Problem 5: Given the following data for a transit tape survey. By applying the compass rule, the corrections, to be applied respectively to the latitude and departure of course AB are a. 1.60m & 0.45m b. 0.45 m & 1.60 m c. 1.76m & 1.50m d. 1.99 m & 0.20 m e. 0.56 m & 1.39 m Length Latitude +N Course Departure +E -S 119.09 -W AB 197.20 372.31 165.03 157.16 BC 372.29 4.54 CD 1.16 164.63 DA 242.50 242.16 12.99 -170.15 Sum 997.04 +372.28 -362.41 +169.17

Problem 5: Given the following data for a transit tape survey. By applying the compass rule, the corrections, to be applied respectively to the latitude and departure of course AB are a. 1.60m & 0.45m b. 0.45 m & 1.60 m c. 1.76m & 1.50m d. 1.99 m & 0.20 m e. 0.56 m & 1.39 m Length Latitude +N Course Departure +E -S 119.09 -W AB 197.20 372.31 165.03 157.16 BC 372.29 4.54 CD 1.16 164.63 DA 242.50 242.16 12.99 -170.15 Sum 997.04 +372.28 -362.41 +169.17

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

Q.3 A level circuit is shown in figure. The survey, needed for local engineering commenced at BM 35, the elevation of new benchmarks 36, 37, 38 and 39 were project, 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 вм 36 0.9 km вм 35 0.8 km Фвм 37 0.8 km 0.6 km BM 39 BM 38 0.7 km
HM 36 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 09 km. DM 35 O8 km DBM 37 08 km 08 km BM 39 BM 38 0.7 km
3. Given the following three wire leveling notes: Station Backsight Stadia BM1 Foresight Stadia Elevation 425.00 1.152 1.117 0.935 0.899 0.718 0.662 TP1 2.784 1.904 2.420 1.537 2.057 1.172 ТР2 2.591 2.210 2.094 1.714 1.599 1.218 BM2 a. Find the difference in elevation of TP2 and BM1, b. Find the elevation of BM2
4-) Complete the benchmark-leveling field notes by computing and recording the HI and elevation for each TP and Benchmark. Do the arithmetic check. STATION BS HI ES ELEVATION STATION BS HI FS ELEVATION BM1 2.367 200.00 BM1 3.784 150.00 1 0.345 1.277 1 1.451 0.278 2 1.128 3.785 2 0.555 2.123 3 0.978 0.426 3 2.354 0.754 4 2.127 0.785 0.750 1.075 BM2 3.945 BM2 2.102 BS FS BS FS
I need Answer of A only A. A tachometric surveying was conducted to obtain some readings forinaccessible points, where it was found that the multiplying and additiveconstants are 98 and 0.1, respectively. The instrument was setup at point C and the readings listed in the table were taken while the staff was held vertically. Calculate the horizontal distance between BM and station D. Also, calculate the reduced level of station D. Figure 1 shows a schematic view of the site.
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
Q3: A: Draw the diagram of traverse classification. B: In closed traverse ABCD. The partial coordinates were known as shown below. Determine its area by DMD method. 04 N Dofland OWODY dienstl it with (nm) dinostim Mem. AB BC CD DA ΔΧ 40 -80 -50 AY 60 100 -70
Following are the observations recorded using a total station between two points Aand B:The zenith angle is 82°32'20", the slope distance is 8335.26 ft. The height of theinstrument and rod recording r are equal. If the elevation of point A is 1431.31 ft.Determine the elevation of point B.
B. Double-Rodded Leveling 3. Complete differential level notes for the information shown and perform the customary arithmetic check. STATION BS HI FS STATION 1.722 BM1 127.378 m 1.722 ΤΡΙΗ 2.623 1.382 TP1L 2.789 1.522 TP2 H 3.089 0.899 TP2 L 3.214 0.714 ТРЗ Н 1.521 2.452 TP3 L 1.632 2.678 ТРА Н 2.333 1.113 TP4 L 2.589 1.523 TP5 H 0.667 2.877 TP5 L 0.986 2.543 3.521 BM2 3.214
Problem 1. From the given data of a three wire leveling as shown in the tabulation: Point Backsight Stadia Foresight Stadia Elevation BMA 100.00 5.25 2.61 4.75 2.10 4.26 1.60 TP 9.95 7.80 9.34 7.21 8.72 6.61 BMB a) Find the difference in elevation of TP, and BMA. b) Find the elevation of BMB-
The readings given in the table below were recorded in a levelling operation from points 1 to 10. Fine the reduce levels by the height of instrument method and apply appropriate checks. The point 10 i a bench mark having elevation of 66.374 m. Determine the loop closure. Station Chainage (m) B.S. I.S. F.S. Remarks 1 0.597 B.M. = 68.233 m %3D 20 2.587 3.132 С.Р. 3 40 1.565 4 60 1.911 5 80 0.376 6 100 2.244 1.522 С.Р. 7 120 3.771 8 140 1.334 1.985 C.P 160 0.601 10 180 2.002
Meet - nzw-pimo-rw anuwwlPx-a_a7HdrtDNnmaFDF2GGA/formResponse none of the above. O To establish a new benchmark on point Busing a levelling instrument. The work had been done between B.M1 and B.M2 to the required point B, and the observations were recorded as in the tables below. Calculate the best value *of elevation for point B B.M1 B B.M2 B BS Point B.M1 T. P1 BS FS 3.305 Remarks RL-34.585 FS 0.405 4.24 3.35 Point Remarks B 30.06 0.672 1.825 2.485 3.315 1.23S 3.715 T. P4 T. PS B.M2 1.795 A 4.01 3.485 Т. Р2 T. P3 3.315 2.78 RL=42.093 B 39.97m 41.97m 46.97m 45.98m O 44.37m O 40.95m O * Reciprocal ranging is employed when one end of a line is inaccessible. O the ends of the line are not visible even from intermediate points. O both the ends are inaccessible. the two ends of a line are not intervisible,O O O O O O O O O O OC