HELP PLEASE WITH THE OBJECTIVES, INTRODUCTION, AND CONCLUSION (ATTACHED PHOTO ARE THE DATA AND RESULTS AND THE COMPUTATION) 3 SENTENCES PLEASEE. THANK YOUU
HELP PLEASE WITH THE OBJECTIVES, INTRODUCTION, AND CONCLUSION (ATTACHED PHOTO ARE THE DATA AND RESULTS AND THE COMPUTATION) 3 SENTENCES PLEASEE. THANK YOUU
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
HELP PLEASE WITH THE OBJECTIVES, INTRODUCTION, AND CONCLUSION (ATTACHED PHOTO ARE THE DATA AND RESULTS AND THE COMPUTATION) 3 SENTENCES PLEASEE. THANK YOUU
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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