.65 m² c. 11, 126.15 m² b. 11, 152 d. 11, 156.20 m² SITUATION 11: A closed traverse has the following data: 39. Find the Course 1 - 2 2 - 3 3- 4 4 5 5 1 - Bearing N 9.27 E S 88.43 E N 86.78E S 5.3E S 72.07 W a 29.67 m c. 38.42 m Distance (m) 58.7 m 27.3 m length of the course 3 - 4. 35.0 m b. 36.98 m d. 35.2 m 40. Compute the length of the course 5 - 1. a. 90.30 m 81.14 m b. 78.96 m d. 63.62 m

.65 m² c. 11, 126.15 m² b. 11, 152 d. 11, 156.20 m² SITUATION 11: A closed traverse has the following data: 39. Find the Course 1 - 2 2 - 3 3- 4 4 5 5 1 - Bearing N 9.27 E S 88.43 E N 86.78E S 5.3E S 72.07 W a 29.67 m c. 38.42 m Distance (m) 58.7 m 27.3 m length of the course 3 - 4. 35.0 m b. 36.98 m d. 35.2 m 40. Compute the length of the course 5 - 1. a. 90.30 m 81.14 m b. 78.96 m d. 63.62 m

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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On the drawing below, the virtual condition of the 4 holes is? O .450 .375 O.395 O .020 D O .430 Ø3.000 .020 0 10 45° 4X Ø.395-.430 Ø.500-.510 Ø.020MABMCM +0.010M ADM Ø2.000 .375-.395 4.000MABM C A 0.005MA B
ANSWER KEY=D 108.1502m∆h 22.4596m
The calcula te Based and lengths shown, On The bearing and lantude departure, linear micclosure Of the traverre, If all station s have 0,953 angle and Pre cicion correction NCY? 104 35' G10, 45 10. o00 NCY) 10,000 ES 30¢'54' 6 47,02 sian'on Bea ing lLength 26°10' 04° 35' l ar 30' 358°18' 306°4! A 285.10 G10.45 720,48 203.00 E 69 7.02 195 30 358 18 203 .00
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