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
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
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
