METHOD OF SECTIONS - PROBLEMS SOLUTION PROBLEM NO.1: Determine the axial forces in members CD, CJ, and IJ. Determine the force of the members CF and CG. 1 m M K 100 kN D M G K L 100 kN Free- Body Diagram METHOD OF SECTIONS - PROBLEMS SOLUTION PROBLEM NO.1: Determine the axial forces in members CD, CJ, and IJ. Determine the force of the members CF and CG. D M G H JK L 100 kN EF, = -Tcp - T cos 45° - T, = 0,
METHOD OF SECTIONS - PROBLEMS SOLUTION PROBLEM NO.1: Determine the axial forces in members CD, CJ, and IJ. Determine the force of the members CF and CG. 1 m M K 100 kN D M G K L 100 kN Free- Body Diagram METHOD OF SECTIONS - PROBLEMS SOLUTION PROBLEM NO.1: Determine the axial forces in members CD, CJ, and IJ. Determine the force of the members CF and CG. D M G H JK L 100 kN EF, = -Tcp - T cos 45° - T, = 0,
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Please help me with the following question. You can use the second photo as reference and please put the whole process and solution because I’m having a difficulty with understanding it. Thank you very much.
FN=78
MN=104
SN=66
![Req: Internal force in each member in KN using both methods
SN KN
a
G
5 kN
4 m
V17)
ΜMN KN
EN KN
4 m
B
-2 m-
-2 m.
1m'1m'lm'l m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F220dff5d-ec46-4bf0-b75f-da33cddc0ddf%2Fbd7c0cfb-8f44-423b-bbd8-6a0c6df2b078%2Fek1g9tf_processed.png&w=3840&q=75)
Transcribed Image Text:Req: Internal force in each member in KN using both methods
SN KN
a
G
5 kN
4 m
V17)
ΜMN KN
EN KN
4 m
B
-2 m-
-2 m.
1m'1m'lm'l m
![METHOD OF SECTIONS - PROBLEMS
SOLUTION PROBLEM NO.1 :
Determine the axial forces in members CD, CJ, and IJ.
Determine the force of the members CF and CG.
1 m
M
G
K
L
100 kN
D
M
G
K L
100 kN
Free- Body Diagram
METHOD OF SECTIONS - PROBLEMS
SOLUTION PROBLEM NO.1:
Determine the axial forces in members CD, CJ, and IJ.
Determine the force of the members CF and CG.
B
F
M
G H
JK
100 kN
EF, = -TcD - TCJ cos 45° – T = 0,
EF, = TCI sin 45° -100 kN = 0,
Tep D
EMpoint / = (1 m)TcD - (3 m)(100 kN) = 0.
%3D
45°
K L
Tco = 300 KN (T);
Το- 141 ΚN (T1;
100 kN
Ty = -400 KN (C)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F220dff5d-ec46-4bf0-b75f-da33cddc0ddf%2Fbd7c0cfb-8f44-423b-bbd8-6a0c6df2b078%2Fhed6dk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:METHOD OF SECTIONS - PROBLEMS
SOLUTION PROBLEM NO.1 :
Determine the axial forces in members CD, CJ, and IJ.
Determine the force of the members CF and CG.
1 m
M
G
K
L
100 kN
D
M
G
K L
100 kN
Free- Body Diagram
METHOD OF SECTIONS - PROBLEMS
SOLUTION PROBLEM NO.1:
Determine the axial forces in members CD, CJ, and IJ.
Determine the force of the members CF and CG.
B
F
M
G H
JK
100 kN
EF, = -TcD - TCJ cos 45° – T = 0,
EF, = TCI sin 45° -100 kN = 0,
Tep D
EMpoint / = (1 m)TcD - (3 m)(100 kN) = 0.
%3D
45°
K L
Tco = 300 KN (T);
Το- 141 ΚN (T1;
100 kN
Ty = -400 KN (C)
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