04-flexao pura - pt
TRANSCRIPT
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Flexo pura
Traduo e adaptao: Victor Franco
Ref.: Mechanics of Materials, Beer, Johnston & DeWolf McGraw-Hill.Mechanics of Materials, R. Hibbeler, Pearsons Education.
Mecnica dos Materiais
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Flexo pura
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Flexo pura: Viga que no troo CD estsujeita a flexo pura devida aos doisbinrios iguais e opostos que actuam nomesmo plano
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Outros esforos que provocam flexo
Foras excntricas : foras axiais cujalinha de aco no passa no centride daseco produzem foras internas
equivalentes a uma fora axial e ummomento.
Foras transversais : Foras
concentradas ou distribudas transversaisproduzem foras internas equivalentes auma fora de corte e um momento.
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Vigas sujeitas a flexo e esforos transversos
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Flexo pura
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Viga simtrica sujeita a flexo pura
==
== ==
M dA y M
dA z M
dAF
x z
x y
x x
0
0
Foras internas:
As foras internas em qualquer secotransversal so equivalentes a um binrio.O momento desse binrio o momento flector M na seco.
y
z
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Deformaes devidas a flexo
Considere-se uma viga de comprimento L.
Aps deformao, o comprimento da superfcieneutra mantm-se L.
Nas outras seces:
( )
( )
max
max / /
'
c
y
c y
y y L
y y L L
y L
x
x
x
=
=
======
=
linearm.)varia(deform.
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Tenses devidas a flexo
Para um material linear e elstico:
linearm.)varia(tensomax=
==
c y
E c y
E
x
m x x
Para equilibrio esttico:
=
===
dA yc
dAc y
dAF x x
max
max
0
0
O primeiro momento de rea emrelao superficie neutra zero.Assim, a superfcie neutra temnecessariamente de passar no
centride da seco transversal.
Para equilibrio esttico:
I y M
c y
S M
I Mc
c I
dA yc
M
dAc y ydA y M
x
x
x
=
=
==
=
=
==
max
max
max2max
max
doSubstituin
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Deformaes numa seco transversal A deformao devida ao momento flector M
quantificada pela curvatura da superfcie neutra
EI M
I Mc
Ec Ec
c
=
==
=
1
11
1
max
max
Embora as seces transversais se mantenhamplanas quando sujeitas a flexo, as deformaes no
plano so diferentes de zero:
= =
= =
y
y
x z
x y
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Momento de inrcia de uma seco - reviso
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Second moments or moments of inertia of anarea with respect to the x and y axes,
== dA x I dA y I y x 22
Evaluation of the integrals is simplified bychoosing d to be a thin strip parallel to one ofthe coordinate axes.
For a rectangular area,
331
0
22 bhbdy ydA y I h
x ===
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Raio de girao:
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Consider area A with moment of inertia I x.Imagine that the area is concentrated in athin strip parallel to the x axis withequivalent I x.
A
I k Ak I x x x x ==
2
k x = radius of gyration with respect tothe x axis
Similarly,
A J k Ak J
A
I k Ak I
OOOO
y y y y
==
==
2
2
222 y xO k k k +=
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Momentos de inrcia de seces
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Propriedades geomtricas das seces normalizadas
secodaflexoaresistencidemdulo
secodainrciademomento
max
==
=
==
c I
W
I
W M
I c M
f
f
HEA, HEB
IPEUPN IPN
Perfis normalizados
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Exemplo 4.2
Um componente de ferro-fundido sujeito
a um Momento de 3 kN.m. Sabendo que E = 165 GPa e desprezando os raios deconcordancia, calcular: (a) as tensesmximas em traco e em compresso,
(b) o raio de curvatura do componentedeformado.
Resoluo:
Calcular a localizao docentride da seco e omomento de inrcia da seco:
( ) +=
= 2d A I I
A A y
Y x
Calcular as tenses mximas:
I Mc
=max
Calcular a curvatura:
EI M
= 1
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Exemplo 4.2 cont.
Tenses mximas em compresso e em traco:
49
49
max
m10868
m038.0mkN3m10868
m022.0mkN3
==
==
=
I c M
I c M
I Mc
B B
A A
MPa0.76+= A
MPa3.131= B
Curvatura:
( )( )49- m10868GPa165mkN3
1
=
= EI
M
m7.47
m1095.201 1-3
=
=
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As tenses devidas a um carregamento axialdescentrado, calculam-se sobrepondo a tensonormal axial com a distribuio linear causadapelo momento flector
( ) ( )
I My
AP
x x x
=
+= flexoaxialnormal
Carrregamento axial descentrado
Fora descentrada
Pd M
PF
== Esta sobreposio s vlida em regime
elstico e para pequenas deformaes, comefeito desprezvel na geometria.
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Exemplo 4.8
A tenso admissivel para a peaem ferro fundido representadana figura, de 30 MPa em
traco e 120 MPa emcompresso.
Determinar a maior fora P quepode ser aplicada pea.
49
23
m10868
m038.0
m103
=
==
I
Y
A
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Exemplo 4.8 Fora descentrada e momento flector:
flectormomento
adescentradfora
m
====
==
PPd M
P
d
028.0
028.0010.0038.0
Tenses crticas e foras mximas:
kN77MPa1201559kN6.79MPa30377
=====+=
PPPP
B
A
kN0.77=P Mxima fora P admssivel:
Sobreposio das tenses:( )( )
( )( ) PPP I
Mc AP
PPP
I
Mc
AP
A B
A A
155910868
038.0028.0103
37710868
022.0028.0
103
93
93
=
==
+=
+
=+=
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Concentrao de tenses
I c M
K = max
Tenso mxima:
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Unsymmetric Bending
Analysis of pure bending has been limitedto members subjected to bending couplesacting in a plane of symmetry.
Will now consider situations in which thebending couples do not act in a plane ofsymmetry.
In general, the neutral axis of the section willnot coincide with the axis of the couple.
Cannot assume that the member will bendin the plane of the couples.
The neutral axis of the cross sectioncoincides with the axis of the couple
Members remain symmetric and bend inthe plane of symmetry.
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Unsymmetric Bending
Wish to determine the conditions underwhich the neutral axis of a cross section
of arbitrary shape coincides with theaxis of the couple as shown.
couple vector must be directed alonga principal centroidal axis
inertiaof product I dA yz
dAc
y zdA z M
yz
m x y
===
===
0or
0
The resultant force and momentfrom the distribution ofelementary forces in the sectionmust satisfy
coupleapplied M M M F z y x ==== 0
neutral axis passes through centroid
=
===
dA y
dAc ydAF m x x
0or
0
defines stress distribution
inertiaof moment I I c
I
dAc y
y M M
zm
m z
===
==
Mor
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Unsymmetric Bending
Superposition is applied to determine stresses inthe most general case of unsymmetric bending.
Resolve the couple vector into components alongthe principle centroidal axes.
sincos M M M M y z ==
Superpose the component stress distributions
y
y
z
z x I
y M
I y M
+=
Along the neutral axis,( ) ( )
tantan
sincos0
y
z
y z y
y
z
z x
I I
z y
I y M
I y M
I
y M
I y M
==
+=+==
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Example 4.08
Determine the angle of the neutral axis.
143.3
30tanin9844.0
in359.5tantan 4
4
=
== y
z I I
o4.72=
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