point is

me intersection of symmetrical summit parabolic curve with gi= 7% and g -9% is at Sta 1+250 with elevation of 220 m. if the vertical increment at L/2 is H= a.bc m, B₁ = 7% L/2 Sta Pl 1+250 eley 220 m Не m₂=-9% L/2 PT Determine the following a. The length of curve is L= b.The distance of highest point from PC, S c. The Elevation of the Highest point is

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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o T ☺ ť In a vertical parabolic curve (g1= +1.2%, g2=-2.0%), :length 320 m, fill all below blanks :Rate of change of grade per station If you know station of PVI (20+18.99), then station of PVC is: (e.g. 12+34.65) If you know elevation of PVT 25.642 m, then :the elevation of PVC is :Elevation of station (19+20) Station of highest point on the curve: (e.g. 12+34.65) :Elevation of highest point on the curve Station of a point has elevation 26.631 m: (e.g. 12+34.65) مر ر ل لأعلى لاستعراض الفلاتر + إضافة شرح. . . ( الحالة )مخصص( >
SITUATION 5: A compound curve has the following data: Di = 3° D2 = 4° Li = 28° L2 = 31° Sta. P.l at 30 + 120.5 13. Determine the length of the common tangent. a. 381.97 m c. 286.48 m b. 174.68 m d. 267.37 m 14. Compute the stationing of P.C.C. a. 30 + 106,97 b. 30 + 110.73 c. 30 + 118.46 d. 30 + 235.45
Given for a Simple Curve, a. R=750m b. I-34 deg c. STA. V = 20+050 Detemine the Elements of the given Simple Curve. (Use 100m-Stations) i. Degree of the Curve ii. Tangent Distance iii. Length of the Chord iv. Length of the Curve v. Extemal Distance vi. Middle of Ordinate vii. Sta. PC and Sta. PT
Problem 1 A simple curve connects two tangents AB and BC with bearings N 75°12' E and S 78°36' E, respectively. If the stationing of P.C. is 10+185.28. and its middle ordinate is 7.46 m, Determine the following: a. Radius of the curve. b. Tangent Distance C. External Distance Long Chord. d. e. Length of the curve. f. Stationing of the vertex g. Degree of the curve using Arc Basis
Q 3: A horizontal curve of a 4-lane undivided highway is built in acut section of side slope 3:1. If the design speed is 100kph, station of P.I=(55+00. Shoulder width = 1.5m, e=8%, q=1.5%, t=2.5sec, g=1.5% and A= 45° 1. Determine the min. lane width that would allow a safe operaion on this curve? 2. Draw the cross section at station (56+00) and at station (70+01)?
4. The broken back curve shown is to be replaced by a compound curve R2 = 1.8R1. T1=250m, T2=350m, Δ1/2=30°, Δ2/2=20°, and short tangent =45m. Determine: a.new values of radii b. new values of tangents (ans: a) R1=1101.928m R2=1983.470m,and m=295.261m and n=349.739m)
Two straights AB and BC meet in an accessible point B and are to be connected by a simple curve of 600 m radius. Two points P and Q were selected on AB and BC respectively, and the following data were obtained: angle APQ = 150 degrees, angle CQP = 130 degrees, PQ = 125 m. If station of P is at 1+600 and using a 30 m full station, solve the following:-Station A -Station C -I -Deflection angle at Sta. 1+220 -Chord lenth at Sta. 1+250
The tangent thru the PC has a direction due to North; and the tangent through the PT has a bearing of N 50° E. It has a radius of 200 m. Stationing of PC is 12+060. % a. Compute the tangent distance of the curve. Answer: T= [tangent] m b. Compute the long chord of the curve. Answer: Lc = [Lchord] m c. A line is drawn from the center of the curve (at the central angle) and intersects the curve at point B. This line, when extended, makes an angle of 62° with the tangent thru PC. What is the stationing of point B? Answer: Sta. B = [StaB]
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6. A compound curve has a central angle of the first curve equal to 40° and a radius of 280m. while the second curve has a degree of curve 5° with a central angle 28°. Compute the length of the common tangent. 7. A compound curve has the following data: I₁ = 28⁰ D₁ = 3⁰ 12 = 31° D₂ = 4° Sta. of P.I. = 30 +120.5 Find the stationing of P.C.C.
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