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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

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

Steel Design (Activate Learning with these NEW titles from Engineering!)
Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN:9781337094740
Author:Segui, William T.
Publisher:Segui, William T.
Chapter4: Compression Members
Section: Chapter Questions
Problem 4.3.3P
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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.

Use method of sections and/or joints.

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
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)
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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