Šituation: Simple curve connects two tangents XY and YZ The survey data are the following: Bearing of XY = N 85' 30' E Bearing of YZS 68' 30' E Station of Vertex / Intersection = 4 + 360.2 Station of Point of Curvature= 4 + 288.4 Given Angle of Intersection (1) = 26° Tangent Distance, T= 71.8m Radius of the Curve, R= 311m Degree of Curve, deg. (arc basis) = 3.68 Degree of Curve, deg. (chord basis) =3.69 External Distance, E= 8.18m Middle Ordinate, M = 7.97m Stations Interval - 20 m Chord/Sub-Chord Arc/Sub-arc distance First Station Next Station Deflection Angle Distance PC (Sta. 4 + 288.4) 4+300 4+320 4+340 PT (Sta. 4 + 360.2) 4+300 4+320 4+340 Complete the table and find a) Chord distance (from PC to PT) b) Length of Curve (from PC to PT) c) Station point of tangency

Šituation: Simple curve connects two tangents XY and YZ The survey data are the following: Bearing of XY = N 85' 30' E Bearing of YZS 68' 30' E Station of Vertex / Intersection = 4 + 360.2 Station of Point of Curvature= 4 + 288.4 Given Angle of Intersection (1) = 26° Tangent Distance, T= 71.8m Radius of the Curve, R= 311m Degree of Curve, deg. (arc basis) = 3.68 Degree of Curve, deg. (chord basis) =3.69 External Distance, E= 8.18m Middle Ordinate, M = 7.97m Stations Interval - 20 m Chord/Sub-Chord Arc/Sub-arc distance First Station Next Station Deflection Angle Distance PC (Sta. 4 + 288.4) 4+300 4+320 4+340 PT (Sta. 4 + 360.2) 4+300 4+320 4+340 Complete the table and find a) Chord distance (from PC to PT) b) Length of Curve (from PC to PT) c) Station point of tangency

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

Two Tangents intersect at chainage 1190m, deflection angle A319 a curve of radius 300m by E setting out Calculate all the necessary deblection angle method if Þeg data for interval is 30m.
Simple Curve A railway curve is to connect intersecting tangents with the following design data: Back tangent = N 20° 25' E Forward tangent = S 49°35' E Degree of curve = 5⁰ Design speed = 70 Kph Station PI = 5+555.55 1.Determine the length of line from PC to Pl. a) 234.67m b) 244.67m 2.Compute the straight distance from PC to PT. a) 402.34m b) 375.60m c) 245.89m d) 327.42m c) 389.78m d) 423.67m 3. If the width of the land intended for the railway is 10m, determine the area of land from PC to PT. a) 4,456.56 sqm b) 3,456,43 sqm c) 5,112.34 sqm d) 4,399.92 sqm 4.How long would a 300m long train able to leave the curve completely. a) 25.67 sec b) 38.06m sec c) 40.34 sec d) 29.67 sec
Given: Stationing B= 18 + 685 Stationing C= 19 + 125 Tangent Lines: AB: Azim= 224° 24' BC: Azim= 264° 29' CD: Azim= 3150 28' A proposed highway curve is to connect these three tangents. Find: Radius of simple curve that connects these tangents. b. Sta PC. c. Length of the curve from PC to PT. a.
Two roads AB & CD distance between them as a circular curve with degree of curvature 30°in point M, calculate: 1-Station B, 2- length of circular curve O O O O 90° A B C D R 30° Point Station E M 103+00 105+00 95+00 1700 с 90 N 1000 2000 1400 * Station B= (100+00), length of circular curve=300m Station B= (101+00),length of circular curve=300m Station B= (101+00), length of circular curve=500m Station B= (100+00),length of circular curve=500m Station B= (100+00),length of circular curve=350m
Two tangents intersect at a chainage of 1190m, the deflection angle being 36o. Calculate all the data necessary for setting out a curve of radius 300m using: Offset from long chord method, Offset from tangent (Radial and Perpendicular both) Offsets from chords produced method. Deflection angle method Length of long chord is 100m and peg interval may be taken as 30m. Make figures with each method seperately.
The length of the curve of a simple curve having a degree of curve 4o is equal to 210 m. Compute the area of the fillet of the curve. Choices: 1442.69 m2 1244.69 m2 1224.69 m2 1424.69 m2
SIMPLE CURVE BEARING OF Point of Curvature - V = N 60° E BEARING OF V - Point of Tangency = S 70° E DEGREE OF CURVE (D) = 05° STATION V = 12 + 234.50 COMPUTE THE FOLLOWING: A. TANGENT DISTANCE B. EXTERNAL DISTANCE C. MIDDLE ORDINATE D. LENGTH OF CURVE E. LONG CHORD F. STATIONING POINT OF CURVATURE AND POINT OF TANGENCY
A simple curve connects two tangents XY and YZ. The survey data are the following: Bearing of XY = N 85 30' E Bearing of YZ =S 68' 30' E Station of Vertex / Intersection = 4 + 360.2 Station of Point of Curvature = 4 + 288.4 Determine the following: Degree of Curve, deg. = Degree of Curve, deg. External Distance, E Middle Ordinate, M (arc basis) (chord basis) m. m. PI M m 1/2 PC PT L/2 L/2 R-m Back Tangent Forward Tangent R R 20 I/2
Q#02: Two tangents intersect at chainage 4100 m. Ø 35 35 35". Carry out all the data necessary for setting out the curve, with R = 1000 m and Peg interval being 15 m by: 1) By deflection angle 2) By offsets from chords produced 3) By offset from tangent (Both cases)
In the figure shown AB = 103.20m with A at Station 10+158.93. The angle VAC is 15O21’ andangle VBC is 18O31’, where C is the point to which a simple curve is to be constructed tangent toline AB.a. Determine the radius of the Curveb. Determine the Stationing of point Cc. Determine the Stationing of PT
Determine: USE 3 Decimal all final answers a) The radius of the curve b) Tangent distance of the curve c) If the vertex V is at station 10+100. find the stationing of PT
1. (Simple Curve) Refer to problem Example 15.6 (textbook Garber and Hoel) and re-do with the following changes to the given information: (a) Intersectional angle for 1 station = 3 degrees (b) Sta of PC is 800+25.58 60 deg. (c) Inscribed angle for entire curve is 800+25.58 60 degrees 50% 25% L 75% PC PT Calculate length of curve. Determine the stations at 25% Lc, 50% Lc, 75% Le (L, is length of the curve) then calculate the chords and deflection angles at these points; show all your solution then summarize the results into a tabulation. Calculate also the following: Middle Ordinate and External Distance. Determine stationing of PT. 2. (Compound Curve) Refer to problem Example 15.7 (textbook Garber and Hoel) and re- do with the following changes to the given information: PC (a) The two simple curves are joined at Sta. 238 + 80 (b) First curve: R1 = 550' & A¡=40deg (c) Second curve: R2 = 450' & A2=25deg R.R radii of simple curves forming compound curve A.4, - intersection angles of…