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resmat2007a


DisciplinaResistência dos Materiais II4.494 materiais116.339 seguidores
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sendo os demais esforc¸os simples nulos.
\u2022 Barras de eixo reto e sec¸a\u2dco transversal circular (cheia) ou anular (coroa circular)
conforme figura 3.27. Barras com estas caracter´\u131sticas sa\u2dco comumente denominadas
de eixos
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd
D = 2R d = 2r D = 2R
Figura 3.27: Sec¸a\u2dco circular e anular
\u2022 Eixos sujeitos a` momento torsor constante conforme figura 3.28.
=
T
T
A B
T
+ DMT
BABA
T
Figura 3.28: Eixo sujeito a` torsor constante
\u2022 Pequenas deformac¸o\u2dces: as sec¸o\u2dces permanecem planas e perpendiculares ao eixo,
com forma e dimenso\u2dces conservadas. As deformac¸o\u2dces sa\u2dco deslocamentos angulares
(a\u2c6ngulos de torc¸a\u2dco), em torno do eixo-x (eixo da barra), de uma sec¸a\u2dco em relac¸a\u2dco a
outra.
O momento torsor, conforme estudado no item 3, esta´ associado a`s tenso\u2dces cisalhantes
\u3c4xy e \u3c4xz. A equac¸a\u2dco 3.16, que confirma esta afirmac¸a\u2dco, e´ reescrita abaixo para facilitar o
trabalho do leitor.
T =
\u222b
A
(z\u3c4xy \u2212 y\u3c4xz) dA (3.34)
Analisando um ponto P (z, y) gene´rico e contido numa sec¸a\u2dco transversal de um eixo
conforme figura 3.29, e´ poss´\u131vel transformar a equac¸a\u2dco 3.34 numa forma mais compacta.
Chamando de \u3c4 a soma vetorial entre \u3c4xy e \u3c4xz e observando figura 3.29 tem-se:
~\u3c4 = ~\u3c4xy + ~\u3c4xz (3.35)
z = \u3c1 cos\u3c6 (3.36)
y = \u3c1 sin\u3c6 (3.37)
\u3c4xy = \u3c4 cos\u3c6 (3.38)
\u3c4xz = \u2212\u3c4 sin\u3c6 (3.39)
55
\u3c4xz
\u3c4xy
z
y
\u3c4
\u3c6\u3c1
\u3c6
z
y
Figura 3.29: Tenso\u2dces cisalhantes na torc¸a\u2dco
Substituindo as equac¸o\u2dces 3.35 a 3.39 na equac¸a\u2dco 3.34 tem-se:
T =
\u222b
A
(\u3c1 cos\u3c6\u3c4 cos\u3c6+ \u3c1 sin\u3c6\u3c4 sin\u3c6) dA
T =
\u222b
A
\u3c1\u3c4(cos2 \u3c6+ sin2 \u3c6) dA
T =
\u222b
A
\u3c1\u3c4 dA (3.40)
A equac¸a\u2dco 3.40 pode ser compreendida como a equac¸a\u2dco 3.34 em coordenadas polares.
Assim, as coordenadas que definem a posic¸a\u2dco do ponto gene´rico P podem ser escritas
como \u3c1 e \u3c6. O pro´ximo passo desta ana´lise e´ definir uma relac¸a\u2dco entre \u3c4 e a coordenada
(\u3c1, \u3c6) do ponto gene´rico P , ou simplesmente: \u3c4 = \u3c4(\u3c1, \u3c6).
3.2.2 Ana´lise de Tenso\u2dces e deformac¸o\u2dces na torc¸a\u2dco
Sejam:
\u2022 \u3b3 a distorc¸a\u2dco angular do \u201creta\u2c6ngulo\u201d abcd, contido em uma superf´\u131cie cil´\u131ndrica de
raio \u3c1 e comprimento dx conforme figura 3.30.
\u2022 d\u3b8 o deslocamento angular (a\u2c6ngulo de torc¸a\u2dco) elementar da sec¸a\u2dco Sd em relac¸a\u2dco a`
sec¸a\u2dco Se conforme figura 3.30.
Da figura 3.30 pode-se escrever:
bb\u2032 = \u3c1d\u3b8 (3.41)
bb\u2032 = \u3b3dx (3.42)
Igualando as equac¸o\u2dces 3.41 e 3.42 tem-se:
\u3b3 = \u3c1
d\u3b8
dx
(3.43)
Da Lei de Hooke tem-se:
56
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd d\u3b8
d\u3b8
2
L
A B
S Se d
d
a b
c
\u3c1 2\u3c1
dx
b
d
a
S Se d
\u3b3
\u3b3
c\u2019x
dx
b\u2019
c
Figura 3.30: Ana´lise das deformac¸o\u2dces na torc¸a\u2dco
\u3c4 = G\u3b3 (3.44)
lembrando que G e´ o mo´dulo de elasticidade transversal.
Substituindo o valor de \u3b3 da equac¸a\u2dco 3.43 na equac¸a\u2dco 3.44 tem-se:
\u3c4 = \u3c1 G
d\u3b8
dx
(3.45)
Como \u3b8 varia linearmente com x (ver figura 3.30), sua derivada com relac¸a\u2dco a x e´
constante e pode-se dizer que:
G
d\u3b8
dx
= constante = K (3.46)
Pode-se concluir enta\u2dco que \u3c4 e´ func¸a\u2dco somente de \u3c1, na\u2dco e´ func¸a\u2dco de \u3c6 (\u3c4 = K\u3c1),
portanto constante em pontos de mesmo \u3c1 ( 0 \u2264 \u3c1 \u2264 R ), para qualquer \u3c6 ( 0 \u2264 \u3c6 \u2264 2pi
) . A variac¸a\u2dco de \u3c4 com \u3c1 e´ linear, conforme mostra a figura 3.31.
\u3c1
o
T T
\u3c4max
Figura 3.31: Variac¸a\u2dco da tensa\u2dco cisalhante em func¸a\u2dco de \u3c1
Para calcular a constante K basta substituir \u3c4 = K\u3c1 na equac¸a\u2dco 3.40:
T =
\u222b
A
\u3c1\u3c4 dA =
\u222b
A
\u3c1K\u3c1 dA = (K
\u222b
A
\u3c12 dA\ufe38 \ufe37\ufe37 \ufe38
Momento de ine´rcia polar: Io
) = K.I0 (3.47)
Logo:
K =
T
Io
(3.48)
e:
\u3c4 =
T
Io
\u3c1 (3.49)
57
A tensa\u2dco cisalhante \u3c4max ma´xima se da´ \u3c1 = R:
\u3c4max =
T
Io
R (3.50)
A raza\u2dco entre Io e R (Wo) e´ chamada de mo´dulo de resiste\u2c6ncia a` torc¸a\u2dco. Enta\u2dco:
\u3c4max =
T
Wo
(3.51)
Da Meca\u2c6nica Geral, o valor de Io para uma sec¸a\u2dco circular e´:
Io =
pi
32
D4(sec,a\u2dco circular) (3.52)
e para sec¸a\u2dco anular, sendo D o dia\u2c6metro de eixo temos:
Io =
pi
32
(D4e \u2212D4i ) =
pi
32
D4e(1\u2212 n4)(sec,a\u2dco anular) (3.53)
para anular sendo De o dia\u2c6metro externo, Di o dia\u2c6metro interno do eixo e n = Di/De
Substituindo os valores de R = D/2 (sec¸a\u2dco circular), R = De/2(sec¸a\u2dco anular) e de Io
das equac¸o\u2dces 3.52 e 3.53, pode-se chegar facilmente a:
\u3c4max =
16T
piD3
(sec,a\u2dco circular) (3.54)
\u3c4max =
16T
piD3
(
1
1\u2212 n4 ) (sec,a\u2dco anular) (3.55)
3.2.3 Ca´lculo do a\u2c6ngulo de torc¸a\u2dco
O a\u2c6ngulo de torc¸a\u2dco (rotac¸a\u2dco relativa) entre duas sec¸o\u2dces distantes de L unidades de com-
primento e´:
\u3b8 =
\u222b L
0
d\u3b8 =
\u222b L
0
\u3b3
\u3c1
dx\ufe38 \ufe37\ufe37 \ufe38
ver eq. 3.43
=
\u222b L
0
Lei de Hooke\ufe37\ufe38\ufe38\ufe37
\u3c4
G
1
\u3c1
dx (3.56)
Substituindo o valor de \u3c4 (equac¸a\u2dco 3.49) a equac¸a\u2dco 3.56 pode ser reescrita como:
\u3b8 =
\u222b L
0
T
Io
\u3c1\ufe38\ufe37\ufe37\ufe38
eq.3.49
1
G \u3c1
dx
\u3b8 =
T L
G Io
(3.57)
3.2.4 Torque Aplicado ao eixo na Transmissa\u2dco de Pote\u2c6ncia
Se um eixo transmite uma pote\u2c6ncia P a uma velocidade angular \u3c9, enta\u2dco ele esta´ sujeito
a um torque (momento de torc¸a\u2dco): T = P/\u3c9.
Justificativa: O trabalho executado pelo momento torsor T , constante, e´:
58
W = T\u3c6 (3.58)
dW = Td\u3c6 (3.59)
(3.60)
onde \u3c6 e´ o deslocamento angular, em radianos. Como pote\u2c6ncia e´ trabalho por unidade
de tempo:
P =
dW
dt
= T
d\u3c6
dt
= T\u3c9 (3.61)
Unidades no SI:
\u2022 Pote\u2c6ncia (P ): watt (1W = 1 Nm/s).
\u2022 Velocidade angular \u3c9 = 2pif : rad/s.
\u2022 Frequ¨e\u2c6ncia f : hertz = Hz
\u2022 Torque (T): Nm.
Se a pote\u2c6ncia for expressa em cavalos-vapor (CV) ou horse-power (hp), enta\u2dco os fatores
de conversa\u2dco para W sa\u2dco, respectivamente:
1 CV = 736 W e 1 hp = 746 W (3.62)
3.2.5 Exerc´\u131cios
1. Dimensionar o eixo de uma ma´quina, de 9 m de comprimento, que transmite 200
CV de pote\u2c6ncia, dados \u3c4 = 21 MPa e G = 85 GPa a uma frequ¨e\u2c6ncia de 120 rpm, e
calcular o correspondente deslocamento angular, adotando:
\u2022 Sec¸a\u2dco circular cheia. Resposta: (D = 142 mm, \u3b8 = 0, 03107 rad);
\u2022 Sec¸a\u2dco anular com d/D = 0,5.
Resposta: (D = 145 mm, \u3b8 = 0, 03048 rad);
2. Calcular o momento de torque ma´ximo admiss´\u131vel e o correspondente a\u2c6ngulo de
torc¸a\u2dco em um eixo de comprimento de 2 m dados \u3c4adm = 80 MPa e G = 85 GPa e
sec¸a\u2dco:
\u2022 Circular, D = 250 mm; Resposta: (T = 245,4 KNm e \u3b8 = 0,01506 rad);
\u2022 Anular, com d = 150 mm e D = 250 mm; Resposta: (T = 213,4 KNm e \u3b8 =
0,01504 rad);
3. Um eixo de ac¸o, sec¸a\u2dco circular com D = 60 mm, gira a uma frequ¨e\u2c6ncia de 250
rpm. Determine a pote\u2c6ncia (em CV) que ele pode transmitir, dado \u3c4 = 80 MPa.
Resposta: (P =120,7 CV)
4. Dimensionar um eixo de sec¸a\u2dco circular que transmite a pote\u2c6ncia de 1800 CV a uma
rotac¸a\u2dco de 250 rpm, para uma tensa\u2dco admiss´\u131vel ao cisalhamento de 85 MPa e para
um a\u2c6ngulo de rotac¸a\u2dco de 1 grau para um comprimento igual a 20 vezes o dia\u2c6metro.
Dado o mo´dulo de elasticidade transversal de 80 GPa. Resposta: (D = 195 mm)
59
5. Determine a raza\u2dco entre os pesos P1 e P2 (por unidade de comprimento) de dois
eixos de mesmo material e sujeitos a um mesmo torque, sendo o eixo-1 de sec¸a\u2dco
circular cheia e o eixo-2 de sec¸a\u2dco anular com d/D = 0,75. Resposta: (P1/P2 =
1,7737)
6. Calcular os dia\u2c6metros externo e interno de um eixo de ac¸o sujeito a um torque de
25 KNm, de modo que a tensa\u2dco ma´xima de cisalhamento seja 84 MPa e o