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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