Determine the force in each member of the Truss shown in the figure and indicate whether the members are in tension or compression. SN Newton 2 m 45 0 2 m

Structural Analysis
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ISBN:9781337630931
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Chapter2: Loads On Structures
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Please 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.

Given:

SN=66

Determine the force in each member of the TruSs shown in
the figure and indicate whether the members are in
tension or compression.
SN Newton
2 m
45 °
2 m
Transcribed Image Text:Determine the force in each member of the TruSs shown in the figure and indicate whether the members are in tension or compression. SN Newton 2 m 45 ° 2 m
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
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