4.791Bm 1571.589°THOT #8.629 Tpt12.1247.2.32Tp-316785,643'1.747'Bm 39тр-2Place data in the form of field notes. Include thearithmetic check. Assuming the average length of eachBS andFS is125ft. determine the order of accuracy.of closure for construction projects.Use the evver2.932'Constractionsite

#2 4.7 8.629 Bm 15 71.589° 12.124 Tp. - 82116 7.2.32 Tp-3 1.747' 5.643' 2.932' Constraction. site Bm 39 тр-г HOT # Place data in the form of field notes. Include the arithmetic check. Assuming the average length of each 125 ft. determine the order of accuracy. Use the erver of closure for construction projects. BS and FS is

#2 4.7 8.629 Bm 15 71.589° 12.124 Tp. - 82116 7.2.32 Tp-3 1.747' 5.643' 2.932' Constraction. site Bm 39 тр-г HOT # Place data in the form of field notes. Include the arithmetic check. Assuming the average length of each 125 ft. determine the order of accuracy. Use the erver of closure for construction projects. BS and FS is

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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GIVEN G=82H=78I=79J=38.5
E =29000Ksi I =750inch^4
a=2 b=5 3xb=15x mm 12xb=60x mm 10xa kN = 20x kN ((3xb)-0,00735)) = (15x- 0,00735) mm
ANSWER KEY: D= 51.8319m∆h= 13.5430mElev Sta. A= 70.8530m
A= 9kN B=9kN/m
Ray= 97.8553kn Rbx= 73.7132kn Rby= 14.14221 kn this must be the answer show solution pls
Q) All answers of second moment of area shall be in \times× 106 mm4 Solve it early I upvote
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.
I cannot understand these For point of contraflexural, 40090300=200×x90300 x=2 m θA=0 (fixed support) θDA=θD-θA=θD=Area of MEI Dia between A and D = 19030012×600×3+600×4+12×300×3-12×200×2 =355090300 θD=0.03931 radianθD= 2.252° Deflection at point D ∆D=31-CosθD ∆D=2.3176×10-3 mm Deflection at point D is 2.3176×10-3 mm
correction: w = 10kn/m
a (deg) = 16.446q0 (kN/m) = 37.632
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