ME 221 Statics
Lecture #29
Sections 6.1 – 6.4
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Homework #10
Chapter 7 problems:
– 5, 8, 19, 21, 24, 26 & 35
• Due Wednesday, November 13
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Homework #11
Chapter 6 problems:
– 2, 3, 6 & 7 – Method of Joints
– 32, 36, 47 & 53 – Method of Sections
– 6.47: E-G should be 0.2m
• Due Friday, November 22
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Quiz #6
Today
Exam #3
Friday, November 15
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Chapter 6: Analysis of Structures
• The knowledge of internal forces is a
fundamental requirement for true design
• The Structural elements are:
Trusses
Frames
Machines
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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.
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Frames
Are designed to support,
prevent or transmit loads
(forces, no moments). At
least one member is not a
two-force member.
Machines
Are designed to transmit force, motion or energy
(forces and moments). Will always have at least one
member with multiple forces.
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Planar Trusses
If the planar truss (2-D) is statically
determinant, then the number of joints is
related to the number of reactions and
internal members by :
2j=m+r
j = number of joints
m = number of internal members
r = number of reactions
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Analysis of Trusses Using the Method of Joints
PBy
PBx
B
We need to solve for:
(1) - Internal forces
FAB, FAC, and FCA
(2) - Reactions
Ax, Ay and Cy
PAY
PAx
Ax
a
A
Ay
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Lecture 29
PCy
b
PCx
C
C
y
9
Joint A:

FAC  FAC iˆ

FAB  FAB (cos aiˆ  sin aˆj )

PA  PAxiˆ  PAy ˆj

A  Ax iˆ  Ay ˆj




FAB  FAC  PA  A  0
PAY
PAx
Ax
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a
A
FAC
Ay
……………….(1)
FAB cosa+FAC+Ax=-Pax
FAB sina+Ay=-Pay
FAB
……………………(2)
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At Joint C:

FCA   FAC iˆ

FCB  FCB ( cos biˆ  sin bˆj )

PC  PCx iˆ  PCy ˆj

C  C y ˆj




FCB  FCA  PC  C  0
FCB
FCA
PCy
b
PCx
C
 FAC  FCB cos b   Pcx ..........(3)
C
FCB sin b  C y   Pcy ................(4)
y
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At Joint B:

FBA   FAB (cos aiˆ  sin aˆj )

FBC   FCB ( cos biˆ  sin bˆj )

PB  PBxiˆ  PBy ˆj



FBC  FBA  PB  0
PBy
PBx
B
FBA
FBC
a
b
 FAB cos a  FCB cos b   PBx ..........(5)
 FAB sin a  FCB sin b   PBy ................(6)
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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
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0 0

1 0
0 0

0 0
0 0

0 1
 PAx 
 P 
 Ay 
  PBx 



P
 By 
  PCx 


 PCy 
13
Example
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Example
6 kN
A
B
3 kN
0.9 m
C
D
E
1.2 m
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1.2 m
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6 kN
A
4
B
4
9 5
15
3 kN
C
16
9
15
16
5
4
4
E
D
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