BALANCE THE TRAVERSE BY TRANSIT RULE AND COMPUTE THE AREA BY DMD METHODCorr'nCorr'nDistanceLineBearingLatitudeDepartureCorrected LatitudeDMDCorrected DepartureLatDepDoubledeg min1 2 s 23 37E 113.352 3 s 25 56 w 93.323 4 s 87 25 w 172.924 5 N 55 46 w 165.885 6 N 51 59E 184.896 1 s 85 51E 160S() E (+)S(-)N(*)W(-)N (*)E (+)W()Area

transit method - in this method the distribution of error in different latitude and departure in…

BALANCE THE TRAVERSE BY TRANSIT RULE AND COMPUTE THE AREA BY DMD METHOD Corr'n Corr'n Line Bearing Distance Corrected Departure Latitude Departure Corrected Latitude DMD Lat Dep Double deg min E (*) w() N (*) E (*) N (+) Area 1 2 s 23 37E 113.35 2 3 s 25 56 w 93.32 3 4 s 87 25 w 172.92 4 5N 55 46 w 165.88 5 6 N 51 59E 184.89 6 1 s 85 51E 160

BALANCE THE TRAVERSE BY TRANSIT RULE AND COMPUTE THE AREA BY DMD METHOD Corr'n Corr'n Line Bearing Distance Corrected Departure Latitude Departure Corrected Latitude DMD Lat Dep Double deg min E () w() N () E (*) N (+) Area 1 2 s 23 37E 113.35 2 3 s 25 56 w 93.32 3 4 s 87 25 w 172.92 4 5N 55 46 w 165.88 5 6 N 51 59E 184.89 6 1 s 85 51E 160

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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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
(a) Briefly explain the main reason the used of transition curves, how the transition curve is used in road geometry design and the purpose of super-elevation in combine road curve design. Q3. (b) Two straights, Al and IB, whose bearings are DDD° AI' Al" and DDD° IB' IB" respectively intersect at point I, are to be connected by a combined curve. The chainage of point / is CCCC m, while the design speed and radius have been fixed at 90 km/h and 550 m respectively. Adopting a rate of change of acceleration of 0.3 m/s. As a civil engineer have been assign to do a trial analysis of the proposed design of horizontal combine curve (T1, T2, T3 and T4) and prepared road curve setting-out data (T1 to T2, T2 to T3, T3 to T4) at 20 m intervals using deflection angle and chord distance method. (c) By adopting the coordinate of intersection point / as (YYYY N, XXXX E). Calculate the coordinate of tangent point for T1, T2, T3 and T4.
6. AREA adjusted DPD. -In the latitudes accompanying and adjusted. departures tabulations are the of BY two tra- verses. ADJUSTED LATITUDE ADJUSTED DEPARTURE LINE +N -S +E 420.75 713.84 296.08 693.05 CD 894.37 405.36 DE 50 1.25 926.54 386.42 884,53 EF 695.04 F.A 190.67 Sums +1691.70 -169 1.70 1812.25 18.12.25 ADJUSTED LATITUDE ADJUSTED DEPARTURE COURSE +N +E 1- 2 299.30 295.58 2- 3 338. 05 223.56 3 - 4 24.52 245. 77 4- 5 5 6 6 - 1 8.05 381.36 230.52 463.09 626. 1 1 309. 87 Sums +949.93 - 949.93 + 772.96 -772.96 By employing the double parallel distance methoa, determine PROBLEM SETS 563' A BCDEE
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
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
Following are the data gathered from the traverse: Lat Dep DPD 2A 399.790 0 565.388565.277 -350.184605.780 -564.112-205.360 -50.882-965.697 Express all answer unitless and with 3 decimals for A to J No new data to sav
Q6) Determine |AB| and coordinates of points C,D,E,F,G,H and X. N 79.08g S3 111.77159 s2 S1 144.3233g E A S4 293.03739 S5 252.1676g B (grad) a (grad) S (m) Y (m) x (m) STATION Ду Дх A 70.984 632.089 B 232.964 687.333 158.090 141.338 D 188.521 E 99.712 F 118.203 G 172.782 H 97.426 116.0859g 2806279g
7. ERROR OF CLOSURE. Given the Eollowing tabulated data for two transit-tape surveys: LATITUDE DEPARTURE LINE LENGTH BEARING +N +E - W s 06°35' W N 17 48'w N 70 59 w N 59°33'E S 39° 1 a'E 1-2 445.66 m 442.72 51.09 2-3 326.73 311.09 99.88 3-4 434.50 1 41.58 410.79 4-5 427.00 216.40 368.10 5-1 301.95 233.66 191.25. Sums 1935.84 m + 669.07 -676.38 +559 35 -561.76 (D) (ENL) (ESL) CEEDI E WD)
Given is Station C with N,E= (1000, 1000) B Line AB CD DE EA D Length: Bearing: 59.55 S 65° 47' E 96.00 $43° 9' E 75.32 542°51″W 117.88 N 58° 6' W 95.96 N 19°54'E Lat -27.23 Dep: 52.96 ماما .05) 40.04 - -55.23 -51.22 G2.28 -100.08 90.23 32.07 Compute for the Northings and Eastings of the stations, and COMPUTE for the AREA by splitting it into MULTIPLE TRIANGLES ABC, ACD, and ADE and SHOW the computation for lengths and bearings of sides CA and DA.
VIII. Compass & Transit Rule 1. Complete the table using Compass Rule Bearing S 56 3 E Course AB BC CD DE EF FG Total: Distance 235.1 243 241.2 235.2 271.4 253.5 N 85 15' E N 23 00' E N 27 28' E S 76 09' W S 32 00' W Lat Dep Cl Cd Adj Lat Adj Dep Adj. L Adj. B
Compute for the unknown Line AB BC CD DE EA LINE AB BC CD DE EA Distance 300 400 250 350 277.96 DISTANCE 250 UNKNOWN 300 UNKNOWN 400 Bearing S 37 30 E S 43 15 W N 73 W UNKNOWN UNKNOWN BEARING N 73 00 W N 12 45 E S 37 30 E N 65 45 E S 43 15 W ■
Find the missing data and complete the table below. DIST (m) 23.324 27.459 ? 69.771 22.361 20.591 20.616 15.811 LINE AB BC CD CD DE EF FG GH HA Missing data: LINE Bearing N 30° 58' E N 56° 53' E ? S 27° 18' W N 26° 34' W S 60° 57' W N 14° 2¹ E N 71° 34' W DISTANCE (M) DIRECTION BEARING ANGLE (°)