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PREGUNTAS DE REPASO 21.1. ¿Cuáles son las tres categorías básicas de procesos de remoción de material? Como organizada en este texto, las tres categorías básicas de los procesos de eliminación de material son (1) el mecanizado convencional, (2) los procesos abrasivos y (3) los procesos no tradicionales. tradicionales .
21.2. ¿En qué se distingue el maquinado de otros procesos de manufactura?
21.3. Identifique algunas de las razones por la que el maquinado es comercial y tecnológicamente tecnológicamente importante . La razón incluir lo siguiente: (1) es aplicable a la mayoría de los materiales; (2) que puede producir una variedad de geometrías de una parte: (3) que puede alcanzar tolerancias más que la mayoría de los procesos, y (4) que puede crear buenos acabados superficiales .
21.4. Mencione los tres procesos de maquinado más comunes . Los tres procesos de maquinado frecuentes son (1) girando, (2) de perforación, y (3) la molienda.. molienda
21.5. ¿Cuáles son las dos categorías básicas de herramientas de corte en maquinado? Dé dos ejemplos de operaciones de maquinado que use cada uno de los tipos de herramientas. Las dos categorías son (1) las herramientas de un solo punto, que se utilizan en operaciones tales como encender y aburrido, y (2) borde múltiple herramientas, utilizadas en operaciones como el fresado y taladrado de corte.
21.6. Identifique los parámetros de una operación de maquinado que se incluyen en el conjunto de las condiciones de corte. Condición de corte incluyen velocidad, avance, profundidad de corte, y abrirle ella o no se utiliza un fluido de corte
21.7. Defina la diferencia entre las operaciones de desbaste primario y las de acabado en maquinado. Una operación de desbaste se utiliza para eliminar grandes cantidades de material rápidamente y para producir una geometría de la pieza cerca de la forma deseada. Una operación de acabado sigue desbaste y se utiliza para conseguir la geometría y acabado de la superficie final.
21.8. ¿Qué es una máquina herramienta? herramienta?
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Una máquina herramienta se puede definir como una máquina de potencia del controlador que las posiciones y se mueve con relación a una herramienta para llevar a cabo el trabajo de mecanizado u otro proceso de conformación de metales.
21.9. ¿Qué es una operación de corte ortogonal ? De corte ortogonal implica el uso de una herramienta en forma de cuña en la que el borde de corte es perpendicular a la dirección de movimiento de velocidad en el material de trabajo.
21.10. ¿Por qué es útil el modelo de corte ortogonal en el análisis del maquinado metálico? De corte ortogonal es útil en el análisis de mecanizado de metales, ya que simplifica el bastante complejo tridimensional situación de mecanizado a dos dimensiones. Además, el utillaje en el modelo ortogonal tiene sólo dos parámetros (ángulo de ataque y ángulo de alivio). Lo cual es una geometría simple que una herramienta de un solo punto.
21.11. Mencione y describa brevemente los cuatro tipos de viruta que se producen en el corte de metales. Los cuatro tipos son (I) discontinua, en la que los chips se forma en segmentos separados; (2) continua, en el que el chip no hace segmento y está formado de un metal dúctil; (3) construida borde de la pizca continua, que es el mismo que (2), excepto que la fricción en la interfaz de la herramienta chip provoca la adhesión de una pequeña porción de materi al de trabajo a la cara de la herramienta rastrillo, y (4) dentadas, que son semi-continua en el sentido de que poseen una apariencia de sierra diente que se produce por una formación de viruta cíclica de alterna de alta tensión de cizallamiento, seguido por la cepa de baja cizalladura.
21.12. Identifique Identif ique las cuatro fuerzas que actúan ac túan sobre la viruta en el modelo de corte metálico ortogonal, pero que no pueden medirse directamente en una operación. Las cuatro fuerzas que actúan sobre el chip son (1) la fuerza de fricción, (2) la fuerza normal a la fricción, (3) la fuerza de corte, y (4) la l a fuerza normal a la fricción. fri cción.
21.15. Describa con palabras qué dice la ecuación de Merchant. La ecuación comerciante establece que el ángulo de plano de corte aumenta cuando se aumenta ángulo de ataque y ángulo de fricción se reduce.
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La potencia requerida en una operación de corte es igual a la fuerza de corte multiplicado por la velocidad de corte. corte. De trabajo.
21.17. ¿Qué es la energía específica en el maquinado de metales? La energía específica es la cantidad de energía requerida para eliminar una unidad de volumen del material
21.18. ¿Qué significa el término efecto de tamaño tamaño en el corte de metales? El efecto de tamaño se refiere al hecho de que los aumentos de energía específicos como la sección transversal son del chip (x w t0 en el corte ortogonal o fxd en giro) disminuye.
21.19. ¿Qué es un termopar herramienta-viruta? herramienta-viruta? Un termopar de chip de herramientas se comprenden del chip herramienta como los dos metales diferentes que forman la unión de termopar: como el - chip de interfaz se calienta durante el corte, un pequeño voltaje se emite desde la unión que se puede medir para indicar la temperatura de corte.
CUESTIONARIO DE OPCIÓN MÚLTIPLE (v) azul 21.1. ¿Cuál de los procesos de manufactura siguientes se clasifica como procesos de remoción de material? (dos respuestas correctas ): e) molido, f ) maquinado. 21.2. ¿La máquina herramienta “torno” se utiliza para realizar cuál siguientes operaciones de manufactura?: manufactura?: e) torneado.
de las
21.3. ¿Con cuál de las formas geométricas siguientes está la operación de taladrado más íntimamente relacionada?: c) agujero redondo 21.4. Si las condiciones condicione s de corte en una operación de torneado son velocidad de corte = 300 ft/min, avance = 0.010 in/rev y profundidad de corte = 0.100 in, ¿cuál de las siguientes es la tasa de remoción de material?: d ) 3.6 in3/min. 21.5. ¿Una operación de desbaste primario involucra generalmente a cuál
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21.6. ¿Cuáles de las siguientes son las características del modelo de corte ortogonal? (tres respuestas mejores): d ) solamente dos dimensiones juegan un papel activo en el análisis, f ) el filo del corte es perpendicular a la dirección de la velocidad del corte y g) los dos elementos de la forma de la herramienta son los ángulos de inclinación y de relieve. 21.7. ¿Cuál de las siguientes es la relación de espesor de viruta?: b) to/tc, donde tc = espesor de la viruta después del corte, to = espesor de la viruta antes del corte, f = avance, d = profundidad y w = ancho del corte. 21.8. ¿Cuál de los cuatro tipos de viruta se podría esperar en una operación de torneado conducida a baja velocidad de corte sobre un material de trabajo frágil? c) discontinua. 21.9. De acuerdo con la ecuación de Merchant, ¿cuál de los siguientes resultados podría tener un incremento en el ángulo de inclinación, si los otros factores permanecen igual (dos mejores respuestas): b) disminución de los requerimientos de potencia, e) incremento en el ángulo del plano de corte? 21.10. Al usar el modelo de corte ortogonal para aproximar una operación de torneado, ¿el espesor de la viruta antes del corte to corresponde a cuál de los siguientes condiciones del torneado? b) avance f 21.11. ¿Cuál de los siguientes metales podría tener generalmente los caballos de fuerza unitarios más bajos en una operación de maquinado? a) aluminio 21.12. ¿Para cuál de los siguientes valores de espesor de viruta antes del corte to esperaría usted que fuera más grande la energía específica? c) 0.12 mm 21.13. ¿Cuál de las siguientes condiciones de corte tiene un efecto mayor en la temperatura de corte? b) ¿velocidad?
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y el corte produce un espesor de viruta deformada de 0.65 mm. Calcule a) el ángulo del plano de corte y b) la deformación cortante para la operación. Solución: (a) r = t o / t c = 0.30/0.65 = 0.4615 φ = tan-1(0.4615 cos 15/(1 - 0.4615 sen 15)) = tan-1(0.5062) = 26.85° (b) deformación de corte γ = cot 26.85 + tan (26.85 - 15) = 1.975 + 0.210 = 2.185
21.2. En el problema 21.1, suponga que el ángulo de inclinación cambiara a a = 0°. Suponiendo que el ángulo de fricción permaneciera igual, determine a) el ángulo plano de corte, b) el espesor de la viruta y c) la deformación cortante cortante para la operación. Solución: del Problema 21.1, ecuación. ecuación. (21.16):
= 15 y = 26.85 . Usando la ecuación Mercante, la
= 45 + /2 - /2; reordenand reordenando, o, = 2(45) + - 2 = 2(45) + α - 2( ) = 90 + 15 – 2(26.85) = 51.3 Ahora, con = 0 y restante en el mismo 51.3 mismo 51.3 , = 45 + 0/2 – 51.3/2 = 19.35 (b) Grosor de la viruta en = 0: t c c = = t o/tan = 0.30/tan 19.35 = 0.854 mm (c) deformación de corte = cot 19.35 + tan (19.35 - 0) = 2.848 + 0.351 = 3.199
21.3. En una operación de corte ortogonal, la herramienta de 0.250 in de ancho tiene un ángulo de inclinación de 5º. El torno se configura para que el espesor de la viruta antes del corte sea de 0.010 in. Después del corte, el espesor de la viruta deformada se mide y tiene un valor de 0.027 in. Calcule a) el ángulo del plano de corte y b) la deformación cortante para la operación. Solución: (a) r = = t o/t c c = = 0.010/0.027 = 0.3701 -1 = tan (0.3701 cos 5/(1 - 0.3701 sin 5)) = tan -1(0.3813) = 20.9 (b) deformación de corte = cot 20.9 + tan (20.9 – 5) = 2.623 + 0.284 = 2.907
21.4. En una operación de torneado, la velocidad de la aguja se configura para proporcionar una velocidad de corte de 1.8 m/s. El avance y profundidad del corte son 0.30 mm y 2.6 mm, respectivamente. El ángulo
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Utilice el modelo de corte ortogonal como una aproximación del proceso de torneado. Solución: (a) r = = t o/t c c = = 0.30/0.49 = 0.612 -1 = tan (0.612 cos 8/(1 – 0.612 sin 8)) = tan -1(0.6628) = 33.6 (b) = cot 33.6 + tan (33.6 - 8) = 1.509 + 0.478 = 1.987 (c) RMR = (1.8 m/s x 10 3 mm/m)(0.3)(2.6) = 1404 mm3/s
21.5. La fuerza de corte y la fuerza de empuje en una operación de corte ortogonal son 1 470 N y 1 589 N, respectivamente. El ángulo de inclinación es de 5°, el ancho del corte es de 5.0 mm, el espesor de la viruta antes del corte es de 0.6 y la relación de espesor de la viruta es de 0.38. Determine a) la resistencia cortante del material de trabajo y b) el coeficiente de fricción en la operación. Solución: (a) = tan-1(0.38 cos 5/(1 - 0.38 sin 5)) = tan -1(0.3916) = 21.38 F s = 1470 cos 21.38 – 1589 sin 21.38 = 789.3 N As = (0.6)(5.0)/sin 21.38 = 3.0/.3646 = 8.23 mm 2 S = 789.3/8.23 = 95.9 N/mm 2 = 95.9 MPa (b) = 45 + /2 - /2; reordenando, = 2(45) + = 2(45) + α - 2( ) = 90 + 5 – 2(21.38) = 52.24 = tan 52.24 = 1.291
-2
21.6. La fuerza de corte y la fuerza de empuje se han medido en una operación de corte ortogonal y son de 300 lb y 291 lb, respectivamente. El ángulo de inclinación es de 10º, el ancho de corte de 0.200 in, el espesor de la viruta antes del corte de 0.015 y la relación de espesor de la viruta de 0.4. Determine a) la resistencia al corte del material de trabajo y b) el coeficiente de fricción de la operación. Solución: = tan-1(0.4 cos 10/(1 - 0.4 sin 10)) = tan -1(0.4233) = 22.94 F s = 300 cos 22.94 - 291sin 22.94 = 162.9 lb. As = (0.015)(0.2)/sin 22.94 = 0.0077 in 2 S = 162.9/0.007 162.9/0.0077 7 = 21,167 lb/in2 = 2(45) + α - 2( ) = 90 + 10 - 2(22.94) = 54.1 = tan 54.1 = 1.38
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del corte es de 0.55. Determine a) el espesor de la viruta después del corte, b) el ángulo de corte, c) el ángulo de fricción, d ) el coeficiente de fricción y e) la deformación cortante. Solución: (a) r = = t o/t c c, t c c = = t o/r = = 0.012/0.55 = 0.022 in (b) = tan-1(0.55 cos 15/(1 - 0.55 sin 15)) = tan -1(0.6194) = 31.8 (c) = 2(45) + α - 2( ) = 90 + 15 - 2(31.8) = 41.5 (d) = tan 41.5 = 0.88 (e) = cot 31.8 + tan(31.8 - 15) = 1.615 + 0.301 = 1.92
21.8. La operación de corte ortogonal descrita en el problema 21.7 involucra un material de trabajo cuya resistencia al corte es de 40 000 lb/in2. Con base en sus respuestas al problema anterior, calcule a) la fuerza cortante, b) la fuerza de corte, c) la fuerza de empuje y d ) la fuerza de fricción. Solución: (a) As = (0.012)(0.10)/sin 31.8 = 0.00228 in 2. F s = AsS = 0.0028(40,000) = 91.2 lb (b) F c c = = 91.2 cos (41.5 - 15)/cos (31.8 + 41.5 -15) = 155 lb (c) F t t = = 91.2 sin (41.5 - 15)/cos (31.8 + 41.5 -15) = 77.2 lb (d) F = = 155 sin 15 - 77.2 cos 15 = 115 lb
21.9. En una operación de corte ortogonal, el ángulo de in clinación es de – 5º, el espesor de la viruta antes del corte es de 0.2 mm y el ancho del corte es de 4.0 mm. La relación de viruta es de 0.4. Determine a) el espesor de la viruta después del corte, b) el ángulo de corte, c) el ángulo de fricción, d ) el coeficiente de fricción y e) la deformación cortante. Solución: (a) r = = t /t t = t /r = 0.2/.4 = 0.5 mm
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fuerza cortante, c) la fuerza de corte y la fuerza de empuje y d ) la fuerza de fricción. Solución: (a) = tan-1(0.5 cos 20/(1 - 0.5 sin 20)) = tan -1(0.5668) = 29.5 (b) As = (0.015)(0.15)/sin 29.5 = 0.00456 in 2. F s = AsS = 0.00456(50,000) = 228 lb (c) = 2(45) + α - 2( ) = 90 + 20 - 2(29.5) = 50.9 = 228 cos (50.9 - 20)/cos (29.5 + 50.9 -20) = 397 lb F c c = F t t = = 228 sin (50.9 - 20)/cos (29.5 + 50.9 -20) = 238 lb (d) F = = 397 sin 20 - 238 cos 20 = 359 lb
21.11. Repite el problema 21.10 excepto porque el ángulo de inclinación se modificó a – 5° 5° y la relación de espesor de la viruta resultante es de 0.35. Solución: (a) = tan-1(0.35 cos( –5)/(1 - 0.35 sin(-5))) = tan -1(0.3384) = 18.7 (b) As = (0.015)(0.15)/sin 18.7 = 0.00702 in 2. F s = AsS = 0.00702(50,000) = 351 lb (c) = 2(45) + α - 2( ) = 90 + (-5) - 2(18.7) = 47.6 F c c = = 351 cos(47.6 - (-5))/cos(18.7 + 47.6 - (-5)) = 665 lb F t t = = 351 sin(47.6 - (-5))/cos(18.7 + 47.6 - (-5)) = 870 lb (d) F = = 665 sin (-5) - 870 cos (-5) = 808 lb
21.12. Una barra de acero de carbono de 7.64 in de diámetro tiene una resistencia a la tensión de 65 000 lb/in2 y una resistencia al corte de 45 000 lb/in2. El diámetro se reduce utilizando una operación de torneado a una velocidad de corte de 400 ft/min. El avance es de 0.011 in/rev y la profundidad de corte es de 0.120 in. El ángulo de inclinación de la herramienta en la dirección del flujo de la viruta es de 13°. Las condiciones de corte dan como resultado una relación de viruta de 0.52. Utilizando el
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F t t = = F ssin (β – α)/cos ( + β – α) F t t = = 264.1 sin (43.3 - 13)/cos (29.8 + 43.3 - 13) = 121 lb (d) μ = tan β = tan 43.3 = 0.942
21.13. Acero al bajo carbono con una resistencia a la tensión de 300 MPa y una resistencia al corte de 220 MPa se corta en una operación de torneado con una velocidad de corte de 3.0 m/s. El avance es de 0.20 mm/rev y la profundidad del corte es de 3.0 mm. El ángulo de inclinación de la herramienta es de 5º en la dirección del flujo de la viruta. La relación de viruta resultante es de 0.45. Utilizando el modelo ortogonal como una aproximación aproximación al torneado, determine a) el ángulo del plano de corte, b) la fuerza de corte, c) la fuerza cortante y la fuerza de avance. Solución: (a) = tan-1(0.45 cos 5/(1 - 0.45 sin 5)) = tan -1(0.4666) = 25.0 (b) As = t ow /sin /sin = (0.2)(3.0)/sin 25.0 = 1.42 mm 2. F s = AsS = 1.42(220) = 312 N (c) = 2(45) + α - 2( ) = 90 + 5 - 2(25.0) = 45.0 = F scos (β – α)/cos ( + β – α) F c c = = 312 cos(45 - 5)/cos(25.0 + 45.0 - 5) = 566 N F c c = = F ssin(β – α)/cos( + β – α) F t t = = 312 sin(45 - 5)/cos(25.0 + 45.0 - 5) = 474 N F t t =
21.14. Una operación de torneado se hace con un ángulo de inclinación de 10º, un avance de 0.010 in/rev y una profundidad de corte de 0.100 in. Se sabe que la resistencia al corte del material de trabajo es de 50 000 lb/in2 y la relación de espesor de la viruta medida después del corte es de 0.40. Determine la fuerza de corte y la fuerza del avance. Use el modelo
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Solución: Comience con la definición de la relación de chips, la ecuación. (21.2): r = = t o/t c c = sin /cos ( - ) Reordenando, r cos cos ( - ) = sin Usando la identidad trigonométrica cos( - ) = cos cos + sin sin r (cos cos + sin sin ) = sin Dividiendo ambos lados por sin , obtenemos r cos /tan + r sin = 1 sin r cos /tan = 1 - r sin Reorganizar, tan = r cos /(1 - r sin ) Q.E.D.
21.16. Demuestre cómo la ecuación 21.4 se deduce a partir de la figura 21.6. Solución: En la figura, figura, = AC /BD = ( AD AD + DC )/ )/BD = AD/BD + DC /BD AD/BD = cot y DC /BD = tan ( - ) asi, = cot + tan ( - ) Q.E.D.
21.17. Deduzca las ecuaciones de fuerza para F, N, FS y Fn (ecuaciones 21.9 a 21.12 en el texto), utilizando el diagrama de fuerzas de la figura 21.11. Solución: Eq. (21.9): En la figura 23.11, construir una línea que comienza en la intersección de F t t y y F c c que que es perpendicular a la fuerza de fricción F . La línea construida es en un ángulo con F c c. El vector F se divide en dos segmentos de línea, una de las cuales = F c c sin y el otro = F t t cos . asi, F = = F c c sin
+ F t t cos .
Q.E.D.
Eq. (21.10): En la figura 23.11, traducir vector N verticalmente hacia arriba hasta que coincide con la línea previamente construido, cuya longitud = F c c cos cos . Siguiente, traducir vector pies hacia la derecha y hacia abajo en un ángulo hasta que su base se
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asi F s (original) = F c c cos cos - F t t sin sin
Q.E.D.
Eq. (21.12): En la figura 23.11, construir una línea desde la intersección de pies y Fc que es perpendicular perpendicular y se cruza con el vector Fn. Vector Vector Fn se divide ahora en dos dos segmentos de línea, una de las cuales = F t t cos cos y el otro = F c c sin sin . Por lo tanto, F n = F c c sin sin + F t t cos cos
Q.E.D.
Potencia y energía en en maquinado m aquinado
21.18. En una operación de torneado de acero inoxidable con una dureza de 200 HB, la velocidad de corte de 200 m/min, el avance de 0.25 mm/rev y la profundidad del corte de 7.5min, ¿cuánta potencia consumirá el torno para llevar a cabo esta operación si su eficiencia mecánica es de 90%? Utilice la tabla 21.2 para obtener o btener el valor de energía específico apropiado. Solución: De la Tabla 21.3, Tabla 21.3, U = = 2.8 N-m/mm3 = 2.8 J/mm3 RMR = vfd = = (200 m/min)(10 3 mm/m)(0.25 mm)(7.5 mm) = 375,000 mm3/min = 6250 mm3/s = (6250 mm3/s)(2.8 J/mm3) = 17,500 J/s = 17,500 W = 17.5 kW P c c = Contabilización de la eficiencia mecánica , P g = 17.5/0.90 = 19.44 kW
21.19. En el problema anterior, calcule los requerimientos de potencia del torno si el avance es de 0.50 mm/rev. Solución: Este es el mismo problema básico que el anterior, excepto que una corrección se debe hacer para el Uso de la figura "efecto tamaño." 21.14, para f = = 0.50 mm, factor de 3 corrección= 0.85. de de la tabla 21.3, U = = 2.8 J/mm . Con el factor de corrección, U = 2.8(0.85) 3 = 2.38 J/mm . RMR = vfd = (200 m/min)(10 3 mm/m)(0.50 mm)(7.5 mm) = 750,000 mm 3/min = 12,500 3/s
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Solucion: de la tabla 21.3, HP u = 0.25 hp/(in3/min) para el aluminio. Dado que la alimentación es mayor que 0.010 in / rev en la tabla, un factor de corrección se debe aplicar en la figura 21.14. Para f = 0,020 in / rev = a, el factor de corrección = 0,9. HP c c = = HP u x RMR, HP g = HP c c/ E = 900 x 12(.020)(0.250) = 54 in 3/min RMR = vfd = = 0.9(0.25)(54) = 12.2 hp HP c c = HP g = 12.2/0.87 = 14.0 hp
21.21. En una operación de maquinado con acero simple al carbono cuya dureza de Brinell es de 275 HB, la velocidad de corte se configura a 200 m/min y la profundidad de corte es de 6.0 mm. El motor del torno consume 25 kW y su eficiencia mecánica es de 90%. Utilizando el valor de energía específica apropiada de la tabla 21.2, determine el avance máximo que se puede obtener en esta operación. Solución: de la tabla 21.3, U = = 2.8 N-m/mm3 = 2.8 J/mm3 RMR = vfd = = (200 m/min)(10 3 mm/m)(6 mm)f = 1200(103)f mm3/min = 20(103)f mm3/s Potencia disponible P c c = = P g E = = 25(103)(0.90) = 22.5 (10 3) = 22,500W = 22,500 N-m/s Potencia requerida P c c = = (2.8 N-m/mm3)( 20 x 10 3) f = 56,000 f (units are N-m/s) N -m/s) Ajuste de potencia disponible = potencia requerida, 22,500 = 56,000 f f = = 22,500/56,000 = 0.402 mm (esto debe ser interpretado como mm / rev para
una operación de torneado)
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0.010, un factor de corrección se debe aplicar. De la figura 21.14, el factor de corrección es 0,7. Por lo tanto, la corrección HPU = 0,7 * 1,3 = 0,91 CV / (pulg 3 / min) RMR = VFD = 375 m / min (12 en / ft) (0,03 in) (0.150 in) = 20.25 in3 / min HPc = (20,25 in 3 / min) (0,91 hp / (in 3 / min)) = 18,43 hp necesaria. En la eficiencia E = 87%, disponible caballos de fuerza = 0,87 (20) = 17.4 hp Desde caballos de fuerza requerida excede la potencia disponible, el trabajo no puede llevarse a cabo en el torno de 20 CV, al menos no a la velocidad de corte especificado de 375 ft / min.
21.23. Suponga que la velocidad de corte de los problemas 21.7 y 21.8 es de 200 ft/min. A partir de sus respuestas en estos problemas, encuentre a) los caballos de fuerza consumidos en la operación, b) la tasa de remoción del material en in3/min, c) los caballos de fuerza unitaria, hp-min/in3, d ) la energía específica (in-lb/in3). Solución: (a) De Problema 21.8, HPc Fc = £ 155 = 155 (200) / 33,000 = 0,94 hp (b) RMR = VFD = (200 x 12) (0,012) (0,100) = 2.88 in3 / min (c) HPU = 0,94 / 2,88 = 0,326 CV / (pulg3 / min) (d) T = 155 (200) /2.88 = 10.764 ft-lb / in3 = 129167 in-lb / in3
21.24. En el problema 21.12, el torno tiene una eficiencia mecánica de 0.83. Determine a) los caballos de fuerza consumidos por la operación de torneado, b)
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21.26. Repite el problema 21.25, excepto porque el avance es de 0.0075 in/rev y el material de trabajo es acero inoxidable (dureza Brinell 240 HB). Solución: (a) De la Tabla 21.3, HPU = 1,0 hp / (in 3 / min) para el acero inoxidable. Dado que la alimentación es menor que 0.010 en / rev en la tabla, un factor de corrección se debe aplicar en la figura 21.14. Para f = 0,0075 en / rev = a, a , el factor de corrección = 1,1. HPc = HPU x RMR RMR = 400 x 12 (0.0075) (0,12) = 4.32 in3 / min HPc = 1,1 (1,0) (4,32) = 4,75 hp (b) HPg = 5,01 / 0,83 = 5,73 CV
21.27. Una operación de torneado se lleva a cabo en aluminio(100 BHN). Las condiciones de corte son las siguientes: velocidad de corte de 5.6 m/s, el avance de 0.25 mm/rev y la profundidad de corte de 2.0 mm. El torno tiene una eficiencia mecánica de 0.85. Con base en los valores de energía específica de la tabla 21.2 determine a) la potencia de corte y b) la potencia bruta en la operación de torneado, en Watts. Solución: (a) De la Tabla 21.3, U = 0,7 Nm / mm 3 para el aluminio. RMR = VFD = 5,6 (10 3) (. 25) (2,0) = 2,8 (10 3) mm3 / s. Pc = U RMR = 0,7 (2,8) (10 3) = 1,96 (103) nm / s = 1.960 W
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tabla 21.2, calcule los caballos de fuerza f uerza del motor si el torno tiene una eficiencia de 85%. b) Con base en los caballos de fuerza, calcule un estimado de la fuerza de corte para la operación de torneado. Use el modelo de corte ortogonal como una aproximación del proceso de torneado. Solución: (a) De la Tabla 21.3, U = Pu = 520000-in lb / in3 de acero de aleación de la dureza especificada. Dado que la alimentación es mayor que 0.010 en / rev en la tabla, un factor de corrección se debe aplicar en la figura 21.14. Para f = 0,015 en / rev = a, el factor de corrección = 0,95. Por lo tanto, U = = 520,000(0.95) = 494,000 in-lb/in 3 = 41,167 ft-lb/in3. 12(.015)(0.125)) = 6.75 in 3/min RMR = 300 x 12(.015)(0.125 P c c = = U RMR = 41,167(6.75) = 277,875 ft-lb/min = 277,875/33,000 = 8.42 hp HP c c = HP g = 8.42/0.85 = 9.9 hp (b) HPc = VFC / 33.000. Reorganizar, Fc = 33.000 (HPC / v) = 33.000 (8,42 / 300)= 926 lb. Compruebe: Uso unidad caballos de fuerza de la Tabla 21.3 en lugar de energía específica. HPU = 1,3 CV / (pulg3 / min). m in). Aplicando el factor de corrección de factor de corrección = 0,95, HPU = 1,235 hp / (in3 / min). 12(.015)(0.125)) = 6.75 in 3/min, igual que antes RMR = 300 x 12(.015)(0.125 HP c c = = 1.235(6.75) = 8.34 hp HP g = 8.34/0.85 = 9.8 hp (b) F c c = = 33,000 (8.3/300) = 913 lb.
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/RMR = 3.61/11.3 = 0.319 hp/(in 3/min) HP u = HP c c / (c) Factor de corrección = 0,85 partir de la Fig. 21.14 a explicar el hecho de que f = 0,015 en / vuelta en lugar de 0,010 en / rev. Tomando este factor de corrección en cuenta, HPU = 0,375 / 0,85 = 0,441 CV / (pulg3 / min) tal y como aparecería en la Tabla 21.3 para una alimentación (a) = 0,010 en / rev.
Temperatura Temperatura de cor te
21.31. Se lleva a cabo un corte ortogonal en un metal cuyo calor específico volumétrico es de 1.0 J/g-oC, una densidad de 2.9 g/cm3 y una difusividad térmica de 0.8 cm2/s. Se utilizan las condiciones de corte siguientes: la velocidad de corte es de 4.5 m/s, el espesor de la viruta sin cortar es de 0.25 mm y el ancho del corte es de 2.2 mm. La fuerza de corte tiene un valor de 1170 N. Utilizando la ecuación de Cook, determi -ne la temperatura de corte si la temperatura ambiente esde 22 °C. Solución: C = = (2.9 g/cm 3)(1.0 J/g- C) = 2.90 J/cm3- C = (2.90x10-3) J/mm3- C K = = 0.8 cm2/s = 80 mm2/s = F c cv U = v RMR = 1175 N x 4.5 m/s/(4500 mm/s x 0.25 mm x 2.2 mm) = 2.135 N-m/mm 3 / T = = 0.4U /( /( ρ ρC C ) x (vt o/K )0.333 = 22 + (0.4 x 2.135 N-m/mm N-m/m m3/(2.90x10-3) J/mm3-C) [4500 mm/s x 0.25 mm/80 T = mm2/s]0.333
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De la tabla 4.1, C = = 0.11 Cal/g- C. De la nota "a" en la parte inferior de la tabla, 1 cal = 4.186 J. Por lo tanto, C = = 0.11(4.186) = 0.460 J/ g- C = (7.87 g/cm 3)(0.46 J/g- C) = 3.62(10-3) J/mm3- C C = De la tabla 4.2, conductividad térmica k = = 0.046 J/s-mm- C De la ec. (4.3), difusividad térmica K = = k / C = 0.046 J/s-mm- C /[(7.87 x 10 -3 g/mm3)(0.46 J/g- C)] = 12.7 mm 2/s K = Usando la ecuación de Cook, t o = f = = 0.25 mm -3 3 = (0.4(2.2)/3.62(1 (0.4(2.2)/3.62(10 0 ))[3(10 )(0.25)/12.7]0.333 = 0.2428(103)(59.06)0.333 T = = 242.8(3.89) = 944.4 C Temperatura final, teniendo en cuenta la temperatura ambiente T = = 20 + 944 = 964 C
21.33. Una operación de corte ortogonal se lleva a cabo con un cierto metal cuyo
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lb/in3-°F. Si el avance f to 0.010 in y el ancho del corte es de 0.100 in, utilice la fórmula de Cook para calcular la temperatura de corte en la operación dado que la temperatura ambiente es de 70 °F. Solution: RMR = vt ow , v = = RMR/t ow = = 1.8/(0.01 x 0.100) = 1800 in/min = 30 in/sec = F c cv = 300(30)/(30 x 0.010 x 0.100) = 300,000 in-lb/in 3. U = v vt ow = / T = = 70 + (0.4 U / C )( )(vt o/K )0.333 = 70 + (0.4 x 300,000/124)(30 x 0.010/0.18) 0.333 = 70 + (968)(1.667) 0.333 = 70 + 1147 = 1217 F
21.36. Una operación de torneado utiliza una velocidad de corte de 200 m/min, un avance de 0.25 mm/rev y una profundi dad de corte de 4.00 mm. La difusividad térmica del material de trabajo es de 20 mm2/s y el calor específico volumétrico es de 3.5 (10-3) J/mm3-°C. Si un termo-acoplador
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TECNO TECNOL L OG OGÍA ÍA DE LAS L AS HERRAMIENTAS DE CORTE PREGUNTAS DE REPASO 23.1. ¿Cuáles son los dos aspectos principales de la tecnología de herramientas de corte? Los dos aspectos principales de la tecnología de la herramienta son: (1) material de la herramienta y (2) la geometría de la herramienta.
23.2. Mencione los tres modos de falla de la herramienta de maquinado. Los tres modos de fallo herramienta son (1) el fracaso de la fractura. (2) el fracaso de la
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de la herramienta. (5) problemas de eliminación de viruta, (6) la degradación de acabado, (7) aumento de potencia, (8) número de piezas, y (9) la longitud del tiempo de corte de la herramienta.
23.8. Identifique tres propiedades deseables de un material para herramienta de corte. Tres propiedades deseables son (1) la dureza para resistir el fracaso de la fractura, (2) dureza en caliente para resistir el fracaso fracas o de la temperatura, y (3) la resistencia al desgaste para prolongar la vida de la herramienta durante durante el desgaste gradual.
23.9. ¿Cuáles son los elementos principales de aleación de los aceros de alta velocidad? Principales ingredientes de aleación en HSS son (1) o bien de tungsteno o una
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de herramientas de acero de alta velocidad; (2) insertos de placas soldadas, que se utiliza para algunos carburos cementados; y (3) sujetado mecánicamente insertos, que se utiliza para la mayoría de los materiales para herramientas duras incluidos los carburos cementados, carburos recubiertos, cermet, cerámica, SPD, y CBN. 23.15. Mencione las dos categorías principales de fluidos para corte de acuerdo con su función.
Las dos categorías funcionales de los fluidos de corte son: (1) refrigerantes y (2) los lubricant lubricantes. es.
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vida útil del fluido antes de su eliminación. Ventajas Ve ntajas de los sistemas sistema s de filtro incluyen la vida del fluido por más tiempo, la reducción de los costes de eliminación,, una mejor higiene, mantenimiento de la máquina herramienta eliminación inferior, y una mayor vida útil de corte.
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23.4. ¿Cuál de los siguientes no es un ingrediente común de las herramientas de corte de carburo cementado? (dos respuestas correctas): a) Al2O3 y c) CrC. 23.5. ¿Cuál de los siguientes efectos sobre los carburos cementados WC-Co tiene
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velocidades de corte. b) En papel logarítmico natural, grafique los resultados a los que llegó en el inciso anterior. A partir de la gráfica, determine los valores de n y C en la ecuación de Taylor de periodo de vida. c) A manera de comparación, calcule los valores de n y C en la ecuación de Taylor resolviendo las ecuaciones
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flanco en función del tiempo. Utilizando 0.020 in de desgaste del flanco como criterio de la falla de la herramienta, determine los tiempos de vida para las dos velocidades de corte. b) En un pedazo de papel logarítmico natural, grafique los resultados que obtuvo en el inciso anterior. A partir de la gráfica, determine los
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