STATION DIRECT DOUBLE MEAN REMARKS DEFL. DEFL. DEFL. Azimuth 111 - 112 = 121° 21' 111 147° 01' 293° 59' R (+) R (+) R (+) 112 116° 11' 232° 22' 113 157° 5' 314° 02' 114 59° 2' 119° 8' L (-) Σ Angle Check (Closed Traverse Only): Misclosure by sum of deflection angles: Misclosure by azimuth calculation: show work for azimuth calculations separately) Note: Add the deflection angles algebraically ( R = + Angle, L = - Angle

STATION DIRECT DOUBLE MEAN REMARKS DEFL. DEFL. DEFL. Azimuth 111 - 112 = 121° 21' 111 147° 01' 293° 59' R (+) R (+) R (+) 112 116° 11' 232° 22' 113 157° 5' 314° 02' 114 59° 2' 119° 8' L (-) Σ Angle Check (Closed Traverse Only): Misclosure by sum of deflection angles: Misclosure by azimuth calculation: show work for azimuth calculations separately) Note: Add the deflection angles algebraically ( R = + Angle, L = - Angle

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

Q avg = 0.181 m^3/s
Given:Upper reading = 1.64 mMiddle reading = 1.57 mLower reading = 1.51 mVertical Middle Angle = 10°32’00”Elev of point A = 333 mH of instrument = 1.2 mK = 100Stadia Constant = 0 Solve the ff.: A. Inclined, horizontal, and vertical distance B. Elevation of point B.
A tower is located on a hilltop at T. Angular readings are taken with a theodolitefrom two instrumentation stations A and B, which are 50 m apart. Given the followingdata, determine the R.L. of hilltop at T and the tower height. Height of instrument (R.L.of instrumental centre) = 10 m. Vertical angles from A to the top and bottom of thetower at T is 35° and 28°(upward) respectively. Whole circle bearing of line AT, ABand BTis 25°, 110°, 5° respectively.
Transit is set up at A observeda stadia intercept of 2.2 m at a verticalangle of +1823’. The stadia interval factor ofthe instrument used is 95.5 and stadiaconstant of 0.3 m. If the height of instrumentis 1.62 m, and the rod reading is 1.95 m.Determine the horizontal distance from thetransit set up A to the rod at point B. a. 201.44 m c. 198.45 b. 189.48 m d. 186.56 m
a (deg) = 16.446q0 (kN/m) = 37.632
Ans. key : 1) 19.45 KPa. 2) 48.63 KPa 3) 1.37 kN/m
Find the equal height of towers at triangulation stations A and C so that line AC will clear theelevated ground between A and C by 6m. with the instrument set at point B, the observedvertical angle to ground station A is 0 0 55’ and the vertical angle to ground station C is 0 0 48’.Elevation of A is 237m. Distance AB is 10km and distance of BC is 8.7km. Yhe instrument set-upat B is 2.7m above the ground. Use hcr= 0.067k ^2
correction: w = 10kn/m
Ray= 97.8553kn Rbx= 73.7132kn Rby= 14.14221 kn this must be the answer show solution pls
Answer key ; 1. 1160kn#2 170 kn -m#3 540 kn and210 kn-m
The following data were obtained by stadia observation: vertical angle, + 12deg 51 min; upper and lower stadia hair reading, 4.311 and 3.730 m, respectively. the rod reading was taken at 3.785m above the stadia point A and the height of instrument set up at point B is 1.620m. If the stadia interval factor is 99.5 and the stadia constant is 0.381m, what is the vertical distance? a.54.098 b.94.780 c.12.620 d.26.364
E =29000Ksi I =750inch^4