ME 221 Statics
Lecture #29
Sections 6.6 – 6.7
ME221
Lecture #29
1
Homework #10
Chapter 7 problems:
– 2, 5, 6, 8, 19, 21, 24, 26 & 35
• Due Today
ME221
Lecture #29
2
Homework #11
Chapter 6 problems:
–
–
–
–
ME221
2, 3, 6 & 7 – Method of Joints
32, 36, 47 & 53 – Method of Sections
68 & 75
Due Wednesday, November 19
Lecture #29
3
Exam #3
Wednesday, November 12
Details on Monday
ME221
Lecture #29
4
Trusses
Composed of slender straight pieces connected
together by frictionless pins where all the loads (no
moments) are applied.
Each member will act as a two-force member
(either in tension or compression).
All the forces acting on a truss member are axial.
ME221
Lecture #29
5
Analysis of Trusses Using the Method of Joints
PBy
PBx
B
We need to solve for:
(1) - Internal forces
PAY
FAB, FAC, and FCA
(2) - Reactions
Ax, Ay and Cy
PAx
Ax
a
A
Ay
ME221
Lecture #29
PCy
b
PCx
C
C
y
6
Using Matrix Notation
1
0
1
 FAB   cos a
 F   sin a
0
0
0
AC

 
cos b 0
 FBC   cos a 0

 =   sin a 0  sin b 0
A
 x  
 1  cos b 0
 Ay   0

  0
0
sin
b
0
C

 y 
Using manual calculations
Look for joints with 2 unknowns
ME221
Lecture #29
-1
0 0

1 0
0 0

0 0
0 0

0 1
 PAx 
 P 
 Ay 
  PBx 



P
 By 
  PCx 


 PCy 
7
Example
6 kN
A
B
3 kN
0.9 m
C
D
E
1.2 m
ME221
1.2 m
Lecture #29
8
6 kN
A
4
B
4
9 5
15
3 kN
C
16
15
16
9
5
4
4
E
D
ME221
Lecture #29
9
Method of Sectioning
If the question is to find internal forces in selected
members of the truss, then one can alternatively
use the method of sectioning.
Example: Determine the force in members FG and FH
12 kN
A
C
12 kN
E
G
I
1.8 m
B
D
F
H
J
4 @ 2.4 m=9.6 m
ME221
Lecture #29
10
12 kN
A
C
12 kN
E
G
FGE
I
1.8 m
FGF
B
D
FHF
F
12 kN
A
C
H
J
12 kN
E
G
FEG
I
FFG
B
D
 Fx
ME221
0
F
 Fy
0
Lecture #29
FFH
1.8 m
H
 M Anypo int
J
0
11