analysis-of-structure.ppt

VECTOR MECHANICS FOR ENGINEERS:
STATICS
Ninth Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
Lecture Notes:
J. Walt Oler
Texas Tech University
CHAPTER
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
6 Analysis of Structures

Vector Mechanics for Engineers: Statics
Ninth
Edition
Contents
6 - 2
Introduction
Definition of a Truss
Simple Trusses
Analysis of Trusses by the Method
of Joints
Joints Under Special Loading
Conditions
Space Trusses
Sample Problem 6.1
Analysis of Trusses by the Method
of Sections
Trusses Made of Several Simple
Trusses
Sample Problem 6.3
Analysis of Frames
Frames Which Cease to be Rigid
When Detached From Their
Supports
Sample Problem 6.4
Machines

Ninth
Edition
Introduction
6 - 3
• For the equilibrium of structures made of several
connected parts, the internal forces as well the external
forces are considered.
• In the interaction between connected parts, Newton’s 3rd
Law states that the forces of action and reaction
between bodies in contact have the same magnitude,
same line of action, and opposite sense.
• Three categories of engineering structures are considered:
a) Frames: contain at least one one multi-force
member, i.e., member acted upon by 3 or more
forces.
b) Trusses: formed from two-force members, i.e.,
straight members with end point connections
c) Machines: structures containing moving parts
designed to transmit and modify forces.

Ninth
Edition
6 - 4
• A truss consists of straight members connected at
joints. No member is continuous through a joint.
• Bolted or welded connections are assumed to be
pinned together. Forces acting at the member ends
reduce to a single force and no couple. Only two-
force members are considered.
• Most structures are made of several trusses joined
together to form a space framework. Each truss
carries those loads which act in its plane and may
be treated as a two-dimensional structure.
• When forces tend to pull the member apart, it is in
tension. When the forces tend to compress the
member, it is in compression.

Ninth
Edition
6 - 5
Members of a truss are slender and not capable of
supporting large lateral loads. Loads must be applied at
the joints.

Ninth
Edition
6 - 6

Ninth
Edition
Simple Trusses
6 - 7
• A rigid truss will not collapse under
the application of a load.
• A simple truss is constructed by
successively adding two members and
one connection to the basic triangular
truss.
• In a simple truss, m = 2n - 3 where
m is the total number of members
and n is the number of joints.

Ninth
Edition
Analysis of Trusses by the Method of Joints
6 - 8
• Dismember the truss and create a freebody
diagram for each member and pin.
• The two forces exerted on each member are
equal, have the same line of action, and
opposite sense.
• Forces exerted by a member on the pins or
joints at its ends are directed along the member
and equal and opposite.
• Conditions of equilibrium on the pins provide
2n equations for 2n unknowns. For a simple
truss, 2n = m + 3. May solve for m member
forces and 3 reaction forces at the supports.
• Conditions for equilibrium for the entire truss
provide 3 additional equations which are not
independent of the pin equations.

Ninth
Edition
Joints Under Special Loading Conditions
6 - 9
• Forces in opposite members intersecting in
two straight lines at a joint are equal.
• The forces in two opposite members are
equal when a load is aligned with a third
member. The third member force is equal
to the load (including zero load).
• The forces in two members connected at a
joint are equal if the members are aligned
and zero otherwise.
• Recognition of joints under special loading
conditions simplifies a truss analysis.

Ninth
Edition
Space Trusses
6 - 10
• An elementary space truss consists of 6 members
connected at 4 joints to form a tetrahedron.
• A simple space truss is formed and can be
extended when 3 new members and 1 joint are
added at the same time.
• Equilibrium for the entire truss provides 6
additional equations which are not independent of
the joint equations.
• In a simple space truss, m = 3n - 6 where m is the
number of members and n is the number of joints.
• Conditions of equilibrium for the joints provide 3n
equations. For a simple truss, 3n = m + 6 and the
equations can be solved for m member forces and
6 support reactions.

Ninth
Edition
Sample Problem 6.1
6 - 11
Using the method of joints, determine
the force in each member of the truss.
SOLUTION:
• Based on a free-body diagram of the
entire truss, solve the 3 equilibrium
equations for the reactions at E and C.
• Joint A is subjected to only two unknown
member forces. Determine these from the
joint equilibrium requirements.
• In succession, determine unknown
member forces at joints D, B, and E from
• All member forces and support reactions
are known at joint C. However, the joint
equilibrium requirements may be applied
to check the results.

Ninth
Edition
Sample Problem 6.1
6 - 12
SOLUTION:
• Based on a free-body diagram of the entire truss,
solve the 3 equilibrium equations for the reactions
at E and C.
       
m
3
m
6
kN
5
m
12
kN
10
0
E
MC






 kN
50
E
 
 x
x C
F 0 0

x
C
 



 y
y C
F kN
50
kN
5
-
kN
10
0

 kN
35
y
C

Ninth
Edition
Sample Problem 6.1
6 - 13
• Joint A is subjected to only two unknown
member forces. Determine these from the
5
3
4
kN
10 AD
AB F
F

 C
F
T
F
AD
AB
kN
5
.
12
kN
5
.
7


• There are now only two unknown member
forces at joint D.
  DA
DE
DA
DB
F
F
F
F
5
3
2


C
F
T
F
DE
DB
kN
15
kN
5
.
12



Ninth
Edition
Sample Problem 6.1
6 - 14
• There are now only two unknown member
forces at joint B. Assume both are in tension.
 
kN
75
.
18
kN
12
kN
5
0 5
4
5
4








BE
BE
y
F
F
F
C
FBE kN
75
.
18

   
kN
25
.
26
75
.
18
kN
5
.
12
kN
5
.
7
0 5
3
5
3








BC
BC
x
F
F
F
T
FBC kN
25
.
26

• There is one unknown member force at joint
E. Assume the member is in tension.
 
kN
75
.
43
kN
75
.
18
kN
15
0 5
3
5
3







EC
EC
x
F
F
F
C
FEC kN
75
.
43


Ninth
Edition
Sample Problem 6.1
6 - 15
• All member forces and support reactions are
known at joint C. However, the joint equilibrium
requirements may be applied to check the results.
   
   
checks
0
75
.
43
35
checks
0
75
.
43
25
.
26
5
4
5
3










y
x
F
F

Ninth
Edition
Analysis of Trusses by the Method of Sections
6 - 16
• When the force in only one member or the
forces in a very few members are desired, the
method of sections works well.
• To determine the force in member BD, pass a
section through the truss as shown and create
a free body diagram for the left side.
• With only three members cut by the section,
the equations for static equilibrium may be
applied to determine the unknown member
forces, including FBD.

Ninth
Edition
Trusses Made of Several Simple Trusses
6 - 17
• Compound trusses are statically
determinant, rigid, and completely
constrained.
3
2 
 n
m
• Truss contains a redundant member
and is statically indeterminate.
3
2 
 n
m
• Necessary but insufficient condition
for a compound truss to be statically
determinant, rigid, and completely
constrained,
n
r
m 2


non-rigid rigid
3
2 
 n
m
• Additional reaction forces may be
necessary for a rigid truss.
4
2 
 n
m

Ninth
Edition
Sample Problem 6.3
6 - 18
Determine the force in members FH,
GH, and GI.
SOLUTION:
• Take the entire truss as a free body.
Apply the conditions for static equilib-
rium to solve for the reactions at A and L.
• Pass a section through members FH,
GH, and GI and take the right-hand
section as a free body.
• Apply the conditions for static
equilibrium to determine the desired
member forces.

Ninth
Edition
Sample Problem 6.3
6 - 19
SOLUTION:
• Take the entire truss as a free body.
Apply the conditions for static equilib-
rium to solve for the reactions at A and L.
        
       



















kN
5
.
12
kN
20
0
kN
5
.
7
m
25
kN
1
m
25
kN
1
m
20
kN
6
m
15
kN
6
m
10
kN
6
m
5
0
A
A
L
F
L
L
M
y
A

Ninth
Edition
Sample Problem 6.3
6 - 20
• Pass a section through members FH, GH, and GI
and take the right-hand section as a free body.
       
kN
13
.
13
0
m
33
.
5
m
5
kN
1
m
10
kN
7.50
0







GI
GI
H
F
F
M
• Apply the conditions for static equilibrium to
determine the desired member forces.
T
FGI kN
13
.
13


Ninth
Edition
Sample Problem 6.3
6 - 21
        
  
kN
82
.
13
0
m
8
cos
m
5
kN
1
m
10
kN
1
m
15
kN
7.5
0
07
.
28
5333
.
0
m
15
m
8
tan













FH
FH
G
F
F
M
GL
FG



C
FFH kN
82
.
13

 
        
kN
371
.
1
0
m
10
cos
m
5
kN
1
m
10
kN
1
0
15
.
43
9375
.
0
m
8
m
5
tan
3
2












GH
GH
L
F
F
M
HI
GI



C
FGH kN
371
.
1


Ninth
Edition
Analysis of Frames
6 - 22
• Frames and machines are structures with at least one
multiforce member. Frames are designed to support loads
and are usually stationary. Machines contain moving parts
and are designed to transmit and modify forces.
• A free body diagram of the complete frame is used to
determine the external forces acting on the frame.
• Internal forces are determined by dismembering the frame
and creating free-body diagrams for each component.
• Forces between connected components are equal, have the
same line of action, and opposite sense.
• Forces on two force members have known lines of action
but unknown magnitude and sense.
• Forces on multiforce members have unknown magnitude
and line of action. They must be represented with two
unknown components.

Ninth
Edition
Frames Which Cease To Be Rigid When Detached From Their Supports
6 - 23
• Some frames may collapse if removed from
their supports. Such frames can not be treated
as rigid bodies.
• A free-body diagram of the complete frame
indicates four unknown force components which
can not be determined from the three equilibrium
conditions.
• The frame must be considered as two distinct, but
related, rigid bodies.
• With equal and opposite reactions at the contact
point between members, the two free-body
diagrams indicate 6 unknown force components.
• Equilibrium requirements for the two rigid
bodies yield 6 independent equations.

Ninth
Edition
Sample Problem 6.4
6 - 24
Members ACE and BCD are
connected by a pin at C and by the
link DE. For the loading shown,
determine the force in link DE and the
components of the force exerted at C
on member BCD.
SOLUTION:
• Create a free-body diagram for the
complete frame and solve for the support
reactions.
• Define a free-body diagram for member
BCD. The force exerted by the link DE
has a known line of action but unknown
magnitude. It is determined by summing
moments about C.
• With the force on the link DE known, the
sum of forces in the x and y directions
may be used to find the force
components at C.
• With member ACE as a free-body,
check the solution by summing
moments about A.

Ninth
Edition
Sample Problem 6.4
6 - 25
SOLUTION:
• Create a free-body diagram for the complete frame
and solve for the support reactions.
N
480
0 


 y
y A
F 
 N
480
y
A
    
mm
160
mm
100
N
480
0 B
M A 





 N
300
B
x
x A
B
F 


 0 

 N
300
x
A


 
07
.
28
tan 150
80
1

Note:

Ninth
Edition
Sample Problem 6.4
6 - 26
• Define a free-body diagram for member
BCD. The force exerted by the link DE has a
known line of action but unknown
magnitude. It is determined by summing
moments about C.
        
N
561
mm
100
N
480
mm
0
6
N
300
mm
250
sin
0







DE
DE
C
F
F
M 
C
FDE N
561

• Sum of forces in the x and y directions may be used to find the force
components at C.
  N
300
cos
N
561
0
N
300
cos
0











x
DE
x
x
C
F
C
F
N
795


x
C
  N
480
sin
N
561
0
N
480
sin
0











y
DE
y
y
C
F
C
F
N
216

y
C

Ninth
Edition
Sample Problem 6.4
6 - 27
• With member ACE as a free-body, check
the solution by summing moments about A.
       
         0
mm
220
795
mm
100
sin
561
mm
300
cos
561
mm
220
mm
100
sin
mm
300
cos














 x
DE
DE
A C
F
F
M
(checks)

Ninth
Edition
Machines
6 - 28
• Machines are structures designed to transmit
and modify forces. Their main purpose is to
transform input forces into output forces.
• Given the magnitude of P, determine the
magnitude of Q.
• Create a free-body diagram of the complete
machine, including the reaction that the wire
exerts.
• The machine is a nonrigid structure. Use
one of the components as a free-body.
• Taking moments about A,
P
b
a
Q
bQ
aP
M A 



 0

analysis-of-structure.ppt

More Related Content

What's hot (20)

Similar to analysis-of-structure.ppt (20)

Recently uploaded (20)

analysis-of-structure.ppt