Question

In the truss shown in figure,compute the forces in each member.

Answer 1

Solution :
We can say that FD,GH and CB are zero force members in the given truss
Joint A :

Applying the conditions of equilibrium
ΣFy=0
-1 – FAC sin30 = 0
FAC = -2kN
Applying the conditions of equilibrium
ΣFx = 0
FAB + FAC cos30 = 0
FAB = 1.7321 Kn
JOINT C :

Applying the conditions of equilibrium
ΣFx = 0
FCE = FCA = -2kN
JOINT B :

Applying the conditions of equilibrium
ΣFy = 0
-1 - FBE sin60 = 0
FBE = -1.1547 kN
Applying the conditions of equilibrium
ΣFx = 0
-FBA + FBEcos60 + FBD = 0
FBD = 2.3094 kN
JOINT D :

Applying the conditions of equilibrium
ΣFy = 0
-1 - FDEsin60 = 0
FDE = -1.1547 kN
Applying the conditions of equilibrium
ΣFx = 0
-FDB - FDEcos60 + FDG = 0
FDG = 1.7321 kN
JOINT E :

Applying the conditions of equilibrium
ΣFy = 0
FED + FEFcos30 + FEBsin30 = 0
FEFcos30 = -(-1.1547)-(-1.1547) x`1/2`
FEF = 2kN
Joint F :

Applying the conditions of equilibrium
ΣFx = 0
FFG = FFE = -2kN
Final answer :

Sr.no.	MEMBER	MAGNITUDE OF FORCE (in kN)	NATURE OF FORCE
1.	AC	2	COMPRESSION
2.	AB	1.7321	TENSION
3.	CB	0	-
4.	CE	2	COMPRESSION
5.	BE	1.1547	COMPRESSION
6.	BD	2.3094	TENSION
7.	DE	1.1547	COMPRESSION
8..	DG	1.7321	TENSION
9.	EF	2	TENSION
10.	EH	4	COMPRESSION
11.	FD	0	-
12.	FG	2	COMPRESSION
13.	GH	0	-