1) This document provides methods for analyzing trusses to determine the forces acting on individual members, including finding zero force members, using equilibrium equations at joints, and using a section cut method. 2) It applies these methods to solve for forces in several example trusses, determining zero force members, forces at various joints using equilibrium, and using section cuts to solve for additional member forces. 3) The key steps shown are finding zero force members, using the equations of equilibrium to solve for forces at joints, and using virtual section cuts combined with equilibrium to determine remaining member forces.

1. 7laha in Truss
Never stop thinking
With: kites Team

2. Brief methods to solve truss problem
method Simple diagram
• Finding zero loading members
• Case1:if we have two
members in the
truss(single/apart
from other members)
without external force
or support reaction
applied to their joint .
the two members the
two members should
be zero force members
• In this case Fdc=Fde=0
F
c
D
Ef
BA

3. Zero force member method
• Second case
• If three members form a joint
in in truss where two members
are collinear and there is not
any external force or support
reaction .
• The third member is zero force
member
• Simple diagram
• ∑Fx=0 fda=0
f E
D
C
A
B
F da
F dc
F de
x

4. Sections method
• Illustration
• Imagine that you cut apart
from the truss , we consider
the members where the cut
happen have external forces
according to the reminder part
so we can use the equilibrium
rules
• example
A
B
G
C D
F E
P
P
Fbc
Fg
c
Fgf
fbc
fgc
fgf
yd
xd
Re

5. Smart example
•
5KN5KN 7.5KN
a
b
N
m c
L
D
K J
E
I
F h
G
4.5KN
2.5m2.5m2.5m2.5m
2.5m
2.5m
2.5m

6. • 1-Determine the force in
members CD , LD and KL
• 2-Determine zero force
members
• 3-Determine the force in the
member FH
• 4-determine the force in the
members JI and ML
• Solution:
• First there are not zero –force
members
• From the equilibrium
• ∑f y=0 so y G+YA=17.5KN
• MA=0soYG=11KN
• ∑f x= 0 so XG=4.5 KN

7. solution
• At point G:
• Θ=45 deg
• So FGF=4.5√2KN
• FGH=6.5KN
• At point H:
• ∑FY=0
• FIH=6.5√2 KN
• ∑FX=0
• FFH=6.5KN
FGH
11KN
4.5KNFGF
θ
θ
FIH
FGH
FfH

8. • 1-when we cut at KJ , JD , ED • ∑FY=0
• FJD=(11-5)*√2 =6√2KN
• ∑FX=0
• FDE-FKJ=10.5 KN
• MJ=0
• FDE*2.5=4.5*7.5
• So FDE=13.5KN
• FKJ=3KN
FED
J
E
FKJ
FJD
5K
N
G 4.5
KN
11K
N

9. • When we cut at LK ,LD and
CD.
• ∑FY=0
• So FLD cos45 =6.5-5
• FLD=2.12KN
• ∑FX=0
• FKL=4.5+1.5+FCD
• ML=0
• So 6.5*5=FCD*2.5
• FCD=13KN
• FKL=19KN
5KN4.5 KN
FLD
FCD
FLK
6.5K
N
A
L

10. FML*cos45=FKL-FLD*cos45 -4.5
so FML=13√2 KN
• FJI*Cos45=FDJ*cos45+FK
J=9kN

11. 7laha
•
• Determine the force in members JI and DE of
the k truss.
• Hint : use sections aa and bb
• Final answer :Fji=10.67KN FDE=10.67KN
L k
J I H
A B C
D
E F G
6K
N
7.5
KN
9K
N
6m6m6m6m6m
4.5m
4.5m
a
a
b
b
M N o P