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HELP WITH OBJECTIVE, INTRODUCTION, AND CONCLUSION (ATTACHED PHOTO ARE THE DATA AND RESULTS AND COMPUTATIONS)IT'S ABOUT MEASUREMENT OF ANGLES AND DIRECTIONS: CLOSED COMPASS TRAVERSE

HELP WITH OBJECTIVE, INTRODUCTION, AND CONCLUSION (ATTACHED PHOTO ARE THE DATA AND RESULTS AND COMPUTATIONS)IT'S ABOUT MEASUREMENT OF ANGLES AND DIRECTIONS: CLOSED COMPASS TRAVERSE

Structural Analysis
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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HELLO PLEASE, HELP WITH OBJECTIVE, INTRODUCTION, AND CONCLUSION (ATTACHED PHOTO ARE THE DATA AND RESULTS AND COMPUTATIONS)IT'S ABOUT MEASUREMENT OF ANGLES AND DIRECTIONS: CLOSED COMPASS TRAVERSE

Observed Bearings Computed Adjusted values Length (m) line Sta Internal Corr Interior For w ard васк Forward Васк Bearing Angle 238 ° 40' Angle Bearing AB 16.50 S30° 40'W N30° 40'E A 59' 239°39' 210 40' 30°40' 15.15 N84° 30 'W B 65° 30' 59' 66°29' 97°9' 277°q' BC S83° 50' E 82 ° 30' 59' 83°29' 180 ° 38' CD 117.35 N02°00'W SO2°15'E C 38 DE 14.92 S89°30'W DUE E AST D 87 ° 15' 59' 86 °14' 268°52' 88°52' E A 60.25 S28° 50' N 28° 00' W E 61° 10' 59' 62°9' 151° 1' 28° 55' Şolution: from A- B-C.· A Corrected ungle: bb Take Eraverse in Anticlockwise , AS we go LA = 2380 H0! +591=2390391! H Included anyle = (Fore beaung ot next line - B.B ot Previous line) 2B = 65°30! + 59' - 66° 29' 82°30' + 59' = 83° 291 H Beauing of line &hould be whole urele beauing. ZD =87° Is! + 59' = 88°14! = 610101 + S9' = 62° 9! Now, LA = Internal dngle at A' = FBot A B - B.B of FA ZE = C1804 30040') - (360°- 28o o') = 238 40' Beaung calautation : . = F.B of BC - B.B of AB b Staion A 8 R is free from local attoaction, so s°beaing = (1800-83050' ) - ( 30° 40') = |65°30! taken at A ss B ale corrock. F.B of CD - BB of BC Now, = (3600 - a°o' ) - (360° - 84° 30') = |82° 30! %3D F-B of AB as a coreck reading 2100 401 - FBot AB LA = = F.B of DE - BB ot CD ニ 1800 + 30°40! = C180°+ 89°30' )-C l800-215') = B.B ot AB = 30° 40! LE = FB ot EA - B.B of DE = (leo°- 28°s0!) -(90°) =161° 10! F.B of BC = 66° 29'+30°Ho' -(4709! ZB ニ Sum ot the interior angle : LA +LB+LC+ LD + LE B. B ot Bc = 180° + 9Tº9' =a77°9', F.B of CD = 83°99'+ 277°9' = [38' Can - 4) x90 = (2x5-A)X 90 = S40° Theoretial sum = B.B of CD - 38'+180° = |1800 38' ニ Error in inteuor angle : 53505! -SHOD ZD = F.Bof DE = 88°1A'+ 18o°38! =268 °sa = E- 4° ss' Correction in each interior angle = B.B of DE : 26 8°sa! - 180° - [88°s 40 55! ZE = FB Đf EA = ba°91 +8 = I° 2' = B.B of EA : 1800-151°I! = 28D55'
Transcribed Image Text:Observed Bearings Computed Adjusted values Length (m) line Sta Internal Corr Interior For w ard васк Forward Васк Bearing Angle 238 ° 40' Angle Bearing AB 16.50 S30° 40'W N30° 40'E A 59' 239°39' 210 40' 30°40' 15.15 N84° 30 'W B 65° 30' 59' 66°29' 97°9' 277°q' BC S83° 50' E 82 ° 30' 59' 83°29' 180 ° 38' CD 117.35 N02°00'W SO2°15'E C 38 DE 14.92 S89°30'W DUE E AST D 87 ° 15' 59' 86 °14' 268°52' 88°52' E A 60.25 S28° 50' N 28° 00' W E 61° 10' 59' 62°9' 151° 1' 28° 55' Şolution: from A- B-C.· A Corrected ungle: bb Take Eraverse in Anticlockwise , AS we go LA = 2380 H0! +591=2390391! H Included anyle = (Fore beaung ot next line - B.B ot Previous line) 2B = 65°30! + 59' - 66° 29' 82°30' + 59' = 83° 291 H Beauing of line &hould be whole urele beauing. ZD =87° Is! + 59' = 88°14! = 610101 + S9' = 62° 9! Now, LA = Internal dngle at A' = FBot A B - B.B of FA ZE = C1804 30040') - (360°- 28o o') = 238 40' Beaung calautation : . = F.B of BC - B.B of AB b Staion A 8 R is free from local attoaction, so s°beaing = (1800-83050' ) - ( 30° 40') = |65°30! taken at A ss B ale corrock. F.B of CD - BB of BC Now, = (3600 - a°o' ) - (360° - 84° 30') = |82° 30! %3D F-B of AB as a coreck reading 2100 401 - FBot AB LA = = F.B of DE - BB ot CD ニ 1800 + 30°40! = C180°+ 89°30' )-C l800-215') = B.B ot AB = 30° 40! LE = FB ot EA - B.B of DE = (leo°- 28°s0!) -(90°) =161° 10! F.B of BC = 66° 29'+30°Ho' -(4709! ZB ニ Sum ot the interior angle : LA +LB+LC+ LD + LE B. B ot Bc = 180° + 9Tº9' =a77°9', F.B of CD = 83°99'+ 277°9' = [38' Can - 4) x90 = (2x5-A)X 90 = S40° Theoretial sum = B.B of CD - 38'+180° = |1800 38' ニ Error in inteuor angle : 53505! -SHOD ZD = F.Bof DE = 88°1A'+ 18o°38! =268 °sa = E- 4° ss' Correction in each interior angle = B.B of DE : 26 8°sa! - 180° - [88°s 40 55! ZE = FB Đf EA = ba°91 +8 = I° 2' = B.B of EA : 1800-151°I! = 28D55'
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