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 memberA4 ft000D450 lbAB4 ftReactionsWF0UnrMipt 04 ftFAB FBA-225 lbs (T)FAD FDA =318 lbs (C)Internal forces FBD=FDB=0 lbs1 Diagnosis (including stability, determinacy, zero-force member)2 Free body Diagram (FBD):3 Reaction forces4 Internal forcesAy=225 lbsCy-225 lbs andCx=450 lbsFBC 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 forces4 Internal forces5 Summary3 m4 mGrF = 0M/ point A = 0Summary:FBC FCB 800 N (T)FGE=FEG-800 N (C)FGc=FcG=500 N (T)Baa4 mEс-4 m1200 NA-400NA-300 ND₂=900N400 ND

m=number of members r=external support reaction j=number of joints To find support reaction and…

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

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Beams

In structural engineering, a beam is a component made of a variety of materials (including steel alloys, and wood) that can resist loads applied laterally to the beam axis. Members, elements, rafters, shafts, and purlins are all terms used to describe beams.

Question

100%

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

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

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