Q3): The following data were collected while running a closed traverse ABCDA. Calculate:1- Coordinates of B (Easting and Northing).NLineABBCCDDA4- Interior angles A and B5- Distance BC and CD.Length (m)330827Sum= 2435.15Azimuth181°25'89°50'355°00'264°18'20"PointABCDA (calculated)=Easting500B1323.167Northing500582.059A=B=Angles

Answered: Image /qna-images/answer/d5217420-d552-4dc0-afdf-2a2c76957048.jpg

Q3): The following data were collected while running a closed traverse ABCDA. Calculate: 1- Coordinates of B (Easting and Northing). ne B C D A 4- Interior angles A and B 5- Distance BC and CD. Length (m) 330 827 Sum= 2435.15 Azimuth 181°25' 89°50' 355°00' 264°18'20" Point A B C D A (calculated) = Easting 500 1323.167 = Northing 500 582.059 A= B= Angles

Q3): The following data were collected while running a closed traverse ABCDA. Calculate: 1- Coordinates of B (Easting and Northing). ne B C D A 4- Interior angles A and B 5- Distance BC and CD. Length (m) 330 827 Sum= 2435.15 Azimuth 181°25' 89°50' 355°00' 264°18'20" Point A B C D A (calculated) = Easting 500 1323.167 = Northing 500 582.059 A= B= Angles

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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Q3): The following data were collected while running a closed traverse ABCDA. Calculate: 1- Coordinates of B (Easting and Northing). N Line AB BC CD DA 4- 5- Distance BC and CD. 6- Coordinates of C (Easting and Northing). Interior angles A and B Length(m) 330 827 Sum= 2435.15 Azimuth 181°25' 89°50' 355°00' 264°18'20" Point A B C D A (calculated) = Easting 500 B 1323.167 Northing 500 582.059 A= B= Angles
Q3): The following data were collected while running a closed traverse ABCDA. Calculate: Line AB BC CD DA 2- Linear misclosure 3- Traverse precision Length(m) 330 866.8846 411.0045 827 Sum=2435.15 Azimuth 181°25' 89°50' 355°00' 264°18'20" Point A B C 1358.7 D 1323.167 A (calculated) 500 Easting 500 491.84 = = Northing 500 170.1 172.62 582.059 500
3. Complete the table and sketch the traverse, assuming no measurements were obtained for line EF. LATITUDE DEPARTURE LINE LENGTH, m BEARING N (+) S() E(+) W(-) АВ N 5°30'8.33" E 490.71 47.27 BC 587.12 608.89 CD 327.41 786.78 DE 1002.76 218.32 EF FA 375.01 544.64 SUM
Use DMD & DPD Method to calculate the Area of the closed traverse whose data is given below. Draw the figure. LINE CORRECTED DEPARTURE CORRECTED DISTANCE 413.53341 596.41168 AB CORRECTED LATITUDE +296.42872 - 246.85339 + 104.50281 154.07815 + 288.33990 BC CD +542.92753 - 597.93576 - 233.33167 606.99919 DA 279.61356
PROBLEM 1. Given in the accompanying tabulation are the adjusted latitude and adjusted departure of a closed traverse. Calculate the area by the DMD method LINE ADJ ADJ LAT DEP AB - 125.055 + 196.387 BC +20.241 + 241.003 CD +180.107 + 65.326 DE +210.169 - 104.519 EF -74.003 - 260.516 FA - 211.459 - 137.681
Bearing Angle Traverse Given the tabulated measurement of the closed traverse as shown. Determine the Area of the Traverse. Line Length Bearing AB 589.060 m N 87°30’ E BC 391.730 m N 07°02’ E CD 291.860 m N 30°30’ E DE 656.980 m S 42°45’ E EF 654.060 m Unknown FA 686.660 m Unknown Note: EXPRESS YOUR ANSWER UPTO 3 DECIMAL PLACES.
. Problem 1 Compute the area within the following traverse using the Coordinate method and CHECK by the DMDS method Balanced Coordinates Course Dep Lat Point X Y AB -226.9846 98.7106 A 709,489.434 709,262.449 709,176.011 709,061.165 709,361.616 708,846.452 G 709,568.788 709,682.105 I 709,563.827 121,072.089 121,170.800 120,992.637 120,797.489 120,807.032 120,287.159 120,244.663 119,715.939 120,610.402 BC -86.4386 -178.1623 CD -114.8457 -195.1486 C DE 300.4512 9.5436 EF -515.1646 -519.8733 日 FG 722.3360 -42.4958 F GH 113.3175 -528.7245 HI -118.2782 894.4631 IA -74.3929 461.6870
AREA MEASUREMENTS W- E U AREA BY DOUBLE MERIDIAN DISTANCE (DMD) Problem 1: In the tabulation are given the adjusted latitudes and adjusted departures of a closed traverse. V Determine the area using DMD method. LINE ADJUSTED LATITUDE ADJUSTED DEPARTURE (+)N (-)S (+)E (-)W Y AB 113.92 388.84 ВС 178.86 285.13 CD 295.88 134.96 DE 164.35 135.39 EF 13.53 251.86 FA 182.95 151.76 SUM +474.74 -474.74 +673.97 -673.97
3. The following data if a closed traverse is shown in the technical description. The area of the closed traverse is 63147.05 sq. m. LINES AB BC CD LAT. a -240.20 -160.00 DA 179.70 Solve for the unknown variables. DEP. 120 с e -220 DMD b d f 8 2A 26460.0 X -86480.0 39534.0
Q3): The following data were collected while running a closed traverse ABCDA. Calculate: 1- Coordinates of B (Easting and Northing). N 2- Linear misclosure Line AB BC CD DA 3- Traverse precision 4- Interior angles A and B 5- Distance BC and CD. 6- Coordinates of C (Easting and Northing). Length(m) 330 827 Sum= 2435.15 Azimuth 181°25' 89°50' 355°00' 264°18'20" Point A B C D A (calculated) Easting 500 B 1323.167 Northing 500 582.059 A= B= Angles
3. The following tabulated data is to be used in determining the area of traverse ABCDE by the DMD method. Determine the total area of the traverse in m?. Adjusted Latitude Adjusted Departure Course +N -S +E -W АВ 558.25 78.34 ВС 102.58 746.38 CD 285.33 319.24 DE 352.05 395.68 EA 23.45 591.60 Sums +660.83 m -660.83 m +1065.62 m -1065.62
Bearing Angle Traverse Given the tabulated measurement of the closed traverse as shown. Determine the area of the traverse using DPD Method. Line Length Bearing AB 413.600 m N 09°30’ W BC 304.380 m N 55°06’ W CD 407.210 m S 57°35’ W DE Unknown S 04°05’ E EA 649.630 m Unknown Note: EXPRESS YOUR ANSWER UPTO 3 DECIMAL PLACES.