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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) \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd 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) \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd \ufffd\ufffd 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 \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd