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Pb 4 Methof of joints: Internal forces in each member 000 4 ft B D -4 ft 450 lb 4 ft 1 Diagnosis (including stability, determinacy, zer. force member) 2 Free body Diagram (FBD): 3 Reaction forces 4 Internal forces

Pb 4 Methof of joints: Internal forces in each member 000 4 ft B D -4 ft 450 lb 4 ft 1 Diagnosis (including stability, determinacy, zer. force member) 2 Free body Diagram (FBD): 3 Reaction forces 4 Internal forces

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 provide the diagnosis, free body diagram, reaction forces and internal forces for the following two figures:
Pb 4 Methof of joints: Internal forces in each member A 4 ft 000 D 450 lb A B 4 ft Reactions W F0 Un r Mipt 0 4 ft FAB FBA-225 lbs (T) FAD FDA =318 lbs (C) Internal forces FBD=FDB=0 lbs 1 Diagnosis (including stability, determinacy, zero- force member) 2 Free body Diagram (FBD): 3 Reaction forces 4 Internal forces Ay=225 lbs Cy-225 lbs and Cx=450 lbs FBC FAB FCB FBA-225 lbs (T) FCD FC 3 18 lbs (T) DC
Transcribed Image Text:Pb 4 Methof of joints: Internal forces in each member A 4 ft 000 D 450 lb A B 4 ft Reactions W F0 Un r Mipt 0 4 ft FAB FBA-225 lbs (T) FAD FDA =318 lbs (C) Internal forces FBD=FDB=0 lbs 1 Diagnosis (including stability, determinacy, zero- force member) 2 Free body Diagram (FBD): 3 Reaction forces 4 Internal forces Ay=225 lbs Cy-225 lbs and Cx=450 lbs FBC FAB FCB FBA-225 lbs (T) FCD FC 3 18 lbs (T) DC
Pb 5 Method of Sections Internal forces in members GE, GC and BC. 1 Diagnosis (including stability, determinacy, zero-force member) 2 Free body Diagram (FBD): 3 Reaction forces 4 Internal forces 5 Summary 3 m 4 m G r F = 0 M/ point A = 0 Summary: FBC FCB 800 N (T) FGE=FEG-800 N (C) FGc=FcG=500 N (T) B a a 4 m E с -4 m 1200 N A-400N A-300 N D₂=900N 400 N D
Transcribed Image Text:Pb 5 Method of Sections Internal forces in members GE, GC and BC. 1 Diagnosis (including stability, determinacy, zero-force member) 2 Free body Diagram (FBD): 3 Reaction forces 4 Internal forces 5 Summary 3 m 4 m G r F = 0 M/ point A = 0 Summary: FBC FCB 800 N (T) FGE=FEG-800 N (C) FGc=FcG=500 N (T) B a a 4 m E с -4 m 1200 N A-400N A-300 N D₂=900N 400 N D
Expert Solution
Step 1

m=number of members

r=external support reaction

j=number of joints 

To find support reaction and force in all members of the truss 

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