chapter06-analysis of structures
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VECTOR MECHANICS FOR ENGINEERS:
STATICS
Tenth Editionin SI Units
Ferdinand P. Beer
E. Russell Johnston, Jr.David F. Mazurek
CHAPTER
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
6 Analysis of Structures
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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 Methodof Sections
Trusses Made of Several SimpleTrusses
Sample Problem 6.3
Analysis of a Frame
Frames Which Cease to be RigidWhen Detached From Their
Supports
Sample Problem 6.4
Machines
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Application
6 - 3
Design of support structures requiresknowing the loads, or forces, that each
member of the structure will
experience.
Functional elements, such as the holding
force of this pliers, can be determined
from concepts in this section.
T
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Introduction
6 - 4
• 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) Trusses : formed from two-force members, i.e.,
straight members with end point connections and
forces that act only at these end points.b) Frames : contain at least one multi-force member,
i.e., member acted upon by 3 or more forces.
c) Machines : structures containing moving parts
designed to transmit and modify forces.
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Definition of a Truss
6 - 5
• 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 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.
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Definition of a Truss
6 - 6
•
Members of a truss are slender and not capable ofsupporting large lateral loads. Loads must be applied
at the joints.
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Definition of a Truss
6 - 7 T I
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Simple Trusses
6 - 8
• 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 triangulartruss.
f ST I n
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Analysis of Trusses by the Method of Joints
6 - 9
• Dismember the truss and create a free body
diagram for each member and pin.
• The two forces exerted on each member are
equal, have the same line of action, andopposite 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 are used to solve for
2 unknown forces at each pin (or joint), giving a
total of 2n solutions, where n=number of joints.
Forces are found by solving for unknown forces
while moving from joint to joint sequentially.
• Conditions for equilibrium for the entire truss
can be used to solve for 3 support reactions.
V t M h i f E i St tiT I n
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Joints Under Special Loading Conditions
6 - 10
• 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.
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Space Trusses
6 - 11
• 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 thenumber 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 and6 support reactions.
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Sample Problem 6.1
6 - 12
Using the method of joints, determine
the force in each member of the truss.
SOLUTION:
• What’s the first step to solving this
problem? Think, then discuss this with a
neighbor.
• DRAW THE FREE BODY DIAGRAMFOR THE ENTIRE TRUSS (always first)
and solve for the 3 support reactions
• Draw this FBD and compare your sketch
with a neighbor. Discuss with each other
any differences.
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Sample Problem 6.1
6 - 13
SOLUTION:
• Based on a free body diagram of the entire truss,solve the 3 equilibrium equations for the reactions
at E and C .
• Looking at the FBD, which “sum of moments”
equation could you apply in order to find one of
the unknown reactions with just this one equation?
• Next, apply the remaining
equilibrium conditions to
find the remaining 2 support
reactions.
m3m6kN5m12kN10
0
E
M C
kN50 E
x x C F 0 0 xC
y y C F kN50kN5-kN100
kN35 yC
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Sample Problem 6.1
6 - 14
DA DE
DA DB
F F
F F
5
32
• We now solve the problem by moving
sequentially from joint to joint and solving
the associated FBD for the unknown forces.
• Which joint should you start with, and why?
Think, then discuss with a neighbor.
• Joints A or C are equally good because each
has only 2 unknown forces. Use joint A and
draw its FBD and find the unknown forces.
• Which joint should you move to next, and why? Discuss.• Joint D, since it has 2 unknowns remaining
(joint B has 3). Draw the FBD and solve.
534
kN10 AD AB
F F
C . F
T . F
AD
AB
kN512
kN57
C F
T . F
DE
DB
kN15
kN512
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Sample Problem 6.1
6 - 15
• There are now only two unknown member
forces at joint B. Assume both are in tension.
• There is one remaining unknown member
force at joint E (or C ). Use joint E andassume the member is in tension.
kN7518
kN512kN505
4
5
4
. F
F . F
BE
BE y
C . F BE kN7518
kN2526
kN7518kN512kN5705
3
5
3
. F
... F F
BC
BC x
T . F BC kN2526
kN7543
kN7518kN1505
3
5
3
. F
. F F
EC
EC x
C . F EC kN7543
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Sample Problem 6.1
6 - 16
• All member forces and support reactions are
known at joint C . However, the joint equilibriumrequirements may be applied to check the results.
checks 0kN7543kN35
checks 0kN7543kN2526
5
4
5
3
. F
.. F
y
x
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Analysis of Trusses by the Method of Sections
6 - 17
• When the force in only one member or the
forces in a very few members are desired, themethod of sections works well.
• To determine the force in member BD, form a
section by “cutting” the truss at n-n 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 F BD.
• A FBD could have been created for the right
side, but why is this a less desirable choice?
Think and discuss.
• Notice that the exposed internal forces
are all assumed to be in tension.
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Analysis of Trusses by the Method of Sections
6 - 18
• Using the left-side FBD, write one
equilibrium equation that can be solved tofind F BD. Check your equation with a
neighbor; resolve any differences between
your answers if you can.
• So, for example, this cut with line p-p is
acceptable.
• Assume that the initial section cut was made
using line k-k . Why would this be a poor
choice? Think, then discuss with a neighbor.
• Notice that any cut may be chosen, solong as the cut creates a separated section.
k
k
p p
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Trusses Made of Several Simple Trusses
6 - 19
• Compound trusses are statically
determinate, rigid, and completelyconstrained.
32 nm
• Truss contains a redundant member
and is statically indeterminate. 32 nm
• Necessary but insufficient condition
for a compound truss to be statically
determinate, rigid, and completely
constrained,
nr m 2
non-rigid rigid
32 nm
• Additional reaction forces may be
necessary for a rigid truss.
42 nm
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Sample Problem 6.3
6 - 20
Determine the force in members FH ,
GH , and GI .
SOLUTION:
• List the steps for solving this problem.
Discuss your list with a neighbor.
1. Draw the FBD for the entire truss.
Apply the equilibrium conditions andsolve for the reactions at A and L.
3. Apply the conditions for static
equilibrium to determine the desired
member forces.
2. Make a cut through members FH ,
GH , and GI and take the right-hand
section as a free body (the left side
would also be good).
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Sample Problem 6.3
6 - 21
SOLUTION:
• Take the entire truss as a free body.
Apply the conditions for static
equilibrium to solve for the reactions at A
and L.
x x
y
y y
A
A F
. A
A L F
. L
L
M
0
kN512
kN200
kN57
m25kN1m25kN1m20
kN6m15kN6m10kN6m50
LAy
Ax
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Sample Problem 6.3
6 - 22
• Make a cut through members FH , GH , and GI and take the right-hand section as a free body.
Draw this FBD.
kN1313
0m335m5kN1m10kN7.50
0
. F
. F
M
GI
GI
H
• Sum of the moments about point H :
T F GI kN13.13
• What is the one equilibrium equation that could be solved to find FGI ? Confirm your answer with
a neighbor.
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Sample Problem 6.3
6 - 23
kN8113
0m8cos
m5kN1m10kN1m15kN7.5
0
072853330 m15
m8tan
. F
F
M
..GL
FG
FH
FH
G
C . F FH kN8113
kN3711
0m15cosm5kN1m10kN1
0
154393750 m8
m5tan
3
2
. F
F
M
.. HI GI
GH
GH
L
C F GH kN371.1
• F FH is shown as its components. What one
equilibrium equation will determine F FH
?
• There are many options for finding FGH at this
point (e.g., S F x=0, S F y=0). Here is one more:
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Analysis of a Frame
6 - 24
• Frames and machines are structures with at least one
multiforce (>2 forces) member. Frames are designed to
support loads and are usually stationary. Machines contain
moving parts and 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 magnitudeand line of action. They must be represented with two
unknown components.
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Frames Which Cease To Be Rigid When Detached From Their Supports
6 - 25
• Some frames may collapse if removed from
their supports. Such frames cannot be treated asrigid bodies.
• A free-body diagram of the complete frame
indicates four unknown force components which
cannot be determined from the three equilibrium
conditions (statically indeterminate).
• 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-bodydiagrams show 6 unknown force components.
• Equilibrium requirements for the two rigid bodies
yield 6 independent equations. Thus, taking the
frame apart made the problem solvable.
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Sample Problem 6.4
6 - 26
Members ACE and BCD are connected
by a pin at C and by the link DE . For
the loading shown, determine the forcein link DE and the components of the
force exerted at C on member BCD.
SOLUTION:
1. Create a free body diagram for thecomplete frame and solve for the
support reactions.
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Sample Problem 6.4
6 - 27
SOLUTION:
1. Create a free-body diagram for the completeframe and solve for the support reactions.
N4800 y y A F N480 y A
mm160mm100 N4800 B M A
N300 B
N300
0
x
x x
A
A B F
N300 x A
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Sample Problem 6.4
6 - 28
SOLUTION (cont.):
2. Create a free body diagram for member BCD (since the problem asked for forces on this
body). Choose the best FBD, then discuss your
choice with a neighbor. Justify your choice.
FDE,x
FDE,y
FDE
FDE
FDE,x
FDE,y
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Sample Problem 6.4
6 - 29
N561
mm100 N480mm06 N300mm250sin0
DE
DE C
F
F M
C F DE N561
• Sum of forces in the x and y directions may be used to find the force
components at C .
N300cos N5610
N300cos0
x
DE x xC
F C F N795 xC
N480sin N5610
N480sin0
y
DE y y
C
F C F
N216 yC
SOLUTION (cont.):
3. Using the best FBD for member BCD, what isthe one equilibrium equation that can directly
find F DE ? Please discuss.
07.28tan150
801
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Sample Problem 6.4
6 - 30
• With member ACE as a free body with no
additional unknown forces, check the
solution by summing moments about A.
0mm220795mm100sin561mm300cos561
mm220mm100sinmm300cos
x DE DE A C F F M
(checks)
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Machines
6 - 31
• Machines are structures designed to transmit
and modify forces. Typically they transform
input forces (P) into output forces (Q).
• Given the magnitude of P, determine the
magnitude of Q.
• Create a free-body diagram of the completemachine, including the reaction that the wire
exerts.
• The machine is a nonrigid structure. Use
one of the components as a free-body.
Discuss why the forces at A are such.
• Sum moments about A,
P b
aQbQaP M A 0
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Tutorials
Chapter 2
•
2.19 2.38 (2.F4) 2.67 2.70 2.121Chapter 3
• 3.4 3.9 3.10 3.49 3.97 3.101 3.114 3.148
Chapter 4
• (4.F1) (4.F4) 4.3 4.8 4.10 4.14 4.15 4.19 4.45
• (4.F7) 4.103 4.105 4.113 4.121
Chapter 5
• (Fig P5.1) (Fig P5.7) (Fig P5.15)
5.20 5.21 5.102 5.103 5.104
Chapter 6
• (Fig P6.2) (Fig P6.5)
Chapter 7• 7.29 7.35 7.36 7.37 7.58
Chapter 8
• (8.F1) (8.F2) (8.F3) (8.F4)
• 8.1 8.2 8.6 8.15 8.19