Advertisement Remove all ads

A Truss is Loaded and Supported as Shown.Determine the Following: (1)Identify the Zero Force Members,If Any (2)Find the Forces in Members Ef,Ed and Fc by Method of Joints. - Engineering Mechanics

Advertisement Remove all ads
Advertisement Remove all ads
Answer in Brief

A truss is loaded and supported as shown.Determine the following:
(1)Identify the zero force members,if any
(2)Find the forces in members EF,ED and FC by method of joints.
(3)Find the forces in members GF,GC and BC by method of sections

Advertisement Remove all ads

Solution

By analysis of truss,we can say that DE is zero force member

METHOD OF JOINTS:

Joint E:
Applying the conditions of equilibrium

ΣFY=0 
FEFsin30-100=0

FEF=200 kN
Applying the conditions of equilibrium

ΣFX=0 
-FEFcos30-FED+200=0
-200cos30+200=FED
FED=26.7949 kN

△FED is congruent to △FCD

∠FCD=∠FED=30° 

JOINT F:

Applying the conditions of equilibrium

ΣFY=0 
FFGsin30-FFCsin30-FFEsin30-100=0 
FFG-FFC-200=200 

FFG-FFC=400 ……….(1)

ΣFX=0 
-FFGcos30-FFCcos30+FFEcos30=0 
Dividing by cos30
FFG +FFC = 200 ………(2) 
Solving (1) and (2)
FFG=300 kN
FFC=-100 kN 

METHOD OF SECTIONS:

In △FED
tan 30 = `(FD)/(DE)`
DE=3m
FD=`sqrt(3)` m
Consider the equilibrium of the truss section

ΣMC=0 
FFGcos 30 x FD + FFGsin30 x CD -100 x CD -100 X CE = 0
3FFG=900 
FFG=300 kN
Applying the conditions of equilibrium

ΣFX=0
-FFGcos30-FcB+200=0 

-300cos30+200=FcB
FcB=-59.8076 kN

ΣFY=0 
FcG+FFGsin30-100-100=0
FcG=50 kN

Member of truss Magnitude of force(kN) Nature of force
BC 59.8076 Compression
GC 50 Tension
GF 300 Tension
FC 100 Compression
ED 26.7949 Tension
EF 200 Tension
Concept: Analysis of Plane Trusses by Using Method of Joints
  Is there an error in this question or solution?
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×