Method of Sections
Method of joints works well when you want to know the forces in all the members.
If we want to know a force in a particular member, the method of sections is more
efficient.
Look at the following truss:
F1
F2
A
F3
B
D
G
E
C
To solve for the reactions at E and G, we use the entire truss as a free-body.
F1
F2
A
F3
B
Gy
D
G
Gx
E
C
Ey
By method of joints we use free-body diagrams of individual nodes.
F1
A
AB
AC
We know that the entire truss is in equilibrium.
If this is the case, then each joint is also in equilibrium.
Well, if this is the case, then a portion of the truss smaller than the entire but larger than
the joints must also be in equilibrium.
Lecture16.doc
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For instance:
F1
F2
A
B
BD
BE
C
CE
E
This section must also be in equilibrium.
So, we can cut the truss into sections and solve for the unknowns.
What is the only limitation of this method?
We can only cut such that there are 3 unknowns because:
F y  0
Fx  0
M  0
If I wanted to find BD what would I do?
BD is only unknown!!
M E  0
What about CE? M B  0
CE only unknown!!!
What about BE? Fy  0
BE only unknown!!
Lecture16.doc
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Examples
1). Given: The truss shown below
28 kips
A
28 kips
C
E
G
I
K
16 kips
10 '
B
D
8'
F
J
H
8'
8'
8'
8'
Find: Determine the force in members EF and GI.
FBD
28 kips
A
28 kips
C
E
G
I
K
16 kips
y
Bx
B
D
F
J
H
By
x
Jy
Fy  0
  M B  0
Fx  0
 28(8)  28(24)  16(10)  J y (32)  0
16  B x  0
 28  28  33  B y  0
J y  33 kips
B x  16 kips
B y  23 kips
Member EF
28 kips
F y  0
A
C
FEG
23  28  FEF  0
FEF  5 kips C
FEF
16 kips
B
D
FDF
23 kips
Lecture16.doc
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Member GI
I
FGI
K
16 kips
FHI
H
FHJ
J
33 kips
  M H  0
33(8)  16(10)  FGI (10)  0
FGI  10.4 kips
FGI  10.4 kips C
Lecture16.doc
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2). Given :
A
P
B
C
D
E
d
F
G
I
J
H
d
K
O
L
M
d
d
N
d
d
Find: Determine the force in members FK and JO
FBD
P
A
F
B
FFG
FFK
C
FGH
H
D
FHI
E
FIJ
J
FJO
  M F  0
  M J  0
FJO (4d )  Pd  0
FFK (4d )  Pd  0
FJO 
1
P Comp
4
Lecture16.doc
FFK 
1
P Tension
4
5
3). Given: the truss shown below
H
3m
I
F
G
D
E
B
3m
K
J
C
L
3m
M
3m
A
16 kN
16 kN
N
6 Panels @ 4m = 24 m
Find: Force in members DF,EF, and EG.
FBD - whole structure =>
Ax = 0
Ay = 0
Member DF (enlarged view)
F
FDF
d

D
B
C
A
tan  
FEG
FEF
16 kN
3
4
  36.87 
  M E  0
16(6)  FDF (4 sin 36.87  )  0
FDF  40 kN T
Member EF
Member EG
  M A  0
  M F  0
FEF (8)  FDF (4 sin 36.87  )  0
16(9)  FEG (4 sin 36.87  )  0
FEF (8)  40(4 sin 36.87  )
FEG  60 kN
FEF  12 kN T
Lecture16.doc
FEG  60 kN C
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