Determine the area of a closed traverse using Double Meridian Distance and Double Parallel Distance AzimuthFrom NorthComputed DepartureComputed LatitudeCourseDistance(+) N(-) S(+) E(-) WCD380.38 m0°50'380.34 m5.53 mDE519.30 m296°06'228.46 m466.35 mEF693.87 m205°54'624.18 m303.08 mFG417.98 m135°20'297.27 m293.83 mGH403.93 m33°27'337.03 m222.65 mHC249.00 m95°39'24.51 m247.79 mSUM2664.46 m+945.83 m -945.96 m+769.8 m-769.43 m

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Answered: Determine the area of a closed traverse…

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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If the Greenwich civil time is 20hr30m15s P.M on june 13, 2019. Find the LMT for place 120degreeW.
0:£) B s, Mt B Traverse.. X o I onder 1o determine a tunnel between staton A and B where the vision was obstrucied between them, a traverse was slected o link between these stations, The field measurements of the raverseas in the table blow: - Line Length(m) | Azimuth DA 500 2000 be 00 oo ) 0o [oor AB ? ? Tength and the bearing of the tunnel (AB) and the angle BAD for stting it out
Exa . 3 D“:l'[::: 1: the following offsets were taken from a chain line to hedge: | Offset B et I B, 120 160 220 280 . 108 136 182 96 BA& TS 63 46 the hedge and the offset by (iv) Trapezoidal (Ci)o;r;;):tcrdxpe area inc.ludcd between the chain line, id ordinate rule, (ii) Average ordinate rule, (iii) Simpson's rule, rule.
degree (a) = 14.01 gshsysujw thnx
Answer Keys: By = 49,111.11 N Vcmax = -9277.78 N +Mcmax = 54,222.22 N-m
Two triangulation towers 120 m apart are on the samehorizontal plane. A surveyor standing and setting upsuccessively the transit at the bases of the towersobserves that the angle of elevation of the taller is twicethat of smaller. At the midway point, he observes that theangles of elevation are complementary angles. The heightof the instruments in all set up are is maintained to be 1.20m from the plane. Determine the height of the smallertower.
The following consecutive readings were taken with a level and a 4mt. leveling staff on acontinuously sloping ground at common intervals of 30 m:0.855(onA),1.545,2.335,3.115,3.825,0.455,1.380,2.055,2.855,3.455,0.585,1.015,1.850,2.755,3.845(on B)Enter the readings on a level field book and work out the staff readings on that, Find theR.L of different points using Collimation Plane Method. RL of the BM is 200.000m.Calculate the slope of the ground.
The slope distance and zenith angle (measured from vertical) between points A and B were observed with a totalstation instrument at ‘A’ as 9585.26 ft and 81o42’20”, respectively. The height o f th e instrument and rod reading were 4.5 ft and 5.7 ft respectively. If the station and elevation of point A is 1238.4 ft respectively, compute theelevation of point B.
If an engineer's transit was set-up at TP with HI=1.60m, sta.F was sighted with RR=2.00m. What would be its stadia intercept?
Q - The subsequent readings were curious about dumpy level with four.zero m leveling team of workers and 20.0 m c program languageperiod. 1.755, 2.505, 3.050, 3.950, 1.750, 2.355, 3.a hundred, three.805, zero.905, 2.0.5 and 3.250. If R.L of first point is 15.zero m. Calculate the R.L of different points by way of H.I approach.
Leveling Methods An instrument at point B is set up in the bottom of a depression withits scope measured 1.40m above ground. The purpose of the station is todetermine the elevations of both points A and C which are located on thepeaks of the valley. Two vertical angles of +11°30’ and +13°15’ are madewith the scopes of the instrument to points A and C respectively.Preliminary survey suggests that points AB is measured with a horizontaldistance of 583m and points BC is measured with a horizontal distance of1,125m. Considering the earth’s curvature and atmospheric refraction,determine the elevation of both A and C if the elevation of the stationis 75m above sea level.
Given the following data for lengths AB (197.20m), BC (372.31m), CD (165.03m), DA (242.50m) and Latitudes AB (-119.09), BC (372.28), CD (-1.16), DA (-242.16) and Departures AB (-157.16), BC (4.54), CD (164.63), DA (-12.99). By applying the transit rule, the adjusted latitude and adjusted departure of course AB, respectively, are • -120.69 m and 157.36 m • -117.49 m and -157.61 m • -120.69 m and -156.71 m • -117.10 m and -157.36 m • -121.08 m and -156.96 m

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