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7laha in Truss
Never stop thinking
With: kites Team
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
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
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
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
• 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
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
• 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
• 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
FML*cos45=FKL-FLD*cos45 -4.5
so FML=13√2 KN
• FJI*Cos45=FDJ*cos45+FK
J=9kN
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

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7laha in truss

  • 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