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. Solve for the internal force in each member in kN using the method of sections and/or method of joints. FN=78MN=104SN=66 METHOD OF SECTIONS - PROBLEMSSOLUTION PROBLEM NO.1 :Determine the axial forces in members CD, CJ, and IJ.Determine the force of the members CF and CG.1 mMGKL100 kNDMGK L100 kNFree- Body DiagramMETHOD OF SECTIONS - PROBLEMSSOLUTION PROBLEM NO.1:Determine the axial forces in members CD, CJ, and IJ.Determine the force of the members CF and CG.BFMG HJK100 kNEF, = -TcD - TCJ cos 45° – T = 0,EF, = TCI sin 45° -100 kN = 0,Tep DEMpoint / = (1 m)TcD - (3 m)(100 kN) = 0.%3D45°K LTco = 300 KN (T);Το- 141 ΚN (T1;100 kNTy = -400 KN (C) SN KNG5 kN4 mN17FEN KNCΜΝ ΚΝ24 mABa-2 m-2 m-1m 1m Im 1 m

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Answered: METHOD OF SECTIONS - PROBLEMS SOLUTION…

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.

Solve for the internal force in each member in kN using the method of sections and/or method of joints.

FN=78

MN=104

SN=66

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)

SN KN
G
5 kN
4 m
N17
F
EN KN
C
ΜΝ ΚΝ
2
4 m
A
B
a
-2 m
-2 m-
1m 1m Im 1 m

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