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Capitulo 7.indd 382

382

(7-48)

(7-49)

Un eje está cargado en flexión y en torsión, de modo que Ma ⫽ 70 N · m, Ta ⫽ 45 N · m, Mm ⫽ 55 N · m y Tm ⫽ 35 N · m. Para el eje, Su ⫽ 700 MPa y Sy ⫽ 560 MPa, y se supone un límite de resistencia a la fatiga completamente corregido de Se ⫽ 210 MPa. Sean Kf ⫽ 2.2 y Kfs ⫽ 1.8. Con un factor de diseño de 2.0, determine el diámetro mínimo aceptable del eje usando el a) criterio de ED-Gerber. b) criterio de ED-elíptico. c) criterio de ED-Soderberg. d) criterio de ED-Goodman. Analice y compare los resultados.

La sección de un eje, que se presenta en la figura, se diseñará con tamaños relativos aproximados de d ⫽ 0.75D y r ⫽ D/20 con el diámetro d conformándose al tamaño de los diámetros interiores de los rodamientos estándares. El eje se hará de acero SAE 2340, tratado térmicamente para obtener resistencias mínimas en el área del hombro de resistencia última a la tensión de 175 kpsi y resistencia a la fluencia de 160 kpsi con una dureza Brinell no menor que 370. En el hombro, el eje se someterá a un momento flexionante completamente reversible de 600 lbf · pulg, acompañado de una torsión uniforme de 400 lbf · pulg. Use un factor de diseño de 2.5 y dimensione el eje para vida infinita.

7-1

7-2

Los problemas marcados con un asterisco (*) están vinculados a problemas en otros capítulos, los cuales se resumen en la tabla 1-1 de la sección 1-16, página 23.

PROBLEMAS

La interferencia mínima, de la ecuación (7-42), debe usarse para determinar la presión mínima para verificar la cantidad máxima de par de torsión de manera que la junta esté diseñada para transmitir sin deslizamiento.

T ⫽ ( /2)fpld 2

T ⫽ Ff d /2 ⫽ f ppd l (d /2)

donde l es la longitud de la maza. Esta fuerza de fricción actúa con un brazo de palanca de d/2 para proporcionar la capacidad de par de torsión de la junta; en consecuencia,

Ff ⫽ f N ⫽ f ( pA) ⫽ f [ p2 (d/2)l] ⫽ fpld

to finito para la interfase será útil siempre que esté garantizado. Un elemento de esfuerzo sobre la superficie de un eje rotatorio experimentará un esfuerzo flexionante completamente reversible en la dirección longitudinal, así como los esfuerzos de compresión estables en las direcciones tangencial y radial. Éste es un elemento de esfuerzo en tres dimensiones. También puede estar presente el esfuerzo cortante debido a la torsión en el eje. Como los esfuerzos debidos al ajuste por presión son de compresión, por lo general la situación de fatiga realmente mejora. Por esta razón, puede ser aceptable simplificar el análisis del eje sin tomar en cuenta los esfuerzos de compresión estables debidos al ajuste por presión. Sin embargo, existe un efecto de concentración del esfuerzo en el esfuerzo de flexión en el eje, cercano a los extremos de la geometría de la maza, y por consiguiente en su uniformidad y rigidez, que puede tener un efecto significativo en el valor específico del factor de la concentración del esfuerzo, lo que hace difícil reportar valores generalizados. Para las primeras estimaciones, los valores típicos no son mayores que 2. La cantidad de par de torsión que puede transmitirse a través de un ajuste de interferencia puede estimarse mediante un análisis simple de fricción en la interfase. La fuerza de fricción es el producto del coeficiente de fricción f y la fuerza normal que actúa en la interfase. La fuerza normal puede representarse mediante el producto de la presión p y el área de la superficie A de interfase. Por lo tanto, la fuerza de fricción Ff es

Capítulo 7 Ejes, flechas y sus componentes

20/03/12 20:08

Capitulo 7.indd 383

Problema 7-4 El material se mueve debajo del rodamiento. Dimensiones en pulgadas.

7-4

Problema 7-3

7-3

Problema 7-2 Sección de un eje que contiene una ranura de alivio rectificada. A menos que se especifique lo contrario, el diámetro en la raíz de la ranura dr ⫽ d ⫺ 2r y aunque la sección de diámetro d esté rectificada, la raíz de la ranura es una superficie maquinada. d

383

A

B

250 mm

C

100 mm

D

F 20⬚

z

3

14

O

8

y

3

14 3

24

3

4 diá.

Engrane 4 3 diá.

A

B

2

x

20°

F

20/03/12 20:08

Un rodillo industrial con engranes, que se muestra en la figura, se impulsa a 300 rpm por una fuerza F que actúa en un círculo de paso de 3 pulg de diámetro. El rodillo ejerce una fuerza normal de 30 lbf/pulg de longitud del rodillo sobre el material que se jala a través de él. El material pasa debajo del rodillo. El coeficiente de fricción es 0.40. Desarrolle los diagramas de momento flexionante y fuerza cortante del eje modelando la fuerza del rodillo como: a) una fuerza concentrada en el centro del rodillo, y b) una fuerza uniformemente distribuida a lo largo del rodillo. Estos diagramas aparecerán en dos planos ortogonales.

TA

El eje giratorio de acero sólido simplemente apoyado en los puntos B y C está impulsado por un engrane (que no se muestra) el cual se une con el engrane recto en D, que tiene un diámetro de paso de 150 mm. La fuerza F del engrane impulsor actúa a un ángulo de presión de 20°. El eje transmite un par de torsión al punto A de TA ⫽ 340 N · m. El eje de acero está maquinado con Sy ⫽ 420 MPa y Sut ⫽ 560 MPa. Usando un factor de seguridad de 2.5, determine el diámetro mínimo permisible de la sección de 250 mm del eje con base en a) un análisis estático de la fluencia con base en la teoría de la energía de distorsión y b) un análisis de falla por fatiga. Para estimar los factores de concentración del esfuerzo suponga radios de filete agudos en los hombros del cojinete.

D

r

Problemas

Capitulo 7.indd 384

7-17

7-7* a 7-16*

1

12

10

1

12

A

1

1 4

4

cuñero

7 8

3-68, 132 3-69, 132 3-70, 132 3-71, 132 3-72, 133 3-73, 133 3-74, 134 3-76, 134 3-77, 134 3-79, 135

7-7*

7-8*

7-9*

7-10*

7-11*

7-12*

7-13*

7-14*

7-15*

7-16*

En el tren de engranes con doble reducción que se muestra en la figura, el eje a está impulsado por un motor unido mediante un cople flexible conectado a la saliente. El motor proporciona un par de torsión de 2 500 lbf · pulg a una velocidad de 1 200 rpm. Los engranes tienen un ángulo de presión de 20°, con los diámetros que se muestran en la figura. Use un acero estirado en frío AISI 1020. Diseñe uno de los ejes (según lo especifique su profesor) con un factor de diseño de 1.5, realizando las siguientes tareas. a) Bosqueje una configuración general del eje; incluya medios para localizar los engranes y cojinetes y para transmitir el par de torsión. b) Realice un análisis de fuerzas para encontrar las fuerzas de reacción del cojinete y genere diagramas de corte y momento flexionante. c) Determine las ubicaciones críticas potenciales para el diseño por esfuerzo.

Problema original, número de página

Número de problema

Para el problema especificado en la tabla, trabaje sobre los resultados del problema original para obtener un diseño preliminar del eje mediante la realización de las siguientes tareas. a) Dibuje un diseño general del eje, incluyendo los medios para localizar los componentes y para transmitir el par de torsión. En este punto son aceptables las estimaciones de las anchuras de los componentes. b) Especifique un material adecuado para el eje. c) Determine los diámetros críticos del eje con base en la vida infinita a la fatiga y con un factor de diseño de 1.5. Verifique la fluencia. d) Haga cualquier otra decisión dimensional necesaria para especificar todos los diámetros y dimensiones axiales. Haga un bosquejo a escala del eje que muestre todas las dimensiones propuestas. e) Verifique las deflexiones en los engranajes y las pendientes de los engranes y cojinetes de modo que satisfagan los límites recomendados en la tabla 7-2. Suponga que es muy poco probable que las deflexiones en las poleas sean críticas. Si alguna de las deflexiones supera los límites recomendados, haga los cambios necesarios para colocarlas dentro de los límites.

1

1 14

En la figura se muestra el diseño propuesto para el rodillo industrial del problema 7-4. Se propone usar cojinetes de lubricación de película hidrodinámica. Todas las superficies están maquinadas, excepto los muñones, que son esmerilados y pulidos. El material es acero 1035 HR. Realice una evaluación del diseño. ¿Es satisfactorio el diseño?

7-6

O

Diseñe un eje para la situación del rodillo industrial del problema 7-4, con un factor de diseño de 2 y una meta de confiabilidad de 0.999 contra falla por fatiga. Planee para un cojinete de bolas a la izquierda y un rodamiento cilíndrico a la derecha. Para deformación emplee un factor de seguridad de 2.

7-5

Capítulo 7 Ejes, flechas y sus componentes

Problema 7-6 Radios de los hombros de los 1 cojinetes 0.030 pulg, los otros 16 pulg. El cuñero tipo trineo tiene 1 una longitud de 3 2 pulg.

384

20/03/12 20:08

Capitulo 7.indd 385

7-19*

Problema 7-18 Radios de los filetes de los hombros en asientos de cojinetes 0.030 pulg, los otros 18 pulg, excepto transición de asiento de cojinete derecho, 14 pulg. El material es acero 1030 HR. Cuñas 38 pulg de 3 pulg de profundidad. ancho por 16 Dimensiones en pulgadas.

7-18

Problema 7-17 Dimensiones en pulgadas.

12

8

20

24

16

A

C

E

3

9

8

2

B

D

F

b

6

c

a

Determine los diámetros críticos del eje con base en la fatiga y los esfuerzos estáticos en las ubicaciones críticas. Tome algunas otras decisiones dimensionales necesarias para especificar todos los diámetros y dimensiones axiales. Bosqueje el eje a escala, y muestre en un dibujo a escala, todas las dimensiones propuestas. Verifique la deflexión en el engrane y las pendientes en el engrane y los cojinetes para satisfacer los límites que se recomiendan en la tabla 7-2. Si alguna de las deflexiones supera los límites que se recomiendan, haga los cambios apropiados para ubicarlas dentro de los límites.

385

3

9

1.875

74

11

1.875 1.574

0.453

6

1.500

20/03/12 20:08

Se propone el eje mostrado en la figura para la aplicación definida en el problema 3-72, página 133. El material es acero AISI 1018 estirado en frío. Los engranes de seguridad están contra los hombros y tienen centros con tornillos para fijarlos en su lugar. Se muestran los centros efectivos de los engranes para la transmisión de fuerza. Los cuñeros se cortan con fresas estándar. Los cojinetes están ajustados a presión contra los hombros. Determine el factor de seguridad mínimo contra la fatiga.

1.574

0.354

8

En la figura se muestra el diseño propuesto de un eje que se usará como eje de entrada a en el problema 7-17. Se planeó usar un cojinete de bolas a la izquierda y un cojinete de rodillos cilíndricos a la derecha. a) Determine el factor de seguridad de la fatiga mínima mediante la evaluación de algunas ubicaciones críticas. Use un criterio de falla por fatiga que se considere típico de los datos de falla, en lugar de uno que se considere conservador. También asegúrese de que el eje no alcanzará fluencia en el primer ciclo de carga. b) Verifique el diseño para su adecuación, respecto de la deformación, de acuerdo con las recomendaciones de la tabla 7-2.

4

g)

f)

e)

d)

Problemas

41

1.75

11

10

2

1.00

Centro del engrane

1.3

9

75

1 080

50

Centro del engrane 300

325

285

42

30

30

Problema 7-23 Dimensiones en pulgadas.

1 4

1 8

cuñero

1 16

R.

2.0 1.000

1 32

2.75

R.

0.15

1.181

8.50 1.70

12.87

3 8

1 8 3 16

R. cuñero

0.485 1.750

0.20

0.1 R.

2.0

1 8

R.

1.40

2.20

1 32

R.

1.181

0.75

El eje que se muestra en la figura, impulsado por un engrane en el cuñero de la derecha, impulsa a un ventilador en el cuñero izquierdo, y está soportado por dos cojinetes de bola con ranura profunda. El eje está hecho de acero estirado en frío AISI 1020. A una velocidad de estado constante, el engrane transmite una carga radial de 230 lbf y una carga tangencial de 633 lbf con un diámetro de paso de 8 pulg. a) Determine los factores de seguridad a la fatiga en algunas ubicaciones potencialmente críticas. b) Verifique que las deflexiones satisfagan los mínimos sugeridos para los cojinetes y engranes.

425

385

50

350

7-23

30

40

Centro del engrane

Continúe el problema 7-21 verificando que las deflexiones satisfagan los mínimos sugeridos para cojinetes y engranes en la tabla 7-2. Si alguna de las deflexiones supera los límites recomendados, haga los cambios necesarios para colocarlas dentro de los límites.

30

400

7-22*

Problema 7-21* Todos los filetes de 2 mm. Dimensiones en mm.

15

Se propone el eje mostrado en la figura para la aplicación definida en el problema 3-73, página 133. El material es acero AISI 1018 estirado en frío. Los engranes de seguridad están contra los hombros y tienen centros con tornillos para fijarlos en su lugar. Se muestran los centros efectivos de los engranes para la transmisión de fuerza. Los cuñeros se cortan con fresas estándar. Los cojinetes se ajustan a presión contra los hombros. Determine el factor de seguridad mínimo contra la fatiga.

17

2.5

14

7-21*

15

1.75

Centro del engrane

Continúe el problema 7-19 verificando que las deflexiones satisfagan los mínimos sugeridos para cojinetes y engranes en la tabla 7-2. Si alguna de las deflexiones supera los límites recomendados, haga los cambios necesarios para colocarlas dentro de los límites.

1

1.3

16

7-20*

1.00

0.5

Capítulo 7 Ejes, flechas y sus componentes

Problema 7-19* 1 pulg. Todos los filetes de 16 Dimensiones en pulgadas.

Capitulo 7.indd 386

386

20/03/12 20:08

Capitulo 7.indd 387

Problema 7-26 Dimensiones en milímetros.

7-26

Problema 7-25 Dimensiones en milímetros.

7-25

Problema 7-24 Dimensiones en milímetros.

7-24

387

55

35

115

155 40 55

10

7 kN

375

45

40

150

35

85

30

60

30

20

175

100

400

50

125

B

CL brg

RA

A

100

T

2.4 kN

100

350 RB

RB

B

75

22.4 kN T = 540 N . m

3.8 kN

100

20/03/12 20:08

Se debe diseñar un eje de acero tratado térmicamente para soportar el engrane recto y el sinfín en voladizo que se muestran en la figura. Un cojinete en A toma carga radial pura. El cojinete en B toma la carga de empuje del sinfín para cualquier dirección de rotación. Las dimensiones y las cargas se presentan en la figura; note que las cargas radiales están en el mismo plano. Haga un diseño completo del eje, incluyendo un bosquejo del eje que muestre todas las dimensiones. Identifique el material y su tratamiento térmico (si es necesario). Proporcione una evaluación de adecuación de su diseño final. La velocidad del eje es de 310 rpm.

A

CL brg

Se debe diseñar un eje para soportar el piñón recto y el engrane helicoidal que se muestran en la figura sobre dos cojinetes espaciados 700 mm entre centros. El cojinete A es cilíndrico de rodillos y sólo tomará carga radial; el cojinete B tomará una carga de empuje de 900 N producida por el engrane helicoidal y su parte de la carga radial. El cojinete en B puede ser de bolas. Las cargas radiales de ambos engranes están en el mismo plano y son de 2.7 kN para el piñón y de 900 N para el engrane. La velocidad del eje es 1 200 rpm. Diseñe y haga un bosquejo a escala del eje, donde se indiquen todos los tamaños de los filetes, cuñeros, hombros y diámetros. Especifique el material y su tratamiento.

Todos los filetes 2 mm

30

30

Un eje de acero AISI 1020 estirado en frío con la geometría que se muestra en la figura, soporta una carga transversal de 7 kN y transmite un par de torsión de 107 N · m. Examine el eje por resistencia y deflexión. Si la mayor inclinación permisible de los cojinetes es de 0.001 rad y en el acoplamiento del engrane 0.0005 rad, ¿cuál es el factor de seguridad que protege contra el daño por distorsión? ¿Cuál es el factor de seguridad que protege contra la falla por fatiga? Si el eje resulta insatisfactorio, ¿qué recomendaría para corregir el problema?

Problemas

7-27

7-33

Problema 7-32 Dimensiones en pulgadas.

2

9 14 15 16

32 lbf 2.000

Un agujero transversal taladrado y escariado se emplea en un eje sólido para sujetar un pasador que coloca y retiene un elemento mecánico, como la maza de un engrane, en posición axial y permite la transmisión del par de torsión. Como un agujero de diámetro pequeño introduce alta concentración de

1

2.763

El eje de acero que se muestra en la figura soporta un engrane de 18 lbf a la izquierda y uno de 32 lbf a la derecha. Estime la primera velocidad crítica debida a las cargas, la velocidad crítica del eje sin las cargas y la velocidad crítica de la combinación.

7-32

2.472

Si se ahueca un eje de diámetro uniforme, ¿se incrementa o disminuye la velocidad crítica?

7-31

18 lbf

Compare la ecuación (7-27) de la frecuencia angular de un eje de dos discos con la ecuación (7-28) y observe que las constantes en las dos ecuaciones son iguales. a) Desarrolle una expresión para la segunda velocidad crítica. b) Estime la segunda velocidad crítica del eje del ejemplo 7-5, incisos a) y b).

7-30

2.000

Demuestre qué tan rápido converge el método de Rayleigh para el eje sólido de diámetro uniforme del problema 7-28, al dividir el eje primero en uno, luego en dos y finalmente en tres elementos.

2

4P, 16T

1

1 2 diá.

7-29

6

2

Un eje con diámetro uniforme de 25 mm tiene una longitud de 600 mm entre cojinetes. a) Encuentre la velocidad crítica mínima del eje. b) Si la meta es incrementar al doble la velocidad crítica, determine el nuevo diámetro. c) ¿Cuál es la velocidad crítica de un modelo del eje a la mitad del tamaño?

4

1 2

El eje falló aquí

7-28

1 8 diá.

3

Un eje con engrane cónico montado en dos cojinetes de bolas de 40 mm serie 02 es impulsado a 1 720 rpm por un motor conectado a través de un acoplamiento flexible. En la figura se presenta el eje, el engrane y los cojinetes. El eje ha tenido problemas (en realidad, dos ejes ya han fallado), y el tiempo de parada de la máquina es tan costoso que se ha decidido que usted rediseñe el eje. Una verificación de dureza de los ejes en la proximidad de la fractura de los ejes mostró un promedio de 198 Bhn de uno y 204 Bhn del otro. Con tanta exactitud como fue posible se estimó que los ejes fallaron a vidas medidas entre 600 000 y 1 200 000 ciclos de operación. Las superficies de los ejes estaban maquinadas, pero no esmeriladas. Los tamaños de los filetes no se midieron, pero corresponden a las recomendaciones para los cojinetes de bolas que se colocaron. Usted sabe que la carga es del tipo pulsante o de impacto, pero no tiene idea de la magnitud, porque el eje impulsa un mecanismo de posicionamiento y las fuerzas 3 3 son de inercia. Los cuñeros tienen un ancho de 8 pulg por 16 pulg de profundidad. El piñón cónico de dientes rectos impulsa un engrane cónico de 48 dientes. Especifique un nuevo eje con suficientes detalles para asegurar una vida larga y sin problemas.

Capítulo 7 Ejes, flechas y sus componentes

Problema 7-27 Dimensiones en pulgadas.

Capitulo 7.indd 388

388

20/03/12 20:08

Capitulo 7.indd 389

389

Un engrane y un eje con diámetro nominal de 35 mm deben ensamblarse con un ajuste de interferencia medio, como se especifica en la tabla 7-9. El engrane tiene una maza, con un diámetro externo de 60 mm, y una longitud total de 50 mm. El eje está hecho de acero AISI 1020 CD, y el engrane de acero completamente endurecido para proporcionar Su ⫽ 700 MPa y Sy ⫽ 600 MPa. a) Especifique las dimensiones con tolerancias para lograr el ajuste deseado entre el eje y orificio del engrane. b) Determine las presiones mínima y máxima que podrían experimentarse en la interfase con las tolerancias especificadas. c) Determine los peores factores de seguridad estáticos que protegen contra la fluencia en el ensamble del eje y el engrane, con base en la teoría de falla por energía de distorsión. d) Determine el par de torsión máximo que se espera transmita la junta sin deslizamiento, es decir, cuando el valor de la presión de interferencia es mínimo para las tolerancias especificadas.

7-42

20/03/12 20:08

Se mide cuidadosamente el diámetro de un eje que resulta ser de 1.5020 pulg. Se selecciona un cojinete con una especificación de catálogo para el diámetro interior de 1.500 a 1.501 pulg. Determine si es una selección aceptable si se desea un ajuste de interferencia localizada.

7-41

Se necesita un perno para formar un pivote de una articulación. Encuentre las dimensiones que se requieren para un pasador y una horquilla de tamaño básico de 45 mm, con ajuste de deslizamiento.

7-38

Se seleccionó un cojinete de bolas con el diámetro interior especificado en el catálogo, de 35.000 mm a 35.020 mm. Especifique los diámetros mínimo y máximo adecuados para el eje a fin de proporcionar un ajuste de interferencia localizada.

Se requiere un ajuste de interferencia para una maza de hierro fundido de un engrane, el cual se montará sobre un eje de acero. Tome las decisiones dimensionales para realizar un ajuste de interferencia medio con tamaño básico de 1.75 pulg.

7-37

7-40

Se necesita un pasador guía para alinear el ensamble de un accesorio de dos piezas. El tamaño nominal del pasador es de 15 mm. Tome las decisiones dimensionales para realizar un ajuste con un juego de ubicación de tamaño básico de 15 mm.

7-36

Es necesario describir una chumacera y su casquillo. El tamaño nominal es de 1.25 pulg. ¿Qué dimensiones se necesitan para un tamaño básico de 1.25 pulg con un ajuste estrecho de operación, si se trata de un ensamble de chumacera y casquillo ligeramente cargado?

Se propone el eje mostrado en el problema 7-21 para la aplicación definida en el problema 3-73, página 133. Especifique una cuña recta para el engrane B, con un factor de seguridad de 1.1.

7-35*

7-39

Se propone el eje mostrado en el problema 7-19 para la aplicación definida en el problema 3-72, página 133. Especifique una cuña recta para el engrane B, con un factor de seguridad de 1.1.

7-34*

esfuerzo, y un diámetro mayor debilita el área que resiste flexión y torsión, investigue la existencia del diámetro de un pasador con efecto adverso mínimo en el eje. Luego formule una regla de diseño. (Sugerencia: Utilice la tabla A-16.)

Problemas

4  K f M m  3  K fsTm

2

B

2

2

2

2

· ¸ ¸¸ ¹

1/3

º½ »° »¾ »° ¼¿



1/3

265.5 N ˜ m

338.4 N ˜ m



(d) DE-Goodman: Eq. (7-8) can be shown to be

ª16n § A B · º « ¨  ¸» «¬ S ¨© Se S y ¸¹ »¼

1/3



1/3

1/3

·º ¸» ¸» ¹¼



ª16(2) § 338.4 265.5 « ¨  d « S ¨ 210 106 560 106 © ¬ d = 27.70 (103) m = 27.70 mm Ans.

(c) DE-Soderberg, Eq. (7-14) can be shown to be

1/3

§ · 2 2 § 16n A2 B 2 ·  338.4   265.5 ¸ ¨ 16(2)  2 ¸ d ¨ 2 2 2 ¨ S ¸ ¨ S Se S y ¸¹ ª 210 106 º ª 560 106 º ¸ ¨ © ¬ ¼ ¬ ¼ ¹ © d = 25.77 (103) m = 25.77 mm Ans.



 º»  »¼

2 1/2

4 > (2.2)(55) @  3 > (1.8)(35) @

(b) DE-elliptic, Eq. (7-12) can be shown to be



2

4 > (2.2)(70) @  3 > (1.8)(45) @

­ ª § 6 ° 8(2)(338.4) « ¨ ª 2(265.5)  210 10 «   d ® 1 1 « 6 6 ¨ ° S  210 10 « ¨ «¬ 338.4  700 10 ¬ © ¯ d = 25.85 (103) m = 25.85 mm Ans.

2

2

4  K f M a  3  K fsTa

A

(a) DE-Gerber, Eq. (7-10):

7-2

1/3

ª16n § A B · º « ¨  ¸» ¬« S © Se Sut ¹ ¼»



Chapter 7 - Rev. A, Page 1/45

This problem has to be done by successive trials, since S e is a function of shaft size. The material is SAE 2340 for which S ut = 175 kpsi, S y = 160 kpsi, and H B • 370.

d

ª16(2) § 338.4 265.5 · º» « ¨ ¸  « S ¨ 210 106 700 106 ¸¹ » © ¬ ¼ d = 27.27 (103) m = 27.27 mm Ans. ________________________________________________________________________ Criterion d (mm) Compared to DE-Gerber DE-Gerber 25.85 DE-Elliptic 25.77 0.31% Lower Less conservative DE-Soderberg 27.70 7.2% Higher More conservative DE-Goodman 27.27 5.5% Higher More conservative ______________________________________________________________________________

7-1

Chapter 7 ka

2.70(175) 0.265

Eq. (6-8), p.282:

0.879(0.799) 0.107

0.90

S e = 0.69 (0.90)(0.5)(175) = 54.3 kpsi dr 0.799 D 1.23 in 0.65 0.65 r = D / 20 = 1.23/20 = 0.062 in

kb

Trial #2: Choose d r = 0.799 in.





Chapter 7 - Rev. A, Page 2/45

1/3 2 2 1/2 ½ ­ ª § 1.46(400) · º ° °16(2.5) « § 1.81(600) · ¸  3¨ ¸ » ¾ dr ® 4¨ 3 « ¨ 160 103 ¸ » ° ° S « ¨© 54.9 10 ¸¹ © ¹ ¼» ¬ ¯ ¿ d r = 0.799 in

We select the DE-ASME Elliptic failure criteria, Eq. (7-12), with d as d r , and M m = T a = 0,

d d r  2r 0.75  2(0.058) 0.808 in d 0.808 1.08 dr 0.75 r 0.058 0.077 dr 0.75 K t = 1.9 Fig. 6-20, p. 295: r = 0.058 in, q = 0.90 Eq. (6-32), p. 295: K f = 1 + 0.90 (1.9 – 1) = 1.81 Fig. A-15-15: K ts = 1.5 Fig. 6-21, p. 296: r = 0.058 in, q s = 0.92 Eq. (6-32), p. 295: K fs = 1 + 0.92 (1.5 – 1) = 1.46

Fig. A-15-14:

0.5 175 87.5 kpsi

0.91

d  2r 0.75D  2 D / 20 0.65D dr 0.75 1.15 in D 0.65 0.65 D 1.15 r 0.058 in 20 20 dr

0.5Sut

0.879(0.75) 0.107

0.69

S e = 0.69 (0.91)(87.5) = 54.9 kpsi

Sec

Eq. (6-18), p. 287:

kb

Eq. (6-20), p. 288:

Trial #1: Choose d r = 0.75 in

Eq. (6-19), p. 287:

1.9, K ts 1.5, q 0.90, qs 0.92 1.81, K fs 1.46 Using Eq. (7-12) produces d r = 0.802 in. Further iteration produces no change. With d r = 0.802 in, 0.802 D 1.23 in 0.65 d 0.75(1.23) 0.92 in

7-3

2.4 2.1

1  0.8(2.7  1)

1  0.9(2.2  1)

Kf

K fs

Eq. (7-15):

c V max

ª§ 32 K f M a ·2 § 16 K fsTm · «¨ ¸  3¨ ¸ 3 3 «¬© S d ¹ © Sd ¹

º » »¼

2 1/2

Chapter 7 - Rev. A, Page 3/45

(a) We will choose to include fatigue stress concentration factors even for the static analysis to avoid localized yielding.

Eq. (6-32):

For sharp fillet radii at the shoulders, from Table 7-1, K t = 2.7, and K ts = 2.2. Examining Figs. 6-20 and 6-21 (pp. 295 and 296 respectively) with Sut 560 MPa, conservatively estimate q = 0.8 and qs 0.9. These estimates can be checked once a specific fillet radius is determined.

F cos 20q(d / 2) = T A , F = 2 T A / ( d cos 20q) = 2(340) / (0.150 cos 20q) = 4824 N. The maximum bending moment will be at point C, with M C = 4824(0.100) = 482.4 N·m. Due to the rotation, the bending is completely reversed, while the torsion is constant. Thus, M a = 482.4 N·m, T m = 340 N·m, M m = T a = 0.

A look at a bearing catalog finds that the next available bore diameter is 0.9375 in. In nominal sizes, we select d = 0.94 in, D = 1.25 in, r = 0.0625 in Ans. ______________________________________________________________________________

?Kf

Kt

With these ratios only slightly different from the previous iteration, we are at the limit of readability of the figures. We will keep the same values as before.

Figs. A-15-14 and A-15-15: d d r  2r 0.799  2(0.062) 0.923 in d 0.923 1.16 d r 0.799 r 0.062 0.078 d r 0.799 16

¬

ka

2

4.51(560) 0.265

0.84

Ans.



kb

1.51d 0.157

1.51 53.4

0.157

0.81

· ¸ ¸ ¹

º » » ¼»

2 1/ 2

Ta 1/3

½ ° ¾ ° ¿

0.

1/3

· ¸ ¸ ¹

q = 0.72 q s = 0.77

K f = 1 + 0.72 (2.7 – 1) = 2.2 K fs = 1 + 0.77 (2.2 – 1) = 1.9 S e = 0.84(0.81)(0.5)(560) = 191 MPa d = 53 mm Ans.

Chapter 7 - Rev. A, Page 4/45

Further iteration does not change the results. _____________________________________________________________________________

Eq. (6-18): Eq. (7-12):

Eq. (6-32):

Iterating with these new estimates,

Fig. (6-20): Fig. (6-21):

Assuming a sharp fillet radius, from Table 7-1, r = 0.02d = 0.02 (53.4) = 1.07 mm.

Eq. (6-20):

With this diameter, we can refine our estimates for k b and q.



2 ­ ª § 2.1(340) °16(2.5) « § 2.4(482.4) · ¸  3¨ 4¨ ® « ¨ 200 106 ¸ ¨ 420 106 S ° ¹ © ¬« © ¯ 0.0534 m 53.4 mm

Selecting the DE-ASME Elliptic criteria, use Eq. (7-12) with M m

S e = 0.84(0.85)(0.5)(560) = 200 MPa

d

2 1/2

4 > (2.4)(482.4) @  3 > (2.1)(340) @ `  ^

§ 16(2.5) ¨ ¨ S  420 106 ©

1/3

¼

S d 3S y ª 2 2 1/ 2 4 K M  3  K fsTm º «  f a »

­° 16n 1/2 ½ ° ª¬ 4( K f M a ) 2  3( K fsTa ) 2 º¼ ¾ ® S S ¯° y ¿°

Sy c V max

d = 0.0430 m = 43.0 mm

d

n

Assume k b = 0.85 for now. Check later once a diameter is known.

(b)

Solving for d,

Eq. (7-16):

7-4

F tan 20

z B

D

y A

128 tan 20

D

46.6 lbf

xz-plane

Bending moment diagram:

240(5.75)  F (11.5)  46.6(14.25) 0 240(5.75)  46.6(14.25) FAy 62.3 lbf 11.5 y 6M A FO (11.5)  46.6(2.75)  240(5.75) 0 240(5.75)  46.6(2.75) FOy 131.1 lbf 11.5

6M O

(a) xy-plane

F

y B

FBz

T

96 lbf

Chapter 7 - Rev. A, Page 5/45

FCz (2) 96(2) 192 lbf ˜ in T 192 128 lbf 1.5 1.5

FCz

240 lbf

30(8) 0.4(240)

FCy

We have a design task of identifying bending moment and torsion diagrams which are preliminary to an industrial roller shaft design. Let point C represent the center of the span of the roller.

( 128) 2  ( 352) 2

100 2  ( 754) 2

375 lbf ˜ in

2

2

131.1x  15 x  1.75  15 x  9.75  62.3 x  11.5 Bending moment diagram:

M xy

(b) xy-plane

Chapter 7 - Rev. A, Page 6/45

1

Torque: The torque is constant from C to B, with a magnitude previously obtained of 192 lbf·in.

MA

MC

Bending moment diagram:

761 lbf ˜ in

0 96(5.75)  FAz (11.5)  128(14.25) 96(5.75)  128(14.25) F 206.6 lbf 11.5 z 6M A 0 FO (11.5)  128(2.75)  96(5.75) 96(5.75)  128(2.75) FOz 17.4 lbf 11.5 z A

6M O

131.1(5.75)  15(5.75  1.75) 2

514 lbf ˜ in

M xy2  M xz2

2

1

Chapter 7 - Rev. A, Page 7/45

This is a design problem, which can have many acceptable designs. See the solution for Prob. 7-17 for an example of the design process. ______________________________________________________________________________

7-5

Torque: The torque rises from 0 to 192 lbf·in linearly across the roller, then is constant to B. Ans. ______________________________________________________________________________

M max = 516 lbf · in at x = 6.25 in

Plot M net (x), 1.75 ” x ” 11.5 in

Let M net

2

17.4 x  6 x  1.75  6 x  9.75  206.6 x  11.5

Bending moment diagram:

M xz

xz-plane

This is reduced from 754 lbf · in found in part (a). The maximum occurs at x 6.12 in rather than C, but it is close enough.

MC

M max = –516 lbf · in and occurs at 6.12 in.

7-6



EIy

0 at x 0

Ÿ

C2

0 y 0 at x 11.5 Ÿ C1 1908.4 lbf ˜ in 3 x = 0: EIy' = 1908.4 x = 11.5: EIy' = –2153.1

y



 EIz

C4

0

1908.42  8.9752

1908.4 lbf ˜ in 3

EIT

Ÿ

At O:

0 From (2),

0 at x

z 0 at x 11.5 Ÿ C3 8.975 lbf ˜ in 3 x = 0: EIz' = 8.975 x = 11.5: EIz' = –683.5

z

(2)

(1)

Chapter 7 - Rev. A, Page 8/45

17.4 2 206.6 3 3 2 x  2 x  1.75  2 x  9.75  x  11.5  C3 2 2 17.4 3 1 1 206.6 4 4 3 x  x  1.75  x  9.75  x  11.5  C3 x  C4 6 2 2 6 EIzc

xz plane (treating z n  )

From (1),



131.1 2 62.3 3 3 2 x  5 x  1.75  5 x  9.75  x  11.5  C1 2 2 131.1 3 5 5 62.3 4 4 3  x  x  1.75  x  9.75  x  11.5  C1 x  C2 6 4 4 6

EIy c 

xy plane, with dy/dx = y'

From Prob. 7-4, integrate M xy and M xz .

If students have access to finite element or beam analysis software, have them model the shaft to check deflections. If not, solve a simpler version of shaft for deflection. The 1 in diameter sections will not affect the deflection results much, so model the 1 in diameter as 1.25 in. Also, ignore the step in AB.

n

T

EIT



I1 in section OCA, ycA 2153.1/ EI1

0.001 1.59 0.000 628



2259 30 106 S / 64 1.254



2259.0 lbf ˜ in 3 (dictates size)

0.000 628 rad

(2153.1) 2  ( 683.5) 2



z cB

z cB / A





2 EI 2

128 2.752



484 EI 2 484









T A 7.09(104 ) rad T B 1.10(103 ) rad

Finite element results: T O 5.47(104 ) rad



0.0%

11.4%

Error in simplified model 3.0%

0.00110 rad



(0.000 803) 2  ( 0.000 751) 2

6

683.5  30 10 S / 64 1.254 30 106 S / 64 0.8754

683.5 , EI1

Crowned teeth must be used.

TB



z cA

xz plane

Chapter 7 - Rev. A, Page 9/45

0.000 751 rad

= –0.000 803 rad (magnitude greater than 0.0005 rad)

Since y' B/A is a cantilever, from Table A-9-1, with I I 2 in section AB Fx( x  2l ) 46.6 ycB / A (2.75)[2.75  2(2.75)] 176.2 / EI 2 2 EI 2 2 EI 2 2153.1 176.2 ? ycB ycA  ycB / A   30 106 S / 64 1.254 30 106 S / 64  0.8754

With I

At gear mesh, B xy plane

At A: Sut

72 kpsi, S y

39.5 kpsi

§ 1 · ¨ ¸ © 0.3 ¹ k d ke kf

1

K fs

1  0.70(1.8  1) 1.56

192 lbf ˜ in

7-17

Chapter 7 - Rev. A, Page 10/45

(a) One possible shaft layout is shown in part (e). Both bearings and the gear will be located against shoulders. The gear and the motor will transmit the torque through the

7-7 through 7-16 These are design problems, which can have many acceptable designs. See the solution for Prob. 7-17 for an example of the design process. ______________________________________________________________________________

The main problem with the design is the undersized shaft overhang with excessive slope at the gear. The use of crowned-teeth in the gears will eliminate this problem. ______________________________________________________________________________

Perform a similar analysis at the profile keyway under the gear.

n = 3.91

1 n

T

a 1/2 2 2 ª1.56(192) º ½° 16 ­° ª1.85(360) º 4  3« ¾ » » 3 ® « S 1 ¯° ¬ 27 500 ¼ ¬ 39 500 ¼ ¿°

Using DE-ASME Elliptic, Eq. (7-11) with M m

0,

360.0 lbf ˜ in

Se 0.869(0.879)(36) 27.5 kpsi D / d = 1.25, r / d = 0.03 Fig. A-15-8: K ts = 1.8 Fig. A-15-9: K t = 2.3 Fig. 6-20: q = 0.65 Fig. 6-21: q s = 0.70 Eq. (6-32): K f 1  0.65(2.3  1) 1.85

kc

kb

0.879

2.70(72) 0.265  0.869

ka 0.107

( 209.3) 2  ( 293) 2 0.5(72) 36 kpsi

M

Sce

At the shoulder at A, x 10.75 in. From Prob. 7-4, M xy 209.3 lbf ˜ in, M xz 293.0 lbf ˜ in, T

Strength

The simplified model yielded reasonable results.

2

 455  12502

1/2



455 lbf

1330 lbf

RA (9)

242(9)

2178 lbf ˜ in

Chapter 7 - Rev. A, Page 11/45

(c) Potential critical locations occur at each stress concentration (shoulders and keyways). To be thorough, the stress at each potentially critical location should be evaluated. For

Shear force, bending moment, and torque diagrams can now be obtained.

M max

Reactions RA and RB , and the load W are all in the same plane. From force and moment balance, RA 1330(2 /11) 242 lbf RB 1330(9 /11) 1088 lbf

1/2



 Wt 2

W

W

2 r

1250 tan 20D

Wt tan I

Wr

Wt T / (d / 2) 2500 / (4 / 2) 1250 lbf The radial component of gear force is related by the pressure angle.

(b) From summing moments around the shaft axis, the tangential transmitted load through the gear will be

keys. The bearings can be lightly pressed onto the shaft. The left bearing will locate the shaft in the housing, while the right bearing will float in the housing.

2.70(68) 0.265

1  0.8(3.0  1)

0.883

2.6

1  0.75(2.14  1) 1.9

0.883(0.816)(0.5)(68)

Se

0.816

(2 / 0.3) 0.107 kb

For estimating kb , guess d

ka

K fs

Kf

24.5 kpsi

2 in.

d 1.57 in

Ans.

d

1/3 1/2 2 1/2 ª ­ª º ª3  2.6 ˜ 2500 2 º ½ º «16(1.5) ° ¬ 4 1.9 ˜ 2178 ¼ ¬ ¼ °»  ¾» « S ® 24 500 68 000 ° °» « ¯ ¿¼ ¬

d

Chapter 7 - Rev. A, Page 12/45

With this diameter, the estimates for notch sensitivity and size factor were conservative, but close enough for a first iteration until deflections are checked. Check yielding with this diameter.

Eq. (7-8):

1/2 2 1/2 ª ­ª º ª3  K T 2 º ½ º «16n °° ¬« 4  K f M a ¼» fs m « » °° » ¬ ¼  « ® ¾» Se Sut « S ° °» ¯° ¿° ¼» ¬«

1/3

Selecting the DE-Goodman criteria for a conservative first design,

Eq. (6-20) Eq. (6-18)

Eq. (6-19):

Eq. (6-32):

M a 2178 lbf ˜ in Tm 2500 lbf ˜ in M m Ta 0 From Table 7-1, estimate stress concentrations for the end-milled keyseat to be K t = 2.14 and K ts = 3.0. For the relatively low strength steel specified (AISI 1020 CD), roughly estimate notch sensitivities of q = 0.75 and q s = 0.80, obtained by observation of Figs. 620 and 6-21, assuming a typical radius at the bottom of the keyseat of r / d = 0.02 (p. 373), and a shaft diameter of up to 3 inches.

(d) At the gear keyway, approximately 9 in from the left end of the shaft, the bending is completely reversed and the torque is steady.

now, we will choose the most likely critical location, by observation of the loading situation, to be in the keyway for the gear. At this point there is a large stress concentration, a large bending moment, and the torque is present. The other locations either have small bending moments, or no torque. The stress concentration for the keyway is highest at the ends. For simplicity, and to be conservative, we will use the maximum bending moment, even though it will have dropped off a little at the end of the keyway.

2

ª§ 32(1.9)(2178) · § 16(2.6)(2500) · º «¨ ¸  3¨ ¸ » 3 3 (1.57) S ¹ © S (1.57) ¹ »¼ ¬«© c S y / V max 57 / 18.4 3.1 Ans.

2 1/ 2

1/2

2 ª§ 32 K f M a ·2 § 16 K fsTm · º «¨ ¸  3¨ ¸ » 3 3 «¬© S d ¹ © S d ¹ »¼

18389 psi

18.4 kpsi

Chapter 7 - Rev. A, Page 13/45

(g) To use a non-crowned gear, the gear slope is recommended to be less than 0.0005 rad. Since all other deflections are acceptable, we will target an increase in diameter only for the long section between the left bearing and the gear. Increasing this diameter from the proposed 1.56 in to 1.75 in, produces a gear slope of  0.000 401 rad. All other deflections are improved as well. ______________________________________________________________________________

(f) Entering this shaft geometry into beam analysis software (or Finite Element software), the following deflections are determined: Left bearing slope: 0.000 532 rad Right bearing slope:  0.000 850 rad Gear slope:  0.000 545 rad Right end of shaft slope:  0.000 850 rad Gear deflection:  0.001 45 in Right end of shaft deflection: 0.005 10 in Comparing these deflections to the recommendations in Table 7-2, everything is within typical range except the gear slope is a little high for an uncrowned gear.

(e) Now estimate other diameters to provide typical shoulder supports for the gear and bearings (p. 372). Also, estimate the gear and bearing widths.

ny

c V max

Eq. (7-15):

c V max

7-18

37.5 kpsi, H B 0.883

137

1 nf

1

Kt K ts

Fig. A-15-9: Fig. A-15-8:

1.6

2.4

Chapter 7 - Rev. A, Page 14/45

Right bearing shoulder The text does not give minimum and maximum shoulder diameters for 03-series bearings (roller). Use D = 1.75 in. r 0.030 D 1.75 0.019, 1.11 d 1.574 d 1.574

Eq. (7-11):

2.1

2 2 ­° ª1.6(2178) º 2 ª 2.1(2500) º ½° 16 4  3« ¾ » » 3 ® « S (1.875 ) ¯° ¬ 24 700 ¼ ¬ 37 500 ¼ ¿° n f = 3.5 Ans.

Se

Eq. (6-18):

§ 1.875 · 0.822 ¨ ¸ © 0.30 ¹ 0.883(0.822)(34.0) 24.7 kpsi

0.107

Kf

kb

1  qs ( K ts  1) 1  0.57(3.0  1) 1  0.51(2.14  1) 1.6

K fs

q 0.51 qs 0.57

Eq. (6-20):

Fig. 6-20: Fig. 6-21: Eq. (6-32):

For an end-mill profile keyway cutter of 0.010 in radius, estimate notch sensitivities.

Left keyway See Table 7-1 for keyway stress concentration factors, Kt 2.14½ ¾ Profile keyway Kts 3.0 ¿

1

kd

kc

ke

2.70(68) 0.265

ka

Eq. (6-19):

68 kpsi, S y

0.5(68) 34.0 kpsi

Sce

Eq. (6-8):

Table A-20 for 1030 HR: Sut

Candidate critical locations for strength: x Left seat keyway x Right bearing shoulder x Right keyway

(a) Use the distortion-energy elliptic failure locus. The torque and moment loadings on the shaft are shown in the solution to Prob. 7-17.

1  0.70(1.6  1) 1.42

§ 0.453 · 2178 ¨ ¸ © 2 ¹

K fs

M

1 nf

1/ 2

2 ª § 1.91(493) · 2 § 1.42(2500) · º 16 «4 ¨ ¸  3¨ ¸ » 3 S (1.574 ) «¬ © 24 700 ¹ © 37 500 ¹ »¼ Ans. n f = 4.2

493 lbf ˜ in

1  0.65(2.4  1) 1.91

Kf

q 0.65 qs 0.70

S d 3S y Ans.

S 1.53 (37 500)

1/2

2 ª§ 32 K f M a ·2 § 16 K fsTm · º «¨ ¸  3¨ ¸ » 3 3 «¬© S d ¹ © S d ¹ »¼ 1/2

c V max 1/2

13 722 psi 13.7 kpsi

ª § 16 2.1 2500 ·2 º ¸ » «3 ¨   3 ¸ » « ¨ S 1.5  ¹ ¼ ¬ ©

ª § 16 K fsTm ·2 º «3 ¨ ¸ » 3 «¬ © S d ¹ »¼

1/2

Chapter 7 - Rev. A, Page 15/45

2 2 ª§ · § · º «¨ 32 1.6  2178 ¸  3 ¨ 16  2.1  2500 ¸ » ¨ S 1.875 3 ¸ » «¨ S 1.875 3 ¸ ¹ © ¹ ¼ ¬© 8791 psi 8.79 kpsi Sy 37.5 ny Ans. 4.3 c 8.79 V max Check in smaller diameter at right end of shaft where only steady torsion exists.

c V max

Yielding Check for yielding at the left keyway, where the completely reversed bending is maximum, and the steady torque is present. Using Eq. (7-15), with M m = T a = 0,

n f = 2.7

nf

Right keyway Use the same stress concentration factors as for the left keyway. There is no bending moment, thus Eq. (7-11) reduces to: 1 16 3K fsTm 16 3(2.1)(2500)

Eq. (7-11):

Fig. 6-20: Fig. 6-21: Eq. (6-32): Sy c V max

37.5 13.7 2.7 Ans.

Fb (3 x 2  b 2  l 2 ) 6 EIl

T T

2.4124(10 6 )(3 x 2  117)

d y AB dx

T BC







n fs

n fs

0.0008 1.8 0.000 434 2

Allowable slope Actual slope

Ans.

0.001 0.000 282 3

3.5

Ans.

n fs 

Chapter 7 - Rev. A, Page 16/45

0.0005 1.6 Ans. 0.000 304 ______________________________________________________________________________

Gear mesh slope: Table 7-2 recommends a minimum relative slope of 0.0005 rad. While we don’t know the slope on the next shaft, we know that it will need to have a larger diameter and be stiffer. At the moment we can say

Right bearing:

Left bearing:

1449(9)(112  92 ) 6(30)(106 )(S / 64)(1.8754 )(11)

Fa 3 x 2  6 xl  2l 2  a 2 6 EIl

Fa 2 l  a2 6 EIl

d yBC dx

Obtain allowable slopes from Table 7-2.

T

At x = l = 11 in:

At x = 0 in:

4.342(10 4 ) rad

1449(2)(3 x 2  2 2  112 ) 6(30)(106 )(S / 64)(1.8754 )(11)

2.823(104 ) rad 3.040(104 ) rad At x = 9 in: To the right of the load, from Table A-9, case 6, p. 1015,

T AB

To the left of the load, from Table A-9, case 6, p. 1015,

(b) One could take pains to model this shaft exactly, using finite element software. However, for the bearings and the gear, the shaft is basically of uniform diameter, 1.875 in. The reductions in diameter at the bearings will change the results insignificantly. Use E = 30 Mpsi for steel.

ny

7-19

1

kd

kc

ke

2.70(64) 0.265

ka

0.897

S ut = 64 kpsi, S y = 54 kpsi Sce 0.5(64) 32.0 kpsi

kb

Se

Eq. (6-20):

Eq. (6-18):

K fs

§ 1.75 · 0.828 ¨ ¸ © 0.30 ¹ 0.897(0.828)(32) 23.8 kpsi

0.107

1  qs ( K ts  1) 1  0.71(3.0  1) 2.4

Chapter 7 - Rev. A, Page 17/45

Keyway at A Assuming r / d = 0.02 for typical end-milled keyway cutter (p. 373), with d = 1.75 in, r = 0.02d = 0.035 in. Table 7-1: K t = 2.14, K ts = 3.0 Fig. 6-20: q = 0.65 Fig. 6-21: q s = 0.71 Eq. (6-32): K f 1  q  K t  1 1  0.65(2.14  1) 1.7

Eq. (6-19):

Table A-20: Eq. (6-8):

The most likely critical locations for fatigue are at locations where the bending moment is high, the cross section is small, stress concentration exists, and torque exists. The twoplane bending moment diagrams, shown in the solution to Prob. 3-72, indicate decreasing moments in both planes to the left of A and to the right of C, with combined values at A and C of M A = 5324 lbf·in and M C = 6750 lbf·in. The torque is constant between A and B, with T = 2819 lbf·in. The most likely critical locations are at the stress concentrations near A and C. The two shoulders near A can be eliminated since the shoulders near C have the same geometry but a higher bending moment. We will consider the following potentially critical locations: x keyway at A x shoulder to the left of C x shoulder to the right of C 3  K fsTm

B

2

3 ª¬ 2.4  2819 º¼ 2

4 ¬ª1.7  5324 ¼º 2

11 718 lbf ˜ in 11.72 kip ˜ in

18102 lbf ˜ in 18.10 kip ˜ in

S d 3 Se ° ¯

Se

kb

K fs

0.107

§ 1.75 · 0.828 ¨ ¸ © 0.30 ¹ 0.897(0.828)(32) 23.8 kpsi

1  qs ( K ts  1) 1  0.76(1.8  1) 1.6

K t = 2.2 K ts = 1.8 q = 0.71 q s = 0.76 K f 1  q  K t  1 1  0.71(2.2  1) 1.9

1 n

3  K fsTm B

2

3 ¬ª1.6  2819 ¼º

2

7812 lbf ˜ in

8  25.65

25.65 kip ˜ in

Chapter 7 - Rev. A, Page 18/45

7.812 kip ˜ in

25 650 lbf ˜ in

2 1/2 ½ ­ ª ° « § 2  7.812  23.8 · »º ° ¸ ®1  1  ¨ ¾ S 1.753  23.8 ° « ©¨  25.65  64 ¹¸ » ° ¼ ¿ ¯ ¬

2

4 ª¬1.9  6750 º¼

2 1/2 ½ ­ 8 A ° ª § 2 BSe · º ° ®1  «1  ¨ ¸ » ¾ 3 S d Se ° « © ASut ¹ » ° ¼ ¿ ¯ ¬

2

4K f Ma A

For convenience, we will use the full value of the bending moment at C, even though it will be slightly less at the shoulder. Using Eq. (7-9) with M m = T a = 0,

Eq. (6-18):

Eq. (6-20):

Fig. A-15-9: Fig. A-15-8: Fig. 6-20: Fig. 6-21: Eq. (6-32):

Shoulder to the left of C r / d = 0.0625 / 1.75 = 0.036, D / d = 2.5 / 1.75 = 1.43

1 n

2 1/2

­ ½ 8 A ° ª § 2 BSe · º ° ®1  «1  ¨ ¸ » ¾ AS » ° ut ¹ ¼ ¬« © ¿ 2 1/2 ½ ­ ª § 8 18.10 2 11.72  23.8 · º ° ° « » ¾ 1 1   ¨ ¸ ® S 1.753  23.8 ° « ¨© 18.10  64 ¸¹ » ° ¼ ¿ ¯ ¬ n = 1.3

2

4K f Ma

A

We will choose the DE-Gerber criteria since this is an analysis problem in which we would like to evaluate typical expectations. Using Eq. (7-9) with M m = T a = 0,

K fs

0.107

1  qs ( K ts  1) 1  0.76(1.7  1) 1.5

K t = 2.0 K ts = 1.7 q = 0.71 q s = 0.76 K f 1  q  K t  1 1  0.71(2.0  1) 1.7

2

½ ° ¾ °¿

2

7324 lbf ˜ in

8  22.95

22.95 kip ˜ in

7.324 kip ˜ in

22 950 lbf ˜ in

2 1/2 ½ ­ ª ° « § 2  7.324  24.5 · »º ° 1 1   ¨ ¸ ® ¾ S 1.33  24.5 ° « ©¨  22.95  64 ¹¸ » ° ¼ ¿ ¯ ¬

º » ¼»

2 1/2

3 ª¬1.5  2819 º¼ 2

4 ¬ª1.7  6750 ¼º

­ 8 A ° ª § 2 BSe · ®1  «1  ¨ ¸ 3 S d Se ° « © ASut ¹ ¯ ¬

n = 0.45

1 n

3  K fsTm

B

2

4K f Ma

A

kb

c V max

2 ª§ · § · «¨ 32 1.7  6750 ¸  3 ¨ 16 1.5  2819 ¸ 3 3 ¨ ¸ ¸ «¨ S 1.3 S 1.3 © ¹ ¹ ¬©

2 ª§ 32 K f M a ·2 § 16 K fsTm · º «¨ »  3 ¸ ¨ ¸ 3 Sd3 ¹ © S d ¹ ¼» ¬«©

1/2

º » » ¼

2 1/2

Chapter 7 - Rev. A, Page 19/45

55 845 psi 55.8 kpsi

Though not explicitly called for in the problem statement, a static check for yielding is especially warranted with such a low fatigue factor of safety. Using Eq. (7-15), with M m = T a = 0,

The critical location is at the shoulder to the right of C, where n = 0.45 and finite life is predicted. Ans.

Eq. (6-20):

§ 1.3 · 0.855 ¨ ¸ © 0.30 ¹ Eq. (6-18): Se 0.897(0.855)(32) 24.5 kpsi For convenience, we will use the full value of the bending moment at C, even though it will be slightly less at the shoulder. Using Eq. (7-9) with M m = T a = 0,

Fig. A-15-9: Fig. A-15-8: Fig. 6-20: Fig. 6-21: Eq. (6-32):

Shoulder to the right of C r / d = 0.0625 / 1.3 = 0.048, D / d = 1.75 / 1.3 = 1.35

n = 0.96

Sy c V max

54 55.8 0.97

7-20

Slope Deflection (rad) (in) 0.00640 0.00000 0.00434 0.00000 0.00260 0.04839 0.01078 0.07517

 slope all

nd  dy / dx old

1/4

(1)(0.01078) 0.0005

1/4

2.15

Left bearing O Right bearing C Left Gear A Right Gear B

Location

Slope (rad) 0.00030 0.00020 0.00012 0.00050

Deflection (in) 0.00000 0.00000 0.00225 0.00350

Chapter 7 - Rev. A, Page 20/45

Multiplying all diameters by 2.15, we obtain the following deflections:

d new d old

Comparing these values to the recommended limits in Table 7-2, we find that they are all out of the desired range. This is not unexpected since the stress analysis of Prob. 7-19 also indicated the shaft is undersized for infinite life. The slope at the right gear is the most excessive, so we will attempt to increase all diameters to bring it into compliance. Using Eq. (7-18) at the right gear,

Left bearing O Right bearing C Left Gear A Right Gear B

Location

For a shaft with significantly varying diameters over its length, we will choose to use shaft analysis software or finite element software to calculate the deflections. Entering the geometry from the shaft as defined in Prob. 7-19, and the loading as defined in Prob. 3-72, the following deflection magnitudes are determined:

It is interesting to note the impact of stress concentration on the acceptability of the proposed design. This problem is linked with several previous problems (see Table 1-1, p. 24) in which the shaft was considered to have a constant diameter of 1.25 in. In each of the previous problems, the 1.25 in diameter was more than adequate for deflection, static, and fatigue considerations. In this problem, even though practically the entire shaft has diameters larger than 1.25 in, the stress concentrations significantly reduce the anticipated fatigue life. ______________________________________________________________________________

This indicates localized yielding is predicted at the stress-concentration, though after localized cold-working it may not be a problem. The finite fatigue life is still likely to be the failure mode that will dictate whether this shaft is acceptable.

n

7-21

S ut = 440 MPa, S y = 370 MPa

K fs

1  qs ( K ts  1) 1  0.72(3.0  1) 2.4

K t = 2.14, K ts = 3.0 q = 0.66 q s = 0.72 K f 1  q  K t  1 1  0.66(2.14  1) 1.8

Fig. A-15-9:

K t = 2.2

Shoulder to the left of B r / d = 2 / 50 = 0.04, D / d = 75 / 50 = 1.5

Table 7-1: Fig. 6-20: Fig. 6-21: Eq. (6-32):

Chapter 7 - Rev. A, Page 21/45

Keyway at A Assuming r / d = 0.02 for typical end-milled keyway cutter (p. 373), with d = 50 mm, r = 0.02d = 1 mm.

Table A-20:

The most likely critical locations for fatigue are at locations where the bending moment is high, the cross section is small, stress concentration exists, and torque exists. The twoplane bending moment diagrams, shown in the solution to Prob. 3-73, indicate both planes have a maximum bending moment at B. At this location, the combined bending moment from both planes is M = 4097 N·m, and the torque is T = 3101 N·m. The shoulder to the right of B will be eliminated since its diameter is only slightly smaller, and there is no torque. Comparing the shoulder to the left of B with the keyway at B, the primary difference between the two is the stress concentration, since they both have essentially the same bending moment, torque, and size. We will check the stress concentration factors for both to determine which is critical.

This brings the slope at the right gear just to the limit for an uncrowned gear, and all other slopes well below the recommended limits. For the gear deflections, the values are below recommended limits as long as the diametral pitch is less than 20. ______________________________________________________________________________ 1  qs ( K ts  1) 1  0.78(1.8  1) 1.6

kb

kc Se

Eq. (6-19): Eq. (6-20):

Eq. (6-18):

0.107

0.818

0.899(0.818)(220) 162 MPa

§ 50 · ¨ ¸ © 7.62 ¹ k d ke 1

0.899

0.5(440) 220 MPa 4.51(440) 0.265

3 ¬ª 2.4  3101 ¼º

2

12 890 N ˜ m

14 750 N ˜ m

c V max

1/2 2 2 ª§ · § · º «¨ 32 1.8  4097 ¸  3 ¨ 16  2.4  3101 ¸ » ¨ S  0.050 3 ¸ » «¨ S  0.050 3 ¸ ¹ © ¹ ¼ ¬©

2 ª§ 32 K f M a ·2 § 16 K fsTm · º «¨ ¸  3¨ ¸ » 3 3 «¬© S d ¹ © S d ¹ »¼

1/2

798 MPa

Chapter 7 - Rev. A, Page 22/45



7.98 108 Pa

Though not explicitly called for in the problem statement, a static check for yielding is especially warranted with such a low fatigue factor of safety. Using Eq. (7-15), with M m = T a = 0,

8 14 750

2

4 ª¬1.8  4097 º¼

2 1/2 ½ ­ 8 A ° ª § 2 BSe · º ° « 1 1   ® ¨ ¸ » ¾ S d 3 Se ° « © ASut ¹ » ° ¼ ¿ ¯ ¬

2

2 1/2 ½ ­ ª 6 º ° « § 2 12 890 162 10 · » ° ¸ ®1  1  ¨ ¾ S  0.0503 162 106 ° « ¨© 14 750  440 106 ¸¹ » ° ¼ ¿ ¯ ¬ n = 0.25 Infinite life is not predicted. Ans.

1 n

3  K fsTm B

2

4K f Ma A

We will choose the DE-Gerber criteria since this is an analysis problem in which we would like to evaluate typical expectations. Using Eq. (7-9) with M m = T a = 0,

Sce ka

Eq. (6-8):

Examination of the stress concentration factors indicates the keyway will be the critical location.

K fs

Fig. A-15-8: K ts = 1.8 Fig. 6-20: q = 0.73 Fig. 6-21: q s = 0.78 Eq. (6-32): K f 1  q  K t  1 1  0.73(2.2  1) 1.9

Sy

c V max

370 798

0.46

7-22

Slope (rad) 0.01445 0.01843 0.00358 0.00366

Deflection (mm) 0.000 0.000 3.761 3.676

nd yold yall

1/4

(1)(3.761) 0.254

1/4

1.96

Left bearing O Right bearing C Left Gear A Right Gear B

Location

Slope (rad) 0.00090 0.00115 0.00022 0.00023

Deflection (mm) 0.000 0.000 0.235 0.230 Chapter 7 - Rev. A, Page 23/45

Multiplying all diameters by 2, we obtain the following deflections:

d new d old

Comparing these values to the recommended limits in Table 7-2, we find that they are all well out of the desired range. This is not unexpected since the stress analysis in Prob. 7-21 also indicated the shaft is undersized for infinite life. The transverse deflection at the left gear is the most excessive, so we will attempt to increase all diameters to bring it into compliance. Using Eq. (7-17) at the left gear, assuming from Table 7-2 an allowable deflection of y all = 0.01 in = 0.254 mm,

Left bearing O Right bearing C Left Gear A Right Gear B

Location

For a shaft with significantly varying diameters over its length, we will choose to use shaft analysis software or finite element software to calculate the deflections. Entering the geometry from the shaft as defined in Prob. 7-21, and the loading as defined in Prob. 3-73, the following deflection magnitudes are determined:

This problem is linked with several previous problems (see Table 1-1, p. 24) in which the shaft was considered to have a constant diameter of 50 mm. The results here are consistent with the previous problems, in which the 50 mm diameter was found to slightly undersized for static, and significantly undersized for fatigue. Though in the current problem much of the shaft has larger than 50 mm diameter, the added contribution of stress concentration limits the fatigue life. ______________________________________________________________________________

This indicates localized yielding is predicted at the stress-concentration. Even without the stress concentration effects, the static factor of safety turns out to be 0.93. Static failure is predicted, rendering this proposed shaft design unacceptable.

n

7-23

Chapter 7 - Rev. A, Page 24/45

Potentially critical locations are identified as follows: x Keyway at C, where the torque is high, the diameter is small, and the keyway creates a stress concentration.

(a) Label the approximate locations of the effective centers of the bearings as A and B, the fan as C, and the gear as D, with axial dimensions as shown. Since there is only one gear, we can combine the radial and tangential gear forces into a single resultant force with an accompanying torque, and handle the statics problem in a single plane. From statics, the resultant reactions at the bearings can be found to be R A = 209.9 lbf and R B = 464.5 lbf. The bending moment and torque diagrams are shown, with the maximum bending moment at D of M D = 209.9(6.98) = 1459 lbf·in and a torque transmitted from D to C of T = 633 (8/2) = 2532 lbf·in. Due to the shaft rotation, the bending stress on any stress element will be completely reversed, while the torsional stress will be steady. Since we do not have any information about the fan, we will ignore any axial load that it would introduce. It would not likely contribute much compared to the bending anyway.

This brings the deflection at the gears just within the limit for a spur gear (assuming P < 10 teeth/in), and all other deflections well below the recommended limits. ______________________________________________________________________________

ka

2.70(68) 0.265 0.883

ny

W

57 / 2 12.9

12.9 kpsi

2.21

S 1.004 / 32

2532 1.00 / 2

Sy / 2

Tr J

Se

Eq. (6-18):

§ 1.75 · 0.828 ¨ ¸ © 0.30 ¹ 0.883(0.828)(34.0) 24.9 kpsi

0.107

1  qs ( K ts  1) 1  0.72(3.0  1) 2.4

Chapter 7 - Rev. A, Page 25/45

We will choose the DE-Gerber criteria since this is an analysis problem in which we would like to evaluate typical expectations. Using Eq. (7-9) with M m = T a = 0,

kb

K fs

K t = 2.14, K ts = 3.0 q = 0.66 q s = 0.72 K f 1  q  K t  1 1  0.66(2.14  1) 1.8

Eq. (6-20):

Table 7-1: Fig. 6-20: Fig. 6-21: Eq. (6-32):

Keyway at D Assuming r / d = 0.02 for typical end-milled keyway cutter (p. 373), with d = 1.75 in, r = 0.02d = 0.035 in.

Eq. (5-3):

W

Keyway at C Since there is only steady torsion here, only a static check needs to be performed. We’ll use the maximum shear stress theory.

Eq. (6-19):

S ut = 68 kpsi, S y = 57 kpsi Sce 0.5(68) 34.0 kpsi

Keyway at D, where the bending moment is maximum, the torque is high, and the keyway creates a stress concentration. Groove at E, where the diameter is smaller than at D, the bending moment is still high, and the groove creates a stress concentration. There is no torque here, though. Shoulder at F, where the diameter is smaller than at D or E, the bending moment is still moderate, and the shoulder creates a stress concentration. There is no torque here, though. The shoulder to the left of D can be eliminated since the change in diameter is very slight, so that the stress concentration will undoubtedly be much less than at D.

Table A-20: Eq. (6-8):

x

x

x

x 2

3 ¬ª 2.4  2532 ¼º

2

2

4 ª¬1.8 1459 º¼

2 1/2 ½ ­ 8 A ° ª § 2 BSe · º ° «   1 1 ® ¨ ¸ » ¾ S d 3 Se ° « © ASut ¹ » ° ¼ ¿ ¯ ¬

3  K fsTm 2

4K f Ma

Se

kb

0.107

§ 1.55 · 0.839 ¨ ¸ © 0.30 ¹ 0.883(0.839)(34) 25.2 kpsi

Fig. A-15-9: Fig. 6-20:

Shoulder at F

2

^

1/2

Ans.

r / d = 0.125 / 1.40 = 0.089, K t = 1.7 q = 0.78

3

1/2

`

2

4122 lbf ˜ in

4.122 kip ˜ in

Chapter 7 - Rev. A, Page 26/45

D / d = 2.0 / 1.40 = 1.43

2 1  ª1   0 º ¬ ¼ S 1.55  25.2

½ ° ¾ °¿

4 ª¬1.8 1115 º¼

2 ­ 8 A ° ª § 2 BSe · º ®1  «1  ¨ ¸ » 3 S d Se ° « © ASut ¹ » ¼ ¯ ¬

8  4.122

4K f Ma

n = 4.47

1 n

B=0

A

Using Eq. (7-9) with M m = T a = T m = 0,

Eq. (6-18):

Eq. (6-20):

r / d = 0.1 / 1.55 = 0.065, D / d = 1.75 / 1.55 = 1.13 Fig. A-15-14: K t = 2.1 Fig. 6-20: q = 0.76 Eq. (6-32): K f 1  q  K t  1 1  0.76(2.1  1) 1.8

Groove at E We will assume Figs. A-15-14 is applicable since the 2 in diameter to the right of the groove is relatively narrow and will likely not allow the stress flow to fully develop. (See Fig.7-9 for the stress flow concept.)

8  5.252

5.252 kip ˜ in

10 525 lbf ˜ in 10.53 kip ˜ in

5252 lbf ˜ in

2 1/2 ½ ­ ª ° « § 2 10.53  24.9 · º» ° ¸ ®1  1  ¨ ¾ S 1.753  24.9 ° « ¨©  5.252  68 ¸¹ » ° ¼ ¿ ¯ ¬ n = 3.59 Ans.

1 n

B

A

kb

Se

Eq. (6-20):

Eq. (6-18):

§ 1.40 · 0.848 ¨ ¸ © 0.30 ¹ 0.883(0.848)(34) 25.5 kpsi

0.107

2

^

1/2

1/2

`

2535 lbf ˜ in

2.535 kip ˜ in

Left bearing A Right bearing B Fan C Gear D

Location

Slope Deflection (rad) (in) 0.000290 0.000000 0.000400 0.000000 0.000290 0.000404 0.000146 0.000928

Chapter 7 - Rev. A, Page 27/45

The deflection problem can readily (though tediously) be solved with singularity functions. For examples, see Ex. 4-7, p. 159, or the solution to Prob. 7-24. Alternatively, shaft analysis software or finite element software may be used. Using any of the methods, the results should be as follows:

Section Diameter Length (in) (in) 1 1.00 2.90 2 1.70 7.77 3 1.40 2.20

(b) The deflection will not be much affected by the details of fillet radii, grooves, and keyways, so these can be ignored. Also, the slight diameter changes, as well as the narrow 2.0 in diameter section, can be neglected. We will model the shaft with the following three sections:

Ans.

2 1  ª1   0 º ¬ ¼ S 1.403  25.5

½ ° ¾ °¿

4 ¬ª1.5  845 ¼º

2 ­ 8 A ° ª § 2 BSe · º ®1  «1  ¨ ¸ » 3 S d Se ° « © ASut ¹ » ¼ ¯ ¬

8  2.535

4K f Ma

n = 5.42

1 n

B=0

A

2

1  q  K t  1 1  0.78(1.7  1) 1.5

Using Eq. (7-9) with M m = T a = T m = 0,

Kf

Eq. (6-32):

7-24

x(m) 0 0.04 0.04 0.1 0.1 0.14 0.14 0.21 0.21 0.275 0.275 0.315

M/I (109 N/m3) 0 2.112 1.2375 3.094 1.932 2.705 2.705 1.623 2.6 0.99 1.6894 0

Plot M/I as a function of x. Slope 52.8 30.942 19.325 –15.457 -24.769 -42.235

Step

–0.8745 –1.162 0 0.977 0.6994

Chapter 7 - Rev. A, Page 28/45

-17.47

-9.312

–34.78

–11.617

–21.86

'Slope

I 35 = (S/64)(0.0354) = 7.366(10-8) m4, I 40 = 1.257(10-7) m4, I 45 = 2.013(10-7) m4

100 140 210 275 315 x(mm) 0 40 M(N · m) 0 155.56 388.89 544.44 326.67 124.44 0

Determine the bending moment at each step.

Statics: Left support: R1 7(315  140) / 315 3.889 kN Right support: R2 7(140) / 315 3.111 kN

Deflection: First we will ignore the steps near the bearings where the bending moments are low. Thus let the 30 mm dia. be 35 mm. Secondly, the 55 mm dia. is very thin, 10 mm. The full bending stresses will not develop at the outer fibers so full stiffness will not develop either. Thus, ignore this step and let the diameter be 45 mm.

Shaft analysis software or finite element software can be utilized if available. Here we will demonstrate how the problem can be simplified and solved using singularity functions.

Comparing these values to the recommended limits in Table 7-2, we find that they are all within the recommended range. ______________________________________________________________________________

Ey

dy dx

2

2 1

 0.977 x  0.21

1

3

1.552 x  0.21  0.3497 x  0.275

2

2

 2.912 x  0.275

3

3

1

x (mm) 0 140 315

T (rad) –0.001 4260 –0.000 1466 0.001 3120 F.E. Model –0.001 4270 –0.000 1467 0.001 3280

Chapter 7 - Rev. A, Page 29/45

Full F.E. Model –0.001 4160 –0.000 1646 0.001 3150

Equation (1) with C 1 = –0.295 25 provides the slopes at the bearings and gear. The following table gives the results in the second column. The third column gives the results from a similar finite element model. The fourth column gives the results of a full model which models the 35 and 55 mm diameter steps.

 C1x  C2 º 109 ¼

2

 C1 º 109 (1) ¼

ª8.8 x 3  0.4373 x  0.04  3.643 x  0.04  0.581 x  0.1 ¬ 3 3 2 1.937 x  0.1  5.797 x  0.14  0.4885 x  0.21

2

 1.162 x  0.1

4.655 x  0.21  0.6994 x  0.275  8.735 x  0.275

2

5.81 x  0.1  17.39 x  0.14

2

1  17.47 x  0.275 º 109 ¼

ª 26.4 x 2  0.8745 x  0.04 1  10.93 x  0.04 ¬

0

Boundary conditions: y = 0 at x = 0 yields C 2 = 0; y = 0 at x = 0.315 m yields C 1 = –0.295 25 N/m2.

E

1

9.312 x  0.21  0.6994 x  0.275

ª52.8 x  0.8745 x  0.04 0  21.86 x  0.04 1  1.162 x  0.1 0 ¬ 1 1 0 11.617 x  0.1  34.78 x  0.14  0.977 x  0.21

Integrate twice:

M /I

The steps and the change of slopes are evaluated in the table. From these, the function M/I can be generated:

 slope all

nd  dy / dx old 1/ 4

(1)(0.0014260) 0.001

1/4

1.093

0 15

40

100 110 140 210 275 300 330 61.9 47.8 60.9 52.0 39.6 17.6 0 0 6 8.5 12.7 20.2 68.1 0 61.9 47.8 61.8 53.1 45.3 39.2 118.0

At x = 210 mm: Eq. (6-19): ka

4.51(470) 0.265

0.883

Chapter 7 - Rev. A, Page 30/45

Table A-20 for AISI 1020 CD steel: S ut = 470 MPa, S y = 390 MPa

x (mm)

V (MPa) 0 22.0 37.0 0 W (MPa) 0 0 V c(MPa) 0 22.0 37.0

Strength: Due to stress concentrations and reduced shaft diameters, there are a number of locations to look at. A table of nominal stresses is given below. Note that torsion is only to the right of the 7 kN load. Using V = 32M/(Sd3) and W = 16T/(Sd3),

This is well within our goal. Have the students try a goal of 0.0005 rad at the gears.

T = –9.30 u 10-4 rad x = 0: x = 140 mm: T = –1.09 u 10-4 rad x = 315 mm: T = 8.65 u 10-4 rad

Repeating the full finite element model results in

20.00 30.00 35.00 40.00 45.00 55.00 Old d, mm New ideal d, mm 21.86 32.79 38.26 43.72 49.19 60.12 Rounded up d, mm 22.00 34.00 40.00 44.00 50.00 62.00

Form a table:

d new d old

To increase the stiffness of the shaft, apply Eq. (7-18) to the most offending deflection (at x = 0) to determine a multiplier to be used for all diameters.

It can be seen that the allowable slopes at the bearings are exceeded. Thus, either the load has to be reduced or the shaft “beefed” up. If the allowable slope is 0.001 rad, then the maximum load should be F max = (0.001/0.001 426)7 = 4.91 kN. With a design factor this would be reduced further.

The main discrepancy between the results is at the gear location (x = 140 mm). The larger value in the full model is caused by the stiffer 55 mm diameter step. As was stated earlier, this step is not as stiff as modeling implicates, so the exact answer is somewhere between the full model and the simplified model which in any event is a small value. As expected, modeling the 30 mm dia. as 35 mm does not affect the results much.

kb (40 / 7.62) 0.107 0.837 S e = 0.883 (0.837)(0.5)(470) = 174 MPa D / d = 45 / 40 = 1.125, r / d = 2 / 40 = 0.05 K ts = 1.4 K t = 1.9 q = 0.75 q s = 0.79 K f = 1 + 0.75(1.9 –1) = 1.68 K f s = 1 + 0.79(1.4 – 1) = 1.32







Chapter 7 - Rev. A, Page 31/45

7-25 and 7-26 With these design tasks each student will travel different paths and almost all details will differ. The important points are x The student gets a blank piece of paper, a statement of function, and some constraints – explicit and implied. At this point in the course, this is a good experience. x It is a good preparation for the capstone design course.

If worse-case is at x = 210 mm, the changes discussed for the slope criterion will improve the strength issue. ______________________________________________________________________________

Check the other locations.

Note that since there is only a steady torque, Eq. (7-11) reduces to essentially the equivalent of the distortion energy failure theory.

n = 2.49



D / d = 30 / 20 = 1.5, r / d = 2 / 20 = 0.1 Fig. A-15-9: K ts = 1.42 Fig. 6-21: q s = 0.79 Eq. (6-32): K f s = 1 + 0.79(1.42 – 1) = 1.33 Eq. (7-11): ª1.33(107) º 1 16 » 3 « 6 n S  3 ¬« 390 10 ¼»

At x = 330 mm: The von Mises stress is the highest but it comes from the steady torque only.



2 2 ­ ª ª1.32(107) º ½° 1 16 ° 1.68(326.67) º « » « » 4 3  ® ¾ n S 0.043 ° « 174 106 » «¬ 390 106 »¼ ° ¼ ¯ ¬ ¿ n 1.98

1/2

Choosing DE-ASME Elliptic to inherently include the yield check, from Eq. (7-11), with M m = T a = 0,

Fig. A-15-8: Fig. A-15-9: Fig. 6-20: Fig. 6-21: Eq. (6-32):

Eq. (6-20): Eq. (6-18):

Take the first half hour, resisting the temptation of putting pencil to paper, and decide what the problem really is. Take another twenty minutes to list several possible remedies. Pick one, and show your instructor how you would implement it.

Also, a fix at the bearing seat lead-in could transfer the problem to the shoulder fillet, and the problem may not be solved.

The students’ initial reaction is that he/she does not know much from the problem statement. Then, slowly the realization sets in that they do know some important things that the designer did not. They knew how it failed, where it failed, and that the design wasn’t good enough; it was close, though.

x x

x

This task was once given as a final exam problem. This problem is a learning experience. Following the task statement, the following guidance was added.

7-28

d

J gE

4l 2Z

S2

(2)

(1)

2

9 § S · § 0.025 · 9.81(207)(10 ) Z ¨ ¸ ¨ ¸ 3 76.5 10 © 0.6 ¹ © 4 ¹

(a) From Eq. (1) and Table A-5

or

2

§ S · § d · gE Z ¨ ¸ ¨ ¸ © l ¹ ©4¹ J

In Eq. (7-22) set Sd4 Sd2 I , A 4 64 to obtain

883 rad/s

Ans.

Chapter 7 - Rev. A, Page 32/45

To many students’ credit, they chose to keep the shaft geometry, and selected a new material to realize about twice the Brinell hardness. ______________________________________________________________________________

7-27

The adequacy of their design must be demonstrated and possibly include a designer’s notebook. x Many of the fundaments of the course, based on this text and this course, are useful. The student will find them useful and notice that he/she is doing it. x Don’t let the students create a time sink for themselves. Tell them how far you want them to go. ______________________________________________________________________________

x

gE

J

4 l

constant 0.6(883) 529.8 529.8 1766 rad/s Ans. 0.3

7-29

207(10 ) Pa,

E







2

AJ l





2



4

4

8

4.909 10

0.3(0.3) 0.6  0.3  0.3

2

w









2

6w y g 6w y 2

1 11

 10

8

5.756 104

 9.81 1.47110

10

5

6

6(207) 109 (1.917) 108 (0.6)

Two elements:

Z1

2 1

1

G11 5

6

620 rad/s

Chapter 7 - Rev. A, Page 33/45

(30% low)

22.53 N

7.65 104 N/m 3



J

7.65 10 4 (0.6)

1.917 108 m 4 ,

1.134 10 m/N  y wG 22.53(1.134) 10 2.555 10 m y 6.528 10 6w y 22.53(2.555) 10 5.756 10 6w y 22.53(6.528) 10 1.47110

Eq. (7-24):

One element:

9

I

883 rad/s

4.909 10 4 m 2 ,

A

From Prob. 7-28, Z

Thus the first critical speed doubles. ______________________________________________________________________________

Z

lZ

Since d / l is the same regardless of the scale,

2

S d







876 rad/s

4







0.1(0.1) 0.62  0.12  0.12















5



3.500 107 m/N

2

7





6

5.460 107 m/N

1.134 106 m/N

6(207) 109 (1.917) 108 (0.6)



0.3(0.1) 0.62  0.32  0.12

 

6(207) 109 (1.917) 108 (0.6)

9

(0.8% low)

6(207) 109 (1.917) 108 (0.6)

 

º» »¼

10

0.5(0.1) 0.62  0.52  0.12

 



4.961 107 m/N

 2.877 10 2(11.265)(1.632) 10 3.677 10







6

11.265(6.379) 10 7  11.265(4.961) 10 7





6.379 107 m/N





3

2

1

2

7

7

7

6

6

2

7

6

7

5

6(207) 109 (1.917) 108 (0.6)

6

7

7

7

9

Chapter 7 - Rev. A, Page 34/45

2

4

6

5



1.277 105 m

2.380 10 m/N   y 7.51 ª¬3.500 10  5.460 10  2.380 10 º¼ 8.516 10 y 7.51 ª¬5.460 10  1.134 10  5.460 10 º¼ 1.672 10 y 7.51 ª¬ 2.380 10  5.460 10  3.500 10 º¼ 8.516 10 6w y 7.51 ª¬8.516 10  1.672 10  8.516 10 º¼ 2.535 10 6w y 7.51^ª¬8.516 10 º¼  ª¬1.672 10 º¼  ª¬8.516 10 º¼ ` 3.189 10

G13



1.632 10 10

w1G11  w2G12



0.3(0.3)  0.62  0.32  0.32

G12 G 32

G 22



0.15(0.15)(0.62  0.152  0.152 ) 6(207) 109 (1.917) 108 (0.6)

ª 2.877 104 9.81 « «¬ 3.677 109

G11 G 33



6(207) 109 (1.917) 108 (0.6)

Three elements:

Z1

6w y 2

 

0.45(0.15) 0.62  0.452  0.152

2(11.265)(1.277) 10 5

y22

y12 6w y

y2

y1

G12 G 21

(c) From Eq. (2),

lZ

G11 G 22

(b) From Eq. (1), we observe that the critical speed is linearly proportional to the diameter. Thus, to double the critical speed, we should double the diameter to d = 50 mm. Ans.

 

ª 2.535 104 9.81 « «¬ 3.189 109

º» »¼

883 rad/s

7-30

( m2G 22  1/ Z 2 )

m1G 21

0

2

2

1 ·§ 1 · § 1 ·§ 1 · § 1 · § 1 ¨ 2 ¸  ¨ 2  2 ¸¨ ¸  ¨ 2 ¸¨ 2 ¸ 0 © Z ¹ © Z1 Z2 ¹ © Z ¹ © Z1 ¹ © Z2 ¹

2 2

m1m2 (G11G 22  G12G 21 )

and it follows that

Z12 Z22

1 1 Ÿ

Z22

1

(2)

(1)

Chapter 7 - Rev. A, Page 35/45

Z12 m1m2 (G11G 22  G12G 21 )

Equate the third terms of Eqs. (1) and (2), which must be identical.

or,

§ 1 1 ·§ 1 1 · ¨ 2  2 ¸¨ 2  2 ¸ 0 © Z Z1 ¹ © Z Z2 ¹

2 1

Eq. (1) has two roots 1 / Z and 1 / Z . Thus

2

§ 1 · § 1 · ¨ 2 ¸  (m1G11  m2G 22 ) ¨ 2 ¸  m1m2 (G11G 22  G12G 21 ) 0 ©Z ¹ © Z1 ¹

Expanding the determinant yields,

m2G12

(m1G11  1/ Z 2 )

(a) For two bodies, Eq. (7-26) is

The result is the same as in Prob. 7-28. The point was to show that convergence is rapid using a static deflection beam equation. The method works because: x If a deflection curve is chosen which meets the boundary conditions of momentfree and deflection-free ends, as in this problem, the strain energy is not very sensitive to the equation used. x Since the static bending equation is available, and meets the moment-free and deflection-free ends, it works. ______________________________________________________________________________

Z1

1

g2

Z1 w1w2 (G11G 22  G12G 21 )

Ans.



I A

S d  d

2 o

2 i

/4

S  do4  di4 / 64

2 2 2 2 1  d o  di  d o  di 16 d o2  di2

1 d o2  di2 4

§d · 1 ¨ i ¸ © do ¹

2

Z1 1  0 to about Z1 1  1

The possible values of di are 0 d di d d o , so the range of the critical speeds is

(1/ 4) d o2

(1/ 4) do2  di2

This means that when a solid shaft is hollowed out, the critical speed increases beyond that of the solid shaft of the same size. By how much?

Z1 v

In Eq. (7-22), for Z 1 , the term I / A appears. For a hollow uniform diameter shaft,



7-32

x M

x M

12 0.5

1 2 3 0.875 1.75 1.625 9 10 11 0.875 0.75 0.625

0 0

8 1

Chapter 7 - Rev. A, Page 36/45

16 0

5 6 7 1.375 1.25 1.125 13 14 15 0.375 0.25 0.125

4 1.5

All steps will be modeled using singularity functions with a spreadsheet. Programming both loads will enable the user to first set the left load to 1, the right load to 0 and calculate G 11 and G 21 . Then set the left load to 0 and the right to 1 to get G 12 and G 22 . The spreadsheet shows the G 11 and G 21 calculation. A table for M / I vs. x is easy to make. First, draw the bending-moment diagram as shown with the data.

or from Z1 to 2 Z1. Ans. ______________________________________________________________________________

7-31

Z2

1 3862 466 rad/s Ans. 124.8 35(55) ª¬ 2.061(3.534)  2.2342 º¼ 108 ______________________________________________________________________________

(b) In Ex. 7-5, part (b), the first critical speed of the two-disk shaft (w 1 = 35 lbf, w 2 = 55 lbf) is Z 1 = 124.8 rad/s. From part (a), using influence coefficients,

Z2

S  2 4 / 64 0.7854 in 4

S  2.472 / 64 1.833 in 4

S  2.7634 / 64 2.861 in 4

4

16 0

1 1 2 2 3 4 5 6 7 8 1.1141 0.4774 0.9547 0.9547 0.8865 0.8183 0.7501 0.6819 0.6137 0.5456

9 9 10 11 12 13 14 14 15 15 0.4774 0.3058 0.2621 0.2185 0.1748 0.1311 0.0874 0.0874 0.0437 0.1592

0 0

Chapter 7 - Rev. A, Page 37/45

From this diagram, one can see where changes in value (steps) and slope occur. Using a spreadsheet, one can form a table of these changes. An example of a step is, at x = 1 in, M/I goes from 0.875/0.7854 = 1.1141 lbf/in3 to 0.875/1.833 = 0.4774 lbf/in3, a step change of 0.4774  1.1141 =  0.6367 lbf/in3. A slope change also occurs at at x = 1 in.

x M/I

x M/I

Divide M by I at the key points x = 0, 1, 2, 9, 14, 15, and 16 in and plot

9 d x d 15 in , I 3

1 d x d 9 in , I 2

0 d x d 1 in and 15 d x d 16 in, I1

The second-area moments are:

A x 1a 1b 2 2 9a 9b 14 14 15a 15b 16

B M 0.875 0.875 1.75 1.75 0.875 0.875 0.25 0.25 0.125 0.125 0

C M/I 1.114085 0.477358 0.954716 0.954716 0.477358 0.305854 0.087387 0.087387 0.043693 0.159155 0.000000

D step 0.000000 -0.636727 0.000000 0.000000 0.000000 -0.171504 0.000000 0.000000 0.000000 0.115461 0.000000

E Slope 1.114085 0.477358 0.477358 -0.068194 -0.068194 -0.043693 -0.043693 -0.043693 -0.043693 -0.159155 -0.159155

1

1 0

1

0

 D7 x  9  F7 x  9  D11 x  15  F11 x  15

0

E2 ( x)  D3 x  1  F3 x  1  F5 x  2

F ' Slope 0.000000 -0.636727 0.000000 -0.545552 0.000000 0.024501 0.000000 0.000000 0.000000 -0.115461 0.000000

1

Z1

386

5.105(104 ) 5.212(109 )

Chapter 7 - Rev. A, Page 38/45

6149 rad/s or 58 720 rev/min

Ans.

18(2.917)(107 )  32(1.627)(107 ) 1.046(105 ) 18(1.627)(107 )  32(2.231)(107 ) 1.007(105 ) 1.093(1010 ), y22 1.014(1010 ) 5.105(104 ), ¦ w y 2 5.212(109 ) Neglecting the shaft, Eq. (7-23) gives

y1 y2 y12 ¦ wy

Integrating twice gives the equation for Ey. Assume the shaft is steel. Boundary conditions y = 0 at x = 0 and at x = 16 inches provide integration constants (C 1 =  4.906 lbf/in and C 2 = 0). Substitution back into the deflection equation at x = 2 and 14 in provides the G ’s. The results are: G 11 = 2.917(10–7) and G 12 = 1.627(10–7). Repeat for F 1 = 0 and F 2 = 1, resulting in G 21 = 1.627(10–7) and G 22 = 2.231(10–7). This can be verified by finite element analysis.

M /I

The equation for M / I in terms of the spreadsheet cell locations is:

1 2 3 4 5 6 7 8 9 10 11 12

The slope for 0 d x d 1 in is 1.1141/1 = 1.1141 lbf/in2, which changes to (0.9547  0.4774)/1 = 0.4774 lbf/in2, a change of 0.4774  1.1141 =  0.6367 lbf/in2. Following this approach, a table is made of all the changes. The table shown indicates the column letters and row numbers for the spreadsheet.

3.072 10 10 , y22

1.615 10 10

9

4

2

9

, ¦ w y 5.37110 ª 3.449 10 º » 4980 rad/s 386 « «¬ 5.37110 »¼

3.449 10

4

5.472 10 7

1

1

7-33

W

K tsW 0

K tsc W 0c

Chapter 7 - Rev. A, Page 39/45

We must not let the basis of the stress concentration factor, as presented, impose a viewpoint on the designer. Table A-16 shows K ts as a decreasing monotonic as a function of a/D. All is not what it seems. Let us change the basis for data presentation to the full section rather than the net section.

  Ÿ Z1  3870 rad/s Ans. Z12 61492 49802 ______________________________________________________________________________

1

Combination: Using Dunkerley’s equation, Eq. (7-32):

A finite element model of the exact shaft gives Z 1 = 5340 rad/s. The simple model is 6.8% low.

Z1

¦w y

5.718 10 7 , G 22

1.27110 5

y12

y1 1.753 10 5 , y2

9.605 10 7 , G12 G 21

G11

The spreadsheet can be easily modified to give

Without the loads, we will model the shaft using 2 elements, one between 0 d x d 9 in, and one between 0 d x d 16 in. As an approximation, we will place their weights at x = 9/2 = 4.5 in, and x = 9 + (16  9)/2 = 12.5 in. From Table A-5, the weight density of steel is J = 0.282 lbf/in3. The weight of the left element is S §S · w1 J ¦ d 2l 0.282 ¨ ¸ ª¬ 22 1  2.4722  8 º¼ 11.7 lbf 4 ©4¹ The right element is §S · w2 0.282 ¨ ¸ ª¬ 2.7632  6  22 1 º¼ 11.0 lbf ©4¹ K ts A

§ 32T · K tsc ¨ 3 ¸ ©SD ¹

Our knowledge of the minima location is

7-34

T r

2819 1.00 / 2

5638 lbf

Failure by shear across the key:

Chapter 7 - Rev. A, Page 40/45

Selecting 1020 CD steel for the key, with S y = 57 kpsi, and using the distortion-energy theory, S sy = 0.577 S y = (0.577)(57) = 32.9 kpsi.

F

From the solution to Prob. 3-72, the torque to be transmitted through the key from the gear to the shaft is T = 2819 lbf·in. From Prob. 7-19, the nominal shaft diameter supporting the gear is 1.00 in. From Table 7-6, a 0.25 in square key is appropriate for a 1.00 in shaft diameter. The force applied to the key is

0.075 d (a / D ) d 0.125 We can form a design rule: In torsion, the pin diameter should be about 1/10 of the shaft diameter, for greatest shaft capacity. However, it is not catastrophic if one forgets the rule. ______________________________________________________________________________

x

K tsc has the following attributes: x It exhibits a minimum; x It changes little over a wide range; x Its minimum is a stationary point minimum at a / D  0.100;

Form a table:

K tsc

32T

S AD 3

Therefore

K ts

W

F A

F

t / 2 l

n

V

Sy

l

2 F /  tl

Ÿ

Ÿ

nF tS sy

l

2 Fn tS y

1.1 5638 0.25  32 900

2  5638 1.1

0.25  57 103

0.754 in

0.870 in

7-35

Ÿ l

2 Fn tS y

 0.014  390 106

2 124 103 1.1

0.0500 m 50.0 mm

Shaft: Eq. (7-37):

Hole: Eq. (7-36):

D max = D + 'D = 15 + 0.018 = 15.018 mm D min = D = 15.000 mm

Ans. Ans.

Choose basic size D = d = 15 mm. From Table 7-9, a locational clearance fit is designated as 15H7/h6. From Table A-11, the tolerance grades are 'D = 0.018 mm and 'd = 0.011 mm. From Table A-12, the fundamental deviation is G F = 0 mm.

D max = D + 'D = 1.75 + 0.0010 = 1.7510 in D min = D = 1.7500 in

Ans. Ans.

W

S sy

F A

S sy

F /  tl

F tl

Failure by crushing:

n

W

Ÿ l

nF tS sy

 0.014  225 10

1.1124 103 6

43.3 mm

Chapter 7 - Rev. A, Page 41/45



0.0433 m

D max = D + 'D = 45 + 0.025 = 45.025 mm D min = D = 45.000 mm

Ans. Ans.

Chapter 7 - Rev. A, Page 42/45

Ans. d max = d + G F = 45.000 + (–0.009) = 44.991 mm Ans. d min = d + G F – 'd = 45.000 + (–0.009) – 0.016 = 44.975 mm ______________________________________________________________________________

Shaft: Eq. (7-37):

Hole: Eq. (7-36):

Choose basic size D = d = 45 mm. From Table 7-9, a sliding fit is designated as H7/g6. From Table A-11, the tolerance grades are 'D = 0.025 mm and 'd = 0.016 mm. From Table A-12, the fundamental deviation is G F = –0.009 mm.

7-38

Shaft: Eq. (7-38):

Hole: Eq. (7-36):

Ans. d max = d + G F = 15.000 + 0 = 15.000 mm Ans. d min = d + G F – 'd = 15.000 + 0 – 0.011 = 14.989 mm ______________________________________________________________________________ 7-37 Choose basic size D = d = 1.75 in. From Table 7-9, a medium drive fit is designated as H7/s6. From Table A-13, the tolerance grades are 'D = 0.0010 in and 'd = 0.0006 in. From Table A-14, the fundamental deviation is G F = 0.0017 in.

7-36

Failure by shear across the key:

3101 124 103 N 0.050 / 2

2 F /  tl

Sy

F

t / 2 l

Ans. d min = d + G F = 1.75 + 0.0017 = 1.7517 in d max = d + G F + 'd = 1.75 + 0.0017 + 0.0006 = 1.7523 in Ans. ______________________________________________________________________________

T r

V

Sy

F A

Select 14 mm square key, 50 mm long, 1020 CD steel. Ans. ______________________________________________________________________________

n

V

Selecting 1020 CD steel for the key, with S y = 390 MPa, and using the distortion-energy theory, S sy = 0.577 S y = 0.577(390) = 225 MPa.

F

The force applied to the key is

From the solution to Prob. 3-73, the torque to be transmitted through the key from the gear to the shaft is T = 3101 N·m. From Prob. 7-21, the nominal shaft diameter supporting the gear is 50 mm. To determine an appropriate key size for the shaft diameter, we can either convert to inches and use Table 7-6, or we can look up standard metric key sizes from the internet or a machine design handbook. It turns out that the recommended metric key for a 50 mm shaft is 14 x 9 mm. Since the problem statement specifies a square key, we will use a 14 x 14 mm key. For comparison, using Table 7-6 as a guide, for d = 50 mm = 1.97 in, a 0.5 in square key is appropriate. This is equivalent to 12.7 mm. A 14 x 14 mm size is conservative, but reasonable after rounding up to standard sizes.

Select ¼-in square key, 7/8 in long, 1020 CD steel. Ans. ______________________________________________________________________________

V

Sy

F / tl

F F A tl S sy S sy

Failure by crushing:

n

W

Shaft: Eq. (7-37):

Hole: Eq. (7-36):

D max = D + 'D = 1.250 + 0.0015 = 1.2515 in D min = D = 1.2500 in Ans. Ans.

Choose basic size D = d = 1.250 in. From Table 7-9, a close running fit is designated as H8/f7. From Table A-13, the tolerance grades are 'D = 0.0015 in and 'd = 0.0010 in. From Table A-14, the fundamental deviation is G F = –0.0010 in.

D max = D + 'D = 35 + 0.025 = 35.025 mm D min = D = 35.000 mm

Shaft: Eq. (7-38):

The bearing bore specifications are within the hole specifications for a locational interference fit. Now find the necessary shaft sizes.

Hole: Eq. (7-36):

Choose basic size D = d = 35 mm. From Table 7-9, a locational interference fit is designated as H7/p6. From Table A-11, the tolerance grades are 'D = 0.025 mm and 'd = 0.016 mm. From Table A-12, the fundamental deviation is G F = 0.026 mm.

7-41

D max = D + 'D = 1.5000 + 0.0010 = 1.5010 in D min = D = 1.5000 in

Shaft: Eq. (7-38):

Chapter 7 - Rev. A, Page 43/45

d min = d + G F = 1.5000 + 0.0010 = 1.5010 in d max = d + G F + 'd = 1.5000 + 0.0010 + 0.0006 = 1.5016 in

The bearing bore specifications exactly match the requirements for a locational interference fit. Now check the shaft.

Hole: Eq. (7-36):

Choose basic size D = d = 1.5 in. From Table 7-9, a locational interference fit is designated as H7/p6. From Table A-13, the tolerance grades are 'D = 0.0010 in and 'd = 0.0006 in. From Table A-14, the fundamental deviation is G F = 0.0010 in.

Ans. d min = d + G F = 35 + 0.026 = 35.026 mm Ans. d max = d + G F + 'd = 35 + 0.026 + 0.016 = 35.042 mm ______________________________________________________________________________

7-40

Ans. d max = d + G F = 1.250 + (–0.0010) = 1.2490 in Ans. d min = d + G F – 'd = 1.250 + (–0.0010) – 0.0010 = 1.2480 in ______________________________________________________________________________

7-39

7-42

pmin

pmax



1/ 2



115 MPa

115 MPa

 V 1V 2  V 22

p













p

p

V r ,hub Eq. (7-46):

1/2

35.1 MPa

115 MPa

Ans.

Ans.

Chapter 7 - Rev. A, Page 44/45

115 MPa

»¼

º»

»¼

º»

§ 602  352 · 115 ¨ 2 234 MPa 2 ¸ © 60  35 ¹ 115 MPa d o2  d 2 do2  d 2

ª¬ ( 115) 2  ( 115)( 115)  ( 115) 2 º¼ S y / V c 390 / 115 3.4 Ans.

V

2 1



 

 

p

V t ,hub

n

Vc

V r ,shaft



2 2 2 2 EG min ª d o  d d  di º « » 2d 3 « d o2  di2 »¼ ¬ 207 109  0.018 ª 602  352 352  0 « 602  0 2 353 «¬

Eq. (7-45):

For the hub:

Eq. (5-13):

Eq. (7-46):



2 2 2 2 EG max ª d o  d d  d i º « » 3 2 2 d o  di 2d « »¼ ¬ 9 2 207 10  0.059 ª 60  352 352  0 « 602  0 2 353 «¬

G min d min  Dmax 35.043  35.025 0.018 mm G max d max  Dmin 35.059  35.000 0.059 mm

(c) For the shaft: Eq. (7-44): V t ,shaft

Eq. (7-40):

Eq. (7-43):

(b) Eq. (7-42):

(a) Basic size is D = d = 35 mm. Table 7-9: H7/s6 is specified for medium drive fit. Table A-11: Tolerance grades are 'D = 0.025 mm and 'd = 0.016 mm. Table A-12: Fundamental deviation is G F 0.043 mm. Eq. (7-36): D max = D + 'D = 35 + 0.025 = 35.025 mm D min = D = 35.000 mm Ans. Eq. (7-38): d min = d + G F = 35 + 0.043 = 35.043 mm Ans. d max = d + G F + 'd = 35 + 0.043 + 0.016 = 35.059 mm

The shaft diameter of 1.5020 in is greater than the maximum allowable diameter of 1.5016 in, and therefore does not meet the specifications for the locational interference fit. Ans. ______________________________________________________________________________

Vc

2 1

 V 1V 2  V 22

1/ 2



ª¬ (234) 2  (234)( 115)  ( 115) 2 º¼

V 1/ 2

308 MPa

2700 N ˜ m Ans.

Chapter 7 - Rev. A, Page 45/45

______________________________________________________________________________

(S / 2)(0.8)(35.1) 106 (0.050)(0.035) 2



n S y / V c 600 / 308 1.9 Ans. (d) A value for the static coefficient of friction for steel to steel can be obtained online or from a physics textbook as approximately f = 0.8. Eq. (7-49) T (S / 2) f pmin ld 2

Eq. (5-13):