Structural Analysis
Method of Sections
If you had everything, where would you keep it?
Review – Method of Joints
All loads are made on the pins/joints
No pin can be analyzed that has more than two
unknowns
We have only two equilibrium conditions: sum
forces in x and y direction
All the forces in the members lie along the
members
2
Method of Sections
Friday, March 19, 2010
Review – Method of Joints
If a member is in compression, it is pushing on
the pin at each end of the member
If a member is in tension, it is pulling on the pin
at each end of the member
A member may carry no force, then it is known
as a zero force member
3
Method of Sections
Friday, March 19, 2010
Method of Sections
4
The second method of truss analysis that
we will consider is called the method of
sections.
The method of sections depends on our
ability to separate the truss into two
separate parts, hence two separate FBDs,
and then perform an analysis on one of the
two parts.
Method of Sections
Friday, March 19, 2010
Method of Sections
5
The method of sections utilizes both force
and moment equilibrium.
The method of sections is often utilized
when we want to know the forces in just a
few members of a complex truss.
We usually divide the truss at the members
we want to know
Method of Sections
Friday, March 19, 2010
Method of Sections
6
Unless we have more information, we will
not separate the truss using more than
three members
The separation must divide the truss into
two completely separate parts
Method of Sections
Friday, March 19, 2010
Method of Sections
• To illustrate the method of sections, we will
analyze the truss shown in following
problem, but this time we will only
determine the forces is members BC, CG,
and GF
7
Method of Sections
Friday, March 19, 2010
Method of Sections
• First we will consider the truss itself,
assuming that we already know the
reactions at A and E
B
C
D
2.0 m
A
G
1.5 m
35 kN
8
F
1.5 m
40 kN
Method of Sections
1.5 m
E
1.5 m
20 kN
25 kN
Friday, March 19, 2010
Method of Sections
• We could use the method of joints to
find the members we are looking for but
the method of sections is lazier
B
C
D
2.0 m
A
G
1.5 m
35 kN
9
F
1.5 m
40 kN
Method of Sections
1.5 m
E
1.5 m
20 kN
25 kN
Friday, March 19, 2010
Method of Sections
• Actually we could have found the
reaction at only one of the supports
since only one of the supports will be in
our final FBD
B
C
D
2.0 m
A
G
1.5 m
10
35 kN
F
1.5 m
Method40
of kN
Sections
1.5 m
E
1.5 m
20 kN
25
kNMarch 19, 2010
Friday,
Method of Sections
• Lets begin by removing members
• Assume that the members are in
compression (your choice)
B
C
D
2.0 m
A
G
1.5 m
11
35 kN
F
1.5 m
Method40
of kN
Sections
1.5 m
E
1.5 m
20 kN
25
kNMarch 19, 2010
Friday,
Method of Sections
• So if we remove BC we have
B
BC C
D
BC
2.0 m
A
G
1.5 m
12
35 kN
F
1.5 m
Method40
of kN
Sections
1.5 m
E
1.5 m
20 kN
25
kNMarch 19, 2010
Friday,
Method of Sections
• Now that isn’t two forces BC, it is the
compressive force in member BC acting
on each pin it is connected to
B
BC C
D
BC
2.0 m
A
G
1.5 m
13
35 kN
F
1.5 m
Method40
of kN
Sections
1.5 m
E
1.5 m
20 kN
25
kNMarch 19, 2010
Friday,
Method of Sections
• Now we do the same thing with member
CG
B
BC C
D
BC
CG
2.0 m
CG
A
G
1.5 m
14
35 kN
F
1.5 m
Method40
of kN
Sections
1.5 m
E
1.5 m
20 kN
25
kNMarch 19, 2010
Friday,
Method of Sections
• And finally with member FG
B
BC C
D
BC
CG
2.0 m
CG
FG
A
G
1.5 m
15
35 kN
FG
F
1.5 m
Method40
of kN
Sections
1.5 m
E
1.5 m
20 kN
25
kNMarch 19, 2010
Friday,
Method of Sections
• Notice that we have effectively
separated the truss into two parts or
sections
B
BC C
D
BC
CG
2.0 m
CG
FG
A
G
1.5 m
16
35 kN
FG
F
1.5 m
Method40
of kN
Sections
1.5 m
E
1.5 m
20 kN
25
kNMarch 19, 2010
Friday,
Method of Sections
• Each of the two parts of the truss must
still be in equilibrium so we can choose
either side and make our calculations
B
BC C
D
BC
CG
2.0 m
CG
FG
A
G
1.5 m
17
35 kN
FG
F
1.5 m
Method40
of kN
Sections
1.5 m
E
1.5 m
20 kN
25
kNMarch 19, 2010
Friday,
Method of Sections
• We will choose the left side
• For the remainder of this analysis, we
can ignore the right side of the truss
B
BC C
D
BC
CG
2.0 m
CG
FG
A
G
1.5 m
18
35 kN
FG
F
1.5 m
Method40
of kN
Sections
1.5 m
E
1.5 m
20 kN
25
kNMarch 19, 2010
Friday,
Method of Sections
• This is the FBD of the section that we
will now work with
B
BC
2.0 m
CG
FG
A
G
1.5 m
19
35 kN
Method40
of kN
Sections
Friday, March 19, 2010
Method of Sections
• Notice that anything that is not directly
connected to the side we are using is
not included in the FBD
B
BC
2.0 m
CG
FG
A
G
1.5 m
20
35 kN
Method40
of kN
Sections
Friday, March 19, 2010
Method of Sections
• We can now use our equilibrium
calculation to solve for the three cut
members
B
BC
2.0 m
CG
FG
A
G
1.5 m
21
35 kN
Method40
of kN
Sections
Friday, March 19, 2010
Method of Sections
• Since we are not looking at a point we
can use all three equations of
equilibrium
B
BC
2.0 m
CG
FG
A
G
1.5 m
22
35 kN
Method40
of kN
Sections
Friday, March 19, 2010
Method of Sections
• We can start by taking the moments
about point G and solving for BC
∑M
B
G
=0
BC
− (1.5m )( 35kN ) + ( 2.0m )( BC ) = 0
2.0 m
CG
FG
A
G
1.5 m
35 kN
23
1.5m )( 35kN )
(
BC =
= 26.25kN
2.0m
BC = 26.25kN C
40 kN
Method of Sections
Friday, March 19, 2010
Method of Sections
• We also could have chosen to sum the
moments about C to solve for FG
B
BC
2.0 m
CG
FG
A
G
1.5 m
35 kN
24
40 kN
Method of Sections
Friday, March 19, 2010
Method of Sections
• The point we choose to sum moments
about doesn’t have to be on the FBD but
it should eliminate two of the unknowns
B
BC
2.0 m
CG
FG
A
G
1.5 m
35 kN
25
40 kN
Method of Sections
Friday, March 19, 2010
Method of Sections
• Now we can sum forces in the ydirection and solve for CG
• CG is on a 3-4-5 triangle
B
26.25 kN
2.0 m
CG
FG
A
G
1.5 m
35 kN
26
40 kN
∑F
y
=0
4
35kN − 40kN − CG = 0
5
5
CG = − 5kN = −6.25kN
4
CG = 6.25kN T
Method of Sections
Friday, March 19, 2010
Method of Sections
• Finally, we can sum the forces in the xdirection and solve for FG
• Notice I have redrawn the FBD using my
new information about CG
B
26.25 kN
2.0 m
6.25 kN
FG
A
G
1.5 m
27
35 kN
40 kN
∑F
x
=0
3
−26.25kN + 6.25kN − FG = 0
5
3
FG = 6.25kN − 26.25kN = −22.5kN
5
FG
=of22.5
T
Method
SectionskN
Friday, March 19, 2010
An Example Problem
• Problem 6-33
28
Method of Sections
Friday, March 19, 2010