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resmat2007a


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a\u2c6ngulo
de torc¸a\u2dco seja de 2, 5 graus para um comprimento de 3 m. Dado G = 84 GPa.
Resposta: (D = 137,5 mm e d = 110,5 mm);
7. No eixo representado na figura 3.32, calcular a tensa\u2dco ma´xima em cada trecho e o
a\u2c6ngulo de torc¸a\u2dco C x A, dados: T1 = 6 KNm, T2 = 8 KNm.
\u2022 AB alum\u131´nio, D1 = 100 mm, G1 = 28 GPa;
\u2022 BC lata\u2dco, D2 = 60 mm, G2 = 35 GPa;
Resposta: (\u3c4AB = 71,3 MPa, \u3c4BC = 141,5 MPa e \u3b8 = 0,1318 rad)
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C
T
B
T
A
0,60m1,0m
2 1
Figura 3.32: Figura do exerc´\u131cio 7
8. No eixo representado na figura 3.33, calcular a tensa\u2dco ma´xima em cada trecho e o
a\u2c6ngulo de torc¸a\u2dco CxA. T1 = 6 KNm, T2 = 9 KNm, G = 84 GPa, D = 100 mm em
AB e D = 76 mm em BC.
Resposta: (\u3c4AB = 15,3 MPa, \u3c4BC = 69,6 MPa e \u3b8 = 0,01163 rad)
T1T2
0,7m
A B C
1,0m
Figura 3.33: Figura do exerc´\u131cio 8
9. O eixo da figura 3.34 tem sec¸a\u2dco circular com 50 mm de dia\u2c6metro, e´ movimentado
pela polia em C a uma rotac¸a\u2dco de 200 rpm e movimenta duas ma´quinas em A (40
CV) e B (25 CV). Calcular a tensa\u2dco ma´xima em cada trecho e o a\u2c6ngulo de torc¸a\u2dco
BxA, dado G = 80 GPa.
Resposta: (\u3c4AC = 57,3 MPa, \u3c4CB = 35,8 MPa e \u3b8 = 0,01611 rad)
10. No exerc´\u131cio 9, qual deveria ser a raza\u2dco entre os dia\u2c6metros D1 em AC e D2 em CB
de modo que a tensa\u2dco ma´xima nos dois trechos seja a mesma. Resposta: (R = 1,17)
60
BCA
1,5m 1,5m
Figura 3.34: Figura do exerc´\u131cio 9
11. Um eixo de ac¸o (veja figura 3.35), dia\u2c6metros D1 = 80 mm em AB e D2 = 60 mm
em BC, esta´ sujeito a dois torques iguais a T nas sec¸o\u2dces B e C. Dado o mo´dulo de
elasticidade transversal de 82 GPa, a tensa\u2dco tangencial admiss´\u131vel de 102 MPa e o
a\u2c6ngulo de torc¸a\u2dco CxA admiss´\u131vel 0, 08 rad, calcular o valor ma´ximo admiss´\u131vel de
T .
Resposta. (T = 3, 913 KNm)
1,0m 1,5m
BA C
T
T
Figura 3.35: Figura do exerc´\u131cio 11
12. Calcular o valor ma´ximo admiss´\u131vel do torque T e os valores correspondentes das
tenso\u2dces ma´ximas e do a\u2c6ngulo de torc¸a\u2dco CxA, dados D = 50 mm em AB e D =
50mm e d = 30 mm em BC, a tensa\u2dco admiss´\u131vel \u3c4 = 80 MPa e o valor de G = 80
GPa.
Resposta: (T = 1,709 KNm, \u3c4AB = 55,7 MPa, \u3c4BC = 80MPa e \u3b8 = 0,001065 rad)
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1,8 T
 T
BA C
60cm90 cm
Figura 3.36: Figura do exerc´\u131cio 12
61
3.2.6 Torc¸a\u2dco em tubos de paredes delgadas
Supondo-se uma barra sujeita a` torc¸a\u2dco tenha sec¸a\u2dco vazada de forma qualquer, com espes-
sura e (constante ou varia´vel. De forma semalhante ao abordado na sec¸a\u2dco 3.28, pode-se
mostrar que as tenso\u2dces cisalhantes sa\u2dco diretamante proporcionais a` dista\u2c6ncia ao centro da
sec¸a\u2dco. Sendo a espessura pequena com relac¸a\u2dco a`s dimenso\u2dces da sec¸a\u2dco, considera-se nestes
casos a tensa\u2dco \u3c4 constante na espessura (podendo variar ao redor da sec¸a\u2dco) conforme
mostra figura 3.37
T
T T
\u3c4
Figura 3.37: Torc¸a\u2dco em tubo de paredes delgadas
Seja um elemento de volume de espessura e1 e e2 e dimenso\u2dces elementares dx (longi-
tudinal) e ds transversal conforme figura 3.38
ba
c
dx
dx
d
a b
c
TT
e11F
e2
F 2
F4 F3
d ds
A
Figura 3.38: Elemento infinitezimal
Sejam \u3c41 e \u3c42 as tenso\u2dces nas faces longitudinais do elemento infinitesimal. Considerando-
se constante estas tenso\u2dces, as correspndentes forc¸as sa\u2dco dadas por:
F1 = \u3c41 e1 dx (3.63)
F2 = \u3c42 e2 dx (3.64)
Obviamente, da condic¸a\u2dco equil´\u131brio escreve-se
F1 = F2 \u21d2 \u3c41 e1 = \u3c42 e2 (3.65)
Como o elemento de volume e´ gene´rico, conclui-se que:
f = \u3c4 e (constante ao redor da sec,a\u2dco) (3.66)
onde f e´ chamado de fluxo de cisalhamento.
Pode-se concluir tambe´m que:
\u2022 e constante \u2192 \u3c4 constante
\u2022 e ma´ximo \u2192 \u3c4 m\u131´nimo
62
\u2022 e m\u131´nimo \u2192 \u3c4 ma´ximo
Fazendo-se o equil´\u131brio de momento com relac¸a\u2dco ao ponto A indicado na figura 3.38
tem-se, admitindo uma variac¸a\u2dco linear da espessura:
\u3c43
(e1 + e2)
2
ds dx = \u3c41 e1 dx ds
\u3c43
(e1 + e2)
2
= f (3.67)
Tomando-se a resultante de forc¸as na face 3 do volume infinitesimal obtem-se
F3 =
f\ufe37 \ufe38\ufe38 \ufe37
\u3c43
(e1 + e2)
2
ds = f ds (3.68)
A equac¸a\u2dco de equil´\u131brio entre forc¸as externas e internas numa sec¸a\u2dco de tubo de paredes
finas, equivalente a` equac¸a\u2dco 3.34 em tubos de sec¸a\u2dco cheia, pode ser obtida fazendo-se o
somato´rio ao longo da linha me´dia da espessura (Lm) dos torques elementar resultantes
(dT = F3) num comprimento ds do so´lido infinitesimal (ver figura 3.39):
r f ds
ds
T
O
Figura 3.39: Equil´\u131\u131brio entre forc¸as internas e externas
T =
\u222b Lm
0
dT
T =
\u222b Lm
0
F3
T =
\u222b Lm
0
r f ds (3.69)
A equac¸a\u2dco pode ser reescrita de forma mais simplificada observando a a´rea me´dia Am
(ver figura 3.39), limitada pela linha me´dia Lm e que o fluxo de cisalhamanto (f) e´ uma
constante na sec¸a\u2dco:
T = f
2Am\ufe37 \ufe38\ufe38 \ufe37\u222b Lm
0
r ds = 2 Am f (3.70)
e observando equac¸a\u2dco 3.66:
\u3c4 =
T
2 e Am
(3.71)
63
A equac¸a\u2dco 3.71 e´ conhecida como primeira fo´rmula de Bredt.
Demonstra-se igualando a energia de deformac¸a\u2dco com o trabalho efetuado pelo torque
T que o angulo de torc¸a\u2dco \u3b8 para um comprimento L de tubo e´:
\u3b8 =
T L
G I
(3.72)
sendo:
I =
4 A2m\u222b Lm
o
ds
e
(3.73)
Para tubos de espessura constante tem-se:
I =
4 A2m e
Lm
(3.74)
e a equac¸a\u2dco 3.72 fica:
\u3b8 =
\u3c4\ufe37 \ufe38\ufe38 \ufe37
T
2 e Am
L Lm
2 Am G
=
\u3c4 L Lm
2 G Am
(3.75)
A equac¸a\u2dco 3.75 e´ conhecida como segunda fo´rmula de Bredt.
3.2.7 Exerc´\u131cios
1. Um tubo de alum\u131´nio (G = 28 GPa) de 1, 0 m de comprimento e sec¸a\u2dco reta\u2c6ngular 60
mm x 100 mm (dimenso\u2dces externas) esta´ sujeito a um torque T = 3 kNm.Determinar
a tensa\u2dco de cisalhamento em cada uma das paredes do tubo e o a\u2c6ngulo de torc¸a\u2dco,
se:
\u2022 a) a espessura e´ constante, igual a 4 mm
\u2022 b)devido a um defeito de fabricac¸a\u2dco duas paredes adjacenetes te\u2c6mespessura 3
mm, e as outras duas te\u2c6m espessura de 5 mm.
Resposta: a) 69, 75 MPa e 0,07044 rad b)93, 0 MPa e 0,07513rad
2. Um tubo circular vazado de espessura 25 mm e dia\u2c6metro interno 225 mm esta´ sujeito
a um torque T = 170, 25 kNm. Calcular as tenso\u2dces ma´xima de cisalhamento no tubo
usando a teoria aproximada da tubos de paredes finas e a teoria exata de torc¸a\u2dco
Resposta 69, 4 MPa e 76, 08 MPa
3. Um tubo fino de sec¸a\u2dco eliptica Esta´ sujeito a um torque T = 5, 67 kNm. Dados
espessura 5 mm, eixo maior = 150 mm, eixo menor = 100 mm e G = 80,5 GPa,
calcular a tensa\u2dco de cisalhamento e o a\u2c6ngulo de torc¸a\u2dco para um comprimento de 1,0
m. Admitindo que o per´\u131metro da el´\u131pse pode ser aproximado por:
P = 1, 5 pi (a+ b)\u2212 pi
\u221a
a b (3.76)
Resposta 52,41 MPa e 0,01147rad
64
4. Calcular o torque ma´ximo admissivel em um tubo de paredes finas de espessura
constante de 1, 5 mm e sec¸a\u2dco representada na figura 3.40 (dimenso\u2dces externas dadas
em mm) para uma tensa\u2dco admissivel ao cisalhamento de 2, 5 MPa.
Resposta 10, 89 Nm
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