Highway curve has R = 1000 ft, I = 25° 00', andPl station = 26 + 40.00 ft.Part ADetermine degree of curvature Da of the curve.Express your answer in degrees to five decimal places.Da =SubmitPart BSubmitV— ΑΣΦL, T = 436.33,221.69 ftRequest AnswerDetermine curve length L and tangent distance T of the curve.Express your answers in feet to two decimal places separated by a comma.Part CCorrectPrevious AnswersSubmitvecVGI ΑΣΦM, E= 23.7,20.40Determine middle ordinate M and external distance E of the curve.Express your answers in feet to two decimal places separated by a comma.?vecPrevious Answers Request Answer?ft

R = 1000 ft I = 25o00'We have to find-(A) Determine degree of curvature Da of the curve.(B)…

Highway curve has R = 1000 ft, I = 25°00', and Pl station = 26 + 40.00 ft. Part A Determine degree of curvature Da of the curve. Express your answer in degrees to five decimal pla Da = Submit Part B Submit ΠΠΑΣΦ Request Answer Determine curve length I and tangent distance T of th Express your answers in feet to two decimal places L, T = 436.33,221.69 ft Part C Previous Answers Correct vec Determine middle ordinate M and external distance E Express your answers in feet to two decimal places ΑΣΦ M. E= 23.7,20.40 vec Submit Previous Answers Request Answer

Highway curve has R = 1000 ft, I = 25°00', and Pl station = 26 + 40.00 ft. Part A Determine degree of curvature Da of the curve. Express your answer in degrees to five decimal pla Da = Submit Part B Submit ΠΠΑΣΦ Request Answer Determine curve length I and tangent distance T of th Express your answers in feet to two decimal places L, T = 436.33,221.69 ft Part C Previous Answers Correct vec Determine middle ordinate M and external distance E Express your answers in feet to two decimal places ΑΣΦ M. E= 23.7,20.40 vec Submit Previous Answers Request Answer

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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SURVEYING (UPVOTE WILL BE GIVEN. PLEASE WRITE THE COMPLETE SOLUTIONS AND DIAGRAMS *IF NECESSARY* LEGIBLY. NO LONG EXPLANATION NEEDED. ANSWER IN 2 DECIMAL PLACES AND BOX THE FINAL ANSWER.) A simple curve with radius of 190m is laid out to have an extemal distance equal to 15.00 meters. If the station of PC is 10+420, Solve the following: a. Degree of curve (D) b. Degree of Ordinate (M) c. Length of the circular curve (Lc) d. Station of PT e. Stationing of a point along the curve which intersects with the line making a deflection angle 10° with the tangent thru the PC.
Two tangents intersect at STA 2+550. A compound curve laid on their tangents has the following data: I1=31deg.;I2=36deg.;D1=3deg.;D2=5deg.; (a) Compute the stationing at PC,PCC and PT
1. Draw a simple horizontal curve label the curve and define each term 2. For the curve data shown calculate deflection angles also tabulate and show all necessary calculationsalso briefly explain layout method based on your resultsI= 42° 16’ 00” D= 4° 0’ 0”, PI station =93+62.71 3- Highway curve with R = 1000ft ,I = 60° 00’ 00”and PI =85+40Complete the curve calculations showing all details
Problem No. 1: Two tangents intersect at STA 2+550. A compound curve laid on their tangents has the following data: I1=31deg.;I2=36deg.;D1=3deg.;D2=5deg.; (a) Compute the stationing at PC,PCC and PT
Situation 1] A circular road is proposed to be laid out with tangent AB whose bearing is N85º54’E intersecting another tangent line BC whose bearing is N48º30’E. If the degree of the curve is 4.5º, using arc basis, determine the following: Tangent distance of the curve, T Long chord of the curve, C Middle ordinate, M External distance, E Length of the road, LC If the station of point B 10+584, determine the station of PT.
A.) Given the set of numbers .14, 19, 19, 1, 2, 7, 10, 24, 10, 11, 23, 24, 3, 4 , 7, 26What is the ratio of the second quartile to the third quartile?1.50.52.01.0 B.) The common tangent BC of a reversed curve is 280.50m and has a bearing of S 47deg 31minE. AB is the tangent of the first curve whose bearing is N 72deg 45min E. CD is a tangent of thesecond curve whose bearing is N 38deg 13min E. A is at the PC while D is at the PT . Theradius of the first curve is 180m . The PI1 is at Sta 12 + 523.37. Find the stationing of the PT12 + 878.1612 + 883.6512 + 893.2412 + 889.54
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
SURVEYING (UPVOTE WILL BE GIVEN. PLEASE WRITE THE COMPLETE SOLUTIONS AND FIGURES *IF NECESSARY* LEGIBLY. NO LONG EXPLANATION NEEDED. ANSWER IN 2 DECIMAL PLACES AND BOX THE FINAL ANSWER.) A simple curve with radius of 190m is laid out to have an extemal distance equal to 15.00 meters. If the station of PC is 10+420, Solve the following: a. Degree of curve (D) b. Degree of Ordinate (M) c. Length of the circular curve (Lc) d. Station of PT e. Stationing of a point along the curve which intersects with the line making a deflection angle 10° with the tangent thru the PC.
Horizontal highway curve: Given PI at 95 + 54.78, Δ = 24°28’, and D = 5°30’, compute thedeflections to the nearest second for the full stations (means stations 94+00, 95+00, 96+00, etc.)as well as for the beginning and end of the curve.
Please Handwrite Answer! Horizontal highway curve: Given PI at 95 + 54.78, Δ = 24°28’, and D = 5°30’, compute thedeflections to the nearest second for the full stations (means stations 94+00, 95+00, 96+00, etc.)as well as for the beginning and end of the curve.
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