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Ninth EditionNinth Edition
VECTOR MECHANICS FOR ENGINEERS:
STATICSSTATICSCHAPTER
Ferdinand P. BeerFerdinand P. Beer
E. Russell Johnston Jr.E. Russell Johnston Jr.
Lecture Notes:Lecture Notes:
Analysis of Structures:
..
Texas Tech UniversityTexas Tech University Trusses
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Introduction For the equilibrium of structures made of several
connected parts, the internal forces as well the external
orces are considered.
In the interaction between connected parts, Newtons 3rd
Law states that theforces of action and reaction
between bodies in contact have the same magnitude,
same line of action, and opposite sense.
a) Frames: contain at least one one multi-force
member, i.e., member acted upon by 3 or more
.
b) Trusses: formed from two-force members, i.e.,
straight members with end point connections
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Definition of a Truss A truss consists of straight members connected at
joints. No member is continuous through a joint.
Most structures are made of several trusses joinedtogether to form a space framework. Each truss
Bolted or welded connections are assumed to be
be treated as a two-dimensional structure.
pinned together. Forces acting at the member ends
reduce to a single force and no couple. Only two-
force members are considered.
When forces tend to pull the member apart, it is in
tension. When the forces tend to compress the
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member, it is in compression.
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Definition of a Truss
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Analysis of Trusses by the Method of J oints Dismember the truss and create a freebodydiagram for each member and pin.
e wo orces exer e on eac mem er 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 on the pins provide
2n equations for 2n unknowns. For a simple
=, .
forces and 3 reaction forces at the supports.
Conditions for equilibrium for the entire truss
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provide 3 additional equations which are not
independent of the pin equations.
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J oints Under Special Loading Conditions Forces in opposite members intersecting intwo straight lines at a joint are equal.
equal when a load is aligned with a thirdmember. The third member force is equal
to the load includin zero load .
The forces in two members connected at a
joint are equal if the members are alignedand zero otherwise.
Recognition of joints under special loading
conditions simplifies a truss analysis.
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Sample Problem 6.1Using the method of joints, determine the force
in each member of the truss.
SOLUTION:
Based on a free-body diagram of the entire
0M =
truss, solve the 3 equilibrium equations for the
reactions atEandC.
( )( ) ( )( ) ( )ft6ft12lb1000ft24lb2000 E+=
= lb000,10E
== xx CF 0 0=xC
++==F lb10 000lb1000-lb20000
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= lb7000yC
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Sample Problem 6.1
member forces. Determine these from the
joint equilibrium requirements.
534
lb2000 ADAB FF ==CF
TF
AD
AB
lb2500
lb1500
=
=
There are now only two unknown memberforces at joint D.
=
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2=
CFDE
DB
lb3000
=
=
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Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth
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Sample Problem 6.1 There are now only two unknown member
forces at joint B. Assume both are in tension.
( )lb3750
250010000 54
5
4
=
==
BE
BEy
FFF
CFBE lb3750=
( ) ( )
lb5250
375025001500053
53
+=
==
BC
BCx
F
FF
TFBC lb5250=
There is one unknown member force at joint
E. Assume the member is in tension.
( )
lb8750
37503000053
53
=
++==
EC
ECx
F
FF
CFEC lb8750=
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Sample Problem 6.1
All member forces and support reactions are
known at joint C. However, the joint equilibrium
re uirements ma be a lied to check the results.
( ) ( )checks087505250
4
53
==
=+= xF
5y
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Analysis of Trusses by the Method of Sections
When the force in only one member or the
forces in a very few members are desired, the
.
To determine the force in memberBD, pass a
section throu h 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.
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Sample Problem 6.3
SOLUTION:
Take the entire truss as a free body.
Apply the conditions for static equilib-
rium to solve for the reactions atA andL.
Pass a section throu h members FH
GH, andGIand take the right-hand
section as a free body.
pp y e con ons or s a c
equilibrium to determine the desired
member forces.
Determine the force in members FH,GH, andGI.
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Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth
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Sample Problem 6.3
SOLUTION:
Take the entire truss as a free bod .
Apply the conditions for static equilib-rium to solve for the reactions atA andL.
== kN6m15kN6m10kN6m50M
( )( ) ( )( ) ( )
=
+
kN5.7
m25kN1m25kN1m20
L
L
=++==
kN5.12
kN200
A
ALFy
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Sample Problem 6.3
Pass a section through members FH, GH, andGI
and take the right-hand section as a free body.
Apply the conditions for static equilibrium to
determine the desired member forces.
( )( ) ( )( ) ( )
kN13.13
0m33.5m5kN1m10kN7.50
0
+=
=
=
GI
H
F
F
M
TFGI kN13.13=
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Sample Problem 6.3
0
07.285333.0m15
m8tan
=
====
GMGL
FG
( )( ) ( )( ) ( )( )( )( ) 0m8cosm5kN1m10kN1m15kN7.5
=+
FHF
.=FHCFFH kN82.13=
( )15.439375.0
m8
m5tan
32
====HI
GI
( )( ) ( )( ) ( )( ) 0m10cosm5kN1m10kN1
0
=++
=GH
L
F
M
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.=GHCFGH kN371.1=