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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.   MN=104 SN=66

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.   MN=104 SN=66

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
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.

 

MN=104

SN=66

METHOD OF JOINTS - PROBLEMS SOLUTION PROBLEM NO.1 Determine the axial forces in members AB and AC of the truss. EF, = A, + B = 0, 3 m EF, = A, - 2 kN = 0, EMpoint = -(6 m) A, - (10 m)(2 kN) = 0. 3 m D Solving yields A, = -3.33 kN, A, = 2 kN, B 5m 5 m- and B = 3.33 kN. 2 kN References: Statics by Bedford Fowler SE Free- Body Diagram METHOD OF JOINTS - PROBLEMS SOLUTION PROBLEM NO.1 Determine the axial forces in members AB and AC of the truss. ,= 2 KN A = 3.33 KN A = 3.33 KN A,= 2 KN TẠC 3 m a =59 3 m TAB D. a = arctan(5/3) B = 3.33kk- 5 m 5m %3D 2 kN 2 AN References: Statics by Bedford Fowler SE B = 3.33kN EF, = TACsina - 3.33 kN = 0, TẠc = 3.89 KN EF, = 2 kN -TAR - TẠC cos a = 0. Ta = O KN AlB
Transcribed Image Text:METHOD OF JOINTS - PROBLEMS SOLUTION PROBLEM NO.1 Determine the axial forces in members AB and AC of the truss. EF, = A, + B = 0, 3 m EF, = A, - 2 kN = 0, EMpoint = -(6 m) A, - (10 m)(2 kN) = 0. 3 m D Solving yields A, = -3.33 kN, A, = 2 kN, B 5m 5 m- and B = 3.33 kN. 2 kN References: Statics by Bedford Fowler SE Free- Body Diagram METHOD OF JOINTS - PROBLEMS SOLUTION PROBLEM NO.1 Determine the axial forces in members AB and AC of the truss. ,= 2 KN A = 3.33 KN A = 3.33 KN A,= 2 KN TẠC 3 m a =59 3 m TAB D. a = arctan(5/3) B = 3.33kk- 5 m 5m %3D 2 kN 2 AN References: Statics by Bedford Fowler SE B = 3.33kN EF, = TACsina - 3.33 kN = 0, TẠc = 3.89 KN EF, = 2 kN -TAR - TẠC cos a = 0. Ta = O KN AlB
d 9 ft 9 ft e a f MN kip 12 ft SN kip h 9 ft 12 ft 12 ft
Transcribed Image Text:d 9 ft 9 ft e a f MN kip 12 ft SN kip h 9 ft 12 ft 12 ft
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