A-2.60% grade meets a +1.70% grade at station 44+25 and elevation 3387.96 ft, 400-ft curve, stakeout at half stations. Tabulate station elevations for an equal-tangent parabolic curve for the data given above. Express your answers in feet to six significant figures separated by commas in order of increasing station numbers. Elev = Submit 17 ΑΣΦ OF Request Answer ↓↑ vec ? ft

A-2.60% grade meets a +1.70% grade at station 44+25 and elevation 3387.96 ft, 400-ft curve, stakeout at half stations. Tabulate station elevations for an equal-tangent parabolic curve for the data given above. Express your answers in feet to six significant figures separated by commas in order of increasing station numbers. Elev = Submit 17 ΑΣΦ OF Request Answer ↓↑ vec ? 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

Following are the data gathered from the traverse: Lat Dep DPD 2A 399.790 0 565.388 565.277 -350.184 605.780 -564.112 -205.360 -50.882 -965.697 Express all answer unitless and with 3 decimals for A to J
Item 1 A-2.5% grade meets a +2.50% grade at station 4 + 200 and elevation 146.657 m, 180-m curve, stakeout at 30-m increments. Part A List station elevations for an equal-tangent parabolic curve for the data given. Give the elevations in order of increasing X. Express your answers in meters to six significant figures separated by commas. Elev = Submit Provide Feedback OF V—| ΑΣΦ| | | vec Request Answer ? m
46 K/s ۱۸ من ۱۸ Q.3 A level circuit is shown in figure. The survey, needed for local engineering project, commenced at BM 35, the elevation of new benchmarks 36, 37, 38 and 39 were determined, and then the level survey was looped back to BM 35, the point of commencement. Fined the adjusted new BMs. ВМ 35 Elevation(m) 119.010(fixed) 35 36 37 38 39 35 123.342 118.206 118.451 120.491 118.933 BM 38 0.9 kg BM 35 DUM 37 08 km 06 km BM 38 سلوان 2 8:11p äc lul 21.03.06
The technical description of a six-corner lot is tabulated in the table below. LINES BEARING DIST LAT DEP DMD DPD ТР-1 15 11 20 3051.74 1-2 33 W 190.30 2-3 33 3 W 300.00 3-4 33 3 W 4-5 89 54 E 302.27 5-6 6. 6-1 W 312.82 a. Plot the given data. b. Calculate the distances of line 3-4 and line 5-6. c. Calculate the area of the lot by DMD.
Question 1 A parcel of lot (trapezoidal) has a technical description as shown in the tabulated data. Lines Bearing A-B S 01-27 E 27.72 B-C S 88-47 W 38.21 C-D N 06-57 E 28.01 D-A N 88-47 E 34.12 [Select] [Select] [Select] Distance(m) [Select] What is the DPD of line CD? in m What is the double area of the line BC? in m² What is the total area of the lot using DPD? m² What is the angle to the right of line B-C Next
Q1:- A lot (ABCDF) with a curved boundary (FA) (whose radius DF-DA-40 m) is cut off by a fence (JK), Compute: - 1- Length and Azimuth of line JK. 2- Coordinate of point K. 3- Cut off area (ABJK). (400, 700) с J Dom B 286 21 go tom 30m (due w) A(500, 500)
Reves Grades of g =2 50% and g +1.50% VPI elovation 856 00 ft at station 22 - 00, and a foxed elevation 800 00 ft at station 20 + 50 Part A Field conditions require a highwity curve to pass through a fxed point Compute a sutate equal tangent vertical curve and station elevations Give the elevations in order of increasing X Express your answers in feet to five significant figures separated by commas VO AE vec Eev Submit Reauest Ansr Nest> Provde Eeedback
Q.3 A level circuit is shown in figure. The survey, needed for local engineering project, commenced at BM 35, the elevation of new benchmarks 36, 37, 38 and 39 were determined, and then the level survey was looped back to BM 35, the point of commencement. Fined the adjusted new BMs. BM 35 Elevation(m) 190.000(fixed) 35 39 186.397 35 190.033 36 37 38 188.242 186.299 188.251 BA 36 09 km. BM 35 O8 km DeM 37 0.8 km 06 km BM 39 BM 38 0.7 km
7. ERROR OF CLOSURE. Given the following tabulated data for two trans it-tape surveys: DEPARTURE LATITUDE LINE LENGTH BEARING +E - W +N s 06 35'w N17 48'w N 70 59'w N 59 33'E S 39*18'E 51.09 99.88 410.79 442.72 1-2 2-3 3-4 445.66 m 326.73 434.50 427.00 301.95 311.09 141.58 216.40 368.10 4-5 233.66 191.25 5-1 +669.07 -676.38 +559.35 -561.76 Sums 1935.84 m (ZNL) E SL) CEED) EWD) (D) LATITUDE DEPARTURE AZIMUTH FROM NORTH COURSE LENGTH -S +E -W 227.57 82 12' 131°46 228 20' 267 43' 352 01 31.17 AB 229.70m 87.02 107.34 97.32 130.55 161.46 BC 120.61 CD 7.19 180. 35 DE EA 160.49 170.16 23.86 171.63 +201.33 -201.55 +324.89 -324.82 Sums 874.03m (NL) (SL) (ZED) (EWD) (D) each traverse determine the linear error of direction of the side of error, and the precision the measurements made. Tabulate all observed and For closure, the of computed values
Compute L, T, E, M, LC, R, and stations of the PC and PT for the highway curve with D,-4'00, I-13 30', and Pl station 19+22.00 ft. Use the arc definition for the highway curve. Part A What are radius R and length L7 Express your answers in feet to two decimal places separated by a comma. 19Η ΑΣΦ. 11 Ivoc R.L- Submit Part B Request Answer What are distances T, E, M, and long chord LC? Express your answers in feet to two decimal places separated by commas. VAX T. E. M, LC= Submit Part C Request Answer ft Ivec ? ft
Q1. Calculate departures and latitudes of the following ABCDE traverse. 3 N A 285.40' •N26°10'E E 5 75 25 E 640.45' 647.02' 5306 W N 0°42' W 720.8 203.00 5 15 30¹ W
Direction: Given the following data on the table. Fill up the needed data. Balance the traverse by transit rule. Solve the area by DPD or DMD. Sketch the traverse. All values must be in decimal places. LIN BEARIN E DISTAN CE (M) AZIMUT COMPUTED ADJUSTED DEPARTURE ADJUST ED DISTAN CE ADJUST DPD ED AZIMUT DOUBLE COMPUTED DEPARTURE ADJUSTED LATITUDE ADJUST ED BEARIN H LATITUDE AREA (from S) H (nearest (nearest minute) minute) +NL -SL +ED -WD +NL -SL +ED -WD AB N 10-00 650.00 BC S 70-29 E CD S 26-28 895.00 315.00 DE S 65-33 875.00 EA N 45-31 410.00 TOTAL LEC : RP (nearest hundred) AREA