adjusted departure adjusted azimuth length departure correction latitude correction latitude 32.100 1263.450 671.390 -1.671 669.719 1070.300 1.334 1071.634 112.430 843.950 780.100 -1.116 778.984 -322.010 0.891 -321.119 167.680 1028.630 219.480 -1.360 218.120 -1004.940 1.086 -1003.854 263.730 1126.360 -1119.620 -1.489 -1121.109 -123.010 1.190 -121.820 304.500 661.110 -544.840 -0.874 -545.714 374.460 0.698 375.158

adjusted departure adjusted azimuth length departure correction latitude correction latitude 32.100 1263.450 671.390 -1.671 669.719 1070.300 1.334 1071.634 112.430 843.950 780.100 -1.116 778.984 -322.010 0.891 -321.119 167.680 1028.630 219.480 -1.360 218.120 -1004.940 1.086 -1003.854 263.730 1126.360 -1119.620 -1.489 -1121.109 -123.010 1.190 -121.820 304.500 661.110 -544.840 -0.874 -545.714 374.460 0.698 375.158

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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Q.1 A line of level is run from B.M.1 (elevation=103.00m) to B.M. 2(elevation=107.00m). The field observations were recorded as given below: Station В.S I.S F. S Remarks 1 1.95 В.М. 1 0.65 3 2.60 1.45 -1.90 -1.60 0.55 6. 0.50 В.М. 2 Reduce the level of point 2, 3, 4 and 5, determine the total error of closure, and adjust the values.
Station BS HI FS Elevation BM-125 920.00 ТР-1 3.11 X 1.27 920.47 ТР-2 Y 926.28 1.11 ТР-3 2.74 2.00 BM-126 1.33 What is the value of Y?
معاينة Find the coordinates of points A, B, C, D and E by adjusting this traverse. Complete all cells in the tables below. GIVEN: Azimuth S5-S2 = 32°47'27" XS2 = - 715.000m YS2 = - 355.000m Complete all tables below and calculate all disclosure errors and their correction values. D A AS2 Page 1 of 4 مشاركة حد د المنطقة التخطيط II Measured angle Clockwise Angular 115°30'25" 225°10'08" 245°40'55" 223°05'05" 260°15'45" 242°51'26" 127°26'30"
Q.1 A line of level is run from B.M .1 (elevation=102.00m) to B.M. 2(elevation=106.00m). The field observations were recorded as given below: Station В. S I.S F. S Remarks 1 1.95 В.М. 1 2 0.65 2.60 1.45 4 -1.90 -1.60 5 0.55 6 0.50 В.М. 2 Reduce the level of point 2, 3, 4 and 5, determine the total error of closure, and adjust the values.
Azimuth Azimuth Fraction) Length Degrees Minutes Seconds Line 10047 86.99 100 28 AB 44 186.18 186 10 BC 259 89 0.00 CD 100.11 125.32 115 12 44 115.21 DE 323 10 323.17 EA 284.548 SUM 856.858 7. Determine the FGCC dassification of the angle measurements in the problem shown here. Interior Angles Degrees Minutes Seconds 80 28 86 10 44 100 01 215 12 4. 58 3. 18 SUM 69
Module 03- Coursework Activity 04 A transit with stadia interval factor of 100.8 was set up at point C on the line between points A and B. The following readings were observed: A B ROD VERTICAL ANGLE +15-25 -09-07 1.972 1.330 UPPER HAIR READINGS MIDDLE 1.854 1.175 1.732 1.010 LOWER If the distance from the instrument to the focus is 0.38m, determine the following: a) Horizontal distance between A and B. b) Difference in elevation between A and B.
What is the equation to convert the Azimuth below to a Bearing? οεε οτε οτε οοε οστ 087 οιτ 097 OSZ Obz οετ οττ οΙΖ Ο 300 310 320 330 340 350 10 20 30 40 50 60 70 120 130 140 150 160 170 180 190 200 210 220 230 240 80
Bearing or Azimuth of Line AB: ____________________ Sta Occ Sta Obs Dist (m) Mean Dist Observed Interior Angle Corr Adjusted Interior Angle Direct Reversed Mean A E 11.55 102°57 B 13.20 B A 13.20 C 17.42 C B 17.42 D 10.00 D C 10.00 E 13.90 E D 13.90 A 11.55
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
A level instrument at a height of 1.320 m has been placed at a station having a Reduced Level (RL) of 112.565 m. The instrument reads -2.835 m on a levelling staff held at the bottom of a bridge deck. The RL (in m) of the bottom of the bridge deck is A. 116.720 B. 116.080 C. 114.080 D. 111.050
In AHJK, PM I| JK as shown. Find HPM. H. M 76 10 25 27' K O HPM = 76°10' O HPM = 25°27' %3D O HPM = 90° O HPM = 78°23' %3D None of the above
2/A- The following data were obtained by stadia observation: zenith angle = 81° 50', s = 2.50 ft. The stadia interval factor is known to be 0.75 and K = 1 ft. Calculate H and V. 02/B- The following table gives the Departure Latitud