Truss Analysis
Method of Sections
ENGR 1310
Truss Analysis
+
y
The Method of Joints will always work to analyze a truss,
but it’s not always the easiest method.
x
P  300 N
B
10 m
C

h  5m
A
R3
R1
E
10 m
10 m
D
G
F
FBC
10 m
C
FCE
R2

FFC
FCG
Slide 2
y
Truss Analysis
Method of Sections
P  300 N
+
x
10 m
B
C
E

R3  0
A
10 m
10 m
10 m
R1  50 N
• Determine External Reactions
using Newton’s 1st Law:
x
D
G
F
F
h  5m
R 2  250 N
 R3  0
 F  R  R  300N  0
 M  (300N  5m)  ( R  30m)  0
y
1
D
2
1
Slide 3
y
Truss Analysis
Method of Sections
+
x
P  300 N
B
10 m
C
E

A
R3
R1  50 N
10 m
10 m
F
10 m
h  5m
D
G
R2  250 N
The first step is to cut a section through the member whose force you want to
obtain and through two other members.
Slide 4
y
Truss Analysis
Method of Sections
+
x
P  300 N
C
B
E

A
10 m
10 m
F
50 N
h  5m
D
G
250 N
Slide 5
+
y
Truss Analysis
Method of Sections
B
x
FBC
FFC
h  5m
A
FFG
10 m
F
M
M
M
A
50 N
  FBC (5m)  FFCy(10m)  0
B
 50 N (5m)  FFCx(5m)  FFCy(5m)  FFG (10m)  0
F
  FBC (5m)  50 N (10m)  0
FBC  100 N
We assumed Tension,  since the sign is negative,
FBC  100 N (C )
Slide 6
y
Truss Analysis
Method of Sections
+
x
B
FBC  100 N (C )
FFC
h  5m
A
FFG
10 m
F
F
 0  50 N  FFCy
FFCy  50 N  50 N (C )
FFCx  FFCy  50 N (C )
FFC  50 2 N (C )
Y
50 N
F
X
 0 FBC  FFCx  FFG
 100 N  50 N  FFG
 FFG  150 N (T )
Slide 7
Truss Analysis
Method of Sections
The other internal forces in the other members can be solved by cutting sections in
those members, setting up the proper equations of equilibrium for the sum of forces
in the X-direction, the sum of forces in the Y-direction, and the sum of moments.
300 N
B
50 2 N
(C )
50 2 N
(C )
50 2 N
(T )
A
50 N (T )
50 N
C
100 N (C )
E
200 N (C )
50 2 N
(C )
250 2 N
(C )
50 2 N
(T )
F
150 N (T )
D
G
250 N (T )
250 N
Slide 8