Question

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

Answer 1

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

METHOD OF JOINTS:

Joint E:
Applying the conditions of equilibrium

ΣF_Y=0 
F_EFsin30-100=0

F_EF=200 kN
Applying the conditions of equilibrium

ΣF_X=0 
-F_EFcos30-F_ED+200=0
-200cos30+200=F_ED
F_ED=26.7949 kN

△FED is congruent to △FCD

∠FCD=∠FED=30° 

JOINT F:

Applying the conditions of equilibrium

ΣF_Y=0 
F_FGsin30-F_FCsin30-F_FEsin30-100=0 
F_FG-F_FC-200=200 

F_FG-F_FC=400 ……….(1)

ΣF_X=0 
-F_FGcos30-F_FCcos30+F_FEcos30=0 
Dividing by cos30
F_FG +F_FC = 200 ………(2) 
Solving (1) and (2)
F_FG=300 kN
F_FC=-100 kN 

METHOD OF SECTIONS:

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

ΣM_C=0 
F_FGcos 30 x FD + F_FGsin30 x CD -100 x CD -100 X CE = 0
3F_FG=900 
F_FG=300 kN
Applying the conditions of equilibrium

ΣF_X=0
-F_FGcos30-F_cB+200=0 

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

ΣF_Y=0 
Fc_G+F_FGsin30-100-100=0
Fc_G=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