Determine the force in each member. MN=104 SN=66
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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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.
Determine the force in each member.
MN=104
SN=66
![1. DETERMINE PHE FORCE (N EACH MEMBERS
QFt
の
104 kip
12ft
Cace kip
9ft
イ2f4
12ft](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F835fff2e-ec06-48f8-aff1-9866fdac17a6%2F4e96c263-cc16-4018-bc2a-8aa3b2c866ef%2Frm634om_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1. DETERMINE PHE FORCE (N EACH MEMBERS
QFt
の
104 kip
12ft
Cace kip
9ft
イ2f4
12ft
![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%2F835fff2e-ec06-48f8-aff1-9866fdac17a6%2F4e96c263-cc16-4018-bc2a-8aa3b2c866ef%2F8ht7jnr_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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